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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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idea
First of all, if n == 1 you should probably return 1. And yes, this recursive function computes 1 + 2^3 + 3^3 + ... + n^3. How do we know that?
Well, take an example like n = 5;
If you add them up => R(5) returns 5^3 + 4^3 + .. + 2^3 + 1.
EXPLANATION:
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: 4a3b+2c =0
1: a +b c+d=0
Solving this triangular system is straigthforward:
a=1/4
b=1/2
c=1/4
d=0.
and finally
S(n) = (n^4+2n³+n²)/4 = n²(n+1)²/4.
Check iteration method of power 2 k=log n
T(n) = 2kT (n/(2k)) + (n+n+n+....+n)
= n + n log n
see on page 31
Lecture number 8 the solution s given
EXAMPLE FROM BOOK:
plz guide us ya working to nahe karne
T(n) = 2kT(n/(2k)) + (n + n + n + · · · + n)  {z }
k times
= 2kT(n/(2k)) + kn
= 2(logn)T(n/(2(logn))) + (log n)n
= 2(logn)T(n/n) + (log n)n
= nT(1) + n log n = n + n log n
Q 2 ka solution???
koi ALLAH ka banda correct solution bta day pleaseeee .........
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