Given two weighted automata, we consider the problem of whether one is big-O of the other, i.e., if the weight of every finite word in the first is not greater than some constant multiple of the weight in the second. We show that the problem is undecidable, even for the instantiation of weighted automata as labelled Markov chains. Moreover, even when it is known that one weighted automaton is big-O of another, the problem of finding or approximating the associated constant is also undecidable. Our positive results show that the big-O problem is polynomial-time solvable for unambiguous automata, coNP-complete for unlabelled weighted automata (i.e., when the alphabet is a single character) and decidable, subject to Schanuel's conjecture, when the language is bounded (i.e., a subset of $w_1^*\dots w_m^*$ for some finite words $w_1,\dots,w_m$) or when the automaton has finite ambiguity. On labelled Markov chains, the problem can be restated as a ratio total variation distance, which, instead of finding the maximum difference between the probabilities of any two events, finds the maximum ratio between the probabilities of any two events. The problem is related to $\epsilon$-differential privacy, for which the optimal constant of the big-O notation is exactly $\exp(\epsilon)$.
翻译:在两个加权自动数据中, 我们考虑一个是否是另一个大的 O, 也就是说, 如果第一个有限的单词的重量不大于第二个的数倍。 我们显示, 问题是不可估量的, 即使是加权的自动数据即刻化, 贴上Markov 链条的标签。 此外, 即使已知一个加权的自动数据是另一个的 O, 找到或接近相关常数的问题也是不可估量的。 我们的积极结果显示, 大- O 的问题对于明确的自动数据来说, 是多元时间的可溶性, 并不大于一个常数的倍数。 我们显示, 这个问题是不可估量的自动数据( 例如, 当字母是一个单一的字符) 问题, 即使是在Schanuel 的推测下, 当语言被捆绑( 例如, 某定数单数的 $_ 1 \\\\\ o\ w_ m $ $ 美元) 时, 问题也是不可估量的 。 当Onalmaton 问题对于一个不精确的常数, 之间, 找到一个最精确的差 。