A Practical Introduction To Data Structures And Algorithm Analysis Third Edition (Java Version)
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This document provides a summary of a textbook on data structures and algorithms. It begins with an introduction to fundamental concepts like abstract data types and common design patterns. It then covers important preliminary topics in mathematics. The main body of the text focuses on analyzing the efficiency of algorithms and implementing various data structures like lists, trees, graphs, and indexes. It also addresses sorting and searching algorithms, file processing, advanced data structures, algorithm analysis techniques, limits of computation, and patterns of algorithms. The textbook is intended as a teaching text for undergraduate courses on data structures or algorithms.
![Sec. 1.5 Further Reading 19
4. It must be composed of a finite number of steps. If the description for the
algorithm were made up of an infinite number of steps, we could never hope
to write it down, nor implement it as a computer program. Most languages for
describing algorithms (including English and “pseudocode”) provide some
way to perform repeated actions, known as iteration. Examples of iteration
in programming languages include the while and for loop constructs of
Java. Iteration allows for short descriptions, with the number of steps actually
performed controlled by the input.
5. It must terminate. In other words, it may not go into an infinite loop.
Programs: We often think of a computer program as an instance, or concrete
representation, of an algorithm in some programming language. In this book,
nearly all of the algorithms are presented in terms of programs, or parts of pro-
grams. Naturally, there are many programs that are instances of the same alg-
orithm, because any modern computer programming language can be used to im-
plement the same collection of algorithms (although some programming languages
can make life easier for the programmer). To simplify presentation throughout
the remainder of the text, I often use the terms “algorithm” and “program” inter-
changeably, despite the fact that they are really separate concepts. By definition,
an algorithm must provide sufficient detail that it can be converted into a program
when needed.
The requirement that an algorithm must terminate means that not all computer
programs meet the technical definition of an algorithm. Your operating system is
one such program. However, you can think of the various tasks for an operating sys-
tem (each with associated inputs and outputs) as individual problems, each solved
by specific algorithms implemented by a part of the operating system program, and
each one of which terminates once its output is produced.
To summarize: A problem is a function or a mapping of inputs to outputs.
An algorithm is a recipe for solving a problem whose steps are concrete and un-
ambiguous. The algorithm must be correct, of finite length, and must terminate for
all inputs. A program is an instantiation of an algorithm in a computer program-
ming language.
1.5 Further Reading
The first authoritative work on data structures and algorithms was the series of
books The Art of Computer Programming by Donald E. Knuth, with Volumes 1
and 3 being most relevant to the study of data structures [Knu97, Knu98]. A mod-
ern encyclopedic approach to data structures and algorithms that should be easy](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-38-320.jpg)
![20 Chap. 1 Data Structures and Algorithms
to understand once you have mastered this book is Algorithms by Robert Sedge-
wick [Sed03]. For an excellent and highly readable (but more advanced) teaching
introduction to algorithms, their design, and their analysis, see Introduction to Al-
gorithms: A Creative Approach by Udi Manber [Man89]. For an advanced, en-
cyclopedic approach, see Introduction to Algorithms by Cormen, Leiserson, and
Rivest [CLRS01]. Steven S. Skiena’s The Algorithm Design Manual [Ski98] pro-
vides pointers to many implementations for data structures and algorithms that are
available on the Web.
For a gentle introduction to ADTs and program specification, see Abstract Data
Types: Their Specification, Representation, and Use by Thomas, Robinson, and
Emms [TRE88].
The claim that all modern programming languages can implement the same
algorithms (stated more precisely, any function that is computable by one program-
ming language is computable by any programming language with certain standard
capabilities) is a key result from computability theory. For an easy introduction to
this field see James L. Hein, Discrete Structures, Logic, and Computability [Hei03].
Much of computer science is devoted to problem solving. Indeed, this is what
attracts many people to the field. How to Solve It by George Pólya [Pól57] is con-
sidered to be the classic work on how to improve your problem-solving abilities. If
you want to be a better student (as well as a better problem solver in general), see
Strategies for Creative Problem Solving by Folger and LeBlanc [FL95], Effective
Problem Solving by Marvin Levine [Lev94], and Problem Solving & Comprehen-
sion by Arthur Whimbey and Jack Lochhead [WL99].
See The Origin of Consciousness in the Breakdown of the Bicameral Mind by
Julian Jaynes [Jay90] for a good discussion on how humans use the concept of
metaphor to handle complexity. More directly related to computer science educa-
tion and programming, see “Cogito, Ergo Sum! Cognitive Processes of Students
Dealing with Data Structures” by Dan Aharoni [Aha00] for a discussion on mov-
ing from programming-context thinking to higher-level (and more design-oriented)
programming-free thinking.
On a more pragmatic level, most people study data structures to write better
programs. If you expect your program to work correctly and efficiently, it must
first be understandable to yourself and your co-workers. Kernighan and Pike’s The
Practice of Programming [KP99] discusses a number of practical issues related to
programming, including good coding and documentation style. For an excellent
(and entertaining!) introduction to the difficulties involved with writing large pro-
grams, read the classic The Mythical Man-Month: Essays on Software Engineering
by Frederick P. Brooks [Bro95].](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-39-320.jpg)
![Sec. 1.6 Exercises 21
If you want to be a successful Java programmer, you need good reference man-
uals close at hand. David Flanagan’s Java in a Nutshell [Fla05] provides a good
reference for those familiar with the basics of the language.
After gaining proficiency in the mechanics of program writing, the next step
is to become proficient in program design. Good design is difficult to learn in any
discipline, and good design for object-oriented software is one of the most difficult
of arts. The novice designer can jump-start the learning process by studying well-
known and well-used design patterns. The classic reference on design patterns
is Design Patterns: Elements of Reusable Object-Oriented Software by Gamma,
Helm, Johnson, and Vlissides [GHJV95] (this is commonly referred to as the “gang
of four” book). Unfortunately, this is an extremely difficult book to understand,
in part because the concepts are inherently difficult. A number of Web sites are
available that discuss design patterns, and which provide study guides for the De-
sign Patterns book. Two other books that discuss object-oriented software design
are Object-Oriented Software Design and Construction with C++ by Dennis Ka-
fura [Kaf98], and Object-Oriented Design Heuristics by Arthur J. Riel [Rie96].
1.6 Exercises
The exercises for this chapter are different from those in the rest of the book. Most
of these exercises are answered in the following chapters. However, you should
not look up the answers in other parts of the book. These exercises are intended to
make you think about some of the issues to be covered later on. Answer them to
the best of your ability with your current knowledge.
1.1 Think of a program you have used that is unacceptably slow. Identify the spe-
cific operations that make the program slow. Identify other basic operations
that the program performs quickly enough.
1.2 Most programming languages have a built-in integer data type. Normally
this representation has a fixed size, thus placing a limit on how large a value
can be stored in an integer variable. Describe a representation for integers
that has no size restriction (other than the limits of the computer’s available
main memory), and thus no practical limit on how large an integer can be
stored. Briefly show how your representation can be used to implement the
operations of addition, multiplication, and exponentiation.
1.3 Define an ADT for character strings. Your ADT should consist of typical
functions that can be performed on strings, with each function defined in](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-40-320.jpg)
![26 Chap. 2 Mathematical Preliminaries
{1, 4} A set composed of the members 1 and 4
{x | x is a positive integer} A set definition using a set former
Example: the set of all positive integers
x ∈ P x is a member of set P
x /
∈ P x is not a member of set P
∅ The null or empty set
|P| Cardinality: size of set P
or number of members for set P
P ⊆ Q, Q ⊇ P Set P is included in set Q,
set P is a subset of set Q,
set Q is a superset of set P
P ∪ Q Set Union:
all elements appearing in P OR Q
P ∩ Q Set Intersection:
all elements appearing in P AND Q
P − Q Set difference:
all elements of set P NOT in set Q
Figure 2.1 Set notation.
|P| = 3 (because P has three members) and |Q| = 2 (because Q has two members).
The union of P and Q, written P ∪ Q, is the set of elements in either P or Q, which
is {2, 3, 5, 10}. The intersection of P and Q, written P ∩ Q, is the set of elements
that appear in both P and Q, which is {5}. The set difference of P and Q, written
P − Q, is the set of elements that occur in P but not in Q, which is {2, 3}. Note
that P ∪ Q = Q ∪ P and that P ∩ Q = Q ∩ P, but in general P − Q 6= Q − P.
In this example, Q − P = {10}. Note that the set {4, 3, 5} is indistinguishable
from set P, because sets have no concept of order. Likewise, set {4, 3, 4, 5} is also
indistinguishable from P, because sets have no concept of duplicate elements.
The powerset of a set S is the set of all possible subsets for S. Consider the set
S = {a, b, c}. The powerset of S is
{∅, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}}.
Sometimes we wish to define a collection of elements with no order (like a
set), but with duplicate-valued elements. Such a collection is called a bag.1 To
distinguish bags from sets, I use square brackets [] around a bag’s elements. For
1
The object referred to here as a bag is sometimes called a multilist. But, I reserve the term
multilist for a list that may contain sublists (see Section 12.1).](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-45-320.jpg)
![Sec. 2.1 Sets and Relations 27
example, bag [3, 4, 5, 4] is distinct from bag [3, 4, 5], while set {3, 4, 5, 4} is
indistinguishable from set {3, 4, 5}. However, bag [3, 4, 5, 4] is indistinguishable
from bag [3, 4, 4, 5].
A sequence is a collection of elements with an order, and which may contain
duplicate-valued elements. A sequence is also sometimes called a tuple or a vec-
tor. In a sequence, there is a 0th element, a 1st element, 2nd element, and so
on. I indicate a sequence by using angle brackets hi to enclose its elements. For
example, h3, 4, 5, 4i is a sequence. Note that sequence h3, 5, 4, 4i is distinct from
sequence h3, 4, 5, 4i, and both are distinct from sequence h3, 4, 5i.
A relation R over set S is a set of ordered pairs from S. As an example of a
relation, if S is {a, b, c}, then
{ha, ci, hb, ci, hc, bi}
is a relation, and
{ha, ai, ha, ci, hb, bi, hb, ci, hc, ci}
is a different relation. If tuple hx, yi is in relation R, we may use the infix notation
xRy. We often use relations such as the less than operator (<) on the natural
numbers, which includes ordered pairs such as h1, 3i and h2, 23i, but not h3, 2i or
h2, 2i. Rather than writing the relationship in terms of ordered pairs, we typically
use an infix notation for such relations, writing 1 < 3.
Define the properties of relations as follows, where R is a binary relation over
set S.
• R is reflexive if aRa for all a ∈ S.
• R is symmetric if whenever aRb, then bRa, for all a, b ∈ S.
• R is antisymmetric if whenever aRb and bRa, then a = b, for all a, b ∈ S.
• R is transitive if whenever aRb and bRc, then aRc, for all a, b, c ∈ S.
As examples, for the natural numbers, < is antisymmetric and transitive; ≤ is
reflexive, antisymmetric, and transitive, and = is reflexive, antisymmetric, and tran-
sitive. For people, the relation “is a sibling of” is symmetric and transitive. If we
define a person to be a sibling of himself, then it is reflexive; if we define a person
not to be a sibling of himself, then it is not reflexive.
R is an equivalence relation on set S if it is reflexive, symmetric, and transitive.
An equivalence relation can be used to partition a set into equivalence classes. If
two elements a and b are equivalent to each other, we write a ≡ b. A partition of
a set S is a collection of subsets that are disjoint from each other and whose union
is S. An equivalence relation on set S partitions the set into subsets whose elements
are equivalent. See Section 6.2 for a discussion on how to represent equivalence
classes on a set. One application for disjoint sets appears in Section 11.5.2.](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-46-320.jpg)
![30 Chap. 2 Mathematical Preliminaries
while n! grows slower than nn (because
√
2πn/en < 1), it grows faster than cn for
any positive integer constant c.
Permutations: A permutation of a sequence S is simply the members of S ar-
ranged in some order. For example, a permutation of the integers 1 through n
would be those values arranged in some order. If the sequence contains n distinct
members, then there are n! different permutations for the sequence. This is because
there are n choices for the first member in the permutation; for each choice of first
member there are n − 1 choices for the second member, and so on. Sometimes
one would like to obtain a random permutation for a sequence, that is, one of the
n! possible permutations is selected in such a way that each permutation has equal
probability of being selected. A simple Java function for generating a random per-
mutation is as follows. Here, the n values of the sequence are stored in positions 0
through n − 1 of array A, function swap(A, i, j) exchanges elements i and
j in array A, and Random(n) returns an integer value in the range 0 to n − 1 (see
the Appendix for more information on swap and Random).
// Randomly permute the values of array "A"
static <E> void permute(E[] A) {
for (int i = A.length; i > 0; i--) // for each i
swap(A, i-1, DSutil.random(i)); // swap A[i-1] with
} // a random element
Boolean variables: A Boolean variable is a variable (of type boolean in Java)
that takes on one of the two values true and false. These two values are often
associated with the values 1 and 0, respectively, although there is no reason why
this needs to be the case. It is poor programming practice to rely on the corre-
spondence between 0 and false, because these are logically distinct objects of
different types.
Floor and ceiling: The floor of x (written ⌊x⌋) takes real value x and returns the
greatest integer ≤ x. For example, ⌊3.4⌋ = 3, as does ⌊3.0⌋, while ⌊−3.4⌋ = −4
and ⌊−3.0⌋ = −3. The ceiling of x (written ⌈x⌉) takes real value x and returns
the least integer ≥ x. For example, ⌈3.4⌉ = 4, as does ⌈4.0⌉, while ⌈−3.4⌉ =
⌈−3.0⌉ = −3.
Modulus operator: The modulus (or mod) function returns the remainder of an
integer division. Sometimes written n mod m in mathematical expressions, the
syntax for the Java modulus operator is n % m. From the definition of remainder,
n mod m is the integer r such that n = qm + r for q an integer, and |r| < |m|.
Therefore, the result of n mod m must be between 0 and m − 1 when n and m are](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-49-320.jpg)
![Sec. 2.8 Further Reading 49
typical car is driven about 12,000 miles per year. If gasoline costs $2/gallon,
then the yearly gas bill is $1200 for the less efficient car and $800 for the
more efficient car. If we ignore issues such as the payback that would be
received if we invested $2000 in a bank, it would take 5 years to make
up the difference in price. At this point, the buyer must decide if price is
the only criterion and if a 5-year payback time is acceptable. Naturally,
a person who drives more will make up the difference more quickly, and
changes in gasoline prices will also greatly affect the outcome.
Example 2.20 When at the supermarket doing the week’s shopping, can
you estimate about how much you will have to pay at the checkout? One
simple way is to round the price of each item to the nearest dollar, and add
this value to a mental running total as you put the item in your shopping
cart. This will likely give an answer within a couple of dollars of the true
total.
2.8 Further Reading
Most of the topics covered in this chapter are considered part of Discrete Math-
ematics. An introduction to this field is Discrete Mathematics with Applications
by Susanna S. Epp [Epp04]. An advanced treatment of many mathematical topics
useful to computer scientists is Concrete Mathematics: A Foundation for Computer
Science by Graham, Knuth, and Patashnik [GKP94].
See “Technically Speaking” from the February 1995 issue of IEEE Spectrum
[Sel95] for a discussion on the standard for indicating units of computer storage
used in this book.
Introduction to Algorithms by Udi Manber [Man89] makes extensive use of
mathematical induction as a technique for developing algorithms.
For more information on recursion, see Thinking Recursively by Eric S. Roberts
[Rob86]. To learn recursion properly, it is worth your while to learn the program-
ming language LISP, even if you never intend to write a LISP program. In particu-
lar, Friedman and Felleisen’s The Little LISPer [FF89] is designed to teach you how
to think recursively as well as teach you LISP. This book is entertaining reading as
well.
A good book on writing mathematical proofs is Daniel Solow’s How to Read
and Do Proofs [Sol90]. To improve your general mathematical problem-solving
abilities, see The Art and Craft of Problem Solving by Paul Zeitz [Zei07]. Zeitz](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-68-320.jpg)
![50 Chap. 2 Mathematical Preliminaries
also discusses the three proof techniques presented in Section 2.6, and the roles of
investigation and argument in problem solving.
For more about estimating techniques, see two Programming Pearls by John
Louis Bentley entitled The Back of the Envelope and The Envelope is Back [Ben84,
Ben00, Ben86, Ben88]. Genius: The Life and Science of Richard Feynman by
James Gleick [Gle92] gives insight into how important back of the envelope calcu-
lation was to the developers of the atomic bomb, and to modern theoretical physics
in general.
2.9 Exercises
2.1 For each relation below, explain why the relation does or does not satisfy
each of the properties reflexive, symmetric, antisymmetric, and transitive.
(a) “isBrotherOf” on the set of people.
(b) “isFatherOf” on the set of people.
(c) The relation R = {hx, yi | x2 + y2 = 1} for real numbers x and y.
(d) The relation R = {hx, yi | x2 = y2} for real numbers x and y.
(e) The relation R = {hx, yi | x mod y = 0} for x, y ∈ {1, 2, 3, 4}.
(f) The empty relation ∅ (i.e., the relation with no ordered pairs for which
it is true) on the set of integers.
(g) The empty relation ∅ (i.e., the relation with no ordered pairs for which
it is true) on the empty set.
2.2 For each of the following relations, either prove that it is an equivalence
relation or prove that it is not an equivalence relation.
(a) For integers a and b, a ≡ b if and only if a + b is even.
(b) For integers a and b, a ≡ b if and only if a + b is odd.
(c) For nonzero rational numbers a and b, a ≡ b if and only if a × b > 0.
(d) For nonzero rational numbers a and b, a ≡ b if and only if a/b is an
integer.
(e) For rational numbers a and b, a ≡ b if and only if a − b is an integer.
(f) For rational numbers a and b, a ≡ b if and only if |a − b| ≤ 2.
2.3 State whether each of the following relations is a partial ordering, and explain
why or why not.
(a) “isFatherOf” on the set of people.
(b) “isAncestorOf” on the set of people.
(c) “isOlderThan” on the set of people.
(d) “isSisterOf” on the set of people.
(e) {ha, bi, ha, ai, hb, ai} on the set {a, b}.](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-69-320.jpg)
![Sec. 3.1 Introduction 59
compiled with the same compiler and run on the same computer under the same
conditions. As much as possible, the same amount of care should be taken in the
programming effort devoted to each program to make the implementations “equally
efficient.” In this sense, all of the factors mentioned above should cancel out of the
comparison because they apply to both algorithms equally.
If you truly wish to understand the running time of an algorithm, there are other
factors that are more appropriate to consider than machine speed, programming
language, compiler, and so forth. Ideally we would measure the running time of
the algorithm under standard benchmark conditions. However, we have no way
to calculate the running time reliably other than to run an implementation of the
algorithm on some computer. The only alternative is to use some other measure as
a surrogate for running time.
Of primary consideration when estimating an algorithm’s performance is the
number of basic operations required by the algorithm to process an input of a
certain size. The terms “basic operations” and “size” are both rather vague and
depend on the algorithm being analyzed. Size is often the number of inputs pro-
cessed. For example, when comparing sorting algorithms, the size of the problem
is typically measured by the number of records to be sorted. A basic operation
must have the property that its time to complete does not depend on the particular
values of its operands. Adding or comparing two integer variables are examples
of basic operations in most programming languages. Summing the contents of an
array containing n integers is not, because the cost depends on the value of n (i.e.,
the size of the input).
Example 3.1 Consider a simple algorithm to solve the problem of finding
the largest value in an array of n integers. The algorithm looks at each
integer in turn, saving the position of the largest value seen so far. This
algorithm is called the largest-value sequential search and is illustrated by
the following Java function:
// Return position of largest value in "A"
static int largest(int[] A) {
int currlarge = 0; // Holds largest element position
for (int i=1; i<A.length; i++) // For each element
if (A[currlarge] < A[i]) // if A[i] is larger
currlarge = i; // remember its position
return currlarge; // Return largest position
}
Here, the size of the problem is n, the number of integers stored in A. The
basic operation is to compare an integer’s value to that of the largest value](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-78-320.jpg)
![74 Chap. 3 Algorithm Analysis
because n grows faster than 2 log n. Thus, n2 is in Ω(2n log n).
3.5 Calculating the Running Time for a Program
This section presents the analysis for several simple code fragments.
Example 3.9 We begin with an analysis of a simple assignment statement
to an integer variable.
a = b;
Because the assignment statement takes constant time, it is Θ(1).
Example 3.10 Consider a simple for loop.
sum = 0;
for (i=1; i<=n; i++)
sum += n;
The first line is Θ(1). The for loop is repeated n times. The third
line takes constant time so, by simplifying rule (4) of Section 3.4.4, the
total cost for executing the two lines making up the for loop is Θ(n). By
rule (3), the cost of the entire code fragment is also Θ(n).
Example 3.11 We now analyze a code fragment with several for loops,
some of which are nested.
sum = 0;
for (j=1; j<=n; j++) // First for loop
for (i=1; i<=j; i++) // is a double loop
sum++;
for (k=0; k<n; k++) // Second for loop
A[k] = k;
This code fragment has three separate statements: the first assignment
statement and the two for loops. Again the assignment statement takes
constant time; call it c1. The second for loop is just like the one in Exam-
ple 3.10 and takes c2n = Θ(n) time.
The first for loop is a double loop and requires a special technique. We
work from the inside of the loop outward. The expression sum++ requires
constant time; call it c3. Because the inner for loop is executed i times, by
simplifying rule (4) it has cost c3i. The outer for loop is executed n times,](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-93-320.jpg)
![78 Chap. 3 Algorithm Analysis
Key
Position 0 2 3 4 5 6 7 8
26 29 36
10 11 12 13 14 15
11 13 21 41 45 51 54
1
56 65 72 77
9
83
40
Figure 3.4 An illustration of binary search on a sorted array of 16 positions.
Consider a search for the position with value K = 45. Binary search first checks
the value at position 7. Because 41 < K, the desired value cannot appear in any
position below 7 in the array. Next, binary search checks the value at position 11.
Because 56 > K, the desired value (if it exists) must be between positions 7
and 11. Position 9 is checked next. Again, its value is too great. The final search
is at position 8, which contains the desired value. Thus, function binary returns
position 8. Alternatively, if K were 44, then the same series of record accesses
would be made. After checking position 8, binary would return a value of n,
indicating that the search is unsuccessful.
// Return the position of an element in sorted array "A"
// with value "K". If "K" is not in "A", return A.length.
static int binary(int[] A, int K) {
int l = -1;
int r = A.length; // l and r are beyond array bounds
while (l+1 != r) { // Stop when l and r meet
int i = (l+r)/2; // Check middle of remaining subarray
if (K < A[i]) r = i; // In left half
if (K == A[i]) return i; // Found it
if (K > A[i]) l = i; // In right half
}
return A.length; // Search value not in A
}
To find the cost of this algorithm in the worst case, we can model the running
time as a recurrence and then find the closed-form solution. Each recursive call
to binary cuts the size of the array approximately in half, so we can model the
worst-case cost as follows, assuming for simplicity that n is a power of two.
T(n) = T(n/2) + 1 for n > 1; T(1) = 1.
If we expand the recurrence, we find that we can do so only log n times before
we reach the base case, and each expansion adds one to the cost. Thus, the closed-
form solution for the recurrence is T(n) = log n.
Function binary is designed to find the (single) occurrence of K and return
its position. A special value is returned if K does not appear in the array. This
algorithm can be modified to implement variations such as returning the position
of the first occurrence of K in the array if multiple occurrences are allowed, and
returning the position of the greatest value less than K when K is not in the array.](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-97-320.jpg)
![Sec. 3.8 Multiple Parameters 83
position i along the x axis). This is why we must always say that function
f(n) is in O(g(n)) in the best, average, or worst case! If we leave off which
class of inputs we are discussing, we cannot know which cost measure we
are referring to for most algorithms.
3.8 Multiple Parameters
Sometimes the proper analysis for an algorithm requires multiple parameters to de-
scribe the cost. To illustrate the concept, consider an algorithm to compute the rank
ordering for counts of all pixel values in a picture. Pictures are often represented by
a two-dimensional array, and a pixel is one cell in the array. The value of a pixel is
either the code value for the color, or a value for the intensity of the picture at that
pixel. Assume that each pixel can take any integer value in the range 0 to C − 1.
The problem is to find the number of pixels of each color value and then sort the
color values with respect to the number of times each value appears in the picture.
Assume that the picture is a rectangle with P pixels. A pseudocode algorithm to
solve the problem follows.
for (i=0; i<C; i++) // Initialize count
count[i] = 0;
for (i=0; i<P; i++) // Look at all of the pixels
count[value(i)]++; // Increment a pixel value count
sort(count); // Sort pixel value counts
In this example, count is an array of size C that stores the number of pixels for
each color value. Function value(i) returns the color value for pixel i.
The time for the first for loop (which initializes count) is based on the num-
ber of colors, C. The time for the second loop (which determines the number of
pixels with each color) is Θ(P). The time for the final line, the call to sort, de-
pends on the cost of the sorting algorithm used. From the discussion of Section 3.6,
we can assume that the sorting algorithm has cost Θ(P log P) if P items are sorted,
thus yielding Θ(P log P) as the total algorithm cost.
Is this a good representation for the cost of this algorithm? What is actu-
ally being sorted? It is not the pixels, but rather the colors. What if C is much
smaller than P? Then the estimate of Θ(P log P) is pessimistic, because much
fewer than P items are being sorted. Instead, we should use P as our analysis vari-
able for steps that look at each pixel, and C as our analysis variable for steps that
look at colors. Then we get Θ(C) for the initialization loop, Θ(P) for the pixel
count loop, and Θ(C log C) for the sorting operation. This yields a total cost of
Θ(P + C log C).](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-102-320.jpg)
![Sec. 3.9 Space Bounds 85
The need to maintain a balance between these opposing goals is what makes the
study of data structures so interesting.
One important aspect of algorithm design is referred to as the space/time trade-
off principle. The space/time tradeoff principle says that one can often achieve a
reduction in time if one is willing to sacrifice space or vice versa. Many programs
can be modified to reduce storage requirements by “packing” or encoding informa-
tion. “Unpacking” or decoding the information requires additional time. Thus, the
resulting program uses less space but runs slower. Conversely, many programs can
be modified to pre-store results or reorganize information to allow faster running
time at the expense of greater storage requirements. Typically, such changes in time
and space are both by a constant factor.
A classic example of a space/time tradeoff is the lookup table. A lookup table
pre-stores the value of a function that would otherwise be computed each time it is
needed. For example, 12! is the greatest value for the factorial function that can be
stored in a 32-bit int variable. If you are writing a program that often computes
factorials, it is likely to be much more time efficient to simply pre-compute the 12
storable values in a table. Whenever the program needs the value of n! for n ≤ 12,
it can simply check the lookup table. (If n > 12, the value is too large to store as
an int variable anyway.) Compared to the time required to compute factorials, it
may be well worth the small amount of additional space needed to store the lookup
table.
Lookup tables can also store approximations for an expensive function such as
sine or cosine. If you compute this function only for exact degrees or are willing
to approximate the answer with the value for the nearest degree, then a lookup
table storing the computation for exact degrees can be used instead of repeatedly
computing the sine function. Note that initially building the lookup table requires
a certain amount of time. Your application must use the lookup table often enough
to make this initialization worthwhile.
Another example of the space/time tradeoff is typical of what a programmer
might encounter when trying to optimize space. Here is a simple code fragment for
sorting an array of integers. We assume that this is a special case where there are n
integers whose values are a permutation of the integers from 0 to n − 1. This is an
example of a Binsort, which is discussed in Section 7.7. Binsort assigns each value
to an array position corresponding to its value.
for (i=0; i<n; i++)
B[A[i]] = A[i];](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-104-320.jpg)
![86 Chap. 3 Algorithm Analysis
This is efficient and requires Θ(n) time. However, it also requires two arrays
of size n. Next is a code fragment that places the permutation in order but does so
within the same array (thus it is an example of an “in place” sort).
for (i=0; i<n; i++)
while (A[i] != i) // Swap element A[i] with A[A[i]]
DSutil.swap(A, i, A[i]);
Function swap(A, i, j) exchanges elements i and j in array A (see the
Appendix). It may not be obvious that the second code fragment actually sorts the
array. To see that this does work, notice that each pass through the for loop will
at least move the integer with value i to its correct position in the array, and that
during this iteration, the value of A[i] must be greater than or equal to i. A total
of at most n swap operations take place, because an integer cannot be moved out
of its correct position once it has been placed there, and each swap operation places
at least one integer in its correct position. Thus, this code fragment has cost Θ(n).
However, it requires more time to run than the first code fragment. On my computer
the second version takes nearly twice as long to run as the first, but it only requires
half the space.
A second principle for the relationship between a program’s space and time
requirements applies to programs that process information stored on disk, as dis-
cussed in Chapter 8 and thereafter. Strangely enough, the disk-based space/time
tradeoff principle is almost the reverse of the space/time tradeoff principle for pro-
grams using main memory.
The disk-based space/time tradeoff principle states that the smaller you can
make your disk storage requirements, the faster your program will run. This is be-
cause the time to read information from disk is enormous compared to computation
time, so almost any amount of additional computation needed to unpack the data is
going to be less than the disk-reading time saved by reducing the storage require-
ments. Naturally this principle does not hold true in all cases, but it is good to keep
in mind when designing programs that process information stored on disk.
3.10 Speeding Up Your Programs
In practice, there is not such a big difference in running time between an algorithm
whose growth rate is Θ(n) and another whose growth rate is Θ(n log n). There is,
however, an enormous difference in running time between algorithms with growth
rates of Θ(n log n) and Θ(n2). As you shall see during the course of your study
of common data structures and algorithms, it is not unusual that a problem whose
obvious solution requires Θ(n2) time also has a solution that requires Θ(n log n)](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-105-320.jpg)
![90 Chap. 3 Algorithm Analysis
Another approach to analysis is simulation. The idea of simulation is to model
the problem with a computer program and then run it to get a result. In the con-
text of algorithm analysis, simulation is distinct from empirical comparison of two
competitors because the purpose of the simulation is to perform analysis that might
otherwise be too difficult. A good example of this appears in Figure 9.8. This fig-
ure shows the cost for inserting or deleting a record from a hash table under two
different assumptions for the policy used to find a free slot in the table. The y axes
is the cost in number of hash table slots evaluated, and the x axes is the percentage
of slots in the table that are full. The mathematical equations for these curves can
be determined, but this is not so easy. A reasonable alternative is to write simple
variations on hashing. By timing the cost of the program for various loading condi-
tions, it is not difficult to construct a plot similar to Figure 9.8. The purpose of this
analysis is not to determine which approach to hashing is most efficient, so we are
not doing empirical comparison of hashing alternatives. Instead, the purpose is to
analyze the proper loading factor that would be used in an efficient hashing system
to balance time cost versus hash table size (space cost).
3.12 Further Reading
Pioneering works on algorithm analysis include The Art of Computer Programming
by Donald E. Knuth [Knu97, Knu98], and The Design and Analysis of Computer
Algorithms by Aho, Hopcroft, and Ullman [AHU74]. The alternate definition for
Ω comes from [AHU83]. The use of the notation “T(n) is in O(f(n))” rather
than the more commonly used “T(n) = O(f(n))” I derive from Brassard and
Bratley [BB96], though certainly this use predates them. A good book to read for
further information on algorithm analysis techniques is Compared to What? by
Gregory J.E. Rawlins [Raw92].
Bentley [Ben88] describes one problem in numerical analysis for which, be-
tween 1945 and 1988, the complexity of the best known algorithm had decreased
from O(n7) to O(n3). For a problem of size n = 64, this is roughly equivalent
to the speedup achieved from all advances in computer hardware during the same
time period.
While the most important aspect of program efficiency is the algorithm, much
improvement can be gained from efficient coding of a program. As cited by Freder-
ick P. Brooks in The Mythical Man-Month [Bro95], an efficient programmer can of-
ten produce programs that run five times faster than an inefficient programmer, even
when neither takes special efforts to speed up their code. For excellent and enjoy-
able essays on improving your coding efficiency, and ways to speed up your code
when it really matters, see the books by Jon Bentley [Ben82, Ben00, Ben88]. The](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-109-320.jpg)
![Sec. 3.13 Exercises 91
situation described in Example 3.18 arose when we were working on the project
reported on in [SU92].
As an interesting aside, writing a correct binary search algorithm is not easy.
Knuth [Knu98] notes that while the first binary search was published in 1946, the
first bug-free algorithm was not published until 1962! Bentley (“Writing Correct
Programs” in [Ben00]) has found that 90% of the computer professionals he tested
could not write a bug-free binary search in two hours.
3.13 Exercises
3.1 For each of the five expressions of Figure 3.1, give the range of values of n
for which that expression is most efficient.
3.2 Graph the following expressions. For each expression, state the range of
values of n for which that expression is the most efficient.
4n2
log3 n 3n
20n 2 log2 n n2/3
3.3 Arrange the following expressions by growth rate from slowest to fastest.
4n2
log3 n n! 3n
20n 2 log2 n n2/3
See Stirling’s approximation in Section 2.2 for help in classifying n!.
3.4 (a) Suppose that a particular algorithm has time complexity T(n) = 3 ×
2n, and that executing an implementation of it on a particular machine
takes t seconds for n inputs. Now suppose that we are presented with a
machine that is 64 times as fast. How many inputs could we process on
the new machine in t seconds?
(b) Suppose that another algorithm has time complexity T(n) = n2, and
that executing an implementation of it on a particular machine takes
t seconds for n inputs. Now suppose that we are presented with a ma-
chine that is 64 times as fast. How many inputs could we process on
the new machine in t seconds?
(c) A third algorithm has time complexity T(n) = 8n. Executing an im-
plementation of it on a particular machine takes t seconds for n inputs.
Given a new machine that is 64 times as fast, how many inputs could
we process in t seconds?
3.5 Hardware vendor XYZ Corp. claims that their latest computer will run 100
times faster than that of their competitor, Prunes, Inc. If the Prunes, Inc.
computer can execute a program on input of size n in one hour, what size](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-110-320.jpg)
![Sec. 3.13 Exercises 93
(b) sum = 0;
for (i=0; i<3; i++)
for (j=0; j<n; j++)
sum++;
(c) sum=0;
for (i=0; i<n*n; i++)
sum++;
(d) for (i=0; i < n-1; i++)
for (j=i+1; j < n; j++) {
tmp = AA[i][j];
AA[i][j] = AA[j][i];
AA[j][i] = tmp;
}
(e) sum = 0;
for (i=1; i<=n; i++)
for (j=1; j<=n; j*=2)
sum++;
(f) sum = 0;
for (i=1; i<=n; i*=2)
for (j=1; j<=n; j++)
sum++;
(g) Assume that array A contains n values, Random takes constant time,
and sort takes n log n steps.
for (i=0; i<n; i++) {
for (j=0; j<n; j++)
A[i] = DSutil.random(n);
sort(A);
}
(h) Assume array A contains a random permutation of the values from 0 to
n − 1.
sum = 0;
for (i=0; i<n; i++)
for (j=0; A[j]!=i; j++)
sum++;
(i) sum = 0;
if (EVEN(n))
for (i=0; i<n; i++)
sum++;
else
sum = sum + n;
3.13 Show that big-Theta notation (Θ) defines an equivalence relation on the set
of functions.](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-112-320.jpg)
![104 Chap. 4 Lists, Stacks, and Queues
/** Array-based list implementation */
class AList<E> implements List<E> {
private static final int defaultSize = 10; // Default size
private int maxSize; // Maximum size of list
private int listSize; // Number of list items now
private int curr; // Position current element
private E[] listArray; // Array holding list elements
/** Constructors */
/** Create a list with the default capacity. */
AList() { this(defaultSize); }
/** Create a new list object.
@param size Max number of elements list can contain. */
@SuppressWarnings("unchecked") // Generic array allocation
AList(int size) {
maxSize = size;
listSize = curr = 0;
listArray = (E[])new Object[size]; // Create listArray
}
public void clear() // Reinitialize the list
{ listSize = curr = 0; } // Simply reinitialize values
/** Insert "it" at current position */
public void insert(E it) {
assert listSize < maxSize : "List capacity exceeded";
for (int i=listSize; i>curr; i--) // Shift elements up
listArray[i] = listArray[i-1]; // to make room
listArray[curr] = it;
listSize++; // Increment list size
}
public void append(E it) { // Append "it"
assert listSize < maxSize : "List capacity exceeded";
listArray[listSize++] = it;
}
public void moveToStart() { curr = 0; } // Reset position
Figure 4.2 An array-based list implementation.
Figure 4.2 shows the array-based list implementation, named AList. AList
inherits from abstract class List and so must implement all of the member func-
tions of List.](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-123-320.jpg)
![Sec. 4.1 Lists 105
// Remove and return the current element.
public E remove() {
if ((curr<0) || (curr>=listSize)) // No current element
return null;
E it = listArray[curr]; // Copy the element
for(int i=curr; i<listSize-1; i++) // Shift them down
listArray[i] = listArray[i+1];
listSize--; // Decrement size
return it;
}
public void moveToEnd() { curr = listSize; } // Reset
public void prev() { if (curr != 0) curr--; } // Back up
public void next() { if (curr < listSize) curr++; } // Next
// Return list size
public int length() { return listSize; }
public int currPos() { // Return current position
return curr;
}
// Set current list position to "pos"
public void moveToPos(int pos) {
assert (pos>=0) && (pos<=listSize) : "Pos out of range";
curr = pos;
}
public E getValue() { // Return current element
assert (curr>=0)&&(curr<listSize) : "No current element";
return listArray[curr];
}
Figure 4.2 (continued)
Class AList’s private portion contains the data members for the array-based
list. These include listArray, the array which holds the list elements. Because
listArray must be allocated at some fixed size, the size of the array must be
known when the list object is created. Note that an optional parameter is declared
for the AList constructor. With this parameter, the user can indicate the maximum
number of elements permitted in the list. If no parameter is given, then it takes the
value DefaultListSize, which is assumed to be a suitably defined constant
value.
Because each list can have a differently sized array, each list must remember
its maximum permitted size. Data member maxSize serves this purpose. At any](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-124-320.jpg)
![Sec. 4.2 Stacks 127
/** Array-based stack implementation */
class AStack<E> implements Stack<E> {
private static final int defaultSize = 10;
private int maxSize; // Maximum size of stack
private int top; // Index for top Object
private E [] listArray; // Array holding stack
// Constructors
AStack() { this(defaultSize); }
@SuppressWarnings("unchecked") // Generic array allocation
AStack(int size) {
maxSize = size;
top = 0;
listArray = (E[])new Object[size]; // Create listArray
}
public void clear() // Reinitialize stack
{ top = 0; }
public void push(E it) { // Push "it" onto stack
assert top != maxSize : "Stack is full";
listArray[top++] = it;
}
public E pop() { // Pop top element
assert top != 0 : "Stack is empty";
return listArray[--top];
}
public E topValue() { // Return top element
assert top != 0 : "Stack is empty";
return listArray[top-1];
}
public int length() { return top; } // Return length
Figure 4.18 Array-based stack class implementation.](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-146-320.jpg)
![138 Chap. 4 Lists, Stacks, and Queues
// Array-based queue implementation
class AQueue<E> implements Queue<E> {
private static final int defaultSize = 10;
private int maxSize; // Maximum size of queue
private int front; // Index of front element
private int rear; // Index of rear element
private E[] listArray; // Array holding queue elements
// Constructors
AQueue() { this(defaultSize); }
@SuppressWarnings("unchecked") // For generic array
AQueue(int size) {
maxSize = size+1;
rear = 0; front = 1;
listArray = (E[])new Object[maxSize]; // Create listArray
}
public void clear() // Reinitialize
{ rear = 0; front = 1; }
public void enqueue(E it) { // Put "it" in queue
assert ((rear+2) % maxSize) != front : "Queue is full";
rear = (rear+1) % maxSize; // Circular increment
listArray[rear] = it;
}
public E dequeue() { // Take element out of queue
assert length() != 0 : "Queue is empty";
E it = listArray[front];
front = (front+1) % maxSize; // Circular increment
return it;
}
public E frontValue() { // Get front value
assert length() != 0 : "Queue is empty";
return listArray[front];
}
public int length() // Return length
{ return ((rear+maxSize) - front + 1) % maxSize; }
Figure 4.25 An array-based queue implementation.](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-157-320.jpg)
![Sec. 4.5 Further Reading 147
4.5 Further Reading
For more discussion on choice of functions used to define the List ADT, see the
work of the Reusable Software Research Group from Ohio State. Their definition
for the List ADT can be found in [SWH93]. More information about designing
such classes can be found in [SW94].
4.6 Exercises
4.1 Assume a list has the following configuration:
h | 2, 23, 15, 5, 9 i.
Write a series of Java statements using the List ADT of Figure 4.1 to delete
the element with value 15.
4.2 Show the list configuration resulting from each series of list operations using
the List ADT of Figure 4.1. Assume that lists L1 and L2 are empty at the
beginning of each series. Show where the current position is in the list.
(a) L1.append(10);
L1.append(20);
L1.append(15);
(b) L2.append(10);
L2.append(20);
L2.append(15);
L2.moveToStart();
L2.insert(39);
L2.next();
L2.insert(12);
4.3 Write a series of Java statements that uses the List ADT of Figure 4.1 to
create a list capable of holding twenty elements and which actually stores the
list with the following configuration:
h 2, 23 | 15, 5, 9 i.
4.4 Using the list ADT of Figure 4.1, write a function to interchange the current
element and the one following it.](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-166-320.jpg)
![182 Chap. 5 Binary Trees
import java.lang.Comparable;
/** Max-heap implementation */
public class MaxHeap<E extends Comparable<? super E>> {
private E[] Heap; // Pointer to the heap array
private int size; // Maximum size of the heap
private int n; // Number of things in heap
public MaxHeap(E[] h, int num, int max)
{ Heap = h; n = num; size = max; buildheap(); }
/** Return current size of the heap */
public int heapsize() { return n; }
/** Is pos a leaf position? */
public boolean isLeaf(int pos)
{ return (pos >= n/2) && (pos < n); }
/** Return position for left child of pos */
public int leftchild(int pos) {
assert pos < n/2 : "Position has no left child";
return 2*pos + 1;
}
/** Return position for right child of pos */
public int rightchild(int pos) {
assert pos < (n-1)/2 : "Position has no right child";
return 2*pos + 2;
}
/** Return position for parent */
public int parent(int pos) {
assert pos > 0 : "Position has no parent";
return (pos-1)/2;
}
/** Heapify contents of Heap */
public void buildheap()
{ for (int i=n/2-1; i>=0; i--) siftdown(i); }
/** Insert into heap */
public void insert(E val) {
assert n < size : "Heap is full";
int curr = n++;
Heap[curr] = val; // Start at end of heap
// Now sift up until curr’s parent’s key > curr’s key
while ((curr != 0) &&
(Heap[curr].compareTo(Heap[parent(curr)]) > 0)) {
DSutil.swap(Heap, curr, parent(curr));
curr = parent(curr);
}
}
Figure 5.19 Max-heap implementation.](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-201-320.jpg)
![Sec. 5.5 Heaps and Priority Queues 183
/** Put element in its correct place */
private void siftdown(int pos) {
assert (pos >= 0) && (pos < n) : "Illegal heap position";
while (!isLeaf(pos)) {
int j = leftchild(pos);
if ((j<(n-1)) && (Heap[j].compareTo(Heap[j+1]) < 0))
j++; // j is now index of child with greater value
if (Heap[pos].compareTo(Heap[j]) >= 0)
return;
DSutil.swap(Heap, pos, j);
pos = j; // Move down
}
}
public E removemax() { // Remove maximum value
assert n > 0 : "Removing from empty heap";
DSutil.swap(Heap, 0, --n); // Swap maximum with last value
if (n != 0) // Not on last element
siftdown(0); // Put new heap root val in correct place
return Heap[n];
}
/** Remove element at specified position */
public E remove(int pos) {
assert (pos >= 0) && (pos < n) : "Illegal heap position";
DSutil.swap(Heap, pos, --n); // Swap with last value
while (Heap[pos].compareTo(Heap[parent(pos)]) > 0) {
DSutil.swap(Heap, pos, parent(pos));
pos = parent(pos);
}
if (n != 0) siftdown(pos); // Push down
return Heap[n];
}
}
Figure 5.19 (continued)
This class definition makes two concessions to the fact that an array-based im-
plementation is used. First, heap nodes are indicated by their logical position within
the heap rather than by a pointer to the node. In practice, the logical heap position
corresponds to the identically numbered physical position in the array. Second, the
constructor takes as input a pointer to the array to be used. This approach provides
the greatest flexibility for using the heap because all data values can be loaded into
the array directly by the client. The advantage of this comes during the heap con-
struction phase, as explained below. The constructor also takes an integer parame-
ter indicating the initial size of the heap (based on the number of elements initially](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-202-320.jpg)
![Sec. 5.6 Huffman Coding Trees 197
Huffman coding does better when there is large variation in the frequencies of
letters. In the particular case of the frequencies shown in Figure 5.31, we can
determine the expected savings from Huffman coding if the actual frequencies of a
coded message match the expected frequencies.
Example 5.11 Because the sum of the frequencies in Figure 5.31 is 306
and E has frequency 120, we expect it to appear 120 times in a message
containing 306 letters. An actual message might or might not meet this
expectation. Letters D, L, and U have code lengths of three, and together
are expected to appear 121 times in 306 letters. Letter C has a code length of
four, and is expected to appear 32 times in 306 letters. Letter M has a code
length of five, and is expected to appear 24 times in 306 letters. Finally,
letters K and Z have code lengths of six, and together are expected to appear
only 9 times in 306 letters. The average expected cost per character is
simply the sum of the cost for each character (ci) times the probability of
its occurring (pi), or
c1p1 + c2p2 + · · · + cnpn.
This can be reorganized as
c1f1 + c2f2 + · · · + cnfn
fT
where fi is the (relative) frequency of letter i and fT is the total for all letter
frequencies. For this set of frequencies, the expected cost per letter is
[(1×120)+(3×121)+(4×32)+(5×24)+(6×9)]/306 = 785/306 ≈ 2.57
A fixed-length code for these eight characters would require log 8 = 3 bits
per letter as opposed to about 2.57 bits per letter for Huffman coding. Thus,
Huffman coding is expected to save about 14% for this set of letters.
Huffman coding for all ASCII symbols should do better than this. The letters of
Figure 5.31 are atypical in that there are too many common letters compared to the
number of rare letters. Huffman coding for all 26 letters would yield an expected
cost of 4.29 bits per letter. The equivalent fixed-length code would require about
five bits. This is somewhat unfair to fixed-length coding because there is actually
room for 32 codes in five bits, but only 26 letters. More generally, Huffman coding
of a typical text file will save around 40% over ASCII coding if we charge ASCII
coding at eight bits per character. Huffman coding for a binary file (such as a](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-216-320.jpg)
![198 Chap. 5 Binary Trees
compiled executable) would have a very different set of distribution frequencies and
so would have a different space savings. Most commercial compression programs
use two or three coding schemes to adjust to different types of files.
In the preceding example, “DEED” was coded in 8 bits, a saving of 33% over
the twelve bits required from a fixed-length coding. However, “MUCK” requires
18 bits, more space than required by the corresponding fixed-length coding. The
problem is that “MUCK” is composed of letters that are not expected to occur
often. If the message does not match the expected frequencies of the letters, than
the length of the encoding will not be as expected either.
5.7 Further Reading
See Shaffer and Brown [SB93] for an example of a tree implementation where an
internal node pointer field stores the value of its child instead of a pointer to its
child when the child is a leaf node.
Many techniques exist for maintaining reasonably balanced BSTs in the face of
an unfriendly series of insert and delete operations. One example is the AVL tree of
Adelson-Velskii and Landis, which is discussed by Knuth [Knu98]. The AVL tree
(see Section 13.2) is actually a BST whose insert and delete routines reorganize the
tree structure so as to guarantee that the subtrees rooted by the children of any node
will differ in height by at most one. Another example is the splay tree [ST85], also
discussed in Section 13.2.
See Bentley’s Programming Pearl “Thanks, Heaps” [Ben85, Ben88] for a good
discussion on the heap data structure and its uses.
The proof of Section 5.6.1 that the Huffman coding tree has minimum external
path weight is from Knuth [Knu97]. For more information on data compression
techniques, see Managing Gigabytes by Witten, Moffat, and Bell [WMB99], and
Codes and Cryptography by Dominic Welsh [Wel88]. Tables 5.23 and 5.24 are
derived from Welsh [Wel88].
5.8 Exercises
5.1 Section 5.1.1 claims that a full binary tree has the highest number of leaf
nodes among all trees with n internal nodes. Prove that this is true.
5.2 Define the degree of a node as the number of its non-empty children. Prove
by induction that the number of degree 2 nodes in any binary tree is one less
than the number of leaves.
5.3 Define the internal path length for a tree as the sum of the depths of all
internal nodes, while the external path length is the sum of the depths of all](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-217-320.jpg)
![Sec. 6.2 The Parent Pointer Implementation 211
/** General Tree class implementation for UNION/FIND */
class ParPtrTree {
private Integer [] array; // Node array
public ParPtrTree(int size) {
array = new Integer[size]; // Create node array
for (int i=0; i<size; i++)
array[i] = null;
}
/** Determine if nodes are in different trees */
public boolean differ(int a, int b) {
Integer root1 = FIND(a); // Find root of node a
Integer root2 = FIND(b); // Find root of node b
return root1 != root2; // Compare roots
}
/** Merge two subtrees */
public void UNION(int a, int b) {
Integer root1 = FIND(a); // Find root of node a
Integer root2 = FIND(b); // Find root of node b
if (root1 != root2) array[root2] = root1; // Merge
}
String print() {
StringBuffer out = new StringBuffer(100);
for (int i=0; i<array.length; i++) {
if (array[i] == null) out.append("-1 ");
else {
Integer temp = array[i];
for (int j=0; j<array.length; j++)
if (temp == array[j]) {
out.append(j + " ");
break;
}
}
}
return out.toString();
}
}
public Integer FIND(Integer curr) {
if (array[curr] == null) return curr; // At root
while (array[curr] != null) curr = array[curr];
return curr;
}
Figure 6.4 General tree implementation using parent pointers for the
UNION/FIND algorithm.](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-230-320.jpg)
![Sec. 6.2 The Parent Pointer Implementation 215
as to which node is set to be the root for the combined tree. In the case
of equivalence pair (E, G), the root of E is D while the root of G is F.
Because F is the root of the larger tree, node D is set to point to F.
Not all equivalences will combine two trees. If edge (F, G) is processed when
the representation is in the state shown in Figure 6.7(c), no change will be made
because F is already the root for G.
The weighted union rule helps to minimize the depth of the tree, but we can do
better than this. Path compression is a method that tends to create extremely shal-
low trees. Path compression takes place while finding the root for a given node X.
Call this root R. Path compression resets the parent of every node on the path from
X to R to point directly to R. This can be implemented by first finding R. A second
pass is then made along the path from X to R, assigning the parent field of each
node encountered to R. Alternatively, a recursive algorithm can be implemented as
follows. This version of FIND not only returns the root of the current node, but
also makes all ancestors of the current node point to the root.
public Integer FIND(Integer curr) {
if (array[curr] == null) return curr; // At root
array[curr] = FIND(array[curr]);
return array[curr];
}
Example 6.4 Figure 6.7(d) shows the result of processing equivalence
pair (H, E) on the the representation shown in Figure 6.7(c) using the stan-
dard weighted union rule without path compression. Figure 6.8 illustrates
the path compression process for the same equivalence pair. After locating
the root for node H, we can perform path compression to make H point
directly to root object A. Likewise, E is set to point directly to its root, F.
Finally, object A is set to point to root object F.
Note that path compression takes place during the FIND operation, not
during the merge operation. In Figure 6.8, this means that nodes B, C, and
H have node A remain as their parent, rather than changing their parent to
be F. While we might prefer to have these nodes point to F, to accomplish
this would require that additional information from the FIND operation be
passed back to the UNION operation. This would not be practical.
Path compression keeps the cost of each FIND operation very close to constant.
To be more precise about what is meant by “very close to constant,” the cost of path
compression for n FIND operations on n nodes (when combined with the weighted](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-234-320.jpg)
![226 Chap. 6 Non-Binary Trees
will use the “)” symbol) to indicate the end of a child list. All leaf nodes are fol-
lowed by a “)” symbol because they have no children. A leaf node that is also the
last child for its parent would indicate this by two or more successive “)” symbols.
Example 6.8 For the general tree of Figure 6.3, we get the sequential
representation
RAC)D)E))BF))) (6.4)
Note that F is followed by three “)” marks, because it is a leaf, the last node
of B’s rightmost subtree, and the last node of R’s rightmost subtree.
Note that this representation for serializing general trees cannot be used for bi-
nary trees. This is because a binary tree is not merely a restricted form of general
tree with at most two children. Every binary tree node has a left and a right child,
though either or both might be empty. For example, the representation of Exam-
ple 6.8 cannot let us distinguish whether node D in Figure 6.17 is the left or right
child of node B.
6.6 Further Reading
The expression log∗
n cited in Section 6.2 is closely related to the inverse of Ack-
ermann’s function. For more information about Ackermann’s function and the cost
of path compression, see Robert E. Tarjan’s paper “On the efficiency of a good but
not linear set merging algorithm” [Tar75]. The article “Data Structures and Algo-
rithms for Disjoint Set Union Problems” by Galil and Italiano [GI91] covers many
aspects of the equivalence class problem.
Foundations of Multidimensional and Metric Data Structures by Hanan Samet
[Sam06] treats various implementations of tree structures in detail within the con-
text of K-ary trees. Samet covers sequential implementations as well as the linked
and array implementations such as those described in this chapter and Chapter 5.
While these books are ostensibly concerned with spatial data structures, many of
the concepts treated are relevant to anyone who must implement tree structures.
6.7 Exercises
6.1 Write an algorithm to determine if two general trees are identical. Make the
algorithm as efficient as you can. Analyze your algorithm’s running time.](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-245-320.jpg)
![238 Chap. 7 Internal Sorting
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Figure 7.1 An illustration of Insertion Sort. Each column shows the array after
the iteration with the indicated value of i in the outer for loop. Values above
the line in each column have been sorted. Arrows indicate the upward motions of
records through the array.
7.2.1 Insertion Sort
Imagine that you have a stack of phone bills from the past two years and that you
wish to organize them by date. A fairly natural way to do this might be to look at
the first two bills and put them in order. Then take the third bill and put it into the
right order with respect to the first two, and so on. As you take each bill, you would
add it to the sorted pile that you have already made. This naturally intuitive process
is the inspiration for our first sorting algorithm, called Insertion Sort. Insertion
Sort iterates through a list of records. Each record is inserted in turn at the correct
position within a sorted list composed of those records already processed. The
following is a Java implementation. The input is an array of n records stored in
array A.
static <E extends Comparable<? super E>> void Sort(E[] A) {
for (int i=1; i<A.length; i++) // Insert i’th record
for (int j=i; (j>0) && (A[j].compareTo(A[j-1])<0); j--)
DSutil.swap(A, j, j-1);
}
Consider the case where inssort is processing the ith record, which has key
value X. The record is moved upward in the array as long as X is less than the
key value immediately above it. As soon as a key value less than or equal to X is
encountered, inssort is done with that record because all records above it in the
array must have smaller keys. Figure 7.1 illustrates how Insertion Sort works.
The body of inssort is made up of two nested for loops. The outer for
loop is executed n − 1 times. The inner for loop is harder to analyze because
the number of times it executes depends on how many keys in positions 1 to i − 1
have a value less than that of the key in position i. In the worst case, each record](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-257-320.jpg)
![240 Chap. 7 Internal Sorting
i=0 1 2 3 4 5 6
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Figure 7.2 An illustration of Bubble Sort. Each column shows the array after
the iteration with the indicated value of i in the outer for loop. Values above the
line in each column have been sorted. Arrows indicate the swaps that take place
during a given iteration.
7.2.2 Bubble Sort
Our next sort is called Bubble Sort. Bubble Sort is often taught to novice pro-
grammers in introductory computer science courses. This is unfortunate, because
Bubble Sort has no redeeming features whatsoever. It is a relatively slow sort, it
is no easier to understand than Insertion Sort, it does not correspond to any in-
tuitive counterpart in “everyday” use, and it has a poor best-case running time.
However, Bubble Sort serves as the basis for a better sort that will be presented in
Section 7.2.3.
Bubble Sort consists of a simple double for loop. The first iteration of the
inner for loop moves through the record array from bottom to top, comparing
adjacent keys. If the lower-indexed key’s value is greater than its higher-indexed
neighbor, then the two values are swapped. Once the smallest value is encountered,
this process will cause it to “bubble” up to the top of the array. The second pass
through the array repeats this process. However, because we know that the smallest
value reached the top of the array on the first pass, there is no need to compare
the top two elements on the second pass. Likewise, each succeeding pass through
the array compares adjacent elements, looking at one less value than the preceding
pass. Figure 7.2 illustrates Bubble Sort. A Java implementation is as follows:
static <E extends Comparable<? super E>> void Sort(E[] A) {
for (int i=0; i<A.length-1; i++) // Bubble up i’th record
for (int j=A.length-1; j>i; j--)
if ((A[j].compareTo(A[j-1]) < 0))
DSutil.swap(A, j, j-1);
}](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-259-320.jpg)
![Sec. 7.2 Three Θ(n2) Sorting Algorithms 241
Determining Bubble Sort’s number of comparisons is easy. Regardless of the
arrangement of the values in the array, the number of comparisons made by the
inner for loop is always i, leading to a total cost of
n
X
i=1
i = Θ(n2).
Bubble Sort’s running time is roughly the same in the best, average, and worst
cases.
The number of swaps required depends on how often a value is less than the
one immediately preceding it in the array. We can expect this to occur for about
half the comparisons in the average case, leading to Θ(n2) for the expected number
of swaps. The actual number of swaps performed by Bubble Sort will be identical
to that performed by Insertion Sort.
7.2.3 Selection Sort
Consider again the problem of sorting a pile of phone bills for the past year. An-
other intuitive approach might be to look through the pile until you find the bill for
January, and pull that out. Then look through the remaining pile until you find the
bill for February, and add that behind January. Proceed through the ever-shrinking
pile of bills to select the next one in order until you are done. This is the inspiration
for our last Θ(n2) sort, called Selection Sort. The ith pass of Selection Sort “se-
lects” the ith smallest key in the array, placing that record into position i. In other
words, Selection Sort first finds the smallest key in an unsorted list, then the second
smallest, and so on. Its unique feature is that there are few record swaps. To find
the next smallest key value requires searching through the entire unsorted portion
of the array, but only one swap is required to put the record in place. Thus, the total
number of swaps required will be n − 1 (we get the last record in place “for free”).
Figure 7.3 illustrates Selection Sort. Below is a Java implementation.
static <E extends Comparable<? super E>> void Sort(E[] A) {
for (int i=0; i<A.length-1; i++) { // Select i’th record
int lowindex = i; // Remember its index
for (int j=A.length-1; j>i; j--) // Find the least value
if (A[j].compareTo(A[lowindex]) < 0)
lowindex = j; // Put it in place
DSutil.swap(A, i, lowindex);
}
}
Selection Sort is essentially a Bubble Sort, except that rather than repeatedly
swapping adjacent values to get the next smallest record into place, we instead](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-260-320.jpg)
![Sec. 7.3 Shellsort 245
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Figure 7.6 An example of Shellsort. Sixteen items are sorted in four passes.
The first pass sorts 8 sublists of size 2 and increment 8. The second pass sorts
4 sublists of size 4 and increment 4. The third pass sorts 2 sublists of size 8 and
increment 2. The fourth pass sorts 1 list of size 16 and increment 1 (a regular
Insertion Sort).
static <E extends Comparable<? super E>> void Sort(E[] A) {
for (int i=A.length/2; i>2; i/=2) // For each increment
for (int j=0; j<i; j++) // Sort each sublist
inssort2(A, j, i);
inssort2(A, 0, 1); // Could call regular inssort here
}
// Modified version of Insertion Sort for varying increments
static <E extends Comparable<? super E>>
void inssort2(E[] A, int start, int incr) {
for (int i=start+incr; i<A.length; i+=incr)
for (int j=i; (j>=incr)&&
(A[j].compareTo(A[j-incr])<0); j-=incr)
DSutil.swap(A, j, j-incr);
}
Shellsort will work correctly regardless of the size of the increments, provided
that the final pass has increment 1 (i.e., provided the final pass is a regular Insertion
Sort). If Shellsort will aways conclude with a regular Insertion Sort, then how
can it be any improvement on Insertion Sort? The expectation is that each of the
(relatively cheap) sublist sorts will make the list “more sorted” than it was before.
It is not necessarily the case that this will be true, but it is almost always true in
practice. When the final Insertion Sort is conducted, the list should be “almost
sorted,” yielding a relatively cheap final Insertion Sort pass.](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-264-320.jpg)
![248 Chap. 7 Internal Sorting
subarrays of size 2, which are in turn merged into subarrays of size 4, and so on.
We need to avoid having each merge operation require a new array. With some
difficulty, an algorithm can be devised that alternates between two arrays. A much
simpler approach is to copy the sorted sublists to the auxiliary array first, and then
merge them back to the original array. Here is a complete implementation for
mergesort following this approach:
static <E extends Comparable<? super E>>
void mergesort(E[] A, E[] temp, int l, int r) {
int mid = (l+r)/2; // Select midpoint
if (l == r) return; // List has one element
mergesort(A, temp, l, mid); // Mergesort first half
mergesort(A, temp, mid+1, r); // Mergesort second half
for (int i=l; i<=r; i++) // Copy subarray to temp
temp[i] = A[i];
// Do the merge operation back to A
int i1 = l; int i2 = mid + 1;
for (int curr=l; curr<=r; curr++) {
if (i1 == mid+1) // Left sublist exhausted
A[curr] = temp[i2++];
else if (i2 > r) // Right sublist exhausted
A[curr] = temp[i1++];
else if (temp[i1].compareTo(temp[i2])<0) // Get smaller
A[curr] = temp[i1++];
else A[curr] = temp[i2++];
}
}
An optimized Mergesort implementation is shown next. It reverses the order of
the second subarray during the initial copy. Now the current positions of the two
subarrays work inwards from the ends, allowing the end of each subarray to act as
a sentinel for the other. Unlike the previous implementation, no test is needed to
check for when one of the two subarrays becomes empty. This version also uses
Insertion Sort to sort small subarrays.](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-267-320.jpg)
![Sec. 7.5 Quicksort 249
static <E extends Comparable<? super E>>
void mergesort(E[] A, E[] temp, int l, int r) {
int i, j, k, mid = (l+r)/2; // Select the midpoint
if (l == r) return; // List has one element
if ((mid-l) >= THRESHOLD) mergesort(A, temp, l, mid);
else inssort(A, l, mid-l+1);
if ((r-mid) > THRESHOLD) mergesort(A, temp, mid+1, r);
else inssort(A, mid+1, r-mid);
// Do the merge operation. First, copy 2 halves to temp.
for (i=l; i<=mid; i++) temp[i] = A[i];
for (j=1; j<=r-mid; j++) temp[r-j+1] = A[j+mid];
// Merge sublists back to array
for (i=l,j=r,k=l; k<=r; k++)
if (temp[i].compareTo(temp[j])<0) A[k] = temp[i++];
else A[k] = temp[j--];
}
Analysis of Mergesort is straightforward, despite the fact that it is a recursive
algorithm. The merging part takes time Θ(i) where i is the total length of the two
subarrays being merged. The array to be sorted is repeatedly split in half until
subarrays of size 1 are reached, at which time they are merged to be of size 2,
these merged to subarrays of size 4, and so on as shown in Figure 7.7. Thus, the
depth of the recursion is log n for n elements (assume for simplicity that n is a
power of two). The first level of recursion can be thought of as working on one
array of size n, the next level working on two arrays of size n/2, the next on four
arrays of size n/4, and so on. The bottom of the recursion has n arrays of size 1.
Thus, n arrays of size 1 are merged (requiring n total steps), n/2 arrays of size 2
(again requiring n total steps), n/4 arrays of size 4, and so on. At each of the log n
levels of recursion, Θ(n) work is done, for a total cost of Θ(n log n). This cost is
unaffected by the relative order of the values being sorted, thus this analysis holds
for the best, average, and worst cases.
7.5 Quicksort
While Mergesort uses the most obvious form of divide and conquer (split the list in
half then sort the halves), it is not the only way that we can break down the sorting
problem. And we saw that doing the merge step for Mergesort when using an array
implementation is not so easy. So prehaps a different divide and conquer strategy
might turn out to be more efficient?
Quicksort is aptly named because, when properly implemented, it is the fastest
known general-purpose in-memory sorting algorithm in the average case. It does
not require the extra array needed by Mergesort, so it is space efficent as well.
Quicksort is widely used, and is typically the algorithm implemented in a library](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-268-320.jpg)
![Sec. 7.5 Quicksort 251
static <E extends Comparable<? super E>>
void qsort(E[] A, int i, int j) { // Quicksort
int pivotindex = findpivot(A, i, j); // Pick a pivot
DSutil.swap(A, pivotindex, j); // Stick pivot at end
// k will be the first position in the right subarray
int k = partition(A, i-1, j, A[j]);
DSutil.swap(A, k, j); // Put pivot in place
if ((k-i) > 1) qsort(A, i, k-1); // Sort left partition
if ((j-k) > 1) qsort(A, k+1, j); // Sort right partition
}
Function partition will move records to the appropriate partition and then
return k, the first position in the right partition. Note that the pivot value is initially
placed at the end of the array (position j). Thus, partition must not affect the
value of array position j. After partitioning, the pivot value is placed in position k,
which is its correct position in the final, sorted array. By doing so, we guarantee
that at least one value (the pivot) will not be processed in the recursive calls to
qsort. Even if a bad pivot is selected, yielding a completely empty partition to
one side of the pivot, the larger partition will contain at most n − 1 elements.
Selecting a pivot can be done in many ways. The simplest is to use the first
key. However, if the input is sorted or reverse sorted, this will produce a poor
partitioning with all values to one side of the pivot. It is better to pick a value
at random, thereby reducing the chance of a bad input order affecting the sort.
Unfortunately, using a random number generator is relatively expensive, and we
can do nearly as well by selecting the middle position in the array. Here is a simple
findpivot function:
static <E extends Comparable<? super E>>
int findpivot(E[] A, int i, int j)
{ return (i+j)/2; }
We now turn to function partition. If we knew in advance how many keys
are less than the pivot, partition could simply copy elements with key values
less than the pivot to the low end of the array, and elements with larger keys to the
high end. Because we do not know in advance how many keys are less than the
pivot, we use a clever algorithm that moves indices inwards from the ends of the
subarray, swapping values as necessary until the two indices meet. Here is a Java
implementation for the partition step:](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-270-320.jpg)
![252 Chap. 7 Internal Sorting
static <E extends Comparable<? super E>>
int partition(E[] A, int l, int r, E pivot) {
do { // Move bounds inward until they meet
while (A[++l].compareTo(pivot)<0);
while ((r!=0) && (A[--r].compareTo(pivot)>0));
DSutil.swap(A, l, r); // Swap out-of-place values
} while (l < r); // Stop when they cross
DSutil.swap(A, l, r); // Reverse last, wasted swap
return l; // Return first position in right partition
}
Figure 7.8 illustrates partition. Initially, variables l and r are immediately
outside the actual bounds of the subarray being partitioned. Each pass through the
outer do loop moves the counters l and r inwards, until eventually they meet.
Note that at each iteration of the inner while loops, the bounds are moved prior
to checking against the pivot value. This ensures that progress is made by each
while loop, even when the two values swapped on the last iteration of the do
loop were equal to the pivot. Also note the check that r > l in the second while
loop. This ensures that r does not run off the low end of the partition in the case
where the pivot is the least value in that partition. Function partition returns
the first index of the right partition so that the subarray bound for the recursive
calls to qsort can be determined. Figure 7.9 illustrates the complete Quicksort
algorithm.
To analyze Quicksort, we first analyze the findpivot and partition
functions operating on a subarray of length k. Clearly, findpivot takes con-
stant time. Function partition contains a do loop with two nested while
loops. The total cost of the partition operation is constrained by how far l and r
can move inwards. In particular, these two bounds variables together can move a
total of s steps for a subarray of length s. However, this does not directly tell us
how much work is done by the nested while loops. The do loop as a whole is
guaranteed to move both l and r inward at least one position on each first pass.
Each while loop moves its variable at least once (except in the special case where
r is at the left edge of the array, but this can happen only once). Thus, we see that
the do loop can be executed at most s times, the total amount of work done moving
l and r is s, and each while loop can fail its test at most s times. The total work
for the entire partition function is therefore Θ(s).
Knowing the cost of findpivot and partition, we can determine the
cost of Quicksort. We begin with a worst-case analysis. The worst case will occur
when the pivot does a poor job of breaking the array, that is, when there are no
elements in one partition, and n − 1 elements in the other. In this case, the divide
and conquer strategy has done a poor job of dividing, so the conquer phase will
work on a subproblem only one less than the size of the original problem. If this](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-271-320.jpg)
![254 Chap. 7 Internal Sorting
happens at each partition step, then the total cost of the algorithm will be
n
X
k=1
k = Θ(n2).
In the worst case, Quicksort is Θ(n2). This is terrible, no better than Bubble
Sort.2 When will this worst case occur? Only when each pivot yields a bad parti-
tioning of the array. If the pivot values are selected at random, then this is extremely
unlikely to happen. When selecting the middle position of the current subarray, it
is still unlikely to happen. It does not take many good partitionings for Quicksort
to work fairly well.
Quicksort’s best case occurs when findpivot always breaks the array into
two equal halves. Quicksort repeatedly splits the array into smaller partitions, as
shown in Figure 7.9. In the best case, the result will be log n levels of partitions,
with the top level having one array of size n, the second level two arrays of size n/2,
the next with four arrays of size n/4, and so on. Thus, at each level, all partition
steps for that level do a total of n work, for an overall cost of n log n work when
Quicksort finds perfect pivots.
Quicksort’s average-case behavior falls somewhere between the extremes of
worst and best case. Average-case analysis considers the cost for all possible ar-
rangements of input, summing the costs and dividing by the number of cases. We
make one reasonable simplifying assumption: At each partition step, the pivot is
equally likely to end in any position in the (sorted) array. In other words, the pivot
is equally likely to break an array into partitions of sizes 0 and n−1, or 1 and n−2,
and so on.
Given this assumption, the average-case cost is computed from the following
equation:
T(n) = cn +
1
n
n−1
X
k=0
[T(k) + T(n − 1 − k)], T(0) = T(1) = c.
This equation is in the form of a recurrence relation. Recurrence relations are
discussed in Chapters 2 and 14, and this one is solved in Section 14.2.4. This
equation says that there is one chance in n that the pivot breaks the array into
subarrays of size 0 and n − 1, one chance in n that the pivot breaks the array into
subarrays of size 1 and n − 2, and so on. The expression “T(k) + T(n − 1 − k)” is
the cost for the two recursive calls to Quicksort on two arrays of size k and n−1−k.
2
The worst insult that I can think of for a sorting algorithm.](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-273-320.jpg)
![Sec. 7.6 Heapsort 257
and fast. The algorithm should take advantage of the fact that sorting is a special-
purpose application in that all of the values to be stored are available at the start.
This means that we do not necessarily need to insert one value at a time into the
tree structure.
Heapsort is based on the heap data structure presented in Section 5.5. Heapsort
has all of the advantages just listed. The complete binary tree is balanced, its array
representation is space efficient, and we can load all values into the tree at once,
taking advantage of the efficient buildheap function. The asymptotic perfor-
mance of Heapsort is Θ(n log n) in the best, average, and worst cases. It is not as
fast as Quicksort in the average case (by a constant factor), but Heapsort has special
properties that will make it particularly useful when sorting data sets too large to fit
in main memory, as discussed in Chapter 8.
A sorting algorithm based on max-heaps is quite straightforward. First we use
the heap building algorithm of Section 5.5 to convert the array into max-heap order.
Then we repeatedly remove the maximum value from the heap, restoring the heap
property each time that we do so, until the heap is empty. Note that each time we
remove the maximum element from the heap, it is placed at the end of the array.
Assume the n elements are stored in array positions 0 through n−1. After removing
the maximum value from the heap and readjusting, the maximum value will now
be placed in position n − 1 of the array. The heap is now considered to be of size
n − 1. Removing the new maximum (root) value places the second largest value
in position n − 2 of the array. At the end of the process, the array will be properly
sorted from least to greatest. This is why Heapsort uses a max-heap rather than
a min-heap as might have been expected. Figure 7.10 illustrates Heapsort. The
complete Java implementation is as follows:
static <E extends Comparable<? super E>>
void heapsort(E[] A) { // Heapsort
MaxHeap<E> H = new MaxHeap<E>(A, A.length, A.length);
for (int i=0; i<A.length; i++) // Now sort
H.removemax(); // Removemax places max at end of heap
}
Because building the heap takes Θ(n) time (see Section 5.5), and because
n deletions of the maximum element each take Θ(log n) time, we see that the en-
tire heapsort operation takes Θ(n log n) time in the worst, average, and best cases.
While typically slower than Quicksort by a constant factor, Heapsort has one spe-
cial advantage over the other sorts studied so far. Building the heap is relatively
cheap, requiring Θ(n) time. Removing the maximum element from the heap re-
quires Θ(log n) time. Thus, if we wish to find the k largest elements in an array,
we can do so in time Θ(n+k log n). If k is small, this is a substantial improvement](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-276-320.jpg)
![Sec. 7.7 Binsort and Radix Sort 259
over the time required to find the k largest elements using one of the other sorting
methods described earlier. One situation where we are able to take advantage of this
concept is in the implementation of Kruskal’s minimum-cost spanning tree (MST)
algorithm of Section 11.5.2. That algorithm requires that edges be visited in as-
cending order (so, use a min-heap), but this process stops as soon as the MST is
complete. Thus, only a relatively small fraction of the edges need be sorted.
7.7 Binsort and Radix Sort
Imagine that for the past year, as you paid your various bills, you then simply piled
all the paperwork onto the top of a table somewhere. Now the year has ended and
its time to sort all of these papers by what the bill was for (phone, electricity, rent,
etc.) and date. A pretty natural approach is to make some space on the floor, and as
you go through the pile of papers, put the phone bills into one pile, the electric bills
into another pile, and so on. Once this initial assignment of bills to piles is done (in
one pass), you can sort each pile by date relatively quickly because they are each
fairly small. This is the basic idea behind a Binsort.
Section 3.9 presented the following code fragment to sort a permutation of the
numbers 0 through n − 1:
for (i=0; i<n; i++)
B[A[i]] = A[i];
Here the key value is used to determine the position for a record in the final sorted
array. This is the most basic example of a Binsort, where key values are used
to assign records to bins. This algorithm is extremely efficient, taking Θ(n) time
regardless of the initial ordering of the keys. This is far better than the performance
of any sorting algorithm that we have seen so far. The only problem is that this
algorithm has limited use because it works only for a permutation of the numbers
from 0 to n − 1.
We can extend this simple Binsort algorithm to be more useful. Because Binsort
must perform direct computation on the key value (as opposed to just asking which
of two records comes first as our previous sorting algorithms did), we will assume
that the records use an integer key type.
The simplest extension is to allow for duplicate values among the keys. This
can be done by turning array slots into arbitrary-length bins by turning B into an
array of linked lists. In this way, all records with key value i can be placed in bin
B[i]. A second extension allows for a key range greater than n. For example,
a set of n records might have keys in the range 1 to 2n. The only requirement is](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-278-320.jpg)
![260 Chap. 7 Internal Sorting
that each possible key value have a corresponding bin in B. The extended Binsort
algorithm is as follows:
static void binsort(Integer A[]) {
List<Integer>[] B = (LList<Integer>[])new LList[MaxKey];
Integer item;
for (int i=0; i<MaxKey; i++)
B[i] = new LList<Integer>();
for (int i=0; i<A.length; i++) B[A[i]].append(A[i]);
for (int i=0; i<MaxKey; i++)
for (B[i].moveToStart();
(item = B[i].getValue()) != null; B[i].next())
output(item);
}
This version of Binsort can sort any collection of records whose key values fall
in the range from 0 to MaxKeyValue−1. The total work required is simply that
needed to place each record into the appropriate bin and then take all of the records
out of the bins. Thus, we need to process each record twice, for Θ(n) work.
Unfortunately, there is a crucial oversight in this analysis. Binsort must also
look at each of the bins to see if it contains a record. The algorithm must process
MaxKeyValue bins, regardless of how many actually hold records. If MaxKey-
Value is small compared to n, then this is not a great expense. Suppose that
MaxKeyValue = n2. In this case, the total amount of work done will be Θ(n +
n2) = Θ(n2). This results in a poor sorting algorithm, and the algorithm becomes
even worse as the disparity between n and MaxKeyValue increases. In addition,
a large key range requires an unacceptably large array B. Thus, even the extended
Binsort is useful only for a limited key range.
A further generalization to Binsort yields a bucket sort. Each bin is associated
with not just one key, but rather a range of key values. A bucket sort assigns records
to bins and then relies on some other sorting technique to sort the records within
each bin. The hope is that the relatively inexpensive bucketing process will put
only a small number of records in each bin, and that a “cleanup sort” within the
bins will then be relatively cheap.
There is a way to keep the number of bins and the related processing small
while allowing the cleanup sort to be based on Binsort. Consider a sequence of
records with keys in the range 0 to 99. If we have ten bins available, we can first
assign records to bins by taking their key value modulo 10. Thus, every key will
be assigned to the bin matching its rightmost decimal digit. We can then take these
records from the bins in order and reassign them to the bins on the basis of their
leftmost (10’s place) digit (define values in the range 0 to 9 to have a leftmost digit](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-279-320.jpg)
![Sec. 7.7 Binsort and Radix Sort 261
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
Result of first pass: 91 1 72 23 84 5 25 27 97 17 67 28
Result of second pass: 1 17
5 23 25 27 28 67 72 84 91 97
First pass
(on right digit)
Second pass
(on left digit)
Initial List: 27 91 1 97 17 23 84 28 72 5 67 25
23
84
5 25
27
28
91 1
72
97 17 67
17
91 97
72
84
1 5
23 25
67
27 28
Figure 7.11 An example of Radix Sort for twelve two-digit numbers in base ten.
Two passes are required to sort the list.
of 0). In other words, assign the ith record from array A to a bin using the formula
A[i]/10. If we now gather the values from the bins in order, the result is a sorted
list. Figure 7.11 illustrates this process.
In this example, we have r = 10 bins and n = 12 keys in the range 0 to r2 − 1.
The total computation is Θ(n), because we look at each record and each bin a
constant number of times. This is a great improvement over the simple Binsort
where the number of bins must be as large as the key range. Note that the example
uses r = 10 so as to make the bin computations easy to visualize: Records were
placed into bins based on the value of first the rightmost and then the leftmost
decimal digits. Any number of bins would have worked. This is an example of a
Radix Sort, so called because the bin computations are based on the radix or the
base of the key values. This sorting algorithm can be extended to any number of
keys in any key range. We simply assign records to bins based on the keys’ digit
values working from the rightmost digit to the leftmost. If there are k digits, then
this requires that we assign keys to bins k times.](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-280-320.jpg)
![262 Chap. 7 Internal Sorting
As with Mergesort, an efficient implementation of Radix Sort is somewhat dif-
ficult to achieve. In particular, we would prefer to sort an array of values and avoid
processing linked lists. If we know how many values will be in each bin, then an
auxiliary array of size r can be used to hold the bins. For example, if during the first
pass the 0 bin will receive three records and the 1 bin will receive five records, then
we could simply reserve the first three array positions for the 0 bin and the next five
array positions for the 1 bin. Exactly this approach is taken by the following Java
implementation. At the end of each pass, the records are copied back to the original
array.
static void radix(Integer[] A, Integer[] B,
int k, int r, int[] count) {
// Count[i] stores number of records in bin[i]
int i, j, rtok;
for (i=0, rtok=1; i<k; i++, rtok*=r) { // For k digits
for (j=0; j<r; j++) count[j] = 0; // Initialize count
// Count the number of records for each bin on this pass
for (j=0; j<A.length; j++) count[(A[j]/rtok)%r]++;
// count[j] will be index in B for last slot of bin j.
for (j=1; j<r; j++) count[j] = count[j-1] + count[j];
// Put records into bins, working from bottom of bin
// Since bins fill from bottom, j counts downwards
for (j=A.length-1; j>=0; j--)
B[--count[(A[j]/rtok)%r]] = A[j];
for (j=0; j<A.length; j++) A[j] = B[j]; // Copy B back
}
}
The first inner for loop initializes array cnt. The second loop counts the
number of records to be assigned to each bin. The third loop sets the values in cnt
to their proper indices within array B. Note that the index stored in cnt[j] is the
last index for bin j; bins are filled from the bottom. The fourth loop assigns the
records to the bins (within array B). The final loop simply copies the records back
to array A to be ready for the next pass. Variable rtoi stores ri for use in bin
computation on the i’th iteration. Figure 7.12 shows how this algorithm processes
the input shown in Figure 7.11.
This algorithm requires k passes over the list of n numbers in base r, with
Θ(n + r) work done at each pass. Thus the total work is Θ(nk + rk). What is
this in terms of n? Because r is the size of the base, it might be rather small.](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-281-320.jpg)
![Sec. 7.7 Binsort and Radix Sort 263
First pass values for Count.
Count array:
Index positions for Array B.
End of Pass 1: Array A.
rtoi = 1.
Second pass values for Count.
rtoi = 10.
Count array:
Index positions for Array B.
End of Pass 2: Array A.
0 1 2 3 4 5 6 7 8 9
0 1 2 3 4 5 6 7 8 9
Initial Input: Array A
91 23 84 25 27 97 17 67 28
72
91 1 97 17 23 84 28 72 5 67 25
27
11 12 12
2 3 4 5 7 7
0
1 5
1 2 3 4 5 6 7 8 9
0
12
10
0 1 2 3 4 5 6 7 8 9
17 23 25 27 28 67 72 84 91 97
5
1
2 1 1 1 2 0 4 1 0
0
2
1
1
1
0
4
1 0
0
2
9
8
7 7
7
7
3
2
Figure 7.12 An example showing function radix applied to the input of Fig-
ure 7.11. Row 1 shows the initial values within the input array. Row 2 shows the
values for array cnt after counting the number of records for each bin. Row 3
shows the index values stored in array cnt. For example, cnt[0] is 0, indicat-
ing no input values are in bin 0. Cnt[1] is 2, indicating that array B positions 0
and 1 will hold the values for bin 1. Cnt[2] is 3, indicating that array B position
2 will hold the (single) value for bin 2. Cnt[7] is 11, indicating that array B
positions 7 through 10 will hold the four values for bin 7. Row 4 shows the results
of the first pass of the Radix Sort. Rows 5 through 7 show the equivalent steps for
the second pass.](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-282-320.jpg)
![Sec. 7.9 Lower Bounds for Sorting 269
Yes No
Yes No Yes No
Yes No Yes No
A[1]<A[0]?
A[2]<A[1]? A[2]<A[1]?
A[1]<A[0]?
A[1]<A[0]?
(Y<X?)
(Z<X?) (Z<Y?)
(Z<X?)
(Z<Y?)
XYZ
ZYX YZX
XYZ
XZY
YXZ
YZX
ZXY
ZYX
YXZ
YXZ
YZX
ZYX
XZY
YXZ
YZX
YZX
ZYX
XYZ
XYZ
XZY
ZXY
XZY
ZXY
XYZ
ZXY XZY
Figure 7.14 Decision tree for Insertion Sort when processing three values la-
beled X, Y, and Z, initially stored at positions 0, 1, and 2, respectively, in input
array A.
initially stored in positions 0, 1, and 2, respectively, of input array A. Consider the
possible outputs. Initially, we know nothing about the final positions of the three
values in the sorted output array. The correct output could be any permutation of
the input values. For three values, there are n! = 6 permutations. Thus, the root
node of the decision tree lists all six permutations that might be the eventual result
of the algorithm.
When n = 3, the first comparison made by Insertion Sort is between the sec-
ond item in the input array (Y) and the first item in the array (X). There are two
possibilities: Either the value of Y is less than that of X, or the value of Y is not
less than that of X. This decision is modeled by the first branch in the tree. If Y is
less than X, then the left branch should be taken and Y must appear before X in the
final output. Only three of the original six permutations have this property, so the
left child of the root lists the three permutations where Y appears before X: YXZ,
YZX, and ZYX. Likewise, if Y were not less than X, then the right branch would
be taken, and only the three permutations in which Y appears after X are possible
outcomes: XYZ, XZY, and ZXY. These are listed in the right child of the root.
Let us assume for the moment that Y is less than X and so the left branch is
taken. In this case, Insertion Sort swaps the two values. At this point the array
stores YXZ. Thus, in Figure 7.14 the left child of the root shows YXZ above the
line. Next, the third value in the array is compared against the second (i.e., Z is](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-288-320.jpg)
![Sec. 7.10 Further Reading 271
• A binary tree of height n can store at most 2n − 1 nodes.
• Equivalently, a tree with n nodes requires at least ⌈log(n + 1)⌉ levels.
What is the minimum number of nodes that must be in the decision tree for any
comparison-based sorting algorithm for n values? Because sorting algorithms are
in the business of determining which unique permutation of the input corresponds
to the sorted list, all sorting algorithms must contain at least one leaf node for
each possible permutation. There are n! permutations for a set of n numbers (see
Section 2.2).
Because there are at least n! nodes in the tree, we know that the tree must
have Ω(log n!) levels. From Stirling’s approximation (Section 2.2), we know log n!
is in Ω(n log n). The decision tree for any comparison-based sorting algorithm
must have nodes Ω(n log n) levels deep. Thus, in the worst case, any such sorting
algorithm must require Ω(n log n) comparisons.
Any sorting algorithm requiring Ω(n log n) comparisons in the worst case re-
quires Ω(n log n) running time in the worst case. Because any sorting algorithm
requires Ω(n log n) running time, the problem of sorting also requires Ω(n log n)
time. We already know of sorting algorithms with O(n log n) running time, so we
can conclude that the problem of sorting requires Θ(n log n) time. As a corol-
lary, we know that no comparison-based sorting algorithm can improve on existing
Θ(n log n) time sorting algorithms by more than a constant factor.
7.10 Further Reading
The definitive reference on sorting is Donald E. Knuth’s Sorting and Searching
[Knu98]. A wealth of details is covered there, including optimal sorts for small
size n and special purpose sorting networks. It is a thorough (although somewhat
dated) treatment on sorting. For an analysis of Quicksort and a thorough survey
on its optimizations, see Robert Sedgewick’s Quicksort [Sed80]. Sedgewick’s Al-
gorithms [Sed03] discusses most of the sorting algorithms described here and pays
special attention to efficient implementation. The optimized Mergesort version of
Section 7.4 comes from Sedgewick.
While Ω(n log n) is the theoretical lower bound in the worst case for sorting,
many times the input is sufficiently well ordered that certain algorithms can take
advantage of this fact to speed the sorting process. A simple example is Insertion
Sort’s best-case running time. Sorting algorithms whose running time is based on
the amount of disorder in the input are called adaptive. For more information on
adaptive sorting algorithms, see “A Survey of Adaptive Sorting Algorithms” by
Estivill-Castro and Wood [ECW92].](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-290-320.jpg)
![274 Chap. 7 Internal Sorting
7.13 Graph f1(n) = n log n, f2(n) = n1.5, and f3(n) = n2 in the range 1 ≤ n ≤
1000 to visually compare their growth rates. Typically, the constant factor
in the running-time expression for an implementation of Insertion Sort will
be less than the constant factors for Shellsort or Quicksort. How many times
greater can the constant factor be for Shellsort to be faster than Insertion Sort
when n = 1000? How many times greater can the constant factor be for
Quicksort to be faster than Insertion Sort when n = 1000?
7.14 Imagine that there exists an algorithm SPLITk that can split a list L of n
elements into k sublists, each containing one or more elements, such that
sublist i contains only elements whose values are less than all elements in
sublist j for i < j <= k. If n < k, then k−n sublists are empty, and the rest
are of length 1. Assume that SPLITk has time complexity O(length of L).
Furthermore, assume that the k lists can be concatenated again in constant
time. Consider the following algorithm:
List SORTk(List L) {
List sub[k]; // To hold the sublists
if (L.length() > 1) {
SPLITk(L, sub); // SPLITk places sublists into sub
for (i=0; i<k; i++)
sub[i] = SORTk(sub[i]); // Sort each sublist
L = concatenation of k sublists in sub;
return L;
}
}
(a) What is the worst-case asymptotic running time for SORTk? Why?
(b) What is the average-case asymptotic running time of SORTk? Why?
7.15 Here is a variation on sorting. The problem is to sort a collection of n nuts
and n bolts by size. It is assumed that for each bolt in the collection, there
is a corresponding nut of the same size, but initially we do not know which
nut goes with which bolt. The differences in size between two nuts or two
bolts can be too small to see by eye, so you cannot rely on comparing the
sizes of two nuts or two bolts directly. Instead, you can only compare the
sizes of a nut and a bolt by attempting to screw one into the other (assume
this comparison to be a constant time operation). This operation tells you
that either the nut is bigger than the bolt, the bolt is bigger than the nut, or
they are the same size. What is the minimum number of comparisons needed
to sort the nuts and bolts in the worst case?](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-293-320.jpg)
![Sec. 8.3 Buffers and Buffer Pools 293
(on disk)
Secondary Storage
2
3
4
5
6
7
8
9
8
3
5
0
1 7
1
Main Memory
(in RAM)
Figure 8.5 An illustration of virtual memory. The complete collection of infor-
mation resides in the slowe, secondary storage (on disk). Those sectors recently
accessed are held in the fast main memory (in RAM). In this example, copies of
Sectors 1, 7, 5, 3, and 8 from secondary storage are currently stored in the main
memory. If a memory access to Sector 9 is received, one of the sectors currently
in main memory must be replaced.
the buffer pool class itself. The first approach is to pass “messages” between the
two. This approach is illustrated by the following abstract class:
/** ADT for buffer pools using the message-passing style */
public interface BufferPoolADT {
/** Copy "sz" bytes from "space" to position "pos" in the
buffered storage */
public void insert(byte[] space, int sz, int pos);
/** Copy "sz" bytes from position "pos" of the buffered
storage to "space". */
public void getbytes(byte[] space, int sz, int pos);
}
This simple class provides an interface with two member functions, insert
and getbytes. The information is passed between the buffer pool user and the
buffer pool through the space parameter. This is storage space, provided by the
bufferpool client and at least sz bytes long, which the buffer pool can take in-
formation from (the insert function) or put information into (the getbytes
function). Parameter pos indicates where the information will be placed in the](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-312-320.jpg)
![294 Chap. 8 File Processing and External Sorting
buffer pool’s logical storage space. Physically, it will actually be copied to the
appropriate byte position in some buffer in the buffer pool.
Example 8.4 Assume each sector of the disk file (and thus each block in
the buffer pool) stores 1024 bytes. Assume that the buffer pool is in the
state shown in Figure 8.5. If the next request is to copy 40 bytes begin-
ning at position 6000 of the file, these bytes should be placed into Sector 5
(whose bytes go from position 5120 to position 6143). Because Sector 5
is currently in the buffer pool, we simply copy the 40 bytes contained in
space to byte positions 880-919. The buffer containing Sector 5 is then
moved to the buffer pool ahead of the buffer containing Sector 1.
The alternative approach is to have the buffer pool provide to the user a direct
pointer to a buffer that contains the necessary information. Such an interface might
look as follows:
/** ADT for buffer pools using the buffer-passing style */
public interface BufferPoolADT {
/** Return pointer to the requested block */
public byte[] getblock(int block);
/** Set the dirty bit for the buffer holding "block" */
public void dirtyblock(int block);
// Tell the size of a buffer
public int blocksize();
};
In this approach, the user of the buffer pool is made aware that the storage
space is divided into blocks of a given size, where each block is the size of a buffer.
The user requests specific blocks from the buffer pool, with a pointer to the buffer
holding the requested block being returned to the user. The user might then read
from or write to this space. If the user writes to the space, the buffer pool must be
informed of this fact. The reason is that, when a given block is to be removed from
the buffer pool, the contents of that block must be written to the backing storage if
it has been modified. If the block has not been modified, then it is unnecessary to
write it out.
Example 8.5 We wish to write 40 bytes beginning at logical position
6000 in the file. Assume that the buffer pool is in the state shown in Fig-
ure 8.5. Using the second ADT, the client would need to know that blocks
(buffers) are of size 1024, and therefore would request access to Sector 5.](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-313-320.jpg)
![296 Chap. 8 File Processing and External Sorting
tents must then issue a readBlock request to read the data from disk into the
buffer, and then a getDataPointer request to gain direct access to the buffer’s
data contents.
/** Improved ADT for buffer pools using the buffer-passing
style. Most user functionality is in the buffer class,
not the buffer pool itself. */
/** A single buffer in the buffer pool */
public interface BufferADT {
/** Read the associated block from disk (if necessary) and
return a pointer to the data */
public byte[] readBlock();
/** Return a pointer to the buffer’s data array
(without reading from disk) */
public byte[] getDataPointer();
/** Flag the buffer’s contents as having changed, so that
flushing the block will write it back to disk */
public void markDirty();
/** Release the block’s access to this buffer. Further
accesses to this buffer are illegal. */
public void releaseBuffer();
}
/** The bufferpool */
public interface BufferPoolADT {
/** Relate a block to a buffer, returning a pointer to a
buffer object */
Buffer acquireBuffer(int block);
}
Clearly, the buffer-passing approach places more obligations on the user of the
buffer pool. These obligations include knowing the size of a block, not corrupting
the buffer pool’s storage space, and informing the buffer pool both when a block
has been modified and when it is no longer needed. So many obligations make this
approach prone to error. An advantage is that there is no need to do an extra copy
step when getting information from the user to the buffer. If the size of the records
stored is small, this is not an important consideration. If the size of the records is
large (especially if the record size and the buffer size are the same, as typically is the
case when implementing B-trees, see Section 10.5), then this efficiency issue might
become important. Note however that the in-memory copy time will always be far
less than the time required to write the contents of a buffer to disk. For applications](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-315-320.jpg)
![Sec. 8.4 The Programmer’s View of Files 297
where disk I/O is the bottleneck for the program, even the time to copy lots of
information between the buffer pool user and the buffer might be inconsequential.
Another advantage to buffer passing is the reduction in unnecessary read operations
for data that will be overwritten anyway.
You should note that the implementations for class BufferPool above are
not generic. Instead, the space parameter and the buffer pointer are declared to
be byte[]. When a class is generic, that means that the record type is arbitrary,
but that the class knows what the record type is. In contrast, using a byte[]
pointer for the space means that not only is the record type arbitrary, but also the
buffer pool does not even know what the user’s record type is. In fact, a given buffer
pool might have many users who store many types of records.
In a buffer pool, the user decides where a given record will be stored but has
no control over the precise mechanism by which data are transfered to the backing
storage. This is in contrast to the memory manager described in Section 12.3 in
which the user passes a record to the manager and has no control at all over where
the record is stored.
8.4 The Programmer’s View of Files
The Java programmer’s logical view of a random access file is a single stream
of bytes. Interaction with a file can be viewed as a communications channel for
issuing one of three instructions: read bytes from the current position in the file,
write bytes to the current position in the file, and move the current position within
the file. You do not normally see how the bytes are stored in sectors, clusters, and
so forth. The mapping from logical to physical addresses is done by the file system,
and sector-level buffering is done automatically by the disk controller.
When processing records in a disk file, the order of access can have a great
effect on I/O time. A random access procedure processes records in an order
independent of their logical order within the file. Sequential access processes
records in order of their logical appearance within the file. Sequential processing
requires less seek time if the physical layout of the disk file matches its logical
layout, as would be expected if the file were created on a disk with a high percentage
of free space.
Following are the primary Java functions for accessing information from ran-
dom access disk files when using class RandomAccessFile.
• RandomAccessFile(String name, String mode): Class con-
structor, opens a disk file for processing.](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-316-320.jpg)
![298 Chap. 8 File Processing and External Sorting
• read(byte[] b): Read some bytes from the current position in the file.
The current position moves forward as the bytes are read.
• write(byte[] b): Write some bytes at the current position in the file
(overwriting the bytes already at that position). The current position moves
forward as the bytes are written.
• seek(long pos): Move the current position in the file. This allows bytes
at other places within the file to be read or written.
• close(): Close a file at the end of processing.
8.5 External Sorting
We now consider the problem of sorting collections of records too large to fit in
main memory. Because the records must reside in peripheral or external memory,
such sorting methods are called external sorts. This is in contrast to the internal
sorts discussed in Chapter 7 which assume that the records to be sorted are stored in
main memory. Sorting large collections of records is central to many applications,
such as processing payrolls and other large business databases. As a consequence,
many external sorting algorithms have been devised. Years ago, sorting algorithm
designers sought to optimize the use of specific hardware configurations, such as
multiple tape or disk drives. Most computing today is done on personal computers
and low-end workstations with relatively powerful CPUs, but only one or at most
two disk drives. The techniques presented here are geared toward optimized pro-
cessing on a single disk drive. This approach allows us to cover the most important
issues in external sorting while skipping many less important machine-dependent
details. Readers who have a need to implement efficient external sorting algorithms
that take advantage of more sophisticated hardware configurations should consult
the references in Section 8.6.
When a collection of records is too large to fit in main memory, the only prac-
tical way to sort it is to read some records from disk, do some rearranging, then
write them back to disk. This process is repeated until the file is sorted, with each
record read perhaps many times. Given the high cost of disk I/O, it should come as
no surprise that the primary goal of an external sorting algorithm is to minimize the
amount of information that must be read from or written to disk. A certain amount
of additional CPU processing can profitably be traded for reduced disk access.
Before discussing external sorting techniques, consider again the basic model
for accessing information from disk. The file to be sorted is viewed by the program-
mer as a sequential series of fixed-size blocks. Assume (for simplicity) that each](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-317-320.jpg)
![310 Chap. 8 File Processing and External Sorting
File Sort 1 Sort 2 Sort 3
Size Memory size (in blocks) Memory size (in blocks)
2 4 16 256 2 4 16
1 0.61 0.27 0.24 0.19 0.10 0.21 0.15 0.13
4,864 2,048 1,792 1,280 256 2,048 1,024 512
4 2.56 1.30 1.19 0.96 0.61 1.15 0.68 0.66*
21,504 10,240 9,216 7,168 3,072 10,240 5,120 2,048
16 11.28 6.12 5.63 4.78 3.36 5.42 3.19 3.10
94,208 49,152 45,056 36,864 20,480 49,152 24,516 12,288
256 220.39 132.47 123.68 110.01 86.66 115.73 69.31 68.71
1,769K 1,048K 983K 852K 589K 1,049K 524K 262K
Figure 8.11 A comparison of three external sorts on a collection of small records
for files of various sizes. Each entry in the table shows time in seconds and total
number of blocks read and written by the program. File sizes are in Megabytes.
For the third sorting algorithm, on file size of 4MB, the time and blocks shown in
the last column are for a 32-way merge. 32 is used instead of 16 because 32 is a
root of the number of blocks in the file (while 16 is not), thus allowing the same
number of runs to be merged at every pass.
merges for R beyond about 4 or 8 runs does not help much because a lot of time is
spent determining which is the next smallest element among the R runs.
We see from this experiment that building large initial runs reduces the running
time to slightly more than one third that of standard Mergesort, depending on file
and memory sizes. Using a multiway merge further cuts the time nearly in half.
In summary, a good external sorting algorithm will seek to do the following:
• Make the initial runs as long as possible.
• At all stages, overlap input, processing, and output as much as possible.
• Use as much working memory as possible. Applying more memory usually
speeds processing. In fact, more memory will have a greater effect than a
faster disk. A faster CPU is unlikely to yield much improvement in running
time for external sorting, because disk I/O speed is the limiting factor.
• If possible, use additional disk drives for more overlapping of processing
with I/O, and to allow for sequential file processing.
8.6 Further Reading
A good general text on file processing is Folk and Zoellig’s File Structures: A
Conceptual Toolkit [FZ98]. A somewhat more advanced discussion on key issues in
file processing is Betty Salzberg’s File Structures: An Analytical Approach [Sal88].](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-329-320.jpg)
![Sec. 8.7 Exercises 311
A great discussion on external sorting methods can be found in Salzberg’s book.
The presentation in this chapter is similar in spirit to Salzberg’s.
For details on disk drive modeling and measurement, see the article by Ruemm-
ler and Wilkes, “An Introduction to Disk Drive Modeling” [RW94]. See Andrew
S. Tanenbaum’s Structured Computer Organization [Tan06] for an introduction to
computer hardware and organization. An excellent, detailed description of mem-
ory and hard disk drives can be found online at “The PC Guide,” by Charles M.
Kozierok [Koz05] (www.pcguide.com). The PC Guide also gives detailed de-
scriptions of the Microsoft Windows and UNIX (Linux) file systems.
See “Outperforming LRU with an Adaptive Replacement Cache Algorithm” by
Megiddo and Modha for an example of a more sophisticated algorithm than LRU
for managing buffer pools.
The snowplow argument comes from Donald E. Knuth’s Sorting and Searching
[Knu98], which also contains a wide variety of external sorting algorithms.
8.7 Exercises
8.1 Computer memory and storage prices change rapidly. Find out what the
current prices are for the media listed in Figure 8.1. Does your information
change any of the basic conclusions regarding disk processing?
8.2 Assume a disk drive from the late 1990s is configured as follows. The to-
tal storage is approximately 675MB divided among 15 surfaces. Each sur-
face has 612 tracks; there are 144 sectors/track, 512 bytes/sector, and 8 sec-
tors/cluster. The disk turns at 3600 rpm. The track-to-track seek time is
20 ms, and the average seek time is 80 ms. Now assume that there is a
360KB file on the disk. On average, how long does it take to read all of the
data in the file? Assume that the first track of the file is randomly placed on
the disk, that the entire file lies on adjacent tracks, and that the file completely
fills each track on which it is found. A seek must be performed each time the
I/O head moves to a new track. Show your calculations.
8.3 Using the specifications for the disk drive given in Exercise 8.2, calculate the
expected time to read one entire track, one sector, and one byte. Show your
calculations.
8.4 Using the disk drive specifications given in Exercise 8.2, calculate the time
required to read a 10MB file assuming
(a) The file is stored on a series of contiguous tracks, as few tracks as pos-
sible.
(b) The file is spread randomly across the disk in 4KB clusters.](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-330-320.jpg)
![318 Chap. 9 Searching
This and the following chapter treat these three approaches in turn. Any of
these approaches are potentially suitable for implementing the Dictionary ADT
introduced in Section 4.4. However, each has different performance characteristics
that make it the method of choice in particular circumstances.
The current chapter considers methods for searching data stored in lists and
tables. A table is simply another term for an array. List in this context means any
list implementation including a linked list or an array. Most of these methods are
appropriate for sequences (i.e., duplicate key values are allowed), although special
techniques applicable to sets are discussed in Section 9.3. The techniques from the
first three sections of this chapter are most appropriate for searching a collection of
records stored in RAM. Section 9.4 discusses hashing, a technique for organizing
data in a table such that the location of each record within the table is a function of
its key value. Hashing is appropriate when records are stored either in RAM or on
disk.
Chapter 10 discusses tree-based methods for organizing information on disk,
including a commonly used file structure called the B-tree. Nearly all programs that
must organize large collections of records stored on disk use some variant of either
hashing or the B-tree. Hashing is practical for only certain access functions (exact-
match queries) and is generally appropriate only when duplicate key values are
not allowed. B-trees are the method of choice for disk-based applications anytime
hashing is not appropriate.
9.1 Searching Unsorted and Sorted Arrays
The simplest form of search has already been presented in Example 3.1: the se-
quential search algorithm. Sequential search on an unsorted list requires Θ(n) time
in the worst case.
How many comparisons does linear search do on average? A major consid-
eration is whether K is in list L at all. We can simplify our analysis by ignoring
everything about the input except the position of K if it is found in L. Thus, we
have n + 1 distinct possible events: That K is in one of positions 0 to n − 1 in
L (each with its own probability), or that it is not in L at all. We can express the
probability that K is not in L as
P(K /
∈ L) = 1 −
n
X
i=1
P(K = L[i])
where P(x) is the probability of event x.](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-337-320.jpg)
![320 Chap. 9 Searching
element in L, that is, we check elements L[j], L[2j], and so on. So long as K is
greater than the values we are checking, we continue on. But when we reach a
value in L greater than K, we do a linear search on the piece of length j − 1 that
we know brackets K if it is in the list.
If mj ≤ n < (m+1)j, then the total cost of this algorithm is at most m+j −1
3-way comparisons. Therefore, the cost to run the algorithm on n items with a jump
of size j is
T(n, j) = m + j − 1 =
n
j
+ j − 1.
What is the best value that we can pick for j? We want to minimize the cost:
min
1≤j≤n
n
j
+ j − 1
Take the derivative and solve for f′(j) = 0 to find the minimum, which is
j =
√
n. In this case, the worst case cost will be roughly 2
√
n.
This example teaches us some lessons about algorithm design. We want to
balance the work done while selecting a sublist with the work done while searching
a sublist. In general, it is a good strategy to make subproblems of equal effort. This
is an example of a divide and conquer algorithm.
What if we extend this idea to three levels? We would first make jumps of
some size j to find a sublist of size j − 1 whose end vlaues bracket value K. Ww
would then work through this sublist by making jumps of some smaller size, say
j1. Finally, once we find a bracketed sublist of size j1 − 1, we would do sequential
search to complete the process.
This probably sounds convoluted to do two levels of jumping to be followed by
a sequential search. While it might make sense to do a two-level algorithm (that is,
jump search jumps to find a sublist and then does sequential search on the sublist),
it almost never seems to make sense to do a three-level algorithm. Instead, when
we go beyond two levels, we nearly always generalize by using recursion. This
leads us to the most commonly used search algorithm for sorted arrays, the binary
search described in Section 3.5.
If we know nothing about the distribution of key values, then binary search
is the best algorithm available for searching a sorted array (see Exercise 9.22).
However, sometimes we do know something about the expected key distribution.
Consider the typical behavior of a person looking up a word in a large dictionary.
Most people certainly do not use sequential search! Typically, people use a mod-
ified form of binary search, at least until they get close to the word that they are](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-339-320.jpg)
![Sec. 9.1 Searching Unsorted and Sorted Arrays 321
looking for. The search generally does not start at the middle of the dictionary. A
person looking for a word starting with ‘S’ generally assumes that entries beginning
with ‘S’ start about three quarters of the way through the dictionary. Thus, he or
she will first open the dictionary about three quarters of the way through and then
make a decision based on what they find as to where to look next. In other words,
people typically use some knowledge about the expected distribution of key values
to “compute” where to look next. This form of “computed” binary search is called
a dictionary search or interpolation search. In a dictionary search, we search L
at a position p that is appropriate to the value of K as follows.
p =
K − L[1]
L[n] − L[1]
The location of a particular key within the key range is translated into the
expected position for the corresponding record in the table, and this position is
checked first. As with binary search, the value of the key found eliminates all
records either above or below that position. The actual value of the key found can
then be used to compute a new position within the remaining range of the table.
The next check is made based on the new computation. This proceeds until either
the desired record is found, or the table is narrowed until no records are left.
A variation on dictionary search is known as Quadratic Binary Search (QBS),
and we will analyze this in detail because its analysis is easier than that of the
general dictionary search. QBS will first compute p and t hen examine L[⌈pn⌉]. If
K L[⌈pn⌉] then QBS will sequentially probe by steps of size
√
n, that is, we step
through
L[⌈pn − i
√
n⌉], i = 1, 2, 3, ...
until we reach a value less than or equal to K. Similarly for K L[⌈pn⌉] we will
step forward by
√
n until we reach a value in L that is greater than K. We are now
within
√
n positions of K. Assume (for now) that it takes a constant number of
comparisons to bracket K within a sublist of size
√
n. We then take this sublist and
repeat the process recursively.
What is the cost for QBS? Note that
√
cn = cn/2, and we will be repeatedly
taking square roots of the current sublist size until we find the item we are looking
for. Because n = 2log n and we can cut log n in half only log log n times, the cost
is Θ(log log n) if the number of probes on jump search is constant.
The number of comparisons needed is
√
n
X
i=1
iP(need exactly i probes)](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-340-320.jpg)
![330 Chap. 9 Searching
The set difference A − B can be implemented in Java using the expression A˜B
(˜ is the symbol for bitwise negation). For larger sets that do not fit into a single
computer word, the equivalent operations can be performed in turn on the series of
words making up the entire bit vector.
This method of computing sets from bit vectors is sometimes applied to doc-
ument retrieval. Consider the problem of picking from a collection of documents
those few which contain selected keywords. For each keyword, the document re-
trieval system stores a bit vector with one bit for each document. If the user wants to
know which documents contain a certain three keywords, the corresponding three
bit vectors are AND’ed together. Those bit positions resulting in a value of 1 cor-
respond to the desired documents. Alternatively, a bit vector can be stored for each
document to indicate those keywords appearing in the document. Such an organiza-
tion is called a signature file. The signatures can be manipulated to find documents
with desired combinations of keywords.
9.4 Hashing
This section presents a completely different approach to searching tables: by direct
access based on key value. The process of finding a record using some computa-
tion to map its key value to a position in the table is called hashing. Most hash-
ing schemes place records in the table in whatever order satisfies the needs of the
address calculation, thus the records are not ordered by value or frequency. The
function that maps key values to positions is called a hash function and is usually
denoted by h. The array that holds the records is called the hash table and will be
denoted by HT. A position in the hash table is also known as a slot. The number
of slots in hash table HT will be denoted by the variable M, with slots numbered
from 0 to M − 1. The goal for a hashing system is to arrange things such that, for
any key value K and some hash function h, i = h(K) is a slot in the table such
that 0 ≤ h(K) M, and we have the key of the record stored at HT[i] equal to
K.
Hashing only works to store sets. That is, hashing cannnot be used for applica-
tions where multiple records with the same key value are permitted. Hashing is not
a good method for answering range searches. In other words, we cannot easily find
all records (if any) whose key values fall within a certain range. Nor can we easily
find the record with the minimum or maximum key value, or visit the records in
key order. Hashing is most appropriate for answering the question, “What record,
if any, has key value K?” For applications where access involves only exact-match
queries, hashing is usually the search method of choice because it is extremely ef-
ficient when implemented correctly. As you will see in this section, however, there](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-349-320.jpg)
![Sec. 9.4 Hashing 331
are many approaches to hashing and it is easy to devise an inefficient implementa-
tion. Hashing is suitable for both in-memory and disk-based searching and is one
of the two most widely used methods for organizing large databases stored on disk
(the other is the B-tree, which is covered in Chapter 10).
As a simple (though unrealistic) example of hashing, consider storing n records,
each with a unique key value in the range 0 to n − 1. In this simple case, a record
with key k can be stored in HT[k], and the hash function is simply h(k) = k. To
find the record with key value k, simply look in HT[k].
Typically, there are many more values in the key range than there are slots in
the hash table. For a more realistic example, suppose that the key can take any
value in the range 0 to 65,535 (i.e., the key is a two-byte unsigned integer), and that
we expect to store approximately 1000 records at any given time. It is impractical
in this situation to use a hash table with 65,536 slots, because most of the slots will
be left empty. Instead, we must devise a hash function that allows us to store the
records in a much smaller table. Because the possible key range is larger than the
size of the table, at least some of the slots must be mapped to from multiple key
values. Given a hash function h and two keys k1 and k2, if h(k1) = β = h(k2)
where β is a slot in the table, then we say that k1 and k2 have a collision at slot β
under hash function h.
Finding a record with key value K in a database organized by hashing follows
a two-step procedure:
1. Compute the table location h(K).
2. Starting with slot h(K), locate the record containing key K using (if neces-
sary) a collision resolution policy.
9.4.1 Hash Functions
Hashing generally takes records whose key values come from a large range and
stores those records in a table with a relatively small number of slots. Collisions
occur when two records hash to the same slot in the table. If we are careful—or
lucky—when selecting a hash function, then the actual number of collisions will
be few. Unfortunately, even under the best of circumstances, collisions are nearly
unavoidable.1 For example, consider a classroom full of students. What is the
1
The exception to this is perfect hashing. Perfect hashing is a system in which records are
hashed such that there are no collisions. A hash function is selected for the specific set of records
being hashed, which requires that the entire collection of records be available before selecting the
hash function. Perfect hashing is efficient because it always finds the record that we are looking
for exactly where the hash function computes it to be, so only one access is required. Selecting
a perfect hash function can be expensive but might be worthwhile when extremely efficient search](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-350-320.jpg)
![334 Chap. 9 Searching
Example 9.6 A good hash function for numerical values is the mid-square
method. The mid-square method squares the key value, and then takes the
middle r bits of the result, giving a value in the range 0 to 2r − 1. This
works well because most or all bits of the key value contribute to the result.
For example, consider records whose keys are 4-digit numbers in base 10.
The goal is to hash these key values to a table of size 100 (i.e., a range of 0
to 99). This range is equivalent to two digits in base 10. That is, r = 2 If
the input is the number 4567, squaring yields an 8-digit number, 20857489.
The middle two digits of this result are 57. All digits (equivalently, all bits
when the number is viewed in binary) contribute to the middle two digits
of the squared value. Thus, the result is not dominated by the distribution
of the bottom digit or the top digit of the original key value.
Example 9.7 Here is a hash function for strings of characters:
int h(String x, int M) {
char ch[];
ch = x.toCharArray();
int xlength = x.length();
int i, sum;
for (sum=0, i=0; ix.length(); i++)
sum += ch[i];
return sum % M;
}
This function sums the ASCII values of the letters in a string. If the hash
table size M is small, this hash function should do a good job of distributing
strings evenly among the hash table slots, because it gives equal weight to
all characters. This is an example of the folding approach to designing a
hash function. Note that the order of the characters in the string has no
effect on the result. A similar method for integers would add the digits of
the key value, assuming that there are enough digits to (1) keep any one
or two digits with bad distribution from skewing the results of the process
and (2) generate a sum much larger than M. As with many other hash
functions, the final step is to apply the modulus operator to the result, using
table size M to generate a value within the table range. If the sum is not
sufficiently large, then the modulus operator will yield a poor distribution.](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-353-320.jpg)
![Sec. 9.4 Hashing 335
For example, because the ASCII value for “A” is 65 and “Z” is 90, sum will
always be in the range 650 to 900 for a string of ten upper case letters. For
a hash table of size 100 or less, a good distribution results with all slots in
the table accepting either two or three of the values in the key range. For a
hash table of size 1000, the distribution is terrible because only slots 650 to
900 can possibly be the home slot for some key value.
Example 9.8 Here is a much better hash function for strings.
long sfold(String s, int M) {
int intLength = s.length() / 4;
long sum = 0;
for (int j = 0; j intLength; j++) {
char c[] = s.substring(j * 4, (j * 4) + 4).toCharArray();
long mult = 1;
for (int k = 0; k c.length; k++) {
sum += c[k] * mult;
mult *= 256;
}
}
char c[] = s.substring(intLength * 4).toCharArray();
long mult = 1;
for (int k = 0; k c.length; k++) {
sum += c[k] * mult;
mult *= 256;
}
return(Math.abs(sum) % M);
}
This function takes a string as input. It processes the string four bytes
at a time, and interprets each of the four-byte chunks as a single (unsigned)
long integer value. The integer values for the four-byte chunks are added
together. In the end, the resulting sum is converted to the range 0 to M − 1
using the modulus operator.2
For example, if the string “aaaabbbb” is passed to sfold, then the first
four bytes (“aaaa”) will be interpreted as the integer value 1,633,771,873
2
Recall from Section 2.2 that the implementation for n mod m on many C++ and Java compilers
will yield a negative number if n is negative. Implementors for hash functions need to be careful that
their hash function does not generate a negative number. This can be avoided either by insuring
that n is positive when computing n mod m, or adding m to the result if n mod m is negative.
All computation in sfold is done using unsigned long values in part to protect against taking the
modulus of an negative number.](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-354-320.jpg)
![340 Chap. 9 Searching
/** Insert record r with key k into HT */
void hashInsert(K k, E r) {
int home; // Home position for r
int pos = home = h(k); // Initial position
for (int i=1; HT[pos] != null; i++) {
pos = (home + p(k, i)) % M; // Next pobe slot
assert HT[pos].key().compareTo(k) != 0 :
Duplicates not allowed;
}
HT[pos] = new KVpairK,E(k, r); // Insert R
}
Figure 9.4 Insertion method for a dictionary implemented by a hash table.
/** Search in hash table HT for the record with key k */
E hashSearch(K k) {
int home; // Home position for k
int pos = home = h(k); // Initial position
for (int i = 1; (HT[pos] != null)
(HT[pos].key().compareTo(k) != 0); i++)
pos = (home + p(k, i)) % M; // Next probe position
if (HT[pos] == null) return null; // K not in hash table
else return HT[pos].value(); // Found it
}
Figure 9.5 Search method for a dictionary implemented by a hash table.
Method hashInsert first checks to see if the home slot for the key is empty.
If the home slot is occupied, then we use the probe function, p(k, i) to locate a free
slot in the table. Function p has two parameters, the key k and a count i for where
in the probe sequence we wish to be. That is, to get the first position in the probe
sequence after the home slot for key K, we call p(K, 1). For the next slot in the
probe sequence, call p(K, 2). Note that the probe function returns an offset from
the original home position, rather than a slot in the hash table. Thus, the for loop
in hashInsert is computing positions in the table at each iteration by adding
the value returned from the probe function to the home position. The ith call to p
returns the ith offset to be used.
Searching in a hash table follows the same probe sequence that was followed
when inserting records. In this way, a record not in its home position can be recov-
ered. A Java implementation for the search procedure is shown in Figure 9.5
Both the insert and the search routines assume that at least one slot on the probe
sequence of every key will be empty. Otherwise, they will continue in an infinite
loop on unsuccessful searches. Thus, the dictionary should keep a count of the](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-359-320.jpg)
![Sec. 9.4 Hashing 343
sequence that cycles through only the even slots. Likewise, the probe sequence
for a key whose home position is in an odd slot will cycle through the odd slots.
Thus, this combination of table size and linear probing constant effectively divides
the records into two sets stored in two disjoint sections of the hash table. So long
as both sections of the table contain the same number of records, this is not really
important. However, just from chance it is likely that one section will become fuller
than the other, leading to more collisions and poorer performance for those records.
The other section would have fewer records, and thus better performance. But the
overall system performance will be degraded, as the additional cost to the side that
is more full outweighs the improved performance of the less-full side.
Constant c must be relatively prime to M to generate a linear probing sequence
that visits all slots in the table (that is, c and M must share no factors). For a hash
table of size M = 10, if c is any one of 1, 3, 7, or 9, then the probe sequence
will visit all slots for any key. When M = 11, any value for c between 1 and 10
generates a probe sequence that visits all slots for every key.
Consider the situation where c = 2 and we wish to insert a record with key k1
such that h(k1) = 3. The probe sequence for k1 is 3, 5, 7, 9, and so on. If another
key k2 has home position at slot 5, then its probe sequence will be 5, 7, 9, and so on.
The probe sequences of k1 and k2 are linked together in a manner that contributes
to clustering. In other words, linear probing with a value of c 1 does not solve
the problem of primary clustering. We would like to find a probe function that does
not link keys together in this way. We would prefer that the probe sequence for k1
after the first step on the sequence should not be identical to the probe sequence of
k2. Instead, their probe sequences should diverge.
The ideal probe function would select the next position on the probe sequence
at random from among the unvisited slots; that is, the probe sequence should be a
random permutation of the hash table positions. Unfortunately, we cannot actually
select the next position in the probe sequence at random, because then we would not
be able to duplicate this same probe sequence when searching for the key. However,
we can do something similar called pseudo-random probing. In pseudo-random
probing, the ith slot in the probe sequence is (h(K) + ri) mod M where ri is the
ith value in a random permutation of the numbers from 1 to M − 1. All insertion
and search operations must use the same random permutation. The probe function
would be
p(K, i) = Perm[i − 1],
where Perm is an array of length M − 1 containing a random permutation of the
values from 1 to M − 1.](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-362-320.jpg)
![344 Chap. 9 Searching
Example 9.9 Consider a table of size M = 101, with Perm[1] = 5,
Perm[2] = 2, and Perm[3] = 32. Assume that we have two keys k1 and
k2 where h(k1) = 30 and h(k2) = 35. The probe sequence for k1 is 30,
then 35, then 32, then 62. The probe sequence for k2 is 35, then 40, then
37, then 67. Thus, while k2 will probe to k1’s home position as its second
choice, the two keys’ probe sequences diverge immediately thereafter.
Another probe function that eliminates primary clustering is called quadratic
probing. Here the probe function is some quadratic function
p(K, i) = c1i2
+ c2i + c3
for some choice of constants c1, c2, and c3. The simplest variation is p(K, i) = i2
(i.e., c1 = 1, c2 = 0, and c3 = 0. Then the ith value in the probe sequence would
be (h(K) + i2) mod M. Under quadratic probing, two keys with different home
positions will have diverging probe sequences.
Example 9.10 Given a hash table of size M = 101, assume for keys k1
and k2 that h(k1) = 30 and h(k2) = 29. The probe sequence for k1 is 30,
then 31, then 34, then 39. The probe sequence for k2 is 29, then 30, then
33, then 38. Thus, while k2 will probe to k1’s home position as its second
choice, the two keys’ probe sequences diverge immediately thereafter.
Unfortunately, quadratic probing has the disadvantage that typically not all hash
table slots will be on the probe sequence. Using p(K, i) = i2 gives particularly in-
consistent results. For many hash table sizes, this probe function will cycle through
a relatively small number of slots. If all slots on that cycle happen to be full, then
the record cannot be inserted at all! For example, if our hash table has three slots,
then records that hash to slot 0 can probe only to slots 0 and 1 (that is, the probe
sequence will never visit slot 2 in the table). Thus, if slots 0 and 1 are full, then
the record cannot be inserted even though the table is not full! A more realistic
example is a table with 105 slots. The probe sequence starting from any given slot
will only visit 23 other slots in the table. If all 24 of these slots should happen to
be full, even if other slots in the table are empty, then the record cannot be inserted
because the probe sequence will continually hit only those same 24 slots.
Fortunately, it is possible to get good results from quadratic probing at low
cost. The right combination of probe function and table size will visit many slots
in the table. In particular, if the hash table size is a prime number and the probe
function is p(K, i) = i2, then at least half the slots in the table will be visited.
Thus, if the table is less than half full, we can be certain that a free slot will be](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-363-320.jpg)
![Sec. 9.5 Further Reading 351
occupied the slot but does so no longer. If a tombstone is encountered when search-
ing through a probe sequence, the search procedure is to continue with the search.
When a tombstone is encountered during insertion, that slot can be used to store the
new record. However, to avoid inserting duplicate keys, it will still be necessary for
the search procedure to follow the probe sequence until a truly empty position has
been found, simply to verify that a duplicate is not in the table. However, the new
record would actually be inserted into the slot of the first tombstone encountered.
The use of tombstones allows searches to work correctly and allows reuse of
deleted slots. However, after a series of intermixed insertion and deletion opera-
tions, some slots will contain tombstones. This will tend to lengthen the average
distance from a record’s home position to the record itself, beyond where it could
be if the tombstones did not exist. A typical database application will first load a
collection of records into the hash table and then progress to a phase of intermixed
insertions and deletions. After the table is loaded with the initial collection of
records, the first few deletions will lengthen the average probe sequence distance
for records (it will add tombstones). Over time, the average distance will reach
an equilibrium point because insertions will tend to decrease the average distance
by filling in tombstone slots. For example, after initially loading records into the
database, the average path distance might be 1.2 (i.e., an average of 0.2 accesses
per search beyond the home position will be required). After a series of insertions
and deletions, this average distance might increase to 1.6 due to tombstones. This
seems like a small increase, but it is three times longer on average beyond the home
position than before deletions.
Two possible solutions to this problem are
1. Do a local reorganization upon deletion to try to shorten the average path
length. For example, after deleting a key, continue to follow the probe se-
quence of that key and swap records further down the probe sequence into
the slot of the recently deleted record (being careful not to remove a key from
its probe sequence). This will not work for all collision resolution policies.
2. Periodically rehash the table by reinserting all records into a new hash table.
Not only will this remove the tombstones, but it also provides an opportunity
to place the most frequently accessed records into their home positions.
9.5 Further Reading
For a comparison of the efficiencies for various self-organizing techniques, see
Bentley and McGeoch, “Amortized Analysis of Self-Organizing Sequential Search
Heuristics” [BM85]. The text compression example of Section 9.2 comes from](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-370-320.jpg)
![352 Chap. 9 Searching
Bentley et al., “A Locally Adaptive Data Compression Scheme” [BSTW86]. For
more on Ziv-Lempel coding, see Data Compression: Methods and Theory by
James A. Storer [Sto88]. Knuth covers self-organizing lists and Zipf distributions
in Volume 3 of The Art of Computer Programming[Knu98].
Introduction to Modern Information Retrieval by Salton and McGill [SM83] is
an excellent source for more information about document retrieval techniques.
See the paper “Practical Minimal Perfect Hash Functions for Large Databases”
by Fox et al. [FHCD92] for an introduction and a good algorithm for perfect hash-
ing.
