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The Knapsack Problem
The classic Knapsack problem is typically put forth as:
A thief breaks into a store and wants to fill his
knapsack with as much value in goods as possible
before making his escape. Given the following list of
items available, what should he take?
• Item A, weighing w
A
pounds and valued at v
A
• Item B, weighing w
B
pounds and valued at v
B
• Item C, weighing w
C
pounds and valued at v
C
•
•
•
The Simplest Versions…
Can items be divided up such that only a portion is taken?
The thief can hold 5 pounds and has to choose from:
3 pounds of gold dust at $379.22/pound
6 pounds of silver dust at $188.89/pound
1/9 pound of platinum dust at $433.25/pound
Are all of the weights or total values identical?
The thief breaks into a ring shop where all of the rings
weight 1oz. He can hold 12 ounces; which should he
take?
A Deceptively Hard Version…
What if each problem has the same price/pound?
This problem reduces to the binpacking problem: we
want to fit as many pounds of material into the knapsack
as possible.
How can we approach this problem?
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Example
The thief breaks into a gold refinery; he can steal from a
selection of raw gold nuggets, each of the same value per
pound. If he can carry 50 pounds, what selection would
maximize the amount he carries out?
47.3 pounds
6.0 pounds
5.2 pounds
36.7 pounds
5.6 pounds
5.2 pounds
25.5 pounds
5.6 pounds
5.0 pounds
16.7 pounds
5.4 pounds
3.2 pounds
8.8 pounds 5.3 pounds
0.25 pounds
An Easier Version...
What if all of the sizes we are working with are
relatively
small
integers? For example, if we could fit 10 pounds
and:
Object A is 2 pounds and worth $40
Object B is 3 pounds and worth $50
Object C is 1 pound and worth $100
Object D is 5 pounds and worth $95
Object E is 3 pounds and worth $30
We can use dynamic programming!
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 Spring '08
 OFRIA
 Algorithms, Dynamic Programming, Force, Mass, pounds

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