A list of items is given, each item has its own value and weight. Items can be placed in a knapsack whose maximum weight limit is W. The problem is to find the weight that is less than or equal to W, and value is maximized.
There are two types of Knapsack problem.
- 0 – 1 Knapsack
- Fractional Knapsack
For the 0 – 1 Knapsack, items cannot be divided into smaller pieces, and for fractional knapsack, items can be broken into smaller pieces.
Here we will discuss the fractional knapsack problem.
The time complexity of this algorithm is O(n Log n).
Input and Output
Input: Maximum weight = 50. List of items with value and weight. {(60, 10), (100, 20), (120, 30)} Output: Maximum value: 240 By taking the items of weight 20 and 30
Algorithm
fractionalKnapsack(weight, itemList, n)
Input − maximum weight of the knapsack, list of items and the number of items
Output: The maximum value obtained.
Begin sort the item list based on the ration of value and weight currentWeight := 0 knapsackVal := 0 for all items i in the list do if currentWeight + weight of item[i] < weight then currentWeight := currentWeight + weight of item[i] knapsackVal := knapsackVal + value of item[i] else remaining := weight – currentWeight knapsackVal “= knapsackVal + value of item[i] * (remaining/weight of item[i]) break the loop done End
EXample
#include <iostream> #include<algorithm> using namespace std; struct item { int value, weight; }; bool cmp(struct item a, struct item b) { //compare item a and item b based on the ration of value and weight double aRatio = (double)a.value / a.weight; double bRatio = (double)b.value / b.weight; return aRatio > bRatio; } double fractionalKnapsack(int weight, item itemList[], int n) { sort(itemList, itemList + n, cmp); //sort item list using compare function int currWeight = 0; // Current weight in knapsack double knapsackVal = 0.0; for (int i = 0; i < n; i++) { //check through all items if (currWeight + itemList[i].weight <= weight) { //when the space is enough for selected item, add it currWeight += itemList[i].weight; knapsackVal += itemList[i].value; }else{ //when no place for whole item, break it into smaller parts int remaining = weight - currWeight; knapsackVal += itemList[i].value * ((double) remaining / itemList[i].weight); break; } } return knapsackVal; } int main() { int weight = 50; // Weight of knapsack item itemList[] = {{60, 10}, {100, 20}, {120, 30}}; int n = 3; cout << "Maximum value: " << fractionalKnapsack(weight, itemList, n); }
Output
Maximum value: 240