Fractional Knapsack Problem

Given the weights and values of n items, we need to put these items in a knapsack of capacity W to get the maximum total value in the knapsack.

In the 0-1 Knapsack problem, we are not allowed to break items. We either take the whole item or don’t take it. 

Input: 
Items as (value, weight) pairs 
arr[] = {{60, 10}, {100, 20}, {120, 30}} 
Knapsack Capacity, W = 50; 

Output:
Maximum possible value = 240 
by taking items of weight 10 and 20 kg and 2/3 fraction 
of 30 kg. Hence total price will be 60+100+(2/3)(120) = 240

In Fractional Knapsack, we can break items for maximizing the total value of knapsack. This problem in which we can break an item is also called the fractional knapsack problem. 

Input : 
Same as above

Output :
Maximum possible value = 240
By taking full items of 10 kg, 20 kg and 
2/3rd of last item of 30 kg

A brute-force solution would be to try all possible subsets with all different fractions but that will be too much time taking. 

An efficient solution is to use the Greedy approach. The basic idea of the greedy approach is to calculate the ratio value/weight for each item and sort the item on basis of this ratio. Then take the item with the highest ratio and add them until we can’t add the next item as a whole and at the end add the next item as much as we can. Which will always be the optimal solution to this problem.

Algorithm:-

1. Calculate the ratio(value/weight) for each item.
2. Sort all the items in decreasing order of the ratio.
3. Initialize res =0, curr_cap = given_cap.
4. Do the following for every item “i” in the sorted order:

while(i.weight){
    if(i.weight<=curr_cap)
    {
        curr_cap -= i.weight;
        res += i.value;
    }
    else
    {
        res += (i.value * (curr_cap/i.weight));
        return res;
    }
}

        5. Return res.

A simple code with our own comparison function can be written as follows, please see the sort function more closely, the third argument to the sort function is our comparison function which sorts the item according to value/weight ratio in non-decreasing order. 
After sorting we need to loop over these items and add them to our knapsack satisfying the above-mentioned criteria.

Below is the implementation of the above idea:

C++

// C/C++ program to solve fractional Knapsack Problem
#include <bits/stdc++.h>

using namespace std;

// Structure for an item which stores weight and
// corresponding value of Item
struct Item {
    int value, weight;

    // Constructor
    Item(int value, int weight)
    {
        this->value = value;
        this->weight = weight;
    }
};

// Comparison function to sort Item according to val/weight
// ratio
bool cmp(struct Item a, struct Item b)
{
    double r1 = (double)a.value / (double)a.weight;
    double r2 = (double)b.value / (double)b.weight;
    return r1 > r2;
}

// Main greedy function to solve problem
double fractionalKnapsack(int W, struct Item arr[], int n)
{
    //    sorting Item on basis of ratio
    sort(arr, arr + n, cmp);

    //    Uncomment to see new order of Items with their
    //    ratio
    /*
    for (int i = 0; i < n; i++)
    {
        cout << arr[i].value << "  " << arr[i].weight << " :
    "
             << ((double)arr[i].value / arr[i].weight) <<
    endl;
    }
    */

    double finalvalue = 0.0; // Result (value in Knapsack)

    // Looping through all Items
    for (int i = 0; i < n; i++) {
        // If adding Item won't overflow, add it completely
        if (arr[i].weight <= W) {
            W -= arr[i].weight;
            finalvalue += arr[i].value;
        }

        // If we can't add current Item, add fractional part
        // of it
        else {
            finalvalue
                += arr[i].value
                   * ((double)W / (double)arr[i].weight);
            break;
        }
    }

    // Returning final value
    return finalvalue;
}

// Driver code
int main()
{
    int W = 50; //    Weight of knapsack
    Item arr[] = { { 60, 10 }, { 100, 20 }, { 120, 30 } };

    int n = sizeof(arr) / sizeof(arr[0]);

    // Function call
    cout << "Maximum value we can obtain = "
         << fractionalKnapsack(W, arr, n);
    return 0;
}

Java

// Java program to solve fractional Knapsack Problem
import java.util.Arrays;
import java.util.Comparator;

// Greedy approach
public class FractionalKnapSack {
    // function to get maximum value
    private static double getMaxValue(int[] wt, int[] val,
                                      int capacity)
    {
        ItemValue[] iVal = new ItemValue[wt.length];

        for (int i = 0; i < wt.length; i++) {
            iVal[i] = new ItemValue(wt[i], val[i], i);
        }

