Given an undirected graph and a number m, the task is to color the given graph with at most m colors such that no two adjacent vertices of the graph are colored with the same color
Note: Here coloring of a graph means the assignment of colors to all vertices
Below is an example of a graph that can be colored with 3 different colors:
Examples:
Input: graph = {0, 1, 1, 1},
{1, 0, 1, 0},
{1, 1, 0, 1},
{1, 0, 1, 0}
Output: Solution Exists: Following are the assigned colors: 1 2 3 2
Explanation: By coloring the vertices with following colors,
adjacent vertices does not have same colorsInput: graph = {1, 1, 1, 1},
{1, 1, 1, 1},
{1, 1, 1, 1},
{1, 1, 1, 1}Output: Solution does not exist
Explanation: No solution exits
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Naive Approach for M-Coloring Problem:
Generate all possible configurations of colors. Since each node can be colored using any of the m available colors, the total number of color configurations possible is mV. After generating a configuration of color, check if the adjacent vertices have the same color or not. If the conditions are met, print the combination.
Time Complexity: O(mV). There is a total O(mV) combination of colors
Auxiliary Space: O(V). The Recursive Stack of graph coloring(…) function will require O(V) space.
M-Coloring Problem using Backtracking:
Assign colors one by one to different vertices, starting from vertex 0. Before assigning a color, check for safety by considering already assigned colors to the adjacent vertices i.e check if the adjacent vertices have the same color or not. If there is any color assignment that does not violate the conditions, mark the color assignment as part of the solution. If no assignment of color is possible then backtrack and return false
Follow the given steps to solve the problem:
- Create a recursive function that takes the graph, current index, number of vertices, and color array.
- If the current index is equal to the number of vertices. Print the color configuration in the color array.
- Assign a color to a vertex from the range (1 to m).
- For every assigned color, check if the configuration is safe, (i.e. check if the adjacent vertices do not have the same color) and recursively call the function with the next index and number of vertices otherwise, return false
- If any recursive function returns true then return true
- If no recursive function returns true then return false
Illustration:
- To color the graph, color each node one by one.
- To color the first node there are 3 choices of colors Red, Green and Blue, so lets take the red color for first node.
- After Red color for first node is fixed then we have made choice for second node in similar manner as we did for first node, then for 3rd node and so on.
- There is one important point to remember. while choosing color for the node, it should not be same as the color of the adjacent node.
Choosing Red For Node 1
- As shown in the above diagram, all the solutions are shown by coloring the first node in Red.
- Let’s choose Green color for the first node and explore the options for the remaining nodes.
Choosing Green for Node 1
- As shown in the above diagram, all the solutions are shown by coloring the first node in Green.
- Let’s choose Blue color for the first node and explore the options for the remaining nodes.
Choosing Blue for Node 1
Below is the implementation of the above approach:
C++
// C++ program for solution of M// Coloring problem using backtracking#include <bits/stdc++.h>using namespace std;// Number of vertices in the graph#define V 4void printSolution(int color[]);/* A utility function to check if the current color assignment is safe for vertex v i.e. checks whether the edge exists or not (i.e, graph[v][i]==1). If exist then checks whether the color to be filled in the new vertex(c is sent in the parameter) is already used by its adjacent vertices(i-->adj vertices) or not (i.e, color[i]==c) */bool isSafe(int v, bool graph[V][V], int color[], int c){ for (int i = 0; i < V; i++) if (graph[v][i] && c == color[i]) return false; return true;}/* A recursive utility functionto solve m coloring problem */bool graphColoringUtil(bool graph[V][V], int m, int color[], int v){ /* base case: If all vertices are assigned a color then return true */ if (v == V) return true; /* Consider this vertex v and try different colors */ for (int c = 1; c <= m; c++) { /* Check if