यह प्रत्येक उपसमुच्चय में k तत्वों के साथ सभी संभावित उपसमुच्चय उत्पन्न करने के लिए एक C++ कार्यक्रम है।
एल्गोरिदम
Begin function PossibleSubSet(char a[], int reqLen, int s, int currLen, bool check[], int l): If currLen > reqLen Return Else if currLen = reqLen Then print the new generated sequence. If s = l Then return no further element is left. For every index there are two options: either proceed with a start as ‘true’ and recursively call PossibleSubSet() with incremented value of ‘currLen’ and ‘s’. Or proceed with a start as ‘false’ and recursively call PossibleSubSet() with only incremented value of ‘s’. End
उदाहरण
#include<iostream> using namespace std; void PossibleSubSet(char a[], int reqLen, int s, int currLen, bool check[], int l) //print the all possible combination of given array set { if(currLen > reqLen) return; else if (currLen == reqLen) { cout<<"\t"; for (int i = 0; i < l; i++) { if (check[i] == true) { cout<<a[i]<<" "; } } cout<<"\n"; return; } if (s == l) { return; } check[s] = true; PossibleSubSet(a, reqLen, s + 1, currLen + 1, check, l); //recursively call PossibleSubSet() with incremented value of ‘currLen’ and ‘s’. check[s] = false; PossibleSubSet(a, reqLen, s + 1, currLen, check, l); //recursively call PossibleSubSet() with only incremented value of ‘s’. } int main() { int i,n,m; bool check[n]; cout<<"Enter the number of elements: "; cin>>n; char a[n]; cout<<"\n"; for(i = 0; i < n; i++) { cout<<"Enter "<<i+1<<" element: "; cin>>a[i]; check[i] = false; } cout<<"\nEnter the length of the subsets required: "; cin>>m; cout<<"\nThe possible combination of length "<<m<<" for the given array set:\n"; PossibleSubSet(a, m, 0, 0, check, n); return 0; }
आउटपुट
Enter the number of elements: 7 Enter 1 element: 7 Enter 2 element: 6 Enter 3 element: 5 Enter 4 element: 4 Enter 5 element: 3 Enter 6 element: 2 Enter 7 element: 1 Enter the length of the subsets required: 6 The possible combination of length 6 for the given array set: 7 6 5 4 3 2 7 6 5 4 3 1 7 6 5 4 2 1 7 6 5 3 2 1 7 6 4 3 2 1 7 5 4 3 2 1 6 5 4 3 2 1