On the Complexity of k-SAT

The k-SAT problem is to determine if a given k-CNF has a satisfying assignment. It is a celebrated open question as to whether it requires exponential time to solve k-SAT for k 3. Here exponential time means 2 n for some >0. In this paper, assuming that, for k 3, k-SAT requires exponential time complexity, we show that the complexity of k-SAT increases as k increases. More precisely, for k 3, define sk=inf{ :there exists 2 n algorithm for solving k-SAT}. Define ETH (Exponential-Time Hypothesis) for k-SAT as follows: for k 3, sk>0. In this paper, we show that sk is increasing infinitely often assuming ETH for k-SAT. Let s∞ be the limit of sk. We will in fact show that sk (1 d/k)s∞ for some constant d>0. We prove this result by bringing together the ideas of critical clauses and the Sparsification Lemma to reduce the satisfiability of a k-CNF to the satisfiability of a disjunction of 2 nk -CNFs in fewer variables for some k k and arbitrarily small >0. We also show that such a disjunction can be computed in time 2 n for arbitrarily small >0.