1. Analysis of Non – Recursive Algorithms

2. Why Analysis of Algorithms?
For a given problem there can be multiple solutions available for the same problem. Analysis of algorithm helps us to find which among them is better in terms of time and space.
For Ex: If you have a stream of numbers coming and you would like to sort it. We might use Insertion sort ahead of bubble sort or selection sort.
This decision can be taken if we would analyze the algorithm.

3. Goal of Analysis of Algorithms
The goal of analysis of algorithms is to compare algorithms mainly in-terms of running time and memory.

4. What is running time analysis?
It is the process of determining how processing time increases as size of the problem (input) increases.
Input size is the number of elements in the input.
Depending on the problem type the size of input may be different for different types.

5. How to Compare Algorithms
Execution Time?
Number of statements executed?
Ideal Solution?
Expressing the running time of an algorithm as a function of input size 'n'. (i.e. f(n) ).
Compare these functions corresponding to running times.
Independent of machine, programming language,etc..

6. Commonly used Rate of Growths
What is Rate of Growth?
The rate at which the running time increases as a function of input is called rate of growth.
Commonly used Rate of Growths
Time Complexity
Name 1
Constant
logn
Logarithmic n
Linear
nlogn
Linear Logarithmic n2
Quadratic n3
Cubic 2n
Exponential

7. Types of Analysis Worst Case :
Defines the input for which algorithm takes huge time.
Input is the one for which algorithm runs slower.
Best Case
Defines the input for which algorithm takes lowest time.
Input is the one for which algorithm runs faster.
Average Case
Provides a prediction about running time of algorithm
Assumes that input is random
Lower_Bound <= Average Time <= Upper_Bound

8. Asymptotic Notation
It is a syntax that is used to represent the upper bound and lower bound.
Big – O Notation
Omega - Ω Notation
Theta - Θ Notation

9. Big – O Notation
This notation gives the tight upper bound of the given function.
It is represented as, f(n) = O (g(n)).
It means at large values of 'n', the upper bound of f(n) is g(n)

10. Big – O Notation c.g(n) f(n) 0<=f(n)<=c.g(n), for all n>=n0
RATE OF GROWTH
0<=f(n)<=c.g(n), for all n>=n0
g(n) is tight upper bound of f(n) n0
INPUT SIZE, n

11. Big – O Notation : Examples
Ex 1 : Find the upper bound of f(n) = 3n + 8
Solution : 3n + 8 <= 4n, for all n>=8
c = 4, n0 = 8
3n+8 = O(n)
Ex 2 : Find the upper bound of f(n) = n2 + 4
Solution : n2 + 4 <= 2n2, for all n>=2
c = 2, n0 = 2
n2 + 4 = O(n2)

12. Big – O Notation : Examples
Ex 3 : Find the upper bound of f(n) = n n2 + 50
Solution : n n <= 2n4, for all n>=11
c = 2, n0 = 11
n n = O(n4)
Ex 4 : Find the upper bound of f(n) = n
Solution : n<= n, for all n>=1
c = 1, n0 = 1
n = O(n)
Ex 5: Find the upper bound of f(n) = 10.
Answer : O(1)

13. Omega - Ω Notation
It gives the tight lower bound of the given algorithm
Represented as, f(n) = Ω(g(n))
It means for larger value of 'n' the tight lower bound of f(n) is g(n).

14. Omega - Ω Notation f(n) c.g(n) 0<=c.g(n)<=f(n), for all n>=n0
RATE OF GROWTH
0<=c.g(n)<=f(n), for all n>=n0
g(n) is tight lower bound of f(n) n0
INPUT SIZE, n

15. Omega - Ω Notation : Examples
Ex 1 : Prove f(n) = 100n + 5 ≠ Ω (n2)
Solution : By definition of Ω(g(n)) = f(n),
We write 0<=c.g(n)<=f(n)
0 <=c.n2 <= 100n
100n+5<=100n+5n, for all n>=1
=> 100n+5<=105n
From 1 and 2,
cn2<=105n
cn2-105n<=0
cn<=105
n<=105/c

