Design and Analysis - 0-1 Knapsack

We discussed the fractional knapsack problem using the greedy approach, earper in this tutorial. It is shown that Greedy approach gives an optimal solution for Fractional Knapsack. However, this chapter will cover 0-1 Knapsack problem using dynamic programming approach and its analysis.

Unpke in fractional knapsack, the items are always stored fully without using the fractional part of them. Its either the item is added to the knapsack or not. That is why, this method is known as the 0-1 Knapsack problem.

Hence, in case of 0-1 Knapsack, the value of x_i can be either 0 or 1, where other constraints remain the same.

0-1 Knapsack cannot be solved by Greedy approach. Greedy approach does not ensure an optimal solution in this method. In many instances, Greedy approach may give an optimal solution.

0-1 Knapsack Algorithm

Problem Statement − A thief is robbing a store and can carry a maximal weight of W into his knapsack. There are n items and weight of i^th item is wi and the profit of selecting this item is pi. What items should the thief take?

Let i be the highest-numbered item in an optimal solution S for W dollars. Then S’ = S − {i} is an optimal solution for W – w_i dollars and the value to the solution S is V_i plus the value of the sub-problem.

We can express this fact in the following formula: define c[i, w] to be the solution for items 1,2, … , i and the maximum weight w.

The algorithm takes the following inputs

The maximum weight W

The number of items n

The two sequences v = <v1, v2, …, vn> and w = <w1, w2, …, wn>

The set of items to take can be deduced from the table, starting at c[n, w] and tracing backwards where the optimal values came from.

If c[i, w] = c[i-1, w], then item i is not part of the solution, and we continue tracing with c[i-1, w]. Otherwise, item i is part of the solution, and we continue tracing with c [i-1, w-W].

Dynamic-0-1-knapsack (v, w, n, W)
for w = 0 to W do
   c[0, w] = 0
for i = 1 to n do
   c[i, 0] = 0
   for w = 1 to W do
      if wi ≤ w then
         if vi + c[i-1, w-wi] then
            c[i, w] = vi + c[i-1, w-wi]
         else c[i, w] = c[i-1, w]
      else
         c[i, w] = c[i-1, w]

The following examples will estabpsh our statement.

Example

Let us consider that the capacity of the knapsack is W = 8 and the items are as shown in the following table.

Solution

Using the greedy approach of 0-1 knapsack, the weight that’s stored in the knapsack would be A+B = 4 with the maximum profit 2 + 4 = 6. But, that solution would not be the optimal solution.

Therefore, dynamic programming must be adopted to solve 0-1 knapsack problems.

Step 1

Construct an adjacency table with maximum weight of knapsack as rows and items with respective weights and profits as columns.

Values to be stored in the table are cumulative profits of the items whose weights do not exceed the maximum weight of the knapsack (designated values of each row)

So we add zeroes to the 0^th row and 0^th column because if the weight of item is 0, then it weighs nothing; if the maximum weight of knapsack is 0, then no item can be added into the knapsack.

The remaining values are filled with the maximum profit achievable with respect to the items and weight per column that can be stored in the knapsack.

The formula to store the profit values is −

$$cleft [ i,w ight ]=maxleft{cleft [ i-1,w-wleft [ i ight ] ight ]+Pleft [ i ight ] ight}$$

By computing all the values using the formula, the table obtained would be −

To find the items to be added in the knapsack, recognize the maximum profit from the table and identify the items that make up the profit, in this example, its {1, 7}.

The optimal solution is {1, 7} with the maximum profit is 12.

Analysis

This algorithm takes Ɵ(n.w) times as table c has (n+1).(w+1) entries, where each entry requires Ɵ(1) time to compute.

Example

Following is the final implementation of 0-1 Knapsack Algorithm using Dynamic Programming Approach.

#include <stdio.h>
#include <string.h>
int findMax(int n1, int n2){
   if(n1>n2) {
      return n1;
   } else {
      return n2;
   }
}
int knapsack(int W, int wt[], int val[], int n){
   int K[n+1][W+1];
   for(int i = 0; i<=n; i++) {
      for(int w = 0; w<=W; w++) {
         if(i == 0 || w == 0) {
            K[i][w] = 0;
         } else if(wt[i-1] <= w) {
            K[i][w] = findMax(val[i-1] + K[i-1][w-wt[i-1]], K[i-1][w]);
         } else {
            K[i][w] = K[i-1][w];
         }
      }
   }
   return K[n][W];
}
int main(){
   int val[5] = {70, 20, 50};
   int wt[5] = {11, 12, 13};
   int W = 30;
   int len = sizeof val / sizeof val[0];
   printf("Maximum Profit achieved with this knapsack: %d", knapsack(W, wt, val, len));
}

Maximum Profit achieved with this knapsack: 120

#include <bits/stdc++.h>
using namespace std;
int max(int a, int b){
   return (a > b) ? a : b;
}
int knapSack(int W, int wt[], int val[], int n){
   int i, w;
   vector<vector<int>> K(n + 1, vector<int>(W + 1));
   for(i = 0; i <= n; i++) {
      for(w = 0; w <= W; w++) {
         if (i == 0 || w == 0)
            K[i][w] = 0;
         else if (wt[i - 1] <= w)
            K[i][w] = max(val[i - 1] + K[i - 1][w - wt[i - 1]], K[i - 1][w]);
         else
            K[i][w] = K[i - 1][w];
      }
   }
   return K[n][W];
}
int main(){
   int val[] = { 70, 20, 50 };
   int wt[] = { 11, 12, 13 };
   int W = 30;
   int n = sizeof(val) / sizeof(val[0]);
   cout << "Maximum Profit achieved with this knapsack: " << knapSack(W, wt, val, n);
   return 0;
}

Maximum Profit achieved with this knapsack: 120

import java.util.*;
import java.lang.*;
pubpc class 01Knapsack {
   pubpc static int findMax(int n1, int n2) {
      if(n1>n2) {
         return n1;
      } else {
         return n2;
      }
   }
   pubpc static int knapsack(int W, int wt[], int val[], int n) {
      int K[][] = new int[n+1][W+1];
      for(int i = 0; i<=n; i++) {
         for(int w = 0; w<=W; w++) {
            if(i == 0 || w == 0) {
               K[i][w] = 0;
            } else if(wt[i-1] <= w) {
               K[i][w] = findMax(val[i-1] + K[i-1][w-wt[i-1]], K[i-1][w]);
            } else {
               K[i][w] = K[i-1][w];
            }
         }
      }
      return K[n][W];
   }
   pubpc static void main(String[] args) {
      int[] val = {70, 20, 50};
      int[] wt = {11, 12, 13};
      int W = 30;
      int len = val.length;
      System.out.print("Maximum Profit achieved with this knapsack: " + knapsack(W, wt, val, len));
   }
}

Maximum Profit achieved with this knapsack: 120

def knapsack(W, wt, val, n):
   K = [[0] * (W+1) for i 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
         epf(wt[i-1] <= w):
            K[i][w] = max(val[i-1] + K[i-1][w-wt[i-1]], K[i-1][w])
         else:
            K[i][w] = K[i-1][w]
   return K[n][W]

val = [70, 20, 50];
wt = [11, 12, 13];
W = 30;
ln = len(val);
profit = knapsack(W, wt, val, ln)
print("Maximum Profit achieved with this knapsack: ")
print(profit)

Maximum Profit achieved with this knapsack: 
120