Finding minimum vertex cover size of a graph using binary search

A vertex cover of an undirected graph is a subset of its vertices such that for every edge (u, v) of the graph, either ‘u’ or ‘v’ is in vertex cover. There may be a lot of vertex covers possible for a graph.

Problem Find the size of the minimum size vertex cover, that is, cardinality of a vertex cover with minimum cardinality, for an undirected connected graph with V vertices and m edges.

Examples:

Input: V = 6, E = 6
		     6
	            /
		   /
		  1 -----5
		 /|\
		3 | \
		\ |  \
		  2   4
Output: Minimum vertex cover size = 2
Consider subset of vertices {1, 2}, every edge 
in above graph is either incident on vertex 1 
or 2. Hence the minimum vertex cover = {1, 2}, 
the size of which is 2.

Input: V = 6, E = 7
		   2 ---- 4 ---- 6
		  /|      |
		1  |      |
		  \|      |
		   3 ---- 5
Output: Minimum vertex cover size = 3
Consider subset of vertices {2, 3, 4}, every
edge in above graph is either incident on 
vertex 2, 3 or 4. Hence the minimum size of
a vertex cover can be 3.

Method 1 (Naive)

We can check in O(E + V) time if a given subset of vertices is a vertex cover or not, using the following algorithm.

Generate all 2V subsets of vertices in graph and
do following for every subset.
  1. edges_covered = 0
  2. for each vertex in current subset
  3.   for all edges emerging out of current vertex
  4.	  if the edge is not already marked visited
  5.          mark the edge visited
  6.          edges_covered++
  7. if edges_covered is equal to total number edges
  8.   return size of current subset

An upper bound on time complexity of this solution is O((E + V) * 2V)



 

Method 2 (Binary Search)

If we generate 2V subsets first by generating VCV subsets, then VC(V-1) subsets, and so on upto VC0 subsets(2V = VCV + VC(V-1) + … + VC1 + VC0). Our objective is now to find the minimum k such that at least one subset of size ‘k’ amongst VCk subsets is a vertex cover [ We know that if minimum size vertex cover is of size k, then there will exist a vertex cover of all sizes more than k. That is, there will be a vertex cover of size k + 1, k + 2, k + 3, …, n. ]

Now let’s imagine a boolean array of size n and call it isCover[]. So if the answer of the question; “Does a vertex cover of size x exist?” is yes, we put a ‘1’ at xth position, otherwise ‘0’.

The array isCover[] will look like:

1 2 3 . . . k . . . n
0 0 0 . . . 1 . . . 1

The array is sorted and hence binary searchable, as no index before k will have a ‘1’, and every index after k(inclusive) will have a ‘1’, so k is the answer.

So we can apply Binary Search to find the minimum size vertex set that covers all edges (this problem is equivalent to finding last 1 in isCover[]). Now the problem is how to generate all subsets of a given size. The idea is to use Gosper’s hack.

What is Gosper’s Hack?
Gosper’s hack is a technique to get the next number with same number of bits set. So we set the first x bits from right and generate next number with x bits set until the number is less than 2V. In this way, we can generate all VCx numbers with x bits set.

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int set = (1 << k) - 1;
int limit = (1 << V);
while (set < limit) 
{
    // Do your stuff with current set
    doStuff(set);
  
    // Gosper's hack:
    int c = set & -set;
    int r = set + c;
    set = (((r^set) >>> 2) / c) | r;
}

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Source : StackExchange

We use gosper’s hack to generate all subsets of size x(0 < x <= V), that is, to check whether we have a '1' or '0' at any index x in isCover[] array.

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// A C++ program to find size of minimum vertex
// cover using Binary Search
#include<bits/stdc++.h>
#define maxn 25
  
using namespace std;
  
// Global array to store the graph
// Note: since the array is global, all the
// elements are 0 initially
bool gr[maxn][maxn];
  
// Returns true if there is a possible subset
// of size 'k' that can be a vertex cover
bool isCover(int V, int k, int E)
{
    // Set has first 'k' bits high initially
    int set = (1 << k) - 1;
  
    int limit = (1 << V);
  
    // to mark the edges covered in each subset
    // of size 'k'
    bool vis[maxn][maxn];
  
    while (set < limit)
    {
        // Reset visited array for every subset
        // of vertices
        memset(vis, 0, sizeof vis);
  
        // set counter for number of edges covered
        // from this subset of vertices to zero
        int cnt = 0;
  
        // selected vertex cover is the the indices
        // where 'set' has its bit high
        for (int j = 1, v = 1 ; j < limit ; j = j << 1, v++)
        {
            if (set & j)
            {
                // Mark all edges emerging out of this
                // vertex visited
                for (int k = 1 ; k <= V ; k++)
                {
                    if (gr[v][k] && !vis[v][k])
                    {
                        vis[v][k] = 1;
                        vis[k][v] = 1;
                        cnt++;
                    }
                }
            }
        }
  
        // If the current subset covers all the edges
        if (cnt == E)
            return true;
  
        // Generate previous combination with k bits high
        // set & -set = (1 << last bit high in set)
        int c = set & -set;
        int r = set + c;
        set = (((r^set) >> 2) / c) | r;
    }
    return false;
}
  
// Returns answer to graph stored in gr[][]
int findMinCover(int n, int m)
{
    // Binary search the answer
    int left = 1, right = n;
    while (right > left)
    {
        int mid = (left + right) >> 1;
        if (isCover(n, mid, m) == false)
            left = mid + 1;
        else
            right = mid;
    }
  
    // at the end of while loop both left and
    // right will be equal,/ as when they are
    // not, the while loop won't exit the minimum
    // size vertex cover = left = right
    return left;
}
  
// Inserts an edge in the graph
void insertEdge(int u, int v)
{
    gr[u][v] = 1;
    gr[v][u] = 1;  // Undirected graph
}
  
// Driver code
int main()
{
    /*
            6
           /
          1 ----- 5   vertex cover = {1, 2}
         /|\
        3 | \
        \ |  \
          2   4   */
    int V = 6, E = 6;
    insertEdge(1, 2);
    insertEdge(2, 3);
    insertEdge(1, 3);
    insertEdge(1, 4);
    insertEdge(1, 5);
    insertEdge(1, 6);
    cout << "Minimum size of a vertex cover = "
         << findMinCover(V, E) << endl;
  
  
    // Let us create another graph
    memset(gr, 0, sizeof gr);
    /*
           2 ---- 4 ---- 6
          /|      |
        1  |      |   vertex cover = {2, 3, 4}
         \ |      |
           3 ---- 5    */
  
    V = 6, E = 7;
    insertEdge(1, 2);
    insertEdge(1, 3);
    insertEdge(2, 3);
    insertEdge(2, 4);
    insertEdge(3, 5);
    insertEdge(4, 5);
    insertEdge(4, 6);
    cout << "Minimum size of a vertex cover = "
         << findMinCover(V, E) << endl;
  
    return 0;
}

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Output:



Minimum size of a vertex cover = 2
Minimum size of a vertex cover = 3

Conclusion:
Time Complexity : O (E * ( VCV/2 + VCV/4 + VCV/8 +…upto VCk ) )

These terms are not more than log(V) in worst case.

Note: Gosper’s hack works for upto V = 31 only, if we take ‘long long int’ instead of ‘int’ it can work upto V = 63.

This article is contributed by Saumye Malhotra .If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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