For further details on the analysis for various collision resolution policies, see
Knuth, Volume 3 [Knu98] and Concrete Mathematics: A Foundation for Computer
Science by Graham, Knuth, and Patashnik [GKP94].
The model of hashing presented in this chapter has been of a fixed-size hash
table. A problem not addressed is what to do when the hash table gets half full and
more records must be inserted. This is the domain of dynamic hashing methods.
A good introduction to this topic is “Dynamic Hashing Schemes” by R.J. Enbody
and H.C. Du [ED88].
9.6 Exercises
9.1 Create a graph showing expected cost versus the probability of an unsuc-
cessful search when performing sequential search (see Section 9.1). What
can you say qualitatively about the rate of increase in expected cost as the
probability of unsuccessful search grows?
9.2 Modify the binary search routine of Section 3.5 to implement interpolation
search. Assume that keys are in the range 1 to 10,000, and that all key values
within the range are equally likely to occur.
9.3 Write an algorithm to find the Kth smallest value in an unsorted array of n
numbers (K = n). Your algorithm should require Θ(n) time in the average
case. Hint: Your algorithm shuld look similar to Quicksort.
9.4 Example 9.9.3 discusses a distribution where the relative frequencies of the
records match the harmonic series. That is, for every occurance of the first
record, the second record will appear half as often, the third will appear one
third as often, the fourth one quarter as often, and so on. The actual prob-
ability for the ith record was defined to be 1/(iHn). Explain why this is
correct.
9.5 Graph the equations T(n) = log2 n and T(n) = n/ loge n. Which gives the
better performance, binary search on a sorted list, or sequential search on a](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-371-320.jpg)
![376 Chap. 10 Indexing
values, but these are used solely as placeholders to guide the search. This means
that internal nodes are significantly different in structure from leaf nodes. Inter-
nal nodes store keys to guide the search, associating each key with a pointer to a
child B+-tree node. Leaf nodes store actual records, or else keys and pointers to
actual records in a separate disk file if the B+-tree is being used purely as an in-
dex. Depending on the size of a record as compared to the size of a key, a leaf
node in a B+-tree of order m might have enough room to store more or less than
m records. The requirement is simply that the leaf nodes store enough records to
remain at least half full. The leaf nodes of a B+-tree are normally linked together
to form a doubly linked list. Thus, the entire collection of records can be traversed
in sorted order by visiting all the leaf nodes on the linked list. Here is a Java-like
pseudocode representation for the B+-tree node interface. Leaf node and internal
node subclasses would implement this base class.
/** Interface for B+ Tree nodes */
public interface BPNodeK,E {
public boolean isLeaf();
public int numrecs();
public K[] keys();
}
An important implementation detail to note is that while Figure 10.16 shows
internal nodes containing three keys and four pointers, class BPNode is slightly
different in that it stores key/pointer pairs. Figure 10.16 shows the B+-tree as it
is traditionally drawn. To simplify implementation in practice, nodes really do
associate a key with each pointer. Each internal node should be assumed to hold
in the leftmost position an additional key that is less than or equal to any possible
key value in the node’s leftmost subtree. B+-tree implementations typically store
an additional dummy record in the leftmost leaf node whose key value is less than
any legal key value.
B+-trees are exceptionally good for range queries. Once the first record in
the range has been found, the rest of the records with keys in the range can be
accessed by sequential processing of the remaining records in the first node, and
then continuing down the linked list of leaf nodes as far as necessary. Figure 10.17
illustrates the B+-tree.
Search in a B+-tree is nearly identical to search in a regular B-tree, except that
the search must always continue to the proper leaf node. Even if the search-key
value is found in an internal node, this is only a placeholder and does not provide
access to the actual record. To find a record with key value 33 in the B+-tree of
Figure 10.17, search begins at the root. The value 33 stored in the root merely
serves as a placeholder, indicating that keys with values greater than or equal to](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-395-320.jpg)
![Sec. 10.5 B-Trees 377
23
30 31 33 45 47
48
48 50 52
10 12 15 18 19 20 21 22
33
18
23
Figure 10.17 Example of a B+
-tree of order four. Internal nodes must store
between two and four children. For this example, the record size is assumed to be
such that leaf nodes store between three and five records.
33 are found in the second subtree. From the second child of the root, the first
branch is taken to reach the leaf node containing the actual record (or a pointer to
the actual record) with key value 33. Here is a pseudocode sketch of the B+-tree
search algorithm:
private E findhelp(BPNodeK,E rt, K k) {
int currec = binaryle(rt.keys(), rt.numrecs(), k);
if (rt.isLeaf())
if ((((BPLeafK,E)rt).keys())[currec].compareTo(k)
== 0)
return ((BPLeafK,E)rt).recs(currec);
else return null;
else
return findhelp(((BPInternalK,E)rt).pointers(currec),
k);
}
B+-tree insertion is similar to B-tree insertion. First, the leaf L that should
contain the record is found. If L is not full, then the new record is added, and no
other B+-tree nodes are affected. If L is already full, split it in two (dividing the
records evenly among the two nodes) and promote a copy of the least-valued key
in the newly formed right node. As with the 2-3 tree, promotion might cause the
parent to split in turn, perhaps eventually leading to splitting the root and causing
the B+-tree to gain a new level. B+-tree insertion keeps all leaf nodes at equal
depth. Figure 10.18 illustrates the insertion process through several examples. Fig-
ure 10.19 shows a Java-like pseudocode sketch of the B+-tree insert algorithm.
To delete record R from the B+-tree, first locate the leaf L that contains R. If L
is more than half full, then we need only remove R, leaving L still at least half full.
This is demonstrated by Figure 10.20.
If deleting a record reduces the number of records in the node below the min-
imum threshold (called an underflow), then we must do something to keep the](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-396-320.jpg)
![380 Chap. 10 Indexing
48
(a)
45 4748 50 52
23
33
18
(b)
18 19 20 21 23 30 31
101215 22 485052
4547
Figure 10.22 Deleting the record with key value 33 from the B+
-tree of Fig-
ure 10.17 via collapsing siblings. (a) The two leftmost leafnodes merge together
to form a single leaf. Unfortunately, the parent node now has only one child.
(b) Because the left subtree has a spare leaf node, that node is passed to the right
subtree. The placeholder values of the root and the right internal node are updated
to reflect the changes. Value 23 moves to the root, and old root value 33 moves to
the rightmost internal node.
/** Delete a record with the given key value, and
return true if the root underflows */
private boolean removehelp(BPNodeK,E rt, K k) {
int currec = binaryle(rt.keys(), rt.numrecs(), k);
if (rt.isLeaf())
if ((((BPLeafK,E)rt).keys()[currec]).compareTo(k)
== 0)
return ((BPLeafK,E)rt).delete(currec);
else return false;
else // Process internal node
if (removehelp(((BPInternalK,E)rt).pointers(currec),
k))
// Child will merge if necessary
return ((BPInternalK,E)rt).underflow(currec);
else return false;
}
Figure 10.23 Java-like pseudocode for the B+
-tree delete algorithm.
together, then the tree loses a level. Figure 10.22 illustrates the node-merge deletion
process. Figure 10.23 shows Java-like pseudocode for the B+-tree delete algorithm.](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-399-320.jpg)
![382 Chap. 10 Indexing
(levels 0 and 1) require relatively little space. If the B-tree is only four levels deep,
then at most two disk fetches (internal nodes at level two and leaves at level three)
are required to reach the pointer to any given record.
As mentioned earlier, a buffer pool should be used to manage nodes of the
B-tree. Several nodes of the tree would typically be in main memory at one time.
The most straightforward approach is to use a standard method such as LRU to
do node replacement. However, sometimes it might be desirable to “lock” certain
nodes such as the root into the buffer pool. In general, if the buffer pool is even
of modest size (say at least twice the depth of the tree), no special techniques for
node replacement will be required because the upper-level nodes will naturally be
accessed frequently.
10.6 Further Reading
For an expanded discussion of the issues touched on in this chapter, see a gen-
eral file processing text such as File Structures: A Conceptual Toolkit by Folk and
Zoellick [FZ98]. In particular, Folk and Zoellick provide a good discussion of
the relationship between primary and secondary indices. The most thorough dis-
cussion on various implementations for the B-tree is the survey article by Comer
[Com79]. Also see [Sal88] for further details on implementing B-trees. See Shaf-
fer and Brown [SB93] for a discussion of buffer pool management strategies for
B+-tree-like data structures.
10.7 Exercises
10.1 Assume that a computer system has disk blocks of 1024 bytes, and that you
are storing records that have 4-byte keys and 4-byte data fields. The records
are sorted and packed sequentially into the disk file.
(a) Assume that a linear index uses 4 bytes to store the key and 4 bytes
to store the block ID for the associated records. What is the greatest
number of records that can be stored in the file if a linear index of size
256KB is used?
(b) What is the greatest number of records that can be stored in the file if
the linear index is also stored on disk (and thus its size is limited only
by the second-level index) when using a second-level index of 1024
bytes (i.e., 256 key values) as illustrated by Figure 10.2? Each element
of the second-level index references the smallest key value for a disk
block of the linear index.](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-401-320.jpg)
![396 Chap. 11 Graphs
class Graphm implements Graph { // Graph: Adjacency matrix
private int[][] matrix; // The edge matrix
private int numEdge; // Number of edges
public int[] Mark; // The mark array
public Graphm() {}
public Graphm(int n) { // Constructor
Init(n);
}
public void Init(int n) {
Mark = new int[n];
matrix = new int[n][n];
numEdge = 0;
}
public int n() { return Mark.length; } // # of vertices
public int e() { return numEdge; } // # of edges
public int first(int v) { // Get v’s first neighbor
for (int i=0; iMark.length; i++)
if (matrix[v][i] != 0) return i;
return Mark.length; // No edge for this vertex
}
public int next(int v, int w) { // Get v’s next edge
for (int i=w+1; iMark.length; i++)
if (matrix[v][i] != 0)
return i;
return Mark.length; // No next edge;
}
public boolean isEdge(int i, int j) // Is (i, j) an edge?
{ return matrix[i][j] != 0; }
// Set edge weight
public void setEdge(int i, int j, int wt) {
assert wt!=0 : Cannot set weight to 0;
if (matrix[i][j] == 0) numEdge++;
matrix[i][j] = wt;
}
public void delEdge(int i, int j) { // Delete edge (i, j)
if (matrix[i][j] != 0) {
matrix[i][j] = 0;
numEdge--;
}
}
public int weight(int i, int j) { // Return edge weight
if (i == j) return 0;
if (matrix[i][j] == 0) return Integer.MAX VALUE;
return matrix[i][j];
}
// Get and set marks
public void setMark(int v, int val) { Mark[v] = val; }
public int getMark(int v) { return Mark[v]; }
}
Figure 11.6 An implementation for the adjacency matrix implementation.](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-415-320.jpg)
![398 Chap. 11 Graphs
// Adjacency list graph implementation
class Graphl implements Graph {
private GraphList[] vertex; // The vertex list
private int numEdge; // Number of edges
public int[] Mark; // The mark array
public Graphl() {}
public Graphl(int n) // Constructor
{ Init(n); }
public void Init(int n) {
Mark = new int[n];
vertex = new GraphList[n];
for (int i=0; in; i++)
vertex[i] = new GraphList();
numEdge = 0;
}
public int n() { return Mark.length; } // # of vertices
public int e() { return numEdge; } // # of edges
public int first(int v) { // Get v’s first neighbor
vertex[v].moveToStart();
Edge it = vertex[v].getValue();
if (it == null) return Mark.length;
else return it.vertex();
}
public boolean isEdge(int v, int w) { // Is (i,j) an edge?
Edge it = vertex[v].getValue();
if ((it != null) (it.vertex() == w)) return true;
for (vertex[v].moveToStart();
vertex[v].currPos() vertex[v].length();
vertex[v].next()) // Check whole list
if (vertex[v].getValue().vertex() == w) return true;
return false;
}
public int next(int v, int w) { // Get next neighbor
Edge it = null;
if (isEdge(v, w)) {
vertex[v].next();
it = vertex[v].getValue();
}
if (it != null)
return it.vertex();
else return Mark.length;
}
Figure 11.7 An implementation for the adjacency list.](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-417-320.jpg)
![Sec. 11.3 Graph Traversals 399
// Store edge weight
public void setEdge(int i, int j, int weight) {
assert weight != 0 : May not set weight to 0;
Edge currEdge = new Edge(j, weight);
if (isEdge(i, j)) { // Edge already exists in graph
vertex[i].remove();
vertex[i].insert(currEdge);
}
else { // Keep neighbors sorted by vertex index
numEdge++;
for (vertex[i].moveToStart();
vertex[i].currPos() vertex[i].length();
vertex[i].next())
if (vertex[i].getValue().vertex() j) break;
vertex[i].insert(currEdge);
}
}
public void delEdge(int i, int j) // Delete edge
{ if (isEdge(i, j)) { vertex[i].remove(); numEdge--; } }
public int weight(int i, int j) { // Return weight of edge
if (i == j) return 0;
if (isEdge(i, j)) return vertex[i].getValue().weight();
else return Integer.MAX VALUE;
}
// Set and get marks
public void setMark(int v, int val) { Mark[v] = val; }
public int getMark(int v) { return Mark[v]; }
}
Figure 11.7 (continued)
of states. Solving the problem may require getting from a specified start state to a
specified goal state by moving between states only through the connections. Typi-
cally, the start and goal states are not directly connected. To solve this problem, the
vertices of the graph must be searched in some organized manner.
Graph traversal algorithms typically begin with a start vertex and attempt to
visit the remaining vertices from there. Graph traversals must deal with a number
of troublesome cases. First, it may not be possible to reach all vertices from the
start vertex. This occurs when the graph is not connected. Second, the graph may
contain cycles, and we must make sure that cycles do not cause the algorithm to go
into an infinite loop.
Graph traversal algorithms can solve both of these problems by maintaining a
mark bit for each vertex on the graph. At the beginning of the algorithm, the mark](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-418-320.jpg)
![406 Chap. 11 Graphs
J1 J2
J3 J4
J5 J7
J6
Figure 11.13 An example graph for topological sort. Seven tasks have depen-
dencies as shown by the directed graph.
static void topsort(Graph G) { // Topological sort: Queue
QueueInteger Q = new AQueueInteger(G.n());
int[] Count = new int[G.n()];
int v;
for (v=0; vG.n(); v++) Count[v] = 0; // Initialize
for (v=0; vG.n(); v++) // Process every edge
for (int w = G.first(v); w G.n(); w = G.next(v, w))
Count[w]++; // Add to v2’s prereq count
for (v=0; vG.n(); v++) // Initialize Queue
if (Count[v] == 0) // V has no prerequisites
Q.enqueue(v);
while (Q.length() 0) { // Process the vertices
v = Q.dequeue().intValue();
printout(v); // PreVisit for Vertex V
for (int w = G.first(v); w G.n(); w = G.next(v, w)) {
Count[w]--; // One less prerequisite
if (Count[w] == 0) // This vertex is now free
Q.enqueue(w);
}
}
}
Figure 11.14 A queue-based topological sort algorithm.
prerequisites are placed on the queue. We then begin processing the queue. When
Vertex V is taken off of the queue, it is printed, and all neighbors of V (that is, all
vertices that have V as a prerequisite) have their counts decremented by one. Any
neighbor whose count is now zero is placed on the queue. If the queue becomes
empty without printing all of the vertices, then the graph contains a cycle (i.e., there
is no possible ordering for the tasks that does not violate some prerequisite). The
printed order for the vertices of the graph in Figure 11.13 using the queue version of
topological sort is J1, J2, J3, J6, J4, J5, J7. Figure 11.14 shows an implementation
for the queue-based topological sort algorithm.](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-425-320.jpg)
![Sec. 11.4 Shortest-Paths Problems 409
// Compute shortest path distances from s, store them in D
static void Dijkstra(Graph G, int s, int[] D) {
for (int i=0; iG.n(); i++) // Initialize
D[i] = Integer.MAX VALUE;
D[s] = 0;
for (int i=0; iG.n(); i++) { // Process the vertices
int v = minVertex(G, D); // Find next-closest vertex
G.setMark(v, VISITED);
if (D[v] == Integer.MAX VALUE) return; // Unreachable
for (int w = G.first(v); w G.n(); w = G.next(v, w))
if (D[w] (D[v] + G.weight(v, w)))
D[w] = D[v] + G.weight(v, w);
}
}
Figure 11.16 An implementation for Dijkstra’s algorithm.
static int minVertex(Graph G, int[] D) {
int v = 0; // Initialize v to any unvisited vertex;
for (int i=0; iG.n(); i++)
if (G.getMark(i) == UNVISITED) { v = i; break; }
for (int i=0; iG.n(); i++) // Now find smallest value
if ((G.getMark(i) == UNVISITED) (D[i] D[v]))
v = i;
return v;
}
Because this scan is done |V| times, and because each edge requires a constant-
time update to D, the total cost for this approach is Θ(|V|2 + |E|) = Θ(|V|2),
because |E| is in O(|V|2).
The second method is to store unprocessed vertices in a min-heap ordered by
distance values. The next-closest vertex can be found in the heap in Θ(log |V|)
time. Every time we modify D(X), we could reorder X in the heap by deleting
and reinserting it. This is an example of a priority queue with priority update, as
described in Section 5.5. To implement true priority updating, we would need to
store with each vertex its array index within the heap. A simpler approach is to
add the new (smaller) distance value for a given vertex as a new record in the heap.
The smallest value for a given vertex currently in the heap will be found first, and
greater distance values found later will be ignored because the vertex will already
be marked as VISITED. The only disadvantage to repeatedly inserting distance
values is that it will raise the number of elements in the heap from Θ(|V|) to Θ(|E|)
in the worst case. The time complexity is Θ((|V| + |E|) log |E|), because for each
edge we must reorder the heap. Because the objects stored on the heap need to
know both their vertex number and their distance, we create a simple class for the
purpose called DijkElem, as follows. DijkElem is quite similar to the Edge
class used by the adjacency list representation.](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-428-320.jpg)
![410 Chap. 11 Graphs
// Dijkstra’s shortest-paths: priority queue version
static void Dijkstra(Graph G, int s, int[] D) {
int v; // The current vertex
DijkElem[] E = new DijkElem[G.e()]; // Heap for edges
E[0] = new DijkElem(s, 0); // Initial vertex
MinHeapDijkElem H = new MinHeapDijkElem(E, 1, G.e());
for (int i=0; iG.n(); i++) // Initialize distance
D[i] = Integer.MAX VALUE;
D[s] = 0;
for (int i=0; iG.n(); i++) { // For each vertex
do { v = (H.removemin()).vertex(); } // Get position
while (G.getMark(v) == VISITED);
G.setMark(v, VISITED);
if (D[v] == Integer.MAX VALUE) return; // Unreachable
for (int w = G.first(v); w G.n(); w = G.next(v, w))
if (D[w] (D[v] + G.weight(v, w))) { // Update D
D[w] = D[v] + G.weight(v, w);
H.insert(new DijkElem(w, D[w]));
}
}
}
Figure 11.17 An implementation for Dijkstra’s algorithm using a priority queue.
import java.lang.Comparable;
class DijkElem implements ComparableDijkElem {
private int vertex;
private int weight;
public DijkElem(int inv, int inw)
{ vertex = inv; weight = inw; }
public DijkElem() {vertex = 0; weight = 0; }
public int key() { return weight; }
public int vertex() { return vertex; }
public int compareTo(DijkElem that) {
if (weight that.key()) return -1;
else if (weight == that.key()) return 0;
else return 1;
}
}
Figure 11.17 shows an implementation for Dijkstra’s algorithm using the prior-
ity queue.
Using MinVertex to scan the vertex list for the minimum value is more ef-
ficient when the graph is dense, that is, when |E| approaches |V|2. Using a prior-
ity queue is more efficient when the graph is sparse because its cost is Θ((|V| +](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-429-320.jpg)
![412 Chap. 11 Graphs
A
9
7 5
B
C
1 2
6
D 2
1
E
F
Figure 11.19 A graph and its MST. All edges appear in the original graph.
Those edges drawn with heavy lines indicate the subset making up the MST. Note
that edge (C, F) could be replaced with edge (D, F) to form a different MST with
equal cost.
11.5.1 Prim’s Algorithm
The first of our two algorithms for finding MSTs is commonly referred to as Prim’s
algorithm. Prim’s algorithm is very simple. Start with any Vertex N in the graph,
setting the MST to be N initially. Pick the least-cost edge connected to N. This
edge connects N to another vertex; call this M. Add Vertex M and Edge (N, M) to
the MST. Next, pick the least-cost edge coming from either N or M to any other
vertex in the graph. Add this edge and the new vertex it reaches to the MST. This
process continues, at each step expanding the MST by selecting the least-cost edge
from a vertex currently in the MST to a vertex not currently in the MST.
Prim’s algorithm is quite similar to Dijkstra’s algorithm for finding the single-
source shortest paths. The primary difference is that we are seeking not the next
closest vertex to the start vertex, but rather the next closest vertex to any vertex
currently in the MST. Thus we replae the lines
if (D[w] (D[v] + G-weight(v, w)))
D[w] = D[v] + G-weight(v, w);
in Djikstra’s algorithm with the lines
if (D[w] G-weight(v, w))
D[w] = G-weight(v, w);
in Prim’s algorithm.
Figure 11.20 shows an implementation for Prim’s algorithm that searches the
distance matrix for the next closest vertex. For each vertex I, when I is processed
by Prim’s algorithm, an edge going to I is added to the MST that we are building.](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-431-320.jpg)
![Sec. 11.5 Minimum-Cost Spanning Trees 413
// Compute a minimal-cost spanning tree
static void Prim(Graph G, int s, int[] D, int[] V) {
for (int i=0; iG.n(); i++) // Initialize
D[i] = Integer.MAX VALUE;
D[s] = 0;
for (int i=0; iG.n(); i++) { // Process the vertices
int v = minVertex(G, D);
G.setMark(v, VISITED);
if (v != s) AddEdgetoMST(V[v], v);
if (D[v] == Integer.MAX VALUE) return; // Unreachable
for (int w = G.first(v); w G.n(); w = G.next(v, w))
if (D[w] G.weight(v, w)) {
D[w] = G.weight(v, w);
V[w] = v;
}
}
}
Figure 11.20 An implementation for Prim’s algorithm.
Array V[I] stores the previously visited vertex that is closest to Vertex I. This
information lets us know which edge goes into the MST when Vertex I is processed.
The implementation of Figure 11.20 also contains calls to AddEdgetoMST to
indicate which edges are actually added to the MST.
Alternatively, we can implement Prim’s algorithm using a priority queue to find
the next closest vertex, as shown in Figure 11.21. As with the priority queue version
of Dijkstra’s algorithm, the heap’s Elem type stores a DijkElem object.
Prim’s algorithm is an example of a greedy algorithm. At each step in the
for loop, we select the least-cost edge that connects some marked vertex to some
unmarked vertex. The algorithm does not otherwise check that the MST really
should include this least-cost edge. This leads to an important question: Does
Prim’s algorithm work correctly? Clearly it generates a spanning tree (because
each pass through the for loop adds one edge and one unmarked vertex to the
spanning tree until all vertices have been added), but does this tree have minimum
cost?
Theorem 11.1 Prim’s algorithm produces a minimum-cost spanning tree.
Proof: We will use a proof by contradiction. Let G = (V, E) be a graph for which
Prim’s algorithm does not generate an MST. Define an ordering on the vertices
according to the order in which they were added by Prim’s algorithm to the MST:
v0, v1, ..., vn−1. Let edge ei connect (vx, vi) for some x i and i ≥ 1. Let ej be the
lowest numbered (first) edge added by Prim’s algorithm such that the set of edges
selected so far cannot be extended to form an MST for G. In other words, ej is the](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-432-320.jpg)
![414 Chap. 11 Graphs
// Prims’s MST algorithm: priority queue version
static void Prim(Graph G, int s, int[] D, int[] V) {
int v; // The current vertex
DijkElem[] E = new DijkElem[G.e()]; // Heap for edges
E[0] = new DijkElem(s, 0); // Initial vertex
MinHeapDijkElem H = new MinHeapDijkElem(E, 1, G.e());
for (int i=0; iG.n(); i++) // Initialize
D[i] = Integer.MAX VALUE; // distances
D[s] = 0;
for (int i=0; iG.n(); i++) { // Now, get distances
do { v = (H.removemin()).vertex(); } // Get position
while (G.getMark(v) == VISITED);
G.setMark(v, VISITED);
if (v != s) AddEdgetoMST(V[v], v); // Add edge to MST
if (D[v] == Integer.MAX VALUE) return; // Unreachable
for (int w = G.first(v); w G.n(); w = G.next(v, w))
if (D[w] G.weight(v, w)) { // Update D
D[w] = G.weight(v, w);
V[w] = v; // Where it came from
H.insert(new DijkElem(w, D[w]));
}
}
}
Figure 11.21 An implementation of Prim’s algorithm using a priority queue.
first edge where Prim’s algorithm “went wrong.” Let T be the “true” MST. Call vp
(p j) the vertex connected by edge ej, that is, ej = (vp, vj).
Because T is a tree, there exists some path in T connecting vp and vj. There
must be some edge e′ in this path connecting vertices vu and vw, with u j and
w ≥ j. Because ej is not part of T, adding edge ej to T forms a cycle. Edge e′ must
be of lower cost than edge ej, because Prim’s algorithm did not generate an MST.
This situation is illustrated in Figure 11.22. However, Prim’s algorithm would have
selected the least-cost edge available. It would have selected e′, not ej. Thus, it is a
contradiction that Prim’s algorithm would have selected the wrong edge, and thus,
Prim’s algorithm must be correct. ✷
Example 11.3 For the graph of Figure 11.19, assume that we begin by
marking Vertex A. From A, the least-cost edge leads to Vertex C. Vertex C
and edge (A, C) are added to the MST. At this point, our candidate edges
connecting the MST (Vertices A and C) with the rest of the graph are (A, E),
(C, B), (C, D), and (C, F). From these choices, the least-cost edge from the
MST is (C, D). So we add Vertex D to the MST. For the next iteration, our
edge choices are (A, E), (C, B), (C, F), and (D, F). Because edges (C, F)](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-433-320.jpg)
![416 Chap. 11 Graphs
MST. The third edge we process is (C, F), which causes the MST contain-
ing Vertices C and D to merge with MST containing Vertices E and F. The
next edge to process is (D, F). But because Vertices D and F are currently
in the same MST, this edge is rejected. The algorithm will continue on to
accept edges (B, C) and (A, C) into the MST.
The edges can be processed in order of weight by using a min-heap. This is
generally faster than sorting the edges first, because in practice we need only visit
a small fraction of the edges before completing the MST. This is an example of
finding only a few smallest elements in a list, as discussed in Section 7.6.
The only tricky part to this algorithm is determining if two vertices belong to
the same equivalence class. Fortunately, the ideal algorithm is available for the
purpose — the UNION/FIND algorithm based on the parent pointer representation
for trees described in Section 6.2. Figure 11.24 shows an implementation for the
algorithm. Class KruskalElem is used to store the edges on the min-heap.
Kruskal’s algorithm is dominated by the time required to process the edges.
The differ and UNION functions are nearly constant in time if path compression
and weighted union is used. Thus, the total cost of the algorithm is Θ(|E| log |E|)
in the worst case, when nearly all edges must be processed before all the edges of
the spanning tree are found and the algorithm can stop. More often the edges of the
spanning tree are the shorter ones,and only about |V| edges must be processed. If
so, the cost is often close to Θ(|V| log |E|) in the average case.
11.6 Further Reading
Many interesting properties of graphs can be investigated by playing with the pro-
grams in the Stanford Graphbase. This is a collection of benchmark databases and
graph processing programs. The Stanford Graphbase is documented in [Knu94].
11.7 Exercises
11.1 Prove by induction that a graph with n vertices has at most n(n−1)/2 edges.
11.2 Prove the following implications regarding free trees.
(a) IF an undirected graph is connected and has no simple cycles, THEN
the graph has |V| − 1 edges.
(b) IF an undirected graph has |V| − 1 edges and no cycles, THEN the
graph is connected.
11.3 (a) Draw the adjacency matrix representation for the graph of Figure 11.25.](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-435-320.jpg)
![418 Chap. 11 Graphs
class KruskalElem implements ComparableKruskalElem {
private int v, w, weight;
public KruskalElem(int inweight, int inv, int inw)
{ weight = inweight; v = inv; w = inw; }
public int v1() { return v; }
public int v2() { return w; }
public int key() { return weight; }
public int compareTo(KruskalElem that) {
if (weight that.key()) return -1;
else if (weight == that.key()) return 0;
else return 1;
}
}
static void Kruskal(Graph G) { // Kruskal’s MST algorithm
ParPtrTree A = new ParPtrTree(G.n()); // Equivalence array
KruskalElem[] E = new KruskalElem[G.e()]; // Minheap array
int edgecnt = 0; // Count of edges
for (int i=0; iG.n(); i++) // Put edges in the array
for (int w = G.first(i); w G.n(); w = G.next(i, w))
E[edgecnt++] = new KruskalElem(G.weight(i, w), i, w);
MinHeapKruskalElem H =
new MinHeapKruskalElem(E, edgecnt, edgecnt);
int numMST = G.n(); // Initially n classes
for (int i=0; numMST1; i++) { // Combine equiv classes
KruskalElem temp = H.removemin(); // Next cheapest
int v = temp.v1(); int u = temp.v2();
if (A.differ(v, u)) { // If in different classes
A.UNION(v, u); // Combine equiv classes
AddEdgetoMST(v, u); // Add this edge to MST
numMST--; // One less MST
}
}
}
Figure 11.24 An implementation for Kruskal’s algorithm.
11.7 Wright a pseudocode algorithm to create a BFS tree for an undirected, con-
nected graph starting at a specified vertex V.
11.8 The BFS topological sort algorithm can report the existence of a cycle if one
is encountered. Modify this algorithm to print the vertices possibly appearing
in cycles (if there are any cycles).
11.9 Explain why, in the worst case, Dijkstra’s algorithm is (asymptotically) as
efficient as any algorithm for finding the shortest path from some vertex I to
another vertex J.](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-437-320.jpg)
![Sec. 12.2 Matrix Representations 427
a00 0 0 0
a10 a11 0 0
a20 a21 a22 0
a30 a31 a32 a33
(a)
a00 a01 a02 a03
0 a11 a12 a13
0 0 a22 a23
0 0 0 a33
(b)
Figure 12.6 Triangular matrices. (a) A lower triangular matrix. (b) An upper
triangular matrix.
12.2 Matrix Representations
Some applications must represent a large, two-dimensional matrix where many of
the elements have a value of zero. One example is the lower triangular matrix that
results from solving systems of simultaneous equations. A lower triangular matrix
stores zero values at positions [r, c] such that r c, as shown in Figure 12.6(a).
Thus, the upper-right triangle of the matrix is always zero. Another example is
the representation of undirected graphs in an adjacency matrix (see Project 11.2).
Because all edges between Vertices i and j go in both directions, there is no need to
store both. Instead we can just store one edge going from the higher-indexed vertex
to the lower-indexed vertex. In this case, only the lower triangle of the matrix can
have non-zero values.
We can take advantage of this fact to save space. Instead of storing n(n + 1)/2
pieces of information in an n × n array, it would save space to use a list of length
n(n + 1)/2. This is only practical if some means can be found to locate within the
list the element that would correspond to position [r, c] in the original matrix.
To derive an equation to do this computation, note that row 0 of the matrix
has one non-zero value, row 1 has two non-zero values, and so on. Thus, row r
is preceded by r rows with a total of
Pr
k=1 k = (r2 + r)/2 non-zero elements.
Adding c to reach the cth position in the rth row yields the following equation to
convert position [r, c] in the original matrix to the correct position in the list.
matrix[r, c] = list[(r2
+ r)/2 + c].
A similar equation can be used to store an upper triangular matrix, that is, a matrix
with zero values at positions [r, c] such that r c, as shown in Figure 12.6(b). For
an n × n upper triangular matrix, the equation would be
matrix[r, c] = list[rn − (r2
+ r)/2 + c].
A more difficult situation arises when the vast majority of values stored in an
n × n matrix are zero, but there is no restriction on which positions are zero and
which are non-zero. This is known as a sparse matrix.](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-446-320.jpg)
![Sec. 12.3 Memory Management 431
/** Memory Manager interface */
interface MemManADT {
/** Store a record and return a handle to it */
public MemHandle insert(byte[] info); // Request space
/** Get back a copy of a stored record */
public byte[] get(MemHandle h); // Retrieve data
/** Release the space associated with a record */
public void release(MemHandle h); // Release space
}
Figure 12.8 A simple ADT for a memory manager.
sage actually stored. Method release allows the client to tell the memory man-
ager to release the space that stores a given message.
When all inserts and releases follow a simple pattern, such as last requested,
first released (stack order), or first requested, first released (queue order), memory
management is fairly easy. We are concerned in this section with the general case
where blocks of any size might be requested and released in any order. This is
known as dynamic storage allocation. One example of dynamic storage allocation
is managing free store for a compiler’s runtime environment, such as the system-
level new operations in Java. Another example is managing main memory in a
multitasking operating system. Here, a program might require a certain amount
of space, and the memory manager must keep track of which programs are using
which parts of the main memory. Yet another example is the file manager for a
disk drive. When a disk file is created, expanded, or deleted, the file manager must
allocate or deallocate disk space.
A block of memory or disk space managed in this way is sometimes referred to
as a heap. The term “heap” is being used here in a different way than the heap data
structure discussed in Section 5.5. Here “heap” refers to the memory controlled by
a dynamic memory management scheme.
In the rest of this section, we first study techniques for dynamic memory man-
agement. We then tackle the issue of what to do when no single block of memory
in the memory pool is large enough to honor a given request.
12.3.1 Dynamic Storage Allocation
For the purpose of dynamic storage allocation, we view memory as a single array
broken into a series of variable-size blocks, where some of the blocks are free and
some are reserved or already allocated. The free blocks are linked together to form](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-450-320.jpg)
![440 Chap. 12 Lists and Arrays Revisited
a
c
d e
f
g h
A
B
C
Freelist
Figure 12.16 Example of LISP list variables, including the system freelist.
int[] p = new int[5];
int[] q = new int[10];
p = q;
While in Java this would be no problem (due to auotmatic garbage collection),
in some languages such as C++, this would be considered bad form because the
original space allocated to p is lost as a result of the third assignment. This space
cannot be used again by the program. Such lost memory is referred to as garbage,
also known as a memory leak. When no program variable points to a block of
space, no future access to that space is possible. Of course, if another variable
had first been assigned to point to p’s space, then reassigning p would not create
garbage.
Some programming languages take a different view towards garbage. In par-
ticular, the LISP programming language uses the multilist representation of Fig-
ure 12.5, and all storage is in the form either of internal nodes with two pointers
or atoms. Figure 12.16 shows a typical collection of LISP structures, headed by
variables named A, B, and C, along with a freelist.
In LISP, list objects are constantly being put together in various ways as tem-
porary variables, and then all reference to them is lost when the object is no longer
needed. Thus, garbage is normal in LISP, and in fact cannot be avoided during
normal processing. When LISP runs out of memory, it resorts to a garbage collec-
tion process to recover the space tied up in garbage. Garbage collection consists of](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-459-320.jpg)
![Sec. 12.4 Further Reading 443
(a)
a
b
4
prev
e
c
curr
(b)
4
6
6
2
2
3
5
1
3
1
5
a
b
c e
Figure 12.18 Example of the Deutsch-Schorr-Waite garbage collection alg-
orithm. (a) The initial multilist structure. (b) The multilist structure of (a) at
the instant when link node 5 is being processed by the garbage collection alg-
orithm. A chain of pointers stretching from variable prev to the head node of the
structure has been (temporarily) created by the garbage collection algorithm.
original condition. This is known as the Deutsch-Schorr-Waite garbage collection
algorithm.
12.4 Further Reading
An introductory text on operating systems covers many topics relating to memory
management issues, including layout of files on disk and caching of information
in main memory. All of the topics covered here on memory management, buffer
pools, and paging are relevant to operating system implementation. For example,
see Operating Systems by William Stallings[Sta05].
For information on LISP, see The Little LISPer by Friedman and Felleisen
[FF89]. Another good LISP reference is Common LISP: The Language by Guy L.
Steele [Ste84]. For information on Emacs, which is both an excellent text editor
and a fully developed programming environment, see the GNU Emacs Manual by
Richard M. Stallman [Sta07]. You can get more information about Java’s garbage
collection system from The Java Programming Language by Ken Arnold and James
Gosling [AG06].](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-462-320.jpg)
![Sec. 13.3 Spatial Data Structures 463
equivalent to the findhelp function of the BST class. Note that KD class private
member D stores the key’s dimension.
private E findhelp(KDNodeE rt, int[] key, int level) {
if (rt == null) return null;
E it = rt.element();
int[] itkey = rt.key();
if ((itkey[0] == key[0]) (itkey[1] == key[1]))
return rt.element();
if (itkey[level] key[level])
return findhelp(rt.left(), key, (level+1)%D);
else
return findhelp(rt.right(), key, (level+1)%D);
}
Inserting a new node into the k-d tree is similar to BST insertion. The k-d tree
search procedure is followed until a null pointer is found, indicating the proper
place to insert the new node.
Example 13.6 Inserting a record at location (10, 50) in the k-d tree of
Figure 13.11 first requires a search to the node containing record B. At this
point, the new record is inserted into B’s left subtree.
Deleting a node from a k-d tree is similar to deleting from a BST, but slightly
harder. As with deleting from a BST, the first step is to find the node (call it N)
to be deleted. It is then necessary to find a descendant of N which can be used to
replace N in the tree. If N has no children, then N is replaced with a null pointer.
Note that if N has one child that in turn has children, we cannot simply assign N’s
parent to point to N’s child as would be done in the BST. To do so would change the
level of all nodes in the subtree, and thus the discriminator used for a search would
also change. The result is that the subtree would no longer be a k-d tree because a
node’s children might now violate the BST property for that discriminator.
Similar to BST deletion, the record stored in N should be replaced either by the
record in N’s right subtree with the least value of N’s discriminator, or by the record
in N’s left subtree with the greatest value for this discriminator. Assume that N was
at an odd level and therefore y is the discriminator. N could then be replaced by the
record in its right subtree with the least y value (call it Ymin). The problem is that
Ymin is not necessarily the leftmost node, as it would be in the BST. A modified
search procedure to find the least y value in the left subtree must be used to find it
instead. The implementation for findmin is shown in Figure 13.12. A recursive
call to the delete routine will then remove Ymin from the tree. Finally, Ymin’s record
is substituted for the record in node N.](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-482-320.jpg)
![464 Chap. 13 Advanced Tree Structures
private KDNodeE
findmin(KDNodeE rt, int descrim, int level) {
KDNodeE temp1, temp2;
int[] key1 = null;
int[] key2 = null;
if (rt == null) return null;
temp1 = findmin(rt.left(), descrim, (level+1)%D);
if (temp1 != null) key1 = temp1.key();
if (descrim != level) {
temp2 = findmin(rt.right(), descrim, (level+1)%D);
if (temp2 != null) key2 = temp2.key();
if ((temp1 == null) || ((temp2 != null)
(key1[descrim] key2[descrim])))
temp1 = temp2;
key1 = key2;
} // Now, temp1 has the smaller value
int[] rtkey = rt.key();
if ((temp1 == null) || (key1[descrim] rtkey[descrim]))
return rt;
else
return temp1;
}
Figure 13.12 The k-d tree findmin method. On levels using the minimum
value’s discriminator, branching is to the left. On other levels, both children’s
subtrees must be visited. Helper function min takes two nodes and a discriminator
as input, and returns the node with the smaller value in that discriminator.
Note that we can replace the node to be deleted with the least-valued node
from the right subtree only if the right subtree exists. If it does not, then a suitable
replacement must be found in the left subtree. Unfortunately, it is not satisfactory
to replace N’s record with the record having the greatest value for the discriminator
in the left subtree, because this new value might be duplicated. If so, then we
would have equal values for the discriminator in N’s left subtree, which violates
the ordering rules for the k-d tree. Fortunately, there is a simple solution to the
problem. We first move the left subtree of node N to become the right subtree (i.e.,
we simply swap the values of N’s left and right child pointers). At this point, we
proceed with the normal deletion process, replacing the record of N to be deleted
with the record containing the least value of the discriminator from what is now
N’s right subtree.
Assume that we want to print out a list of all records that are within a certain
distance d of a given point P. We will use Euclidean distance, that is, point P is
defined to be within distance d of point N if1
q
(Px − Nx)2 + (Py − Ny)2 ≤ d.
1
A more efficient computation is (Px − Nx)2
+ (Py − Ny)2
≤ d2
. This avoids performing a
square root function.](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-483-320.jpg)
![Sec. 13.3 Spatial Data Structures 467
private void rshelp(KDNodeE rt, int[] point,
int radius, int lev) {
if (rt == null) return;
int[] rtkey = rt.key();
if (InCircle(point, radius, rtkey))
System.out.println(rt.element());
if (rtkey[lev] (point[lev] - radius))
rshelp(rt.left(), point, radius, (lev+1)%D);
if (rtkey[lev] (point[lev] + radius))
rshelp(rt.right(), point, radius, (lev+1)%D);
}
Figure 13.15 The k-d tree region search method.
tains more than a single data point, then the region is split into four equal quadrants.
The corresponding PR quadtree then contains an internal node and four subtrees,
each subtree representing a single quadrant of the region, which might in turn be
split into subquadrants. Each internal node of a PR quadtree represents a single
split of the two-dimensional region. The four quadrants of the region (or equiva-
lently, the corresponding subtrees) are designated (in order) NW, NE, SW, and SE.
Each quadrant containing more than a single point would in turn be recursively di-
vided into subquadrants until each leaf of the corresponding PR quadtree contains
at most one point.
For example, consider the region of Figure 13.16(a) and the corresponding
PR quadtree in Figure 13.16(b). The decomposition process demands a fixed key
range. In this example, the region is assumed to be of size 128 × 128. Note that the
internal nodes of the PR quadtree are used solely to indicate decomposition of the
region; internal nodes do not store data records. Because the decomposition lines
are predetermined (i.e, key-space decomposition is used), the PR quadtree is a trie.
Search for a record matching point Q in the PR quadtree is straightforward.
Beginning at the root, we continuously branch to the quadrant that contains Q until
our search reaches a leaf node. If the root is a leaf, then just check to see if the
node’s data record matches point Q. If the root is an internal node, proceed to
the child that contains the search coordinate. For example, the NW quadrant of
Figure 13.16 contains points whose x and y values each fall in the range 0 to 63.
The NE quadrant contains points whose x value falls in the range 64 to 127, and
whose y value falls in the range 0 to 63. If the root’s child is a leaf node, then that
child is checked to see if Q has been found. If the child is another internal node, the
search process continues through the tree until a leaf node is found. If this leaf node
stores a record whose position matches Q then the query is successful; otherwise Q
is not in the tree.](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-486-320.jpg)
![Sec. 13.4 Further Reading 473
Spatial data structures can also be used to store line object, rectangle object,
or objects of arbitrary shape (such as polygons in two dimensions or polyhedra in
three dimensions). A simple, yet effective, data structure for storing rectangles or
arbitrary polygonal shapes can be derived from the PR quadtree. Pick a threshold
value c, and subdivide any region into four quadrants if it contains more than c
objects. A special case must be dealt with when more than c object intersect.
Some of the most interesting developments in spatial data structures have to
do with adapting them for disk-based applications. However, all such disk-based
implementations boil down to storing the spatial data structure within some variant
on either B-trees or hashing.
13.4 Further Reading
PATRICIA tries and other trie implementations are discussed in Information Re-
trieval: Data Structures Algorithms, Frakes and Baeza-Yates, eds. [FBY92].
See Knuth [Knu97] for a discussion of the AVL tree. For further reading on
splay trees, see “Self-adjusting Binary Search” by Sleator and Tarjan [ST85].
The world of spatial data structures is rich and rapidly evolving. For a good
introduction, see Foundations of Multidimensional and Metric Data Structures by
Hanan Samet [Sam06]. This is also the best reference for more information on
the PR quadtree. The k-d tree was invented by John Louis Bentley. For further
information on the k-d tree, in addition to [Sam06], see [Ben75]. For information
on using a quadtree to store arbitrary polygonal objects, see [SH92].
For a discussion on the relative space requirements for two-way versus multi-
way branching, see “A Generalized Comparison of Quadtree and Bintree Storage
Requirements” by Shaffer, Juvvadi, and Heath [SJH93].
Closely related to spatial data structures are data structures for storing multidi-
mensional data (which might not necessarily be spatial in nature). A popular data
structure for storing such data is the R-tree, originally proposed by Guttman [Gut84].
13.5 Exercises
13.1 Show the binary trie (as illustrated by Figure 13.1) for the following collec-
tion of values: 42, 12, 100, 10, 50, 31, 7, 11, 99.
13.2 Show the PAT trie (as illustrated by Figure 13.3) for the following collection
of values: 42, 12, 100, 10, 50, 31, 7, 11, 99.
13.3 Write the insertion routine for a binary trie as shown in Figure 13.1.
13.4 Write the deletion routine for a binary trie as shown in Figure 13.1.](https://image.slidesharecdn.com/apracticalintroductiontodatastructuresandalgorithmanalysisthirdeditionjavaversion-230805172128-4a683686/85/A-Practical-Introduction-To-Data-Structures-And-Algorithm-Analysis-Third-Edition-Java-Version-492-320.jpg)
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A Practical Introduction To Data Structures And Algorithm Analysis Third Edition (Java Version)
- 1. A Practical Introduction to Data Structures and Algorithm Analysis Third Edition (Java Version) Clifford A. Shaffer Department of Computer Science Virginia Tech Blacksburg, VA 24061 October 7, 2010 Copyright c 2009-2010 by Clifford A. Shaffer. This document is made freely available for educational and other non-commercial use. You may make copies of this file and redistribute it without charge. You may extract portions of this document provided that the front page, including the title, author, and this notice are included. Any commercial use of this document requires the written consent of the author. The author can be reached at [email protected]. Further information about this text is available at http://people.cs.vt.edu/˜shaffer/Book/
- 2. Contents Preface xiii I Preliminaries 1 1 Data Structures and Algorithms 3 1.1 A Philosophy of Data Structures 4 1.1.1 The Need for Data Structures 4 1.1.2 Costs and Benefits 6 1.2 Abstract Data Types and Data Structures 8 1.3 Design Patterns 12 1.3.1 Flyweight 13 1.3.2 Visitor 14 1.3.3 Composite 15 1.3.4 Strategy 16 1.4 Problems, Algorithms, and Programs 17 1.5 Further Reading 19 1.6 Exercises 21 2 Mathematical Preliminaries 25 2.1 Sets and Relations 25 2.2 Miscellaneous Notation 29 2.3 Logarithms 31 2.4 Summations and Recurrences 33 iii
- 3. iv Contents 2.5 Recursion 36 2.6 Mathematical Proof Techniques 39 2.6.1 Direct Proof 40 2.6.2 Proof by Contradiction 40 2.6.3 Proof by Mathematical Induction 41 2.7 Estimating 47 2.8 Further Reading 49 2.9 Exercises 50 3 Algorithm Analysis 57 3.1 Introduction 57 3.2 Best, Worst, and Average Cases 63 3.3 A Faster Computer, or a Faster Algorithm? 65 3.4 Asymptotic Analysis 67 3.4.1 Upper Bounds 68 3.4.2 Lower Bounds 70 3.4.3 Θ Notation 71 3.4.4 Simplifying Rules 72 3.4.5 Classifying Functions 73 3.5 Calculating the Running Time for a Program 74 3.6 Analyzing Problems 79 3.7 Common Misunderstandings 81 3.8 Multiple Parameters 83 3.9 Space Bounds 84 3.10 Speeding Up Your Programs 86 3.11 Empirical Analysis 89 3.12 Further Reading 90 3.13 Exercises 91 3.14 Projects 95 II Fundamental Data Structures 97 4 Lists, Stacks, and Queues 99
- 4. Contents v 4.1 Lists 100 4.1.1 Array-Based List Implementation 103 4.1.2 Linked Lists 106 4.1.3 Comparison of List Implementations 117 4.1.4 Element Implementations 119 4.1.5 Doubly Linked Lists 120 4.2 Stacks 125 4.2.1 Array-Based Stacks 125 4.2.2 Linked Stacks 128 4.2.3 Comparison of Array-Based and Linked Stacks 129 4.2.4 Implementing Recursion 129 4.3 Queues 133 4.3.1 Array-Based Queues 134 4.3.2 Linked Queues 137 4.3.3 Comparison of Array-Based and Linked Queues 140 4.4 Dictionaries and Comparators 140 4.5 Further Reading 147 4.6 Exercises 147 4.7 Projects 150 5 Binary Trees 153 5.1 Definitions and Properties 153 5.1.1 The Full Binary Tree Theorem 156 5.1.2 A Binary Tree Node ADT 157 5.2 Binary Tree Traversals 158 5.3 Binary Tree Node Implementations 162 5.3.1 Pointer-Based Node Implementations 163 5.3.2 Space Requirements 169 5.3.3 Array Implementation for Complete Binary Trees 170 5.4 Binary Search Trees 171 5.5 Heaps and Priority Queues 180 5.6 Huffman Coding Trees 187 5.6.1 Building Huffman Coding Trees 189
- 5. vi Contents 5.6.2 Assigning and Using Huffman Codes 195 5.7 Further Reading 198 5.8 Exercises 198 5.9 Projects 202 6 Non-Binary Trees 205 6.1 General Tree Definitions and Terminology 205 6.1.1 An ADT for General Tree Nodes 206 6.1.2 General Tree Traversals 207 6.2 The Parent Pointer Implementation 208 6.3 General Tree Implementations 216 6.3.1 List of Children 217 6.3.2 The Left-Child/Right-Sibling Implementation 218 6.3.3 Dynamic Node Implementations 218 6.3.4 Dynamic “Left-Child/Right-Sibling” Implementation 220 6.4 K-ary Trees 221 6.5 Sequential Tree Implementations 223 6.6 Further Reading 226 6.7 Exercises 226 6.8 Projects 230 III Sorting and Searching 233 7 Internal Sorting 235 7.1 Sorting Terminology and Notation 236 7.2 Three Θ(n2) Sorting Algorithms 237 7.2.1 Insertion Sort 238 7.2.2 Bubble Sort 240 7.2.3 Selection Sort 241 7.2.4 The Cost of Exchange Sorting 243 7.3 Shellsort 244 7.4 Mergesort 246 7.5 Quicksort 249
- 6. Contents vii 7.6 Heapsort 256 7.7 Binsort and Radix Sort 259 7.8 An Empirical Comparison of Sorting Algorithms 265 7.9 Lower Bounds for Sorting 267 7.10 Further Reading 271 7.11 Exercises 272 7.12 Projects 275 8 File Processing and External Sorting 279 8.1 Primary versus Secondary Storage 280 8.2 Disk Drives 282 8.2.1 Disk Drive Architecture 283 8.2.2 Disk Access Costs 286 8.3 Buffers and Buffer Pools 289 8.4 The Programmer’s View of Files 297 8.5 External Sorting 298 8.5.1 Simple Approaches to External Sorting 301 8.5.2 Replacement Selection 304 8.5.3 Multiway Merging 307 8.6 Further Reading 310 8.7 Exercises 311 8.8 Projects 315 9 Searching 317 9.1 Searching Unsorted and Sorted Arrays 318 9.2 Self-Organizing Lists 324 9.3 Bit Vectors for Representing Sets 329 9.4 Hashing 330 9.4.1 Hash Functions 331 9.4.2 Open Hashing 336 9.4.3 Closed Hashing 337 9.4.4 Analysis of Closed Hashing 346 9.4.5 Deletion 350
- 7. viii Contents 9.5 Further Reading 351 9.6 Exercises 352 9.7 Projects 355 10 Indexing 357 10.1 Linear Indexing 359 10.2 ISAM 361 10.3 Tree-based Indexing 364 10.4 2-3 Trees 366 10.5 B-Trees 372 10.5.1 B+-Trees 375 10.5.2 B-Tree Analysis 381 10.6 Further Reading 382 10.7 Exercises 382 10.8 Projects 385 IV Advanced Data Structures 387 11 Graphs 389 11.1 Terminology and Representations 390 11.2 Graph Implementations 394 11.3 Graph Traversals 397 11.3.1 Depth-First Search 400 11.3.2 Breadth-First Search 401 11.3.3 Topological Sort 405 11.4 Shortest-Paths Problems 407 11.4.1 Single-Source Shortest Paths 407 11.5 Minimum-Cost Spanning Trees 411 11.5.1 Prim’s Algorithm 412 11.5.2 Kruskal’s Algorithm 415 11.6 Further Reading 416 11.7 Exercises 416 11.8 Projects 420
- 8. Contents ix 12 Lists and Arrays Revisited 423 12.1 Multilists 423 12.2 Matrix Representations 427 12.3 Memory Management 430 12.3.1 Dynamic Storage Allocation 431 12.3.2 Failure Policies and Garbage Collection 438 12.4 Further Reading 443 12.5 Exercises 444 12.6 Projects 445 13 Advanced Tree Structures 447 13.1 Tries 447 13.2 Balanced Trees 452 13.2.1 The AVL Tree 453 13.2.2 The Splay Tree 455 13.3 Spatial Data Structures 459 13.3.1 The K-D Tree 461 13.3.2 The PR quadtree 466 13.3.3 Other Point Data Structures 471 13.3.4 Other Spatial Data Structures 471 13.4 Further Reading 473 13.5 Exercises 473 13.6 Projects 475 V Theory of Algorithms 479 14 Analysis Techniques 481 14.1 Summation Techniques 482 14.2 Recurrence Relations 487 14.2.1 Estimating Upper and Lower Bounds 488 14.2.2 Expanding Recurrences 491 14.2.3 Divide and Conquer Recurrences 493 14.2.4 Average-Case Analysis of Quicksort 495
- 9. x Contents 14.3 Amortized Analysis 497 14.4 Further Reading 500 14.5 Exercises 500 14.6 Projects 505 15 Lower Bounds 507 15.1 Introduction to Lower Bounds Proofs 508 15.2 Lower Bounds on Searching Lists 510 15.2.1 Searching in Unsorted Lists 510 15.2.2 Searching in Sorted Lists 512 15.3 Finding the Maximum Value 513 15.4 Adversarial Lower Bounds Proofs 515 15.5 State Space Lower Bounds Proofs 518 15.6 Finding the ith Best Element 522 15.7 Optimal Sorting 524 15.8 Further Reading 527 15.9 Exercises 527 15.10Projects 529 16 Patterns of Algorithms 531 16.1 Greedy Algorithms 531 16.2 Dynamic Programming 532 16.2.1 Knapsack Problem 533 16.2.2 All-Pairs Shortest Paths 534 16.3 Randomized Algorithms 536 16.3.1 Skip Lists 538 16.4 Numerical Algorithms 543 16.4.1 Exponentiation 544 16.4.2 Largest Common Factor 545 16.4.3 Matrix Multiplication 546 16.4.4 Random Numbers 548 16.4.5 Fast Fourier Transform 549 16.5 Further Reading 553
- 10. Contents xi 16.6 Exercises 553 16.7 Projects 554 17 Limits to Computation 555 17.1 Reductions 556 17.2 Hard Problems 561 17.2.1 The Theory of NP-Completeness 562 17.2.2 NP-Completeness Proofs 567 17.2.3 Coping with NP-Complete Problems 571 17.3 Impossible Problems 575 17.3.1 Uncountability 576 17.3.2 The Halting Problem Is Unsolvable 579 17.4 Further Reading 583 17.5 Exercises 583 17.6 Projects 586 Bibliography 587 Index 593
- 12. Preface We study data structures so that we can learn to write more efficient programs. But why must programs be efficient when new computers are faster every year? The reason is that our ambitions grow with our capabilities. Instead of rendering effi- ciency needs obsolete, the modern revolution in computing power and storage ca- pability merely raises the efficiency stakes as we computerize more complex tasks. The quest for program efficiency need not and should not conflict with sound design and clear coding. Creating efficient programs has little to do with “program- ming tricks” but rather is based on good organization of information and good al- gorithms. A programmer who has not mastered the basic principles of clear design is not likely to write efficient programs. Conversely, “software engineering” cannot be used as an excuse to justify inefficient performance. Generality in design can and should be achieved without sacrificing performance, but this can only be done if the designer understands how to measure performance and does so as an integral part of the design and implementation process. Most computer science curricula recognize that good programming skills begin with a strong emphasis on funda- mental software engineering principles. Then, once a programmer has learned the principles of clear program design and implementation, the next step is to study the effects of data organization and algorithms on program efficiency. Approach: This book describes many techniques for representing data. These techniques are presented within the context of the following principles: 1. Each data structure and each algorithm has costs and benefits. Practitioners need a thorough understanding of how to assess costs and benefits to be able to adapt to new design challenges. This requires an understanding of the principles of algorithm analysis, and also an appreciation for the significant effects of the physical medium employed (e.g., data stored on disk versus main memory). xiii
- 13. xiv Preface 2. Related to costs and benefits is the notion of tradeoffs. For example, it is quite common to reduce time requirements at the expense of an increase in space requirements, or vice versa. Programmers face tradeoff issues regularly in all phases of software design and implementation, so the concept must become deeply ingrained. 3. Programmers should know enough about common practice to avoid rein- venting the wheel. Thus, programmers need to learn the commonly used data structures, their related algorithms, and the most frequently encountered design patterns found in programming. 4. Data structures follow needs. Programmers must learn to assess application needs first, then find a data structure with matching capabilities. To do this requires competence in principles 1, 2, and 3. As I have taught data structures through the years, I have found that design issues have played an ever greater role in my courses. This can be traced through the various editions of this textbook by the increasing coverage for design patterns and generic interfaces. The first edition had no mention of design patterns. The second edition had limited coverage of a few example patterns, and introduced the dictionary ADT and comparator classes. With the third edition, there is explicit coverage of some design patterns that are encountered when programming the basic data structures and algorithms covered in the book. Using the Book in Class: Data structures and algorithms textbooks tend to fall into one of two categories: teaching texts or encyclopedias. Books that attempt to do both usually fail at both. This book is intended as a teaching text. I believe it is more important for a practitioner to understand the principles required to select or design the data structure that will best solve some problem than it is to memorize a lot of textbook implementations. Hence, I have designed this as a teaching text that covers most standard data structures, but not all. A few data structures that are not widely adopted are included to illustrate important principles. Some relatively new data structures that should become widely used in the future are included. Within an undergraduate program, this textbook is designed for use in either an advanced lower division (sophomore or junior level) data structures course, or for a senior level algorithms course. New material has been added in the third edition to support its use in an algorithms course. Normally, this text would be used in a course beyond the standard freshman level “CS2” course that often serves as the ini- tial introduction to data structures. Readers of this book should have programming experience, typically two semesters or the equivalent of a structured programming language such as Pascal or C, and including at least some exposure to Java. Read- ers who are already familiar with recursion will have an advantage. Students of
- 14. Preface xv data structures will also benefit from having first completed a good course in Dis- crete Mathematics. Nonetheless, Chapter 2 attempts to give a reasonably complete survey of the prerequisite mathematical topics at the level necessary to understand their use in this book. Readers may wish to refer back to the appropriate sections as needed when encountering unfamiliar mathematical material. A sophomore-level class where students have only a little background in basic data structures or analysis (that is, background equivalent to what would be had from a traditional CS2 course) might cover Chapters 1-11 in detail, as well as se- lected topics from Chapter 13. That is how I use the book for my own sophomore- level class. Students with greater background might cover Chapter 1, skip most of Chapter 2 except for reference, briefly cover Chapters 3 and 4, and then cover chapters 5-12 in detail. Again, only certain topics from Chapter 13 might be cov- ered, depending on the programming assignments selected by the instructor. A senior-level algorithms course would focus on Chapters 11 and 14-17. Chapter 13 is intended in part as a source for larger programming exercises. I recommend that all students taking a data structures course be required to im- plement some advanced tree structure, or another dynamic structure of comparable difficulty such as the skip list or sparse matrix representations of Chapter 12. None of these data structures are significantly more difficult to implement than the binary search tree, and any of them should be within a student’s ability after completing Chapter 5. While I have attempted to arrange the presentation in an order that makes sense, instructors should feel free to rearrange the topics as they see fit. The book has been written so that once the reader has mastered Chapters 1-6, the remaining material has relatively few dependencies. Clearly, external sorting depends on understand- ing internal sorting and disk files. Section 6.2 on the UNION/FIND algorithm is used in Kruskal’s Minimum-Cost Spanning Tree algorithm. Section 9.2 on self- organizing lists mentions the buffer replacement schemes covered in Section 8.3. Chapter 14 draws on examples from throughout the book. Section 17.2 relies on knowledge of graphs. Otherwise, most topics depend only on material presented earlier within the same chapter. Most chapters end with a section entitled “Further Reading.” These sections are not comprehensive lists of references on the topics presented. Rather, I include books and articles that, in my opinion, may prove exceptionally informative or entertaining to the reader. In some cases I include references to works that should become familiar to any well-rounded computer scientist. Use of Java: The programming examples are written in JavaTM . As with any programming language, Java has both advantages and disadvantages. Java is a
- 15. xvi Preface small language. There usually is only one way to do something, and this has the happy tendency of encouraging a programmer toward clarity when used correctly. In this respect, it is superior to C or C++. Java serves nicely for defining and using most traditional data structures such as lists and trees. On the other hand, Java is quite poor when used to do file processing, being both cumbersome and inefficient. It is also a poor language when fine control of memory is required. As an example, applications requiring memory management, such as those discussed in Section 12.3, are difficult to write in Java. Since I wish to stick to a single language throughout the text, like any programmer I must take the bad along with the good. The most important issue is to get the ideas across, whether or not those ideas are natural to a particular language of discourse. Most programmers will use a variety of programming languages throughout their career, and the concepts described in this book should prove useful in a variety of circumstances. I do not wish to discourage those unfamiliar with Java from reading this book. I have attempted to make the examples as clear as possible while maintaining the advantages of Java. Java is used here strictly as a tool to illustrate data structures concepts. Fortunately, Java is an easy language for C or Pascal programmers to read with a minimal amount of study of the syntax related to object-oriented program- ming. In particular, I make use of Java’s support for hiding implementation details, including features such as classes, private class members, and interfaces. These features of the language support the crucial concept of separating logical design, as embodied in the abstract data type, from physical implementation as embodied in the data structure. I make no attempt to teach Java within the text. An Appendix is provided that describes the Java syntax and concepts necessary to understand the program examples. I also provide the actual Java code used in the text through anonymous FTP. Inheritance, a key feature of object-oriented programming, is used only spar- ingly in the code examples. Inheritance is an important tool that helps programmers avoid duplication, and thus minimize bugs. From a pedagogical standpoint, how- ever, inheritance often makes code examples harder to understand since it tends to spread the description for one logical unit among several classes. Thus, some of my class definitions for objects such as tree or list nodes do not take full advantage of possible inheritance from earlier code examples. This does not mean that a pro- grammer should do likewise. Avoiding code duplication and minimizing errors are important goals. Treat the programming examples as illustrations of data structure principles, but do not copy them directly into your own programs.
- 16. Preface xvii My Java implementations serve to provide concrete illustrations of data struc- ture principles. They are not meant to be a series of commercial-quality Java class implementations. The code examples provide less parameter checking than is sound programming practice for commercial programmers. Some parameter checking is included in the form of calls to methods in class Assert. These methods are modeled after the C standard library function assert. Method Assert.notFalse takes a Boolean expression. If this expression evaluates to false, then the program terminates immediately. Method Assert.notNull takes a reference to class Object, and terminates the program if the value of the reference is null. (To be precise, these functions throw an IllegalArgument- Exception, which typically results in terminating the program unless the pro- grammer takes action to handle the exception.) Terminating a program when a function receives a bad parameter is generally considered undesirable in real pro- grams, but is quite adequate for understanding how a data structure is meant to op- erate. In real programming applications, Java’s exception handling features should be used to deal with input data errors. I make a distinction in the text between “Java implementations” and “pseu- docode.” Code labeled as a Java implementation has actually been compiled and tested on one or more Java compilers. Pseudocode examples often conform closely to Java syntax, but typically contain one or more lines of higher level description. Pseudocode is used where I perceived a greater pedagogical advantage to a simpler, but less precise, description. Exercises and Projects: Proper implementation and anaysis of data structures cannot be learned simply by reading a book. You must practice by implementing real programs, constantly comparing different techniques to see what really works best in a given situation. One of the most important aspects of a course in data structures is that it is where students really learn to program using pointers and dynamic memory alloca- tion, by implementing data structures such as linked lists and trees. Its also where students truly learn recursion. In our curriculum, this is the first course where students do significant design, because it often requires real data structures to mo- tivate significant design exercises. Finally, the fundamental differences between memory-based and disk-based data access cannot be appreciated without practical programming experience. For all of these reasons, a data structures course cannot succeed without a significant programming component. In our department, the data structures course is arguably the most difficult programming course in the curricu- lum.
- 17. xviii Preface Students should also work problems to develop their analytical abilities. I pro- vide over 400 exercises and suggestions for programming projects. I urge readers to take advantage of them. Contacting the Author and Supplementary Materials: A book such as this is sure to contain errors and have room for improvement. I welcome bug reports and constructive criticism. I can be reached by electronic mail via the Internet at [email protected]. Alternatively, comments can be mailed to Cliff Shaffer Department of Computer Science Virginia Tech Blacksburg, VA 24061 A set of transparency masters for use in conjunction with this book can be ob- tained via the WWW at http://www.cs.vt.edu/˜shaffer/book.html. The Java code examples are also available this site. Online Web pages for Virginia Tech’s sophomore-level data structures class can be found at URL http://courses.cs.vt.edu/˜cs3114 This book was originally typeset by the author with L A TEX. The bibliography was prepared using BIBTEX. The index was prepared using makeindex. The figures were mostly drawn with Xfig. Figures 3.1 and 9.8 were partially created using Mathematica. Acknowledgments: It takes a lot of help from a lot of people to make a book. I wish to acknowledge a few of those who helped to make this book possible. I apologize for the inevitable omissions. Virginia Tech helped make this whole thing possible through sabbatical re- search leave during Fall 1994, enabling me to get the project off the ground. My de- partment heads during the time I have written the various editions of this book, Den- nis Kafura and Jack Carroll, provided unwavering moral support for this project. Mike Keenan, Lenny Heath, and Jeff Shaffer provided valuable input on early ver- sions of the chapters. I also wish to thank Lenny Heath for many years of stimulat- ing discussions about algorithms and analysis (and how to teach both to students). Steve Edwards deserves special thanks for spending so much time helping me on various redesigns of the C++ and Java code versions for the second and third edi- tions, and many hours of discussion on the principles of program design. Thanks to Layne Watson for his help with Mathematica, and to Bo Begole, Philip Isenhour, Jeff Nielsen, and Craig Struble for much technical assistance. Thanks to Bill Mc- Quain, Mark Abrams and Dennis Kafura for answering lots of silly questions about C++ and Java.
- 18. Preface xix I am truly indebted to the many reviewers of the various editions of this manu- script. For the first edition these reviewers included J. David Bezek (University of Evansville), Douglas Campbell (Brigham Young University), Karen Davis (Univer- sity of Cincinnati), Vijay Kumar Garg (University of Texas – Austin), Jim Miller (University of Kansas), Bruce Maxim (University of Michigan – Dearborn), Jeff Parker (Agile Networks/Harvard), Dana Richards (George Mason University), Jack Tan (University of Houston), and Lixin Tao (Concordia University). Without their help, this book would contain many more technical errors and many fewer insights. For the second edition, I wish to thank these reviewers: Gurdip Singh (Kansas State University), Peter Allen (Columbia University), Robin Hill (University of Wyoming), Norman Jacobson (University of California – Irvine), Ben Keller (East- ern Michigan University), and Ken Bosworth (Idaho State University). In addition, I wish to thank Neil Stewart and Frank J. Thesen for their comments and ideas for improvement. Third edition reviewers included Randall Lechlitner (University of Houstin, Clear Lake) and Brian C. Hipp (York Technical College). I thank them for their comments. Without the hard work of many people at Prentice Hall, none of this would be possible. Authors simply do not create printer-ready books on their own. Foremost thanks go to Kate Hargett, Petra Rector, Laura Steele, and Alan Apt, my editors over the years. My production editors, Irwin Zucker for the second edition, Kath- leen Caren for the original C++ version, and Ed DeFelippis for the Java version, kept everything moving smoothly during that horrible rush at the end. Thanks to Bill Zobrist and Bruce Gregory (I think) for getting me into this in the first place. Others at Prentice Hall who helped me along the way include Truly Donovan, Linda Behrens, and Phyllis Bregman. I am sure I owe thanks to many others at Prentice Hall for their help in ways that I am not even aware of. I wish to express my appreciation to Hanan Samet for teaching me about data structures. I learned much of the philosophy presented here from him as well, though he is not responsible for any problems with the result. Thanks to my wife Terry, for her love and support, and to my daughters Irena and Kate for pleasant diversions from working too hard. Finally, and most importantly, to all of the data structures students over the years who have taught me what is important and what should be skipped in a data structures course, and the many new insights they have provided. This book is dedicated to them. Clifford A. Shaffer Blacksburg, Virginia
- 22. 1 Data Structures and Algorithms How many cities with more than 250,000 people lie within 500 miles of Dallas, Texas? How many people in my company make over $100,000 per year? Can we connect all of our telephone customers with less than 1,000 miles of cable? To answer questions like these, it is not enough to have the necessary information. We must organize that information in a way that allows us to find the answers in time to satisfy our needs. Representing information is fundamental to computer science. The primary purpose of most computer programs is not to perform calculations, but to store and retrieve information — usually as fast as possible. For this reason, the study of data structures and the algorithms that manipulate them is at the heart of computer science. And that is what this book is about — helping you to understand how to structure information to support efficient processing. This book has three primary goals. The first is to present the commonly used data structures. These form a programmer’s basic data structure “toolkit.” For many problems, some data structure in the toolkit provides a good solution. The second goal is to introduce the idea of tradeoffs and reinforce the concept that there are costs and benefits associated with every data structure. This is done by describing, for each data structure, the amount of space and time required for typical operations. The third goal is to teach how to measure the effectiveness of a data structure or algorithm. Only through such measurement can you determine which data structure in your toolkit is most appropriate for a new problem. The techniques presented also allow you to judge the merits of new data structures that you or others might invent. There are often many approaches to solving a problem. How do we choose between them? At the heart of computer program design are two (sometimes con- flicting) goals: 3
- 23. 4 Chap. 1 Data Structures and Algorithms 1. To design an algorithm that is easy to understand, code, and debug. 2. To design an algorithm that makes efficient use of the computer’s resources. Ideally, the resulting program is true to both of these goals. We might say that such a program is “elegant.” While the algorithms and program code examples pre- sented here attempt to be elegant in this sense, it is not the purpose of this book to explicitly treat issues related to goal (1). These are primarily concerns of the disci- pline of Software Engineering. Rather, this book is mostly about issues relating to goal (2). How do we measure efficiency? Chapter 3 describes a method for evaluating the efficiency of an algorithm or computer program, called asymptotic analysis. Asymptotic analysis also allows you to measure the inherent difficulty of a problem. The remaining chapters use asymptotic analysis techniques for every algorithm presented. This allows you to see how each algorithm compares to other algorithms for solving the same problem in terms of its efficiency. This first chapter sets the stage for what is to follow, by presenting some higher- order issues related to the selection and use of data structures. We first examine the process by which a designer selects a data structure appropriate to the task at hand. We then consider the role of abstraction in program design. We briefly consider the concept of a design pattern and see some examples. The chapter ends with an exploration of the relationship between problems, algorithms, and programs. 1.1 A Philosophy of Data Structures 1.1.1 The Need for Data Structures You might think that with ever more powerful computers, program efficiency is becoming less important. After all, processor speed and memory size still seem to double every couple of years. Won’t any efficiency problem we might have today be solved by tomorrow’s hardware? As we develop more powerful computers, our history so far has always been to use additional computing power to tackle more complex problems, be it in the form of more sophisticated user interfaces, bigger problem sizes, or new problems previously deemed computationally infeasible. More complex problems demand more computation, making the need for efficient programs even greater. Worse yet, as tasks become more complex, they become less like our everyday experience. Today’s computer scientists must be trained to have a thorough understanding of the principles behind efficient program design, because their ordinary life experiences often do not apply when designing computer programs.
- 24. Sec. 1.1 A Philosophy of Data Structures 5 In the most general sense, a data structure is any data representation and its associated operations. Even an integer or floating point number stored on the com- puter can be viewed as a simple data structure. More typically, a data structure is meant to be an organization or structuring for a collection of data items. A sorted list of integers stored in an array is an example of such a structuring. Given sufficient space to store a collection of data items, it is always possible to search for specified items within the collection, print or otherwise process the data items in any desired order, or modify the value of any particular data item. Thus, it is possible to perform all necessary operations on any data structure. However, using the proper data structure can make the difference between a program running in a few seconds and one requiring many days. A solution is said to be efficient if it solves the problem within the required resource constraints. Examples of resource constraints include the total space available to store the data — possibly divided into separate main memory and disk space constraints — and the time allowed to perform each subtask. A solution is sometimes said to be efficient if it requires fewer resources than known alternatives, regardless of whether it meets any particular requirements. The cost of a solution is the amount of resources that the solution consumes. Most often, cost is measured in terms of one key resource such as time, with the implied assumption that the solution meets the other resource constraints. It should go without saying that people write programs to solve problems. How- ever, it is crucial to keep this truism in mind when selecting a data structure to solve a particular problem. Only by first analyzing the problem to determine the perfor- mance goals that must be achieved can there be any hope of selecting the right data structure for the job. Poor program designers ignore this analysis step and apply a data structure that they are familiar with but which is inappropriate to the problem. The result is typically a slow program. Conversely, there is no sense in adopting a complex representation to “improve” a program that can meet its performance goals when implemented using a simpler design. When selecting a data structure to solve a problem, you should follow these steps. 1. Analyze your problem to determine the basic operations that must be sup- ported. Examples of basic operations include inserting a data item into the data structure, deleting a data item from the data structure, and finding a specified data item. 2. Quantify the resource constraints for each operation. 3. Select the data structure that best meets these requirements.
- 25. 6 Chap. 1 Data Structures and Algorithms This three-step approach to selecting a data structure operationalizes a data- centered view of the design process. The first concern is for the data and the op- erations to be performed on them, the next concern is the representation for those data, and the final concern is the implementation of that representation. Resource constraints on certain key operations, such as search, inserting data records, and deleting data records, normally drive the data structure selection pro- cess. Many issues relating to the relative importance of these operations are ad- dressed by the following three questions, which you should ask yourself whenever you must choose a data structure: • Are all data items inserted into the data structure at the beginning, or are insertions interspersed with other operations? • Can data items be deleted? • Are all data items processed in some well-defined order, or is search for specific data items allowed? Typically, interspersing insertions with other operations, allowing deletion, and supporting search for data items all require more complex representations. 1.1.2 Costs and Benefits Each data structure has associated costs and benefits. In practice, it is hardly ever true that one data structure is better than another for use in all situations. If one data structure or algorithm is superior to another in all respects, the inferior one will usually have long been forgotten. For nearly every data structure and algorithm presented in this book, you will see examples of where it is the best choice. Some of the examples might surprise you. A data structure requires a certain amount of space for each data item it stores, a certain amount of time to perform a single basic operation, and a certain amount of programming effort. Each problem has constraints on available space and time. Each solution to a problem makes use of the basic operations in some relative pro- portion, and the data structure selection process must account for this. Only after a careful analysis of your problem’s characteristics can you determine the best data structure for the task. Example 1.1 A bank must support many types of transactions with its customers, but we will examine a simple model where customers wish to open accounts, close accounts, and add money or withdraw money from accounts. We can consider this problem at two distinct levels: (1) the re- quirements for the physical infrastructure and workflow process that the
- 26. Sec. 1.1 A Philosophy of Data Structures 7 bank uses in its interactions with its customers, and (2) the requirements for the database system that manages the accounts. The typical customer opens and closes accounts far less often than he or she accesses the account. Customers are willing to wait many minutes while accounts are created or deleted but are typically not willing to wait more than a brief time for individual account transactions such as a deposit or withdrawal. These observations can be considered as informal specifica- tions for the time constraints on the problem. It is common practice for banks to provide two tiers of service. Hu- man tellers or automated teller machines (ATMs) support customer access to account balances and updates such as deposits and withdrawals. Spe- cial service representatives are typically provided (during restricted hours) to handle opening and closing accounts. Teller and ATM transactions are expected to take little time. Opening or closing an account can take much longer (perhaps up to an hour from the customer’s perspective). From a database perspective, we see that ATM transactions do not mod- ify the database significantly. For simplicity, assume that if money is added or removed, this transaction simply changes the value stored in an account record. Adding a new account to the database is allowed to take several minutes. Deleting an account need have no time constraint, because from the customer’s point of view all that matters is that all the money be re- turned (equivalent to a withdrawal). From the bank’s point of view, the account record might be removed from the database system after business hours, or at the end of the monthly account cycle. When considering the choice of data structure to use in the database system that manages customer accounts, we see that a data structure that has little concern for the cost of deletion, but is highly efficient for search and moderately efficient for insertion, should meet the resource constraints imposed by this problem. Records are accessible by unique account number (sometimes called an exact-match query). One data structure that meets these requirements is the hash table described in Chapter 9.4. Hash tables allow for extremely fast exact-match search. A record can be modified quickly when the modification does not affect its space requirements. Hash tables also support efficient insertion of new records. While deletions can also be supported efficiently, too many deletions lead to some degradation in performance for the remaining operations. However, the hash table can be reorganized periodically to restore the system to peak efficiency. Such reorganization can occur offline so as not to affect ATM transactions.
- 27. 8 Chap. 1 Data Structures and Algorithms Example 1.2 A company is developing a database system containing in- formation about cities and towns in the United States. There are many thousands of cities and towns, and the database program should allow users to find information about a particular place by name (another example of an exact-match query). Users should also be able to find all places that match a particular value or range of values for attributes such as location or population size. This is known as a range query. A reasonable database system must answer queries quickly enough to satisfy the patience of a typical user. For an exact-match query, a few sec- onds is satisfactory. If the database is meant to support range queries that can return many cities that match the query specification, the entire opera- tion may be allowed to take longer, perhaps on the order of a minute. To meet this requirement, it will be necessary to support operations that pro- cess range queries efficiently by processing all cities in the range as a batch, rather than as a series of operations on individual cities. The hash table suggested in the previous example is inappropriate for implementing our city database, because it cannot perform efficient range queries. The B+-tree of Section 10.5.1 supports large databases, insertion and deletion of data records, and range queries. However, a simple linear in- dex as described in Section 10.1 would be more appropriate if the database is created once, and then never changed, such as an atlas distributed on a CD-ROM. 1.2 Abstract Data Types and Data Structures The previous section used the terms “data item” and “data structure” without prop- erly defining them. This section presents terminology and motivates the design process embodied in the three-step approach to selecting a data structure. This mo- tivation stems from the need to manage the tremendous complexity of computer programs. A type is a collection of values. For example, the Boolean type consists of the values true and false. The integers also form a type. An integer is a simple type because its values contain no subparts. A bank account record will typically contain several pieces of information such as name, address, account number, and account balance. Such a record is an example of an aggregate type or composite type. A data item is a piece of information or a record whose value is drawn from a type. A data item is said to be a member of a type.
- 28. Sec. 1.2 Abstract Data Types and Data Structures 9 A data type is a type together with a collection of operations to manipulate the type. For example, an integer variable is a member of the integer data type. Addition is an example of an operation on the integer data type. A distinction should be made between the logical concept of a data type and its physical implementation in a computer program. For example, there are two tra- ditional implementations for the list data type: the linked list and the array-based list. The list data type can therefore be implemented using a linked list or an ar- ray. Even the term “array” is ambiguous in that it can refer either to a data type or an implementation. “Array” is commonly used in computer programming to mean a contiguous block of memory locations, where each memory location stores one fixed-length data item. By this meaning, an array is a physical data structure. However, array can also mean a logical data type composed of a (typically ho- mogeneous) collection of data items, with each data item identified by an index number. It is possible to implement arrays in many different ways. For exam- ple, Section 12.2 describes the data structure used to implement a sparse matrix, a large two-dimensional array that stores only a relatively few non-zero values. This implementation is quite different from the physical representation of an array as contiguous memory locations. An abstract data type (ADT) is the realization of a data type as a software component. The interface of the ADT is defined in terms of a type and a set of operations on that type. The behavior of each operation is determined by its inputs and outputs. An ADT does not specify how the data type is implemented. These implementation details are hidden from the user of the ADT and protected from outside access, a concept referred to as encapsulation. A data structure is the implementation for an ADT. In an object-oriented lan- guage such as Java, an ADT and its implementation together make up a class. Each operation associated with the ADT is implemented by a member function or method. The variables that define the space required by a data item are referred to as data members. An object is an instance of a class, that is, something that is created and takes up storage during the execution of a computer program. The term “data structure” often refers to data stored in a computer’s main mem- ory. The related term file structure often refers to the organization of data on peripheral storage, such as a disk drive or CD-ROM. Example 1.3 The mathematical concept of an integer, along with opera- tions that manipulate integers, form a data type. The Java int variable type is a physical representation of the abstract integer. The int variable type, along with the operations that act on an int variable, form an ADT. Un-
- 29. 10 Chap. 1 Data Structures and Algorithms fortunately, the int implementation is not completely true to the abstract integer, as there are limitations on the range of values an int variable can store. If these limitations prove unacceptable, then some other represen- tation for the ADT “integer” must be devised, and a new implementation must be used for the associated operations. Example 1.4 An ADT for a list of integers might specify the following operations: • Insert a new integer at a particular position in the list. • Return true if the list is empty. • Reinitialize the list. • Return the number of integers currently in the list. • Delete the integer at a particular position in the list. From this description, the input and output of each operation should be clear, but the implementation for lists has not been specified. One application that makes use of some ADT might use particular member functions of that ADT more than a second application, or the two applications might have different time requirements for the various operations. These differences in the requirements of applications are the reason why a given ADT might be supported by more than one implementation. Example 1.5 Two popular implementations for large disk-based database applications are hashing (Section 9.4) and the B+-tree (Section 10.5). Both support efficient insertion and deletion of records, and both support exact- match queries. However, hashing is more efficient than the B+-tree for exact-match queries. On the other hand, the B+-tree can perform range queries efficiently, while hashing is hopelessly inefficient for range queries. Thus, if the database application limits searches to exact-match queries, hashing is preferred. On the other hand, if the application requires support for range queries, the B+-tree is preferred. Despite these performance is- sues, both implementations solve versions of the same problem: updating and searching a large collection of records. The concept of an ADT can help us to focus on key issues even in non-com-ut- ing applications.
- 30. Sec. 1.2 Abstract Data Types and Data Structures 11 Example 1.6 When operating a car, the primary activities are steering, accelerating, and braking. On nearly all passenger cars, you steer by turn- ing the steering wheel, accelerate by pushing the gas pedal, and brake by pushing the brake pedal. This design for cars can be viewed as an ADT with operations “steer,” “accelerate,” and “brake.” Two cars might imple- ment these operations in radically different ways, say with different types of engine, or front- versus rear-wheel drive. Yet, most drivers can oper- ate many different cars because the ADT presents a uniform method of operation that does not require the driver to understand the specifics of any particular engine or drive design. These differences are deliberately hidden. The concept of an ADT is one instance of an important principle that must be understood by any successful computer scientist: managing complexity through abstraction. A central theme of computer science is complexity and techniques for handling it. Humans deal with complexity by assigning a label to an assembly of objects or concepts and then manipulating the label in place of the assembly. Cognitive psychologists call such a label a metaphor. A particular label might be related to other pieces of information or other labels. This collection can in turn be given a label, forming a hierarchy of concepts and labels. This hierarchy of labels allows us to focus on important issues while ignoring unnecessary details. Example 1.7 We apply the label “hard drive” to a collection of hardware that manipulates data on a particular type of storage device, and we ap- ply the label “CPU” to the hardware that controls execution of computer instructions. These and other labels are gathered together under the label “computer.” Because even small home computers have millions of compo- nents, some form of abstraction is necessary to comprehend how a com- puter operates. Consider how you might go about the process of designing a complex computer program that implements and manipulates an ADT. The ADT is implemented in one part of the program by a particular data structure. While designing those parts of the program that use the ADT, you can think in terms of operations on the data type without concern for the data structure’s implementation. Without this ability to simplify your thinking about a complex program, you would have no hope of understanding or implementing it.
- 31. 12 Chap. 1 Data Structures and Algorithms Example 1.8 Consider the design for a relatively simple database system stored on disk. Typically, records on disk in such a program are accessed through a buffer pool (see Section 8.3) rather than directly. Variable length records might use a memory manager (see Section 12.3) to find an appro- priate location within the disk file to place the record. Multiple index struc- tures (see Chapter 10) will typically be used to access records in various ways. Thus, we have a chain of classes, each with its own responsibili- ties and access privileges. A database query from a user is implemented by searching an index structure. This index requests access to the record by means of a request to the buffer pool. If a record is being inserted or deleted, such a request goes through the memory manager, which in turn interacts with the buffer pool to gain access to the disk file. A program such as this is far too complex for nearly any human programmer to keep all of the details in his or her head at once. The only way to design and imple- ment such a program is through proper use of abstraction and metaphors. In object-oriented programming, such abstraction is handled using classes. Data types have both a logical and a physical form. The definition of the data type in terms of an ADT is its logical form. The implementation of the data type as a data structure is its physical form. Figure 1.1 illustrates this relationship between logical and physical forms for data types. When you implement an ADT, you are dealing with the physical form of the associated data type. When you use an ADT elsewhere in your program, you are concerned with the associated data type’s logical form. Some sections of this book focus on physical implementations for a given data structure. Other sections use the logical ADT for the data type in the context of a higher-level task. Example 1.9 A particular Java environment might provide a library that includes a list class. The logical form of the list is defined by the public functions, their inputs, and their outputs that define the class. This might be all that you know about the list class implementation, and this should be all you need to know. Within the class, a variety of physical implementations for lists is possible. Several are described in Section 4.1. 1.3 Design Patterns At a higher level of abstraction than ADTs are abstractions for describing the design of programs — that is, the interactions of objects and classes. Experienced software
- 32. Sec. 1.3 Design Patterns 13 Data Type ADT: • Type • Operations Data Items: Logical Form Physical Form Data Items: Data Structure: • Storage Space • Subroutines Figure 1.1 The relationship between data items, abstract data types, and data structures. The ADT defines the logical form of the data type. The data structure implements the physical form of the data type. designers learn and reuse various techniques for combining software components. Such techniques are sometimes referred to as design patterns. A design pattern embodies and generalizes important design concepts for a recurring problem. A primary goal of design patterns is to quickly transfer the knowledge gained by expert designers to newer programmers. Another goal is to allow for efficient communication between programmers. Its much easier to discuss a design issue when you share a vocabulary relevant to the topic. Specific design patterns emerge from the discovery that a particular design problem appears repeatedly in many contexts. They are meant to solve real prob- lems. Design patterns are a bit like generics: They describe the structure for a design solution, with the details filled in for any given problem. Design patterns are a bit like data structures: Each one provides costs and benefits, which implies that tradeoffs are possible. Therefore, a given design pattern might have variations on its application to match the various tradeoffs inherent in a given situation. The rest of this section introduces a few simple design patterns that are used later in the book. 1.3.1 Flyweight The Flyweight design pattern is meant to solve the following problem. You have an application with many objects. Some of these objects are identical in the in- formation that they contain, and the role that they play. But they must be reached from various places, and conceptually they really are distinct objects. Because so much information is shared, we would like to take advantage of the opportunity to reduce memory cost by sharing space. An example comes from representing the
- 33. 14 Chap. 1 Data Structures and Algorithms layout for a document. The letter “C” might reasonably be represented by an object that describes that character’s strokes and bounding box. However, we don’t want to create a separate “C” object everywhere in the document that a “C” appears. The solution is to allocate a single copy of the shared representation for “C” ob- ject. Then, every place in the document that needs a “C” in a given font, size, and typeface will reference this single copy. The various instances of references to “C” are called flyweights. A flyweight includes the reference to the shared information, and might include additional information specific to that instance. We could imagine describing the layout of text on a page by using a tree struc- ture. The root of the tree is a node representing the page. The page has multiple child nodes, one for each column. The column nodes have child nodes for each row. And the rows have child nodes for each character. These representations for characters are the flyweights. The flyweight includes the reference to the shared shape information, and might contain additional information specific to that in- stance. For example, each instance for “C” will contain a reference to the shared information about strokes and shapes, and it might also contain the exact location for that instance of the character on the page. Flyweights are used in the implementation for the PR quadtree data structure for storing collections of point objects, described in Section 13.3. In a PR quadtree, we again have a tree with leaf nodes. Many of these leaf nodes (the ones that represent empty areas) contain the same information. These identical nodes can be implemented using the Flyweight design pattern for better memory efficiency. 1.3.2 Visitor Given a tree of objects to describe a page layout, we might wish to perform some activity on every node in the tree. Section 5.2 discusses tree traversal, which is the process of visiting every node in the tree in a defined order. A simple example for our text composition application might be to count the number of nodes in the tree that represents the page. At another time, we might wish to print a listing of all the nodes for debugging purposes. We could write a separate traversal function for each such activity that we in- tend to perform on the tree. A better approach would be to write a generic traversal function, and pass in the activity to be performed at each node. This organization constitutes the visitor design pattern. The visitor design pattern is used in Sec- tions 5.2 (tree traversal) and 11.3 (graph traversal).
- 34. Sec. 1.3 Design Patterns 15 1.3.3 Composite There are two fundamental approaches to dealing with the relationship between a collection of actions and a hierarchy of object types. First consider the typical procedural approach. Say we have a base class for page layout entities, with a subclass hierarchy to define specific subtypes (page, columns, rows, figures, char- acters, etc.). And say there are actions to be performed on a collection of such objects (such as rendering the objects to the screen). The procedural design ap- proach is for each action to be implemented as a method that takes as a parameter a pointer to the base class type. Each such method will traverse through the collec- tion of objects, visiting each object in turn. Each method contains something like a case statement that defines the details of the action for each subclass in the collec- tion (e.g., page, column, row, character). We can cut the code down some by using the visitor design pattern so that we only need to write the traversal once, and then write a visitor subroutine for each action that might be applied to the collection of objects. But each such visitor subroutine must still contain logic for dealing with each of the possible subclasses. In our page composition application, there are only a few activities that we would like to perform on the page representation. We might render the objects in full detail. Or we might want a “rough draft” rendering that prints only the bound- ing boxes of the objects. If we come up with a new activity to apply to the collection of objects, we do not need to change any of the code that implements the existing activities. But adding new activities won’t happen often for this application. In contrast, there could be many object types, and we might frequently add new ob- ject types to our implementation. Unfortunately, adding a new object type requires that we modify each activity, and the subroutines implementing the activities get rather long case statements to distinguish the behavior of the many subclasses. An alternative design is to have each object subclass in the hierarchy embody the action for each of the various activities that might be performed. Each subclass will have code to perform each activity (such as full rendering or bounding box rendering). Then, if we wish to apply the activity to the collection, we simply call the first object in the collection and specify the action (as a method call on that object). In the case of our page layout and its hierarchical collection of objects, those objects that contain other objects (such as a row objects that contains letters) will call the appropriate method for each child. If we want to add a new activity with this organization, we have to change the code for every subclass. But this is relatively rare for our text compositing application. In contrast, adding a new object into the subclass hierarchy (which for this application is far more likely than adding a new rendering function) is easy. Adding a new subclass does not require changing
- 35. 16 Chap. 1 Data Structures and Algorithms any of the existing subclasses. It merely requires that we define the behavior of each activity that can be performed on that subclass. This second design approach of burying the functional activity in the subclasses is called the Composite design pattern. A detailed example for using the Composite design pattern is presented in Section 5.3.1. 1.3.4 Strategy Our final example of a design pattern lets us encapsulate and make interchangeable a set of alternative actions that might be performed as part of some larger activity. Again continuing our text compositing example, each output device that we wish to render to will require its own function for doing the actual rendering. That is, the objects will be broken down into constituent pixels or strokes, but the actual mechanics of rendering a pixel or stroke will depend on the output device. We don’t want to build this rendering functionality into the object subclasses. Instead, we want to pass to the subroutine performing the rendering action a method or class that does the appropriate rendering details for that output device. That is, we wish to hand to the object the appropriate “strategy” for accomplishing the details of the rendering task. Thus, we call this approach the Strategy design pattern. The Strategy design pattern will be discussed further in Chapter 7. There, a sorting function is given a class (called a comparator) that understands how to ex- tract and compare the key values for records to be sorted. In this way, the sorting function does not need to know any details of how its record type is implemented. One of the biggest challenges to understanding design patterns is that many of them appear to be pretty much the same. For example, you might be confused about the difference between the composite pattern and the visitor pattern. The distinction is that the composite design pattern is about whether to give control of the traversal process to the nodes of the tree or to the tree itself. Both approaches can make use of the visitor design pattern to avoid rewriting the traversal function many times, by encapsulating the activity performed at each node. But isn’t the strategy design pattern doing the same thing? The difference be- tween the visitor pattern and the strategy pattern is more subtle. Here the difference is primarily one of intent and focus. In both the strategy design pattern and the visi- tor design pattern, an activity is being passed in as a parameter. The strategy design pattern is focused on encapsulating an activity that is part of a larger process, so that different ways of performing that activity can be substituted. The visitor de- sign pattern is focused on encapsulating an activity that will be performed on all members of a collection so that completely different activities can be substituted within a generic method that accesses all of the collection members.
- 36. Sec. 1.4 Problems, Algorithms, and Programs 17 1.4 Problems, Algorithms, and Programs Programmers commonly deal with problems, algorithms, and computer programs. These are three distinct concepts. Problems: As your intuition would suggest, a problem is a task to be performed. It is best thought of in terms of inputs and matching outputs. A problem definition should not include any constraints on how the problem is to be solved. The solution method should be developed only after the problem is precisely defined and thor- oughly understood. However, a problem definition should include constraints on the resources that may be consumed by any acceptable solution. For any problem to be solved by a computer, there are always such constraints, whether stated or implied. For example, any computer program may use only the main memory and disk space available, and it must run in a “reasonable” amount of time. Problems can be viewed as functions in the mathematical sense. A function is a matching between inputs (the domain) and outputs (the range). An input to a function might be a single value or a collection of information. The values making up an input are called the parameters of the function. A specific selection of values for the parameters is called an instance of the problem. For example, the input parameter to a sorting function might be an array of integers. A particular array of integers, with a given size and specific values for each position in the array, would be an instance of the sorting problem. Different instances might generate the same output. However, any problem instance must always result in the same output every time the function is computed using that particular input. This concept of all problems behaving like mathematical functions might not match your intuition for the behavior of computer programs. You might know of programs to which you can give the same input value on two separate occasions, and two different outputs will result. For example, if you type “date” to a typical UNIX command line prompt, you will get the current date. Naturally the date will be different on different days, even though the same command is given. However, there is obviously more to the input for the date program than the command that you type to run the program. The date program computes a function. In other words, on any particular day there can only be a single answer returned by a properly running date program on a completely specified input. For all computer programs, the output is completely determined by the program’s full set of inputs. Even a “random number generator” is completely determined by its inputs (although some random number generating systems appear to get around this by accepting a random input from a physical process beyond the user’s control). The relationship between programs and functions is explored further in Section 17.3.
- 37. 18 Chap. 1 Data Structures and Algorithms Algorithms: An algorithm is a method or a process followed to solve a problem. If the problem is viewed as a function, then an algorithm is an implementation for the function that transforms an input to the corresponding output. A problem can be solved by many different algorithms. A given algorithm solves only one problem (i.e., computes a particular function). This book covers many problems, and for several of these problems I present more than one algorithm. For the important problem of sorting I present nearly a dozen algorithms! The advantage of knowing several solutions to a problem is that solution A might be more efficient than solution B for a specific variation of the problem, or for a specific class of inputs to the problem, while solution B might be more efficient than A for another variation or class of inputs. For example, one sorting algorithm might be the best for sorting a small collection of integers, another might be the best for sorting a large collection of integers, and a third might be the best for sorting a collection of variable-length strings. By definition, an algorithm possesses several properties. Something can only be called an algorithm to solve a particular problem if it has all of the following properties. 1. It must be correct. In other words, it must compute the desired function, converting each input to the correct output. Note that every algorithm im- plements some function Because every algorithm maps every input to some output (even if that output is a system crash). At issue here is whether a given algorithm implements the intended function. 2. It is composed of a series of concrete steps. Concrete means that the action described by that step is completely understood — and doable — by the person or machine that must perform the algorithm. Each step must also be doable in a finite amount of time. Thus, the algorithm gives us a “recipe” for solving the problem by performing a series of steps, where each such step is within our capacity to perform. The ability to perform a step can depend on who or what is intended to execute the recipe. For example, the steps of a cookie recipe in a cookbook might be considered sufficiently concrete for instructing a human cook, but not for programming an automated cookie- making factory. 3. There can be no ambiguity as to which step will be performed next. Often it is the next step of the algorithm description. Selection (e.g., the if statements in Java) is normally a part of any language for describing algorithms. Selec- tion allows a choice for which step will be performed next, but the selection process is unambiguous at the time when the choice is made.
- 38. Sec. 1.5 Further Reading 19 4. It must be composed of a finite number of steps. If the description for the algorithm were made up of an infinite number of steps, we could never hope to write it down, nor implement it as a computer program. Most languages for describing algorithms (including English and “pseudocode”) provide some way to perform repeated actions, known as iteration. Examples of iteration in programming languages include the while and for loop constructs of Java. Iteration allows for short descriptions, with the number of steps actually performed controlled by the input. 5. It must terminate. In other words, it may not go into an infinite loop. Programs: We often think of a computer program as an instance, or concrete representation, of an algorithm in some programming language. In this book, nearly all of the algorithms are presented in terms of programs, or parts of pro- grams. Naturally, there are many programs that are instances of the same alg- orithm, because any modern computer programming language can be used to im- plement the same collection of algorithms (although some programming languages can make life easier for the programmer). To simplify presentation throughout the remainder of the text, I often use the terms “algorithm” and “program” inter- changeably, despite the fact that they are really separate concepts. By definition, an algorithm must provide sufficient detail that it can be converted into a program when needed. The requirement that an algorithm must terminate means that not all computer programs meet the technical definition of an algorithm. Your operating system is one such program. However, you can think of the various tasks for an operating sys- tem (each with associated inputs and outputs) as individual problems, each solved by specific algorithms implemented by a part of the operating system program, and each one of which terminates once its output is produced. To summarize: A problem is a function or a mapping of inputs to outputs. An algorithm is a recipe for solving a problem whose steps are concrete and un- ambiguous. The algorithm must be correct, of finite length, and must terminate for all inputs. A program is an instantiation of an algorithm in a computer program- ming language. 1.5 Further Reading The first authoritative work on data structures and algorithms was the series of books The Art of Computer Programming by Donald E. Knuth, with Volumes 1 and 3 being most relevant to the study of data structures [Knu97, Knu98]. A mod- ern encyclopedic approach to data structures and algorithms that should be easy
- 39. 20 Chap. 1 Data Structures and Algorithms to understand once you have mastered this book is Algorithms by Robert Sedge- wick [Sed03]. For an excellent and highly readable (but more advanced) teaching introduction to algorithms, their design, and their analysis, see Introduction to Al- gorithms: A Creative Approach by Udi Manber [Man89]. For an advanced, en- cyclopedic approach, see Introduction to Algorithms by Cormen, Leiserson, and Rivest [CLRS01]. Steven S. Skiena’s The Algorithm Design Manual [Ski98] pro- vides pointers to many implementations for data structures and algorithms that are available on the Web. For a gentle introduction to ADTs and program specification, see Abstract Data Types: Their Specification, Representation, and Use by Thomas, Robinson, and Emms [TRE88]. The claim that all modern programming languages can implement the same algorithms (stated more precisely, any function that is computable by one program- ming language is computable by any programming language with certain standard capabilities) is a key result from computability theory. For an easy introduction to this field see James L. Hein, Discrete Structures, Logic, and Computability [Hei03]. Much of computer science is devoted to problem solving. Indeed, this is what attracts many people to the field. How to Solve It by George Pólya [Pól57] is con- sidered to be the classic work on how to improve your problem-solving abilities. If you want to be a better student (as well as a better problem solver in general), see Strategies for Creative Problem Solving by Folger and LeBlanc [FL95], Effective Problem Solving by Marvin Levine [Lev94], and Problem Solving & Comprehen- sion by Arthur Whimbey and Jack Lochhead [WL99]. See The Origin of Consciousness in the Breakdown of the Bicameral Mind by Julian Jaynes [Jay90] for a good discussion on how humans use the concept of metaphor to handle complexity. More directly related to computer science educa- tion and programming, see “Cogito, Ergo Sum! Cognitive Processes of Students Dealing with Data Structures” by Dan Aharoni [Aha00] for a discussion on mov- ing from programming-context thinking to higher-level (and more design-oriented) programming-free thinking. On a more pragmatic level, most people study data structures to write better programs. If you expect your program to work correctly and efficiently, it must first be understandable to yourself and your co-workers. Kernighan and Pike’s The Practice of Programming [KP99] discusses a number of practical issues related to programming, including good coding and documentation style. For an excellent (and entertaining!) introduction to the difficulties involved with writing large pro- grams, read the classic The Mythical Man-Month: Essays on Software Engineering by Frederick P. Brooks [Bro95].
- 40. Sec. 1.6 Exercises 21 If you want to be a successful Java programmer, you need good reference man- uals close at hand. David Flanagan’s Java in a Nutshell [Fla05] provides a good reference for those familiar with the basics of the language. After gaining proficiency in the mechanics of program writing, the next step is to become proficient in program design. Good design is difficult to learn in any discipline, and good design for object-oriented software is one of the most difficult of arts. The novice designer can jump-start the learning process by studying well- known and well-used design patterns. The classic reference on design patterns is Design Patterns: Elements of Reusable Object-Oriented Software by Gamma, Helm, Johnson, and Vlissides [GHJV95] (this is commonly referred to as the “gang of four” book). Unfortunately, this is an extremely difficult book to understand, in part because the concepts are inherently difficult. A number of Web sites are available that discuss design patterns, and which provide study guides for the De- sign Patterns book. Two other books that discuss object-oriented software design are Object-Oriented Software Design and Construction with C++ by Dennis Ka- fura [Kaf98], and Object-Oriented Design Heuristics by Arthur J. Riel [Rie96]. 1.6 Exercises The exercises for this chapter are different from those in the rest of the book. Most of these exercises are answered in the following chapters. However, you should not look up the answers in other parts of the book. These exercises are intended to make you think about some of the issues to be covered later on. Answer them to the best of your ability with your current knowledge. 1.1 Think of a program you have used that is unacceptably slow. Identify the spe- cific operations that make the program slow. Identify other basic operations that the program performs quickly enough. 1.2 Most programming languages have a built-in integer data type. Normally this representation has a fixed size, thus placing a limit on how large a value can be stored in an integer variable. Describe a representation for integers that has no size restriction (other than the limits of the computer’s available main memory), and thus no practical limit on how large an integer can be stored. Briefly show how your representation can be used to implement the operations of addition, multiplication, and exponentiation. 1.3 Define an ADT for character strings. Your ADT should consist of typical functions that can be performed on strings, with each function defined in
- 41. 22 Chap. 1 Data Structures and Algorithms terms of its input and output. Then define two different physical representa- tions for strings. 1.4 Define an ADT for a list of integers. First, decide what functionality your ADT should provide. Example 1.4 should give you some ideas. Then, spec- ify your ADT in Java in the form of an abstract class declaration, showing the functions, their parameters, and their return types. 1.5 Briefly describe how integer variables are typically represented on a com- puter. (Look up one’s complement and two’s complement arithmetic in an introductory computer science textbook if you are not familiar with these.) Why does this representation for integers qualify as a data structure as de- fined in Section 1.2? 1.6 Define an ADT for a two-dimensional array of integers. Specify precisely the basic operations that can be performed on such arrays. Next, imagine an application that stores an array with 1000 rows and 1000 columns, where less than 10,000 of the array values are non-zero. Describe two different imple- mentations for such arrays that would be more space efficient than a standard two-dimensional array implementation requiring one million positions. 1.7 You have been assigned to implement a sorting program. The goal is to make this program general purpose, in that you don’t want to define in advance what record or key types are used. Describe ways to generalize a simple sorting algorithm (such as insertion sort, or any other sort you are familiar with) to support this generalization. 1.8 You have been assigned to implement a simple seqential search on an array. The problem is that you want the search to be as general as possible. This means that you need to support arbitrary record and key types. Describe ways to generalize the search function to support this goal. Consider the possibility that the function will be used multiple times in the same program, on differing record types. Consider the possibility that the function will need to be used on different keys (possibly with the same or different types) of the same record. For example, a student data record might be searched by zip code, by name, by salary, or by GPA. 1.9 Does every problem have an algorithm? 1.10 Does every algorithm have a Java program? 1.11 Consider the design for a spelling checker program meant to run on a home computer. The spelling checker should be able to handle quickly a document of less than twenty pages. Assume that the spelling checker comes with a dictionary of about 20,000 words. What primitive operations must be imple- mented on the dictionary, and what is a reasonable time constraint for each operation?
- 42. Sec. 1.6 Exercises 23 1.12 Imagine that you have been hired to design a database service containing information about cities and towns in the United States, as described in Ex- ample 1.2. Suggest two possible implementations for the database. 1.13 Imagine that you are given an array of records that is sorted with respect to some key field contained in each record. Give two different algorithms for searching the array to find the record with a specified key value. Which one do you consider “better” and why? 1.14 How would you go about comparing two proposed algorithms for sorting an array of integers? In particular, (a) What would be appropriate measures of cost to use as a basis for com- paring the two sorting algorithms? (b) What tests or analysis would you conduct to determine how the two algorithms perform under these cost measures? 1.15 A common problem for compilers and text editors is to determine if the parentheses (or other brackets) in a string are balanced and properly nested. For example, the string “((())())()” contains properly nested pairs of paren- theses, but the string “)()(” does not; and the string “())” does not contain properly matching parentheses. (a) Give an algorithm that returns true if a string contains properly nested and balanced parentheses, and false if otherwise. Hint: At no time while scanning a legal string from left to right will you have encoun- tered more right parentheses than left parentheses. (b) Give an algorithm that returns the position in the string of the first of- fending parenthesis if the string is not properly nested and balanced. That is, if an excess right parenthesis is found, return its position; if there are too many left parentheses, return the position of the first ex- cess left parenthesis. Return −1 if the string is properly balanced and nested. 1.16 A graph consists of a set of objects (called vertices) and a set of edges, where each edge connects two vertices. Any given pair of vertices can be connected by only one edge. Describe at least two different ways to represent the con- nections defined by the vertices and edges of a graph. 1.17 Imagine that you are a shipping clerk for a large company. You have just been handed about 1000 invoices, each of which is a single sheet of paper with a large number in the upper right corner. The invoices must be sorted by this number, in order from lowest to highest. Write down as many different approaches to sorting the invoices as you can think of.
- 43. 24 Chap. 1 Data Structures and Algorithms 1.18 Imagine that you are a programmer who must write a function to sort an array of about 1000 integers from lowest value to highest value. Write down at least five approaches to sorting the array. Do not write algorithms in Java or pseudocode. Just write a sentence or two for each approach to describe how it would work. 1.19 Think of an algorithm to find the maximum value in an (unsorted) array. Now, think of an algorithm to find the second largest value in the array. Which is harder to implement? Which takes more time to run (as measured by the number of comparisons performed)? Now, think of an algorithm to find the third largest value. Finally, think of an algorithm to find the middle value. Which is the most difficult of these problems to solve? 1.20 An unsorted list of integers allows for constant-time insert simply by adding a new integer at the end of the list. Unfortunately, searching for the integer with key value X requires a sequential search through the unsorted list until you find X, which on average requires looking at half the list. On the other hand, a sorted array-based list of n integers can be searched in log n time by using a binary search. Unfortunately, inserting a new integer requires a lot of time because many integers might be shifted in the array if we want to keep it sorted. How might data be organized to support both insertion and search in log n time?
- 44. 2 Mathematical Preliminaries This chapter presents mathematical notation, background, and techniques used throughout the book. This material is provided primarily for review and reference. You might wish to return to the relevant sections when you encounter unfamiliar notation or mathematical techniques in later chapters. Section 2.7 on estimating might be unfamiliar to many readers. Estimating is not a mathematical technique, but rather a general engineering skill. It is enor- mously useful to computer scientists doing design work, because any proposed solution whose estimated resource requirements fall well outside the problem’s re- source constraints can be discarded immediately. 2.1 Sets and Relations The concept of a set in the mathematical sense has wide application in computer science. The notations and techniques of set theory are commonly used when de- scribing and implementing algorithms because the abstractions associated with sets often help to clarify and simplify algorithm design. A set is a collection of distinguishable members or elements. The members are typically drawn from some larger population known as the base type. Each member of a set is either a primitive element of the base type or is a set itself. There is no concept of duplication in a set. Each value from the base type is either in the set or not in the set. For example, a set named P might be the three integers 7, 11, and 42. In this case, P’s members are 7, 11, and 42, and the base type is integer. Figure 2.1 shows the symbols commonly used to express sets and their rela- tionships. Here are some examples of this notation in use. First define two sets, P and Q. P = {2, 3, 5}, Q = {5, 10}. 25
- 45. 26 Chap. 2 Mathematical Preliminaries {1, 4} A set composed of the members 1 and 4 {x | x is a positive integer} A set definition using a set former Example: the set of all positive integers x ∈ P x is a member of set P x / ∈ P x is not a member of set P ∅ The null or empty set |P| Cardinality: size of set P or number of members for set P P ⊆ Q, Q ⊇ P Set P is included in set Q, set P is a subset of set Q, set Q is a superset of set P P ∪ Q Set Union: all elements appearing in P OR Q P ∩ Q Set Intersection: all elements appearing in P AND Q P − Q Set difference: all elements of set P NOT in set Q Figure 2.1 Set notation. |P| = 3 (because P has three members) and |Q| = 2 (because Q has two members). The union of P and Q, written P ∪ Q, is the set of elements in either P or Q, which is {2, 3, 5, 10}. The intersection of P and Q, written P ∩ Q, is the set of elements that appear in both P and Q, which is {5}. The set difference of P and Q, written P − Q, is the set of elements that occur in P but not in Q, which is {2, 3}. Note that P ∪ Q = Q ∪ P and that P ∩ Q = Q ∩ P, but in general P − Q 6= Q − P. In this example, Q − P = {10}. Note that the set {4, 3, 5} is indistinguishable from set P, because sets have no concept of order. Likewise, set {4, 3, 4, 5} is also indistinguishable from P, because sets have no concept of duplicate elements. The powerset of a set S is the set of all possible subsets for S. Consider the set S = {a, b, c}. The powerset of S is {∅, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}}. Sometimes we wish to define a collection of elements with no order (like a set), but with duplicate-valued elements. Such a collection is called a bag.1 To distinguish bags from sets, I use square brackets [] around a bag’s elements. For 1 The object referred to here as a bag is sometimes called a multilist. But, I reserve the term multilist for a list that may contain sublists (see Section 12.1).
- 46. Sec. 2.1 Sets and Relations 27 example, bag [3, 4, 5, 4] is distinct from bag [3, 4, 5], while set {3, 4, 5, 4} is indistinguishable from set {3, 4, 5}. However, bag [3, 4, 5, 4] is indistinguishable from bag [3, 4, 4, 5]. A sequence is a collection of elements with an order, and which may contain duplicate-valued elements. A sequence is also sometimes called a tuple or a vec- tor. In a sequence, there is a 0th element, a 1st element, 2nd element, and so on. I indicate a sequence by using angle brackets hi to enclose its elements. For example, h3, 4, 5, 4i is a sequence. Note that sequence h3, 5, 4, 4i is distinct from sequence h3, 4, 5, 4i, and both are distinct from sequence h3, 4, 5i. A relation R over set S is a set of ordered pairs from S. As an example of a relation, if S is {a, b, c}, then {ha, ci, hb, ci, hc, bi} is a relation, and {ha, ai, ha, ci, hb, bi, hb, ci, hc, ci} is a different relation. If tuple hx, yi is in relation R, we may use the infix notation xRy. We often use relations such as the less than operator (<) on the natural numbers, which includes ordered pairs such as h1, 3i and h2, 23i, but not h3, 2i or h2, 2i. Rather than writing the relationship in terms of ordered pairs, we typically use an infix notation for such relations, writing 1 < 3. Define the properties of relations as follows, where R is a binary relation over set S. • R is reflexive if aRa for all a ∈ S. • R is symmetric if whenever aRb, then bRa, for all a, b ∈ S. • R is antisymmetric if whenever aRb and bRa, then a = b, for all a, b ∈ S. • R is transitive if whenever aRb and bRc, then aRc, for all a, b, c ∈ S. As examples, for the natural numbers, < is antisymmetric and transitive; ≤ is reflexive, antisymmetric, and transitive, and = is reflexive, antisymmetric, and tran- sitive. For people, the relation “is a sibling of” is symmetric and transitive. If we define a person to be a sibling of himself, then it is reflexive; if we define a person not to be a sibling of himself, then it is not reflexive. R is an equivalence relation on set S if it is reflexive, symmetric, and transitive. An equivalence relation can be used to partition a set into equivalence classes. If two elements a and b are equivalent to each other, we write a ≡ b. A partition of a set S is a collection of subsets that are disjoint from each other and whose union is S. An equivalence relation on set S partitions the set into subsets whose elements are equivalent. See Section 6.2 for a discussion on how to represent equivalence classes on a set. One application for disjoint sets appears in Section 11.5.2.
- 47. 28 Chap. 2 Mathematical Preliminaries Example 2.1 For the integers, = is an equivalence relation that partitions each element into a distinct subset. In other words, for any integer a, three things are true. 1. a = a, 2. if a = b then b = a, and 3. if a = b and b = c, then a = c. Of course, for distinct integers a, b, and c there are never cases where a = b, b = a, or b = c. So the claims that = is symmetric and transitive are vacuously true (there are never examples in the relation where these events occur). But becausethe requirements for symmetry and transitivity are not violated, the relation is symmetric and transitive. Example 2.2 If we clarify the definition of sibling to mean that a person is a sibling of him- or herself, then the sibling relation is an equivalence relation that partitions the set of people. Example 2.3 We can use the modulus function (defined in the next sec- tion) to define an equivalence relation. For the set of integers, use the mod- ulus function to define a binary relation such that two numbers x and y are in the relation if and only if x mod m = y mod m. Thus, for m = 4, h1, 5i is in the relation because 1 mod 4 = 5 mod 4. We see that modulus used in this way defines an equivalence relation on the integers, and this re- lation can be used to partition the integers into m equivalence classes. This relation is an equivalence relation because 1. x mod m = x mod m for all x; 2. if x mod m = y mod m, then y mod m = x mod m; and 3. if x mod m = y mod m and y mod m = z mod m, then x mod m = z mod m. A binary relation is called a partial order if it is antisymmetric and transitive.2 The set on which the partial order is defined is called a partially ordered set or a poset. Elements x and y of a set are comparable under a given relation if either 2 Not all authors use this definition for partial order. I have seen at least three significantly different definitions in the literature. I have selected the one that lets < and ≤ both define partial orders on the integers, becausethis seems the most natural to me.
- 48. Sec. 2.2 Miscellaneous Notation 29 xRy or yRx. If every pair of distinct elements in a partial order are comparable, then the order is called a total order or linear order. Example 2.4 For the integers, the relations < and ≤ both define partial orders. Operation < is a total order because, for every pair of integers x and y such that x 6= y, either x < y or y < x. Likewise, ≤ is a total order because, for every pair of integers x and y such that x 6= y, either x ≤ y or y ≤ x. Example 2.5 For the powerset of the integers, the subset operator defines a partial order (because it is antisymmetric and transitive). For example, {1, 2} ⊆ {1, 2, 3}. However, sets {1, 2} and {1, 3} are not comparable by the subset operator, because neither is a subset of the other. Therefore, the subset operator does not define a total order on the powerset of the integers. 2.2 Miscellaneous Notation Units of measure: I use the following notation for units of measure. “B” will be used as an abbreviation for bytes, “b” for bits, “KB” for kilobytes (210 = 1024 bytes), “MB” for megabytes (220 bytes), “GB” for gigabytes (230 bytes), and “ms” for milliseconds (a millisecond is 1 1000 of a second). Spaces are not placed be- tween the number and the unit abbreviation when a power of two is intended. Thus a disk drive of size 25 gigabytes (where a gigabyte is intended as 230 bytes) will be written as “25GB.” Spaces are used when a decimal value is intended. An amount of 2000 bits would therefore be written “2 Kb” while “2Kb” represents 2048 bits. 2000 milliseconds is written as 2000 ms. Note that in this book large amounts of storage are nearly always measured in powers of two and times in powers of ten. Factorial function: The factorial function, written n! for n an integer greater than 0, is the product of the integers between 1 and n, inclusive. Thus, 5! = 1 · 2 · 3 · 4 · 5 = 120. As a special case, 0! = 1. The factorial function grows quickly as n becomes larger. Because computing the factorial function directly is a time-consuming process, it can be useful to have an equation that provides a good approximation. Stirling’s approximation states that n! ≈ √ 2πn(n e )n, where e ≈ 2.71828 (e is the base for the system of natural logarithms).3 Thus we see that 3 The symbol “≈” means “approximately equal.”
- 49. 30 Chap. 2 Mathematical Preliminaries while n! grows slower than nn (because √ 2πn/en < 1), it grows faster than cn for any positive integer constant c. Permutations: A permutation of a sequence S is simply the members of S ar- ranged in some order. For example, a permutation of the integers 1 through n would be those values arranged in some order. If the sequence contains n distinct members, then there are n! different permutations for the sequence. This is because there are n choices for the first member in the permutation; for each choice of first member there are n − 1 choices for the second member, and so on. Sometimes one would like to obtain a random permutation for a sequence, that is, one of the n! possible permutations is selected in such a way that each permutation has equal probability of being selected. A simple Java function for generating a random per- mutation is as follows. Here, the n values of the sequence are stored in positions 0 through n − 1 of array A, function swap(A, i, j) exchanges elements i and j in array A, and Random(n) returns an integer value in the range 0 to n − 1 (see the Appendix for more information on swap and Random). // Randomly permute the values of array "A" static <E> void permute(E[] A) { for (int i = A.length; i > 0; i--) // for each i swap(A, i-1, DSutil.random(i)); // swap A[i-1] with } // a random element Boolean variables: A Boolean variable is a variable (of type boolean in Java) that takes on one of the two values true and false. These two values are often associated with the values 1 and 0, respectively, although there is no reason why this needs to be the case. It is poor programming practice to rely on the corre- spondence between 0 and false, because these are logically distinct objects of different types. Floor and ceiling: The floor of x (written ⌊x⌋) takes real value x and returns the greatest integer ≤ x. For example, ⌊3.4⌋ = 3, as does ⌊3.0⌋, while ⌊−3.4⌋ = −4 and ⌊−3.0⌋ = −3. The ceiling of x (written ⌈x⌉) takes real value x and returns the least integer ≥ x. For example, ⌈3.4⌉ = 4, as does ⌈4.0⌉, while ⌈−3.4⌉ = ⌈−3.0⌉ = −3. Modulus operator: The modulus (or mod) function returns the remainder of an integer division. Sometimes written n mod m in mathematical expressions, the syntax for the Java modulus operator is n % m. From the definition of remainder, n mod m is the integer r such that n = qm + r for q an integer, and |r| < |m|. Therefore, the result of n mod m must be between 0 and m − 1 when n and m are
- 50. Sec. 2.3 Logarithms 31 positive integers. For example, 5 mod 3 = 2; 25 mod 3 = 1, 5 mod 7 = 5, and 5 mod 5 = 0. Unfortunately, there is more than one way to assign values to q and r, de- pending on how integer division is interpreted. The most common mathematical definition computes the mod function as n mod m = n − m⌊n/m⌋. In this case, −3 mod 5 = 2. However, Java and C++ compilers typically use the underly- ing processor’s machine instruction for computing integer arithmetic. On many computers this is done by truncating the resulting fraction, meaning n mod m = n − m(trunc(n/m)). Under this definition, −3 mod 5 = −3. Unfortunately, for many applications this is not what the user wants or expects. For example, many hash systems will perform some computation on a record’s key value and then take the result modulo the hash table size. The expectation here would be that the result is a legal index into the hash table, not a negative number. Implementors of hash functions must either insure that the result of the computation is always postive, or else add the hash table size to the result of the modulo function when that result is negative. 2.3 Logarithms A logarithm of base b for value y is the power to which b is raised to get y. Nor- mally, this is written as logb y = x. Thus, if logb y = x then bx = y, and blogby = y. Logarithms are used frequently by programmers. Here are two typical uses. Example 2.6 Many programs require an encoding for a collection of ob- jects. What is the minimum number of bits needed to represent n distinct code values? The answer is ⌈log2 n⌉ bits. For example, if you have 1000 codes to store, you will require at least ⌈log2 1000⌉ = 10 bits to have 1000 different codes (10 bits provide 1024 distinct code values). Example 2.7 Consider the binary search algorithm for finding a given value within an array sorted by value from lowest to highest. Binary search first looks at the middle element and determines if the value being searched for is in the upper half or the lower half of the array. The algorithm then continues splitting the appropriate subarray in half until the desired value is found. (Binary search is described in more detail in Section 3.5.) How many times can an array of size n be split in half until only one element remains in the final subarray? The answer is ⌈log2 n⌉ times.
- 51. 32 Chap. 2 Mathematical Preliminaries In this book, nearly all logarithms used have a base of two. This is because data structures and algorithms most often divide things in half, or store codes with binary bits. Whenever you see the notation log n in this book, either log2 n is meant or else the term is being used asymptotically and the actual base does not matter. If any base for the logarithm other than two is intended, then the base will be shown explicitly. Logarithms have the following properties, for any positive values of m, n, and r, and any positive integers a and b. 1. log(nm) = log n + log m. 2. log(n/m) = log n − log m. 3. log(nr) = r log n. 4. loga n = logb n/ logb a. The first two properties state that the logarithm of two numbers multiplied (or divided) can be found by adding (or subtracting) the logarithms of the two num- bers.4 Property (3) is simply an extension of property (1). Property (4) tells us that, for variable n and any two integer constants a and b, loga n and logb n differ by the constant factor logb a, regardless of the value of n. Most runtime analyses in this book are of a type that ignores constant factors in costs. Property (4) says that such analyses need not be concerned with the base of the logarithm, because this can change the total cost only by a constant factor. Note that 2log n = n. When discussing logarithms, exponents often lead to confusion. Property (3) tells us that log n2 = 2 log n. How do we indicate the square of the logarithm (as opposed to the logarithm of n2)? This could be written as (log n)2, but it is traditional to use log2 n. On the other hand, we might want to take the logarithm of the logarithm of n. This is written log log n. A special notation is used in the rare case where we would like to know how many times we must take the log of a number before we reach a value ≤ 1. This quantity is written log∗ n. For example, log∗ 1024 = 4 because log 1024 = 10, log 10 ≈ 3.33, log 3.33 ≈ 1.74, and log 1.74 < 1, which is a total of 4 log opera- tions. 4 These properties are the idea behind the slide rule. Adding two numbers can be viewed as joining two lengths together and measuring their combined length. Multiplication is not so easily done. However, if the numbers are first converted to the lengths of their logarithms, then those lengths can be added and the inverse logarithm of the resulting length gives the answer for the multiplication (this is simply logarithm property (1)). A slide rule measures the length of the logarithm for the numbers, lets you slide bars representing these lengths to add up the total length, and finally converts this total length to the correct numeric answer by taking the inverse of the logarithm for the result.
