Knapsack problem

1
Knapsack problem
Data Structures & Algorithms

2
Given some items, pack the knapsack to get
the maximum total value. Each item has some
weight and some value. Total weight that we can
carry is no more than some fixed number W.
So we must consider weights of items as well as
their values.
Item # Weight Value
1 1 8
2 3 6
3 5 5
Knapsack problem

3
Knapsack problem
There are two versions of the problem:
1. “0-1 knapsack problem”
 Items are indivisible; you either take an item or not. Some
special instances can be solved with dynamic programming
1. “Fractional knapsack problem”
 Items are divisible: you can take any fraction of an item

4
 Given a knapsack with maximum capacity W, and
a set S consisting of n items
 Each item i has some weight wi and benefit value
bi (all wi and W are integer values)
 Problem: How to pack the knapsack to achieve
maximum total value of packed items?
0-1 Knapsack problem

5
 Problem, in other words, is to find
∑∑ ∈∈
≤
Ti
i
Ti
i Wwb subject tomax
0-1 Knapsack problem
The problem is called a “0-1” problem,
because each item must be entirely
accepted or rejected.

6
Let’s first solve this problem with a
straightforward algorithm
 Since there are n items, there are 2n
possible
combinations of items.
 We go through all combinations and find the one
with maximum value and with total weight less or
equal to W
 Running time will be O(2n
)
0-1 Knapsack problem:
brute-force approach

7
 We can do better with an algorithm based on
dynamic programming
 We need to carefully identify the subproblems
0-1 Knapsack problem:
dynamic programming approach

8
Defining a Subproblem

9
 We can do better with an algorithm based on
dynamic programming
 We need to carefully identify the subproblems
Let’s try this:
If items are labeled 1..n, then a subproblem
would be to find an optimal solution for
Sk = {items labeled 1, 2, .. k}

10
If items are labeled 1..n, then a subproblem would be
to find an optimal solution for Sk = {items labeled
1, 2, .. k}
 This is a reasonable subproblem definition.
 The question is: can we describe the final solution
(Sn ) in terms of subproblems (Sk)?
 Unfortunately, we can’t do that.

11
Max weight: W = 20
For S4:
Total weight: 14
Maximum benefit: 20
w1 =2
b1 =3
w2 =4
b2 =5
w3 =5
b3 =8
w4 =3
b4 =4 wi bi
10
85
54
43
32
Weight Benefit
9
Item
#
4
3
2
1
5
S4
S5
w1 =2
b1 =3
w2 =4
b2 =5
w3 =5
b3 =8
w5 =9
b5 =10
For S5:
Total weight: 20
Maximum benefit: 26
Solution for S4 is
not part of the
solution for S !!!
?

12
 As we have seen, the solution for S4 is not part of the
solution for S5
 So our definition of a subproblem is flawed and we
need another one!

13

14
 Let’s add another parameter: w, which will represent
the maximum weight for each subset of items
 The subproblem then will be to compute V[k,w], i.e.,
to find an optimal solution for Sk = {items labeled 1,
2, .. k} in a knapsack of size w

15
 The subproblem will then be to compute V[k,w], i.e.,
to find an optimal solution for Sk = {items labeled 1,
2, .. k} in a knapsack of size w
 Assuming knowing V[i, j], where i=0,1, 2, … k-1,
j=0,1,2, …w, how to derive V[k,w]?
Recursive Formula for
subproblems

16
It means, that the best subset of Sk that has total
weight w is:
1) the best subset of Sk-1 that has total weight ≤ w, or
2) the best subset of Sk-1 that has total weight ≤ w-wk plus the
item k



+−−−
>−
=
else}],1[],,1[max{
if],1[
],[
kk
k
bwwkVwkV
wwwkV
wkV
Recursive formula for subproblems:
Recursive Formula for
subproblems (continued)

17
Recursive Formula
The best subset of Sk that has the total weight ≤ w,
either contains item k or not.
First case: wk>w. Item k can’t be part of the solution,
since if it was, the total weight would be > w, which
is unacceptable.
Second case: wk ≤ w. Then the item k can be in the
solution, and we choose the case with greater value.



