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CS502 Assignment No. 02 Solution & Discussion Due Date: Dec 07, 2015
Assignment No. 02 CS502: Fundamentals of Algorithms Due Date:07/12/2015 Instructions Please read the following instructions carefully before submitting assignment:
It should be clear that your assignment will not get any credit (zero marks) if:
Objective
The objective of this assignment is to enable students:
Submission
You are required to submit your solution through LMS as MS Word document.
For any query about the assignment, contact at CS502@vu.edu.pk GOOD LUCK


Question 1:
Consider the following recursive algorithm for computing the sum of the first n squares: Sum(n) = 12 + 22 + . . . + n2.
Algorithm: SUM(n) if n = 1 return 1 else return SUM(n − 1) + n ∗ n
Write recurrence relation for above algorithm and solve it using Iteration Method.
Question 2:
In Divide and conquer strategy, three main steps are performed:
Write an algorithm to find minimum number from a given array of size ‘n’ using divide and conquer approach.


Lectures Covered: This assignment covers first 15 Lectures. Deadline: Your assignment must be uploaded/submitted at or before 07 Dec, 2015. 
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please koi ye to btyn na knsa solution kis question ka hy???????
plzzzzzzzzzz jaldi sy bta dyn knsa solution kis question ka hy???
or knsa solution theek hy???
part 1 ist ka ....
auer part 2 scnd ka .........scnd ka answr mjhe cnfrm nhe :)
KYA YE THEAK HEIN ??????PLEASE TELL ME
Assignment No. 02
ID=MC140401807
Question 1:
Consider the following recursive algorithm for computing the sum of the first n squares:
Sum(n) = 1^{2} + 2^{2} + . . . + n^{2}.
Algorithm: SUM(n)
if n = 1 return 1
else return SUM (n1) + n _{*} n
Write recurrence relation for above algorithm and solve it using Iteration Method.
Answer:
Iteration method of power 2 k=log n
T(n) = 2kT (n/(2k)) + (n+n+n+....+n)
= n + n log n
Question 2:
In Divide and conquer strategy, three main steps are performed:
Write an algorithm to find minimum number from a given array of size ‘n’ using divide and conquer approach.
Answer: Denoting S(n) the sum of the first n cubes, S(n) must be a polynomial of the fourth degree in n, let
S(n) = an^4+bn³+cn²+dn
This is because
1) S(0)= 0, so there is no independent term
2) When computing S(n)S(n1), which must equal n³, you get a polynomial of the third degree, by cancellation of the quartic term:
S (n)S(n1) = a(n^4(n1)^4)+b(n³(n1)³)+c(n²(n1)²)+d(n(n1)).
Developing and simplifying
a(4n³6n²+4n1)+b(3n²3n+1)+c(2n1)+d = n³.
Let us identify the coefficients.
n³: 4a = 1
n²: 6a + 3b = 0
n: 4a  3b + 2c = 0
1:  a + b – c + d = 0
Solving this triangular system is straightforward:
a= 1/4
b= 1/2
c= 1/4
d= 0
So,
S(n) = (n^4+2n³+n²)/4 = n²(n+1)²/4.
YE theak hy ya galat ?
Part1
T(n)=T(n1)+n*n as we know that T(n1)=T(n2)+n*n so we put this value
T(n)=T(n2)+n*n+n*n =T(n2)+2(n*n)
T(n)= T(n2)+2(n*n) T(n2)=T(n3)+n*n so we put this value
T(n)= T(n3)+2(n*n)+n*n= T(n3)+3(n*n)
T(n3)+3(n*n) T(n3)=T(n4)+n*n so we put this value
T(n)=T(n4)+3(n*n)+n*n=T(n4)+4(n*n)
..........
T(n)= T(nk)+k(n*n)
If we set n=k
Then
T(n)= T(nn)+n(n*n)=n^3
Part2:
SELECT(Array A,int p ,int r,int k=1)
if(p=r)
then return A[p]
else x CHOOSE_PIVOT(A,p,r)
q PARTITION (A,p,r,x)
rankx qp+1
if k= rankx
then return x
if k< rankx
then return SELECT(A, p ,q1, k)
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