On the Interplay
Between Babai and Černý’s Conjectures
Abstract
Motivated by the Babai conjecture and the Černý conjecture, we study the reset thresholds of automata with the transition monoid equal to the full monoid of transformations of the state set. For automata with states in this class, we prove that the reset thresholds are upperbounded by and can attain the value . In addition, we study diameters of the pair digraphs of permutation automata and construct state permutation automata with diameter .
1 Background and Overview
^{0}^{0}footnotetext: A short version of this work has been presented at the conference DLT 2017.Completely reachable automata, i.e., deterministic finite automata in which every nonempty subset of the state set occurs as the image of the whole state set under the action of a suitable input word, appeared in the study of descriptional complexity of formal languages [26] and in relation to the Černý conjecture [13]. In [6] an emphasis has been made on automata in this class with minimal transition monoid size. In the present paper we focus on automata being in a sense the extreme opposites of those studied in [6], namely, on automata of maximal transition monoid size. In other words, we consider automata with full transition monoid, i.e., transition monoid equal to the full monoid of transformations of the state set; clearly, automata with this property are completely reachable. There are several reasons justifying special attention to automata with full transition monoid. First, as observed in [6], the membership problem for this class of automata is decidable in polynomial time (of the size of the input automaton) while the complexity of membership in the class of all completely reachable automata still remains unknown. Second, this class contains automata that correspond to Brzozowski’s most complex regular languages [7] and to other regular languages that play a distinguished role in descriptive complexity analysis. Finally, and most importantly from our viewpoint, automata with full transition monoid are synchronizing and their synchronization issues constitute a sort of meeting point for two famous open problems—the Babai conjecture and the Černý conjecture. Next, we recall these conjectures and outline the contribution of the present paper in view of these problems.
1.1. The Babai Conjecture.
Let be a set of generators of a finite group . The Cayley graph consists of as the set of vertices and the edges for all , . The diameter of is the maximum among the lengths of shortest paths between any two vertices. In group theory terms, the diameter of is the smallest such that every can be represented as , where and for all . The diameter of is the maximal diameter of among all generating sets of . The notion of group diameter is related to the growth rate in groups, expander graphs, random walks on groups and their mixing times, see, e.g., [33, 23]. Recently, the following conjecture received significant attention:
Conjecture 1 (Babai [4])
The diameter of each nonabelian finite simple group does not exceed , where the implied constant is absolute.
Note that for the case of the symmetric group , this conjecture readily implies . (The group is not simple but for it contains a nonabelian simple subgroup of index 2.)
The Babai conjecture was proved for various classes of groups, but despite intensive research effort it remains open, see [22] for an overview. In the case of , a recent breakthrough gives only a quasipolynomial upper bound, namely, , and it relies on the Classification of Finite Simple Groups [22]. It is even more astonishing if we compare it to the best known lower bound in this case: for the classical set of generators consisting of the transposition and the full cycle , every permutation in can be expressed as a product of at most (asymptotically) generators [40].
1.2. The Černý Conjecture.
Recall that a deterministic finite state automaton (DFA) is a triple^{1}^{1}1As initial and final states play no role in our considerations, we omit them. , where is a finite set of states, is a finite set of input symbols called the alphabet, and is a function called the transition function. A word is a sequence of letters from the alphabet. The length of a word is the number of its letters. We can look at as the result of the action of the letter at the state . We extend this action to the action of words over on denoting, for any word and any state , the state resulting in successive applications of the letters of from left to right by . For a subset , we write for the set .
A DFA is called synchronizing if there exist a word and a state such that . Any such word is called a synchronizing or reset word. The minimum length of reset words for is called the reset threshold of and is denoted by . Synchronizing automata appear in various branches of mathematics and are related to synchronizing codes [5], part orienting problems [27, 28], substitution systems [16], primitive sets of matrices [19], synchronizing groups [3], convex optimization [20], and consensus theory [11].
If the conjecture holds true, then the value is optimal, since for every there exists an state automaton with the reset threshold equal to [9].
The Černý conjecture has gained a lot of attention in automata theory. It has been shown to hold true in various special classes[14, 24, 31, 36, 21, 35], but in the general case, it remains open for already half a century. For more than 30 years, the best upper bound was , obtained in [30, 15] and independently in [25]. Recently, a small improvement on this bound has been reported in [37]: the new bound is still cubic in but improves the coefficient at by . A survey on synchronizing automata and the Černý conjecture can be found in [38].
