Check whether a given graph is Bipartite or not - GeeksforGeeks

Check whether a given graph is Bipartite or not - GeeksforGeeks
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.

Graph BFS:

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).

1. Assign RED color to the source vertex (putting into set U).
2. Color all the neighbors with BLUE color (putting into set V).
3. Color all neighbor’s neighbor with RED color (putting into set U).
4. This way, assign color to all vertices such that it satisfies all the constraints of m way coloring problem where m = 2.
5. While assigning colors, if we find a neighbor which is colored with same color as current vertex, then the graph cannot be colored with 2 vertices (or graph is not Bipartite)
http://likemyblogger.blogspot.com/2015/08/mj-23-bipartite-graph.html

    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;

    }

BFS Java Code: http://www.sanfoundry.com/java-program-check-whether-graph-bipartite-using-bfs/
http://karmaandcoding.blogspot.com/2012/03/bipartite-graph-check.html
http://www.fgdsb.com/2015/01/03/check-whether-a-graph-is-bipartite-or-not/
DFS:
http://algs4.cs.princeton.edu/41undirected/Bipartite.java.html
http://www.sanfoundry.com/cpp-program-graph-bipartite-dfs/

    private boolean isBipartite;   // is the graph bipartite?
    private boolean[] color;       // color[v] gives vertices on one side of bipartition
    private boolean[] marked;      // marked[v] = true if v has been visited in DFS
    private int[] edgeTo;          // edgeTo[v] = last edge on path to v
    private Stack<Integer> cycle;  // odd-length cycle
    public Bipartite(Graph G) {
        isBipartite = true;
        color  = new boolean[G.V()];
        marked = new boolean[G.V()];
        edgeTo = new int[G.V()];

        for (int v = 0; v < G.V(); v++) {
            if (!marked[v]) {
                dfs(G, v);
            }
        }
        assert check(G);
    }

    private void dfs(Graph G, int v) { 
        marked[v] = true;
        for (int w : G.adj(v)) {

            // short circuit if odd-length cycle found
            if (cycle != null) return;

            // found uncolored vertex, so recur
            if (!marked[w]) {
                edgeTo[w] = v;
                color[w] = !color[v];
                dfs(G, w);
            } 

            // if v-w create an odd-length cycle, find it
            else if (color[w] == color[v]) {
                isBipartite = false;
                cycle = new Stack<Integer>();
                cycle.push(w);  // don't need this unless you want to include start vertex twice
                for (int x = v; x != w; x = edgeTo[x]) {
                    cycle.push(x);
                }
                cycle.push(w);
            }
        }
    }

http://www.keithschwarz.com/interesting/code/bipartite-verify/BipartiteVerify.java.html
    public static <T> boolean isBipartite(UndirectedGraph<T> g) {
        Map<T, Boolean> parityTable = new HashMap<T, Boolean>();
        for (T node: g)
            if (!parityTable.containsKey(node) && 
                !dfsExplore(node, g, parityTable, true))
                return false;
        return true;
    }
    private static <T> boolean dfsExplore(T node, UndirectedGraph<T> g,
                                          Map<T, Boolean> parityTable,
                                          boolean parity) {
        if (parityTable.containsKey(node))
            return parityTable.get(node).equals(parity);
        parityTable.put(node, parity);
        for (T endpoint: g.edgesFrom(node))
            if (!dfsExplore(endpoint, g, parityTable, !parity))
                return false;
        return true;
    }
http://www.sanfoundry.com/java-program-check-whether-graph-bipartite-using-2-color-algorithm/
http://www.cnblogs.com/EdwardLiu/p/6552027.html

input friends relations{{1,2}, {2,3}, {3,4}}
把人分成两拨，每拨人互相不认识，
所以应该是group1{1,3}, group2{2,4}

这道题应该是how to bipartite a graph

Taken from GeeksforGeeks

Following is a simple algorithm to find out whether a given graph is Birpartite or not using Breadth First Search (BFS) :-

1. Assign RED color to the source vertex (putting into set U).
2. Color all the neighbors with BLUE color (putting into set V).
3. Color all neighbor’s neighbor with RED color (putting into set U).
4. This way, assign color to all vertices such that it satisfies all the constraints of m way coloring problem where m = 2.
5. While assigning colors, if we find a neighbor which is colored with same color as current vertex, then the graph cannot be colored with 2 vertices (or graph is not Bipartite).

Also, NOTE :-

-> It is possible to color a cycle graph with even cycle using two colors.

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

EDIT :-

If a graph is not connected, it may have more than one bipartition. You need to check all those components separately with the algorithm as mentioned above.

So, for various disconnected sub-graph of the same graph, you need to perform this bipartition check on all of them separately using the same algorithm discussed above. All of those various disconnected sub-graph of the same graph will account for its own set of bipartition.

And, the graph will be termed bipartite, IF AND ONLY IF, each of its connected components are proved to be bipartite .

 6     HashSet<Integer> list1 = new HashSet<Integer>();
 7     HashSet<Integer> list2 = new HashSet<Integer>();
 8     
 9     public void bfs(int[][] relations) {
10         HashMap<Integer, HashSet<Integer>> graph = new HashMap<Integer, HashSet<Integer>>();
11         for (int[] each : relations) {
12             if (!graph.containsKey(each[0]))
13                 graph.put(each[0], new HashSet<Integer>());
14             if (!graph.containsKey(each[1]))
15                 graph.put(each[1], new HashSet<Integer>());
16             graph.get(each[0]).add(each[1]);
17             graph.get(each[1]).add(each[0]);
18         }
19         
20         
21         Queue<Integer> queue = new LinkedList<Integer>();
22         queue.offer(relations[0][0]);
23         list1.add(relations[0][0]);
24         HashSet<Integer> visited = new HashSet<Integer>();
25         visited.add(relations[0][0]);
26         int count = 1;
27         while (!queue.isEmpty()) {
28             int size = queue.size();
29             for (int i=0; i<size; i++) {
30                 int person = queue.poll();
31                 HashSet<Integer> friends = graph.get(person);
32                 for (int each : friends) {
33                     if (list1.contains(each)&&list1.contains(person) || list2.contains(each)&&list2.contains(person)) {
34                         list1.clear();
35                         list2.clear();
36                         return;
37                     }
38                         
39                     if (!visited.contains(each)) {
40                         if (count%2 == 1) list2.add(each);
41                         else list1.add(each);
42                         queue.offer(each);
43                         visited.add(each);
44                     }
45                 }
46             }
47             count++;
48         }
49     }

Graph Coloring | Set 1 (Introduction and Applications)
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