assignment of color c to v is fine*/ if (isSafe(v, graph, color, c)) { color[v] = c; /* recur to assign colors to rest of the vertices */ if (graphColoringUtil(graph, m, color, v + 1) == true) return true; /* If assigning color c doesn't lead to a solution then remove it */ color[v] = 0; } } /* If no color can be assigned to this vertex then return false */ return false;}/* This function solves the m Coloring problem using Backtracking. It mainly uses graphColoringUtil() to solve the problem. It returns false if the m colors cannot be assigned, otherwise return true and prints assignments of colors to all vertices. Please note that there may be more than one solutions, this function prints one of the feasible solutions.*/bool graphColoring(bool graph[V][V], int m){ // Initialize all color values as 0. // This initialization is needed // correct functioning of isSafe() int color[V]; for (int i = 0; i < V; i++) color[i] = 0; // Call graphColoringUtil() for vertex 0 if (graphColoringUtil(graph, m, color, 0) == false) { cout << "Solution does not exist"; return false; } // Print the solution printSolution(color); return true;}/* A utility function to print solution */void printSolution(int color[]){ cout << "Solution Exists:" << " Following are the assigned colors" << "\n"; for (int i = 0; i < V; i++) cout << " " << color[i] << " "; cout << "\n";}// Driver codeint main(){ /* Create following graph and test whether it is 3 colorable (3)---(2) | / | | / | | / | (0)---(1) */ bool graph[V][V] = { { 0, 1, 1, 1 }, { 1, 0, 1, 0 }, { 1, 1, 0, 1 }, { 1, 0, 1, 0 }, }; // Number of colors int m = 3; // Function call graphColoring(graph, m); return 0;}// This code is contributed by Shivani |
C
// C program for solution of M// Coloring problem using backtracking#include <stdbool.h>#include <stdio.h>// Number of vertices in the graph#define V 4void printSolution(int color[]);/* A utility function to check if the current color assignment is safe for vertex v i.e. checks whether the edge exists or not (i.e, graph[v][i]==1). If exist then checks whether the color to be filled in the new vertex(c is sent in the parameter) is already used by its adjacent vertices(i-->adj vertices) or not (i.e, color[i]==c) */bool isSafe(int v, bool graph[V][V], int color[], int c){ for (int i = 0; i < V; i++) if (graph[v][i] && c == color[i]) return false; return true;}/* A recursive utility functionto solve m coloring problem */bool graphColoringUtil(bool graph[V][V], int m, int color[], int v){ /* base case: If all vertices are assigned a color then return true */ if (v == V) return true; /* Consider this vertex v and try different colors */ for (int c = 1; c <= m; c++) { /* Check if assignment of color c to v is fine*/ if (isSafe(v, graph, color, c)) { color[v] = c; /* recur to assign colors to rest of the vertices */ if (graphColoringUtil(graph, m, color, v + 1) == true) return true; /* If assigning color c doesn't lead to a solution then remove it */ color[v] = 0; } } /* If no color can be assigned to this vertex then return false */ return false;}/* This function solves the m Coloring problem using Backtracking. It mainly uses graphColoringUtil() to solve the problem. It returns false if the m colors cannot be assigned, otherwise return true and prints assignments of colors to all vertices. Please note that there may be more than one solutions, this function prints one of the feasible solutions.*/bool graphColoring(bool graph[V][V], int m){ // Initialize all color values as 0. // This initialization is needed // correct functioning of isSafe() int color[V]; for (int i = 0; i < V; i++) color[i] = 0; // Call graphColoringUtil() for vertex 0 if (graphColoringUtil(graph, m, color, 0) == false) { printf("Solution does not exist"); return false; } // Print the solution printSolution(color); return true;}/* A utility function to print solution */void printSolution(int color[]){ printf("Solution Exists:" " Following are the assigned colors \n"); for (int i = 0; i < V; i++) printf(" %d ", color[i]); printf("\n");}// Driver codeint main(){ /* Create following graph and test whether it is 3 colorable (3)---(2) | / | | / | | / | (0)---(1) */ bool graph[V][V] = { { 0, 1, 1, 1 }, { 1, 0, 1, 0 }, { 1, 1, 0, 1 }, { 1, 0, 1, 0 }, }; int m = 3; // Number of colors // Function call graphColoring(graph, m); return 0;} |
Java
