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:-
- 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:
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 |
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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