Lecture 3 insertion sort and complexity analysis

Lecture 3 : Analysis of
Algorithms & Insertion Sort
Jayavignesh T
Asst Professor
SENSE

Time Complexity
• Amount of computer time required by an algorithm
to run on completion
• Difficult to compute time complexity in terms of
physically clocked time.
• Drawbacks of measuring running time in-terms of
seconds, millisecond etc are
– Dependence of speed of a underlying hardware
– Number of other programs running (System load)
– Dependence of compiler used in generating
machine code

How to calculate running time then?
• Time complexity given in terms of FREQUENCY
COUNT
• Count denoting number of times of execution of
statement.
For (i=0; i <n;i++) { // St1 : 1, St 2 : n+1 , St 3 : n times
sum = sum + a[i]; // n times
}
3n + 2 ; O(n) neglecting constants and lower order terms

for (i=0; i < n ; i ++) // 1 ; n+1 ; n times
{
for (j=0; j < n ; j ++) // n ; n(n+1) ; n(n)
{
c[i][j] = a[i][j] + b[i][j];
}
}
3n2+4n+ 2 = O(n2)

Time Complexity
• Number of steps required by an algorithm varies
with the size of the problem it is solving.
• Normally expressed as order of magnitude
– eg O(n2)
– Size of problem doubles then the algorithm will
take 4 times as many steps to complete

• All Algorithms run longer on larger inputs
• Algorithm’s efficiency - f(n)
• Identify the most important operations of the
algorithm – BASIC Operation
• Basic operation – contributing to most of total
running time
• Compute the number of times basic operation is
executed (mostly in inner loop)
Ex : Sorting Algorithms – Comparison (< >)
Matrix Multiplication, Polynomial evaluation – Arithmetic Operations ( *, +)
= (assignment), ==(equality) etc..

Lecture 3 insertion sort and complexity analysis

Order of Growth of Algorithm
• Measuring the performance of an algorithm
in relation with input size n
• Cannot says it equals n2 , but it grows like n2

Rate of Growth of Algorithm as fn of i/p size

Determination of Complexities
• How do you determine the running time of
piece of code?
Ans : Depends on the kinds of statements used

1. Sequence of Statements
Statement 1;
Statement 2;
…
…
Statement k;
• Independent statement in a piece of code and not an
unrolled loop
• Total Time : Adding the time for all statements.
• Total Time = Time (Statement 1) + Time (Statement 2) + … +
Time (Statement k)
• Each statement – simple (basic operations) – Time constant –
Total time is also constant O(1)

1 (Constant Time)
• When instructions of program are executed once or at
most only a few times , then the running time
complexity of such algorithm is known as constant time.
• It is independent of the problem size.
• It is represented as O(1).
• For example, linear search best case complexity is O(1)

Log n (Logarithmic)
• The running time of the algorithm in which large
problem is solved by transforming into smaller sizes sub
problems is said to be Logarithmic in nature.
• Becomes slightly slower as n grows.
• It does not process all the data element of input size n.
• The running time does not double until n increases to
n2.
• It is represented as O(log n).
• For example binary search algorithm running time
complexity is O(log n).

2.For loops
for (i=0; i<N;i++)
{
Sequence of statements
}
• Loop executes N times, Sequence of statements also executes N
times.
• Total time for the for loop = N*O(1) = O(N)

3.If-then-else statements
If(cond) {
Sequence of statements 1
}
Else
{
Sequence of statements 2
}
• Either Sequence 1 or Sequence 2 will execute.
• Worst Case Time is slowest of two possibilities
– Max { time (sequence 1), time (sequence 2) }
– If Sequence 1 is O(N) and Sequence 2 is O(1), Worst case time for
if-then-else would be O(N)

n (Linear)
• The complete set of instruction is executed once for each
input i.e input of size n is processed.
• It is represented as O(n).
• This is the best option to be used when the whole input has
to be processed.
• In this situation time requirement increases directly with the
size of the problem.
• For example linear search Worst case complexity is O(n).

4.Nested Loops
For (i=0;i<N;i++){
for(j=0;j<M;j++){
sequence of statements;
}
}
Total Complexity = O(N*M)
= O(N2)

5.Statement with function calls
• for (j=0; j<N; j++) g(N); has complexity O(N2)
– Loop executes N times
– g(N) has complexity O(N)

n2 (Quadratic)
• Running time of an algorithm is quadratic in nature
when it process all pairs of data items.
• Such algorithm will have two nested loops.
• For input size n, running time will be O(n2).
• Practically this is useful for problem with small input
size or elementary sorting problems.
• In this situation time requirement increases fast with
the size of the problem.
• For example insertion sort running time complexity is
O(n2).

Prob1. Calculate worst-case complexity!
• Nested Loop + Non-nested loop
for (i=0;i<N;i++){
for(j=0;j<N;j++){
}
}
for(k=0;k<N;j++){
}
• O(N2), O(N) = O(max(N2,N) = O(N2)

Prob 2.Calculate worst-case complexity!
• Nested Loop
for (i=0;i<N;i++){
for(j=i;j<N;j++){
}
}
• N+ (N-1) + (N-2) + …. + 1 = N(N+1)/2 = O(N2)

Approaches of Designing Algorithms
• Incremental Approach
• Insertion sort
– In each iteration one more element joins the sorted array
• Divide and Conquer Approach
– Recursively break down into 2 or more sub problems until it
becomes easy to solve. Solutions are combined to give
solution to original problem
• Merge Sort
• Quick Sort

Insertion Sort
3 4 6 8 9 7 2 5 1
1 nj 
 i
Strategy
• Start empty handed
• Insert a card in the right position
of the already sorted hand
• Continue until all the cards are
Inserted/sorted

Insertion Sort – Tracing Input

Analysis – Insertion Sort
• Assume that the i th line takes time ci , which is a constant.
(Since the third line is a comment, it takes no time.)
• For j = 2, 3, . . . , n, let tj be the number of times that the
while loop test is executed for that value of j .
• Note that when a for or while loop exits in the usual way -
due to the test in the loop header - the test is executed
one time more than the loop body.

Analysis – Insertion Sort – Running time

Divide-and-Conquer
• The most-well known algorithm design strategy:
1. Divide instance of problem into two or more smaller
instances
2. Solve smaller instances recursively
3. Obtain solution to original (larger) instance by
combining these solutions
• Type of recurrence relation

Divide-and-Conquer Technique (cont.)

Lecture 3 insertion sort and complexity analysis

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