lecture09-knapsack

8/7/2019 lecture09-knapsack

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CS 312: Algorithm Analysis

Lecture #9: Knapsack

Objectives

Remember the form of a generic greedyalgorithm

Define the Knapsack Problem

Learn how to formulate the problem inalgebraic terms

Get exposed to the formulation as a linearprogram (for future reference)

Work through an example

Generic Greedy Alg.

Function greedy (C:set): set{C is the set of candidates}

S {S will be the solution}

while (C != & !solution(S))

x select(C)

C C \ {x}

if feasible(S{x}) then

S S {x}

if solution(S) then return S

else there are no solutions

The Knapsack Problem

Given n objects withweight wi and value vi

Knapsack can only hold atotal weight of W

Fill Knapsack

Maximize total value

Dont exceed W

The Knapsack ProblemMathematically, represent a selectionfrom n objects by a vector x

Fix Order

of Objects

0

0

0

0

0

1

means

and

0

1

0

0

0

1

means{ , }

and

0

0

0

0

0

75.

means

The Knapsack ProblemMathematically, represent a selectionfrom n objects by a vector x

Fix Order

of Objects

=

6

5

4

3

2

1

x

x

x

x

x

x

x

10 ix

=

6

5

4

3

2

1

w

w

w

w

w

w

w

Corresponding

Object Weights

=

6

5

4

3

2

1

v

v

v

v

v

v

v

Corresponding

Object Values


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The problem is written mathematically as

Wxw

xvx

T

T

max

subject to

where vi>0, wi>0, and 0


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Knapsack

function knapsack (w[1n], v[1n], W) : array [1n]for i = 1 to n do x[i] 0weight 0while weight < W do

i the best remaining objectif weight + w[i]


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Example: Let W=100 and consider

Selection Function Ideas Most valuable Lightest weight Highest value/unit weight

=

50

40

30

20

10

w

=

60

40

66

30

20

v

=

2.1

1

2.2

5.1

2

/wv

=

0

0

0

0

0

x weight = 0




=

50

40

30

20

10

w

=

60

40

66

30

20

v

=

2.1

1

2.2

5.1

2

/wv

=

0

0

0

0

0

x weight = 0




=

50

40

30

20

10

w

=

60

40

66

30

20

v

=

2.1

1

2.2

5.1

2

/wv

=

00

1

0

0

x weight = 30

=

50

40

20

10

w

=

60

40

30

20

v




=

2.1

1

2.2

5.1

2

/wv

=

00

1

0

0

x weight = 30

=

50

40

20

10

w

=

60

40

30

20

v




=

2.1

1

2.2

5.1

2

/wv

=

0

0

1

0

0

x weight = 30

=

50

40

20

10

w

=

60

40

30

20

v




=

2.1

1

2.2

5.1

2

/wv

=

1

0

1

0

0

x weight = 80


5/8

5

=

40

20

10

w

=

1

0

1

0

0

x

=

40

30

20

v




=

2.1

1

2.2

5.1

2

/wv

weight = 80

=

1

0

1

0

0

x

=

40

30

20

v

=

40

20

10

w




=

2.1

1

2.2

5.1

2

/wv

weight = 80

=

15.0

1

0

0

x

=

40

30

20

v

=

40

20

10

w




=

2.1

1

2.2

5.1

2

/wv

weight = 100

=

15.0

1

0

0

x

=

30

20

v

=

20

10

w




=

2.1

1

2.2

5.1

2

/wv

weight = 100Score 1 = 146


=

0

0

0

0

0

x

=

60

40

66

30

20

v

=

50

40

30

20

10

w



=

2.1

1

2.2

5.1

2

/wv



=

0

1

1

1

1

x

=

60

40

66

30

20

v

=

50

40

30

20

10

w



=

2.1

1

2.2

5.1

2

/wv



6/8


7/8

7

=

50

40

20

w

=

2.1

1

2.2

5.1

2

/wv


Score 1 = 146Score 2 = 156

=

0

0

1

0

1

x

=

60

40

66

30

20

v



weight = 40

=

2.1

1

2.2

5.1

2

/wv

=

50

40

20

w


Score 1 = 146Score 2 = 156

=

0

0

1

0

1

x

=

60

40

66

30

20

v



weight = 40

=

2.1

1

2.2

5.1

2

/wv

=

50

40

w


Score 1 = 146Score 2 = 156

=

00

1

1

1

x

=

60

40

66

30

20

v



weight = 60

=

2.1

1

2.2

5.1

2

/wv

=

50

40

w


Score 1 = 146Score 2 = 156

=

00

1

1

1

x

=

60

40

66

30

20

v



weight = 60

=

2.1

1

2.2

5.1

2

/wv

=

8.0

0

1

1

1

x

=

40

w


Score 1 = 146Score 2 = 156

=

60

40

66

30

20

v



weight = 100

=

8.0

0

1

1

1

x

=

2.1

1

2.2

5.1

2

/wv

=

40

w


Score 1 = 146Score 2 = 156Score 3 = 164

=

60

40

66

30

20

v



weight = 100


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Score 1 = 146Score 2 = 156Score 3 = 164


Theorem: If objects are selected in order of decreasing

vi/wi, then algorithmknapsack finds an optimal solution.

Proof: Work through proofIn book on page 204.

Efficiency: Show that theAlgorithm exhibits timecomplexity that grows inO(nlogn).

Assignment

HW #6.5 Question 6.18

Read 7.1, 7.2

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