An improvement to a recent upper bound for synchronizing words of finite automata

TL;DR

This paper improves the upper bound on the length of the shortest reset word in finite automata from approximately 0.1664 to 0.1654, refining the theoretical understanding of automata synchronization.

Contribution

It introduces a modified approach and a new counting argument that yields a tighter upper bound on reset word length.

Findings

Upper bound improved to 0.1654

New counting argument developed

Refinement of previous theoretical bounds

Abstract

It has been known since the 60's that any complete discrete $n$-state automaton admits a reset word of length not exceeding $\alpha n^3+o(n^3)$ for some absolute constant $\alpha$. J.-E. Pin and P. Frankl proved this statement with $\alpha=1/6=0.1666...$ in 1982, and this bound remained best known until 2017, when M. Szyku\l{}a decreased its value to $\alpha\approx0.1664$. In this note, we present a modification to the latest approach and develop a different counting argument which leads to a more substantial improvement of $\alpha\leqslant 0.1654$.

Equations

$$l+\min \left\{\frac{(r+1)(r+2)}{2},2(s_1+2s_2+3s_3+\dots +r{s}_r)+2r\right\}.$$

$$\frac{7}{48}n^{3}+2{\sum}\limits_{r=\rho}^{\lfloor n/2\rfloor}\min\left\{\frac{r^{2}}{4},1s_{1}+\ldots+rs_{r}\right\}+3n^{2}.$$

$$\varphi(s_{1},\ldots,s_{k}):={\sum}\limits_{r=\rho}^{k}\min\left\{\frac{r^{2}}{4},1s_{1}+\ldots+rs_{r}\right\}\leqslant\frac{15\,625\,n^{3}}{1\,597\,536}+o(n^{3}).$$

$$\left(\frac{7}{48}+\frac{2\cdot 15625}{1597536}\right)n^3+o(n^3)$$

$$1s_{1}+\ldots+\tau s_{\tau}\leqslant\tau^{2}/4\,\,\,\,\,\,\,\,\mbox{for all $\tau\in\{\rho,\ldots,k\}$}$$

$$\alpha :=1s_1+\dots +t{s}_t>t^2/4\text{mbox}and1s_1+\dots +r{s}_r⩽r^2/4\text{mbox}forallr<t.$$

$${\sigma }_{\beta }^{\prime }={\sigma }_{\beta }-\epsilon ,{\sigma }_{\nu }^{\prime }={\sigma }_{\nu }+\epsilon (\nu -\beta +1),{\sigma }_{\nu +1}^{\prime }={\sigma }_{\nu +1}-\epsilon (\nu -\beta )$$

$$\phi (\sigma )=\sum _{r=\rho }^k(k-r+1)r{\sigma }_r$$

$${\sigma }_r=0.5-0.25/r\text{mbox}forallr\in \{\beta +2,\dots ,\gamma -1\}.$$

$${\sigma }_{\beta }+{\sigma }_{\beta +1}⩾0.25(\beta +1).$$

$$\beta ⩾0.25(\beta +1)+0.5(\gamma -\beta -2)-0.25\ln n.$$

$$\phi (\sigma )⩽\psi (\beta ,\gamma ):=(-2{\beta }^3-4{\gamma }^3+3{\gamma }^{2}n+6{\gamma }^2+6\gamma +2)/24.$$

Full text

1. Introduction

Let $\mathcal{A}=(Q,\Sigma,\delta)$ be a deterministic finite automaton, where $Q$ is a finite set of states, $\Sigma$ is a finite alphabet, and $\delta:Q\times\Sigma\to Q$ is a transition function, which assigns a mapping $Q\to Q$ to every letter of $\Sigma$. This function naturally extends to an action $Q\times\Sigma^{*}\to Q$ of the free monoid $\Sigma^{*}$ on $Q$, and this action is still denoted by $\delta$. For a subset $S\subseteq Q$ and a word $w\in\Sigma^{*}$, we define $S\cdot w$ as the set of all images $s\cdot w$ of elements $s\in S$ under $w$. The cardinality of $Q\cdot w$ is called the rank of a word $w$, and the rank of an automaton is defined as the smallest possible rank of a word. An automaton $\mathcal{A}$ of rank one is called synchronizing, and the length of the shortest rank-one words is called the reset threshold of $\mathcal{A}$ and denoted by $\operatorname{rt}(\mathcal{A})$.