+−−−
>−
=
else}],1[],,1[max{
if],1[
],[
kk
k
bwwkVwkV
wwwkV
wkV

18
for w = 0 to W
V[0,w] = 0
for i = 1 to n
V[i,0] = 0
for i = 1 to n
for w = 0 to W
if wi <= w // item i can be part of the solution
if bi + V[i-1,w-wi] > V[i-1,w]
V[i,w] = bi + V[i-1,w- wi]
else
V[i,w] = V[i-1,w]
else V[i,w] = V[i-1,w] // wi > w
0-1 Knapsack Algorithm

19
for w = 0 to W
V[0,w] = 0
for i = 1 to n
V[i,0] = 0
for i = 1 to n
for w = 0 to W
< the rest of the code >
What is the running time of this
algorithm?
O(W)
O(W)
Repeat n times
O(n*W)
Remember that the brute-force algorithm
takes O(2n
)
Running time

20
Let’s run our algorithm on the
following data:
n = 4 (# of elements)
W = 5 (max weight)
Elements (weight, benefit):
(2,3), (3,4), (4,5), (5,6)
Example

21
for w = 0 to W
V[0,w] = 0
0 0 0 0 000
1
2
3
4 50 1 2 3
4
iW
Example (2)

22
for i = 1 to n
V[i,0] = 0
0
0
0
0
0 0 0 0 000
1
2
3
4 50 1 2 3
4
iW
Example (3)

23
V[i,w] = bi + V[i-1,w- wi]
else
V[i,w] = V[i-1,w]
0
Items:
1: (2,3)
2: (3,4)
3: (4,5)
4: (5,6)
0
i=1
bi=3
wi=2
w=1
w-wi =-1
0 0 0 0 000
1
2
3
4 50 1 2 3
4
iW
0
0
0
Example (4)

24
Items:
1: (2,3)
2: (3,4)
3: (4,5)
4: (5,6)
300
0
0
0
0 0 0 0 000
1
2
3
4 50 1 2 3
4
iW
i=1
bi=3
wi=2
w=2
w-wi =0
V[i,w] = bi + V[i-1,w- wi]
else
V[i,w] = V[i-1,w]
Example (5)

25
Items:
1: (2,3)
2: (3,4)
3: (4,5)
4: (5,6)
300
0
0
0
0 0 0 0 000
1
2
3
4 50 1 2 3
4
iW
i=1
bi=3
wi=2
w=3
w-wi =1
V[i,w] = bi + V[i-1,w- wi]
else
V[i,w] = V[i-1,w]
3
Example (6)

26
Items:
1: (2,3)
2: (3,4)
3: (4,5)
4: (5,6)
300
0
0
0
0 0 0 0 000
1
2
3
4 50 1 2 3
4
iW
i=1
bi=3
wi=2
w=4
w-wi =2
V[i,w] = bi + V[i-1,w- wi]
else
V[i,w] = V[i-1,w]
3 3
Example (7)

27
Items:
1: (2,3)
2: (3,4)
3: (4,5)
4: (5,6)
300
0
0
0
0 0 0 0 000
1
2
3
4 50 1 2 3
4
iW
i=1
bi=3
wi=2
w=5
w-wi =3
V[i,w] = bi + V[i-1,w- wi]
else
V[i,w] = V[i-1,w]
3 3 3
Example (8)

28
Items:
1: (2,3)
2: (3,4)
3: (4,5)
4: (5,6)
00
0
0
0
0 0 0 0 000
1
2
3
4 50 1 2 3
4
iW
i=2
bi=4
wi=3
w=1
w-wi =-2
3 3 3 3
0
V[i,w] = bi + V[i-1,w- wi]
else
V[i,w] = V[i-1,w]
Example (9)

29
Items:
1: (2,3)
2: (3,4)
3: (4,5)
4: (5,6)
00
0
0
0
0 0 0 0 000
1
2
3
4 50 1 2 3
4
iW
i=2
bi=4
wi=3
w=2
w-wi =-1
3 3 3 3
3
V[i,w] = bi + V[i-1,w- wi]
else
V[i,w] = V[i-1,w]
0
Example (10)

30
Items:
1: (2,3)
2: (3,4)
3: (4,5)
4: (5,6)
00
0
0
0
0 0 0 0 000
1
2
3
4 50 1 2 3
4
iW
i=2
bi=4
wi=3
w=3
w-wi =0
3 3 3 3
0
V[i,w] = bi + V[i-1,w- wi]
else
V[i,w] = V[i-1,w]
43
Example (11)

31
Items:
1: (2,3)
2: (3,4)
3: (4,5)
4: (5,6)
00
0
0
0
0 0 0 0 000
1
2
3
4 50 1 2 3
4
iW
i=2
bi=4
wi=3
w=4
w-wi =1
3 3 3 3
0
V[i,w] = bi + V[i-1,w- wi]
else
V[i,w] = V[i-1,w]
43 4
Example (12)

