Reversible Languages Having Finitely Many Reduced Automata

TL;DR

This paper characterizes reversible regular languages with finitely many reduced reversible automata, providing an efficient decision procedure and advancing understanding of reversible automata's structural properties.

Contribution

It introduces a forbidden pattern characterization for these languages and offers an NL decision procedure, addressing an open problem in reversible automata theory.

Findings

Characterization of reversible languages with finitely many reduced automata

Introduction of a forbidden pattern for recognition

Efficient NL decision procedure for the class

Abstract

Reversible forms of computations are often interesting from an energy efficiency point of view. When the computation device in question is an automaton, it is known that the minimal reversible automaton recognizing a given language is not necessarily unique, moreover, there are languages having arbitrarily large reversible recognizers possessing no nontrivial reversible congruence. However, the exact characterization of this class of languages was open. In this paper we give a forbidden pattern capturing the reversible regular languages having only finitely many reduced reversible automata, allowing an efficient (NL) decision procedure.

Equations

$$p\Theta q\iff (p=q)\text{ or }(p^{\ast }=q^{\ast }\in Z).$$

$$r_{\_}0,(a_{\_}1,e_{\_}1),r_{\_}1,(a_{\_}2,e_{\_}2),\dots ,(a_{\_}\ell ,e_{\_}\ell ),r_{\_}\ell$$

$$(q,u)\cdot a=\{\begin{array}{ll}(qa,ua)& ifqais\; not\; an\; \infty -state,\\ (qa,u)& otherwise.\end{array}$$

$$(q^{\prime },i)b=\{\begin{array}{ll}(q^{\prime }b,(i+1)\text{mod}k)& if{q}^{\prime }=qandb=a\\ (q^{\prime }b,i)& otherwise.\end{array}$$

Full text

1 Introduction

Landauer’s principle [13] states that any logically irreversible manipulation of information – such as the merging of two computation paths – is accompanied by a corresponding entropy increase of non-information-bearing degrees of freedom in the processing apparatus. In practice, this can be read as “merging two computation paths generates heat”, though it has been demonstrated [23] that the entropy cost can be taken in e.g. angular momentum. Being a principle in physics, there is some debate regarding its validity, challenged [6, 20, 21] and defended [5, 4, 12] several times recently. In any case, the study of reversible computations, in which distinct computation paths never merge, looks appealing. In the context of quantum computing, one allows only reversible logic gates [19]. For classical Turing machines, it is known that each deterministic machine can be simulated by a reversible one, using the same amount of space [3, 14]. Hence, each regular language is accepted by a reversible two-way deterministic finite automaton (also shown in [9]).

In the case of classical, i.e. one-way automata, the situation is different: not all regular languages can be recognized by a reversible automaton, not even if we allow partial automata (that is, trap states can be removed, thus the transition function being a partial one). Those languages that can be recognized by a reversible one are called reversible languages. It is clear that one has to allow being partially defined at least since otherwise exactly the regular group languages (those languages in whose minimal automata each letter induces a permutation) would be reversible.

Several variants of reversible automata were defined and studied [22, 2, 18]. The variant we work with (partial deterministic automata with a single initial state and and arbitrary set of final states) have been treated in [10, 11, 7, 17, 16, 15]. In particular, in [7] the class of reversible languages is characterized by means of a forbidden pattern in the minimal automaton of the language in question, and an algorithm is provided to compute a minimal reversible automaton for a reversible language, given its minimal automaton. Here “minimal” means minimizing the number of states, and the minimal reversible automaton is shown to be not necessarily unique. In [17], the notion of reduced reversible automata is introduced: a reversible automaton is reduced if it is trim (all of its states are accessible and coaccessible), and none of its nontrivial factor automata is reversible. The authors characterize the class of those reversible languages (again, by developing a forbidden pattern) having a unique reduced reversible automaton (up to isomorphism), and leave open the problem to find a characterization of the class of those reversible languages having finitely many reduced reversible automata (up to isomorphism).

In this paper we solve this open problem of [17], by also developing a forbidden pattern characterization which allows an $\mathbf{NL}$ algorithm.

