🤖 AI Summary
研究通过定义一个新的整数函数B(S)来确定有限自动机中子集S的最短扩展词长度，解决了如何有效认证和计算非欧拉自动机中状态子集扩展性的问题。
📝 Abstract
Let $\mathcal A$ be a complete deterministic finite automaton on a state set $Q$ of size $n$ with $k$ letters, and for a proper nonempty subset $S$ of $Q$ let $\mathrm{minext}(S)$ be the length of a shortest word $u$ with $|Su^{-1}|>|S|$, where $Su^{-1}=\{q: q\cdot u\in S\}$. To each state $q$ attach the integer $β^{\ast}_q=\sum_{t=1}^{n-1}k^{\,n-1-t}(\mathrm{indeg}_t(q)-k^{t})$, where $\mathrm{indeg}_t(q)$ counts the pairs $(p,u)$ with $|u|=t$ and $p\cdot u=q$, and let $B(S)=\sum_{q\in S}β^{\ast}_q$. On every synchronizing automaton, $B(S)\ge0$ implies $\mathrm{minext}(S)\le n-1$, so, as $B(Q)=0$, one of $S$ and $Q\setminus S$ extends within $n-1$; when $B(S)>0$ no hypothesis is needed. Kari's Eulerian extension lemma is the case $β^{\ast}=0$, and $β^{\ast}$, like every member of the family $\sum_{t=1}^{n-1}c_tσ_t$, $c_t>0$, vanishes identically if and only if the automaton is Eulerian, where $σ_t(S)=\sum_{q\in S}(\mathrm{indeg}_t(q)-k^{t})$. On strongly connected automata $σ_t(S)/k^{t}$ has Cesàro limit $n\,e(S)/e(Q)-|S|$ for Friedman's weight $e$; that limit certifies singletons but no larger subset in general. The hypothesis $B(S)\ge0$ cannot be relaxed by one integer unit, nor can the constant $n-1$ be improved. A second-moment test on the sizes $|Su^{-1}|$ certifies 60 to 95 percent of the subsets with $B(S)<0$ at $n\le7$. Along non-Eulerian automata whose words of length $n-1$ merge a fraction of the state pairs bounded below, with $\max_q\mathrm{indeg}_{n-1}(q)=o(nk^{n-1})$, it certifies all but a vanishing share of them. The functional $B$ certifies half of the subsets outside $\{B=0\}$. At each subset size coprime to $n$ ($n\ge4$) some synchronizing Eulerian binary automaton attains the constant $n-1$; whether only there is open. No reset bound follows: Černý's automata have subsets not extending within $n-1$.