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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:
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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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Our main purpose here discussion not just Solution
We are here with you hands in hands to facilitate your learning and do not appreciate the idea of copying or replicating solutions.
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.
can u plz explain it? i hve problem in algo
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.
first we should find the recurrence relation of sum of the first n squares
1^2 + 2^2 + . . . + n^2.
T(N)= T(n2)+n^2
there is mistake in above recurrence relation. pls find it.
now it is easy to solve with iteration method.
is it comlete solution .?
Dear Students Don’t wait for solution post your problems here and discuss ... after discussion a perfect solution will come in a result. So, Start it now, replies here give your comments according to your knowledge and understandings....
I think asey hi ho ga
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
any one got the complet solution
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