The document discusses design and analysis algorithms, focusing on recurrences and recursion with examples and problem solutions. It highlights the advantages and disadvantages of recursion, as well as providing a detailed explanation of binary search with corresponding recurrence relations. The document includes code examples illustrating recursive and iterative implementations.

1. Design and
Analysis
Algorithm

2. The Recurrence
and its example and
problem solutions
* Presented by BSCS-E1 (girls)
(Roll no 4,5,7,9,15,19, 20)
Designed by Sara Yousaf

3. For example:
Fact(4)=24
• Solution:
4*fact(3)
4*3*fact(2)
4*3*2*fact(1)
4*3*2*1=24
• In recursion one function called itself, the number of parameters
never
changes only the parameter value changes
• The process in which the function calls itself is called
recursion
• The corresponding function is called the recursive function.
THE RECURSION

4. The Recursion
Advantages:
• Recursion adds clarity to the code.
• It can reduce time complexity
• Reduce time to debug
• It is also used for analyzing and
optimizing the algorithm
Disadvantages:
• Uses more memory
• Recursion becomes slow
• A recursive program has a greater
space
requirement than an iterative program
as each function in the program
remains
in the stack until the base case is
reached

5. THE RECURRENCE
• Recurrence is an equation or an in-
equality that describes the function in
terms of its smaller inputs
• Recurrences arise when an algorithm
contain recursive calls to itself.
The simplest form is as follows:
T(n)= T(n-1)+n

7. The Recurrence
• Binary Search:
T(n)=T(n/2)+1
Input size
T(n)=T(n/2)+1
Reducing of input
size
Signifies a function
1 2 3 4 5 6 7 8
1 2 3 4 5 6 7 8
2 3 5 7 9 10 11 12
lo hi
mid
Mid =[(lo+hi)/2]
Exp;
mid pt= 4 approx

8. The Recurrence
• Binary Search:
Binary Search (A, lo, hi, x)
 if(lo>hi)
return false
mid (lo+hi)/2
 If x=A[mid]
return True
 If(x<A[mid])
Binary-Search(A,lo,mid-1,x)
 If(x>A[mid])
Binary-Search(A,mid+1,hi,x)
Example:
A[8]={1,2,3,4,5,7,9,11} midpoint
-lo=1 ,hi=8,x=7
mid=4, lo=5 , hi=8
mid=6, A[mid]=x Found!
midpoint
1 2 3 4 5 6 7 8
1 2 3 4 5 7 9 11
1 2 3 4 5 6 7 8
1 2 3 4 5 7 9 11

9. T
H
E
R
E
C
U
R
R
E
N
C
E
Exp-1:
• Simple code
void fun(int n)
{
if(n>0){
for(i=0;i<n;i++)
{
print(n);{
fun(n-1)}
}
T(n)=T(n-1)+2n+2
T(n)=T(n-1)+n
n
1
n+1
n
n-1
Exp-2:
int binarySearch(intarr[],int n , int key)
{
int low=0;
int high=n-1;
while(low<=high){
int mid=(low+high)/2;
if(arr[mid]==key) {
return mid; }
else if (arr[mid]<key) {
low=mid+1; }
else {
high= mid-1; } }
return -1; }
//O(1)
//O(1)
//O(1)
//O(log n)
//O(1)
//O(1)
//O(1)
//O(1)
//O(1)
//O(1)
T(n)=O(1)+)O(1)+O(logn)*(O(1)+O(1)+O(1))+O(1)
T(n)=O(1)+O(1)+O(logn)+O(1)+O(1)
T(n)=O(logn)