# Check whether a given graph is Bipartite or not

• Difficulty Level : Medium
• Last Updated : 13 Oct, 2022

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:
One approach is to check whether the graph is 2-colorable or not using backtracking algorithm m coloring problem
Following is a simple algorithm to find out whether a given graph is Bipartite 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 neighborâ€™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)

Recommended Practice

## C++

 // C++ program to find out whether a// given graph is Bipartite or not#include #include #define V 4 using namespace std; // This function returns true if graph// G[V][V] is Bipartite, else falsebool isBipartite(int G[][V], int src){    // Create a color array to store colors    // assigned to all vertices. 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[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    queue 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();         // Return false if there is a self-loop        if (G[u][u] == 1)        return false;         // 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;} // 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, 0) ? cout << "Yes" : cout << "No";    return 0;}

## Java

 // 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 vertices.        // 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; iq = new LinkedList();        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();             // Return false if there is a self-loop            if (G[u][u] == 1)                return false;             // Find all non-colored adjacent vertices            for (int v=0; v

## Python3

 # Python program to find out whether a# given graph is Bipartite or not class Graph():     def __init__(self, V):        self.V = V        self.graph = [[0 for column in range(V)] \                                for row in range(V)]     # This function returns true if graph G[V][V]    # is Bipartite, else false    def isBipartite(self, src):         # Create a color array to store colors        # assigned to all vertices. 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.        colorArr = [-1] * self.V         # Assign first color to source        colorArr[src] = 1         # Create a queue (FIFO) of vertex numbers and        # enqueue source vertex for BFS traversal        queue = []        queue.append(src)         # Run while there are vertices in queue        # (Similar to BFS)        while queue:             u = queue.pop()             # Return false if there is a self-loop            if self.graph[u][u] == 1:                return False;             for v in range(self.V):                 # An edge from u to v exists and destination                # v is not colored                if self.graph[u][v] == 1 and colorArr[v] == -1:                     # Assign alternate color to this                    # adjacent v of u                    colorArr[v] = 1 - colorArr[u]                    queue.append(v)                 # An edge from u to v exists and destination                # v is colored with same color as u                elif self.graph[u][v] == 1 and 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 functiong = Graph(4)g.graph = [[0, 1, 0, 1],            [1, 0, 1, 0],            [0, 1, 0, 1],            [1, 0, 1, 0]            ]             print ("Yes" if g.isBipartite(0) else "No") # This code is contributed by Divyanshu Mehta

## C#

 // C# program to find out whether// a given graph is Bipartite or notusing System;using System.Collections.Generic; class GFG{    readonly static int V = 4; // No. of Vertices     // This function returns true if    // graph G[V,V] is Bipartite, else false    bool isBipartite(int [,]G, int src)    {        // Create a color array to store        // colors assigned to all vertices.        // 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        Listq = new List();        q.Add(src);         // Run while there are vertices        // in queue (Similar to BFS)        while (q.Count != 0)        {            // Dequeue a vertex from queue            int u = q[0];            q.RemoveAt(0);             // Return false if there is a self-loop            if (G[u, u] == 1)                return false;             // 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 Code    public static void Main(String[] args)    {        int [,]G = {{0, 1, 0, 1},                    {1, 0, 1, 0},                    {0, 1, 0, 1},                    {1, 0, 1, 0}};        GFG b = new GFG();        if (b.isBipartite(G, 0))            Console.WriteLine("Yes");        else            Console.WriteLine("No");    }} // This code is contributed by Rajput-Ji

## Javascript



Output:

Yes

Time Complexity : O(V*V) as adjacency matrix is used for graph but can be made O(V+E) by using adjacency list

Space Complexity : O(V) due to queue and color vector.

The above algorithm works only if the graph is 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 Bipartite. Note that the Bipartite condition says all edges should be from one set to another.

We can extend the above code to handle cases when a graph is not connected. The idea is repeatedly called above method for all not yet visited vertices.

