CS702: Advance Algorithm Analysis and Design

Assignment No.2 MS (CS), Fall 2014 Maximum Points: 50

20+20+9=49

Question No. 1 (20 Points) 20

Use Brute Force Method to find an optimal solution for the 0-1 Knapsack problem.Item weight value knapsack capacity W = 201 2 32 3 43 4 54 5 8

59 10

12345

345810

23459

Capacity = 20

Dynamic Programming Method:

Keep = 0

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 0

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 0

Keep = 1

Keep = 1

Keep = 1

Keep = 0

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 0

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 0

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 0

Keep = 0

Keep = 0

Keep = 0

Keep = 0

Keep = 0

Keep = 0

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Keep = 1

Let W = 20Final Solution: V[5, 20] = 26Item selected: [1, 3, 4, 5]

Brute Force Method:

ItemsSubsetTotal weightTotal Value

1

00

2{1}23

3{2}34

4{3}45

5{4}58

6{5}910

7{1,2}57

8{1,3}68

9{1,4}711

10{1,5}1113

11{2,3}79

12{2,4}812

13{2,5}1214

14{3,4}913

15{3,5}1315

16{4,5}1418

17{1,2,3}912

18{1,2,4}1015

19{1,2,5}1417

20{1,3,4}1116

21{1,3,5}1518

22{2,3,4}1217

23{2,3,5}1619

24{2,4,5}1722

25{3,4,5}1823

26{1,2,3,4}1420

27{1,2,3,5}1822

28{1,3,4,5}2026

29{1,2,4,5}1825

30{2,3,4,5}2127 not feasible

31{1,2,3,4,5}2330 not feasible

Question No. 2 (20 Points) 20

You have been given four matrices A1.A2.A3.A4 with p0=5, p1=4, p2=6, p3=2 and p4=7. You have to find m [1, 4].

Solution:

Main Diagonal:

Computing m [1, 2]:

Computing m [2, 3]:

Computing m [3, 4]:

Computing m [1, 3]:

Computing m [2, 4]:

Computing m [1, 4]:

Final Cost Matrix:012088158

048104

084

0

Order of Computation:15810

269

37

4

K,s Values Leading Minimum m[i, j]0113

022

03

0

Representing Order using Binary Tree

The above computation shows that the minimum cost for multiplying those four matrices is 158.

The optimal order for multiplication is, for.

158 for m[1,4]. and (1,3,4,5)giving maximum value

Question No. 3 (10 Points) 9

Read the attached research paper entitled The Automatic Construction ofPure Recurrence Relations, with this assignment very carefully. Answer the following question:

a. Which method is used for the construction of Pure Recurrence Relation?

Abstract:To generalize the hyper geometric functions we construct the method of pure recurrence relations, an improvement of a technique by Fasenmeyer suitable for automatic computer programming.

Geometrical interpretation and constructive proof for the existence of a pure recurrence relation for functions of the form:

(1)

Where a, s, b s are polynomials in; k, n and possibly other atoms;

We will denote these atoms as; k, y. However these polynomials must be linearly dependent on k and n and their coefficients must be integers.

The functionsinclude most of the generalized hyper geometric functions:

(2)

Whereis the usual Pochhammer symbol such that

Where

A pure recurrence relation for (i) as a function of n is an expression of the form:

(3)

The coefficients of the termsare rational expressions in y. r is the order of the recurrence relation.

Problem of the construction of a recurrence relation in two parts:

1)

Finding V, the set of integer pairssuch that is a term of the recurrence relation.2) computing the coefficients

If the structure of the recurrence relation is known, if V is known, then thecan be found as solutions of a system of linear equations with polynomials as coefficients.

The method involves rewriting eachas

(4)

Whereis define as

(5)

The are rational expression. We can rewrite (3) as:

(6)

The nominator of the coefficient of.after expansion to a common denominator must vanish for each value of k. We call this nominator a polynomial in, with as coefficients polynomials in y and the, the associated polynomial of (3). To yield simple equations, it is wisest to get them by a judicious mixture of employing specific values of k and equating coefficients of powers of k to zero. More details can be found in [1, 2]. From now on we concentrate on finding the structure of the recurrence relation.