Upper bounds for reset thresholds of synchronizing automata were a topic of extensive research in the last 50 years, and one of the main goals of this study is a famous conjecture stating that $\operatorname{rt}(\mathcal{A})\leqslant(n-1)^{2}$ for any synchronizing $n$-state automaton $\mathcal{A}$; this statement was considered many years ago by different authors and became known as the Černý conjecture (see a historical survey in [12]). There are a lot of progress on this question for different special classes of automata [6, 8, 10], but the general version of the Černý conjecture remains wide open. The cubic upper bounds on the reset threshold, that is, inequalities of the form $\operatorname{rt}(\mathcal{A})\leqslant\alpha n^{3}+o(n^{3})$ for some fixed $\alpha$, have been known since 1966, see [7]. After a series of improvements [1, 2, 5, 6], the progress stuck for 35 years on the celebrated $\alpha=1/6=0.1666...$ bound of J.-E. Pin and P. Frankl [3, 6]. In 2011, A. Trahtman [11] discovered an idea of how to find a relatively short word of rank at most $n/2$, and M. Szykuła [9] combined it with a neat linear algebraic argument and finally improved the upper bound to $\alpha\approx 0.1664$ in 2017. The purpose of this note is to modify the approach of [9] and get a more substantial improvement of $\alpha\leqslant 0.1654$.

2. Modifying the method

From now on, we denote by $\mathcal{A}=(Q,\Sigma,\delta)$ a synchronizing automaton with $n$ states, and we define the corank of a word $w\in\Sigma^{*}$ as $n-\operatorname{rk}w$. Our aim is to give a relevant modification of the following proposition, which plays a crucial role in [9].

Theorem 1**.**

(Lemma 2 in [9].) Let $A$ and $S$ be subsets of $Q$ satisfying $\varnothing\subsetneq A\subsetneq S$. Suppose that there is a word $w\in\Sigma^{*}$ such that $A\nsubseteq S\cdot w$. Then there exists a word $w$ of length at most $n-|A|$ satisfying either (1) $A\nsubseteq S\cdot w$ or (2) $|S\cdot w|<|S|$.

In [9], a successive application of Theorem 1 was used to construct a word $\omega$ that satisfies $A\nsubseteq S\cdot\omega$. At the first glance, the following theorem may look like a mere reformulation of this technique avoiding a direct mention of a successive application. However, it gives a more explicit description of a desired word $\omega$ which will give a significant improvement on the bound of [9] later in this note.

Theorem 2**.**

Let $u\in\Sigma^{*}$ be a word of length $l$ and corank $r\in[1,n/2-1]$. Assume that, for an integer $\lambda$, there exists a word $v$ of length $\lambda$ such that $\operatorname{rk}v\leqslant\operatorname{rk}v^{\prime}$ for any word $v^{\prime}$ of length at most $\lambda+2r$. Then there is a word of length at most $l+\lambda+2r$ and corank at least $r+1$.

Proof.

For a state $\sigma$ in $Q\cdot u$, we denote by $\sigma\cdot u^{-1}$ the preimage of $\sigma$ under the mapping $q\to q\cdot u$. Let $A$ be the union of all those preimages $\sigma\cdot u^{-1}$ which are singleton sets; according to Lemma 7 in [9], one has $|A|\geqslant n-2r$.

Now we want to find a word $w$ of length at most $2r$ such that $A\nsubseteq Q\cdot v\cdot w$, which would allow us to find an element $a\in A$ satisfying $a\notin Q\cdot v\cdot w$, which would imply $a\cdot u\notin Q\cdot v\cdot w\cdot u$ and thus lead to a desired conclusion $Q\cdot u\supsetneq Q\cdot v\cdot w\cdot u$. Such a word $w$ is found immediately if $A$ and $S:=Q\cdot v$ satisfy the assumptions of Theorem 1, because the second possibility of its conclusion means that $|Q\cdot v\cdot w|<|Q\cdot v|$ and contradicts the assumption of the current theorem.

As to the assumptions of Theorem 1, the one in the second sentence holds because our automaton is synchronizing. In particular, there should be a letter $b\in\Sigma$ such that $A\cdot b\neq A$. So if $A$ was equal to $Q\cdot v$, then we could have taken $w=b$ and proceed as in the previous paragraph, and, similarly, if $A$ was not a subset of $Q\cdot v$ at all, we could have taken $w$ to be the empty word and do the same thing. ∎

One more theorem is needed before we can proceed to counting — we cannot improve on the Pin–Frankl bound without using the Pin–Frankl bound.

Theorem 3**.**

[3, 6]** Let $u\in\Sigma^{*}$ be a word of length $l$ and corank $r\leqslant n-2$. Then there is a word of length at most $l+(r+1)(r+2)/2$ and corank at least $r+1$.

3. Counting

As Theorem 2 suggests, we are going to study the gaps between the smallest lengths of words with ranks taking consecutive pairs of values. Formally speaking, we denote by $\lambda_{i}$ the smallest length of a word with corank at least $i\in\{0,\ldots,n-1\}$; we obviously have $0=\lambda_{0}\leqslant\ldots\leqslant\lambda_{n-1}=\operatorname{rt}(\mathcal{A})$. We also write $\lambda_{n}=+\infty$ and define $\rho$ as the smallest corank satisfying $\lambda_{\rho+1}-\lambda_{\rho}>n$.