## C++

 // C++ program to find out whether// a given graph is Bipartite or not.// It works for disconnected graph also.#include  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 a    // nd enqueue source vertex for BFS traversal    queue 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();         // Return false if there is a self-loop        if (G[u][u] == 1)            return false;         // 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    // vertices. 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[V];    for (int i = 0; i < V; ++i)        colorArr[i] = -1;     // This code is to handle disconnected graph    for (int i = 0; i < V; i++)        if (colorArr[i] == -1)            if (isBipartiteUtil(G, i, colorArr) == false)                return false;     return true;} // Driver codeint 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;}

## Java

 // JAVA Code to check whether a given// graph is Bipartite or notimport java.util.*; class Bipartite {     public static int V = 4;     // This function returns true if graph    // G[V][V] is Bipartite, else false    public static boolean    isBipartiteUtil(int G[][], int src, int colorArr[])    {        colorArr[src] = 1;         // Create a queue (FIFO) of vertex numbers and        // enqueue source vertex for BFS traversal        LinkedList q = new LinkedList();        q.add(src);         // Run while there are vertices in queue        // (Similar to BFS)        while (!q.isEmpty()) {            // Dequeue a vertex from queue            // ( Refer http://goo.gl/35oz8 )            int u = q.getFirst();            q.pop();             // Return false if there is a self-loop            if (G[u][u] == 1)                return false;             // 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.push(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;    }     // Returns true if G[][] is Bipartite, else false    public static boolean isBipartite(int G[][])    {        // Create a color array to store colors assigned        // to all vertices. 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;         // This code is to handle disconnected graph        for (int i = 0; i < V; i++)            if (colorArr[i] == -1)                if (isBipartiteUtil(G, i, colorArr)                    == false)                    return false;         return true;    }     /* Driver code*/    public static void main(String[] args)    {        int G[][] = { { 0, 1, 0, 1 },                      { 1, 0, 1, 0 },                      { 0, 1, 0, 1 },                      { 1, 0, 1, 0 } };         if (isBipartite(G))            System.out.println("Yes");        else            System.out.println("No");    }} // This code is contributed by Arnav Kr. Mandal.

## Python3

 # Python3 program to find out whether a# given graph is Bipartite or not  class Graph():     def __init__(self, V):        self.V = V        self.graph = [[0 for column in range(V)]                      for row in range(V)]         self.colorArr = [-1 for i in range(self.V)]     # This function returns true if graph G[V][V]    # is Bipartite, else false    def isBipartiteUtil(self, src):         # Create a color array to store colors        # assigned to all vertices. Vertex        # number is used as index in this array.        # The value '-1' of self.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.         # Assign first color to source         # Create a queue (FIFO) of vertex numbers and        # enqueue source vertex for BFS traversal        queue = []        queue.append(src)         # Run while there are vertices in queue        # (Similar to BFS)        while queue:             u = queue.pop()             # Return false if there is a self-loop            if self.graph[u][u] == 1:                return False             for v in range(self.V):                 # An edge from u to v exists and                # destination v is not colored                if (self.graph[u][v] == 1 and                        self.colorArr[v] == -1):                     # Assign alternate color to                    # this adjacent v of u                    self.colorArr[v] = 1 - self.colorArr[u]                    queue.append(v)                 # An edge from u to v exists and destination                # v is colored with same color as u                elif (self.graph[u][v] == 1 and                      self.colorArr[v] == self.colorArr[u]):                    return False         # If we reach here, then all adjacent        # vertices can be colored with alternate        # color        return True     def isBipartite(self):        self.colorArr = [-1 for i in range(self.V)]        for i in range(self.V):            if self.colorArr[i] == -1:                if not self.isBipartiteUtil(i):                    return False        return True  # Driver Codeg = Graph(4)g.graph = [[0, 1, 0, 1],           [1, 0, 1, 0],           [0, 1, 0, 1],           [1, 0, 1, 0]] print ("Yes" if g.isBipartite() else "No") # This code is contributed by Anshuman Sharma

