On Two Algorithmic Problems about Synchronizing Automata

Abstract

Under the assumption P = N P, we prove that two natural problems from the theory of synchronizing automata cannot be solved in polynomial time. The first problem is to decide whether a given reachable partial automaton is synchronizing. The second one is, given an n-state binary complete synchronizing automaton, to compute its reset threshold within performance ratio less than d ln (n) for a specific constant d > 0.

Exact synchronization in partial deterministic automata

“…Namely, checking whether a given strongly connected PFA is exactly synchronizing can be done in polynomial time, and for strongly connected exactly synchronizing PFAs with n states, there exits a cubic in n upper bound on the minimum length of exactly synchronizing words-both these facts readily follow from a result in [32]. However, Berlinkov [3] has shown that in the absence of strong connectivity, testing a given PFA for exact synchronization becomes PSPACE-complete; he has also constructed is a series of n-state PFAs whose shortest exactly synchronizing words have length of magnitude 2 Ω(n) . Thus, for the general case, problems related to exact synchronization are no less complicated than those related to careful synchronization.…”

Strong Inapproximability of the Shortest Reset Word

“…Later, Gerbush and Heeringa [11] used the log n−approximation hardness of SetCover [9] to prove that O(log n)-approximation of the shortest reset word is NP-hard. Finally, Berlinkov [7] extended their result to hold even for the binary alphabet, and conjectured that a polynomial time O(log n)-approximation algorithm exists. We refute the conjecture by showing that, for every constant ε > 0, no polynomial time n 1−ε −approximation is possible unless P = NP.…”

“…Our construction will use Σ = {0, 1, 2}, which can be then reduced to the binary alphabet using the method of Berlinkov [7]. It is based on encoding every letter in binary and adding some intermediate states.…”

“…Lemma 2.1 (Lemma 7 of [7]) Suppose SynAppx({0, 1}, n α ) can be solved in polynomial time for some α ∈ (0, 1), then so can be SynAppx(Σ, O(n α )) for any Σ of constant size.…”

“…Proof: As shown in Lemma 7 of [7], given an n-state automaton A over an alphabet Σ, one can efficiently construct an automaton B onñ := 2|Σ|n states over the binary alphabet, such that Syn(A)t ≤ Syn(B) ≤ t(1 + Syn(A)), where t = log 2 |Σ| + 1. Then, if we can approximate Syn(B) within a factor ofñ α , we can compute in polynomial time an x such that Syn(B) ≤ x ≤ Syn(B)ñ α .…”