- 52. Sec. 2.4 Summations and Recurrences 33 2.4 Summations and Recurrences Most programs contain loop constructs. When analyzing running time costs for programs with loops, we need to add up the costs for each time the loop is executed. This is an example of a summation. Summations are simply the sum of costs for some function applied to a range of parameter values. Summations are typically written with the following “Sigma” notation: n X i=1 f(i). This notation indicates that we are summing the value of f(i) over some range of (integer) values. The parameter to the expression and its initial value are indicated below the P symbol. Here, the notation i = 1 indicates that the parameter is i and that it begins with the value 1. At the top of the P symbol is the expression n. This indicates the maximum value for the parameter i. Thus, this notation means to sum the values of f(i) as i ranges from 1 through n. This can also be written f(1) + f(2) + · · · + f(n − 1) + f(n). Within a sentence, Sigma notation is typeset as Pn i=1 f(i). Given a summation, you often wish to replace it with a direct equation with the same value as the summation. This is known as a closed-form solution, and the process of replacing the summation with its closed-form solution is known as solv- ing the summation. For example, the summation Pn i=1 1 is simply the expression “1” summed n times (remember that i ranges from 1 to n). Because the sum of n 1s is n, the closed-form solution is n. The following is a list of useful summations, along with their closed-form solutions. n X i=1 i = n(n + 1) 2 . (2.1) n X i=1 i2 = 2n3 + 3n2 + n 6 = n(2n + 1)(n + 1) 6 . (2.2) log n X i=1 n = n log n. (2.3) ∞ X i=0 ai = 1 1 − a for 0 < a < 1. (2.4)
- 53. 34 Chap. 2 Mathematical Preliminaries n X i=0 ai = an+1 − 1 a − 1 for a 6= 1. (2.5) As special cases to Equation 2.5, n X i=1 1 2i = 1 − 1 2n , (2.6) and n X i=0 2i = 2n+1 − 1. (2.7) As a corollary to Equation 2.7, log n X i=0 2i = 2log n+1 − 1 = 2n − 1. (2.8) Finally, n X i=1 i 2i = 2 − n + 2 2n . (2.9) The sum of reciprocals from 1 to n, called the Harmonic Series and written Hn, has a value between loge n and loge n+1. To be more precise, as n grows, the summation grows closer to Hn ≈ loge n + γ + 1 2n , (2.10) where γ is Euler’s constant and has the value 0.5772... Most of these equalities can be proved easily by mathematical induction (see Section 2.6.3). Unfortunately, induction does not help us derive a closed-form solu- tion. It only confirms when a proposed closed-form solution is correct. Techniques for deriving closed-form solutions are discussed in Section 14.1. The running time for a recursive algorithm is most easily expressed by a recur- sive expression because the total time for the recursive algorithm includes the time to run the recursive call(s). A recurrence relation defines a function by means of an expression that includes one or more (smaller) instances of itself. A classic example is the recursive definition for the factorial function: n! = (n − 1)! · n for n > 1; 1! = 0! = 1. Another standard example of a recurrence is the Fibonacci sequence: Fib(n) = Fib(n − 1) + Fib(n − 2) for n > 2; Fib(1) = Fib(2) = 1.
- 54. Sec. 2.4 Summations and Recurrences 35 From this definition we see that the first seven numbers of the Fibonacci sequence are 1, 1, 2, 3, 5, 8, and 13. Notice that this definition contains two parts: the general definition for Fib(n) and the base cases for Fib(1) and Fib(2). Likewise, the definition for factorial contains a recursive part and base cases. Recurrence relations are often used to model the cost of recursive functions. For example, the number of multiplications required by function fact of Section 2.5 for an input of size n will be zero when n = 0 or n = 1 (the base cases), and it will be one plus the cost of calling fact on a value of n − 1. This can be defined using the following recurrence: T(n) = T(n − 1) + 1 for n > 1; T(0) = T(1) = 0. As with summations, we typically wish to replace the recurrence relation with a closed-form solution. One approach is to expand the recurrence by replacing any occurrences of T on the right-hand side with its definition. Example 2.8 If we expand the recurrence T(n) = T(n − 1) + 1, we get T(n) = T(n − 1) + 1 = (T(n − 2) + 1) + 1. We can expand the recurrence as many steps as we like, but the goal is to detect some pattern that will permit us to rewrite the recurrence in terms of a summation. In this example, we might notice that (T(n − 2) + 1) + 1 = T(n − 2) + 2 and if we expand the recurrence again, we get T(n) = T(n − 2) + 2 = T(n − 3) + 1 + 2 = T(n − 3) + 3 which generalizes to the pattern T(n) = T(n − i) + i. We might conclude that T(n) = T(n − (n − 1)) + (n − 1) = T(1) + n − 1 = n − 1.
- 55. 36 Chap. 2 Mathematical Preliminaries Because we have merely guessed at a pattern and not actually proved that this is the correct closed form solution, we can use an induction proof to complete the process (see Example 2.13). Example 2.9 A slightly more complicated recurrence is T(n) = T(n − 1) + n; T(1) = 1. Expanding this recurrence a few steps, we get T(n) = T(n − 1) + n = T(n − 2) + (n − 1) + n = T(n − 3) + (n − 2) + (n − 1) + n. We should then observe that this recurrence appears to have a pattern that leads to T(n) = T(n − (n − 1)) + (n − (n − 2)) + · · · + (n − 1) + n = 1 + 2 + · · · + (n − 1) + n. This is equivalent to the summation Pn i=1 i, for which we already know the closed-form solution. Techniques to find closed-form solutions for recurrence relations are discussed in Section 14.2. Prior to Chapter 14, recurrence relations are used infrequently in this book, and the corresponding closed-form solution and an explanation for how it was derived will be supplied at the time of use. 2.5 Recursion An algorithm is recursive if it calls itself to do part of its work. For this approach to be successful, the “call to itself” must be on a smaller problem then the one originally attempted. In general, a recursive algorithm must have two parts: the base case, which handles a simple input that can be solved without resorting to a recursive call, and the recursive part which contains one or more recursive calls to the algorithm where the parameters are in some sense “closer” to the base case than those of the original call. Here is a recursive Java function to compute the
- 56. Sec. 2.5 Recursion 37 factorial of n. A trace of fact’s execution for a small value of n is presented in Section 4.2.4. static long fact(int n) { // Compute n! recursively // fact(20) is the largest value that fits in a long assert (n >= 0) && (n <= 20) : "n out of range"; if (n <= 1) return 1; // Base case: return base solution return n * fact(n-1); // Recursive call for n > 1 } The first two lines of the function constitute the base cases. If n ≤ 1, then one of the base cases computes a solution for the problem. If n > 1, then fact calls a function that knows how to find the factorial of n − 1. Of course, the function that knows how to compute the factorial of n − 1 happens to be fact itself. But we should not think too hard about this while writing the algorithm. The design for recursive algorithms can always be approached in this way. First write the base cases. Then think about solving the problem by combining the results of one or more smaller — but similar — subproblems. If the algorithm you write is correct, then certainly you can rely on it (recursively) to solve the smaller subproblems. The secret to success is: Do not worry about how the recursive call solves the subproblem. Simply accept that it will solve it correctly, and use this result to in turn correctly solve the original problem. What could be simpler? Recursion has no counterpart in everyday problem solving. The concept can be difficult to grasp because it requires you to think about problems in a new way. To use recursion effectively, it is necessary to train yourself to stop analyzing the recursive process beyond the recursive call. The subproblems will take care of themselves. You just worry about the base cases and how to recombine the sub- problems. The recursive version of the factorial function might seem unnecessarily com- plicated to you because the same effect can be achieved by using a while loop. Here is another example of recursion, based on a famous puzzle called “Towers of Hanoi.” The natural algorithm to solve this problem has multiple recursive calls. It cannot be rewritten easily using while loops. The Towers of Hanoi puzzle begins with three poles and n rings, where all rings start on the leftmost pole (labeled Pole 1). The rings each have a different size, and are stacked in order of decreasing size with the largest ring at the bottom, as shown in Figure 2.2.a. The problem is to move the rings from the leftmost pole to the rightmost pole (labeled Pole 3) in a series of steps. At each step the top ring on some pole is moved to another pole. There is one limitation on where rings may be moved: A ring can never be moved on top of a smaller ring.
- 57. 38 Chap. 2 Mathematical Preliminaries (a) (b) Figure 2.2 Towers of Hanoi example. (a) The initial conditions for a problem with six rings. (b) A necessary intermediate step on the road to a solution. How can you solve this problem? It is easy if you don’t think too hard about the details. Instead, consider that all rings are to be moved from Pole 1 to Pole 3. It is not possible to do this without first moving the bottom (largest) ring to Pole 3. To do that, Pole 3 must be empty, and only the bottom ring can be on Pole 1. The remaining n − 1 rings must be stacked up in order on Pole 2, as shown in Figure 2.2.b. How can you do this? Assume that a function X is available to solve the problem of moving the top n − 1 rings from Pole 1 to Pole 2. Then move the bottom ring from Pole 1 to Pole 3. Finally, again use function X to move the remaining n − 1 rings from Pole 2 to Pole 3. In both cases, “function X” is simply the Towers of Hanoi function called on a smaller version of the problem. The secret to success is relying on the Towers of Hanoi algorithm to do the work for you. You need not be concerned about the gory details of how the Towers of Hanoi subproblem will be solved. That will take care of itself provided that two things are done. First, there must be a base case (what to do if there is only one ring) so that the recursive process will not go on forever. Second, the recursive call to Towers of Hanoi can only be used to solve a smaller problem, and then only one of the proper form (one that meets the original definition for the Towers of Hanoi problem, assuming appropriate renaming of the poles). Here is a Java implementation for the recursive Towers of Hanoi algorithm. Function move(start, goal) takes the top ring from Pole start and moves it to Pole goal. If the move function were to print the values of its parameters, then the result of calling TOH would be a list of ring-moving instructions that solves the problem.
- 58. Sec. 2.6 Mathematical Proof Techniques 39 static void TOH(int n, Pole start, Pole goal, Pole temp) { if (n == 0) return; // Base case TOH(n-1, start, temp, goal); // Recursive call: n-1 rings move(start, goal); // Move bottom disk to goal TOH(n-1, temp, goal, start); // Recursive call: n-1 rings } Those who are unfamiliar with recursion might find it hard to accept that it is used primarily as a tool for simplifying the design and description of algorithms. A recursive algorithm usually does not yield the most efficient computer program for solving the problem because recursion involves function calls, which are typi- cally more expensive than other alternatives such as a while loop. However, the recursive approach usually provides an algorithm that is reasonably efficient in the sense discussed in Chapter 3. (But not always! See Exercise 2.11.) If necessary, the clear, recursive solution can later be modified to yield a faster implementation, as described in Section 4.2.4. Many data structures are naturally recursive, in that they can be defined as be- ing made up of self-similar parts. Tree structures are an example of this. Thus, the algorithms to manipulate such data structures are often presented recursively. Many searching and sorting algorithms are based on a strategy of divide and con- quer. That is, a solution is found by breaking the problem into smaller (similar) subproblems, solving the subproblems, then combining the subproblem solutions to form the solution to the original problem. This process is often implemented using recursion. Thus, recursion plays an important role throughout this book, and many more examples of recursive functions will be given. 2.6 Mathematical Proof Techniques Solving any problem has two distinct parts: the investigation and the argument. Students are too used to seeing only the argument in their textbooks and lectures. But to be successful in school (and in life after school), one needs to be good at both, and to understand the differences between these two phases of the process. To solve the problem, you must investigate successfully. That means engaging the problem, and working through until you find a solution. Then, to give the answer to your client (whether that “client” be your instructor when writing answers on a homework assignment or exam, or a written report to your boss), you need to be able to make the argument in a way that gets the solution across clearly and succinctly. The argument phase involves good technical writing skills — the ability to make a clear, logical argument. Being conversant with standard proof techniques can help you in this process. Knowing how to write a good proof helps in many ways. First, it clarifies your
- 59. 40 Chap. 2 Mathematical Preliminaries thought process, which in turn clarifies your explanations. Second, if you use one of the standard proof structures such as proof by contradiction or an induction proof, then both you and your reader are working from a shared understanding of that structure. That makes for less complexity to your reader to understand your proof, because the reader need not decode the structure of your argument from scratch. This section briefly introduces three commonly used proof techniques: (i) de- duction, or direct proof; (ii) proof by contradiction, and (iii) proof by mathematical induction. 2.6.1 Direct Proof In general, a direct proof is just a “logical explanation.” A direct proof is some- times referred to as an argument by deduction. This is simply an argument in terms of logic. Often written in English with words such as “if ... then,” it could also be written with logic notation such as “P ⇒ Q.” Even if we don’t wish to use symbolic logic notation, we can still take advantage of fundamental theorems of logic to structure our arguments. For example, if we want to prove that P and Q are equivalent, we can first prove P ⇒ Q and then prove Q ⇒ P. In some domains, proofs are essentially a series of state changes from a start state to an end state. Formal predicate logic can be viewed in this way, with the vari- ous “rules of logic” being used to make the changes from one formula or combining a couple of formulas to make a new formula on the route to the destination. Sym- bolic manipulations to solve integration problems in introductory calculus classes are similar in spirit, as are high school geometry proofs. 2.6.2 Proof by Contradiction The simplest way to disprove a theorem or statement is to find a counterexample to the theorem. Unfortunately, no number of examples supporting a theorem is sufficient to prove that the theorem is correct. However, there is an approach that is vaguely similar to disproving by counterexample, called Proof by Contradiction. To prove a theorem by contradiction, we first assume that the theorem is false. We then find a logical contradiction stemming from this assumption. If the logic used to find the contradiction is correct, then the only way to resolve the contradiction is to recognize that the assumption that the theorem is false must be incorrect. That is, we conclude that the theorem must be true. Example 2.10 Here is a simple proof by contradiction. Theorem 2.1 There is no largest integer.
- 60. Sec. 2.6 Mathematical Proof Techniques 41 Proof: Proof by contradiction. Step 1. Contrary assumption: Assume that there is a largest integer. Call it B (for “biggest”). Step 2. Show this assumption leads to a contradiction: Consider C = B + 1. C is an integer because it is the sum of two integers. Also, C > B, which means that B is not the largest integer after all. Thus, we have reached a contradiction. The only flaw in our reasoning is the initial assumption that the theorem is false. Thus, we conclude that the theorem is correct. ✷ A related proof technique is proving the contrapositive. We can prove that P ⇒ Q by proving (not Q) ⇒ (not P). 2.6.3 Proof by Mathematical Induction Mathematical induction is much like recursion. It is applicable to a wide variety of theorems. Induction also provides a useful way to think about algorithm design, because it encourages you to think about solving a problem by building up from simple subproblems. Induction can help to prove that a recursive function produces the correct result. Within the context of algorithm analysis, one of the most important uses for mathematical induction is as a method to test a hypothesis. As explained in Sec- tion 2.4, when seeking a closed-form solution for a summation or recurrence we might first guess or otherwise acquire evidence that a particular formula is the cor- rect solution. If the formula is indeed correct, it is often an easy matter to prove that fact with an induction proof. Let Thrm be a theorem to prove, and express Thrm in terms of a positive integer parameter n. Mathematical induction states that Thrm is true for any value of parameter n (for n ≥ c, where c is some constant) if the following two conditions are true: 1. Base Case: Thrm holds for n = c, and 2. Induction Step: If Thrm holds for n − 1, then Thrm holds for n. Proving the base case is usually easy, typically requiring that some small value such as 1 be substituted for n in the theorem and applying simple algebra or logic as necessary to verify the theorem. Proving the induction step is sometimes easy, and sometimes difficult. An alternative formulation of the induction step is known as strong induction. The induction step for strong induction is: 2a. Induction Step: If Thrm holds for all k, c ≤ k < n, then Thrm holds for n.
- 61. 42 Chap. 2 Mathematical Preliminaries Proving either variant of the induction step (in conjunction with verifying the base case) yields a satisfactory proof by mathematical induction. The two conditions that make up the induction proof combine to demonstrate that Thrm holds for n = 2 as an extension of the fact that Thrm holds for n = 1. This fact, combined again with condition (2) or (2a), indicates that Thrm also holds for n = 3, and so on. Thus, Thrm holds for all values of n (larger than the base cases) once the two conditions have been proved. What makes mathematical induction so powerful (and so mystifying to most people at first) is that we can take advantage of the assumption that Thrm holds for all values less than n to help us prove that Thrm holds for n. This is known as the induction hypothesis. Having this assumption to work with makes the induction step easier to prove than tackling the original theorem itself. Being able to rely on the induction hypothesis provides extra information that we can bring to bear on the problem. There are important similarities between recursion and induction. Both are anchored on one or more base cases. A recursive function relies on the ability to call itself to get the answer for smaller instances of the problem. Likewise, induction proofs rely on the truth of the induction hypothesis to prove the theorem. The induction hypothesis does not come out of thin air. It is true if and only if the theorem itself is true, and therefore is reliable within the proof context. Using the induction hypothesis it do work is exactly the same as using a recursive call to do work. Example 2.11 Here is a sample proof by mathematical induction. Call the sum of the first n positive integers S(n). Theorem 2.2 S(n) = n(n + 1)/2. Proof: The proof is by mathematical induction. 1. Check the base case. For n = 1, verify that S(1) = 1(1+1)/2. S(1) is simply the sum of the first positive number, which is 1. Because 1(1 + 1)/2 = 1, the formula is correct for the base case. 2. State the induction hypothesis. The induction hypothesis is S(n − 1) = n−1 X i=1 i = (n − 1)((n − 1) + 1) 2 = (n − 1)(n) 2 . 3. Use the assumption from the induction hypothesis for n − 1 to show that the result is true for n. The induction hypothesis states
- 62. Sec. 2.6 Mathematical Proof Techniques 43 that S(n − 1) = (n − 1)(n)/2, and because S(n) = S(n − 1) + n, we can substitute for S(n − 1) to get n X i=1 i = n−1 X i=1 i ! + n = (n − 1)(n) 2 + n = n2 − n + 2n 2 = n(n + 1) 2 . Thus, by mathematical induction, S(n) = n X i=1 i = n(n + 1)/2. ✷ Note carefully what took place in this example. First we cast S(n) in terms of a smaller occurrence of the problem: S(n) = S(n − 1) + n. This is important because once S(n − 1) comes into the picture, we can use the induction hypothesis to replace S(n − 1)) with (n − 1)(n)/2. From here, it is simple algebra to prove that S(n − 1) + n equals the right-hand side of the original theorem. Example 2.12 Here is another simple proof by induction that illustrates choosing the proper variable for induction. We wish to prove by induction that the sum of the first n positive odd numbers is n2. First we need a way to describe the nth odd number, which is simply 2n − 1. This also allows us to cast the theorem as a summation. Theorem 2.3 Pn i=1(2i − 1) = n2. Proof: The base case of n = 1 yields 1 = 12, which is true. The induction hypothesis is n−1 X i=1 (2i − 1) = (n − 1)2 . We now use the induction hypothesis to show that the theorem holds true for n. The sum of the first n odd numbers is simply the sum of the first n − 1 odd numbers plus the nth odd number. In the second line below, we will use the induction hypothesis to replace the partial summation (shown in brackets in the first line) with its closed-form solution. After that, algebra
- 63. 44 Chap. 2 Mathematical Preliminaries takes care of the rest. n X i=1 (2i − 1) = "n−1 X i=1 (2i − 1) # + 2n − 1 = (n − 1)2 + 2n − 1 = n2 − 2n + 1 + 2n − 1 = n2 . Thus, by mathematical induction, Pn i=1(2i − 1) = n2. ✷ Example 2.13 This example shows how we can use induction to prove that a proposed closed-form solution for a recurrence relation is correct. Theorem 2.4 The recurrence relation T(n) = T(n−1)+1; T(1) = 0 has closed-form solution T(n) = n − 1. Proof: To prove the base case, we observe that T(1) = 1 − 1 = 0. The induction hypothesis is that T(n − 1) = n − 2. Combining the definition of the recurrence with the induction hypothesis, we see immediately that T(n) = T(n − 1) + 1 = n − 2 + 1 = n − 1 for n > 1. Thus, we have proved the theorem correct by mathematical induction. ✷ Example 2.14 This example uses induction without involving summa- tions or other equations. It also illustrates a more flexible use of base cases. Theorem 2.5 2c / and 5c / stamps can be used to form any value (for values ≥ 4). Proof: The theorem defines the problem for values ≥ 4 because it does not hold for the values 1 and 3. Using 4 as the base case, a value of 4c / can be made from two 2c / stamps. The induction hypothesis is that a value of n−1 can be made from some combination of 2c / and 5c / stamps. We now use the induction hypothesis to show how to get the value n from 2c / and 5c / stamps. Either the makeup for value n − 1 includes a 5c / stamp, or it does not. If so,
- 64. Sec. 2.6 Mathematical Proof Techniques 45 then replace a 5c / stamp with three 2c / stamps. If not, then the makeup must have included at least two 2c / stamps (because it is at least of size 4 and contains only 2c / stamps). In this case, replace two of the 2c / stamps with a single 5c / stamp. In either case, we now have a value of n made up of 2c / and 5c / stamps. Thus, by mathematical induction, the theorem is correct. ✷ Example 2.15 Here is an example using strong induction. Theorem 2.6 For n > 1, n is divisible by some prime number. Proof: For the base case, choose n = 2. 2 is divisible by the prime num- ber 2. The induction hypothesis is that any value a, 2 ≤ a < n, is divisible by some prime number. There are now two cases to consider when proving the theorem for n. If n is a prime number, then n is divisible by itself. If n is not a prime number, then n = a × b for a and b, both integers less than n but greater than 1. The induction hypothesis tells us that a is divisible by some prime number. That same prime number must also divide n. Thus, by mathematical induction, the theorem is correct. ✷ Our next example of mathematical induction proves a theorem from geometry. It also illustrates a standard technique of induction proof where we take n objects and remove some object to use the induction hypothesis. Example 2.16 Define a two-coloring for a set of regions as a way of as- signing one of two colors to each region such that no two regions sharing a side have the same color. For example, a chessboard is two-colored. Fig- ure 2.3 shows a two-coloring for the plane with three lines. We will assume that the two colors to be used are black and white. Theorem 2.7 The set of regions formed by n infinite lines in the plane can be two-colored. Proof: Consider the base case of a single infinite line in the plane. This line splits the plane into two regions. One region can be colored black and the other white to get a valid two-coloring. The induction hypothesis is that the set of regions formed by n − 1 infinite lines can be two-colored. To prove the theorem for n, consider the set of regions formed by the n − 1 lines remaining when any one of the n lines is removed. By the induction hy- pothesis, this set of regions can be two-colored. Now, put the nth line back. This splits the plane into two half-planes, each of which (independently) has a valid two-coloring inherited from the two-coloring of the plane with
- 65. 46 Chap. 2 Mathematical Preliminaries Figure 2.3 A two-coloring for the regions formed by three lines in the plane. n − 1 lines. Unfortunately, the regions newly split by the nth line violate the rule for a two-coloring. Take all regions on one side of the nth line and reverse their coloring (after doing so, this half-plane is still two-colored). Those regions split by the nth line are now properly two-colored, Because the part of the region to one side of the line is now black and the region to the other side is now white. Thus, by mathematical induction, the entire plane is two-colored. ✷ Compare the proof of Theorem 2.7 with that of Theorem 2.5. For Theorem 2.5, we took a collection of stamps of size n − 1 (which, by the induction hypothesis, must have the desired property) and from that “built” a collection of size n that has the desired property. We therefore proved the existence of some collection of stamps of size n with the desired property. For Theorem 2.7 we must prove that any collection of n lines has the desired property. Thus, our strategy is to take an arbitrary collection of n lines, and “re- duce” it so that we have a set of lines that must have the desired property because it matches the induction hypothesis. From there, we merely need to show that re- versing the original reduction process preserves the desired property. In contrast, consider what would be required if we attempted to “build” from a set of lines of size n − 1 to one of size n. We would have great difficulty justifying that all possible collections of n lines are covered by our building process. By reducing from an arbitrary collection of n lines to something less, we avoid this problem. This section’s final example shows how induction can be used to prove that a recursive function produces the correct result.
- 66. Sec. 2.7 Estimating 47 Example 2.17 We would like to prove that function fact does indeed compute the factorial function. There are two distinct steps to such a proof. The first is to prove that the function always terminates. The second is to prove that the function returns the correct value. Theorem 2.8 Function fact will terminate for any value of n. Proof: For the base case, we observe that fact will terminate directly whenever n ≤ 0. The induction hypothesis is that fact will terminate for n − 1. For n, we have two possibilities. One possibility is that n ≥ 12. In that case, fact will terminate directly because it will fail its assertion test. Otherwise, fact will make a recursive call to fact(n-1). By the induction hypothesis, fact(n-1) must terminate. ✷ Theorem 2.9 Function fact does compute the factorial function for any value in the range 0 to 12. Proof: To prove the base case, observe that when n = 0 or n = 1, fact(n) returns the correct value of 1. The induction hypothesis is that fact(n-1) returns the correct value of (n − 1)!. For any value n within the legal range, fact(n) returns n ∗ fact(n-1). By the induction hy- pothesis, fact(n-1) = (n − 1)!, and because n ∗ (n − 1)! = n!, we have proved that fact(n) produces the correct result. ✷ We can use a similar process to prove many recursive programs correct. The general form is to show that the base cases perform correctly, and then to use the induction hypothesis to show that the recursive step also produces the correct result. Prior to this, we must prove that the function always terminates, which might also be done using an induction proof. 2.7 Estimating One of the most useful life skills that you can gain from your computer science training is knowing how to perform quick estimates. This is sometimes known as “back of the napkin” or “back of the envelope” calculation. Both nicknames sug- gest that only a rough estimate is produced. Estimation techniques are a standard part of engineering curricula but are often neglected in computer science. Estima- tion is no substitute for rigorous, detailed analysis of a problem, but it can serve to indicate when a rigorous analysis is warranted: If the initial estimate indicates that the solution is unworkable, then further analysis is probably unnecessary. Estimating can be formalized by the following three-step process: 1. Determine the major parameters that affect the problem.
- 67. 48 Chap. 2 Mathematical Preliminaries 2. Derive an equation that relates the parameters to the problem. 3. Select values for the parameters, and apply the equation to yield an estimated solution. When doing estimations, a good way to reassure yourself that the estimate is reasonable is to do it in two different ways. In general, if you want to know what comes out of a system, you can either try to estimate that directly, or you can estimate what goes into the system (assuming that what goes in must later come out). If both approaches (independently) give similar answers, then this should build confidence in the estimate. When calculating, be sure that your units match. For example, do not add feet and pounds. Verify that the result is in the correct units. Always keep in mind that the output of a calculation is only as good as its input. The more uncertain your valuation for the input parameters in Step 3, the more uncertain the output value. However, back of the envelope calculations are often meant only to get an answer within an order of magnitude, or perhaps within a factor of two. Before doing an estimate, you should decide on acceptable error bounds, such as within 10%, within a factor of two, and so forth. Once you are confident that an estimate falls within your error bounds, leave it alone! Do not try to get a more precise estimate than necessary for your purpose. Example 2.18 How many library bookcases does it take to store books containing one million pages? I estimate that a 500-page book requires one inch on the library shelf (for example, look at the size of this book), yielding about 200 feet of shelf space for one million pages. If a shelf is 4 feet wide, then 50 shelves are required. If a bookcase contains 5 shelves, this yields about 10 library bookcases. To reach this conclusion, I estimated the number of pages per inch, the width of a shelf, and the number of shelves in a bookcase. None of my estimates are likely to be precise, but I feel confident that my answer is correct to within a factor of two. (After writing this, I went to Virginia Tech’s library and looked at some real bookcases. They were only about 3 feet wide, but typically had 7 shelves for a total of 21 shelf-feet. So I was correct to within 10% on bookcase capacity, far better than I expected or needed. One of my selected values was too high, and the other too low, which canceled out the errors.) Example 2.19 Is it more economical to buy a car that gets 20 miles per gallon, or one that gets 30 miles per gallon but costs $2000 more? The
- 68. Sec. 2.8 Further Reading 49 typical car is driven about 12,000 miles per year. If gasoline costs $2/gallon, then the yearly gas bill is $1200 for the less efficient car and $800 for the more efficient car. If we ignore issues such as the payback that would be received if we invested $2000 in a bank, it would take 5 years to make up the difference in price. At this point, the buyer must decide if price is the only criterion and if a 5-year payback time is acceptable. Naturally, a person who drives more will make up the difference more quickly, and changes in gasoline prices will also greatly affect the outcome. Example 2.20 When at the supermarket doing the week’s shopping, can you estimate about how much you will have to pay at the checkout? One simple way is to round the price of each item to the nearest dollar, and add this value to a mental running total as you put the item in your shopping cart. This will likely give an answer within a couple of dollars of the true total. 2.8 Further Reading Most of the topics covered in this chapter are considered part of Discrete Math- ematics. An introduction to this field is Discrete Mathematics with Applications by Susanna S. Epp [Epp04]. An advanced treatment of many mathematical topics useful to computer scientists is Concrete Mathematics: A Foundation for Computer Science by Graham, Knuth, and Patashnik [GKP94]. See “Technically Speaking” from the February 1995 issue of IEEE Spectrum [Sel95] for a discussion on the standard for indicating units of computer storage used in this book. Introduction to Algorithms by Udi Manber [Man89] makes extensive use of mathematical induction as a technique for developing algorithms. For more information on recursion, see Thinking Recursively by Eric S. Roberts [Rob86]. To learn recursion properly, it is worth your while to learn the program- ming language LISP, even if you never intend to write a LISP program. In particu- lar, Friedman and Felleisen’s The Little LISPer [FF89] is designed to teach you how to think recursively as well as teach you LISP. This book is entertaining reading as well. A good book on writing mathematical proofs is Daniel Solow’s How to Read and Do Proofs [Sol90]. To improve your general mathematical problem-solving abilities, see The Art and Craft of Problem Solving by Paul Zeitz [Zei07]. Zeitz
- 69. 50 Chap. 2 Mathematical Preliminaries also discusses the three proof techniques presented in Section 2.6, and the roles of investigation and argument in problem solving. For more about estimating techniques, see two Programming Pearls by John Louis Bentley entitled The Back of the Envelope and The Envelope is Back [Ben84, Ben00, Ben86, Ben88]. Genius: The Life and Science of Richard Feynman by James Gleick [Gle92] gives insight into how important back of the envelope calcu- lation was to the developers of the atomic bomb, and to modern theoretical physics in general. 2.9 Exercises 2.1 For each relation below, explain why the relation does or does not satisfy each of the properties reflexive, symmetric, antisymmetric, and transitive. (a) “isBrotherOf” on the set of people. (b) “isFatherOf” on the set of people. (c) The relation R = {hx, yi | x2 + y2 = 1} for real numbers x and y. (d) The relation R = {hx, yi | x2 = y2} for real numbers x and y. (e) The relation R = {hx, yi | x mod y = 0} for x, y ∈ {1, 2, 3, 4}. (f) The empty relation ∅ (i.e., the relation with no ordered pairs for which it is true) on the set of integers. (g) The empty relation ∅ (i.e., the relation with no ordered pairs for which it is true) on the empty set. 2.2 For each of the following relations, either prove that it is an equivalence relation or prove that it is not an equivalence relation. (a) For integers a and b, a ≡ b if and only if a + b is even. (b) For integers a and b, a ≡ b if and only if a + b is odd. (c) For nonzero rational numbers a and b, a ≡ b if and only if a × b > 0. (d) For nonzero rational numbers a and b, a ≡ b if and only if a/b is an integer. (e) For rational numbers a and b, a ≡ b if and only if a − b is an integer. (f) For rational numbers a and b, a ≡ b if and only if |a − b| ≤ 2. 2.3 State whether each of the following relations is a partial ordering, and explain why or why not. (a) “isFatherOf” on the set of people. (b) “isAncestorOf” on the set of people. (c) “isOlderThan” on the set of people. (d) “isSisterOf” on the set of people. (e) {ha, bi, ha, ai, hb, ai} on the set {a, b}.
- 70. Sec. 2.9 Exercises 51 (f) {h2, 1i, h1, 3i, h2, 3i} on the set {1, 2, 3}. 2.4 How many total orderings can be defined on a set with n elements? Explain your answer. 2.5 Define an ADT for a set of integers (remember that a set has no concept of duplicate elements, and has no concept of order). Your ADT should consist of the functions that can be performed on a set to control its membership, check the size, check if a given element is in the set, and so on. Each function should be defined in terms of its input and output. 2.6 Define an ADT for a bag of integers (remember that a bag may contain du- plicates, and has no concept of order). Your ADT should consist of the func- tions that can be performed on a bag to control its membership, check the size, check if a given element is in the set, and so on. Each function should be defined in terms of its input and output. 2.7 Define an ADT for a sequence of integers (remember that a sequence may contain duplicates, and supports the concept of position for its elements). Your ADT should consist of the functions that can be performed on a se- quence to control its membership, check the size, check if a given element is in the set, and so on. Each function should be defined in terms of its input and output. 2.8 An investor places $30,000 into a stock fund. 10 years later the account has a value of $69,000. Using logarithms and anti-logarithms, present a formula for calculating the average annual rate of increase. Then use your formula to determine the average annual growth rate for this fund. 2.9 Rewrite the factorial function of Section 2.5 without using recursion. 2.10 Rewrite the for loop for the random permutation generator of Section 2.2 as a recursive function. 2.11 Here is a simple recursive function to compute the Fibonacci sequence: // Recursive Fibonacci generator static long fibr(int n) { // fibr(91) is the largest value that fits in a long assert (n > 0) && (n <= 91) : "n out of range"; if ((n == 1) || (n == 2)) return 1; // Base case return fibr(n-1) + fibr(n-2); // Recursive call } This algorithm turns out to be very slow, calling Fibr a total of Fib(n) times. Contrast this with the following iterative algorithm:
- 71. 52 Chap. 2 Mathematical Preliminaries // Iterative Fibonacci generator static long fibi(int n) { // fibr(91) is the largest value that fits in a long assert (n > 0) && (n <= 91) : "n out of range"; long curr, prev, past; if ((n == 1) || (n == 2)) return 1; curr = prev = 1; // curr holds current Fib value for (int i=3; i<=n; i++) { // Compute next value past = prev; // past holds fibi(i-2) prev = curr; // prev holds fibi(i-1) curr = past + prev; // curr now holds fibi(i) } return curr; } Function Fibi executes the for loop n − 2 times. (a) Which version is easier to understand? Why? (b) Explain why Fibr is so much slower than Fibi. 2.12 Write a recursive function to solve a generalization of the Towers of Hanoi problem where each ring may begin on any pole so long as no ring sits on top of a smaller ring. 2.13 Consider the following function: static void foo (double val) { if (val != 0.0) foo(val/2.0); } This function makes progress towards the base case on every recursive call. In theory (that is, if double variables acted like true real numbers), would this function ever terminate for input val a nonzero number? In practice (an actual computer implementation), will it terminate? 2.14 Write a function to print all of the permutations for the elements of an array containing n distinct integer values. 2.15 Write a recursive algorithm to print all of the subsets for the set of the first n positive integers. 2.16 The Largest Common Factor (LCF) for two positive integers n and m is the largest integer that divides both n and m evenly. LCF(n, m) is at least one, and at most m, assuming that n ≥ m. Over two thousand years ago, Euclid provided an efficient algorithm based on the observation that, when n mod m 6= 0, LCF(n, m) = GCD(m, n mod m). Use this fact to write two algorithms to find LCF for two positive integers. The first version should compute the value iteratively. The second version should compute the value using recursion.
- 72. Sec. 2.9 Exercises 53 2.17 Prove by contradiction that the number of primes is infinite. 2.18 (a) Use induction to show that n2 − n is always even. (b) Give a direct proof in one or two sentences that n2 − n is always even. (c) Show that n3 − n is always divisible by three. (d) Is n5 − n aways divisible by 5? Explain your answer. 2.19 Prove that √ 2 is irrational. 2.20 Explain why n X i=1 i = n X i=1 (n − i + 1) = n−1 X i=0 (n − i). 2.21 Prove Equation 2.2 using mathematical induction. 2.22 Prove Equation 2.6 using mathematical induction. 2.23 Prove Equation 2.7 using mathematical induction. 2.24 Find a closed-form solution and prove (using induction) that your solution is correct for the summation n X i=1 3i . 2.25 Prove that the sum of the first n even numbers is n2 + n (a) indirectly by assuming that the sum of the first n odd numbers is n2. (b) directly by mathematical induction. 2.26 Give a closed-form formula for the summation Pn i=a i where a is an integer between 1 and n. 2.27 Prove that Fib(n) < (5 3 )n. 2.28 Prove, for n ≥ 1, that n X i=1 i3 = n2(n + 1)2 4 . 2.29 The following theorem is called the Pigeonhole Principle. Theorem 2.10 When n + 1 pigeons roost in n holes, there must be some hole containing at least two pigeons. (a) Prove the Pigeonhole Principle using proof by contradiction. (b) Prove the Pigeonhole Principle using mathematical induction. 2.30 For this problem, you will consider arrangements of infinite lines in the plane such that three or more lines never intersect at a single point. (a) Give a recurrence relation that expresses the number of regions formed by n lines, and explain why your recurrence is correct.
- 73. 54 Chap. 2 Mathematical Preliminaries (b) Give the summation that results from expanding your recurrence. (c) Give a closed-form solution for the summation. 2.31 Prove (using induction) that the recurrence T(n) = T(n − 1)+n; T(1) = 1 has as its closed-form solution T(n) = n(n + 1)/2. 2.32 Expand the following recurrence to help you find a closed-form solution, and then use induction to prove your answer is correct. T(n) = 2T(n − 1) + 1 for n > 0; T(0) = 0. 2.33 Expand the following recurrence to help you find a closed-form solution, and then use induction to prove your answer is correct. T(n) = T(n − 1) + 3n + 1 for n > 0; T(0) = 1. 2.34 Assume that an n-bit integer (represented by standard binary notation) takes any value in the range 0 to 2n − 1 with equal probability. (a) For each bit position, what is the probability of its value being 1 and what is the probability of its value being 0? (b) What is the average number of “1” bits for an n-bit random number? (c) What is the expected value for the position of the leftmost “1” bit? In other words, how many positions on average must we examine when moving from left to right before encountering a “1” bit? Show the appropriate summation. 2.35 What is the total volume of your body in liters (or, if you prefer, gallons)? 2.36 An art historian has a database of 20,000 full-screen color images. (a) About how much space will this require? How many CD-ROMs would be required to store the database? (A CD-ROM holds about 600MB of data). Be sure to explain all assumptions you made to derive your answer. (b) Now, assume that you have access to a good image compression tech- nique that can store the images in only 1/10 of the space required for an uncompressed image. Will the entire database fit onto a single CD- ROM if the images are compressed? 2.37 How many cubic miles of water flow out of the mouth of the Mississippi River each day? DO NOT look up the answer or any supplemental facts. Be sure to describe all assumptions made in arriving at your answer.
- 74. Sec. 2.9 Exercises 55 2.38 When buying a home mortgage, you often have the option of paying some money in advance (called “discount points”) to get a lower interest rate. As- sume that you have the choice between two 15-year mortgages: one at 8%, and the other at 73 4 % with an up-front charge of 1% of the mortgage value. How long would it take to recover the 1% charge when you take the mort- gage at the lower rate? As a second, more precise estimate, how long would it take to recover the charge plus the interest you would have received if you had invested the equivalent of the 1% charge in the bank at 5% interest while paying the higher rate? DO NOT use a calculator to help you answer this question. 2.39 When you build a new house, you sometimes get a “construction loan” which is a temporary line of credit out of which you pay construction costs as they occur. At the end of the construction period, you then replace the construc- tion loan with a regular mortgage on the house. During the construction loan, you only pay each month for the interest charged against the actual amount borrowed so far. Assume that your house construction project starts at the beginning of April, and is complete at the end of six months. Assume that the total construction cost will be $300,000 with the costs occurring at the be- ginning of each month in $50,000 increments. The construction loan charges 6% interest. Estimate the total interest payments that must be paid over the life of the construction loan. 2.40 Here are some questions that test your working knowledge of how fast com- puters operate. Is disk drive access time normally measured in milliseconds (thousandths of a second) or microseconds (millionths of a second)? Does your RAM memory access a word in more or less than one microsecond? How many instructions can your CPU execute in one year if the machine is left running at full speed all the time? DO NOT use paper or a calculator to derive your answers. 2.41 Does your home contain enough books to total one million pages? How many total pages are stored in your school library building? 2.42 How many words are in this book? 2.43 How many hours are one million seconds? How many days? Answer these questions doing all arithmetic in your head. 2.44 How many cities and towns are there in the United States? 2.45 How many steps would it take to walk from Boston to San Francisco? 2.46 A man begins a car trip to visit his in-laws. The total distance is 60 miles, and he starts off at a speed of 60 miles per hour. After driving exactly 1 mile, he loses some of his enthusiasm for the journey, and (instantaneously) slows
- 75. 56 Chap. 2 Mathematical Preliminaries down to 59 miles per hour. After traveling another mile, he again slows to 58 miles per hour. This continues, progressively slowing by 1 mile per hour for each mile traveled until the trip is complete. (a) How long does it take the man to reach his in-laws? (b) How long would the trip take in the continuous case where the speed smoothly diminishes with the distance yet to travel?
- 76. 3 Algorithm Analysis How long will it take to process the company payroll once we complete our planned merger? Should I buy a new payroll program from vendor X or vendor Y? If a particular program is slow, is it badly implemented or is it solving a hard problem? Questions like these ask us to consider the difficulty of a problem, or the relative efficiency of two or more approaches to solving a problem. This chapter introduces the motivation, basic notation, and fundamental tech- niques of algorithm analysis. We focus on a methodology known as asymptotic algorithm analysis, or simply asymptotic analysis. Asymptotic analysis attempts to estimate the resource consumption of an algorithm. It allows us to compare the relative costs of two or more algorithms for solving the same problem. Asymptotic analysis also gives algorithm designers a tool for estimating whether a proposed solution is likely to meet the resource constraints for a problem before they imple- ment an actual program. After reading this chapter, you should understand • the concept of a growth rate, the rate at which the cost of an algorithm grows as the size of its input grows; • the concept of upper and lower bounds for a growth rate, and how to estimate these bounds for a simple program, algorithm, or problem; and • the difference between the cost of an algorithm (or program) and the cost of a problem. The chapter concludes with a brief discussion of the practical difficulties encoun- tered when empirically measuring the cost of a program, and some principles for code tuning to improve program efficiency. 3.1 Introduction How do you compare two algorithms for solving some problem in terms of effi- ciency? One way is to implement both algorithms as computer programs and then 57
- 77. 58 Chap. 3 Algorithm Analysis run them on a suitable range of inputs, measuring how much of the resources in question each program uses. This approach is often unsatisfactory for four reasons. First, there is the effort involved in programming and testing two algorithms when at best you want to keep only one. Second, when empirically comparing two al- gorithms there is always the chance that one of the programs was “better written” than the other, and that the relative qualities of the underlying algorithms are not truly represented by their implementations. This is especially likely to occur when the programmer has a bias regarding the algorithms. Third, the choice of empirical test cases might unfairly favor one algorithm. Fourth, you could find that even the better of the two algorithms does not fall within your resource budget. In that case you must begin the entire process again with yet another program implementing a new algorithm. But, how would you know if any algorithm can meet the resource budget? Perhaps the problem is simply too difficult for any implementation to be within budget. These problems can often be avoided by using asymptotic analysis. Asymp- totic analysis measures the efficiency of an algorithm, or its implementation as a program, as the input size becomes large. It is actually an estimating technique and does not tell us anything about the relative merits of two programs where one is always “slightly faster” than the other. However, asymptotic analysis has proved useful to computer scientists who must determine if a particular algorithm is worth considering for implementation. The critical resource for a program is most often its running time. However, you cannot pay attention to running time alone. You must also be concerned with other factors such as the space required to run the program (both main memory and disk space). Typically you will analyze the time required for an algorithm (or the instantiation of an algorithm in the form of a program), and the space required for a data structure. Many factors affect the running time of a program. Some relate to the environ- ment in which the program is compiled and run. Such factors include the speed of the computer’s CPU, bus, and peripheral hardware. Competition with other users for the computer’s resources can make a program slow to a crawl. The program- ming language and the quality of code generated by a particular compiler can have a significant effect. The “coding efficiency” of the programmer who converts the algorithm to a program can have a tremendous impact as well. If you need to get a program working within time and space constraints on a particular computer, all of these factors can be relevant. Yet, none of these factors address the differences between two algorithms or data structures. To be fair, pro- grams derived from two algorithms for solving the same problem should both be
- 78. Sec. 3.1 Introduction 59 compiled with the same compiler and run on the same computer under the same conditions. As much as possible, the same amount of care should be taken in the programming effort devoted to each program to make the implementations “equally efficient.” In this sense, all of the factors mentioned above should cancel out of the comparison because they apply to both algorithms equally. If you truly wish to understand the running time of an algorithm, there are other factors that are more appropriate to consider than machine speed, programming language, compiler, and so forth. Ideally we would measure the running time of the algorithm under standard benchmark conditions. However, we have no way to calculate the running time reliably other than to run an implementation of the algorithm on some computer. The only alternative is to use some other measure as a surrogate for running time. Of primary consideration when estimating an algorithm’s performance is the number of basic operations required by the algorithm to process an input of a certain size. The terms “basic operations” and “size” are both rather vague and depend on the algorithm being analyzed. Size is often the number of inputs pro- cessed. For example, when comparing sorting algorithms, the size of the problem is typically measured by the number of records to be sorted. A basic operation must have the property that its time to complete does not depend on the particular values of its operands. Adding or comparing two integer variables are examples of basic operations in most programming languages. Summing the contents of an array containing n integers is not, because the cost depends on the value of n (i.e., the size of the input). Example 3.1 Consider a simple algorithm to solve the problem of finding the largest value in an array of n integers. The algorithm looks at each integer in turn, saving the position of the largest value seen so far. This algorithm is called the largest-value sequential search and is illustrated by the following Java function: // Return position of largest value in "A" static int largest(int[] A) { int currlarge = 0; // Holds largest element position for (int i=1; i<A.length; i++) // For each element if (A[currlarge] < A[i]) // if A[i] is larger currlarge = i; // remember its position return currlarge; // Return largest position } Here, the size of the problem is n, the number of integers stored in A. The basic operation is to compare an integer’s value to that of the largest value
- 79. 60 Chap. 3 Algorithm Analysis seen so far. It is reasonable to assume that it takes a fixed amount of time to do one such comparison, regardless of the value of the two integers or their positions in the array. Because the most important factor affecting running time is normally size of the input, for a given input size n we often express the time T to run the algorithm as a function of n, written as T(n). We will always assume T(n) is a non-negative value. Let us call c the amount of time required to compare two integers in function largest. We do not care right now what the precise value of c might be. Nor are we concerned with the time required to increment vari- able i because this must be done for each value in the array, or the time for the actual assignment when a larger value is found, or the little bit of extra time taken to initialize currlarge. We just want a reasonable ap- proximation for the time taken to execute the algorithm. The total time to run largest is therefore approximately cn, Because we must make n comparisons, with each comparison costing c time. We say that function largest (and the largest-value sequential search algorithm in general) has a running time expressed by the equation T(n) = cn. This equation describes the growth rate for the running time of the largest- value sequential search algorithm. Example 3.2 The running time of a statement that assigns the first value of an integer array to a variable is simply the time required to copy the value of the first array value. We can assume this assignment takes a constant amount of time regardless of the value. Let us call c1 the amount of time necessary to copy an integer. No matter how large the array on a typical computer (given reasonable conditions for memory and array size), the time to copy the value from the first position of the array is always c1. Thus, the equation for this algorithm is simply T(n) = c1, indicating that the size of the input n has no effect on the running time. This is called a constant running time.
- 80. Sec. 3.1 Introduction 61 Example 3.3 Consider the following Java code: sum = 0; for (i=1; i<=n; i++) for (j=1; j<=n; j++) sum++; What is the running time for this code fragment? Clearly it takes longer to run when n is larger. The basic operation in this example is the increment operation for variable sum. We can assume that incrementing takes constant time; call this time c2. (We can ignore the time required to initialize sum, and to increment the loop counters i and j. In practice, these costs can safely be bundled into time c2.) The total number of increment operations is n2. Thus, we say that the running time is T(n) = c2n2. The growth rate for an algorithm is the rate at which the cost of the algorithm grows as the size of its input grows. Figure 3.1 shows a graph for six equations, each meant to describe the running time for a particular program or algorithm. A variety of growth rates representative of typical algorithms are shown. The two equations labeled 10n and 20n are graphed by straight lines. A growth rate of cn (for c any positive constant) is often referred to as a linear growth rate or running time. This means that as the value of n grows, the running time of the algorithm grows in the same proportion. Doubling the value of n roughly doubles the running time. An algorithm whose running-time equation has a highest-order term containing a factor of n2 is said to have a quadratic growth rate. In Figure 3.1, the line labeled 2n2 represents a quadratic growth rate. The line labeled 2n represents an exponential growth rate. This name comes from the fact that n appears in the exponent. The line labeled n! is also growing exponentially. As you can see from Figure 3.1, the difference between an algorithm whose running time has cost T(n) = 10n and another with cost T(n) = 2n2 becomes tremendous as n grows. For n > 5, the algorithm with running time T(n) = 2n2 is already much slower. This is despite the fact that 10n has a greater constant factor than 2n2. Comparing the two curves marked 20n and 2n2 shows that changing the constant factor for one of the equations only shifts the point at which the two curves cross. For n > 10, the algorithm with cost T(n) = 2n2 is slower than the algorithm with cost T(n) = 20n. This graph also shows that the equation T(n) = 5n log n grows somewhat more quickly than both T(n) = 10n and T(n) = 20n, but not nearly so quickly as the equation T(n) = 2n2. For constants a, b > 1, na grows faster than either logb n or log nb. Finally, algorithms with cost T(n) = 2n or T(n) = n! are prohibitively expensive for even modest values of n. Note that for constants a, b ≥ 1, an grows faster than nb.
- 81. 62 Chap. 3 Algorithm Analysis 0 100 200 300 400 10n 20n 2n2 5n log n 2n n! 0 5 10 15 0 10 20 30 40 50 Input size n 10n 20n 5n log n 2n2 2n n! 0 200 400 600 800 1000 1200 1400 Figure 3.1 Two views of a graph illustrating the growth rates for six equations. The bottom view shows in detail the lower-left portion of the top view. The hor- izontal axis represents input size. The vertical axis can represent time, space, or any other measure of cost.
- 82. Sec. 3.2 Best, Worst, and Average Cases 63 n log log n log n n n log n n2 n3 2n 16 2 4 24 2 · 24 = 25 28 212 216 256 3 8 28 8 · 28 = 211 216 224 2256 1024 ≈ 3.3 10 210 10 · 210 ≈ 213 220 230 21024 64K 4 16 216 16 · 216 = 220 232 248 264K 1M ≈ 4.3 20 220 20 · 220 ≈ 224 240 260 21M 1G ≈ 4.9 30 230 30 · 230 ≈ 235 260 290 21G Figure 3.2 Costs for growth rates representative of most computer algorithms. We can get some further insight into relative growth rates for various algorithms from Figure 3.2. Most of the growth rates that appear in typical algorithms are shown, along with some representative input sizes. Once again, we see that the growth rate has a tremendous effect on the resources consumed by an algorithm. 3.2 Best, Worst, and Average Cases Consider the problem of finding the factorial of n. For this problem, there is only one input of a given “size” (that is, there is only a single instance of size n for each value of n). Now consider our largest-value sequential search algorithm of Example 3.1, which always examines every array value. This algorithm works on many inputs of a given size n. That is, there are many possible arrays of any given size. However, no matter what array the algorithm looks at, its cost will always be the same in that it always looks at every element in the array one time. For some algorithms, different inputs of a given size require different amounts of time. For example, consider the problem of searching an array containing n integers to find the one with a particular value K (assume that K appears exactly once in the array). The sequential search algorithm begins at the first position in the array and looks at each value in turn until K is found. Once K is found, the algorithm stops. This is different from the largest-value sequential search algorithm of Example 3.1, which always examines every array value. There is a wide range of possible running times for the sequential search alg- orithm. The first integer in the array could have value K, and so only one integer is examined. In this case the running time is short. This is the best case for this algorithm, because it is not possible for sequential search to look at less than one value. Alternatively, if the last position in the array contains K, then the running time is relatively long, because the algorithm must examine n values. This is the worst case for this algorithm, because sequential search never looks at more than
- 83. 64 Chap. 3 Algorithm Analysis n values. If we implement sequential search as a program and run it many times on many different arrays of size n, or search for many different values of K within the same array, we expect the algorithm on average to go halfway through the array before finding the value we seek. On average, the algorithm examines about n/2 values. We call this the average case for this algorithm. When analyzing an algorithm, should we study the best, worst, or average case? Normally we are not interested in the best case, because this might happen only rarely and generally is too optimistic for a fair characterization of the algorithm’s running time. In other words, analysis based on the best case is not likely to be representative of the behavior of the algorithm. However, there are rare instances where a best-case analysis is useful — in particular, when the best case has high probability of occurring. In Chapter 7 you will see some examples where taking advantage of the best-case running time for one sorting algorithm makes a second more efficient. How about the worst case? The advantage to analyzing the worst case is that you know for certain that the algorithm must perform at least that well. This is es- pecially important for real-time applications, such as for the computers that monitor an air traffic control system. Here, it would not be acceptable to use an algorithm that can handle n airplanes quickly enough most of the time, but which fails to perform quickly enough when all n airplanes are coming from the same direction. For other applications — particularly when we wish to aggregate the cost of running the program many times on many different inputs — worst-case analy- sis might not be a representative measure of the algorithm’s performance. Often we prefer to know the average-case running time. This means that we would like to know the typical behavior of the algorithm on inputs of size n. Unfortunately, average-case analysis is not always possible. Average-case analysis first requires that we understand how the actual inputs to the program (and their costs) are dis- tributed with respect to the set of all possible inputs to the program. For example, it was stated previously that the sequential search algorithm on average examines half of the array values. This is only true if the element with value K is equally likely to appear in any position in the array. If this assumption is not correct, then the algorithm does not necessarily examine half of the array values in the average case. See Section 9.2 for further discussion regarding the effects of data distribution on the sequential search algorithm. The characteristics of a data distribution have a significant effect on many search algorithms, such as those based on hashing (Section 9.4) and search trees (e.g., see Section 5.4). Incorrect assumptions about data distribution can have dis-
- 84. Sec. 3.3 A Faster Computer, or a Faster Algorithm? 65 astrous consequences on a program’s space or time performance. Unusual data distributions can also be used to advantage, as shown in Section 9.2. In summary, for real-time applications we are likely to prefer a worst-case anal- ysis of an algorithm. Otherwise, we often desire an average-case analysis if we know enough about the distribution of our input to compute the average case. If not, then we must resort to worst-case analysis. 3.3 A Faster Computer, or a Faster Algorithm? Imagine that you have a problem to solve, and you know of an algorithm whose running time is proportional to n2. Unfortunately, the resulting program takes ten times too long to run. If you replace your current computer with a new one that is ten times faster, will the n2 algorithm become acceptable? If the problem size remains the same, then perhaps the faster computer will allow you to get your work done quickly enough even with an algorithm having a high growth rate. But a funny thing happens to most people who get a faster computer. They don’t run the same problem faster. They run a bigger problem! Say that on your old computer you were content to sort 10,000 records because that could be done by the computer during your lunch break. On your new computer you might hope to sort 100,000 records in the same time. You won’t be back from lunch any sooner, so you are better off solving a larger problem. And because the new machine is ten times faster, you would like to sort ten times as many records. If your algorithm’s growth rate is linear (i.e., if the equation that describes the running time on input size n is T(n) = cn for some constant c), then 100,000 records on the new machine will be sorted in the same time as 10,000 records on the old machine. If the algorithm’s growth rate is greater than cn, such as c1n2, then you will not be able to do a problem ten times the size in the same amount of time on a machine that is ten times faster. How much larger a problem can be solved in a given amount of time by a faster computer? Assume that the new machine is ten times faster than the old. Say that the old machine could solve a problem of size n in an hour. What is the largest problem that the new machine can solve in one hour? Figure 3.3 shows how large a problem can be solved on the two machines for the five running-time functions from Figure 3.1. This table illustrates many important points. The first two equations are both linear; only the value of the constant factor has changed. In both cases, the machine that is ten times faster gives an increase in problem size by a factor of ten. In other words, while the value of the constant does affect the absolute size of the problem that can be solved in a fixed amount of time, it does not affect the improvement in
- 85. 66 Chap. 3 Algorithm Analysis f(n) n n′ Change n′/n 10n 1000 10, 000 n′ = 10n 10 20n 500 5000 n′ = 10n 10 5n log n 250 1842 √ 10n < n′ < 10n 7.37 2n2 70 223 n′ = √ 10n 3.16 2n 13 16 n′ = n + 3 −− Figure 3.3 The increase in problem size that can be run in a fixed period of time on a computer that is ten times faster. The first column lists the right-hand sides for each of the five growth rate equations of Figure 3.1. For the purpose of this example, arbitrarily assume that the old machine can run 10,000 basic operations in one hour. The second column shows the maximum value for n that can be run in 10,000 basic operations on the old machine. The third column shows the value for n′ , the new maximum size for the problem that can be run in the same time on the new machine that is ten times faster. Variable n′ is the greatest size for the problem that can run in 100,000 basic operations. The fourth column shows how the size of n changed to become n′ on the new machine. The fifth column shows the increase in the problem size as the ratio of n′ to n. problem size (as a proportion to the original size) gained by a faster computer. This relationship holds true regardless of the algorithm’s growth rate: Constant factors never affect the relative improvement gained by a faster computer. An algorithm with time equation T(n) = 2n2 does not receive nearly as great an improvement from the faster machine as an algorithm with linear growth rate. Instead of an improvement by a factor of ten, the improvement is only the square root of that: √ 10 ≈ 3.16. Thus, the algorithm with higher growth rate not only solves a smaller problem in a given time in the first place, it also receives less of a speedup from a faster computer. As computers get ever faster, the disparity in problem sizes becomes ever greater. The algorithm with growth rate T(n) = 5n log n improves by a greater amount than the one with quadratic growth rate, but not by as great an amount as the algo- rithms with linear growth rates. Note that something special happens in the case of the algorithm whose running time grows exponentially. In Figure 3.1, the curve for the algorithm whose time is proportional to 2n goes up very quickly. In Figure 3.3, the increase in problem size on the machine ten times as fast is shown to be about n + 3 (to be precise, it is n + log2 10). The increase in problem size for an algorithm with exponential growth rate is by a constant addition, not by a multiplicative factor. Because the old value of n was 13, the new problem size is 16. If next year you buy another
- 86. Sec. 3.4 Asymptotic Analysis 67 computer ten times faster yet, then the new computer (100 times faster than the original computer) will only run a problem of size 19. If you had a second program whose growth rate is 2n and for which the original computer could run a problem of size 1000 in an hour, than a machine ten times faster can run a problem only of size 1003 in an hour! Thus, an exponential growth rate is radically different than the other growth rates shown in Figure 3.3. The significance of this difference is explored in Chapter 17. Instead of buying a faster computer, consider what happens if you replace an algorithm whose running time is proportional to n2 with a new algorithm whose running time is proportional to n log n. In the graph of Figure 3.1, a fixed amount of time would appear as a horizontal line. If the line for the amount of time available to solve your problem is above the point at which the curves for the two growth rates in question meet, then the algorithm whose running time grows less quickly is faster. An algorithm with running time T(n) = n2 requires 1024 × 1024 = 1, 048, 576 time steps for an input of size n = 1024. An algorithm with running time T(n) = n log n requires 1024 × 10 = 10, 240 time steps for an input of size n = 1024, which is an improvement of much more than a factor of ten when compared to the algorithm with running time T(n) = n2. Because n2 > 10n log n whenever n > 58, if the typical problem size is larger than 58 for this example, then you would be much better off changing algorithms instead of buying a computer ten times faster. Furthermore, when you do buy a faster computer, an algorithm with a slower growth rate provides a greater benefit in terms of larger problem size that can run in a certain time on the new computer. 3.4 Asymptotic Analysis Despite the larger constant for the curve labeled 10n in Figure 3.1, the curve labeled 2n2 crosses it at the relatively small value of n = 5. What if we double the value of the constant in front of the linear equation? As shown in the graph, the curve labeled 20n is surpassed by the curve labeled 2n2 once n = 10. The additional factor of two for the linear growth rate does not much matter; it only doubles the x-coordinate for the intersection point. In general, changes to a constant factor in either equation only shift where the two curves cross, not whether the two curves cross. When you buy a faster computer or a faster compiler, the new problem size that can be run in a given amount of time for a given growth rate is larger by the same factor, regardless of the constant on the running-time equation. The time curves for two algorithms with different growth rates still cross, regardless of their running-time equation constants. For these reasons, we usually ignore the constants
- 87. 68 Chap. 3 Algorithm Analysis when we want an estimate of the running time or other resource requirements of an algorithm. This simplifies the analysis and keeps us thinking about the most important aspect: the growth rate. This is called asymptotic algorithm analysis. To be precise, asymptotic analysis refers to the study of an algorithm as the input size “gets big” or reaches a limit (in the calculus sense). However, it has proved to be so useful to ignore all constant factors that asymptotic analysis is used for most algorithm comparisons. It is not always reasonable to ignore the constants. When comparing algorithms meant to run on small values of n, the constant can have a large effect. For exam- ple, if the problem is to sort a collection of exactly five records, then an algorithm designed for sorting thousands of records is probably not appropriate, even if its asymptotic analysis indicates good performance. There are rare cases where the constants for two algorithms under comparison can differ by a factor of 1000 or more, making the one with lower growth rate impractical for most purposes due to its large constant. Asymptotic analysis is a form of “back of the envelope” esti- mation for algorithm resource consumption. It provides a simplified model of the running time or other resource needs of an algorithm. This simplification usually helps you understand the behavior of your algorithms. Just be aware of the limi- tations to asymptotic analysis in the rare situation where the constant is important. 3.4.1 Upper Bounds Several terms are used to describe the running-time equation for an algorithm. These terms — and their associated symbols — indicate precisely what aspect of the algorithm’s behavior is being described. One is the upper bound for the growth of the algorithm’s running time. It indicates the upper or highest growth rate that the algorithm can have. To make any statement about the upper bound of an algorithm, we must be making it about some class of inputs of size n. We measure this upper bound nearly always on the best-case, average-case, or worst-case inputs. Thus, we cannot say, “this algorithm has an upper bound to its growth rate of n2.” We must say something like, “this algorithm has an upper bound to its growth rate of n2 in the average case.” Because the phrase “has an upper bound to its growth rate of f(n)” is long and often used when discussing algorithms, we adopt a special notation, called big-Oh notation. If the upper bound for an algorithm’s growth rate (for, say, the worst case) is f(n), then we would write that this algorithm is “in the set O(f(n))in the worst case” (or just “in O(f(n))in the worst case”). For example, if n2 grows as
- 88. Sec. 3.4 Asymptotic Analysis 69 fast as T(n) (the running time of our algorithm) for the worst-case input, we would say the algorithm is “in O(n2) in the worst case.” The following is a precise definition for an upper bound. T(n) represents the true running time of the algorithm. f(n) is some expression for the upper bound. For T(n) a non-negatively valued function, T(n) is in set O(f(n)) if there exist two positive constants c and n0 such that T(n) ≤ cf(n) for all n > n0. Constant n0 is the smallest value of n for which the claim of an upper bound holds true. Usually n0 is small, such as 1, but does not need to be. You must also be able to pick some constant c, but it is irrelevant what the value for c actually is. In other words, the definition says that for all inputs of the type in question (such as the worst case for all inputs of size n) that are large enough (i.e., n > n0), the algorithm always executes in less than cf(n) steps for some constant c. Example 3.4 Consider the sequential search algorithm for finding a spec- ified value in an array of integers. If visiting and examining one value in the array requires cs steps where cs is a positive number, and if the value we search for has equal probability of appearing in any position in the ar- ray, then in the average case T(n) = csn/2. For all values of n > 1, csn/2 ≤ csn. Therefore, by the definition, T(n) is in O(n) for n0 = 1 and c = cs. Example 3.5 For a particular algorithm, T(n) = c1n2 + c2n in the av- erage case where c1 and c2 are positive numbers. Then, c1n2 + c2n ≤ c1n2 + c2n2 ≤ (c1 + c2)n2 for all n > 1. So, T(n) ≤ cn2 for c = c1 + c2, and n0 = 1. Therefore, T(n) is in O(n2) by the definition. Example 3.6 Assigning the value from the first position of an array to a variable takes constant time regardless of the size of the array. Thus, T(n) = c (for the best, worst, and average cases). We could say in this case that T(n) is in O(c). However, it is traditional to say that an algorithm whose running time has a constant upper bound is in O(1). Just knowing that something is in O(f(n)) says only how bad things can get. Perhaps things are not nearly so bad. Because we know sequential search is in O(n) in the worst case, it is also true to say that sequential search is in O(n2). But
- 89. 70 Chap. 3 Algorithm Analysis sequential search is practical for large n, in a way that is not true for some other algorithms in O(n2). We always seek to define the running time of an algorithm with the tightest (lowest) possible upper bound. Thus, we prefer to say that sequen- tial search is in O(n). This also explains why the phrase “is in O(f(n))” or the notation “∈ O(f(n))” is used instead of “is O(f(n))” or “= O(f(n)).” There is no strict equality to the use of big-Oh notation. O(n) is in O(n2), but O(n2) is not in O(n). 3.4.2 Lower Bounds Big-Oh notation describes an upper bound. In other words, big-Oh notation states a claim about the greatest amount of some resource (usually time) that is required by an algorithm for some class of inputs of size n (typically the worst such input, the average of all possible inputs, or the best such input). Similar notation is used to describe the least amount of a resource that an alg- orithm needs for some class of input. Like big-Oh notation, this is a measure of the algorithm’s growth rate. Like big-Oh notation, it works for any resource, but we most often measure the least amount of time required. And again, like big-Oh no- tation, we are measuring the resource required for some particular class of inputs: the worst-, average-, or best-case input of size n. The lower bound for an algorithm (or a problem, as explained later) is denoted by the symbol Ω, pronounced “big-Omega” or just “Omega.” The following defi- nition for Ω is symmetric with the definition of big-Oh. For T(n) a non-negatively valued function, T(n) is in set Ω(g(n)) if there exist two positive constants c and n0 such that T(n) ≥ cg(n) for all n > n0.1 1 An alternate (non-equivalent) definition for Ω is T(n) is in the set Ω(g(n)) if there exists a positive constant c such that T(n) ≥ cg(n) for an infinite number of values for n. This definition says that for an “interesting” number of cases, the algorithm takes at least cg(n) time. Note that this definition is not symmetric with the definition of big-Oh. For g(n) to be a lower bound, this definition does not require that T(n) ≥ cg(n) for all values of n greater than some constant. It only requires that this happen often enough, in particular that it happen for an infinite number of values for n. Motivation for this alternate definition can be found in the following example. Assume a particular algorithm has the following behavior: T(n) = n for all odd n ≥ 1 n2 /100 for all even n ≥ 0 From this definition, n2 /100 ≥ 1 100 n2 for all even n ≥ 0. So, T(n) ≥ cn2 for an infinite number of values of n (i.e., for all even n) for c = 1/100. Therefore, T(n) is in Ω(n2 ) by the definition.
- 90. Sec. 3.4 Asymptotic Analysis 71 Example 3.7 Assume T(n) = c1n2 + c2n for c1 and c2 > 0. Then, c1n2 + c2n ≥ c1n2 for all n > 1. So, T(n) ≥ cn2 for c = c1 and n0 = 1. Therefore, T(n) is in Ω(n2) by the definition. It is also true that the equation of Example 3.7 is in Ω(n). However, as with big-Oh notation, we wish to get the “tightest” (for Ω notation, the largest) bound possible. Thus, we prefer to say that this running time is in Ω(n2). Recall the sequential search algorithm to find a value K within an array of integers. In the average and worst cases this algorithm is in Ω(n), because in both the average and worst cases we must examine at least cn values (where c is 1/2 in the average case and 1 in the worst case). 3.4.3 Θ Notation The definitions for big-Oh and Ω give us ways to describe the upper bound for an algorithm (if we can find an equation for the maximum cost of a particular class of inputs of size n) and the lower bound for an algorithm (if we can find an equation for the minimum cost for a particular class of inputs of size n). When the upper and lower bounds are the same within a constant factor, we indicate this by using Θ (big-Theta) notation. An algorithm is said to be Θ(h(n)) if it is in O(h(n)) and it is in Ω(h(n)). Note that we drop the word “in” for Θ notation, because there is a strict equality for two equations with the same Θ. In other words, if f(n) is Θ(g(n)), then g(n) is Θ(f(n)). Because the sequential search algorithm is both in O(n) and in Ω(n) in the average case, we say it is Θ(n) in the average case. Given an algebraic equation describing the time requirement for an algorithm, the upper and lower bounds always meet. That is because in some sense we have a perfect analysis for the algorithm, embodied by the running-time equation. For For this equation for T(n), it is true that all inputs of size n take at least cn time. But an infinite number of inputs of size n take cn2 time, so we would like to say that the algorithm is in Ω(n2 ). Unfortunately, using our first definition will yield a lower bound of Ω(n) because it is not possible to pick constants c and n0 such that T(n) ≥ cn2 for all n > n0. The alternative definition does result in a lower bound of Ω(n2 ) for this algorithm, which seems to fit common sense more closely. Fortu- nately, few real algorithms or computer programs display the pathological behavior of this example. Our first definition for Ω generally yields the expected result. As you can see from this discussion, asymptotic bounds notation is not a law of nature. It is merely a powerful modeling tool used to describe the behavior of algorithms.
- 91. 72 Chap. 3 Algorithm Analysis many algorithms (or their instantiations as programs), it is easy to come up with the equation that defines their runtime behavior. Most algorithms presented in this book are well understood and we can almost always give a Θ analysis for them. However, Chapter 17 discusses a whole class of algorithms for which we have no Θ analysis, just some unsatisfying big-Oh and Ω analyses. Exercise 3.14 presents a short, simple program fragment for which nobody currently knows the true upper or lower bounds. While some textbooks and programmers will casually say that an algorithm is “order of” or “big-Oh” of some cost function, it is generally better to use Θ notation rather than big-Oh notation whenever we have sufficient knowledge about an alg- orithm to be sure that the upper and lower bounds indeed match. Throughout this book, Θ notation will be used in preference to big-Oh notation whenever our state of knowledge makes that possible. Limitations on our ability to analyze certain algorithms may require use of big-Oh or Ω notations. In rare occasions when the discussion is explicitly about the upper or lower bound of a problem or algorithm, the corresponding notation will be used in preference to Θ notation. 3.4.4 Simplifying Rules Once you determine the running-time equation for an algorithm, it really is a simple matter to derive the big-Oh, Ω, and Θ expressions from the equation. You do not need to resort to the formal definitions of asymptotic analysis. Instead, you can use the following rules to determine the simplest form. 1. If f(n) is in O(g(n)) and g(n) is in O(h(n)), then f(n) is in O(h(n)). 2. If f(n) is in O(kg(n)) for any constant k > 0, then f(n) is in O(g(n)). 3. If f1(n) is in O(g1(n)) and f2(n) is in O(g2(n)), then f1(n) + f2(n) is in O(max(g1(n), g2(n))). 4. If f1(n) is in O(g1(n)) and f2(n) is in O(g2(n)), then f1(n)f2(n) is in O(g1(n)g2(n)). The first rule says that if some function g(n) is an upper bound for your cost function, then any upper bound for g(n) is also an upper bound for your cost func- tion. A similar property holds true for Ω notation: If g(n) is a lower bound for your cost function, then any lower bound for g(n) is also a lower bound for your cost function. Likewise for Θ notation. The significance of rule (2) is that you can ignore any multiplicative constants in your equations when using big-Oh notation. This rule also holds true for Ω and Θ notations.
- 92. Sec. 3.4 Asymptotic Analysis 73 Rule (3) says that given two parts of a program run in sequence (whether two statements or two sections of code), you need consider only the more expensive part. This rule applies to Ω and Θ notations as well: For both, you need consider only the more expensive part. Rule (4) is used to analyze simple loops in programs. If some action is repeated some number of times, and each repetition has the same cost, then the total cost is the cost of the action multiplied by the number of times that the action takes place. This rule applies to Ω and Θ notations as well. Taking the first three rules collectively, you can ignore all constants and all lower-order terms to determine the asymptotic growth rate for any cost function. The advantages and dangers of ignoring constants were discussed near the begin- ning of this section. Ignoring lower-order terms is reasonable when performing an asymptotic analysis. The higher-order terms soon swamp the lower-order terms in their contribution to the total cost as n becomes larger. Thus, if T(n) = 3n4 +5n2, then T(n) is in O(n4). The n2 term contributes relatively little to the total cost. Throughout the rest of this book, these simplifying rules are used when dis- cussing the cost for a program or algorithm. 3.4.5 Classifying Functions Given functions f(n) and g(n) whose growth rates are expressed as algebraic equa- tions, we might like to determine if one grows faster than the other. The best way to do this is to take the limit of the two functions as n grows towards infinity, lim n→∞ f(n) g(n) . If the limit goes to ∞, then f(n) is in Ω(g(n)) because f(n) grows faster. If the limit goes to zero, then f(n) is in O(g(n)) because g(n) grows faster. If the limit goes to some constant other than zero, then f(n) = Θ(g(n)) because both grow at the same rate. Example 3.8 If f(n) = 2n log n and g(n) = n2, is f(n) in O(g(n)), Ω(g(n)), or Θ(g(n))? Because n2 2n log n = n 2 log n , we easily see that lim n→∞ n2 2n log n = ∞
- 93. 74 Chap. 3 Algorithm Analysis because n grows faster than 2 log n. Thus, n2 is in Ω(2n log n). 3.5 Calculating the Running Time for a Program This section presents the analysis for several simple code fragments. Example 3.9 We begin with an analysis of a simple assignment statement to an integer variable. a = b; Because the assignment statement takes constant time, it is Θ(1). Example 3.10 Consider a simple for loop. sum = 0; for (i=1; i<=n; i++) sum += n; The first line is Θ(1). The for loop is repeated n times. The third line takes constant time so, by simplifying rule (4) of Section 3.4.4, the total cost for executing the two lines making up the for loop is Θ(n). By rule (3), the cost of the entire code fragment is also Θ(n). Example 3.11 We now analyze a code fragment with several for loops, some of which are nested. sum = 0; for (j=1; j<=n; j++) // First for loop for (i=1; i<=j; i++) // is a double loop sum++; for (k=0; k<n; k++) // Second for loop A[k] = k; This code fragment has three separate statements: the first assignment statement and the two for loops. Again the assignment statement takes constant time; call it c1. The second for loop is just like the one in Exam- ple 3.10 and takes c2n = Θ(n) time. The first for loop is a double loop and requires a special technique. We work from the inside of the loop outward. The expression sum++ requires constant time; call it c3. Because the inner for loop is executed i times, by simplifying rule (4) it has cost c3i. The outer for loop is executed n times,
- 94. Sec. 3.5 Calculating the Running Time for a Program 75 but each time the cost of the inner loop is different because it costs c3i with i changing each time. You should see that for the first execution of the outer loop, i is 1. For the second execution of the outer loop, i is 2. Each time through the outer loop, i becomes one greater, until the last time through the loop when i = n. Thus, the total cost of the loop is c3 times the sum of the integers 1 through n. From Equation 2.1, we know that n X i=1 i = n(n + 1) 2 , which is Θ(n2). By simplifying rule (3), Θ(c1 + c2n + c3n2) is simply Θ(n2). Example 3.12 Compare the asymptotic analysis for the following two code fragments: sum1 = 0; for (i=1; i<=n; i++) // First double loop for (j=1; j<=n; j++) // do n times sum1++; sum2 = 0; for (i=1; i<=n; i++) // Second double loop for (j=1; j<=i; j++) // do i times sum2++; In the first double loop, the inner for loop always executes n times. Because the outer loop executes n times, it should be obvious that the state- ment sum1++ is executed precisely n2 times. The second loop is similar to the one analyzed in the previous example, with cost Pn j=1 j. This is ap- proximately 1 2 n2. Thus, both double loops cost Θ(n2), though the second requires about half the time of the first. Example 3.13 Not all doubly nested for loops are Θ(n2). The follow- ing pair of nested loops illustrates this fact.
- 95. 76 Chap. 3 Algorithm Analysis sum1 = 0; for (k=1; k<=n; k*=2) // Do log n times for (j=1; j<=n; j++) // Do n times sum1++; sum2 = 0; for (k=1; k<=n; k*=2) // Do log n times for (j=1; j<=k; j++) // Do k times sum2++; When analyzing these two code fragments, we will assume that n is a power of two. The first code fragment has its outer for loop executed log n + 1 times because on each iteration k is multiplied by two until it reaches n. Because the inner loop always executes n times, the total cost for the first code fragment can be expressed as Plog n i=0 n. Note that a variable substitution takes place here to create the summation, with k = 2i. From Equation 2.3, the solution for this summation is Θ(n log n). In the second code fragment, the outer loop is also executed log n + 1 times. The inner loop has cost k, which doubles each time. The summation can be expressed as Plog n i=0 2i where n is assumed to be a power of two and again k = 2i. From Equation 2.8, we know that this summation is simply Θ(n). What about other control statements? While loops are analyzed in a manner similar to for loops. The cost of an if statement in the worst case is the greater of the costs for the then and else clauses. This is also true for the average case, assuming that the size of n does not affect the probability of executing one of the clauses (which is usually, but not necessarily, true). For switch statements, the worst-case cost is that of the most expensive branch. For subroutine calls, simply add the cost of executing the subroutine. There are rare situations in which the probability for executing the various branches of an if or switch statement are functions of the input size. For exam- ple, for input of size n, the then clause of an if statement might be executed with probability 1/n. An example would be an if statement that executes the then clause only for the smallest of n values. To perform an average-case analysis for such programs, we cannot simply count the cost of the if statement as being the cost of the more expensive branch. In such situations, the technique of amortized analysis (see Section 14.3) can come to the rescue. Determining the execution time of a recursive subroutine can be difficult. The running time for a recursive subroutine is typically best expressed by a recurrence relation. For example, the recursive factorial function fact of Section 2.5 calls itself with a value one less than its input value. The result of this recursive call is
- 96. Sec. 3.5 Calculating the Running Time for a Program 77 then multiplied by the input value, which takes constant time. Thus, the cost of the factorial function, if we wish to measure cost in terms of the number of multi- plication operations, is one more than the number of multiplications made by the recursive call on the smaller input. Because the base case does no multiplications, its cost is zero. Thus, the running time for this function can be expressed as T(n) = T(n − 1) + 1 for n > 1; T(1) = 0. We know from Examples 2.8 and 2.13 that the closed-form solution for this recur- rence relation is Θ(n). The final example of algorithm analysis for this section will compare two algo- rithms for performing search in an array. Earlier, we determined that the running time for sequential search on an array where the search value K is equally likely to appear in any location is Θ(n) in both the average and worst cases. We would like to compare this running time to that required to perform a binary search on an array whose values are stored in order from lowest to highest. Binary search begins by examining the value in the middle position of the ar- ray; call this position mid and the corresponding value kmid. If kmid = K, then processing can stop immediately. This is unlikely to be the case, however. Fortu- nately, knowing the middle value provides useful information that can help guide the search process. In particular, if kmid > K, then you know that the value K cannot appear in the array at any position greater than mid. Thus, you can elim- inate future search in the upper half of the array. Conversely, if kmid < K, then you know that you can ignore all positions in the array less than mid. Either way, half of the positions are eliminated from further consideration. Binary search next looks at the middle position in that part of the array where value K may exist. The value at this position again allows us to eliminate half of the remaining positions from consideration. This process repeats until either the desired value is found, or there are no positions remaining in the array that might contain the value K. Figure 3.4 illustrates the binary search method. Here is a Java implementation for binary search:
- 97. 78 Chap. 3 Algorithm Analysis Key Position 0 2 3 4 5 6 7 8 26 29 36 10 11 12 13 14 15 11 13 21 41 45 51 54 1 56 65 72 77 9 83 40 Figure 3.4 An illustration of binary search on a sorted array of 16 positions. Consider a search for the position with value K = 45. Binary search first checks the value at position 7. Because 41 < K, the desired value cannot appear in any position below 7 in the array. Next, binary search checks the value at position 11. Because 56 > K, the desired value (if it exists) must be between positions 7 and 11. Position 9 is checked next. Again, its value is too great. The final search is at position 8, which contains the desired value. Thus, function binary returns position 8. Alternatively, if K were 44, then the same series of record accesses would be made. After checking position 8, binary would return a value of n, indicating that the search is unsuccessful. // Return the position of an element in sorted array "A" // with value "K". If "K" is not in "A", return A.length. static int binary(int[] A, int K) { int l = -1; int r = A.length; // l and r are beyond array bounds while (l+1 != r) { // Stop when l and r meet int i = (l+r)/2; // Check middle of remaining subarray if (K < A[i]) r = i; // In left half if (K == A[i]) return i; // Found it if (K > A[i]) l = i; // In right half } return A.length; // Search value not in A } To find the cost of this algorithm in the worst case, we can model the running time as a recurrence and then find the closed-form solution. Each recursive call to binary cuts the size of the array approximately in half, so we can model the worst-case cost as follows, assuming for simplicity that n is a power of two. T(n) = T(n/2) + 1 for n > 1; T(1) = 1. If we expand the recurrence, we find that we can do so only log n times before we reach the base case, and each expansion adds one to the cost. Thus, the closed- form solution for the recurrence is T(n) = log n. Function binary is designed to find the (single) occurrence of K and return its position. A special value is returned if K does not appear in the array. This algorithm can be modified to implement variations such as returning the position of the first occurrence of K in the array if multiple occurrences are allowed, and returning the position of the greatest value less than K when K is not in the array.
- 98. Sec. 3.6 Analyzing Problems 79 Comparing sequential search to binary search, we see that as n grows, the Θ(n) running time for sequential search in the average and worst cases quickly becomes much greater than the Θ(log n) running time for binary search. Taken in isolation, binary search appears to be much more efficient than sequential search. This is despite the fact that the constant factor for binary search is greater than that for sequential search, because the calculation for the next search position in binary search is more expensive than just incrementing the current position, as sequential search does. Note however that the running time for sequential search will be roughly the same regardless of whether or not the array values are stored in order. In contrast, binary search requires that the array values be ordered from lowest to highest. De- pending on the context in which binary search is to be used, this requirement for a sorted array could be detrimental to the running time of a complete program, be- cause maintaining the values in sorted order requires to greater cost when inserting new elements into the array. This is an example of a tradeoff between the advan- tage of binary search during search and the disadvantage related to maintaining a sorted array. Only in the context of the complete problem to be solved can we know whether the advantage outweighs the disadvantage. 3.6 Analyzing Problems You most often use the techniques of “algorithm” analysis to analyze an algorithm, or the instantiation of an algorithm as a program. You can also use these same techniques to analyze the cost of a problem. It should make sense to you to say that the upper bound for a problem cannot be worse than the upper bound for the best algorithm that we know for that problem. But what does it mean to give a lower bound for a problem? Consider a graph of cost over all inputs of a given size n for some algorithm for a given problem. Define A to be the collection of all algorithms that solve the problem (theoretically, there an infinite number of such algorithms). Now, consider the collection of all the graphs for all of the (infinitely many) algorithms in A. The worst case lower bound is the least of all the highest points on all the graphs. It is much easier to show that an algorithm (or program) is in Ω(f(n)) than it is to show that a problem is in Ω(f(n)). For a problem to be in Ω(f(n)) means that every algorithm that solves the problem is in Ω(f(n)), even algorithms that we have not thought of! So far all of our examples of algorithm analysis give “obvious” results, with big-Oh always matching Ω. To understand how big-Oh, Ω, and Θ notations are
- 99. 80 Chap. 3 Algorithm Analysis properly used to describe our understanding of a problem or an algorithm, it is best to consider an example where you do not already know a lot about the problem. Let us look ahead to analyzing the problem of sorting to see how this process works. What is the least possible cost for any sorting algorithm in the worst case? The algorithm must at least look at every element in the input, just to determine that the input is truly sorted. It is also possible that each of the n values must be moved to another location in the sorted output. Thus, any sorting algorithm must take at least cn time. For many problems, this observation that each of the n inputs must be looked at leads to an easy Ω(n) lower bound. In your previous study of computer science, you have probably seen an example of a sorting algorithm whose running time is in O(n2) in the worst case. The simple Bubble Sort and Insertion Sort algorithms typically given as examples in a first year programming course have worst case running times in O(n2). Thus, the problem of sorting can be said to have an upper bound in O(n2). How do we close the gap between Ω(n) and O(n2)? Can there be a better sorting algorithm? If you can think of no algorithm whose worst-case growth rate is better than O(n2), and if you have discovered no analysis technique to show that the least cost for the problem of sorting in the worst case is greater than Ω(n), then you cannot know for sure whether or not there is a better algorithm. Chapter 7 presents sorting algorithms whose running time is in O(n log n) for the worst case. This greatly narrows the gap. Witht his new knowledge, we now have a lower bound in Ω(n) and an upper bound in O(n log n). Should we search for a faster algorithm? Many have tried, without success. Fortunately (or perhaps unfortunately?), Chapter 7 also includes a proof that any sorting algorithm must have running time in Ω(n log n) in the worst case.2 This proof is one of the most important results in the field of algorithm analysis, and it means that no sorting algorithm can possibly run faster than cn log n for the worst-case input of size n. Thus, we can conclude that the problem of sorting is Θ(n log n) in the worst case, because the upper and lower bounds have met. Knowing the lower bound for a problem does not give you a good algorithm. But it does help you to know when to stop looking. If the lower bound for the problem matches the upper bound for the algorithm (within a constant factor), then we know that we can find an algorithm that is better only by a constant factor. 2 While it is fortunate to know the truth, it is unfortunate that sorting is Θ(n log n) rather than Θ(n)!
- 100. Sec. 3.7 Common Misunderstandings 81 3.7 Common Misunderstandings Asymptotic analysis is one of the most intellectually difficult topics that undergrad- uate computer science majors are confronted with. Most people find growth rates and asymptotic analysis confusing and so develop misconceptions about either the concepts or the terminology. It helps to know what the standard points of confusion are, in hopes of avoiding them. One problem with differentiating the concepts of upper and lower bounds is that, for most algorithms that you will encounter, it is easy to recognize the true growth rate for that algorithm. Given complete knowledge about a cost function, the upper and lower bound for that cost function are always the same. Thus, the distinction between an upper and a lower bound is only worthwhile when you have incomplete knowledge about the thing being measured. If this distinction is still not clear, reread Section 3.6. We use Θ-notation to indicate that there is no meaningful difference between what we know about the growth rates of the upper and lower bound (which is usually the case for simple algorithms). It is a common mistake to confuse the concepts of upper bound or lower bound on the one hand, and worst case or best case on the other. The best, worst, or average cases each give us a concrete instance that we can apply to an algorithm description to get a cost measure. The upper and lower bounds describe our under- standing of the growth rate for that cost measure. So to define the growth rate for an algorithm or problem, we need to determine what we are measuring (the best, worst, or average case) and also our description for what we know about the growth rate of that cost measure (big-Oh, Ω, or Θ). The upper bound for an algorithm is not the same as the worst case for that algorithm for a given input of size n. What is being bounded is not the actual cost (which you can determine for a given value of n), but rather the growth rate for the cost. There cannot be a growth rate for a single point, such as a particular value of n. The growth rate applies to the change in cost as a change in input size occurs. Likewise, the lower bound is not the same as the best case for a given size n. Another common misconception is thinking that the best case for an algorithm occurs when the input size is as small as possible, or that the worst case occurs when the input size is as large as possible. What is correct is that best- and worse- case instances exist for each possible size of input. That is, for all inputs of a given size, say i, one (or more) of the inputs of size i is the best and one (or more) of the inputs of size i is the worst. Often (but not always!), we can characterize the best input case for an arbitrary size, and we can characterize the worst input case for an arbitrary size. Ideally, we can determine the growth rate for the best, worst, and average cases as the input size grows.
- 101. 82 Chap. 3 Algorithm Analysis Example 3.14 What is the growth rate of the best case for sequential search? For any array of size n, the best case occurs when the value we are looking for appears in the first position of the array. This is true regard- less of the size of the array. Thus, the best case (for arbitrary size n) occurs when the desired value is in the first of n positions, and its cost is 1. It is not correct to say that the best case occurs when n = 1. Example 3.15 Imagine drawing a graph to show the cost of finding the maximum value among n values, as n grows. That is, the x axis would be n, and the y value would be the cost. Of course, this is a diagonal line going up to the right, as n increases (you might want to sketch this graph for yourself before reading further). Now, imagine the graph showing the cost for each instance of the prob- lem of finding the maximum value among (say) 20 elements in an array. The first position along the x axis of the graph might correspond to having the maximum element in the first position of the array. The second position along the x axis of the graph might correspond to having the maximum el- ement in the second position of the array, and so on. Of course, the cost is always 20. Therefore, the graph would be a horizontal line with value 20. You should sketch this graph for yourself. Now, let’s switch to the problem of doing a sequential search for a given value in an array. Think about the graph showing all the problem instances of size 20. The first problem instance might be when the value we search for is in the first position of the array. This has cost 1. The second problem instance might be when the value we search for is in the second position of the array. This has cost 2. And so on. If we arrange the problem instances of size 20 from least expensive on the left to most expensive on the right, we see that the graph forms a diagonal line from lower left (with value 0) to upper right (with value 20). Sketch this graph for yourself. Finally, lets consider the cost for performing sequential search as the size of the array n gets bigger. What will this graph look like? Unfortu- nately, there’s not one simple answer, as there was for finding the maximum value. The shape of this graph depends on whether we are considering the best case cost (that would be a horizontal line with value 1), the worst case cost (that would be a diagonal line with value i at position i along the x axis), or the average cost (that would be a a diagonal line with value i/2 at
- 102. Sec. 3.8 Multiple Parameters 83 position i along the x axis). This is why we must always say that function f(n) is in O(g(n)) in the best, average, or worst case! If we leave off which class of inputs we are discussing, we cannot know which cost measure we are referring to for most algorithms. 3.8 Multiple Parameters Sometimes the proper analysis for an algorithm requires multiple parameters to de- scribe the cost. To illustrate the concept, consider an algorithm to compute the rank ordering for counts of all pixel values in a picture. Pictures are often represented by a two-dimensional array, and a pixel is one cell in the array. The value of a pixel is either the code value for the color, or a value for the intensity of the picture at that pixel. Assume that each pixel can take any integer value in the range 0 to C − 1. The problem is to find the number of pixels of each color value and then sort the color values with respect to the number of times each value appears in the picture. Assume that the picture is a rectangle with P pixels. A pseudocode algorithm to solve the problem follows. for (i=0; i<C; i++) // Initialize count count[i] = 0; for (i=0; i<P; i++) // Look at all of the pixels count[value(i)]++; // Increment a pixel value count sort(count); // Sort pixel value counts In this example, count is an array of size C that stores the number of pixels for each color value. Function value(i) returns the color value for pixel i. The time for the first for loop (which initializes count) is based on the num- ber of colors, C. The time for the second loop (which determines the number of pixels with each color) is Θ(P). The time for the final line, the call to sort, de- pends on the cost of the sorting algorithm used. From the discussion of Section 3.6, we can assume that the sorting algorithm has cost Θ(P log P) if P items are sorted, thus yielding Θ(P log P) as the total algorithm cost. Is this a good representation for the cost of this algorithm? What is actu- ally being sorted? It is not the pixels, but rather the colors. What if C is much smaller than P? Then the estimate of Θ(P log P) is pessimistic, because much fewer than P items are being sorted. Instead, we should use P as our analysis vari- able for steps that look at each pixel, and C as our analysis variable for steps that look at colors. Then we get Θ(C) for the initialization loop, Θ(P) for the pixel count loop, and Θ(C log C) for the sorting operation. This yields a total cost of Θ(P + C log C).
- 103. 84 Chap. 3 Algorithm Analysis Why can we not simply use the value of C for input size and say that the cost of the algorithm is Θ(C log C)? Because, C is typically much less than P. For example, a picture might have 1000 × 1000 pixels and a range of 256 possible colors. So, P is one million, which is much larger than C log C. But, if P is smaller, or C larger (even if it is still less than P), then C log C can become the larger quantity. Thus, neither variable should be ignored. 3.9 Space Bounds Besides time, space is the other computing resource that is commonly of concern to programmers. Just as computers have become much faster over the years, they have also received greater allotments of memory. Even so, the amount of available disk space or main memory can be significant constraints for algorithm designers. The analysis techniques used to measure space requirements are similar to those used to measure time requirements. However, while time requirements are nor- mally measured for an algorithm that manipulates a particular data structure, space requirements are normally determined for the data structure itself. The concepts of asymptotic analysis for growth rates on input size apply completely to measuring space requirements. Example 3.16 What are the space requirements for an array of n inte- gers? If each integer requires c bytes, then the array requires cn bytes, which is Θ(n). Example 3.17 Imagine that we want to keep track of friendships between n people. We can do this with an array of size n × n. Each row of the array represents the friends of an individual, with the columns indicating who has that individual as a friend. For example, if person j is a friend of person i, then we place a mark in column j of row i in the array. Likewise, we should also place a mark in column i of row j if we assume that friendship works both ways. For n people, the total size of the array is Θ(n2). A data structure’s primary purpose is to store data in a way that allows efficient access to those data. To provide efficient access, it may be necessary to store addi- tional information about where the data are within the data structure. For example, each node of a linked list must store a pointer to the next value on the list. All such information stored in addition to the actual data values is referred to as overhead. Ideally, overhead should be kept to a minimum while allowing maximum access.
- 104. Sec. 3.9 Space Bounds 85 The need to maintain a balance between these opposing goals is what makes the study of data structures so interesting. One important aspect of algorithm design is referred to as the space/time trade- off principle. The space/time tradeoff principle says that one can often achieve a reduction in time if one is willing to sacrifice space or vice versa. Many programs can be modified to reduce storage requirements by “packing” or encoding informa- tion. “Unpacking” or decoding the information requires additional time. Thus, the resulting program uses less space but runs slower. Conversely, many programs can be modified to pre-store results or reorganize information to allow faster running time at the expense of greater storage requirements. Typically, such changes in time and space are both by a constant factor. A classic example of a space/time tradeoff is the lookup table. A lookup table pre-stores the value of a function that would otherwise be computed each time it is needed. For example, 12! is the greatest value for the factorial function that can be stored in a 32-bit int variable. If you are writing a program that often computes factorials, it is likely to be much more time efficient to simply pre-compute the 12 storable values in a table. Whenever the program needs the value of n! for n ≤ 12, it can simply check the lookup table. (If n > 12, the value is too large to store as an int variable anyway.) Compared to the time required to compute factorials, it may be well worth the small amount of additional space needed to store the lookup table. Lookup tables can also store approximations for an expensive function such as sine or cosine. If you compute this function only for exact degrees or are willing to approximate the answer with the value for the nearest degree, then a lookup table storing the computation for exact degrees can be used instead of repeatedly computing the sine function. Note that initially building the lookup table requires a certain amount of time. Your application must use the lookup table often enough to make this initialization worthwhile. Another example of the space/time tradeoff is typical of what a programmer might encounter when trying to optimize space. Here is a simple code fragment for sorting an array of integers. We assume that this is a special case where there are n integers whose values are a permutation of the integers from 0 to n − 1. This is an example of a Binsort, which is discussed in Section 7.7. Binsort assigns each value to an array position corresponding to its value. for (i=0; i<n; i++) B[A[i]] = A[i];
- 105. 86 Chap. 3 Algorithm Analysis This is efficient and requires Θ(n) time. However, it also requires two arrays of size n. Next is a code fragment that places the permutation in order but does so within the same array (thus it is an example of an “in place” sort). for (i=0; i<n; i++) while (A[i] != i) // Swap element A[i] with A[A[i]] DSutil.swap(A, i, A[i]); Function swap(A, i, j) exchanges elements i and j in array A (see the Appendix). It may not be obvious that the second code fragment actually sorts the array. To see that this does work, notice that each pass through the for loop will at least move the integer with value i to its correct position in the array, and that during this iteration, the value of A[i] must be greater than or equal to i. A total of at most n swap operations take place, because an integer cannot be moved out of its correct position once it has been placed there, and each swap operation places at least one integer in its correct position. Thus, this code fragment has cost Θ(n). However, it requires more time to run than the first code fragment. On my computer the second version takes nearly twice as long to run as the first, but it only requires half the space. A second principle for the relationship between a program’s space and time requirements applies to programs that process information stored on disk, as dis- cussed in Chapter 8 and thereafter. Strangely enough, the disk-based space/time tradeoff principle is almost the reverse of the space/time tradeoff principle for pro- grams using main memory. The disk-based space/time tradeoff principle states that the smaller you can make your disk storage requirements, the faster your program will run. This is be- cause the time to read information from disk is enormous compared to computation time, so almost any amount of additional computation needed to unpack the data is going to be less than the disk-reading time saved by reducing the storage require- ments. Naturally this principle does not hold true in all cases, but it is good to keep in mind when designing programs that process information stored on disk. 3.10 Speeding Up Your Programs In practice, there is not such a big difference in running time between an algorithm whose growth rate is Θ(n) and another whose growth rate is Θ(n log n). There is, however, an enormous difference in running time between algorithms with growth rates of Θ(n log n) and Θ(n2). As you shall see during the course of your study of common data structures and algorithms, it is not unusual that a problem whose obvious solution requires Θ(n2) time also has a solution that requires Θ(n log n)
- 106. Sec. 3.10 Speeding Up Your Programs 87 time. Examples include sorting and searching, two of the most important computer problems. Example 3.18 The following is a true story. A few years ago, one of my graduate students had a big problem. His thesis work involved several intricate operations on a large database. He was now working on the final step. “Dr. Shaffer,” he said, “I am running this program and it seems to be taking a long time.” After examining the algorithm we realized that its running time was Θ(n2), and that it would likely take one to two weeks to complete. Even if we could keep the computer running uninterrupted for that long, he was hoping to complete his thesis and graduate before then. Fortunately, we realized that there was a fairly easy way to convert the algorithm so that its running time was Θ(n log n). By the next day he had modified the program. It ran in only a few hours, and he finished his thesis on time. While not nearly so important as changing an algorithm to reduce its growth rate, “code tuning” can also lead to dramatic improvements in running time. Code tuning is the art of hand-optimizing a program to run faster or require less storage. For many programs, code tuning can reduce running time by a factor of ten, or cut the storage requirements by a factor of two or more. I once tuned a critical function in a program — without changing its basic algorithm — to achieve a factor of 200 speedup. To get this speedup, however, I did make major changes in the representation of the information, converting from a symbolic coding scheme to a numeric coding scheme on which I was able to do direct computation. Here are some suggestions for ways to speed up your programs by code tuning. The most important thing to realize is that most statements in a program do not have much effect on the running time of that program. There are normally just a few key subroutines, possibly even key lines of code within the key subroutines, that account for most of the running time. There is little point to cutting in half the running time of a subroutine that accounts for only 1% of the total running time. Focus your attention on those parts of the program that have the most impact. When tuning code, it is important to gather good timing statistics. Many com- pilers and operating systems include profilers and other special tools to help gather information on both time and space use. These are invaluable when trying to make a program more efficient, because they can tell you where to invest your effort. A lot of code tuning is based on the principle of avoiding work rather than speeding up work. A common situation occurs when we can test for a condition
- 107. 88 Chap. 3 Algorithm Analysis that lets us skip some work. However, such a test is never completely free. Care must be taken that the cost of the test does not exceed the amount of work saved. While one test might be cheaper than the work potentially saved, the test must always be made and the work can be avoided only some fraction of the time. Example 3.19 A common operation in computer graphics applications is to find which among a set of complex objects contains a given point in space. Many useful data structures and algorithms have been developed to deal with variations of this problem. Most such implementations involve the following tuning step. Directly testing whether a given complex ob- ject contains the point in question is relatively expensive. Instead, we can screen for whether the point is contained within a bounding box for the object. The bounding box is simply the smallest rectangle (usually defined to have sides perpendicular to the x and y axes) that contains the object. If the point is not in the bounding box, then it cannot be in the object. If the point is in the bounding box, only then would we conduct the full com- parison of the object versus the point. Note that if the point is outside the bounding box, we saved time because the bounding box test is cheaper than the comparison of the full object versus the point. But if the point is inside the bounding box, then that test is redundant because we still have to com- pare the point against the object. Typically the amount of work avoided by making this test is greater than the cost of making the test on every object. Example 3.20 Section 7.2.3 presents a sorting algorithm named Selec- tion Sort. The chief distinguishing characteristic of this algorithm is that it requires relatively few swaps of records stored in the array being sorted. However, it sometimes performs an unnecessary swap operation, specifi- cally when it tries to swap a record with itself. This work could be avoided by testing whether the two indices being swapped are the same. However, this event has insufficient chance of occurring. Because the cost of the test is high enough compared to the work saved when the test is successful, adding the test typically will slow down the program rather than speed it up. Be careful not to use tricks that make the program unreadable. Most code tun- ing is simply cleaning up a carelessly written program, not taking a clear program and adding tricks. In particular, you should develop an appreciation for the capa-
- 108. Sec. 3.11 Empirical Analysis 89 bilities of modern compilers to make extremely good optimizations of expressions. “Optimization of expressions” here means a rearrangement of arithmetic or logical expressions to run more efficiently. Be careful not to damage the compiler’s ability to do such optimizations for you in an effort to optimize the expression yourself. Always check that your “optimizations” really do improve the program by running the program before and after the change on a suitable benchmark set of input. Many times I have been wrong about the positive effects of code tuning in my own pro- grams. Most often I am wrong when I try to optimize an expression. It is hard to do better than the compiler. The greatest time and space improvements come from a better data structure or algorithm. The final thought for this section is First tune the algorithm, then tune the code. 3.11 Empirical Analysis This chapter has focused on asymptotic analysis. This is an analytic tool, whereby we model the key aspects of an algorithm to determine the growth rate of the alg- orithm as the input size grows. As pointed out previously, there are many limita- tions to this approach. These include the effects at small problem size, determining the finer distinctions between algorithms with the same growth rate, and the inher- ent difficulty of doing mathematical modeling for more complex problems. An alternative to analytical approaches are empirical approaches. The most ob- vious empirical approach is simply to run two competitors and see which performs better. In this way we might overcome the deficiencies of analytical approaches. Be warned that comparative timing of programs is a difficult business, often subject to experimental errors arising from uncontrolled factors (system load, the language or compiler used, etc.). The most important point is not to be biased in favor of one of the programs. If you are biased, this is certain to be reflected in the timings. One look at competing software or hardware vendors’ advertisements should convince you of this. The most common pitfall when writing two programs to compare their performance is that one receives more code-tuning effort than the other. As mentioned in Section 3.10, code tuning can often reduce running time by a factor of ten. If the running times for two programs differ by a constant factor regardless of input size (i.e., their growth rates are the same), then differences in code tuning might account for any difference in running time. Be suspicious of empirical comparisons in this situation.
- 109. 90 Chap. 3 Algorithm Analysis Another approach to analysis is simulation. The idea of simulation is to model the problem with a computer program and then run it to get a result. In the con- text of algorithm analysis, simulation is distinct from empirical comparison of two competitors because the purpose of the simulation is to perform analysis that might otherwise be too difficult. A good example of this appears in Figure 9.8. This fig- ure shows the cost for inserting or deleting a record from a hash table under two different assumptions for the policy used to find a free slot in the table. The y axes is the cost in number of hash table slots evaluated, and the x axes is the percentage of slots in the table that are full. The mathematical equations for these curves can be determined, but this is not so easy. A reasonable alternative is to write simple variations on hashing. By timing the cost of the program for various loading condi- tions, it is not difficult to construct a plot similar to Figure 9.8. The purpose of this analysis is not to determine which approach to hashing is most efficient, so we are not doing empirical comparison of hashing alternatives. Instead, the purpose is to analyze the proper loading factor that would be used in an efficient hashing system to balance time cost versus hash table size (space cost). 3.12 Further Reading Pioneering works on algorithm analysis include The Art of Computer Programming by Donald E. Knuth [Knu97, Knu98], and The Design and Analysis of Computer Algorithms by Aho, Hopcroft, and Ullman [AHU74]. The alternate definition for Ω comes from [AHU83]. The use of the notation “T(n) is in O(f(n))” rather than the more commonly used “T(n) = O(f(n))” I derive from Brassard and Bratley [BB96], though certainly this use predates them. A good book to read for further information on algorithm analysis techniques is Compared to What? by Gregory J.E. Rawlins [Raw92]. Bentley [Ben88] describes one problem in numerical analysis for which, be- tween 1945 and 1988, the complexity of the best known algorithm had decreased from O(n7) to O(n3). For a problem of size n = 64, this is roughly equivalent to the speedup achieved from all advances in computer hardware during the same time period. While the most important aspect of program efficiency is the algorithm, much improvement can be gained from efficient coding of a program. As cited by Freder- ick P. Brooks in The Mythical Man-Month [Bro95], an efficient programmer can of- ten produce programs that run five times faster than an inefficient programmer, even when neither takes special efforts to speed up their code. For excellent and enjoy- able essays on improving your coding efficiency, and ways to speed up your code when it really matters, see the books by Jon Bentley [Ben82, Ben00, Ben88]. The
- 110. Sec. 3.13 Exercises 91 situation described in Example 3.18 arose when we were working on the project reported on in [SU92]. As an interesting aside, writing a correct binary search algorithm is not easy. Knuth [Knu98] notes that while the first binary search was published in 1946, the first bug-free algorithm was not published until 1962! Bentley (“Writing Correct Programs” in [Ben00]) has found that 90% of the computer professionals he tested could not write a bug-free binary search in two hours. 3.13 Exercises 3.1 For each of the five expressions of Figure 3.1, give the range of values of n for which that expression is most efficient. 3.2 Graph the following expressions. For each expression, state the range of values of n for which that expression is the most efficient. 4n2 log3 n 3n 20n 2 log2 n n2/3 3.3 Arrange the following expressions by growth rate from slowest to fastest. 4n2 log3 n n! 3n 20n 2 log2 n n2/3 See Stirling’s approximation in Section 2.2 for help in classifying n!. 3.4 (a) Suppose that a particular algorithm has time complexity T(n) = 3 × 2n, and that executing an implementation of it on a particular machine takes t seconds for n inputs. Now suppose that we are presented with a machine that is 64 times as fast. How many inputs could we process on the new machine in t seconds? (b) Suppose that another algorithm has time complexity T(n) = n2, and that executing an implementation of it on a particular machine takes t seconds for n inputs. Now suppose that we are presented with a ma- chine that is 64 times as fast. How many inputs could we process on the new machine in t seconds? (c) A third algorithm has time complexity T(n) = 8n. Executing an im- plementation of it on a particular machine takes t seconds for n inputs. Given a new machine that is 64 times as fast, how many inputs could we process in t seconds? 3.5 Hardware vendor XYZ Corp. claims that their latest computer will run 100 times faster than that of their competitor, Prunes, Inc. If the Prunes, Inc. computer can execute a program on input of size n in one hour, what size
- 111. 92 Chap. 3 Algorithm Analysis input can XYZ’s computer execute in one hour for each algorithm with the following growth rate equations? n n2 n3 2n 3.6 (a) Find a growth rate that squares the run time when we double the input size. That is, if T(n) = X, then T(2n) = x2 (b) Find a growth rate that cubes the run time when we double the input size. That is, if T(n) = X, then T(2n) = x3 3.7 Using the definition of big-Oh, show that 1 is in O(1) and that 1 is in O(n). 3.8 Using the definitions of big-Oh and Ω, find the upper and lower bounds for the following expressions. Be sure to state appropriate values for c and n0. (a) c1n (b) c2n3 + c3 (c) c4n log n + c5n (d) c62n + c7n6 3.9 (a) What is the smallest integer k such that √ n = O(nk)? (b) What is the smallest integer k such that n log n = O(nk)? 3.10 (a) Is 2n = Θ(3n)? Explain why or why not. (b) Is 2n = Θ(3n)? Explain why or why not. 3.11 For each of the following pairs of functions, either f(n) is in O(g(n)), f(n) is in Ω(g(n)), or f(n) = Θ(g(n)). For each pair, determine which relation- ship is correct. Justify your answer, using the method of limits discussed in Section 3.4.5. (a) f(n) = log n2; g(n) = log n + 5. (b) f(n) = √ n; g(n) = log n2. (c) f(n) = log2 n; g(n) = log n. (d) f(n) = n; g(n) = log2n. (e) f(n) = n log n + n; g(n) = log n. (f) f(n) = log n2; g(n) = (log n)2. (g) f(n) = 10; g(n) = log 10. (h) f(n) = 2n; g(n) = 10n2. (i) f(n) = 2n; g(n) = n log n. (j) f(n) = 2n; g(n) = 3n. (k) f(n) = 2n; g(n) = nn. 3.12 Determine Θ for the following code fragments in the average case. Assume that all variables are of type int. (a) a = b + c; d = a + e;
- 112. Sec. 3.13 Exercises 93 (b) sum = 0; for (i=0; i<3; i++) for (j=0; j<n; j++) sum++; (c) sum=0; for (i=0; i<n*n; i++) sum++; (d) for (i=0; i < n-1; i++) for (j=i+1; j < n; j++) { tmp = AA[i][j]; AA[i][j] = AA[j][i]; AA[j][i] = tmp; } (e) sum = 0; for (i=1; i<=n; i++) for (j=1; j<=n; j*=2) sum++; (f) sum = 0; for (i=1; i<=n; i*=2) for (j=1; j<=n; j++) sum++; (g) Assume that array A contains n values, Random takes constant time, and sort takes n log n steps. for (i=0; i<n; i++) { for (j=0; j<n; j++) A[i] = DSutil.random(n); sort(A); } (h) Assume array A contains a random permutation of the values from 0 to n − 1. sum = 0; for (i=0; i<n; i++) for (j=0; A[j]!=i; j++) sum++; (i) sum = 0; if (EVEN(n)) for (i=0; i<n; i++) sum++; else sum = sum + n; 3.13 Show that big-Theta notation (Θ) defines an equivalence relation on the set of functions.
- 113. 94 Chap. 3 Algorithm Analysis 3.14 Give the best lower bound that you can for the following code fragment, as a function of the initial value of n. while (n > 1) if (ODD(n)) n = 3 * n + 1; else n = n / 2; Do you think that the upper bound is likely to be the same as the answer you gave for the lower bound? 3.15 Does every algorithm have a Θ running-time equation? In other words, are the upper and lower bounds for the running time (on any specified class of inputs) always the same? 3.16 Does every problem for which there exists some algorithm have a Θ running- time equation? In other words, for every problem, and for any specified class of inputs, is there some algorithm whose upper bound is equal to the problem’s lower bound? 3.17 Given an array storing integers ordered by value, modify the binary search routine to return the position of the first integer with value K in the situation where K can appear multiple times in the array. Be sure that your algorithm is Θ(log n), that is, do not resort to sequential search once an occurrence of K is found. 3.18 Given an array storing integers ordered by value, modify the binary search routine to return the position of the integer with the greatest value less than K when K itself does not appear in the array. Return ERROR if the least value in the array is greater than K. 3.19 Modify the binary search routine to support search in an array of infinite size. In particular, you are given as input a sorted array and a key value K to search for. Call n the position of the smallest value in the array that is equal to or larger than X. Provide an algorithm that can determine n in O(log n) comparisons in the worst case. Explain why your algorithm meets the required time bound. 3.20 It is possible to change the way that we pick the dividing point in a binary search, and still get a working search routine. However, where we pick the dividing point could affect the performance of the algorithm. (a) If we change the dividing point computation in function binary from i = (l + r)/2 to i = (l + ((r − l)/3)), what will the worst-case run- ning time be in asymptotic terms? If the difference is only a constant time factor, how much slower or faster will the modified program be compared to the original version of binary?
- 114. Sec. 3.14 Projects 95 (b) If we change the dividing point computation in function binary from i = (l + r)/2 to i = r − 2, what will the worst-case running time be in asymptotic terms? If the difference is only a constant time factor, how much slower or faster will the modified program be compared to the original version of binary? 3.21 Design an algorithm to assemble a jigsaw puzzle. Assume that each piece has four sides, and that each piece’s final orientation is known (top, bottom, etc.). Assume that you have available a function bool compare(Piece a, Piece b, Side ad) that can tell, in constant time, whether piece a connects to piece b on a’s side ad and b’s opposite side bd. The input to your algorithm should consist of an n × m array of random pieces, along with dimensions n and m. The algorithm should put the pieces in their correct positions in the array. Your algorithm should be as efficient as possible in the asymptotic sense. Write a summation for the running time of your algorithm on n pieces, and then derive a closed-form solution for the summation. 3.22 Can the average case cost for an algorithm be worse than the worst case cost? Can it be better than the best case cost? Explain why or why not. 3.23 Prove that if an algorithm is Θ(f(n)) in the average case, then it is Ω(f(n)) in the worst case. 3.24 Prove that if an algorithm is Θ(f(n)) in the average case, then it is O(f(n)) in the best case. 3.14 Projects 3.1 Imagine that you are trying to store 32 boolean values, and must access them frequently. Compare the time required to access Boolean values stored al- ternatively as a single bit field, a character, a short integer, or a long integer. There are two things to be careful of when writing your program. First, be sure that your program does enough variable accesses to make meaningful measurements. A single access is much smaller than the measurement rate for all four methods. Second, be sure that your program spends as much time as possible doing variable accesses rather than other things such as calling timing functions or incrementing for loop counters. 3.2 Implement sequential search and binary search algorithms on your computer. Run timings for each algorithm on arrays of size n = 10i for i ranging from 1 to as large a value as your computer’s memory and compiler will allow. For both algorithms, store the values 0 through n − 1 in order in the array, and
- 115. 96 Chap. 3 Algorithm Analysis use a variety of random search values in the range 0 to n − 1 on each size n. Graph the resulting times. When is sequential search faster than binary search for a sorted array? 3.3 Implement a program that runs and gives timings for the two Fibonacci se- quence functions provided in Exercise 2.11. Graph the resulting running times for as many values of n as your computer can handle.
- 116. PART II Fundamental Data Structures 97
- 118. 4 Lists, Stacks, and Queues If your program needs to store a few things — numbers, payroll records, or job descriptions for example — the simplest and most effective approach might be to put them in a list. Only when you have to organize or search through a large number of things do more sophisticated data structures usually become necessary. (We will study how to organize and search through medium amounts of data in Chapters 5, 7, and 9, and discuss how to deal with large amounts of data in Chapters 8–10.) Many applications don’t require any form of search, and they do not require that any ordering be placed on the objects being stored. Some applications require processing in a strict chronological order, perhaps processing objects in the order that they arrived, or perhaps processing objects in the reverse of the order that they arrived. For all these situations, a simple list structure is appropriate. This chapter describes representations for lists in general, as well as two impor- tant list-like structures called the stack and the queue. Along with presenting these fundamental data structures, the other goals of the chapter are to: (1) Give examples of separating a logical representation in the form of an ADT from a physical im- plementation for a data structure. (2) Illustrate the use of asymptotic analysis in the context of some simple operations that you might already be familiar with. In this way you can begin to see how asymptotic analysis works, without the complica- tions that arise when analyzing more sophisticated algorithms and data structures. (3) Introduce the concept and use of dictionaries and comparator classes. We begin by defining an ADT for lists in Section 4.1. Two implementations for the list ADT — the array-based list and the linked list — are covered in detail and their relative merits discussed. Sections 4.2 and 4.3 cover stacks and queues, re- spectively. Sample implementations for each of these data structures are presented. Section 4.4 presents the Dictionary ADT for storing and retrieving data, which sets a context for implementing search structures such as the Binary Search Tree of Section 5.4. 99
- 119. 100 Chap. 4 Lists, Stacks, and Queues 4.1 Lists We all have an intuitive understanding of what we mean by a “list,” so our first step is to define precisely what is meant so that this intuitive understanding can eventually be converted into a concrete data structure and its operations. Perhaps the most important concept related to lists is that of position. In other words, we perceive that there is a first element in the list, a second element, and so on. Thus, we should view a list as embodying the mathematical concepts of a sequence, as defined in Section 2.1. We define a list to be a finite, ordered sequence of data items known as ele- ments. “Ordered” in this definition means that each element has a position in the list. (We will not use “ordered” in this context to mean that the list is sorted.) Each list element has a data type. In the simple list implementations discussed in this chapter, all elements of the list have the same data type, although there is no con- ceptual objection to lists whose elements have differing data types if the application requires it (see Section 12.1). The operations defined as part of the list ADT do not depend on the elemental data type. For example, the list ADT can be used for lists of integers, lists of characters, lists of payroll records, even lists of lists. A list is said to be empty when it contains no elements. The number of ele- ments currently stored is called the length of the list. The beginning of the list is called the head, the end of the list is called the tail. There might or might not be some relationship between the value of an element and its position in the list. For example, sorted lists have their elements positioned in ascending order of value, while unsorted lists have no particular relationship between element values and positions. This section will consider only unsorted lists. Chapters 7 and 9 treat the problems of how to create and search sorted lists efficiently. When presenting the contents of a list, we use the same notation as was in- troduced for sequences in Section 2.1. To be consistent with Java array indexing, the first position on the list is denoted as 0. Thus, if there are n elements in the list, they are given positions 0 through n − 1 as ha0, a1, ..., an−1i. The subscript indicates an element’s position within the list. Using this notation, the empty list would appear as hi. Before selecting a list implementation, a program designer should first consider what basic operations the implementation must support. Our common intuition about lists tells us that a list should be able to grow and shrink in size as we insert and remove elements. We should be able to insert and remove elements from any- where in the list. We should be able to gain access to any element’s value, either to read it or to change it. We must be able to create and clear (or reinitialize) lists. It is also convenient to access the next or previous element from the “current” one.
- 120. Sec. 4.1 Lists 101 The next step is to define the ADT for a list object in terms of a set of operations on that object. We will use the Java notation of an interface to formally define the list ADT. Interface List defines the member functions that any list implementation inheriting from it must support, along with their parameters and return types. We increase the flexibility of the list ADT by writing it as a Java generic. True to the notion of an ADT, an interface does not specify how operations are implemented. Two complete implementations are presented later in this sec- tion, both of which use the same list ADT to define their operations, but they are considerably different in approaches and in their space/time tradeoffs. Figure 4.1 presents our list ADT. Class List is a generic with one type pa- rameter, named E. E serves as a placeholder for whatever element type the user would like to store in a list. The comments given in Figure 4.1 describe precisely what each member function is intended to do. However, some explanation of the basic design is in order. Given that we wish to support the concept of a sequence, with access to any position in the list, the need for many of the member functions such as insert and moveToPos is clear. The key design decision embodied in this ADT is support for the concept of a current position. For example, member moveToStart sets the current position to be the first element on the list, while methods next and prev move the current position to the next and previous ele- ments, respectively. The intention is that any implementation for this ADT support the concept of a current position. The current position is where any action such as insertion or deletion will take place. Since insertions take place at the current position, and since we want to be able to insert to the front or the back of the list as well as anywhere in between, there are actuall n + 1 possible “current positions” when there are n elements in the list. It is helpful to modify our list display notation to show the position of the current element. I will use a vertical bar, such as h20, 23 | 12, 15i to indicate the list of four elements, with the current position being to the right of the bar at element 12. Given this configuration, calling insert with value 10 will change the list to be h20, 23 | 10, 12, 15i. If you examine Figure 4.1, you should find that the list member functions pro- vided allow you to build a list with elements in any desired order, and to access any desired position in the list. You might have noticed that the clear method is not necessary, in that it could be implemented by means of the other member functions in the same asymptotic time. It is included merely for convenience. Method getValue returns a reference to the current element. If there is no current value, then null is returned. A list can be iterated through as shown in this following code fragment.
- 121. 102 Chap. 4 Lists, Stacks, and Queues /** List ADT */ public interface List<E> { /** Remove all contents from the list, so it is once again empty. Client is responsible for reclaiming storage used by the list elements. */ public void clear(); /** Insert an element at the current location. The client is responsible for ensuring that the list’s capacity is not exceeded. @param item The element to be inserted. */ public void insert(E item); /** Append an element at the end of the list. The client is responsible for ensuring that the list’s capacity is not exceeded. @param item The element to be appended. */ public void append(E item); /** Remove and return the current element. @return The element that was removed. */ public E remove(); /** Set the current position to the start of the list */ public void moveToStart(); /** Set the current position to the end of the list */ public void moveToEnd(); /** Move the current position one step left. No change if already at beginning. */ public void prev(); /** Move the current position one step right. No change if already at end. */ public void next(); /** @return The number of elements in the list. */ public int length(); /** @return The position of the current element. */ public int currPos(); /** Set current position. @param pos The position to make current. */ public void moveToPos(int pos); /** @return The current element. */ public E getValue(); } Figure 4.1 The Java interface for a list.
- 122. Sec. 4.1 Lists 103 for (L.moveToStart(); L.currPos()<L.length(); L.next()) { it = L.getValue(); doSomething(it); } In this example, each element of the list in turn is stored in it, and passed to the doSomething function. The loop terminates when the current position reaches the end of the list. The list class declaration presented here is just one of many possible interpreta- tions for lists. Figure 4.1 provides most of the operations that one naturally expects to perform on lists and serves to illustrate the issues relevant to implementing the list data structure. As an example of using the list ADT, we can create a function to return true if there is an occurrence of a given integer in the list, and false otherwise. The find method needs no knowledge about the specific list imple- mentation, just the list ADT. /** @return True if "k" is in list "L", false otherwise */ public static boolean find(List<Integer> L, int k) { int it; for (L.moveToStart(); L.currPos()<L.length(); L.next()) { it = L.getValue(); if (k == it) return true; // Found k } return false; // k not found } While this implementation for find could be written as a generic with re- spect to the element type, it would still be limited in its ability to handle different data types stored on the list. In particular, it only works when the description for the object being searched for (k in the function) is of the same type as the objects them- selves. A more typical situation is that we are searching for a record that contains a key field who’s value matches k. Similar functions to find and return a composite element based on a key value can be created using the list implementation, but to do so requires some agreement between the list ADT and the find function on the concept of a key. This topic will be discussed in Section 4.4. 4.1.1 Array-Based List Implementation There are two standard approaches to implementing lists, the array-based list, and the linked list. This section discusses the array-based approach. The linked list is presented in Section 4.1.2. Time and space efficiency comparisons for the two are discussed in Section 4.1.3.
- 123. 104 Chap. 4 Lists, Stacks, and Queues /** Array-based list implementation */ class AList<E> implements List<E> { private static final int defaultSize = 10; // Default size private int maxSize; // Maximum size of list private int listSize; // Number of list items now private int curr; // Position current element private E[] listArray; // Array holding list elements /** Constructors */ /** Create a list with the default capacity. */ AList() { this(defaultSize); } /** Create a new list object. @param size Max number of elements list can contain. */ @SuppressWarnings("unchecked") // Generic array allocation AList(int size) { maxSize = size; listSize = curr = 0; listArray = (E[])new Object[size]; // Create listArray } public void clear() // Reinitialize the list { listSize = curr = 0; } // Simply reinitialize values /** Insert "it" at current position */ public void insert(E it) { assert listSize < maxSize : "List capacity exceeded"; for (int i=listSize; i>curr; i--) // Shift elements up listArray[i] = listArray[i-1]; // to make room listArray[curr] = it; listSize++; // Increment list size } public void append(E it) { // Append "it" assert listSize < maxSize : "List capacity exceeded"; listArray[listSize++] = it; } public void moveToStart() { curr = 0; } // Reset position Figure 4.2 An array-based list implementation. Figure 4.2 shows the array-based list implementation, named AList. AList inherits from abstract class List and so must implement all of the member func- tions of List.
