A maxmin problem on finite automata

We solve the following problem proposed by H. Straubing. Given a two letter alphabet A, what is the maximal number of states f(n) of the minimal automaton of a subset of An, the set of all words of length n. We give an explicit formula to compute f(n) and we show that 1 = lim n→∞ nf(n)/2n ≤ limn→∞ nf(n)/2 n = 2. The purpose of this note is to solve the following question, raised by H. Straubing. Let A = {a, b} be a two-letter alphabet. For each finite language L, denote by s(L) the number of states of the minimal (deterministic) automaton of L, and put f(n) = max{s(L) | L ⊂ A} The problem is to compute f(n) and to give, if possible, an asymptotic equivalent. We first recall some definitions (see [1] for more details). An automaton A = (Q,A, · , q0, F ) consists of a (finite) set of states Q, a finite set of letters A, an initial state q0 ∈ Q, a set of final states F ⊂ Q, and a partial function Q × A → Q denoted by (q, a) → q · a. This function is extended to a (partial) function Q×A → Q, called the transition function, by the rules (a) for every q ∈ Q, q · 1 = q, (b) for every q ∈ Q, u ∈ A and a ∈ A, q · (ua) = (q ·u)· a if (q ·u) and (q ·u)· a are defined, and q · (ua) is undefined otherwise. If the transition function is a total function, A is a complete automaton and it is uncomplete otherwise. The language accepted by A is the set L(A) = {u ∈ A | q0 ·u ∈ F} A state q is accessible (resp. coaccessible ) if q0 ·u = q (respectively q ·u ∈ F ) for some word u ∈ A. Two states q and q are equivalent (in A) if, for every word u ∈ A, ∗This research was supported by the “Programme de Recherches Coordonnées Mathématiques et Informatique”.