A new algorithm for testing if a regular language is locally threshold testable

A new algorithm is presented for testing if a regular language is locally threshold testable. The new algorithm is slower than existing algorithms, but its correctness proof is shorter. The proof idea is to restate the problem in Presburger arithmetic. A language L ⊆ A * is called locally threshold testable (LTT) if it is a Boolean combination of languages of the form: a) words that have w ∈ A * as a prefix; or b) words that have w ∈ A * as a suffix; or c) words that have w ∈ A * as an infix at least n times. For instance, the language a + b + a + b + is locally threshold testable, as witnessed by: " words that have a as a prefix, have ab as an infix exactly two times, and have ba as an infix exactly one time ". By Gaifman's theorem, locally threshold testable is equivalent to being definable in first-order logic with the successor relation on positions. This paper presents a new proof of the following theorem: Theorem 1 It is decidable if a regular language L is locally threshold testable. As observed by Beauquier and Pin in [1,2], the above problem is decidable by a result of Thérien and Weiss [8]. A polynomial time algorithm – with respect to the minimal deterministic automaton for L – was presented by Pin in [4,5]. The approach of [8,2,1,4,5] uses semigroup theory; the crux is that a language is locally threshold testable if and only if its syntactic semigroup satisfies certain swapping conditions. However, proving that the swapping conditions are sufficient requires involved combinatorics. The proof in this paper does not use semigroups, only Parikh images and Presburger arithmetic. The point is to provide an alternative proof of decidability, even if the algorithm is quite slow (several exponentials).