/* Java program for solution of M Coloring problem using backtracking */public class mColoringProblem { final int V = 4; int color[]; /* A utility function to check if the current color assignment is safe for vertex v */ boolean isSafe(int v, int graph[][], int color[], int c) { for (int i = 0; i < V; i++) if (graph[v][i] == 1 && c == color[i]) return false; return true; } /* A recursive utility function to solve m coloring problem */ boolean graphColoringUtil(int graph[][], int m, int color[], int v) { /* base case: If all vertices are assigned a color then return true */ if (v == V) return true; /* Consider this vertex v and try different colors */ for (int c = 1; c <= m; c++) { /* Check if assignment of color c to v is fine*/ if (isSafe(v, graph, color, c)) { color[v] = c; /* recur to assign colors to rest of the vertices */ if (graphColoringUtil(graph, m, color, v + 1)) return true; /* If assigning color c doesn't lead to a solution then remove it */ color[v] = 0; } } /* If no color can be assigned to this vertex then return false */ return false; } /* This function solves the m Coloring problem using Backtracking. It mainly uses graphColoringUtil() to solve the problem. It returns false if the m colors cannot be assigned, otherwise return true and prints assignments of colors to all vertices. Please note that there may be more than one solutions, this function prints one of the feasible solutions.*/ boolean graphColoring(int graph[][], int m) { // Initialize all color values as 0. This // initialization is needed correct // functioning of isSafe() color = new int[V]; for (int i = 0; i < V; i++) color[i] = 0; // Call graphColoringUtil() for vertex 0 if (!graphColoringUtil(graph, m, color, 0)) { System.out.println("Solution does not exist"); return false; } // Print the solution printSolution(color); return true; } /* A utility function to print solution */ void printSolution(int color[]) { System.out.println("Solution Exists: Following" + " are the assigned colors"); for (int i = 0; i < V; i++) System.out.print(" " + color[i] + " "); System.out.println(); } // Driver code public static void main(String args[]) { mColoringProblem Coloring = new mColoringProblem(); /* Create following graph and test whether it is 3 colorable (3)---(2) | / | | / | | / | (0)---(1) */ int graph[][] = { { 0, 1, 1, 1 }, { 1, 0, 1, 0 }, { 1, 1, 0, 1 }, { 1, 0, 1, 0 }, }; int m = 3; // Number of colors // Function call Coloring.graphColoring(graph, m); }}// This code is contributed by Abhishek Shankhadhar |
Python3
# Python3 program for solution of M Coloring# problem using backtrackingclass Graph(): def __init__(self, vertices): self.V = vertices self.graph = [[0 for column in range(vertices)] for row in range(vertices)] # A utility function to check # if the current color assignment # is safe for vertex v def isSafe(self, v, colour, c): for i in range(self.V): if self.graph[v][i] == 1 and colour[i] == c: return False return True # A recursive utility function to solve m # coloring problem def graphColourUtil(self, m, colour, v): if v == self.V: return True for c in range(1, m + 1): if self.isSafe(v, colour, c) == True: colour[v] = c if self.graphColourUtil(m, colour, v + 1) == True: return True colour[v] = 0 def graphColouring(self, m): colour = [0] * self.V if self.graphColourUtil(m, colour, 0) == None: return False # Print the solution print("Solution exist and Following are the assigned colours:") for c in colour: print(c, end=' ') return True# Driver Codeif __name__ == '__main__': g = Graph(4) g.graph = [[0, 1, 1, 1], [1, 0, 1, 0], [1, 1, 0, 1], [1, 0, 1, 0]] m = 3 # Function call g.graphColouring(m)# This code is contributed by Divyanshu Mehta |
C#
/* C# program for solution of M Coloring problemusing backtracking */using System;class GFG { readonly int V = 4; int[] color; /* A utility function to check if the current color assignment is safe for vertex v */ bool isSafe(int v, int[, ] graph, int[] color, int c) { for (int i = 0; i < V; i++) if (graph[v, i] == 1 && c == color[i]) return false; return true; } /* A recursive utility function to solve m coloring problem */ bool graphColoringUtil(int[, ] graph, int m, int[] color, int v) { /* base case: If all vertices are assigned a color then return true */ if (v == V) return true; /* Consider this vertex v and try different colors */ for (int c = 1; c <= m; c++) { /* Check if assignment of color c to v is fine*/ if (isSafe(v, graph, color, c)) { color[v] = c; /* recur to assign colors to rest of the vertices */ if (graphColoringUtil(graph, m, color, v + 1)) return true; /* If assigning color c doesn't lead to a solution then remove it */ color[v] = 0; } } /* If no color can be assigned to this vertex then return false */ return false; } /* This function solves the m Coloring problem using Backtracking. It mainly uses graphColoringUtil() to solve the problem. It returns false if the m colors cannot be assigned, otherwise return true and prints assignments of colors to all vertices. Please note that there may be more than one solutions, this function prints one of the feasible solutions.