# Finding Chromatic Number w/ Brute Force Algorithm

• November 15th, 2011, 05:45 PM
thecrazytaco
Finding Chromatic Number w/ Brute Force Algorithm
Hello, I'm new to the forums here and really in a bind.

I'm trying to implement this pseudocode:

Code :

```//graph G, vertices n, vertices are numbered 0, 1 . . n-1 //G is stored in adjacency list or adjacency matrix p //colors are stored in array q //q[i] has color of vertex i, initially all 0   //colors G using minimum number of colors, returns chromatic number int color() { for(i = 1 to n) { //if G can be colored using i colors starting at vertex 0 if(color(0,i)) { chromatic number is i, return it } } }   //colors G using m colors starting at vertex v boolean color(v,m) { if(v exceeds n - 1) all vertices have been colored, success else { for(i = 1 to m) { assign color i to vertex v   check whether colors of v and its neighbors conflict   if(colors do not conflict) { color the rest of G using m colors starting at vertex v+1 done by recursive call color(v+1,m)   if(the rest of G can be colored) all vertices have been colored, success } } assign color 0 to vertex v and fail, this happens when none of the m colors can be assigned to vertex v } }   //in color() method, chromatic numbers can be tried in ascending order //(i = 1 to n) or descending order (i = n to 1), and both orders can be //used to find a range of possible values of the chromatic number```

I thought that my implementation was working because I tested it with a small sample graph and got the correct output. However, when testing with large graphs this algorithm was solving them in seconds, but is supposed to have a runtime of hours. I think that I may be missing a recursive call somewhere, or maybe am breaking out of the function too early somehow.

Anyway, here is my implementation of the above pseudocode. I am sincerely grateful for any help that you guys can provide. Thanks :D

Explanation of my variables:

this.size = the number of vertices in the graph
this.vertexColors = 1 dimensional array with this.size number of elements. holds
a chromatic number for each vertex
this.inputMatrix = a this.size x this.size element 2d matrix. only holds 0s and 1s.
the 0s represent no line between vertices, and a 1 represents a line

Example:
vertex
1 2 3 4

vertex 1 0 0 0 0
vertex 2 0 0 1 0
vertex 3 0 0 0 0
vertex 4 0 0 0 0

This represents a line connecting vertex 2 and 3. A 1 in row 3, column 2 would mean the same thing. It is also not necessary to have a 1 in both of those to represent the line, just one will work.

My implementation:

Code java:

``` //colors graph using minimum number of colors, returns chromatic number public int colorAscending() { for(int i = 1; i < this.size + 1; i++) { //if graph can be colored using i colors starting at //vertex 0 if(colorAscending(0,i)) { return i; } } //syntactically necessary return statement. should never get here return 0; }             //v = current vertex, m = number of colors to try //colors graph using m colors starting at vertex v private boolean colorAscending(int v, int m) { if(v > this.size - 1) { //all vertices have been colored, success return true; } else { for(int i = 1; i < m+1; i++) { //assign color i to vertex v this.vertexColors[v] = i;   //check whether colors of v and its neighbors conflict if(!checkForConflict(v)) { if(colorAscending(v+1,m)) { return true; } else { return false; } } } //assign color 0 to vertex v and fail this.vertexColors[v] = 0; return false; } }           //checks if vertices adjacent to v have the same color or not private boolean checkForConflict(int vertex) { for(int i = 0; i < this.size; i++) { if((this.inputMatrix[i][vertex] == 1) && (this.vertexColors[i] == this.vertexColors[vertex])) { return true; } } return false; }```
• November 15th, 2011, 07:10 PM
Junky
Re: Finding Chromatic Number w/ Brute Force Algorithm
• November 16th, 2011, 06:35 AM
Mr.777
Re: Finding Chromatic Number w/ Brute Force Algorithm
Can you explain the flow of your implementation. Which function calls first and next and next?
This implementation is taking us no where.