## C#

 // C# Code to check whether a given// graph is Bipartite or notusing System;using System.Collections.Generic; class GFG {    public static int V = 4;     // This function returns true if graph    // G[V,V] is Bipartite, else false    public static bool isBipartiteUtil(int[, ] G, int src,                                       int[] colorArr)    {        colorArr[src] = 1;         // Create a queue (FIFO) of vertex numbers and        // enqueue source vertex for BFS traversal        Queue q = new Queue();        q.Enqueue(src);         // Run while there are vertices in queue        // (Similar to BFS)        while (q.Count != 0) {            // Dequeue a vertex from queue            // ( Refer http://goo.gl/35oz8 )            int u = q.Peek();            q.Dequeue();             // Return false if there is a self-loop            if (G[u, u] == 1)                return false;             // 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.Enqueue(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;    }     // Returns true if G[,] is Bipartite,    // else false    public static bool isBipartite(int[, ] G)    {        // Create a color array to store        // colors assigned to all vertices.        // 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;         // This code is to handle disconnected graph        for (int i = 0; i < V; i++)            if (colorArr[i] == -1)                if (isBipartiteUtil(G, i, colorArr)                    == false)                    return false;         return true;    }     // Driver Code    public static void Main(String[] args)    {        int[, ] G = { { 0, 1, 0, 1 },                      { 1, 0, 1, 0 },                      { 0, 1, 0, 1 },                      { 1, 0, 1, 0 } };         if (isBipartite(G))            Console.WriteLine("Yes");        else            Console.WriteLine("No");    }} // This code is contributed by Rajput-Ji

## Javascript



Output:

Yes

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).

If Graph is represented using Adjacency List .Time Complexity will be O(V+E).

Works for connected as well as disconnected graph.

## C++

 #include using namespace std; bool isBipartite(int V, vector adj[]){    // vector to store colour of vertex    // assigning all to -1 i.e. uncoloured    // colours are either 0 or 1      // for understanding take 0 as red and 1 as blue    vector col(V, -1);     // queue for BFS storing {vertex , colour}    queue > q;         //loop incase graph is not connected    for (int i = 0; i < V; i++) {             //if not coloured        if (col[i] == -1) {                     //colouring with 0 i.e. red            q.push({ i, 0 });            col[i] = 0;                       while (!q.empty()) {                pair p = q.front();                q.pop();                                 //current vertex                int v = p.first;                  //colour of current vertex                int c = p.second;                                   //traversing vertexes connected to current vertex                for (int j : adj[v]) {                                         //if already coloured with parent vertex color                      //then bipartite graph is not possible                    if (col[j] == c)                        return 0;                                         //if uncoloured                    if (col[j] == -1) {                      //colouring with opposite color to that of parent                        col[j] = (c) ? 0 : 1;                        q.push({ j, col[j] });                    }                }            }        }    }    //if all vertexes are coloured such that      //no two connected vertex have same colours    return 1;}  // { Driver Code Starts.int main(){     int V, E;    V = 4 , E = 8;      //adjacency list for storing graph    vector adj[V];      adj[0] = {1,3};      adj[1] = {0,2};      adj[2] = {1,3};      adj[3] = {0,2};            bool ans = isBipartite(V, adj);    //returns 1 if bipartite graph is possible      if (ans)        cout << "Yes\n";    //returns 0 if bipartite graph is not possible      else        cout << "No\n";     return 0;} // code Contributed By Devendra Kolhe

## Python3

 def isBipartite(V, adj):    # vector to store colour of vertex    # assigning all to -1 i.e. uncoloured    # colours are either 0 or 1    # for understanding take 0 as red and 1 as blue    col = [-1]*(V)       # queue for BFS storing {vertex , colour}    q = []       #loop incase graph is not connected    for i in range(V):               # if not coloured        if (col[i] == -1):                       # colouring with 0 i.e. red            q.append([i, 0])            col[i] = 0                       while len(q) != 0:                p = q[0]                q.pop(0)                               # current vertex                v = p[0]                                 # colour of current vertex                c = p[1]                                   # traversing vertexes connected to current vertex                for j in adj[v]:                                       # if already coloured with parent vertex color                    # then bipartite graph is not possible                    if (col[j] == c):                        return False                                       # if uncoloured                    if (col[j] == -1):                                               # colouring with opposite color to that of parent                        if c == 1:                            col[j] = 0                        else:                            col[j] = 1                        q.append([j, col[j]])         # if all vertexes are coloured such that    # no two connected vertex have same colours    return True V, E = 4, 8 # adjacency list for storing graphadj = []adj.append([1,3])adj.append([0,2])adj.append([1,3])adj.append([0,2])  ans = isBipartite(V, adj) # returns 1 if bipartite graph is possibleif (ans):    print("Yes")     # returns 0 if bipartite graph is not possibleelse:    print("No")         # This code is contributed by divyesh072019.

