{"id":26267,"date":"2017-10-26T20:45:17","date_gmt":"2017-10-26T15:15:17","guid":{"rendered":"https:\/\/www.wikitechy.com\/technology\/?p=26267"},"modified":"2017-10-26T20:45:17","modified_gmt":"2017-10-26T15:15:17","slug":"java-algorithm-check-whether-given-graph-bipartite-not","status":"publish","type":"post","link":"https:\/\/www.wikitechy.com\/technology\/java-algorithm-check-whether-given-graph-bipartite-not\/","title":{"rendered":"Java Algorithm – Check whether a given graph is Bipartite or not"},"content":{"rendered":"

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. <\/span>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.<\/p>\n

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.<\/p>\n

It is not possible to color a cycle graph with odd cycle using two colors.<\/p>\n

Algorithm to check if a graph is Bipartite:<\/em><\/strong>
\nOne approach is to check whether the graph is 2-colorable or not using backtracking algorithm m coloring problem.
\nFollowing is a simple algorithm to find out whether a given graph is Birpartite or not using Breadth First Search (BFS).
\n1. Assign RED color to the source vertex (putting into set U).
\n2. Color all the neighbors with BLUE color (putting into set V).
\n3. Color all neighbor\u2019s neighbor with RED color (putting into set U).
\n4. This way, assign color to all vertices such that it satisfies all the constraints of m way coloring problem where m = 2.
\n5. 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)<\/p>\n[ad type=”banner”]\n

Java Programming:<\/strong><\/p>\n

<\/div> 
\/\/ Java program to find out whether a given graph is Bipartite or notimport 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<\/code><\/pre> <\/div>\n
Output:<\/strong><\/p>\n
Yes<\/pre>\n
The above algorithm works only if the graph is strongly connected. In above code, we always start with source 0 and assume that vertices are visited from it. One important observation is a graph with no edges is also Bipiartite. Note that the Bipartite condition says all edges should be from one set to another.<\/p>\n
We can extend the above code to handle cases when a graph is not connected. The idea is repeatedly call above method for all not yet visited vertices.<\/p>\n[ad type=”banner”]\n
C++ Programming:<\/strong><\/p>\n
 
 <\/div> 
\/\/ C++ program to find out whether a given graph is Bipartite or not.\/\/ It works for disconnected graph also.#include <bits\/stdc++.h>using namespace std; const int V = 4; \/\/ This function returns true if graph G[V][V] is Bipartite,\/\/ else falsebool isBipartiteUtil(int G[][V], int src, int colorArr[]){    colorArr[src] = 1;     \/\/ Create a queue (FIFO) of vertex numbers and enqueue    \/\/ source vertex for BFS traversal    queue <int> q;    q.push(src);     \/\/ Run while there are vertices in queue (Similar to BFS)    while (!q.empty())    {        \/\/ Dequeue a vertex from queue ( Refer http:\/\/goo.gl\/35oz8 )        int u = q.front();        q.pop();          \/\/ 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] && colorArr[v] == -1)            {                \/\/ Assign alternate color to this                \/\/ adjacent v of u                colorArr[v] = 1 - colorArr[u];                q.push(v);            }             \/\/ An edge from u to v exists and destination            \/\/ v is colored with same color as u            else if (G[u][v] && colorArr[v] == colorArr[u])                return false;        }    }     \/\/ If we reach here, then all adjacent vertices can    \/\/ be colored with alternate color    return true;} \/\/ Returns true if G[][] is Bipartite, else falsebool isBipartite(int G[][V]){    \/\/ 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    \/\/ ndicate 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[V];    for (int i = 0; i < V; ++i)        colorArr[i] = -1;     \/\/ This code is to handle disconnected graoh    for (int i = 0; i < V; i++)      if (colorArr[i] == -1)        if (isBipartiteUtil(G, i, colorArr) == false)           return false;      return true;} \/\/ Driver program to test above functionint main(){    int G[][V] = {{0, 1, 0, 1},        {1, 0, 1, 0},        {0, 1, 0, 1},        {1, 0, 1, 0}    };     isBipartite(G) ? cout << "Yes" : cout << "No";    return 0;}<\/code><\/pre> <\/div>\n
Output:<\/strong><\/p>\n
Yes<\/pre>\n
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).<\/p>\n[ad type=”banner”]\n","protected":false},"excerpt":{"rendered":"
Java Algorithm – Check whether a given graph is Bipartite or not – Graph Algorithm – A Bipartite Graph is a graph whose vertices can be divided<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,74168,73906,2139],"tags":[78028,78022,78026,78029,78023,78021,78030,78065,78024,78031,78025,78027,78032],"class_list":["post-26267","post","type-post","status-publish","format-standard","hentry","category-coding","category-dfs-and-bfs","category-graph-algorithms","category-java","tag-bipartite-graph-algorithm","tag-bipartite-graph-algorithm-using-dfs","tag-bipartite-graph-applications","tag-bipartite-graph-c","tag-bipartite-graph-code-in-c","tag-bipartite-graph-example","tag-bipartite-graph-example-ppt","tag-bipartite-graph-geeksforgeeks","tag-bipartite-graph-matching","tag-bipartite-graph-pdf","tag-check-if-a-graph-is-bipartite-using-bfs","tag-complete-bipartite-graph","tag-tripartite-graph"],"_links":{"self":[{"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/posts\/26267","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/comments?post=26267"}],"version-history":[{"count":0,"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/posts\/26267\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/media?parent=26267"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/categories?post=26267"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/tags?post=26267"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}