Decidability of the Equivalence of Multi-Letter Quantum Finite Automata

TL;DR

This paper investigates the decidability of equivalence for multi-letter quantum finite automata, providing a characterization of when two such automata are equivalent and designing an efficient polynomial-time algorithm for this problem.

Contribution

It establishes a decidability criterion for multi-letter QFA equivalence and introduces a polynomial-time algorithm, extending previous results for measure-once QFAs.

Findings

Two multi-letter QFAs are equivalent if and only if they are $(n^2m^{k-1}-m^{k-1}+k)$-equivalent.

The worst-case time complexity for checking equivalence is exponential.

A polynomial-time algorithm with complexity $O(m^{2k-1}n^{8}+km^kn^{6})$ is proposed for equivalence checking.

Abstract

Multi-letter {\it quantum finite automata} (QFAs) were a quantum variant of classical {\it one-way multi-head finite automata} (J. Hromkovi\v{c}, Acta Informatica 19 (1983) 377-384), and it has been shown that this new one-way QFAs (multi-letter QFAs) can accept with no error some regular languages $(a+b)^{*}b$ that are unacceptable by the previous one-way QFAs. In this paper, we study the decidability of the equivalence of multi-letter QFAs, and the main technical contributions are as follows: (1) We show that any two automata, a $k_{1}$-letter QFA ${\cal A}_1$ and a $k_{2}$-letter QFA ${\cal A}_2$, over the same input alphabet $\Sigma$ are equivalent if and only if they are $(n^2m^{k-1}-m^{k-1}+k)$-equivalent, where $m=|\Sigma|$ is the cardinality of $\Sigma$, $k=\max(k_{1},k_{2})$, and $n=n_{1}+n_{2}$, with $n_{1}$ and $n_{2}$ being the numbers of states of ${\cal A}_{1}$ and ${\cal A}_{2}$, respectively. When $k=1$, we obtain the decidability of equivalence of measure-once QFAs in the literature. It is worth mentioning that our technical method is essentially different from that for the decidability of the case of single input alphabet (i.e., $m=1$). (2) However, if we determine the equivalence of multi-letter QFAs by checking all strings of length not more than $ n^2m^{k-1}-m^{k-1}+k$, then the worst time complexity is exponential, i.e., $O(n^6m^{n^2m^{k-1}-m^{k-1}+2k-1})$. Therefore, we design a polynomial-time $O(m^{2k-1}n^{8}+km^kn^{6})$ algorithm for determining the equivalence of any two multi-letter QFAs. Here, the time complexity is concerning the number of states in the multi-letter QFAs, and $k$ is thought of as a constant.