SOLUTION AT Australian Expert Writers

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6.4 Knapsack Problem

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Knapsack Problem

Knapsack problem.

■ Given n items and a “knapsack“ such that

– item i weighs wi > 0 and has value vi > 0, and

– the knapsack has the capacity of total weight W.

■ Goal: fill the knapsack so as to maximize the total value, i.e., find a

subset S of the set of n items such that Σi∈Svi is maximized subject

to Σi∈Swi ≤ W.

Ex: Optimal solution for this one?

The optimal solution is items 3, 4

which has the total value 40. 1

Value

18

22

28

1

Weight

5 6

6 2

7

Item

1 2 3 4 5

W = 11

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Side Note: 0-1 Knapsack Problem

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Greedy: repeatedly add an item with the maximum ratio vi/wi, subject

to the capacity limit of W.

Q. Is this greedy optimal?

A. No!

This greedy finds items 5, 2, 1 which achieves only value 35.

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Knapsack Problem: Greedy ?

1

Value

18

22

28

1

Weight

5 6

6 2

7

Item

1 2 3 4 5

W = 11

Value/Weight

1

3.6

3.66…

3 4

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Dynamic Programming: As We Did Before

Notation. OPT(i) = max total value of the items in an optimal subset of

items 1, 2, .., i .

To compute OPT(i):

■ Case 1: item i is not in the optimal solution.

– OPT(i) is computed from the remaining items 1, 2, …, i-1

– So, OPT(i) = OPT(i-1)

■ Case 2: item i is in the optimal solution.

– OPT(i) is the sum of vi and the total value of an optimal solution

from the remaining items 1, 2, …, i-1

– So, OPT(i) = vi + OPT(i-1)

■ So, OPT(i) = max OPT(i-1), vi + OPT(i-1) = vi + OPT(i-1) Not right!

■ What’s wrong? Accepting an item i changes the available capacity of

the knapsack, which is one of the variables defining the problem. So,

we need to redefine the subproblems to involve the available capacity.

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Dynamic Programming: Now Differently (and Correctly)

Notation. OPT(i, w) = max total value of items in an optimal subset of

items 1, 2, .., i subject to the total weight limit w.

■ Case 1: item i is not in the optimal solution.

– OPT(i, w) is computed from the remaining items 1, 2, …, i-1

subject to the total weight limit w.

■ Case 2: item i is in the optimal solution (possible only if wi ≤ w).

– new weight limit = w – wi

– OPT(i, w) is the sum of vi and the total value of an optimal solution

from the remaining items 1, 2, …, i-1 subject to the total weight

limit w – wi.

OPT(i, w) =

0 if i = 0

OPT(i -1, w) if wi > w

max OPT(i -1, w), vi + OPT(i -1, w – wi ) otherwise

Optimal substructure:

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Knapsack. Fill up an array M[0..n][0..W]. (Assume integer weights.)

■ (We have a two-variable OPT(i, w), so we need a two-dimensional

array.)

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Input: W, w1,…,wn, v1,…,vn

Global array: M[0..n][0..W]

for w = 0 to W

M[0, w] = 0

for i = 1 to n

for w = 0 to W

if (wi > w)

M[i,w] = M[i-1,w]

else

M[i,w] = maxM[i-1,w], vi + M[i-1,w-wi]

return M[n, W]

Knapsack Algorithm: Bottom-Up

OPT(i, w) =

0

OPT(i -1, w)

if i = 0

if wi > w

max OPT(i -1, w), vi + OPT(i -1, w – wi ) otherwise

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Knapsack Algorithm: Tracing

i

φ

1, 2

1, 2, 3

1, 2, 3, 4

1

1, 2, 3, 4, 5

0 0 0 0 0 0 0

1 0 1 1 1 1 1

2 0 1 6 6 6 6

3 0 1 7 7 7 7

4 0 1 7 7 7 7

5 0 7

18

18

1

18

6 0 7

19

22

1

22

7 0 7

24

24

1

28

8 0 7

25

28

1

29

9 0 7

25

29

1

34

10

0 7

25

29

1

35

11

0 7

25

40

1

40

w

1

Value

18

22

28

1

Weight

5 6

6 2

7

Item

1 2 3 4 5

W = 11

value = 22 + 18 = 40 (for 4, 3 )

for w = 0 to W

M[0, w] = 0

for i = 1 to n

for w = 0 to W

if (wi > w)

M[i, w] = M[i-1, w]

else

M[i, w] = maxM[i-1, w],

vi + M[i-1, w-wi]

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Knapsack Problem: Running Time

Running time. Θ(n W).

■ Polynomial in n and W, but see the note below.

Note. This is not a polynomial algorithm.

– There are two inputs: items (i.e., their values and weights) and

knapsack capacity.

– n, number of items, is not an input but its cardinality.

So, the runtime has nothing to do with the numeric values of

the values of items.

– W, capacity, is the numeric value of an input, i.e., limit on max total

weight.

So, if W is very large, the runtime is very long.

– Computationally, we say the running time Θ(n W) = Θ(n 2b), where b

is the length of the binary representation of W. So, exponential!

This kind of running time is called pseudo-polynomial. (More

on this later when we are studying NP.)

Side Note: Pseudo-polynomial Runtime

Def. Pseudo-polynomial: polynomial in the magnitude of the input

values but not in the number of bits needed to represent the values.

“A numeric algorithm runs in pseudo-polynomial time if its running

time is polynomial in the numeric value of the input (which is

exponential in the length of the input — its number of digits). “ –

Wikipedia

Compare: “really”-polynomial runtime – examples:

– Dijkstra’s shortest-path finding algorithm: O(m logn), where m is

the number of edges and n is the number of vertices, either of

which has nothing to do with the weights of the edges.

– Weighted interval scheduling algorithm: O(n), where n is the

number of jobs, which has nothing to do with the weights of the

jobs.

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Knapsack: Finding a Solution

To find an optimal subset of items in addition to the optimal total value

of them.

Idea (Bookkeeping). Record the “winning case” (item i included or not

subject to weight limit w) in the array entry M[i,w] at each step of

the iteration, and then backtrack from M[n,W] to collect items that

are included.

Idea (Post-processing). Trace the array M[0..n][0..W] backward

recursively starting with the last element M[n][W].

Exercise!

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Side Note: Knapsack Problem — Greedy (Again)

Q. What if we allow part of an item to be included? What difference

would it make?

■ (Note: In this case we call it the fractional knapsack problem; in

contrast, we call the original one the 0-1 knapsack problem.)

Q. Does greedy work?

■ Greedy choice strategy: include the item with the largest

value/weight ratio first.

A. Yes, greedy works! The greedy finds items item 5, 2/3 of item 4 ,

achieving 42.666… which is the optimal value.

Running-time:

1

Value

18

22

28

1

Weight

5 6

6 2

7

Item

1 2 3 4 5

W = 11

Value/Weight

1

3.6

3.66…

3 4

O(n logn) for sorting

the items by the ratio

value/weight.

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