2 Notation

We assume the reader has some knowledge in automata and formal language theory (see e.g. [8]).

In this paper we consider deterministic partial automata with a single initial state and an arbitrary set of final states. That is, an automaton, or DFA, is a tuple $M=(Q,\Sigma,\delta,q_{\_}0,F)$ with $Q$ being the finite set of states, $q_{\_}0\in Q$ the initial state, $F\subseteq Q$ the set of final or accepting states, $\Sigma$ the finite, nonempty input alphabet of symbols or letters and $\delta:Q\times\Sigma\to Q$ the partial transition function which is extended in the usual way to a partial function also denoted by $\delta:Q\times\Sigma^{*}\to Q$ with $\delta(q,\varepsilon)=q$ for the empty word $\varepsilon$ and $\delta(q,wa)=\delta(\delta(q,w),a)$ if $\delta(q,w)$ is defined and undefined otherwise. When $\delta$ is understood from the context, we write $q\cdot w$ or $qw$ for $\delta(q,w)$ in order to ease notation. When $M$ is an automaton and $q$ is a state of $M$, then the language recognized by $M$ from $q$ is $L(M,q)=\{w\in\Sigma^{*}:qw\in F\}$. The language recognized by $M$ is $L(M)=L(M,q_{\_}0)$. A language is called regular or recognizable if some automaton recognizes it.

When $p$ and $q$ are states of the automata $M$ and $N$, respectively, we say that $p$ and $q$ are equivalent, denoted $p\equiv q$, if $L(M,p)=L(N,q)$. (When $M$ or $N$ is unclear from the context, we may write $(M,p)\equiv(N,q)$.) The automata $M$ and $N$ are said to be equivalent if their initial states are equivalent.

A state $q$ of $M$ is useful if it is reachable ($q_{\_}0w=q$ for some $w$) and productive ($qw\in F$ for some $w$). A DFA is trim if it only has useful states. When $L(M)$ is nonempty, one can erase the non-useful states of $M$: the resulting automaton will be trim and equivalent to $M$ (and may be partially defined even if $M$ is totally defined, if $M$ has a trap state $q$ for which $L(M,q)=\emptyset$). An equivalence relation $\Theta$ on the state set $Q$ is a congruence of $M$ if $p\Theta q$ implies both $p\in F~{}\Leftrightarrow~{}q\in F$ (that is, $F$ is saturated by $\Theta$) and $pa\Theta qa$ for each $a\in\Sigma$ (that is, $\Theta$ is compatible with the action). In particular, in any $\Theta$-class, $pa$ is defined if and only if so is $qa$.

Clearly the identity relation $\Delta_{\_}Q$ on $Q$ is always a congruence, the trivial congruence. A trim automaton is reduced if it has no nontrivial congruence. When $\Theta$ is a congruence of $M$ and $p~{}\Theta~{}q$ are states falling into the same $\Theta$-class, then $p\equiv q$. Given $M$ and a congruence $\Theta$ on $M$, the factor automaton of $M$ is $M/\Theta~{}=~{}(Q/\Theta,\Sigma,\delta/\Theta,q_{\_}0/\Theta,F/\Theta)$ where $p/\Theta$ denotes the $\Theta$-class of $p$, $X/\Theta$ denotes the set $\{p/\Theta:~{}p\in X\}$ of $\Theta$-classes for a set $X\subseteq Q$ and $\delta(q/\Theta,a)=\delta(q,a)/\Theta$ if $\delta(q,a)$ is defined, and is undefined otherwise.

Then, for each $p\in Q$ the states $p$ and $p/\Theta$ are equivalent, thus any automaton is equivalent to each of its factor automata. It is also known that for any automaton $M$ there is a unique (up to isomorphism, i.e. modulo renaming states) equivalent reduced automaton, the one we get by trimming $M$, then factoring the useful part of $M$ by the language equivalence relation $p\Theta_{\_}Mq~{}\Leftrightarrow~{}p\equiv q$.