- 124. Sec. 4.1 Lists 105 // Remove and return the current element. public E remove() { if ((curr<0) || (curr>=listSize)) // No current element return null; E it = listArray[curr]; // Copy the element for(int i=curr; i<listSize-1; i++) // Shift them down listArray[i] = listArray[i+1]; listSize--; // Decrement size return it; } public void moveToEnd() { curr = listSize; } // Reset public void prev() { if (curr != 0) curr--; } // Back up public void next() { if (curr < listSize) curr++; } // Next // Return list size public int length() { return listSize; } public int currPos() { // Return current position return curr; } // Set current list position to "pos" public void moveToPos(int pos) { assert (pos>=0) && (pos<=listSize) : "Pos out of range"; curr = pos; } public E getValue() { // Return current element assert (curr>=0)&&(curr<listSize) : "No current element"; return listArray[curr]; } Figure 4.2 (continued) Class AList’s private portion contains the data members for the array-based list. These include listArray, the array which holds the list elements. Because listArray must be allocated at some fixed size, the size of the array must be known when the list object is created. Note that an optional parameter is declared for the AList constructor. With this parameter, the user can indicate the maximum number of elements permitted in the list. If no parameter is given, then it takes the value DefaultListSize, which is assumed to be a suitably defined constant value. Because each list can have a differently sized array, each list must remember its maximum permitted size. Data member maxSize serves this purpose. At any
- 125. 106 Chap. 4 Lists, Stacks, and Queues given time the list actually holds some number of elements that can be less than the maximum allowed by the array. This value is stored in listSize. Data member curr stores the current position. Because listArray, maxSize, listSize, and curr are all declared to be private, they may only be accessed by methods of Class AList. Class AList stores the list elements in contiguous array positions. Array po- sitions correspond to list positions. In other words, the element at position i in the list is stored at array cell i. The head of the list is always at position 0. This makes random access to any element in the list quite easy. Given some position in the list, the value of the element in that position can be accessed directly. Thus, access to any element using the moveToPos method followed by the getValue method takes Θ(1) time. Because the array-based list implementation is defined to store list elements in contiguous cells of the array, the insert, append, and remove methods must maintain this property. Inserting or removing elements at the tail of the list is easy, and the append operation takes Θ(1) time. However, if we wish to insert an element at the head of the list, all elements currently in the list must shift one position toward the tail to make room, as illustrated by Figure 4.3. This process takes Θ(n) time if there are n elements already in the list. If we wish to insert at position i within a list of n elements, then n − i elements must shift toward the tail. Removing an element from the head of the list is similar in that all remaining elements in the array must shift toward the head by one position to fill in the gap. To remove the element at position i, n − i − 1 elements must shift toward the head. In the average case, insertion or removal requires moving half of the elements, which is Θ(n). Most of the other member functions for Class AList simply access the current list element or move the current position. Such operations all require Θ(1) time. Aside from insert and remove, the only other operations that might require more than constant time are the constructor, the destructor, and clear. These three member functions each make use of the system free-store operation new. As discussed further in Section 4.1.2, system free-store operations can be expensive. 4.1.2 Linked Lists The second traditional approach to implementing lists makes use of pointers and is usually called a linked list. The linked list uses dynamic memory allocation, that is, it allocates memory for new list elements as needed. A linked list is made up of a series of objects, called the nodes of the list. Because a list node is a distinct object (as opposed to simply a cell in an array), it is
- 126. Sec. 4.1 Lists 107 Insert 23: 12 20 3 8 13 12 20 8 3 3 8 20 12 13 23 13 (a) (b) (c) 5 0 1 2 4 4 3 2 1 0 1 2 3 4 5 5 0 3 Figure 4.3 Inserting an element at the head of an array-based list requires shift- ing all existing elements in the array by one position toward the tail. (a) A list containing five elements before inserting an element with value 23. (b) The list after shifting all existing elements one position to the right. (c) The list after 23 has been inserted in array position 0. Shading indicates the unused part of the array. good practice to make a separate list node class. An additional benefit to creating a list node class is that it can be reused by the linked implementations for the stack and queue data structures presented later in this chapter. Figure 4.4 shows the implementation for list nodes, called the Link class. Objects in the Link class contain an element field to store the element value, and a next field to store a pointer to the next node on the list. The list built from such nodes is called a singly linked list, or a one-way list, because each list node has a single pointer to the next node on the list. The Link class is quite simple. There are two forms for its constructor, one with an initial element value and one without. Because the Link class is also used by the stack and queue implementations presented later, its data members are made public. While technically this is breaking encapsulation, in practice the Link class should be implemented as a private class of the linked list (or stack or queue) implementation, and thus not visible to the rest of the program. Figure 4.5(a) shows a graphical depiction for a linked list storing four integers. The value stored in a pointer variable is indicated by an arrow “pointing” to some- thing. Java uses the special symbol null for a pointer value that points nowhere, such as for the last list node’s next field. A null pointer is indicated graphically by a diagonal slash through a pointer variable’s box. The vertical line between the nodes labeled 23 and 12 in Figure 4.5(a) indicates the current position. The first link node of the list is accessed from a pointer named head. To speed access to the end of the list, in particular to allow the append method to
- 127. 108 Chap. 4 Lists, Stacks, and Queues class Link<E> { // Singly linked list node private E element; // Value for this node private Link<E> next; // Pointer to next node in list // Constructors Link(E it, Link<E> nextval) { element = it; next = nextval; } Link(Link<E> nextval) { next = nextval; } Link<E> next() { return next; } Link<E> setNext(Link<E> nextval) { return next = nextval; } E element() { return element; } E setElement(E it) { return element = it; } } // class Link Figure 4.4 A simple singly linked list node implementation. tail head 20 23 12 15 (a) head tail 15 12 10 23 20 (b) curr curr Figure 4.5 Illustration of a faulty linked-list implementation where curr points directly to the current node. (a) Linked list prior to inserting element with value 10. (b) Desired effect of inserting element with value 10. be performed in constant time, a pointer named tail is also kept to the last link of the list. The position of the current element is indicated by another pointer, named curr. Finally, because there is no simple way to compute the length of the list simply from these three pointers, the list length must be stored explicitly, and updated by every operation that modifies the list size. The value cnt stores the length of the list. Note that LList’s constructor maintains the optional parameter for minimum list size introduced for Class AList. This is done simply to keep the calls to the constructor the same for both variants. Because the linked list class does not need
- 128. Sec. 4.1 Lists 109 to declare a fixed-size array when the list is created, this parameter is unnecessary for linked lists. It is ignored by the implementation. A key design decision for the linked list implementation is how to represent the current position. The most reasonable choices appear to be a pointer to the current element. But there is a big advantage to making curr point to the element preceding the current element. Figure 4.5(a) shows the list’s curr pointer pointing to the current element. The vertical line between the nodes containing 23 and 12 indicates the logical position of the current element. Consider what happens if we wish to insert a new node with value 10 into the list. The result should be as shown in Figure 4.5(b). However, there is a problem. To “splice” the list node containing the new element into the list, the list node storing 23 must have its next pointer changed to point to the new node. Unfortunately, there is no convenient access to the node preceding the one pointed to by curr. However, if curr points directlyto the preceding element, there is no difficulty in adding a new element after curr. See Exercise 4.5 for further discussion of why making curr point directly to the current element fails. Unfortunately, we encounter a number of problems when the list is empty, or when the current position is at an end of the list. In particular, when the list is empty we have no element for head, tail, and curr to point to. One solution is to implement a number of special cases in the implementations for insert and remove. This increases code complexity, making it harder to understand, and thus increases the chance of introducing a programming bug. These special cases can be eliminated by implementing linked lists with a spe- cial header node as the first node of the list. This header node is a link node like any other, but its value is ignored and it is not considered to be an actual element of the list. The header node saves coding effort because we no longer need to consider special cases for empty lists or when the current position is at one end of the list. The cost of this simplification is the space for the header node. However, there are space savings due to smaller code size, because statements to handle the special cases are omitted. In practice, this reduction in code size typically saves more space than that required for the header node, depending on the number of lists created. Figure 4.6 shows the state of an initialized or empty list when using a header node. Figure 4.7 shows the insertion example of Figure 4.5 using a header node and the convention that curr points to the node preceding the current node. Figure 4.8 shows the definition for the linked list class, named LList. Class LList inherits from the abstract list class and thus must implement all of Class List’s member functions.
- 129. 110 Chap. 4 Lists, Stacks, and Queues tail head curr Figure 4.6 Initial state of a linked list when using a header node. tail curr head 20 23 12 15 (a) head tail 20 23 10 12 (b) 15 curr Figure 4.7 Insertion using a header node, with curr pointing one node head of the current element. (a) Linked list before insertion. The current node contains 12. (b) Linked list after inserting the node containing 10. Implementation for most member functions of the list class is straightfor- ward. However, insert and remove should be studied carefully. Inserting a new element is a three-step process. First, the new list node is created and the new element is stored into it. Second, the next field of the new list node is assigned to point to the current node (the one (after) the node that curr points to). Third, the next field of node pointed to by curr is assigned to point to the newly inserted node. The following line in the insert method of Figure 4.8 actually does all three of these steps. curr.setNext(new Link<E>(it, curr.next())); Operator new creates the new link node and calls the constructor for the Link class, which takes two parameters. The first is the element. The second is the value to be placed in the list node’s next field, in this case “curr.next().” Figure 4.9 illustrates this three-step process. Once the new node is added, tail is pushed forward if the new element was added to the end of the list. Insertion requires Θ(1) time. Removing a node from the linked list requires only that the appropriate pointer be redirected around the node to be deleted. This memory eventually be reclaimed by the garbage collector. The following lines from the remove method of Fig- ure 4.8 do precisely this.
- 130. Sec. 4.1 Lists 111 // Linked list implementation class LList<E> implements List<E> { private Link<E> head; // Pointer to list header private Link<E> tail; // Pointer to last element protected Link<E> curr; // Access to current element int cnt; // Size of list //Constructors LList(int size) { this(); } // Constructor -- Ignore size LList() { curr = tail = head = new Link<E>(null); // Create header cnt = 0; } public void clear() { // Remove all elements head.setNext(null); // Drop access to links curr = tail = head = new Link<E>(null); // Create header cnt = 0; } // Insert "it" at current position public void insert(E it) { curr.setNext(new Link<E>(it, curr.next())); if (tail == curr) tail = curr.next(); // New tail cnt++; } public void append(E it) { // Append "it" to list tail = tail.setNext(new Link<E>(it, null)); cnt++; } public void moveToStart() // Set curr at list start { curr = head; } // Remove and return current element public E remove() { if (curr.next() == null) return null; // Nothing to remove E it = curr.next().element(); // Remember value if (tail == curr.next()) tail = curr; // Removed last curr.setNext(curr.next().next()); // Remove from list cnt--; // Decrement count return it; // Return value } Figure 4.8 A linked list implementation.
- 131. 112 Chap. 4 Lists, Stacks, and Queues public void moveToEnd() // Set curr at list end { curr = tail; } // Move curr one step left; no change if already at front public void prev() { if (curr == head) return; // No previous element Link<E> temp = head; // March down list until we find the previous element while (temp.next() != curr) temp = temp.next(); curr = temp; } // Move curr one step right; no change if already at end public void next() { if (curr != tail) { curr = curr.next(); } } public int length() { return cnt; } // Return the position of the current element public int currPos() { Link<E> temp = head; int i; for (i=0; curr != temp; i++) temp = temp.next(); return i; } // Move down list to "pos" position public void moveToPos(int pos) { assert (pos>=0) && (pos<=cnt) : "Position out of range"; curr = head; for(int i=0; i<pos; i++) curr = curr.next(); } public E getValue() { // Return current element if(curr.next() == null) return null; return curr.next().element(); } Figure 4.8 (continued)
- 132. Sec. 4.1 Lists 113 curr ... ... (a) ... ... 3 (b) 1 2 curr 23 12 Insert 10: 10 23 12 10 Figure 4.9 The linked list insertion process. (a) The linked list before insertion. (b) The linked list after insertion. 1 marks the element field of the new link node. 2 marks the next field of the new link node, which is set to point to what used to be the current node (the node with value 12). 3 marks the next field of the node preceding the current position. It used to point to the node containing 12; now it points to the new node containing 10. E it = curr.next().element(); // Remember value curr.setNext(curr.next().next()); // Remove from list Figure 4.10 illustrates the remove method. Removing an element requires Θ(1) time. Method next simply moves curr one position toward the tail of the list, which takes Θ(1) time. Method prev moves curr one position toward the head of the list, but its implementation is more difficult. In a singly linked list, there is no pointer to the previous node. Thus, the only alternative is to march down the list from the beginning until we reach the current node (being sure always to remember the node before it, because that is what we really want). This takes Θ(n) time in the average and worst cases. Implementation of method moveToPos is similar in that finding the ith position requires marching down i positions from the head of the list, taking Θ(i) time. Implementations for the remaining operations are straightforward, each requir- ing Θ(1) time.
- 133. 114 Chap. 4 Lists, Stacks, and Queues 2 ... ... ... ... (a) 1 (b) it curr 23 12 12 10 10 23 curr Figure 4.10 The linked list removal process. (a) The linked list before removing the node with value 10. (b) The linked list after removal. 1 marks the list node being removed. it is set to point to the element. 2 marks the next field of the preceding list node, which is set to point to the node following the one being deleted. Freelists The new operator is relatively expensive to use. Section 12.3 discusses how a general-purpose memory manager can be implemented. The problem is that free- store routines must be capable of handling requests to and from free store with no particular pattern, as well as requests of vastly different sizes. Garbage collection is also expensive. Most compilers today provide reasonable implementations for their free-store operators. However, the requirement that free-store managers be able to handle any pattern of new operations, combined with unpredictable freeing of space by the garbage collector, makes them inefficient compared to what might be implemented for more controlled patterns of memory access. The conditions under which list nodes are created and deleted in a linked list implementation allow the Link class programmer to provide simple but efficient memory management routines in place of the system-level free-store operators. Instead of making repeated calls to new, the Link class can handle its own freelist. A freelist holds those list nodes that are not currently being used. When a node is deleted from a linked list, it is placed at the head of the freelist. When a new element is to be added to a linked list, the freelist is checked to see if a list node is available. If so, the node is taken from the freelist. If the freelist is empty, the standard new operator must then be called.
- 134. Sec. 4.1 Lists 115 Freelists are particularly useful for linked lists that periodically grow and then shrink. The freelist will never grow larger than the largest size yet reached by the linked list. Requests for new nodes (after the list has shrunk) can be handled by the freelist. One approach to implementing freelists would be to create two new methods to handle requesting and freeing nodes. This requires that the user’s code, such as the linked list class implementation of Figure 4.8, be modified to call these freelist op- erators. In the implementation shown here, the link class is augmented with meth- ods get and release. Figure 4.11 shows the reimplementation for the Link class to support these methods. Note how simple they are, because they need only remove and add an element to the front of the freelist, respectively. The freelist methods get and release both run in Θ(1) time, except in the case where the freelist is exhausted and the new operator must be called. In Figure 4.11, you should note the use of the static definition for the freelist header. The purpose of the keyword static is to create a single variable shared among all instances of the Link nodes. We want only a single freelist for all Link nodes of a given type. A program might create multiple lists. If they are all of the same type (that is, their element types are the same), then they can and should share the same freelist. This will happen with the implementation of Figure 4.11. If lists are created that have different element types, because this code is implemented with templates, the need for different list implementations will be discovered by the compiler at compile time. Separate versions of the list class will be generated for each element type. Thus, each element type will also get its own separate copy of the Link class. And each distinct Link class implementation will get a separate freelist. Let’s consider a more complex situation where we want separate freelists, but we don’t know this until runtime. For example, perhaps we are using lists to store variable-length strings. A given node gets space allocated for a string of a specific length. We’d like to be able to reuse the nodes by placing them on a freelist, instead of relying on the system free store operators. But we cannot reuse a given list node unless the size allocated for its string matches the length of the string we want to store in this node. Thus, each specific node size must be stored on its own freelist. That is, the nodes for strings of length one are stored on one freelist, the nodes for strings of length two on another freelist, and so on. In this way, its easy to find a node of the proper size. Unfortunately, we don’t know in advance what the various sizes of strings will be. If we want to support long strings, then there could be thousands of different node sizes, many of which are not used in any particular run of the program. Thus,
- 135. 116 Chap. 4 Lists, Stacks, and Queues // Singly linked list node with freelist support class Link<E> { private E element; // Value for this node private Link<E> next; // Point to next node in list // Constructors Link(E it, Link<E> nextval) { element = it; next = nextval; } Link(Link<E> nextval) { next = nextval; } Link<E> next() { return next; } Link<E> setNext(Link<E> nextval) { return next = nextval; } E element() { return element; } E setElement(E it) { return element = it; } // Extensions to support freelists static Link freelist = null; // Freelist for the class // Get new link static <E> Link<E> get(E it, Link<E> nextval) { if (freelist == null) return new Link<E>(it, nextval); // Get from "new" Link<E> temp = freelist; // Get from freelist freelist = freelist.next(); temp.setElement(it); temp.setNext(nextval); return temp; } void release() { // Return Link to freelist element = null; // Drop reference to the element next = freelist; freelist = this; } } // class Link Figure 4.11 Implementation for the Link class with a freelist. The static declaration for member freelist means that all Link class objects share the same freelist pointer variable instead of each object storing its own copy.
- 136. Sec. 4.1 Lists 117 we do not want to allocate freelists in advance for each potential node length. On the other hand, we need to make sure no more than one copy of the freelist for a given size is created. We can modify our freelist free-store methods to request the appropriate freel- ist. This access function will search for the proper freelist. If it exists, that freelist is used. If not, that freelist is created. 4.1.3 Comparison of List Implementations Now that you have seen two substantially different implementations for lists, it is natural to ask which is better. In particular, if you must implement a list for some task, which implementation should you choose? Array-based lists have the disadvantage that their size must be predetermined before the array can be allocated. Array-based lists cannot grow beyond their pre- determined size. Whenever the list contains only a few elements, a substantial amount of space might be tied up in a largely empty array. Linked lists have the advantage that they only need space for the objects actually on the list. There is no limit to the number of elements on a linked list, as long as there is free-store memory available. The amount of space required by a linked list is Θ(n), while the space required by the array-based list implementation is Ω(n), but can be greater. Array-based lists have the advantage that there is no wasted space for an indi- vidual element. Linked lists require that a pointer be added to every list node. If the element size is small, then the overhead for links can be a significant fraction of the total storage. When the array for the array-based list is completely filled, there is no storage overhead. The array-based list will then be more space efficient, by a constant factor, than the linked implementation. A simple formula can be used to determine whether the array-based list or linked list implementation will be more space efficient in a particular situation. Call n the number of elements currently in the list, P the size of a pointer in stor- age units (typically four bytes), E the size of a data element in storage units (this could be anything, from one bit for a Boolean variable on up to thousands of bytes or more for complex records), and D the maximum number of list elements that can be stored in the array. The amount of space required for the array-based list is DE, regardless of the number of elements actually stored in the list at any given time. The amount of space required for the linked list is n(P + E). The smaller of these expressions for a given value n determines the more space-efficient imple- mentation for n elements. In general, the linked implementation requires less space than the array-based implementation when relatively few elements are in the list. Conversely, the array-based implementation becomes more space efficient when
- 137. 118 Chap. 4 Lists, Stacks, and Queues the array is close to full. Using the equation, we can solve for n to determine the break-even point beyond which the array-based implementation is more space efficient in any particular situation. This occurs when n > DE/(P + E). If P = E, then the break-even point is at D/2. This would happen if the element field is either a four-byte int value or a pointer, and the next field is a typical four- byte pointer. That is, the array-based implementation would be more efficient (if the link field and the element field are the same size) whenever the array is more than half full. As a rule of thumb, linked lists are better when implementing lists whose num- ber of elements varies widely or is unknown. Array-based lists are generally more space efficient when the user knows in advance approximately how large the list will become. Array-based lists are faster for random access by position. Positions can easily be adjusted forwards or backwards by the next and prev methods. These opera- tions always take Θ(1) time. In contrast, singly linked lists have no explicit access to the previous element, and access by position requires that we march down the list from the front (or the current position) to the specified position. Both of these operations require Θ(n) time in the average and worst cases, if we assume that each position on the list is equally likely to be accessed on any call to prev or movetoPos. Given a pointer to a suitable location in the list, the insert and remove methods for linked lists require only Θ(1) time. Array-based lists must shift the re- mainder of the list up or down within the array. This requires Θ(n) time in the aver- age and worst cases. For many applications, the time to insert and delete elements dominates all other operations. For this reason, linked lists are often preferred to array-based lists. When implementing the array-based list, an implementor could allow the size of the array to grow and shrink depending on the number of elements that are ac- tually stored. This data structure is known as a dynamic array. For example, the Java Vector class implements a dynamic array. Dynamic arrays allow the pro- grammer to get around the limitation on the standard array that its size cannot be changed once the array has been created. This also means that space need not be allocated to the dynamic array until it is to be used. The disadvantage of this ap- proach is that it takes time to deal with space adjustments on the array. Each time the array grows in size, its contents must be copied. A good implementation of the dynamic array will grow and shrink the array in such a way as to keep the overall
- 138. Sec. 4.1 Lists 119 cost for a series of insert/delete operations relatively inexpensive, even though an occasional insert/delete operation might be expensive. To analyze the cost of dy- namic array operations, we need to use a technique known as amortized analysis, which is discussed in Section 14.3. 4.1.4 Element Implementations List users must decide whether they wish to store a copy of any given element on each list that contains it. For small elements such as an integer, this makes sense. If the elements are payroll records, it might be desirable for the list node to store a pointer to the record rather than store a copy of the record itself. This change would allow multiple list nodes (or other data structures) to point to the same record, rather than make repeated copies of the record. Not only might this save space, but it also means that a modification to an element’s value is automatically reflected at all locations where it is referenced. The disadvantage of storing a pointer to each ele- ment is that the pointer requires space of its own. If elements are never duplicated, then this additional space adds unnecessary overhead. Java most naturally stores references to objects, meaning that only a single copy of an object such as a payroll record will be maintained, even if it is on multiple lists. Whether it is more advantageous to use references to shared elements or sepa- rate copies depends on the intended application. In general, the larger the elements and the more they are duplicated, the more likely that references to shared elements is the better approach. A second issue faced by implementors of a list class (or any other data structure that stores a collection of user-defined data elements) is whether the elements stored are all required to be of the same type. This is known as homogeneity in a data structure. In some applications, the user would like to define the class of the data element that is stored on a given list, and then never permit objects of a different class to be stored on that same list. In other applications, the user would like to permit the objects stored on a single list to be of differing types. For the list implementations presented in this section, the compiler requires that all objects stored on the list be of the same type. Besides Java generics, there are other techniques that implementors of a list class can use to ensure that the element type for a given list remains fixed, while still permitting different lists to store different element types. One approach is to store an object of the appropriate type in the header node of the list (perhaps an object of the appropriate type is supplied as a parameter to the list constructor), and then check that all insert operations on that list use the same element type.
- 139. 120 Chap. 4 Lists, Stacks, and Queues head 20 23 curr 12 15 tail Figure 4.12 A doubly linked list. The third issue that users of the list implementations must face is primarily of concern when programming in languages that do not support automatic garbage collection. That is how to deal with the memory of the objects stored on the list when the list is deleted or the clear method is called. The list destructor and the clear method are problematic in that there is a potential that they will be misused, thus causing a memory leak. Deleting listArray in the array-based implementation, or deleting a link node in the linked list implementation, might remove the only reference to an object, leaving its memory space inaccessible. Unfortunately, there is no way for the list implementation to know whether a given object is pointed to in another part of the program or not. Thus, the user of the list must be responsible for deleting these objects when that is appropriate. 4.1.5 Doubly Linked Lists The singly linked list presented in Section 4.1.2 allows for direct access from a list node only to the next node in the list. A doubly linked list allows convenient access from a list node to the next node and also to the preceding node on the list. The doubly linked list node accomplishes this in the obvious way by storing two pointers: one to the node following it (as in the singly linked list), and a second pointer to the node preceding it. The most common reason to use a doubly linked list is because it is easier to implement than a singly linked list. While the code for the doubly linked implementation is a little longer than for the singly linked version, it tends to be a bit more “obvious” in its intention, and so easier to implement and debug. Figure 4.12 illustrates the doubly linked list concept. Like our singly linked list implementation, the doubly linked list implementa- tion makes use of a header node. We also add a tailer node to the end of the list. The tailer is similar to the header, in that it is a node that contains no value, and it always exists. When the doubly linked list is initialized, the header and tailer nodes are created. Data member head points to the header node, and tail points to the tailer node. The purpose of these nodes is to simplify the insert, append, and remove methods by eliminating all need for special-case code when the list is empty. Whether a list implementation is doubly or singly linked should be hidden from the List class user. Figure 4.13 shows the complete implementation for a Link
- 140. Sec. 4.1 Lists 121 class DLink<E> { // Doubly linked list node private E element; // Value for this node private DLink<E> next; // Pointer to next node in list private DLink<E> prev; // Pointer to previous node // Constructors DLink(E it, DLink<E> n, DLink<E> p) { element = it; next = n; prev = p; } DLink(DLink<E> n, DLink<E> p) { next = n; prev = p; } DLink<E> next() { return next; } DLink<E> setNext(DLink<E> nextval) { return next = nextval; } DLink<E> prev() { return prev; } DLink<E> setPrev(DLink<E> prevval) { return prev = prevval; } E element() { return element; } E setElement(E it) { return element = it; } } // class DLink Figure 4.13 Doubly linked list node implementation with a freelist. class to be used with doubly linked lists. This code is a little longer than that for the singly linked list node implementation. Not only do the doubly linked list nodes have an extra data member, but the constructors are a little more intelligent. When a new node is being added to a doubly linked list, the neighboring nodes in the list must also point back to the newly added node. Thus, the constructors will check to see if the next or prev fields are non-null. When they are, the node being pointed to will have its appropriate link field modified to point back to the new node being added to the list. This simplifies the doubly linked list insert method. Figure 4.14 shows the implementation for the insert, append, remove, and prev doubly linked list methods. The class declaration and the remaining member functions for the doubly linked list class are nearly identical to those of Figures 4.8. The insert method is especially simple for our doubly linked list implemen- tation, because most of the work is done by the node’s constructor. Figure 4.15 shows the list before and after insertion of a node with value 10. The following line of code from Figure 4.14 does the actual work. curr.setNext(new DLink<E>(it, curr.next(), curr)); The three parameters to the new operator allow the list node class constructor to set the element, prev, and next fields, respectively, for the new link node. The new operator returns a pointer to the newly created node. The node constructor also updates the next field of the node that the new node’s prev field points to. It
- 141. 122 Chap. 4 Lists, Stacks, and Queues // Insert "it" at current position public void insert(E it) { curr.setNext(new DLink<E>(it, curr.next(), curr)); if (curr.next().next() != null) curr.next().next().setPrev(curr.next()); if (tail == curr) tail = curr.next(); // New tail cnt++; } public void append(E it) { // Append "it" to list tail.setNext(new DLink<E>(it, null, tail)); tail = tail.next(); cnt++; } // Remove and return first element in right partition public E remove() { if (curr.next() == null) return null; // Nothing to remove E it = curr.next().element(); // Remember value if (curr.next().next() != null) curr.next().next().setPrev(curr); else tail = curr; // Removed last Object: set tail curr.setNext(curr.next().next()); // Remove from list cnt--; // Decrement the count return it; // Return value removed } // Move curr one step left; no change if at front public void prev() { if (curr != head) curr = curr.prev(); } Figure 4.14 Implementations for doubly linked list class insert, append, remove, and prev methods. then updates the prev field of the node that the new node’s next field points to. The existance of the header and tailer nodes mean that there are no special cases to be worried about when inserting into an empty list. The List class append method from Figure 4.14 is also simple. Here is the crucial line of code. tail.setNext(new DLink<E>(it, null, tail)); Again, the Link class constructor sets the element, prev, and next fields of the node when the new operator is executed, and it also sets the appropriate pointers in the previous last node and the list tailer node.
- 142. Sec. 4.1 Lists 123 ... 12 23 5 ... 20 (a) ... 20 4 curr Insert 10: 10 ... 23 12 1 10 3 2 (b) curr Figure 4.15 Insertion for doubly linked lists. The labels 1 , 2 , and 3 cor- respond to assignments done by the linked list node constructor. 4 marks the assignment to curr->next. 5 marks the assignment to the prev pointer of the node following the newly inserted node. ... 20 curr ... 23 12 ... ... 20 12 curr (b) (a) 23 it Figure 4.16 Doubly linked list removal. Pointer ltemp is set to point to the current node. Then the nodes to either side of the node being removed have their pointers adjusted.
- 143. 124 Chap. 4 Lists, Stacks, and Queues Method remove (illustrated by Figure 4.16) is straightforward, though the code is somewhat longer. First, the variable it is assigned the value being re- moved, Note that we must separate the element, which is returned to the caller, from the link object. The following lines then adjust the pointers. E it = curr.next().element(); // Remember value if (curr.next().next() != null) curr.next().next().setPrev(curr); else tail = curr; // Removed last Object: set tail curr.setNext(curr.next().next()); // Remove from list The first line stores the value of the node being removed. The second line makes the next node’s prev pointer point to the left of the node being removed. If necessary, tail is updated. Finally, the next field of the node preceding the one being deleted is adjusted. The final steps of method remove are to update the list length and to return the value of the deleted element. The only disadvantage of the doubly linked list as compared to the singly linked list is the additional space used. The doubly linked list requires two pointers per node, and so in the implementation presented it requires twice as much overhead as the singly linked list. Example 4.1 There is a space-saving technique that can be employed to eliminate the additional space requirement, though it will complicate the implementation and be somewhat slower. Thus, the technique is an example of a space/time tradeoff. It is based on observing that, if we store the sum of two values, then we can get either value back by subtracting the other. That is, if we store a + b in variable c, then b = c − a and a = c − b. Of course, to recover one of the values out of the stored summation, the other value must be supplied. A pointer to the first node in the list, along with the value of one of its two link fields, will allow access to all of the remaining nodes of the list in order. This is because the pointer to the node must be the same as the value of the following node’s prev pointer, as well as the previous node’s next pointer. It is possible to move down the list breaking apart the summed link fields as though you were opening a zipper. Details for implementing this variation are left as an exercise. The principle behind this technique is worth remembering, because it has many applications. The following code fragment will swap the contents of two variables without using a temporary variable (at the cost of three arithmetic operations). a = a + b; b = a - b; // Now b contains original value of a a = a - b; // Now a contains original value of b
- 144. Sec. 4.2 Stacks 125 A similar effect can be had by using the exclusive-or operator. This fact is widely used in computer graphics. A region of the computer screen can be highlighted by XORing the outline of a box around it. XORing the box outline a second time restores the original contents of the screen. 4.2 Stacks The stack is a list-like structure in which elements may be inserted or removed from only one end. While this restriction makes stacks less flexible than lists, it also makes stacks both efficient (for those operations they can do) and easy to im- plement. Many applications require only the limited form of insert and remove operations that stacks provide. In such cases, it is more efficient to use the sim- pler stack data structure rather than the generic list. For example, the freelist of Section 4.1.2 is really a stack. Despite their restrictions, stacks have many uses. Thus, a special vocabulary for stacks has developed. Accountants used stacks long before the invention of the computer. They called the stack a “LIFO” list, which stands for “Last-In, First- Out.” Note that one implication of the LIFO policy is that stacks remove elements in reverse order of their arrival. It is traditional to call the accessible element of the stack the top element. El- ements are not said to be inserted; instead they are pushed onto the stack. When removed, an element is said to be popped from the stack. Figure 4.17 shows a sample stack ADT. As with lists, there are many variations on stack implementation. The two ap- proaches presented here are array-based and linked stacks, which are analogous to array-based and linked lists, respectively. 4.2.1 Array-Based Stacks Figure 4.18 shows a complete implementation for the array-based stack class. As with the array-based list implementation, listArray must be declared of fixed size when the stack is created. In the stack constructor, size serves to indicate this size. Method top acts somewhat like a current position value (because the “current” position is always at the top of the stack), as well as indicating the number of elements currently in the stack. The array-based stack implementation is essentially a simplified version of the array-based list. The only important design decision to be made is which end of the array should represent the top of the stack. One choice is to make the top be
- 145. 126 Chap. 4 Lists, Stacks, and Queues /** Stack ADT */ public interface Stack<E> { /** Reinitialize the stack. The user is responsible for reclaiming the storage used by the stack elements. */ public void clear(); /** Push an element onto the top of the stack. @param it The element being pushed onto the stack. */ public void push(E it); /** Remove and return the element at the top of the stack. @return The element at the top of the stack. */ public E pop(); /** @return A copy of the top element. */ public E topValue(); /** @return The number of elements in the stack. */ public int length(); }; Figure 4.17 The Java interface for a stack. at position 0 in the array. In terms of list functions, all insert and remove operations would then be on the element in position 0. This implementation is inefficient, because now every push or pop operation will require that all elements currently in the stack be shifted one position in the array, for a cost of Θ(n) if there are n elements. The other choice is have the top element be at position n − 1 when there are n elements in the stack. In other words, as elements are pushed onto the stack, they are appended to the tail of the list. Method pop removes the tail element. In this case, the cost for each push or pop operation is only Θ(1). For the implementation of Figure 4.18, top is defined to be the array index of the first free position in the stack. Thus, an empty stack has top set to 0, the first available free position in the array. (Alternatively, top could have been defined to be the index for the top element in the stack, rather than the first free position. If this had been done, the empty list would initialize top as −1.) Methods push and pop simply place an element into, or remove an element from, the array position indicated by top. Because top is assumed to be at the first free position, push first inserts its value into the top position and then increments top, while pop first decrements top and then removes the top element.
- 146. Sec. 4.2 Stacks 127 /** Array-based stack implementation */ class AStack<E> implements Stack<E> { private static final int defaultSize = 10; private int maxSize; // Maximum size of stack private int top; // Index for top Object private E [] listArray; // Array holding stack // Constructors AStack() { this(defaultSize); } @SuppressWarnings("unchecked") // Generic array allocation AStack(int size) { maxSize = size; top = 0; listArray = (E[])new Object[size]; // Create listArray } public void clear() // Reinitialize stack { top = 0; } public void push(E it) { // Push "it" onto stack assert top != maxSize : "Stack is full"; listArray[top++] = it; } public E pop() { // Pop top element assert top != 0 : "Stack is empty"; return listArray[--top]; } public E topValue() { // Return top element assert top != 0 : "Stack is empty"; return listArray[top-1]; } public int length() { return top; } // Return length Figure 4.18 Array-based stack class implementation.
- 147. 128 Chap. 4 Lists, Stacks, and Queues // Linked stack implementation class LStack<E> implements Stack<E> { private Link<E> top; // Pointer to first element private int size; // Number of elements //Constructors public LStack() { top = null; size = 0; } public LStack(int size) { top = null; size = 0; } // Reinitialize stack public void clear() { top = null; size = 0; } public void push(E it) { // Put "it" on stack top = new Link<E>(it, top); size++; } public E pop() { // Remove "it" from stack assert top != null : "Stack is empty"; E it = top.element(); top = top.next(); size--; return it; } public E topValue() { // Return top value assert top != null : "Stack is empty"; return top.element(); } public int length() { return size; } // Return length Figure 4.19 Linked stack class implementation. 4.2.2 Linked Stacks The linked stack implementation is a simplified version of the linked list imple- mentation. The freelist of Section 4.1.2 is an example of a linked stack. Elements are inserted and removed only from the head of the list. The header node is not used because no special-case code is required for lists of zero or one elements. Fig- ure 4.19 shows the complete class implementation for the linked stack. The only data member is top, a pointer to the first (top) link node of the stack. Method push first modifies the next field of the newly created link node to point to the top of the stack and then sets top to point to the new link node. Method pop is also quite simple. The variable temp stores the value of the top node, while ltemp keeps a link to the top node as it is removed from the stack. The stack is updated by setting top to point to the next element in the stack. The old top node
- 148. Sec. 4.2 Stacks 129 top1 top2 Figure 4.20 Two stacks implemented within in a single array, both growing toward the middle. is then returned to free store (or the freelist), and the element value is returned as the value of the pop method. 4.2.3 Comparison of Array-Based and Linked Stacks All operations for the array-based and linked stack implementations take constant time, so from a time efficiency perspective, neither has a significant advantage. Another basis for comparison is the total space required. The analysis is similar to that done for list implementations. The array-based stack must declare a fixed-size array initially, and some of that space is wasted whenever the stack is not full. The linked stack can shrink and grow but requires the overhead of a link field for every element. When multiple stacks are to be implemented, it is possible to take advantage of the one-way growth of the array-based stack. This can be done by using a single array to store two stacks. One stack grows inward from each end as illustrated by Figure 4.20, hopefully leading to less wasted space. However, this only works well when the space requirements of the two stacks are inversely correlated. In other words, ideally when one stack grows, the other will shrink. This is particularly effective when elements are taken from one stack and given to the other. If instead both stacks grow at the same time, then the free space in the middle of the array will be exhausted quickly. 4.2.4 Implementing Recursion Perhaps the most common computer application that uses stacks is not even visible to its users. This is the implementation of subroutine calls in most programming language runtime environments. A subroutine call is normally implemented by placing necessary information about the subroutine (including the return address, parameters, and local variables) onto a stack. This information is called an ac- tivation record. Further subroutine calls add to the stack. Each return from a subroutine pops the top activation record off the stack. Figure 4.21 illustrates the implementation of the recursive factorial function of Section 2.5 from the runtime environment’s point of view.
- 149. 130 Chap. 4 Lists, Stacks, and Queues β2 β2 β2 β β1 β1 β1 β3 β β1 β 4 4 3 2 3 4 Call fact(1) Call fact(2) Call fact(3) Call fact(4) Return 1 4 3 Return 2 4 Return 6 Return 24 β β β1 β β Currptr Currptr Currptr Currptr Currptr Currptr Currptr Currptr Currptr Currptr Currptr Currptr Currptr Currptr Currptr Currptr n n n n n n n n n Figure 4.21 Implementing recursion with a stack. β values indicate the address of the program instruction to return to after completing the current function call. On each recursive function call to fact (from Section 2.5), both the return ad- dress and the current value of n must be saved. Each return from fact pops the top activation record off the stack. Consider what happens when we call fact with the value 4. We use β to indicate the address of the program instruction where the call to fact is made. Thus, the stack must first store the address β, and the value 4 is passed to fact. Next, a recursive call to fact is made, this time with value 3. We will name the program address from which the call is made β1. The address β1, along with the current value for n (which is 4), is saved on the stack. Function fact is invoked with input parameter 3. In similar manner, another recursive call is made with input parameter 2, re- quiring that the address from which the call is made (say β2) and the current value for n (which is 3) are stored on the stack. A final recursive call with input parame-
- 150. Sec. 4.2 Stacks 131 ter 1 is made, requiring that the stack store the calling address (say β3) and current value (which is 2). At this point, we have reached the base case for fact, and so the recursion begins to unwind. Each return from fact involves popping the stored value for n from the stack, along with the return address from the function call. The return value for fact is multiplied by the restored value for n, and the result is returned. Because an activation record must be created and placed onto the stack for each subroutine call, making subroutine calls is a relatively expensive operation. While recursion is often used to make implementation easy and clear, sometimes you might want to eliminate the overhead imposed by the recursive function calls. In some cases, such as the factorial function of Section 2.5, recursion can easily be replaced by iteration. Example 4.2 As a simple example of replacing recursion with a stack, consider the following non-recursive version of the factorial function. static long fact(int n) { // Compute n! // To fit n! in a long variable, require n < 21 assert (n >= 0) && (n <= 20) : "n out of range"; // Make a stack just big enough Stack<Integer> S = new AStack<Integer>(n); while (n > 1) S.push(n--); long result = 1; while (S.length() > 0) result = result * S.pop(); return result; } Here, we simply push successively smaller values of n onto the stack un- til the base case is reached, then repeatedly pop off the stored values and multiply them into the result. In practice, an iterative form of the factorial function would be both simpler and faster than the version shown in Example 4.2. Unfortunately, it is not always possible to replace recursion with iteration. Recursion, or some imitation of it, is necessary when implementing algorithms that require multiple branching such as in the Towers of Hanoi algorithm, or when traversing a binary tree. The Mergesort and Quicksort algorithms of Chapter 7 are also examples in which recursion is required. Fortunately, it is always possible to imitate recursion with a stack. Let us now turn to a non-recursive version of the Towers of Hanoi function, which cannot be done iteratively.
- 151. 132 Chap. 4 Lists, Stacks, and Queues Example 4.3 The TOH function shown in Figure 2.2 makes two recursive calls: one to move n − 1 rings off the bottom ring, and another to move these n − 1 rings back to the goal pole. We can eliminate the recursion by using a stack to store a representation of the three operations that TOH must perform: two recursive calls and a move operation. To do so, we must first come up with a representation of the various operations, implemented as a class whose objects will be stored on the stack. Here is such a class. class TOHobj { public operation op; public int num; public Pole start, goal, temp; TOHobj(operation o, int n, Pole s, Pole g, Pole t) { op = o; num = n; start = s; goal = g; temp = t; } TOHobj(operation o, Pole s, Pole g) // MOVE { op = o; start = s; goal = g; } } Class TOHobj stores five values: an operation field (indicating either a move or a new TOH operation), the number of rings, and the three poles. Note that the move operation actually needs only to store information about two poles. Thus, there are two constructors: one to store the state when imitating a recursive call, and one to store the state for a move operation. The non-recursive version of TOH can now be presented.
- 152. Sec. 4.3 Queues 133 static void TOH(int n, Pole start, Pole goal, Pole temp) { // Make a stack just big enough Stack<TOHobj> S = new AStack<TOHobj>(2*n+1); S.push(new TOHobj(operation.TOH, n, start, goal, temp)); while (S.length() > 0) { TOHobj it = S.pop(); // Get next task if (it.op == operation.MOVE) // Do a move move(it.start, it.goal); else if (it.num > 0) { // Imitate TOH recursive // solution (in reverse) S.push(new TOHobj(operation.TOH, it.num-1, it.temp, it.goal, it.start)); S.push(new TOHobj(operation.MOVE, it.start, it.goal)); // A move to do S.push(new TOHobj(operation.TOH, it.num-1, it.start, it.temp, it.goal)); } } } We first define an enumerated type called TOHop, with two values MOVE and TOH, to indicate calls to the move function and recursive calls to TOH, respectively. Note that an array-based stack is used, because we know that the stack will need to store exactly 2n + 1 elements. The new version of TOH begins by placing on the stack a description of the initial problem of n rings. The rest of the function is simply a while loop that pops the stack and executes the appropriate operation. In the case of a TOH operation (for n > 0), we store on the stack representations for the three operations executed by the recursive version. However, these operations must be placed on the stack in reverse order, so that they will be popped off in the correct order. Some “naturally recursive” applications lend themselves to efficient implemen- tation with a stack, because the amount of information needed to describe a sub- problem is small. For example, Section 7.5 discusses a stack-based implementation for Quicksort. 4.3 Queues Like the stack, the queue is a list-like structure that provides restricted access to its elements. Queue elements may only be inserted at the back (called an enqueue
- 153. 134 Chap. 4 Lists, Stacks, and Queues /** Queue ADT */ public interface Queue<E> { /** Reinitialize the queue. The user is responsible for reclaiming the storage used by the queue elements. */ public void clear(); /** Place an element at the rear of the queue. @param The element being enqueued. */ public void enqueue(E it); /** Remove and return element at the front of the queue. @return The element at the front of the queue. */ public E dequeue(); /** @return The front element. */ public E frontValue(); /** @return The number of elements in the queue. */ public int length(); } Figure 4.22 The Java interface for a queue. operation) and removed from the front (called a dequeue operation). Queues oper- ate like standing in line at a movie theater ticket counter.1 If nobody cheats, then newcomers go to the back of the line. The person at the front of the line is the next to be served. Thus, queues release their elements in order of arrival. Accountants have used queues since long before the existence of computers. They call a queue a “FIFO” list, which stands for “First-In, First-Out.” Figure 4.22 shows a sample queue ADT. This section presents two implementations for queues: the array-based queue and the linked queue. 4.3.1 Array-Based Queues The array-based queue is somewhat tricky to implement effectively. A simple con- version of the array-based list implementation is not efficient. Assume that there are n elements in the queue. By analogy to the array-based list implementation, we could require that all elements of the queue be stored in the first n positions of the array. If we choose the rear element of the queue to be in position 0, then dequeue operations require only Θ(1) time because the front ele- ment of the queue (the one being removed) is the last element in the array. However, 1 In Britain, a line of people is called a “queue,” and getting into line to wait for service is called “queuing up.”
- 154. Sec. 4.3 Queues 135 front rear 20 5 12 17 (a) rear (b) 12 17 3 30 4 front Figure 4.23 After repeated use, elements in the array-based queue will drift to the back of the array. (a) The queue after the initial four numbers 20, 5, 12, and 17 have been inserted. (b) The queue after elements 20 and 5 are deleted, following which 3, 30, and 4 are inserted. enqueue operations will require Θ(n) time, because the n elements currently in the queue must each be shifted one position in the array. If instead we chose the rear element of the queue to be in position n − 1, then an enqueue operation is equivalent to an append operation on a list. This requires only Θ(1) time. But now, a dequeue operation requires Θ(n) time, because all of the elements must be shifted down by one position to retain the property that the remaining n − 1 queue elements reside in the first n − 1 positions of the array. A far more efficient implementation can be obtained by relaxing the require- ment that all elements of the queue must be in the first n positions of the array. We will still require that the queue be stored be in contiguous array positions, but the contents of the queue will be permitted to drift within the array, as illustrated by Figure 4.23. Now, both the enqueue and the dequeue operations can be performed in Θ(1) time because no other elements in the queue need be moved. This implementation raises a new problem. Assume that the front element of the queue is initially at position 0, and that elements are added to successively higher-numbered positions in the array. When elements are removed from the queue, the front index increases. Over time, the entire queue will drift toward the higher-numbered positions in the array. Once an element is inserted into the highest-numbered position in the array, the queue has run out of space. This hap- pens despite the fact that there might be free positions at the low end of the array where elements have previously been removed from the queue. The “drifting queue” problem can be solved by pretending that the array is circular and so allow the queue to continue directly from the highest-numbered position in the array to the lowest-numbered position. This is easily implemented
- 155. 136 Chap. 4 Lists, Stacks, and Queues rear front rear (a) (b) 20 5 12 17 12 17 3 30 4 front Figure 4.24 The circular queue with array positions increasing in the clockwise direction. (a) The queue after the initial four numbers 20, 5, 12, and 17 have been inserted. (b) The queue after elements 20 and 5 are deleted, following which 3, 30, and 4 are inserted. through use of the modulus operator (denoted by % in Java). In this way, positions in the array are numbered from 0 through size−1, and position size−1 is de- fined to immediately precede position 0 (which is equivalent to position size % size). Figure 4.24 illustrates this solution. There remains one more serious, though subtle, problem to the array-based queue implementation. How can we recognize when the queue is empty or full? Assume that front stores the array index for the front element in the queue, and rear stores the array index for the rear element. If both front and rear have the same position, then with this scheme there must be one element in the queue. Thus, an empty queue would be recognized by having rear be one less than front (tak- ing into account the fact that the queue is circular, so position size−1 is actually considered to be one less than position 0). But what if the queue is completely full? In other words, what is the situation when a queue with n array positions available contains n elements? In this case, if the front element is in position 0, then the rear element is in position size−1. But this means that the value for rear is one less than the value for front when the circular nature of the queue is taken into account. In other words, the full queue is indistinguishable from the empty queue! You might think that the problem is in the assumption about front and rear being defined to store the array indices of the front and rear elements, respectively, and that some modification in this definition will allow a solution. Unfortunately, the problem cannot be remedied by a simple change to the definition for front and rear, because of the number of conditions or states that the queue can be in. Ignoring the actual position of the first element, and ignoring the actual values of the elements stored in the queue, how many different states are there? There can
- 156. Sec. 4.3 Queues 137 be no elements in the queue, one element, two, and so on. At most there can be n elements in the queue if there are n array positions. This means that there are n + 1 different states for the queue (0 through n elements are possible). If the value of front is fixed, then n+1 different values for rear are needed to distinguish among the n+1 states. However, there are only n possible values for rear unless we invent a special case for, say, empty queues. This is an example of the Pigeonhole Principle defined in Exercise 2.29. The Pigeonhole Principle states that, given n pigeonholes and n + 1 pigeons, when all of the pigeons go into the holes we can be sure that at least one hole contains more than one pigeon. In similar manner, we can be sure that two of the n + 1 states are indistinguishable by their relative values of front and rear. We must seek some other way to distinguish full from empty queues. One obvious solution is to keep an explicit count of the number of elements in the queue, or at least a Boolean variable that indicates whether the queue is empty or not. Another solution is to make the array be of size n + 1, and only allow n elements to be stored. Which of these solutions to adopt is purely a matter of the implementor’s taste in such affairs. My choice is to use an array of size n + 1. Figure 4.25 shows an array-based queue implementation. listArray holds the queue elements, and as usual, the queue constructor allows an optional param- eter to set the maximum size of the queue. The array as created is actually large enough to hold one element more than the queue will allow, so that empty queues can be distinguished from full queues. Method size is used to control the circular motion of the queue (it is the base for the modulus operator). Method rear is set to the position of the rear element. In this implementation, the front of the queue is defined to be toward the lower numbered positions in the array (in the counter-clockwise direction in Fig- ure 4.24), and the rear is defined to be toward the higher-numbered positions. Thus, enqueue increments the rear pointer (modulus size), and dequeue increments the front pointer. Implementation of all member functions is straightforward. 4.3.2 Linked Queues The linked queue implementation is a straightforward adaptation of the linked list. Figure 4.26 shows the linked queue class declaration. Methods front and rear are pointers to the front and rear queue elements, respectively. We will use a header link node, which allows for a simpler implementation of the enqueue operation by avoiding any special cases when the queue is empty. On initialization, the front and rear pointers will point to the header node, and front will always point to the header node while rear points to the true last link node in the queue. Method
- 157. 138 Chap. 4 Lists, Stacks, and Queues // Array-based queue implementation class AQueue<E> implements Queue<E> { private static final int defaultSize = 10; private int maxSize; // Maximum size of queue private int front; // Index of front element private int rear; // Index of rear element private E[] listArray; // Array holding queue elements // Constructors AQueue() { this(defaultSize); } @SuppressWarnings("unchecked") // For generic array AQueue(int size) { maxSize = size+1; rear = 0; front = 1; listArray = (E[])new Object[maxSize]; // Create listArray } public void clear() // Reinitialize { rear = 0; front = 1; } public void enqueue(E it) { // Put "it" in queue assert ((rear+2) % maxSize) != front : "Queue is full"; rear = (rear+1) % maxSize; // Circular increment listArray[rear] = it; } public E dequeue() { // Take element out of queue assert length() != 0 : "Queue is empty"; E it = listArray[front]; front = (front+1) % maxSize; // Circular increment return it; } public E frontValue() { // Get front value assert length() != 0 : "Queue is empty"; return listArray[front]; } public int length() // Return length { return ((rear+maxSize) - front + 1) % maxSize; } Figure 4.25 An array-based queue implementation.
- 158. Sec. 4.3 Queues 139 // Linked queue implementation class LQueue<E> implements Queue<E> { private Link<E> front; // Pointer to front queue node private Link<E> rear; // Pointer to rear queuenode int size; // Number of elements in queue // Constructors public LQueue() { init(); } public LQueue(int size) { init(); } // Ignore size private void init() { // Initialize queue front = rear = new Link<E>(null); size = 0; } public void clear() { init(); } // Reinitialize queue public void enqueue(E it) { // Put element on rear rear.setNext(new Link<E>(it, null)); rear = rear.next(); size++; } public E dequeue() { // remove element from front assert size != 0 : "Queue is empty"; E it = front.next().element(); // Store dequeued value front.setNext(front.next().next()); // Advance front if (front.next() == null) rear = front; // Last Object size--; return it; // Return Object } public E frontValue() { // Get front element assert size != 0 : "Queue is empty"; return front.next().element(); } public int length() { return size; } // Return length Figure 4.26 Linked queue class implementation.
- 159. 140 Chap. 4 Lists, Stacks, and Queues enqueue places the new element in a link node at the end of the linked list (i.e., the node that rear points to) and then advances rear to point to the new link node. Method dequeue grabs the first element of the list removes it. 4.3.3 Comparison of Array-Based and Linked Queues All member functions for both the array-based and linked queue implementations require constant time. The space comparison issues are the same as for the equiva- lent stack implementations. Unlike the array-based stack implementation, there is no convenient way to store two queues in the same array, unless items are always transferred directly from one queue to the other. 4.4 Dictionaries and Comparators The most common objective of computer programs is to store and retrieve data. Much of this book is about efficient ways to organize collections of data records so that they can be stored and retrieved quickly. In this section we describe a simple interface for such a collection, called a dictionary. The dictionary ADT provides operations for storing records, finding records, and removing records from the collection. This ADT gives us a standard basis for comparing various data structures. Before we can discuss the interface for a dictionary, we must first define the concepts of a key and comparable objects. If we want to search for a given record in a database, how should we describe what we are looking for? A database record could simply be a number, or it could be quite complicated, such as a payroll record with many fields of varying types. We do not want to describe what we are looking for by detailing and matching the entire contents of the record. If we knew every- thing about the record already, we probably would not need to look for it. Instead, we typically define what record we want in terms of a key value. For example, if searching for payroll records, we might wish to search for the record that matches a particular ID number. In this example the ID number is the search key. To implement the search function, we require that keys be comparable. At a minimum, we must be able to take two keys and reliably determine whether they are equal or not. That is enough to enable a sequential search through a database of records and find one that matches a given key. However, we typically would like for the keys to define a total order (see Section 2.1), which means that we can tell which of two keys is greater than the other. Using key types with total orderings gives the database implementor the opportunity to organize a collection of records in a way that makes searching more efficient. An example is storing the
- 160. Sec. 4.4 Dictionaries and Comparators 141 /** The Dictionary abstract class. */ public interface Dictionary<K, E> { /** Reinitialize dictionary */ public void clear(); /** Insert a record @param k The key for the record being inserted. @param e The record being inserted. */ public void insert(K k, E e); /** Remove and return a record. @param k The key of the record to be removed. @return A maching record. If multiple records match "k", remove an arbitrary one. Return null if no record with key "k" exists. */ public E remove(K k); /** Remove and return an arbitrary record from dictionary. @return the record removed, or null if none exists. */ public E removeAny(); /** @return A record matching "k" (null if none exists). If multiple records match, return an arbitrary one. */ public E find(K k); /** @return the number of records in the dictionary. */ public int size(); }; Figure 4.27 The abstract class definition for a simple dictionary. records in sorted order in an array, which permits a binary search. Fortunately, in practice most fields of most records consist of simple data types with natural total orders. For example, integers, floats, doubles, and character strings all are totally ordered. Ordering fields that are naturally multi-dimensional, such as a point in two or three dimensions, present special opportunities if we wish to take advantage of their multidimensional nature. This problem is addressed in Section 13.3. Figure 4.27 shows the definition for a simple abstract dictionary class. The methods insert and find are the heart of the class. Method insert takes a record and inserts it into the dictionary. Method find takes a key value and returns some record from the dictionary whose key matches the one provided. If there are multiple records in the dictionary with that key value, there is no requirement as to which one is returned. Method clear simply reinitializes the dictionary. The remove method is similar to find, except that it also deletes the record returned from the dictionary.
- 161. 142 Chap. 4 Lists, Stacks, and Queues Once again, if there are multiple records in the dictionary that match the desired key, there is no requirement as to which one actually is removed and returned. Method size returns the number of elements in the dictionary. The remaining Method is removeAny. This is similar to remove, except that it does not take a key value. Instead, it removes an arbitrary record from the dictionary, if one exists. The purpose of this method is to allow a user the ability to iterate over all elements in the dictionary (of course, the dictionary will become empty in the process). Without the removeAny method, a dictionary user could not get at a record of the dictionary that he didn’t already know the key value for. With the removeAny method, the user can process all records in the dictionary as shown in the following code fragment. while (dict.size() > 0) { it = dict.removeAny(); doSomething(it); } There are other approaches that might seem more natural for iterating though a dictionary, such as using a “first” and a “next” function. But not all data structures that we want to use to implement a dictionary are able to do “first” efficiently. For example, a hash table implementation cannot efficiently locate the record in the table with the smallest key value. By using RemoveAny, we have a mechanism that provides generic access. Given a database storing records of a particular type, we might want to search for records in multiple ways. For example, we might want to store payroll records in one dictionary that allows us to search by ID, and also store those same records in a second dictionary that allows us to search by name. Figure 4.28 shows the definition for a payroll record. The Payroll class has multiple fields, each of which might be used as a search key. Simply by varying the type for the key, and looking at the appropriate field in each record, we can define a dictionary whose search key is the ID field, another whose search key is the name field, and a third whose search key is the address field of the Payroll class. Figure 4.29 shows an example where Payroll objects are stored in two separate dictionaries, one using the ID field as the key and the other using the name field as the key. A fundamental operation on a dictionary is to find a record that matches a given key. This raises the issue of how to extract the key from a record. This could be done by providing to the dictionary some class that knows how to extract the key from a record. Unfortunately, this solution does not work in all situations, because there are record types for which it is not possible to write the key extraction
- 162. Sec. 4.4 Dictionaries and Comparators 143 // A simple payroll entry with ID, name, address fields class Payroll { private Integer ID; private String name; private String address; // Constructor Payroll(int inID, String inname, String inaddr) { ID = inID; name = inname; address = inaddr; } // Data member access functions public Integer getID() { return ID; } public String getname() { return name; } public String getaddr() { return address; } } Figure 4.28 A payroll record implementation. // IDdict organizes Payroll records by ID Dictionary<Integer, Payroll> IDdict = new UALdictionary<Integer, Payroll>(); // namedict organizes Payroll records by name Dictionary<String, Payroll> namedict = new UALdictionary<String, Payroll>(); Payroll foo1 = new Payroll(5, "Joe", "Anytown"); Payroll foo2 = new Payroll(10, "John", "Mytown"); IDdict.insert(foo1.getID(), foo1); IDdict.insert(foo2.getID(), foo2); namedict.insert(foo1.getname(), foo1); namedict.insert(foo2.getname(), foo2); Payroll findfoo1 = IDdict.find(5); Payroll findfoo2 = namedict.find("John"); Figure 4.29 A dictionary search example. Here, payroll records are stored in two dictionaries, one organized by ID and the other organized by name. Both dictionaries are implemented with an unsorted array-based list.
- 163. 144 Chap. 4 Lists, Stacks, and Queues // Container for a key-value pair class KVpair<K, E> { private K k; private E e; // Constructors KVpair() { k = null; e = null; } KVpair(K kval, E eval) { k = kval; e = eval; } // Data member access functions public K key() { return k; } public E value() { return e; } } Figure 4.30 Implementation for a class representing a key-value pair. method.2 The fundamental issue is that the key value for a record is not an intrinsic property of the record’s class, or of any field within the class. The key for a record is actually a property of the context in which the record is used. An alternative is to explicitly store the key associated with a given record, as a separate field in the dictionary. That is, each entry in the dictionary will contain both a record and its associated key. Such entries will be known as key-value pairs. It is typical that storing the key explicitly duplicates some field in the record. How- ever, keys tend to be much smaller than records, so this additional space overhead will not be great. A simple class for representing key-value pairs is shown in Fig- ure 4.30. The insert method of the dictionary class supports the key-value pair implementation because it takes two parameters, a record and its associated key for that dictionary. Now that we have defined the dictionary ADT and settled on the design ap- proach of storing key-value pairs for our dictionary entries, we are ready to consider ways to implement it. Two possibilities would be to use an array-based or linked list. Figure 4.31 shows an implementation for the dictionary using an (unsorted) array-based list. 2 One example of such a situation occurs when we have a collection of records that describe books in a library. One of the fields for such a record might be a list of subject keywords, where the typical record stores a few keywords. Our dictionary might be implemented as a list of records sorted by keyword. If a book contains three keywords, it would appear three times on the list, once for each associated keyword. However, given the record, there is no simple way to determine which keyword on the keyword list triggered this appearance of the record. Thus, we cannot write a function that extracts the key from such a record.
- 164. Sec. 4.4 Dictionaries and Comparators 145 /** Dictionary implemented by unsorted array-based list. */ class UALdictionary<K, E> implements Dictionary<K, E> { private static final int defaultSize = 10; // Default size private AList<KVpair<K,E>> list; // To store dictionary // Constructors UALdictionary() { this(defaultSize); } UALdictionary(int sz) { list = new AList<KVpair<K, E>>(sz); } public void clear() { list.clear(); } // Reinitialize /** Insert an element: append to list */ public void insert(K k, E e) { KVpair<K,E> temp = new KVpair<K,E>(k, e); list.append(temp); } /** Use sequential search to find the element to remove */ public E remove(K k) { E temp = find(k); if (temp != null) list.remove(); return temp; } /** Remove the last element */ public E removeAny() { if (size() != 0) { list.moveToEnd(); list.prev(); KVpair<K,E> e = list.remove(); return e.value(); } else return null; } // Find "k" using sequential search public E find(K k) { for(list.moveToStart(); list.currPos() < list.length(); list.next()) { KVpair<K,E> temp = list.getValue(); if (k == temp.key()) return temp.value(); } return null; // "k" does not appear in dictionary } Figure 4.31 A dictionary implemented with an unsorted array-based list.
- 165. 146 Chap. 4 Lists, Stacks, and Queues public int size() // Return list size { return list.length(); } } Figure 4.31 (continued) Examining class UALdict (UAL stands for “unsorted array-based list), we can easily see that insert is a constant time operation, because it simply inserts the new record at the end of the list. However, find, and remove both require Θ(n) time in the average and worst cases, because we need to do a sequential search. Method remove in particular must touch every record in the list, because once the desired record is found, the remaining records must be shifted down in the list to fill the gap. Method removeAny removes the last record from the list, so this is a constant-time operation. As an alternative, we could implement the dictionary using a linked list. The implementation would be quite similar to that shown in Figure 4.31, and the cost of the functions should be the same asymptotically. Another alternative would be to implement the dictionary with a sorted list. The advantage of this approach would be that we might be able to speed up the find operation by using a binary search. To do so, first we must define a variation on the List ADT to support sorted lists. A sorted list is somewhat different from an unsorted list in that it cannot permit the user to control where elements get inserted. Thus, the insert method must be quite different in a sorted list than in an unsorted list. Likewise, the user cannot be permitted to append elements onto the list. For these reasons, a sorted list cannot be implemented with straightforward inheritance from the List ADT. The cost for find in a sorted list is Θ(log n) for a list of length n. This is a great improvement over the cost of find in an unsorted list. Unfortunately, the cost of insert changes from constant time in the unsorted list to Θ(n) time in the sorted list. Whether the sorted list implementation for the dictionary ADT is more or less efficient than the unsorted list implementation depends on the relative number of insert and find operations to be performed. If many more find operations than insert operations are used, then it might be worth using a sorted list to implement the dictionary. In both cases, remove requires Θ(n) time in the worst and average cases. Even if we used binary search to cut down on the time to find the record prior to removal, we would still need to shift down the remaining records in the list to fill the gap left by the remove operation.
- 166. Sec. 4.5 Further Reading 147 4.5 Further Reading For more discussion on choice of functions used to define the List ADT, see the work of the Reusable Software Research Group from Ohio State. Their definition for the List ADT can be found in [SWH93]. More information about designing such classes can be found in [SW94]. 4.6 Exercises 4.1 Assume a list has the following configuration: h | 2, 23, 15, 5, 9 i. Write a series of Java statements using the List ADT of Figure 4.1 to delete the element with value 15. 4.2 Show the list configuration resulting from each series of list operations using the List ADT of Figure 4.1. Assume that lists L1 and L2 are empty at the beginning of each series. Show where the current position is in the list. (a) L1.append(10); L1.append(20); L1.append(15); (b) L2.append(10); L2.append(20); L2.append(15); L2.moveToStart(); L2.insert(39); L2.next(); L2.insert(12); 4.3 Write a series of Java statements that uses the List ADT of Figure 4.1 to create a list capable of holding twenty elements and which actually stores the list with the following configuration: h 2, 23 | 15, 5, 9 i. 4.4 Using the list ADT of Figure 4.1, write a function to interchange the current element and the one following it.
- 167. 148 Chap. 4 Lists, Stacks, and Queues 4.5 In the linked list implementation presented in Section 4.1.2, the current po- sition is implemented using a pointer to the element ahead of the logical current node. The more “natural” approach might seem to be to have curr point directly to the node containing the current element. However, if this was done, then the pointer of the node preceeding the current one cannot be updated properly because there is no access to this node from curr. An alternative is to add a new node after the current element, copy the value of the current element to this new node, and then insert the new value into the old current node. (a) What happens if curr is at the end of the list already? Is there still a way to make this work? Is the resulting code simpler or more complex than the implementation of Section 4.1.2? (b) Will deletion always work in constant time if curr points directly to the current node? 4.6 Add to the LList class implementation a member function to reverse the order of the elements on the list. Your algorithm should run in Θ(n) time for a list of n elements. 4.7 Write a function to merge two linked lists. The input lists have their elements in sorted order, from smallest to highest. The output list should also be sorted from highest to lowest. Your algorithm should run in linear time on the length of the output list. 4.8 A circular linked list is one in which the next field for the last link node of the list points to the first link node of the list. This can be useful when you wish to have a relative positioning for elements, but no concept of an absolute first or last position. (a) Modify the code of Figure 4.8 to implement circular singly linked lists. (b) Modify the code of Figure 4.14 to implement circular doubly linked lists. 4.9 Section 4.1.3 states “the space required by the array-based list implementa- tion is Ω(n), but can be greater.” Explain why this is so. 4.10 Section 4.1.3 presents an equation for determining the break-even point for the space requirements of two implementations of lists. The variables are D, E, P, and n. What are the dimensional units for each variable? Show that both sides of the equation balance in terms of their dimensional units. 4.11 Use the space equation of Section 4.1.3 to determine the break-even point for an array-based list and linked list implementation for lists when the sizes for the data field, a pointer, and the array-based list’s array are as specified.
- 168. Sec. 4.6 Exercises 149 (a) The data field is eight bytes, a pointer is four bytes, and the array holds twenty elements. (b) The data field is two bytes, a pointer is four bytes, and the array holds thirty elements. (c) The data field is one byte, a pointer is four bytes, and the array holds thirty elements. (d) The data field is 32 bytes, a pointer is four bytes, and the array holds forty elements. 4.12 Determine the size of an int variable, a double variable, and a pointer on your computer. (a) Calculate the break-even point, as a function of n, beyond which the array-based list is more space efficient than the linked list for lists whose elements are of type int. (b) Calculate the break-even point, as a function of n, beyond which the array-based list is more space efficient than the linked list for lists whose elements are of type double. 4.13 Modify the code of Figure 4.18 to implement two stacks sharing the same array, as shown in Figure 4.20. 4.14 Modify the array-based queue definition of Figure 4.25 to use a separate Boolean member to keep track of whether the queue is empty, rather than require that one array position remain empty. 4.15 A palindrome is a string that reads the same forwards as backwards. Using only a fixed number of stacks and queues, the stack and queue ADT func- tions, and a fixed number of int and char variables, write an algorithm to determine if a string is a palindrome. Assume that the string is read from standard input one character at a time. The algorithm should output true or false as appropriate. 4.16 Reimplement function fibr from Exercise 2.11, using a stack to replace the recursive call as described in Section 4.2.4. 4.17 Write a recursive algorithm to compute the value of the recurrence relation T(n) = T(⌈n/2⌉) + T(⌊n/2⌋) + n; T(1) = 1. Then, rewrite your algorithm to simulate the recursive calls with a stack. 4.18 Let Q be a non-empty queue, and let S be an empty stack. Using only the stack and queue ADT functions and a single element variable X, write an algorithm to reverse the order of the elements in Q.
- 169. 150 Chap. 4 Lists, Stacks, and Queues 4.19 A common problem for compilers and text editors is to determine if the parentheses (or other brackets) in a string are balanced and properly nested. For example, the string “((())())()” contains properly nested pairs of paren- theses, but the string “)()(” does not, and the string “())” does not contain properly matching parentheses. (a) Give an algorithm that returns true if a string contains properly nested and balanced parentheses, and false otherwise. Use a stack to keep track of the number of left parentheses seen so far. Hint: At no time while scanning a legal string from left to right will you have encoun- tered more right parentheses than left parentheses. (b) Give an algorithm that returns the position in the string of the first of- fending parenthesis if the string is not properly nested and balanced. That is, if an excess right parenthesis is found, return its position; if there are too many left parentheses, return the position of the first ex- cess left parenthesis. Return −1 if the string is properly balanced and nested. Use a stack to keep track of the number and positions of left parentheses seen so far. 4.20 Imagine that you are designing an application where you need to perform the operations Insert, Delete Maximum, and Delete Minimum. For this application, the cost of inserting is not important, because it can be done off-line prior to startup of the time-critical section, but the performance of the two deletion operations are critical. Repeated deletions of either kind must work as fast as possible. Suggest a data structure that can support this application, and justify your suggestion. What is the time complexity for each of the three key operations? 4.21 Write a function that reverses the order of an array of n items. 4.7 Projects 4.1 A deque (pronounced “deck”) is like a queue, except that items may be added and removed from both the front and the rear. Write either an array-based or linked implementation for the deque. 4.2 One solution to the problem of running out of space for an array-based list implementation is to replace the array with a larger array whenever the origi- nal array overflows. A good rule that leads to an implementation that is both space and time efficient is to double the current size of the array when there is an overflow. Reimplement the array-based List class of Figure 4.2 to support this array-doubling rule.
- 170. Sec. 4.7 Projects 151 4.3 Use singly linked lists to implement integers of unlimited size. Each node of the list should store one digit of the integer. You should implement addition, subtraction, multiplication, and exponentiation operations. Limit exponents to be positive integers. What is the asymptotic running time for each of your operations, expressed in terms of the number of digits for the two operands of each function? 4.4 Implement doubly linked lists by storing the sum of the next and prev pointers in a single pointer variable as described in Example 4.1. 4.5 Implement a city database using unordered lists. Each database record con- tains the name of the city (a string of arbitrary length) and the coordinates of the city expressed as integer x and y coordinates. Your database should allow records to be inserted, deleted by name or coordinate, and searched by name or coordinate. Another operation that should be supported is to print all records within a given distance of a specified point. Implement the database using an array-based list implementation, and then a linked list im- plementation. Collect running time statistics for each operation in both im- plementations. What are your conclusions about the relative advantages and disadvantages of the two implementations? Would storing records on the list in alphabetical order by city name speed any of the operations? Would keeping the list in alphabetical order slow any of the operations? 4.6 Modify the code of Figure 4.18 to support storing variable-length strings of at most 255 characters. The stack array should have type char. A string is represented by a series of characters (one character per stack element), with the length of the string stored in the stack element immediately above the string itself, as illustrated by Figure 4.32. The push operation would store an element requiring i storage units in the i positions beginning with the current value of top and store the size in the position i storage units above top. The value of top would then be reset above the newly inserted element. The pop operation need only look at the size value stored in position top−1 and then pop off the appropriate number of units. You may store the string on the stack in reverse order if you prefer, provided that when it is popped from the stack, it is returned in its proper order. 4.7 Implement a collection of freelists for variable-length strings, as described at the end of Section 4.1.2. For each such freelist, you will need an access function to get it if it exists, and implement it if it does not. A major de- sign consideration is how to organize the collection of freelists, which are distinguished by the length of the strings. Essentially, what is needed is a dictionary of freelists, organized by string lengths.
- 171. 152 Chap. 4 Lists, Stacks, and Queues top = 10 ‘a’ ‘b’ ‘c’ 3 ‘h’ ‘e’ ‘l’ ‘o’ 5 0 1 2 3 4 5 6 7 8 9 10 ‘l’ Figure 4.32 An array-based stack storing variable-length strings. Each position stores either one character or the length of the string immediately to the left of it in the stack. 4.8 Define an ADT for a bag (see Section 2.1) and create an array-based imple- mentation for bags. Be sure that your bag ADT does not rely in any way on knowing or controlling the position of an element. Then, implement the dictionary ADT of Figure 4.27 using your bag implementation. 4.9 Implement the dictionary ADT of Figure 4.27 using an unsorted linked list as defined by class LList in Figure 4.8. Make the implementation as efficient as you can, given the restriction that your implementation must use the un- sorted linked list and its access operations to implement the dictionary. State the asymptotic time requirements for each function member of the dictionary ADT under your implementation. 4.10 Implement the dictionary ADT of Figure 4.27 based on stacks. Your imple- mentation should declare and use two stacks. 4.11 Implement the dictionary ADT of Figure 4.27 based on queues. Your imple- mentation should declare and use two queues.
- 172. 5 Binary Trees The list representations of Chapter 4 have a fundamental limitation: Either search or insert can be made efficient, but not both at the same time. Tree structures permit both efficient access and update to large collections of data. Binary trees in particular are widely used and relatively easy to implement. But binary trees are useful for many things besides searching. Just a few examples of applications that trees can speed up include prioritizing jobs, describing mathematical expressions and the syntactic elements of computer programs, or organizing the information needed to drive data compression algorithms. This chapter begins by presenting definitions and some key properties of bi- nary trees. Section 5.2 discusses how to process all nodes of the binary tree in an organized manner. Section 5.3 presents various methods for implementing binary trees and their nodes. Sections 5.4 through 5.6 present three examples of binary trees used in specific applications: the Binary Search Tree (BST) for implementing dictionaries, heaps for implementing priority queues, and Huffman coding trees for text compression. The BST, heap, and Huffman coding tree each have distinctive features that affect their implementation and use. 5.1 Definitions and Properties A binary tree is made up of a finite set of elements called nodes. This set either is empty or consists of a node called the root together with two binary trees, called the left and right subtrees, which are disjoint from each other and from the root. (Disjoint means that they have no nodes in common.) The roots of these subtrees are children of the root. There is an edge from a node to each of its children, and a node is said to be the parent of its children. If n1, n2, ..., nk is a sequence of nodes in the tree such that ni is the parent of ni+1 for 1 ≤ i < k, then this sequence is called a path from n1 to nk. The length 153
- 173. 154 Chap. 5 Binary Trees G I E F A C B D H Figure 5.1 An example binary tree. Node A is the root. Nodes B and C are A’s children. Nodes B and D together form a subtree. Node B has two children: Its left child is the empty tree and its right child is D. Nodes A, C, and E are ancestors of G. Nodes D, E, and F make up level 2 of the tree; node A is at level 0. The edges from A to C to E to G form a path of length 3. Nodes D, G, H, and I are leaves. Nodes A, B, C, E, and F are internal nodes. The depth of I is 3. The height of this tree is 4. of the path is k − 1. If there is a path from node R to node M, then R is an ancestor of M, and M is a descendant of R. Thus, all nodes in the tree are descendants of the root of the tree, while the root is the ancestor of all nodes. The depth of a node M in the tree is the length of the path from the root of the tree to M. The height of a tree is one more than the depth of the deepest node in the tree. All nodes of depth d are at level d in the tree. The root is the only node at level 0, and its depth is 0. A leaf node is any node that has two empty children. An internal node is any node that has at least one non-empty child. Figure 5.1 illustrates the various terms used to identify parts of a binary tree. Figure 5.2 illustrates an important point regarding the structure of binary trees. Because all binary tree nodes have two children (one or both of which might be empty), the two binary trees of Figure 5.2 are not the same. Two restricted forms of binary tree are sufficiently important to warrant special names. Each node in a full binary tree is either (1) an internal node with exactly two non-empty children or (2) a leaf. A complete binary tree has a restricted shape obtained by starting at the root and filling the tree by levels from left to right. In the complete binary tree of height d, all levels except possibly level d−1 are completely full. The bottom level has its nodes filled in from the left side.
- 174. Sec. 5.1 Definitions and Properties 155 (b) (d) (c) (a) B EMPTY EMPTY A A A B B B A Figure 5.2 Two different binary trees. (a) A binary tree whose root has a non- empty left child. (b) A binary tree whose root has a non-empty right child. (c) The binary tree of (a) with the missing right child made explicit. (d) The binary tree of (b) with the missing left child made explicit. (a) (b) Figure 5.3 Examples of full and complete binary trees. (a) This tree is full (but not complete). (b) This tree is complete (but not full). Figure 5.3 illustrates the differences between full and complete binary trees.1 There is no particular relationship between these two tree shapes; that is, the tree of Figure 5.3(a) is full but not complete while the tree of Figure 5.3(b) is complete but not full. The heap data structure (Section 5.5) is an example of a complete binary tree. The Huffman coding tree (Section 5.6) is an example of a full binary tree. 1 While these definitions for full and complete binary tree are the ones most commonly used, they are not universal. Some textbooks even reverse these definitions! Because the common meaning of the words “full” and “complete” are quite similar, there is little that you can do to distinguish between them other than to memorize the definitions. Here is a memory aid that you might find useful: “Complete” is a wider word than “full,” and complete binary trees tend to be wider than full binary trees because each level of a complete binary tree is as wide as possible.
- 175. 156 Chap. 5 Binary Trees Any number of internal nodes Figure 5.4 A tree containing n internal nodes and a single leaf. 5.1.1 The Full Binary Tree Theorem Some binary tree implementations store data only at the leaf nodes, using the inter- nal nodes to provide structure to the tree. More generally, binary tree implementa- tions might require some amount of space for internal nodes, and a different amount for leaf nodes. Thus, to analyze the space required by such implementations, it is useful to know the minimum and maximum fraction of the nodes that are leaves in a tree containing n internal nodes. Unfortunately, this fraction is not fixed. A binary tree of n internal nodes might have only one leaf. This occurs when the internal nodes are arranged in a chain ending in a single leaf as shown in Figure 5.4. In this case, the number of leaves is low because each internal node has only one non-empty child. To find an upper bound on the number of leaves for a tree of n internal nodes, first note that the upper bound will occur when each internal node has two non-empty children, that is, when the tree is full. However, this observation does not tell what shape of tree will yield the highest percentage of non-empty leaves. It turns out not to matter, because all full binary trees with n internal nodes have the same number of leaves. This fact allows us to compute the space requirements for a full binary tree implementation whose leaves require a different amount of space from its internal nodes. Theorem 5.1 Full Binary Tree Theorem: The number of leaves in a non-empty full binary tree is one more than the number of internal nodes. Proof: The proof is by mathematical induction on n, the number of internal nodes. This is an example of an induction proof where we reduce from an arbitrary in- stance of size n to an instance of size n − 1 that meets the induction hypothesis. • Base Cases: The non-empty tree with zero internal nodes has one leaf node. A full binary tree with one internal node has two leaf nodes. Thus, the base cases for n = 0 and n = 1 conform to the theorem. • Induction Hypothesis: Assume that any full binary tree T containing n − 1 internal nodes has n leaves.
- 176. Sec. 5.1 Definitions and Properties 157 • Induction Step: Given tree T with n internal nodes, select an internal node I whose children are both leaf nodes. Remove both of I’s children, making I a leaf node. Call the new tree T′ . T′ has n − 1 internal nodes. From the induction hypothesis, T′ has n leaves. Now, restore I’s two children. We once again have tree T with n internal nodes. How many leaves does T have? Because T′ has n leaves, adding the two children yields n+2. However, node I counted as one of the leaves in T′ and has now become an internal node. Thus, tree T has n + 1 leaf nodes and n internal nodes. By mathematical induction the theorem holds for all values of n ≥ 0. ✷ When analyzing the space requirements for a binary tree implementation, it is useful to know how many empty subtrees a tree contains. A simple extension of the Full Binary Tree Theorem tells us exactly how many empty subtrees there are in any binary tree, whether full or not. Here are two approaches to proving the following theorem, and each suggests a useful way of thinking about binary trees. Theorem 5.2 The number of empty subtrees in a non-empty binary tree is one more than the number of nodes in the tree. Proof 1: Take an arbitrary binary tree T and replace every empty subtree with a leaf node. Call the new tree T′ . All nodes originally in T will be internal nodes in T′ (because even the leaf nodes of T have children in T′ ). T′ is a full binary tree, because every internal node of T now must have two children in T′ , and each leaf node in T must have two children in T′ (the leaves just added). The Full Binary Tree Theorem tells us that the number of leaves in a full binary tree is one more than the number of internal nodes. Thus, the number of new leaves that were added to create T′ is one more than the number of nodes in T. Each leaf node in T′ corresponds to an empty subtree in T. Thus, the number of empty subtrees in T is one more than the number of nodes in T. ✷ Proof 2: By definition, every node in binary tree T has two children, for a total of 2n children in a tree of n nodes. Every node except the root node has one parent, for a total of n − 1 nodes with parents. In other words, there are n − 1 non-empty children. Because the total number of children is 2n, the remaining n + 1 children must be empty. ✷ 5.1.2 A Binary Tree Node ADT Just as we developed a generic list ADT on which to build specialized list im- plementations, we would like to define a generic binary tree ADT based on those
- 177. 158 Chap. 5 Binary Trees /** ADT for binary tree nodes */ public interface BinNode<E> { /** Return and set the element value */ public E element(); public void setElement(E v); /** Return the left child */ public BinNode<E> left(); /** Return the right child */ public BinNode<E> right(); /** Return true if this is a leaf node */ public boolean isLeaf(); } Figure 5.5 A binary tree node ADT. aspects of binary trees that are relevent to all applications. For example, we must be able to initialize a binary tree, and we might wish to determine if the tree is empty. Some activities might be unique to the application. For example, we might wish to combine two binary trees by making their roots be the children of a new root node. Other activities are centered around the nodes. For example, we might need access to the left or right child of a node, or to the node’s data value. Clearly there are activities that relate to nodes (e.g., reach a node’s child or get a node’s value), and activities that relate to trees (e.g., tree initialization). This indicates that nodes and trees should be implemented as separate classes. For now, we concentrate on the class to implement binary tree nodes. This class will be used by some of the binary tree structures presented later. Figure 5.5 shows an interface for binary tree nodes, called BinNode. Class BinNode is a generic with parameter E, which is the type for the data record stored in a node. Member functions are provided that set or return the element value, return a reference to the left child, return a reference to the right child, or indicate whether the node is a leaf. 5.2 Binary Tree Traversals Often we wish to process a binary tree by “visiting” each of its nodes, each time performing a specific action such as printing the contents of the node. Any process for visiting all of the nodes in some order is called a traversal. Any traversal that lists every node in the tree exactly once is called an enumeration of the tree’s nodes. Some applications do not require that the nodes be visited in any particular order as long as each node is visited precisely once. For other applications, nodes
- 178. Sec. 5.2 Binary Tree Traversals 159 must be visited in an order that preserves some relationship. For example, we might wish to make sure that we visit any given node before we visit its children. This is called a preorder traversal. Example 5.1 The preorder enumeration for the tree of Figure 5.1 is ABDCEGFHI. The first node printed is the root. Then all nodes of the left subtree are printed (in preorder) before any node of the right subtree. Alternatively, we might wish to visit each node only after we visit its children (and their subtrees). For example, this would be necessary if we wish to return all nodes in the tree to free store. We would like to delete the children of a node before deleting the node itself. But to do that requires that the children’s children be deleted first, and so on. This is called a postorder traversal. Example 5.2 The postorder enumeration for the binary tree of Figure 5.1 is DBGEHIFCA. An inorder traversal first visits the left child (including its entire subtree), then visits the node, and finally visits the right child (including its entire subtree). The binary search tree of Section 5.4 makes use of this traversal. Example 5.3 The inorder enumeration for the binary tree of Figure 5.1 is BDAGECHFI. A traversal routine is naturally written as a recursive function. Its input parame- ter is a pointer to a node which we will call root because each node can be viewed as the root of a some subtree. The initial call to the traversal function passes in a pointer to the root node of the tree. The traversal function visits root and its chil- dren (if any) in the desired order. For example, a preorder traversal specifies that root be visited before its children. This can easily be implemented as follows. void preorder(BinNode rt) // rt is the root of the subtree { if (rt == null) return; // Empty subtree visit(rt); preorder(rt.left()); preorder(rt.right()); }
- 179. 160 Chap. 5 Binary Trees Function preorder first checks that the tree is not empty (if it is, then the traversal is done and preorder simply returns). Otherwise, preorder makes a call to visit, which processes the root node (i.e., prints the value or performs some computation as required by the application). Function preorder is then called recursively on the left subtree, which will visit all nodes in that subtree. Finally, preorder is called on the right subtree, visiting all remaining nodes in the tree. Postorder and inorder traversals are similar. They simply change the order in which the node and its children are visited, as appropriate. An important design decision regarding the implementation of any recursive function on trees is when to check for an empty subtree. Function preorder first checks to see if the value for root is null. If not, it will recursively call itself on the left and right children of root. In other words, preorder makes no attempt to avoid calling itself on an empty child. Some programmers attempt an alternate design in which the left and right pointers of the current node are checked so that the recursive call is made only on non-empty children. Such a design typically looks as follows: void preorder2(BinNode rt) { visit(rt); if (rt.left() != null) preorder(rt.left()); if (rt.right() != null) preorder(rt.right()); } At first it might appear that preorder2 is more efficient than preorder, because it makes only half as many recursive calls. (Why?) On the other hand, preorder2 must access the left and right child pointers twice as often. The net result is little or no performance improvement. In reality, the design of preorder2 is generally inferior to that of preorder for two reasons. First, while it is not apparent in this simple example, for more complex traversals it can become awkward to place the check for the null pointer in the calling code. Even in this simple example we had to write two tests for null, rather than one in the first design. The second and more important problem with preorder2 is that it tends to be error prone. While preorder2 insures that no recursive calls will be made on empty subtrees, it will fail if the initial call passes in a null pointer. This would occur if the original tree is empty. To avoid the bug, either preorder2 needs an additional test for a null pointer at the beginning (making the subsequent tests redundant after all), or the caller of preorder2 has a hidden obligation to pass in a non-empty tree, which is unreliable design. The net result is that many programmers forget to test for the possibility that the empty tree
- 180. Sec. 5.2 Binary Tree Traversals 161 is being traversed. By using the first design, which explicitly supports processing of empty subtrees, the problem is avoided. Another issue to consider when designing a traversal is how to define the visitor function that is to be executed on every node. One approach is simply to write a new version of the traversal for each such visitor function as needed. The disad- vantage to this is that whatever function does the traversal must have access to the BinNode class. It is probably better design to permit only the tree class to have access to the BinNode class. Another approach is for the tree class to supply a generic traversal function which takes the visitor as a function parameter. This is known as the visitor design pattern. A major contraint on this approach is that the signature for all visitor functions, that is, their return type and parameters, must be fixed in advance. Thus, the designer of the generic traversal function must be able to adequately judge what parameters and return type will likely be needed by potential visitor functions. Properly handling information flow between parts of a program can often be a significant design challenge. This issue tends to be particularly confusing when dealing with recursive functions such as tree traversals. In general, we can run into trouble either with passing in the correct information needed by the function to do its work, or with returning information to the recursive function’s caller. We will see many examples throughout the book that illustrate methods for passing infor- mation in and out of recursive functions as they traverse a tree structure. Before leaving this section, we will study a few simple examples. Let us first consider the simple case where a computation requires that we com- municate information back up the tree to the end user. Example 5.4 We wish to write a function that counts the number of nodes in a binary tree. The key insight is that the total count for any (non-empty) subtree is one for the root plus the counts for the left and right subtrees. We can implement the function as follows. int count(BinNode rt) { if (rt == null) return 0; // Nothing to count return 1 + count(rt.left()) + count(rt.right()); } Another problem that occurs when recursively processing data collections is controlling which members of the collection will be visited. For example, some tree “traversals” might in fact visit only some tree nodes, while avoiding processing of others. Exercise 5.20 must solve exactly this problem in the context of a binary search tree. It must visit only those children of a given node that might possibly
- 181. 162 Chap. 5 Binary Trees 20 50 40 75 20 to 40 Figure 5.6 To be a binary search tree, the left child of the node with value 40 must have a value between 20 and 40. fall within a given range of values. Fortunately, it requires only a simple local calcuation to determine which child(ren) to visit. A more difficult situation is illustrated by the following problem. Given an arbitrary binary tree we wish to determine if, for every node A, are all nodes in A’s left subtree less than the value of A, and are all nodes in A’s right subtree greater than the value of A? (This happens to be the definition for a binary search tree, described in Section 5.4.) Unfortunately, to make this decision we need to know some context that is not available just by looking at the node’s parent or children. As shown by Figure 5.6, it is not enough to verify that A’s left child has a value less than that of A, and that A’s right child has a greater value. Nor is it enough to verify that A has a value consistent with that of its parent. In fact, we need to know information about what range of values is legal for a given node. That information might come from any ancestor for the node. Thus, relevent range information must be passed down the tree. We can implement this function as follows. boolean checkBST(BSTNode<Integer,Integer> root, Integer low, Integer high) { if (root == null) return true; // Empty subtree Integer rootkey = root.key(); if ((rootkey < low) || (rootkey > high)) return false; // Out of range if (!checkBST(root.left(), low, rootkey)) return false; // Left side failed return checkBST(root.right(), rootkey, high); } 5.3 Binary Tree Node Implementations In this section we will examine ways to implement binary tree nodes. We begin with some options for pointer-based binary tree node implementations. Then comes a
- 182. Sec. 5.3 Binary Tree Node Implementations 163 discussion on techniques for determining the space requirements for a given binary tree node implementation. The section concludes with an introduction to the array- based implementation for complete binary trees. 5.3.1 Pointer-Based Node Implementations By definition, all binary tree nodes have two children, though one or both children can be empty. Binary tree nodes normally contain a value field, with the type of the field depending on the application. The most common node implementation includes a value field and pointers to the two children. Figure 5.7 shows a simple implementation for the BinNode abstract class, which we will name BSTNode. Class BSTNode includes a data member of type element, (which is the second generic parameter) for the element type. To sup- port search structures such as the Binary Search Tree, an additional field is included, with corresponding access methods, store a key value (whose purpose is explained in Section 4.4). Its type is determined by the first generic parameter, named K. Ev- ery BSTNode object also has two pointers, one to its left child and another to its right child. Figure 5.8 shows an illustration of the BSTNode implementation. Some programmers find it convenient to add a pointer to the node’s parent, allowing easy upward movement in the tree. Using a parent pointer is somewhat analogous to adding a link to the previous node in a doubly linked list. In practice, the parent pointer is almost always unnecessary and adds to the space overhead for the tree implementation. It is not just a problem that parent pointers take space. More importantly, many uses of the parent pointer are driven by improper under- standing of recursion and so indicate poor programming. If you are inclined toward using a parent pointer, consider if there is a more efficient implementation possible. An important decision in the design of a pointer-based node implementation is whether the same class definition will be used for leaves and internal nodes. Using the same class for both will simplify the implementation, but might be an inefficient use of space. Some applications require data values only for the leaves. Other applications require one type of value for the leaves and another for the in- ternal nodes. Examples include the binary trie of Section 13.1, the PR quadtree of Section 13.3, the Huffman coding tree of Section 5.6, and the expression tree illus- trated by Figure 5.9. By definition, only internal nodes have non-empty children. If we use the same node implementation for both internal and leaf nodes, then both must store the child pointers. But it seems wasteful to store child pointers in the leaf nodes. Thus, there are many reasons why it can save space to have separate implementations for internal and leaf nodes.
- 183. 164 Chap. 5 Binary Trees /** Binary tree node implementation: Pointers to children */ class BSTNode<K,E> implements BinNode<E> { private K key; // Key for this node private E element; // Element for this node private BSTNode<K,E> left; // Pointer to left child private BSTNode<K,E> right; // Pointer to right child /** Constructors */ public BSTNode() {left = right = null; } public BSTNode(K k, E val) { left = right = null; key = k; element = val; } public BSTNode(K k, E val, BSTNode<K,E> l, BSTNode<K,E> r) { left = l; right = r; key = k; element = val; } /** Return and set the key value */ public K key() { return key; } public void setKey(K k) { key = k; } /** Return and set the element value */ public E element() { return element; } public void setElement(E v) { element = v; } /** Return and set the left child */ public BSTNode<K,E> left() { return left; } public void setLeft(BSTNode<K,E> p) { left = p; } /** Return and set the right child */ public BSTNode<K,E> right() { return right; } public void setRight(BSTNode<K,E> p) { right = p; } /** Return true if this is a leaf node */ public boolean isLeaf() { return (left == null) && (right == null); } } Figure 5.7 A binary tree node class implementation. As an example of a tree that stores different information at the leaf and inter- nal nodes, consider the expression tree illustrated by Figure 5.9. The expression tree represents an algebraic expression composed of binary operators such as ad- dition, subtraction, multiplication, and division. Internal nodes store operators, while the leaves store operands. The tree of Figure 5.9 represents the expression 4x(2x + a) − c. The storage requirements for a leaf in an expression tree are quite different from those of an internal node. Internal nodes store one of a small set of operators, so internal nodes could store either a small code identifying the operator or a single byte for the operator’s character symbol. In contrast, leaves must store variable names or numbers. Thus, the leaf node value field must be considerably
- 184. Sec. 5.3 Binary Tree Node Implementations 165 A C G H E D B F I Figure 5.8 Illustration of a typical pointer-based binary tree implementation, where each node stores two child pointers and a value. ∗ + ∗ 4 x ∗ x 2 − c a Figure 5.9 An expression tree for 4x(2x + a) − c. larger to handle the wider range of possible values. At the same time, leaf nodes need not store child pointers. Java allows us to differentiate leaf from internal nodes through the use of class inheritance. As we have seen with lists and with class BinNode, a base class provides a general definition for an object, and a subclass modifies a base class to add more detail. A base class can be declared for nodes in general, with subclasses defined for the internal and leaf nodes. The base class of Figure 5.10 is named VarBinNode. It includes a virtual member function named isLeaf, which indi- cates the node type. Subclasses for the internal and leaf node types each implement isLeaf. Internal nodes store child pointers of the base class type; they do not dis- tinguish their children’s actual subclass. Whenever a node is examined, its version of isLeaf indicates the node’s subclass.
- 185. 166 Chap. 5 Binary Trees public interface VarBinNode { public boolean isLeaf(); } class VarLeafNode implements VarBinNode { // Leaf node private String operand; // Operand value public VarLeafNode(String val) { operand = val; } public boolean isLeaf() { return true; } public String value() { return operand; } }; /** Internal node */ class VarIntlNode implements VarBinNode { private VarBinNode left; // Left child private VarBinNode right; // Right child private Character operator; // Operator value public VarIntlNode(Character op, VarBinNode l, VarBinNode r) { operator = op; left = l; right = r; } public boolean isLeaf() { return false; } public VarBinNode leftchild() { return left; } public VarBinNode rightchild() { return right; } public Character value() { return operator; } } /** Preorder traversal */ public static void traverse(VarBinNode rt) { if (rt == null) return; // Nothing to visit if (rt.isLeaf()) // Process leaf node Visit.VisitLeafNode(((VarLeafNode)rt).value()); else { // Process internal node Visit.VisitInternalNode(((VarIntlNode)rt).value()); traverse(((VarIntlNode)rt).leftchild()); traverse(((VarIntlNode)rt).rightchild()); } } Figure 5.10 An implementation for separate internal and leaf node representa- tions using Java class inheritance and virtual functions.
- 186. Sec. 5.3 Binary Tree Node Implementations 167 Figure 5.10 presents a sample node implementation. It includes two classes derived from class VarBinNode, named LeafNode and IntlNode. Class IntlNode accesses its children through pointers of type VarBinNode. Function traverse illustrates the use of these classes. Where traverse calls method “root.isLeaf(),” Java’s runtime environment determines which subclass this particular instance of root happens to be and calls that subclass’s version of isLeaf. Method isLeaf then provides the actual node type to its caller. The other member functions for the derived subclasses are accessed by type-casting the base class pointer as appropriate, as shown in function traverse. There is another approach that we can take to represent separate leaf and inter- nal nodes, also using a virtual base class and separate node classes for the two types. This is to implement nodes using the composite design pattern. This approach is noticeably different from the one of Figure 5.10 in that the node classes themselves implement the functionality of traverse. Figure 5.11 shows the implementa- tion. Here, base class VarBinNode declares a member function traverse that each subclass must implement. Each subclass then implements its own appropriate behavior for its role in a traversal. The whole traversal process is called by invoking traverse on the root node, which in turn invokes traverse on its children. When comparing the implementations of Figures 5.10 and 5.11, each has ad- vantages and disadvantages. The first does not require that the node classes know about the traverse function. With this approach, it is easy to add new methods to the tree class that do other traversals or other operations on nodes of the tree. However, we see that traverse in Figure 5.10 does need to be familiar with each node subclass. Adding a new node subclass would therefore require modifications to the traverse function. In contrast, the approach of Figure 5.11 requires that any new operation on the tree that requires a traversal also be implemented in the node subclasses. On the other hand, the approach of Figure 5.11 avoids the need for the traverse function to know anything about the distinct abilities of the node subclasses. Those subclasses handle the responsibility of performing a traversal on themselves. A secondary benefit is that there is no need for traverse to explic- itly enumerate all of the different node subclasses, directing appropriate action for each. With only two node classes this is a minor point. But if there were many such subclasses, this could become a bigger problem. A disadvantage is that the traversal operation must not be called on a null pointer, because there is no object to catch the call. This problem could be avoided by using a flyweight to implement empty nodes. Typically, the version of Figure 5.10 would be preferred in this example if traverse is a member function of the tree class, and if the node subclasses are
- 187. 168 Chap. 5 Binary Trees public interface VarBinNode { public boolean isLeaf(); public void traverse(); } class VarLeafNode implements VarBinNode { // Leaf node private String operand; // Operand value public VarLeafNode(String val) { operand = val; } public boolean isLeaf() { return true; } public String value() { return operand; } public void traverse() { Visit.VisitLeafNode(operand); } } class VarIntlNode implements VarBinNode { // Internal node private VarBinNode left; // Left child private VarBinNode right; // Right child private Character operator; // Operator value public VarIntlNode(Character op, VarBinNode l, VarBinNode r) { operator = op; left = l; right = r; } public boolean isLeaf() { return false; } public VarBinNode leftchild() { return left; } public VarBinNode rightchild() { return right; } public Character value() { return operator; } public void traverse() { Visit.VisitInternalNode(operator); if (left != null) left.traverse(); if (right != null) right.traverse(); } } /** Preorder traversal */ public static void traverse(VarBinNode rt) { if (rt != null) rt.traverse(); } Figure 5.11 A second implementation for separate internal and leaf node repre- sentations using Java class inheritance and virtual functions using the composite design pattern. Here, the functionality of traverse is embedded into the node subclasses.
- 188. Sec. 5.3 Binary Tree Node Implementations 169 hidden from users of that tree class. On the other hand, if the nodes are objects that have meaning to users of the tree separate from their existence as nodes in the tree, then the version of Figure 5.11 might be preferred because hiding the internal behavior of the nodes becomes more important. 5.3.2 Space Requirements This section presents techniques for calculating the amount of overhead required by a binary tree implementation. Recall that overhead is the amount of space necessary to maintain the data structure. In other words, it is any space not used to store data records. The amount of overhead depends on several factors including which nodes store data values (all nodes, or just the leaves), whether the leaves store child pointers, and whether the tree is a full binary tree. In a simple pointer-based implementation for the binary tree such as that of Figure 5.7, every node has two pointers to its children (even when the children are null). This implementation requires total space amounting to n(2P + D) for a tree of n nodes. Here, P stands for the amount of space required by a pointer, and D stands for the amount of space required by a data value. The total overhead space will be 2Pn for the entire tree. Thus, the overhead fraction will be 2P/(2P + D). The actual value for this expression depends on the relative size of pointers versus data fields. If we arbitrarily assume that P = D, then a full tree has about two thirds of its total space taken up in overhead. Worse yet, Theorem 5.2 tells us that about half of the pointers are “wasted” null values that serve only to indicate tree structure, but which do not provide access to new data. If only leaves store data values, then the fraction of total space devoted to over- head depends on whether the tree is full. If the tree is not full, then conceivably there might only be one leaf node at the end of a series of internal nodes. Thus, the overhead can be an arbitrarily high percentage for non-full binary trees. The overhead fraction drops as the tree becomes closer to full, being lowest when the tree is truly full. In this case, about one half of the nodes are internal. Great savings can be had by eliminating the pointers from leaf nodes in full binary trees. Because about half of the nodes are leaves and half internal nodes, and because only internal nodes now have overhead, the overhead fraction in this case will be approximately n 2 (2P) n 2 (2P) + Dn = P P + D .
- 189. 170 Chap. 5 Binary Trees If P = D, the overhead drops to about one half of the total space. However, if only leaf nodes store useful information, the overhead fraction for this implementation is actually three quarters of the total space, because half of the “data” space is unused. If a full binary tree needs to store data only at the leaf nodes, a better implemen- tation would have the internal nodes store two pointers and no data field while the leaf nodes store only a data field. This implementation requires 2Pn + D(n + 1) units of space. If P = D, then the overhead is about 2P/(2P +D) = 2/3. It might seem counter-intuitive that the overhead ratio has gone up while the total amount of space has gone down. The reason is because we have changed our definition of “data” to refer only to what is stored in the leaf nodes, so while the overhead fraction is higher, it is from a total storage requirement that is lower. There is one serious flaw with this analysis. When using separate implemen- tations for internal and leaf nodes, there must be a way to distinguish between the node types. When separate node types are implemented via Java subclasses, the runtime environment stores information with each object allowing it to deter- mine, for example, the correct subclass to use when the isLeaf virtual function is called. Thus, each node requires additional space. Only one bit is truly necessary to distinguish the two possibilities. In rare applications where space is a critical resource, implementors can often find a spare bit within the node’s value field in which to store the node type indicator. An alternative is to use a spare bit within a node pointer to indicate node type. For example, this is often possible when the compiler requires that structures and objects start on word boundaries, leaving the last bit of a pointer value always zero. Thus, this bit can be used to store the node- type flag and is reset to zero before the pointer is dereferenced. Another alternative when the leaf value field is smaller than a pointer is to replace the pointer to a leaf with that leaf’s value. When space is limited, such techniques can make the differ- ence between success and failure. In any other situation, such “bit packing” tricks should be avoided because they are difficult to debug and understand at best, and are often machine dependent at worst.2 5.3.3 Array Implementation for Complete Binary Trees The previous section points out that a large fraction of the space in a typical binary tree node implementation is devoted to structural overhead, not to storing data. 2 In the early to mid 1980s, I worked on a Geographic Information System that stored spatial data in quadtrees (see Section 13.3). At the time space was a critical resource, so we used a bit-packing approach where we stored the nodetype flag as the last bit in the parent node’s pointer. This worked perfectly on various 32-bit workstations. Unfortunately, in those days IBM PC-compatibles used 16-bit pointers. We never did figure out how to port our code to the 16-bit machine.
- 190. Sec. 5.4 Binary Search Trees 171 This section presents a simple, compact implementation for complete binary trees. Recall that complete binary trees have all levels except the bottom filled out com- pletely, and the bottom level has all of its nodes filled in from left to right. Thus, a complete binary tree of n nodes has only one possible shape. You might think that a complete binary tree is such an unusual occurrence that there is no reason to develop a special implementation for it. However, the complete binary tree has practical uses, the most important being the heap data structure discussed in Sec- tion 5.5. Heaps are often used to implement priority queues (Section 5.5) and for external sorting algorithms (Section 8.5.2). We begin by assigning numbers to the node positions in the complete binary tree, level by level, from left to right as shown in Figure 5.12(a). An array can store the tree’s data values efficiently, placing each data value in the array position corresponding to that node’s position within the tree. Figure 5.12(b) lists the array indices for the children, parent, and siblings of each node in Figure 5.12(a). From Figure 5.12(b), you should see a pattern regarding the positions of a node’s relatives within the array. Simple formulae can be derived for calculating the array index for each relative of a node r from r’s index. No explicit pointers are necessary to reach a node’s left or right child. This means there is no overhead to the array implementation if the array is selected to be of size n for a tree of n nodes. The formulae for calculating the array indices of the various relatives of a node are as follows. The total number of nodes in the tree is n. The index of the node in question is r, which must fall in the range 0 to n − 1. • Parent(r) = ⌊(r − 1)/2⌋ if r 6= 0. • Left child(r) = 2r + 1 if 2r + 1 < n. • Right child(r) = 2r + 2 if 2r + 2 < n. • Left sibling(r) = r − 1 if r is even. • Right sibling(r) = r + 1 if r is odd and r + 1 < n. 5.4 Binary Search Trees Section 4.4 presented the dictionary ADT, along with dictionary implementations based on sorted and unsorted lists. When implementing the dictionary with an unsorted list, inserting a new record into the dictionary can be performed quickly by putting it at the end of the list. However, searching an unsorted list for a particular record requires Θ(n) time in the average case. For a large database, this is probably much too slow. Alternatively, the records can be stored in a sorted list. If the list is implemented using a linked list, then no speedup to the search operation will result from storing the records in sorted order. On the other hand, if we use a sorted
- 191. 172 Chap. 5 Binary Trees 5 6 8 9 10 11 7 (a) 4 0 1 3 2 Position 0 1 2 3 4 5 6 7 8 9 10 11 Parent – 0 0 1 1 2 2 3 3 4 4 5 Left Child 1 3 5 7 9 11 – – – – – – Right Child 2 4 6 8 10 – – – – – – – Left Sibling – – 1 – 3 – 5 – 7 – 9 – Right Sibling – 2 – 4 – 6 – 8 – 10 – – (b) Figure 5.12 A complete binary tree and its array implementation. (a) The com- plete binary tree with twelve nodes. Each node has been labeled with its position in the tree. (b) The positions for the relatives of each node. A dash indicates that the relative does not exist. array-based list to implement the dictionary, then binary search can be used to find a record in only Θ(log n) time. However, insertion will now require Θ(n) time on average because, once the proper location for the new record in the sorted list has been found, many records might be shifted to make room for the new record. Is there some way to organize a collection of records so that inserting records and searching for records can both be done quickly? This section presents the binary search tree (BST), which allows an improved solution to this problem. A BST is a binary tree that conforms to the following condition, known as the Binary Search Tree Property: All nodes stored in the left subtree of a node whose key value is K have key values less than K. All nodes stored in the right subtree of a node whose key value is K have key values greater than or equal to K. Figure 5.13 shows two BSTs for a collection of values. One consequence of the Binary Search Tree Property is that if the BST nodes are printed using an inorder traversal (see Section 5.2), the resulting enumeration will be in sorted order from lowest to highest.
- 192. Sec. 5.4 Binary Search Trees 173 7 2 32 42 40 120 7 42 (a) 37 42 (b) 24 120 42 24 2 32 37 40 Figure 5.13 Two Binary Search Trees for a collection of values. Tree (a) results if values are inserted in the order 37, 24, 42, 7, 2, 40, 42, 32, 120. Tree (b) results if the same values are inserted in the order 120, 42, 42, 7, 2, 32, 37, 24, 40. Figure 5.14 shows a class declaration for the BST that implements the dictio- nary ADT. The public member functions include those required by the dictionary ADT, along with a constructor and destructor. To find a record with key value K in a BST, begin at the root. If the root stores a record with key value K, then the search is over. If not, then we must search deeper in the tree. What makes the BST efficient during search is that we need search only one of the node’s two subtrees. If K is less than the root node’s key value, we search only the left subtree. If K is greater than the root node’s key value, we search only the right subtree. This process continues until a record with key value K is found, or we reach a leaf node. If we reach a leaf node without encountering K, then no record exists in the BST whose key value is K. Example 5.5 Consider searching for the node with key value 32 in the tree of Figure 5.13(a). Because 32 is less than the root value of 37, the search proceeds to the left subtree. Because 32 is greater than 24, we search in 24’s right subtree. At this point the node containing 32 is found. If the search value were 35, the same path would be followed to the node containing 32. Because this node has no children, we know that 35 is not in the BST. Notice that in Figure 5.14, public member function find calls a private mem- ber function named findhelp. Method find takes the search key as an explicit parameter and its BST as an implicit parameter, along with space to place a copy of
- 193. 174 Chap. 5 Binary Trees import java.lang.Comparable; /** Binary Search Tree implementation for Dictionary ADT */ class BST<K extends Comparable<? super K>, E> implements Dictionary<K, E> { private BSTNode<K,E> root; // Root of the BST int nodecount; // Number of nodes in the BST /** Constructor */ BST() { root = null; nodecount = 0; } /** Reinitialize tree */ public void clear() { root = null; nodecount = 0; } /** Insert a record into the tree. @param k Key value of the record. @param e The record to insert. */ public void insert(K k, E e) { root = inserthelp(root, k, e); nodecount++; } /** Remove a record from the tree. @param k Key value of record to remove. @return Record removed, or null if there is none. */ public E remove(K k) { E temp = findhelp(root, k); // First find it if (temp != null) { root = removehelp(root, k); // Now remove it nodecount--; } return temp; } Figure 5.14 Class declaration for the binary search tree. the record if it is found. However, the find operation is most easily implemented as a recursive function whose parameters are the root of a BST subtree, the search key, and space for the element once it is found. Member findhelp is of the desired form for this recursive subroutine and is implemented as follows: private E findhelp(BSTNode<K,E> rt, K k) { if (rt == null) return null; if (rt.key().compareTo(k) > 0) return findhelp(rt.left(), k); else if (rt.key().compareTo(k) == 0) return rt.element(); else return findhelp(rt.right(), k); }
- 194. Sec. 5.4 Binary Search Trees 175 /** Remove and return the root node from the dictionary. @return The record removed, null if tree is empty. */ public E removeAny() { if (root != null) { E temp = root.element(); root = removehelp(root, root.key()); nodecount--; return temp; } else return null; } /** @return Record with key value k, null if none exist. @param k The key value to find. */ public E find(K k) { return findhelp(root, k); } /** @return The number of records in the dictionary. */ public int size() { return nodecount; } } Figure 5.14 (continued) Once the desired record is found, it is passed up the chain of recursive calls to findhelp in the third parameter, “it.” The return value for the function (true or false, depending on whether a suitable element has been found) is also simply passed back up the chain of recursive calls. Inserting a record with key value K requires that we first find where that record would have been if it were in the tree. This takes us to either a leaf node, or to an internal node with no child in the appropriate direction.3 Call this node R ′. We then add a new node containing the new record as a child of R ′. Figure 5.15 illustrates this operation. The value 35 is added as the right child of the node with value 32. Here is the implementation for inserthelp: private BSTNode<K,E> inserthelp(BSTNode<K,E> rt, K k, E e) { if (rt == null) return new BSTNode<K,E>(k, e); if (rt.key().compareTo(k) > 0) rt.setLeft(inserthelp(rt.left(), k, e)); else rt.setRight(inserthelp(rt.right(), k, e)); return rt; } 3 This assumes that no node has a key value equal to the one being inserted. If we find a node that duplicates the key value to be inserted, we have two options. If the application does not allow nodes with equal keys, then this insertion should be treated as an error (or ignored). If duplicate keys are allowed, our convention will be to insert the duplicate in the right subtree.
- 195. 176 Chap. 5 Binary Trees 37 24 2 32 35 42 40 42 120 7 Figure 5.15 An example of BST insertion. A record with value 35 is inserted into the BST of Figure 5.13(a). The node with value 32 becomes the parent of the new node containing 35. You should pay careful attention to the implementation for inserthelp. Note that inserthelp returns a pointer to a BSTNode. What is being returned is a subtree identical to the old subtree, except that it has been modified to contain the new record being inserted. Each node along a path from the root to the parent of the new node added to the tree will have its appropriate child pointer assigned to it. Except for the last node in the path, none of these nodes will actually change their child’s pointer value. In that sense, many of the assignments seem redundant. However, the cost of these additional assignments is worth paying to keep the inser- tion process simple. The alternative is to check if a given assignment is necessary, which is probably more expensive than the assignment! The shape of a BST depends on the order in which elements are inserted. A new element is added to the BST as a new leaf node, potentially increasing the depth of the tree. Figure 5.13 illustrates two BSTs for a collection of values. It is possible for the BST containing n nodes to be a chain of nodes with height n. This would happen if, for example, all elements were inserted in sorted order. In general, it is preferable for a BST to be as shallow as possible. This keeps the average cost of a BST operation low. Removing a node from a BST is a bit trickier than inserting a node, but it is not complicated if all of the possible cases are considered individually. Before tackling the general node removal process, let us first discuss how to remove from a given subtree the node with the smallest key value. This routine will be used later by the general node removal function. To remove the node with the minimum key value from a subtree, first find that node by continuously moving down the left link until there is no further left link to follow. Call this node S. To remove S, simply have the parent of S change its pointer to point to the right child of S. We know that S has no left child (because if S did have a left child, S would not be the node with
- 196. Sec. 5.4 Binary Search Trees 177 9 5 20 5 10 subroot Figure 5.16 An example of deleting the node with minimum value. In this tree, the node with minimum value, 5, is the left child of the root. Thus, the root’s left pointer is changed to point to 5’s right child. minimum key value). Thus, changing the pointer as described will maintain a BST, with S removed. The code for this method, named deletemin, is as follows: private BSTNode<K,E> deletemin(BSTNode<K,E> rt) { if (rt.left() == null) return rt.right(); else { rt.setLeft(deletemin(rt.left())); return rt; } } Example 5.6 Figure 5.16 illustrates the deletemin process. Beginning at the root node with value 10, deletemin follows the left link until there is no further left link, in this case reaching the node with value 5. The node with value 10 is changed to point to the right child of the node containing the minimum value. This is indicated in Figure 5.16 by a dashed line. A pointer to the node containing the minimum-valued element is stored in pa- rameter S. The return value of the deletemin method is the subtree of the cur- rent node with the minimum-valued node in the subtree removed. As with method inserthelp, each node on the path back to the root has its left child pointer reassigned to the subtree resulting from its call to the deletemin method. Removing a node with given key value R from the BST requires that we first find R and then remove it from the tree. So, the first part of the remove operation is a search to find R. Once R is found, there are several possibilities. If R has no children, then R’s parent has its pointer set to null. If R has one child, then R’s
- 197. 178 Chap. 5 Binary Trees 37 40 24 7 32 42 40 42 120 2 Figure 5.17 An example of removing the value 37 from the BST. The node containing this value has two children. We replace value 37 with the least value from the node’s right subtree, in this case 40. parent has its pointer set to R’s child (similar to deletemin). The problem comes if R has two children. One simple approach, though expensive, is to set R’s parent to point to one of R’s subtrees, and then reinsert the remaining subtree’s nodes one at a time. A better alternative is to find a value in one of the subtrees that can replace the value in R. Thus, the question becomes: Which value can substitute for the one being re- moved? It cannot be any arbitrary value, because we must preserve the BST prop- erty without making major changes to the structure of the tree. Which value is most like the one being removed? The answer is the least key value greater than (or equal to) the one being removed, or else the greatest key value less than the one being removed. If either of these values replace the one being removed, then the BST property is maintained. Example 5.7 Assume that we wish to remove the value 37 from the BST of Figure 5.13(a). Instead of removing the root node, we remove the node with the least value in the right subtree (using the deletemin operation). This value can then replace the value in the root. In this example we first remove the node with value 40, because it contains the least value in the right subtree. We then substitute 40 as the new value for the root node. Figure 5.17 illustrates this process. When duplicate node values do not appear in the tree, it makes no difference whether the replacement is the greatest value from the left subtree or the least value from the right subtree. If duplicates are stored, then we must select the replacement from the right subtree. To see why, call the greatest value in the left subtree G. If multiple nodes in the left subtree have value G, selecting G as the replacement value for the root of the subtree will result in a tree with equal values to the left of
- 198. Sec. 5.4 Binary Search Trees 179 /** Remove a node with key value k @return The tree with the node removed */ private BSTNode<K,E> removehelp(BSTNode<K,E> rt, K k) { if (rt == null) return null; if (rt.key().compareTo(k) > 0) rt.setLeft(removehelp(rt.left(), k)); else if (rt.key().compareTo(k) < 0) rt.setRight(removehelp(rt.right(), k)); else { // Found it if (rt.left() == null) return rt.right(); else if (rt.right() == null) return rt.left(); else { // Two children BSTNode<K,E> temp = getmin(rt.right()); rt.setElement(temp.element()); rt.setKey(temp.key()); rt.setRight(deletemin(rt.right())); } } return rt; } Figure 5.18 Implementation for the BST removehelp method. the node now containing G. Precisely this situation occurs if we replace value 120 with the greatest value in the left subtree of Figure 5.13(b). Selecting the least value from the right subtree does not have a similar problem, because it does not violate the Binary Search Tree Property if equal values appear in the right subtree. From the above, we see that if we want to remove the record stored in a node with two children, then we simply call deletemin on the node’s right subtree and substitute the record returned for the record being removed. Figure 5.18 shows is the code for removehelp. The cost for findhelp and inserthelp is the depth of the node found or inserted. The cost for removehelp is the depth of the node being removed, or in the case when this node has two children, the depth of the node with smallest value in its right subtree. Thus, in the worst case, the cost of any one of these operations is the depth of the deepest node in the tree. This is why it is desirable to keep BSTs balanced, that is, with least possible height. If a binary tree is bal- anced, then the height for a tree of n nodes is approximately log n. However, if the tree is completely unbalanced, for example in the shape of a linked list, then the height for a tree of n nodes can be as great as n. Thus, a balanced BST will in the average case have operations costing Θ(log n), while a badly unbalanced BST can have operations in the worst case costing Θ(n). Consider the situation where we
- 199. 180 Chap. 5 Binary Trees construct a BST of n nodes by inserting the nodes one at a time. If we are fortunate to have them arrive in an order that results in a balanced tree (a “random” order is likely to be good enough for this purpose), then each insertion will cost on average Θ(log n), for a total cost of Θ(n log n). However, if the nodes are inserted in order of increasing value, then the resulting tree will be a chain of height n. The cost of insertion in this case will be Pn i=1 i, which is Θ(n2). Traversing a BST costs Θ(n) regardless of the shape of the tree. Each node is visited exactly once, and each child pointer is followed exactly once. Below is an example traversal, named printhelp. It performs an inorder traversal on the BST to print the node values in ascending order. private void printhelp(BSTNode<K,E> rt) { if (rt == null) return; printhelp(rt.left()); printVisit(rt.element()); printhelp(rt.right()); } While the BST is simple to implement and efficient when the tree is balanced, the possibility of its being unbalanced is a serious liability. There are techniques for organizing a BST to guarantee good performance. Two examples are the AVL tree and the splay tree of Section 13.2. Other search trees are guaranteed to remain balanced, such as the 2-3 tree of Section 10.4. 5.5 Heaps and Priority Queues There are many situations, both in real life and in computing applications, where we wish to choose the next “most important” from a collection of people, tasks, or objects. For example, doctors in a hospital emergency room often choose to see next the “most critical” patient rather than the one who arrived first. When scheduling programs for execution in a multitasking operating system, at any given moment there might be several programs (usually called jobs) ready to run. The next job selected is the one with the highest priority. Priority is indicated by a particular value associated with the job (and might change while the job remains in the wait list). When a collection of objects is organized by importance or priority, we call this a priority queue. A normal queue data structure will not implement a prior- ity queue efficiently because search for the element with highest priority will take Θ(n) time. A list, whether sorted or not, will also require Θ(n) time for either in- sertion or removal. A BST that organizes records by priority could be used, with the total of n inserts and n remove operations requiring Θ(n log n) time in the average
- 200. Sec. 5.5 Heaps and Priority Queues 181 case. However, there is always the possibility that the BST will become unbal- anced, leading to bad performance. Instead, we would like to find a data structure that is guaranteed to have good performance for this special application. This section presents the heap4 data structure. A heap is defined by two prop- erties. First, it is a complete binary tree, so heaps are nearly always implemented using the array representation for complete binary trees presented in Section 5.3.3. Second, the values stored in a heap are partially ordered. This means that there is a relationship between the value stored at any node and the values of its children. There are two variants of the heap, depending on the definition of this relationship. A max-heap has the property that every node stores a value that is greater than or equal to the value of either of its children. Because the root has a value greater than or equal to its children, which in turn have values greater than or equal to their children, the root stores the maximum of all values in the tree. A min-heap has the property that every node stores a value that is less than or equal to that of its children. Because the root has a value less than or equal to its children, which in turn have values less than or equal to their children, the root stores the minimum of all values in the tree. Note that there is no necessary relationship between the value of a node and that of its sibling in either the min-heap or the max-heap. For example, it is possible that the values for all nodes in the left subtree of the root are greater than the values for every node of the right subtree. We can contrast BSTs and heaps by the strength of their ordering relationships. A BST defines a total order on its nodes in that, given the positions for any two nodes in the tree, the one to the “left” (equivalently, the one appearing earlier in an inorder traversal) has a smaller key value than the one to the “right.” In contrast, a heap implements a partial order. Given their positions, we can determine the relative order for the key values of two nodes in the heap only if one is a descendent of the other. Min-heaps and max-heaps both have their uses. For example, the Heapsort of Section 7.6 uses the max-heap, while the Replacement Selection algorithm of Section 8.5.2 uses a min-heap. The examples in the rest of this section will use a max-heap. Be careful not to confuse the logical representation of a heap with its physical implementation by means of the array-based complete binary tree. The two are not synonymous because the logical view of the heap is actually a tree structure, while the typical physical implementation uses an array. Figure 5.19 shows an implementation for heaps. The class is a generic with one type parameter. E defines the type for the data elements stored in the heap. 4 The term “heap” is also sometimes used to refer to a memory pool. See Section 12.3.
- 201. 182 Chap. 5 Binary Trees import java.lang.Comparable; /** Max-heap implementation */ public class MaxHeap<E extends Comparable<? super E>> { private E[] Heap; // Pointer to the heap array private int size; // Maximum size of the heap private int n; // Number of things in heap public MaxHeap(E[] h, int num, int max) { Heap = h; n = num; size = max; buildheap(); } /** Return current size of the heap */ public int heapsize() { return n; } /** Is pos a leaf position? */ public boolean isLeaf(int pos) { return (pos >= n/2) && (pos < n); } /** Return position for left child of pos */ public int leftchild(int pos) { assert pos < n/2 : "Position has no left child"; return 2*pos + 1; } /** Return position for right child of pos */ public int rightchild(int pos) { assert pos < (n-1)/2 : "Position has no right child"; return 2*pos + 2; } /** Return position for parent */ public int parent(int pos) { assert pos > 0 : "Position has no parent"; return (pos-1)/2; } /** Heapify contents of Heap */ public void buildheap() { for (int i=n/2-1; i>=0; i--) siftdown(i); } /** Insert into heap */ public void insert(E val) { assert n < size : "Heap is full"; int curr = n++; Heap[curr] = val; // Start at end of heap // Now sift up until curr’s parent’s key > curr’s key while ((curr != 0) && (Heap[curr].compareTo(Heap[parent(curr)]) > 0)) { DSutil.swap(Heap, curr, parent(curr)); curr = parent(curr); } } Figure 5.19 Max-heap implementation.
- 202. Sec. 5.5 Heaps and Priority Queues 183 /** Put element in its correct place */ private void siftdown(int pos) { assert (pos >= 0) && (pos < n) : "Illegal heap position"; while (!isLeaf(pos)) { int j = leftchild(pos); if ((j<(n-1)) && (Heap[j].compareTo(Heap[j+1]) < 0)) j++; // j is now index of child with greater value if (Heap[pos].compareTo(Heap[j]) >= 0) return; DSutil.swap(Heap, pos, j); pos = j; // Move down } } public E removemax() { // Remove maximum value assert n > 0 : "Removing from empty heap"; DSutil.swap(Heap, 0, --n); // Swap maximum with last value if (n != 0) // Not on last element siftdown(0); // Put new heap root val in correct place return Heap[n]; } /** Remove element at specified position */ public E remove(int pos) { assert (pos >= 0) && (pos < n) : "Illegal heap position"; DSutil.swap(Heap, pos, --n); // Swap with last value while (Heap[pos].compareTo(Heap[parent(pos)]) > 0) { DSutil.swap(Heap, pos, parent(pos)); pos = parent(pos); } if (n != 0) siftdown(pos); // Push down return Heap[n]; } } Figure 5.19 (continued) This class definition makes two concessions to the fact that an array-based im- plementation is used. First, heap nodes are indicated by their logical position within the heap rather than by a pointer to the node. In practice, the logical heap position corresponds to the identically numbered physical position in the array. Second, the constructor takes as input a pointer to the array to be used. This approach provides the greatest flexibility for using the heap because all data values can be loaded into the array directly by the client. The advantage of this comes during the heap con- struction phase, as explained below. The constructor also takes an integer parame- ter indicating the initial size of the heap (based on the number of elements initially
- 203. 184 Chap. 5 Binary Trees loaded into the array) and a second integer parameter indicating the maximum size allowed for the heap (the size of the array). Method heapsize returns the current size of the heap. H.isLeaf(pos) will return true if position pos is a leaf in heap H, and false otherwise. Mem- bers leftchild, rightchild, and parent return the position (actually, the array index) for the left child, right child, and parent of the position passed, respec- tively. One way to build a heap is to insert the elements one at a time. Method insert will insert a new element V into the heap. You might expect the heap insertion pro- cess to be similar to the insert function for a BST, starting at the root and working down through the heap. However, this approach is not likely to work because the heap must maintain the shape of a complete binary tree. Equivalently, if the heap takes up the first n positions of its array prior to the call to insert, it must take up the first n + 1 positions after. To accomplish this, insert first places V at po- sition n of the array. Of course, V is unlikely to be in the correct position. To move V to the right place, it is compared to its parent’s value. If the value of V is less than or equal to the value of its parent, then it is in the correct place and the insert routine is finished. If the value of V is greater than that of its parent, then the two elements swap positions. From here, the process of comparing V to its (current) parent continues until V reaches its correct position. Each call to insert takes Θ(log n) time in the worst case, because the value being inserted can move at most the distance from the bottom of the tree to the top of the tree. Thus, the time to insert n values into the heap, if we insert them one at a time, will be Θ(n log n) in the worst case. If all n values are available at the beginning of the building process, we can build the heap faster than just inserting the values into the heap one by one. Con- sider Figure 5.20(a), which shows one series of exchanges that will result in a heap. Note that this figure shows the input in its logical form as a complete binary tree, but you should realize that these values are physically stored in an array. All ex- changes are between a node and one of its children. The heap is formed as a result of this exchange process. The array for the right-hand tree of Figure 5.20(a) would appear as follows: 7 4 6 1 2 3 5 Figure 5.20(b) shows an alternate series of exchanges that also forms a heap, but much more efficiently. From this example, it is clear that the heap for any given set of numbers is not unique, and we see that some rearrangements of the input values require fewer exchanges than others to build the heap. So, how do we pick the best rearrangement?
