0/1 Knapsack Problem to print all possible solutions
Given weights and profits of N items, put these items in a knapsack of capacity W. The task is to print all possible solutions to the problem in such a way that there are no remaining items left whose weight is less than the remaining capacity of the knapsack. Also, compute the maximum profit.
Examples:
Input: Profits[] = {60, 100, 120, 50}
Weights[] = {10, 20, 30, 40}, W = 40
Output:
10: 60, 20: 100,
10: 60, 30: 120,
Maximum Profit = 180
Explanation:
Maximum profit from all the possible solutions is 180
Input: Profits[] = {60, 100, 120, 50}
Weights[] = {10, 20, 30, 40}, W = 50
Output:
10: 60, 20: 100,
10: 60, 30: 120,
20: 100, 30: 120,
Maximum Profit = 220
Explanation:
Maximum profit from all the possible solutions is 220
Approach: The idea is to make pairs for the weight and the profits of the items and then try out all permutations of the array and including the weights until their is no such item whose weight is less than the remaining capacity of the knapsack. Meanwhile after including an item increment the profit for that solution by the profit of that item.
Below is the implementation of the above approach:
C++
// C++ implementation to print all // the possible solutions of the // 0/1 Knapsack problem #include <bits/stdc++.h> using namespace std; // Utility function to find the // maximum of the two elements int max( int a, int b) { return (a > b) ? a : b; } // Function to find the all the // possible solutions of the // 0/1 knapSack problem int knapSack( int W, vector< int > wt, vector< int > val, int n) { // Mapping weights with Profits map< int , int > umap; set<vector<pair< int , int >>> set_sol; // Making Pairs and inserting // into the map for ( int i = 0; i < n; i++) { umap.insert({ wt[i], val[i] }); } int result = INT_MIN; int remaining_weight; int sum = 0; // Loop to iterate over all the // possible permutations of array do { sum = 0; // Initially bag will be empty remaining_weight = W; vector<pair< int , int >> possible; // Loop to fill up the bag // until there is no weight // such which is less than // remaining weight of the // 0-1 knapSack for ( int i = 0; i < n; i++) { if (wt[i] <= remaining_weight) { remaining_weight -= wt[i]; auto itr = umap.find(wt[i]); sum += (itr->second); possible.push_back({itr->first, itr->second }); } } sort(possible.begin(), possible.end()); if (sum > result) { result = sum; } if (set_sol.find(possible) == set_sol.end()){ for ( auto sol: possible){ cout << sol.first << ": " << sol.second << ", " ; } cout << endl; set_sol.insert(possible); } } while ( next_permutation(wt.begin(), wt.end())); return result; } // Driver Code int main() { vector< int > val{ 60, 100, 120 }; vector< int > wt{ 10, 20, 30 }; int W = 50; int n = val.size(); int maximum = knapSack(W, wt, val, n); cout << "Maximum Profit = " ; cout << maximum; return 0; } |
Python3
# Python3 implementation to print all # the possible solutions of the # 0/1 Knapsack problem INT_MIN = - 2147483648 def nextPermutation(nums: list ) - > None : """ Do not return anything, modify nums in-place instead. """ if sorted (nums,reverse = True ) = = nums: return None n = len (nums) brk_point = - 1 for pos in range (n - 1 , 0 , - 1 ): if nums[pos]>nums[pos - 1 ]: brk_point = pos break else : nums.sort() return replace_with = - 1 for j in range (brk_point,n): if nums[j]>nums[brk_point - 1 ]: replace_with = j else : break nums[replace_with],nums[brk_point - 1 ] = nums[brk_point - 1 ],nums[replace_with] nums[brk_point:] = sorted (nums[brk_point:]) return nums # Function to find the all the # possible solutions of the # 0/1 knapSack problem def knapSack(W, wt, val, n): # Mapping weights with Profits umap = dict () set_sol = set () # Making Pairs and inserting # o the map for i in range (n) : umap[wt[i]] = val[i] result = INT_MIN remaining_weight = 0 sum = 0 # Loop to iterate over all the # possible permutations of array while True : sum = 0 # Initially bag will be empty remaining_weight = W possible = [] # Loop to fill up the bag # until there is no weight # such which is less than # remaining weight of the # 0-1 knapSack for i in range (n) : if (wt[i] < = remaining_weight) : remaining_weight - = wt[i] sum + = (umap[wt[i]]) possible.append((wt[i], umap[wt[i]]) ) possible.sort() if ( sum > result) : result = sum if ( tuple (possible) not in set_sol): for sol in possible: print (sol[ 0 ], ": " , sol[ 1 ], ", " ,end = '') print () set_sol.add( tuple (possible)) if not nextPermutation(wt): break return result # Driver Code if __name__ = = '__main__' : val = [ 60 , 100 , 120 ] wt = [ 10 , 20 , 30 ] W = 50 n = len (val) maximum = knapSack(W, wt, val, n) print ( "Maximum Profit =" ,maximum) #This code was contributed by Amartya Ghosh |
10: 60, 20: 100, 10: 60, 30: 120, 20: 100, 30: 120, Maximum Profit = 220