Any automaton can be seen as an edge-labeled multigraph and thus its strongly connected components, or SCCs, are well-defined classes of its states: the states $p$ and $q$ belong to the same SCC if $pu=q$ and $qv=p$ for some words $u,v\in\Sigma^{*}$. Clearly, this is an equivalence relation. We call an SCC trivial if it consists of a single state $p$ with $pu\neq p$ for any nonempty word $u$ (that is, if it contains absolutely no edges, not even loops), and nontrivial otherwise.

3 Reversible languages

An automaton $M=(Q,\Sigma,\delta,q_{\_}0,F)$ is reversible if $pa=qa$ implies $p=q$ for each $p,q\in Q$ and $a\in\Sigma$. A language $L\subseteq\Sigma^{*}$ is reversible if it is recognizable by some reversible automaton. A reversible congruence of a reversible automaton $M$ is a congruence $\Theta$ of $M$ such that the factor automaton $M/\Theta$ is also reversible. The automaton $M$ is a reduced reversible automaton if it has no nontrivial reversible congruence.

It is known [7] that a language is reversible if and only if its minimal automaton has no distinct states $p\neq q$, a letter $a$ and a word $w$ such that $pa=qa$ and $paw=w$ (the forbidden pattern is depicted on Figure 1). Equivalently, for any state $r$ belonging to a nontrivial component, and letter $a$, the set $\{p\in Q:pa=r\}$ has to have at most one element.

Contrary to the general case of regular languages, there can be more than one reduced reversible automata, up to isomorphism, recognizing the same (reversible) language. For example, see Figure 2 of [17].

In that example the minimal automaton (depicted on Subfigure (a)) is not reversible and there are two nonisomorphic reversible reduced automata recognizing the same language (with four states). We note that in this particular example there are actually an infinite number of reduced reversible automata, recognizing the same language. In [17] the set of states of a minimal automaton was partitioned into two classes: the irreversible states are such states which are reachable from some distinct states $p\neq q$ with the same word $w$, while the reversible states are those which are not irreversible. For example, in the case of Figure 2, $q$ is the only irreversible state. One of the results of [17] is that if there exists an irreversible state $p$ which is reachable from a nontrivial SCC of the automaton (allowing the case when $p$ itself belongs to a nontrivial SCC, as $q$ does in the the example), then there exist an infinite number of nonisomorphic reduced reversible automata, each recognizing the same (reversible) language. A natural question is to precisely characterize the class of these reversible languages.

4 Result

In this section we give a forbidden pattern characterization for those reversible languages having a finite number of reduced reversible automata, up to isomorphism.

For this part, let us fix a reversible language $L$. Let $M=(Q^{*},\Sigma,\delta^{*},q_{\_}0^{*},F^{*})$ be the minimal automaton of $L$. We partition the states of $M$ into classes as follows: a state $q$ is a…

• •

$1$-state if there exists exactly one word $u$ with $q_{\_}0^{*}u=q$;

• •

$\infty$-state if there exist infinitely many words $u$ with $q_{\_}0^{*}u=q$;

• •

$\oplus$-state if it is neither a $1$-state nor an $\infty$-state

and orthogonally, $q$ is an…

• •

irreversible state if there are distinct states $p_{\_}1^{*}\neq p_{\_}2^{*}\in Q^{*}$ and a word $u$ such that $p_{\_}1^{*}u=p_{\_}2^{*}u=q$;

• •

reversible if it is not irreversible.

As an example, consider Figure 3.

Here, states $q_{\_}0$, $q_{\_}1$ and $q_{\_}3$ are $1$-states, reachable by the words $\varepsilon$, $a$ and $c$, respectively; $q_{\_}2$ and $q_{\_}4$ are $\oplus$-states as they are reachable by $\{b,d\}$ and $\{aa,bb,db,ca\}$, respectively and $q_{\_}5$ is a $\infty$-state, reachable by words of the form $(b+d)ab^{*}$. Moreover, $q_{\_}4$ is the only irreversible state (as $q_{\_}1a=q_{\_}3a$).

