Data Structure and Algorithms course slides

CSCI 2309
Data Structure
and Algorithms
Dr. Ing- Wazen SHBAIR
wazen.shbair@gmail.com
DSA.WEEK1.Lecture3
2023 - 2024
1

Analysis of Algorithms
Algorithm
Input Output
© 2014 Goodrich, Tamassia, Goldwasser 2
Presentation for use with the textbook Data Structures and
Algorithms in Java, 6th edition, by M. T. Goodrich, R. Tamassia,
and M. H. Goldwasser, Wiley, 2014

Analysis of Algorithms 3
Running Time
q Most algorithms transform
input objects into output
objects.
q The running time of an
algorithm typically grows
with the input size.
q Average case time is often
difficult to determine.
q We focus on the worst case
running time.
n Easier to analyze
n Crucial to applications such as
games, finance and robotics
0
20
40
60
80
100
120
Running
Time
1000 2000 3000 4000
Input Size
best case
average case
worst case
© 2014 Goodrich, Tamassia, Goldwasser

Experimental
Studies
q Write a program
implementing the
algorithm
q Run the program with
inputs of varying size
and composition,
noting the time
needed:
q Plot the results
0
1000
2000
3000
4000
5000
6000
7000
8000
9000
0 50 100
Input Size
Time
(ms)

Limitations of Experiments
q It is necessary to implement the
algorithm, which may be difficult
q Results may not be indicative of the
running time on other inputs not included
in the experiment.
q In order to compare two algorithms, the
same hardware and software
environments must be used

Theoretical Analysis
q Uses a high-level description of the
algorithm instead of an implementation
q Characterizes running time as a
function of the input size, n
q Takes into account all possible inputs
q Allows us to evaluate the speed of an
algorithm independent of the
hardware/software environment

Pseudocode
q High-level description of an algorithm
q More structured than English prose
q Less detailed than a program
q Preferred notation for describing
algorithms
q Hides program design issues

Pseudocode Details
q Control flow
n if … then … [else …]
n while … do …
n repeat … until …
n for … do …
n Indentation replaces braces
q Method declaration
Algorithm method (arg [, arg…])
Input …
Output …
q Method call
method (arg [, arg…])
q Return value
return expression
q Expressions:
¬Assignment
= Equality testing
n2 Superscripts and other
mathematical
formatting allowed

The Random Access Machine (RAM) Model
A RAM consists of
q A CPU
q An potentially unbounded bank
of memory cells, each of which
can hold an arbitrary number or
character
q Memory cells are numbered and
accessing any cell in memory
takes unit time
0
1
2

Seven Important Functions
q Seven functions that
often appear in algorithm
analysis:
n Constant » 1
n Logarithmic » log n
n Linear » n
n N-Log-N » n log n
n Quadratic » n2
n Cubic » n3
n Exponential » 2n
q In a log-log chart, the
slope of the line
corresponds to the
growth rate
1E+0
1E+2
1E+4
1E+6
1E+8
1E+10
1E+12
1E+14
1E+16
1E+18
1E+20
1E+22
1E+24
1E+26
1E+28
1E+30
1E+0 1E+2 1E+4 1E+6 1E+8 1E+10
n
T
(n
)
Cubic
Quadratic
Linear

Functions Graphed
Using “Normal” Scale
g(n) = 2n
g(n) = 1
g(n) = lg n
g(n) = n lg n
g(n) = n
g(n) = n2
g(n) = n3
Slide by Matt Stallmann
included with permission.

