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.
Input:
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.
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: 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. V [i, j] = max {V [i  1, j], v_{i} + V [i  1, j w_{i}] The value of V [3, 7] was computed as follows: V [3, 7] = max {V [3  1, 7], v_{3} + V [3  1, 7  w_{3}] = max {V [2, 7], 18 + V [2, 7  5]} = max {7, 18 + 6} = 24 Finally, the output is: The maximum value of items in the knapsack is 40, the bottomright entry). The dynamic programming approach can now be coded as the following algorithm: Algorithm of Knapsack ProblemKNAPSACK (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 (w_{i}≤ w & v_{i} + V [i1, w  w_{i}]> V [i 1,W]) 7. then V [i, W] ← v_{i} + V [i  1, w  w_{i}] 8. else V [i, W] ← V [i  1, w]
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