        // sorting items by value;
        Arrays.sort(iVal, new Comparator<ItemValue>() {
            @Override
            public int compare(ItemValue o1, ItemValue o2)
            {
                return o2.cost.compareTo(o1.cost);
            }
        });

        double totalValue = 0d;

        for (ItemValue i : iVal) {

            int curWt = (int)i.wt;
            int curVal = (int)i.val;

            if (capacity - curWt >= 0) {
                // this weight can be picked while
                capacity = capacity - curWt;
                totalValue += curVal;
            }
            else {
                // item cant be picked whole
                double fraction
                    = ((double)capacity / (double)curWt);
                totalValue += (curVal * fraction);
                capacity
                    = (int)(capacity - (curWt * fraction));
                break;
            }
        }

        return totalValue;
    }

    // item value class
    static class ItemValue {
        Double cost;
        double wt, val, ind;

        // item value function
        public ItemValue(int wt, int val, int ind)
        {
            this.wt = wt;
            this.val = val;
            this.ind = ind;
            cost = new Double((double)val / (double)wt);
        }
    }

    // Driver code
    public static void main(String[] args)
    {
        int[] wt = { 10, 40, 20, 30 };
        int[] val = { 60, 40, 100, 120 };
        int capacity = 50;

        double maxValue = getMaxValue(wt, val, capacity);

        // Function call
        System.out.println("Maximum value we can obtain = "
                           + maxValue);
    }
}

Python3

# Python3 program to solve fractional
# Knapsack Problem


class ItemValue:

    """Item Value DataClass"""

    def __init__(self, wt, val, ind):
        self.wt = wt
        self.val = val
        self.ind = ind
        self.cost = val // wt

    def __lt__(self, other):
        return self.cost < other.cost

# Greedy Approach


class FractionalKnapSack:

    """Time Complexity O(n log n)"""
    @staticmethod
    def getMaxValue(wt, val, capacity):
        """function to get maximum value """
        iVal = []
        for i in range(len(wt)):
            iVal.append(ItemValue(wt[i], val[i], i))

        # sorting items by value
        iVal.sort(reverse=True)

        totalValue = 0
        for i in iVal:
            curWt = int(i.wt)
            curVal = int(i.val)
            if capacity - curWt >= 0:
                capacity -= curWt
                totalValue += curVal
            else:
                fraction = capacity / curWt
                totalValue += curVal * fraction
                capacity = int(capacity - (curWt * fraction))
                break
        return totalValue


# Driver Code
if __name__ == "__main__":
    wt = [10, 40, 20, 30]
    val = [60, 40, 100, 120]
    capacity = 50

    # Function call
    maxValue = FractionalKnapSack.getMaxValue(wt, val, capacity)
    print("Maximum value in Knapsack =", maxValue)

# This code is contributed by vibhu4agarwal

C#

// C# program to solve fractional Knapsack Problem
using System;
using System.Collections;

class GFG {

    // Class for an item which stores weight and
    // corresponding value of Item
    class item {
        public int value;
        public int weight;

        public item(int value, int weight)
        {
            this.value = value;
            this.weight = weight;
        }
    }

    // Comparison function to sort Item according
    // to val/weight ratio
    class cprCompare : IComparer {
        public int Compare(Object x, Object y)
        {
            item item1 = (item)x;
            item item2 = (item)y;
            double cpr1 = (double)item1.value
                          / (double)item1.weight;
            double cpr2 = (double)item2.value
                          / (double)item2.weight;

            if (cpr1 < cpr2)
                return 1;

            return cpr1 > cpr2 ? -1 : 0;
        }
    }

    // Main greedy function to solve problem
    static double FracKnapSack(item[] items, int w)
    {

        // Sort items based on cost per units
        cprCompare cmp = new cprCompare();
        Array.Sort(items, cmp);

        // Traverse items, if it can fit,
        // take it all, else take fraction
        double totalVal = 0f;
        int currW = 0;

        foreach(item i in items)
        {
            float remaining = w - currW;

            // If the whole item can be
            // taken, take it
            if (i.weight <= remaining) {
                totalVal += (double)i.value;
                currW += i.weight;
            }

            // dd fraction until we run out of space
            else {
                if (remaining == 0)
                    break;

                double fraction
                    = remaining / (double)i.weight;
                totalVal += fraction * (double)i.value;
                currW += (int)(fraction * (double)i.weight);
            }
        }
        return totalVal;
    }

    // Driver code
    static void Main(string[] args)
    {
        item[] arr = { new item(60, 10), new item(100, 20),
                       new item(120, 30) };

        Console.WriteLine("Maximum value we can obtain = "
                          + FracKnapSack(arr, 50));
    }
}

// This code is contributed by Mohamed Adel

Output

Maximum value we can obtain = 240

As main time taking step is sorting, the whole problem can be solved in O(n log n) only. 

Time Complexity: O(nlogn)
Auxiliary Space: O(1)

This article is contributed by Utkarsh Trivedi.

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