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.
Please Discuss here about this assignment.Thanks
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
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