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 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)
C++
// C++ program to find out whether a // given graph is Bipartite or not #include <iostream> #include <queue> #define V 4 using namespace std; // This function returns true if graph // G[V][V] is Bipartite, else false bool 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 <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(); // 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 function int 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 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 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 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(); // 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 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 |
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 function g = 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 not using 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 List<int>q = new List<int>(); 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
<script> // Javascript program to find out whether // a given graph is Bipartite or not let V = 4; // No. of Vertices // This function returns true if // graph G[V][V] is Bipartite, else false function isBipartite(G,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. let colorArr = new Array(V); for (let 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 let q = []; q.push(src); // Run while there are vertices in queue (Similar to BFS) while (q.length != 0) { // Dequeue a vertex from queue let u = q.shift(); // Return false if there is a self-loop if (G[u][u] == 1) return false; // Find all non-colored adjacent vertices for (let 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; } // Driver program to test above function let G=[[0, 1, 0, 1], [1, 0, 1, 0], [0, 1, 0, 1], [1, 0, 1, 0]]; if (isBipartite(G, 0)) document.write("Yes"); else document.write("No"); // This code is contributed by avanitrachhadiya2155 </script> |
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
Auxiliary Space: 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 <bits/stdc++.h> using namespace std; const int V = 4; // This function returns true if // graph G[V][V] is Bipartite, else false bool 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<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(); // 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 false bool 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 code int 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 not import 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<Integer> q = new LinkedList<Integer>(); 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 Code g = 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 not using 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<int> q = new Queue<int>(); 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
<script> // Javascript Code to check whether a given // graph is Bipartite or not var V = 4; // This function returns true if graph // G[V,V] is Bipartite, else false function isBipartiteUtil(G, src, colorArr) { colorArr[src] = 1; // Create a queue (FIFO) of vertex numbers and // enqueue source vertex for BFS traversal var q = []; q.push(src); // Run while there are vertices in queue // (Similar to BFS) while (q.length != 0) { // Dequeue a vertex from queue // ( Refer http://goo.gl/35oz8 ) var u = q[0]; q.shift(); // Return false if there is a self-loop if (G[u, u] == 1) return false; // Find all non-colored adjacent vertices for (var 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 function isBipartite(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. var colorArr = Array(V); for (var i = 0; i < V; ++i) colorArr[i] = -1; // This code is to handle disconnected graph for (var i = 0; i < V; i++) if (colorArr[i] == -1) if (isBipartiteUtil(G, i, colorArr) == false) return false; return true; } // Driver Code var G = [ [ 0, 1, 0, 1 ], [ 1, 0, 1, 0 ], [ 0, 1, 0, 1 ], [ 1, 0, 1, 0 ] ]; if (isBipartite(G)) document.write("Yes"); else document.write("No"); // This code is contributed by rrrtnx. </script> |
Output
Yes
Time complexity: O(V+E).
Auxiliary Space: O(V), because we have a V-size array.
If Graph is represented using Adjacency List .Time Complexity will be O(V+E).
Works for connected as well as disconnected graph.
C++
#include <bits/stdc++.h> using namespace std; bool isBipartite(int V, vector<int> 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<int> col(V, -1); // queue for BFS storing {vertex , colour} queue<pair<int, int> > 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<int, int> 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<int> 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 |
Java
import java.util.*; public class GFG{ static class Pair{ int first, second; Pair(int f, int s){ first = f; second = s; } } static boolean isBipartite(int V, ArrayList<ArrayList<Integer>> 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 int col[] = new int[V]; Arrays.fill(col, -1); // queue for BFS storing {vertex , colour} Queue<Pair> q = new LinkedList<Pair>(); //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.add(new Pair(i, 0)); col[i] = 0; while (!q.isEmpty()) { Pair p = q.peek(); q.poll(); //current vertex int v = p.first; // colour of current vertex int c = p.second; // traversing vertexes connected to current vertex for (int j : adj.get(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 col[j] = (c==1) ? 0 : 1; q.add(new Pair(j, col[j])); } } } } } // if all vertexes are coloured such that // no two connected vertex have same colours return true; } // Driver Code Starts. public static void main(String args[]) { int V, E; V = 4 ; E = 8; // adjacency list for storing graph ArrayList<ArrayList<Integer>> adj = new ArrayList<ArrayList<Integer>>(); for(int i = 0; i < V; i++){ adj.add(new ArrayList<Integer>()); } adj.get(0).add(1); adj.get(0).add(3); adj.get(1).add(0); adj.get(1).add(2); adj.get(2).add(1); adj.get(2).add(3); adj.get(3).add(0); adj.get(3).add(2); boolean ans = isBipartite(V, adj); // returns 1 if bipartite graph is possible if (ans) System.out.println("Yes"); // returns 0 if bipartite graph is not possible else System.out.println("No"); } } // This code is contributed by adityapande88. |