Observation 4**.**

We have $\lambda_{\rho}<n^{2}$.

Further, we set $\delta_{j}=\lambda_{j+1}-\lambda_{j}$ for any $j\in\{0,\ldots,\rho\}$, and, for any integer $r\leqslant n/2$, we define the quantity $s_{r}$ as the number of those $j\in\{0,\ldots,\rho\}$ which satisfy $\delta_{j}\in\{2r-1,2r\}$. Let us translate Theorems 2 and 3 to this language.

Theorem 5**.**

Let $u\in\Sigma^{*}$ be a word of length $l$ and corank $r\in[1,n/2-1]$. Then there is a word of corank at least $r+1$ and length not exceeding

[TABLE]

Proof.

We are allowed to put the first argument of $\min$ by Theorem 3 immediately. Further, let us pick a word $v$ of corank $\tau$ and length $\lambda_{\tau}$, where $\tau$ is the minimal index for which $\delta_{\tau}$ exceeds $2r$. The length of $v$ does not exceed the sum of all the $\delta_{j}$’s not exceeding $2r$, which is at most $2(s_{1}+2s_{2}+3s_{3}+\ldots+rs_{r})$. Also, we cannot get a word of rank less than $\operatorname{rk}v$ unless we take $\delta_{\tau}>2r$ more letters than $v$ has — therefore, we can apply Theorem 2 and justify the second argument of $\min$. ∎

Corollary 6**.**

The reset threshold of $\mathcal{A}$ does not exceed

[TABLE]

Proof.

We use Observation 4 to get a word of corank at least $\rho$ and length at most $n^{2}$, then we upgrade it to a word of rank $\leqslant\lceil n/2\rceil$ by a successive application of Theorem 5, and then we construct a synchronizing word by Theorem 3 (the cost of this last step is $\leqslant 7n^{3}/48$ additional letters). Also, the expressions under the minimum were simplified by isolating the $O(n^{2})$ terms in the last summand. ∎

The following statement is going to complete the proof of our main result.

Proposition 7**.**

Let $n$ and $\rho<n/2$ be positive integers, let $k=\lfloor n/2\rfloor$. Let $s_{1},\ldots,s_{k}$ be nonnegative real numbers satisfying $s_{1}+\ldots+s_{k}\leqslant\rho$. Then

[TABLE]

The numbers $s_{1},\ldots,s_{k}$ appearing in Corollary 6 are clearly nonnegative and have the sum not exceeding $\rho$, so we can apply Proposition 7 and get

[TABLE]

or $0.1654n^{3}+\operatorname{O}(1)$ as an upper bound for the reset threshold of $\mathcal{A}$.

4. Proving Proposition 7

The last section is devoted to a solution of the optimization problem appearing in Proposition 7. First, we restrict our attention to a certain special case.

Claim 8**.**

It is sufficient to prove Proposition 7 under the additional assumptions of $s_{1}=\ldots=s_{\rho-1}=0$ and

[TABLE]

(where the latter says that the minimum is always attained at the second argument).

Proof.

Let us define $s_{r}^{\prime}=0$ for $r<\rho$, $s_{r}^{\prime}=s_{r}$ for $r>\rho$, and $s_{\rho}^{\prime}=(1s_{1}+\ldots+\rho s_{\rho})/\rho$. The new values are nonnegative and sum to at most $s_{1}+\ldots+s_{k}\leqslant\rho$, so they satisfy the assumptions of Proposition 7. Also, we have $1s_{1}+\ldots+rs_{r}=1s_{1}^{\prime}+\ldots+rs_{r}^{\prime}$ for all $r\geqslant\rho$, so the arguments of the minimuma do not change, and we can pass to $(s_{1}^{\prime},\ldots,s_{k}^{\prime})$ without loss of generality.

Now, for a tuple $s=(s_{1},\ldots,s_{k})$ not satisfying one of the conditions (8), we define $t(s)$ as the smallest $\tau$ for which it fails. In other words, we have

[TABLE]

Then we set $s_{t}^{\prime}=s_{t}+t/4-\alpha/t$ and $s_{r}^{\prime}=s_{r}$ for $r\notin\{\tau,\tau+1\}$, and also, if $t\neq k$, we define $s_{t+1}^{\prime}=s_{t+1}-t/4+\alpha/t$. The values $s_{r}^{\prime}$ are again nonnegative and sum to at most $s_{1}+\ldots+s_{k}\leqslant\rho$, so they satisfy the assumptions of Proposition 7. Also, we have $\varphi(s^{\prime})\geqslant\varphi(s)$ because the summands with $r\in[\rho,t]$ did not change while the $(t+1)$st and later summands could not have decreased. Finally, we note that, even if $s^{\prime}$ still does not satisfy some of the conditions (8), we still can prove the second statement of the current theorem by induction because $t(s^{\prime})>t(s)$.∎

