Luca Trevisan Problem Set 6 Solutions

2) We prove that Half-Clique is NP-hard by reducing 3-SAT to it. The reduction is very similar to the one from 3-SAT to Clique described in CLR. We start from an instance φ = C1 ∧ C2 ∧ . . . ∧ Ck of 3-SAT. Without loss of generality we can suppose that for r = 1, . . . , k Cr contains three different literals lr 1, l r 2, l r 3. Now we have to construct a graph that has a clique of size |V |/2 if and only if φ is satisfiable. For each clause Cr = lr 1 ∨ lr 2 ∨ lr 3 we put the vertices vr 1, vr 2, vr 3 in G and we add an edge between two vertices vs i , v r j if the following conditions are met. 1) vs i and v r j are in different triplets (r 6= s) 2) their correpsonding literals are consistent (lr i is not the negation of l s j) Finally we add the vertices z1, . . . , zk to V and we connect every zi to all the other vertices in the graph. Thus G has 4k vertices. It is easy to show that the above construction can be done in polynomial time with respect to k. It remains to show that it is indeed a reduction.