In order to make the relationship between the Černý and the Babai conjectures more visible, we borrow from [2] the idea of restating the former in terms similar to those used in the formulation of the latter. Let be the full transformation monoid of an element set . A transformation is a constant if there exists such that for all we have . We can state the Černý conjecture as follows: for every set of transformations , if the submonoid generated by contains a constant, then there exists a constant such that , where and for all . It is easy to see that this formulation is equivalent to the original one by treating the letters of an automaton as the transformations of its state set since reset words precisely correspond to constant transformations.
1.3. Our Contributions.
The first part of our paper is devoted to the following hybrid Babai–Černý problem^{2}^{2}2During the preparation of this paper we discovered that the same question was also posed in [34, Conjecture 3], though its connection with Babai’s problem was not registered there.: given a set of generators of the full transformation monoid , what is the length of the shortest product with which is equal to a constant? Namely, we are interested in the bounds on that depend only on . The hybrid Babai–Černý problem is a special case of the Černý problem. Indeed, it is a restriction to the class of DFAs with the transition monoid, i.e., the transformation monoid generated by the actions of letters, equal to . Of course, the general cubic upper bound is valid, but not the lower bound, since the Černý automata do not belong to this class (even though they are completely reachable, see [6]). In Section 2 we establish that the growth rate of is , more precisely, we show that . We also present the exact values of for small values of resulting from our computational experiments. Our contribution can be also seen as a progress towards resolution of Conjecture 3 from [34].
The second part of our paper is devoted to a “local” version of the Babai problem where we restrict our attention to the action on the set of (unordered) pairs. Let be a set of permutations from . The pair digraph consists of pairs as the set of vertices and the edges for all and . The diameter of , denoted , is the maximum among the lengths of shortest (directed) paths between any two vertices. We study the behavior of in terms of . The problem comes from analysis of certain aspects of Markov chains and group theory [17], but our interest in it is mainly motivated by its importance for the theory of synchronizing automata. Indeed, every synchronizing automaton must have a letter , say, whose action merges a pair of states. Thus, one can construct a reset word for by successively moving pairs of states to a pair merged by . If possesses sufficiently many letters acting as permutations (as automata with the full transition monoid do), one can move pairs by these permutations, and hence, upper bounds on the diameter of the corresponding pair digraph induce upper bounds on .
Clearly, for all . In Section 3 we establish the lower bound on by presenting a series of examples with only two generators for every odd .
1.4. Related Work.
The diameters of groups and semigroups constitute a relatively well studied topic. A general discussion on diameters and growth rates of groups can be found in [23]. Various results about the diameter of and its submonoids are described in [34, 29]. The length of the shortest representation of a constant (including the case of partially defined transformations) is typically studied in the framework of synchronizing automata, see [38, 39, 1].
2 Automata with Full Transition Monoid
2.1. Naïve Construction.
Recall that, on the one hand, the Černý automata from [9] have two letters of which one acts as a cyclic permutation and the other fixes all states, except one, which is mapped to the next element in the cyclic order defined by the cyclic permutation. On the other hand, the extremal case of the Babai conjecture for is composed of a cyclic permutation and the transformation which fixes all elements except two, which are neighbors in the cyclic order defined by the cyclic permutation. Therefore, one could wonder if a combination of these transformations could result in a DFA with both large reset threshold and full transition monoid.
The construction is defined as follows. There are states and three letters , , and . The letter acts as a cyclic permutation on the states, following their indices. The letter fixes all states, except , which is mapped to by . The letter fixes all states, except and , for some , which are swapped by . The resulting automaton is shown in Fig. 1. We notice that if we remove the letter , we obtain the automaton from the Černý family providing the largest currently existing lower bound in the Černý problem, and if we remove the letter , we obtain a generating set of the group providing the largest currently existing lower bound in the Babai problem for . Also observe that in the case where , our automaton is nothing but Brzozowski’s “Universal Witness” [7] recognizing the most complex regular language, i.e., the language witnessing at once practically all tight lower bounds found for the state complexity of various operations with regular languages, see [7, Theorem 6]. The next result shows that, however, the reset threshold of the automaton is upperbounded by , while, as we show later, among automata with full transition monoid there exist ones whose reset threshold is a quadratic function of their state number.
Theorem 2.1
The automaton has a reset word of length at most .
Proof
Recall that we aim to show that, for each , the automaton has a reset word of length at most . It is easy to see that the word of length resets the automaton , so that we assume that in the rest of the proof.
We construct a word letterbyletter in several rounds, starting with the empty word. The main parameter in our construction is the current image of the state set of under the action of the word constructed so far; let stand for this image. (Thus, we have at the beginning, and becomes a singleton at the end of the process.) It is quite helpful to visualize as the set whose states bear certain tokens. If one colors states covered by tokens lightgray, then Fig. 1 represents the initial position while Fig. 2 shows some intermediate situation.
