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# Thread: Finding Chromatic Number w/ Brute Force Algorithm

1. ## 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:

```//graph G, vertices n, vertices are numbered 0, 1 . . n-1
//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

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:

```    //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;
}```

3. ## 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.