The knapsack problem or rucksack problem is a problem in combinatorial optimization: Given a set of items, each with a weight and a value, determine the number of each item to include in a collection so that the total weight is less than or equal to a given limit and the total value is as large as possible. It derives its name from the problem faced by someone who is constrained by a fixed-size knapsackand must fill it with the most valuable items.
0/1 Knapsack Problem: Dynamic Programming Approach:
Knapsack Problem:
Knapsack is basically means bag. A bag of given capacity.
We want to pack n items in your luggage.
- The ith item is worth vi dollars and weight wi pounds.
- Take as valuable a load as possible, but cannot exceed W pounds.
- vi wi W are integers.
0/1 Knapsack Problem: Dynamic Programming Approach:
Knapsack Problem:
Knapsack is basically means bag. A bag of given capacity.
We want to pack n items in your luggage.
- The ith item is worth vi dollars and weight wi pounds.
- Take as valuable a load as possible, but cannot exceed W pounds.
- vi wi W are integers.
0/1 Knapsack Problem: Dynamic Programming Approach:
Knapsack Problem:
Knapsack is basically means bag. A bag of given capacity.
We want to pack n items in your luggage. - The ith item is worth vi dollars and weight wi pounds.
- Take as valuable a load as possible, but cannot exceed W pounds.
- vi wi W are integers.
0/1 Knapsack Problem: Dynamic Programming Approach:
Knapsack Problem:
Knapsack is basically means bag. A bag of given capacity.
We want to pack n items in your luggage. - The ith item is worth vi dollars and weight wi pounds.
- Take as valuable a load as possible, but cannot exceed W pounds.
- vi wi W are integers.
- Knapsack of capacity
- List (Array) of weight and their corresponding value.
Output: To maximize profit and minimize weight in capacity.
The knapsack problem where we have to pack the knapsack with maximum value in such a manner that the total weight of the items should not be greater than the capacity of the knapsack.
Knapsack problem can be further divided into two parts:
1. Fractional Knapsack: Fractional knapsack problem can be solved by Greedy Strategy where as 0 /1 problem is not.
It cannot be solved by Dynamic Programming Approach.
0/1 Knapsack Problem:
In this item cannot be broken which means thief should take the item as a whole or should leave it. That's why it is called 0/1 knapsack Problem.
- Each item is taken or not taken.
- Cannot take a fractional amount of an item taken or take an item more than once.
- It cannot be solved by the Greedy Approach because it is enable to fill the knapsack to capacity.
- Greedy Approach doesn't ensure an Optimal Solution.
Example of 0/1 Knapsack Problem:
Example: The maximum weight the knapsack can hold is W is 11. There are five items to choose from. Their weights and values are presented in the following table:
![Example of 0/1 Knapsack Problem:](https://static.javatpoint.com/tutorial/daa/images/example-of-0-1-knapsack-problem.png)
The [i, j] entry here will be V [i, j], the best value obtainable using the first "i" rows of items if the maximum capacity were j. We begin by initialization and first row.
![Example of 0/1 Knapsack Problem:](https://static.javatpoint.com/tutorial/daa/images/example-of-0-1-knapsack-problem2.png)
V [i, j] = max {V [i - 1, j], vi + V [i - 1, j -wi]
![Example of 0/1 Knapsack Problem](https://static.javatpoint.com/tutorial/daa/images/example-of-0-1-knapsack-problem3.png)
![Example of 0/1 Knapsack Problem](https://static.javatpoint.com/tutorial/daa/images/example-of-0-1-knapsack-problem4.png)
The value of V [3, 7] was computed as follows:
V [3, 7] = max {V [3 - 1, 7], v3 + V [3 - 1, 7 - w3] = max {V [2, 7], 18 + V [2, 7 - 5]} = max {7, 18 + 6} = 24
![Example of 0/1 Knapsack Problem](https://static.javatpoint.com/tutorial/daa/images/example-of-0-1-knapsack-problem6.png)
Finally, the output is:
![Example of 0/1 Knapsack Problem](https://static.javatpoint.com/tutorial/daa/images/example-of-0-1-knapsack-problem7.png)
The maximum value of items in the knapsack is 40, the bottom-right entry). The dynamic programming approach can now be coded as the following algorithm:
Algorithm of Knapsack Problem
KNAPSACK (n, W) 1. for w = 0, W 2. do V [0,w] ← 0 3. for i=0, n 4. do V [i, 0] ← 0 5. for w = 0, W 6. do if (wi≤ w & vi + V [i-1, w - wi]> V [i -1,W]) 7. then V [i, W] ← vi + V [i - 1, w - wi] 8. else V [i, W] ← V [i - 1, w]
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