We note that our notion of irreversible states is not exactly the same as in [17]: what we call irreversible states are those states which belong to the “irreversible part” of the automaton in the terms of [17]. There, a state $q$ is called irreversible only if $p_{\_}1a=p_{\_}2a=q$ for some distinct pair $p_{\_}1\neq p_{\_}2$ of states and letter $a$.

Clearly, a state is an $\infty$-state iff it can be reached from some nontrivial SCC of $M$. Now we define the set $Z\subseteq Q^{*}$ of zig-zag states111The coined term “zig-zag” originates from an earlier version of Figure 5 on which forward edges had a “northeast” direction while backward edges had a “southeast” direction. as follows: $Z$ is the least set $X$ satisfying the following conditions:

All the $\infty$-states belong to $X$. 2. 2.

If $q\in X$ and $a\in\Sigma$ is a letter with $q\cdot a$ being defined, then $q\cdot a\in X$. 3. 3.

If $q$ is a $\oplus$-state and $a\in\Sigma$ is a letter with $q\cdot a\in X$, then $q\in X$.

The main result of the paper is the following:

Theorem 1**.**

There are only finitely many reduced reversible automata recognizing $L$ if and only if every zig-zag state of $M$ is reversible.

We break the proof up into several parts.

4.1 When all the zig-zag states are reversible

In this part we show that whenever all the zig-zag states are reversible, and $N$ is a trim reversible automaton recognizing $L$, then there is a reversible congruence $\Theta$ on $N$ such that $N/\Theta$ has a bounded number of states (the bound in question is computable from the minimal automaton $M$). So let us assume that there is no irreversible zig-zag state in $M$ and let $N=(Q,\Sigma,\delta,q_{\_}0,F)$ be a trim reversible automaton recognizing $L$. Then, for each $q\in Q$ there exists a unique state $q^{*}\in Q^{*}$ with $q\equiv q^{*}$, and the function $q\mapsto q^{*}$ is a homomorphism.

Now let us define the relation $\Theta$ on $Q$ as follows:

[TABLE]

Lemma 1**.**

The relation $\Theta$ is a reversible congruence on $N$.

Proof.

It is clear that $\Theta$ is an equivalence relation: reflexivity and symmetry are trivial, and $p~{}\Theta~{}q~{}\Theta~{}r$ either entails $p=q$ or $q=r$ (in which case $p~{}\Theta~{}r$ is obvious) or that $p^{*}=q^{*}=r^{*}\in Z$ (and then, $p~{}\Theta~{}r$ also holds).

Now if $p~{}\Theta~{}q$ and $p\cdot a$ is defined, then we have to show that $q\cdot a$ is also defined and $p\cdot a~{}\Theta~{}q\cdot a$. This is clear if $p=q$. Otherwise we have $p^{*}=q^{*}$ is a zig-zag state, thus $(p\cdot a)^{*}=(q\cdot a)^{*}$ as starring is a homomorphism from $N$ to $M$ (thus in particular, $q\cdot a$ is defined). As $p^{*}\in Z$ and $Z$ is closed under action, we have that $p^{*}\cdot a=(p\cdot a)^{*}$ is also a zig-zag state, thus $p\cdot a~{}\Theta~{}q\cdot a$ indeed holds and $\Theta$ is a congruence on $N$.