16. Theta - Θ Notation
This notation decides whether the upper bound and lower bound of a given function is same or not.
It is defined as,
Θ(g(n)) = f(n), if there exists positive constants c1, c2 and n0 such that,
0 <= c1.g(n) <= f(n) <= c2.g(n), for-all, n>=n0

17. Theta - Θ Notation
c2g(n)
f(n)
RATE OF GROWTH
c1g(n) no
INPUT SIZE, n

18. Theta - Θ Notation : Examples
Ex 1: Find Θ bound for f(n) = n2/2 – n/2
Solution :
n2/5 <= n2/2 – n/2 <= n2, for-all, n>=1
c1=1/5, c2=1.
And g(n) = n2
f(n) = Θ(g(n))
Ex 2: Prove that n ≠ Θ(n2)
c1n2 <= n <= c2n2, Holds true only for n<=1/c1
Hence, n ≠ Θ(n2)

19. Guidelines for Asymptotic Analysis
Loops : The running time of a loop is at most, the running time of statements inside the loop.
for(i=1; i<=n; i++)
X = X + 2; //Constant Time, c
Total Time = c*n = O(n)

20. Guidelines for Asymptotic Analysis
Nested Loops : The total running time is the product of sizes of all loops.
for(i=1; i<=n; i++){
X = X + 2; //Constant Time, c1
for(j=1; j<=n; j++) {
M = M -2; //Constant Time, c2 }
Total Time = c2*n*n = O(n2)

21. Guidelines for Asymptotic Analysis
Consecutive statements : Add the time complexities of each statements.
for(i=0;i<n;i++)
printf(“Hello....\n”); //Constant Time, c
for(i=1; i<=n; i++){
X = X + 2; //Constant Time, c1
for(j=1; j<=n; j++) {
M = M -2; //Constant Time, c2 }
Total Time = n*c + c2*n*n = O(n2)

22. Guidelines for Asymptotic Analysis
Logarithmic Complexity
for(i=1; i<=n; i=i*2)
X = X + 2; //Constant Time, c
Total Time = O(log2n)

23. Commonly used Logarithms and Summations
logxy = ylogx
logxy = log x + log y
log logn = log(log n)
log(x/y) = log(x) - log(y)
alogbx = xlogba
logk n=(logn)k
N = N*(N+1)/2
1+x+x2+x xn = (xn+1 – 1) / (x-1)
1+1/2+1/ /n ≈ logn

24. Analyse the following function
void function(int n) {
int i, count = 0;
for(i=0; i*i <=n ; i++)
count++; }
Answer : O(√n)

25. What is the complexity of following function
void function(int n) {
int i, j, k, count=0;
for(i=n/2; i<=n ; i++)
for(j=1; j+n/2 <=n; j++)
for(k=1;k<=n;k=k*2)
count++; }
Answer : O(n2logn)

26. What is the complexity of following function
void function(int n) {
int i, j, k, count=0;
for(i=n/2; i<=n ; i++)
for(j=1; j <=n; j=j*2)
for(k=1;k<=n;k=k*2)
count++; }
Answer : O(nlog2n)

27. What is the complexity of following function
void function(int n) {
if(n==1) return;
for(i=1; i<=n; i++)
for(j=1; j<=n; j++)
printf(“*”);
break; }
Answer : O(n)

28. What is the complexity of following function
void function(int n) {
float sum=0;
for(i=1; i<=n; i++)
for(j=1;j<=n;j=j*2)
printf(“Hello World\n”); }
Answer : O(nlogn)

29. What is the complexity of following function
void function(int n) {
float sum=0;
for(i=1; i<=n/3; i++)
for(j=1;j<=n;j=j+5)
printf(“Hello World\n”); }
Answer : O(n2)

30. What is the complexity of following function
void function(int n) {
int i=1;
while(i<=n)
int j = n;
while(j>0)
j = j / 2;
i = i*2; }
Answer : O(log2n)