## Javascript



Output

Yes

Time Complexity: O(V+E)

Auxiliary Space: O(V)

Exercise:
1. Can DFS algorithm be used to check the bipartite-ness of a graph? If yes, how?
Solution :

## C++

 // C++ program to find out whether a given graph is Bipartite or not.// Using recursion.#include  using namespace std;#define V 4  bool colorGraph(int G[][V],int color[],int pos, int c){         if(color[pos] != -1 && color[pos] !=c)        return false;             // color this pos as c and all its neighbours and 1-c    color[pos] = c;    bool ans = true;    for(int i=0;i

## Java

 // Java program to find out whether// a given graph is Bipartite or not.// Using recursion.class GFG{    static final int V = 4;     static boolean colorGraph(int G[][],                              int color[],                              int pos, int c)    {        if (color[pos] != -1 &&            color[pos] != c)            return false;         // color this pos as c and        // all its neighbours as 1-c        color[pos] = c;        boolean ans = true;        for (int i = 0; i < V; i++)        {            if (G[pos][i] == 1)            {                if (color[i] == -1)                    ans &= colorGraph(G, color, i, 1 - c);                 if (color[i] != -1 && color[i] != 1 - c)                    return false;            }            if (!ans)                return false;        }        return true;    }     static boolean isBipartite(int G[][])    {        int[] color = new int[V];        for (int i = 0; i < V; i++)            color[i] = -1;         // start is vertex 0;        int pos = 0;             // two colors 1 and 0        return colorGraph(G, color, pos, 1);    }     // Driver Code    public static void main(String[] args)    {        int G[][] = { { 0, 1, 0, 1 },                      { 1, 0, 1, 0 },                      { 0, 1, 0, 1 },                      { 1, 0, 1, 0 } };         if (isBipartite(G))            System.out.print("Yes");        else            System.out.print("No");    }} // This code is contributed by Rajput-Ji

## Python3

 # Python3 program to find out whether a given# graph is Bipartite or not using recursion.V = 4 def colorGraph(G, color, pos, c):         if color[pos] != -1 and color[pos] != c:        return False             # color this pos as c and all its neighbours and 1-c    color[pos] = c    ans = True    for i in range(0, V):        if G[pos][i]:            if color[i] == -1:                ans &= colorGraph(G, color, i, 1-c)                             if color[i] !=-1 and color[i] != 1-c:                return False                  if not ans:            return False          return True  def isBipartite(G):         color = [-1] * V             #start is vertex 0    pos = 0    # two colors 1 and 0    return colorGraph(G, color, pos, 1) if __name__ == "__main__":      G = [[0, 1, 0, 1],         [1, 0, 1, 0],         [0, 1, 0, 1],         [1, 0, 1, 0]]          if isBipartite(G): print("Yes")    else: print("No") # This code is contributed by Rituraj Jain

## C#

 // C# program to find out whether// a given graph is Bipartite or not.// Using recursion.using System; class GFG{    static readonly int V = 4;     static bool colorGraph(int [,]G,                           int []color,                           int pos, int c)    {        if (color[pos] != -1 &&            color[pos] != c)            return false;         // color this pos as c and        // all its neighbours as 1-c        color[pos] = c;        bool ans = true;        for (int i = 0; i < V; i++)        {            if (G[pos, i] == 1)            {                if (color[i] == -1)                    ans &= colorGraph(G, color, i, 1 - c);                 if (color[i] != -1 && color[i] != 1 - c)                    return false;            }            if (!ans)                return false;        }        return true;    }     static bool isBipartite(int [,]G)    {        int[] color = new int[V];        for (int i = 0; i < V; i++)            color[i] = -1;         // start is vertex 0;        int pos = 0;             // two colors 1 and 0        return colorGraph(G, color, pos, 1);    }     // Driver Code    public static void Main(String[] args)    {        int [,]G = {{ 0, 1, 0, 1 },                    { 1, 0, 1, 0 },                    { 0, 1, 0, 1 },                    { 1, 0, 1, 0 }};         if (isBipartite(G))            Console.Write("Yes");        else            Console.Write("No");    }} // This code is contributed by 29AjayKumar

## Javascript



Output

Yes

Time Complexity: O(V+E)

Auxiliary Space: O(V)

References:
http://en.wikipedia.org/wiki/Graph_coloring
http://en.wikipedia.org/wiki/Bipartite_graph