Cygan, M and Dell, H and Lokshtanov, D and Marx, Dániel and Nederlof, J (2016) On Problems as Hard as CNF-SAT. ACM TRANSACTIONS ON ALGORITHMS, 12 (3). p. 41. ISSN 1549-6325 10.1145/2925416

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Abstract

The field of exact exponential time algorithms for non-deterministic polynomial-time hard problems has thrived since the mid-2000s. While exhaustive search remains asymptotically the fastest known algorithm for some basic problems, non-trivial exponential time algorithms have been found for a myriad of problems, including GRAPH COLORING, HAMILTONIAN PATH, DOMINATING SET, and 3-CNF-SAT. In some instances, improving these algorithms further seems to be out of reach. The CNF-SAT problem is the canonical example of a problem for which the trivial exhaustive search algorithm runs in time O(2(n)), where n is the number of variables in the input formula. While there exist non-trivial algorithms for CNF-SAT that run in time o(2(n)), no algorithm was able to improve the growth rate 2 to a smaller constant, and hence it is natural to conjecture that 2 is the optimal growth rate. The strong exponential time hypothesis (SETH) by Impagliazzo and Paturi [JCSS 2001] goes a little bit further and asserts that, for every epsilon < 1, there is a (large) integer k such that k-CNF-SAT cannot be computed in time 2(epsilon n). In this article, we show that, for every epsilon < 1, the problems HITTING SET, SET SPLITTING, and NAE-SAT cannot be computed in time O(2(epsilon n)) unless SETH fails. Here n is the number of elements or variables in the input. For these problems, we actually get an equivalence to SETH in a certain sense. We conjecture that SETH implies a similar statement for SET COVER and prove that, under this assumption, the fastest known algorithms for STEINER TREE, CONNECTED VERTEX COVER, SET PARTITIONING, and the pseudo-polynomial time algorithm for SUBSET SUM cannot be significantly improved. Finally, we justify our assumption about the hardness of SET COVER by showing that the parity of the number of solutions to SET COVER cannot be computed in time O(2(epsilon n)) for any epsilon < 1 unless SETH fails.

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