To see that $\Theta$ is a reversible congruence, assume $p\cdot a~{}\Theta~{}q\cdot a$. We have to show that $p~{}\Theta~{}q$. If $p\cdot a=q\cdot a$, then $p=q$ (thus $p~{}\Theta~{}q$), since $N$ is a reversible automaton. Otherwise, let $p\cdot a\neq q\cdot a$ (hence $p\neq q$) and $(p\cdot a)^{*}=(q\cdot a)^{*}\in Z$. By assumption on $M$, this state $(p\cdot a)^{*}$ is reversible. Thus, as $p^{*}\cdot a=q^{*}\cdot a=(p\cdot a)^{*}$, we get that $p^{*}=q^{*}$. Hence to show $p^{*}~{}\Theta~{}q^{*}$ it suffices to show that it is also a zig-zag state. If $p^{*}$ is a $\infty$-state, then it is a zig-zag state by definition of $Z$. Also, if $p^{*}$ is a $\oplus$-state, then it is still a zig-zag state (as $p^{*}\cdot a$ is a zig-zag state, we can apply Condition 3 in the definition of $Z$). Finally, observe that $p^{*}$ cannot be a $1$-state since $p$ and $q$ are different (reachable) states of $N$, hence there are words $u\neq v$ with $q_{\_}0u=p$ and $q_{\_}0v=q$. For these words, by starring being a homomorphism we get that $q_{\_}0^{*}u=q_{\_}0^{*}v=p^{*}$($=q^{*}$), thus $p^{*}$ is reachable by at least two distinct words.

Hence, $\Theta$ is indeed a reversible congruence. ∎

To conclude this case observe the following facts:

• •

For each non-$\infty$-state $p^{*}$ there exist a finite number $u_{\_}1,\ldots,u_{\_}{n(p^{*})}$ of words leading into $p^{*}$ in $M$, thus there can be at most $n(p^{*})$ states in $N$ which are equivalent to $p^{*}$ (since $N$ is trim). This bound $n(p^{*})$ is computable from $M$.

• •

If $p^{*}$ is a $\infty$-state of $M$, then $p^{*}$ is a zig-zag state, thus all the states of $N$ equivalent to $p^{*}$ are collapsed into a single class of $\Theta$. For these states, let us define the value $n(p^{*})$ to be $1$.

Hence, $n=\mathop{\sum}\limits_{\_}{p^{*}\in Q^{*}}n(p^{*})$ is a (finite, computable) upper bound on the number of states in the factor automaton $N/\Theta$ (hence it is an upper bound for the number of states in any reduced reversible automaton recognizing $L$ as in that case $\Theta$ has to be the trivial congruence). Thus we have proved the first part of Theorem 1:

Theorem 2**.**

If all the zig-zag states of $M$ are reversible, then there is a finite upper bound for the number of states of any reduced reversible automata recognizing $L$. Hence, in that case there exists only a finite number of nonisomorphic reduced reversible automata recognizing $L$.

4.2 When there is an irreversible zig-zag state in $M$

In this part let us assume that $M$ has an irreversible zig-zag state. We will start from an arbitrary reduced reversible automaton $N$ recognizing $L$, and then “blow it up” to some arbitrarily large equivalent reduced reversible automaton. Before giving the construction, we illustrate the process in the case of Figure 3 (there, $q_{\_}4$ is an irreversible zig-zag state).

On Figure 4 (a), we have a reversible automaton $N$. On (b), we replace the state $q_{\_}5$ having a loop by a cycle of length $5$ (and we also duplicate the state $q_{\_}2$ – we can do that since $q_{\_}2$ is not a $1$-state). Then, the automaton $N^{\prime}$ is a reduced reversible automaton, whenever the length of the cycle (which is now chosen to be $5$) is a prime number [17].

Indeed, any reversible congruence $\Theta$ on $N^{\prime}$ collapses equivalent states only. Suppose e.g. $\Theta$ collapses $q_{\_}{51}$ and $q_{\_}{53}$. Applying $b$ we get that $q_{\_}{52}$ and $q_{\_}{54}$ also get collapsed, and so $q_{\_}{53}$ and $q_{\_}{55}$, and $q_{\_}{54}$ and $q_{\_}{51}$ – thus, all the $q_{\_}{5x}$ states fall into a single $\Theta$-class then. As $\Theta$ is assumed to be reversible, it has to collapse $q_{\_}2$ and $q_{\_}2^{\prime}$ as well; applying $b$ also the states $q_{\_}4$ and $q_{\_}4^{\prime}$ have to be collapsed and finally, applying again reversibility we get that $q_{\_}1$ and $q_{\_}3$ should be collapsed but this cannot happen as they are not equivalent states. This reasoning works for any choice of different copies of $q_{\_}5$ (as far as the number chosen is a prime), thus $N^{\prime}$ has only the trivial reversible congruence and is a reduced reversible automaton.

