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Tighter Connections Between Formula-SAT and Shaving Logs


Bringmann,  Karl
Algorithms and Complexity, MPI for Informatics, Max Planck Society;

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(Preprint), 426KB

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Abboud, A., & Bringmann, K. (2018). Tighter Connections Between Formula-SAT and Shaving Logs. Retrieved from

A noticeable fraction of Algorithms papers in the last few decades improve the running time of well-known algorithms for fundamental problems by logarithmic factors. For example, the $O(n^2)$ dynamic programming solution to the Longest Common Subsequence problem (LCS) was improved to $O(n^2/\log^2 n)$ in several ways and using a variety of ingenious tricks. This line of research, also known as "the art of shaving log factors", lacks a tool for proving negative results. Specifically, how can we show that it is unlikely that LCS can be solved in time $O(n^2/\log^3 n)$? Perhaps the only approach for such results was suggested in a recent paper of Abboud, Hansen, Vassilevska W. and Williams (STOC'16). The authors blame the hardness of shaving logs on the hardness of solving satisfiability on Boolean formulas (Formula-SAT) faster than exhaustive search. They show that an $O(n^2/\log^{1000} n)$ algorithm for LCS would imply a major advance in circuit lower bounds. Whether this approach can lead to tighter barriers was unclear. In this paper, we push this approach to its limit and, in particular, prove that a well-known barrier from complexity theory stands in the way for shaving five additional log factors for fundamental combinatorial problems. For LCS, regular expression pattern matching, as well as the Fr\'echet distance problem from Computational Geometry, we show that an $O(n^2/\log^{7+\varepsilon} n)$ runtime would imply new Formula-SAT algorithms. Our main result is a reduction from SAT on formulas of size $s$ over $n$ variables to LCS on sequences of length $N=2^{n/2} \cdot s^{1+o(1)}$. Our reduction is essentially as efficient as possible, and it greatly improves the previously known reduction for LCS with $N=2^{n/2} \cdot s^c$, for some $c \geq 100$.