Boolean formula for a graph colouring algorithm - algorithm

Question

I have a graph which has 5 vertexes.
Graph g1 = new Graph(5);
g1.addEdge(0, 1);
g1.addEdge(0, 2);
g1.addEdge(1, 2);
g1.addEdge(1, 3);
g1.addEdge(3, 2);
g1.addEdge(3, 4);
g1.addEdge(4, 2);
System.out.println("Coloring of graph 1");
g1.greedyColoring();
I need to express the problem of coloring this graph as a boolean expression.
Suppose a, b, and c are three colours, and the literal ai means vertex i has color a. Then the above graph could be colored like this:
a0, b1, c2, a3, b4
How can I get a boolean formula that, when satisfied, provides a solution for coloring this graph?

Answer 1

You are looking for coloring all vertices in a graph, each with one out of three available colors, such that no two adjacent nodes have the same color.
There are thus three types of conditions:
1. Two adjacent nodes cannot have the same color
So for instance, the edge (0, 1) will imply these three constraints:
Node 0 and 1 cannot both have color a
Node 0 and 1 cannot both have color b
Node 0 and 1 cannot both have color c
Translated into a boolean expression:
¬a0 ∨ ¬a1
¬b0 ∨ ¬b1
¬c0 ∨ ¬c1
You would need to generate such triplets of disjunctions for all edges. So in total you'll have 3 x 7 = 21 boolean disjunctions:
¬a0 ∨ ¬a1 ¬a0 ∨ ¬a2 ¬a1 ∨ ¬a2 ¬a1 ∨ ¬a3 ¬a3 ∨ ¬a2 ¬a3 ∨ ¬a4 ¬a4 ∨ ¬a2
¬b0 ∨ ¬b1 ¬b0 ∨ ¬b2 ¬b1 ∨ ¬b2 ¬b1 ∨ ¬b3 ¬b3 ∨ ¬b2 ¬b3 ∨ ¬b4 ¬b4 ∨ ¬b2
¬c0 ∨ ¬c1 ¬c0 ∨ ¬c2 ¬c1 ∨ ¬c2 ¬c1 ∨ ¬c3 ¬c3 ∨ ¬c2 ¬c3 ∨ ¬c4 ¬c4 ∨ ¬c2
2. All nodes must get a color
So for instance, for node 0 we will have this constraint:
Node 0 must have color a, b, or c
Translated into a boolean expression:
a0 ∨ b0 ∨ c0
You would need to do the same for all nodes, so in total you'll have 5 such expressions:
a0 ∨ b0 ∨ c0
a1 ∨ b1 ∨ c1
a2 ∨ b2 ∨ c2
a3 ∨ b3 ∨ c3
a4 ∨ b4 ∨ c4
3. No nodes can get more than one color
So for instance, for node 0 we will have:
Node 0 cannot have both color a and b
Node 0 cannot have both color a and c
Node 0 cannot have both color b and c
Translated into a boolean expression:
¬a0 ∨ ¬b0
¬a0 ∨ ¬c0
¬b0 ∨ ¬c0
You would need to do the same for all nodes, so in total you'll have 3 * 5 = 15 such expressions for this type:
¬a0 ∨ ¬b0 ¬a1 ∨ ¬b1 ¬a2 ∨ ¬b2 ¬a3 ∨ ¬b3 ¬a4 ∨ ¬b4
¬a0 ∨ ¬c0 ¬a1 ∨ ¬c1 ¬a2 ∨ ¬c2 ¬a3 ∨ ¬c3 ¬a4 ∨ ¬c4
¬b0 ∨ ¬c0 ¬b1 ∨ ¬c1 ¬b2 ∨ ¬c2 ¬b3 ∨ ¬c3 ¬b4 ∨ ¬c4
Conclusion
All the above disjunctions (there are 21 + 5 + 15 = 41 of them) need to be true (conjugated). Such a problem is a Boolean satisfiability problem, and more particularly 3-SAT, and is an NP-Complete problem.
Code to produce the boolean expressions
The following code assumes that the Graph object will exposes a nodes list where each node has an id and neighbors.
The disjunctions are output as strings, each on a separate line:
Graph g1 = new Graph(5);
g1.addEdge(0, 1);
g1.addEdge(0, 2);
g1.addEdge(1, 2);
g1.addEdge(1, 3);
g1.addEdge(3, 2);
g1.addEdge(3, 4);
g1.addEdge(4, 2);
char colors[] = {'a', 'b', 'c'};
// Type 1
for (Node node : g1.nodes) {
for (Node neighbor : node.neighbors) {
for (char color : colors) {
System.out.println(String.format("¬%1$c%2$d ∨ ¬%1$c%3$d", color, node.id, neighbor.id));
}
}
}
// Type 2
for (Node node : g1.nodes) {
String expr[] = new String[colors.length];
int i = 0;
for (char color : colors) {
expr[i++] = String.format("%s%d", color, node.id);
}
System.out.println(String.join(" ∨ ", expr));
}
// Type 3
for (Node node : g1.nodes) {
for (char color1 : colors) {
for (char color2 : colors) {
if (color1 < color2) {
System.out.println(String.format("¬%1$c%3$d ∨ ¬%2$c%3$d", color1, color2, node.id));
}
}
}
}
See it run on repl.it.

Answer 2

You get an equation for each edge and connect them with and (ci is the color used for vertex i):
c0 != c1 && c0 != c2 && c1 != c2 && c1 != c3 && c2 != c3 && c2 != c4 && c3 != c4
The boolean formula only checks if you have found a valid coloring for the graph, not if the number of colors is minimized.