Find if an undirected graph contains an independent set of a given size in C++

C++Server Side ProgrammingProgramming

Concept

With respect of a given undirected graph, verify if it contains an independent set of size l. If there exists an independent set of size l print ‘Yes’, else print ‘No’. It should be noted that an independent set in a graph is defined as a set of vertices which are not directly connected to each other.

Input

L = 4,
graph = [[1, 0, 1, 0, 0],
[0, 1, 1, 0, 0],[1, 1, 1, 1, 1],
[0, 0, 1, 1, 0],[0, 0, 1, 0, 1]];

Output

Yes

The above graph contains an independent set of size 4 (vertices 0, 1, 3, 4 are not directly connected to each other). Hence the output is ‘Yes’.

Input

L = 4,
graph =[[1, 1, 1, 0, 0],[1, 1, 1, 0, 0],[1, 1, 1, 1, 1],[0, 0, 1, 1, 0],[0, 0, 1, 0, 1]];

Output

No

In the diagram, the above graph doesn’t contain an independent set of size 4. Hence output is ‘No’.

Method

  • At first, initialize a variable sol with boolean False value.
  • Determine all the possible sets of vertices of size L from the given graph.
  • It has been seen that if an independent set of size l is found, change the value of sol to True and return.
  • Otherwise continue checking for other possible sets.
  • At last, if sol is True, print ‘Yes’ else print ‘No’.

Example

 Live Demo

// C++ code to check if a given graph
// contains an independent set of size k
#include <bits/stdc++.h>
using namespace std;
// Shows function prototype
bool check1(int[][5], vector<int>&, int);
// Shows function to construct a set of given size l
bool func(int graph1[][5], vector<int>&arr1,
int l, int index1, bool sol1[]){
   // Verify if the selected set is independent or not.
   // Used to change the value of sol to True and return
   // if it is independent
      if (l == 0){
         if (check1(graph1, arr1, arr1.size())){
            sol1[0] = true;
            return true;
         }
      }
      else{
         // Now set of size l can be formed even if we don't
         // include the vertex at current index.
         if (index1 >= l){
            vector<int> newvec(arr1.begin(), arr1.end());
            newvec.push_back(index1);
            return (func(graph1, newvec, l - 1,
            index1 - 1, sol1) or
            func(graph1, arr1, l, index1 - 1, sol1));
         }
            // Now set of size l cannot be formed if we don't
         // include the vertex at current index.
         else{
            arr1.push_back(index1);
            return func(graph1, arr1, l - 1,
            index1 - 1, sol1);
         }
      }
   }
   // Shows function to verify if the given set is
   // independent or not
   // arr --> set of size l (contains the
   // index of included vertex)
   bool check1(int graph1[][5], vector<int>&arr1, int n1){
      // Verify if each vertex is connected to any other
         // vertex in the set or not
      for (int i = 0; i < n1; i++)
         for (int j = i + 1; j < n1; j++)
            if (graph1[arr1[i]][arr1[j]] == 1)
            return false;
      return true;
}
// Driver Code
int main(){
   int graph1[][5] = {{1, 0, 1, 0, 0},{0, 1, 1, 0, 0},{1, 1, 1, 1, 1},{0, 0, 1, 1, 0},
      {0, 0, 1, 0, 1}};
      int l = 4;
      vector<int> arr1; // Empty set
      bool sol1[] = {false};
      int n1 = sizeof(graph1) /
      sizeof(graph1[0]);
      func(graph1, arr1, l, n1 - 1, sol1);
      if (sol1[0])
         cout << "Yes" << endl;
      else
         cout << "No" << endl;
      return 0;
}

Output

Yes
raja
Published on 24-Jul-2020 11:15:33
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