- 204. Sec. 5.5 Heaps and Priority Queues 185 (a) 6 (b) 4 5 6 7 5 7 4 2 3 2 2 6 6 3 5 1 3 7 5 4 2 1 3 7 4 1 1 Figure 5.20 Two series of exchanges to build a max-heap. (a) This heap is built by a series of nine exchanges in the order (4-2), (4-1), (2-1), (5-2), (5-4), (6-3), (6-5), (7-5), (7-6). (b) This heap is built by a series of four exchanges in the order (5-2), (7-3), (7-1), (6-1). One good algorithm stems from induction. Suppose that the left and right sub- trees of the root are already heaps, and R is the name of the element at the root. This situation is illustrated by Figure 5.21. In this case there are two possibilities. (1) R has a value greater than or equal to its two children. In this case, construction is complete. (2) R has a value less than one or both of its children. In this case, R should be exchanged with the child that has greater value. The result will be a heap, except that R might still be less than one or both of its (new) children. In this case, we simply continue the process of “pushing down” R until it reaches a level where it is greater than its children, or is a leaf node. This process is implemented by the private method siftdown of the heap class. The siftdown operation is illustrated by Figure 5.22. This approach assumes that the subtrees are already heaps, suggesting that a complete algorithm can be obtained by visiting the nodes in some order such that the children of a node are visited before the node itself. One simple way to do this is simply to work from the high index of the array to the low index. Actually, the build process need not visit the leaf nodes (they can never move down because they are already at the bottom), so the building algorithm can start in the middle of the
- 205. 186 Chap. 5 Binary Trees R H1 H2 Figure 5.21 Final stage in the heap-building algorithm. Both subtrees of node R are heaps. All that remains is to push R down to its proper level in the heap. (a) (b) (c) 5 1 7 7 5 1 7 5 6 4 2 4 3 6 2 6 3 4 2 1 3 Figure 5.22 The siftdown operation. The subtrees of the root are assumed to be heaps. (a) The partially completed heap. (b) Values 1 and 7 are swapped. (c) Values 1 and 6 are swapped to form the final heap. array, with the first internal node. The exchanges shown in Figure 5.20(b) result from this process. Method buildHeap implements the building algorithm. What is the cost of buildHeap? Clearly it is the sum of the costs for the calls to siftdown. Each siftdown operation can cost at most the number of levels it takes for the node being sifted to reach the bottom of the tree. In any complete tree, approximately half of the nodes are leaves and so cannot be moved downward at all. One quarter of the nodes are one level above the leaves, and so their elements can move down at most one level. At each step up the tree we get half the number of nodes as were at the previous level, and an additional height of one. The maximum sum of total distances that elements can go is therefore log n X i=1 (i − 1) n 2i = n 2 log n X i=1 i − 1 2i−1 . From Equation 2.9 we know that this summation has a closed-form solution of approximately 2, so this algorithm takes Θ(n) time in the worst case. This is far better than building the heap one element at a time, which would cost Θ(n log n) in the worst case. It is also faster than the Θ(n log n) average-case time and Θ(n2) worst-case time required to build the BST.
- 206. Sec. 5.6 Huffman Coding Trees 187 Removing the maximum (root) value from a heap containing n elements re- quires that we maintain the complete binary tree shape, and that the remaining n − 1 node values conform to the heap property. We can maintain the proper shape by moving the element in the last position in the heap (the current last element in the array) to the root position. We now consider the heap to be one element smaller. Unfortunately, the new root value is probably not the maximum value in the new heap. This problem is easily solved by using siftdown to reorder the heap. Be- cause the heap is log n levels deep, the cost of deleting the maximum element is Θ(log n) in the average and worst cases. The heap is a natural implementation for the priority queues discussed at the beginning of this section. Jobs can be added to the heap (using their priority value as the ordering key) when needed. Method removemax can be called whenever a new job is to be executed. Some applications of priority queues require the ability to change the priority of an object already stored in the queue. This might require that the object’s position in the heap representation be updated. Unfortunately, a max-heap is not efficient when searching for an arbitrary value; it is only good for finding the maximum value. However, if we already know the index for an object within the heap, it is a simple matter to update its priority (including changing its position to maintain the heap property) or remove it. The remove method takes as input the position of the node to be removed from the heap. A typical implementation for priority queues requiring updating of priorities will need to use an auxiliary data structure that supports efficient search for objects (such as a BST). Records in the auxiliary data structure will store the object’s heap index, so that the object can be deleted from the heap and reinserted with its new priority (see Project 5.5). Sections 11.4.1 and 11.5.1 present applications for a priority queue with priority updating. 5.6 Huffman Coding Trees The space/time tradeoff principle from Section 3.9 suggests that one can often gain an improvement in space requirements in exchange for a penalty in running time. There are many situations where this is a desirable tradeoff. A typical example is storing files on disk. If the files are not actively used, the owner might wish to compress them to save space. Later, they can be uncompressed for use, which costs some time, but only once. We often represent a set of items in a computer program by assigning a unique code to each item. For example, the standard ASCII coding scheme assigns a unique eight-bit value to each character. It takes a certain minimum number of bits to provide unique codes for each character. For example, it takes ⌈log 128⌉ or
- 207. 188 Chap. 5 Binary Trees Letter Frequency Letter Frequency A 77 N 67 B 17 O 67 C 32 P 20 D 42 Q 5 E 120 R 59 F 24 S 67 G 17 T 85 H 50 U 37 I 76 V 12 J 4 W 22 K 7 X 4 L 42 Y 22 M 24 Z 2 Figure 5.23 Relative frequencies for the 26 letters of the alphabet as they ap- pear in a selected set of English documents. “Frequency” represents the expected frequency of occurrence per 1000 letters, ignoring case. seven bits to provide the 128 unique codes needed to represent the 128 symbols of the ASCII character set.5 The requirement for ⌈log n⌉ bits to represent n unique code values assumes that all codes will be the same length, as are ASCII codes. This is called a fixed-length coding scheme. If all characters were used equally often, then a fixed-length coding scheme is the most space efficient method. However, you are probably aware that not all characters are used equally often. Figure 5.23 shows the relative frequencies of the letters of the alphabet. From this table we can see that the letter ‘E’ appears about 60 times more often than the letter ‘Z.’ In normal ASCII, the words “DEED” and “MUCK” require the same amount of space (four bytes). It would seem that words such as “DEED,” which are composed of relatively common letters, should be storable in less space than words such as “MUCK,” which are composed of relatively uncommon letters. If some characters are used more frequently than others, is it possible to take advantage of this fact and somehow assign them shorter codes? The price could be that other characters require longer codes, but this might be worthwhile if such characters appear rarely enough. This concept is at the heart of file compression 5 The ASCII standard is eight bits, not seven, even though there are only 128 characters repre- sented. The eighth bit is used either to check for transmission errors, or to support extended ASCII codes with an additional 128 characters.
- 208. Sec. 5.6 Huffman Coding Trees 189 Letter C D E K L M U Z Frequency 32 42 120 7 42 24 37 2 Figure 5.24 The relative frequencies for eight selected letters. techniques in common use today. The next section presents one such approach to assigning variable-length codes, called Huffman coding. While it is not commonly used in its simplest form for file compression (there are better methods), Huffman coding gives the flavor of such coding schemes. 5.6.1 Building Huffman Coding Trees Huffman coding assigns codes to characters such that the length of the code de- pends on the relative frequency or weight of the corresponding character. Thus, it is a variable-length code. If the estimated frequencies for letters match the actual frequency found in an encoded message, then the length of that message will typi- cally be less than if a fixed-length code had been used. The Huffman code for each letter is derived from a full binary tree called the Huffman coding tree, or simply the Huffman tree. Each leaf of the Huffman tree corresponds to a letter, and we define the weight of the leaf node to be the weight (frequency) of its associated letter. The goal is to build a tree with the minimum external path weight. Define the weighted path length of a leaf to be its weight times its depth. The binary tree with minimum external path weight is the one with the minimum sum of weighted path lengths for the given set of leaves. A letter with high weight should have low depth, so that it will count the least against the total path length. As a result, another letter might be pushed deeper in the tree if it has less weight. The process of building the Huffman tree for n letters is quite simple. First, create a collection of n initial Huffman trees, each of which is a single leaf node containing one of the letters. Put the n partial trees onto a min-heap (a priority queue) organized by weight (frequency). Next, remove the first two trees (the ones with lowest weight) from the heap. Join these two trees together to create a new tree whose root has the two trees as children, and whose weight is the sum of the weights of the two trees. Put this new tree back on the heap. This process is repeated until all of the partial Huffman trees have been combined into one. Example 5.8 Figure 5.25 illustrates part of the Huffman tree construction process for the eight letters of Figure 5.24. Ranking D and L arbitrarily by alphabetical order, the letters are ordered by frequency as
- 209. 190 Chap. 5 Binary Trees Step 1: Step 2: 9 Step 3: Step 4: 65 Step 5: 42 32 C 65 33 9 E 79 L 24 L 120 37 42 C 42 32 24 U D E 2 7 K Z M 9 9 24 37 U 42 D 42 M 32 120 C L E M C U D 2 Z 7 24 32 37 42 42 L K 120 E 2 7 K M C 32 37 42 42 24 L Z D 120 E U 120 2 Z 7 K 37 42 D U 2 Z 7 33 33 M K Figure 5.25 The first five steps of the building process for a sample Huffman tree.
- 210. Sec. 5.6 Huffman Coding Trees 191 306 0 1 E 0 79 0 1 37 U 42 1 107 0 42 1 65 0 C 1 0 1 9 0 1 2 Z 7 D L M K 32 33 24 120 186 Figure 5.26 A Huffman tree for the letters of Figure 5.24. Letter Z K M C U D L E Frequency 2 7 24 32 37 42 42 120 Because the first two letters on the list are Z and K, they are selected to be the first trees joined together. They become the children of a root node with weight 9. Thus, a tree whose root has weight 9 is placed back on the list, where it takes up the first position. The next step is to take values 9 and 24 off the heap (corresponding to the partial tree with two leaf nodes built in the last step, and the partial tree storing the letter M, respectively) and join them together. The resulting root node has weight 33, and so this tree is placed back into the heap. Its priority will be between the trees with values 32 (for letter C) and 37 (for letter U). This process continues until a tree whose root has weight 306 is built. This tree is shown in Figure 5.26. Figure 5.27 shows an implementation for Huffman tree nodes. This implemen- tation is similar to the VarBinNode implementation of Figure 5.10. There is an abstract base class, named HuffNode, and two subclasses, named LeafNode and IntlNode. This implementation reflects the fact that leaf and internal nodes contain distinctly different information. Figure 5.28 shows the implementation for the Huffman tree. Nodes of the tree store key-value pairs (see Figure 4.30), where the key is the weight and the value is the character. Figure 5.29 shows the Java code for the tree-building process.
- 211. 192 Chap. 5 Binary Trees /** Binary tree node implementation with just an element field (no key) and pointers to children */ class HuffNode<E> implements BinNode<E> { private E element; // Element for this node private HuffNode<E> left; // Pointer to left child private HuffNode<E> right; // Pointer to right child /** Constructors */ public HuffNode() {left = right = null; } public HuffNode(E val) { left = right = null; element = val; } public HuffNode(E val, HuffNode<E> l, HuffNode<E> r) { left = l; right = r; element = val; } /** Return and set the element value */ public E element() { return element; } public E setElement(E v) { return element = v; } /** Return and set the left child */ public HuffNode<E> left() { return left; } public HuffNode<E> setLeft(HuffNode<E> p) { return left = p; } /** Return and set the right child */ public HuffNode<E> right() { return right; } public HuffNode<E> setRight(HuffNode<E> p) { return right = p; } /** Return true if this is a leaf node */ public boolean isLeaf() { return (left == null) && (right == null); } } Figure 5.27 Implementation for Huffman tree nodes. Internal nodes and leaf nodes are represented by separate classes, each derived from an abstract base class. Huffman tree building is an example of a greedy algorithm. At each step, the algorithm makes a “greedy” decision to merge the two subtrees with least weight. This makes the algorithm simple, but does it give the desired result? This sec- tion concludes with a proof that the Huffman tree indeed gives the most efficient arrangement for the set of letters. The proof requires the following lemma. Lemma 5.1 For any Huffman tree built by function buildHuff containing at least two letters, the two letters with least frequency are stored in siblings nodes whose depth is at least as deep as any other leaf nodes in the tree.
- 212. Sec. 5.6 Huffman Coding Trees 193 class HuffTree { // A Huffman coding tree private HuffNode<LettFreq> root; // Root of the tree public HuffTree(LettFreq val) { root = new HuffNode<LettFreq>(val); } public HuffTree(LettFreq val, HuffTree l, HuffTree r) { root = new HuffNode<LettFreq>(val, l.root(), r.root()); } public HuffNode<LettFreq> root() { return root; } public int weight() // Weight of tree is weight of root { return root.element().weight(); } } Figure 5.28 Class declarations for the Huffman tree. Proof: Call the two letters with least frequency l1 and l2. They must be siblings because buildHuff selects them in the first step of the construction process. Assume that l1 and l2 are not the deepest nodes in the tree. In this case, the Huffman tree must either look as shown in Figure 5.30, or in some sense be symmetrical to this. For this situation to occur, the parent of l1 and l2, labeled V, must have greater weight than the node labeled X. Otherwise, function buildHuff would have selected node V in place of node X as the child of node U. However, this is impossible because l1 and l2 are the letters with least frequency. ✷ Theorem 5.3 Function buildHuff builds the Huffman tree with the minimum external path weight for the given set of letters. Proof: The proof is by induction on n, the number of letters. • Base Case: For n = 2, the Huffman tree must have the minimum external path weight because there are only two possible trees, each with identical weighted path lengths for the two leaves. • Induction Hypothesis: Assume that any tree created by buildHuff that contains n − 1 leaves has minimum external path length. • Induction Step: Given a Huffman tree T built by buildHuff with n leaves, n ≥ 2, suppose that w1 ≤ w2 ≤ · · · ≤ wn where w1 to wn are the weights of the letters. Call V the parent of the letters with frequencies w1 and w2. From the lemma, we know that the leaf nodes containing the letters with frequencies w1 and w2 are as deep as any nodes in T. If any other leaf nodes in the tree were deeper, we could reduce their weighted path length by swapping them with w1 or w2. But the lemma tells us that no such deeper nodes exist. Call T′ the Huffman tree that is identical to T except that node
- 213. 194 Chap. 5 Binary Trees // Build a Huffman tree from list hufflist static HuffTree buildTree(List<HuffTree> hufflist) { HuffTree tmp1, tmp2, tmp3; LettFreq tmpnode; for(hufflist.moveToPos(1); hufflist.length() > 1; hufflist.moveToPos(1)) { // While at least two items left hufflist.moveToStart(); tmp1 = hufflist.remove(); tmp2 = hufflist.remove(); tmpnode = new LettFreq(tmp1.weight() + tmp2.weight()); tmp3 = new HuffTree(tmpnode, tmp1, tmp2); // return to the list in sorted order for (hufflist.moveToStart(); hufflist.currPos() < hufflist.length(); hufflist.next()) if (tmp3.weight() <= hufflist.getValue().weight()) { hufflist.insert(tmp3); break; } // Put in list if (hufflist.currPos() >= hufflist.length()) hufflist.append(tmp3); // This is heaviest value } hufflist.moveToStart(); // This is only tree on list return hufflist.remove(); // Return the tree } Figure 5.29 Implementation for the Huffman tree construction function. buildHuff takes as input fl, the min-heap of partial Huffman trees, which initially are single leaf nodes as shown in Step 1 of Figure 5.25. The body of function buildTree consists mainly of a for loop. On each iteration of the for loop, the first two partial trees are taken off the heap and placed in variables temp1 and temp2. A tree is created (temp3) such that the left and right subtrees are temp1 and temp2, respectively. Finally, temp3 is returned to fl. l1 X V l2 U Figure 5.30 An impossible Huffman tree, showing the situation where the two nodes with least weight, l1 and l2, are not the deepest nodes in the tree. Triangles represent subtrees.
- 214. Sec. 5.6 Huffman Coding Trees 195 Letter Freq Code Bits C 32 1110 4 D 42 101 3 E 120 0 1 K 7 111101 6 L 42 110 3 M 24 11111 5 U 37 100 3 Z 2 111100 6 Figure 5.31 The Huffman codes for the letters of Figure 5.24. V is replaced with a leaf node V ′ whose weight is w1 + w2. By the induction hypothesis, T′ has minimum external path length. Returning the children to V ′ restores tree T, which must also have minimum external path length. Thus by mathematical induction, function buildHuff creates the Huffman tree with minimum external path length. ✷ 5.6.2 Assigning and Using Huffman Codes Once the Huffman tree has been constructed, it is an easy matter to assign codes to individual letters. Beginning at the root, we assign either a ‘0’ or a ‘1’ to each edge in the tree. ‘0’ is assigned to edges connecting a node with its left child, and ‘1’ to edges connecting a node with its right child. This process is illustrated by Figure 5.26. The Huffman code for a letter is simply a binary number determined by the path from the root to the leaf corresponding to that letter. Thus, the code for E is ‘0’ because the path from the root to the leaf node for E takes a single left branch. The code for K is ‘111101’ because the path to the node for K takes four right branches, then a left, and finally one last right. Figure 5.31 lists the codes for all eight letters. Given codes for the letters, it is a simple matter to use these codes to encode a text message. We simply replace each letter in the string with its binary code. A lookup table can be used for this purpose. Example 5.9 Using the code generated by our example Huffman tree, the word “DEED” is represented by the bit string “10100101” and the word “MUCK” is represented by the bit string “111111001110111101.” Decoding the message is done by looking at the bits in the coded string from left to right until a letter is decoded. This can be done by using the Huffman tree in
- 215. 196 Chap. 5 Binary Trees a reverse process from that used to generate the codes. Decoding a bit string begins at the root of the tree. We take branches depending on the bit value — left for ‘0’ and right for ‘1’ — until reaching a leaf node. This leaf contains the first character in the message. We then process the next bit in the code restarting at the root to begin the next character. Example 5.10 To decode the bit string “1011001110111101” we begin at the root of the tree and take a right branch for the first bit which is ‘1.’ Because the next bit is a ‘0’ we take a left branch. We then take another right branch (for the third bit ‘1’), arriving at the leaf node corresponding to the letter D. Thus, the first letter of the coded word is D. We then begin again at the root of the tree to process the fourth bit, which is a ‘1.’ Taking a right branch, then two left branches (for the next two bits which are ‘0’), we reach the leaf node corresponding to the letter U. Thus, the second letter is U. In similar manner we complete the decoding process to find that the last two letters are C and K, spelling the word “DUCK.” A set of codes is said to meet the prefix property if no code in the set is the prefix of another. The prefix property guarantees that there will be no ambiguity in how a bit string is decoded. In other words, once we reach the last bit of a code during the decoding process, we know which letter it is the code for. Huffman codes certainly have the prefix property because any prefix for a code would correspond to an internal node, while all codes correspond to leaf nodes. For example, the code for M is ‘11111.’ Taking five right branches in the Huffman tree of Figure 5.26 brings us to the leaf node containing M. We can be sure that no letter can have code ‘111’ because this corresponds to an internal node of the tree, and the tree-building process places letters only at the leaf nodes. How efficient is Huffman coding? In theory, it is an optimal coding method whenever the true frequencies are known, and the frequency of a letter is indepen- dent of the context of that letter in the message. In practice, the frequencies of letters do change depending on context. For example, while E is the most com- monly used letter of the alphabet in English documents, T is more common as the first letter of a word. This is why most commercial compression utilities do not use Huffman coding as their primary coding method, but instead use techniques that take advantage of the context for the letters. Another factor that affects the compression efficiency of Huffman coding is the relative frequencies of the letters. Some frequency patterns will save no space as compared to fixed-length codes; others can result in great compression. In general,
- 216. Sec. 5.6 Huffman Coding Trees 197 Huffman coding does better when there is large variation in the frequencies of letters. In the particular case of the frequencies shown in Figure 5.31, we can determine the expected savings from Huffman coding if the actual frequencies of a coded message match the expected frequencies. Example 5.11 Because the sum of the frequencies in Figure 5.31 is 306 and E has frequency 120, we expect it to appear 120 times in a message containing 306 letters. An actual message might or might not meet this expectation. Letters D, L, and U have code lengths of three, and together are expected to appear 121 times in 306 letters. Letter C has a code length of four, and is expected to appear 32 times in 306 letters. Letter M has a code length of five, and is expected to appear 24 times in 306 letters. Finally, letters K and Z have code lengths of six, and together are expected to appear only 9 times in 306 letters. The average expected cost per character is simply the sum of the cost for each character (ci) times the probability of its occurring (pi), or c1p1 + c2p2 + · · · + cnpn. This can be reorganized as c1f1 + c2f2 + · · · + cnfn fT where fi is the (relative) frequency of letter i and fT is the total for all letter frequencies. For this set of frequencies, the expected cost per letter is [(1×120)+(3×121)+(4×32)+(5×24)+(6×9)]/306 = 785/306 ≈ 2.57 A fixed-length code for these eight characters would require log 8 = 3 bits per letter as opposed to about 2.57 bits per letter for Huffman coding. Thus, Huffman coding is expected to save about 14% for this set of letters. Huffman coding for all ASCII symbols should do better than this. The letters of Figure 5.31 are atypical in that there are too many common letters compared to the number of rare letters. Huffman coding for all 26 letters would yield an expected cost of 4.29 bits per letter. The equivalent fixed-length code would require about five bits. This is somewhat unfair to fixed-length coding because there is actually room for 32 codes in five bits, but only 26 letters. More generally, Huffman coding of a typical text file will save around 40% over ASCII coding if we charge ASCII coding at eight bits per character. Huffman coding for a binary file (such as a
- 217. 198 Chap. 5 Binary Trees compiled executable) would have a very different set of distribution frequencies and so would have a different space savings. Most commercial compression programs use two or three coding schemes to adjust to different types of files. In the preceding example, “DEED” was coded in 8 bits, a saving of 33% over the twelve bits required from a fixed-length coding. However, “MUCK” requires 18 bits, more space than required by the corresponding fixed-length coding. The problem is that “MUCK” is composed of letters that are not expected to occur often. If the message does not match the expected frequencies of the letters, than the length of the encoding will not be as expected either. 5.7 Further Reading See Shaffer and Brown [SB93] for an example of a tree implementation where an internal node pointer field stores the value of its child instead of a pointer to its child when the child is a leaf node. Many techniques exist for maintaining reasonably balanced BSTs in the face of an unfriendly series of insert and delete operations. One example is the AVL tree of Adelson-Velskii and Landis, which is discussed by Knuth [Knu98]. The AVL tree (see Section 13.2) is actually a BST whose insert and delete routines reorganize the tree structure so as to guarantee that the subtrees rooted by the children of any node will differ in height by at most one. Another example is the splay tree [ST85], also discussed in Section 13.2. See Bentley’s Programming Pearl “Thanks, Heaps” [Ben85, Ben88] for a good discussion on the heap data structure and its uses. The proof of Section 5.6.1 that the Huffman coding tree has minimum external path weight is from Knuth [Knu97]. For more information on data compression techniques, see Managing Gigabytes by Witten, Moffat, and Bell [WMB99], and Codes and Cryptography by Dominic Welsh [Wel88]. Tables 5.23 and 5.24 are derived from Welsh [Wel88]. 5.8 Exercises 5.1 Section 5.1.1 claims that a full binary tree has the highest number of leaf nodes among all trees with n internal nodes. Prove that this is true. 5.2 Define the degree of a node as the number of its non-empty children. Prove by induction that the number of degree 2 nodes in any binary tree is one less than the number of leaves. 5.3 Define the internal path length for a tree as the sum of the depths of all internal nodes, while the external path length is the sum of the depths of all
- 218. Sec. 5.8 Exercises 199 leaf nodes in the tree. Prove by induction that if tree T is a full binary tree with n internal nodes, I is T’s internal path length, and E is T’s external path length, then E = I + 2n for n ≥ 0. 5.4 Explain why function preorder2 from Section 5.2 makes half as many recursive calls as function preorder. Explain why it makes twice as many accesses to left and right children. 5.5 (a) Modify the preorder traversal of Section 5.2 to perform an inorder traversal of a binary tree. (b) Modify the preorder traversal of Section 5.2 to perform a postorder traversal of a binary tree. 5.6 Write a recursive function named search that takes as input the pointer to the root of a binary tree (not a BST!) and a value K, and returns true if value K appears in the tree and false otherwise. 5.7 Write an algorithm that takes as input the pointer to the root of a binary tree and prints the node values of the tree in level order. Level order first prints the root, then all nodes of level 1, then all nodes of level 2, and so on. Hint: Preorder traversals make use of a stack through recursive calls. Consider making use of another data structure to help implement the level- order traversal. 5.8 Write a recursive function that returns the height of a binary tree. 5.9 Write a recursive function that returns a count of the number of leaf nodes in a binary tree. 5.10 Assume that a given BST stores integer values in its nodes. Write a recursive function that sums the values of all nodes in the tree. 5.11 Assume that a given BST stores integer values in its nodes. Write a recursive function that traverses a binary tree, and prints the value of every node who’s grandparent has a value that is a multiple of five. 5.12 Write a recursive function that traverses a binary tree, and prints the value of every node which has at least four great-grandchildren. 5.13 Compute the overhead fraction for each of the following full binary tree im- plementations. (a) All nodes store data, two child pointers, and a parent pointer. The data field requires four bytes and each pointer requires four bytes. (b) All nodes store data and two child pointers. The data field requires sixteen bytes and each pointer requires four bytes. (c) All nodes store data and a parent pointer, and internal nodes store two child pointers. The data field requires eight bytes and each pointer re- quires four bytes.
- 219. 200 Chap. 5 Binary Trees (d) Only leaf nodes store data; internal nodes store two child pointers. The data field requires eight bytes and each pointer requires four bytes. 5.14 Why is the BST Property defined so that nodes with values equal to the value of the root appear only in the right subtree, rather than allow equal-valued nodes to appear in either subtree? 5.15 (a) Show the BST that results from inserting the values 15, 20, 25, 18, 16, 5, and 7 (in that order). (b) Show the enumerations for the tree of (a) that result from doing a pre- order traversal, an inorder traversal, and a postorder traversal. 5.16 Draw the BST that results from adding the value 5 to the BST shown in Figure 5.13(a). 5.17 Draw the BST that results from deleting the value 7 from the BST of Fig- ure 5.13(b). 5.18 Write a function that prints out the node values for a BST in sorted order from highest to lowest. 5.19 Write a recursive function named smallcount that, given the pointer to the root of a BST and a key K, returns the number of nodes having key values less than or equal to K. Function smallcount should visit as few nodes in the BST as possible. 5.20 Write a recursive function named printRange that, given the pointer to the root to a BST, a low key value, and a high key value, prints in sorted order all records whose key values fall between the two given keys. Function printRange should visit as few nodes in the BST as possible. 5.21 Describe a simple modification to the BST that will allow it to easily support finding the Kth smallest value in Θ(log n) average case time. Then write a pseudo-code function for finding the Kth smallest value in your modified BST. 5.22 What are the minimum and maximum number of elements in a heap of height h? 5.23 Where in a max-heap might the smallest element reside? 5.24 Show the max-heap that results from running buildHeap on the following values stored in an array: 10 5 12 3 2 1 8 7 9 4 5.25 (a) Show the heap that results from deleting the maximum value from the max-heap of Figure 5.20b. (b) Show the heap that results from deleting the element with value 5 from the max-heap of Figure 5.20b.
- 220. Sec. 5.8 Exercises 201 5.26 Revise the heap definition of Figure 5.19 to implement a min-heap. The member function removemax should be replaced by a new function called removemin. 5.27 Build the Huffman coding tree and determine the codes for the following set of letters and weights: Letter A B C D E F G H I J K L Frequency 2 3 5 7 11 13 17 19 23 31 37 41 What is the expected length in bits of a message containing n characters for this frequency distribution? 5.28 What will the Huffman coding tree look like for a set of sixteen characters all with equal weight? What is the average code length for a letter in this case? How does this differ from the smallest possible fixed length code for sixteen characters? 5.29 A set of characters with varying weights is assigned Huffman codes. If one of the characters is assigned code 001, then, (a) Describe all codes that cannot have been assigned. (b) Describe all codes that must have been assigned. 5.30 Assume that a sample alphabet has the following weights: Letter Q Z F M T S O E Frequency 2 3 10 10 10 15 20 30 (a) For this alphabet, what is the worst-case number of bits required by the Huffman code for a string of n letters? What string(s) have the worst- case performance? (b) For this alphabet, what is the best-case number of bits required by the Huffman code for a string of n letters? What string(s) have the best- case performance? (c) What is the average number of bits required by a character using the Huffman code for this alphabet? 5.31 You must keep track of some data. Your options are: (1) A linked-list maintained in sorted order. (2) A linked-list of unsorted records. (3) A binary search tree. (4) An array-based list maintained in sorted order. (5) An array-based list of unsorted records.
- 221. 202 Chap. 5 Binary Trees For each of the following scenarios, which of these choices would be best? Explain your answer. (a) The records are guaranteed to arrive already sorted from lowest to high- est (i.e., whenever a record is inserted, its key value will always be greater than that of the last record inserted). A total of 1000 inserts will be interspersed with 1000 searches. (b) The records arrive with values having a uniform random distribution (so the BST is likely to be well balanced). 1,000,000 insertions are performed, followed by 10 searches. (c) The records arrive with values having a uniform random distribution (so the BST is likely to be well balanced). 1000 insertions are interspersed with 1000 searches. (d) The records arrive with values having a uniform random distribution (so the BST is likely to be well balanced). 1000 insertions are performed, followed by 1,000,000 searches. 5.9 Projects 5.1 Reimplement the composite design for the binary tree node class of Fig- ure 5.11 using a flyweight in place of null pointers to empty nodes. 5.2 One way to deal with the “problem” of null pointers in binary trees is to use that space for some other purpose. One example is the threaded binary tree. Extending the node implementation of Figure 5.7, the threaded binary tree stores with each node two additional bit fields that indicate if the child pointers lc and rc are regular pointers to child nodes or threads. If lc is not a pointer to a non-empty child (i.e., if it would be null in a regular binary tree), then it instead stores a pointer to the inorder predecessor of that node. The inorder predecessor is the node that would be printed immediately before the current node in an inorder traversal. If rc is not a pointer to a child, then it instead stores a pointer to the node’s inorder successor. The inorder successor is the node that would be printed immediately after the current node in an inorder traversal. The main advantage of threaded binary trees is that operations such as inorder traversal can be implemented without using recursion or a stack. Reimplement the BST as a threaded binary tree, and include a non-recursive version of the preorder traversal 5.3 Implement a city database using a BST to store the database records. Each database record contains the name of the city (a string of arbitrary length) and the coordinates of the city expressed as integer x- and y-coordinates.
- 222. Sec. 5.9 Projects 203 The BST should be organized by city name. Your database should allow records to be inserted, deleted by name or coordinate, and searched by name or coordinate. Another operation that should be supported is to print all records within a given distance of a specified point. Collect running-time statistics for each operation. Which operations can be implemented reason- ably efficiently (i.e., in Θ(log n) time in the average case) using a BST? Can the database system be made more efficient by using one or more additional BSTs to organize the records by location? 5.4 Create a binary tree ADT that includes generic traversal methods that take a visitor, as described in Section 5.2. Write functions count and BSTcheck of Section 5.2 as visitors to be used with the generic traversal method. 5.5 Implement a priority queue class based on the max-heap class implementa- tion of Figure 5.19. The following methods should be supported for manip- ulating the priority queue: void enqueue(int ObjectID, int priority); int dequeue(); void changeweight(int ObjectID, int newPriority); Method enqueue inserts a new object into the priority queue with ID num- ber ObjectID and priority priority. Method dequeue removes the object with highest priority from the priority queue and returns its object ID. Method changeweight changes the priority of the object with ID number ObjectID to be newPriority. The type for Elem should be a class that stores the object ID and the priority for that object. You will need a mech- anism for finding the position of the desired object within the heap. Use an array, storing the object with ObjectID i in position i. (Be sure in your testing to keep the ObjectIDs within the array bounds.) You must also modify the heap implementation to store the object’s position in the auxil- iary array so that updates to objects in the heap can be updated as well in the array. 5.6 The Huffman coding tree function buildHuff of Figure 5.29 manipulates a sorted list. This could result in a Θ(n2) algorithm, because placing an inter- mediate Huffman tree on the list could take Θ(n) time. Revise this algorithm to use a priority queue based on a min-heap instead of a list. 5.7 Complete the implementation of the Huffman coding tree, building on the code presented in Section 5.6. Include a function to compute and store in a table the codes for each letter, and functions to encode and decode messages. This project can be further extended to support file compression. To do so requires adding two steps: (1) Read through the input file to generate actual
- 223. 204 Chap. 5 Binary Trees frequencies for all letters in the file; and (2) store a representation for the Huffman tree at the beginning of the encoded output file to be used by the decoding function. If you have trouble with devising such a representation, see Section 6.5.
- 224. 6 Non-Binary Trees Many organizations are hierarchical in nature, such as the military and most busi- nesses. Consider a company with a president and some number of vice presidents who report to the president. Each vice president has some number of direct sub- ordinates, and so on. If we wanted to model this company with a data structure, it would be natural to think of the president in the root node of a tree, the vice presi- dents at level 1, and their subordinates at lower levels in the tree as we go down the organizational hierarchy. Because the number of vice presidents is likely to be more than two, this com- pany’s organization cannot easily be represented by a binary tree. We need instead to use a tree whose nodes have an arbitrary number of children. Unfortunately, when we permit trees to have nodes with an arbrary number of children, they be- come much harder to implement than binary trees. We consider such trees in this chapter. To distinguish them from the more commonly used binary tree, we use the term general tree. Section 6.1 presents general tree terminology. Section 6.2 presents a simple representation for solving the important problem of processing equivalence classes. Several pointer-based implementations for general trees are covered in Section 6.3. Aside from general trees and binary trees, there are also uses for trees whose in- ternal nodes have a fixed number K of children where K is something other than two. Such trees are known as K-ary trees. Section 6.4 generalizes the properties of binary trees to K-ary trees. Sequential representations, useful for applications such as storing trees on disk, are covered in Section 6.5. 6.1 General Tree Definitions and Terminology A tree T is a finite set of one or more nodes such that there is one designated node R, called the root of T. If the set (T−{R}) is not empty, these nodes are partitioned 205
- 225. 206 Chap. 6 Non-Binary Trees S1 S2 Children of V Subtree rooted at V Siblings of V Ancestors of V R Root Parent of V P V C3 C1 C2 Figure 6.1 Notation for general trees. Node P is the parent of nodes V, S1, and S2. Thus, V, S1, and S2 are children of P. Nodes R and P are ancestors of V. Nodes V, S1, and S2 are called siblings. The oval surrounds the subtree having V as its root. into n ≥ 0 disjoint subsets T0, T1, ..., Tn−1, each of which is a tree, and whose roots R1, R2, ..., Rn, respectively, are children of R. The subsets Ti (0 ≤ i < n) are said to be subtrees of T. These subtrees are ordered in that Ti is said to come before Tj if i < j. By convention, the subtrees are arranged from left to right with subtree T0 called the leftmost child of R. A node’s out degree is the number of children for that node. A forest is a collection of one or more trees. Figure 6.1 presents further tree notation generalized from the notation for binary trees presented in Chapter 5. Each node in a tree has precisely one parent, except for the root, which has no parent. From this observation, it immediately follows that a tree with n nodes must have n − 1 edges because each node, aside from the root, has one edge connecting that node to its parent. 6.1.1 An ADT for General Tree Nodes Before discussing general tree implementations, we should first make precise what operations such implementations must support. Any implementation must be able to initialize a tree. Given a tree, we need access to the root of that tree. There must be some way to access the children of a node. In the case of the ADT for binary tree nodes, this was done by providing member functions that give explicit
- 226. Sec. 6.1 General Tree Definitions and Terminology 207 /** General tree ADT */ interface GenTree<E> { public void clear(); // Clear the tree public GTNode<E> root(); // Return the root // Make the tree have a new root, give first child and sib public void newroot(E value, GTNode<E> first, GTNode<E> sib); public void newleftchild(E value); // Add left child } Figure 6.2 The general tree node and general tree classes. access to the left and right child pointers. Unfortunately, because we do not know in advance how many children a given node will have in the general tree, we cannot give explicit functions to access each child. An alternative must be found that works for any number of children. One choice would be to provide a function that takes as its parameter the index for the desired child. That combined with a function that returns the number of children for a given node would support the ability to access any node or process all children of a node. Unfortunately, this view of access tends to bias the choice for node implementations in favor of an array-based approach, because these functions favor random access to a list of children. In practice, an implementation based on a linked list is often preferred. An alternative is to provide access to the first (or leftmost) child of a node, and to provide access to the next (or right) sibling of a node. Figure 6.2 shows class declarations for general trees and their nodes. Based on these two access functions, the children of a node can be traversed like a list. Trying to find the next sibling of the rightmost sibling would return null. 6.1.2 General Tree Traversals In Section 5.2, three tree traversals were presented for binary trees: preorder, pos- torder, and inorder. For general trees, preorder and postorder traversals are defined with meanings similar to their binary tree counterparts. Preorder traversal of a gen- eral tree first visits the root of the tree, then performs a preorder traversal of each subtree from left to right. A postorder traversal of a general tree performs a pos- torder traversal of the root’s subtrees from left to right, then visits the root. Inorder traversal does not have a natural definition for the general tree, because there is no particular number of children for an internal node. An arbitrary definition — such as visit the leftmost subtree in inorder, then the root, then visit the remaining sub- trees in inorder — can be invented. However, inorder traversals are generally not useful with general trees.
- 227. 208 Chap. 6 Non-Binary Trees B D E F C A R Figure 6.3 An example of a general tree. Example 6.1 A preorder traversal of the tree in Figure 6.3 visits the nodes in order RACDEBF. A postorder traversal of this tree visits the nodes in order CDEAFBR. To perform a preorder traversal, it is necessary to visit each of the children for a given node (say R) from left to right. This is accomplished by starting at R’s leftmost child (call it T). From T, we can move to T’s right sibling, and then to that node’s right sibling, and so on. Using the ADT of Figure 6.2, here is an implementation to print the nodes of a general tree in preorder Note the for loop at the end, which processes the list of children by beginning with the leftmost child, then repeatedly moving to the next child until calling next returns null. /** Preorder traversal for general trees */ static <E> void preorder(GTNode<E> rt) { PrintNode(rt); if (!rt.isLeaf()) { GTNode<E> temp = rt.leftmostChild(); while (temp != null) { preorder(temp); temp = temp.rightSibling(); } } } 6.2 The Parent Pointer Implementation Perhaps the simplest general tree implementation is to store for each node only a pointer to that node’s parent. We will call this the parent pointer implementation.
- 228. Sec. 6.2 The Parent Pointer Implementation 209 Clearly this implementation is not general purpose, because it is inadequate for such important operations as finding the leftmost child or the right sibling for a node. Thus, it may seem to be a poor idea to implement a general tree in this way. However, the parent pointer implementation stores precisely the information required to answer the following, useful question: “Given two nodes, are they in the same tree?” To answer the question, we need only follow the series of parent pointers from each node to its respective root. If both nodes reach the same root, then they must be in the same tree. If the roots are different, then the two nodes are not in the same tree. The process of finding the ultimate root for a given node we will call FIND. The parent pointer representation is most often used to maintain a collection of disjoint sets. Two disjoint sets share no members in common (their intersection is empty). A collection of disjoint sets partitions some objects such that every object is in exactly one of the disjoint sets. There are two basic operations that we wish to support: (1) determine if two objects are in the same set, and (2) merge two sets together. Because two merged sets are united, the merging operation is usually called UNION and the whole process of determining if two objects are in the same set and then merging the sets goes by the name “UNION/FIND.” To implement UNION/FIND, we represent each disjoint set with a separate general tree. Two objects are in the same disjoint set if they are in the same tree. Every node of the tree (except for the root) has precisely one parent. Thus, each node requires the same space to represent it. The collection of objects is typically stored in an array, where each element of the array corresponds to one object, and each element stores the object’s value. The objects also correspond to nodes in the various disjoint trees (one tree for each disjoint set), so we also store the parent value with each object in the array. Those nodes that are the roots of their respective trees store an appropriate indicator. Note that this representation means that a single array is being used to implement a collection of trees. This makes it easy to merge trees together with UNION operations. Figure 6.4 shows the parent pointer implementation for the general tree, called ParPtrTree. This class is greatly simplified from the declarations of Figure 6.2 because we need only a subset of the general tree operations. Instead of implement- ing a separate node class, ParPtrTree simply stores an array where each array element corresponds to a node of the tree. Each position i of the array stores the value for node i and the array position for the parent of node i. Class ParPtrTree is given two new functions, differ and UNION. Function differ checks if two
- 229. 210 Chap. 6 Non-Binary Trees objects are in different sets, and function UNION merges two sets together. A pri- vate function FIND is used to find the ultimate root for an object. An application using the UNION/FIND operations should store a set of n ob- jects, where each object is assigned a unique index in the range 0 to n − 1. The indices refer to the corresponding parent pointers in the array. Class ParPtrTree creates and initializes the UNION/FIND array, and functions differ and UNION take array indices as inputs. Figure 6.5 illustrates the parent pointer implementation. Note that the nodes can appear in any order within the array, and the array can store up to n separate trees. For example, Figure 6.5 shows two trees stored in the same array. Thus, a single array can store a collection of items distributed among an arbitrary (and changing) number of disjoint subsets. Consider the problem of assigning the members of a set to disjoint subsets called equivalence classes. Recall from Section 2.1 that an equivalence relation is reflexive, symmetric, and transitive. Thus, if objects A and B are equivalent, and objects B and C are equivalent, we must be able to recognize that objects A and C are also equivalent. There are many practical uses for disjoint sets and representing equivalences. For example, consider Figure 6.6 which shows a graph of ten nodes labeled A through J. Notice that for nodes A through I, there is some series of edges that connects any pair of the nodes, but node J is disconnected from the rest of the nodes. Such a graph might be used to represent connections such as wires be- tween components on a circuit board, or roads between cities. We can consider two nodes of the graph to be equivalent if there is a path between them. Thus, nodes A, H, and E would be equivalent in Figure 6.6, but J is not equivalent to any other. A subset of equivalent (connected) edges in a graph is called a connected component. The goal is to quickly classify the objects into disjoint sets that corre- spond to the connected components. Another application for UNION/FIND occurs in Kruskal’s algorithm for computing the minimal cost spanning tree for a graph (Section 11.5.2). The input to the UNION/FIND algorithm is typically a series of equivalence pairs. In the case of the connected components example, the equivalence pairs would simply be the set of edges in the graph. An equivalence pair might say that object C is equivalent to object A. If so, C and A are placed in the same subset. If a later equivalence relates A and B, then by implication C is also equivalent to B. Thus, an equivalence pair may cause two subsets to merge, each of which contains several objects.
- 230. Sec. 6.2 The Parent Pointer Implementation 211 /** General Tree class implementation for UNION/FIND */ class ParPtrTree { private Integer [] array; // Node array public ParPtrTree(int size) { array = new Integer[size]; // Create node array for (int i=0; i<size; i++) array[i] = null; } /** Determine if nodes are in different trees */ public boolean differ(int a, int b) { Integer root1 = FIND(a); // Find root of node a Integer root2 = FIND(b); // Find root of node b return root1 != root2; // Compare roots } /** Merge two subtrees */ public void UNION(int a, int b) { Integer root1 = FIND(a); // Find root of node a Integer root2 = FIND(b); // Find root of node b if (root1 != root2) array[root2] = root1; // Merge } String print() { StringBuffer out = new StringBuffer(100); for (int i=0; i<array.length; i++) { if (array[i] == null) out.append("-1 "); else { Integer temp = array[i]; for (int j=0; j<array.length; j++) if (temp == array[j]) { out.append(j + " "); break; } } } return out.toString(); } } public Integer FIND(Integer curr) { if (array[curr] == null) return curr; // At root while (array[curr] != null) curr = array[curr]; return curr; } Figure 6.4 General tree implementation using parent pointers for the UNION/FIND algorithm.
- 231. 212 Chap. 6 Non-Binary Trees C D F W X Y Parent’s Index 1 1 1 2 E D C B A R Label Z W Z Y X F 0 0 7 7 7 B A E R Node Index 0 1 2 3 4 5 6 7 8 9 10 Figure 6.5 The parent pointer array implementation. Each node corresponds to a position in the node array stores its value and a pointer to its parent. The parent pointers are represented by the position in the array of the parent. The root of any tree stores ROOT, represented graphically by a slash in the “Parent’s Index” box. This figure shows two trees stored in the same parent pointer array, one rooted at R, and the other rooted at W. B E D G F J A C I H Figure 6.6 A graph with two connected components. Equivalence classes can be managed efficiently with the UNION/FIND alg- orithm. Initially, each object is at the root of its own tree. An equivalence pair is processed by checking to see if both objects of the pair are in the same tree us- ing function differ. If they are in the same tree, then no change need be made because the objects are already in the same equivalence class. Otherwise, the two equivalence classes should be merged by the UNION function. Example 6.2 As an example of solving the equivalence class problem, consider the graph of Figure 6.6. Initially, we assume that each node of the graph is in a distinct equivalence class. This is represented by storing each as the root of its own tree. Figure 6.7(a) shows this initial configuration
- 232. Sec. 6.2 The Parent Pointer Implementation 213 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 (a) (b) (c) (d) 0 1 2 3 4 5 6 7 8 9 B H G A C F A F C G H F A B C G D E H J J D E D B E J B C D E F G H B C D F G H G B C D E G H A A A E F I B C D J I I E H J F A I I I I J J 0 0 5 5 0 3 5 2 3 2 0 0 5 3 2 5 5 A B D E F G H I C 5 5 5 J Figure 6.7 An example of equivalence processing. (a) Initial configuration for the ten nodes of the graph in Figure 6.6. The nodes are placed into ten independent equivalence classes. (b) The result of processing five edges: (A, B), (C, H), (G, F), (D, E), and (I, F). (c) The result of processing two more edges: (H, A) and (E, G). (d) The result of processing edge (H, E).
- 233. 214 Chap. 6 Non-Binary Trees using the parent pointer array representation. Now, consider what happens when equivalence relationship (A, B) is processed. The root of the tree containing A is A, and the root of the tree containing B is B. To make them equivalent, one of these two nodes is set to be the parent of the other. In this case it is irrelevant which points to which, so we arbitrarily select the first in alphabetical order to be the root. This is represented in the parent pointer array by setting the parent field of B (the node in array position 1 of the array) to store a pointer to A. Equivalence pairs (C, H), (G, F), and (D, E) are processed in similar fashion. When processing the equivalence pair (I, F), because I and F are both their own roots, I is set to point to F. Note that this also makes G equivalent to I. The result of processing these five equivalences is shown in Figure 6.7(b). The parent pointer representation places no limit on the number of nodes that can share a parent. To make equivalence processing as efficient as possible, the distance from each node to the root of its respective tree should be as small as possible. Thus, we would like to keep the height of the trees small when merging two equivalence classes together. Ideally, each tree would have all nodes pointing directly to the root. Achieving this goal all the time would require too much ad- ditional processing to be worth the effort, so we must settle for getting as close as possible. A low-cost approach to reducing the height is to be smart about how two trees are joined together. One simple technique, called the weighted union rule, joins the tree with fewer nodes to the tree with more nodes by making the smaller tree’s root point to the root of the bigger tree. This will limit the total depth of the tree to O(log n), because the depth of nodes only in the smaller tree will now increase by one, and the depth of the deepest node in the combined tree can only be at most one deeper than the deepest node before the trees were combined. The total number of nodes in the combined tree is therefore at least twice the number in the smaller subtree. Thus, the depth of any node can be increased at most log n times when n equivalences are processed. Example 6.3 When processing equivalence pair (I, F) in Figure 6.7(b), F is the root of a tree with two nodes while I is the root of a tree with only one node. Thus, I is set to point to F rather than the other way around. Figure 6.7(c) shows the result of processing two more equivalence pairs: (H, A) and (E, G). For the first pair, the root for H is C while the root for A is itself. Both trees contain two nodes, so it is an arbitrary decision
- 234. Sec. 6.2 The Parent Pointer Implementation 215 as to which node is set to be the root for the combined tree. In the case of equivalence pair (E, G), the root of E is D while the root of G is F. Because F is the root of the larger tree, node D is set to point to F. Not all equivalences will combine two trees. If edge (F, G) is processed when the representation is in the state shown in Figure 6.7(c), no change will be made because F is already the root for G. The weighted union rule helps to minimize the depth of the tree, but we can do better than this. Path compression is a method that tends to create extremely shal- low trees. Path compression takes place while finding the root for a given node X. Call this root R. Path compression resets the parent of every node on the path from X to R to point directly to R. This can be implemented by first finding R. A second pass is then made along the path from X to R, assigning the parent field of each node encountered to R. Alternatively, a recursive algorithm can be implemented as follows. This version of FIND not only returns the root of the current node, but also makes all ancestors of the current node point to the root. public Integer FIND(Integer curr) { if (array[curr] == null) return curr; // At root array[curr] = FIND(array[curr]); return array[curr]; } Example 6.4 Figure 6.7(d) shows the result of processing equivalence pair (H, E) on the the representation shown in Figure 6.7(c) using the stan- dard weighted union rule without path compression. Figure 6.8 illustrates the path compression process for the same equivalence pair. After locating the root for node H, we can perform path compression to make H point directly to root object A. Likewise, E is set to point directly to its root, F. Finally, object A is set to point to root object F. Note that path compression takes place during the FIND operation, not during the merge operation. In Figure 6.8, this means that nodes B, C, and H have node A remain as their parent, rather than changing their parent to be F. While we might prefer to have these nodes point to F, to accomplish this would require that additional information from the FIND operation be passed back to the UNION operation. This would not be practical. Path compression keeps the cost of each FIND operation very close to constant. To be more precise about what is meant by “very close to constant,” the cost of path compression for n FIND operations on n nodes (when combined with the weighted
- 235. 216 Chap. 6 Non-Binary Trees 5 0 0 5 5 5 0 5 A B C D J 9 8 7 6 5 4 3 2 1 0 A B G E E F G H I J C H I D F Figure 6.8 An example of path compression, showing the result of processing equivalence pair (H, E) on the representation of Figure 6.7(c). union rule for joining sets) is Θ(n log∗ n). The notation “log∗ n” means the number of times that the log of n must be taken before n ≤ 1. For example, log∗ 65536 is 4 because log 65536 = 16, log 16 = 4, log 4 = 2, and finally log 2 = 1. Thus, log∗ n grows very slowly, so the cost for a series of n FIND operations is very close to n. Note that this does not mean that the tree resulting from processing n equiva- lence pairs necessarily has depth Θ(log∗ n). One can devise a series of equivalence operations that yields Θ(log n) depth for the resulting tree. However, many of the equivalences in such a series will look only at the roots of the trees being merged, requiring little processing time. The total amount of processing time required for n operations will be Θ(n log∗ n), yielding nearly constant time for each equivalence operation. This is an example of the technique of amortized analysis, discussed further in Section 14.3. 6.3 General Tree Implementations We now tackle the problem of devising an implementation for general trees that al- lows efficient processing of all member functions of the ADT shown in Figure 6.2. This section presents several approaches to implementing general trees. Each im- plementation yields advantages and disadvantages in the amount of space required to store a node and the relative ease with which key operations can be performed. General tree implementations should place no restriction on how many children a node may have. In some applications, once a node is created the number of children never changes. In such cases, a fixed amount of space can be allocated for the node when it is created, based on the number of children for the node. Matters become more complicated if children can be added to or deleted from a node, requiring that the node’s space allocation be adjusted accordingly.
- 236. Sec. 6.3 General Tree Implementations 217 R B A C D E F Index Val Par 0 1 2 3 4 5 6 7 R A C B D F E 0 1 0 1 3 1 3 2 4 6 5 1 Figure 6.9 The “list of children” implementation for general trees. The column of numbers to the left of the node array labels the array indices. The column labeled “Val” stores node values. The column labeled “Par” stores pointers to the parents. For clarity, these pointers are shown as array indices. The last column stores pointers to the linked list of children for each internal node. Each element on the linked list stores a pointer to one of the node’s children (shown as the array index of the target node). 6.3.1 List of Children Our first attempt to create a general tree implementation is called the “list of chil- dren” implementation for general trees. It simply stores with each internal node a linked list of its children, in order from left to right. This is illustrated by Figure 6.9. The “list of children” implementation stores the tree nodes in an array. Each node contains a value, a pointer to its parent, and a pointer to a linked list of the node’s children, stored from left to right. Each list element contains a pointer to one child. Thus, the leftmost child of a node can be found directly because it is the first element in the linked list. However, to find the right sibling for a node is more difficult. Consider the case of a node M and its parent P. To find M’s right sibling, we must move down the child list of P until the linked list element storing the pointer to M has been found. Going one step further takes us to the linked list element that stores a pointer to M’s right sibling. Thus, in the worst case, to find M’s right sibling requires that all children of M’s parent be searched. Combining trees using this representation is difficult if each tree is stored in a separate node array. If the nodes of both trees are stored in a single node array, then
- 237. 218 Chap. 6 Non-Binary Trees R′ Left Val ParRight R B A C D E F X X 7 1 1 1 0 2 1 3 4 R A B C D E F 5 R′ 8 6 2 0 Figure 6.10 The “left-child/right-sibling” implementation. adding tree T as a subtree of node R is done by simply adding the root of T to R’s list of children. 6.3.2 The Left-Child/Right-Sibling Implementation With the “list of children” implementation, it is difficult to access a node’s right sibling. Figure 6.10 presents an improvement. Here, each node stores its value and pointers to its parent, leftmost child, and right sibling. Thus, each of the basic ADT operations can be implemented by reading a value directly from the node. If two trees are stored within the same node array, then adding one as the subtree of the other simply requires setting three pointers. Combining trees in this way is illustrated by Figure 6.11. This implementation is more space efficient than the “list of children” implementation, and each node requires a fixed amount of space in the node array. 6.3.3 Dynamic Node Implementations The two general tree implementations just described use an array to store the col- lection of nodes. In contrast, our standard implementation for binary trees stores each node as a separate dynamic object containing its value and pointers to its two children. Unfortunately, nodes of a general tree can have any number of children, and this number may change during the life of the node. A general tree node imple- mentation must support these properties. One solution is simply to limit the number
- 238. Sec. 6.3 General Tree Implementations 219 0 1 1 1 7 2 0 R′ X R B A D E F R Left Val Par Right C 1 8 3 A 2 6 B C 4 D 5 E F X 0 7 R′ Figure 6.11 Combining two trees that use the “left-child/right-sibling” imple- mentation. The subtree rooted at R in Figure 6.10 now becomes the first child of R′ . Three pointers are adjusted in the node array: The left-child field of R′ now points to node R, while the right-sibling field for R points to node X. The parent field of node R points to node R′ . of children permitted for any node and allocate pointers for exactly that number of children. There are two major objections to this. First, it places an undesirable limit on the number of children, which makes certain trees unrepresentable by this implementation. Second, this might be extremely wasteful of space because most nodes will have far fewer children and thus leave some pointer positions empty. The alternative is to allocate variable space for each node. There are two basic approaches. One is to allocate an array of child pointers as part of the node. In essence, each node stores an array-based list of child pointers. Figure 6.12 illus- trates the concept. This approach assumes that the number of children is known when the node is created, which is true for some applications but not for others. It also works best if the number of children does not change. If the number of children does change (especially if it increases), then some special recovery mech- anism must be provided to support a change in the size of the child pointer array. One possibility is to allocate a new node of the correct size from free store and re- turn the old copy of the node to free store for later reuse. This works especially well in a language with built-in garbage collection such as Java. For example, assume that a node M initially has two children, and that space for two child pointers is al- located when M is created. If a third child is added to M, space for a new node with three child pointers can be allocated, the contents of M is copied over to the new
- 239. 220 Chap. 6 Non-Binary Trees Val Size (b) (a) F B D E C A R R 2 A 3 B 1 C 0 D 0 E 0 F 0 Figure 6.12 A dynamic general tree representation with fixed-size arrays for the child pointers. (a) The general tree. (b) The tree representation. For each node, the first field stores the node value while the second field stores the size of the child pointer array. space, and the old space is then returned to free store. As an alternative to relying on the system’s garbage collector, a memory manager for variable size storage units can be implemented, as described in Section 12.3. Another possibility is to use a collection of free lists, one for each array size, as described in Section 4.1.2. Note in Figure 6.12 that the current number of children for each node is stored explicitly in a size field. The child pointers are stored in an array with size elements. Another approach that is more flexible, but which requires more space, is to store a linked list of child pointers with each node as illustrated by Figure 6.13. This implementation is essentially the same as the “list of children” implementation of Section 6.3.1, but with dynamically allocated nodes rather than storing the nodes in an array. 6.3.4 Dynamic “Left-Child/Right-Sibling” Implementation The “left-child/right-sibling” implementation of Section 6.3.2 stores a fixed number of pointers with each node. This can be readily adapted to a dynamic implemen- tation. In essence, we substitute a binary tree for a general tree. Each node of the “left-child/right-sibling” implementation points to two “children” in a new binary tree structure. The left child of this new structure is the node’s first child in the general tree. The right child is the node’s right sibling. We can easily extend this conversion to a forest of general trees, because the roots of the trees can be con- sidered siblings. Converting from a forest of general trees to a single binary tree is illustrated by Figure 6.14. Here we simply include links from each node to its right sibling and remove links to all children except the leftmost child. Figure 6.15 shows how this might look in an implementation with two pointers at each node.
- 240. Sec. 6.4 K-ary Trees 221 (b) (a) B F E D R C A R B A C D E F Figure 6.13 A dynamic general tree representation with linked lists of child pointers. (a) The general tree. (b) The tree representation. (a) root (b) Figure 6.14 Converting from a forest of general trees to a single binary tree. Each node stores pointers to its left child and right sibling. The tree roots are assumed to be siblings for the purpose of converting. Compared with the implementation illustrated by Figure 6.13 which requires over- head of three pointers/node, the implementation of Figure 6.15 only requires two pointers per node. Because each node of the general tree now contains a fixed number of pointers, and because each function of the general tree ADT can now be implemented ef- ficiently, the dynamic “left-child/right-sibling” implementation is preferred to the other general tree implementations described in Sections 6.3.1 to 6.3.3. 6.4 K-ary Trees K-ary trees are trees whose internal nodes all have exactly K children. Thus, a full binary tree is a 2-ary tree. The PR quadtree discussed in Section 13.3 is an
- 241. 222 Chap. 6 Non-Binary Trees (a) B F E D R C A A R B C D F E (b) Figure 6.15 A general tree converted to the dynamic “left-child/right-sibling” representation. Compared to the representation of Figure 6.13, this representation requires less space. (a) (b) Figure 6.16 Full and complete 3-ary trees. (a) This tree is full (but not complete). (b) This tree is complete (but not full). example of a 4-ary tree. Because K-ary tree nodes have a fixed number of children, unlike general trees, they are relatively easy to implement. In general, K-ary trees bear many similarities to binary trees, and similar implementations can be used for K-ary tree nodes. Note that as K becomes large, the potential number of null pointers grows, and the difference between the required sizes for internal nodes and leaf nodes increases. Thus, as K becomes larger, the need to choose different implementations for the internal and leaf nodes becomes more pressing. Full and complete K-ary trees are analogous to full and complete binary trees, respectively. Figure 6.16 shows full and complete K-ary trees for K = 3. In practice, most applications of K-ary trees limit them to be either full or complete. Many of the properties of binary trees extend to K-ary trees. Equivalent theo- rems to those in Section 5.1.1 regarding the number of NULL pointers in a K-ary tree and the relationship between the number of leaves and the number of internal
- 242. Sec. 6.5 Sequential Tree Implementations 223 nodes in a K-ary tree can be derived. We can also store a complete K-ary tree in an array, similar to the approach shown in Section 5.3.3. 6.5 Sequential Tree Implementations Next we consider a fundamentally different approach to implementing trees. The goal is to store a series of node values with the minimum information needed to reconstruct the tree structure. This approach, known as a sequential tree imple- mentation, has the advantage of saving space because no pointers are stored. It has the disadvantage that accessing any node in the tree requires sequentially process- ing all nodes that appear before it in the node list. In other words, node access must start at the beginning of the node list, processing nodes sequentially in whatever order they are stored until the desired node is reached. Thus, one primary virtue of the other implementations discussed in this section is lost: efficient access (typi- cally Θ(log n) time) to arbitrary nodes in the tree. Sequential tree implementations are ideal for archiving trees on disk for later use because they save space, and the tree structure can be reconstructed as needed for later processing. Seqential tree implementations can also be used to serialize a tree structure. Serialization is the process of storing an object as a series of bytes, typically so that the data structure can be transmitted between computers. This capability is important when using data structures in a distributed processing environment. A sequential tree implementation stores the node values as they would be enu- merated by a preorder traversal, along with sufficient information to describe the tree’s shape. If the tree has restricted form, for example if it is a full binary tree, then less information about structure typically needs to be stored. A general tree, because it has the most flexible shape, tends to require the most additional shape information. There are many possible sequential tree implementation schemes. We will begin by describing methods appropriate to binary trees, then generalize to an implementation appropriate to a general tree structure. Because every node of a binary tree is either a leaf or has two (possibly empty) children, we can take advantage of this fact to implicitly represent the tree’s struc- ture. The most straightforward sequential tree implementation lists every node value as it would be enumerated by a preorder traversal. Unfortunately, the node values alone do not provide enough information to recover the shape of the tree. In particular, as we read the series of node values, we do not know when a leaf node has been reached. However, we can treat all non-empty nodes as internal nodes with two (possibly empty) children. Only null values will be interpreted as leaf nodes, and these can be listed explicitly. Such an augmented node list provides enough information to recover the tree structure.
- 243. 224 Chap. 6 Non-Binary Trees G I E F A C B D H Figure 6.17 Sample binary tree for sequential tree implementation examples. Example 6.5 For the binary tree of Figure 6.17, the corresponding se- quential representation would be as follows (assuming that ‘/’ stands for null): AB/D//CEG///FH//I// (6.1) To reconstruct the tree structure from this node list, we begin by setting node A to be the root. A’s left child will be node B. Node B’s left child is a null pointer, so node D must be B’s right child. Node D has two null children, so node C must be the right child of node A. To illustrate the difficulty involved in using the sequential tree representation for processing, consider searching for the right child of the root node. We must first move sequentially through the node list of the left subtree. Only at this point do we reach the value of the root’s right child. Clearly the sequential representation is space efficient, but not time efficient for descending through the tree along some arbitrary path. Assume that each node value takes a constant amount of space. An example would be if the node value is a positive integer and null is indicated by the value zero. From the Full Binary Tree Theorem of Section 5.1.1, we know that the size of the node list will be about twice the number of nodes (i.e., the overhead fraction is 1/2). The extra space is required by the null pointers. We should be able to store the node list more compactly. However, any sequential implementation must recognize when a leaf node has been reached, that is, a leaf node indicates the end of a subtree. One way to do this is to explicitly list with each node whether it is an internal node or a leaf. If a node X is an internal node, then we know that its
- 244. Sec. 6.5 Sequential Tree Implementations 225 two children (which may be subtrees) immediately follow X in the node list. If X is a leaf node, then the next node in the list is the right child of some ancestor of X, not the right child of X. In particular, the next node will be the child of X’s most recent ancestor that has not yet seen its right child. However, this assumes that each internal node does in fact have two children, in other words, that the tree is full. Empty children must be indicated in the node list explicitly. Assume that internal nodes are marked with a prime (′) and that leaf nodes show no mark. Empty children of internal nodes are indicated by ‘/’, but the (empty) children of leaf nodes are not represented at all. Note that a full binary tree stores no null values with this implementation, and so requires less overhead. Example 6.6 We can represent the tree of Figure 6.17 as follows: A′ B′ /DC′ E′ G/F′ HI (6.2) Note that slashes are needed for the empty children because this is not a full binary tree. Storing n bits can be a considerable savings over storing n null values. In Example 6.6, each node is shown with a mark if it is internal, or no mark if it is a leaf. This requires that each node value has space to store the mark bit. This might be true if, for example, the node value were stored as a 4-byte integer but the range of the values sored was small enough so that not all bits are used. An example would be if all node values must be positive. Then the high-order (sign) bit of the integer value could be used as the mark bit. Another approach is to store a separate bit vector to represent the status of each node. In this case, each node of the tree corresponds to one bit in the bit vector. A value of ‘1’ could indicate an internal node, and ‘0’ could indicate a leaf node. Example 6.7 The bit vector for the tree if Figure 6.17 would be 11001100100 (6.3) Storing general trees by means of a sequential implementation requires that more explicit structural information be included with the node list. Not only must the general tree implementation indicate whether a node is leaf or internal, it must also indicate how many children the node has. Alternatively, the implementation can indicate when a node’s child list has come to an end. The next example dis- penses with marks for internal or leaf nodes. Instead it includes a special mark (we
- 245. 226 Chap. 6 Non-Binary Trees will use the “)” symbol) to indicate the end of a child list. All leaf nodes are fol- lowed by a “)” symbol because they have no children. A leaf node that is also the last child for its parent would indicate this by two or more successive “)” symbols. Example 6.8 For the general tree of Figure 6.3, we get the sequential representation RAC)D)E))BF))) (6.4) Note that F is followed by three “)” marks, because it is a leaf, the last node of B’s rightmost subtree, and the last node of R’s rightmost subtree. Note that this representation for serializing general trees cannot be used for bi- nary trees. This is because a binary tree is not merely a restricted form of general tree with at most two children. Every binary tree node has a left and a right child, though either or both might be empty. For example, the representation of Exam- ple 6.8 cannot let us distinguish whether node D in Figure 6.17 is the left or right child of node B. 6.6 Further Reading The expression log∗ n cited in Section 6.2 is closely related to the inverse of Ack- ermann’s function. For more information about Ackermann’s function and the cost of path compression, see Robert E. Tarjan’s paper “On the efficiency of a good but not linear set merging algorithm” [Tar75]. The article “Data Structures and Algo- rithms for Disjoint Set Union Problems” by Galil and Italiano [GI91] covers many aspects of the equivalence class problem. Foundations of Multidimensional and Metric Data Structures by Hanan Samet [Sam06] treats various implementations of tree structures in detail within the con- text of K-ary trees. Samet covers sequential implementations as well as the linked and array implementations such as those described in this chapter and Chapter 5. While these books are ostensibly concerned with spatial data structures, many of the concepts treated are relevant to anyone who must implement tree structures. 6.7 Exercises 6.1 Write an algorithm to determine if two general trees are identical. Make the algorithm as efficient as you can. Analyze your algorithm’s running time.
- 246. Sec. 6.7 Exercises 227 6.2 Write an algorithm to determine if two binary trees are identical when the ordering of the subtrees for a node is ignored. For example, if a tree has root node with value R, left child with value A and right child with value B, this would be considered identical to another tree with root node value R, left child value B, and right child value A. Make the algorithm as efficient as you can. Analyze your algorithm’s running time. How much harder would it be to make this algorthm work on a general tree? 6.3 Write a postorder traversal function for general trees, similar to the preorder traversal function named print given in Section 6.1.2. 6.4 Write a function that takes as input a general tree and returns the number of nodes in that tree. Write your function to use the GenTree and GTNode ADTs of Figure 6.2. 6.5 Describe how to implement the weighted union rule efficiently. In particular, describe what information must be stored with each node and how this infor- mation is updated when two trees are merged. Modify the implementation of Figure 6.4 to support the weighted union rule. 6.6 A potential alternatative to the weighted union rule for combining two trees is the height union rule. The height union rule requires that the root of the tree with greater height become the root of the union. Explain why the height union rule can lead to worse average time behavior than the weighted union rule. 6.7 Using the weighted union rule and path compression, show the array for the parent pointer implementation that results from the following series of equivalences on a set of objects indexed by the values 0 through 15. Initially, each element in the set should be in a separate equivalence class. When two trees to be merged are the same size, make the root with greater index value be the child of the root with lesser index value. (0, 2) (1, 2) (3, 4) (3, 1) (3, 5) (9, 11) (12, 14) (3, 9) (4, 14) (6, 7) (8, 10) (8, 7) (7, 0) (10, 15) (10, 13) 6.8 Using the weighted union rule and path compression, show the array for the parent pointer implementation that results from the following series of equivalences on a set of objects indexed by the values 0 through 15. Initially, each element in the set should be in a separate equivalence class. When two trees to be merged are the same size, make the root with greater index value be the child of the root with lesser index value. (2, 3) (4, 5) (6, 5) (3, 5) (1, 0) (7, 8) (1, 8) (3, 8) (9, 10) (11, 14) (11, 10) (12, 13) (11, 13) (14, 1)
- 247. 228 Chap. 6 Non-Binary Trees 6.9 Devise a series of equivalence statements for a collection of sixteen items that yields a tree of height 5 when both the weighted union rule and path compression are used. What is the total number of parent pointers followed to perform this series? 6.10 One alternative to path compression that gives similar performance gains is called path halving. In path halving, when the path is traversed from the node to the root, we make the grandparent of every other node i on the path the new parent of i. Write a version of FIND that implements path halving. Your FIND operation should work as you move up the tree, rather than require the two passes needed by path compression. 6.11 Analyze the fraction of overhead required by the “list of children” imple- mentation, the “left-child/right-sibling” implementation, and the two linked implementations of Section 6.3.3. How do these implementations compare in space efficiency? 6.12 Using the general tree ADT of Figure 6.2, write a function that takes as input the root of a general tree and returns a binary tree generated by the conversion process illustrated by Figure 6.14. 6.13 Use mathematical induction to prove that the number of leaves in a non- empty full K-ary tree is (K − 1)n + 1, where n is the number of internal nodes. 6.14 Derive the formulae for computing the relatives of a non-empty complete K-ary tree node stored in the complete tree representation of Section 5.3.3. 6.15 Find the overhead fraction for a full K-ary tree implementation with space requirements as follows: (a) All nodes store data, K child pointers, and a parent pointer. The data field requires four bytes and each pointer requires four bytes. (b) All nodes store data and K child pointers. The data field requires six- teen bytes and each pointer requires four bytes. (c) All nodes store data and a parent pointer, and internal nodes store K child pointers. The data field requires eight bytes and each pointer re- quires four bytes. (d) Only leaf nodes store data; only internal nodes store K child pointers. The data field requires four bytes and each pointer requires two bytes. 6.16 (a) Write out the sequential representation for Figure 6.18 using the coding illustrated by Example 6.5. (b) Write out the sequential representation for Figure 6.18 using the coding illustrated by Example 6.6.
- 248. Sec. 6.7 Exercises 229 C A B F E H I D G Figure 6.18 A sample tree for Exercise 6.16. 6.17 Draw the binary tree representing the following sequential representation for binary trees illustrated by Example 6.5: ABD//E//C/F// 6.18 Draw the binary tree representing the following sequential representation for binary trees illustrated by Example 6.6: A′ /B′ /C′ D′ G/E Show the bit vector for leaf and internal nodes (as illustrated by Example 6.7) for this tree. 6.19 Draw the general tree represented by the following sequential representation for general trees illustrated by Example 6.8: XPC)Q)RV)M)))) 6.20 (a) Write a function to decode the sequential representation for binary trees illustrated by Example 6.5. The input should be the sequential repre- sentation and the output should be a pointer to the root of the resulting binary tree. (b) Write a function to decode the sequential representation for full binary trees illustrated by Example 6.6. The input should be the sequential representation and the output should be a pointer to the root of the re- sulting binary tree. (c) Write a function to decode the sequential representation for general trees illustrated by Example 6.8. The input should be the sequential representation and the output should be a pointer to the root of the re- sulting general tree. 6.21 Devise a sequential representation for Huffman coding trees suitable for use as part of a file compression utility (see Project 5.7).
- 249. 230 Chap. 6 Non-Binary Trees 6.8 Projects 6.1 Write classes that implement the general tree class declarations of Figure 6.2 using the dynamic “left-child/right-sibling” representation described in Sec- tion 6.3.4. 6.2 Write classes that implement the general tree class declarations of Figure 6.2 using the linked general tree implementation with child pointer arrays of Fig- ure 6.12. Your implementation should support only fixed-size nodes that do not change their number of children once they are created. Then, reim- plement these classes with the linked list of children representation of Fig- ure 6.13. How do the two implementations compare in space and time effi- ciency and ease of implementation? 6.3 Write classes that implement the general tree class declarations of Figure 6.2 using the linked general tree implementation with child pointer arrays of Fig- ure 6.12. Your implementation must be able to support changes in the num- ber of children for a node. When created, a node should be allocated with only enough space to store its initial set of children. Whenever a new child is added to a node such that the array overflows, allocate a new array from free store that can store twice as many children. 6.4 Implement a BST file archiver. Your program should take a BST created in main memory using the implementation of Figure 5.14 and write it out to disk using one of the sequential representations of Section 6.5. It should also be able to read in disk files using your sequential representation and create the equivalent main memory representation. 6.5 Use the UNION/FIND algorithm to implement a solution to the following problem. Given a set of points represented by their xy-coordinates, assign the points to clusters. Any two points are defined to be in the same cluster if they are within a specified distance d of each other. For the purpose of this problem, clustering is an equivalence relationship. In other words, points A, B, and C are defined to be in the same cluster if the distance between A and B is less than d and the distance between A and C is also less than d, even if the distance between B and C is greater than d. To solve the problem, compute the distance between each pair of points, using the equivalence processing algorithm to merge clusters whenever two points are within the specified distance. What is the asymptotic complexity of this algorithm? Where is the bottleneck in processing? 6.6 In this project, you will run some empirical tests to deterimine if some vari- ations on path compression in the UNION/FIND algorithm will lead to im-
- 250. Sec. 6.8 Projects 231 proved performance. You should compare the following four implementa- tions: (a) Standard UNION/FIND with path compression and weighted union. (b) Path compression and weighted union, except that path compression is done after the UNION, instead of during the FIND operation. That is, make all nodes along the paths traversed in both trees point directly to the root of the larger tree. (c) Weighted union and a simplified form of path compression. At the end of every FIND operation, make the node point to its tree’s root (but don’t change the pointers for other nodes along the path). (d) Weighted union and a simplified form of path compression. Both nodes in the equivalence will be set to point directly to the root of the larger tree after the UNION operation. For example, consider processing the equivalence (A, B) where A′ is the root of A and B′ is the root of B. Assume the tree with root A′ is bigger than the tree with root B′ . At the end of the UNION/FIND operation, nodes A, B, and B′ will all point directly to A′ .
- 252. PART III Sorting and Searching 233
- 254. 7 Internal Sorting We sort many things in our everyday lives: A handful of cards when playing Bridge; bills and other piles of paper; jars of spices; and so on. And we have many intuitive strategies that we can use to do the sorting, depending on how many objects we have to sort and how hard they are to move around. Sorting is also one of the most frequently performed computing tasks. We might sort the records in a database so that we can search the collection efficiently. We might sort the records by zip code so that we can print and mail them more cheaply. We might use sorting as an intrinsic part of an algorithm to solve some other problem, such as when computing the minimum-cost spanning tree (see Section 11.5). Because sorting is so important, naturally it has been studied intensively and many algorithms have been devised. Some of these algorithms are straightforward adaptations of schemes we use in everyday life. Others are totally alien to how hu- mans do things, having been invented to sort thousands or even millions of records stored on the computer. After years of study, there are still unsolved problems related to sorting. New algorithms are still being developed and refined for special- purpose applications. While introducing this central problem in computer science, this chapter has a secondary purpose of illustrating many important issues in algorithm design and analysis. The collection of sorting algorithms presented will illustate that divide- and-conquer is a powerful approach to solving a problem, and that there are multi- ple ways to do the dividing. Mergesort divides a list in half. Quicksort divides a list into big values and small values. And Radix Sort divides the problem by working on one digit of the key at a time. Sorting algorithms will be used to illustrate a wide variety of analysis tech- niques in this chapter. We’ll find that it is possible for an algorithm to have an average case whose growth rate is significantly smaller than its worse case (Quick- sort). We’ll see how it is possible to speed up sorting algorithms (both Shellsort 235
- 255. 236 Chap. 7 Internal Sorting and Quicksort) by taking advantage of the best case behavior of another algorithm (Insertion sort). We’ll see several examples of how we can tune an algorithm for better performance. We’ll see that special case behavior by some algorithms makes them the best solution for special niche applications (Heapsort). Sorting provides an example of a significant technique for analyzing the lower bound for a problem. Sorting will also be used to motivate the introduction to file processing presented in Chapter 8. The present chapter covers several standard algorithms appropriate for sorting a collection of records that fit in the computer’s main memory. It begins with a dis- cussion of three simple, but relatively slow, algorithms requiring Θ(n2) time in the average and worst cases. Several algorithms with considerably better performance are then presented, some with Θ(n log n) worst-case running time. The final sort- ing method presented requires only Θ(n) worst-case time under special conditions. The chapter concludes with a proof that sorting in general requires Ω(n log n) time in the worst case. 7.1 Sorting Terminology and Notation Except where noted otherwise, input to the sorting algorithms presented in this chapter is a collection of records stored in an array. Records are compared to one another by means of a comparator class, as introduced in Section 4.4. To simplify the discussion we will assume that each record has a key field whose value is ex- tracted from the record by the comparator. The key method of the comparator class is prior, which returns true when its first argument should appear prior to its sec- ond argument in the sorted list. We also assume that for every record type there is a swap function that can interchange the contents of two records in the array (see the Appendix). Given a set of records r1, r2, ..., rn with key values k1, k2, ..., kn, the Sorting Problem is to arrange the records into any order s such that records rs1 , rs2 , ..., rsn have keys obeying the property ks1 ≤ ks2 ≤ ... ≤ ksn . In other words, the sorting problem is to arrange a set of records so that the values of their key fields are in non-decreasing order. As defined, the Sorting Problem allows input with two or more records that have the same key value. Certain applications require that input not contain duplicate key values. The sorting algorithms presented in this chapter and in Chapter 8 can handle duplicate key values unless noted otherwise. When duplicate key values are allowed, there might be an implicit ordering to the duplicates, typically based on their order of occurrence within the input. It might be desirable to maintain this initial ordering among duplicates. A sorting
- 256. Sec. 7.2 Three Θ(n2) Sorting Algorithms 237 algorithm is said to be stable if it does not change the relative ordering of records with identical key values. Many, but not all, of the sorting algorithms presented in this chapter are stable, or can be made stable with minor changes. When comparing two sorting algorithms, the most straightforward approach would seem to be simply program both and measure their running times. An ex- ample of such timings is presented in Figure 7.13. However, such a comparison can be misleading because the running time for many sorting algorithms depends on specifics of the input values. In particular, the number of records, the size of the keys and the records, the allowable range of the key values, and the amount by which the input records are “out of order” can all greatly affect the relative running times for sorting algorithms. When analyzing sorting algorithms, it is traditional to measure the number of comparisons made between keys. This measure is usually closely related to the running time for the algorithm and has the advantage of being machine and data- type independent. However, in some cases records might be so large that their physical movement might take a significant fraction of the total running time. If so, it might be appropriate to measure the number of swap operations performed by the algorithm. In most applications we can assume that all records and keys are of fixed length, and that a single comparison or a single swap operation requires a constant amount of time regardless of which keys are involved. Some special situations “change the rules” for comparing sorting algorithms. For example, an application with records or keys having widely varying length (such as sorting a sequence of variable length strings) will benefit from a special-purpose sorting technique. Some applications require that a small number of records be sorted, but that the sort be performed frequently. An example would be an application that repeatedly sorts groups of five numbers. In such cases, the constants in the runtime equations that are usually ignored in an asymptotic analysis now become crucial. Finally, some situations require that a sorting algorithm use as little memory as possible. We will note which sorting algorithms require significant extra memory beyond the input array. 7.2 Three Θ(n2 ) Sorting Algorithms This section presents three simple sorting algorithms. While easy to understand and implement, we will soon see that they are unacceptably slow when there are many records to sort. Nonetheless, there are situations where one of these simple algorithms is the best tool for the job.
- 257. 238 Chap. 7 Internal Sorting i=1 3 4 5 6 42 20 17 13 28 14 23 15 20 42 17 13 28 14 23 15 2 17 20 42 13 28 14 23 15 13 17 20 42 28 14 23 13 17 20 28 42 14 23 13 14 17 20 28 42 23 13 14 17 20 23 28 42 13 14 15 17 20 23 28 42 7 15 15 15 15 Figure 7.1 An illustration of Insertion Sort. Each column shows the array after the iteration with the indicated value of i in the outer for loop. Values above the line in each column have been sorted. Arrows indicate the upward motions of records through the array. 7.2.1 Insertion Sort Imagine that you have a stack of phone bills from the past two years and that you wish to organize them by date. A fairly natural way to do this might be to look at the first two bills and put them in order. Then take the third bill and put it into the right order with respect to the first two, and so on. As you take each bill, you would add it to the sorted pile that you have already made. This naturally intuitive process is the inspiration for our first sorting algorithm, called Insertion Sort. Insertion Sort iterates through a list of records. Each record is inserted in turn at the correct position within a sorted list composed of those records already processed. The following is a Java implementation. The input is an array of n records stored in array A. static <E extends Comparable<? super E>> void Sort(E[] A) { for (int i=1; i<A.length; i++) // Insert i’th record for (int j=i; (j>0) && (A[j].compareTo(A[j-1])<0); j--) DSutil.swap(A, j, j-1); } Consider the case where inssort is processing the ith record, which has key value X. The record is moved upward in the array as long as X is less than the key value immediately above it. As soon as a key value less than or equal to X is encountered, inssort is done with that record because all records above it in the array must have smaller keys. Figure 7.1 illustrates how Insertion Sort works. The body of inssort is made up of two nested for loops. The outer for loop is executed n − 1 times. The inner for loop is harder to analyze because the number of times it executes depends on how many keys in positions 1 to i − 1 have a value less than that of the key in position i. In the worst case, each record
- 258. Sec. 7.2 Three Θ(n2) Sorting Algorithms 239 must make its way to the top of the array. This would occur if the keys are initially arranged from highest to lowest, in the reverse of sorted order. In this case, the number of comparisons will be one the first time through the for loop, two the second time, and so on. Thus, the total number of comparisons will be n X i=2 i = Θ(n2). In contrast, consider the best-case cost. This occurs when the keys begin in sorted order from lowest to highest. In this case, every pass through the inner for loop will fail immediately, and no values will be moved. The total number of comparisons will be n − 1, which is the number of times the outer for loop executes. Thus, the cost for Insertion Sort in the best case is Θ(n). While the best case is significantly faster than the worst case, the worst case is usually a more reliable indication of the “typical” running time. However, there are situations where we can expect the input to be in sorted or nearly sorted order. One example is when an already sorted list is slightly disordered; restoring sorted order using Insertion Sort might be a good idea if we know that the disordering is slight. Examples of algorithms that take advantage of Insertion Sort’s best-case running time are the Shellsort algorithm of Section 7.3 and the Quicksort algorithm of Section 7.5. What is the average-case cost of Insertion Sort? When record i is processed, the number of times through the inner for loop depends on how far “out of order” the record is. In particular, the inner for loop is executed once for each key greater than the key of record i that appears in array positions 0 through i−1. For example, in the leftmost column of Figure 7.1 the value 15 is preceded by five values greater than 15. Each such occurrence is called an inversion. The number of inversions (i.e., the number of values greater than a given value that occur prior to it in the array) will determine the number of comparisons and swaps that must take place. We need to determine what the average number of inversions will be for the record in position i. We expect on average that half of the keys in the first i − 1 array positions will have a value greater than that of the key at position i. Thus, the average case should be about half the cost of the worst case, which is still Θ(n2). So, the average case is no better than the worst case in asymptotic complexity. Counting comparisons or swaps yields similar results because each time through the inner for loop yields both a comparison and a swap, except the last (i.e., the comparison that fails the inner for loop’s test), which has no swap. Thus, the number of swaps for the entire sort operation is n − 1 less than the number of comparisons. This is 0 in the best case, and Θ(n2) in the average and worst cases.
- 259. 240 Chap. 7 Internal Sorting i=0 1 2 3 4 5 6 42 20 17 13 28 14 23 13 42 20 17 14 28 15 13 14 42 20 17 15 28 23 13 14 15 42 20 17 23 28 13 14 15 17 42 20 23 28 13 14 15 17 20 42 23 28 13 14 15 17 20 23 42 13 14 15 17 20 23 28 42 23 28 15 Figure 7.2 An illustration of Bubble Sort. Each column shows the array after the iteration with the indicated value of i in the outer for loop. Values above the line in each column have been sorted. Arrows indicate the swaps that take place during a given iteration. 7.2.2 Bubble Sort Our next sort is called Bubble Sort. Bubble Sort is often taught to novice pro- grammers in introductory computer science courses. This is unfortunate, because Bubble Sort has no redeeming features whatsoever. It is a relatively slow sort, it is no easier to understand than Insertion Sort, it does not correspond to any in- tuitive counterpart in “everyday” use, and it has a poor best-case running time. However, Bubble Sort serves as the basis for a better sort that will be presented in Section 7.2.3. Bubble Sort consists of a simple double for loop. The first iteration of the inner for loop moves through the record array from bottom to top, comparing adjacent keys. If the lower-indexed key’s value is greater than its higher-indexed neighbor, then the two values are swapped. Once the smallest value is encountered, this process will cause it to “bubble” up to the top of the array. The second pass through the array repeats this process. However, because we know that the smallest value reached the top of the array on the first pass, there is no need to compare the top two elements on the second pass. Likewise, each succeeding pass through the array compares adjacent elements, looking at one less value than the preceding pass. Figure 7.2 illustrates Bubble Sort. A Java implementation is as follows: static <E extends Comparable<? super E>> void Sort(E[] A) { for (int i=0; i<A.length-1; i++) // Bubble up i’th record for (int j=A.length-1; j>i; j--) if ((A[j].compareTo(A[j-1]) < 0)) DSutil.swap(A, j, j-1); }
- 260. Sec. 7.2 Three Θ(n2) Sorting Algorithms 241 Determining Bubble Sort’s number of comparisons is easy. Regardless of the arrangement of the values in the array, the number of comparisons made by the inner for loop is always i, leading to a total cost of n X i=1 i = Θ(n2). Bubble Sort’s running time is roughly the same in the best, average, and worst cases. The number of swaps required depends on how often a value is less than the one immediately preceding it in the array. We can expect this to occur for about half the comparisons in the average case, leading to Θ(n2) for the expected number of swaps. The actual number of swaps performed by Bubble Sort will be identical to that performed by Insertion Sort. 7.2.3 Selection Sort Consider again the problem of sorting a pile of phone bills for the past year. An- other intuitive approach might be to look through the pile until you find the bill for January, and pull that out. Then look through the remaining pile until you find the bill for February, and add that behind January. Proceed through the ever-shrinking pile of bills to select the next one in order until you are done. This is the inspiration for our last Θ(n2) sort, called Selection Sort. The ith pass of Selection Sort “se- lects” the ith smallest key in the array, placing that record into position i. In other words, Selection Sort first finds the smallest key in an unsorted list, then the second smallest, and so on. Its unique feature is that there are few record swaps. To find the next smallest key value requires searching through the entire unsorted portion of the array, but only one swap is required to put the record in place. Thus, the total number of swaps required will be n − 1 (we get the last record in place “for free”). Figure 7.3 illustrates Selection Sort. Below is a Java implementation. static <E extends Comparable<? super E>> void Sort(E[] A) { for (int i=0; i<A.length-1; i++) { // Select i’th record int lowindex = i; // Remember its index for (int j=A.length-1; j>i; j--) // Find the least value if (A[j].compareTo(A[lowindex]) < 0) lowindex = j; // Put it in place DSutil.swap(A, i, lowindex); } } Selection Sort is essentially a Bubble Sort, except that rather than repeatedly swapping adjacent values to get the next smallest record into place, we instead
- 261. 242 Chap. 7 Internal Sorting i=0 1 2 3 4 5 6 42 20 17 13 28 14 23 15 13 20 17 42 28 14 23 15 13 14 17 42 28 20 23 15 13 14 15 42 28 20 23 17 13 14 15 17 28 20 23 42 13 14 15 17 20 28 23 42 13 14 15 17 20 23 28 42 13 14 15 17 20 23 28 42 Figure 7.3 An example of Selection Sort. Each column shows the array after the iteration with the indicated value of i in the outer for loop. Numbers above the line in each column have been sorted and are in their final positions. Key = 42 Key = 5 Key = 42 Key = 5 (a) (b) Key = 23 Key = 10 Key = 23 Key = 10 Figure 7.4 An example of swapping pointers to records. (a) A series of four records. The record with key value 42 comes before the record with key value 5. (b) The four records after the top two pointers have been swapped. Now the record with key value 5 comes before the record with key value 42. remember the position of the element to be selected and do one swap at the end. Thus, the number of comparisons is still Θ(n2), but the number of swaps is much less than that required by bubble sort. Selection Sort is particularly advantageous when the cost to do a swap is high, for example, when the elements are long strings or other large records. Selection Sort is more efficient than Bubble Sort (by a constant factor) in most other situations as well. There is another approach to keeping the cost of swapping records low that can be used by any sorting algorithm even when the records are large. This is to have each element of the array store a pointer to a record rather than store the record itself. In this implementation, a swap operation need only exchange the pointer values; the records themselves do not move. This technique is illustrated by Figure 7.4. Additional space is needed to store the pointers, but the return is a faster swap operation.
- 262. Sec. 7.2 Three Θ(n2) Sorting Algorithms 243 Insertion Bubble Selection Comparisons: Best Case Θ(n) Θ(n2) Θ(n2) Average Case Θ(n2) Θ(n2) Θ(n2) Worst Case Θ(n2) Θ(n2) Θ(n2) Swaps: Best Case 0 0 Θ(n) Average Case Θ(n2) Θ(n2) Θ(n) Worst Case Θ(n2) Θ(n2) Θ(n) Figure 7.5 A comparison of the asymptotic complexities for three simple sorting algorithms. 7.2.4 The Cost of Exchange Sorting Figure 7.5 summarizes the cost of Insertion, Bubble, and Selection Sort in terms of their required number of comparisons and swaps1 in the best, average, and worst cases. The running time for each of these sorts is Θ(n2) in the average and worst cases. The remaining sorting algorithms presented in this chapter are significantly bet- ter than these three under typical conditions. But before continuing on, it is instruc- tive to investigate what makes these three sorts so slow. The crucial bottleneck is that only adjacent records are compared. Thus, comparisons and moves (in all but Selection Sort) are by single steps. Swapping adjacent records is called an ex- change. Thus, these sorts are sometimes referred to as exchange sorts. The cost of any exchange sort can be at best the total number of steps that the records in the array must move to reach their “correct” location (i.e., the number of inversions for each record). What is the average number of inversions? Consider a list L containing n val- ues. Define LR to be L in reverse. L has n(n−1)/2 distinct pairs of values, each of which could potentially be an inversion. Each such pair must either be an inversion in L or in LR. Thus, the total number of inversions in L and LR together is exactly n(n−1)/2 for an average of n(n−1)/4 per list. We therefore know with certainty that any sorting algorithm which limits comparisons to adjacent items will cost at least n(n − 1)/4 = Ω(n2) in the average case. 1 There is a slight anomaly with Selection Sort. The supposed advantage for Selection Sort is its low number of swaps required, yet Selection Sort’s best-case number of swaps is worse than that for Insertion Sort or Bubble Sort. This is because the implementation given for Selection Sort does not avoid a swap in the case where record i is already in position i. The reason is that it usually takes more time to repeatedly check for this situation than would be saved by avoiding such swaps.
- 263. 244 Chap. 7 Internal Sorting 7.3 Shellsort The next sort we consider is called Shellsort, named after its inventor, D.L. Shell. It is also sometimes called the diminishing increment sort. Unlike Insertion and Selection Sort, there is no real life intuitive equivalent to Shellsort. Unlike the exchange sorts, Shellsort makes comparisons and swaps between non-adjacent el- ements. Shellsort also exploits the best-case performance of Insertion Sort. Shell- sort’s strategy is to make the list “mostly sorted” so that a final Insertion Sort can finish the job. When properly implemented, Shellsort will give substantially better performance than Θ(n2) in the worst case. Shellsort uses a process that forms the basis for many of the sorts presented in the following sections: Break the list into sublists, sort them, then recombine the sublists. Shellsort breaks the array of elements into “virtual” sublists. Each sublist is sorted using an Insertion Sort. Another group of sublists is then chosen and sorted, and so on. During each iteration, Shellsort breaks the list into disjoint sublists so that each element in a sublist is a fixed number of positions apart. For example, let us as- sume for convenience that n, the number of values to be sorted, is a power of two. One possible implementation of Shellsort will begin by breaking the list into n/2 sublists of 2 elements each, where the array index of the 2 elements in each sublist differs by n/2. If there are 16 elements in the array indexed from 0 to 15, there would initially be 8 sublists of 2 elements each. The first sublist would be the ele- ments in positions 0 and 8, the second in positions 1 and 9, and so on. Each list of two elements is sorted using Insertion Sort. The second pass of Shellsort looks at fewer, bigger lists. For our example the second pass would have n/4 lists of size 4, with the elements in the list being n/4 positions apart. Thus, the second pass would have as its first sublist the 4 elements in positions 0, 4, 8, and 12; the second sublist would have elements in positions 1, 5, 9, and 13; and so on. Each sublist of four elements would also be sorted using an Insertion Sort. The third pass would be made on two lists, one consisting of the odd positions and the other consisting of the even positions. The culminating pass in this example would be a “normal” Insertion Sort of all elements. Figure 7.6 illustrates the process for an array of 16 values where the sizes of the increments (the distances between elements on the successive passes) are 8, 4, 2, and 1. Below is a Java implementation for Shellsort.
- 264. Sec. 7.3 Shellsort 245 59 20 17 13 28 14 23 83 36 98 59 15 23 14 28 13 11 20 36 28 14 11 13 36 20 17 15 98 36 20 28 15 23 14 17 13 11 11 13 14 15 17 20 23 28 36 41 42 59 65 70 83 98 11 70 65 41 42 15 83 42 41 65 70 17 98 98 42 83 59 41 23 70 65 65 83 59 70 42 41 Figure 7.6 An example of Shellsort. Sixteen items are sorted in four passes. The first pass sorts 8 sublists of size 2 and increment 8. The second pass sorts 4 sublists of size 4 and increment 4. The third pass sorts 2 sublists of size 8 and increment 2. The fourth pass sorts 1 list of size 16 and increment 1 (a regular Insertion Sort). static <E extends Comparable<? super E>> void Sort(E[] A) { for (int i=A.length/2; i>2; i/=2) // For each increment for (int j=0; j<i; j++) // Sort each sublist inssort2(A, j, i); inssort2(A, 0, 1); // Could call regular inssort here } // Modified version of Insertion Sort for varying increments static <E extends Comparable<? super E>> void inssort2(E[] A, int start, int incr) { for (int i=start+incr; i<A.length; i+=incr) for (int j=i; (j>=incr)&& (A[j].compareTo(A[j-incr])<0); j-=incr) DSutil.swap(A, j, j-incr); } Shellsort will work correctly regardless of the size of the increments, provided that the final pass has increment 1 (i.e., provided the final pass is a regular Insertion Sort). If Shellsort will aways conclude with a regular Insertion Sort, then how can it be any improvement on Insertion Sort? The expectation is that each of the (relatively cheap) sublist sorts will make the list “more sorted” than it was before. It is not necessarily the case that this will be true, but it is almost always true in practice. When the final Insertion Sort is conducted, the list should be “almost sorted,” yielding a relatively cheap final Insertion Sort pass.
- 265. 246 Chap. 7 Internal Sorting Some choices for increments will make Shellsort run more efficiently than oth- ers. In particular, the choice of increments described above (2k, 2k−1, ..., 2, 1) turns out to be relatively inefficient. A better choice is the following series based on division by three: (..., 121, 40, 13, 4, 1). The analysis of Shellsort is difficult, so we must accept without proof that the average-case performance of Shellsort (for “divisions by three” increments) is O(n1.5). Other choices for the increment series can reduce this upper bound some- what. Thus, Shellsort is substantially better than Insertion Sort, or any of the Θ(n2) sorts presented in Section 7.2. In fact, Shellsort is competitive with the asymptoti- cally better sorts to be presented whenever n is of medium size. Shellsort illustrates how we can sometimes exploit the special properties of an algorithm (in this case Insertion Sort) even if in general that algorithm is unacceptably slow. 7.4 Mergesort A natural approach to problem solving is divide and conquer. In terms of sorting, we might consider breaking the list to be sorted into pieces, process the pieces, and then put them back together somehow. A simple way to do this would be to split the list in half, sort the halves, and then merge the sorted halves together. This is the idea behind Mergesort. Mergesort is one of the simplest sorting algorithms conceptually, and has good performance both in the asymptotic sense and in empirical running time. Supris- ingly, even though it is based on a simple concept, it is relatively difficult to im- plement in practice. Figure 7.7 illustrates Mergesort. A pseudocode sketch of Mergesort is as follows: List mergesort(List inlist) { if (inlist.length() <= 1) return inlist;; List l1 = half of the items from inlist; List l2 = other half of the items from inlist; return merge(mergesort(l1), mergesort(l2)); } Before discussing how to implement Mergesort, we will first examine the merge function. Merging two sorted sublists is quite simple. Function merge examines the first element of each sublist and picks the smaller value as the smallest element overall. This smaller value is removed from its sublist and placed into the output list. Merging continues in this way, comparing the front elements of the sublists and continually appending the smaller to the output list until no more input elements remain.
- 266. Sec. 7.4 Mergesort 247 36 20 17 13 28 14 23 15 28 23 15 14 36 20 17 13 20 36 13 17 14 28 15 23 13 14 15 17 20 23 28 36 Figure 7.7 An illustration of Mergesort. The first row shows eight numbers that are to be sorted. Mergesort will recursively subdivide the list into sublists of one element each, then recombine the sublists. The second row shows the four sublists of size 2 created by the first merging pass. The third row shows the two sublists of size 4 created by the next merging pass on the sublists of row 2. The last row shows the final sorted list created by merging the two sublists of row 3. Implementing Mergesort presents a number of technical difficulties. The first decision is how to represent the lists. Mergesort lends itself well to sorting a singly linked list because merging does not require random access to the list elements. Thus, Mergesort is the method of choice when the input is in the form of a linked list. Implementing merge for linked lists is straightforward, because we need only remove items from the front of the input lists and append items to the output list. Breaking the input list into two equal halves presents some difficulty. Ideally we would just break the lists into front and back halves. However, even if we know the length of the list in advance, it would still be necessary to traverse halfway down the linked list to reach the beginning of the second half. A simpler method, which does not rely on knowing the length of the list in advance, assigns elements of the input list alternating between the two sublists. The first element is assigned to the first sublist, the second element to the second sublist, the third to first sublist, the fourth to the second sublist, and so on. This requires one complete pass through the input list to build the sublists. When the input to Mergesort is an array, splitting input into two subarrays is easy if we know the array bounds. Merging is also easy if we merge the subarrays into a second array. Note that this approach requires twice the amount of space as any of the sorting methods presented so far, which is a serious disadvantage for Mergesort. It is possible to merge the subarrays without using a second array, but this is extremely difficult to do efficiently and is not really practical. Merging the two subarrays into a second array, while simple to implement, presents another dif- ficulty. The merge process ends with the sorted list in the auxiliary array. Consider how the recursive nature of Mergesort breaks the original array into subarrays, as shown in Figure 7.7. Mergesort is recursively called until subarrays of size 1 have been created, requiring log n levels of recursion. These subarrays are merged into
- 267. 248 Chap. 7 Internal Sorting subarrays of size 2, which are in turn merged into subarrays of size 4, and so on. We need to avoid having each merge operation require a new array. With some difficulty, an algorithm can be devised that alternates between two arrays. A much simpler approach is to copy the sorted sublists to the auxiliary array first, and then merge them back to the original array. Here is a complete implementation for mergesort following this approach: static <E extends Comparable<? super E>> void mergesort(E[] A, E[] temp, int l, int r) { int mid = (l+r)/2; // Select midpoint if (l == r) return; // List has one element mergesort(A, temp, l, mid); // Mergesort first half mergesort(A, temp, mid+1, r); // Mergesort second half for (int i=l; i<=r; i++) // Copy subarray to temp temp[i] = A[i]; // Do the merge operation back to A int i1 = l; int i2 = mid + 1; for (int curr=l; curr<=r; curr++) { if (i1 == mid+1) // Left sublist exhausted A[curr] = temp[i2++]; else if (i2 > r) // Right sublist exhausted A[curr] = temp[i1++]; else if (temp[i1].compareTo(temp[i2])<0) // Get smaller A[curr] = temp[i1++]; else A[curr] = temp[i2++]; } } An optimized Mergesort implementation is shown next. It reverses the order of the second subarray during the initial copy. Now the current positions of the two subarrays work inwards from the ends, allowing the end of each subarray to act as a sentinel for the other. Unlike the previous implementation, no test is needed to check for when one of the two subarrays becomes empty. This version also uses Insertion Sort to sort small subarrays.
- 268. Sec. 7.5 Quicksort 249 static <E extends Comparable<? super E>> void mergesort(E[] A, E[] temp, int l, int r) { int i, j, k, mid = (l+r)/2; // Select the midpoint if (l == r) return; // List has one element if ((mid-l) >= THRESHOLD) mergesort(A, temp, l, mid); else inssort(A, l, mid-l+1); if ((r-mid) > THRESHOLD) mergesort(A, temp, mid+1, r); else inssort(A, mid+1, r-mid); // Do the merge operation. First, copy 2 halves to temp. for (i=l; i<=mid; i++) temp[i] = A[i]; for (j=1; j<=r-mid; j++) temp[r-j+1] = A[j+mid]; // Merge sublists back to array for (i=l,j=r,k=l; k<=r; k++) if (temp[i].compareTo(temp[j])<0) A[k] = temp[i++]; else A[k] = temp[j--]; } Analysis of Mergesort is straightforward, despite the fact that it is a recursive algorithm. The merging part takes time Θ(i) where i is the total length of the two subarrays being merged. The array to be sorted is repeatedly split in half until subarrays of size 1 are reached, at which time they are merged to be of size 2, these merged to subarrays of size 4, and so on as shown in Figure 7.7. Thus, the depth of the recursion is log n for n elements (assume for simplicity that n is a power of two). The first level of recursion can be thought of as working on one array of size n, the next level working on two arrays of size n/2, the next on four arrays of size n/4, and so on. The bottom of the recursion has n arrays of size 1. Thus, n arrays of size 1 are merged (requiring n total steps), n/2 arrays of size 2 (again requiring n total steps), n/4 arrays of size 4, and so on. At each of the log n levels of recursion, Θ(n) work is done, for a total cost of Θ(n log n). This cost is unaffected by the relative order of the values being sorted, thus this analysis holds for the best, average, and worst cases. 7.5 Quicksort While Mergesort uses the most obvious form of divide and conquer (split the list in half then sort the halves), it is not the only way that we can break down the sorting problem. And we saw that doing the merge step for Mergesort when using an array implementation is not so easy. So prehaps a different divide and conquer strategy might turn out to be more efficient? Quicksort is aptly named because, when properly implemented, it is the fastest known general-purpose in-memory sorting algorithm in the average case. It does not require the extra array needed by Mergesort, so it is space efficent as well. Quicksort is widely used, and is typically the algorithm implemented in a library
- 269. 250 Chap. 7 Internal Sorting sort routine such as the UNIX qsort function. Interestingly, Quicksort is ham- pered by exceedingly poor worst-case performance, thus making it inappropriate for certain applications. Before we get to Quicksort, consider for a moment the practicality of using a Binary Search Tree for sorting. You could insert all of the values to be sorted into the BST one by one, then traverse the completed tree using an inorder traversal. The output would form a sorted list. This approach has a number of drawbacks, including the extra space required by BST pointers and the amount of time required to insert nodes into the tree. However, this method introduces some interesting ideas. First, the root of the BST (i.e., the first node inserted) splits the list into two sublits: The left subtree contains those values in the list less than the root value while the right subtree contains those values in the list greater than or equal to the root value. Thus, the BST implicitly implements a “divide and conquer” approach to sorting the left and right subtrees. Quicksort implements this concept in a much more efficient way. Quicksort first selects a value called the pivot. Assume that the input array contains k values less than the pivot. The records are then rearranged in such a way that the k values less than the pivot are placed in the first, or leftmost, k positions in the array, and the values greater than or equal to the pivot are placed in the last, or rightmost, n − k positions. This is called a partition of the array. The values placed in a given partition need not (and typically will not) be sorted with respect to each other. All that is required is that all values end up in the correct partition. The pivot value itself is placed in position k. Quicksort then proceeds to sort the resulting subarrays now on either side of the pivot, one of size k and the other of size n − k − 1. How are these values sorted? Because Quicksort is such a good algorithm, using Quicksort on the subarrays would be appropriate. Unlike some of the sorts that we have seen earlier in this chapter, Quicksort might not seem very “natural” in that it is not an approach that a person is likely to use to sort real objects. But it should not be too suprising that a really efficient sort for huge numbers of abstract objects on a computer would be rather different from our experiences with sorting a relatively few physical objects. The Java code for Quicksort is as follows. Parameters i and j define the left and right indices, respectively, for the subarray being sorted. The initial call to Quicksort would be qsort(array, 0, n-1).
- 270. Sec. 7.5 Quicksort 251 static <E extends Comparable<? super E>> void qsort(E[] A, int i, int j) { // Quicksort int pivotindex = findpivot(A, i, j); // Pick a pivot DSutil.swap(A, pivotindex, j); // Stick pivot at end // k will be the first position in the right subarray int k = partition(A, i-1, j, A[j]); DSutil.swap(A, k, j); // Put pivot in place if ((k-i) > 1) qsort(A, i, k-1); // Sort left partition if ((j-k) > 1) qsort(A, k+1, j); // Sort right partition } Function partition will move records to the appropriate partition and then return k, the first position in the right partition. Note that the pivot value is initially placed at the end of the array (position j). Thus, partition must not affect the value of array position j. After partitioning, the pivot value is placed in position k, which is its correct position in the final, sorted array. By doing so, we guarantee that at least one value (the pivot) will not be processed in the recursive calls to qsort. Even if a bad pivot is selected, yielding a completely empty partition to one side of the pivot, the larger partition will contain at most n − 1 elements. Selecting a pivot can be done in many ways. The simplest is to use the first key. However, if the input is sorted or reverse sorted, this will produce a poor partitioning with all values to one side of the pivot. It is better to pick a value at random, thereby reducing the chance of a bad input order affecting the sort. Unfortunately, using a random number generator is relatively expensive, and we can do nearly as well by selecting the middle position in the array. Here is a simple findpivot function: static <E extends Comparable<? super E>> int findpivot(E[] A, int i, int j) { return (i+j)/2; } We now turn to function partition. If we knew in advance how many keys are less than the pivot, partition could simply copy elements with key values less than the pivot to the low end of the array, and elements with larger keys to the high end. Because we do not know in advance how many keys are less than the pivot, we use a clever algorithm that moves indices inwards from the ends of the subarray, swapping values as necessary until the two indices meet. Here is a Java implementation for the partition step:
- 271. 252 Chap. 7 Internal Sorting static <E extends Comparable<? super E>> int partition(E[] A, int l, int r, E pivot) { do { // Move bounds inward until they meet while (A[++l].compareTo(pivot)<0); while ((r!=0) && (A[--r].compareTo(pivot)>0)); DSutil.swap(A, l, r); // Swap out-of-place values } while (l < r); // Stop when they cross DSutil.swap(A, l, r); // Reverse last, wasted swap return l; // Return first position in right partition } Figure 7.8 illustrates partition. Initially, variables l and r are immediately outside the actual bounds of the subarray being partitioned. Each pass through the outer do loop moves the counters l and r inwards, until eventually they meet. Note that at each iteration of the inner while loops, the bounds are moved prior to checking against the pivot value. This ensures that progress is made by each while loop, even when the two values swapped on the last iteration of the do loop were equal to the pivot. Also note the check that r > l in the second while loop. This ensures that r does not run off the low end of the partition in the case where the pivot is the least value in that partition. Function partition returns the first index of the right partition so that the subarray bound for the recursive calls to qsort can be determined. Figure 7.9 illustrates the complete Quicksort algorithm. To analyze Quicksort, we first analyze the findpivot and partition functions operating on a subarray of length k. Clearly, findpivot takes con- stant time. Function partition contains a do loop with two nested while loops. The total cost of the partition operation is constrained by how far l and r can move inwards. In particular, these two bounds variables together can move a total of s steps for a subarray of length s. However, this does not directly tell us how much work is done by the nested while loops. The do loop as a whole is guaranteed to move both l and r inward at least one position on each first pass. Each while loop moves its variable at least once (except in the special case where r is at the left edge of the array, but this can happen only once). Thus, we see that the do loop can be executed at most s times, the total amount of work done moving l and r is s, and each while loop can fail its test at most s times. The total work for the entire partition function is therefore Θ(s). Knowing the cost of findpivot and partition, we can determine the cost of Quicksort. We begin with a worst-case analysis. The worst case will occur when the pivot does a poor job of breaking the array, that is, when there are no elements in one partition, and n − 1 elements in the other. In this case, the divide and conquer strategy has done a poor job of dividing, so the conquer phase will work on a subproblem only one less than the size of the original problem. If this
- 272. Sec. 7.5 Quicksort 253 Pass 1 Swap 1 Pass 2 Swap 2 Pass 3 72 6 57 88 85 42 83 73 48 60 l r 72 6 57 88 85 42 83 73 48 60 48 6 57 88 85 42 83 73 72 60 r 48 6 57 88 85 42 83 73 72 60 l 48 6 57 42 85 88 83 73 72 60 r l 48 6 57 42 88 83 73 72 60 Initial l l r r 85 l,r Figure 7.8 The Quicksort partition step. The first row shows the initial positions for a collection of ten key values. The pivot value is 60, which has been swapped to the end of the array. The do loop makes three iterations, each time moving counters l and r inwards until they meet in the third pass. In the end, the left partition contains four values and the right partition contains six values. Function qsort will place the pivot value into position 4. Pivot = 6 Pivot = 73 Pivot = 57 Final Sorted Array Pivot = 60 Pivot = 88 42 57 48 57 6 42 48 57 60 72 73 83 85 88 Pivot = 42 Pivot = 85 6 57 88 60 42 83 73 48 85 85 72 73 83 88 60 42 57 6 48 6 48 42 42 48 85 83 88 85 83 72 73 85 88 83 72 Figure 7.9 An illustration of Quicksort.
- 273. 254 Chap. 7 Internal Sorting happens at each partition step, then the total cost of the algorithm will be n X k=1 k = Θ(n2). In the worst case, Quicksort is Θ(n2). This is terrible, no better than Bubble Sort.2 When will this worst case occur? Only when each pivot yields a bad parti- tioning of the array. If the pivot values are selected at random, then this is extremely unlikely to happen. When selecting the middle position of the current subarray, it is still unlikely to happen. It does not take many good partitionings for Quicksort to work fairly well. Quicksort’s best case occurs when findpivot always breaks the array into two equal halves. Quicksort repeatedly splits the array into smaller partitions, as shown in Figure 7.9. In the best case, the result will be log n levels of partitions, with the top level having one array of size n, the second level two arrays of size n/2, the next with four arrays of size n/4, and so on. Thus, at each level, all partition steps for that level do a total of n work, for an overall cost of n log n work when Quicksort finds perfect pivots. Quicksort’s average-case behavior falls somewhere between the extremes of worst and best case. Average-case analysis considers the cost for all possible ar- rangements of input, summing the costs and dividing by the number of cases. We make one reasonable simplifying assumption: At each partition step, the pivot is equally likely to end in any position in the (sorted) array. In other words, the pivot is equally likely to break an array into partitions of sizes 0 and n−1, or 1 and n−2, and so on. Given this assumption, the average-case cost is computed from the following equation: T(n) = cn + 1 n n−1 X k=0 [T(k) + T(n − 1 − k)], T(0) = T(1) = c. This equation is in the form of a recurrence relation. Recurrence relations are discussed in Chapters 2 and 14, and this one is solved in Section 14.2.4. This equation says that there is one chance in n that the pivot breaks the array into subarrays of size 0 and n − 1, one chance in n that the pivot breaks the array into subarrays of size 1 and n − 2, and so on. The expression “T(k) + T(n − 1 − k)” is the cost for the two recursive calls to Quicksort on two arrays of size k and n−1−k. 2 The worst insult that I can think of for a sorting algorithm.
- 274. Sec. 7.5 Quicksort 255 The initial cn term is the cost of doing the findpivot and partition steps, for some constant c. The closed-form solution to this recurrence relation is Θ(n log n). Thus, Quicksort has average-case cost Θ(n log n). This is an unusual situation that the average case cost and the worst case cost have asymptotically different growth rates. Consider what “average case” actually means. We compute an average cost for inputs of size n by summing up for every possible input of size n the product of the running time cost of that input times the probability that that input will occur. To simplify things, we assumed that every permutation is equally likely to occur. Thus, finding the average means summing up the cost for every permutation and dividing by the number of inputs (n!). We know that some of these n! inputs cost O(n2). But the sum of all the permutation costs has to be (n!)(O(n log n)). Given the extremely high cost of the worst inputs, there must be very few of them. In fact, there cannot be a constant fraction of the inputs with cost O(n2). Even, say, 1% of the inputs with cost O(n2) would lead to an average cost of O(n2). Thus, as n grows, the fraction of inputs with high cost must be going toward a limit of zero. We can conclude that Quicksort will always have good behavior if we can avoid those very few bad input permutations. The running time for Quicksort can be improved (by a constant factor), and much study has gone into optimizing this algorithm. The most obvious place for improvement is the findpivot function. Quicksort’s worst case arises when the pivot does a poor job of splitting the array into equal size subarrays. If we are willing to do more work searching for a better pivot, the effects of a bad pivot can be decreased or even eliminated. One good choice is to use the “median of three” algorithm, which uses as a pivot the middle of three randomly selected values. Using a random number generator to choose the positions is relatively expensive, so a common compromise is to look at the first, middle, and last positions of the current subarray. However, our simple findpivot function that takes the middle value as its pivot has the virtue of making it highly unlikely to get a bad input by chance, and it is quite cheap to implement. This is in sharp contrast to selecting the first or last element as the pivot, which would yield bad performance for many permutations that are nearly sorted or nearly reverse sorted. A significant improvement can be gained by recognizing that Quicksort is rel- atively slow when n is small. This might not seem to be relevant if most of the time we sort large arrays, nor should it matter how long Quicksort takes in the rare instance when a small array is sorted because it will be fast anyway. But you should notice that Quicksort itself sorts many, many small arrays! This happens as a natural by-product of the divide and conquer approach.
- 275. 256 Chap. 7 Internal Sorting A simple improvement might then be to replace Quicksort with a faster sort for small numbers, say Insertion Sort or Selection Sort. However, there is an even better — and still simpler — optimization. When Quicksort partitions are below a certain size, do nothing! The values within that partition will be out of order. However, we do know that all values in the array to the left of the partition are smaller than all values in the partition. All values in the array to the right of the partition are greater than all values in the partition. Thus, even if Quicksort only gets the values to “nearly” the right locations, the array will be close to sorted. This is an ideal situation in which to take advantage of the best-case performance of Insertion Sort. The final step is a single call to Insertion Sort to process the entire array, putting the elements into final sorted order. Empirical testing shows that the subarrays should be left unordered whenever they get down to nine or fewer elements. The last speedup to be considered reduces the cost of making recursive calls. Quicksort is inherently recursive, because each Quicksort operation must sort two sublists. Thus, there is no simple way to turn Quicksort into an iterative algorithm. However, Quicksort can be implemented using a stack to imitate recursion, as the amount of information that must be stored is small. We need not store copies of a subarray, only the subarray bounds. Furthermore, the stack depth can be kept small if care is taken on the order in which Quicksort’s recursive calls are executed. We can also place the code for findpivot and partition inline to eliminate the remaining function calls. Note however that by not processing sublists of size nine or less as suggested above, about three quarters of the function calls will already have been eliminated. Thus, eliminating the remaining function calls will yield only a modest speedup. 7.6 Heapsort Our discussion of Quicksort began by considering the practicality of using a binary search tree for sorting. The BST requires more space than the other sorting meth- ods and will be slower than Quicksort or Mergesort due to the relative expense of inserting values into the tree. There is also the possibility that the BST might be un- balanced, leading to a Θ(n2) worst-case running time. Subtree balance in the BST is closely related to Quicksort’s partition step. Quicksort’s pivot serves roughly the same purpose as the BST root value in that the left partition (subtree) stores val- ues less than the pivot (root) value, while the right partition (subtree) stores values greater than or equal to the pivot (root). A good sorting algorithm can be devised based on a tree structure more suited to the purpose. In particular, we would like the tree to be balanced, space efficient,
- 276. Sec. 7.6 Heapsort 257 and fast. The algorithm should take advantage of the fact that sorting is a special- purpose application in that all of the values to be stored are available at the start. This means that we do not necessarily need to insert one value at a time into the tree structure. Heapsort is based on the heap data structure presented in Section 5.5. Heapsort has all of the advantages just listed. The complete binary tree is balanced, its array representation is space efficient, and we can load all values into the tree at once, taking advantage of the efficient buildheap function. The asymptotic perfor- mance of Heapsort is Θ(n log n) in the best, average, and worst cases. It is not as fast as Quicksort in the average case (by a constant factor), but Heapsort has special properties that will make it particularly useful when sorting data sets too large to fit in main memory, as discussed in Chapter 8. A sorting algorithm based on max-heaps is quite straightforward. First we use the heap building algorithm of Section 5.5 to convert the array into max-heap order. Then we repeatedly remove the maximum value from the heap, restoring the heap property each time that we do so, until the heap is empty. Note that each time we remove the maximum element from the heap, it is placed at the end of the array. Assume the n elements are stored in array positions 0 through n−1. After removing the maximum value from the heap and readjusting, the maximum value will now be placed in position n − 1 of the array. The heap is now considered to be of size n − 1. Removing the new maximum (root) value places the second largest value in position n − 2 of the array. At the end of the process, the array will be properly sorted from least to greatest. This is why Heapsort uses a max-heap rather than a min-heap as might have been expected. Figure 7.10 illustrates Heapsort. The complete Java implementation is as follows: static <E extends Comparable<? super E>> void heapsort(E[] A) { // Heapsort MaxHeap<E> H = new MaxHeap<E>(A, A.length, A.length); for (int i=0; i<A.length; i++) // Now sort H.removemax(); // Removemax places max at end of heap } Because building the heap takes Θ(n) time (see Section 5.5), and because n deletions of the maximum element each take Θ(log n) time, we see that the en- tire heapsort operation takes Θ(n log n) time in the worst, average, and best cases. While typically slower than Quicksort by a constant factor, Heapsort has one spe- cial advantage over the other sorts studied so far. Building the heap is relatively cheap, requiring Θ(n) time. Removing the maximum element from the heap re- quires Θ(log n) time. Thus, if we wish to find the k largest elements in an array, we can do so in time Θ(n+k log n). If k is small, this is a substantial improvement
- 277. 258 Chap. 7 Internal Sorting Original Numbers Build Heap Remove 88 Remove 85 Remove 83 73 88 60 48 60 48 72 6 48 60 42 57 72 60 6 42 48 6 60 42 48 6 57 85 85 72 42 83 72 73 42 57 6 72 57 73 83 73 57 83 72 6 60 42 48 83 85 88 57 73 73 57 72 60 42 6 88 83 48 85 73 72 60 42 57 88 83 6 85 48 88 85 83 72 73 57 6 42 60 48 6 57 88 60 42 83 48 85 73 72 88 85 83 73 Figure 7.10 An illustration of Heapsort. The top row shows the values in their original order. The second row shows the values after building the heap. The third row shows the result of the first removefirst operation on key value 88. Note that 88 is now at the end of the array. The fourth row shows the result of the second removefirst operation on key value 85. The fifth row shows the result of the third removefirst operation on key value 83. At this point, the last three positions of the array hold the three greatest values in sorted order. Heapsort continues in this manner until the entire array is sorted.
- 278. Sec. 7.7 Binsort and Radix Sort 259 over the time required to find the k largest elements using one of the other sorting methods described earlier. One situation where we are able to take advantage of this concept is in the implementation of Kruskal’s minimum-cost spanning tree (MST) algorithm of Section 11.5.2. That algorithm requires that edges be visited in as- cending order (so, use a min-heap), but this process stops as soon as the MST is complete. Thus, only a relatively small fraction of the edges need be sorted. 7.7 Binsort and Radix Sort Imagine that for the past year, as you paid your various bills, you then simply piled all the paperwork onto the top of a table somewhere. Now the year has ended and its time to sort all of these papers by what the bill was for (phone, electricity, rent, etc.) and date. A pretty natural approach is to make some space on the floor, and as you go through the pile of papers, put the phone bills into one pile, the electric bills into another pile, and so on. Once this initial assignment of bills to piles is done (in one pass), you can sort each pile by date relatively quickly because they are each fairly small. This is the basic idea behind a Binsort. Section 3.9 presented the following code fragment to sort a permutation of the numbers 0 through n − 1: for (i=0; i<n; i++) B[A[i]] = A[i]; Here the key value is used to determine the position for a record in the final sorted array. This is the most basic example of a Binsort, where key values are used to assign records to bins. This algorithm is extremely efficient, taking Θ(n) time regardless of the initial ordering of the keys. This is far better than the performance of any sorting algorithm that we have seen so far. The only problem is that this algorithm has limited use because it works only for a permutation of the numbers from 0 to n − 1. We can extend this simple Binsort algorithm to be more useful. Because Binsort must perform direct computation on the key value (as opposed to just asking which of two records comes first as our previous sorting algorithms did), we will assume that the records use an integer key type. The simplest extension is to allow for duplicate values among the keys. This can be done by turning array slots into arbitrary-length bins by turning B into an array of linked lists. In this way, all records with key value i can be placed in bin B[i]. A second extension allows for a key range greater than n. For example, a set of n records might have keys in the range 1 to 2n. The only requirement is
- 279. 260 Chap. 7 Internal Sorting that each possible key value have a corresponding bin in B. The extended Binsort algorithm is as follows: static void binsort(Integer A[]) { List<Integer>[] B = (LList<Integer>[])new LList[MaxKey]; Integer item; for (int i=0; i<MaxKey; i++) B[i] = new LList<Integer>(); for (int i=0; i<A.length; i++) B[A[i]].append(A[i]); for (int i=0; i<MaxKey; i++) for (B[i].moveToStart(); (item = B[i].getValue()) != null; B[i].next()) output(item); } This version of Binsort can sort any collection of records whose key values fall in the range from 0 to MaxKeyValue−1. The total work required is simply that needed to place each record into the appropriate bin and then take all of the records out of the bins. Thus, we need to process each record twice, for Θ(n) work. Unfortunately, there is a crucial oversight in this analysis. Binsort must also look at each of the bins to see if it contains a record. The algorithm must process MaxKeyValue bins, regardless of how many actually hold records. If MaxKey- Value is small compared to n, then this is not a great expense. Suppose that MaxKeyValue = n2. In this case, the total amount of work done will be Θ(n + n2) = Θ(n2). This results in a poor sorting algorithm, and the algorithm becomes even worse as the disparity between n and MaxKeyValue increases. In addition, a large key range requires an unacceptably large array B. Thus, even the extended Binsort is useful only for a limited key range. A further generalization to Binsort yields a bucket sort. Each bin is associated with not just one key, but rather a range of key values. A bucket sort assigns records to bins and then relies on some other sorting technique to sort the records within each bin. The hope is that the relatively inexpensive bucketing process will put only a small number of records in each bin, and that a “cleanup sort” within the bins will then be relatively cheap. There is a way to keep the number of bins and the related processing small while allowing the cleanup sort to be based on Binsort. Consider a sequence of records with keys in the range 0 to 99. If we have ten bins available, we can first assign records to bins by taking their key value modulo 10. Thus, every key will be assigned to the bin matching its rightmost decimal digit. We can then take these records from the bins in order and reassign them to the bins on the basis of their leftmost (10’s place) digit (define values in the range 0 to 9 to have a leftmost digit
- 280. Sec. 7.7 Binsort and Radix Sort 261 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 Result of first pass: 91 1 72 23 84 5 25 27 97 17 67 28 Result of second pass: 1 17 5 23 25 27 28 67 72 84 91 97 First pass (on right digit) Second pass (on left digit) Initial List: 27 91 1 97 17 23 84 28 72 5 67 25 23 84 5 25 27 28 91 1 72 97 17 67 17 91 97 72 84 1 5 23 25 67 27 28 Figure 7.11 An example of Radix Sort for twelve two-digit numbers in base ten. Two passes are required to sort the list. of 0). In other words, assign the ith record from array A to a bin using the formula A[i]/10. If we now gather the values from the bins in order, the result is a sorted list. Figure 7.11 illustrates this process. In this example, we have r = 10 bins and n = 12 keys in the range 0 to r2 − 1. The total computation is Θ(n), because we look at each record and each bin a constant number of times. This is a great improvement over the simple Binsort where the number of bins must be as large as the key range. Note that the example uses r = 10 so as to make the bin computations easy to visualize: Records were placed into bins based on the value of first the rightmost and then the leftmost decimal digits. Any number of bins would have worked. This is an example of a Radix Sort, so called because the bin computations are based on the radix or the base of the key values. This sorting algorithm can be extended to any number of keys in any key range. We simply assign records to bins based on the keys’ digit values working from the rightmost digit to the leftmost. If there are k digits, then this requires that we assign keys to bins k times.
