One Idea

In my first post I promised to give a polynomial-time procedure to compute the probability that an infinite word (chosen uniformly at random) is accepted by a given unambiguous Büchi automaton. For example, let's take this Büchi automaton again: Let's call the probability $x_1, x_2, x_3$, depending on the start state: \[ \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} := \begin{pmatrix} \text{Pr(state $1$ accepts a random word)}\\ \text{Pr(state $2$ accepts a random word)}\\ \text{Pr(state $3$ accepts a random word)} \end{pmatrix} \] I showed in the first post that $x_1, x_2, x_3$ are nonzero in this example. In order to compute the $x_i$ we set up a linear system of equations and solve it. We have \[ x_1 \ = \ \frac12 \cdot x_{1,a} + \frac12 \cdot x_{1,b}\,, \] where $x_{1,a}$ and $x_{1,b}$ denote the (conditional) probabilities that state $1$ accepts a random word, under the condition that the word starts with $a$ and $b$, respectively. $x_{1,b} = 0$ because state $

Computer science is about leveraging mathematical structure, for instance, in order to accelerate algorithms. More often than we'd like we can't accelerate much: For instance, checking two NFAs for language inclusion or equivalence is PSPACE-complete. Even the problem whether an NFA accepts all words is PSPACE-complete. For DFAs, those problems are easy. This means that for NFAs we can't do much better than determinizing (using the subset construction and incurring an exponential blowup) and solving the resulting DFA problem. In this post I'll discuss an automata problem that seems to call for determinization and is, in fact, PSPACE-complete for NFAs. But there is an efficient algorithm for unambiguous automata. Here is an example of an unambiguous automaton : The notion of accepting states is not used in this post. By unambiguousness I just mean that the automaton has no diamonds, that is, for any states $q, r$, any word $w$ labels at most one path from $q$ to