We prove that for any distinct $x,y \in \{0,1\}^n$, there is a deterministic finite automaton with $\widetilde{O}(n^{1/3})$ states that accepts $x$ but not $y$. This improves Robson's 1989 upper bound of $\widetilde{O}(n^{2/5})$.
此内容由惯性聚合(RSS阅读器)自动聚合整理，仅供阅读参考。 原文来自 — 版权归原作者所有。