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CS502 Assignment No. 02 Solution & Discussion Due Date: Dec 07, 2015
Assignment No. 02
CS502: Fundamentals of Algorithms
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Consider the following recursive algorithm for computing the sum of the first n squares:
Sum(n) = 12 + 22 + . . . + n2.
if n = 1 return 1
else return SUM(n − 1) + n ∗ n
Write recurrence relation for above algorithm and solve it using Iteration Method.
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 Discuss here about this assignment.Thanks
Our main purpose here discussion not just Solution
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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
S(n) the sum of the first
S(n) must be a polynomial of the fourth degree in
S(n) = an^4+bn³+cn²+dn.
This is because
S(0)= 0, so there is no independent term,
2) When computing
S(n)-S(n-1), which must equal
n³, you get a polynomial of the third degree, by cancellation of the quartic term:
S(n)-S(n-1) = a(n^4-(n-1)^4)+b(n³-(n-1)³)+c(n²-(n-1)²)+d(n-(n-1)).
Developing and simplifying,
a(4n³-6n²+4n-1)+b(3n²-3n+1)+c(2n-1)+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 straigthforward:
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.
there is mistake in above recurrence relation. pls find it.
now it is easy to solve with iteration method.
is it comlete solution .?
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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
any one got the complet solution