When a letter is applied to , the token that covers , say, moves to the state ; in more visual terms, the token “slides” along the arrow representing the transition . If two tokens arrive at the same state, which happens whenever both and bear tokens and the letter is applied, we remove one of the tokens.
In the course of our construction, rounds of two sorts alternate: merging, in which only ’s and ’s are applied to , and pairing, in which only ’s and ’s are applied. We call a state from isolated if both its neighbor states (with respect to the cyclic order defined by ) are not in . A merging round starts whenever and has at most one isolated state, and it lasts while contains nonisolated states; a pairing round starts whenever and all states in are isolated, and it lasts while contains more than one isolated state. Every round consists of a number of steps, in each of which we choose a letter, append the chosen letter to the word and update the set by applying the letter to it. The choice is done according to one of the two following rules (M) and (P) used during merging and pairing rounds, respectively:

is chosen whenever ; otherwise is chosen;

is chosen whenever , but (so that is isolated); otherwise is chosen.
Clearly, at the beginning no state is isolated, and hence, the first round of our construction must be merging. It amounts to an immediate calculation to see that by the end of the first round, we have and .
Now we are going to verify two claims.
Claim 1. If before any of the next merging rounds, then at the end of the round.
We say that two neighbor states form an isolated couple if each of these states has exactly one neighbor in .
Claim 2. If before a pairing round, then at the end of the round is partitioned in either isolated couples (if is even) or isolated couples and one isolated state (if is odd).
First we prove Claim 2. The distance from to is . We order tokens that cover the states in according to the distance from their states to the state : the th token is the one that covers the state with the distance to , where . Now consider the evolution of the set under the choice of letters according to the rule (P). Clearly, the first choices are all ’s. After that the first token reaches . Since the action of translates the set , without affecting distances between its states, all states in remain isolated at this point. In particular, is isolated, and hence, (P) forces the letter to be applied. This moves the first token “backwards” to the state while all other tokens keep their positions. The next letter to be applied is , and its application moves all tokens one step “forwards” so that the token from returns to . Clearly, the distance from the state that holds the second token to becomes after these two moves. If the state remains isolated, another application of is invoked, followed by another application of , and this results in a further decrement of the distance from the state that holds the second token to . Eventually, after the suffix is appended to , the second token reaches the state . At this moment, the third token (if it exists) covers a state with distance to whence form an isolated couple in . The two tokens covering these states will then remain adjacent till the end of the round.
If or , we are done. If , we proceed in the same way. Namely, the next choices are all ’s. After that the third token reaches . Except the first two, all other tokens remain isolated. Now (P) forces and to be alternatively chosen times each. This makes the third token shuffle between and , while the fourth and the next tokens move steps “forwards”. After that form yet another isolated couple in , etc.
We have shown that at the end of the round, the set indeed consists of either isolated couples (if is even) or isolated couples and one isolated state (if is odd). Moreover, the suffix appended to during the round is of the form
(1) 
The letter occurs in this suffix times if is even and times if is odd, and the number of occurrences of is less than that of . Since , we conclude that the length of the suffix (1) is less than .
Now it is easy to prove Claim 1. In view of Claim 2, at the beginning of the round, the set consists of either isolated couples (if is even) or isolated couples and one isolated state (if is odd). If is an isolated couple, we say that is its left state. Now we order isolated couples in according to the distance from their left states to the state : the th couple is the one with the distance from its left state to , where . Consider the evolution of the set under the choice of letters according to the rule (M). The first choices are all ’s. After that the tokens that initially covered the states of the first isolated couple arrive at the states and , and hence, (M) forces the letter to be applied. This application removes the token from and does not change anything else. The state then becomes isolated. The next choices are again all ’s, and the successive applications of these ’s bring tokens that initially covered the states of the second isolated couple to the states and . Then, again, is chosen, removing the token from and creating yet another isolated state in , etc. At the end of the round, exactly one token from each isolated couple is removed and all remaining states are isolated. The number of these states is if is even or if is odd; in short, , as claimed.
Moreover, the suffix appended to during the round is of the form
(2) 
The letter occurs in this suffix times and the letter occurs times, whence the length of the suffix (2) is less than .
