# Reducing 3CNF to Clique: Why do we omit negated literals?

I have an example for a reduction of 3CNF to Clique, there is one thing I don't get about it, hopefully you could clarify it. The reduction works like this:

Construct a graph G = (V, E) as follows:

Vertices: Each literal corresponds to a vertex.

Edges: All vertices are connected with an edge except the vertices of the same clause and vertices with negated literals.

Why is it important that that negated literals will not be connected? How would that effect the reduction?

• What do you think? What have you tried? Have you tried running the reduction by hand on a few sample problem instances (a few 3SAT instances) to see what would happen? We expect you to make a serious effort before asking. This question is one where you could easily discover the answer on your own by just trying a few examples. – D.W. Jan 31 '14 at 3:09
• @D.W. I have tried, a none trivial 3CNF formula that can not be satisfied have 8 clauses, it not very easy to deduce what really happens there. – aviran Jan 31 '14 at 19:53

The goal of the reduction is to enforce the fact that a clique of the graph of size $|Clauses|$ is exactly a valid instanciation of the variables satisfying all clauses of the formula.
So by not connecting variables of the same clause together, you force a clique of size $|Clauses|$ to pick one variable in each clause, setting it to true.
However, you don't want this instanciation to contradict itself: you cannot have $x$ and $\neg x$ true at the same time. For instance the formula with two clauses $x\wedge \neg x$ should not be satisfiable. So you forbid a clique to contain a variable and its negation, by not connecting them together. Therefore, a clique has to be coherent, so it always represents a valid instanciation. Notice that it can be only a partial instanciation, in which case you can choose anything for the remaining variables.