AI 中文总结
该研究针对下推自动机的转折复杂度，证明了有界语言相关的转折数量判定问题可解，其转折数随输入长度线性增长，且一般情况与有界语言的转折特性存在差异。
AI 中文摘要
下推自动机计算中的转折（reversal）指的是栈存储高度从上升阶段切换至下降阶段的操作。给定一个下推自动机，我们首先针对其语言中的每个字符串，研究其接受计算中转折的最小数量（弱度量）。我们证明，对于任意给定的k≥0，判断一个下推自动机是否以常数个转折接受有界语言、是否以k个转折接受有界语言是可判定的；这与一般情况形成对比——一般情况下除了以0个转折接受（该问题可判定）外，其余此类问题均不可判定。此外，我们证明当接受有界语言所需的转折数量不受任何常数限制时，其会随输入长度线性增长，这同样与一般情况形成对比：一般情况下，对于每个非负整数k，都存在一种语言，其接受所需的转折数量为log的k次复合的阶。我们还证明，当考虑所有接受计算的代价（接受度量）时，即使不限制为有界语言，转折数量的线性下界（不受任何常数限制）仍然成立。
英文摘要
A turn or reversal in a computation of a pushdown automaton is a switch from a phase in which the height of the pushdown store increases to a phase in which it decreases. Given a pushdown automaton, we first consider, for each string in its language, the minimum number of turns made in accepting computations (weak measure). We prove that it is decidable whether a pushdown automaton accepts a bounded language in a constant number of turns and whether it accepts a bounded language in k turns, for any given k>=0. This is in contrast to the general case, in which these problems are known to be undecidable, with the exception of acceptance in 0 turns, which is decidable. Furthermore, we prove that when the number of turns sufficient to accept a bounded language is not limited by any constants, it linearly grows with respect to the input length. Also this is in contrast with the general case where, for each nonnegative k, there exists a language for which the number of turns necessary and sufficient is of the order of log^(k), the k times composition of the logarithm with itself. We also prove that, when the costs of all accepting computations are taken into account (accept measure), a linear lower bound for the number of turns, if not limited by any constants, holds even removing the restriction to bounded languages.
CommentsIn Proceedings AFL 2026, arXiv:2608.23071
Journal refEPTCS 451, 2026, pp. 261-274
