Fractional Knapsack Problem

Given the weights and values of N items, in the form of {value, weight} put these items in a knapsack of capacity W to get the maximum total value in the knapsack. In Fractional Knapsack, we can break items for maximizing the total value of the knapsack

Note: 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: arr[] = {{60, 10}, {100, 20}, {120, 30}}, W = 50
Output: 240 
Explanation: 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

Input:  arr[] = {{500, 30}}, W = 10
Output: 166.667

Naive Approach: To solve the problem follow the below idea:

Try all possible subsets with all different fractions.

Algorithm for Knapsack problem using Greedy Strategy:

Algorithm Fr-Knapsack (M,n) 
/*Input: Knapsack capacity = M. n is the number of available items with associated profits P[1: n] and weights W[1: n]. These items are such that P[i]/W[i]>=P[i+1]/W[i +1] where 1 ≤i≤n.

Output: S[1: n] is the fixed size solution vector. S[i] gives the fractional part xi, of item i placed into a knapsack, 0 ≤ xi ≤1 and 1 ≤ i ≤ n. */
{
for (i:=1; i <= n ;i++)

{

S[i] :=0.0; // Initialization of a solution vector.
}

balance:= M; /* 'balance' describes the remaining capacity of a knapsack. Initially it is equal to the knapsack capacity M. */

for(i:=1; i <= n ;i++)

{
if(W [i] > balance ) / *Insufficient capacity of a knapsack */
break;

S[i] :=1.0; // add whole item i.
balance := balance - W[i]; /*Update the remaining capacity after adding item i with weight W[i] into a knapsack.*/
} 
f(i <= n) /*Due to insufficient capacity add fractional part of item i in the knapsack. */
S[i]:= (balance / W[i] ) ;
return S;
}

Fractional Knapsack Problem using Greedy algorithm:

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 the 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.

Follow the given steps to solve the problem using the above approach:

• Calculate the ratio(value/weight) for each item.
• Sort all the items in decreasing order of the ratio.
• Initialize res =0, curr_cap = given_cap.
• Do the following for every item “i” in the sorted order:
  • If the weight of the current item is less than or equal to the remaining capacity then add the value of that item into the result
  • Else add the current item as much as we can and break out of the loop.
• Return res.

Below is the implementation of the above approach:

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
static 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);

    double finalvalue = 0.0;

    // 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;
    Item arr[] = { { 60, 10 }, { 100, 20 }, { 120, 30 } };

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

    // Function call
    cout << fractionalKnapsack(W, arr, N);
    return 0;
}

Java

// Java program to solve fractional Knapsack Problem

import java.io.*;
import java.util.Arrays;
import java.util.Comparator;

// Greedy approach
class FractionalKnapSack {

    // Function to get maximum value
    private static double getMaxValue(ItemValue[] arr,
                                      int capacity)
    {
        // Sorting items by value/weight ratio;
        Arrays.sort(arr, new Comparator<ItemValue>() {
            @Override
            public int compare(ItemValue item1,
                               ItemValue item2)
            {
                double cpr1
                    = new Double((double)item1.value
                                 / (double)item1.weight);
                double cpr2
                    = new Double((double)item2.value
                                 / (double)item2.weight);

                if (cpr1 < cpr2)
                    return 1;
                else
                    return -1;
            }
        });

        double totalValue = 0d;

        for (ItemValue i : arr) {

            int curWt = (int)i.weight;
            int curVal = (int)i.value;

            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 {

        int value, weight;

        // Item value function
        public ItemValue(int val, int wt)
        {
            this.weight = wt;
            this.value = val;
        }
    }

    // Driver code
    public static void main(String[] args)
    {

        ItemValue[] arr = { new ItemValue(60, 10),
                            new ItemValue(100, 20),
                            new ItemValue(120, 30) };

        int capacity = 50;

        double maxValue = getMaxValue(arr, capacity);

        // Function call
        System.out.println(maxValue);
    }
}

Python3

# Structure for an item which stores weight and
# corresponding value of Item
class Item:
    def __init__(self, value, weight):
        self.value = value
        self.weight = weight

# Main greedy function to solve problem
def fractionalKnapsack(W, arr):

    # Sorting Item on basis of ratio
    arr.sort(key=lambda x: (x.value/x.weight), reverse=True)

    # Result(value in Knapsack)
    finalvalue = 0.0

    # Looping through all Items
    for item in arr:

        # If adding Item won't overflow,
        # add it completely
        if item.weight <= W:
            W -= item.weight
            finalvalue += item.value

        # If we can't add current Item,
        # add fractional part of it
        else:
            finalvalue += item.value * W / item.weight
            break

    # Returning final value
    return finalvalue


# Driver Code
if __name__ == "__main__":

    W = 50
    arr = [Item(60, 10), Item(100, 20), Item(120, 30)]

    # Function call
    max_val = fractionalKnapsack(W, arr)
    print(max_val)

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) };

        // Function call
        Console.WriteLine(FracKnapSack(arr, 50));
    }
}

// This code is contributed by Mohamed Adel

Javascript

// JavaScript program to solve fractional Knapsack Problem

// Structure for an item which stores weight and
// corresponding value of Item
class Item {
    constructor(value, weight) {
        this.value = value;
        this.weight = weight;
    }
}

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

// Main greedy function to solve problem
function fractionalKnapsack(W, arr) {
    // Sorting Item on basis of ratio
    arr.sort(cmp);

    let finalvalue = 0.0;

    // Looping through all items
    for (let i = 0; i < arr.length; 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 * (W / arr[i].weight);
            break;
        }
    }

    // Returning final value
    return finalvalue;
}

// Driver code
let W = 50;
let arr = [new Item(60, 10), new Item(100, 20), new Item(120, 30)];

console.log(fractionalKnapsack(W, arr));

// This code is contributed by lokeshpotta20

Output

Maximum value we can obtain = 240

Time Complexity: O(N * log N)
Auxiliary Space: O(N)

This article is contributed by Utkarsh Trivedi.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.