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 graph adj = [] 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 possible if (ans): print("Yes") # returns 0 if bipartite graph is not possible else: print("No") # This code is contributed by divyesh072019. |
C#
using System; using System.Collections.Generic; class GFG { static bool isBipartite(int V, List<List<int>> 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 int[] col = new int[V]; Array.Fill(col, -1); // queue for BFS storing {vertex , colour} List<Tuple<int,int>> q = new List<Tuple<int,int>>(); //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.Add(new Tuple<int,int>(i, 0)); col[i] = 0; while (q.Count > 0) { Tuple<int,int> p = q[0]; q.RemoveAt(0); //current vertex int v = p.Item1; // colour of current vertex int c = p.Item2; // traversing vertexes connected to current vertex foreach(int 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 col[j] = (c==1) ? 0 : 1; q.Add(new Tuple<int,int>(j, col[j])); } } } } } // if all vertexes are coloured such that // no two connected vertex have same colours return true; } static void Main() { int V; V = 4 ; // adjacency list for storing graph List<List<int>> adj = new List<List<int>>(); for(int i = 0; i < V; i++){ adj.Add(new List<int>()); } adj[0].Add(1); adj[0].Add(3); adj[1].Add(0); adj[1].Add(2); adj[2].Add(1); adj[2].Add(3); adj[3].Add(0); adj[3].Add(2); bool ans = isBipartite(V, adj); // returns 1 if bipartite graph is possible if (ans) Console.WriteLine("Yes"); // returns 0 if bipartite graph is not possible else Console.WriteLine("No"); } } // This code is contributed by decode2207. |
Javascript
<script> class Pair { constructor(f,s) { this.first = f; this.second = s; } } function 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 let col = new Array(V); for(let i = 0; i < V; i++) col[i] = -1; // queue for BFS storing {vertex , colour} let q = []; //loop incase graph is not connected for (let i = 0; i < V; i++) { // if not coloured if (col[i] == -1) { // colouring with 0 i.e. red q.push(new Pair(i, 0)); col[i] = 0; while (q.length!=0) { let p = q[0]; q.shift(); //current vertex let v = p.first; // colour of current vertex let c = p.second; // traversing vertexes connected to current vertex for (let j of 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 col[j] = (c==1) ? 0 : 1; q.push(new Pair(j, col[j])); } } } } } // if all vertexes are coloured such that // no two connected vertex have same colours return true; } // Driver Code Starts. let V, E; V = 4 ; E = 8; // adjacency list for storing graph let adj = []; for(let i = 0; i < V; i++){ adj.push([]); } adj[0].push(1); adj[0].push(3); adj[1].push(0); adj[1].push(2); adj[2].push(1); adj[2].push(3); adj[3].push(0); adj[3].push(2); let ans = isBipartite(V, adj); // returns 1 if bipartite graph is possible if (ans) document.write("Yes"); // returns 0 if bipartite graph is not possible else document.write("No"); // This code is contributed by patel2127 </script> |
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?
Explanation:
The algorithm can be described as follows:
Step 1: To perform a DFS traversal, we require a starting node and an array to track visited nodes.
However, instead of using a visited array, we will utilize a color array where each node is initially
assigned the value -1 to indicate that it has not been colored yet.
Step 2: In the DFS function call, it is important to pass the assigned color value and store it in the color array.
We will attempt to color the nodes using 0 and 1, although alternative colors can also be chosen.
Starting with color 0 is suggested, but starting with color 1 is also acceptable; the key is to ensure that adjacent nodes
have the opposite color to their current node.
Step 3: During the DFS traversal, we explore uncolored neighbors by utilizing the adjacency list.
For each uncolored node encountered, we assign it the opposite color of its current node.
Step 4: If, at any point, we encounter an adjacent node in the adjacency list that has already been colored and shares
the same color as the current node, it implies that coloring is not possible.
Consequently, we return false, indicating that the given graph is not bipartite. Otherwise, we continue returning true.
Solution :
C++
// C++ program to find out whether a given graph is Bipartite or not. // Using recursion. #include <iostream> 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<V;i++){ if(G[pos][i]){ 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; } bool isBipartite(int G[][V]){ int color[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); } int 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; } // This code is contributed By Mudit Verma |
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
<script> // JavaScript program to find out whether // a given graph is Bipartite or not. // Using recursion. var V = 4; function colorGraph(G, color, pos, 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; var ans = true; for (var 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; } function isBipartite(G) { var color = new Array(V).fill(0); for (var i = 0; i < V; i++) color[i] = -1; // start is vertex 0; var pos = 0; // two colors 1 and 0 return colorGraph(G, color, pos, 1); } // Driver Code var G = [ [0, 1, 0, 1], [1, 0, 1, 0], [0, 1, 0, 1], [1, 0, 1, 0], ]; if (isBipartite(G)) document.write("Yes"); else document.write("No"); // This code is contributed by rdtank. </script> |
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
This article is compiled by Aashish Barnwal. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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