On the Probability of Being Synchronizable

We prove that a random automaton with n states and any fixed non-singleton alphabet is synchronizing with high probability. Moreover, we also prove that the convergence rate is exactly 1−Θ( 1 n ) as conjectured by Cameron [3] for the most interesting binary alphabet case. Finally, we describe a deterministic algorithm which decides whether a given random automaton is synchronizing in linear expected time. 1 Synchronizing automata Suppose A is a complete deterministic finite automaton whose input alphabet is A and whose state set is Q. The automaton A is called synchronizing if there exists a word w ∈ A whose action resets A, that is, w leaves the automaton in one particular state no matter at which state in Q it is applied: q.w = q.w for all q, q ∈ Q. Any such word w is called a reset word of A. For a brief introduction to the theory of synchronizing automata we refer reader to the survey [13]. Synchronizing automata serve as transparent and natural models of errorresistant systems in many applications (coding theory, robotics, testing of reactive systems) and also reveal interesting connections with symbolic dynamics and other parts of mathematics. We take an example from [1]. Imagine that you are in a dungeon consisting of a number of interconnected caves, all of which appear identical. Each cave has a common number of one-way doors of different colours through which you may leave; these lead to passages to other caves. There is one more door in each cave; in one cave the extra door leads to freedom, in all the others to instant death. You have a map of the dungeon with the escape door identified, but you do not know in which cave you are. If you are lucky, there is a sequence of doors through which you may pass which takes you to the escape cave from any starting point. The result of this paper is very positive; we prove that for an uniformly at random chosen dungeon (automaton) there is a life-saving sequence (reset word) with probability 1 − O( 1 n0.5c ) where n is the number of caves (states) and c is the number of colours (letters). Moreover, we prove that the convergence rate is tight for the most interesting 2-colour case, thus confirming Peter Cameron’s conjecture from [3]. Up to recently, the best results in this direction were much weaker: in [9] was proved that random 4-letter automata are synchronizing with probability p for a specific constant p > 0; in [8] was proved that if a random automaton with n states has at least 72 ln(n) letters then it is almost surely synchronizing. Recently, Nicaud [7] has shown (independently) by a completely different pure combinatoric techniques that a random n-state automaton with 2 letters is synchronizing with probability 1 − O(n 1 8). Our results give a much better convergence rate. 2 The probability of being synchronizable Let Q stand for {1, 2, . . . n} and Σn for the probability space of all unambiguous maps from Q to Q with the uniform probability distribution. Throughout this section let A = 〈Q, {a, b}〉 be a random automaton, that is, maps a and b are chosen independently at random from Σn. The underlying digraph of A = 〈Q,Σ〉 is a digraph denoted by UG(A) whose vertex set is Q and whose edge multi set is {(q, q.a) | q ∈ Q, a ∈ Σ}. In other words, the underlying digraph of an automaton is obtained by erasing all labels from the arrows of the automaton. Given a letter x ∈ Σ, the underlying digraph of x is the underlying digraph of the automaton Ax = 〈Q, {x}〉 where the transition function is the restriction of the original transition function to the letter x. Clearly each directed graph with n vertices and constant out-degree 1 corresponds to the unique map from Σn whence we can mean Σn as the probability space with the uniform distribution on all directed graphs with constant out-degree 1. Theorem 1. The probability of being synchronizable for 2-letter random automata with n states equals 1−Θ( 1 n ). Proof. Since synchronizing automata are necessary weakly connected, the following lemma gives the lower bound of the theorem. Lemma 1. The probability that A is not weakly connected is at least Ω( 1 n ). Proof. Let us count the number of automata having exactly one disconnected loop, that is the state having only (two) incoming arrows from itself. Such automata can be counted as follows. We first choose the state p of a disconnected loop in n ways. The transitions for this state is defined in the unique way. The number of ways to define transitions for any other state q is 1(n− 2) + (n− 2)(n− 1) = n(n− 2) because if a maps q to q then b can map q to any state except {p, q}; if a does not map q to {p, q} then b can map q to any state except {p, q}. Thus the probability of being such automata is equal