### Strong Refutation Heuristics for Random k-SAT

AMIN COJA-OGHLAN, ANDREAS GOERDT, ANDRÉ LANKA
2006 Combinatorics, probability & computing
A simple first moment argument shows that in a randomly chosen k-SAT formula with m clauses over n boolean variables, the fraction of satisfiable clauses is at most 1−2 −k +o(1) as m/n → ∞ almost surely. In this paper, we deal with the corresponding algorithmic strong refutation problem: given a random k-SAT formula, can we find a certificate that the fraction of satisfiable clauses is at most 1 − 2 −k + o(1) in polynomial time? We present heuristics based on spectral techniques that in the
more » ... k = 3, m ≥ ln(n) 6 n 3/2 and in the case k = 4, m ≥ Cn 2 find such certificates almost surely. Our methods also apply to a variety of further problems such as hypergraph coloring. m = (2n) k p. (Thus, in this paper, clauses are order k-tuples, and we allow for multiple occurrences of literals in a clause. Several slightly different models exist, but the differences are only of technical relevance.) The combinatorial structure of random k-SAT formulas has attracted considerable attention. Friedgut [11] has shown that Form n,k,p exhibits a sharp threshold behavior: there exist numbers c k = c k (n) such that Form n,k,p is satisfiable almost surely if m < (1 − ε)c k n, whereas Form n,k,p is unsatisfiable almost surely if m > (1 + ε)c k n. The asymptotic behavior of c k as k → ∞ has been determined by Achlioptas and Peres [2]. Moreover, a simple first moment argument shows that the maximum number of clauses of Form n,k,p that can be satisfied by any assignment is at most (1 − 2 k + o(1))m as m/n → ∞. More precise results have been obtained by Achlioptas, Naor, and Peres [1] . With respect to proof complexity, various types of resolution proofs for the non-existence of satisfying assignments in Form n,k,p have been investigated. Ben-Sasson [4] has shown that tree-like resolution proofs to refute Form n,k,p almost surely have size exp(Ω(n/∆ 1/(k−2)+ε )), where ∆ = n k−1 p and 0 < ε < 1/2 is an arbitrary constant. Hence, tree-like resolution proofs are of exponential length even if the expected number of clauses is n k−1−ε (i.e. p = n −ε−k/2 ). Furthermore, [4, Theorem 2.24] shows that general resolution proofs for the nonexistence of satisfying assignments of Form n,k,p almost surely have super polynomial size if p ≤ n −k/2−δ (δ > 0 constant). Goerdt and Krivelevich [16] have suggested a heuristic that uses spectral techniques for refuting Form n,4,p with p = ln(n) 7 n −2 (i.e. the expected number of clauses is m = ln(n) 7 n 2 ). No efficient resolution-based refutation heuristic is known for this range of p; in fact, tree-like resolution proofs are of exponential length by the aforementioned results. Removing the polylogarithmic factor, Feige and Ofek [8] and (independently) Coja-Oghlan, Goerdt, Lanka, and Schädlich [6] have shown that spectral techniques can be used to refute Form n,4,p if p ≥ Cn −2 for a sufficiently large constant C > 0. Moreover, Feige and Ofek [9] have shown that a heuristic that combines spectral techniques with extracting and refuting a XOR formula from Form n,3,p can refute Form n,3,p for p ≥ Cn −3/2 (i.e. m = Cn 3/2 ). This result improves on previous work by Friedman and Goerdt [12] , and Goerdt and Lanka [15] . We emphasize that in all of the above cases, the values of p to which the refutation heuristics apply exceed the threshold when Form n,k,p actually becomes unsatisfiable almost surely by at least a factor of n (k−2)/2 . The new aspect in the present paper is that we deal with strong refutation heuristics. That is, our aim are heuristics that on input Form n,k,p almost surely certify that not more than a (1 − 2 −k + ε)-fraction of the clauses can be satisfied, for any ε > 0. This aspect has not (at least not explicitly) been studied in the aforementioned references. For instance, resolution proofs cannot provide strong refutation. Moreover, the spectral heuristics studied so far [8, 9, 6, 15, 12, 16] only certify that every assignment leaves a o(1)-fraction of the clauses unsatisfied. With respect to MAX 3-SAT, we have the following result.