Primitive Operations
q Basic computations
performed by an algorithm
q Identifiable in pseudocode
q Largely independent from the
programming language
q Exact definition not important
(we will see why later)
q Assumed to take a constant
amount of time in the RAM
model
q Examples:
n Evaluating an
expression
n Assigning a value
to a variable
n Indexing into an
array
n Calling a method
n Returning from a
method

Counting Primitive Operations
q By inspecting the pseudocode, we can determine the maximum
number of primitive operations executed by an algorithm, as a
function of the input size
q Step 3: 2 ops, 4: 2 ops, 5: 2n ops,
6: 2n ops, 7: 0 to n ops, 8: 1 op

Estimating Running Time
q Algorithm arrayMax executes 5n + 5 primitive
operations in the worst case, 4n + 5 in the best
case. Define:
a = Time taken by the fastest primitive operation
b = Time taken by the slowest primitive operation
q Let T(n) be worst-case time of arrayMax. Then
a (4n + 5) £ T(n) £ b(5n + 5)
q Hence, the running time T(n) is bounded by two
linear functions

Growth Rate of Running Time
q Changing the hardware/ software
environment
n Affects T(n) by a constant factor, but
n Does not alter the growth rate of T(n)
q The linear growth rate of the running
time T(n) is an intrinsic property of
algorithm arrayMax

Why Growth Rate Matters
if runtime
is...
time for n + 1 time for 2 n time for 4 n
c lg n c lg (n + 1) c (lg n + 1) c(lg n + 2)
c n c (n + 1) 2c n 4c n
c n lg n
~ c n lg n
+ c n
2c n lg n +
2cn
4c n lg n +
4cn
c n2 ~ c n2 + 2c n 4c n2 16c n2
c n3 ~ c n3 + 3c n2 8c n3 64c n3
c 2n c 2 n+1 c 2 2n c 2 4n
runtime
quadruples
when
problem
size doubles

Comparison of Two Algorithms
insertion sort is
n2 / 4
merge sort is
2 n lg n
sort a million items?
insertion sort takes
roughly 70 hours
while
merge sort takes
roughly 40 seconds
This is a slow machine, but if
100 x as fast then it’s 40 minutes
versus less than 0.5 seconds

Constant Factors
q The growth rate is
not affected by
n constant factors or
n lower-order terms
q Examples
n 102n + 105 is a linear
function
n 105n2 + 108n is a
quadratic function
1E+0
1E+2
1E+4
1E+6
1E+8
1E+10
1E+12
1E+14
1E+16
1E+18
1E+20
1E+22
1E+24
1E+26
1E+0 1E+2 1E+4 1E+6 1E+8 1E+10
n
T
(n
)
Quadratic
Quadratic
Linear
Linear

Big-Oh Notation
q Given functions f(n) and
g(n), we say that f(n) is
O(g(n)) if there are
positive constants
c and n0 such that
f(n) £ cg(n) for n ³ n0
q Example: 2n + 10 is O(n)
n 2n + 10 £ cn
n (c - 2) n ³ 10
n n ³ 10/(c - 2)
n Pick c = 3 and n0 = 10
1
10
100
1,000
10,000
1 10 100 1,000
n
3n
2n+10
n

Big-Oh Example
q Example: the function
n2 is not O(n)
n n2 £ cn
n n £ c
n The above inequality
cannot be satisfied
since c must be a
constant
1
10
100
1,000
10,000
100,000
1,000,000
1 10 100 1,000
n
n^2
100n
10n
n

More Big-Oh Examples
q 7n - 2
7n-2 is O(n)
need c > 0 and n0 ³ 1 such that 7 n - 2 £ c n for n ³ n0
this is true for c = 7 and n0 = 1
q 3 n3 + 20 n2 + 5
3 n3 + 20 n2 + 5 is O(n3)
need c > 0 and n0 ³ 1 such that 3 n3 + 20 n2 + 5 £ c n3 for n ³ n0
q 3 log n + 5
3 log n + 5 is O(log n)
need c > 0 and n0 ³ 1 such that 3 log n + 5 £ c log n for n ³ n0

Big-Oh and Growth Rate
q The big-Oh notation gives an upper bound on the
growth rate of a function
q The statement “f(n) is O(g(n))” means that the growth
rate of f(n) is no more than the growth rate of g(n)
q We can use the big-Oh notation to rank functions
according to their growth rate
f(n) is O(g(n)) g(n) is O(f(n))
g(n) grows more Yes No
f(n) grows more No Yes
Same growth Yes Yes