Claim 1, together with the observation we made about the first merging round, readily implies that the number of merging rounds is at most . Since merging and pairing rounds alternate, the total number of rounds is upperbounded by . As observed after the proofs of Claims 1 and 2, a suffix of length less than is appended to the current word during each round. Clearly, at the end of the process, becomes a reset word for , and by the construction the length of is less than . ∎
2.2. Random Sampling and Exhaustive Search.
Every DFA with the transition monoid necessarily has permutation letters that generate the whole symmetric group and a letter of rank (i.e., a letter whose image has elements). It is a well known fact that the converse is also true, i.e., the transition monoid of any automaton with permutation letters generating and a letter of rank is equal to , see, e.g., [18, Theorem 3.1.3].
Relying on a grouptheoretic result by Dixon [12], Cameron [8] observed that an automaton formed by two permutation letters taken uniformly at random and an arbitrary nonpermutation letter is synchronizing with high probability. We give an extension by using another nontrivial grouptheoretical result, namely, the following theorem by Friedman et al. [17]:
Theorem 2.2
For every and there is a constant such that for permutations of taken uniformly at random, the following property holds with probability tending to 1 as : for any two tuples of distinct elements in , there is a product of less than of the ’s which maps the first tuple to the second.
Corollary 1
There is a constant such that the reset threshold of an state automaton with two random permutation letters and an arbitrary nonpermutation letter does not exceed with probability that tends to 1 as .
Proof
Let stand for the automaton in the formulation of the corollary. We let be the nonpermutation letter and assume that the two permutation letters in have the property of Theorem 2.2 for with some constant . By Theorem 2.2 this assumption holds true with probability that tends to 1 as .
There exists two different states such that . The set contains less than elements. If , then is a reset word for . If , take two different states . By , there is a product of less than of the permutation letters such that for . Now . If , is a reset word for . If , we apply the same argument to a pair of different states in . Clearly, the process results in a reset word in at most steps while the suffix appended at each step is of length at most . Hence the length of the reset word constructed this way is at most .∎
Corollary 1 indicates that one can hardly discover an state automaton with the transition monoid equal to and sufficiently large reset threshold by a random sampling. Therefore, we performed an exhaustive search among all automata with two permutation letters generating and one letter of rank . Our computational results are summarized in Table 1.
Number of states  2  3  4  5  6  7 
Reset threshold  1  4  8  14  19  27 
Table 1: The largest reset thresholds of state automata two permutation letters generating and one letter of rank
As grows, the reset thresholds of the obtained examples become much smaller than . We were unable to derive a series of state threeletter automata with the transition monoid and quadratically growing reset thresholds. We suspect that the reset threshold of automata in this class is .
In the case of unbounded alphabet, for every , we present an state automaton with the transition monoid such that . The state set of is and the input alphabet consists of letters . The transition function is defined as follows:
Simply speaking, every letter for swaps the states and and fixes the other states. The letter brings both and to and fixes the other states. The automaton is depicted in Fig. 3.
Recall that a state of an DFA is said to be a sink state (or zero) if for every input letter . It is known that every state synchronizing automaton with zero can be reset by a word of length , cf. [31]. To show that this upper bound is tight for each , Rystsov [31] constructed an state and letter synchronizing automaton with zero which cannot be reset by any word of length less than . In fact, our automaton is a slight modification of as the latter automaton is nothing but without the letter .
Theorem 2.3
For every , the automaton has as its transition monoid and .
Proof
The letters generate because the product is a full cycle and any full cycle together with any transposition generates . Since the letter has rank , it together with generates .
The automaton is synchronizing because so is the restricted automaton , and because every reset word for resets as well. It remains to verify that the length of any reset word for must be at least . Let be a reset word of minimum length for . Since is the only nonpermutation letter, we must have for some such that . This is only possible when whence . Consider the function from the set of all nonempty subsets of into the set of nonnegative integers defined as follows: if , then . Clearly, and . For any set and any letter , we have since each letter only exchanges two adjacent states or maps and to . Thus, when we apply the word letterbyletter, the value of after the application of the prefix of of length cannot be less than . Hence, to reach the value , we need at least letters.∎
2.3. Upper Bound on the Reset Threshold.
We now provide a quadratic upper bound on the reset words of automata with the transition monoid equal to . Our proof is inspired by the method of Rystsov [32] adapted to our case.
Let be a DFA. Given a proper nonempty subset and a word over , we say that can be extended by if the cardinality of the set is greater than . Now assume that and the transition monoid of coincides with the full transformation monoid . Then there is a letter of rank . The set consists of a unique state, which is called the excluded state for and is denoted by . Furthermore, the set contains a unique state such that for some ; this state is called the duplicate state for and is denoted by . We notice that a nonempty subset can be extended by if and only if and . Moreover, if a word is a product of permutation letters, can be extended by the word if and only if and . To better understand which extensions are possible, we construct a series of directed graphs (digraphs) , , with the set as the vertex set.