*/ bool graphColoring(int[, ] graph, int m) { // Initialize all color values as 0. This // initialization is needed correct functioning // of isSafe() color = new int[V]; for (int i = 0; i < V; i++) color[i] = 0; // Call graphColoringUtil() for vertex 0 if (!graphColoringUtil(graph, m, color, 0)) { Console.WriteLine("Solution does not exist"); return false; } // Print the solution printSolution(color); return true; } /* A utility function to print solution */ void printSolution(int[] color) { Console.WriteLine("Solution Exists: Following" + " are the assigned colors"); for (int i = 0; i < V; i++) Console.Write(" " + color[i] + " "); Console.WriteLine(); } // Driver Code public static void Main(String[] args) { GFG Coloring = new GFG(); /* Create following graph and test whether it is 3 colorable (3)---(2) | / | | / | | / | (0)---(1) */ int[, ] graph = { { 0, 1, 1, 1 }, { 1, 0, 1, 0 }, { 1, 1, 0, 1 }, { 1, 0, 1, 0 } }; int m = 3; // Number of colors // Function call Coloring.graphColoring(graph, m); }}// This code is contributed by PrinciRaj1992 |
Javascript
<script>/* JavaScript program for solution of M Coloring problem using backtracking */let V = 4;let color;/* A utility function to check if the current color assignment is safe for vertex v */function isSafe(v,graph,color,c){ for (let i = 0; i < V; i++) if ( graph[v][i] == 1 && c == color[i]) return false; return true;}/* A recursive utility function to solve m coloring problem */function graphColoringUtil(graph,m,color,v){ /* base case: If all vertices are assigned a color then return true */ if (v == V) return true; /* Consider this vertex v and try different colors */ for (let c = 1; c <= m; c++) { /* Check if assignment of color c to v is fine*/ if (isSafe(v, graph, color, c)) { color[v] = c; /* recur to assign colors to rest of the vertices */ if ( graphColoringUtil( graph, m, color, v + 1)) return true; /* If assigning color c doesn't lead to a solution then remove it */ color[v] = 0; } } /* If no color can be assigned to this vertex then return false */ return false;}/* This function solves the m Coloring problem using Backtracking. It mainly uses graphColoringUtil() to solve the problem. It returns false if the m colors cannot be assigned, otherwise return true and prints assignments of colors to all vertices. Please note that there may be more than one solutions, this function prints one of the feasible solutions.*/function graphColoring(graph,m){ // Initialize all color values as 0. This // initialization is needed correct // functioning of isSafe() color = new Array(V); for (let i = 0; i < V; i++) color[i] = 0; // Call graphColoringUtil() for vertex 0 if ( !graphColoringUtil( graph, m, color, 0)) { document.write( "Solution does not exist<br>"); return false; } // Print the solution printSolution(color); return true;}/* A utility function to print solution */function printSolution(color){ document.write( "Solution Exists: Following" + " are the assigned colors<br>"); for (let i = 0; i < V; i++) document.write(" " + color[i] + " "); document.write("<br>");} // driver program to test above function/* Create following graph and test whether it is 3 colorable (3)---(2) | / | | / | | / | (0)---(1) */ let graph = [ [ 0, 1, 1, 1 ], [ 1, 0, 1, 0 ], [ 1, 1, 0, 1 ], [ 1, 0, 1, 0 ], ]; let m = 3; // Number of colors graphColoring(graph, m); // This code is contributed by ab2127</script> |
Output
Solution Exists: Following are the assigned colors 1 2 3 2
Time Complexity: O(mV). There is a total of O(mV) combinations of colors. The upper bound time complexity remains the same but the average time taken will be less.
Auxiliary Space: O(V). The recursive Stack of the graph coloring function will require O(V) space.
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