From now on, we assume that the conditions $s_{1}=\ldots=s_{\rho-1}=0$ and (8) hold; we also recall that $s_{r}\geqslant 0$ and $s_{1}+\ldots+s_{k}\leqslant\rho$. We call the set of all tuples $(s_{1},\ldots,s_{k})$ satisfying these conditions a feasible set; Claim 8 allows us to restrict Proposition 7 to it. The feasible set is compact (for any fixed $\rho$), so the function $\varphi$ should have a maximum point $\sigma=(0,\ldots,0,\sigma_{\rho},\ldots,\sigma_{k})$. Let $\beta$ and $\gamma$ be, respectively, the minimal and maximal indices $i$ satisfying $\sigma_{i}\neq 0$.

Claim 9**.**

We have $1\sigma_{1}+\ldots+r\sigma_{r}=r^{2}/4$ for all $r\in\{\beta+1,\ldots,\gamma-1\}$.

Proof.

Assume the converse and find the maximal $\nu\in\{\beta+1,\ldots,\gamma-1\}$ for which $1s_{1}+\ldots+\nu s_{\nu}<\nu^{2}/4$. Then we pick a sufficiently small $\varepsilon>0$ and define

[TABLE]

and $\sigma^{\prime}_{r}=\sigma_{r}$ for $r\notin\{\beta,\nu,\nu+1\}$. The tuple $\sigma^{\prime}=(\sigma_{1}^{\prime},\ldots,\sigma_{k}^{\prime})$ sums to $\sigma_{1}+\ldots+\sigma_{k}\leqslant\rho$, and its coordinates are nonnegative for a sufficiently small $\varepsilon$. Further, the sum as in (8) could not have increased for $\tau<\nu$; the same sum but with $\tau=\nu$ has changed by something proportional to $\varepsilon$ and so it could not have overcome $\nu^{2}/4$. Finally, such a sum with $\tau>\nu$ did not change as we can check directly, so the tuple $\sigma^{\prime}$ belongs to the feasible set. Finally, we have

[TABLE]

if $\sigma$ is in the feasible set, and one can get the inequality $\varphi(\sigma^{\prime})>\varphi(\sigma)$, which contradicts the maximality of $\sigma$. (Such an inequality can be deduced by a straightforward computation, but let us point it out that it follows from the strict concavity of the sequence of the coefficients of $\sigma_{\rho},\ldots,\sigma_{k}$ in the above expression for $\varphi$.) ∎

Now we are going to employ Claim 9 to complete the proof of Proposition 7. First, we have $r\sigma_{r}=r^{2}/4-(r-1)^{2}/4$ or

[TABLE]

Secondly, we have $\beta\sigma_{\beta}+(\beta+1)\sigma_{\beta+1}=(\beta+1)^{2}/4$ or

[TABLE]

Summing the inequalities (4) over all $r\in\{\beta+2,\ldots,\gamma-1\}$ and adding the inequality (4), we get at most $\sigma_{1}+\ldots+\sigma_{k}\leqslant\rho\leqslant\beta$ on the left-hand side, or

[TABLE]

Now we estimate $\varphi(\sigma)$ using Claim 9 again. Clearly, the summands corresponding to $r<\beta$ are zero in the definition (7) of $\varphi$, the summands with $r\in\{\beta,\ldots,\gamma\}$ cannot exceed $0.25r^{2}$, and the summands with $r>\gamma$ are at most $0.25\gamma^{2}$. We get $\varphi(\sigma)\leqslant\sum_{r=\beta}^{\gamma}0.25r^{2}+0.25(0.5n-\gamma)\gamma^{2}$, or

[TABLE]

It remains to maximize $\psi(\beta,\gamma)$ subject to $0\leqslant\beta\leqslant\gamma\leqslant 0.5n$ and (4); these inequalities define a quadrilateral $\Delta$ with vertices $(0,0)$, $(0,0.5\ln n+1.5)$, $(0.5n,0.5n)$, $(0.2n-0.2\ln n-0.6,0.5n)$. Being strongly monotone in $\beta$, the function $\psi$ cannot have a maximum inside $\Delta$, so it remains to perform a basic calculus task and maximize $\psi$ on the edges. The computation shows that the maximum of $\psi$ on $\Delta$ is attained at the point $(25n/129,125n/258)+o(n)$ and confirms that it is equal to the right-hand side of (7). Therefore, the proof of our main result is complete.

As a final remark, let us note that we were not interested to optimize the $o(n^{3})$ term of our bound (3), but a direct computation of the previous paragraph gives an $O(n^{2}\log n)$ estimate of this term. A more careful application of Claim 9 would make it $O(n^{2})$, with explicit and reasonably small coefficients of the powers of $n$.