The digraph has the set as its edge set. Let be the set of permutation letters of . Notice that generates the symmetric group . By we denote the set of words of length at most over the letters in . The digraph for has the edge set . The digraphs , , form a sort of stratification for the graph with the edge set ; the latter digraph has been studied in [32] and [6] (in the context of arbitrary completely reachable automata). Observe that none of the digraphs , , have loops.
Recall that a digraph is said to be strongly connected if for every pair of its vertices, there exists a directed path from the first vertex to the second. We need the two following lemmas.
Lemma 1
If the digraph is strongly connected, then every proper nonempty subset in can be extended by a word of length at most .
Proof
let be a proper nonempty subset in . If is strongly connected, there exists an edge that connects and in the sense that while . As , there exists a word such that . Then and , whence the word extends and has length at most .∎
Lemma 2
The digraph is strongly connected.
Proof
We start with showing that the digraph contains an oriented cycle.
Consider the underlying digraph of the automaton , i.e., the digraph with the vertex set and the edge set . This digraph is strongly connected since generates the whole symmetric group . Therefore, for every , there exists a directed path in from to . If one takes such a path of minimum length, it does not traverse any vertex in more than once, whence the length of the path is at most . Thus, the word belongs to and the pair is an outgoing edge of the vertex in the digraph . We see that every state in has an outgoing edge in . Now, we can walk along the edges of , starting at , which has the outgoing edge , until we reach an already visited state, thus getting an oriented cycle in the graph.
If is a digraph, we say that a vertex is reachable from a vertex if either or there is a directed path from to . The mutual reachability relation is an equivalence on the set , and the digraphs induced on the classes of the mutual reachability relation are either strongly connected or singletons (i.e., digraphs with 1 vertex and no edge). Slightly abusing terminology, we call these induced digraphs (including singletons) the strongly connected components of the digraph .
Consider the strongly connected components of the digraph and let denote their vertex sets. Without any loss we may assume that . Observe that since contains an oriented cycle which is not a loop whence at least one strongly connected component is nonsingleton. (Recall that digraphs of the form are loopless.) If , then already the digraph is strongly connected, and we are done. Otherwise we analyze the evolution of the partition of with into strongly connected components under the action of the letters in . Since generates the symmetric group , it cannot preserve any nontrivial partition of . Thus, there is a nonsingleton component among and a letter in whose action sends two elements of to different components, i.e. and for some .
By the definition of the sets , if , then . Therefore each edge from that connects some vertices in , , translates into an edge from that connects the images of these vertices in . Therefore, the digraphs of induced on the sets are either strongly connected or singletons. In particular, the digraph of induced on is strongly connected. Since and , the digraph of induced on the set also is strongly connected. This implies that the number of strongly connected components in is less than . If is not yet strongly connected, the same reasoning applied to its strongly connected components, shows that the number of strongly connected components in is less than , etc.
Since at each step the number of strongly connected components is reduced at least by , we conclude that we reach a strongly connected digraph in at most steps. Therefore, is strongly connected.∎
Theorem 2.4
Let be an state automaton with the transition monoid equal to . The reset threshold of is at most .
Proof
Let be a letter of rank and . We extend the set by , getting a subset with . Lemmas 1 and 2 imply that proper nonempty subsets in can be extended by words of length at most . Starting with , we extend subsets until we reach the full state set. Let be the word of length at most used for the th of these extensions and let be the number of the extensions. Observe that . Clearly, the word resets and has the length at most .∎
Remark 1
Let be an state DFA that has a letter of rank , and let be the subgroup of the symmetric group generated by the permutation letters from . Our proof of Theorem 2.4 actually works in the case if is a 2transitive group, that is, acts transitively on the set of ordered pairs of .
3 Bounds on the Diameter of the Pair Digraph
In this section we present a lower bound on the largest diameter of the pair digraph for . We proceed by presenting subsets for every odd whose diameter is . In order to simplify the presentation, we mostly use automata terminology and describe the corresponding examples as a family of automata (the input letters of form the subset ). We let and denote pairs of states such as simply by .
The automaton shown in Fig. 4 is the first of the family . The digraph of pairs of its states is shown in Fig. 5. One can verify that the shortest word mapping to has length 15.
The automata of the family are obtained recursively, starting with . From , we construct . The effect of the letters is the same for the states in and . The effect of the letters and at the states , , and is defined as follows: the letters mapping and to themselves in exchange with and with respectively in . The other letter maps and to themselves and , to and respectively . The result is shown in Fig. 6 (for ), in which stands for .