- 281. 262 Chap. 7 Internal Sorting As with Mergesort, an efficient implementation of Radix Sort is somewhat dif- ficult to achieve. In particular, we would prefer to sort an array of values and avoid processing linked lists. If we know how many values will be in each bin, then an auxiliary array of size r can be used to hold the bins. For example, if during the first pass the 0 bin will receive three records and the 1 bin will receive five records, then we could simply reserve the first three array positions for the 0 bin and the next five array positions for the 1 bin. Exactly this approach is taken by the following Java implementation. At the end of each pass, the records are copied back to the original array. static void radix(Integer[] A, Integer[] B, int k, int r, int[] count) { // Count[i] stores number of records in bin[i] int i, j, rtok; for (i=0, rtok=1; i<k; i++, rtok*=r) { // For k digits for (j=0; j<r; j++) count[j] = 0; // Initialize count // Count the number of records for each bin on this pass for (j=0; j<A.length; j++) count[(A[j]/rtok)%r]++; // count[j] will be index in B for last slot of bin j. for (j=1; j<r; j++) count[j] = count[j-1] + count[j]; // Put records into bins, working from bottom of bin // Since bins fill from bottom, j counts downwards for (j=A.length-1; j>=0; j--) B[--count[(A[j]/rtok)%r]] = A[j]; for (j=0; j<A.length; j++) A[j] = B[j]; // Copy B back } } The first inner for loop initializes array cnt. The second loop counts the number of records to be assigned to each bin. The third loop sets the values in cnt to their proper indices within array B. Note that the index stored in cnt[j] is the last index for bin j; bins are filled from the bottom. The fourth loop assigns the records to the bins (within array B). The final loop simply copies the records back to array A to be ready for the next pass. Variable rtoi stores ri for use in bin computation on the i’th iteration. Figure 7.12 shows how this algorithm processes the input shown in Figure 7.11. This algorithm requires k passes over the list of n numbers in base r, with Θ(n + r) work done at each pass. Thus the total work is Θ(nk + rk). What is this in terms of n? Because r is the size of the base, it might be rather small.
- 282. Sec. 7.7 Binsort and Radix Sort 263 First pass values for Count. Count array: Index positions for Array B. End of Pass 1: Array A. rtoi = 1. Second pass values for Count. rtoi = 10. Count array: Index positions for Array B. End of Pass 2: Array A. 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 Initial Input: Array A 91 23 84 25 27 97 17 67 28 72 91 1 97 17 23 84 28 72 5 67 25 27 11 12 12 2 3 4 5 7 7 0 1 5 1 2 3 4 5 6 7 8 9 0 12 10 0 1 2 3 4 5 6 7 8 9 17 23 25 27 28 67 72 84 91 97 5 1 2 1 1 1 2 0 4 1 0 0 2 1 1 1 0 4 1 0 0 2 9 8 7 7 7 7 3 2 Figure 7.12 An example showing function radix applied to the input of Fig- ure 7.11. Row 1 shows the initial values within the input array. Row 2 shows the values for array cnt after counting the number of records for each bin. Row 3 shows the index values stored in array cnt. For example, cnt[0] is 0, indicat- ing no input values are in bin 0. Cnt[1] is 2, indicating that array B positions 0 and 1 will hold the values for bin 1. Cnt[2] is 3, indicating that array B position 2 will hold the (single) value for bin 2. Cnt[7] is 11, indicating that array B positions 7 through 10 will hold the four values for bin 7. Row 4 shows the results of the first pass of the Radix Sort. Rows 5 through 7 show the equivalent steps for the second pass.
- 283. 264 Chap. 7 Internal Sorting One could use base 2 or 10. Base 26 would be appropriate for sorting character strings. For now, we will treat r as a constant value and ignore it for the purpose of determining asymptotic complexity. Variable k is related to the key range: It is the maximum number of digits that a key may have in base r. In some applications we can determine k to be of limited size and so might wish to consider it a constant. In this case, Radix Sort is Θ(n) in the best, average, and worst cases, making it the sort with best asymptotic complexity that we have studied. Is it a reasonable assumption to treat k as a constant? Or is there some rela- tionship between k and n? If the key range is limited and duplicate key values are common, there might be no relationship between k and n. To make this distinction clear, use N to denote the number of distinct key values used by the n records. Thus, N ≤ n. Because it takes a minimum of logr N base r digits to represent N distinct key values, we know that k ≥ logr N. Now, consider the situation in which no keys are duplicated. If there are n unique keys (n = N), then it requires n distinct code values to represent them. Thus, k ≥ logr n. Because it requires at least Ω(log n) digits (within a constant factor) to distinguish between the n distinct keys, k is in Ω(log n). This yields an asymptotic complexity of Ω(n log n) for Radix Sort to process n distinct key values. It is possible that the key range is much larger; logr n bits is merely the best case possible for n distinct values. Thus, the logr n estimate for k could be overly optimistic. The moral of this analysis is that, for the general case of n distinct key values, Radix Sort is at best a Ω(n log n) sorting algorithm. Radix Sort can be much improved by making base r be as large as possible. Consider the case of an integer key value. Set r = 2i for some i. In other words, the value of r is related to the number of bits of the key processed on each pass. Each time the number of bits is doubled, the number of passes is cut in half. When processing an integer key value, setting r = 256 allows the key to be processed one byte at a time. Processing a 32-bit key requires only four passes. It is not unrea- sonable on most computers to use r = 216 = 64K, resulting in only two passes for a 32-bit key. Of course, this requires a cnt array of size 64K. Performance will be good only if the number of records is close to 64K or greater. In other words, the number of records must be large compared to the key size for Radix Sort to be efficient. In many sorting applications, Radix Sort can be tuned in this way to give good performance. Radix Sort depends on the ability to make a fixed number of multiway choices based on a digit value, as well as random access to the bins. Thus, Radix Sort might be difficult to implement for certain key types. For example, if the keys
- 284. Sec. 7.8 An Empirical Comparison of Sorting Algorithms 265 are real numbers or arbitrary length strings, then some care will be necessary in implementation. In particular, Radix Sort will need to be careful about deciding when the “last digit” has been found to distinguish among real numbers, or the last character in variable length strings. Implementing the concept of Radix Sort with the trie data structure (Section 13.1) is most appropriate for these situations. At this point, the perceptive reader might begin to question our earlier assump- tion that key comparison takes constant time. If the keys are “normal integer” values stored in, say, an integer variable, what is the size of this variable compared to n? In fact, it is almost certain that 32 (the number of bits in a standard int vari- able) is greater than log n for any practical computation. In this sense, comparison of two long integers requires Ω(log n) work. Computers normally do arithmetic in units of a particular size, such as a 32-bit word. Regardless of the size of the variables, comparisons use this native word size and require a constant amount of time. In practice, comparisons of two 32-bit values take constant time, even though 32 is much greater than log n. To some extent the truth of the proposition that there are constant time operations (such as integer comparison) is in the eye of the beholder. At the gate level of computer architecture, individual bits are compared. However, constant time comparison for integers is true in practice on most computers, and we rely on such assumptions as the basis for our analyses. In contrast, Radix Sort must do several arithmetic calculations on key values (each requiring constant time), where the number of such calculations is proportional to the key length. Thus, Radix Sort truly does Ω(n log n) work to process n distinct key values. 7.8 An Empirical Comparison of Sorting Algorithms Which sorting algorithm is fastest? Asymptotic complexity analysis lets us distin- guish between Θ(n2) and Θ(n log n) algorithms, but it does not help distinguish between algorithms with the same asymptotic complexity. Nor does asymptotic analysis say anything about which algorithm is best for sorting small lists. For answers to these questions, we can turn to empirical testing. Figure 7.13 shows timing results for actual implementations of the sorting algo- rithms presented in this chapter. The algorithms compared include Insertion Sort, Bubble Sort, Selection Sort, Shellsort, Quicksort, Mergesort, Heapsort and Radix Sort. Shellsort shows both the basic version from Section 7.3 and another with increments based on division by three. Mergesort shows both the basic implemen- tation from Section 7.4 and the optimized version with calls to Insertion Sort for lists of length below nine. For Quicksort, two versions are compared: the basic implementation from Section 7.5 and an optimized version that does not partition
- 285. 266 Chap. 7 Internal Sorting Sort 10 100 1K 10K 100K 1M Up Down Insertion .00023 .007 0.66 64.98 7381.0 674420 0.04 129.05 Bubble .00035 .020 2.25 277.94 27691.0 2820680 70.64 108.69 Selection .00039 .012 0.69 72.47 7356.0 780000 69.76 69.58 Shell .00034 .008 0.14 1.99 30.2 554 0.44 0.79 Shell/O .00034 .008 0.12 1.91 29.0 530 0.36 0.64 Merge .00050 .010 0.12 1.61 19.3 219 0.83 0.79 Merge/O .00024 .007 0.10 1.31 17.2 197 0.47 0.66 Quick .00048 .008 0.11 1.37 15.7 162 0.37 0.40 Quick/O .00031 .006 0.09 1.14 13.6 143 0.32 0.36 Heap .00050 .011 0.16 2.08 26.7 391 1.57 1.56 Heap/O .00033 .007 0.11 1.61 20.8 334 1.01 1.04 Radix/4 .00838 .081 0.79 7.99 79.9 808 7.97 7.97 Radix/8 .00799 .044 0.40 3.99 40.0 404 4.00 3.99 Figure 7.13 Empirical comparison of sorting algorithms run on a 3.4-GHz Intel Pentium 4 CPU running Linux. Shellsort, Quicksort, Mergesort, and Heapsort each are shown with regular and optimized versions. Radix Sort is shown for 4- and 8-bit-per-pass versions. All times shown are milliseconds. sublists below length nine. The first Heapsort version uses the class definitions from Section 5.5. The second version removes all the class definitions and operates directly on the array using inlined code for all access functions. In all cases, the values sorted are random 32-bit numbers. The input to each algorithm is a random array of integers. This affects the timing for some of the sorting algorithms. For example, Selection Sort is not being used to best advantage because the record size is small, so it does not get the best possible showing. The Radix Sort implementation certainly takes advantage of this key range in that it does not look at more digits than necessary. On the other hand, it was not optimized to use bit shifting instead of division, even though the bases used would permit this. The various sorting algorithms are shown for lists of sizes 10, 100, 1000, 10,000, 100,000, and 1,000,000. The final two columns of each figure show the performance for the algorithms when run on inputs of size 10,000 where the num- bers are in ascending (sorted) and descending (reverse sorted) order, respectively. These columns demonstrate best-case performance for some algorithms and worst- case performance for others. These columns also show that for some algorithms, the order of input has little effect. These figures show a number of interesting results. As expected, the O(n2) sorts are quite poor performers for large arrays. Insertion Sort is by far the best of this group, unless the array is already reverse sorted. Shellsort is clearly superior to any of these O(n2) sorts for lists of even 100 elements. Optimized Quicksort is clearly the best overall algorithm for all but lists of 10 elements. Even for small
- 286. Sec. 7.9 Lower Bounds for Sorting 267 arrays, optimized Quicksort performs well because it does one partition step be- fore calling Insertion Sort. Compared to the other O(n log n) sorts, unoptimized Heapsort is quite slow due to the overhead of the class structure. When all of this is stripped away and the algorithm is implemented to manipulate an array directly, it is still somewhat slower than mergesort. In general, optimizating the various algorithms makes a noticible improvement for larger array sizes. Overall, Radix Sort is a surprisingly poor performer. If the code had been tuned to use bit shifting of the key value, it would likely improve substantially; but this would seriously limit the range of element types that the sort could support. 7.9 Lower Bounds for Sorting This book contains many analyses for algorithms. These analyses generally define the upper and lower bounds for algorithms in their worst and average cases. For most of the algorithms presented so far, analysis is easy. This section considers a more difficult task — an analysis for the cost of a problem as opposed to an algorithm. The upper bound for a problem can be defined as the asymptotic cost of the fastest known algorithm. The lower bound defines the best possible efficiency for any algorithm that solves the problem, including algorithms not yet invented. Once the upper and lower bounds for the problem meet, we know that no future algorithm can possibly be (asymptotically) more efficient. A simple estimate for a problem’s lower bound can be obtained by measuring the size of the input that must be read and the output that must be written. Certainly no algorithm can be more efficient than the problem’s I/O time. From this we see that the sorting problem cannot be solved by any algorithm in less than Ω(n) time because it takes at least n steps to read and write the n values to be sorted. Based on our current knowledge of sorting algorithms and the size of the input, we know that the problem of sorting is bounded by Ω(n) and O(n log n). Computer scientists have spent much time devising efficient general-purpose sorting algorithms, but no one has ever found one that is faster than O(n log n) in the worst or average cases. Should we keep searching for a faster sorting algorithm? Or can we prove that there is no faster sorting algorithm by finding a tighter lower bound? This section presents one of the most important and most useful proofs in com- puter science: No sorting algorithm based on key comparisons can possibly be faster than Ω(n log n) in the worst case. This proof is important for three reasons. First, knowing that widely used sorting algorithms are asymptotically optimal is re- assuring. In particular, it means that you need not bang your head against the wall searching for an O(n) sorting algorithm (or at least not one in any way based on key
- 287. 268 Chap. 7 Internal Sorting comparisons). Second, this proof is one of the few non-trivial lower-bounds proofs that we have for any problem; that is, this proof provides one of the relatively few instances where our lower bound is tighter than simply measuring the size of the input and output. As such, it provides a useful model for proving lower bounds on other problems. Finally, knowing a lower bound for sorting gives us a lower bound in turn for other problems whose solution could be used as the basis for a sorting algorithm. The process of deriving asymptotic bounds for one problem from the asymptotic bounds of another is called a reduction, a concept further explored in Chapter 17. Except for the Radix Sort and Binsort, all of the sorting algorithms presented in this chapter make decisions based on the direct comparison of two key values. For example, Insertion Sort sequentially compares the value to be inserted into the sorted list until a comparison against the next value in the list fails. In contrast, Radix Sort has no direct comparison of key values. All decisions are based on the value of specific digits in the key value, so it is possible to take approaches to sorting that do not involve key comparisons. Of course, Radix Sort in the end does not provide a more efficient sorting algorithm than comparison-based sorting. Thus, empirical evidence suggests that comparison-based sorting is a good approach.3 The proof that any comparison sort requires Ω(n log n) comparisons in the worst case is structured as follows. First, you will see how comparison decisions can be modeled as the branches in a binary tree. This means that any sorting alg- orithm based on comparisons can be viewed as a binary tree whose nodes corre- spond to the results of making comparisons. Next, the minimum number of leaves in the resulting tree is shown to be the factorial of n. Finally, the minimum depth of a tree with n! leaves is shown to be in Ω(n log n). Before presenting the proof of an Ω(n log n) lower bound for sorting, we first must define the concept of a decision tree. A decision tree is a binary tree that can model the processing for any algorithm that makes decisions. Each (binary) decision is represented by a branch in the tree. For the purpose of modeling sorting algorithms, we count all comparisons of key values as decisions. If two keys are compared and the first is less than the second, then this is modeled as a left branch in the decision tree. In the case where the first value is greater than the second, the algorithm takes the right branch. Figure 7.14 shows the decision tree that models Insertion Sort on three input values. The first input value is labeled X, the second Y, and the third Z. They are 3 The truth is stronger than this statement implies. In reality, Radix Sort relies on comparisons as well and so can be modeled by the technique used in this section. The result is an Ω(n log n) bound in the general case even for algorithms that look like Radix Sort.
- 288. Sec. 7.9 Lower Bounds for Sorting 269 Yes No Yes No Yes No Yes No Yes No A[1]<A[0]? A[2]<A[1]? A[2]<A[1]? A[1]<A[0]? A[1]<A[0]? (Y<X?) (Z<X?) (Z<Y?) (Z<X?) (Z<Y?) XYZ ZYX YZX XYZ XZY YXZ YZX ZXY ZYX YXZ YXZ YZX ZYX XZY YXZ YZX YZX ZYX XYZ XYZ XZY ZXY XZY ZXY XYZ ZXY XZY Figure 7.14 Decision tree for Insertion Sort when processing three values la- beled X, Y, and Z, initially stored at positions 0, 1, and 2, respectively, in input array A. initially stored in positions 0, 1, and 2, respectively, of input array A. Consider the possible outputs. Initially, we know nothing about the final positions of the three values in the sorted output array. The correct output could be any permutation of the input values. For three values, there are n! = 6 permutations. Thus, the root node of the decision tree lists all six permutations that might be the eventual result of the algorithm. When n = 3, the first comparison made by Insertion Sort is between the sec- ond item in the input array (Y) and the first item in the array (X). There are two possibilities: Either the value of Y is less than that of X, or the value of Y is not less than that of X. This decision is modeled by the first branch in the tree. If Y is less than X, then the left branch should be taken and Y must appear before X in the final output. Only three of the original six permutations have this property, so the left child of the root lists the three permutations where Y appears before X: YXZ, YZX, and ZYX. Likewise, if Y were not less than X, then the right branch would be taken, and only the three permutations in which Y appears after X are possible outcomes: XYZ, XZY, and ZXY. These are listed in the right child of the root. Let us assume for the moment that Y is less than X and so the left branch is taken. In this case, Insertion Sort swaps the two values. At this point the array stores YXZ. Thus, in Figure 7.14 the left child of the root shows YXZ above the line. Next, the third value in the array is compared against the second (i.e., Z is
- 289. 270 Chap. 7 Internal Sorting compared with X). Again, there are two possibilities. If Z is less than X, then these items should be swapped (the left branch). If Z is not less than X, then Insertion Sort is complete (the right branch). Note that the right branch reaches a leaf node, and that this leaf node contains only permutation YXZ. This means that only permutation YXZ can be the outcome based on the results of the decisions taken to reach this node. In other words, Insertion Sort has “found” the single permutation of the original input that yields a sorted list. Likewise, if the second decision resulted in taking the left branch, a third comparison, regardless of the outcome, yields nodes in the decision tree with only single permutations. Again, Insertion Sort has “found” the correct permutation that yields a sorted list. Any sorting algorithm based on comparisons can be modeled by a decision tree in this way, regardless of the size of the input. Thus, all sorting algorithms can be viewed as algorithms to “find” the correct permutation of the input that yields a sorted list. Each algorithm based on comparisons can be viewed as proceeding by making branches in the tree based on the results of key comparisons, and each algorithm can terminate once a node with a single permutation has been reached. How is the worst-case cost of an algorithm expressed by the decision tree? The decision tree shows the decisions made by an algorithm for all possible inputs of a given size. Each path through the tree from the root to a leaf is one possible series of decisions taken by the algorithm. The depth of the deepest node represents the longest series of decisions required by the algorithm to reach an answer. There are many comparison-based sorting algorithms, and each will be mod- eled by a different decision tree. Some decision trees might be well-balanced, oth- ers might be unbalanced. Some trees will have more nodes than others (those with more nodes might be making “unnecessary” comparisons). In fact, a poor sorting algorithm might have an arbitrarily large number of nodes in its decision tree, with leaves of arbitrary depth. There is no limit to how slow the “worst” possible sort- ing algorithm could be. However, we are interested here in knowing what the best sorting algorithm could have as its minimum cost in the worst case. In other words, we would like to know what is the smallest depth possible for the deepest node in the tree for any sorting algorithm. The smallest depth of the deepest node will depend on the number of nodes in the tree. Clearly we would like to “push up” the nodes in the tree, but there is limited room at the top. A tree of height 1 can only store one node (the root); the tree of height 2 can store three nodes; the tree of height 3 can store seven nodes, and so on. Here are some important facts worth remembering:
- 290. Sec. 7.10 Further Reading 271 • A binary tree of height n can store at most 2n − 1 nodes. • Equivalently, a tree with n nodes requires at least ⌈log(n + 1)⌉ levels. What is the minimum number of nodes that must be in the decision tree for any comparison-based sorting algorithm for n values? Because sorting algorithms are in the business of determining which unique permutation of the input corresponds to the sorted list, all sorting algorithms must contain at least one leaf node for each possible permutation. There are n! permutations for a set of n numbers (see Section 2.2). Because there are at least n! nodes in the tree, we know that the tree must have Ω(log n!) levels. From Stirling’s approximation (Section 2.2), we know log n! is in Ω(n log n). The decision tree for any comparison-based sorting algorithm must have nodes Ω(n log n) levels deep. Thus, in the worst case, any such sorting algorithm must require Ω(n log n) comparisons. Any sorting algorithm requiring Ω(n log n) comparisons in the worst case re- quires Ω(n log n) running time in the worst case. Because any sorting algorithm requires Ω(n log n) running time, the problem of sorting also requires Ω(n log n) time. We already know of sorting algorithms with O(n log n) running time, so we can conclude that the problem of sorting requires Θ(n log n) time. As a corol- lary, we know that no comparison-based sorting algorithm can improve on existing Θ(n log n) time sorting algorithms by more than a constant factor. 7.10 Further Reading The definitive reference on sorting is Donald E. Knuth’s Sorting and Searching [Knu98]. A wealth of details is covered there, including optimal sorts for small size n and special purpose sorting networks. It is a thorough (although somewhat dated) treatment on sorting. For an analysis of Quicksort and a thorough survey on its optimizations, see Robert Sedgewick’s Quicksort [Sed80]. Sedgewick’s Al- gorithms [Sed03] discusses most of the sorting algorithms described here and pays special attention to efficient implementation. The optimized Mergesort version of Section 7.4 comes from Sedgewick. While Ω(n log n) is the theoretical lower bound in the worst case for sorting, many times the input is sufficiently well ordered that certain algorithms can take advantage of this fact to speed the sorting process. A simple example is Insertion Sort’s best-case running time. Sorting algorithms whose running time is based on the amount of disorder in the input are called adaptive. For more information on adaptive sorting algorithms, see “A Survey of Adaptive Sorting Algorithms” by Estivill-Castro and Wood [ECW92].
- 291. 272 Chap. 7 Internal Sorting 7.11 Exercises 7.1 Using induction, prove that Insertion Sort will always produce a sorted array. 7.2 Write an Insertion Sort algorithm for integer key values. However, here’s the catch: The input is a stack (not an array), and the only variables that your algorithm may use are a fixed number of integers and a fixed number of stacks. The algorithm should return a stack containing the records in sorted order (with the least value being at the top of the stack). Your algorithm should be Θ(n2) in the worst case. 7.3 The Bubble Sort implementation has the following inner for loop: for (int j=n-1; j>i; j--) Consider the effect of replacing this with the following statement: for (int j=n-1; j>0; j--) Would the new implementation work correctly? Would the change affect the asymptotic complexity of the algorithm? How would the change affect the running time of the algorithm? 7.4 When implementing Insertion Sort, a binary search could be used to locate the position within the first i − 1 elements of the array into which element i should be inserted. How would this affect the number of comparisons re- quired? How would using such a binary search affect the asymptotic running time for Insertion Sort? 7.5 Figure 7.5 shows the best-case number of swaps for Selection Sort as Θ(n). This is because the algorithm does not check to see if the ith record is already in the ith position; that is, it might perform unnecessary swaps. (a) Modify the algorithm so that it does not make unnecessary swaps. (b) What is your prediction regarding whether this modification actually improves the running time? (c) Write two programs to compare the actual running times of the origi- nal Selection Sort and the modified algorithm. Which one is actually faster? 7.6 Recall that a sorting algorithm is said to be stable if the original ordering for duplicate keys is preserved. Of the sorting algorithms Insertion Sort, Bub- ble Sort, Selection Sort, Shellsort, Quicksort, Mergesort, Heapsort, Binsort, and Radix Sort, which of these are stable, and which are not? For each one, describe either why it is or is not stable. If a minor change to the implemen- tation would make it stable, describe the change.
- 292. Sec. 7.11 Exercises 273 7.7 Recall that a sorting algorithm is said to be stable if the original ordering for duplicate keys is preserved. We can make any algorithm stable if we alter the input keys so that (potentially) duplicate key values are made unique in a way that the first occurance of the original duplicate value is less than the second occurance, which in turn is less than the third, and so on. In the worst case, it is possible that all n input records have the same key value. Give an algorithm to modify the key values such that every modified key value is unique, the resulting key values give the same sort order as the original keys, the result is stable (in that the duplicate original key values remain in their original order), and the process of altering the keys is done in linear time using only a constant amount of additional space. 7.8 The discussion of Quicksort in Section 7.5 described using a stack instead of recursion to reduce the number of function calls made. (a) How deep can the stack get in the worst case? (b) Quicksort makes two recursive calls. The algorithm could be changed to make these two calls in a specific order. In what order should the two calls be made, and how does this affect how deep the stack can become? 7.9 Give a permutation for the values 0 through 7 that will cause Quicksort (as implemented in Section 7.5) to have its worst case behavior. 7.10 Assume L is an array, length(L) returns the number of records in the array, and qsort(L, i, j) sorts the records of L from i to j (leaving the records sorted in L) using the Quicksort algorithm. What is the average- case time complexity for each of the following code fragments? (a) for (i=0; i<L.length; i++) qsort(L, 0, i); (b) for (i=0; i<L.length; i++) qsort(L, 0, L.length-1); 7.11 Modify Quicksort to find the smallest K values in an array of records. Your output should be the array modified so that the K smallest values are sorted in the first K positions of the array. Your algorithm should do the minimum amount of work necessary. 7.12 Modify Quicksort to sort a sequence of variable-length strings stored one after the other in a character array, with a second array (storing pointers to strings) used to index the strings. Your function should modify the index array so that the first pointer points to the beginning of the lowest valued string, and so on.
- 293. 274 Chap. 7 Internal Sorting 7.13 Graph f1(n) = n log n, f2(n) = n1.5, and f3(n) = n2 in the range 1 ≤ n ≤ 1000 to visually compare their growth rates. Typically, the constant factor in the running-time expression for an implementation of Insertion Sort will be less than the constant factors for Shellsort or Quicksort. How many times greater can the constant factor be for Shellsort to be faster than Insertion Sort when n = 1000? How many times greater can the constant factor be for Quicksort to be faster than Insertion Sort when n = 1000? 7.14 Imagine that there exists an algorithm SPLITk that can split a list L of n elements into k sublists, each containing one or more elements, such that sublist i contains only elements whose values are less than all elements in sublist j for i < j <= k. If n < k, then k−n sublists are empty, and the rest are of length 1. Assume that SPLITk has time complexity O(length of L). Furthermore, assume that the k lists can be concatenated again in constant time. Consider the following algorithm: List SORTk(List L) { List sub[k]; // To hold the sublists if (L.length() > 1) { SPLITk(L, sub); // SPLITk places sublists into sub for (i=0; i<k; i++) sub[i] = SORTk(sub[i]); // Sort each sublist L = concatenation of k sublists in sub; return L; } } (a) What is the worst-case asymptotic running time for SORTk? Why? (b) What is the average-case asymptotic running time of SORTk? Why? 7.15 Here is a variation on sorting. The problem is to sort a collection of n nuts and n bolts by size. It is assumed that for each bolt in the collection, there is a corresponding nut of the same size, but initially we do not know which nut goes with which bolt. The differences in size between two nuts or two bolts can be too small to see by eye, so you cannot rely on comparing the sizes of two nuts or two bolts directly. Instead, you can only compare the sizes of a nut and a bolt by attempting to screw one into the other (assume this comparison to be a constant time operation). This operation tells you that either the nut is bigger than the bolt, the bolt is bigger than the nut, or they are the same size. What is the minimum number of comparisons needed to sort the nuts and bolts in the worst case?
- 294. Sec. 7.12 Projects 275 7.16 (a) Devise an algorithm to sort three numbers. It should make as few com- parisons as possible. How many comparisons and swaps are required in the best, worst, and average cases? (b) Devise an algorithm to sort five numbers. It should make as few com- parisons as possible. How many comparisons and swaps are required in the best, worst, and average cases? (c) Devise an algorithm to sort eight numbers. It should make as few com- parisons as possible. How many comparisons and swaps are required in the best, worst, and average cases? 7.17 Devise an efficient algorithm to sort a set of numbers with values in the range 0 to 30,000. There are no duplicates. Keep memory requirements to a mini- mum. 7.18 Which of the following operations are best implemented by first sorting the list of numbers? For each operation, briefly describe an algorithm to imple- ment it, and state the algorithm’s asymptotic complexity. (a) Find the minimum value. (b) Find the maximum value. (c) Compute the arithmetic mean. (d) Find the median (i.e., the middle value). (e) Find the mode (i.e., the value that appears the most times). 7.19 Consider a recursive Mergesort implementation that calls Insertion Sort on sublists smaller than some threshold. If there are n calls to Mergesort, how many calls will there be to Insertion Sort? Why? 7.20 Implement Mergesort for the case where the input is a linked list. 7.21 Counting sort (assuming the input key values are integers in the range 0 to m − 1) works by counting the number of records with each key value in the first pass, and then uses this information to place the records in order in a second pass. Write an implementation of counting sort (see the implementa- tion of radix sort for some ideas). What can we say about the relative values of m and n for this to be effective? If m < n, what is the running time of this algorithm? 7.22 Use an argument similar to that given in Section 7.9 to prove that log n is a worst-case lower bound for the problem of searching for a given value in a sorted array containing n elements. 7.12 Projects 7.1 One possible improvement for Bubble Sort would be to add a flag variable and a test that determines if an exchange was made during the current iter- ation. If no exchange was made, then the list is sorted and so the algorithm
- 295. 276 Chap. 7 Internal Sorting can stop early. This makes the best case performance become O(n) (because if the list is already sorted, then no iterations will take place on the first pass, and the sort will stop right there). Modify the Bubble Sort implementation to add this flag and test. Compare the modified implementation on a range of inputs to determine if it does or does not improve performance in practice. 7.2 Starting with the Java code for Quicksort given in this chapter, write a series of Quicksort implementations to test the following optimizations on a wide range of input data sizes. Try these optimizations in various combinations to try and develop the fastest possible Quicksort implementation that you can. (a) Look at more values when selecting a pivot. (b) Do not make a recursive call to qsort when the list size falls below a given threshold, and use Insertion Sort to complete the sorting process. Test various values for the threshold size. (c) Eliminate recursion by using a stack and inline functions. 7.3 Write your own collection of sorting programs to implement the algorithms described in this chapter, and compare their running times. Be sure to im- plement optimized versions, trying to make each program as fast as possible. Do you get the same relative timings as shown in Figure 7.13? If not, why do you think this happened? How do your results compare with those of your classmates? What does this say about the difficulty of doing empirical timing studies? 7.4 Perform a study of Shellsort, using different increments. Compare the ver- sion shown in Section 7.3, where each increment is half the previous one, with others. In particular, try implementing “division by 3” where the incre- ments on a list of length n will be n/3, n/9, etc. Do other increment schemes work as well? 7.5 The implementation for Mergesort given in Section 7.4 takes an array as in- put and sorts that array. At the beginning of Section 7.4 there is a simple pseudocode implementation for sorting a linked list using Mergesort. Im- plement both a linked list-based version of Mergesort and the array-based version of Mergesort, and compare their running times. 7.6 Radix Sort is typically implemented to support only a radix that is a power of two. This allows for a direct conversion from the radix to some number of bits in an integer key value. For example, if the radix is 16, then a 32-bit key will be processed in 8 steps of 4 bits each. This can lead to a more effi- cient implementation because bit shifting can replace the division operations shown in the implementation of Section 7.7. Reimplement the Radix Sort
- 296. Sec. 7.12 Projects 277 code given in Section 7.7 to use bit shifting in place of division. Compare the running time of the old and new Radix Sort implementations. 7.7 It has been proposed that heapsort can be optimized by altering the heap’s siftdown function. Call the value being sifted down X. Siftdown does two comparisons per level: First the children of X are compared, then the winner is compared to X. If X is too small, it is swapped with its larger child and the process repeated. The proposed optimization dispenses with the test against X. Instead, the larger child automatically replaces X, until X reaches the bottom level of the heap. At this point, X might be too large to remain in that position. This is corrected by repeatedly comparing X with its parent and swapping as necessary to “bubble” it up to its proper level. The claim is that this process will save a number of comparisons because most nodes when sifted down end up near the bottom of the tree anyway. Implement both versions of siftdown, and do an empirical study to compare their running times.
- 298. 8 File Processing and External Sorting Earlier chapters presented basic data structures and algorithms that operate on data stored in main memory. Some applications require that large amounts of informa- tion be stored and processed — so much information that it cannot all fit into main memory. In that case, the information must reside on disk and be brought into main memory selectively for processing. You probably already realize that main memory access is much faster than ac- cess to data stored on disk or other storage devices. The relative difference in access times is so great that efficient disk-based programs require a different approach to algorithm design than most programmers are used to. As a result, many program- mers do a poor job when it comes to file processing applications. This chapter presents the fundamental issues relating to the design of algo- rithms and data structures for disk-based applications.1 We begin with a descrip- tion of the significant differences between primary memory and secondary storage. Section 8.2 discusses the physical aspects of disk drives. Section 8.3 presents ba- sic methods for managing buffer pools. Section 8.4 discusses the Java model for random access to data stored on disk. Section 8.5 discusses the basic principles for sorting collections of records too large to fit in main memory. 1 Computer technology changes rapidly. I provide examples of disk drive specifications and other hardware performance numbers that are reasonably up to date as of the time when the book was written. When you read it, the numbers might seem out of date. However, the basic principles do not change. The approximate ratios for time, space, and cost between memory and disk have remained surprisingly steady for over 20 years. 279
- 299. 280 Chap. 8 File Processing and External Sorting Medium 1996 1997 2000 2004 2006 2008 RAM $45.00 7.00 1.500 0.3500 0.1500 0.0339 Disk 0.25 0.10 0.010 0.0010 0.0005 0.0001 Flash drive – – – 0.1000 0.0900 0.0029 Floppy 0.50 0.36 0.250 0.2500 – – Tape 0.03 0.01 0.001 0.0003 – – Figure 8.1 Price comparison table for some writeable electronic data storage media in common use. Prices are in US Dollars/MB. 8.1 Primary versus Secondary Storage Computer storage devices are typically classified into primary or main memory and secondary or peripheral storage. Primary memory usually refers to Random Access Memory (RAM), while secondary storage refers to devices such as hard disk drives, removeable “flash” drives, floppy disks, CDs, DVDs, and tape drives. Primary memory also includes registers, cache, and video memories, but we will ignore them for this discussion because their existence does not affect the principal differences between primary and secondary memory. Along with a faster CPU, every new model of computer seems to come with more main memory. As memory size continues to increase, is it possible that rel- atively slow disk storage will be unnecessary? Probably not, because the desire to store and process larger files grows at least as fast as main memory size. Prices for both main memory and peripheral storage devices have dropped dramatically in re- cent years, as demonstrated by Figure 8.1. However, the cost for disk drive storage per megabyte is about two orders of magnitude less than RAM and has been for many years. There is now a wide range of removable media available for transfering data or storing data offline in relative safety. These include floppy disks (now largely obsolete), writable CDs and DVDs, “flash” drives, and magnetic tape. Optical storage such as CDs and DVDs costs roughly half the price of hard disk drive space per megabyte, and have become practical for use as backup storage within the past few years. Tape used to be much cheaper than other media, and was the preferred means of backup. “Flash” drives cost the most per megabyte, but due to their storage capacity and flexibility, have now replaced floppy disks as the primary storage device for transferring data between computer when direct network transfer is not available. Secondary storage devices have at least two other advantages over RAM mem- ory. Perhaps most importantly, disk and tape files are persistent, meaning that
- 300. Sec. 8.1 Primary versus Secondary Storage 281 they are not erased from disk and tape when the power is turned off. In contrast, RAM used for main memory is usually volatile — all information is lost with the power. A second advantage is that floppy disks, CD-ROMs, and “flash” drives can easily be transferred between computers. This provides a convenient way to take information from one computer to another. In exchange for reduced storage costs, persistence, and portability, secondary storage devices pay a penalty in terms of increased access time. While not all ac- cesses to disk take the same amount of time (more on this later), the typical time required to access a byte of storage from a disk drive in 2007 is around 9 ms (i.e., 9 thousandths of a second). This might not seem slow, but compared to the time required to access a byte from main memory, this is fantastically slow. Typical access time from standard personal computer RAM in 2007 is about 5-10 nanosec- onds (i.e., 5-10 billionths of a second). Thus, the time to access a byte of data from a disk drive is about six orders of magnitude greater than that required to access a byte from main memory. While disk drive and RAM access times are both decreas- ing, they have done so at roughly the same rate. The relative speeds have remained the same for over twenty-five years, in that the difference in access time between RAM and a disk drive has remained in the range between a factor of 100,000 and 1,000,000. To gain some intuition for the significance of this speed difference, consider the time that it might take for you to look up the entry for disk drives in the index of this book, and then turn to the appropriate page. Call this your “primary memory” access time. If it takes you about 20 seconds to perform this access, then an access taking 500,000 times longer would require months. It is interesting to note that while processing speeds have increased dramat- ically, and hardware prices have dropped dramatically, disk and memory access times have improved by less than an order of magnitude over the past ten years. However, the situation is really much better than that modest speedup would sug- gest. During the same time period, the size of both disk and main memory has increased by about three orders of magnitude. Thus, the access times have actually decreased in the face of a massive increase in the density of these storage devices. Due to the relatively slow access time for data on disk as compared to main memory, great care is required to create efficient applications that process disk- based information. The million-to-one ratio of disk access time versus main mem- ory access time makes the following rule of paramount importance when designing disk-based applications: Minimize the number of disk accesses!
- 301. 282 Chap. 8 File Processing and External Sorting There are generally two approaches to minimizing disk accesses. The first is to arrange information so that if you do access data from secondary memory, you will get what you need in as few accesses as possible, and preferably on the first access. File structure is the term used for a data structure that organizes data stored in secondary memory. File structures should be organized so as to minimize the required number of disk accesses. The other way to minimize disk accesses is to arrange information so that each disk access retrieves additional data that can be used to minimize the need for future accesses, that is, to guess accurately what information will be needed later and retrieve it from disk now, if this can be done cheaply. As you shall see, there is little or no difference in the time required to read several hundred contiguous bytes from disk as compared to reading one byte, so this strategy is indeed practical. One way to minimize disk accesses is to compress the information stored on disk. Section 3.9 discusses the space/time tradeoff in which space requirements can be reduced if you are willing to sacrifice time. However, the disk-based space/time tradeoff principle stated that the smaller you can make your disk storage require- ments, the faster your program will run. This is because the time to read informa- tion from disk is enormous compared to computation time, so almost any amount of additional computation to unpack the data is going to be less than the disk read time saved by reducing the storage requirements. This is precisely what happens when files are compressed. CPU time is required to uncompress information, but this time is likely to be much less than the time saved by reducing the number of bytes read from disk. Current file compression programs are not designed to allow random access to parts of a compressed file, so the disk-based space/time tradeoff principle cannot easily be taken advantage of in normal processing using commer- cial disk compression utilities. However, in the future disk drive controllers might automatically compress and decompress files stored on disk, thus taking advan- tage of the disk-based space/time tradeoff principle to save both space and time. Many cartridge tape drives (which must process data sequentially) automatically compress and decompress information during I/O. 8.2 Disk Drives A Java programmer views a random access file stored on disk as a contiguous series of bytes, with those bytes possibly combining to form data records. This is called the logical file. The physical file actually stored on disk is usually not a contiguous series of bytes. It could well be in pieces spread all over the disk. The file manager, a part of the operating system, is responsible for taking requests for data from a logical file and mapping those requests to the physical location
- 302. Sec. 8.2 Disk Drives 283 of the data on disk. Likewise, when writing to a particular logical byte position with respect to the beginning of the file, this position must be converted by the file manager into the corresponding physical location on the disk. To gain some appreciation for the the approximate time costs for these operations, you need to understand the physical structure and basic workings of a disk drive. Disk drives are often referred to as direct access storage devices. This means that it takes roughly equal time to access any record in the file. This is in contrast to sequential access storage devices such as tape drives, which require the tape reader to process data from the beginning of the tape until the desired position has been reached. As you will see, the disk drive is only approximately direct access: At any given time, some records are more quickly accessible than others. 8.2.1 Disk Drive Architecture A hard disk drive is composed of one or more round platters, stacked one on top of another and attached to a central spindle. Platters spin continuously at a constant rate. Each usable surface of each platter is assigned a read/write head or I/O head through which data are read or written, somewhat like the arrangement of a phonograph player’s arm “reading” sound from a phonograph record. Unlike a phonograph needle, the disk read/write head does not actually touch the surface of a hard disk. Instead, it remains slightly above the surface, and any contact during normal operation would damage the disk. This distance is very small, much smaller than the height of a dust particle. It can be likened to a 5000-kilometer airplane trip across the United States, with the plane flying at a height of one meter! A hard disk drive typically has several platters and several read/write heads, as shown in Figure 8.2(a). Each head is attached to an arm, which connects to the boom. The boom moves all of the heads in or out together. When the heads are in some position over the platters, there are data on each platter directly accessible to each head. The data on a single platter that are accessible to any one position of the head for that platter are collectively called a track, that is, all data on a platter that are a fixed distance from the spindle, as shown in Figure 8.2(b). The collection of all tracks that are a fixed distance from the spindle is called a cylinder. Thus, a cylinder is all of the data that can be read when the arms are in a particular position. Each track is subdivided into sectors. Between each sector there are intersec- tor gaps in which no data are stored. These gaps allow the read head to recognize the end of a sector. Note that each sector contains the same amount of data. Be- cause the outer tracks have greater length, they contain fewer bits per inch than do the inner tracks. Thus, about half of the potential storage space is wasted, because only the innermost tracks are stored at the highest possible data density. This ar-
- 303. 284 Chap. 8 File Processing and External Sorting (b) Heads Platters (arm) Boom (a) Track Read/Write Spindle Figure 8.2 (a) A typical disk drive arranged as a stack of platters. (b) One track on a disk drive platter. (a) (b) Intersector Gaps Sectors Bits of Data Figure 8.3 The organization of a disk platter. Dots indicate density of informa- tion. (a) Nominal arrangement of tracks showing decreasing data density when moving outward from the center of the disk. (b) A “zoned” arrangement with the sector size and density periodically reset in tracks further away from the center. rangement is illustrated by Figure 8.3a. Disk drives today actually group tracks into “zones” such that the tracks in the innermost zone adjust their data density going out to maintain the same radial data density, then the tracks of the next zone reset the data density to make better use of their storage ability, and so on. This arrangement is shown in Figure 8.3b. In contrast to the physical layout of a hard disk, a CD-ROM consists of a single spiral track. Bits of information along the track are equally spaced, so the informa- tion density is the same at both the outer and inner portions of the track. To keep
- 304. Sec. 8.2 Disk Drives 285 the information flow at a constant rate along the spiral, the drive must speed up the rate of disk spin as the I/O head moves toward the center of the disk. This makes for a more complicated and slower mechanism. Three separate steps take place when reading a particular byte or series of bytes of data from a hard disk. First, the I/O head moves so that it is positioned over the track containing the data. This movement is called a seek. Second, the sector containing the data rotates to come under the head. When in use the disk is always spinning. At the time of this writing, typical disk spin rates are 7200 rotations per minute (rpm). The time spent waiting for the desired sector to come under the I/O head is called rotational delay or rotational latency. The third step is the actual transfer (i.e., reading or writing) of data. It takes relatively little time to read information once the first byte is positioned under the I/O head, simply the amount of time required for it all to move under the head. In fact, disk drives are designed not to read one byte of data, but rather to read an entire sector of data at each request. Thus, a sector is the minimum amount of data that can be read or written at one time. In general, it is desirable to keep all sectors for a file together on as few tracks as possible. This desire stems from two assumptions: 1. Seek time is slow (it is typically the most expensive part of an I/O operation), and 2. If one sector of the file is read, the next sector will probably soon be read. Assumption (2) is called locality of reference, a concept that comes up frequently in computer applications. Contiguous sectors are often grouped to form a cluster. A cluster is the smallest unit of allocation for a file, so all files are a multiple of the cluster size. The cluster size is determined by the operating system. The file manager keeps track of which clusters make up each file. In Microsoft Windows systems, there is a designated portion of the disk called the File Allocation Table, which stores information about which sectors belong to which file. In contrast, UNIX does not use clusters. The smallest unit of file allocation and the smallest unit that can be read/written is a sector, which in UNIX terminology is called a block. UNIX maintains information about file organization in certain disk blocks called i-nodes. A group of physically contiguous clusters from the same file is called an extent. Ideally, all clusters making up a file will be contiguous on the disk (i.e., the file will consist of one extent), so as to minimize seek time required to access different portions of the file. If the disk is nearly full when a file is created, there might not be an extent available that is large enough to hold the new file. Furthermore, if a file
- 305. 286 Chap. 8 File Processing and External Sorting grows, there might not be free space physically adjacent. Thus, a file might consist of several extents widely spaced on the disk. The fuller the disk, and the more that files on the disk change, the worse this file fragmentation (and the resulting seek time) becomes. File fragmentation leads to a noticeable degradation in performance as additional seeks are required to access data. Another type of problem arises when the file’s logical record size does not match the sector size. If the sector size is not a multiple of the record size (or vice versa), records will not fit evenly within a sector. For example, a sector might be 2048 bytes long, and a logical record 100 bytes. This leaves room to store 20 records with 48 bytes left over. Either the extra space is wasted, or else records are allowed to cross sector boundaries. If a record crosses a sector boundary, two disk accesses might be required to read it. If the space is left empty instead, such wasted space is called internal fragmentation. A second example of internal fragmentation occurs at cluster boundaries. Files whose size is not an even multiple of the cluster size must waste some space at the end of the last cluster. The worst case will occur when file size modulo cluster size is one (for example, a file of 4097 bytes and a cluster of 4096 bytes). Thus, cluster size is a tradeoff between large files processed sequentially (where a large cluster size is desirable to minimize seeks) and small files (where small clusters are desirable to minimize wasted storage). Every disk drive organization requires that some disk space be used to organize the sectors, clusters, and so forth. The layout of sectors within a track is illustrated by Figure 8.4. Typical information that must be stored on the disk itself includes the File Allocation Table, sector headers that contain address marks and informa- tion about the condition (whether usable or not) for each sector, and gaps between sectors. The sector header also contains error detection codes to help verify that the data have not been corrupted. This is why most disk drives have a “nominal” size that is greater than the actual amount of user data that can be stored on the drive. The difference is the amount of space required to organize the information on the disk. Additional space will be lost due to fragmentation. 8.2.2 Disk Access Costs The primary cost when accessing information on disk is normally the seek time. This assumes of course that a seek is necessary. When reading a file in sequential order (if the sectors comprising the file are contiguous on disk), little seeking is necessary. However, when accessing a random disk sector, seek time becomes the dominant cost for the data access. While the actual seek time is highly variable, depending on the distance between the track where the I/O head currently is and
- 306. Sec. 8.2 Disk Drives 287 Intersector Gap Data Sector Header Data Sector Sector Header Sector Intrasector Gap Figure 8.4 An illustration of sector gaps within a track. Each sector begins with a sector header containing the sector address and an error detection code for the contents of that sector. The sector header is followed by a small intrasector gap, followed in turn by the sector data. Each sector is separated from the next sector by a larger intersector gap. the track where the head is moving to, we will consider only two numbers. One is the track-to-track cost, or the minimum time necessary to move from a track to an adjacent track. This is appropriate when you want to analyze access times for files that are well placed on the disk. The second number is the average seek time for a random access. These two numbers are often provided by disk manufacturers. A typical example is the 120GB Western Digital WD Caviar SE serial ATA drive. The manufacturer’s specifications indicate that the track-to-track time is 2.0 ms and the average seek time is 9.0 ms. For many years, typical rotation speed for disk drives was 3600 rpm, or one rotation every 16.7 ms. Most disk drives today have a rotation speed of 7200 rpm, or 8.3 ms per rotation. When reading a sector at random, you can expect that the disk will need to rotate halfway around to bring the desired sector under the I/O head, or 4.2 ms for a 7200-rpm disk drive. Once under the I/O head, a sector of data can be transferred as fast as that sector rotates under the head. If an entire track is to be read, then it will require one rotation (8.3 ms at 7200 rpm) to move the full track under the head. If only part of the track is to be read, then proportionately less time will be required. For example, if there are 16K sectors on the track and one sector is to be read, this will require a trivial amount of time (1/16K of a rotation). Example 8.1 Assume that an older disk drive has a total (nominal) capac- ity of 16.8GB spread among 10 platters, yielding 1.68GB/platter. Each plat- ter contains 13,085 tracks and each track contains (after formating) 256 sec- tors of 512 bytes/sector. Track-to-track seek time is 2.2 ms and average seek time for random access is 9.5 ms. Assume the operating system maintains a cluster size of 8 sectors per cluster (4KB), yielding 32 clusters per track. The disk rotation rate is 5400 rpm (11.1 ms per rotation). Based on this information we can estimate the cost for various file processing operations.
- 307. 288 Chap. 8 File Processing and External Sorting How much time is required to read the track? On average, it will require half a rotation to bring the first sector of the track under the I/O head, and then one complete rotation to read the track. How long will it take to read a file of 1MB divided into 2048 sector- sized (512 byte) records? This file will be stored in 256 clusters, because each cluster holds 8 sectors. The answer to the question depends in large measure on how the file is stored on the disk, that is, whether it is all to- gether or broken into multiple extents. We will calculate both cases to see how much difference this makes. If the file is stored so as to fill all of the sectors of eight adjacent tracks, then the cost to read the first sector will be the time to seek to the first track (assuming this requires a random seek), then a wait for the initial rotational delay, and then the time to read. This requires 9.5 + 11.1 × 1.5 = 26.2 ms. At this point, because we assume that the next seven tracks require only a track-to-track seek because they are adjacent, each requires 2.2 + 11.1 × 1.5 = 18.9 ms. The total time required is therefore 26.2ms + 7 × 18.9ms = 158.5ms. If the file’s clusters are spread randomly across the disk, then we must perform a seek for each cluster, followed by the time for rotational delay. Once the first sector of the cluster comes under the I/O head, very little time is needed to read the cluster because only 8/256 of the track needs to rotate under the head, for a total time of about 5.9 ms for latency and read time. Thus, the total time required is about 256(9.5 + 5.9) ≈ 3942ms or close to 4 seconds. This is much longer than the time required when the file is all together on disk! This example illustrates why it is important to keep disk files from be- coming fragmented, and why so-called “disk defragmenters” can speed up file processing time. File fragmentation happens most commonly when the disk is nearly full and the file manager must search for free space whenever a file is created or changed.
- 308. Sec. 8.3 Buffers and Buffer Pools 289 8.3 Buffers and Buffer Pools Given the specifications of the disk drive from Example 8.1, we find that it takes about 9.5+11.1×1.5 = 26.2 ms to read one track of data on average. It takes about 9.5+11.1/2+(1/256)×11.1 = 15.1 ms on average to read a single sector of data. This is a good savings (slightly over half the time), but less than 1% of the data on the track are read. If we want to read only a single byte, it would save us effectively no time over that required to read an entire sector. For this reason, nearly all disk drives automatically read or write an entire sector’s worth of information whenever the disk is accessed, even when only one byte of information is requested. Once a sector is read, its information is stored in main memory. This is known as buffering or caching the information. If the next disk request is to that same sector, then it is not necessary to read from disk again because the information is already stored in main memory. Buffering is an example of one method for minimizing disk accesses mentioned at the beginning of the chapter: Bring off additional information from disk to satisfy future requests. If information from files were accessed at random, then the chance that two consecutive disk requests are to the same sector would be low. However, in practice most disk requests are close to the location (in the logical file at least) of the previous request. This means that the probability of the next request “hitting the cache” is much higher than chance would indicate. This principle explains one reason why average access times for new disk drives are lower than in the past. Not only is the hardware faster, but information is also now stored using better algorithms and larger caches that minimize the number of times information needs to be fetched from disk. This same concept is also used to store parts of programs in faster memory within the CPU, using the CPU cache that is prevalent in modern microprocessors. Sector-level buffering is normally provided by the operating system and is of- ten built directly into the disk drive controller hardware. Most operating systems maintain at least two buffers, one for input and one for output. Consider what would happen if there were only one buffer during a byte-by-byte copy operation. The sector containing the first byte would be read into the I/O buffer. The output operation would need to destroy the contents of the single I/O buffer to write this byte. Then the buffer would need to be filled again from disk for the second byte, only to be destroyed during output. The simple solution to this problem is to keep one buffer for input, and a second for output. Most disk drive controllers operate independently from the CPU once an I/O request is received. This is useful because the CPU can typically execute millions of instructions during the time required for a single I/O operation. A technique that
- 309. 290 Chap. 8 File Processing and External Sorting takes maximum advantage of this microparallelism is double buffering. Imagine that a file is being processed sequentially. While the first sector is being read, the CPU cannot process that information and so must wait or find something else to do in the meantime. Once the first sector is read, the CPU can start processing while the disk drive (in parallel) begins reading the second sector. If the time required for the CPU to process a sector is approximately the same as the time required by the disk controller to read a sector, it might be possible to keep the CPU continuously fed with data from the file. The same concept can also be applied to output, writing one sector to disk while the CPU is writing to a second output buffer in memory. Thus, in computers that support double buffering, it pays to have at least two input buffers and two output buffers available. Caching information in memory is such a good idea that it is usually extended to multiple buffers. The operating system or an application program might store many buffers of information taken from some backing storage such as a disk file. This process of using buffers as an intermediary between a user and a disk file is called buffering the file. The information stored in a buffer is often called a page, and the collection of buffers is called a buffer pool. The goal of the buffer pool is to increase the amount of information stored in memory in hopes of increasing the likelihood that new information requests can be satisfied from the buffer pool rather than requiring new information to be read from disk. As long as there is an unused buffer available in the buffer pool, new informa- tion can be read in from disk on demand. When an application continues to read new information from disk, eventually all of the buffers in the buffer pool will be- come full. Once this happens, some decision must be made about what information in the buffer pool will be sacrificed to make room for newly requested information. When replacing information contained in the buffer pool, the goal is to select a buffer that has “unnecessary” information, that is, that information least likely to be requested again. Because the buffer pool cannot know for certain what the pattern of future requests will look like, a decision based on some heuristic, or best guess, must be used. There are several approaches to making this decision. One heuristic is “first-in, first-out” (FIFO). This scheme simply orders the buffers in a queue. The buffer at the front of the queue is used next to store new information and then placed at the end of the queue. In this way, the buffer to be replaced is the one that has held its information the longest, in hopes that this in- formation is no longer needed. This is a reasonable assumption when processing moves along the file at some steady pace in roughly sequential order. However, many programs work with certain key pieces of information over and over again, and the importance of information has little to do with how long ago the informa-
- 310. Sec. 8.3 Buffers and Buffer Pools 291 tion was first accessed. Typically it is more important to know how many times the information has been accessed, or how recently the information was last accessed. Another approach is called “least frequently used” (LFU). LFU tracks the num- ber of accesses to each buffer in the buffer pool. When a buffer must be reused, the buffer that has been accessed the fewest number of times is considered to contain the “least important” information, and so it is used next. LFU, while it seems in- tuitively reasonable, has many drawbacks. First, it is necessary to store and update access counts for each buffer. Second, what was referenced many times in the past might now be irrelevant. Thus, some time mechanism where counts “expire” is often desirable. This also avoids the problem of buffers that slowly build up big counts because they get used just often enough to avoid being replaced. An alter- native is to maintain counts for all sectors ever read, not just the sectors currently in the buffer pool. The third approach is called “least recently used” (LRU). LRU simply keeps the buffers in a list. Whenever information in a buffer is accessed, this buffer is brought to the front of the list. When new information must be read, the buffer at the back of the list (the one least recently used) is taken and its “old” information is either discarded or written to disk, as appropriate. This is an easily implemented approx- imation to LFU and is often the method of choice for managing buffer pools unless special knowledge about information access patterns for an application suggests a special-purpose buffer management scheme. The main purpose of a buffer pool is to minimize disk I/O. When the contents of a block are modified, we could write the updated information to disk immediately. But what if the block is changed again? If we write the block’s contents after every change, that might be a lot of disk write operations that can be avoided. It is more efficient to wait until either the file is to be closed, or the buffer containing that block is flushed from the buffer pool. When a buffer’s contents are to be replaced in the buffer pool, we only want to write the contents to disk if it is necessary. That would be necessary only if the contents have changed since the block was read in originally from the file. The way to insure that the block is written when necessary, but only when necessary, is to maintain a boolean variable with the buffer (often referred to as the dirty bit) that is turned on when the buffer’s contents are modified by the client. At the time when the block is flushed from the buffer pool, it is written to disk if and only if the dirty bit has been turned on. Modern operating systems support virtual memory. Virtual memory is a tech- nique that allows the programmer to pretend that there is more of the faster main memory (such as RAM) than actually exists. This is done by means of a buffer pool
- 311. 292 Chap. 8 File Processing and External Sorting reading blocks stored on slower, secondary memory (such as on the disk drive). The disk stores the complete contents of the virtual memory. Blocks are read into main memory as demanded by memory accesses. Naturally, programs using virtual memory techniques are slower than programs whose data are stored completely in main memory. The advantage is reduced programmer effort because a good virtual memory system provides the appearance of larger main memory without modifying the program. Example 8.2 Consider a virtual memory whose size is ten sectors, and which has a buffer pool of five buffers associated with it. We will use a LRU replacement scheme. The following series of memory requests occurs. 9017668135171 After the first five requests, the buffer pool will store the sectors in the order 6, 7, 1, 0, 9. Because Sector 6 is already at the front, the next request can be answered without reading new data from disk or reordering the buffers. The request to Sector 8 requires emptying the contents of the least recently used buffer, which contains Sector 9. The request to Sector 1 brings the buffer holding Sector 1’s contents back to the front. Processing the remaining requests results in the buffer pool as shown in Figure 8.5. Example 8.3 Figure 8.5 illustrates a buffer pool of five blocks mediating a virtual memory of ten blocks. At any given moment, up to five sectors of information can be in main memory. Assume that Sectors 1, 7, 5, 3, and 8 are curently in the buffer pool, stored in this order, and that we use the LRU buffer replacement strategy. If a request for Sector 9 is then received, then one sector currently in the buffer pool must be replaced. Because the buffer containing Sector 8 is the least recently used buffer, its contents will be copied back to disk at Sector 8. The contents of Sector 9 are then copied into this buffer, and it is moved to the front of the buffer pool (leaving the buffer containing Sector 3 as the new least-recently used buffer). If the next memory request were to Sector 5, no data would need to be read from disk. Instead, the buffer containing Sector 5 would be moved to the front of the buffer pool. When implementing buffer pools, there are two basic approaches that can be taken regarding the transfer of information between the user of the buffer pool and
- 312. Sec. 8.3 Buffers and Buffer Pools 293 (on disk) Secondary Storage 2 3 4 5 6 7 8 9 8 3 5 0 1 7 1 Main Memory (in RAM) Figure 8.5 An illustration of virtual memory. The complete collection of infor- mation resides in the slowe, secondary storage (on disk). Those sectors recently accessed are held in the fast main memory (in RAM). In this example, copies of Sectors 1, 7, 5, 3, and 8 from secondary storage are currently stored in the main memory. If a memory access to Sector 9 is received, one of the sectors currently in main memory must be replaced. the buffer pool class itself. The first approach is to pass “messages” between the two. This approach is illustrated by the following abstract class: /** ADT for buffer pools using the message-passing style */ public interface BufferPoolADT { /** Copy "sz" bytes from "space" to position "pos" in the buffered storage */ public void insert(byte[] space, int sz, int pos); /** Copy "sz" bytes from position "pos" of the buffered storage to "space". */ public void getbytes(byte[] space, int sz, int pos); } This simple class provides an interface with two member functions, insert and getbytes. The information is passed between the buffer pool user and the buffer pool through the space parameter. This is storage space, provided by the bufferpool client and at least sz bytes long, which the buffer pool can take in- formation from (the insert function) or put information into (the getbytes function). Parameter pos indicates where the information will be placed in the
- 313. 294 Chap. 8 File Processing and External Sorting buffer pool’s logical storage space. Physically, it will actually be copied to the appropriate byte position in some buffer in the buffer pool. Example 8.4 Assume each sector of the disk file (and thus each block in the buffer pool) stores 1024 bytes. Assume that the buffer pool is in the state shown in Figure 8.5. If the next request is to copy 40 bytes begin- ning at position 6000 of the file, these bytes should be placed into Sector 5 (whose bytes go from position 5120 to position 6143). Because Sector 5 is currently in the buffer pool, we simply copy the 40 bytes contained in space to byte positions 880-919. The buffer containing Sector 5 is then moved to the buffer pool ahead of the buffer containing Sector 1. The alternative approach is to have the buffer pool provide to the user a direct pointer to a buffer that contains the necessary information. Such an interface might look as follows: /** ADT for buffer pools using the buffer-passing style */ public interface BufferPoolADT { /** Return pointer to the requested block */ public byte[] getblock(int block); /** Set the dirty bit for the buffer holding "block" */ public void dirtyblock(int block); // Tell the size of a buffer public int blocksize(); }; In this approach, the user of the buffer pool is made aware that the storage space is divided into blocks of a given size, where each block is the size of a buffer. The user requests specific blocks from the buffer pool, with a pointer to the buffer holding the requested block being returned to the user. The user might then read from or write to this space. If the user writes to the space, the buffer pool must be informed of this fact. The reason is that, when a given block is to be removed from the buffer pool, the contents of that block must be written to the backing storage if it has been modified. If the block has not been modified, then it is unnecessary to write it out. Example 8.5 We wish to write 40 bytes beginning at logical position 6000 in the file. Assume that the buffer pool is in the state shown in Fig- ure 8.5. Using the second ADT, the client would need to know that blocks (buffers) are of size 1024, and therefore would request access to Sector 5.
- 314. Sec. 8.3 Buffers and Buffer Pools 295 A pointer to the buffer containing Sector 5 would be retuned by the call to getblock. The client would then copy 40 bytes to positions 880-919 of the buffer, and call dirtyblock to warn the buffer pool that the contents of this block have been modified. A further problem with the second ADT is the risk of stale pointers. When the buffer pool user is given a pointer to some buffer space at time T1, that pointer does indeed refer to the desired data at that time. As further requests are made to the buffer pool, it is possible that the data in any given buffer will be removed and replaced with new data. If the buffer pool user at a later time T2 then refers to the data refered to by the pointer given at time T1, it is possible that the data are no longer valid because the buffer contents have been replaced in the meantime. Thus the pointer into the buffer pool’s memory has become “stale.” To guarantee that a pointer is not stale, it should not be used if intervening requests to the buffer pool have taken place. We can solve this problem by introducing the concept of a user (or possibly multiple users) gaining access to a buffer, and then releasing the buffer when done. We will add method acquireBuffer and releaseBuffer for this purpose. Method acquireBuffer takes a block ID as input and returns a pointer to the buffer that will be used to store this block. The buffer pool will keep a count of the number of requests currently active for this block. Method releaseBuffer will reduce the count of active users for the associated block. Buffers associated with active blocks will not be eligible for flushing from the buffer pool. This will lead to a problem if the client neglects to release active blocks when they are no longer needed. There would also be a problem if there were more total active blocks than buffers in the buffer pool. However, the buffer pool should always be initialized to include more buffers than will ever be active at one time. An additional problem with both ADTs presented so far comes when the user intends to completely overwrite the contents of a block, and does not need to read in the old contents already on disk. However, the buffer pool cannot in general know whether the user wishes to use the old contents or not. This is especially true with the message-passing approach where a given message might overwrite only part of the block. In this case, the block will be read into memory even when not needed, and then its contents will be overwritten. This inefficiency can be avoided (at least in the buffer-passing version) by sep- arating the assignment of blocks to buffers from actually reading in data for the block. In particular, the following revised buffer-passing ADT does not actually read data in the acquireBuffer method. Users who wish to see the old con-
- 315. 296 Chap. 8 File Processing and External Sorting tents must then issue a readBlock request to read the data from disk into the buffer, and then a getDataPointer request to gain direct access to the buffer’s data contents. /** Improved ADT for buffer pools using the buffer-passing style. Most user functionality is in the buffer class, not the buffer pool itself. */ /** A single buffer in the buffer pool */ public interface BufferADT { /** Read the associated block from disk (if necessary) and return a pointer to the data */ public byte[] readBlock(); /** Return a pointer to the buffer’s data array (without reading from disk) */ public byte[] getDataPointer(); /** Flag the buffer’s contents as having changed, so that flushing the block will write it back to disk */ public void markDirty(); /** Release the block’s access to this buffer. Further accesses to this buffer are illegal. */ public void releaseBuffer(); } /** The bufferpool */ public interface BufferPoolADT { /** Relate a block to a buffer, returning a pointer to a buffer object */ Buffer acquireBuffer(int block); } Clearly, the buffer-passing approach places more obligations on the user of the buffer pool. These obligations include knowing the size of a block, not corrupting the buffer pool’s storage space, and informing the buffer pool both when a block has been modified and when it is no longer needed. So many obligations make this approach prone to error. An advantage is that there is no need to do an extra copy step when getting information from the user to the buffer. If the size of the records stored is small, this is not an important consideration. If the size of the records is large (especially if the record size and the buffer size are the same, as typically is the case when implementing B-trees, see Section 10.5), then this efficiency issue might become important. Note however that the in-memory copy time will always be far less than the time required to write the contents of a buffer to disk. For applications
- 316. Sec. 8.4 The Programmer’s View of Files 297 where disk I/O is the bottleneck for the program, even the time to copy lots of information between the buffer pool user and the buffer might be inconsequential. Another advantage to buffer passing is the reduction in unnecessary read operations for data that will be overwritten anyway. You should note that the implementations for class BufferPool above are not generic. Instead, the space parameter and the buffer pointer are declared to be byte[]. When a class is generic, that means that the record type is arbitrary, but that the class knows what the record type is. In contrast, using a byte[] pointer for the space means that not only is the record type arbitrary, but also the buffer pool does not even know what the user’s record type is. In fact, a given buffer pool might have many users who store many types of records. In a buffer pool, the user decides where a given record will be stored but has no control over the precise mechanism by which data are transfered to the backing storage. This is in contrast to the memory manager described in Section 12.3 in which the user passes a record to the manager and has no control at all over where the record is stored. 8.4 The Programmer’s View of Files The Java programmer’s logical view of a random access file is a single stream of bytes. Interaction with a file can be viewed as a communications channel for issuing one of three instructions: read bytes from the current position in the file, write bytes to the current position in the file, and move the current position within the file. You do not normally see how the bytes are stored in sectors, clusters, and so forth. The mapping from logical to physical addresses is done by the file system, and sector-level buffering is done automatically by the disk controller. When processing records in a disk file, the order of access can have a great effect on I/O time. A random access procedure processes records in an order independent of their logical order within the file. Sequential access processes records in order of their logical appearance within the file. Sequential processing requires less seek time if the physical layout of the disk file matches its logical layout, as would be expected if the file were created on a disk with a high percentage of free space. Following are the primary Java functions for accessing information from ran- dom access disk files when using class RandomAccessFile. • RandomAccessFile(String name, String mode): Class con- structor, opens a disk file for processing.
- 317. 298 Chap. 8 File Processing and External Sorting • read(byte[] b): Read some bytes from the current position in the file. The current position moves forward as the bytes are read. • write(byte[] b): Write some bytes at the current position in the file (overwriting the bytes already at that position). The current position moves forward as the bytes are written. • seek(long pos): Move the current position in the file. This allows bytes at other places within the file to be read or written. • close(): Close a file at the end of processing. 8.5 External Sorting We now consider the problem of sorting collections of records too large to fit in main memory. Because the records must reside in peripheral or external memory, such sorting methods are called external sorts. This is in contrast to the internal sorts discussed in Chapter 7 which assume that the records to be sorted are stored in main memory. Sorting large collections of records is central to many applications, such as processing payrolls and other large business databases. As a consequence, many external sorting algorithms have been devised. Years ago, sorting algorithm designers sought to optimize the use of specific hardware configurations, such as multiple tape or disk drives. Most computing today is done on personal computers and low-end workstations with relatively powerful CPUs, but only one or at most two disk drives. The techniques presented here are geared toward optimized pro- cessing on a single disk drive. This approach allows us to cover the most important issues in external sorting while skipping many less important machine-dependent details. Readers who have a need to implement efficient external sorting algorithms that take advantage of more sophisticated hardware configurations should consult the references in Section 8.6. When a collection of records is too large to fit in main memory, the only prac- tical way to sort it is to read some records from disk, do some rearranging, then write them back to disk. This process is repeated until the file is sorted, with each record read perhaps many times. Given the high cost of disk I/O, it should come as no surprise that the primary goal of an external sorting algorithm is to minimize the amount of information that must be read from or written to disk. A certain amount of additional CPU processing can profitably be traded for reduced disk access. Before discussing external sorting techniques, consider again the basic model for accessing information from disk. The file to be sorted is viewed by the program- mer as a sequential series of fixed-size blocks. Assume (for simplicity) that each
- 318. Sec. 8.5 External Sorting 299 block contains the same number of fixed-size data records. Depending on the ap- plication, a record might be only a few bytes — composed of little or nothing more than the key — or might be hundreds of bytes with a relatively small key field. Records are assumed not to cross block boundaries. These assumptions can be relaxed for special-purpose sorting applications, but ignoring such complications makes the principles clearer. As explained in Section 8.2, a sector is the basic unit of I/O. In other words, all disk reads and writes are for one or more complete sectors. Sector sizes are typically a power of two, in the range 512 to 16K bytes, depending on the operating system and the size and speed of the disk drive. The block size used for external sorting algorithms should be equal to or a multiple of the sector size. Under this model, a sorting algorithm reads a block of data into a buffer in main memory, performs some processing on it, and at some future time writes it back to disk. From Section 8.1 we see that reading or writing a block from disk takes on the order of one million times longer than a memory access. Based on this fact, we can reasonably expect that the records contained in a single block can be sorted by an internal sorting algorithm such as Quicksort in less time than is required to read or write the block. Under good conditions, reading from a file in sequential order is more efficient than reading blocks in random order. Given the significant impact of seek time on disk access, it might seem obvious that sequential processing is faster. However, it is important to understand precisely under what circumstances sequential file processing is actually faster than random access, because it affects our approach to designing an external sorting algorithm. Efficient sequential access relies on seek time being kept to a minimum. The first requirement is that the blocks making up a file are in fact stored on disk in sequential order and close together, preferably filling a small number of contiguous tracks. At the very least, the number of extents making up the file should be small. Users typically do not have much control over the layout of their file on disk, but writing a file all at once in sequential order to a disk drive with a high percentage of free space increases the likelihood of such an arrangement. The second requirement is that the disk drive’s I/O head remain positioned over the file throughout sequential processing. This will not happen if there is competition of any kind for the I/O head. For example, on a multi-user timeshared computer the sorting process might compete for the I/O head with the processes of other users. Even when the sorting process has sole control of the I/O head, it is still likely that sequential processing will not be efficient. Imagine the situation where all processing is done on a single disk drive, with the typical arrangement
- 319. 300 Chap. 8 File Processing and External Sorting of a single bank of read/write heads that move together over a stack of platters. If the sorting process involves reading from an input file, alternated with writing to an output file, then the I/O head will continuously seek between the input file and the output file. Similarly, if two input files are being processed simultaneously (such as during a merge process), then the I/O head will continuously seek between these two files. The moral is that, with a single disk drive, there often is no such thing as effi- cient sequential processing of a data file. Thus, a sorting algorithm might be more efficient if it performs a smaller number of non-sequential disk operations rather than a larger number of logically sequential disk operations that require a large number of seeks in practice. As mentioned previously, the record size might be quite large compared to the size of the key. For example, payroll entries for a large business might each store hundreds of bytes of information including the name, ID, address, and job title for each employee. The sort key might be the ID number, requiring only a few bytes. The simplest sorting algorithm might be to process such records as a whole, reading the entire record whenever it is processed. However, this will greatly increase the amount of I/O required, because only a relatively few records will fit into a single disk block. Another alternative is to do a key sort. Under this method, the keys are all read and stored together in an index file, where each key is stored along with a pointer indicating the position of the corresponding record in the original data file. The key and pointer combination should be substantially smaller than the size of the original record; thus, the index file will be much smaller than the complete data file. The index file will then be sorted, requiring much less I/O because the index records are smaller than the complete records. Once the index file is sorted, it is possible to reorder the records in the original database file. This is typically not done for two reasons. First, reading the records in sorted order from the record file requires a random access for each record. This can take a substantial amount of time and is only of value if the complete collection of records needs to be viewed or processed in sorted order (as opposed to a search for selected records). Second, database systems typically allow searches to be done on multiple keys. For example, today’s processing might be done in order of ID numbers. Tomorrow, the boss might want information sorted by salary. Thus, there might be no single “sorted” order for the full record. Instead, multiple index files are often maintained, one for each sort key. These ideas are explored further in Chapter 10.
- 320. Sec. 8.5 External Sorting 301 8.5.1 Simple Approaches to External Sorting If your operating system supports virtual memory, the simplest “external” sort is to read the entire file into virtual memory and run an internal sorting method such as Quicksort. This approach allows the virtual memory manager to use its normal buffer pool mechanism to control disk accesses. Unfortunately, this might not al- ways be a viable option. One potential drawback is that the size of virtual memory is usually limited to something much smaller than the disk space available. Thus, your input file might not fit into virtual memory. Limited virtual memory can be overcome by adapting an internal sorting method to make use of your own buffer pool. A more general problem with adapting an internal sorting algorithm to external sorting is that it is not likely to be as efficient as designing a new algorithm with the specific goal of minimizing disk fetches. Consider the simple adaptation of Quicksort to use a buffer pool. Quicksort begins by processing the entire array of records, with the first partition step moving indices inward from the two ends. This can be implemented efficiently using a buffer pool. However, the next step is to process each of the subarrays, followed by processing of sub-subarrays, and so on. As the subarrays get smaller, processing quickly approaches random access to the disk drive. Even with maximum use of the buffer pool, Quicksort still must read and write each record log n times on average. We can do much better. Finally, even if the virtual memory manager can give good performance using a standard quicksort, wthi will come at the cost of using a lot of the system’s working memory, which will mean that the system cannot use this space for other work. Better methods can save time while also using less memory. Our approach to external sorting is derived from the Mergesort algorithm. The simplest form of external Mergesort performs a series of sequential passes over the records, merging larger and larger sublists on each pass. The first pass merges sublists of size 1 into sublists of size 2; the second pass merges the sublists of size 2 into sublists of size 4; and so on. A sorted sublist is called a run. Thus, each pass is merging pairs of runs to form longer runs. Each pass copies the contents of the file to another file. Here is a sketch of the algorithm, as illustrated by Figure 8.6. 1. Split the original file into two equal-sized run files. 2. Read one block from each run file into input buffers. 3. Take the first record from each input buffer, and write a run of length two to an output buffer in sorted order. 4. Take the next record from each input buffer, and write a run of length two to a second output buffer in sorted order.
- 321. 302 Chap. 8 File Processing and External Sorting Runs of length 4 Runs of length 2 36 15 23 20 13 14 15 23 36 17 28 20 13 14 14 13 Runs of length 1 15 36 28 17 23 17 20 28 Figure 8.6 A simple external Mergesort algorithm. Input records are divided equally between two input files. The first runs from each input file are merged and placed into the first output file. The second runs from each input file are merged and placed in the second output file. Merging alternates between the two output files until the input files are empty. The roles of input and output files are then reversed, allowing the runlength to be doubled with each pass. 5. Repeat until finished, alternating output between the two output run buffers. Whenever the end of an input block is reached, read the next block from the appropriate input file. When an output buffer is full, write it to the appropriate output file. 6. Repeat steps 2 through 5, using the original output files as input files. On the second pass, the first two records of each input run file are already in sorted order. Thus, these two runs may be merged and output as a single run of four elements. 7. Each pass through the run files provides larger and larger runs until only one run remains. Example 8.6 Using the input of Figure 8.6, we first create runs of length one split between two input files. We then process these two input files sequentially, making runs of length two. The first run has the values 20 and 36, which are output to the first output file. The next run has 13 and 17, which is ouput to the second file. The run 14, 28 is sent to the first file, then run 15, 23 is sent to the second file, and so on. Once this pass has completed, the roles of the input files and output files are reversed. The next pass will merge runs of length two into runs of length four. Runs 20, 36 and 13, 17 are merged to send 13, 17, 20, 36 to the first output file. Then runs 14, 28 and 15, 23 are merged to send run 14, 15, 23, 28 to the second output file. In the final pass, these runs are merged to form the final run 13, 14, 15, 17, 20, 23, 28, 36. This algorithm can easily take advantage of the double buffering techniques described in Section 8.3. Note that the various passes read the input run files se-
- 322. Sec. 8.5 External Sorting 303 quentially and write the output run files sequentially. For sequential processing and double buffering to be effective, however, it is necessary that there be a separate I/O head available for each file. This typically means that each of the input and output files must be on separate disk drives, requiring a total of four disk drives for maximum efficiency. The external Mergesort algorithm just described requires that log n passes be made to sort a file of n records. Thus, each record must be read from disk and written to disk log n times. The number of passes can be significantly reduced by observing that it is not necessary to use Mergesort on small runs. A simple modi- fication is to read in a block of data, sort it in memory (perhaps using Quicksort), and then output it as a single sorted run. Example 8.7 Assume that we have blocks of size 4KB, and records are eight bytes with four bytes of data and a 4-byte key. Thus, each block con- tains 512 records. Standard Mergesort would require nine passes to gener- ate runs of 512 records, whereas processing each block as a unit can be done in one pass with an internal sort. These runs can then be merged by Merge- sort. Standard Mergesort requires eighteen passes to process 256K records. Using an internal sort to create initial runs of 512 records reduces this to one initial pass to create the runs and nine merge passes to put them all together, approximately half as many passes. We can extend this concept to improve performance even further. Available main memory is usually much more than one block in size. If we process larger initial runs, then the number of passes required by Mergesort is further reduced. For example, most modern computers can provide tens or even hundreds of megabytes of RAM to the sorting program. If all of this memory (excepting a small amount for buffers and local variables) is devoted to building initial runs as large as possible, then quite large files can be processed in few passes. The next section presents a technique for producing large runs, typically twice as large as could fit directly into main memory. Another way to reduce the number of passes required is to increase the number of runs that are merged together during each pass. While the standard Mergesort algorithm merges two runs at a time, there is no reason why merging needs to be limited in this way. Section 8.5.3 discusses the technique of multiway merging. Over the years, many variants on external sorting have been presented, but all are based on the following two steps: 1. Break the file into large initial runs.
- 323. 304 Chap. 8 File Processing and External Sorting 2. Merge the runs together to form a single sorted file. 8.5.2 Replacement Selection This section treats the problem of creating initial runs as large as possible from a disk file, assuming a fixed amount of RAM is available for processing. As men- tioned previously, a simple approach is to allocate as much RAM as possible to a large array, fill this array from disk, and sort the array using Quicksort. Thus, if the size of memory available for the array is M records, then the input file can be broken into initial runs of length M. A better approach is to use an algorithm called replacement selection that, on average, creates runs of 2M records in length. Re- placement selection is actually a slight variation on the Heapsort algorithm. The fact that Heapsort is slower than Quicksort is irrelevant in this context because I/O time will dominate the total running time of any reasonable external sorting alg- orithm. Building longer initial runs will reduce the total I/O time required. Replacement selection views RAM as consisting of an array of size M in ad- dition to an input buffer and an output buffer. (Additional I/O buffers might be desirable if the operating system supports double buffering, because replacement selection does sequential processing on both its input and its output.) Imagine that the input and output files are streams of records. Replacement selection takes the next record in sequential order from the input stream when needed, and outputs runs one record at a time to the output stream. Buffering is used so that disk I/O is performed one block at a time. A block of records is initially read and held in the input buffer. Replacement selection removes records from the input buffer one at a time until the buffer is empty. At this point the next block of records is read in. Output to a buffer is similar: Once the buffer fills up it is written to disk as a unit. This process is illustrated by Figure 8.7. Replacement selection works as follows. Assume that the main processing is done in an array of size M records. 1. Fill the array from disk. Set LAST = M − 1. 2. Build a min-heap. (Recall that a min-heap is defined such that the record at each node has a key value less than the key values of its children.) 3. Repeat until the array is empty: (a) Send the record with the minimum key value (the root) to the output buffer. (b) Let R be the next record in the input buffer. If R’s key value is greater than the key value just output ... i. Then place R at the root.
- 324. Sec. 8.5 External Sorting 305 Input Buffer Output Buffer File Input Run File Output RAM Figure 8.7 Overview of replacement selection. Input records are processed se- quentially. Initially RAM is filled with M records. As records are processed, they are written to an output buffer. When this buffer becomes full, it is written to disk. Meanwhile, as replacement selection needs records, it reads them from the input buffer. Whenever this buffer becomes empty, the next block of records is read from disk. ii. Else replace the root with the record in array position LAST, and place R at position LAST. Set LAST = LAST − 1. (c) Sift down the root to reorder the heap. When the test at step 3(b) is successful, a new record is added to the heap, eventually to be output as part of the run. As long as records coming from the input file have key values greater than the last key value output to the run, they can be safely added to the heap. Records with smaller key values cannot be output as part of the current run because they would not be in sorted order. Such values must be stored somewhere for future processing as part of another run. However, because the heap will shrink by one element in this case, there is now a free space where the last element of the heap used to be! Thus, replacement selection will slowly shrink the heap and at the same time use the discarded heap space to store records for the next run. Once the first run is complete (i.e., the heap becomes empty), the array will be filled with records ready to be processed for the second run. Figure 8.8 illustrates part of a run being created by replacement selection. It should be clear that the minimum length of a run will be M records if the size of the heap is M, because at least those records originally in the heap will be part of the run. Under good conditions (e.g., if the input is sorted), then an arbitrarily long run is possible. In fact, the entire file could be processed as one run. If conditions are bad (e.g., if the input is reverse sorted), then runs of only size M result. What is the expected length of a run generated by replacement selection? It can be deduced from an analogy called the snowplow argument. Imagine that a snowplow is going around a circular track during a heavy, but steady, snowstorm. After the plow has been around at least once, snow on the track must be as follows. Immediately behind the plow, the track is empty because it was just plowed. The greatest level of snow on the track is immediately in front of the plow, because
- 325. 306 Chap. 8 File Processing and External Sorting Input Memory Output 16 12 29 16 14 19 21 25 29 56 31 14 35 25 31 21 40 29 56 21 40 56 40 21 25 31 29 16 12 56 40 31 25 19 21 25 21 56 31 40 19 19 19 21 25 31 56 40 29 14 Figure 8.8 Replacement selection example. After building the heap, root value 12 is output and incoming value 16 replaces it. Value 16 is output next, replaced with incoming value 29. The heap is reordered, with 19 rising to the root. Value 19 is output next. Incoming value 14 is too small for this run and is placed at end of the array, moving value 40 to the root. Reordering the heap results in 21 rising to the root, which is output next.
- 326. Sec. 8.5 External Sorting 307 Existing snow Future snow Falling Snow Snowplow Movement Start time T Figure 8.9 The snowplow analogy showing the action during one revolution of the snowplow. A circular track is laid out straight for purposes of illustration, and is shown in cross section. At any time T, the most snow is directly in front of the snowplow. As the plow moves around the track, the same amount of snow is always in front of the plow. As the plow moves forward, less of this is snow that was in the track at time T; more is snow that has fallen since. this is the place least recently plowed. At any instant, there is a certain amount of snow S on the track. Snow is constantly falling throughout the track at a steady rate, with some snow falling “in front” of the plow and some “behind” the plow. (On a circular track, everything is actually “in front” of the plow, but Figure 8.9 illustrates the idea.) During the next revolution of the plow, all snow S on the track is removed, plus half of what falls. Because everything is assumed to be in steady state, after one revolution S snow is still on the track, so 2S snow must fall during a revolution, and 2S snow is removed during a revolution (leaving S snow behind). At the beginning of replacement selection, nearly all values coming from the input file are greater (i.e., “in front of the plow”) than the latest key value output for this run, because the run’s initial key values should be small. As the run progresses, the latest key value output becomes greater and so new key values coming from the input file are more likely to be too small (i.e., “after the plow”); such records go to the bottom of the array. The total length of the run is expected to be twice the size of the array. Of course, this assumes that incoming key values are evenly distributed within the key range (in terms of the snowplow analogy, we assume that snow falls evenly throughout the track). Sorted and reverse sorted inputs do not meet this expectation and so change the length of the run. 8.5.3 Multiway Merging The second stage of a typical external sorting algorithm merges the runs created by the first stage. Assume that we have R runs to merge. If a simple two-way merge is used, then R runs (regardless of their sizes) will require log R passes through the file. While R should be much less than the total number of records (because
- 327. 308 Chap. 8 File Processing and External Sorting Input Runs 12 20 ... 18 23 6 7 ... 15 5 10 ... 5 6 7 10 12 ... Output Buffer Figure 8.10 Illustration of multiway merge. The first value in each input run is examined and the smallest sent to the output. This value is removed from the input and the process repeated. In this example, values 5, 6, and 12 are compared first. Value 5 is removed from the first run and sent to the output. Values 10, 6, and 12 will be compared next. After the first five values have been output, the “current” value of each block is the one underlined. the initial runs should each contain many records), we would like to reduce still further the number of passes required to merge the runs together. Note that two- way merging does not make good use of available memory. Because merging is a sequential process on the two runs, only one block of records per run need be in memory at a time. Keeping more than one block of a run in memory at any time will not reduce the disk I/O required by the merge process. Thus, most of the space just used by the heap for replacement selection (typically many blocks in length) is not being used by the merge process. We can make better use of this space and at the same time greatly reduce the number of passes needed to merge the runs if we merge several runs at a time. Multiway merging is similar to two-way merging. If we have B runs to merge, with a block from each run available in memory, then the B-way merge algorithm simply looks at B values (the frontmost value for each input run) and selects the smallest one to output. This value is removed from its run, and the process is repeated. When the current block for any run is exhausted, the next block from that run is read from disk. Figure 8.10 illustrates a multiway merge. Conceptually, multiway merge assumes that each run is stored in a separate file. However, this is not necessary in practice. We only need to know the position of each run within a single file, and use seek to move to the appropriate block when- ever we need new data from a particular run. Naturally, this approach destroys the ability to do sequential processing on the input file. However, if all runs were stored on a single disk drive, then processing would not be truely sequential anyway be- cause the I/O head would be alternating between the runs. Thus, multiway merging
- 328. Sec. 8.5 External Sorting 309 replaces several (potentially) sequential passes with a single random access pass. If the processing would not be sequential anyway (such as when all processing is on a single disk drive), no time is lost by doing so. Multiway merging can greatly reduce the number of passes required. If there is room in memory to store one block for each run, then all runs can be merged in a single pass. Thus, replacement selection can build initial runs in one pass, and multiway merging can merge all runs in one pass, yielding a total cost of two passes. However, for truly large files, there might be too many runs for each to get a block in memory. If there is room to allocate B blocks for a B-way merge, and the number of runs R is greater than B, then it will be necessary to do multiple merge passes. In other words, the first B runs are merged, then the next B, and so on. These super-runs are then merged by subsequent passes, B super-runs at a time. How big a file can be merged in one pass? Assuming B blocks were allocated to the heap for replacement selection (resulting in runs of average length 2B blocks), followed by a B-way merge, we can process on average a file of size 2B2 blocks in a single multiway merge. 2Bk+1 blocks on average can be processed in k B- way merges. To gain some appreciation for how quickly this grows, assume that we have available 0.5MB of working memory, and that a block is 4KB, yielding 128 blocks in working memory. The average run size is 1MB (twice the working memory size). In one pass, 128 runs can be merged. Thus, a file of size 128MB can, on average, be processed in two passes (one to build the runs, one to do the merge) with only 0.5MB of working memory. A larger block size would reduce the size of the file that can be processed in one merge pass for a fixed-size working memory; a smaller block size or larger working memory would increase the file size that can be processed in one merge pass. With 0.5MB of working memory and 4KB blocks, a file of size 16 gigabytes could be processed in two merge passes, which is big enough for most applications. Thus, this is a very effective algorithm for single disk drive external sorting. Figure 8.11 shows a comparison of the running time to sort various-sized files for the following implementations: (1) standard Mergesort with two input runs and two output runs, (2) two-way Mergesort with large initial runs (limited by the size of available memory), and (3) R-way Mergesort performed after generating large initial runs. In each case, the file was composed of a series of four-byte records (a two-byte key and a two-byte data value), or 256K records per megabyte of file size. We can see from this table that using even a modest memory size (two blocks) to create initial runs results in a tremendous savings in time. Doing 4-way merges of the runs provides another considerable speedup, however largescale multi-way
- 329. 310 Chap. 8 File Processing and External Sorting File Sort 1 Sort 2 Sort 3 Size Memory size (in blocks) Memory size (in blocks) 2 4 16 256 2 4 16 1 0.61 0.27 0.24 0.19 0.10 0.21 0.15 0.13 4,864 2,048 1,792 1,280 256 2,048 1,024 512 4 2.56 1.30 1.19 0.96 0.61 1.15 0.68 0.66* 21,504 10,240 9,216 7,168 3,072 10,240 5,120 2,048 16 11.28 6.12 5.63 4.78 3.36 5.42 3.19 3.10 94,208 49,152 45,056 36,864 20,480 49,152 24,516 12,288 256 220.39 132.47 123.68 110.01 86.66 115.73 69.31 68.71 1,769K 1,048K 983K 852K 589K 1,049K 524K 262K Figure 8.11 A comparison of three external sorts on a collection of small records for files of various sizes. Each entry in the table shows time in seconds and total number of blocks read and written by the program. File sizes are in Megabytes. For the third sorting algorithm, on file size of 4MB, the time and blocks shown in the last column are for a 32-way merge. 32 is used instead of 16 because 32 is a root of the number of blocks in the file (while 16 is not), thus allowing the same number of runs to be merged at every pass. merges for R beyond about 4 or 8 runs does not help much because a lot of time is spent determining which is the next smallest element among the R runs. We see from this experiment that building large initial runs reduces the running time to slightly more than one third that of standard Mergesort, depending on file and memory sizes. Using a multiway merge further cuts the time nearly in half. In summary, a good external sorting algorithm will seek to do the following: • Make the initial runs as long as possible. • At all stages, overlap input, processing, and output as much as possible. • Use as much working memory as possible. Applying more memory usually speeds processing. In fact, more memory will have a greater effect than a faster disk. A faster CPU is unlikely to yield much improvement in running time for external sorting, because disk I/O speed is the limiting factor. • If possible, use additional disk drives for more overlapping of processing with I/O, and to allow for sequential file processing. 8.6 Further Reading A good general text on file processing is Folk and Zoellig’s File Structures: A Conceptual Toolkit [FZ98]. A somewhat more advanced discussion on key issues in file processing is Betty Salzberg’s File Structures: An Analytical Approach [Sal88].
- 330. Sec. 8.7 Exercises 311 A great discussion on external sorting methods can be found in Salzberg’s book. The presentation in this chapter is similar in spirit to Salzberg’s. For details on disk drive modeling and measurement, see the article by Ruemm- ler and Wilkes, “An Introduction to Disk Drive Modeling” [RW94]. See Andrew S. Tanenbaum’s Structured Computer Organization [Tan06] for an introduction to computer hardware and organization. An excellent, detailed description of mem- ory and hard disk drives can be found online at “The PC Guide,” by Charles M. Kozierok [Koz05] (www.pcguide.com). The PC Guide also gives detailed de- scriptions of the Microsoft Windows and UNIX (Linux) file systems. See “Outperforming LRU with an Adaptive Replacement Cache Algorithm” by Megiddo and Modha for an example of a more sophisticated algorithm than LRU for managing buffer pools. The snowplow argument comes from Donald E. Knuth’s Sorting and Searching [Knu98], which also contains a wide variety of external sorting algorithms. 8.7 Exercises 8.1 Computer memory and storage prices change rapidly. Find out what the current prices are for the media listed in Figure 8.1. Does your information change any of the basic conclusions regarding disk processing? 8.2 Assume a disk drive from the late 1990s is configured as follows. The to- tal storage is approximately 675MB divided among 15 surfaces. Each sur- face has 612 tracks; there are 144 sectors/track, 512 bytes/sector, and 8 sec- tors/cluster. The disk turns at 3600 rpm. The track-to-track seek time is 20 ms, and the average seek time is 80 ms. Now assume that there is a 360KB file on the disk. On average, how long does it take to read all of the data in the file? Assume that the first track of the file is randomly placed on the disk, that the entire file lies on adjacent tracks, and that the file completely fills each track on which it is found. A seek must be performed each time the I/O head moves to a new track. Show your calculations. 8.3 Using the specifications for the disk drive given in Exercise 8.2, calculate the expected time to read one entire track, one sector, and one byte. Show your calculations. 8.4 Using the disk drive specifications given in Exercise 8.2, calculate the time required to read a 10MB file assuming (a) The file is stored on a series of contiguous tracks, as few tracks as pos- sible. (b) The file is spread randomly across the disk in 4KB clusters.
- 331. 312 Chap. 8 File Processing and External Sorting Show your calculations. 8.5 Assume that a disk drive is configured as follows. The total storage is ap- proximately 1033MB divided among 15 surfaces. Each surface has 2100 tracks, there are 64 sectors/track, 512 bytes/sector, and 8 sectors/cluster. The disk turns at 7200 rpm. The track-to-track seek time is 3 ms, and the average seek time is 20 ms. Now assume that there is a 512KB file on the disk. On average, how long does it take to read all of the data on the file? Assume that the first track of the file is randomly placed on the disk, that the entire file lies on contiguous tracks, and that the file completely fills each track on which it is found. Show your calculations. 8.6 Using the specifications for the disk drive given in Exercise 8.5, calculate the expected time to read one entire track, one sector, and one byte. Show your calculations. 8.7 Using the disk drive specifications given in Exercise 8.5, calculate the time required to read a 10MB file assuming (a) The file is stored on a series of contiguous tracks, as few tracks as pos- sible. (b) The file is spread randomly across the disk in 4KB clusters. Show your calculations. 8.8 A typical disk drive from 2004 has the following specifications.2 The total storage is approximately 120GB on 6 platter surfaces or 20GB/platter. Each platter has 16K tracks with 2560 sectors/track (a sector holds 512 bytes) and 16 sectors/cluster. The disk turns at 7200 rpm. The track-to-track seek time is 2.0 ms, and the average seek time is 10.0 ms. Now assume that there is a 6MB file on the disk. On average, how long does it take to read all of the data on the file? Assume that the first track of the file is randomly placed on the disk, that the entire file lies on contiguous tracks, and that the file completely fills each track on which it is found. Show your calculations. 8.9 Using the specifications for the disk drive given in Exercise 8.8, calculate the expected time to read one entire track, one sector, and one byte. Show your calculations. 8.10 Using the disk drive specifications given in Exercise 8.8, calculate the time required to read a 10MB file assuming 2 To make the exercise doable, this specification is completely ficticious with respect to the track and sector layout. While sectors do have 512 bytes, and while the number of platters and amount of data per track is plausible, the reality is that all modern drives use a zoned organization to keep the data density from inside to outside of the disk reasonably high. The rest of the numbers are typical for a drive from 2004.
- 332. Sec. 8.7 Exercises 313 (a) The file is stored on a series of contiguous tracks, as few tracks as pos- sible. (b) The file is spread randomly across the disk in 8KB clusters. Show your calculations. 8.11 At the end of 2004, the fastest disk drive I could find specifications for was the Maxtor Atlas. This drive had a nominal capacity of 73.4GB using 4 plat- ters (8 surfaces) or 9.175GB/surface. Assume there are 16,384 tracks with an average of 1170 sectors/track and 512 bytes/sector.3 The disk turns at 15,000 rpm. The track-to-track seek time is 0.4 ms and the average seek time is 3.6 ms. How long will it take on average to read a 6MB file, assuming that the first track of the file is randomly placed on the disk, that the entire file lies on contiguous tracks, and that the file completely fills each track on which it is found. Show your calculations. 8.12 Using the specifications for the disk drive given in Exercise 8.11, calculate the expected time to read one entire track, one sector, and one byte. Show your calculations. 8.13 Using the disk drive specifications given in Exercise 8.11, calculate the time required to read a 10MB file assuming (a) The file is stored on a series of contiguous tracks, as few tracks as pos- sible. (b) The file is spread randomly across the disk in 8KB clusters. Show your calculations. 8.14 Prove that two tracks selected at random from a disk are separated on average by one third the number of tracks on the disk. 8.15 Assume that a file contains one million records sorted by key value. A query to the file returns a single record containing the requested key value. Files are stored on disk in sectors each containing 100 records. Assume that the average time to read a sector selected at random is 10.0 ms. In contrast, it takes only 2.0 ms to read the sector adjacent to the current position of the I/O head. The “batch” algorithm for processing queries is to first sort the queries by order of appearance in the file, and then read the entire file sequentially, processing all queries in sequential order as the file is read. This algorithm implies that the queries must all be available before processing begins. The “interactive” algorithm is to process each query in order of its arrival, search- ing for the requested sector each time (unless by chance two queries in a row are to the same sector). Carefully define under what conditions the batch method is more efficient than the interactive method. 3 Again, this track layout does does not account for the zoned arrangement on modern disk drives.
- 333. 314 Chap. 8 File Processing and External Sorting 8.16 Assume that a virtual memory is managed using a buffer pool. The buffer pool contains five buffers and each buffer stores one block of data. Memory accesses are by block ID. Assume the following series of memory accesses takes place: 5 2 5 12 3 6 5 9 3 2 4 1 5 9 8 15 3 7 2 5 9 10 4 6 8 5 For each of the following buffer pool replacement strategies, show the con- tents of the buffer pool at the end of the series, and indicate how many times a block was found in the buffer pool (instead of being read into memory). Assume that the buffer pool is initially empty. (a) First-in, first out. (b) Least frequently used (with counts kept only for blocks currently in memory, counts for a page are lost when that page is removed, and the oldest item with the smallest count is removed when there is a tie). (c) Least frequently used (with counts kept for all blocks, and the oldest item with the smallest count is removed when there is a tie). (d) Least recently used. (e) Most recently used (replace the block that was most recently accessed). 8.17 Suppose that a record is 32 bytes, a block is 1024 bytes (thus, there are 32 records per block), and that working memory is 1MB (there is also addi- tional space available for I/O buffers, program variables, etc.). What is the expected size for the largest file that can be merged using replacement selec- tion followed by a single pass of multiway merge? Explain how you got your answer. 8.18 Assume that working memory size is 256KB broken into blocks of 8192 bytes (there is also additional space available for I/O buffers, program vari- ables, etc.). What is the expected size for the largest file that can be merged using replacement selection followed by two passes of multiway merge? Ex- plain how you got your answer. 8.19 Prove or disprove the following proposition: Given space in memory for a heap of M records, replacement selection will completely sort a file if no record in the file is preceded by M or more keys of greater value. 8.20 Imagine a database containing ten million records, with each record being 100 bytes long. Provide an estimate of the time it would take (in seconds) to sort the database on a typical workstation. 8.21 Assume that a company has a computer configuration satisfactory for pro- cessing their monthly payroll. Further assume that the bottleneck in payroll
- 334. Sec. 8.8 Projects 315 processing is a sorting operation on all of the employee records, and that an external sorting algorithm is used. The company’s payroll program is so good that it plans to hire out its services to do payroll processing for other companies. The president has an offer from a second company with 100 times as many employees. She realizes that her computer is not up to the job of sorting 100 times as many records in an acceptable amount of time. Describe what impact each of the following modifications to the computing system is likely to have in terms of reducing the time required to process the larger payroll database. (a) A factor of two speedup to the CPU. (b) A factor of two speedup to disk I/O time. (c) A factor of two speedup to main memory access time. (d) A factor of two increase to main memory size. 8.22 How can the external sorting algorithm described in this chapter be extended to handle variable-length records? 8.8 Projects 8.1 For a database application, assume it takes 10 ms to read a block from disk, 1 ms to search for a record in a block stored in memory, and that there is room in memory for a buffer pool of 5 blocks. Requests come in for records, with the request specifying which block contains the record. If a block is accessed, there is a 10% probability for each of the next ten requests that the request will be to the same block. What will be the expected performance improvement for each of the following modifications to the system? (a) Get a CPU that is twice as fast. (b) Get a disk drive that is twice as fast. (c) Get enough memory to double the buffer pool size. Write a simulation to analyze this problem. 8.2 Pictures are typically stored as an array, row by row, on disk. Consider the case where the picture has 16 colors. Thus, each pixel can be represented us- ing 4 bits. If you allow 8 bits per pixel, no processing is required to unpack the pixels (because a pixel corresponds to a byte, the lowest level of address- ing on most machines). If you pack two pixels per byte, space is saved but the pixels must be unpacked. Which takes more time to read from disk and access every pixel of the image: 8 bits per pixel, or 4 bits per pixel with 2 pixels per byte? Program both and compare the times.
- 335. 316 Chap. 8 File Processing and External Sorting 8.3 Implement a disk-based buffer pool class based on the LRU buffer pool re- placement strategy. Disk blocks are numbered consecutively from the begin- ning of the file with the first block numbered as 0. Assume that blocks are 4096 bytes in size, with the first 4 bytes used to store the block ID corre- sponding to that buffer. Use the first BufferPool abstract class given in Section 8.3 as the basis for your implementation. 8.4 Implement an external sort based on replacement selection and multiway merging as described in this chapter. Test your program both on files with small records and on files with large records. For what size record do you find that key sorting would be worthwhile? 8.5 Implement a quicksort for large files on disk by replacing all array access in the normal quicksort application with access to a virtual array implemented using a buffer pool. That is, whenever a record in the array would be read or written by quicksort, use a call to a buffer pool function instead. Compare the running time of this implementation with implementations for external sorting based on mergesort as described in this chapter. 8.6 Section 8.5.1 suggests that an easy modification to the basic 2-way mergesort is to read in a large chunk of data into main memory, sort it with quicksort, and write it out for initial runs. Then, a standard 2-way merge is used in a series of passes to merge the runs together. However, this makes use of only two blocks of working memory at a time. Each block read is essentially random access, because the various files are read in an unknown order, even though each of the input and output files is processed sequentially on each pass. A possible improvement would be, on the merge passes, to divide working memory into four equal sections. One section is allocated to each of the two input files and two output files. All reads during merge passes would be in full sections, rather than single blocks. While the total number of blocks read and written would be the same as a regular 2-way mergsort, it is possible that this would speed processing because a series of blocks that are logically adjacent in the various input and output files would be read/written each time. Implement this variation, and compare its running time against a standard series of 2-way merge passes that read/write only a single block at a time. Before beginning implementation, write down your hypothesis on how the running time will be affected by this change. After implementing, did you find that this change has any meaningful effect on performance?
- 336. 9 Searching Organizing and retrieving information is at the heart of most computer applica- tions, and searching is surely the most frequently performed of all computing tasks. Search can be viewed abstractly as a process to determine if an element with a par- ticular value is a member of a particular set. The more common view of searching is an attempt to find the record within a collection of records that has a particular key value, or those records in a collection whose key values meet some criterion such as falling within a range of values. We can define searching formally as follows. Suppose k1, k2, ... kn are distinct keys, and that we have a collection L of n records of the form (k1, I1), (k2, I2), ..., (kn, In) where Ij is information associated with key kj for 1 ≤ j ≤ n. Given a particular key value K, the search problem is to locate the record (kj, Ij) in L such that kj = K (if one exists). Searching is a systematic method for locating the record (or records) with key value kj = K. A successful search is one in which a record with key kj = K is found. An unsuccessful search is one in which no record with kj = K is found (and no such record exists). An exact-match query is a search for the record whose key value matches a specified key value. A range query is a search for all records whose key value falls within a specified range of key values. We can categorize search algorithms into three general approaches: 1. Sequential and list methods. 2. Direct access by key value (hashing). 3. Tree indexing methods. 317
- 337. 318 Chap. 9 Searching This and the following chapter treat these three approaches in turn. Any of these approaches are potentially suitable for implementing the Dictionary ADT introduced in Section 4.4. However, each has different performance characteristics that make it the method of choice in particular circumstances. The current chapter considers methods for searching data stored in lists and tables. A table is simply another term for an array. List in this context means any list implementation including a linked list or an array. Most of these methods are appropriate for sequences (i.e., duplicate key values are allowed), although special techniques applicable to sets are discussed in Section 9.3. The techniques from the first three sections of this chapter are most appropriate for searching a collection of records stored in RAM. Section 9.4 discusses hashing, a technique for organizing data in a table such that the location of each record within the table is a function of its key value. Hashing is appropriate when records are stored either in RAM or on disk. Chapter 10 discusses tree-based methods for organizing information on disk, including a commonly used file structure called the B-tree. Nearly all programs that must organize large collections of records stored on disk use some variant of either hashing or the B-tree. Hashing is practical for only certain access functions (exact- match queries) and is generally appropriate only when duplicate key values are not allowed. B-trees are the method of choice for disk-based applications anytime hashing is not appropriate. 9.1 Searching Unsorted and Sorted Arrays The simplest form of search has already been presented in Example 3.1: the se- quential search algorithm. Sequential search on an unsorted list requires Θ(n) time in the worst case. How many comparisons does linear search do on average? A major consid- eration is whether K is in list L at all. We can simplify our analysis by ignoring everything about the input except the position of K if it is found in L. Thus, we have n + 1 distinct possible events: That K is in one of positions 0 to n − 1 in L (each with its own probability), or that it is not in L at all. We can express the probability that K is not in L as P(K / ∈ L) = 1 − n X i=1 P(K = L[i]) where P(x) is the probability of event x.
- 338. Sec. 9.1 Searching Unsorted and Sorted Arrays 319 Let pi be the probability that K is in position i of L. When K is not in L, sequential search will require n comparisons. Let p0 be the probability that K is not in L. Then the average cost T(n) will be T(n) = np0 + n X i=1 ipi. What happens to the equation if we assume all the pi’s are equal (except p0)? T(n) = p0n + n X i=1 ip = p0n + p n X i=1 i = p0n + p n(n + 1) 2 = p0n + 1 − p0 n n(n + 1) 2 = n + 1 + p0(n − 1) 2 Depending on the value of p0, n+1 2 ≤ T(n) ≤ n. For large collections of records that are searched repeatedly, sequential search is unacceptably slow. One way to reduce search time is to preprocess the records by sorting them. Given a sorted array, an obvious improvement over simple linear search is to test if the current element in L is greater than K. If it is, then we know that K cannot appear later in the array, and we can quit the search early. But this still does not improve the worst-case cost of the algorithm. We can also observe that if we look first at position 1 in sorted array L and find that K is bigger, then we rule out positions 0 as well as position 1. Because more is often better, what if we look at position 2 in L and find that K is bigger yet? This rules out positions 0, 1, and 2 with one comparison. What if we carry this to the extreme and look first at the last position in L and find that K is bigger? Then we know in one comparison that K is not in L. This is very useful to know, but what is wrong with this approach? While we learn a lot sometimes (in one comparison we might learn that K is not in the list), usually we learn only a little bit (that the last element is not K). The question then becomes: What is the right amount to jump? This leads us to an algorithm known as Jump Search. For some value j, we check every j’th
- 339. 320 Chap. 9 Searching element in L, that is, we check elements L[j], L[2j], and so on. So long as K is greater than the values we are checking, we continue on. But when we reach a value in L greater than K, we do a linear search on the piece of length j − 1 that we know brackets K if it is in the list. If mj ≤ n < (m+1)j, then the total cost of this algorithm is at most m+j −1 3-way comparisons. Therefore, the cost to run the algorithm on n items with a jump of size j is T(n, j) = m + j − 1 = n j + j − 1. What is the best value that we can pick for j? We want to minimize the cost: min 1≤j≤n n j + j − 1 Take the derivative and solve for f′(j) = 0 to find the minimum, which is j = √ n. In this case, the worst case cost will be roughly 2 √ n. This example teaches us some lessons about algorithm design. We want to balance the work done while selecting a sublist with the work done while searching a sublist. In general, it is a good strategy to make subproblems of equal effort. This is an example of a divide and conquer algorithm. What if we extend this idea to three levels? We would first make jumps of some size j to find a sublist of size j − 1 whose end vlaues bracket value K. Ww would then work through this sublist by making jumps of some smaller size, say j1. Finally, once we find a bracketed sublist of size j1 − 1, we would do sequential search to complete the process. This probably sounds convoluted to do two levels of jumping to be followed by a sequential search. While it might make sense to do a two-level algorithm (that is, jump search jumps to find a sublist and then does sequential search on the sublist), it almost never seems to make sense to do a three-level algorithm. Instead, when we go beyond two levels, we nearly always generalize by using recursion. This leads us to the most commonly used search algorithm for sorted arrays, the binary search described in Section 3.5. If we know nothing about the distribution of key values, then binary search is the best algorithm available for searching a sorted array (see Exercise 9.22). However, sometimes we do know something about the expected key distribution. Consider the typical behavior of a person looking up a word in a large dictionary. Most people certainly do not use sequential search! Typically, people use a mod- ified form of binary search, at least until they get close to the word that they are
- 340. Sec. 9.1 Searching Unsorted and Sorted Arrays 321 looking for. The search generally does not start at the middle of the dictionary. A person looking for a word starting with ‘S’ generally assumes that entries beginning with ‘S’ start about three quarters of the way through the dictionary. Thus, he or she will first open the dictionary about three quarters of the way through and then make a decision based on what they find as to where to look next. In other words, people typically use some knowledge about the expected distribution of key values to “compute” where to look next. This form of “computed” binary search is called a dictionary search or interpolation search. In a dictionary search, we search L at a position p that is appropriate to the value of K as follows. p = K − L[1] L[n] − L[1] The location of a particular key within the key range is translated into the expected position for the corresponding record in the table, and this position is checked first. As with binary search, the value of the key found eliminates all records either above or below that position. The actual value of the key found can then be used to compute a new position within the remaining range of the table. The next check is made based on the new computation. This proceeds until either the desired record is found, or the table is narrowed until no records are left. A variation on dictionary search is known as Quadratic Binary Search (QBS), and we will analyze this in detail because its analysis is easier than that of the general dictionary search. QBS will first compute p and t hen examine L[⌈pn⌉]. If K L[⌈pn⌉] then QBS will sequentially probe by steps of size √ n, that is, we step through L[⌈pn − i √ n⌉], i = 1, 2, 3, ... until we reach a value less than or equal to K. Similarly for K L[⌈pn⌉] we will step forward by √ n until we reach a value in L that is greater than K. We are now within √ n positions of K. Assume (for now) that it takes a constant number of comparisons to bracket K within a sublist of size √ n. We then take this sublist and repeat the process recursively. What is the cost for QBS? Note that √ cn = cn/2, and we will be repeatedly taking square roots of the current sublist size until we find the item we are looking for. Because n = 2log n and we can cut log n in half only log log n times, the cost is Θ(log log n) if the number of probes on jump search is constant. The number of comparisons needed is √ n X i=1 iP(need exactly i probes)
- 341. 322 Chap. 9 Searching = 1P1 + 2P2 + 3P3 + · · · + √ nP√ n This is equal to: √ n X i=1 P(need at least i probes) = 1 + (1 − P1) + (1 − P1 − P2) + · · · + P√ n = (P1 + ... + P√ n) + (P2 + ... + P√ n) + (P3 + ... + P√ n) + · · · = 1P1 + 2P2 + 3P3 + · · · + √ nP√ n We require at least two probes to set the bounds, so the cost is 2 + √ n X i=3 P(need at least i probes). We now make take advantage of a useful fact known as Čebyšev’s Inequality. Čebyšev’s inequality states that P(need i probes), or Pi, is Pi ≤ p(1 − p)n (i − 2)2n ≤ 1 4(i − 2)2 because p(1 − p) ≤ 1/4. This assumes uniformly distributed data. Thus, the expected number of probes is 2 + √ n X i=3 1 4(i − 2)2 2 + 1 4 ∞ X i=1 1 i2 = 2 + 1 4 π 6 ≈ 2.4112 Is this better than binary search? Theoretically yes, because a cost of O(log log n) grows slower than a cost of O(log n). However, we have a situation here which illustrates the limits to the model of asymptotic complexity in some practical sit- uations. Yes, c1 log n does grow faster than c2 log log n. In fact, its exponentially faster! But even so, for practical input sizes, the absolute cost difference is fairly small. Thus, the constant factors might play a role. Let’s compare lg lg n to lg n. n lg n lg lg n Difference 16 4 2 2 256 8 3 2.7 216 16 4 4 232 32 5 6.4
- 342. Sec. 9.1 Searching Unsorted and Sorted Arrays 323 It is not always practical to reduce an algorithm’s growth rate. There is a prac- ticality window for every problem, in that we have a practical limit to how big an input we wish to solve for. If our problem size never grows too big, it might not matter if we can reduce the cost by an extra log factor, because the constant factors in the two algorithms might overwhelm the savings. For our two algorithms, let us look further and check the actual number of comparisons used. For binary search, we need about log n − 1 total comparisons. Quadratic binary search requires about 2.4 lg lg n comparisons. If we incorporate this observation into our table, we get a different picture about the relative differ- ences. n lg n − 1 2.4 lg lg n Difference 16 3 4.8 worse 256 7 7.2 ≈ same 64K 15 9.6 1.6 232 31 12 2.6 But we still are not done. This is only a count of comparisons! Binary search is inherently much simpler than QBS, because binary search only needs to calcu- late the midpoint position of the array before each comparison, while quadratic binary search must calculate an interpolation point which is more expensive. So the constant factors for QBS are even higher. Not only are the constant factors worse on average, but QBS is far more depen- dent than binary search on good data distribution to perform well. For example, imagine that you are searching a telephone directory for the name “Young.” Nor- mally you would look near the back of the book. If you found a name beginning with ‘Z,’ you might look just a little ways toward the front. If the next name you find also begins with ’Z,‘ you would look a little further toward the front. If this particular telephone directory were unusual in that half of the entries begin with ‘Z,’ then you would need to move toward the front many times, each time eliminating relatively few records from the search. In the extreme, the performance of interpo- lation search might not be much better than sequential search if the distribution of key values is badly calculated. While it turns out that QBS is not a practical algorithm, this is not a typical situation. Fortunately, algorithm growth rates are usually well behaved, so that as- ymptotic algorithm analysis nearly always gives us a practical indication for which of two algorithms is better.
- 343. 324 Chap. 9 Searching 9.2 Self-Organizing Lists While lists are most commonly ordered by key value, this is not the only viable option. Another approach to organizing lists to speed search is to order the records by expected frequency of access. While the benefits might not be as great as when oganized by key value, the cost to organize by frequency of access is much cheaper, and thus is of use in some situations. Assume that we know, for each key ki, the probability pi that the record with key ki will be requested. Assume also that the list is ordered so that the most frequently requested record is first, then the next most frequently requested record, and so on. Search in the list will be done sequentially, beginning with the first position. Over the course of many searches, the expected number of comparisons required for one search is Cn = 1p1 + 2p2 + ... + npn. In other words, the cost to access the first record is one (because one key value is looked at), and the probability of this occurring is p1. The cost to access the second record is two (because we must look at the first and the second records’ key values), with probability p2, and so on. For n records, assuming that all searches are for records that actually exist, the probabilities p1 through pn must sum to one. Certain probability distributions give easily computed results. Example 9.1 Calculate the expected cost to search a list when each record has equal chance of being accessed (the classic sequential search through an unsorted list). Setting pi = 1/n yields Cn = n X i=1 i/n = (n + 1)/2. This result matches our expectation that half the records will be accessed on average by normal sequential search. If the records truly have equal access probabilities, then ordering records by frequency yields no benefit. We saw in Section 9.1 the more general case where we must conside the probability (labeled p0) that the search key does not match that for any record in the array. In that case, in accordance with our general formula, we get n + 1 + p0(n − 1) 2 . Thus, n+1 2 ≤ Cn ≤ n, depending on the value of p0.
- 344. Sec. 9.2 Self-Organizing Lists 325 A geometric probability distribution can yield quite different results. Example 9.2 Calculate the expected cost for searching a list ordered by frequency when the probabilities are defined as pi = 1/2i if 1 ≤ i ≤ n − 1 1/2n−1 if i = n. Then, Cn ≈ n X i=1 (i/2i ) ≈ 2. For this example, the expected number of accesses is a constant. This is because the probability for accessing the first record is high, the second is much lower but still much higher than for record three, and so on. This shows that for some probability distributions, ordering the list by frequency can yield an efficient search technique. In many search applications, real access patterns follow a rule of thumb called the 80/20 rule. The 80/20 rule says that 80% of the record accesses are to 20% of the records. The values of 80 and 20 are only estimates; every application has its own values. However, behavior of this nature occurs surprisingly often in practice (which explains the success of caching techniques widely used by disk drive and CPU manufacturers for speeding access to data stored in slower memory; see the discussion on buffer pools in Section 8.3). When the 80/20 rule applies, we can expect reasonable search performance from a list ordered by frequency of access. Example 9.3 The 80/20 rule is an example of a Zipf distribution. Nat- urally occurring distributions often follow a Zipf distribution. Examples include the observed frequency for the use of words in a natural language such as English, and the size of the population for cities (i.e., view the relative proportions for the populations as equivalent to the “frequency of use”). Zipf distributions are related to the Harmonic Series defined in Equa- tion 2.10. Define the Zipf frequency for item i in the distribution for n records as 1/(iHn) (see Exercise 9.4). The expected cost for the series whose members follow this Zipf distribution will be Cn = n X i=1 i/iHn = n/Hn ≈ n/ loge n.
- 345. 326 Chap. 9 Searching When a frequency distribution follows the 80/20 rule, the average search looks at about one tenth of the records in a table ordered by frequency. In most applications, we have no means of knowing in advance the frequencies of access for the data records. To complicate matters further, certain records might be accessed frequently for a brief period of time, and then rarely thereafter. Thus, the probability of access for records might change over time (in most database systems, this is to be expected). Self-organizing lists seek to solve both of these problems. Self-organizing lists modify the order of records within the list based on the actual pattern of record access. Self-organizing lists use a heuristic for deciding how to to reorder the list. These heuristics are similar to the rules for managing buffer pools (see Section 8.3). In fact, a buffer pool is a form of self-organizing list. Ordering the buffer pool by expected frequency of access is a good strategy, because typically we must search the contents of the buffers to determine if the desired information is already in main memory. When ordered by frequency of access, the buffer at the end of the list will be the one most appropriate for reuse when a new page of information must be read. Below are three traditional heuristics for managing self-organizing lists: 1. The most obvious way to keep a list ordered by frequency would be to store a count of accesses to each record and always maintain records in this or- der. This method will be referred to as count. Count is similar to the least frequently used buffer replacement strategy. Whenever a record is accessed, it might move toward the front of the list if its number of accesses becomes greater than a record preceding it. Thus, count will store the records in the order of frequency that has actually occurred so far. Besides requiring space for the access counts, count does not react well to changing frequency of access over time. Once a record has been accessed a large number of times under the frequency count system, it will remain near the front of the list regardless of further access history. 2. Bring a record to the front of the list when it is found, pushing all the other records back one position. This is analogous to the least recently used buffer replacement strategy and is called move-to-front. This heuristic is easy to implement if the records are stored using a linked list. When records are stored in an array, bringing a record forward from near the end of the array will result in a large number of records changing position. Move-to-front’s cost is bounded in the sense that it requires at most twice the number of ac- cesses required by the optimal static ordering for n records when at least
- 346. Sec. 9.2 Self-Organizing Lists 327 n searches are performed. In other words, if we had known the series of (at least n) searches in advance and had stored the records in order of frequency so as to minimize the total cost for these accesses, this cost would be at least half the cost required by the move-to-front heuristic. (This will be proved using amortized analysis in Section 14.3.) Finally, move-to-front responds well to local changes in frequency of access, in that if a record is frequently accessed for a brief period of time it will be near the front of the list during that period of access. Move-to-front does poorly when the records are pro- cessed in sequential order, especially if that sequential order is then repeated multiple times. 3. Swap any record found with the record immediately preceding it in the list. This heuristic is called transpose. Transpose is good for list implementations based on either linked lists or arrays. Frequently used records will, over time, move to the front of the list. Records that were once frequently accessed but are no longer used will slowly drift toward the back. Thus, it appears to have good properties with respect to changing frequency of access. Unfortunately, there are some pathological sequences of access that can make transpose perform poorly. Consider the case where the last record of the list (call it X) is accessed. This record is then swapped with the next-to-last record (call it Y), making Y the last record. If Y is now accessed, it swaps with X. A repeated series of accesses alternating between X and Y will continually search to the end of the list, because neither record will ever make progress toward the front. However, such pathological cases are unusual in practice. Example 9.4 Assume that we have eight records, with key values A to H, and that they are initially placed in alphabetical order. Now, consider the result of applying the following access pattern: F D F G E G F A D F G E. If the list is organized by the count heuristic, the final list resulting from these accesses will be F G D E A B C H, and the total cost for the twelve accesses will be 45 comparisons. (Assume that when a record’s frequency count goes up, it moves forward in the list to become the last record with that value for its frequency count. After the first two accesses, F will be the first record and D will be the second.) If the list is organized by the move-to-front heuristic, then the final list will be E G F D A B C H,
- 347. 328 Chap. 9 Searching and the total number of comparisons required is 54. Finally, if the list is organized by the transpose heuristic, then the final list will be A B F D G E C H, and the total number of comparisons required is 62. While self-organizing lists do not generally perform as well as search trees or a sorted list, both of which require O(log n) search time, there are many situations in which self-organizing lists prove a valuable tool. Obviously they have an advantage over sorted lists in that they need not be sorted. This means that the cost to insert a new record is low, which could more than make up for the higher search cost when insertions are frequent. Self-organizing lists are simpler to implement than search trees and are likely to be more efficient for small lists. Nor do they require additional space. Finally, in the case of an application where sequential search is “almost” fast enough, changing an unsorted list to a self-organizing list might speed the application enough at a minor cost in additional code. As an example of applying self-organizing lists, consider an algorithm for com- pressing and transmitting messages. The list is self-organized by the move-to-front rule. Transmission is in the form of words and numbers, by the following rules: 1. If the word has been seen before, transmit the current position of the word in the list. Move the word to the front of the list. 2. If the word is seen for the first time, transmit the word. Place the word at the front of the list. Both the sender and the receiver keep track of the position of words in the list in the same way (using the move-to-front rule), so they agree on the meaning of the numbers that encode repeated occurrences of words. For example, consider the following example message to be transmitted (for simplicity, ignore case in letters). The car on the left hit the car I left. The first three words have not been seen before, so they must be sent as full words. The fourth word is the second appearance of “the,” which at this point is the third word in the list. Thus, we only need to transmit the position value “3.” The next two words have not yet been seen, so must be sent as full words. The seventh word is the third appearance of “the,” which coincidentally is again in the third position. The eighth word is the second appearance of “car,” which is now in the fifth position of the list. “I” is a new word, and the last word “left” is now in the fifth position. Thus the entire transmission would be
- 348. Sec. 9.3 Bit Vectors for Representing Sets 329 0 1 2 3 4 5 6 7 8 9 10 11 12 15 0 0 0 0 0 1 0 1 0 0 1 1 1 0 1 13 14 0 Figure 9.1 The bit table for the set of primes in the range 0 to 15. The bit at position i is set to 1 if and only if i is prime. The car on 3 left hit 3 5 I 5. This approach to compression is similar in spirit to Ziv-Lempel coding, which is a class of coding algorithms commonly used in file compression utilities. Ziv- Lempel coding will replace repeated occurrences of strings with a pointer to the location in the file of the first occurrence of the string. The codes are stored in a self-organizing list in order to speed up the time required to search for a string that has previously been seen. 9.3 Bit Vectors for Representing Sets Determining whether a value is a member of a particular set is a special case of searching for keys in a sequence of records. Thus, any of the search methods discussed in this book can be used to check for set membership. However, we can also take advantage of the restricted circumstances imposed by this problem to develop another representation. In the case where the set elements fall within a limited key range, we can repre- sent the set using a bit array with a bit position allocated for each potential member. Those members actually in the set store a value of 1 in their corresponding bit; those members not in the set store a value of 0 in their corresponding bit. For example, consider the set of primes between 0 and 15. Figure 9.1 shows the corresponding bit table. To determine if a particular value is prime, we simply check the corre- sponding bit. This representation scheme is called a bit vector or a bitmap. The mark array used in several of the graph algorithms of Chapter 11 is an example of such a set representation. If the set fits within a single computer word, then set union, intersection, and difference can be performed by logical bitwise operations. The union of sets A and B is the bitwise OR function (whose symbol is | in Java). The intersection of sets A and B is the bitwise AND function (whose symbol is in Java). For example, if we would like to compute the set of numbers between 0 and 15 that are both prime and odd numbers, we need only compute the expression 0011010100010100 0101010101010101.