32
Items:
1: (2,3)
2: (3,4)
3: (4,5)
4: (5,6)
00
0
0
0
0 0 0 0 000
1
2
3
4 50 1 2 3
4
iW
i=2
bi=4
wi=3
w=5
w-wi =2
3 3 3 3
0
V[i,w] = bi + V[i-1,w- wi]
else
V[i,w] = V[i-1,w]
73 4 4
Example (13)

33
Items:
1: (2,3)
2: (3,4)
3: (4,5)
4: (5,6)
00
0
0
0
0 0 0 0 000
1
2
3
4 50 1 2 3
4
iW
i=3
bi=5
wi=4
w= 1..3
3 3 3 3
0 3 4 4
V[i,w] = bi + V[i-1,w- wi]
else
V[i,w] = V[i-1,w]
7
3 40
Example (14)

34
Items:
1: (2,3)
2: (3,4)
3: (4,5)
4: (5,6)
00
0
0
0
0 0 0 0 000
1
2
3
4 50 1 2 3
4
iW
i=3
bi=5
wi=4
w= 4
w- wi=0
3 3 3 3
0 3 4 4 7
0 3 4 5
V[i,w] = bi + V[i-1,w- wi]
else
V[i,w] = V[i-1,w]
Example (15)

35
Items:
1: (2,3)
2: (3,4)
3: (4,5)
4: (5,6)
00
0
0
0
0 0 0 0 000
1
2
3
4 50 1 2 3
4
iW
i=3
bi=5
wi=4
w= 5
w- wi=1
3 3 3 3
0 3 4 4 7
0 3 4
V[i,w] = bi + V[i-1,w- wi]
else
V[i,w] = V[i-1,w]
5 7
Example (16)

36
Items:
1: (2,3)
2: (3,4)
3: (4,5)
4: (5,6)
00
0
0
0
0 0 0 0 000
1
2
3
4 50 1 2 3
4
iW
i=4
bi=6
wi=5
w= 1..4
3 3 3 3
0 3 4 4
V[i,w] = bi + V[i-1,w- wi]
else
V[i,w] = V[i-1,w]
7
3 40
70 3 4 5
5
Example (17)

37
Items:
1: (2,3)
2: (3,4)
3: (4,5)
4: (5,6)
00
0
0
0
0 0 0 0 000
1
2
3
4 50 1 2 3
4
iW
i=4
bi=6
wi=5
w= 5
w- wi=0
3 3 3 3
0 3 4 4 7
0 3 4
V[i,w] = bi + V[i-1,w- wi]
else
V[i,w] = V[i-1,w]
5
7
7
0 3 4 5
Example (18)

38
Exercise
P303 8.2.1 (a).
How to find out which items are in the optimal subset?

39
Comments
 This algorithm only finds the max possible value
that can be carried in the knapsack
» i.e., the value in V[n,W]
 To know the items that make this maximum value,
an addition to this algorithm is necessary

40
 All of the information we need is in the table.
 V[n,W] is the maximal value of items that can be
placed in the Knapsack.
 Let i=n and k=W
if V[i,k] ≠ V[i−1,k] then
mark the ith
item as in the knapsack
i = i−1, k = k-wi
else
i = i−1 // Assume the ith
item is not in the knapsack
// Could it be in the optimally packed
knapsack?
How to find actual Knapsack
Items

41
Items:
1: (2,3)
2: (3,4)
3: (4,5)
4: (5,6)
00
0
0
0
0 0 0 0 000
1
2
3
4 50 1 2 3
4
iW i=4
k= 5
bi=6
wi=5
V[i,k] = 7
V[i−1,k] =7
3 3 3 3
0 3 4 4 7
0 3 4
i=n, k=W
while i,k > 0
mark the ith
i = i−1, k = k-wi
else
i = i−1
5 7
0 3 4 5 7
Finding the Items

42
Items:
1: (2,3)
2: (3,4)
3: (4,5)
4: (5,6)
00
0
0
0
0 0 0 0 000
1
2
3
4 50 1 2 3
4
iW i=4
k= 5
bi=6
wi=5
V[i,k] = 7
V[i−1,k] =7
3 3 3 3
0 3 4 4 7
0 3 4
i=n, k=W
while i,k > 0
mark the ith
i = i−1, k = k-wi
else
i = i−1
5 7
0 3 4 5 7
Finding the Items (2)

