In this tutorial we will be learning about 0 1 Knapsack problem. In this dynamic programming problem we have n items each with an associated weight and value (benefit or profit). The objective is to fill the knapsack with items such that we have a maximum profit without crossing the weight limit of the knapsack. Since this is a 0 1 knapsack problem hence we can either take an entire item or reject it completely. We can not break an item and fill the knapsack.
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Point to remember
- In this problem we have a Knapsack that has a weight limit W.
- There are items i1, i2, .., in each having weight w1, w2, … wn and some benefit (value or profit) associated with it v1, v2, .. vn
- Our objective is to maximise the benefit such that the total weight inside the knapsack is at most W.
- Since this is a 0 1 Knapsack problem algorithm so, we can either take an entire item or reject it completely. We can not break an item and fill the knapsack.
Problem
Assume that we have a knapsack with max weight capacity W = 5
Our objective is to fill the knapsack with items such that the benefit (value or profit) is maximum.
Our objective is to fill the knapsack with items such that the benefit (value or profit) is maximum.
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Following table contains the items along with their value and weight.
item i | 1 | 2 | 3 | 4 |
value val | 100 | 20 | 60 | 40 |
weight wt | 3 | 2 | 4 | 1 |
Total items n = 4
Total capacity of the knapsack W = 5
Total capacity of the knapsack W = 5
Now we create a value table V[i,w] where, i denotes number of items and w denotes the weight of the items.
Rows denote the items and columns denote the weight.
As there are 4 items so, we have 5 rows from 0 to 4.
And the weight limit of the knapsack is W = 5 so, we have 6 columns from 0 to 5
Rows denote the items and columns denote the weight.
As there are 4 items so, we have 5 rows from 0 to 4.
And the weight limit of the knapsack is W = 5 so, we have 6 columns from 0 to 5
We fill the first row i = 0 with 0. This means when 0 item is considered weight is 0.
Then we fill the first column w = 0 with 0. This means when weight is 0 then items considered is 0.
Rule to fill the V[i,w] table.
After calculation, the value table V
V[i,w] | w = 0 | 1 | 2 | 3 | 4 | 5 |
i = 0 | 0 | 0 | 0 | 0 | 0 | 0 |
1 | 0 | 0 | 0 | 100 | 100 | 100 |
2 | 0 | 0 | 20 | 100 | 100 | 120 |
3 | 0 | 0 | 20 | 100 | 100 | 120 |
4 | 0 | 40 | 40 | 100 | 140 | 140 |
Maximum value earned
Max Value = V[n,W]
= V[4,5]
= 140
Max Value = V[n,W]
= V[4,5]
= 140
So, items we are putting inside the knapsack are 4 and 1.
C Code
Time complexity
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Time complexity of 0 1 Knapsack problem is O(nW) where, n is the number of items and W is the capacity of knapsack.
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