- 349. 330 Chap. 9 Searching The set difference A − B can be implemented in Java using the expression A˜B (˜ is the symbol for bitwise negation). For larger sets that do not fit into a single computer word, the equivalent operations can be performed in turn on the series of words making up the entire bit vector. This method of computing sets from bit vectors is sometimes applied to doc- ument retrieval. Consider the problem of picking from a collection of documents those few which contain selected keywords. For each keyword, the document re- trieval system stores a bit vector with one bit for each document. If the user wants to know which documents contain a certain three keywords, the corresponding three bit vectors are AND’ed together. Those bit positions resulting in a value of 1 cor- respond to the desired documents. Alternatively, a bit vector can be stored for each document to indicate those keywords appearing in the document. Such an organiza- tion is called a signature file. The signatures can be manipulated to find documents with desired combinations of keywords. 9.4 Hashing This section presents a completely different approach to searching tables: by direct access based on key value. The process of finding a record using some computa- tion to map its key value to a position in the table is called hashing. Most hash- ing schemes place records in the table in whatever order satisfies the needs of the address calculation, thus the records are not ordered by value or frequency. The function that maps key values to positions is called a hash function and is usually denoted by h. The array that holds the records is called the hash table and will be denoted by HT. A position in the hash table is also known as a slot. The number of slots in hash table HT will be denoted by the variable M, with slots numbered from 0 to M − 1. The goal for a hashing system is to arrange things such that, for any key value K and some hash function h, i = h(K) is a slot in the table such that 0 ≤ h(K) M, and we have the key of the record stored at HT[i] equal to K. Hashing only works to store sets. That is, hashing cannnot be used for applica- tions where multiple records with the same key value are permitted. Hashing is not a good method for answering range searches. In other words, we cannot easily find all records (if any) whose key values fall within a certain range. Nor can we easily find the record with the minimum or maximum key value, or visit the records in key order. Hashing is most appropriate for answering the question, “What record, if any, has key value K?” For applications where access involves only exact-match queries, hashing is usually the search method of choice because it is extremely ef- ficient when implemented correctly. As you will see in this section, however, there
- 350. Sec. 9.4 Hashing 331 are many approaches to hashing and it is easy to devise an inefficient implementa- tion. Hashing is suitable for both in-memory and disk-based searching and is one of the two most widely used methods for organizing large databases stored on disk (the other is the B-tree, which is covered in Chapter 10). As a simple (though unrealistic) example of hashing, consider storing n records, each with a unique key value in the range 0 to n − 1. In this simple case, a record with key k can be stored in HT[k], and the hash function is simply h(k) = k. To find the record with key value k, simply look in HT[k]. Typically, there are many more values in the key range than there are slots in the hash table. For a more realistic example, suppose that the key can take any value in the range 0 to 65,535 (i.e., the key is a two-byte unsigned integer), and that we expect to store approximately 1000 records at any given time. It is impractical in this situation to use a hash table with 65,536 slots, because most of the slots will be left empty. Instead, we must devise a hash function that allows us to store the records in a much smaller table. Because the possible key range is larger than the size of the table, at least some of the slots must be mapped to from multiple key values. Given a hash function h and two keys k1 and k2, if h(k1) = β = h(k2) where β is a slot in the table, then we say that k1 and k2 have a collision at slot β under hash function h. Finding a record with key value K in a database organized by hashing follows a two-step procedure: 1. Compute the table location h(K). 2. Starting with slot h(K), locate the record containing key K using (if neces- sary) a collision resolution policy. 9.4.1 Hash Functions Hashing generally takes records whose key values come from a large range and stores those records in a table with a relatively small number of slots. Collisions occur when two records hash to the same slot in the table. If we are careful—or lucky—when selecting a hash function, then the actual number of collisions will be few. Unfortunately, even under the best of circumstances, collisions are nearly unavoidable.1 For example, consider a classroom full of students. What is the 1 The exception to this is perfect hashing. Perfect hashing is a system in which records are hashed such that there are no collisions. A hash function is selected for the specific set of records being hashed, which requires that the entire collection of records be available before selecting the hash function. Perfect hashing is efficient because it always finds the record that we are looking for exactly where the hash function computes it to be, so only one access is required. Selecting a perfect hash function can be expensive but might be worthwhile when extremely efficient search
- 351. 332 Chap. 9 Searching probability that some pair of students shares the same birthday (i.e., the same day of the year, not necessarily the same year)? If there are 23 students, then the odds are about even that two will share a birthday. This is despite the fact that there are 365 days in which students can have birthdays (ignoring leap years), on most of which no student in the class has a birthday. With more students, the probability of a shared birthday increases. The mapping of students to days based on their birthday is similar to assigning records to slots in a table (of size 365) using the birthday as a hash function. Note that this observation tells us nothing about which students share a birthday, or on which days of the year shared birthdays fall. To be practical, a database organized by hashing must store records in a hash table that is not so large that it wastes space. Typically, this means that the hash table will be around half full. Because collisions are extremely likely to occur under these conditions (by chance, any record inserted into a table that is half full will have a collision half of the time), ddoes this mean that we need not worry about the ability of a hash function to avoid collisions? Absolutely not. The difference between a good hash function and a bad hash function makes a big difference in practice. Technically, any function that maps all possible key values to a slot in the hash table is a hash function. In the extreme case, even a function that maps all records to the same slot is a hash function, but it does nothing to help us find records during a search operation. We would like to pick a hash function that stores the actual records in the collec- tion such that each slot in the hash table has equal probablility of being filled. Un- fortunately, we normally have no control over the key values of the actual records, so how well any particular hash function does this depends on the distribution of the keys within the allowable key range. In some cases, incoming data are well distributed across their key range. For example, if the input is a set of random numbers selected uniformly from the key range, any hash function that assigns the key range so that each slot in the hash table receives an equal share of the range will likely also distribute the input records uniformly within the table. However, in many applications the incoming records are highly clustered or otherwise poorly distributed. When input records are not well distributed throughout the key range it can be difficult to devise a hash function that does a good job of distributing the records throughout the table, especially if the input distribution is not known in advance. There are many reasons why data values might be poorly distributed. performance is required. An example is searching for data on a read-only CD. Here the database will never change, the time for each access is expensive, and the database designer can build the hash table before issuing the CD.
- 352. Sec. 9.4 Hashing 333 1. Natural distributions are geometric. For example, consider the populations of the 100 largest cities in the United States. If you plot these populations on a numberline, most of them will be clustered toward the low side, with a few outliers on the high side. This is an example of a Zipf distribution (see Section 9.2). Viewed the other way, the home town for a given person is far more likely to be a particular large city than a particular small town. 2. Collected data are likely to be skewed in some way. Field samples might be rounded to, say, the nearest 5 (i.e., all numbers end in 5 or 0). 3. If the input is a collection of common English words, the beginning letter will be poorly distributed. Note that in each of these examples, either high- or low-order bits of the key are poorly distributed. When designing hash functions, we are generally faced with one of two situa- tions. 1. We know nothing about the distribution of the incoming keys. In this case, we wish to select a hash function that evenly distributes the key range across the hash table, while avoiding obvious opportunities for clustering such as hash functions that are sensitive to the high- or low-order bits of the key value. 2. We know something about the distribution of the incoming keys. In this case, we should use a distribution-dependent hash function that avoids assigning clusters of related key values to the same hash table slot. For example, if hashing English words, we should not hash on the value of the first character because this is likely to be unevenly distributed. Below are several examples of hash functions that illustrate these points. Example 9.5 Consider the following hash function used to hash integers to a table of sixteen slots: int h(int x) { return(x % 16); } The value returned by this hash function depends solely on the least significant four bits of the key. Because these bits are likely to be poorly distributed (as an example, a high percentage of the keys might be even numbers, which means that the low order bit is zero), the result will also be poorly distributed. This example shows that the size of the table M can have a big effect on the performance of a hash system because this value is typically used as the modulus.
- 353. 334 Chap. 9 Searching Example 9.6 A good hash function for numerical values is the mid-square method. The mid-square method squares the key value, and then takes the middle r bits of the result, giving a value in the range 0 to 2r − 1. This works well because most or all bits of the key value contribute to the result. For example, consider records whose keys are 4-digit numbers in base 10. The goal is to hash these key values to a table of size 100 (i.e., a range of 0 to 99). This range is equivalent to two digits in base 10. That is, r = 2 If the input is the number 4567, squaring yields an 8-digit number, 20857489. The middle two digits of this result are 57. All digits (equivalently, all bits when the number is viewed in binary) contribute to the middle two digits of the squared value. Thus, the result is not dominated by the distribution of the bottom digit or the top digit of the original key value. Example 9.7 Here is a hash function for strings of characters: int h(String x, int M) { char ch[]; ch = x.toCharArray(); int xlength = x.length(); int i, sum; for (sum=0, i=0; ix.length(); i++) sum += ch[i]; return sum % M; } This function sums the ASCII values of the letters in a string. If the hash table size M is small, this hash function should do a good job of distributing strings evenly among the hash table slots, because it gives equal weight to all characters. This is an example of the folding approach to designing a hash function. Note that the order of the characters in the string has no effect on the result. A similar method for integers would add the digits of the key value, assuming that there are enough digits to (1) keep any one or two digits with bad distribution from skewing the results of the process and (2) generate a sum much larger than M. As with many other hash functions, the final step is to apply the modulus operator to the result, using table size M to generate a value within the table range. If the sum is not sufficiently large, then the modulus operator will yield a poor distribution.
- 354. Sec. 9.4 Hashing 335 For example, because the ASCII value for “A” is 65 and “Z” is 90, sum will always be in the range 650 to 900 for a string of ten upper case letters. For a hash table of size 100 or less, a good distribution results with all slots in the table accepting either two or three of the values in the key range. For a hash table of size 1000, the distribution is terrible because only slots 650 to 900 can possibly be the home slot for some key value. Example 9.8 Here is a much better hash function for strings. long sfold(String s, int M) { int intLength = s.length() / 4; long sum = 0; for (int j = 0; j intLength; j++) { char c[] = s.substring(j * 4, (j * 4) + 4).toCharArray(); long mult = 1; for (int k = 0; k c.length; k++) { sum += c[k] * mult; mult *= 256; } } char c[] = s.substring(intLength * 4).toCharArray(); long mult = 1; for (int k = 0; k c.length; k++) { sum += c[k] * mult; mult *= 256; } return(Math.abs(sum) % M); } This function takes a string as input. It processes the string four bytes at a time, and interprets each of the four-byte chunks as a single (unsigned) long integer value. The integer values for the four-byte chunks are added together. In the end, the resulting sum is converted to the range 0 to M − 1 using the modulus operator.2 For example, if the string “aaaabbbb” is passed to sfold, then the first four bytes (“aaaa”) will be interpreted as the integer value 1,633,771,873 2 Recall from Section 2.2 that the implementation for n mod m on many C++ and Java compilers will yield a negative number if n is negative. Implementors for hash functions need to be careful that their hash function does not generate a negative number. This can be avoided either by insuring that n is positive when computing n mod m, or adding m to the result if n mod m is negative. All computation in sfold is done using unsigned long values in part to protect against taking the modulus of an negative number.
- 355. 336 Chap. 9 Searching and the next four bytes (“bbbb”) will be interpreted as the integer value 1,650,614,882. Their sum is 3,284,386,755 (when treated as an unsigned integer). If the table size is 101 then the modulus function will cause this key to hash to slot 75 in the table. Note that for any sufficiently long string, the sum for the integer quantities will typically cause a 32-bit integer to overflow (thus losing some of the high-order bits) because the resulting values are so large. But this causes no problems when the goal is to compute a hash function. 9.4.2 Open Hashing While the goal of a hash function is to minimize collisions, collisions are normally unavoidable in practice. Thus, hashing implementations must include some form of collision resolution policy. Collision resolution techniques can be broken into two classes: open hashing (also called separate chaining) and closed hashing (also called open addressing).3 The difference between the two has to do with whether collisions are stored outside the table (open hashing), or whether collisions result in storing one of the records at another slot in the table (closed hashing). Open hashing is treated in this section, and closed hashing in Section 9.4.3. The simplest form of open hashing defines each slot in the hash table to be the head of a linked list. All records that hash to a particular slot are placed on that slot’s linked list. Figure 9.2 illustrates a hash table where each slot stores one record and a link pointer to the rest of the list. Records within a slot’s list can be ordered in several ways: by insertion order, by key value order, or by frequency-of-access order. Ordering the list by key value provides an advantage in the case of an unsuccessful search, because we know to stop searching the list once we encounter a key that is greater than the one being searched for. If records on the list are unordered or ordered by frequency, then an unsuccessful search will need to visit every record on the list. Given a table of size M storing N records, the hash function will (ideally) spread the records evenly among the M positions in the table, yielding on average N/M records for each list. Assuming that the table has more slots than there are records to be stored, we can hope that few slots will contain more than one record. In the case where a list is empty or has only one record, a search requires only one access to the list. Thus, the average cost for hashing should be Θ(1). However, if clustering causes many records to hash to only a few of the slots, then the cost to 3 Yes, it is confusing when “open hashing” means the opposite of “open addressing,” but unfortu- nately, that is the way it is.
- 356. Sec. 9.4 Hashing 337 0 1 2 3 4 5 6 7 8 9 9530 1057 2007 1000 3013 9879 9877 Figure 9.2 An illustration of open hashing for seven numbers stored in a ten-slot hash table using the hash function h(K) = K mod 10. The numbers are inserted in the order 9877, 2007, 1000, 9530, 3013, 9879, and 1057. Two of the values hash to slot 0, one value hashes to slot 2, three of the values hash to slot 7, and one value hashes to slot 9. access a record will be much higher because many elements on the linked list must be searched. Open hashing is most appropriate when the hash table is kept in main memory, with the lists implemented by a standard in-memory linked list. Storing an open hash table on disk in an efficient way is difficult, because members of a given linked list might be stored on different disk blocks. This would result in multiple disk accesses when searching for a particular key value, which defeats the purpose of using hashing. There are similarities between open hashing and Binsort. One way to view open hashing is that each record is simply placed in a bin. While multiple records may hash to the same bin, this initial binning should still greatly reduce the number of records accessed by a search operation. In a similar fashion, a simple Binsort reduces the number of records in each bin to a small number that can be sorted in some other way. 9.4.3 Closed Hashing Closed hashing stores all records directly in the hash table. Each record R with key value kR has a home position that is h(kR), the slot computed by the hash function. If R is to be inserted and another record already occupies R’s home position, then
- 357. 338 Chap. 9 Searching 0 1 2 3 4 Overflow Table Hash 9877 2007 3013 9879 1057 9530 1000 Figure 9.3 An illustration of bucket hashing for seven numbers stored in a five- bucket hash table using the hash function h(K) = K mod 5. Each bucket con- tains two slots. The numbers are inserted in the order 9877, 2007, 1000, 9530, 3013, 9879, and 1057. Two of the values hash to bucket 0, three values hash to bucket 2, one value hashes to bucket 3, and one value hashes to bucket 4. Because bucket 2 cannot hold three values, the third one ends up in the overflow bucket. R will be stored at some other slot in the table. It is the business of the collision resolution policy to determine which slot that will be. Naturally, the same policy must be followed during search as during insertion, so that any record not found in its home position can be recovered by repeating the collision resolution process. Bucket Hashing One implementation for closed hashing groups hash table slots into buckets. The M slots of the hash table are divided into B buckets, with each bucket consisting of M/B slots. The hash function assigns each record to the first slot within one of the buckets. If this slot is already occupied, then the bucket slots are searched sequentially until an open slot is found. If a bucket is entirely full, then the record is stored in an overflow bucket of infinite capacity at the end of the table. All buckets share the same overflow bucket. A good implementation will use a hash function that distributes the records evenly among the buckets so that as few records as possible go into the overflow bucket. Figure 9.3 illustrates bucket hashing. When searching for a record, the first step is to hash the key to determine which bucket should contain the record. The records in this bucket are then searched. If
- 358. Sec. 9.4 Hashing 339 the desired key value is not found and the bucket still has free slots, then the search is complete. If the bucket is full, then it is possible that the desired record is stored in the overflow bucket. In this case, the overflow bucket must be searched until the record is found or all records in the overflow bucket have been checked. If many records are in the overflow bucket, this will be an expensive process. A simple variation on bucket hashing is to hash a key value to some slot in the hash table as though bucketing were not being used. If the home position is full, then the collision resolution process is to move down through the table toward the end of the bucket will searching for a free slot in which to store the record. If the bottom of the bucket is reached, then the collision resolution routine wraps around to the top of the bucket to continue the search for an open slot. For example, assume that buckets contain eight records, with the first bucket consisting of slots 0 through 7. If a record is hashed to slot 5, the collision resolution process will attempt to insert the record into the table in the order 5, 6, 7, 0, 1, 2, 3, and finally 4. If all slots in this bucket are full, then the record is assigned to the overflow bucket. The advantage of this approach is that initial collisions are reduced, Because any slot can be a home position rather than just the first slot in the bucket. Bucket methods are good for implementing hash tables stored on disk, because the bucket size can be set to the size of a disk block. Whenever search or insertion occurs, the entire bucket is read into memory. Because the entire bucket is then in memory, processing an insert or search operation requires only one disk access, unless the bucket is full. If the bucket is full, then the overflow bucket must be retrieved from disk as well. Naturally, overflow should be kept small to minimize unnecessary disk accesses. Linear Probing We now turn to the most commonly used form of hashing: closed hashing with no bucketing, and a collision resolution policy that can potentially use any slot in the hash table. During insertion, the goal of collision resolution is to find a free slot in the hash table when the home position for the record is already occupied. We can view any collision resolution method as generating a sequence of hash table slots that can potentially hold the record. The first slot in the sequence will be the home position for the key. If the home position is occupied, then the collision resolution policy goes to the next slot in the sequence. If this is occupied as well, then another slot must be found, and so on. This sequence of slots is known as the probe sequence, and it is generated by some probe function that we will call p. The insert function is shown in Figure 9.4. Insertion works as follows:
- 359. 340 Chap. 9 Searching /** Insert record r with key k into HT */ void hashInsert(K k, E r) { int home; // Home position for r int pos = home = h(k); // Initial position for (int i=1; HT[pos] != null; i++) { pos = (home + p(k, i)) % M; // Next pobe slot assert HT[pos].key().compareTo(k) != 0 : Duplicates not allowed; } HT[pos] = new KVpairK,E(k, r); // Insert R } Figure 9.4 Insertion method for a dictionary implemented by a hash table. /** Search in hash table HT for the record with key k */ E hashSearch(K k) { int home; // Home position for k int pos = home = h(k); // Initial position for (int i = 1; (HT[pos] != null) (HT[pos].key().compareTo(k) != 0); i++) pos = (home + p(k, i)) % M; // Next probe position if (HT[pos] == null) return null; // K not in hash table else return HT[pos].value(); // Found it } Figure 9.5 Search method for a dictionary implemented by a hash table. Method hashInsert first checks to see if the home slot for the key is empty. If the home slot is occupied, then we use the probe function, p(k, i) to locate a free slot in the table. Function p has two parameters, the key k and a count i for where in the probe sequence we wish to be. That is, to get the first position in the probe sequence after the home slot for key K, we call p(K, 1). For the next slot in the probe sequence, call p(K, 2). Note that the probe function returns an offset from the original home position, rather than a slot in the hash table. Thus, the for loop in hashInsert is computing positions in the table at each iteration by adding the value returned from the probe function to the home position. The ith call to p returns the ith offset to be used. Searching in a hash table follows the same probe sequence that was followed when inserting records. In this way, a record not in its home position can be recov- ered. A Java implementation for the search procedure is shown in Figure 9.5 Both the insert and the search routines assume that at least one slot on the probe sequence of every key will be empty. Otherwise, they will continue in an infinite loop on unsuccessful searches. Thus, the dictionary should keep a count of the
- 360. Sec. 9.4 Hashing 341 number of records stored, and refuse to insert into a table that has only one free slot. The discussion on bucket hashing presented a simple method of collision reso- lution. If the home position for the record is occupied, then move down the bucket until a free slot is found. This is an example of a technique for collision resolution known as linear probing. The probe function for simple linear probing is p(K, i) = i. That is, the ith offset on the probe sequence is just i, meaning that the ith step is simply to move down i slots in the table. Once the bottom of the table is reached, the probe sequence wraps around to the beginning of the table. Linear probing has the virtue that all slots in the table will be candidates for inserting a new record before the probe sequence returns to the home position. It is important to understand that, while linear probing is probably the first idea that comes to mind when considering collision resolution policies, it is not the only one possible. Probe function p allows us many options for how to do collision resolution. In fact, linear probing is one of the worst collision resolution methods. The main problem is illustrated by Figure 9.6. Here, we see a hash table of ten slots used to store four-digit numbers, with hash function h(K) = K mod 10. In Figure 9.6(a), five numbers have been placed in the table, leaving five slots remaining. The ideal behavior for a collision resolution mechanism is that each empty slot in the table will have equal probability of receiving the next record inserted (assum- ing that every slot in the table has equal probability of being hashed to initially). In this example, the hash function gives each slot (roughly) equal probability of being the home position for the next key. However, consider what happens to the next record if its key has its home position at slot 0. Linear probing will send the record to slot 2. The same will happen to records whose home position is at slot 1. A record with home position at slot 2 will remain in slot 2. Thus, the probability is 3/10 that the next record inserted will end up in slot 2. In a similar manner, records hashing to slots 7 or 8 will end up in slot 9. However, only records hashing to slot 3 will be stored in slot 3, yielding one chance in ten of this happening. Like- wise, there is only one chance in ten that the next record will be stored in slot 4, one chance in ten for slot 5, and one chance in ten for slot 6. Thus, the resulting probabilities are not equal. To make matters worse, if the next record ends up in slot 9 (which already has a higher than normal chance of happening), then the following record will end up
- 361. 342 Chap. 9 Searching 0 1 2 4 3 5 6 7 9 0 1 2 3 4 5 6 7 8 9 8 9050 1001 9877 9050 1001 9877 2037 1059 2037 (a) (b) Figure 9.6 Example of problems with linear probing. (a) Four values are inserted in the order 1001, 9050, 9877, and 2037 using hash function h(K) = K mod 10. (b) The value 1059 is added to the hash table. in slot 2 with probability 6/10. This is illustrated by Figure 9.6(b). This tendency of linear probing to cluster items together is known as primary clustering. Small clusters tend to merge into big clusters, making the problem worse. The objection to primary clustering is that it leads to long probe sequences. Improved Collision Resolution Methods How can we avoid primary clustering? One possible improvement might be to use linear probing, but to skip slots by a constant c other than 1. This would make the probe function p(K, i) = ci, and so the ith slot in the probe sequence will be (h(K) + ic) mod M. In this way, records with adjacent home positions will not follow the same probe sequence. For example, if we were to skip by twos, then our offsets from the home slot would be 2, then 4, then 6, and so on. One quality of a good probe sequence is that it will cycle through all slots in the hash table before returning to the home position. Clearly linear probing (which “skips” slots by one each time) does this. Unfortunately, not all values for c will make this happen. For example, if c = 2 and the table contains an even number of slots, then any key whose home position is in an even slot will have a probe
- 362. Sec. 9.4 Hashing 343 sequence that cycles through only the even slots. Likewise, the probe sequence for a key whose home position is in an odd slot will cycle through the odd slots. Thus, this combination of table size and linear probing constant effectively divides the records into two sets stored in two disjoint sections of the hash table. So long as both sections of the table contain the same number of records, this is not really important. However, just from chance it is likely that one section will become fuller than the other, leading to more collisions and poorer performance for those records. The other section would have fewer records, and thus better performance. But the overall system performance will be degraded, as the additional cost to the side that is more full outweighs the improved performance of the less-full side. Constant c must be relatively prime to M to generate a linear probing sequence that visits all slots in the table (that is, c and M must share no factors). For a hash table of size M = 10, if c is any one of 1, 3, 7, or 9, then the probe sequence will visit all slots for any key. When M = 11, any value for c between 1 and 10 generates a probe sequence that visits all slots for every key. Consider the situation where c = 2 and we wish to insert a record with key k1 such that h(k1) = 3. The probe sequence for k1 is 3, 5, 7, 9, and so on. If another key k2 has home position at slot 5, then its probe sequence will be 5, 7, 9, and so on. The probe sequences of k1 and k2 are linked together in a manner that contributes to clustering. In other words, linear probing with a value of c 1 does not solve the problem of primary clustering. We would like to find a probe function that does not link keys together in this way. We would prefer that the probe sequence for k1 after the first step on the sequence should not be identical to the probe sequence of k2. Instead, their probe sequences should diverge. The ideal probe function would select the next position on the probe sequence at random from among the unvisited slots; that is, the probe sequence should be a random permutation of the hash table positions. Unfortunately, we cannot actually select the next position in the probe sequence at random, because then we would not be able to duplicate this same probe sequence when searching for the key. However, we can do something similar called pseudo-random probing. In pseudo-random probing, the ith slot in the probe sequence is (h(K) + ri) mod M where ri is the ith value in a random permutation of the numbers from 1 to M − 1. All insertion and search operations must use the same random permutation. The probe function would be p(K, i) = Perm[i − 1], where Perm is an array of length M − 1 containing a random permutation of the values from 1 to M − 1.
- 363. 344 Chap. 9 Searching Example 9.9 Consider a table of size M = 101, with Perm[1] = 5, Perm[2] = 2, and Perm[3] = 32. Assume that we have two keys k1 and k2 where h(k1) = 30 and h(k2) = 35. The probe sequence for k1 is 30, then 35, then 32, then 62. The probe sequence for k2 is 35, then 40, then 37, then 67. Thus, while k2 will probe to k1’s home position as its second choice, the two keys’ probe sequences diverge immediately thereafter. Another probe function that eliminates primary clustering is called quadratic probing. Here the probe function is some quadratic function p(K, i) = c1i2 + c2i + c3 for some choice of constants c1, c2, and c3. The simplest variation is p(K, i) = i2 (i.e., c1 = 1, c2 = 0, and c3 = 0. Then the ith value in the probe sequence would be (h(K) + i2) mod M. Under quadratic probing, two keys with different home positions will have diverging probe sequences. Example 9.10 Given a hash table of size M = 101, assume for keys k1 and k2 that h(k1) = 30 and h(k2) = 29. The probe sequence for k1 is 30, then 31, then 34, then 39. The probe sequence for k2 is 29, then 30, then 33, then 38. Thus, while k2 will probe to k1’s home position as its second choice, the two keys’ probe sequences diverge immediately thereafter. Unfortunately, quadratic probing has the disadvantage that typically not all hash table slots will be on the probe sequence. Using p(K, i) = i2 gives particularly in- consistent results. For many hash table sizes, this probe function will cycle through a relatively small number of slots. If all slots on that cycle happen to be full, then the record cannot be inserted at all! For example, if our hash table has three slots, then records that hash to slot 0 can probe only to slots 0 and 1 (that is, the probe sequence will never visit slot 2 in the table). Thus, if slots 0 and 1 are full, then the record cannot be inserted even though the table is not full! A more realistic example is a table with 105 slots. The probe sequence starting from any given slot will only visit 23 other slots in the table. If all 24 of these slots should happen to be full, even if other slots in the table are empty, then the record cannot be inserted because the probe sequence will continually hit only those same 24 slots. Fortunately, it is possible to get good results from quadratic probing at low cost. The right combination of probe function and table size will visit many slots in the table. In particular, if the hash table size is a prime number and the probe function is p(K, i) = i2, then at least half the slots in the table will be visited. Thus, if the table is less than half full, we can be certain that a free slot will be
- 364. Sec. 9.4 Hashing 345 found. Alternatively, if the hash table size is a power of two and the probe function is p(K, i) = (i2 + i)/2, then every slot in the table will be visited by the probe function. Both pseudo-random probing and quadratic probing eliminate primary cluster- ing, which is the problem of keys sharing substantial segments of a probe sequence. If two keys hash to the same home position, however, then they will always follow the same probe sequence for every collision resolution method that we have seen so far. The probe sequences generated by pseudo-random and quadratic probing (for example) are entirely a function of the home position, not the original key value. This is because function p ignores its input parameter K for these collision resolu- tion methods. If the hash function generates a cluster at a particular home position, then the cluster remains under pseudo-random and quadratic probing. This problem is called secondary clustering. To avoid secondary clustering, we need to have the probe sequence make use of the original key value in its decision-making process. A simple technique for doing this is to return to linear probing by a constant step size for the probe function, but to have that constant be determined by a second hash function, h2. Thus, the probe sequence would be of the form p(K, i) = i∗ h2(K). This method is called double hashing. Example 9.11 Assume a hash table has size M = 101, and that there are three keys k1, k2, and k3 with h(k1) = 30, h(k2) = 28, h(k3) = 30, h2(k1) = 2, h2(k2) = 5, and h2(k3) = 5. Then, the probe sequence for k1 will be 30, 32, 34, 36, and so on. The probe sequence for k2 will be 28, 33, 38, 43, and so on. The probe sequence for k3 will be 30, 35, 40, 45, and so on. Thus, none of the keys share substantial portions of the same probe sequence. Of course, if a fourth key k4 has h(k4) = 28 and h2(k4) = 2, then it will follow the same probe sequence as k1. Pseudo- random or quadratic probing can be combined with double hashing to solve this problem. A good implementation of double hashing should ensure that all of the probe sequence constants are relatively prime to the table size M. This can be achieved easily. One way is to select M to be a prime number, and have h2 return a value in the range 1 ≤ h2(K) ≤ M − 1. Another way is to set M = 2m for some value m and have h2 return an odd value between 1 and 2m. Figure 9.7 shows an implementation of the dictionary ADT by means of a hash table. The simplest hash function is used, with collision resolution by linear prob- ing, as the basis for the structure of a hash table implementation. A suggested
- 365. 346 Chap. 9 Searching project at the end of this chapter asks you to improve the implementation with other hash functions and collision resolution policies. 9.4.4 Analysis of Closed Hashing How efficient is hashing? We can measure hashing performance in terms of the number of record accesses required when performing an operation. The primary operations of concern are insertion, deletion, and search. It is useful to distinguish between successful and unsuccessful searches. Before a record can be deleted, it must be found. Thus, the number of accesses required to delete a record is equiv- alent to the number required to successfully search for it. To insert a record, an empty slot along the record’s probe sequence must be found. This is equivalent to an unsuccessful search for the record (recall that a successful search for the record during insertion should generate an error because two records with the same key are not allowed to be stored in the table). When the hash table is empty, the first record inserted will always find its home position free. Thus, it will require only one record access to find a free slot. If all records are stored in their home positions, then successful searches will also require only one record access. As the table begins to fill up, the probability that a record can be inserted in its home position decreases. If a record hashes to an occupied slot, then the collision resolution policy must locate another slot in which to store it. Finding records not stored in their home position also requires additional record accesses as the record is searched for along its probe sequence. As the table fills up, more and more records are likely to be located ever further from their home positions. From this discussion, we see that the expected cost of hashing is a function of how full the table is. Define the load factor for the table as α = N/M, where N is the number of records currently in the table. An estimate of the expected cost for an insertion (or an unsuccessful search) can be derived analytically as a function of α in the case where we assume that the probe sequence follows a random permutation of the slots in the hash table. Assuming that every slot in the table has equal probability of being the home slot for the next record, the probability of finding the home position occupied is α. The probability of finding both the home position occupied and the next slot on the probe sequence occupied is N(N−1) M(M−1) . The probability of i collisions is N(N − 1) · · · (N − i + 1) M(M − 1) · · · (M − i + 1) .
- 366. Sec. 9.4 Hashing 347 /** Dictionary implemented using hashing. */ class HashDictionaryK extends Comparable? super K, E implements DictionaryK, E { private static final int defaultSize = 10; private HashTableK,E T; // The hash table private int count; // # of records now in table private int maxsize; // Maximum size of dictionary HashDictionary() { this(defaultSize); } HashDictionary(int sz) { T = new HashTableK,E(sz); count = 0; maxsize = sz; } public void clear() { /** Reinitialize */ T = new HashTableK,E(maxsize); count = 0; } public void insert(K k, E e) { /** Insert an element */ assert count maxsize : Hash table is full; T.hashInsert(k, e); count++; } public E remove(K k) { /** Remove an element */ E temp = T.hashRemove(k); if (temp != null) count--; return temp; } public E removeAny() { /** Remove some element. */ if (count != 0) { count--; return T.hashRemoveAny(); } else return null; } /** Find a record with key value k */ public E find(K k) { return T.hashSearch(k); } /** Return number of values in the hash table */ public int size() { return count; } } Figure 9.7 A partial implementation for the dictionary ADT using a hash table.
- 367. 348 Chap. 9 Searching If N and M are large, then this is approximately (N/M)i. The expected number of probes is one plus the sum over i ≥ 1 of the probability of i collisions, which is approximately 1 + ∞ X i=1 (N/M)i = 1/(1 − α). The cost for a successful search (or a deletion) has the same cost as originally inserting that record. However, the expected value for the insertion cost depends on the value of α not at the time of deletion, but rather at the time of the original insertion. We can derive an estimate of this cost (essentially an average over all the insertion costs) by integrating from 0 to the current value of α, yielding a result of 1 α Z α 0 1 1 − x dx = 1 α loge 1 1 − α . It is important to realize that these equations represent the expected cost for operations using the unrealistic assumption that the probe sequence is based on a random permutation of the slots in the hash table (thus avoiding all expense result- ing from clustering). Thus, these costs are lower-bound estimates in the average case. The true average cost under linear probing is 1 2 (1+1/(1−α)2) for insertions or unsuccessful searches and 1 2 (1+1/(1−α)) for deletions or successful searches. Proofs for these results can be found in the references cited in Section 9.5. Figure 9.8 shows the graphs of these four equations to help you visualize the expected performance of hashing based on the load factor. The two solid lines show the costs in the case of a “random” probe sequence for (1) insertion or unsuccessful search and (2) deletion or successful search. As expected, the cost for insertion or unsuccessful search grows faster, because these operations typically search further down the probe sequence. The two dashed lines show equivalent costs for linear probing. As expected, the cost of linear probing grows faster than the cost for “random” probing. From Figure 9.8 we see that the cost for hashing when the table is not too full is typically close to one record access. This is extraordinarily efficient, much better than binary search which requires log n record accesses. As α increases, so does the expected cost. For small values of α, the expected cost is low. It remains below two until the hash table is about half full. When the table is nearly empty, adding a new record to the table does not increase the cost of future search operations by much. However, the additional search cost caused by each additional insertion increases rapidly once the table becomes half full. Based on this analysis, the rule of thumb is to design a hashing system so that the hash table never gets above half
- 368. Sec. 9.4 Hashing 349 1 2 3 4 5 Delete Insert 0 .2 .4 .6 .8 1.0 Figure 9.8 Growth of expected record accesses with α. The horizontal axis is the value for α, the vertical axis is the expected number of accesses to the hash table. Solid lines show the cost for “random” probing (a theoretical lower bound on the cost), while dashed lines show the cost for linear probing (a relatively poor collision resolution strategy). The two leftmost lines show the cost for insertion (equivalently, unsuccessful search); the two rightmost lines show the cost for dele- tion (equivalently, successful search). full. Beyond that point performance will degrade rapidly. This requires that the implementor have some idea of how many records are likely to be in the table at maximum loading, and select the table size accordingly. You might notice that a recommendation to never let a hash table become more than half full contradicts the disk-based space/time tradeoff principle, which strives to minimize disk space to increase information density. Hashing represents an un- usual situation in that there is no benefit to be expected from locality of reference. In a sense, the hashing system implementor does everything possible to eliminate the effects of locality of reference! Given the disk block containing the last record accessed, the chance of the next record access coming to the same disk block is no better than random chance in a well-designed hash system. This is because a good hashing implementation breaks up relationships between search keys. Instead of improving performance by taking advantage of locality of reference, hashing trades increased hash table space for an improved chance that the record will be in its home position. Thus, the more space available for the hash table, the more efficient hashing should be.
- 369. 350 Chap. 9 Searching Depending on the pattern of record accesses, it might be possible to reduce the expected cost of access even in the face of collisions. Recall the 80/20 rule: 80% of the accesses will come to 20% of the data. In other words, some records are accessed more frequently. If two records hash to the same home position, which would be better placed in the home position, and which in a slot further down the probe sequence? The answer is that the record with higher frequency of access should be placed in the home position, because this will reduce the total number of record accesses. Ideally, records along a probe sequence will be ordered by their frequency of access. One approach to approximating this goal is to modify the order of records along the probe sequence whenever a record is accessed. If a search is made to a record that is not in its home position, a self-organizing list heuristic can be used. For example, if the linear probing collision resolution policy is used, then whenever a record is located that is not in its home position, it can be swapped with the record preceding it in the probe sequence. That other record will now be further from its home position, but hopefully it will be accessed less frequently. Note that this approach will not work for the other collision resolution policies presented in this section, because swapping a pair of records to improve access to one might remove the other from its probe sequence. Another approach is to keep access counts for records and periodically rehash the entire table. The records should be inserted into the hash table in frequency order, ensuring that records that were frequently accessed during the last series of requests have the best chance of being near their home positions. 9.4.5 Deletion When deleting records from a hash table, there are two important considerations. 1. Deleting a record must not hinder later searches. In other words, the search process must still pass through the newly emptied slot to reach records whose probe sequence passed through this slot. Thus, the delete process cannot simply mark the slot as empty, because this will isolate records further down the probe sequence. For example, in Figure 9.6(a), keys 9877 and 2037 both hash to slot 7. Key 2037 is placed in slot 8 by the collision resolution policy. If 9877 is deleted from the table, a search for 2037 must still pass through Slot 7 as it probes to slot 8. 2. We do not want to make positions in the hash table unusable because of deletion. The freed slot should be available to a future insertion. Both of these problems can be resolved by placing a special mark in place of the deleted record, called a tombstone. The tombstone indicates that a record once
- 370. Sec. 9.5 Further Reading 351 occupied the slot but does so no longer. If a tombstone is encountered when search- ing through a probe sequence, the search procedure is to continue with the search. When a tombstone is encountered during insertion, that slot can be used to store the new record. However, to avoid inserting duplicate keys, it will still be necessary for the search procedure to follow the probe sequence until a truly empty position has been found, simply to verify that a duplicate is not in the table. However, the new record would actually be inserted into the slot of the first tombstone encountered. The use of tombstones allows searches to work correctly and allows reuse of deleted slots. However, after a series of intermixed insertion and deletion opera- tions, some slots will contain tombstones. This will tend to lengthen the average distance from a record’s home position to the record itself, beyond where it could be if the tombstones did not exist. A typical database application will first load a collection of records into the hash table and then progress to a phase of intermixed insertions and deletions. After the table is loaded with the initial collection of records, the first few deletions will lengthen the average probe sequence distance for records (it will add tombstones). Over time, the average distance will reach an equilibrium point because insertions will tend to decrease the average distance by filling in tombstone slots. For example, after initially loading records into the database, the average path distance might be 1.2 (i.e., an average of 0.2 accesses per search beyond the home position will be required). After a series of insertions and deletions, this average distance might increase to 1.6 due to tombstones. This seems like a small increase, but it is three times longer on average beyond the home position than before deletions. Two possible solutions to this problem are 1. Do a local reorganization upon deletion to try to shorten the average path length. For example, after deleting a key, continue to follow the probe se- quence of that key and swap records further down the probe sequence into the slot of the recently deleted record (being careful not to remove a key from its probe sequence). This will not work for all collision resolution policies. 2. Periodically rehash the table by reinserting all records into a new hash table. Not only will this remove the tombstones, but it also provides an opportunity to place the most frequently accessed records into their home positions. 9.5 Further Reading For a comparison of the efficiencies for various self-organizing techniques, see Bentley and McGeoch, “Amortized Analysis of Self-Organizing Sequential Search Heuristics” [BM85]. The text compression example of Section 9.2 comes from
- 371. 352 Chap. 9 Searching Bentley et al., “A Locally Adaptive Data Compression Scheme” [BSTW86]. For more on Ziv-Lempel coding, see Data Compression: Methods and Theory by James A. Storer [Sto88]. Knuth covers self-organizing lists and Zipf distributions in Volume 3 of The Art of Computer Programming[Knu98]. Introduction to Modern Information Retrieval by Salton and McGill [SM83] is an excellent source for more information about document retrieval techniques. See the paper “Practical Minimal Perfect Hash Functions for Large Databases” by Fox et al. [FHCD92] for an introduction and a good algorithm for perfect hash- ing. For further details on the analysis for various collision resolution policies, see Knuth, Volume 3 [Knu98] and Concrete Mathematics: A Foundation for Computer Science by Graham, Knuth, and Patashnik [GKP94]. The model of hashing presented in this chapter has been of a fixed-size hash table. A problem not addressed is what to do when the hash table gets half full and more records must be inserted. This is the domain of dynamic hashing methods. A good introduction to this topic is “Dynamic Hashing Schemes” by R.J. Enbody and H.C. Du [ED88]. 9.6 Exercises 9.1 Create a graph showing expected cost versus the probability of an unsuc- cessful search when performing sequential search (see Section 9.1). What can you say qualitatively about the rate of increase in expected cost as the probability of unsuccessful search grows? 9.2 Modify the binary search routine of Section 3.5 to implement interpolation search. Assume that keys are in the range 1 to 10,000, and that all key values within the range are equally likely to occur. 9.3 Write an algorithm to find the Kth smallest value in an unsorted array of n numbers (K = n). Your algorithm should require Θ(n) time in the average case. Hint: Your algorithm shuld look similar to Quicksort. 9.4 Example 9.9.3 discusses a distribution where the relative frequencies of the records match the harmonic series. That is, for every occurance of the first record, the second record will appear half as often, the third will appear one third as often, the fourth one quarter as often, and so on. The actual prob- ability for the ith record was defined to be 1/(iHn). Explain why this is correct. 9.5 Graph the equations T(n) = log2 n and T(n) = n/ loge n. Which gives the better performance, binary search on a sorted list, or sequential search on a
- 372. Sec. 9.6 Exercises 353 list ordered by frequency where the frequence conforms to a Zipf distribu- tion? Characterize the difference in running times. 9.6 Assume that the values A through H are stored in a self-organizing list, ini- tially in ascending order. Consider the three self-organizing list heuristics: count, move-to-front, and transpose. For count, assume that the record is moved ahead in the list passing over any other record that its count is now greater than. For each, show the resulting list and the total number of com- parisons required resulting from the following series of accesses: D H H G H E G H G H E C E H G. 9.7 For each of the three self-organizing list heuristics (count, move-to-front, and transpose), describe a series of record accesses for which it would require the greatest number of comparisons of the three. 9.8 Write an algorithm to implement the frequency count self-organizing list heuristic, assuming that the list is implemented using an array. In particu- lar, write a function FreqCount that takes as input a value to be searched for and which adjusts the list appropriately. If the value is not already in the list, add it to the end of the list with a frequency count of one. 9.9 Write an algorithm to implement the move-to-front self-organizing list heuri- stic, assuming that the list is implemented using an array. In particular, write a function MoveToFront that takes as input a value to be searched for and which adjusts the list appropriately. If the value is not already in the list, add it to the beginning of the list. 9.10 Write an algorithm to implement the transpose self-organizing list heuristic, assuming that the list is implemented using an array. In particular, write a function transpose that takes as input a value to be searched for and which adjusts the list appropriately. If the value is not already in the list, add it to the end of the list. 9.11 Write functions for computing union, intersection, and set difference on ar- bitrarily long bit vectors used to represent set membership as described in Section 9.3. Assume that for each operation both vectors are of equal length. 9.12 Compute the probabilities for the following situations. These probabilities can be computed analytically, or you may write a computer program to gen- erate the probabilities by simulation. (a) Out of a group of 23 students, what is the probability that 2 students share the same birthday? (b) Out of a group of 100 students, what is the probability that 3 students share the same birthday?
- 373. 354 Chap. 9 Searching (c) How many students must be in the class for the probability to be at least 50% that there are 2 who share a birthday in the same month? 9.13 Assume that you are hashing key K to a hash table of n slots (indexed from 0 to n − 1). For each of the following functions h(K), is the function ac- ceptable as a hash function (i.e., would the hash program work correctly for both insertions and searches), and if so, is it a good hash function? Function Random(n) returns a random integer between 0 and n − 1, inclusive. (a) h(k) = k/n where k and n are integers. (b) h(k) = 1. (c) h(k) = (k + Random(n)) mod n. (d) h(k) = k mod n where n is a prime number. 9.14 Assume that you have a seven-slot closed hash table (the slots are numbered 0 through 6). Show the final hash table that would result if you used the hash function h(k) = k mod 7 and linear probing on this list of numbers: 3, 12, 9, 2. After inserting the record with key value 2, list for each empty slot the probability that it will be the next one filled. 9.15 Assume that you have a ten-slot closed hash table (the slots are numbered 0 through 9). Show the final hash table that would result if you used the hash function h(k) = k mod 10 and quadratic probing on this list of numbers: 3, 12, 9, 2, 79, 46. After inserting the record with key value 46, list for each empty slot the probability that it will be the next one filled. 9.16 Assume that you have a ten-slot closed hash table (the slots are numbered 0 through 9). Show the final hash table that would result if you used the hash function h(k) = k mod 10 and pseudo-random probing on this list of numbers: 3, 12, 9, 2, 79, 44. The permutation of offsets to be used by the pseudo-random probing will be: 5, 9, 2, 1, 4, 8, 6, 3, 7. After inserting the record with key value 44, list for each empty slot the probability that it will be the next one filled. 9.17 What is the result of running sfold from Section 9.4.1 on the following strings? Assume a hash table size of 101 slots. (a) HELLO WORLD (b) NOW HEAR THIS (c) HEAR THIS NOW 9.18 Using closed hashing, with double hashing to resolve collisions, insert the following keys into a hash table of thirteen slots (the slots are numbered 0 through 12). The hash functions to be used are H1 and H2, defined be- low. You should show the hash table after all eight keys have been inserted.
- 374. Sec. 9.7 Projects 355 Be sure to indicate how you are using H1 and H2 to do the hashing. Func- tion Rev(k) reverses the decimal digits of k, for example, Rev(37) = 73; Rev(7) = 7. H1(k) = k mod 13. H2(k) = (Rev(k + 1) mod 11). Keys: 2, 8, 31, 20, 19, 18, 53, 27. 9.19 Write an algorithm for a deletion function for hash tables that replaces the record with a special value indicating a tombstone. Modify the functions hashInsert and hashSearch to work correctly with tombstones. 9.20 Consider the following permutation for the numbers 1 to 6: 2, 4, 6, 1, 3, 5. Analyze what will happen if this permutation is used by an implementation of pseudo-random probing on a hash table of size seven. Will this permutation solve the problem of primary clustering? What does this say about selecting a permutation for use when implementing pseudo-random probing? 9.7 Projects 9.1 Implement a binary search and the quadratic binary search of Section 9.1. Run your implementations over a large range of problem sizes, timing the results for each algorithm. Graph and compare these timing results. 9.2 Implement the three self-organizing list heuristics count, move-to-front, and transpose. Compare the cost for running the three heuristics on various input data. The cost metric should be the total number of comparisons required when searching the list. It is important to compare the heuristics using input data for which self-organizing lists are reasonable, that is, on frequency dis- tributions that are uneven. One good approach is to read text files. The list should store individual words in the text file. Begin with an empty list, as was done for the text compression example of Section 9.2. Each time a word is encountered in the text file, search for it in the self-organizing list. If the word is found, reorder the list as appropriate. If the word is not in the list, add it to the end of the list and then reorder as appropriate. 9.3 Implement the text compression system described in Section 9.2. 9.4 Implement a system for managing document retrieval. Your system should have the ability to insert (abstract references to) documents into the system, associate keywords with a given document, and to search for documents with specified keywords.
- 375. 356 Chap. 9 Searching 9.5 Implement a database stored on disk using bucket hashing. Define records to be 128 bytes long with a 4-byte key and 120 bytes of data. The remaining 4 bytes are available for you to store necessary information to support the hash table. A bucket in the hash table will be 1024 bytes long, so each bucket has space for 8 records. The hash table should consist of 27 buckets (total space for 216 records with slots indexed by positions 0 to 215) followed by the overflow bucket at record position 216 in the file. The hash function for key value K should be K mod 213. (Note that this means the last three slots in the table will not be home positions for any record.) The collision resolution function should be linear probing with wrap-around within the bucket. For example, if a record is hashed to slot 5, the collision resolution process will attempt to insert the record into the table in the order 5, 6, 7, 0, 1, 2, 3, and finally 4. If a bucket is full, the record should be placed in the overflow section at the end of the file. Your hash table should implement the dictionary ADT of Section 4.4. When you do your testing, assume that the system is meant to store about 100 or so records at a time. 9.6 Implement the dictionary ADT of Section 4.4 by means of a hash table with linear probing as the collision resolution policy. You might wish to begin with the code of Figure 9.7. Using empirical simulation, determine the cost of insert and delete as α grows (i.e., reconstruct the dashed lines of Fig- ure 9.8). Then, repeat the experiment using quadratic probing and pseudo- random probing. What can you say about the relative performance of these three collision resolution policies?
- 376. 10 Indexing Many large-scale computing applications are centered around datasets that are too large to fit into main memory. The classic example is a large database of records with multiple search keys, requiring the ability to insert, delete, and search for records. Hashing provides outstanding performance for such situations, but only in the limited case in which all searches are of the form “find the record with key value K.” Unfortunately, many applications require more general search capabili- ties. One example is a range query search for all records whose key lies within some range. Other queries might involve visiting all records in order of their key value, or finding the record with the greatest key value. Hash tables are not organized to support any of these queries efficiently. This chapter introduces file structures used to organize a large collection of records stored on disk. Such file structures support efficient insertion, deletion, and search operations, for exact-match queries, range queries, and largest/smallest key value searches. Before discussing such file structures, we must become familiar with some ba- sic file-processing terminology. An entry-sequenced file stores records in the order that they were added to the file. Entry-sequenced files are the disk-based equivalent to an unsorted list and so do not support efficient search. The natural solution is to sort the records by order of the search key. However, a typical database, such as a collection of employee or customer records maintained by a business, might con- tain multiple search keys. To answer a question about a particular customer might require a search on the name of the customer. Businesses often wish to sort and output the records by zip code order for a bulk mailing. Government paperwork might require the ability to search by Social Security number. Thus, there might not be a single “correct” order in which to store the records. Indexing is the process of associating a key with the location of a correspond- ing data record. Section 8.5 discussed the concept of a key sort, in which an index 357
- 377. 358 Chap. 10 Indexing file is created whose records consist of key/pointer pairs. Here, each key is asso- ciated with a pointer to a complete record in the main database file. The index file could be sorted or organized using a tree structure, thereby imposing a logical or- der on the records without physically rearranging them. One database might have several associated index files, each supporting efficient access through a different key field. Each record of a database normally has a unique identifier, called the primary key. For example, the primary key for a set of personnel records might be the Social Security number or ID number for the individual. Unfortunately, the ID number is generally an inconvenient value on which to perform a search because the searcher is unlikely to know it. Instead, the searcher might know the desired employee’s name. Alternatively, the searcher might be interested in finding all employees whose salary is in a certain range. If these are typical search requests to the database, then the name and salary fields deserve separate indices. However, key values in the name and salary indices are not likely to be unique. A key field such as salary, where a particular key value might be duplicated in multiple records, is called a secondary key. Most searches are performed using a secondary key. The secondary key index (or more simply, secondary index) will associate a secondary key value with the primary key of each record having that secondary key value. At this point, the full database might be searched directly for the record with that primary key, or there might be a primary key index (or primary index) that relates each primary key value with a pointer to the actual record on disk. In the latter case, only the primary index provides the location of the actual record on disk, while the secondary indices refer to the primary index. Indexing is an important technique for organizing large databases, and many indexing methods have been developed. Direct access through hashing is discussed in Section 9.4. A simple list sorted by key value can also serve as an index to the record file. Indexing disk files by sorted lists are discussed in the following section. Unfortunately, a sorted list does not perform well for insert and delete operations. A third approach to indexing is the tree index. Trees are typically used to or- ganize large databases that must support record insertion, deletion, and key range searches. Section 10.2 briefly describes ISAM, a tentative step toward solving the problem of storing a large database that must support insertion and deletion of records. Its shortcomings help to illustrate the value of tree indexing techniques. Section 10.3 introduces the basic issues related to tree indexing. Section 10.4 in- troduces the 2-3 tree, a balanced tree structure that is a simple form of the B-tree covered in Section 10.5. B-trees are the most widely used indexing method for large disk-based databases, and many variations have been invented. Section 10.5
- 378. Sec. 10.1 Linear Indexing 359 Linear Index Database Records 42 73 98 52 37 52 98 37 42 73 Figure 10.1 Linear indexing for variable-length records. Each record in the index file is of fixed length and contains a pointer to the beginning of the corre- sponding record in the database file. begins with a discussion of the variant normally referred to simply as a “B-tree.” Section 10.5.1 presents the most widely implemented variant, the B+-tree. 10.1 Linear Indexing A linear index is an index file organized as a sequence of key/pointer pairs where the keys are in sorted order and the pointers either (1) point to the position of the complete record on disk, (2) point to the position of the primary key in the primary index, or (3) are actually the value of the primary key. Depending on its size, a linear index might be stored in main memory or on disk. A linear index provides a number of advantages. It provides convenient access to variable-length database records, because each entry in the index file contains a fixed-length key field and a fixed-length pointer to the beginning of a (variable-length) record as shown in Figure 10.1. A linear index also allows for efficient search and random access to database records, becase it is amenable to binary search. If the database contains enough records, the linear index might be too large to store in main memory. This makes binary search of the index more expensive because many disk accesses would typically be required by the search process. One solution to this problem is to store a second-level linear index in main memory that indicates which disk block in the index file stores a desired key. For example, the linear index on disk might reside in a series of 1024-byte blocks. If each key/pointer pair in the linear index requires 8 bytes, then 128 keys are stored per block. The second-level index, stored in main memory, consists of a simple table storing the value of the key in the first position of each block in the linear index file. This arrangement is shown in Figure 10.2. If the linear index requires 1024 disk blocks (1MB), the second-level index contains only 1024 entries, one per disk block. To find which disk block contains a desired search key value, first search through the
- 379. 360 Chap. 10 Indexing 1 2003 5894 Second Level Index 1 2001 5894 9942 10528 10984 Linear Index: Disk Blocks 5688 2003 10528 Figure 10.2 A simple two-level linear index. The linear index is stored on disk. The smaller, second-level index is stored in main memory. Each element in the second-level index stores the first key value in the corresponding disk block of the index file. In this example, the first disk block of the linear index stores keys in the range 1 to 2001, and the second disk block stores keys in the range 2003 to 5688. Thus, the first entry of the second-level index is key value 1 (the first key in the first block of the linear index), while the second entry of the second-level index is key value 2003. Jones Smith Zukowski AA10 AX33 ZQ99 AB12 AX35 AB39 ZX45 FF37 Figure 10.3 A two-dimensional linear index. Each row lists the primary keys associated with a particular secondary key value. In this example, the secondary key is a name. The primary key is a unique four-character code. 1024-entry table to find the greatest value less than or equal to the search key. This directs the search to the proper block in the index file, which is then read into memory. At this point, a binary search within this block will produce a pointer to the actual record in the database. Because the second-level index is stored in main memory, accessing a record by this method requires two disk reads: one from the index file and one from the database file for the actual record. Every time a record is inserted to or deleted from the database, all associated secondary indices must be updated. Updates to a linear index are expensive, be- cause the entire contents of the array might be shifted by one position. Another problem is that multiple records with the same secondary key each duplicate that key value within the index. When the secondary key field has many duplicates, such as when it has a limited range (e.g., a field to indicate job category from among a small number of possible job categories), this duplication might waste considerable space. One improvement on the simple sorted array is a two-dimensional array where each row corresponds to a secondary key value. A row contains the primary keys
- 380. Sec. 10.2 ISAM 361 whose records have the indicated secondary key value. Figure 10.3 illustrates this approach. Now there is no duplication of secondary key values, possibly yielding a considerable space savings. The cost of insertion and deletion is reduced, because only one row of the table need be adjusted. Note that a new row is added to the array when a new secondary key value is added. This might lead to moving many records, but this will happen infrequently in applications suited to using this arrangement. A drawback to this approach is that the array must be of fixed size, which imposes an upper limit on the number of primary keys that might be associated with a particular secondary key. Furthermore, those secondary keys with fewer records than the width of the array will waste the remainder of their row. A better approach is to have a one-dimensional array of secondary key values, where each secondary key is associated with a linked list. This works well if the index is stored in main memory, but not so well when it is stored on disk because the linked list for a given key might be scattered across several disk blocks. Consider a large database of employee records. If the primary key is the em- ployee’s ID number and the secondary key is the employee’s name, then each record in the name index associates a name with one or more ID numbers. The ID number index in turn associates an ID number with a unique pointer to the full record on disk. The secondary key index in such an organization is also known as an inverted list or inverted file. It is inverted in that searches work backwards from the secondary key to the primary key to the actual data record. It is called a list because each secondary key value has (conceptually) a list of primary keys as- sociated with it. Figure 10.4 illustrates this arrangement. Here, we have last names as the secondary key. The primary key is a four-character unique identifier. Figure 10.5 shows a better approach to storing inverted lists. An array of sec- ondary key values is shown as before. Associated with each secondary key is a pointer to an array of primary keys. The primary key array uses a linked-list im- plementation. This approach combines the storage for all of the secondary key lists into a single array, probably saving space. Each record in this array consists of a primary key value and a pointer to the next element on the list. It is easy to insert and delete secondary keys from this array, making this a good implementation for disk-based inverted files. 10.2 ISAM How do we handle large databases that require frequent update? The main prob- lem with the linear index is that it is a single, large array that does not lend itself to updates because a single update can require changing the position of every key
- 381. 362 Chap. 10 Indexing Jones Smith Zukowski Primary Key AA10 AB12 AB39 FF37 AX33 AX35 ZX45 ZQ99 Secondary Key Figure 10.4 Illustration of an inverted list. Each secondary key value is stored in the secondary key list. Each secondary key value on the list has a pointer to a list of the primary keys whose associated records have that secondary key value. Index 0 1 3 Primary Key Next AA10 AX33 ZX45 ZQ99 AB12 AB39 AX35 FF37 4 6 5 7 2 Key Jones Smith Zukowski 0 1 2 3 4 5 6 7 Secondary Figure 10.5 An inverted list implemented as an array of secondary keys and combined lists of primary keys. Each record in the secondary key array contains a pointer to a record in the primary key array. The next field of the primary key array indicates the next record with that secondary key value.
- 382. Sec. 10.2 ISAM 363 Cylinder Overflow Cylinder Overflow Index Cylinder Keys In-memory Table of Cylinder 1 Cylinder 2 Records Records Cylinder Index System Overflow Cylinder Figure 10.6 Illustration of the ISAM indexing system. in the index. Inverted lists reduce this problem, but they are only suitable for sec- ondary key indices with many fewer secondary key values than records. The linear index would perform well as a primary key index if it could somehow be broken into pieces such that individual updates affect only a part of the index. This con- cept will be pursued throughout the rest of this chapter, eventually culminating in the B+-tree, the most widely used indexing method today. But first, we begin by studying ISAM, an early attempt to solve the problem of large databases requiring frequent update. Its weaknesses help to illustrate why the B+-tree works so well. Before the invention of effective tree indexing schemes, a variety of disk-based indexing methods were in use. All were rather cumbersome, largely because no adequate method for handling updates was known. Typically, updates would cause the index to degrade in performance. ISAM is one example of such an index and was widely used by IBM prior to adoption of the B-tree. ISAM is based on a modified form of the linear index, as illustrated by Fig- ure 10.6. Records are stored in sorted order by primary key. The disk file is divided among a number of cylinders on disk.1 Each cylindar holds a section of the list in sorted order. Initially, each cylinder is not filled to capacity, and the extra space is set aside in the cylinder overflow. In memory is a table listing the lowest key value stored in each cylinder of the file. Each cylinder contains a table listing the lowest 1 Recall from Section 8.2.1 that a cylinder is all of the tracks readable from a particular placement of the heads on the multiple platters of a disk drive.
- 383. 364 Chap. 10 Indexing key value for each block in that cylinder, called the cylinder index. When new records are inserted, they are placed in the correct cylinder’s overflow area (in ef- fect, a cylinder acts as a bucket). If a cylinder’s overflow area fills completely, then a system-wide overflow area is used. Search proceeds by determining the proper cylinder from the system-wide table kept in main memory. The cylinder’s block table is brought in from disk and consulted to determine the correct block. If the record is found in that block, then the search is complete. Otherwise, the cylin- der’s overflow area is searched. If that is full, and the record is not found, then the system-wide overflow is searched. After initial construction of the database, so long as no new records are inserted or deleted, access is efficient because it rquires only two disk fetches. The first disk fetch recovers the block table for the desired cylinder. The second disk fetch recovers the block that, under good conditions, contains the record. After many inserts, the overflow list becomes too long, resulting in significant search time as the cylinder overflow area fills up. Under extreme conditions, many searches might eventually lead to the system overflow area. The “solution” to this problem is to periodically reorganize the entire database. This means rebalancing the records among the cylinders, sorting the records within each cylinder, and updating both the system index table and the within-cylinder block table. Such reorganization was typical of database systems during the 1960s and would normally be done each night or weekly. 10.3 Tree-based Indexing Linear indexing is efficient when the database is static, that is, when records are inserted and deleted rarely or never. ISAM is adequate for a limited number of updates, but not for frequent changes. Because it has essentially two levels of indexing, ISAM will also break down for a truly large database where the number of cylinders is too great for the top-level index to fit in main memory. In their most general form, database applications have the following character- istics: 1. Large sets of records are frequently updated. 2. Search is by one or a combination of several keys. 3. Key range queries or min/max queries are used. For such databases, a better organization must be found. One approach would be to use the binary search tree (BST) to store primary and secondary key indices. BSTs can store duplicate key values, they provide efficient insertion and deletion as well as efficient search, and they can perform efficient range queries. When there
- 384. Sec. 10.3 Tree-based Indexing 365 is enough main memory, the BST is a viable option for implementing both primary and secondary key indices. Unfortunately, the BST can become unbalanced. Even under relatively good conditions, the depth of leaf nodes can easily vary by a factor of two. This might not be a significant concern when the tree is stored in main memory because the time required is still Θ(log n) for search and update. When the tree is stored on disk, however, the depth of nodes in the tree becomes crucial. Every time a BST node B is visited, it is necessary to visit all nodes along the path from the root to B. Each node on this path must be retrieved from disk. Each disk access returns a block of information. If a node is on the same block as its parent, then the cost to find that node is trivial once its parent is in main memory. Thus, it is desirable to keep subtrees together on the same block. Unfortunately, many times a node is not on the same block as its parent. Thus, each access to a BST node could potentially require that another block to be read from disk. Using a buffer pool to store multiple blocks in memory can mitigate disk access problems if BST accesses display good locality of reference. But a buffer pool cannot eliminate disk I/O entirely. The problem becomes greater if the BST is unbalanced, because nodes deep in the tree have the potential of causing many disk blocks to be read. Thus, there are two significant issues that must be addressed to have efficient search from a disk-based BST. The first is how to keep the tree balanced. The second is how to arrange the nodes on blocks so as to keep the number of blocks encountered on any path from the root to the leaves at a minimum. We could select a scheme for balancing the BST and allocating BST nodes to blocks in a way that minimizes disk I/O, as illustrated by Figure 10.7. However, maintaining such a scheme in the face of insertions and deletions is difficult. In particular, the tree should remain balanced when an update takes place, but doing so might require much reorganization. Each update should affect only a disk few blocks, or its cost will be too high. As you can see from Figure 10.8, adopting a rule such as requiring the BST to be complete can cause a great deal of rearranging of data within the tree. We can solve these problems by selecting another tree structure that automat- ically remains balanced after updates, and which is amenable to storing in blocks. There are a number of widely used balanced tree data structures, and there are also techniques for keeping BSTs balanced. Examples are the AVL and splay trees dis- cussed in Section 13.2. As an alternative, Section 10.4 presents the 2-3 tree, which has the property that its leaves are always at the same level. The main reason for discussing the 2-3 tree here in preference to the other balanced search trees is that
- 385. 366 Chap. 10 Indexing Figure 10.7 Breaking the BST into blocks. The BST is divided among disk blocks, each with space for three nodes. The path from the root to any leaf is contained on two blocks. 5 3 2 4 6 3 5 7 (a) (b) 7 4 2 6 1 Figure 10.8 An attempt to rebalance a BST after insertion can be expensive. (a) A BST with six nodes in the shape of a complete binary tree. (b) A node with value 1 is inserted into the BST of (a). To maintain both the complete binary tree shape and the BST property, a major reorganization of the tree is required. it naturally leads to the B-tree of Section 10.5, which is by far the most widely used indexing method today. 10.4 2-3 Trees This section presents a data structure called the 2-3 tree. The 2-3 tree is not a binary tree, but instead its shape obeys the following definition: 1. A node contains one or two keys. 2. Every internal node has either two children (if it contains one key) or three children (if it contains two keys). Hence the name. 3. All leaves are at the same level in the tree, so the tree is always height bal- anced. In addition to these shape properties, the 2-3 tree has a search tree property analogous to that of a BST. For every node, the values of all descendants in the left subtree are less than the value of the first key, while values in the center subtree
- 386. Sec. 10.4 2-3 Trees 367 33 23 30 48 18 12 20 21 31 24 15 45 10 47 52 50 Figure 10.9 A 2-3 tree. are greater than or equal to the value of the first key. If there is a right subtree (equivalently, if the node stores two keys), then the values of all descendants in the center subtree are less than the value of the second key, while values in the right subtree are greater than or equal to the value of the second key. To maintain these shape and search properties requires that special action be taken when nodes are inserted and deleted. The 2-3 tree has the advantage over the BST in that the 2-3 tree can be kept height balanced at relatively low cost. Figure 10.9 illustrates the 2-3 tree. Nodes are indicated as rectangular boxes with two key fields. (These nodes actually would contain complete records or point- ers to complete records, but the figures will show oly the keys.) Internal nodes with only two children have an empty right key field. Leaf nodes might contain either one or two keys. Figure 10.10 is a class declaration for the 2-3 tree node. Note that this sample declaration does not distinguish between leaf and internal nodes and so is space inefficient, because leaf nodes store three pointers each. The techniques of Section 5.3.1 can be applied here to implement separate internal and leaf node types. From the defining rules for 2-3 trees we can derive relationships between the number of nodes in the tree and the depth of the tree. A 2-3 tree of height k has at least 2k−1 leaves, because if every internal node has two children it degenerates to the shape of a complete binary tree. A 2-3 tree of height k has at most 3k−1 leaves, because each internal node can have at most three children. Searching for a value in a 2-3 tree is similar to searching in a BST. Search begins at the root. If the root does not contain the search key K, then the search progresses to the only subtree that can possibly contain K. The value(s) stored in the root node determine which is the correct subtree. For example, if searching for the value 30 in the tree of Figure 10.9, we begin with the root node. Because 30 is between 18 and 33, it can only be in the middle subtree. Searching the middle child of the root node yields the desired record. If searching for 15, then the first step is again to search the root node. Because 15 is less than 18, the first (left) branch is taken. At the next level, we take the second branch to the leaf node containing 15.
- 387. 368 Chap. 10 Indexing /** 2-3 tree node implementation */ class TTNodeK extends Comparable? super K,E { private K lkey; // The node’s left key private E lval; // The left record private K rkey; // The node’s right key private E rval; // The right record private TTNodeK,E left; // Pointer to left child private TTNodeK,E center; // Pointer to middle child private TTNodeK,E right; // Pointer to right child public TTNode() { center = left = right = null; } public TTNode(K lk, E lv, K rk, E rv, TTNodeK,E p1, TTNodeK,E p2, TTNodeK,E p3) { lkey = lk; rkey = rk; lval = lv; rval = rv; left = p1; center = p2; right = p3; } public boolean isLeaf() { return left == null; } public TTNodeK,E lchild() { return left; } public TTNodeK,E rchild() { return right; } public TTNodeK,E cchild() { return center; } public K lkey() { return lkey; } // Left key public E lval() { return lval; } // Left value public K rkey() { return rkey; } // Right key public E rval() { return rval; } // Right value public void setLeft(K k, E e) { lkey = k; lval = e; } public void setRight(K k, E e) { rkey = k; rval = e; } public void setLeftChild(TTNodeK,E it) { left = it; } public void setCenterChild(TTNodeK,E it) { center = it; } public void setRightChild(TTNodeK,E it) { right = it; } Figure 10.10 The 2-3 tree node implementation. If the search key were 16, then upon encountering the leaf containing 15 we would find that the search key is not in the tree. Below is an implementation for the 2-3 tree search method.
- 388. Sec. 10.4 2-3 Trees 369 12 10 20 21 33 23 30 24 31 50 18 45 48 47 52 15 15 14 Figure 10.11 Simple insert into the 2-3 tree of Figure 10.9. The value 14 is inserted into the tree at the leaf node containing 15. Because there is room in the node for a second key, it is simply added to the left position with 15 moved to the right position. private E findhelp(TTNodeK,E root, K k) { if (root == null) return null; // val not found if (k.compareTo(root.lkey()) == 0) return root.lval(); if ((root.rkey() != null) (k.compareTo(root.rkey()) == 0)) return root.rval(); if (k.compareTo(root.lkey()) 0) // Search left return findhelp(root.lchild(), k); else if (root.rkey() == null) // Search center return findhelp(root.cchild(), k); else if (k.compareTo(root.rkey()) 0) // Search center return findhelp(root.cchild(), k); else return findhelp(root.rchild(), k); // Search right } Insertion into a 2-3 tree is similar to insertion into a BST to the extent that the new record is placed in the appropriate leaf node. Unlike BST insertion, a new child is not created to hold the record being inserted, that is, the 2-3 tree does not grow downward. The first step is to find the leaf node that would contain the record if it were in the tree. If this leaf node contains only one value, then the new record can be added to that node with no further modification to the tree, as illustrated in Figure 10.11. In this example, a record with key value 14 is inserted. Searching from the root, we come to the leaf node that stores 15. We add 14 as the left value (pushing the record with key 15 to the rightmost position). If we insert the new record into a leaf node L that already contains two records, then more space must be created. Consider the two records of node L and the record to be inserted without further concern for which two were already in L and which is the new record. The first step is to split L into two nodes. Thus, a new node — call it L′ — must be created from free store. L receives the record with the least of the three key values. L′ receives the greatest of the three. The record
- 389. 370 Chap. 10 Indexing 33 15 23 30 48 52 45 47 50 55 10 12 18 20 21 24 31 Figure 10.12 A simple node-splitting insert for a 2-3 tree. The value 55 is added to the 2-3 tree of Figure 10.9. This makes the node containing values 50 and 52 split, promoting value 52 to the parent node. with the middle of the three key value is passed up to the parent node along with a pointer to L′ . This is called a promotion. The promoted key is then inserted into the parent. If the parent currently contains only one record (and thus has only two children), then the promoted record and the pointer to L′ are simply added to the parent node. If the parent is full, then the split-and-promote process is repeated. Figure 10.12 illustrates a simple promotion. Figure 10.13 illustrates what happens when promotions require the root to split, adding a new level to the tree. In either case, all leaf nodes continue to have equal depth. Figures 10.14 and 10.15 present an implementation for the insertion process. Note that inserthelp of Figure 10.14 takes three parameters. The first is a pointer to the root of the current subtree, named rt. The second is the key for the record to be inserted, and the third is the record itself. The return value for inserthelp is a pointer to a 2-3 tree node. If rt is unchanged, then a pointer to rt is returned. If rt is changed (due to the insertion causing the node to split), then a pointer to the new subtree root is returned, with the key value and record value in the leftmost fields, and a pointer to the (single) subtree in the center pointer field. This revised node will then be added to the paren, as illustrated in Figure 10.13. When deleting a record from the 2-3 tree, there are three cases to consider. The simplest occurs when the record is to be removed from a leaf node containing two records. In this case, the record is simply removed, and no other nodes are affected. The second case occurs when the only record in a leaf node is to be removed. The third case occurs when a record is to be removed from an internal node. In both the second and the third cases, the deleted record is replaced with another that can take its place while maintaining the correct order, similar to removing a node from a BST. If the tree is sparse enough, there is no such record available that will allow all nodes to still maintain at least one record. In this situation, sibling nodes are merged together. The delete operation for the 2-3 tree is excessively complex and will not be described further. Instead, a complete discussion of deletion will be
- 390. Sec. 10.4 2-3 Trees 371 23 20 (a) (b) (c) 30 20 24 31 21 24 31 21 19 19 12 10 19 24 30 31 33 45 47 50 52 23 18 20 21 48 15 30 23 33 18 Figure 10.13 Example of inserting a record that causes the 2-3 tree root to split. (a) The value 19 is added to the 2-3 tree of Figure 10.9. This causes the node containing 20 and 21 to split, promoting 20. (b) This in turn causes the internal node containing 23 and 30 to split, promoting 23. (c) Finally, the root node splits, promoting 23 to become the left record in the new root. The result is that the tree becomes one level higher. postponed until the next section, where it can be generalized for a particular variant of the B-tree. The 2-3 tree insert and delete routines do not add new nodes at the bottom of the tree. Instead they cause leaf nodes to split or merge, possibly causing a ripple effect moving up the tree to the root. If necessary the root will split, causing a new root node to be created and making the tree one level deeper. On deletion, if the last two children of the root merge, then the root node is removed and the tree will lose a level. In either case, all leaf nodes are always at the same level. When all leaf nodes are at the same level, we say that a tree is height balanced. Because the 2-3 tree is height balanced, and every internal node has at least two children, we know that the maximum depth of the tree is log n. Thus, all 2-3 tree insert, find, and delete operations require Θ(log n) time.
- 391. 372 Chap. 10 Indexing private TTNodeK,E inserthelp(TTNodeK,E rt, K k, E e) { TTNodeK,E retval; if (rt == null) // Empty tree: create a leaf node for root return new TTNodeK,E(k, e, null, null, null, null, null); if (rt.isLeaf()) // At leaf node: insert here return rt.add(new TTNodeK,E(k, e, null, null, null, null, null)); // Add to internal node if (k.compareTo(rt.lkey()) 0) { // Insert left retval = inserthelp(rt.lchild(), k, e); if (retval == rt.lchild()) return rt; else return rt.add(retval); } else if((rt.rkey() == null) || (k.compareTo(rt.rkey()) 0)) { retval = inserthelp(rt.cchild(), k, e); if (retval == rt.cchild()) return rt; else return rt.add(retval); } else { // Insert right retval = inserthelp(rt.cchild(), k, e); if (retval == rt.cchild()) return rt; else return rt.add(retval); } } Figure 10.14 The 2-3 tree insert routine. 10.5 B-Trees This section presents the B-tree. B-trees are usually attributed to R. Bayer and E. McCreight who described the B-tree in a 1972 paper. By 1979, B-trees had re- placed virtually all large-file access methods other than hashing. B-trees, or some variant of B-trees, are the standard file organization for applications requiring inser- tion, deletion, and key range searches. B-trees address effectively all of the major problems encountered when implementing disk-based search trees: 1. B-trees are always height balanced, with all leaf nodes at the same level. 2. Update and search operations affect only a few disk blocks. The fewer the number of disk blocks affected, the less disk I/O is required. 3. B-trees keep related records (that is, records with similar key values) on the same disk block, which helps to minimize disk I/O on searches due to locality of reference.
- 392. Sec. 10.5 B-Trees 373 /** Add a new key/value pair to the node. There might be a subtree associated with the record being added. This information comes in the form of a 2-3 tree node with one key and a (possibly null) subtree through the center pointer field. */ public TTNodeK,E add(TTNodeK,E it) { if (rkey == null) { // Only one key, add here if (lkey.compareTo(it.lkey()) 0) { rkey = it.lkey(); rval = it.lval(); right = center; center = it.cchild(); } else { rkey = lkey; rval = lval; right = center; lkey = it.lkey(); lval = it.lval(); center = it.cchild(); } return this; } else if (lkey.compareTo(it.lkey()) = 0) { // Add left center = new TTNodeK,E(rkey, rval, null, null, center, right, null); rkey = null; rval = null; right = null; it.setLeftChild(left); left = it; return this; } else if (rkey.compareTo(it.lkey()) 0) { // Add center it.setCenterChild(new TTNodeK,E(rkey, rval, null, null, it.cchild(), right, null)); it.setLeftChild(this); rkey = null; rval = null; right = null; return it; } else { // Add right TTNodeK,E N1 = new TTNodeK,E(rkey, rval, null, null, this, it, null); it.setLeftChild(right); right = null; rkey = null; rval = null; return N1; } } Figure 10.15 The 2-3 tree node add method.
- 393. 374 Chap. 10 Indexing 20 12 18 21 23 30 31 38 47 10 15 24 33 45 48 50 52 60 Figure 10.16 A B-tree of order four. 4. B-trees guarantee that every node in the tree will be full at least to a certain minimum percentage. This improves space efficiency while reducing the typical number of disk fetches necessary during a search or update operation. A B-tree of order m is defined to have the following shape properties: • The root is either a leaf or has at least two children. • Each internal node, except for the root, has between ⌈m/2⌉ and m children. • All leaves are at the same level in the tree, so the tree is always height bal- anced. The B-tree is a generalization of the 2-3 tree. Put another way, a 2-3 tree is a B-tree of order three. Normally, the size of a node in the B-tree is chosen to fill a disk block. A B-tree node implementation typically allows 100 or more children. Thus, a B-tree node is equivalent to a disk block, and a “pointer” value stored in the tree is actually the number of the block containing the child node (usually interpreted as an offset from the beginning of the corresponding disk file). In a typical application, B-tree block I/O will be managed using a buffer pool and a block-replacement scheme such as LRU (see Section 8.3). Figure 10.16 shows a B-tree of order four. Each node contains up to three keys, and internal nodes have up to four children. Search in a B-tree is a generalization of search in a 2-3 tree. It is an alternating two-step process, beginning with the root node of the B-tree. 1. Perform a binary search on the records in the current node. If a record with the search key is found, then return that record. If the current node is a leaf node and the key is not found, then report an unsuccessful search. 2. Otherwise, follow the proper branch and repeat the process. For example, consider a search for the record with key value 47 in the tree of Figure 10.16. The root node is examined and the second (right) branch taken. After
- 394. Sec. 10.5 B-Trees 375 examining the node at level 1, the third branch is taken to the next level to arrive at the leaf node containing a record with key value 47. B-tree insertion is a generalization of 2-3 tree insertion. The first step is to find the leaf node that should contain the key to be inserted, space permitting. If there is room in this node, then insert the key. If there is not, then split the node into two and promote the middle key to the parent. If the parent becomes full, then it is split in turn, and its middle key promoted. Note that this insertion process is guaranteed to keep all nodes at least half full. For example, when we attempt to insert into a full internal node of a B-tree of order four, there will now be five children that must be dealt with. The node is split into two nodes containing two keys each, thus retaining the B-tree property. The middle of the five children is promoted to its parent. 10.5.1 B+ -Trees The previous section mentioned that B-trees are universally used to implement large-scale disk-based systems. Actually, the B-tree as described in the previ- ous section is almost never implemented, nor is the 2-3 tree as described in Sec- tion 10.4. What is most commonly implemented is a variant of the B-tree, called the B+-tree. When greater efficiency is required, a more complicated variant known as the B∗-tree is used. When data are static, it is an extremely efficient way to search. The problem is those pesky inserts and deletes. Imagine that we want to keep the idea of storing a sorted list, but make it more flexible by breaking the list into manageable chunks that are more easily updated. How might we do that? First, we need to decide how big the chunks should be. Since the data are on disk, it seems reasonable to store a chunk that is the size of a disk block, or a small multiple of the disk block size. We could insert a new record with a chunk that hasn’t filled its block. But what if the chunk fills up the entire block that contains it? We could just split it in half. What if we want to delete a record? We could just take the deleted record out of the chunk, but we might not want a lot of near-empty chunks. So we could put adjacent chunks together if they have only a small amount of data between them. Or we could shuffle data between adjacent chunks that together contain more data. The big problem would be how to find the desired chunk when processing a record with a given key. Perhaps some sort of tree-like structure could be used to locate the appropriate chunk. These ideas are exactly what motivate the B+-tree. The B+-tree is essentially a mechanism for managing a list broken into chunks. The most significant difference between the B+-tree and the BST or the 2-3 tree is that the B+-tree stores records only at the leaf nodes. Internal nodes store key
- 395. 376 Chap. 10 Indexing values, but these are used solely as placeholders to guide the search. This means that internal nodes are significantly different in structure from leaf nodes. Inter- nal nodes store keys to guide the search, associating each key with a pointer to a child B+-tree node. Leaf nodes store actual records, or else keys and pointers to actual records in a separate disk file if the B+-tree is being used purely as an in- dex. Depending on the size of a record as compared to the size of a key, a leaf node in a B+-tree of order m might have enough room to store more or less than m records. The requirement is simply that the leaf nodes store enough records to remain at least half full. The leaf nodes of a B+-tree are normally linked together to form a doubly linked list. Thus, the entire collection of records can be traversed in sorted order by visiting all the leaf nodes on the linked list. Here is a Java-like pseudocode representation for the B+-tree node interface. Leaf node and internal node subclasses would implement this base class. /** Interface for B+ Tree nodes */ public interface BPNodeK,E { public boolean isLeaf(); public int numrecs(); public K[] keys(); } An important implementation detail to note is that while Figure 10.16 shows internal nodes containing three keys and four pointers, class BPNode is slightly different in that it stores key/pointer pairs. Figure 10.16 shows the B+-tree as it is traditionally drawn. To simplify implementation in practice, nodes really do associate a key with each pointer. Each internal node should be assumed to hold in the leftmost position an additional key that is less than or equal to any possible key value in the node’s leftmost subtree. B+-tree implementations typically store an additional dummy record in the leftmost leaf node whose key value is less than any legal key value. B+-trees are exceptionally good for range queries. Once the first record in the range has been found, the rest of the records with keys in the range can be accessed by sequential processing of the remaining records in the first node, and then continuing down the linked list of leaf nodes as far as necessary. Figure 10.17 illustrates the B+-tree. Search in a B+-tree is nearly identical to search in a regular B-tree, except that the search must always continue to the proper leaf node. Even if the search-key value is found in an internal node, this is only a placeholder and does not provide access to the actual record. To find a record with key value 33 in the B+-tree of Figure 10.17, search begins at the root. The value 33 stored in the root merely serves as a placeholder, indicating that keys with values greater than or equal to
- 396. Sec. 10.5 B-Trees 377 23 30 31 33 45 47 48 48 50 52 10 12 15 18 19 20 21 22 33 18 23 Figure 10.17 Example of a B+ -tree of order four. Internal nodes must store between two and four children. For this example, the record size is assumed to be such that leaf nodes store between three and five records. 33 are found in the second subtree. From the second child of the root, the first branch is taken to reach the leaf node containing the actual record (or a pointer to the actual record) with key value 33. Here is a pseudocode sketch of the B+-tree search algorithm: private E findhelp(BPNodeK,E rt, K k) { int currec = binaryle(rt.keys(), rt.numrecs(), k); if (rt.isLeaf()) if ((((BPLeafK,E)rt).keys())[currec].compareTo(k) == 0) return ((BPLeafK,E)rt).recs(currec); else return null; else return findhelp(((BPInternalK,E)rt).pointers(currec), k); } B+-tree insertion is similar to B-tree insertion. First, the leaf L that should contain the record is found. If L is not full, then the new record is added, and no other B+-tree nodes are affected. If L is already full, split it in two (dividing the records evenly among the two nodes) and promote a copy of the least-valued key in the newly formed right node. As with the 2-3 tree, promotion might cause the parent to split in turn, perhaps eventually leading to splitting the root and causing the B+-tree to gain a new level. B+-tree insertion keeps all leaf nodes at equal depth. Figure 10.18 illustrates the insertion process through several examples. Fig- ure 10.19 shows a Java-like pseudocode sketch of the B+-tree insert algorithm. To delete record R from the B+-tree, first locate the leaf L that contains R. If L is more than half full, then we need only remove R, leaving L still at least half full. This is demonstrated by Figure 10.20. If deleting a record reduces the number of records in the node below the min- imum threshold (called an underflow), then we must do something to keep the
- 397. 378 Chap. 10 Indexing 33 (b) (a) 1012 233348 10 23 33 50 12 48 33 18 (c) 33 23 48 18 (d) 48 1012 18 20 2123 31 33 45 47 48 50 15 52 12 18 20 21 23 30 31 33 45 47 10 15 48 50 52 Figure 10.18 Examples of B+ -tree insertion. (a) A B+ -tree containing five records. (b) The result of inserting a record with key value 50 into the tree of (a). The leaf node splits, causing creation of the first internal node. (c) The B+ -tree of (b) after further insertions. (d) The result of inserting a record with key value 30 into the tree of (c). The second leaf node splits, which causes the internal node to split in turn, creating a new root. private BPNodeK,E inserthelp(BPNodeK,E rt, K k, E e) { BPNodeK,E retval; if (rt.isLeaf()) // At leaf node: insert here return ((BPLeafK,E)rt).add(k, e); // Add to internal node int currec = binaryle(rt.keys(), rt.numrecs(), k); BPNodeK,E temp = inserthelp( ((BPInternalK,E)root).pointers(currec), k, e); if (temp != ((BPInternalK,E)rt).pointers(currec)) return ((BPInternalK,E)rt).add((BPInternalK,E)temp); else return rt; } Figure 10.19 A Java-like pseudocode sketch of the B+ -tree insert algorithm.
- 398. Sec. 10.5 B-Trees 379 33 23 48 18 101215 23 30 31 19 2021 22 47 33 45 48 50 52 Figure 10.20 Simple deletion from a B+ -tree. The record with key value 18 is removed from the tree of Figure 10.17. Note that even though 18 is also a place- holder used to direct search in the parent node, that value need not be removed from internal nodes even if no record in the tree has key value 18. Thus, the leftmost node at level one in this example retains the key with value 18 after the record with key value 18 has been removed from the second leaf node. 33 19 48 23 101518 19 20 21 22 33 45 47 23 30 31 48 50 52 Figure 10.21 Deletion from the B+ -tree of Figure 10.17 via borrowing from a sibling. The key with value 12 is deleted from the leftmost leaf, causing the record with key value 18 to shift to the leftmost leaf to take its place. Note that the parent must be updated to properly indicate the key range within the subtrees. In this example, the parent node has its leftmost key value changed to 19. node sufficiently full. The first choice is to look at the node’s adjacent siblings to determine if they have a spare record that can be used to fill the gap. If so, then enough records are transferred from the sibling so that both nodes have about the same number of records. This is done so as to delay as long as possible the next time when a delete causes this node to underflow again. This process might require that the parent node has its placeholder key value revised to reflect the true first key value in each node. Figure 10.21 illustrates the process. If neither sibling can lend a record to the underfull node (call it N), then N must give its records to a sibling and be removed from the tree. There is certainly room to do this, because the sibling is at most half full (remember that it had no records to contribute to the current node), and N has become less than half full because it is underflowing. This merge process combines two subtrees of the parent, which might cause it to underflow in turn. If the last two children of the root merge
- 399. 380 Chap. 10 Indexing 48 (a) 45 4748 50 52 23 33 18 (b) 18 19 20 21 23 30 31 101215 22 485052 4547 Figure 10.22 Deleting the record with key value 33 from the B+ -tree of Fig- ure 10.17 via collapsing siblings. (a) The two leftmost leafnodes merge together to form a single leaf. Unfortunately, the parent node now has only one child. (b) Because the left subtree has a spare leaf node, that node is passed to the right subtree. The placeholder values of the root and the right internal node are updated to reflect the changes. Value 23 moves to the root, and old root value 33 moves to the rightmost internal node. /** Delete a record with the given key value, and return true if the root underflows */ private boolean removehelp(BPNodeK,E rt, K k) { int currec = binaryle(rt.keys(), rt.numrecs(), k); if (rt.isLeaf()) if ((((BPLeafK,E)rt).keys()[currec]).compareTo(k) == 0) return ((BPLeafK,E)rt).delete(currec); else return false; else // Process internal node if (removehelp(((BPInternalK,E)rt).pointers(currec), k)) // Child will merge if necessary return ((BPInternalK,E)rt).underflow(currec); else return false; } Figure 10.23 Java-like pseudocode for the B+ -tree delete algorithm. together, then the tree loses a level. Figure 10.22 illustrates the node-merge deletion process. Figure 10.23 shows Java-like pseudocode for the B+-tree delete algorithm.
- 400. Sec. 10.5 B-Trees 381 The B+-tree requires that all nodes be at least half full (except for the root). Thus, the storage utilization must be at least 50%. This is satisfactory for many implementations, but note that keeping nodes fuller will result both in less space required (because there is less empty space in the disk file) and in more efficient processing (fewer blocks on average will be read into memory because the amount of information in each block is greater). Because B-trees have become so popular, many algorithm designers have tried to improve B-tree performance. One method for doing so is to use the B+-tree variant known as the B∗-tree. The B∗-tree is identical to the B+-tree, except for the rules used to split and merge nodes. Instead of splitting a node in half when it overflows, the B∗-tree gives some records to its neighboring sibling, if possible. If the sibling is also full, then these two nodes split into three. Similarly, when a node underflows, it is combined with its two siblings, and the total reduced to two nodes. Thus, the nodes are always at least two thirds full.2 10.5.2 B-Tree Analysis The asymptotic cost of search, insertion, and deletion of records from B-trees, B+-trees, and B∗-trees is Θ(log n) where n is the total number of records in the tree. However, the base of the log is the (average) branching factor of the tree. Typical database applications use extremely high branching factors, perhaps 100 or more. Thus, in practice the B-tree and its variants are extremely shallow. As an illustration, consider a B+-tree of order 100 and leaf nodes that contain up to 100 records. A one-level B+-tree can have at most 100 records. A two-level B+-tree must have at least 100 records (2 leaves with 50 records each). It has at most 10,000 records (100 leaves with 100 records each). A three-level B+-tree must have at least 5000 records (two second-level nodes with 50 children contain- ing 50 records each) and at most one million records (100 second-level nodes with 100 full children each). A four-level B+-tree must have at least 250,000 records and at most 100 million records. Thus, it would require an extremely large database to generate a B+-tree of more than four levels. We can reduce the number of disk fetches required for the B-tree even more by using the following methods. First, the upper levels of the tree can be stored in main memory at all times. Because the tree branches so quickly, the top two levels 2 This concept can be extended further if higher space utilization is required. However, the update routines become much more complicated. I once worked on a project where we implemented 3-for-4 node split and merge routines. This gave better performance than the 2-for-3 node split and merge routines of the B∗ -tree. However, the spitting and merging routines were so complicated that even their author could no longer understand them once they were completed!
- 401. 382 Chap. 10 Indexing (levels 0 and 1) require relatively little space. If the B-tree is only four levels deep, then at most two disk fetches (internal nodes at level two and leaves at level three) are required to reach the pointer to any given record. As mentioned earlier, a buffer pool should be used to manage nodes of the B-tree. Several nodes of the tree would typically be in main memory at one time. The most straightforward approach is to use a standard method such as LRU to do node replacement. However, sometimes it might be desirable to “lock” certain nodes such as the root into the buffer pool. In general, if the buffer pool is even of modest size (say at least twice the depth of the tree), no special techniques for node replacement will be required because the upper-level nodes will naturally be accessed frequently. 10.6 Further Reading For an expanded discussion of the issues touched on in this chapter, see a gen- eral file processing text such as File Structures: A Conceptual Toolkit by Folk and Zoellick [FZ98]. In particular, Folk and Zoellick provide a good discussion of the relationship between primary and secondary indices. The most thorough dis- cussion on various implementations for the B-tree is the survey article by Comer [Com79]. Also see [Sal88] for further details on implementing B-trees. See Shaf- fer and Brown [SB93] for a discussion of buffer pool management strategies for B+-tree-like data structures. 10.7 Exercises 10.1 Assume that a computer system has disk blocks of 1024 bytes, and that you are storing records that have 4-byte keys and 4-byte data fields. The records are sorted and packed sequentially into the disk file. (a) Assume that a linear index uses 4 bytes to store the key and 4 bytes to store the block ID for the associated records. What is the greatest number of records that can be stored in the file if a linear index of size 256KB is used? (b) What is the greatest number of records that can be stored in the file if the linear index is also stored on disk (and thus its size is limited only by the second-level index) when using a second-level index of 1024 bytes (i.e., 256 key values) as illustrated by Figure 10.2? Each element of the second-level index references the smallest key value for a disk block of the linear index.
- 402. Sec. 10.7 Exercises 383 10.2 Assume that a computer system has disk blocks of 4096 bytes, and that you are storing records that have 4-byte keys and 64-byte data fields. The records are sorted and packed sequentially into the disk file. (a) Assume that a linear index uses 4 bytes to store the key and 4 bytes to store the block ID for the associated records. What is the greatest number of records that can be stored in the file if a linear index of size 2MB is used? (b) What is the greatest number of records that can be stored in the file if the linear index is also stored on disk (and thus its size is limited only by the second-level index) when using a second-level index of 4096 bytes (i.e., 1024 key values) as illustrated by Figure 10.2? Each element of the second-level index references the smallest key value for a disk block of the linear index. 10.3 Modify the function binary of Section 3.5 so as to support variable-length records with fixed-length keys indexed by a simple linear index as illustrated by Figure 10.1. 10.4 Assume that a database stores records consisting of a 2-byte integer key and a variable-length data field consisting of a string. Show the linear index (as illustrated by Figure 10.1) for the following collection of records: 397 Hello world! 82 XYZ 1038 This string is rather long 1037 This is shorter 42 ABC 2222 Hello new world! 10.5 Each of the following series of records consists of a four-digit primary key (with no duplicates) and a four-character secondary key (with many dupli- cates). 3456 DEER 2398 DEER 2926 DUCK 9737 DEER 7739 GOAT 9279 DUCK 1111 FROG 8133 DEER 7183 DUCK 7186 FROG
- 403. 384 Chap. 10 Indexing (a) Show the inverted list (as illustrated by Figure 10.4) for this collection of records. (b) Show the improved inverted list (as illustrated by Figure 10.5) for this collection of records. 10.6 Under what conditions will ISAM be more efficient than a B+-tree imple- mentation? 10.7 Prove that the number of leaf nodes in a 2-3 tree with k levels is between 2k−1 and 3k−1. 10.8 Show the result of inserting the values 55 and 46 into the 2-3 tree of Fig- ure 10.9. 10.9 You are given a series of records whose keys are letters. The records arrive in the following order: C, S, D, T, A, M, P, I, B, W, N, G, U, R, K, E, H, O, L, J. Show the 2-3 tree that results from inserting these records. 10.10 You are given a series of records whose keys are letters. The records are inserted in the following order: C, S, D, T, A, M, P, I, B, W, N, G, U, R, K, E, H, O, L, J. Show the tree that results from inserting these records when the 2-3 tree is modified to be a 2-3+ tree, that is, the internal nodes act only as placeholders. Assume that the leaf nodes are capable of holding up to two records. 10.11 Show the result of inserting the value 55 into the B-tree of Figure 10.16. 10.12 Show the result of inserting the values 1, 2, 3, 4, 5, and 6 (in that order) into the B+-tree of Figure 10.17. 10.13 Show the result of deleting the values 18, 19, and 20 (in that order) from the B+-tree of Figure 10.22b. 10.14 You are given a series of records whose keys are letters. The records are inserted in the following order: C, S, D, T, A, M, P, I, B, W, N, G, U, R, K, E, H, O, L, J. Show the B+-tree of order four that results from inserting these records. Assume that the leaf nodes are capable of storing up to three records. 10.15 Assume that you have a B+-tree whose internal nodes can store up to 100 children and whose leaf nodes can store up to 15 records. What are the minimum and maximum number of records that can be stored by the B+-tree for 1, 2, 3, 4, and 5 levels? 10.16 Assume that you have a B+-tree whose internal nodes can store up to 50 children and whose leaf nodes can store up to 50 records. What are the minimum and maximum number of records that can be stored by the B+-tree for 1, 2, 3, 4, and 5 levels?
- 404. Sec. 10.8 Projects 385 10.8 Projects 10.1 Implement a two-level linear index for variable-length records as illustrated by Figures 10.1 and 10.2. Assume that disk blocks are 1024 bytes in length. Records in the database file should typically range between 20 and 200 bytes, including a 4-byte key value. Each record of the index file should store a key value and the byte offset in the database file for the first byte of the corresponding record. The top-level index (stored in memory) should be a simple array storing the lowest key value on the corresponding block in the index file. 10.2 Implement the 2-3+ tree, that is, a 2-3 tree where the internal nodes act only as placeholders. Your 2-3+ tree should implement the dictionary interface of Section 4.4. 10.3 Implement the dictionary ADT of Section 4.4 for a large file stored on disk by means of the B+-tree of Section 10.5. Assume that disk blocks are 1024 bytes, and thus both leaf nodes and internal nodes are also 1024 bytes. Records should store a 4-byte (int) key value and a 60-byte data field. Inter- nal nodes should store key value/pointer pairs where the “pointer” is actually the block number on disk for the child node. Both internal nodes and leaf nodes will need room to store various information such as a count of the records stored on that node, and a pointer to the next node on that level. Thus, leaf nodes will store 15 records, and internal nodes will have room to store about 120 to 125 children depending on how you implement them. Use a buffer pool (Section 8.3) to manage access to the nodes stored on disk.
- 406. PART IV Advanced Data Structures 387
- 408. 11 Graphs Graphs provide the ultimate in data structure flexibility. Graphs can model both real-world systems and abstract problems, so they are used in hundreds of appli- cations. Here is a small sampling of the range of problems that graphs are applied to. 1. Modeling connectivity in computer and communications networks. 2. Representing a map as a set of locations with distances between locations; used to compute shortest routes between locations. 3. Modeling flow capacities in transportation networks. 4. Finding a path from a starting condition to a goal condition; for example, in artificial intelligence problem solving. 5. Modeling computer algorithms, showing transitions from one program state to another. 6. Finding an acceptable order for finishing subtasks in a complex activity, such as constructing large buildings. 7. Modeling relationships such as family trees, business or military organiza- tions, and scientific taxonomies. We begin in Section 11.1 with some basic graph terminology and then define two fundamental representations for graphs, the adjacency matrix and adjacency list. Section 11.2 presents a graph ADT and simple implementations based on the adjacency matrix and adjacency list. Section 11.3 presents the two most commonly used graph traversal algorithms, called depth-first and breadth-first search, with application to topological sorting. Section 11.4 presents algorithms for solving some problems related to finding shortest routes in a graph. Finally, Section 11.5 presents algorithms for finding the minimum-cost spanning tree, useful for deter- mining lowest-cost connectivity in a network. Besides being useful and interesting in their own right, these algorithms illustrate the use of some data structures pre- sented in earlier chapters. 389
- 409. 390 Chap. 11 Graphs (b) (c) 0 3 4 1 2 7 1 2 3 4 (a) 1 Figure 11.1 Examples of graphs and terminology. (a) A graph. (b) A directed graph (digraph). (c) A labeled (directed) graph with weights associated with the edges. In this example, there is a simple path from Vertex 0 to Vertex 3 containing Vertices 0, 1, and 3. Vertices 0, 1, 3, 2, 4, and 1 also form a path, but not a simple path because Vertex 1 appears twice. Vertices 1, 3, 2, 4, and 1 form a simple cycle. 11.1 Terminology and Representations A graph G = (V, E) consists of a set of vertices V and a set of edges E, such that each edge in E is a connection between a pair of vertices in V.1 The number of vertices is written |V|, and the number of edges is written |E|. |E| can range from zero to a maximum of |V|2 − |V|. A graph with relatively few edges is called sparse, while a graph with many edges is called dense. A graph containing all possible edges is said to be complete. A graph with edges directed from one vertex to another (as in Figure 11.1(b)) is called a directed graph or digraph. A graph whose edges are not directed is called an undirected graph (as illustrated by Figure 11.1(a)). A graph with labels associated with its vertices (as in Figure 11.1(c)) is called a labeled graph. Two vertices are adjacent if they are joined by an edge. Such vertices are also called neighbors. An edge connecting Vertices U and V is written (U, V). Such an edge is said to be incident on Vertices U and V. Associated with each edge may be a cost or weight. Graphs whose edges have weights (as in Figure 11.1(c)) are said to be weighted. A sequence of vertices v1, v2, ..., vn forms a path of length n − 1 if there exist edges from vi to vi+1 for 1 ≤ i n. A path is simple if all vertices on the path are distinct. The length of a path is the number of edges it contains. A cycle is a path of length three or more that connects some vertex v1 to itself. A cycle is simple if the path is simple, except for the first and last vertices being the same. 1 Some graph applications require that a given pair of vertices can have multiple edges connecting them, or that a vertex can have an edge to itself. However, the applications discussed in this book do not require either of these special cases, so for simplicity we will assume that they cannot occur.
- 410. Sec. 11.1 Terminology and Representations 391 0 2 4 1 3 6 5 7 Figure 11.2 An undirected graph with three connected components. Vertices 0, 1, 2, 3, and 4 form one connected component. Vertices 5 and 6 form a second connected component. Vertex 7 by itself forms a third connected component. A subgraph S is formed from graph G by selecting a subset Vs of G’s vertices and a subset Es of G’s edges such that for every edge E in Es, both of its vertices are in Vs. An undirected graph is connected if there is at least one path from any vertex to any other. The maximally connected subgraphs of an undirected graph are called connected components. For example, Figure 11.2 shows an undirected graph with three connected components. A graph without cycles is called acyclic. Thus, a directed graph without cycles is called a directed acyclic graph or DAG. A free tree is a connected, undirected graph with no simple cycles. An equiv- alent definition is that a free tree is connected and has |V| − 1 edges. There are two commonly used methods for representing graphs. The adja- cency matrix is illustrated by Figure 11.3(b). The adjacency matrix for a graph is a |V| × |V| array. Assume that |V| = n and that the vertices are labeled from v0 through vn−1. Row i of the adjacency matrix contains entries for Vertex vi. Column j in row i is marked if there is an edge from vi to vj and is not marked oth- erwise. Thus, the adjacency matrix requires one bit at each position. Alternatively, if we wish to associate a number with each edge, such as the weight or distance between two vertices, then each matrix position must store that number. In either case, the space requirements for the adjacency matrix are Θ(|V|2). The second common representation for graphs is the adjacency list, illustrated by Figure 11.3(c). The adjacency list is an array of linked lists. The array is |V| items long, with position i storing a pointer to the linked list of edges for Ver- tex vi. This linked list represents the edges by the vertices that are adjacent to Vertex vi. The adjacency list is therefore a generalization of the “list of children” representation for trees described in Section 6.3.1.
- 411. 392 Chap. 11 Graphs (a) (b) 0 4 2 3 0 1 2 3 4 0 1 2 3 4 1 1 1 1 1 1 1 (c) 0 1 2 3 4 1 3 4 2 1 4 Figure 11.3 Two graph representations. (a) A directed graph. (b) The adjacency matrix for the graph of (a). (c) The adjacency list for the graph of (a). Example 11.1 The entry for Vertex 0 in Figure 11.3(c) stores 1 and 4 because there are two edges in the graph leaving Vertex 0, with one going to Vertex 1 and one going to Vertex 4. The list for Vertex 2 stores an entry for Vertex 4 because there is an edge from Vertex 2 to Vertex 4, but no entry for Vertex 3 because this edge comes into Vertex 2 rather than going out. The storage requirements for the adjacency list depend on both the number of edges and the number of vertices in the graph. There must be an array entry for each vertex (even if the vertex is not adjacent to any other vertex and thus has no elements on its linked list), and each edge must appear on one of the lists. Thus, the cost is Θ(|V| + |E|). Both the adjacency matrix and the adjacency list can be used to store directed or undirected graphs. Each edge of an undirected graph connecting Vertices U and V is represented by two directed edges: one from U to V and one from V to U. Figure 11.4 illustrates the use of the adjacency matrix and the adjacency list for undirected graphs.
- 412. Sec. 11.1 Terminology and Representations 393 (a) (b) (c) 0 1 2 3 0 1 2 3 4 0 1 2 3 4 1 1 1 1 1 1 1 1 1 1 1 1 0 1 3 4 1 0 1 0 4 3 4 2 1 4 2 4 3 2 Figure 11.4 Using the graph representations for undirected graphs. (a) An undi- rected graph. (b) The adjacency matrix for the graph of (a). (c) The adjacency list for the graph of (a). Which graph representation is more space efficient depends on the number of edges in the graph. The adjacency list stores information only for those edges that actually appear in the graph, while the adjacency matrix requires space for each potential edge, whether it exists or not. However, the adjacency matrix requires no overhead for pointers, which can be a substantial cost, especially if the only information stored for an edge is one bit to indicate its existence. As the graph be- comes denser, the adjacency matrix becomes relatively more space efficient. Sparse graphs are likely to have their adjacency list representation be more space efficient. Example 11.2 Assume that the vertex index requires two bytes, a pointer requires four bytes, and an edge weight requires two bytes. Then the adja- cency matrix for the graph of Figure 11.3 requires 2|V2 | = 50 bytes while the adjacency list requires 4|V| + 6|E| = 56 bytes. For the graph of Fig- ure 11.4, the adjacency matrix requires the same space as before, while the adjacency list requires 4|V| + 6|E| = 92 bytes (Because there are now 12 edges instead of 6).
- 413. 394 Chap. 11 Graphs interface Graph { // Graph class ADT public void Init(int n); // Initialize to n vertices public int n(); // Number of vertices public int e(); // Number of edges public int first(int v); // Get v’s first neighbor public int next(int v, int w); // Get v’s next neighbor public void setEdge(int i, int j, int wght); // Set weight public void delEdge(int i, int j); // Delete edge (i, j) public boolean isEdge(int i, int j); // If (i,j) an edge? public int weight(int i, int j); // Return edge weight public void setMark(int v, int val); // Set Mark for v public int getMark(int v); // Get Mark for v } Figure 11.5 A graph ADT. This ADT assumes that the number of vertices is fixed when the graph is created, but that edges can be added and removed. It also supports a mark array to aid graph traversal algorithms. The adjacency matrix often requires a higher asymptotic cost for an algorithm than would result if the adjacency list were used. The reason is that it is common for a graph algorithm to visit each neighbor of each vertex. Using the adjacency list, only the actual edges connecting a vertex to its neighbors are examined. However, the adjacency matrix must look at each of its |V| potential edges, yielding a total cost of Θ(|V2 |) time when the algorithm might otherwise require only Θ(|V|+|E|) time. This is a considerable disadvantage when the graph is sparse, but not when the graph is closer to full. 11.2 Graph Implementations We next turn to the problem of implementing a graph class. Figure 11.5 shows an abstract class defining an ADT for graphs. Vertices are defined by an integer index value. In other words, there is a Vertex 0, Vertex 1, and so on. We can assume that a graph application stores any additional information of interest about a given vertex elsewhere, such as a name or application-dependent value. Note that this ADT is not implemented using a template, because it is the Graph class users’ responsibility to maintain information related to the vertices themselves. The Graph class has no knowledge of the type or content of the information associated with a vertex, only the index number for that vertex. Abstract class Graph has methods to return the number of vertices and edges (methods n and e, respectively). Function weight returns the weight of a given edge, with that edge identified by its two incident vertices. For example, calling weight(0, 4) on the graph of Figure 11.1 (c) would return 4. If no such edge
- 414. Sec. 11.2 Graph Implementations 395 exists, the weight is defined to be 0. So calling weight(0, 2) on the graph of Figure 11.1 (c) would return 0. Functions setEdge and delEdge set the weight of an edge and remove an edge from the graph, respectively. Again, an edge is identified by its two incident vertices. setEdge does not permit the user to set the weight to be 0, because this value is used to indicate a non-existent edge, nor are negative edge weights per- mitted. Functions getMark and setMark get and set, respectively, a requested value in the Mark array (described below) for Vertex V. Nearly every graph algorithm presented in this chapter will require visits to all neighbors of a given vertex. Two methods are provided to support this. They work in a manner similar to linked list access functions. Function first takes as input a vertex V, and returns the edge to the first neighbor for V (we assume the neighbor list is sorted by vertex number). Function next takes as input Vertices V1 and V2 and returns the index for the vertex forming the next edge with V1 after V2 on V1’s edge list. Function next will return a value of n = |V| once the end of the edge list for V1 has been reached. The following line appears in many graph algorithms: for (w = G=first(v); w G-n(); w = G-next(v,w)) This for loop gets the first neighbor of v, then works through the remaining neigh- bors of v until a value equal to G-n() is returned, signaling that all neighbors of v have been visited. For example, first(1) in Figure 11.4 would return 0. next(1, 0) would return 3. next(0, 3) would return 4. next(1, 4) would return 5, which is not a vertex in the graph. It is reasonably straightforward to implement our graph and edge ADTs using either the adjacency list or adjacency matrix. The sample implementations pre- sented here do not address the issue of how the graph is actually created. The user of these implementations must add functionality for this purpose, perhaps reading the graph description from a file. The graph can be built up by using the setEdge function provided by the ADT. Figure 11.6 shows an implementation for the adjacency matrix. Array Mark stores the information manipulated by the setMark and getMark functions. The edge matrix is implemented as an integer array of size n × n for a graph of n ver- tices. Position (i, j) in the matrix stores the weight for edge (i, j) if it exists. A weight of zero for edge (i, j) is used to indicate that no edge connects Vertices i and j. Given a vertex V, function first locates the position in matrix of the first edge (if any) of V by beginning with edge (V, 0) and scanning through row V until an edge is found. If no edge is incident on V, then first returns n.
- 415. 396 Chap. 11 Graphs class Graphm implements Graph { // Graph: Adjacency matrix private int[][] matrix; // The edge matrix private int numEdge; // Number of edges public int[] Mark; // The mark array public Graphm() {} public Graphm(int n) { // Constructor Init(n); } public void Init(int n) { Mark = new int[n]; matrix = new int[n][n]; numEdge = 0; } public int n() { return Mark.length; } // # of vertices public int e() { return numEdge; } // # of edges public int first(int v) { // Get v’s first neighbor for (int i=0; iMark.length; i++) if (matrix[v][i] != 0) return i; return Mark.length; // No edge for this vertex } public int next(int v, int w) { // Get v’s next edge for (int i=w+1; iMark.length; i++) if (matrix[v][i] != 0) return i; return Mark.length; // No next edge; } public boolean isEdge(int i, int j) // Is (i, j) an edge? { return matrix[i][j] != 0; } // Set edge weight public void setEdge(int i, int j, int wt) { assert wt!=0 : Cannot set weight to 0; if (matrix[i][j] == 0) numEdge++; matrix[i][j] = wt; } public void delEdge(int i, int j) { // Delete edge (i, j) if (matrix[i][j] != 0) { matrix[i][j] = 0; numEdge--; } } public int weight(int i, int j) { // Return edge weight if (i == j) return 0; if (matrix[i][j] == 0) return Integer.MAX VALUE; return matrix[i][j]; } // Get and set marks public void setMark(int v, int val) { Mark[v] = val; } public int getMark(int v) { return Mark[v]; } } Figure 11.6 An implementation for the adjacency matrix implementation.
- 416. Sec. 11.3 Graph Traversals 397 Function next locates the edge following edge (i, j) (if any) by continuing down the row of Vertex i starting at position j + 1, looking for an edge. If no such edge exists, next returns n. Functions setEdge and delEdge adjust the appropriate value in the array. Function weight returns the value stored in the appropriate position in the array. Figure 11.7 presents an implementation of the adjacency list representation for graphs. Its main data structure is an array of linked lists, one linked list for each vertex. These linked lists store objects of type Edge, which merely stores the index for the vertex pointed to by the edge, along with the weight of the edge. // Edge class for Adjacency List graph representation class Edge { private int vert, wt; public Edge(int v, int w) // Constructor { vert = v; wt = w; } public int vertex() { return vert; } public int weight() { return wt; } } Implementation for Graphl member functions is straightforward in principle, with the key functions being setEdge, delEdge, and weight. The simplest implementation would start at the beginning of the adjacency list and move along it until the desired vertex has been found. However, many graph algorithms work by taking advantage of the first and next functions to process all edges extending from a given vertex in turn. Thus, there is a significant time savings if setEdge, delEdge, and weight first check to see if the desired edge is the current one on the relevant linked list. The implementation of Figure 11.7 does exactly this. 11.3 Graph Traversals Often it is useful to visit the vertices of a graph in some specific order based on the graph’s topology. This is known as a graph traversal and is similar in concept to a tree traversal. Recall that tree traversals visit every node exactly once, in some specified order such as preorder, inorder, or postorder. Multiple tree traversals exist because various applications require the nodes to be visited in a particular order. For example, to print a BST’s nodes in ascending order requires an inorder traver- sal as opposed to some other traversal. Standard graph traversal orders also exist. Each is appropriate for solving certain problems. For example, many problems in artificial intelligence programming are modeled using graphs. The problem domain may consist of a large collection of states, with connections between various pairs
- 417. 398 Chap. 11 Graphs // Adjacency list graph implementation class Graphl implements Graph { private GraphList[] vertex; // The vertex list private int numEdge; // Number of edges public int[] Mark; // The mark array public Graphl() {} public Graphl(int n) // Constructor { Init(n); } public void Init(int n) { Mark = new int[n]; vertex = new GraphList[n]; for (int i=0; in; i++) vertex[i] = new GraphList(); numEdge = 0; } public int n() { return Mark.length; } // # of vertices public int e() { return numEdge; } // # of edges public int first(int v) { // Get v’s first neighbor vertex[v].moveToStart(); Edge it = vertex[v].getValue(); if (it == null) return Mark.length; else return it.vertex(); } public boolean isEdge(int v, int w) { // Is (i,j) an edge? Edge it = vertex[v].getValue(); if ((it != null) (it.vertex() == w)) return true; for (vertex[v].moveToStart(); vertex[v].currPos() vertex[v].length(); vertex[v].next()) // Check whole list if (vertex[v].getValue().vertex() == w) return true; return false; } public int next(int v, int w) { // Get next neighbor Edge it = null; if (isEdge(v, w)) { vertex[v].next(); it = vertex[v].getValue(); } if (it != null) return it.vertex(); else return Mark.length; } Figure 11.7 An implementation for the adjacency list.
- 418. Sec. 11.3 Graph Traversals 399 // Store edge weight public void setEdge(int i, int j, int weight) { assert weight != 0 : May not set weight to 0; Edge currEdge = new Edge(j, weight); if (isEdge(i, j)) { // Edge already exists in graph vertex[i].remove(); vertex[i].insert(currEdge); } else { // Keep neighbors sorted by vertex index numEdge++; for (vertex[i].moveToStart(); vertex[i].currPos() vertex[i].length(); vertex[i].next()) if (vertex[i].getValue().vertex() j) break; vertex[i].insert(currEdge); } } public void delEdge(int i, int j) // Delete edge { if (isEdge(i, j)) { vertex[i].remove(); numEdge--; } } public int weight(int i, int j) { // Return weight of edge if (i == j) return 0; if (isEdge(i, j)) return vertex[i].getValue().weight(); else return Integer.MAX VALUE; } // Set and get marks public void setMark(int v, int val) { Mark[v] = val; } public int getMark(int v) { return Mark[v]; } } Figure 11.7 (continued) of states. Solving the problem may require getting from a specified start state to a specified goal state by moving between states only through the connections. Typi- cally, the start and goal states are not directly connected. To solve this problem, the vertices of the graph must be searched in some organized manner. Graph traversal algorithms typically begin with a start vertex and attempt to visit the remaining vertices from there. Graph traversals must deal with a number of troublesome cases. First, it may not be possible to reach all vertices from the start vertex. This occurs when the graph is not connected. Second, the graph may contain cycles, and we must make sure that cycles do not cause the algorithm to go into an infinite loop. Graph traversal algorithms can solve both of these problems by maintaining a mark bit for each vertex on the graph. At the beginning of the algorithm, the mark
- 419. 400 Chap. 11 Graphs bit for all vertices is cleared. The mark bit for a vertex is set when the vertex is first visited during the traversal. If a marked vertex is encountered during traversal, it is not visited a second time. This keeps the program from going into an infinite loop when it encounters a cycle. Once the traversal algorithm completes, we can check to see if all vertices have been processed by checking the mark bit array. If not all vertices are marked, we can continue the traversal from another unmarked vertex. Note that this process works regardless of whether the graph is directed or undirected. To ensure visiting all vertices, graphTraverse could be called as follows on a graph G: void graphTraverse(Graph G) { int v; for (v=0; vG.n(); v++) G.setMark(v, UNVISITED); // Initialize for (v=0; vG.n(); v++) if (G.getMark(v) == UNVISITED) doTraverse(G, v); } Function “doTraverse” might be implemented by using one of the graph traver- sals described in this section. 11.3.1 Depth-First Search The first method of organized graph traversal is called depth-first search (DFS). Whenever a vertex V is visited during the search, DFS will recursively visit all of V’s unvisited neighbors. Equivalently, DFS will add all edges leading out of v to a stack. The next vertex to be visited is determined by popping the stack and following that edge. The effect is to follow one branch through the graph to its conclusion, then it will back up and follow another branch, and so on. The DFS process can be used to define a depth-first search tree. This tree is composed of the edges that were followed to any new (unvisited) vertex during the traversal and leaves out the edges that lead to already visited vertices. DFS can be applied to directed or undirected graphs. Here is an implementation for the DFS algorithm: static void DFS(Graph G, int v) { // Depth first search PreVisit(G, v); // Take appropriate action G.setMark(v, VISITED); for (int w = G.first(v); w G.n() ; w = G.next(v, w)) if (G.getMark(w) == UNVISITED) DFS(G, w); PostVisit(G, v); // Take appropriate action }
- 420. Sec. 11.3 Graph Traversals 401 (a) (b) A B D F A B C D F E C E Figure 11.8 (a) A graph. (b) The depth-first search tree for the graph when starting at Vertex A. This implementation contains calls to functions PreVisit and PostVisit. These functions specify what activity should take place during the search. Just as a preorder tree traversal requires action before the subtrees are visited, some graph traversals require that a vertex be processed before ones further along in the DFS. Alternatively, some applications require activity after the remaining vertices are processed; hence the call to function PostVisit. This would be a natural opportunity to make use of the visitor design pattern described in Section 1.3.2. Figure 11.8 shows a graph and its corresponding depth-first search tree. Fig- ure 11.9 illustrates the DFS process for the graph of Figure 11.8(a). DFS processes each edge once in a directed graph. In an undirected graph, DFS processes each edge from both directions. Each vertex must be visited, but only once, so the total cost is Θ(|V| + |E|). 11.3.2 Breadth-First Search Our second graph traversal algorithm is known as a breadth-first search (BFS). BFS examines all vertices connected to the start vertex before visiting vertices fur- ther away. BFS is implemented similarly to DFS, except that a queue replaces the recursion stack. Note that if the graph is a tree and the start vertex is at the root, BFS is equivalent to visiting vertices level by level from top to bottom. Fig- ure 11.10 provides an implementation for the BFS algorithm. Figure 11.11 shows a graph and the corresponding breadth-first search tree. Figure 11.12 illustrates the BFS process for the graph of Figure 11.11(a).
- 421. 402 Chap. 11 Graphs Call DFS on A Mark B Process (B, C) Process (B, F) Print (B, F) and call DFS on F Process (F, E) Print (F, E) and call DFS on E Done with B Pop B Mark A Process (A, C) Print (A, C) and call DFS on C Mark F Process (F, B) Process (F, C) Process (F, D) Print (F, D) and call DFS on D Mark E Process (E, A) Process (E, F) Pop E Continue with C Process (C, E) Process (C, F) Pop C Mark C Process (C, A) Process (C, B) Print (C, B) and call DFS on C Mark D Done with F Pop F Continue with A Process (A, E) Pop A DFS complete Pop D Process (D, C) Process (D, F) E F B C A A F B C A C A F B C A C A D F B C A A B C A B C A F B C A Figure 11.9 A detailed illustration of the DFS process for the graph of Fig- ure 11.8(a) starting at Vertex A. The steps leading to each change in the recursion stack are described.
- 422. Sec. 11.3 Graph Traversals 403 // Breadth first (queue-based) search static void BFS(Graph G, int start) { QueueInteger Q = new AQueueInteger(G.n()); Q.enqueue(start); G.setMark(start, VISITED); while (Q.length() 0) { // Process each vertex on Q int v = Q.dequeue(); PreVisit(G, v); // Take appropriate action for (int w = G.first(v); w G.n(); w = G.next(v, w)) if (G.getMark(w) == UNVISITED) { // Put neighbors on Q G.setMark(w, VISITED); Q.enqueue(w); } PostVisit(G, v); // Take appropriate action } } Figure 11.10 Implementation for the breadth-first graph traversal algorithm (a) (b) B C A C B D D F E E A F Figure 11.11 (a) A graph. (b) The breadth-first search tree for the graph when starting at Vertex A.
- 423. 404 Chap. 11 Graphs Initial call to BFS on A. Mark A and put on the queue. Dequeue A. Process (A, C). Mark and enqueue C. Print (A, C). Process (A, E). Mark and enqueue E. Print(A, E). Dequeue C. Process (C, A). Ignore. Process (C, B). Mark and enqueue B. Print (C, B). Process (C, D). Mark and enqueue D. Print (C, D). Process (C, F). Mark and enqueue F. Print (C, F). Dequeue E. Process (E, A). Ignore. Process (E, F). Ignore. Dequeue B. Process (B, C). Ignore. Process (B, F). Ignore. Dequeue D. Process (D, C). Ignore. Process (D, F). Ignore. Dequeue F. Process (F, B). Ignore. Process (F, C). Ignore. Process (F, D). Ignore. BFS is complete. A E B D F D F C E B D F F Figure 11.12 A detailed illustration of the BFS process for the graph of Fig- ure 11.11(a) starting at Vertex A. The steps leading to each change in the queue are described.
- 424. Sec. 11.3 Graph Traversals 405 11.3.3 Topological Sort Assume that we need to schedule a series of tasks, such as classes or construction jobs, where we cannot start one task until after its prerequisites are completed. We wish to organize the tasks into a linear order that allows us to complete them one at a time without violating any prerequisites. We can model the problem using a DAG. The graph is directed because one task is a prerequisite of another — the vertices have a directed relationship. It is acyclic because a cycle would indicate a conflicting series of prerequisites that could not be completed without violating at least one prerequisite. The process of laying out the vertices of a DAG in a linear order to meet the prerequisite rules is called a topological sort. Figure 11.13 illustrates the problem. An acceptable topological sort for this example is J1, J2, J3, J4, J5, J6, J7. A topological sort may be found by performing a DFS on the graph. When a vertex is visited, no action is taken (i.e., function PreVisit does nothing). When the recursion pops back to that vertex, function PostVisit prints the vertex. This yields a topological sort in reverse order. It does not matter where the sort starts, as long as all vertices are visited in the end. Here is an implementation for the DFS-based algorithm. static void topsort(Graph G) { // Recursive topological sort for (int i=0; iG.n(); i++) // Initialize Mark array G.setMark(i, UNVISITED); for (int i=0; iG.n(); i++) // Process all vertices if (G.getMark(i) == UNVISITED) tophelp(G, i); // Recursive helper function } // Topsort helper function static void tophelp(Graph G, int v) { G.setMark(v, VISITED); for (int w = G.first(v); w G.n(); w = G.next(v, w)) if (G.getMark(w) == UNVISITED) tophelp(G, w); printout(v); // PostVisit for Vertex v } Using this algorithm starting at J1 and visiting adjacent neighbors in alphabetic order, vertices of the graph in Figure 11.13 are printed out in the order J7, J5, J4, J6, J2, J3, J1. When reversed, this yields the legal topological sort J1, J3, J2, J6, J4, J5, J7. We can also implement topological sort using a queue instead of recursion. To do so, we first visit all edges, counting the number of edges that lead to each vertex (i.e., count the number of prerequisites for each vertex). All vertices with no
- 425. 406 Chap. 11 Graphs J1 J2 J3 J4 J5 J7 J6 Figure 11.13 An example graph for topological sort. Seven tasks have depen- dencies as shown by the directed graph. static void topsort(Graph G) { // Topological sort: Queue QueueInteger Q = new AQueueInteger(G.n()); int[] Count = new int[G.n()]; int v; for (v=0; vG.n(); v++) Count[v] = 0; // Initialize for (v=0; vG.n(); v++) // Process every edge for (int w = G.first(v); w G.n(); w = G.next(v, w)) Count[w]++; // Add to v2’s prereq count for (v=0; vG.n(); v++) // Initialize Queue if (Count[v] == 0) // V has no prerequisites Q.enqueue(v); while (Q.length() 0) { // Process the vertices v = Q.dequeue().intValue(); printout(v); // PreVisit for Vertex V for (int w = G.first(v); w G.n(); w = G.next(v, w)) { Count[w]--; // One less prerequisite if (Count[w] == 0) // This vertex is now free Q.enqueue(w); } } } Figure 11.14 A queue-based topological sort algorithm. prerequisites are placed on the queue. We then begin processing the queue. When Vertex V is taken off of the queue, it is printed, and all neighbors of V (that is, all vertices that have V as a prerequisite) have their counts decremented by one. Any neighbor whose count is now zero is placed on the queue. If the queue becomes empty without printing all of the vertices, then the graph contains a cycle (i.e., there is no possible ordering for the tasks that does not violate some prerequisite). The printed order for the vertices of the graph in Figure 11.13 using the queue version of topological sort is J1, J2, J3, J6, J4, J5, J7. Figure 11.14 shows an implementation for the queue-based topological sort algorithm.
