Added Graph Coloring Problem code in Backtracking folder

Author: Karnika06

Backtracking/Graph_Coloring_Problem.cpp

@@ -0,0 +1,160 @@
+/* C++ implementation of Graph Coloring Problem */
+#include <bits/stdc++.h>
+using namespace std;
+
+//Function for generating a next color
+/*x[1],...,x[k-1] have been assigned integer values 
+in the range [1,m] such that adjacent vertices have
+distinct integers*/
+//A value for x[k] is determined in the range [0,m]
+/*x[k] is assigned the next highest numbered color while
+maintaining distinctness from the adjacent vertices of vertex k.
+If no such color exists, then x[k] is 0.*/
+
+void nextvalue(int x[50],int g[50][50],int k,int n,int m)
+{
+	int j;
+ 	while(1)
+ 	{
+   		x[k] = (x[k]+1) % (m+1);    			//next higher color
+   		if(x[k] == 0)             				//all colors have been used
+      		return;
+      		
+      	//checking if this color is distinct from adjacent colors	
+   		for(j=1; j<=n; j++)         
+   		{                          
+     		if(g[k][j] != 0 && (x[k] == x[j]))	//if (k,j) is an edge and if adjacent vertices have the same color
+        		break; 
+   		}
+     	if(j == n+1)           //new color found       
+       		return;                     
+  	}
+}
+
+//Finding all m-colorings of a given graph
+/*All assignments of 1,2,...,m to the vertices of the graph such that 
+adjacent vertices are assigned distinct integers are printed*/
+//k is the index of the next vertex to color
+//set the array x[] to zero, and then invoke mcoloring(1)
+void mcoloring(int x[50],int g[50][50],int k,int n,int m)
+{
+	int i;
+  	do								//loop for generating all legal assignments for x[k]
+ 	{                         
+   		nextvalue(x, g, k,n ,m);    //Assign to x[k] a legal color
+		     
+   		if(x[k] == 0)             	//No new color possible
+     		return; 
+   		
+		if(k == n)                //Atmost m colors have been used to color the n vertices
+   		{
+    		for(i=1 ;i<=n; i++)
+      			cout<<"V"<<i<<": "<<x[i]<<"\t";     
+      		cout<<"\n";
+   		}
+  		else
+		mcoloring(x, g, k+1, n, m); 
+		
+ 	}while(k<n+1); 
+}
+
+//Main function
+int main()
+{
+	int n,ne,u,v,i,j,x[50],g[50][50],m;
+	 
+	cout<<"\nEnter number of vertices: ";
+	cin>>n;				//reading number of vertices of the graph from user
+	 
+	cout<<"\nEnter number of edges: ";
+	cin>>ne;			//reading number of edges of the graph from user
+  
+  	m = n-1;
+  	cout<<"\nThe maximum possible colours that can be assigned is "<<m<<endl;
+  	
+  	//The graph is represented by its boolean adjacency matrix G[1:n,1:n]
+  	for(i=0; i<=n; i++)
+    	for(j=0; j<=n; j++)
+       		g[i][j] = 0;
+       
+  	for(i=1; i<=ne; i++)
+	{
+	 	cout<<"\nEnter the first terminal vertex of edge "<<i<<": ";
+	 	cin>>u;
+	 	
+	 	cout<<"\nEnter the second terminal vertex of edge "<<i<<": ";
+	 	cin>>v;
+	 	
+		g[u][v] = 1;
+   		g[v][u] = 1;
+  	}
+   
+   	for(i=0; i<=n; i++)
+      	x[i] = 0;
+   
+	cout<<endl<<"The colouring possibilities are:"<<endl;
+    mcoloring(x ,g ,1 ,n ,m);
+}
+
+/*
+Input-
+
+Number of vertices= 4
+Number of edges= 4
+
+Adjacency matrix:
+
+	0	1	0	1
+	1	0	1	0
+	0	1	0	1
+	1	0	1	0
+
+Output-
+
+Enter number of vertices: 4
+
+Enter number of edges: 4
+
+The maximum possible colours that can be assigned is 3
+
+Enter the first terminal vertex of edge 1: 1
+
+Enter the second terminal vertex of edge 1: 2
+
+Enter the first terminal vertex of edge 2: 2
+
+Enter the second terminal vertex of edge 2: 3
+
+Enter the first terminal vertex of edge 3: 3
+
+Enter the second terminal vertex of edge 3: 4
+
+Enter the first terminal vertex of edge 4: 1
+
+Enter the second terminal vertex of edge 4: 4
+
+The colouring possibilities are:
+V1: 1   V2: 2   V3: 1   V4: 2
+V1: 1   V2: 2   V3: 1   V4: 3
+V1: 1   V2: 2   V3: 3   V4: 2
+V1: 1   V2: 3   V3: 1   V4: 2
+V1: 1   V2: 3   V3: 1   V4: 3
+V1: 1   V2: 3   V3: 2   V4: 3
+V1: 2   V2: 1   V3: 2   V4: 1
+V1: 2   V2: 1   V3: 2   V4: 3
+V1: 2   V2: 1   V3: 3   V4: 1
+V1: 2   V2: 3   V3: 1   V4: 3
+V1: 2   V2: 3   V3: 2   V4: 1
+V1: 2   V2: 3   V3: 2   V4: 3
+V1: 3   V2: 1   V3: 2   V4: 1
+V1: 3   V2: 1   V3: 3   V4: 1
+V1: 3   V2: 1   V3: 3   V4: 2
+V1: 3   V2: 2   V3: 1   V4: 2
+V1: 3   V2: 2   V3: 3   V4: 1
+V1: 3   V2: 2   V3: 3   V4: 2
+
+Time Complexity: O(m^V)
+Space Complexity: O(V)
+
+where V= number of vertices and m= maximum possible colors
+*/