ON OPTIMAL INVERTERS

2014 ◽  
Vol 20 (1) ◽  
pp. 1-23 ◽  
Author(s):  
YIJIA CHEN ◽  
JÖRG FLUM

AbstractLeonid Levin showed that every algorithm computing a function has an optimal inverter. Recently, we applied his result in various contexts: existence of optimal acceptors, existence of hard sequences for algorithms and proof systems, proofs of Gödel’s incompleteness theorems, analysis of the complexity of the clique problem assuming the nonuniform Exponential Time Hypothesis. We present all these applications here. Even though a simple diagonalization yields Levin’s result, we believe that it is worthwhile to be aware of the explicit result. The purpose of this survey is to convince the reader of our view.

2020 ◽  
Vol 34 (09) ◽  
pp. 13700-13703
Author(s):  
Nikhil Vyas ◽  
Ryan Williams

All known SAT-solving paradigms (backtracking, local search, and the polynomial method) only yield a 2n(1−1/O(k)) time algorithm for solving k-SAT in the worst case, where the big-O constant is independent of k. For this reason, it has been hypothesized that k-SAT cannot be solved in worst-case 2n(1−f(k)/k) time, for any unbounded ƒ : ℕ → ℕ. This hypothesis has been called the “Super-Strong Exponential Time Hypothesis” (Super Strong ETH), modeled after the ETH and the Strong ETH. We prove two results concerning the Super-Strong ETH:1. It has also been hypothesized that k-SAT is hard to solve for randomly chosen instances near the “critical threshold”, where the clause-to-variable ratio is 2k ln 2 −Θ(1). We give a randomized algorithm which refutes the Super-Strong ETH for the case of random k-SAT and planted k-SAT for any clause-to-variable ratio. In particular, given any random k-SAT instance F with n variables and m clauses, our algorithm decides satisfiability for F in 2n(1−Ω( log k)/k) time, with high probability (over the choice of the formula and the randomness of the algorithm). It turns out that a well-known algorithm from the literature on SAT algorithms does the job: the PPZ algorithm of Paturi, Pudlak, and Zane (1998).2. The Unique k-SAT problem is the special case where there is at most one satisfying assignment. It is natural to hypothesize that the worst-case (exponential-time) complexity of Unique k-SAT is substantially less than that of k-SAT. Improving prior reductions, we show the time complexities of Unique k-SAT and k-SAT are very tightly related: if Unique k-SAT is in 2n(1−f(k)/k) time for an unbounded f, then k-SAT is in 2n(1−f(k)(1−ɛ)/k) time for every ɛ > 0. Thus, refuting Super Strong ETH in the unique solution case would refute Super Strong ETH in general.


2015 ◽  
pp. 467-521 ◽  
Author(s):  
Marek Cygan ◽  
Fedor V. Fomin ◽  
Łukasz Kowalik ◽  
Daniel Lokshtanov ◽  
Dániel Marx ◽  
...  

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