- 426. Sec. 11.4 Shortest-Paths Problems 407 5 20 2 10 D B A 3 11 E C 15 Figure 11.15 Example graph for shortest-path definitions. 11.4 Shortest-Paths Problems On a road map, a road connecting two towns is typically labeled with its distance. We can model a road network as a directed graph whose edges are labeled with real numbers. These numbers represent the distance (or other cost metric, such as travel time) between two vertices. These labels may be called weights, costs, or distances, depending on the application. Given such a graph, a typical problem is to find the total length of the shortest path between two specified vertices. This is not a trivial problem, because the shortest path may not be along the edge (if any) connecting two vertices, but rather may be along a path involving one or more intermediate vertices. For example, in Figure 11.15, the cost of the path from A to B to D is 15. The cost of the edge directly from A to D is 20. The cost of the path from A to C to B to D is 10. Thus, the shortest path from A to D is 10 (not along the edge connecting A to D). We use the notation d(A, D) = 10 to indicate that the shortest distance from A to D is 10. In Figure 11.15, there is no path from E to B, so we set d(E, B) = ∞. We define w(A, D) = 20 to be the weight of edge (A, D), that is, the weight of the direct connection from A to D. Because there is no edge from E to B, w(E, B) = ∞. Note that w(D, A) = ∞ because the graph of Figure 11.15 is directed. We assume that all weights are positive. 11.4.1 Single-Source Shortest Paths This section presents an algorithm to solve the single-source shortest-paths prob- lem. Given Vertex S in Graph G, find a shortest path from S to every other vertex in G. We might want only the shortest path between two vertices, S and T. How- ever in the worst case, while finding the shortest path from S to T, we might find the shortest paths from S to every other vertex as well. So there is no better alg- orithm (in the worst case) for finding the shortest path to a single vertex than to find shortest paths to all vertices. The algorithm described here will only compute the
- 427. 408 Chap. 11 Graphs distance to every such vertex, rather than recording the actual path. Recording the path requires modifications to the algorithm that are left as an exercise. Computer networks provide an application for the single-source shortest-paths problem. The goal is to find the cheapest way for one computer to broadcast a message to all other computers on the network. The network can be modeled by a graph with edge weights indicating time or cost to send a message to a neighboring computer. For unweighted graphs (or whenever all edges have the same cost), the single- source shortest paths can be found using a simple breadth-first search. When weights are added, BFS will not give the correct answer. One approach to solving this problem when the edges have differing weights might be to process the vertices in a fixed order. Label the vertices v0 to vn−1, with S = v0. When processing Vertex v1, we take the edge connecting v0 and v1. When processing v2, we consider the shortest distance from v0 to v2 and compare that to the shortest distance from v0 to v1 to v2. When processing Vertex vi, we consider the shortest path for Vertices v0 through vi−1 that have already been processed. Unfortunately, the true shortest path to vi might go through Vertex vj for j i. Such a path will not be considered by this algorithm. However, the problem would not occur if we process the vertices in order of distance from S. Assume that we have processed in order of distance from S to the first i − 1 vertices that are closest to S; call this set of vertices S. We are now about to process the ith closest vertex; call it X. A shortest path from S to X must have its next-to-last vertex in S. Thus, d(S, X) = min U∈S (d(S, U) + w(U, X)). In other words, the shortest path from S to X is the minimum over all paths that go from S to U, then have an edge from U to X, where U is some vertex in S. This solution is usually referred to as Dijkstra’s algorithm. It works by main- taining a distance estimate D(X) for all vertices X in V. The elements of D are ini- tialized to the value INFINITE. Vertices are processed in order of distance from S. Whenever a vertex V is processed, D(X) is updated for every neighbor X of V. Figure 11.16 shows an implementation for Dijkstra’s algorithm. At the end, array D will contain the shortest distance values. There are two reasonable solutions to the key issue of finding the unvisited vertex with minimum distance value during each pass through the main for loop. The first method is simply to scan through the list of |V| vertices searching for the minimum value, as follows:
- 428. Sec. 11.4 Shortest-Paths Problems 409 // Compute shortest path distances from s, store them in D static void Dijkstra(Graph G, int s, int[] D) { for (int i=0; iG.n(); i++) // Initialize D[i] = Integer.MAX VALUE; D[s] = 0; for (int i=0; iG.n(); i++) { // Process the vertices int v = minVertex(G, D); // Find next-closest vertex G.setMark(v, VISITED); if (D[v] == Integer.MAX VALUE) return; // Unreachable for (int w = G.first(v); w G.n(); w = G.next(v, w)) if (D[w] (D[v] + G.weight(v, w))) D[w] = D[v] + G.weight(v, w); } } Figure 11.16 An implementation for Dijkstra’s algorithm. static int minVertex(Graph G, int[] D) { int v = 0; // Initialize v to any unvisited vertex; for (int i=0; iG.n(); i++) if (G.getMark(i) == UNVISITED) { v = i; break; } for (int i=0; iG.n(); i++) // Now find smallest value if ((G.getMark(i) == UNVISITED) (D[i] D[v])) v = i; return v; } Because this scan is done |V| times, and because each edge requires a constant- time update to D, the total cost for this approach is Θ(|V|2 + |E|) = Θ(|V|2), because |E| is in O(|V|2). The second method is to store unprocessed vertices in a min-heap ordered by distance values. The next-closest vertex can be found in the heap in Θ(log |V|) time. Every time we modify D(X), we could reorder X in the heap by deleting and reinserting it. This is an example of a priority queue with priority update, as described in Section 5.5. To implement true priority updating, we would need to store with each vertex its array index within the heap. A simpler approach is to add the new (smaller) distance value for a given vertex as a new record in the heap. The smallest value for a given vertex currently in the heap will be found first, and greater distance values found later will be ignored because the vertex will already be marked as VISITED. The only disadvantage to repeatedly inserting distance values is that it will raise the number of elements in the heap from Θ(|V|) to Θ(|E|) in the worst case. The time complexity is Θ((|V| + |E|) log |E|), because for each edge we must reorder the heap. Because the objects stored on the heap need to know both their vertex number and their distance, we create a simple class for the purpose called DijkElem, as follows. DijkElem is quite similar to the Edge class used by the adjacency list representation.
- 429. 410 Chap. 11 Graphs // Dijkstra’s shortest-paths: priority queue version static void Dijkstra(Graph G, int s, int[] D) { int v; // The current vertex DijkElem[] E = new DijkElem[G.e()]; // Heap for edges E[0] = new DijkElem(s, 0); // Initial vertex MinHeapDijkElem H = new MinHeapDijkElem(E, 1, G.e()); for (int i=0; iG.n(); i++) // Initialize distance D[i] = Integer.MAX VALUE; D[s] = 0; for (int i=0; iG.n(); i++) { // For each vertex do { v = (H.removemin()).vertex(); } // Get position while (G.getMark(v) == VISITED); G.setMark(v, VISITED); if (D[v] == Integer.MAX VALUE) return; // Unreachable for (int w = G.first(v); w G.n(); w = G.next(v, w)) if (D[w] (D[v] + G.weight(v, w))) { // Update D D[w] = D[v] + G.weight(v, w); H.insert(new DijkElem(w, D[w])); } } } Figure 11.17 An implementation for Dijkstra’s algorithm using a priority queue. import java.lang.Comparable; class DijkElem implements ComparableDijkElem { private int vertex; private int weight; public DijkElem(int inv, int inw) { vertex = inv; weight = inw; } public DijkElem() {vertex = 0; weight = 0; } public int key() { return weight; } public int vertex() { return vertex; } public int compareTo(DijkElem that) { if (weight that.key()) return -1; else if (weight == that.key()) return 0; else return 1; } } Figure 11.17 shows an implementation for Dijkstra’s algorithm using the prior- ity queue. Using MinVertex to scan the vertex list for the minimum value is more ef- ficient when the graph is dense, that is, when |E| approaches |V|2. Using a prior- ity queue is more efficient when the graph is sparse because its cost is Θ((|V| +
- 430. Sec. 11.5 Minimum-Cost Spanning Trees 411 A B C D E Initial 0 ∞ ∞ ∞ ∞ Process A 0 10 3 20 ∞ Process C 0 5 3 20 18 Process B 0 5 3 10 18 Process D 0 5 3 10 18 Process E 0 5 3 10 18 Figure 11.18 A listing for the progress of Dijkstra’s algorithm operating on the graph of Figure 11.15. The start vertex is A. |E|) log |E|). However, when the graph is dense, this cost can become as great as Θ(|V|2 log |E|) = Θ(|V |2 log |V |). Figure 11.18 illustrates Dijkstra’s algorithm. The start vertex is A. All vertices except A have an initial value of ∞. After processing Vertex A, its neighbors have their D estimates updated to be the direct distance from A. After processing C (the closest vertex to A), Vertices B and E are updated to reflect the shortest path through C. The remaining vertices are processed in order B, D, and E. 11.5 Minimum-Cost Spanning Trees This section presents two algorithms for determining the minimum-cost spanning tree (MST) for a graph. The MST problem takes as input a connected, undirected graph G, where each edge has a distance or weight measure attached. The MST is the graph containing the vertices of G along with the subset of G’s edges that (1) has minimum total cost as measured by summing the values for all of the edges in the subset, and (2) keeps the vertices connected. Applications where a solution to this problem is useful include soldering the shortest set of wires needed to connect a set of terminals on a circuit board, and connecting a set of cities by telephone lines in such a way as to require the least amount of cable. The MST contains no cycles. If a proposed set of edges did have a cycle, a cheaper MST could be had by removing any one of the edges in the cycle. Thus, the MST is a free tree with |V|−1 edges. The name “minimum-cost spanning tree” comes from the fact that the required set of edges forms a tree, it spans the vertices (i.e., it connects them together), and it has minimum cost. Figure 11.19 shows the MST for an example graph.
- 431. 412 Chap. 11 Graphs A 9 7 5 B C 1 2 6 D 2 1 E F Figure 11.19 A graph and its MST. All edges appear in the original graph. Those edges drawn with heavy lines indicate the subset making up the MST. Note that edge (C, F) could be replaced with edge (D, F) to form a different MST with equal cost. 11.5.1 Prim’s Algorithm The first of our two algorithms for finding MSTs is commonly referred to as Prim’s algorithm. Prim’s algorithm is very simple. Start with any Vertex N in the graph, setting the MST to be N initially. Pick the least-cost edge connected to N. This edge connects N to another vertex; call this M. Add Vertex M and Edge (N, M) to the MST. Next, pick the least-cost edge coming from either N or M to any other vertex in the graph. Add this edge and the new vertex it reaches to the MST. This process continues, at each step expanding the MST by selecting the least-cost edge from a vertex currently in the MST to a vertex not currently in the MST. Prim’s algorithm is quite similar to Dijkstra’s algorithm for finding the single- source shortest paths. The primary difference is that we are seeking not the next closest vertex to the start vertex, but rather the next closest vertex to any vertex currently in the MST. Thus we replae the lines if (D[w] (D[v] + G-weight(v, w))) D[w] = D[v] + G-weight(v, w); in Djikstra’s algorithm with the lines if (D[w] G-weight(v, w)) D[w] = G-weight(v, w); in Prim’s algorithm. Figure 11.20 shows an implementation for Prim’s algorithm that searches the distance matrix for the next closest vertex. For each vertex I, when I is processed by Prim’s algorithm, an edge going to I is added to the MST that we are building.
- 432. Sec. 11.5 Minimum-Cost Spanning Trees 413 // Compute a minimal-cost spanning tree static void Prim(Graph G, int s, int[] D, int[] V) { for (int i=0; iG.n(); i++) // Initialize D[i] = Integer.MAX VALUE; D[s] = 0; for (int i=0; iG.n(); i++) { // Process the vertices int v = minVertex(G, D); G.setMark(v, VISITED); if (v != s) AddEdgetoMST(V[v], v); if (D[v] == Integer.MAX VALUE) return; // Unreachable for (int w = G.first(v); w G.n(); w = G.next(v, w)) if (D[w] G.weight(v, w)) { D[w] = G.weight(v, w); V[w] = v; } } } Figure 11.20 An implementation for Prim’s algorithm. Array V[I] stores the previously visited vertex that is closest to Vertex I. This information lets us know which edge goes into the MST when Vertex I is processed. The implementation of Figure 11.20 also contains calls to AddEdgetoMST to indicate which edges are actually added to the MST. Alternatively, we can implement Prim’s algorithm using a priority queue to find the next closest vertex, as shown in Figure 11.21. As with the priority queue version of Dijkstra’s algorithm, the heap’s Elem type stores a DijkElem object. Prim’s algorithm is an example of a greedy algorithm. At each step in the for loop, we select the least-cost edge that connects some marked vertex to some unmarked vertex. The algorithm does not otherwise check that the MST really should include this least-cost edge. This leads to an important question: Does Prim’s algorithm work correctly? Clearly it generates a spanning tree (because each pass through the for loop adds one edge and one unmarked vertex to the spanning tree until all vertices have been added), but does this tree have minimum cost? Theorem 11.1 Prim’s algorithm produces a minimum-cost spanning tree. Proof: We will use a proof by contradiction. Let G = (V, E) be a graph for which Prim’s algorithm does not generate an MST. Define an ordering on the vertices according to the order in which they were added by Prim’s algorithm to the MST: v0, v1, ..., vn−1. Let edge ei connect (vx, vi) for some x i and i ≥ 1. Let ej be the lowest numbered (first) edge added by Prim’s algorithm such that the set of edges selected so far cannot be extended to form an MST for G. In other words, ej is the
- 433. 414 Chap. 11 Graphs // Prims’s MST algorithm: priority queue version static void Prim(Graph G, int s, int[] D, int[] V) { int v; // The current vertex DijkElem[] E = new DijkElem[G.e()]; // Heap for edges E[0] = new DijkElem(s, 0); // Initial vertex MinHeapDijkElem H = new MinHeapDijkElem(E, 1, G.e()); for (int i=0; iG.n(); i++) // Initialize D[i] = Integer.MAX VALUE; // distances D[s] = 0; for (int i=0; iG.n(); i++) { // Now, get distances do { v = (H.removemin()).vertex(); } // Get position while (G.getMark(v) == VISITED); G.setMark(v, VISITED); if (v != s) AddEdgetoMST(V[v], v); // Add edge to MST if (D[v] == Integer.MAX VALUE) return; // Unreachable for (int w = G.first(v); w G.n(); w = G.next(v, w)) if (D[w] G.weight(v, w)) { // Update D D[w] = G.weight(v, w); V[w] = v; // Where it came from H.insert(new DijkElem(w, D[w])); } } } Figure 11.21 An implementation of Prim’s algorithm using a priority queue. first edge where Prim’s algorithm “went wrong.” Let T be the “true” MST. Call vp (p j) the vertex connected by edge ej, that is, ej = (vp, vj). Because T is a tree, there exists some path in T connecting vp and vj. There must be some edge e′ in this path connecting vertices vu and vw, with u j and w ≥ j. Because ej is not part of T, adding edge ej to T forms a cycle. Edge e′ must be of lower cost than edge ej, because Prim’s algorithm did not generate an MST. This situation is illustrated in Figure 11.22. However, Prim’s algorithm would have selected the least-cost edge available. It would have selected e′, not ej. Thus, it is a contradiction that Prim’s algorithm would have selected the wrong edge, and thus, Prim’s algorithm must be correct. ✷ Example 11.3 For the graph of Figure 11.19, assume that we begin by marking Vertex A. From A, the least-cost edge leads to Vertex C. Vertex C and edge (A, C) are added to the MST. At this point, our candidate edges connecting the MST (Vertices A and C) with the rest of the graph are (A, E), (C, B), (C, D), and (C, F). From these choices, the least-cost edge from the MST is (C, D). So we add Vertex D to the MST. For the next iteration, our edge choices are (A, E), (C, B), (C, F), and (D, F). Because edges (C, F)
- 434. Sec. 11.5 Minimum-Cost Spanning Trees 415 Marked Unmarked Vertices vi, i ≥ j Vertices vi, i j “correct” edge e′ ej Prim’s edge vw vj vp vu Figure 11.22 Prim’s MST algorithm proof. The left oval contains that portion of the graph where Prim’s MST and the “true” MST T agree. The right oval contains the rest of the graph. The two portions of the graph are connected by (at least) edges ej (selected by Prim’s algorithm to be in the MST) and e′ (the “correct” edge to be placed in the MST). Note that the path from vw to vj cannot include any marked vertex vi, i ≤ j, because to do so would form a cycle. and (D, F) happen to have equal cost, it is an arbitrary decision as to which gets selected. Let’s pick (C, F). The next step marks Vertex E and adds edge (F, E) to the MST. Following in this manner, Vertex B (through edge (C, B)) is marked. At this point, the algorithm terminates. 11.5.2 Kruskal’s Algorithm Our next MST algorithm is commonly referred to as Kruskal’s algorithm. Kruskal’s algorithm is also a simple, greedy algorithm. We first partition the set of vertices into |V| equivalence classes (see Section 6.2), each consisting of one vertex. We then process the edges in order of weight. An edge is added to the MST, and the two equivalence classes combined, if the edge connects two vertices in different equivalence classes. This process is repeated until only one equivalence class re- mains. Example 11.4 Figure 11.23 shows the first three steps of Kruskal’s Alg- orithm for the graph of Figure 11.19. Edge (C, D) has the least cost, and because C and D are currently in separate MSTs, they are combined. We next select edge (E, F) to process, and combine these vertices into a single
- 435. 416 Chap. 11 Graphs MST. The third edge we process is (C, F), which causes the MST contain- ing Vertices C and D to merge with MST containing Vertices E and F. The next edge to process is (D, F). But because Vertices D and F are currently in the same MST, this edge is rejected. The algorithm will continue on to accept edges (B, C) and (A, C) into the MST. The edges can be processed in order of weight by using a min-heap. This is generally faster than sorting the edges first, because in practice we need only visit a small fraction of the edges before completing the MST. This is an example of finding only a few smallest elements in a list, as discussed in Section 7.6. The only tricky part to this algorithm is determining if two vertices belong to the same equivalence class. Fortunately, the ideal algorithm is available for the purpose — the UNION/FIND algorithm based on the parent pointer representation for trees described in Section 6.2. Figure 11.24 shows an implementation for the algorithm. Class KruskalElem is used to store the edges on the min-heap. Kruskal’s algorithm is dominated by the time required to process the edges. The differ and UNION functions are nearly constant in time if path compression and weighted union is used. Thus, the total cost of the algorithm is Θ(|E| log |E|) in the worst case, when nearly all edges must be processed before all the edges of the spanning tree are found and the algorithm can stop. More often the edges of the spanning tree are the shorter ones,and only about |V| edges must be processed. If so, the cost is often close to Θ(|V| log |E|) in the average case. 11.6 Further Reading Many interesting properties of graphs can be investigated by playing with the pro- grams in the Stanford Graphbase. This is a collection of benchmark databases and graph processing programs. The Stanford Graphbase is documented in [Knu94]. 11.7 Exercises 11.1 Prove by induction that a graph with n vertices has at most n(n−1)/2 edges. 11.2 Prove the following implications regarding free trees. (a) IF an undirected graph is connected and has no simple cycles, THEN the graph has |V| − 1 edges. (b) IF an undirected graph has |V| − 1 edges and no cycles, THEN the graph is connected. 11.3 (a) Draw the adjacency matrix representation for the graph of Figure 11.25.
- 436. Sec. 11.7 Exercises 417 Initial Step 1 A B C 1 D E F Step 2 Process edge (E, F) 1 1 Step 3 Process edge (C, F) B 1 2 E 1 F Process edge (C, D) A A B D E F C C D B C D E A F Figure 11.23 Illustration of the first three steps of Kruskal’s MST algorithm as applied to the graph of Figure 11.19. (b) Draw the adjacency list representation for the same graph. (c) If a pointer requires four bytes, a vertex label requires two bytes, and an edge weight requires two bytes, which representation requires more space for this graph? (d) If a pointer requires four bytes, a vertex label requires one byte, and an edge weight requires two bytes, which representation requires more space for this graph? 11.4 Show the DFS tree for the graph of Figure 11.25, starting at Vertex 1. 11.5 Wright a pseudocode algorithm to create a DFS tree for an undirected, con- nected graph starting at a specified vertex V. 11.6 Show the BFS tree for the graph of Figure 11.25, starting at Vertex 1.
- 437. 418 Chap. 11 Graphs class KruskalElem implements ComparableKruskalElem { private int v, w, weight; public KruskalElem(int inweight, int inv, int inw) { weight = inweight; v = inv; w = inw; } public int v1() { return v; } public int v2() { return w; } public int key() { return weight; } public int compareTo(KruskalElem that) { if (weight that.key()) return -1; else if (weight == that.key()) return 0; else return 1; } } static void Kruskal(Graph G) { // Kruskal’s MST algorithm ParPtrTree A = new ParPtrTree(G.n()); // Equivalence array KruskalElem[] E = new KruskalElem[G.e()]; // Minheap array int edgecnt = 0; // Count of edges for (int i=0; iG.n(); i++) // Put edges in the array for (int w = G.first(i); w G.n(); w = G.next(i, w)) E[edgecnt++] = new KruskalElem(G.weight(i, w), i, w); MinHeapKruskalElem H = new MinHeapKruskalElem(E, edgecnt, edgecnt); int numMST = G.n(); // Initially n classes for (int i=0; numMST1; i++) { // Combine equiv classes KruskalElem temp = H.removemin(); // Next cheapest int v = temp.v1(); int u = temp.v2(); if (A.differ(v, u)) { // If in different classes A.UNION(v, u); // Combine equiv classes AddEdgetoMST(v, u); // Add this edge to MST numMST--; // One less MST } } } Figure 11.24 An implementation for Kruskal’s algorithm. 11.7 Wright a pseudocode algorithm to create a BFS tree for an undirected, con- nected graph starting at a specified vertex V. 11.8 The BFS topological sort algorithm can report the existence of a cycle if one is encountered. Modify this algorithm to print the vertices possibly appearing in cycles (if there are any cycles). 11.9 Explain why, in the worst case, Dijkstra’s algorithm is (asymptotically) as efficient as any algorithm for finding the shortest path from some vertex I to another vertex J.
- 438. Sec. 11.7 Exercises 419 2 5 4 20 10 3 6 11 3 3 15 5 10 2 1 Figure 11.25 Example graph for Chapter 11 exercises. 11.10 Show the shortest paths generated by running Dijkstra’s shortest-paths alg- orithm on the graph of Figure 11.25, beginning at Vertex 4. Show the D values as each vertex is processed, as in Figure 11.18. 11.11 Modify the algorithm for single-source shortest paths to actually store and return the shortest paths rather than just compute the distances. 11.12 The root of a DAG is a vertex R such that every vertex of the DAG can be reached by a directed path from R. Write an algorithm that takes a directed graph as input and determines the root (if there is one) for the graph. The running time of your algorithm should be Θ(|V| + |E|). 11.13 Write an algorithm to find the longest path in a DAG, where the length of the path is measured by the number of edges that it contains. What is the asymptotic complexity of your algorithm? 11.14 Write an algorithm to determine whether a directed graph of |V| vertices contains a cycle. Your algorithm should run in Θ(|V| + |E|) time. 11.15 Write an algorithm to determine whether an undirected graph of |V| vertices contains a cycle. Your algorithm should run in Θ(|V|) time. 11.16 The single-destination shortest-paths problem for a directed graph is to find the shortest path from every vertex to a specified vertex V. Write an algorithm to solve the single-destination shortest-paths problem. 11.17 List the order in which the edges of the graph in Figure 11.25 are visited when running Prim’s MST algorithm starting at Vertex 3. Show the final MST. 11.18 List the order in which the edges of the graph in Figure 11.25 are visited when running Kruskal’s MST algorithm. Each time an edge is added to the MST, show the result on the equivalence array, (e.g., show the array as in Figure 6.7). 11.19 Write an algorithm to find a maximum cost spanning tree, that is, the span- ning tree with highest possible cost.
- 439. 420 Chap. 11 Graphs 11.20 When can Prim’s and Kruskal’s algorithms yield different MSTs? 11.21 Prove that, if the costs for the edges of Graph G are distinct, then only one MST exists for G. 11.22 Does either Prim’s or Kruskal’s algorithm work if there are negative edge weights? 11.23 Consider the collection of edges selected by Dijkstra’s algorithm as the short- est paths to the graph’s vertices from the start vertex. Do these edges form a spanning tree (not necessarily of minimum cost)? Do these edges form an MST? Explain why or why not. 11.24 Prove that a tree is a bipartite graph. 11.25 Prove that any tree can be two-colored. 11.26 Write an algorithm that deterimines if an arbitrary undirected graph is a bi- partite graph. If the graph is bipartite, then your algorithm should also iden- tify the vertices as to which of the two partitions each belongs to. 11.8 Projects 11.1 Design a format for storing graphs in files. Then implement two functions: one to read a graph from a file and the other to write a graph to a file. Test your functions by implementing a complete MST program that reads an undi- rected graph in from a file, constructs the MST, and then writes to a second file the graph representing the MST. 11.2 An undirected graph need not explicitly store two separate directed edges to represent a single undirected edge. An alternative would be to store only a single undirected edge (I, J) to connect Vertices I and J. However, what if the user asks for edge (J, I)? We can solve this problem by consistently storing the edge such that the lesser of I and J always comes first. Thus, if we have an edge connecting Vertices 5 and 3, requests for edge (5, 3) and (3, 5) both map to (3, 5) because 3 5. Looking at the adacency matrix, we notice that only the lower triangle of the array is used. Thus we could cut the space required by the adjacency matrix from |V|2 positions to |V|(|V|−1)/2 positions. Read Section 12.2 on triangular matrices. The reimplement the adjacency matrix representation of Figure 11.6 to implement undirected graphs using a triangular array. 11.3 While the underlying implementation (whether adjacency matrix or adja- cency list) is hidden behind the graph ADT, these two implementations can have an impact on the efficiency of the resulting program. For Dijkstra’s shortest paths algorithm, two different implementations were given in Sec- tion 11.4.1 that provide diffent ways for determining the next closest vertex
- 440. Sec. 11.8 Projects 421 at each iteration of the algorithm. The relative costs of these two variants depend on who sparse or dense the graph is. They might also depend on whether the graph is implemented using an adjacency list or adjacency ma- trix. Design and implement a study to compare the effects on performance for three variables: (i) the two graph representations (adjacency list and adja- cency matrix); (ii) the two implementations for Djikstra’s shortest paths alg- orithm (searching the table of vertex distances or using a priority queue to track the distances), and (iii) sparse versus dense graphs. Be sure to test your implementations on a variety of graphs that are sufficiently large to generate meaningful times. 11.4 The example implementations for DFS and BFS show calls to functions PreVisit and PostVisit. Better is to implement BFS and DFS using the visitor design pattern, with the visitor functions being passed in as either function or template parameters. Reimplement the BFS and DFS functions to make use of the visitor design pattern. 11.5 Write a program to label the connected components for an undirected graph. In other words, all vertices of the first component are given the first com- ponent’s label, all vertices of the second component are given the second component’s label, and so on. Your algorithm should work by defining any two vertices connected by an edge to be members of the same equivalence class. Once all of the edges have been processed, all vertices in a given equiv- alence class will be connected. Use the UNION/FIND implementation from Section 6.2 to implement equivalence classes.
- 442. 12 Lists and Arrays Revisited Simple lists and arrays are the right tool for the many applications. Other situations require support for operations that cannot be implemented efficiently by the stan- dard list representations of Chapter 4. This chapter presents advanced implementa- tions for lists and arrays that overcome some of the problems of simple linked list and contiguous array representations. A wide range of topics are covered, whose unifying thread is that the data structures are all list- or array-like. This chapter should also serve to reinforce the concept of logical representation versus physical implementation, as some of the “list” implementations have quite different organi- zations internally. Section 12.1 describes a series of representations for multilists, which are lists that may contain sublists. Section 12.2 discusses representations for implementing sparse matrices, large matrices where most of the elements have zero values. Sec- tion 12.3 discusses memory management techniques, which are essentially a way of allocating variable-length sections from a large array. 12.1 Multilists Recall from Chapter 4 that a list is a finite, ordered sequence of items of the form hx0, x1, ..., xn−1i where n ≥ 0. We can represent the empty list by null or hi. In Chapter 4 we assumed that all list elements had the same data type. In this section, we extend the definition of lists to allow elements to be arbitrary in nature. In general, list elements are one of two types. 1. An atom, which is a data record of some type such as a number, symbol, or string. 2. Another list, which is called a sublist. 423
- 443. 424 Chap. 12 Lists and Arrays Revisited x1 y1 a1 a2 y3 z1 z2 x4 Figure 12.1 Example of a multilist represented by a tree. L2 L1 a b c d e L3 Figure 12.2 Example of a reentrant multilist. The shape of the structure is a DAG (all edges point downward). A list containing sublists will be written as hx1, hy1, ha1, a2i, y3i, hz1, z2i, x4i. In this example, the list has four elements. The second element is the sublist hy1, ha1, a2i, y3i and the third is the sublist hz1, z2i. The sublist hy1, ha1, a2i, y3i itself contains a sublist. If a list L has one or more sublists, we call L a multi- list. Lists with no sublists are often referred to as linear lists or chains. Note that this definition for multilist fits well with our definition of sets from Definition 2.1, where a set’s members can be either primitive elements or sets. We can restrict the sublists of a multilist in various ways, depending on whether the multilist should have the form of a tree, a DAG, or a generic graph. A pure list is a list structure whose graph corresponds to a tree, such as in Figure 12.1. In other words, there is exactly one path from the root to any node, which is equivalent to saying that no object may appear more than once in the list. In the pure list, each pair of angle brackets corresponds to an internal node of the tree. The members of the list correspond to the children for the node. Atoms on the list correspond to leaf nodes. A reentrant list is a list structure whose graph corresponds to a DAG. Nodes might be accessible from the root by more than one path, which is equivalent to saying that objects (including sublists) may appear multiple times in the list as long as no cycles are formed. All edges point downward, from the node representing a list or sublist to its elements. Figure 12.2 illustrates a reentrant list. To write out
- 444. Sec. 12.1 Multilists 425 L1 L2 L4 b d a c L3 Figure 12.3 Example of a cyclic list. The shape of the structure is a directed graph. this list in bracket notation, we can duplicate nodes as necessary. Thus, the bracket notation for the list of Figure 12.2 could be written hhha, bii, hha, bi, ci, hc, d, ei, heii. For convenience, we will adopt a convention of allowing sublists and atoms to be labeled, such as “L1:”. Whenever a label is repeated, the element corresponding to that label will be substituted when we write out the list. Thus, the bracket notation for the list of Figure 12.2 could be written hhL1:ha, bii, hL1, L2:ci, hL2, d, L3:ei, hL3ii. A cyclic list is a list structure whose graph corresponds to any directed graph, possibly containing cycles. Figure 12.3 illustrates such a list. Labels are required to write this in bracket notation. Here is the notation for the list of Figure 12.3. hL1:hL2:ha, L1ii, hL2, L3:bi, hL3, c, di, L4:hL4ii. Multilists can be implemented in a number of ways. Most of these should be familiar from implementations suggested earlier in the book for list, tree, and graph data structures. One simple approach is to use an array representation. This works well for chains with fixed-length elements, equivalent to the simple array-based list of Chap- ter 4. We can view nested sublists as variable-length elements. To use this ap- proach, we require some indication of the beginning and end of each sublist. In essence, we are using a sequential tree implementation as discussed in Section 6.5. This should be no surprise, because the pure list is equivalent to a general tree struc- ture. Unfortunately, as with any sequential representation, access to the nth sublist must be done sequentially from the beginning of the list. Because pure lists are equivalent to trees, we can also use linked allocation methods to support direct access to the list of children. Simple linear lists are
- 445. 426 Chap. 12 Lists and Arrays Revisited root − y1 − + y3 + a2 + z1 x4 z2 + a1 − x1 + + + + Figure 12.4 Linked representation for the pure list of Figure 12.1. The first field in each link node stores a tag bit. If the tag bit stores “+,” then the data field stores an atom. If the tag bit stores “−,” then the data field stores a pointer to a sublist. root B C D A Figure 12.5 LISP-like linked representation for the cyclic multilist of Fig- ure 12.3. Each link node stores two pointers. A pointer either points to an atom, or to another link node. Link nodes are represented by two boxes, and atoms by circles. represented by linked lists. Pure lists can be represented as linked lists with an additional tag field to indicate whether the node is an atom or a sublist. If it is a sublist, the data field points to the first element on the sublist. This is illustrated by Figure 12.4. Another approach is to represent all list elements with link nodes storing two pointer fields, except for atoms. Atoms just contain data. This is the system used by the programming language LISP. Figure 12.5 illustrates this representation. Either the pointer contains a tag bit to identify what it points to, or the object being pointed to stores a tag bit to identify itself. Tags distinguish atoms from list nodes. This implementation can easily support reentrant and cyclic lists, because non-atoms can point to any other node.
- 446. Sec. 12.2 Matrix Representations 427 a00 0 0 0 a10 a11 0 0 a20 a21 a22 0 a30 a31 a32 a33 (a) a00 a01 a02 a03 0 a11 a12 a13 0 0 a22 a23 0 0 0 a33 (b) Figure 12.6 Triangular matrices. (a) A lower triangular matrix. (b) An upper triangular matrix. 12.2 Matrix Representations Some applications must represent a large, two-dimensional matrix where many of the elements have a value of zero. One example is the lower triangular matrix that results from solving systems of simultaneous equations. A lower triangular matrix stores zero values at positions [r, c] such that r c, as shown in Figure 12.6(a). Thus, the upper-right triangle of the matrix is always zero. Another example is the representation of undirected graphs in an adjacency matrix (see Project 11.2). Because all edges between Vertices i and j go in both directions, there is no need to store both. Instead we can just store one edge going from the higher-indexed vertex to the lower-indexed vertex. In this case, only the lower triangle of the matrix can have non-zero values. We can take advantage of this fact to save space. Instead of storing n(n + 1)/2 pieces of information in an n × n array, it would save space to use a list of length n(n + 1)/2. This is only practical if some means can be found to locate within the list the element that would correspond to position [r, c] in the original matrix. To derive an equation to do this computation, note that row 0 of the matrix has one non-zero value, row 1 has two non-zero values, and so on. Thus, row r is preceded by r rows with a total of Pr k=1 k = (r2 + r)/2 non-zero elements. Adding c to reach the cth position in the rth row yields the following equation to convert position [r, c] in the original matrix to the correct position in the list. matrix[r, c] = list[(r2 + r)/2 + c]. A similar equation can be used to store an upper triangular matrix, that is, a matrix with zero values at positions [r, c] such that r c, as shown in Figure 12.6(b). For an n × n upper triangular matrix, the equation would be matrix[r, c] = list[rn − (r2 + r)/2 + c]. A more difficult situation arises when the vast majority of values stored in an n × n matrix are zero, but there is no restriction on which positions are zero and which are non-zero. This is known as a sparse matrix.
- 447. 428 Chap. 12 Lists and Arrays Revisited A40 A71 A00 A06 0 1 4 7 A01 A11 A13 A76 0 1 3 6 Cols Rows Figure 12.7 The orthogonal list sparse matrix representation. One approach to representing a sparse matrix is to concatenate (or otherwise combine) the row and column coordinates into a single value and use this as a key in a hash table. Thus, if we want to know the value of a particular position in the matrix, we search the hash table for the appropriate key. If a value for this position is not found, it is assumed to be zero. This is an ideal approach when all queries to the matrix are in terms of access by specified position. However, if we wish to find the first non-zero element in a given row, or the next non-zero element below the current one in a given column, then the hash table requires us to check sequentially through all possible positions in some row or column. Another approach is to implement the matrix as an orthogonal list, as illus- trated in Figure 12.7. Here we have a list of row headers, each of which contains a pointer to a list of matrix records. A second list of column headers also con- tains pointers to matrix records. Each non-zero matrix element stores pointers to its non-zero neighbors in the row, both following and preceding it. Each non-zero
- 448. Sec. 12.2 Matrix Representations 429 element also stores pointers to its non-zero neighbors following and preceding it in the column. Thus, each non-zero element stores its own value, its position within the matrix, and four pointers. Non-zero elements are found by traversing a row or column list. Note that the first non-zero element in a given row could be in any col- umn; likewise, the neighboring non-zero element in any row or column list could be at any (higher) row or column in the array. Thus, each non-zero element must also store its row and column position explicitly. To find if a particular position in the matrix contains a non-zero element, we traverse the appropriate row or column list. For example, when looking for the element at Row 7 and Column 1, we can traverse the list either for Row 7 or for Column 1. When traversing a row or column list, if we come to an element with the correct position, then its value is non-zero. If we encounter an element with a higher position, then the element we are looking for is not in the sparse matrix. In this case, the element’s value is zero. For example, when traversing the list for Row 7 in the matrix of Figure 12.7, we first reach the element at Row 7 and Column 1. If this is what we are looking for, then the search can stop. If we are looking for the element at Row 7 and Column 2, then the search proceeds along the Row 7 list to next reach the element at Column 6. At this point we know that no element at Row 7 and Column 2 is stored in the sparse matrix. Insertion and deletion can be performed by working in a similar way to insert or delete elements within the appropriate row and column lists. Each non-zero element stored in the sparse matrix representation takes much more space than an element stored in a simple n × n matrix. When is the sparse matrix more space efficient than the standard representation? To calculate this, we need to determine how much space the standard matrix requires, and how much the sparse matrix requires. The size of the sparse matrix depends on the number of non- zero elements, while the size of the standard matrix representation does not vary. We need to know the (relative) sizes of a pointer and a data value. For simplicity, our calculation will ignore the space taken up by the row and column header (which is not much affected by the number of elements in the sparse array). As an example, assume that a data value, a row or column index, and a pointer each require four bytes. An n × m matrix requires 4nm bytes. The sparse matrix requires 28 bytes per non-zero element (four pointers, two array indices, and one data value). If we set X to be the percentage of non-zero elements, we can solve for the value of X below which the sparse matrix representation is more space efficient. Using the equation 28mnX = 4mn
- 449. 430 Chap. 12 Lists and Arrays Revisited and solving for X, we find that the sparse matrix using this implementation is more space efficient when X 1/7, that is, when less than about 14% of the elements are non-zero. Different values for the relative sizes of data values, pointers, or matrix indices can lead to a different break-even point for the two implementations. The time required to process a sparse matrix depends on the number of non- zero elements stored. When searching for an element, the cost is the number of elements preceding the desired element on its row or column list. The cost for operations such as adding two matrices should be Θ(n + m) in the worst case when the one matrix stores n non-zero elements and the other stores m non-zero elements. 12.3 Memory Management Most of the data structure implementations described in this book store and access objects of uniform size, such as integers stored in a list or a tree. A few simple methods have been described for storing variable-size records in an array or a stack. This section discusses memory management techniques for the general problem of handling space requests of variable size. The basic model for memory management is that we have a (large) block of contiguous memory locations, which we will call the memory pool. Periodically, memory requests are issued for some amount of space in the pool. The memory manager must find a contiguous block of locations of at least the requested size from somewhere within the memory pool. Honoring such a request is called a memory allocation. The memory manager will typically return some piece of information that permits the user to recover the data that were just stored. This piece of information is called a handle. Previously allocated memory might be returned to the memory manager at some future time. This is called a memory deallocation. We can define an ADT for the memory manager as shown in Figure 12.8. The user of the MemManager ADT provides a pointer (in parameter space) to space that holds some message to be stored or retrieved. This is similar to the basic file read/write methods presented in Section 8.4. The fundamental idea is that the client gives messages to the memory manager for safe keeping. The memory manager returns a “receipt” for the message in the form of a MemHandle object. The client holds the MemHandle until it wishes to get the message back. Method insert lets the client tell the memory manager the length and con- tents of the message to be stored. This ADT assumes that the memory manager will remember the length of the message associated with a given handle, thus method get does not include a length parameter but instead returns the length of the mes-
- 450. Sec. 12.3 Memory Management 431 /** Memory Manager interface */ interface MemManADT { /** Store a record and return a handle to it */ public MemHandle insert(byte[] info); // Request space /** Get back a copy of a stored record */ public byte[] get(MemHandle h); // Retrieve data /** Release the space associated with a record */ public void release(MemHandle h); // Release space } Figure 12.8 A simple ADT for a memory manager. sage actually stored. Method release allows the client to tell the memory man- ager to release the space that stores a given message. When all inserts and releases follow a simple pattern, such as last requested, first released (stack order), or first requested, first released (queue order), memory management is fairly easy. We are concerned in this section with the general case where blocks of any size might be requested and released in any order. This is known as dynamic storage allocation. One example of dynamic storage allocation is managing free store for a compiler’s runtime environment, such as the system- level new operations in Java. Another example is managing main memory in a multitasking operating system. Here, a program might require a certain amount of space, and the memory manager must keep track of which programs are using which parts of the main memory. Yet another example is the file manager for a disk drive. When a disk file is created, expanded, or deleted, the file manager must allocate or deallocate disk space. A block of memory or disk space managed in this way is sometimes referred to as a heap. The term “heap” is being used here in a different way than the heap data structure discussed in Section 5.5. Here “heap” refers to the memory controlled by a dynamic memory management scheme. In the rest of this section, we first study techniques for dynamic memory man- agement. We then tackle the issue of what to do when no single block of memory in the memory pool is large enough to honor a given request. 12.3.1 Dynamic Storage Allocation For the purpose of dynamic storage allocation, we view memory as a single array broken into a series of variable-size blocks, where some of the blocks are free and some are reserved or already allocated. The free blocks are linked together to form
- 451. 432 Chap. 12 Lists and Arrays Revisited Figure 12.9 Dynamic storage allocation model. Memory is made up of a series of variable-size blocks, some allocated and some free. In this example, shaded areas represent memory currently allocated and unshaded areas represent unused memory available for future allocation. Small block: External fragmentation Unused space in allocated block: Internal fragmentation Figure 12.10 An illustration of internal and external fragmentation. a freelist used for servicing future memory requests. Figure 12.9 illustrates the situation that can arise after a series of memory allocations and deallocations. When a memory request is received by the memory manager, some block on the freelist must be found that is large enough to service the request. If no such block is found, then the memory manager must resort to a failure policy such as discussed in Section 12.3.2. If there is a request for m words, and no block exists of exactly size m, then a larger block must be used instead. One possibility in this case is that the entire block is given away to the memory allocation request. This might be desirable when the size of the block is only slightly larger than the request. This is because saving a tiny block that is too small to be useful for a future memory request might not be worthwhile. Alternatively, for a free block of size k, with k m, up to k − m space may be retained by the memory manager to form a new free block, while the rest is used to service the request. Memory managers can suffer from two types of fragmentation. External frag- mentation occurs when a series of memory requests result in lots of small free blocks, no one of which is useful for servicing typical requests. Internal fragmen- tation occurs when more than m words are allocated to a request for m words, wasting free storage. This is equivalent to the internal fragmentation that occurs when files are allocated in multiples of the cluster size. The difference between internal and external fragmentation is illustrated by Figure 12.10. Some memory management schemes sacrifice space to internal fragmentation to make memory management easier (and perhaps reduce external fragmentation). For example, external fragmentation does not happen in file management systems
- 452. Sec. 12.3 Memory Management 433 that allocate file space in clusters. Another example of sacrificing space to inter- nal fragmentation so as to simplify memory management is the buddy method described later in this section. The process of searching the memory pool for a block large enough to service the request, possibly reserving the remaining space as a free block, is referred to as a sequential fit method. Sequential-Fit Methods Sequential-fit methods attempt to find a “good” block to service a storage request. The three sequential-fit methods described here assume that the free blocks are organized into a doubly linked list, as illustrated by Figure 12.11. There are two basic approaches to implementing the freelist. The simpler ap- proach is to store the freelist separately from the memory pool. In other words, a simple linked-list implementation such as described in Chapter 4 can be used, where each node of the linked list contains a pointer to a single free block in the memory pool. This is fine if there is space available for the linked list itself, sepa- rate from the memory pool. The second approach to storing the freelist is more complicated but saves space. Because the free space is free, it can be used by the memory manager to help it do its job; that is, the memory manager temporarily “borrows” space within the free blocks to maintain its doubly linked list. To do so, each unallocated block must be large enough to hold these pointers. In addition, it is usually worthwhile to let the memory manager add a few bytes of space to each reserved block for its own purposes. In other words, a request for m bytes of space might result in slightly more than m bytes being allocated by the memory manager, with the extra bytes used by the memory manager itself rather than the requester. We will assume that all memory blocks are organized as shown in Figure 12.12, with space for tags and linked list pointers. Here, free and reserved blocks are distinguished by a tag bit at both the beginning and the end of the block, for reasons that will be explained. In addition, both free and reserved blocks have a size indicator immediately after the tag bit at the beginning of the block to indicate how large the block is. Free blocks have a second size indicator immediately preceding the tag bit at the end of the block. Finally, free blocks have left and right pointers to their neighbors in the free block list. The information fields associated with each block permit the memory manager to allocate and deallocate blocks as needed. When a request comes in for m words of storage, the memory manager searches the linked list of free blocks until it finds a “suitable” block for allocation. How it determines which block is suitable will
- 453. 434 Chap. 12 Lists and Arrays Revisited Figure 12.11 A doubly linked list of free blocks as seen by the memory manager. Shaded areas represent allocated memory. Unshaded areas are part of the freelist. + Tag Llink Size Tag (a) k Size (b) Tag Size Rlink + − k − Tag k Figure 12.12 Blocks as seen by the memory manager. Each block includes additional information such as freelist link pointers, start and end tags, and a size field. (a) The layout for a free block. The beginning of the block contains the tag bit field, the block size field, and two pointers for the freelist. The end of the block contains a second tag field and a second block size field. (b) A reserved block of k bytes. The memory manager adds to these k bytes an additional tag bit field and block size field at the beginning of the block, and a second tag field at the end of the block. be discussed below. If the block contains exactly m words (plus space for the tag and size fields), then it is removed from the freelist. If the block (of size k) is large enough, then the remaining k − m words are reserved as a block on the freelist, in the current location. When a block F is freed, it must be merged into the freelist. If we do not care about merging adjacent free blocks, then this is a simple insertion into the doubly linked list of free blocks. However, we would like to merge adjacent blocks, because this allows the memory manager to serve requests of the largest possible size. Merging is easily done due to the tag and size fields stored at the ends of each block, as illustrated by Figure 12.13. Here, the memory manager first checks the unit of memory immediately preceding block F to see if the preceding block (call it P) is also free. If it is, then the memory unit before P’s tag bit stores the size of P, thus indicating the position for the beginning of the block in memory. P can
- 454. Sec. 12.3 Memory Management 435 − − + P S F k k Figure 12.13 Adding block F to the freelist. The word immediately preceding the start of F in the memory pool stores the tag bit of the preceding block P. If P is free, merge F into P. We find the end of F by using F’s size field. The word following the end of F is the tag field for block S. If S is free, merge it into F. then simply have its size extended to include block F. If block P is not free, then we just add block F to the freelist. Finally, we also check the bit following the end of block F. If this bit indicates that the following block (call it S) is free, then S is removed from the freelist and the size of F is extended appropriately. We now consider how a “suitable” free block is selected to service a memory request. To illustrate the process, assume there are four blocks on the freelist of sizes 500, 700, 650, and 900 (in that order). Assume that a request is made for 600 units of storage. For our examples, we ignore the overhead imposed for the tag, link, and size fields discussed above. The simplest method for selecting a block would be to move down the free block list until a block of size at least 600 is found. Any remaining space in this block is left on the freelist. If we begin at the beginning of the list and work down to the first free block at least as large as 600, we select the block of size 700. Because this approach selects the first block with enough space, it is called first fit. A simple variation that will improve performance is, instead of always beginning at the head of the freelist, remember the last position reached in the previous search and start from there. When the end of the freelist is reached, search begins again at the head of the freelist. This modification reduces the number of unnecessary searches through small blocks that were passed over by previous requests. There is a potential disadvantage to first fit: It might “waste” larger blocks by breaking them up, and so they will not be available for large requests later. A strategy that avoids using large blocks unnecessarily is called best fit. Best fit looks at the entire list and picks the smallest block that is at least as large as the request (i.e., the “best” or closest fit to the request). Continuing with the preceding example, the best fit for a request of 600 units is the block of size 650, leaving a
- 455. 436 Chap. 12 Lists and Arrays Revisited remainder of size 50. Best fit has the disadvantage that it requires that the entire list be searched. Another problem is that the remaining portion of the best-fit block is likely to be small, and thus useless for future requests. In other words, best fit tends to maximize problems of external fragmentation while it minimizes the chance of not being able to service an occasional large request. A strategy contrary to best fit might make sense because it tends to minimize the effects of external fragmentation. This is called worst fit, which always allocates the largest block on the list hoping that the remainder of the block will be useful for servicing a future request. In our example, the worst fit is the block of size 900, leaving a remainder of size 300. If there are a few unusually large requests, this approach will have less chance of servicing them. If requests generally tend to be of the same size, then this might be an effective strategy. Like best fit, worst fit requires searching the entire freelist at each memory request to find the largest block. Alternatively, the freelist can be ordered from largest to smallest free block, possibly by using a priority queue implementation. Which strategy is best? It depends on the expected types of memory requests. If the requests are of widely ranging size, best fit might work well. If the requests tend to be of similar size, with rare large and small requests, first or worst fit might work well. Unfortunately, there are always request patterns that one of the three sequential fit methods will service, but which the other two will not be able to service. For example, if the series of requests 600, 650, 900, 500, 100 is made to a freelist containing blocks 500, 700, 650, 900 (in that order), the requests can all be serviced by first fit, but not by best fit. Alternatively, the series of requests 600, 500, 700, 900 can be serviced by best fit but not by first fit on this same freelist. Buddy Methods Sequential-fit methods rely on a linked list of free blocks, which must be searched for a suitable block at each memory request. Thus, the time to find a suitable free block would be Θ(n) in the worst case for a freelist containing n blocks. Merging adjacent free blocks is somewhat complicated. Finally, we must either use addi- tional space for the linked list, or use space within the memory pool to support the memory manager operations. In the second option, both free and reserved blocks require tag and size fields. Fields in free blocks do not cost any space (because they are stored in memory that is not otherwise being used), but fields in reserved blocks create additional overhead. The buddy system solves most of these problems. Searching for a block of the proper size is efficient, merging adjacent free blocks is simple, and no tag or other information fields need be stored within reserved blocks. The buddy system
- 456. Sec. 12.3 Memory Management 437 assumes that memory is of size 2N for some integer N. Both free and reserved blocks will always be of size 2k for k ≤ N. At any given time, there might be both free and reserved blocks of various sizes. The buddy system keeps a separate list for free blocks of each size. There can be at most N such lists, because there can only be N distinct block sizes. When a request comes in for m words, we first determine the smallest value of k such that 2k ≥ m. A block of size 2k is selected from the free list for that block size if one exists. The buddy system does not worry about internal fragmentation: The entire block of size 2k is allocated. If no block of size 2k exists, the next larger block is located. This block is split in half (repeatedly if necessary) until the desired block of size 2k is created. Any other blocks generated as a by-product of this splitting process are placed on the appropriate freelists. The disadvantage of the buddy system is that it allows internal fragmentation. For example, a request for 257 words will require a block of size 512. The primary advantages of the buddy system are (1) there is less external fragmentation; (2) search for a block of the right size is cheaper than, say, best fit because we need only find the first available block on the block list for blocks of size 2k; and (3) merging adjacent free blocks is easy. The reason why this method is called the buddy system is because of the way that merging takes place. The buddy for any block of size 2k is another block of the same size, and with the same address (i.e., the byte position in memory, read as a binary value) except that the kth bit is reversed. For example, the block of size 8 with beginning address 0000 in Figure 12.14(a) has buddy with address 1000. Likewise, in Figure 12.14(b), the block of size 4 with address 0000 has buddy 0100. If free blocks are sorted by address value, the buddy can be found by searching the correct block size list. Merging simply requires that the address for the combined buddies be moved to the freelist for the next larger block size. Other Memory Allocation Methods In addition to sequential-fit and buddy methods, there are many ad hoc approaches to memory management. If the application is sufficiently complex, it might be desirable to break available memory into several memory zones, each with a differ- ent memory management scheme. For example, some zones might have a simple memory access pattern of first-in, first-out. This zone can therefore be managed ef- ficiently by using a simple stack. Another zone might allocate only records of fixed size, and so can be managed with a simple freelist as described in Section 4.1.2. Other zones might need one of the general-purpose memory allocation methods
- 457. 438 Chap. 12 Lists and Arrays Revisited (a) (b) 0000 1000 0000 0100 1000 1100 Buddies Buddies Buddies Figure 12.14 Example of the buddy system. (a) Blocks of size 8. (b) Blocks of size 4. discussed in this section. The advantage of zones is that some portions of memory can be managed more efficiently. The disadvantage is that one zone might fill up while other zones have excess memory if the zone sizes are chosen poorly. Another approach to memory management is to impose a standard size on all memory requests. We have seen an example of this concept already in disk file management, where all files are allocated in multiples of the cluster size. This approach leads to internal fragmentation, but managing files composed of clusters is easier than managing arbitrarily sized files. The cluster scheme also allows us to relax the restriction that the memory request be serviced by a contiguous block of memory. Most disk file managers and operating system main memory managers work on a cluster or page system. Block management is usually done with a buffer pool to allocate available blocks in main memory efficiently. 12.3.2 Failure Policies and Garbage Collection At some point during processing, a memory manager could encounter a request for memory that it cannot satisfy. In some situations, there might be nothing that can be done: There simply might not be enough free memory to service the request, and the application may require that the request be serviced immediately. In this case, the memory manager has no option but to return an error, which could in turn lead to a failure of the application program. However, in many cases there are alternatives to simply returning an error. The possible options are referred to collectively as failure policies. In some cases, there might be sufficient free memory to satisfy the request, but it is scattered among small blocks. This can happen when using a sequential-
- 458. Sec. 12.3 Memory Management 439 Memory Block Handle Figure 12.15 Using handles for dynamic memory management. The memory manager returns the address of the handle in response to a memory request. The handle stores the address of the actual memory block. In this way, the memory block might be moved (with its address updated in the handle) without disrupting the application program. fit memory allocation method, where external fragmentation has led to a series of small blocks that collectively could service the request. In this case, it might be possible to compact memory by moving the reserved blocks around so that the free space is collected into a single block. A problem with this approach is that the application must somehow be able to deal with the fact that all of its data have now been moved to different locations. If the application program relies on the absolute positions of the data in any way, this would be disastrous. One approach for dealing with this problem is the use of handles. A handle is a second level of indirection to a memory location. The memory allocation routine does not return a pointer to the block of storage, but rather a pointer to a variable that in turn points to the storage. This variable is the handle. The handle never moves its position, but the position of the block might be moved and the value of the handle updated. Figure 12.15 illustrates the concept. Another failure policy that might work in some applications is to defer the memory request until sufficient memory becomes available. For example, a multi- tasking operating system could adopt the strategy of not allowing a process to run until there is sufficient memory available. While such a delay might be annoying to the user, it is better than halting the entire system. The assumption here is that other processes will eventually terminate, freeing memory. Another option might be to allocate more memory to the memory manager. In a zoned memory allocation system where the memory manager is part of a larger system, this might be a viable option. In a Java program that implements its own memory manager, it might be possible to get more memory from the system-level new operator, such as is done by the freelist of Section 4.1.2. The last failure policy that we will consider is garbage collection. Consider the following series of statements.
- 459. 440 Chap. 12 Lists and Arrays Revisited a c d e f g h A B C Freelist Figure 12.16 Example of LISP list variables, including the system freelist. int[] p = new int[5]; int[] q = new int[10]; p = q; While in Java this would be no problem (due to auotmatic garbage collection), in some languages such as C++, this would be considered bad form because the original space allocated to p is lost as a result of the third assignment. This space cannot be used again by the program. Such lost memory is referred to as garbage, also known as a memory leak. When no program variable points to a block of space, no future access to that space is possible. Of course, if another variable had first been assigned to point to p’s space, then reassigning p would not create garbage. Some programming languages take a different view towards garbage. In par- ticular, the LISP programming language uses the multilist representation of Fig- ure 12.5, and all storage is in the form either of internal nodes with two pointers or atoms. Figure 12.16 shows a typical collection of LISP structures, headed by variables named A, B, and C, along with a freelist. In LISP, list objects are constantly being put together in various ways as tem- porary variables, and then all reference to them is lost when the object is no longer needed. Thus, garbage is normal in LISP, and in fact cannot be avoided during normal processing. When LISP runs out of memory, it resorts to a garbage collec- tion process to recover the space tied up in garbage. Garbage collection consists of
- 460. Sec. 12.3 Memory Management 441 examining the managed memory pool to determine which parts are still being used and which parts are garbage. In particular, a list is kept of all program variables, and any memory locations not reachable from one of these variables are considered to be garbage. When the garbage collector executes, all unused memory locations are placed in free store for future access. This approach has the advantage that it allows for easy collection of garbage. It has the disadvantage, from a user’s point of view, that every so often the system must halt while it performs garbage collec- tion. For example, garbage collection is noticeable in the Emacs text editor, which is normally implemented in LISP. Occasionally the user must wait for a moment while the memory management system performs garbage collection. The Java programming language also makes use of garbage collection. As in LISP, it is common practice in Java to allocate dynamic memory as needed, and to later drop all references to that memory. The garbage collector is responsible for reclaiming such unused space as necessary. This might require extra time when running the program, but it makes life considerably easier for the programmer. In contrast, many large applications written in C++ (even commonly used commercial software) contain memory leaks that will in time cause the program to fail. Several algorithms have been used for garbage collection. One is the reference count algorithm. Here, every dynamically allocated memory block includes space for a count field. Whenever a pointer is directed to a memory block, the reference count is increased. Whenever a pointer is directed away from a memory block, the reference count is decreased. If the count ever becomes zero, then the memory block is considered garbage and is immediately placed in free store. This approach has the advantage that it does not require an explicit garbage collection phase, be- cause information is put in free store immediately when it becomes garbage. The reference count algorithm is used by the UNIX file system. Files can have multiple names, called links. The file system keeps a count of the number of links to each file. Whenever a file is “deleted,” in actuality its link field is simply reduced by one. If there is another link to the file, then no space is recovered by the file system. Whenever the number of links goes to zero, the file’s space becomes available for reuse. Reference counts have several major disadvantages. First, a reference count must be maintained for each memory object. This works well when the objects are large, such as a file. However, it will not work well in a system such as LISP where the memory objects typically consist of two pointers or a value (an atom). Another major problem occurs when garbage contains cycles. Consider Figure 12.17. Here each memory object is pointed to once, but the collection of objects is still garbage because no pointer points to the collection. Thus, reference counts only work when
- 461. 442 Chap. 12 Lists and Arrays Revisited g h Figure 12.17 Garbage cycle example. All memory elements in the cycle have non-zero reference counts because each element has one pointer to it, even though the entire cycle is garbage. the memory objects are linked together without cycles, such as the UNIX file sys- tem where files can only be organized as a DAG. Another approach to garbage collection is the mark/sweep strategy. Here, each memory object needs only a single mark bit rather than a reference counter field. When free store is exhausted, a separate garbage collection phase takes place as follows. 1. Clear all mark bits. 2. Perform depth-first search (DFS) following pointers from each variable on the system’s list of variables. Each memory element encountered during the DFS has its mark bit turned on. 3. A “sweep” is made through the memory pool, visiting all elements. Un- marked elements are considered garbage and placed in free store. The advantages of the mark/sweep approach are that it needs less space than is necessary for reference counts, and it works for cycles. However, there is a major disadvantage. This is a “hidden” space requirement needed to do the processing. DFS is a recursive algorithm: Either it must be implemented recursively, in which case the compiler’s runtime system maintains a stack, or else the memory manager can maintain its own stack. What happens if all memory is contained in a single linked list? Then the depth of the recursion (or the size of the stack) is the number of memory cells! Unfortunately, the space for the DFS stack must be available at the worst conceivable time, that is, when free memory has been exhausted. Fortunately, a clever technique allows DFS to be performed without requiring additional space for a stack. Instead, the structure being traversed is used to hold the stack. At each step deeper into the traversal, instead of storing a pointer on the stack, we “borrow” the pointer being followed. This pointer is set to point back to the node we just came from in the previous step, as illustrated by Figure 12.18. Each borrowed pointer stores an additional bit to tell us whether we came down the left branch or the right branch of the link node being pointed to. At any given instant we have passed down only one path from the root, and we can follow the trail of pointers back up. As we return (equivalent to popping the recursion stack), we set the pointer back to its original position so as to return the structure to its
- 462. Sec. 12.4 Further Reading 443 (a) a b 4 prev e c curr (b) 4 6 6 2 2 3 5 1 3 1 5 a b c e Figure 12.18 Example of the Deutsch-Schorr-Waite garbage collection alg- orithm. (a) The initial multilist structure. (b) The multilist structure of (a) at the instant when link node 5 is being processed by the garbage collection alg- orithm. A chain of pointers stretching from variable prev to the head node of the structure has been (temporarily) created by the garbage collection algorithm. original condition. This is known as the Deutsch-Schorr-Waite garbage collection algorithm. 12.4 Further Reading An introductory text on operating systems covers many topics relating to memory management issues, including layout of files on disk and caching of information in main memory. All of the topics covered here on memory management, buffer pools, and paging are relevant to operating system implementation. For example, see Operating Systems by William Stallings[Sta05]. For information on LISP, see The Little LISPer by Friedman and Felleisen [FF89]. Another good LISP reference is Common LISP: The Language by Guy L. Steele [Ste84]. For information on Emacs, which is both an excellent text editor and a fully developed programming environment, see the GNU Emacs Manual by Richard M. Stallman [Sta07]. You can get more information about Java’s garbage collection system from The Java Programming Language by Ken Arnold and James Gosling [AG06].
- 463. 444 Chap. 12 Lists and Arrays Revisited (c) (b) (a) a b d e c a L1 L1 L2 L4 a b d c L3 Figure 12.19 Some example multilists. 12.5 Exercises 12.1 For each of the following bracket notation desriptions, draw the equivalent multilist in graphical form such as shown in Figure 12.2. (a) ha, b, hc, d, ei, hf, hgi, hii (b) ha, b, hc, d, L1:ei, L1i (c) hL1:a, L1, hL2:bi, L2, hL1ii 12.2 (a) Show the bracket notation for the list of Figure 12.19(a). (b) Show the bracket notation for the list of Figure 12.19(b). (c) Show the bracket notation for the list of Figure 12.19(c). 12.3 Given the linked representation of a pure list such as hx1, hy1, y2, hz1, z2i, y4i, hw1, w2i, x4i, write an in-place reversal algorithm to reverse the sublists at all levels in- cluding the topmost level. For this example, the result would be a linked representation corresponding to hx4, hw2, w1i, hy4, hz2, z1i, y2, y1i, x1i. 12.4 What fraction of the values in a matrix must be zero for the sparse matrix representation of Section 12.2 to be more space efficient than the standard two-dimensional matrix representation when data values require eight bytes, array indices require two bytes, and pointers require four bytes? 12.5 Write a function to add an element at a given position to the sparse matrix representation of Section 12.2. 12.6 Write a function to delete an element from a given position in the sparse matrix representation of Section 12.2. 12.7 Write a function to transpose a sparse matrix as represented in Section 12.2. 12.8 Write a function to add two sparse matrices as represented in Section 12.2.
- 464. Sec. 12.6 Projects 445 12.9 Write memory manager allocation and deallocation routines for the situation where all requests and releases follow a last-requested, first-released (stack) order. 12.10 Write memory manager allocation and deallocation routines for the situation where all requests and releases follow a last-requested, last-released (queue) order. 12.11 Show the result of allocating the following blocks from a memory pool of size 1000 using first fit for each series of block requests. State if a given request cannot be satisfied. (a) Take 300 (call this block A), take 500, release A, take 200, take 300. (b) Take 200 (call this block A), take 500, release A, take 200, take 300. (c) Take 500 (call this block A), take 300, release A, take 300, take 200. 12.12 Show the result of allocating the following blocks from a memory pool of size 1000 using best fit for each series of block requests. State if a given request cannot be satisfied. (a) Take 300 (call this block A), take 500, release A, take 200, take 300. (b) Take 200 (call this block A), take 500, release A, take 200, take 300. (c) Take 500 (call this block A), take 300, release A, take 300, take 200. 12.13 Show the result of allocating the following blocks from a memory pool of size 1000 using worst fit for each series of block requests. State if a given request cannot be satisfied. (a) Take 300 (call this block A), take 500, release A, take 200, take 300. (b) Take 200 (call this block A), take 500, release A, take 200, take 300. (c) Take 500 (call this block A), take 300, release A, take 300, take 200. 12.14 Assume that the memory pool contains three blocks of free storage. Their sizes are 1300, 2000, and 1000. Give examples of storage requests for which (a) first-fit allocation will work, but not best fit or worst fit. (b) best-fit allocation will work, but not first fit or worst fit. (c) worst-fit allocation will work, but not first fit or best fit. 12.6 Projects 12.1 Implement the sparse matrix representation of Section 12.2. Your implemen- tation should support the following operations on the matrix: • insert an element at a given position, • delete an element from a given position, • return the value of the element at a given position, • take the transpose of a matrix, and
- 465. 446 Chap. 12 Lists and Arrays Revisited • add two matrices. 12.2 Implement the MemManager ADT shown at the beginning of Section 12.3. Use a separate linked list to implement the freelist. Your implementation should work for any of the three sequential-fit methods: first fit, best fit, and worst fit. Test your system empirically to determine under what conditions each method performs well. 12.3 Implement the MemManager ADT shown at the beginning of Section 12.3. Do not use separate memory for the free list, but instead embed the free list into the memory pool as shown in Figure 12.12. Your implementation should work for any of the three sequential-fit methods: first fit, best fit, and worst fit. Test your system empirically to determine under what conditions each method performs well. 12.4 Implement the MemManager ADT shown at the beginning of Section 12.3 using the buddy method of Section 12.3.1. Your system should support requests for blocks of a specified size and release of previously requested blocks. 12.5 Implement the Deutsch-Schorr-Waite garbage collection algorithm that is il- lustrated by Figure 12.18.
- 466. 13 Advanced Tree Structures This chapter introduces several tree structures designed for use in specialized appli- cations. The trie of Section 13.1 is commonly used to store strings and is suitable for storing and searching collections of strings. It also serves to illustrate the con- cept of a key space decomposition. The AVL tree and splay tree of Section 13.2 are variants on the BST. They are examples of self-balancing search trees and have guaranteed good performance regardless of the insertion order for records. An introduction to several spatial data structures used to organize point data by xy- coordinates is presented in Section 13.3. Descriptions of the fundamental operations are given for each data structure. Because an important goal for this chapter is to provide material for class program- ming projects, detailed implementations are left for the reader. 13.1 Tries Recall that the shape of a BST is determined by the order in which its data records are inserted. One permutation of the records might yield a balanced tree while another might yield an unbalanced tree in the shape of a linked list. The reason is that the value of the key stored in the root node splits the key range into two parts: those key values less than the root’s key value, and those key values greater than the root’s key value. Depending on the relationship between the root node’s key value and the distribution of the key values for the other records in the the tree, the resulting BST might be balanced or unbalanced. Thus, the BST is an example of a data structure whose organization is based on an object space decomposition, so called because the decomposition of the key range is driven by the objects (i.e., the key values of the data records) stored in the tree. The alternative to object space decomposition is to predefine the splitting posi- tion within the key range for each node in the tree. In other words, the root could be 447
- 467. 448 Chap. 13 Advanced Tree Structures predefined to split the key range into two equal halves, regardless of the particular values or order of insertion for the data records. Those records with keys in the lower half of the key range will be stored in the left subtree, while those records with keys in the upper half of the key range will be stored in the right subtree. While such a decomposition rule will not necessarily result in a balanced tree (the tree will be unbalanced if the records are not well distributed within the key range), at least the shape of the tree will not depend on the order of key insertion. Further- more, the depth of the tree will be limited by the resolution of the key range; that is, the depth of the tree can never be greater than the number of bits required to store a key value. For example, if the keys are integers in the range 0 to 1023, then the resolution for the key is ten bits. Thus, two keys might be identical only until the tenth bit. In the worst case, two keys will follow the same path in the tree only until the tenth branch. As a result, the tree will never be more than ten levels deep. In contrast, a BST containing n records could be as much as n levels deep. Decomposition based on a predetermined subdivision of the key range is called key space decomposition. In computer graphics, a related technique is known as image space decomposition, and this term is sometimes applied to data structures based on key space decomposition as well. Any data structure based on key space decomposition is called a trie. Folklore has it that “trie” comes from “retrieval.” Unfortunately, that would imply that the word is pronounced “tree,” which would lead to confusion with regular use of the word “tree.” “Trie” is actually pronounced as “try.” Like the B+-tree, a trie stores data records only in leaf nodes. Internal nodes serve as placeholders to direct the search process. Figure 13.1 illustrates the trie concept. Upper and lower bounds must be imposed on the key values so that we can compute the middle of the key range. Because the largest value inserted in this example is 120, a range from 0 to 127 is assumed, as 128 is the smallest power of two greater than 120. The binary value of the key determines whether to select the left or right branch at any given point during the search. The most significant bit determines the branch direction at the root. Figure 13.1 shows a binary trie, so called because in this example the trie structure is based on the value of the key interpreted as a binary number, which results in a binary tree. The Huffman coding tree of Section 5.6 is another example of a binary trie. All data values in the Huffman tree are at the leaves, and each branch splits the range of possible letter codes in half. The Huffman codes are actually derived from the letter positions within the trie. These are examples of binary tries, but tries can be built with any branching factor. Normally the branching factor is determined by the alphabet used. For
- 468. Sec. 13.1 Tries 449 0 0 0 2 7 1 24 1 1 1 120 0 0 1 1 32 0 0 1 40 42 37 0 0 0 Figure 13.1 The binary trie for the collection of values 2, 7, 24, 31, 37, 40, 42, 120. All data values are stored in the leaf nodes. Edges are labeled with the value of the bit used to determine the branching direction of each node. The binary form of the key value determines the path to the record, assuming that each key is represented as a 7-bit value representing a number in the range 0 to 127. binary numbers, the alphabet is {0, 1} and a binary trie results. Other alphabets lead to other branching factors. One application for tries is storing a dictionary of words. Such a trie will be referred to as an alphabet trie. For simplicity, our examples will ignore case in letters. We add a special character ($) to the 26 standard English letters. The $ character is used to represent the end of a string. Thus, the branching factor for each node is (up to) 27. Once constructed, the alphabet trie is used to determine if a given word is in the dictionary. Consider searching for a word in the alphabet trie of Figure 13.2. The first letter of the search word determines which branch to take from the root, the second letter determines which branch to take at the next level, and so on. Only the letters that lead to a word are shown as branches. In Figure 13.2(b) the leaf nodes of the trie store a copy of the actual words, while in Figure 13.2(a) the word is built up from the letters associated with each branch. One way to implement a node of the alphabet trie is as an array of 27 pointers indexed by letter. Because most nodes have branches to only a small fraction of the possible letters in the alphabet, an alternate implementation is to use a linked list of pointers to the child nodes, as in Figure 6.9. The depth of a leaf node in the alphabet trie of Figure 13.2(b) has little to do with the number of nodes in the trie. Rather, a node’s depth depends on the number of characters required to distinguish this node’s word from any other. For example, if the words “anteater” and “antelope” are both stored in the trie, it is not until the fifth letter that the two words can be distinguished. Thus, these words must be stored at least as deep as level five. In general, the limiting factor on the depth of nodes in the alphabet trie is the length of the words stored.
- 469. 450 Chap. 13 Advanced Tree Structures e l u o (b) ant e l chicken d u deer duck g h l o horse goose goldfish goat antelope (a) n t $ a t e r $ o p e $ c h i c k n $ e r $ c k $ g o a $ l d f i s h $ r s e a d t o h $ e e s e $ a n t $ a c e o a anteater Figure 13.2 Two variations on the alphabet trie representation for a set of ten words. (a) Each node contains a set of links corresponding to single letters, and each letter in the set of words has a corresponding link. “$” is used to indicate the end of a word. Internal nodes direct the search and also spell out the word one letter per link. The word need not be stored explicitly. “$” is needed to recognize the existence of words that are prefixes to other words, such as ‘ant’ in this example. (b) Here the trie extends only far enough to discriminate between the words. Leaf nodes of the trie each store a complete word; internal nodes merely direct the search.
- 470. Sec. 13.1 Tries 451 1xxxxxx 0 120 01xxxxx 00xxxxx 2 3 0101xxx 4 24 4 5 010101x 2 7 32 37 40 42 000xxxx 0xxxxxx 1 Figure 13.3 The PAT trie for the collection of values 2, 7, 24, 32, 37, 40, 42, 120. Contrast this with the binary trie of Figure 13.1. In the PAT trie, all data values are stored in the leaf nodes, while internal nodes store the bit position used to determine the branching decision, assuming that each key is represented as a 7- bit value representing a number in the range 0 to 127. Some of the branches in this PAT trie have been labeled to indicate the binary representation for all values in that subtree. For example, all values in the left subtree of the node labeled 0 must have value 0xxxxxx (where x means that bit can be either a 0 or a 1). All nodes in the right subtree of the node labeled 3 must have value 0101xxx. However, we can skip branching on bit 2 for this subtree because all values currently stored have a value of 0 for that bit. Poor balance and clumping can result when certain prefixes are heavily used. For example, an alphabet trie storing the common words in the English language would have many words in the “th” branch of the tree, but none in the “zq” branch. Any multiway branching trie can be replaced with a binary trie by replacing the original trie’s alphabet with an equivalent binary code. Alternatively, we can use the techniques of Section 6.3.4 for converting a general tree to a binary tree without modifying the alphabet. The trie implementations illustrated by Figures 13.1 and 13.2 are potentially quite inefficient as certain key sets might lead to a large number of nodes with only a single child. A variant on trie implementation is known as PATRICIA, which stands for “Practical Algorithm To Retrieve Information Coded In Alphanumeric.” In the case of a binary alphabet, a PATRICIA trie (referred to hereafter as a PAT trie) is a full binary tree that stores data records in the leaf nodes. Internal nodes store only the position within the key’s bit pattern that is used to decide on the next branching point. In this way, internal nodes with single children (equivalently, bit positions within the key that do not distinguish any of the keys within the current subtree) are eliminated. A PAT trie corresponding to the values of Figure 13.1 is shown in Figure 13.3.
- 471. 452 Chap. 13 Advanced Tree Structures Example 13.1 When searching for the value 7 (0000111 in binary) in the PAT trie of Figure 13.3, the root node indicates that bit position 0 (the leftmost bit) is checked first. Because the 0th bit for value 7 is 0, take the left branch. At level 1, branch depending on the value of bit 1, which again is 0. At level 2, branch depending on the value of bit 2, which again is 0. At level 3, the index stored in the node is 4. This means that bit 4 of the key is checked next. (The value of bit 3 is irrelevant, because all values stored in that subtree have the same value at bit position 3.) Thus, the single branch that extends from the equivalent node in Figure 13.1 is just skipped. For key value 7, bit 4 has value 1, so the rightmost branch is taken. Because this leads to a leaf node, the search key is compared against the key stored in that node. If they match, then the desired record has been found. Note that during the search process, only a single bit of the search key is com- pared at each internal node. This is significant, because the search key could be quite large. Search in the PAT trie requires only a single full-key comparison, which takes place once a leaf node has been reached. Example 13.2 Consider the situation where we need to store a library of DNA sequences. A DNA sequence is a series of letters, usually many thou- sands of characters long, with the string coming from an alphabet of only four letters that stand for the four amino acids making up a DNA strand. Similar DNA seqences might have long sections of ther string that are iden- tical. The PAT trie woudl avoid making multiple full key comparisons when searching for a specific sequence. 13.2 Balanced Trees We have noted several times that the BST has a high risk of becoming unbalanced, resulting in excessively expensive search and update operations. One solution to this problem is to adopt another search tree structure such as the 2-3 tree. An al- ternative is to modify the BST access functions in some way to guarantee that the tree performs well. This is an appealing concept, and it works well for heaps, whose access functions maintain the heap in the shape of a complete binary tree. Unfortunately, requiring that the BST always be in the shape of a complete binary tree requires excessive modification to the tree during update, as discussed in Sec- tion 10.3.
- 472. Sec. 13.2 Balanced Trees 453 If we are willing to weaken the balance requirements, we can come up with alternative update routines that perform well both in terms of cost for the update and in balence for the resulting tree structure. The AVL tree works in this way, using insertion and deletion routines altered from those of the BST to ensure that, for every node, the depths of the left and right subtrees differ by at most one. The AVL tree is described in Section 13.2.1. A different approach to improving the performance of the BST is to not require that the tree always be balanced, but rather to expend some effort toward making the BST more balanced every time it is accessed. This is a little like the idea of path compression used by the UNION/FIND algorithm presented in Section 6.2. One example of such a compromise is called the splay tree. The splay tree is described in Section 13.2.2. 13.2.1 The AVL Tree The AVL tree (named for its inventors Adelson-Velskii and Landis) should be viewed as a BST with the following additional property: For every node, the heights of its left and right subtrees differ by at most 1. As long as the tree maintains this property, if the tree contains n nodes, then it has a depth of at most O(log n). As a result, search for any node will cost O(log n), and if the updates can be done in time proportional to the depth of the node inserted or deleted, then updates will also cost O(log n), even in the worst case. The key to making the AVL tree work is to make the proper alterations to the insert and delete routines so as to maintain the balance property. Of course, to be practical, we must be able to implement the revised update routines in Θ(log n) time. Consider what happens when we insert a node with key value 5, as shown in Figure 13.4. The tree on the left meets the AVL tree balance requirements. After the insertion, two nodes no longer meet the requirements. Because the original tree met the balance requirement, nodes in the new tree can only be unbalanced by a difference of at most 2 in the subtrees. For the bottommost unbalanced node, call it S, there are 4 cases: 1. The extra node is in the left child of the left child of S. 2. The extra node is in the right child of the left child of S. 3. The extra node is in the left child of the right child of S. 4. The extra node is in the right child of the right child of S. Cases 1 and 4 are symmetrical, as are cases 2 and 3. Note also that the unbalanced nodes must be on the path from the root to the newly inserted node.
- 473. 454 Chap. 13 Advanced Tree Structures 7 2 32 42 40 120 37 42 24 7 2 32 42 40 120 37 42 24 5 Figure 13.4 Example of an insert operation that violates the AVL tree balance property. Prior to the insert operation, all nodes of the tree are balanced (i.e., the depths of the left and right subtrees for every node differ by at most one). After inserting the node with value 5, the nodes with values 7 and 24 are no longer balanced. S X C X S B B C A (a) A (b) Figure 13.5 A single rotation in an AVL tree. This operation occurs when the excess node (in subtree A) is in the left child of the left child of the unbalanced node labeled S. By rearranging the nodes as shown, we preserve the BST property, as well as rebalance the tree to preserve the AVL tree balance property. The case where the excess node is in the right child of the right child of the unbalanced node is handled in the same way. Our problem now is how to balance the tree in O(log n) time. It turns out that we can do this using a series of local operations known as rotations. Cases 1 and 4 can be fixed using a single rotation, as shown in Figure 13.5. Cases 2 and 3 can be fixed using a double rotation, as shown in Figure 13.6. The AVL tree insert algorithm begins with a normal BST insert. Then as the re- cursion unwinds up the tree, we perform the appropriate rotation on any node that is
- 474. Sec. 13.2 Balanced Trees 455 Y S Y X A B S C D X C A (a) B D (b) Figure 13.6 A double rotation in an AVL tree. This operation occurs when the excess node (in subtree B) is in the right child of the left child of the unbalanced node labeled S. By rearranging the nodes as shown, we preserve the BST property, as well as rebalance the tree to preserve the AVL tree balance property. The case where the excess node is in the left child of the right child of S is handled in the same way. found to be unbalanced. Deletion is similar; however consideration for unbalanced nodes must begin at the level of the deletemin operation. Example 13.3 In Figure 13.4 (b), the bottom-most unbalanced node has value 7. The excess node (with value 5) is in the right subtree of the left child of 7, so we have an example of Case 2. This requires a double rotation to fix. After the rotation, 5 becomes the left child of 24, 2 becomes the left child of 5, and 7 becomes the right child of 5. 13.2.2 The Splay Tree Like the AVL tree, the splay tree is not actually a data structure per se but rather is a collection of rules for improving the performance of a BST. These rules govern modifications made to the BST whenever a search, insert, or delete operation is performed. Their purpose is to provide guarantees on the time required by a se- ries of operations, thereby avoiding the worst-case linear time behavior of standard BST operations. No single operation in the splay tree is guaranteed to be efficient. Instead, the splay tree access rules guarantee that a series of m operations will take
- 475. 456 Chap. 13 Advanced Tree Structures O(m log n) time for a tree of n nodes whenever m ≥ n. Thus, a single insert or search operation could take O(n) time. However, m such operations are guaranteed to require a total of O(m log n) time, for an average cost of O(log n) per access operation. This is a desirable performance guarantee for any search-tree structure. Unlike the AVL tree, the splay tree is not guaranteed to be height balanced. What is guaranteed is that the total cost of the entire series of accesses will be cheap. Ultimately, it is the cost of the series of operations that matters, not whether the tree is balanced. Maintaining balance is really done only for the sake of reaching this time efficiency goal. The splay tree access functions operate in a manner reminiscent of the move- to-front rule for self-organizing lists from Section 9.2, and of the path compres- sion technique for managing parent-pointer trees from Section 6.2. These access functions tend to make the tree more balanced, but an individual access will not necessarily result in a more balanced tree. Whenever a node S is accessed (e.g., when S is inserted, deleted, or is the goal of a search), the splay tree performs a process called splaying. Splaying moves S to the root of the BST. When S is being deleted, splaying moves the parent of S to the root. As in the AVL tree, a splay of node S consists of a series of rotations. A rotation moves S higher in the tree by adjusting its position with respect to its parent and grandparent. A side effect of the rotations is a tendency to balance the tree. There are three types of rotation. A single rotation is performed only if S is a child of the root node. The single rotation is illustrated by Figure 13.7. It basically switches S with its parent in a way that retains the BST property. While Figure 13.7 is slightly different from Figure 13.5, in fact the splay tree single rotation is identical to the AVL tree single rotation. Unlike the AVL tree, the splay tree requires two types of double rotation. Dou- ble rotations involve S, its parent (call it P), and S’s grandparent (call it G). The effect of a double rotation is to move S up two levels in the tree. The first double rotation is called a zigzag rotation. It takes place when either of the following two conditions are met: 1. S is the left child of P, and P is the right child of G. 2. S is the right child of P, and P is the left child of G. In other words, a zigzag rotation is used when G, P, and S form a zigzag. The zigzag rotation is illustrated by Figure 13.8. The other double rotation is known as a zigzig rotation. A zigzig rotation takes place when either of the following two conditions are met:
- 476. Sec. 13.2 Balanced Trees 457 P S (a) C S A P A B B C (b) Figure 13.7 Splay tree single rotation. This rotation takes place only when the node being splayed is a child of the root. Here, node S is promoted to the root, rotating with node P. Because the value of S is less than the value of P, P must become S’s right child. The positions of subtrees A, B, and C are altered as appropriate to maintain the BST property, but the contents of these subtrees remains unchanged. (a) The original tree with P as the parent. (b) The tree after a rotation takes place. Performing a single rotation a second time will return the tree to its original shape. Equivalently, if (b) is the initial configuration of the tree (i.e., S is at the root and P is its right child), then (a) shows the result of a single rotation to splay P to the root. (a) (b) S G S P A B G C P C D A B D Figure 13.8 Splay tree zigzag rotation. (a) The original tree with S, P, and G in zigzag formation. (b) The tree after the rotation takes place. The positions of subtrees A, B, C, and D are altered as appropriate to maintain the BST property.
- 477. 458 Chap. 13 Advanced Tree Structures (a) S (b) C D B G B A C S A P G D P Figure 13.9 Splay tree zigzig rotation. (a) The original tree with S, P, and G in zigzig formation. (b) The tree after the rotation takes place. The positions of subtrees A, B, C, and D are altered as appropriate to maintain the BST property. 1. S is the left child of P, which is in turn the left child of G. 2. S is the right child of P, which is in turn the right child of G. Thus, a zigzig rotation takes place in those situations where a zigzag rotation is not appropriate. The zigzig rotation is illustrated by Figure 13.9. While Figure 13.9 appears somewhat different from Figure 13.6, in fact the zigzig rotation is identical to the AVL tree double rotation. Note that zigzag rotations tend to make the tree more balanced, because they bring subtrees B and C up one level while moving subtree D down one level. The result is often a reduction of the tree’s height by one. Zigzig promotions do not typically reduce the height of the tree; they merely bring the newly accessed record toward the root. Splaying node S involves a series of double rotations until S reaches either the root or the child of the root. Then, if necessary, a single rotation makes S the root. This process tends to rebalance the tree. In any case, it will make frequently accessed nodes stay near the top of the tree, resulting in reduced access cost. Proof that the splay tree does in fact meet the guarantee of O(m log n) is beyond the scope of this book. Such a proof can be found in the references in Section 13.4. Example 13.4 Consider a search for value 89 in the splay tree of Fig- ure 13.10(a). The splay tree’s search operation is identical to searching in a BST. However, once the value has been found, it is splayed to the root. Three rotations are required in this example. The first is a zigzig rotation,
- 478. Sec. 13.3 Spatial Data Structures 459 whose result is shown in Figure 13.10(b). The second is a zigzag rotation, whose result is shown in Figure 13.10(c). The final step is a single rotation resulting in the tree of Figure 13.10(d). Notice that the splaying process has made the tree shallower. 13.3 Spatial Data Structures All of the search trees discussed so far — BSTs, AVL trees, splay trees, 2-3 trees, B-trees, and tries — are designed for searching on a one-dimensional key. A typical example is an integer key, whose one-dimensional range can be visualized as a number line. These various tree structures can be viewed as dividing this one- dimensional numberline into pieces. Some databases require support for multiple keys, that is, records can be searched based on any one of several keys. Typically, each such key has its own one- dimensional index, and any given search query searches one of these independent indices as appropriate. A multidimensional search key presents a rather different concept. Imagine that we have a database of city records, where each city has a name and an xy- coordinate. A BST or splay tree provides good performance for searches on city name, which is a one-dimensional key. Separate BSTs could be used to index the x- and y-coordinates. This would allow us to insert and delete cities, and locate them by name or by one coordinate. However, search on one of the two coordinates is not a natural way to view search in a two-dimensional space. Another option is to combine the xy-coordinates into a single key, say by concatenating the two coor- dinates, and index cities by the resulting key in a BST. That would allow search by coordinate, but would not allow for efficient two-dimensional range queries such as searching for all cities within a given distance of a specified point. The problem is that the BST only works well for one-dimensional keys, while a coordinate is a two-dimensional key where neither dimension is more important than the other. Multidimensional range queries are the defining feature of a spatial applica- tion. Because a coordinate gives a position in space, it is called a spatial attribute. To implement spatial applications efficiently requires the use of spatial data struc- tures. Spatial data structures store data objects organized by position and are an important class of data structures used in geographic information systems, com- puter graphics, robotics, and many other fields. This section presents two spatial data structures for storing point data in two or more dimensions. They are the k-d tree and the PR quadtree. The k-d tree is a
- 479. 460 Chap. 13 Advanced Tree Structures S 25 42 99 G P S 75 17 G 99 18 72 42 75 (a) (b) 18 72 89 (c) (d) 17 P S 25 18 72 42 89 92 18 72 42 75 99 89 17 25 92 99 75 92 25 89 P 17 92 Figure 13.10 Example of splaying after performing a search in a splay tree. After finding the node with key value 89, that node is splayed to the root by per- forming three rotations. (a) The original splay tree. (b) The result of performing a zigzig rotation on the node with key value 89 in the tree of (a). (c) The result of performing a zigzag rotation on the node with key value 89 in the tree of (b). (d) The result of performing a single rotation on the node with key value 89 in the tree of (c). If the search had been for 91, the search would have been unsuccessful with the node storing key value 89 being that last one visited. In that case, the same splay operations would take place.
- 480. Sec. 13.3 Spatial Data Structures 461 natural extension of the BST to multiple dimensions. It is a binary tree whose split- ting decisions alternate among the key dimensions. Like the BST, the k-d tree uses object space decomposition. The PR quadtree uses key space decomposition and so is a form of trie. It is a binary tree only for one-dimensional keys (in which case it is a trie with a binary alphabet). For d dimensions it has 2d branches. Thus, in two dimensions, the PR quadtree has four branches (hence the name “quadtree”), split- ting space into four equal-sized quadrants at each branch. Section 13.3.3 briefly mentions two other variations on these data structures, the bintree and the point quadtree. These four structures cover all four combinations of object versus key space decomposition on the one hand, and multi-level binary versus 2d-way branch- ing on the other. Section 13.3.4 briefly discusses spatial data structures for storing other types of spatial data. 13.3.1 The K-D Tree The k-d tree is a modification to the BST that allows for efficient processing of multidimensional keys. The k-d tree differs from the BST in that each level of the k-d tree makes branching decisions based on a particular search key associated with that level, called the discriminator. We define the discriminator at level i to be i mod k for k dimensions. For example, assume that we store data organized by xy-coordinates. In this case, k is 2 (there are two coordinates), with the x- coordinate field arbitrarily designated key 0, and the y-coordinate field designated key 1. At each level, the discriminator alternates between x and y. Thus, a node N at level 0 (the root) would have in its left subtree only nodes whose x values are less than Nx (because x is search key 0, and 0 mod 2 = 0). The right subtree would contain nodes whose x values are greater than Nx. A node M at level 1 would have in its left subtree only nodes whose y values are less than My. There is no restriction on the relative values of Mx and the x values of M’s descendants, because branching decisions made at M are based solely on the y coordinate. Figure 13.11 shows an example of how a collection of two-dimensional points would be stored in a k-d tree. In Figure 13.11 the region containing the points is (arbitrarily) restricted to a 128 × 128 square, and each internal node splits the search space. Each split is shown by a line, vertical for nodes with x discriminators and horizontal for nodes with y discriminators. The root node splits the space into two parts; its children further subdivide the space into smaller parts. The children’s split lines do not cross the root’s split line. Thus, each node in the k-d tree helps to decompose the space into rectangles that show the extent of where nodes can fall in the various subtrees.
- 481. 462 Chap. 13 Advanced Tree Structures B A D C (a) E x y y x B (15, 70) A (40, 45) C (70, 10) D (69, 50) F (85, 90) (b) E (66, 85) F Figure 13.11 Example of a k-d tree. (a) The k-d tree decomposition for a 128 × 128-unit region containing seven data points. (b) The k-d tree for the region of (a). Searching a k-d tree for the record with a specified xy-coordinate is like search- ing a BST, except that each level of the k-d tree is associated with a particular dis- criminator. Example 13.5 Consider searching the k-d tree for a record located at P = (69, 50). First compare P with the point stored at the root (record A in Figure 13.11). If P matches the location of A, then the search is successful. In this example the positions do not match (A’s location (40, 45) is not the same as (69, 50)), so the search must continue. The x value of A is compared with that of P to determine in which direction to branch. Because Ax’s value of 40 is less than P’s x value of 69, we branch to the right subtree (all cities with x value greater than or equal to 40 are in the right subtree). Ay does not affect the decision on which way to branch at this level. At the second level, P does not match record C’s position, so another branch must be taken. However, at this level we branch based on the relative y values of point P and record C (because 1 mod 2 = 1, which corresponds to the y-coordinate). Because Cy’s value of 10 is less than Py’s value of 50, we branch to the right. At this point, P is compared against the position of D. A match is made and the search is successful. As with a BST, if the search process reaches a null pointer, then the search point is not contained in the tree. Here is an implementation for k-d tree search,
- 482. Sec. 13.3 Spatial Data Structures 463 equivalent to the findhelp function of the BST class. Note that KD class private member D stores the key’s dimension. private E findhelp(KDNodeE rt, int[] key, int level) { if (rt == null) return null; E it = rt.element(); int[] itkey = rt.key(); if ((itkey[0] == key[0]) (itkey[1] == key[1])) return rt.element(); if (itkey[level] key[level]) return findhelp(rt.left(), key, (level+1)%D); else return findhelp(rt.right(), key, (level+1)%D); } Inserting a new node into the k-d tree is similar to BST insertion. The k-d tree search procedure is followed until a null pointer is found, indicating the proper place to insert the new node. Example 13.6 Inserting a record at location (10, 50) in the k-d tree of Figure 13.11 first requires a search to the node containing record B. At this point, the new record is inserted into B’s left subtree. Deleting a node from a k-d tree is similar to deleting from a BST, but slightly harder. As with deleting from a BST, the first step is to find the node (call it N) to be deleted. It is then necessary to find a descendant of N which can be used to replace N in the tree. If N has no children, then N is replaced with a null pointer. Note that if N has one child that in turn has children, we cannot simply assign N’s parent to point to N’s child as would be done in the BST. To do so would change the level of all nodes in the subtree, and thus the discriminator used for a search would also change. The result is that the subtree would no longer be a k-d tree because a node’s children might now violate the BST property for that discriminator. Similar to BST deletion, the record stored in N should be replaced either by the record in N’s right subtree with the least value of N’s discriminator, or by the record in N’s left subtree with the greatest value for this discriminator. Assume that N was at an odd level and therefore y is the discriminator. N could then be replaced by the record in its right subtree with the least y value (call it Ymin). The problem is that Ymin is not necessarily the leftmost node, as it would be in the BST. A modified search procedure to find the least y value in the left subtree must be used to find it instead. The implementation for findmin is shown in Figure 13.12. A recursive call to the delete routine will then remove Ymin from the tree. Finally, Ymin’s record is substituted for the record in node N.
- 483. 464 Chap. 13 Advanced Tree Structures private KDNodeE findmin(KDNodeE rt, int descrim, int level) { KDNodeE temp1, temp2; int[] key1 = null; int[] key2 = null; if (rt == null) return null; temp1 = findmin(rt.left(), descrim, (level+1)%D); if (temp1 != null) key1 = temp1.key(); if (descrim != level) { temp2 = findmin(rt.right(), descrim, (level+1)%D); if (temp2 != null) key2 = temp2.key(); if ((temp1 == null) || ((temp2 != null) (key1[descrim] key2[descrim]))) temp1 = temp2; key1 = key2; } // Now, temp1 has the smaller value int[] rtkey = rt.key(); if ((temp1 == null) || (key1[descrim] rtkey[descrim])) return rt; else return temp1; } Figure 13.12 The k-d tree findmin method. On levels using the minimum value’s discriminator, branching is to the left. On other levels, both children’s subtrees must be visited. Helper function min takes two nodes and a discriminator as input, and returns the node with the smaller value in that discriminator. Note that we can replace the node to be deleted with the least-valued node from the right subtree only if the right subtree exists. If it does not, then a suitable replacement must be found in the left subtree. Unfortunately, it is not satisfactory to replace N’s record with the record having the greatest value for the discriminator in the left subtree, because this new value might be duplicated. If so, then we would have equal values for the discriminator in N’s left subtree, which violates the ordering rules for the k-d tree. Fortunately, there is a simple solution to the problem. We first move the left subtree of node N to become the right subtree (i.e., we simply swap the values of N’s left and right child pointers). At this point, we proceed with the normal deletion process, replacing the record of N to be deleted with the record containing the least value of the discriminator from what is now N’s right subtree. Assume that we want to print out a list of all records that are within a certain distance d of a given point P. We will use Euclidean distance, that is, point P is defined to be within distance d of point N if1 q (Px − Nx)2 + (Py − Ny)2 ≤ d. 1 A more efficient computation is (Px − Nx)2 + (Py − Ny)2 ≤ d2 . This avoids performing a square root function.
- 484. Sec. 13.3 Spatial Data Structures 465 A C Figure 13.13 Function InCircle must check the Euclidean distance between a record and the query point. It is possible for a record A to have x- and y- coordinates each within the query distance of the query point C, yet have A itself lie outside the query circle. If the search process reaches a node whose key value for the discriminator is more than d above the corresponding value in the search key, then it is not possible that any record in the right subtree can be within distance d of the search key be- cause all key values in that dimension are always too great. Similarly, if the current node’s key value in the discriminator is d less than that for the search key value, then no record in the left subtree can be within the radius. In such cases, the sub- tree in question need not be searched, potentially saving much time. In the average case, the number of nodes that must be visited during a range query is linear on the number of data records that fall within the query circle. Example 13.7 Find all cities in the k-d tree of Figure 13.14 within 25 units of the point (25, 65). The search begins with the root node, which contains record A. Because (40, 45) is exactly 25 units from the search point, it will be reported. The search procedure then determines which branches of the tree to take. The search circle extends to both the left and the right of A’s (vertical) dividing line, so both branches of the tree must be searched. The left subtree is processed first. Here, record B is checked and found to fall within the search circle. Because the node storing B has no children, pro- cessing of the left subtree is complete. Processing of A’s right subtree now begins. The coordinates of record C are checked and found not to fall within the circle. Thus, it should not be reported. However, it is possible that cities within C’s subtrees could fall within the search circle even if C does not. As C is at level 1, the discriminator at this level is the y-coordinate. Be- cause 65 − 25 10, no record in C’s left subtree (i.e., records above C) could possibly be in the search circle. Thus, C’s left subtree (if it had one) need not be searched. However, cities in C’s right subtree could fall within
- 485. 466 Chap. 13 Advanced Tree Structures B A D C E F Figure 13.14 Searching in the k-d treeof Figure 13.11. (a) The k-d tree decom- position for a 128×128-unit region containing seven data points. (b) The k-d tree for the region of (a). the circle. Thus, search proceeds to the node containing record D. Again, D is outside the search circle. Because 25 + 25 69, no record in D’s right subtree could be within the search circle. Thus, only D’s left subtree need be searched. This leads to comparing record E’s coordinates against the search circle. Record E falls outside the search circle, and processing is complete. So we see that we only search subtrees whose rectangles fall within the search circle. Figure 13.15 shows an implementation for the region search method. When a node is visited, function InCircle is used to check the Euclidean distance between the node’s record and the query point. It is not enough to simply check that the differences between the x- and y-coordinates are each less than the query distances because the the record could still be outside the search circle, as illustrated by Figure 13.13. 13.3.2 The PR quadtree In the Point-Region Quadtree (hereafter referred to as the PR quadtree) each node either has exactly four children or is a leaf. That is, the PR quadtree is a full four- way branching (4-ary) tree in shape. The PR quadtree represents a collection of data points in two dimensions by decomposing the region containing the data points into four equal quadrants, subquadrants, and so on, until no leaf node contains more than a single point. In other words, if a region contains zero or one data points, then it is represented by a PR quadtree consisting of a single leaf node. If the region con-
- 486. Sec. 13.3 Spatial Data Structures 467 private void rshelp(KDNodeE rt, int[] point, int radius, int lev) { if (rt == null) return; int[] rtkey = rt.key(); if (InCircle(point, radius, rtkey)) System.out.println(rt.element()); if (rtkey[lev] (point[lev] - radius)) rshelp(rt.left(), point, radius, (lev+1)%D); if (rtkey[lev] (point[lev] + radius)) rshelp(rt.right(), point, radius, (lev+1)%D); } Figure 13.15 The k-d tree region search method. tains more than a single data point, then the region is split into four equal quadrants. The corresponding PR quadtree then contains an internal node and four subtrees, each subtree representing a single quadrant of the region, which might in turn be split into subquadrants. Each internal node of a PR quadtree represents a single split of the two-dimensional region. The four quadrants of the region (or equiva- lently, the corresponding subtrees) are designated (in order) NW, NE, SW, and SE. Each quadrant containing more than a single point would in turn be recursively di- vided into subquadrants until each leaf of the corresponding PR quadtree contains at most one point. For example, consider the region of Figure 13.16(a) and the corresponding PR quadtree in Figure 13.16(b). The decomposition process demands a fixed key range. In this example, the region is assumed to be of size 128 × 128. Note that the internal nodes of the PR quadtree are used solely to indicate decomposition of the region; internal nodes do not store data records. Because the decomposition lines are predetermined (i.e, key-space decomposition is used), the PR quadtree is a trie. Search for a record matching point Q in the PR quadtree is straightforward. Beginning at the root, we continuously branch to the quadrant that contains Q until our search reaches a leaf node. If the root is a leaf, then just check to see if the node’s data record matches point Q. If the root is an internal node, proceed to the child that contains the search coordinate. For example, the NW quadrant of Figure 13.16 contains points whose x and y values each fall in the range 0 to 63. The NE quadrant contains points whose x value falls in the range 64 to 127, and whose y value falls in the range 0 to 63. If the root’s child is a leaf node, then that child is checked to see if Q has been found. If the child is another internal node, the search process continues through the tree until a leaf node is found. If this leaf node stores a record whose position matches Q then the query is successful; otherwise Q is not in the tree.
- 487. 468 Chap. 13 Advanced Tree Structures (a) 0 0 127 127 A D C B E F (b) nw (40,45) A C (69,50) (70, 10) ne sw B se (15,70) (55,80) (80, 90) E F D Figure 13.16 Example of a PR quadtree. (a) A map of data points. We de- fine the region to be square with origin at the upper-left-hand corner and sides of length 128. (b) The PR quadtree for the points in (a). (a) also shows the block decomposition imposed by the PR quadtree for this region. Inserting record P into the PR quadtree is performed by first locating the leaf node that contains the location of P. If this leaf node is empty, then P is stored at this leaf. If the leaf already contains P (or a record with P’s coordinates), then a duplicate record should be reported. If the leaf node already contains another record, then the node must be repeatedly decomposed until the existing record and P fall into different leaf nodes. Figure 13.17 shows an example of such an insertion. Deleting a record P is performed by first locating the node N of the PR quadtree that contains P. Node N is then changed to be empty. The next step is to look at N’s three siblings. N and its siblings must be merged together to form a single node N ′ if only one point is contained among them. This merging process continues until some level is reached at which at least two points are contained in the subtrees rep- resented by node N ′ and its siblings. For example, if point C is to be deleted from the PR quadtree representing Figure 13.17(b), the resulting node must be merged with its siblings, and that larger node again merged with its siblings to restore the PR quadtree to the decomposition of Figure 13.17(a). Region search is easily performed with the PR quadtree. To locate all points within radius r of query point Q, begin at the root. If the root is an empty leaf node, then no data points are found. If the root is a leaf containing a data record, then the location of the data point is examined to determine if it falls within the circle. If the root is an internal node, then the process is performed recursively, but only on those subtrees containing some part of the search circle.
- 488. Sec. 13.3 Spatial Data Structures 469 C A B (a) (b) B A Figure 13.17 PR quadtree insertion example. (a) The initial PR quadtree con- taining two data points. (b) The result of inserting point C. The block contain- ing A must be decomposed into four subblocks. Points A and C would still be in the same block if only one subdivision takes place, so a second decomposition is required to separate them. Let us now consider how structure of the PR quadtree affects the design of its node representation. The PR quadtree is actually a trie (as defined in Section 13.1). Decomposition takes place at the mid-points for internal nodes, regardless of where the data points actually fall. The placement of the data points does determine whether a decomposition for a node takes place, but not where the decomposi- tion for the node takes place. Internal nodes of the PR quadtree are quite different from leaf nodes, in that internal nodes have children (leaf nodes do not) and leaf nodes have data fields (internal nodes do not). Thus, it is likely to be beneficial to represent internal nodes differently from leaf nodes. Finally, there is the fact that approximately half of the leaf nodes will contain no data field. Another issue to consider is: How does a routine traversing the PR quadtree get the coordinates for the square represented by the current PR quadtree node? One possibility is to store with each node its spatial description (such as upper-left corner and width). However, this will take a lot of space — perhaps as much as the space needed for the data records, depending on what information is being stored. Another possibility is to pass in the coordinates when the recursive call is made. For example, consider the search process. Initially, the search visits the root node of the tree, which has origin at (0, 0), and whose width is the full size of the space being covered. When the appropriate child is visited, it is a simple matter for the search routine to determine the origin for the child, and the width of the square is simply half that of the parent. Not only does passing in the size and position infor-
- 489. 470 Chap. 13 Advanced Tree Structures mation for a node save considerable space, but avoiding storing such information in the nodes we enables a good design choice for empty leaf nodes, as discussed next. How should we represent empty leaf nodes? On average, half of the leaf nodes in a PR quadtree are empty (i.e., do not store a data point). One implementation option is to use a null pointer in internal nodes to represent empty nodes. This will solve the problem of excessive space requirements. There is an unfortunate side effect that using a null pointer requires the PR quadtree processing meth- ods to understand this convention. In other words, you are breaking encapsulation on the node representation because the tree now must know things about how the nodes are implemented. This is not too horrible for this particular application, be- cause the node class can be considered private to the tree class, in which case the node implementation is completely invisible to the outside world. However, it is undesirable if there is another reasonable alternative. Fortunately, there is a good alternative. It is called the Flyweight design pattern. In the PR quadtree, a flyweight is a single empty leaf node that is reused in all places where an empty leaf node is needed. You simply have all of the internal nodes with empty leaf children point to the same node object. This node object is created once at the beginning of the program, and is never removed. The node class recognizes from the pointer value that the flyweight is being accessed, and acts accordingly. Note that when using the Flyweight design pattern, you cannot store coordi- nates for the node in the node. This is an example of the concept of intrinsic versus extrinsic state. Intrinsic state for an object is state information stored in the ob- ject. If you stored the coordinates for a node in the node object, those coordinates would be intrinsic state. Extrinsic state is state information about an object stored elsewhere in the environment, such as in global variables or passed to the method. If your recursive calls that process the tree pass in the coordinates for the current node, then the coordinates will be extrinsic state. A flyweight can have in its intrin- sic state only information that is accurate for all instances of the flyweight. Clearly coordinates do not qualify, because each empty leaf node has its own location. So, if you want to use a flyweight, you must pass in coordinates. Another design choice is: Who controls the work, the node class or the tree class? For example, on an insert operation, you could have the tree class control the flow down the tree, looking at (querying) the nodes to see their type and reacting accordingly. This is the approach used by the BST implementation in Section 5.4. An alternate approach is to have the node class do the work. That is, you have an insert method for the nodes. If the node is internal, it passes the city record to the appropriate child (recursively). If the node is a flyweight, it replaces itself with a
- 490. Sec. 13.3 Spatial Data Structures 471 new leaf node. If the node is a full node, it replaces itself with a subtree. This is an example of the Composite design pattern, discussed in Section 5.3.1. 13.3.3 Other Point Data Structures The differences between the k-d tree and the PR quadtree illustrate many of the design choices encountered when creating spatial data structures. The k-d tree pro- vides an object space decomposition of the region, while the PR quadtree provides a key space decomposition (thus, it is a trie). The k-d tree stores records at all nodes, while the PR quadtree stores records only at the leaf nodes. Finally, the two trees have different structures. The k-d tree is a binary tree, while the PR quadtree is a full tree with 2d branches (in the two-dimensional case, 22 = 4). Consider the extension of this concept to three dimensions. A k-d tree for three dimen- sions would alternate the discriminator through the x, y, and z dimensions. The three-dimensional equivalent of the PR quadtree would be a tree with 23 or eight branches. Such a tree is called an octree. We can also devise a binary trie based on a key space decomposition in each dimension, or a quadtree that uses the two-dimensional equivalent to an object space decomposition. The bintree is a binary trie that uses keyspace decomposition and alternates discriminators at each level in a manner similar to the k-d tree. The bintree for the points of Figure 13.11 is shown in Figure 13.18. Alternatively, we can use a four-way decomposition of space centered on the data points. The tree resulting from such a decomposition is called a point quadtree. The point quadtree for the data points of Figure 13.11 is shown in Figure 13.19. 13.3.4 Other Spatial Data Structures This section has barely scratched the surface of the field of spatial data structures. By now dozens of distinct spatial data structures have been invented, many with variations and alternate implementations. Spatial data structures exist for storing many forms of spatial data other than points. The most important distinctions be- tween are the tree structure (binary or not, regular decompositions or not) and the decomposition rule used to decide when the data contained within a region is so complex that the region must be subdivided. Perhaps the best known spatial data structure is the “region quadtree” for stor- ing images where the pixel values tend to be blocky, such as a map of the countries of the world. The region quadtree uses a four-way regular decomposition scheme similar to the PR quadtree. The decompostion rule is simply to divide any node containing pixels of more than one color or value.
- 491. 472 Chap. 13 Advanced Tree Structures x y A x B y C D E F x y E F D A B C (a) (b) Figure 13.18 An example of the bintree, a binary tree using key space decom- position and discriminators rotating among the dimensions. Compare this with the k-d tree of Figure 13.11 and the PR quadtree of Figure 13.16. 127 0 0 127 (a) B A C F E D (b) C nw sw ne se D A B E F Figure 13.19 An example of the point quadtree, a 4-ary tree using object space decomposition. Compare this with the PR quadtree of Figure 13.11.
- 492. Sec. 13.4 Further Reading 473 Spatial data structures can also be used to store line object, rectangle object, or objects of arbitrary shape (such as polygons in two dimensions or polyhedra in three dimensions). A simple, yet effective, data structure for storing rectangles or arbitrary polygonal shapes can be derived from the PR quadtree. Pick a threshold value c, and subdivide any region into four quadrants if it contains more than c objects. A special case must be dealt with when more than c object intersect. Some of the most interesting developments in spatial data structures have to do with adapting them for disk-based applications. However, all such disk-based implementations boil down to storing the spatial data structure within some variant on either B-trees or hashing. 13.4 Further Reading PATRICIA tries and other trie implementations are discussed in Information Re- trieval: Data Structures Algorithms, Frakes and Baeza-Yates, eds. [FBY92]. See Knuth [Knu97] for a discussion of the AVL tree. For further reading on splay trees, see “Self-adjusting Binary Search” by Sleator and Tarjan [ST85]. The world of spatial data structures is rich and rapidly evolving. For a good introduction, see Foundations of Multidimensional and Metric Data Structures by Hanan Samet [Sam06]. This is also the best reference for more information on the PR quadtree. The k-d tree was invented by John Louis Bentley. For further information on the k-d tree, in addition to [Sam06], see [Ben75]. For information on using a quadtree to store arbitrary polygonal objects, see [SH92]. For a discussion on the relative space requirements for two-way versus multi- way branching, see “A Generalized Comparison of Quadtree and Bintree Storage Requirements” by Shaffer, Juvvadi, and Heath [SJH93]. Closely related to spatial data structures are data structures for storing multidi- mensional data (which might not necessarily be spatial in nature). A popular data structure for storing such data is the R-tree, originally proposed by Guttman [Gut84]. 13.5 Exercises 13.1 Show the binary trie (as illustrated by Figure 13.1) for the following collec- tion of values: 42, 12, 100, 10, 50, 31, 7, 11, 99. 13.2 Show the PAT trie (as illustrated by Figure 13.3) for the following collection of values: 42, 12, 100, 10, 50, 31, 7, 11, 99. 13.3 Write the insertion routine for a binary trie as shown in Figure 13.1. 13.4 Write the deletion routine for a binary trie as shown in Figure 13.1.
- 493. 474 Chap. 13 Advanced Tree Structures 13.5 (a) Show the result (including appropriate rotations) of inserting the value 39 into the AVL tree on the left in Figure 13.4. (b) Show the result (including appropriate rotations) of inserting the value 300 into the AVL tree on the left in Figure 13.4. (c) Show the result (including appropriate rotations) of inserting the value 50 into the AVL tree on the left in Figure 13.4. (d) Show the result (including appropriate rotations) of inserting the value 1 into the AVL tree on the left in Figure 13.4. 13.6 Show the splay tree that results from searching for value 75 in the splay tree of Figure 13.10(d). 13.7 Show the splay tree that results from searching for value 18 in the splay tree of Figure 13.10(d). 13.8 Some applications do not permit storing two records with duplicate key val- ues. In such a case, an attempt to insert a duplicate-keyed record into a tree structure such as a splay tree should result in a failure on insert. What is the appropriate action to take in a splay tree implementation when the insert routine is called with a duplicate-keyed record? 13.9 Show the result of deleting point A from the k-d tree of Figure 13.11. 13.10 (a) Show the result of building a k-d tree from the following points (in- serted in the order given). A (20, 20), B (10, 30), C (25, 50), D (35, 25), E (30, 45), F (30, 35), G (55, 40), H (45, 35), I (50, 30). (b) Show the result of deleting point A from the tree you built in part (a). 13.11 (a) Show the result of deleting F from the PR quadtree of Figure 13.16. (b) Show the result of deleting records E and F from the PR quadtree of Figure 13.16. 13.12 (a) Show the result of building a PR quadtree from the following points (inserted in the order given). Assume the tree is representing a space of 64 by 64 units. A (20, 20), B (10, 30), C (25, 50), D (35, 25), E (30, 45), F (30, 35), G (45, 25), H (45, 30), I (50, 30). (b) Show the result of deleting point C from the tree you built in part (a). (c) Show the result of deleting point F from the resulting tree in part (b). 13.13 On average, how many leaf nodes of a PR quadtree will typically be empty? Explain why. 13.14 When performing a region search on a PR quadtree, we need only search those subtrees of an internal node whose corresponding square falls within the query circle. This is most easily computed by comparing the x and y ranges of the query circle against the x and y ranges of the square corre- sponding to the subtree. However, as illustrated by Figure 13.13, the x and y ranges might overlap without the circle actually intersecting the square. Write a function that accurately determines if a circle and a square intersect.
- 494. Sec. 13.6 Projects 475 13.15 (a) Show the result of building a bintree from the following points (inserted in the order given). Assume the tree is representing a space of 64 by 64 units. A (20, 20), B (10, 30), C (25, 50), D (35, 25), E (30, 45), F (30, 35), G (45, 25), H (45, 30), I (50, 30). (b) Show the result of deleting point C from the tree you built in part (a). (c) Show the result of deleting point F from the resulting tree in part (b). 13.16 Compare the trees constructed for Exercises 12 and 15 in terms of the number of internal nodes, full leaf nodes, empty leaf nodes, and total depths of the two trees. 13.17 Show the result of building a point quadtree from the following points (in- serted in the order given). Assume the tree is representing a space of 64 by 64 units. A (20, 20), B (10, 30), C (25, 50), D (35, 25), E (30, 45), F (31, 35), G (45, 26), H (44, 30), I (50, 30). 13.6 Projects 13.1 Use the trie data structure to devise a program to sort variable-length strings. The program’s running time should be proportional to the total number of letters in all of the strings. Note that some strings might be very long while most are short. 13.2 Define the set of suffix strings for a string S to be S, S without its first char- acter, S without its first two characters, and so on. For example, the complete set of suffix strings for “HELLO” would be {HELLO, ELLO, LLO, LO, O}. A suffix tree is a PAT trie that contains all of the suffix strings for a given string, and associates each suffix with the complete string. The advantage of a suffix tree is that it allows a search for strings using “wildcards.” For example, the search key “TH*” means to find all strings with “TH” as the first two characters. This can easily be done with a regular trie. Searching for “*TH” is not efficient in a regular trie, but it is efficient in a suffix tree. Implement the suffix tree for a dictionary of words or phrases, with support for wildcard search. 13.3 Revise the BST class of Section 5.4 to use the AVL tree rotations. Your new implementation should not modify the original BST class ADT. Compare your AVL tree against an implementation of the standard BST over a wide variety of input data. Under what conditions does the splay tree actually save time?
- 495. 476 Chap. 13 Advanced Tree Structures 13.4 Revise the BST class of Section 5.4 to use the splay tree rotations. Your new implementation should not modify the original BST class ADT. Compare your splay tree against an implementation of the standard BST over a wide variety of input data. Under what conditions does the splay tree actually save time? 13.5 Implement a city database using the k-d tree. Each database record contains the name of the city (a string of arbitrary length) and the coordinates of the city expressed as integer x- and y-coordinates. Your database should allow records to be inserted, deleted by name or coordinate, and searched by name or coordinate. You should also support region queries, that is, a request to print all records within a given distance of a specified point. 13.6 Implement a city database using the PR quadtree. Each database record con- tains the name of the city (a string of arbitrary length) and the coordinates of the city expressed as integer x- and y-coordinates. Your database should allow records to be inserted, deleted by name or coordinate, and searched by name or coordinate. You should also support region queries, that is, a request to print all records within a given distance of a specified point. 13.7 Implement a city database using the bintree. Each database record contains the name of the city (a string of arbitrary length) and the coordinates of the city expressed as integer x- and y-coordinates. Your database should allow records to be inserted, deleted by name or coordinate, and searched by name or coordinate. You should also support region queries, that is, a request to print all records within a given distance of a specified point. 13.8 Implement a city database using the point quadtree. Each database record contains the name of the city (a string of arbitrary length) and the coordinates of the city expressed as integer x- and y-coordinates. Your database should allow records to be inserted, deleted by name or coordinate, and searched by name or coordinate. You should also support region queries, that is, a request to print all records within a given distance of a specified point. 13.9 Use the PR quadtree to implement an efficient solution to Problem 6.5. That is, store the set of points in a PR quadtree. For each point, the PR quadtree is used to find those points within distance D that should be equivalenced. What is the asymptotic complexity of this solution? 13.10 Select any two of the point representations described in this chapter (i.e., the k-d tree, the PR quadtree, the bintree, and the point quadtree). Implement your two choices and compare them over a wide range of data sets. Describe which is easier to implement, which appears to be more space efficient, and which appears to be more time efficient.
- 496. Sec. 13.6 Projects 477 13.11 Implement a representation for a collection of (two dimensional) rectangles using a quadtree based on regular decomposition. Assume that the space being represented is a square whose width and height are some power of two. Rectangles are assumed to have integer coordinates and integer width and height. Pick some value c, and use as a decomposition rule that a region is subdivided into four equal-sized regions whenever it contains more that c rectangles. A special case occurs if all of these rectangles intersect at some point within the current region (because decomposing such a node would never reach termination). In this situation, the node simply stores pointers to more than c rectangles. Try your representation on data sets of rectangles with varying values of c.
- 498. PART V Theory of Algorithms 479
- 500. 14 Analysis Techniques This book contains many examples of asymptotic analysis of the time requirements for algorithms and the space requirements for data structures. Often it is easy to invent an equation to model the behavior of the algorithm or data structure in ques- tion, and also easy to derive a closed-form solution for the equation should it con- tain a recurrence or summation. Sometimes an analysis proves more difficult. It may take a clever insight to de- rive the right model, such as the snowplow argument for analyzing the average run length resulting from Replacement Selection (Section 8.5.2). In this case, once the snowplow argument is understood, the resulting equations are simple. Sometimes, developing the model is straightforward but analyzing the resulting equations is not. An example is the average-case analysis for Quicksort. The equation given in Section 7.5 simply enumerates all possible cases for the pivot position, sum- ming corresponding costs for the recursive calls to Quicksort. However, deriving a closed-form solution for the resulting recurrence relation is not as easy. Many iterative algorithms require that we compute a summation to determine the cost of a loop. Techniques for finding closed-form solutions to summations are presented in Section 14.1. Time requirements for many algorithms based on recursion are best modeled by recurrence relations. A discussion of techniques for solving recurrences is provided in Section 14.2. These sections extend the introduc- tion to summations and recurrences provided in Section 2.4, so the reader should already be familiar with that material. Section 14.3 provides an introduction to the topic of amortized analysis. Am- ortized analysis deals with the cost of a series of operations. Perhaps a single opera- tion in the series has high cost, but as a result the cost of the remaining operations is limited in such a way that the entire series can be done efficiently. Amortized anal- ysis has been used successfully to analyze several of the algorithms presented in 481
- 501. 482 Chap. 14 Analysis Techniques this book, including the cost of a series of UNION/FIND operations (Section 6.2), the cost of a series of splay tree operations (Section 13.2), and the cost of a series of operations on self-organizing lists (Section 9.2). Section 14.3 discusses the topic in more detail. 14.1 Summation Techniques We begin our study of techniques for finding the closed-form solution to a summa- tion by considering the simple example n X i=1 i. In Section 2.6.3 it was proved by induction that this summation has the well-known closed form n(n + 1)/2. But while induction is a good technique for proving that a proposed closed-form expression is correct, how do we find a candidate closed- form expression to test in the first place? Let us try to approach this summation from first principles, as though we had never seen it before. A good place to begin analyzing a summation it is to give an estimate of its value for a given n. Observe that the biggest term for this summation is n, and there are n terms being summed up. So the total must be less than n2. Actually, most terms are much less than n, and the sizes of the terms grows linearly. If we were to draw a picture with bars for the size of the terms, their heights would form a line, and we could enclose them in a box n units wide and n units high. It is easy to see from this that a closer estimate for the summation is about (n2)/2. Having this estimate in hand helps us when trying to determine an exact closed-form solution, because we will hopefully recognize if our proposed solution is badly wrong. Let us now consider some ways that we might hit upon an exact value for the closed form solution to this summation. One particularly clever approach we can take is to observe that we can “pair up” the first and last terms, the second and (n − 1)th terms, and so on. Each pair sums to n + 1. The number of pairs is n/2. Thus, the solution is n(n + 1)/2. This is pretty, and there’s no doubt about it being correct. The problem is that it is not a useful technique for solving many other summations. Now let us try to do something a bit more general. We already recognized that, because the largest term is n and there are n terms, the summation is less than n2. If we are lucky, the closed form solution is a polynomial. Using that as a working assumption, we can invoke a technique called guess-and-test. We will guess that the closed-form solution for this summation is a polynomial of the form
- 502. Sec. 14.1 Summation Techniques 483 c1n2 + c2n + c3 for some constants c1, c2, and c3. If this is the case, we can plug in the answers to small cases of the summation to solve for the coefficients. For this example, substituting 0, 1, and 2 for n leads to three simultaneous equations. Because the summation when n = 0 is just 0, c3 must be 0. For n = 1 and n = 2 we get the two equations c1 + c2 = 1 4c1 + 2c2 = 3, which in turn yield c1 = 1/2 and c2 = 1/2. Thus, if the closed-form solution for the summation is a polynomial, it can only be 1/2n2 + 1/2n + 0 which is more commonly written n(n + 1) 2 . At this point, we still must do the “test” part of the guess-and-test approach. We can use an induction proof to verify whether our candidate closed-form solu- tion is correct. In this case it is indeed correct, as shown by Example 2.11. The induction proof is necessary because our initial assumption that the solution is a simple polynomial could be wrong. For example, it might have been possible that the true solution includes a logarithmic term, such as c1n2 +c2n log n. The process shown here is essentially fitting a curve to a fixed number of points. Because there is always an n-degree polynomial that fits n + 1 points, we had not done enough work to be sure that we to know the true equation without the induction proof. Guess-and-test is useful whenever the solution is a polynomial expression. In particular, similar reasoning can be used to solve for Pn i=1 i2, or more generally Pn i=1 ic for c any positive integer. Why is this not a universal approach to solving summations? Because many summations do not have a polynomial as their closed form solution. A more general approach is based on the subtract-and-guess or divide-and- guess strategies. One form of subtract-and-guess is known as the shifting method. The shifting method subtracts the summation from a variation on the summation. The variation selected for the subtraction should be one that makes most of the terms cancel out. To solve sum f, we pick a known function g and find a pattern in terms of f(n) − g(n) or f(n)/g(n).
- 503. 484 Chap. 14 Analysis Techniques Example 14.1 Find the closed form solution for Pn i=1 i using the divide- and-guess approach. We will try two example functions to illustrate the divide-and-guess method: dividing by n and dividing by f(n − 1). Our goal is to find patterns that we can use to guess a closed-form expression as our candidate for testing with an induction proof. To aid us in finding such patterns, we can construct a table showing the first few numbers of each function, and the result of dividing one by the other, as follows. n 1 2 3 4 5 6 7 8 9 10 f(n) 1 3 6 10 15 21 28 36 46 57 n 1 2 3 4 5 6 7 8 9 10 f(n)/n 2/2 3/2 4/2 5/2 6/2 7/2 8/2 9/2 10/2 11/2 f(n − 1) 0 1 3 6 10 15 21 28 36 46 f(n)/f(n − 1) 3/1 4/2 5/3 6/4 7/5 8/6 9/7 10/8 11/9 Dividing by both n and f(n − 1) happen to give us useful patterns to work with. f(n) n = n+1 2 , and f(n) f(n−1) = n+1 n−1 . Of course, lots of other approaches do not work. For example, f(n)−n = f(n−1). Knowing that f(n) = f(n−1)+n is not useful for determining the closed form solution to this summation. Or consider f(n) − f(n − 1) = n. Again, knowing that f(n) = f(n − 1) + n is not useful. Finding the right combination of equations can be like finding a needle in a haystack. In our first example, we can see directly what the closed-form solution should be. f(n) n = n + 1 2 Obviously, f(n) = n(n + 1)/2. Dividing f(n) by f(n − 1) does not give so obvious a result, but it provides another useful illustration. f(n) f(n − 1) = n + 1 n − 1 f(n)(n − 1) = (n + 1)f(n − 1) f(n)(n − 1) = (n + 1)(f(n) − n) nf(n) − f(n) = nf(n) + f(n) − n2 − n 2f(n) = n2 + n = n(n + 1) f(n) = n(n + 1) 2
- 504. Sec. 14.1 Summation Techniques 485 Once again, we still do not have a proof that f(n) = n(n+1)/2. Why? Because we did not prove that f(n)/n = (n + 1)/2 nor that f(n)/f(n − 1) = (n + 1)(n − 1). We merely hypothesized patterns from looking at a few terms. Fortunately, it is easy to check our hypothesis with induction. Example 14.2 Solve the summation n X i=1 1/2i . We will begin by writing out a table of the first few values of the summation and see if we can detect a pattern. n 1 2 3 4 f(n) 1 2 3 4 7 8 15 16 g(n) 1 1 1 1 g(n) − f(n) 1 2 1 4 1 8 1 16 By direct inspection of the second line of the table, we might recognize the pattern f(n) = 2n−1 2n . A simple induction proof can then prove that this always holds true. Alternatively, consider if we hadn’t noticed the pattern for the form of f(n). We might observe that f(n) appears to be reach- ing an asymptote at one. In which case, we might consider looking at the difference between f(n) and t