43
Items:
1: (2,3)
2: (3,4)
3: (4,5)
4: (5,6)
00
0
0
0
0 0 0 0 000
1
2
3
4 50 1 2 3
4
iW i=3
k= 5
bi=5
wi=4
V[i,k] = 7
V[i−1,k] =7
3 3 3 3
0 3 4 4 7
0 3 4
i=n, k=W
while i,k > 0
mark the ith
i = i−1, k = k-wi
else
i = i−1
5 7
0 3 4 5 7

44
Items:
1: (2,3)
2: (3,4)
3: (4,5)
4: (5,6)
00
0
0
0
0 0 0 0 000
1
2
3
4 50 1 2 3
4
iW i=2
k= 5
bi=4
wi=3
V[i,k] = 7
V[i−1,k] =3
k − wi=2
3 3 3 3
0 3 4 4 7
0 3 4
i=n, k=W
while i,k > 0
mark the ith
i = i−1, k = k-wi
else
i = i−1
5 7
0 3 4 5 7
7

45
Items:
1: (2,3)
2: (3,4)
3: (4,5)
4: (5,6)
00
0
0
0
0 0 0 0 000
1
2
3
4 50 1 2 3
4
iW i=1
k= 2
bi=3
wi=2
V[i,k] = 3
V[i−1,k] =0
k − wi=0
3 3 3 3
0 3 4 4 7
0 3 4
i=n, k=W
while i,k > 0
mark the ith
i = i−1, k = k-wi
else
i = i−1
5 7
0 3 4 5 7
3

46
Items:
1: (2,3)
2: (3,4)
3: (4,5)
4: (5,6)
00
0
0
0
0 0 0 0 000
1
2
3
4 50 1 2 3
4
iW
3 3 3 3
0 3 4 4 7
0 3 4
i=n, k=W
while i,k > 0
mark the nth
i = i−1, k = k-wi
else
i = i−1
5 7
0 3 4 5 7
i=0
k= 0
The optimal
knapsack
should contain
{1, 2}

47
Items:
1: (2,3)
2: (3,4)
3: (4,5)
4: (5,6)
00
0
0
0
0 0 0 0 000
1
2
3
4 50 1 2 3
4
iW
3 3 3 3
0 3 4 4 7
0 3 4
i=n, k=W
while i,k > 0
mark the nth
i = i−1, k = k-wi
else
i = i−1
5 7
0 3 4 5 7
The optimal
knapsack
should contain
{1, 2}
7
3

48
Memorization (Memory Function Method)
 Goal:
» Solve only subproblems that are necessary and solve it only once
 Memorization is another way to deal with overlapping subproblems
in dynamic programming
 With memorization, we implement the algorithm recursively:
» If we encounter a new subproblem, we compute and store the solution.
» If we encounter a subproblem we have seen, we look up the answer
 Most useful when the algorithm is easiest to implement recursively
» Especially if we do not need solutions to all subproblems.

49
for i = 1 to n
for w = 1 to W
V[i,w] = -1
for w = 0 to W
V[0,w] = 0
for i = 1 to n
V[i,0] = 0
MFKnapsack(i, w)
if V[i,w] < 0
if w < wi
value = MFKnapsack(i-1, w)
else
value = max(MFKnapsack(i-1, w),
bi + MFKnapsack(i-1, w-wi))
V[i,w] = value
return V[i,w]
0-1 Knapsack Memory Function Algorithm

50
 Dynamic programming is a useful technique of
solving certain kind of problems
 When the solution can be recursively described in
terms of partial solutions, we can store these
partial solutions and re-use them as necessary
(memorization)
 Running time of dynamic programming algorithm
vs. naïve algorithm:
» 0-1 Knapsack problem: O(W*n) vs. O(2n
)
Conclusion

52
Brute-Force Approach
Design and Analysis of Algorithms - Chapter 8 52

53
Dynamic-Programming Approach
 (1) SMaxV(0) = 0
 (2) MaxV(0) = 0
 (3) for i=1 to n
 (4) SMaxV(i) = max(SmaxV(i-1)+xi, 0)
 (5) MaxV(i) = max(MaxV(i-1), SMaxV(i))
 (6) return MaxV(n)
 Run the algorithm on the following example instance:
» 30, 40, -100, 10, 20, 50, -60, 90, -180, 100
53

Knapsack problem

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