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CS502 Assignment No. 02 Solution & Discussion Due Date: Dec 07, 2015

CS502 - Fundamentals of Algorithms Assignment No. 02 Solution Fall 2015 Due Date Dec 07, 2015

Assignment No. 02
Semester: Fall 2015

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:

  • The assignment is submitted after due date.
  • The submitted assignment is other than .doc file.
  • The submitted assignment does NOT open or file is corrupted.
  • The assignment is copied (from other student or ditto copy from any other source).

 

Objective

 

The objective of this assignment is to enable students:

 

  • Write and solve recurrence relations of recursive algorithms using iteration method
  • Design algorithm using Divide and conquer approach

 

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:

 

  1. 1.      Divide: Divides the problem into a small number of pieces
  2. 2.      Conquer: Solves each piece by applying divide and conquer to it recursively
  3. 3.      Combine: Combines/merges the pieces together into a global solution.

 

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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Replies to This Discussion

ye solution theak hy ?

Very nice you have done

++msn .fb sy mera answr chori kar liya -_-  .. ye lo complete solution :P duaon mein yaad rakhna

Attachments:

yeh ist questtion ka ans hai kia

God Bless u Dear

ye sahi hy kia?

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) = 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.

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:

 

  1. 1.      Divide: Divides the problem into a small number of pieces
  2. 2.      Conquer: Solves each piece by applying divide and conquer to it recursively
  3. 3.      Combine: Combines/merges the pieces together into a global solution.

 

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(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 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(n-1)+n*n as we know that T(n-1)=T(n-2)+n*n so we put this value
T(n)=T(n-2)+n*n+n*n =T(n-2)+2(n*n)
T(n)= T(n-2)+2(n*n) T(n-2)=T(n-3)+n*n so we put this value
T(n)= T(n-3)+2(n*n)+n*n= T(n-3)+3(n*n)
T(n-3)+3(n*n) T(n-3)=T(n-4)+n*n so we put this value
T(n)=T(n-4)+3(n*n)+n*n=T(n-4)+4(n*n)
..........
T(n)= T(n-k)+k(n*n)
If we set n=k
Then
T(n)= T(n-n)+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  q-p+1
if k= rankx
then return x
if k< rankx
then return SELECT(A, p ,q-1, k)

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