Boolean formula for a graph vertex-colouring algorithm

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?

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));
            }
        }
    }
}