Big-Oh Rules
q If is f(n) a polynomial of degree d, then f(n) is
O(nd), i.e.,
1. Drop lower-order terms
2. Drop constant factors
q Use the smallest possible class of functions
n Say “2n is O(n)” instead of “2n is O(n2)”
q Use the simplest expression of the class
n Say “3n + 5 is O(n)” instead of “3n + 5 is O(3n)”

Asymptotic Algorithm Analysis
q The asymptotic analysis of an algorithm determines
the running time in big-Oh notation
q To perform the asymptotic analysis
n We find the worst-case number of primitive operations
executed as a function of the input size
n We express this function with big-Oh notation
q Example:
n We say that algorithm arrayMax “runs in O(n) time”
q Since constant factors and lower-order terms are
eventually dropped anyhow, we can disregard them
when counting primitive operations

Computing Prefix Averages
q We further illustrate
asymptotic analysis with
two algorithms for prefix
averages
q The i-th prefix average of
an array X is average of the
first (i + 1) elements of X:
A[i] = (X[0] + X[1] + … + X[i])/(i+1)
q Computing the array A of
prefix averages of another
array X has applications to
financial analysis
0
5
10
15
20
25
30
35
1 2 3 4 5 6 7
X
A

Prefix Averages (Quadratic)
The following algorithm computes prefix averages in
quadratic time by applying the definition

Arithmetic Progression
q The running time of
prefixAverage1 is
O(1 + 2 + …+ n)
q The sum of the first n
integers is n(n + 1) / 2
n There is a simple visual
proof of this fact
q Thus, algorithm
prefixAverage1 runs in
O(n2) time
0
1
2
3
4
5
6
7
1 2 3 4 5 6

Prefix Averages 2 (Linear)
The following algorithm uses a running summation to
improve the efficiency
Algorithm prefixAverage2 runs in O(n) time!

Math you need to Review
q Properties of powers:
a(b+c) = aba c
abc = (ab)c
ab /ac = a(b-c)
b = a log
a
b
bc = a c*log
a
b
q Properties of logarithms:
logb(xy) = logbx + logby
logb (x/y) = logbx - logby
logbxa = alogbx
logba = logxa/logxb
q Summations
q Powers
q Logarithms
q Proof techniques
q Basic probability
© 2014 Goodrich, Tamassia, Goldwasser Analysis of Algorithms 29

Relatives of Big-Oh
big-Omega
n f(n) is W(g(n)) if there is a constant c > 0
and an integer constant n0 ³ 1 such that
f(n) ³ c g(n) for n ³ n0
big-Theta
n f(n) is Q(g(n)) if there are constants c’ > 0 and
c’’ > 0 and an integer constant n0 ³ 1 such that
c’g(n) £ f(n) £ c’’g(n) for n ³ n0

Intuition for Asymptotic
Notation
big-Oh
n f(n) is O(g(n)) if f(n) is asymptotically
less than or equal to g(n)
big-Omega
n f(n) is W(g(n)) if f(n) is asymptotically
greater than or equal to g(n)
big-Theta
n f(n) is Q(g(n)) if f(n) is asymptotically
equal to g(n)

Example Uses of the
Relatives of Big-Oh
f(n) is Q(g(n)) if it is W(n2) and O(n2). We have already seen the former,
for the latter recall that f(n) is O(g(n)) if there is a constant c > 0 and an
integer constant n0 ³ 1 such that f(n) < c g(n) for n ³ n0
Let c = 5 and n0 = 1
n 5n2 is Q(n2)
f(n) is W(g(n)) if there is a constant c > 0 and an integer constant n0 ³ 1
such that f(n) ³ c g(n) for n ³ n0
let c = 1 and n0 = 1
n 5n2 is W(n)
f(n) is W(g(n)) if there is a constant c > 0 and an integer constant n0 ³ 1
such that f(n) ³ c g(n) for n ³ n0
let c = 5 and n0 = 1
n 5n2 is W(n2)

Data Structure and Algorithms course slides

Recommended

More Related Content

Similar to Data Structure and Algorithms course slides (20)

Recently uploaded (20)

Data Structure and Algorithms course slides