The equivalence problem of multitape finite automata
Theoretical Computer Science
A polynomial-time algorithm for the equivalence of probabilistic automata
SIAM Journal on Computing
The unsolvability of the Equivalence Problem for Λ-Free nondeterministic generalized machines
Journal of the ACM (JACM)
Matching is as easy as matrix inversion
STOC '87 Proceedings of the nineteenth annual ACM symposium on Theory of computing
Automata, Languages, and Machines
Automata, Languages, and Machines
More on Noncommutative Polynomial Identity Testing
CCC '05 Proceedings of the 20th Annual IEEE Conference on Computational Complexity
Derandomizing the Isolation Lemma and Lower Bounds for Circuit Size
APPROX '08 / RANDOM '08 Proceedings of the 11th international workshop, APPROX 2008, and 12th international workshop, RANDOM 2008 on Approximation, Randomization and Combinatorial Optimization: Algorithms and Techniques
Two-sided finite-state transductions
SWCT '63 Proceedings of the 1963 Proceedings of the Fourth Annual Symposium on Switching Circuit Theory and Logical Design
Elements of Automata Theory
Finite automata and their decision problems
IBM Journal of Research and Development
| Hi-index | 0.00 |
The decidability of determining equivalence of deterministic multitape automata (or transducers) was a longstanding open problem until it was resolved by Harju and Karhumäki in the early 1990s. Their proof of decidability yields a co-NP upper bound, but apparently not much more is known about the complexity of the problem. In this paper we give an alternative proof of decidability, which follows the basic strategy of Harju and Karhumäki but replaces their use of group theory with results on matrix algebras. From our proof we obtain a simple randomised algorithm for deciding equivalence of deterministic multitape automata, as well as automata with transition weights in the field of rational numbers. The algorithm involves only matrix exponentiation and runs in polynomial time for each fixed number of tapes. If the two input automata are inequivalent then the algorithm outputs a word on which they differ.