*Check whether a given graph is Bipartite or not

A Bipartite Graph is a graph whose vertices can be divided into two independent sets, U and V such that every edge (u, v) either connects a vertex from U to V or a vertex from V to U. In other words, for every edge (u, v), either u belongs to U and v to V, or u belongs to V and v to U. We can also say that there is no edge that connects vertices of same set.

A bipartite graph is possible if the graph coloring is possible using two colors such that vertices in a set are colored with the same color. Note that it is possible to color a cycle graph with even cycle using two colors. For example, see the following graph.

It is not possible to color a cycle graph with odd cycle using two colors.

Algorithm to check if a graph is Bipartite:

// Java program to find out whether a given graph is Bipartite or not
import java.util.*;
import java.lang.*;
import java.io.*;
 
class Bipartite
{
    final static int V = 4; // No. of Vertices
 
    // This function returns true if graph G[V][V] is Bipartite, else false
    boolean isBipartite(int G[][],int src)
    {
        // Create a color array to store colors assigned to all veritces.
        // Vertex number is used as index in this array. The value '-1'
        // of  colorArr[i] is used to indicate that no color is assigned
        // to vertex 'i'.  The value 1 is used to indicate first color
        // is assigned and value 0 indicates second color is assigned.
        int colorArr[] = new int[V];
        for (int i=0; i<V; ++i)
            colorArr[i] = -1;
 
        // Assign first color to source
        colorArr[src] = 1;
 
        // Create a queue (FIFO) of vertex numbers and enqueue
        // source vertex for BFS traversal
        LinkedList<Integer>q = new LinkedList<Integer>();
        q.add(src);
 
        // Run while there are vertices in queue (Similar to BFS)
        while (q.size() != 0)
        {
 
            // Dequeue a vertex from queue
            int u = q.poll();
 
            // Find all non-colored adjacent vertices
            for (int v=0; v<V; ++v)
            {
                // An edge from u to v exists and destination v is
                // not colored
                if (G[u][v]==1 && colorArr[v]==-1)
                {
                    // Assign alternate color to this adjacent v of u
                    colorArr[v] = 1-colorArr[u];
                    q.add(v);
                }
 
                // An edge from u to v exists and destination v is
                // colored with same color as u
                else if (G[u][v]==1 && colorArr[v]==colorArr[u])
                    return false;
            }
        }
        // If we reach here, then all adjacent vertices can
        //  be colored with alternate color
        return true;
    }
 
    // Driver program to test above function
    public static void main (String[] args)
    {
        int G[][] = {{0, 1, 0, 1},
            {1, 0, 1, 0},
            {0, 1, 0, 1},
            {1, 0, 1, 0}
        };
        Bipartite b = new Bipartite();
        if (b.isBipartite(G, 0))
           System.out.println("Yes");
        else
           System.out.println("No");
    }
}
 
// Contributed by Aakash Hasija

Output:

Yes

Refer this for C implementation of the same.

Time Complexity of the above approach is same as that Breadth First Search. In above implementation is O(V^2) where V is number of vertices. If graph is represented using adjacency list, then the complexity becomes O(V+E).