Knapsack Problem Basics

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The Problem: Given a set of items where each item contains a weight and value, determine the number of each 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.

Pseudo code for Knapsack Problem


  1. Values(array v)
  2. Weights(array w)
  3. Number of distinct items(n)
  4. Capacity(W)
for j from 0 to W do:
    m[0, j] := 0
for i from 1 to n do:
    for j from 0 to W do:
        if w[i] > j then:
            m[i, j] := m[i-1, j]
            m[i, j] := max(m[i-1, j], m[i-1, j-w[i]] + v[i])

A simple implementation of the above pseudo code using Python:

def knapSack(W, wt, val, n):
    K = [[0 for x in range(W+1)] for x in range(n+1)]
    for i in range(n+1):
        for w in range(W+1):
            if i==0 or w==0:
                K[i][w] = 0
            elif wt[i-1] <= w:
                K[i][w] = max(val[i-1] + K[i-1][w-wt[i-1]],  K[i-1][w])
                K[i][w] = K[i-1][w]
    return K[n][W]
val = [60, 100, 120]
wt = [10, 20, 30]
W = 50
n = len(val)
print(knapSack(W, wt, val, n))

Running the code: Save this in a file named

$ python

Time Complexity of the above code: O(nW) where n is the number of items and W is the capacity of knapsack.

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