The careful reader might realize that we actually followed in this reasoning a zig-zag path from the $\infty$-state $q_{\_}5$ to the irreversible state $q_{\_}4$ during the above reasoning. In this part we show that this approach can always be generalized whenever there exists an irreversible zig-zag state in $M$.

By the definition of the zig-zag states, if there exists some irreversible zig-zag state, then there is a sequence

[TABLE]

such that $a_{\_}i\in\Sigma$ and $e_{\_}i\in\{+,-\}$ for each $1\leq i\leq\ell$, and $r_{\_}i\in Q^{*}$ are states of $M$ for each $0\leq i\leq\ell$, moreover,

1. (i)

if $e_{\_}i=+$ then $r_{\_}{i-1}\cdot a_{\_}i=r_{\_}i$ (denoted by $r_{\_}{i-1}\mathop{\longrightarrow}\limits^{a_{\_}i}r_{\_}i$ in the examples and patterns), 2. (ii)

if $e_{\_}i=-$ then $r_{\_}i\cdot a_{\_}{i}=r_{\_}{i-1}$ and $r_{\_}i$ is a $\oplus$-state (denoted by $r_{\_}{i-1}\mathop{\longleftarrow}\limits^{a_{\_}i}r_{\_}i$), 3. (iii)

$r_{\_}\ell$ is an irreversible state, 4. (iv)

$r_{\_}0$ is an $\infty$-state.

Let us choose such a sequence of minimal length. Then, by minimality,

• •

the states $r_{\_}i$ are pairwise different,

• •

all the states $r_{\_}i$, $1\leq i<\ell$ are reversible $\oplus$-states.

To see that all the states are $\oplus$-states (them being pairwise different reversible states is obvious), observe that if $r_{\_}i$ is an $\infty$-state for $0<i$, then $r_{\_}i,\ldots,r_{\_}\ell$ is a shorter such sequence. Hence all the states $r_{\_}i$, $0<i$ are either $1$-states or $\oplus$-states. We show by induction that all of them are $\oplus$-states. The claim holds for $i=1$ as $e_{\_}1=+$ would imply that $r_{\_}1$ should be an $\infty$-state which cannot happen thus $e_{\_}1=-$, hence $r_{\_}1$ is an $\oplus$-state. Now if $r_{\_}i$ is an $\oplus$-state, then either $r_{\_}{i+1}=r_{\_}ia_{\_}{i+1}$ (if $e_{\_}{i+1}=+$) which implies that $r_{\_}{i+1}$ cannot be a $1$-state (thus it is an $\oplus$-state), or $r_{\_}{i+1}$ is an $\oplus$-state (if $e_{\_}{i+1}=-$, applying ii)).

Now we extend the above sequence in both directions as follows.

First, $r_{\_}0$ being an $\infty$-state implies that there exists a state $p_{\_}0$ belonging to a nontrivial SCC of $M$ (that is, $p_{\_}0w=p_{\_}0$ for some nonempty word $w$) from which $r_{\_}0$ is reachable. That is, there is a word $b_{\_}1b_{\_}2\ldots b_{\_}m$ and states $p_{\_}1,\ldots,p_{\_}m$ with $p_{\_}m=r_{\_}0$ and $p_{\_}ib_{\_}{i+1}=p_{\_}{i+1}$ for each $0\leq i<m$.

Second, $r_{\_}\ell$ being an irreversible state implies that there exist different states $s$ and $s^{\prime}$ of $M$ from which $r_{\_}\ell$ is reachable by the same (nonempty) word. That is, there is a word $c_{\_}1c_{\_}2\ldots c_{\_}n$ and states $s_{\_}1,s_{\_}2,\ldots,s_{\_}n=r_{\_}\ell$ such that $sc_{\_}1=s^{\prime}c_{\_}1=s_{\_}1$ and $s_{\_}ic_{\_}{i+1}=s_{\_}{i+1}$ for each $1\leq i<n$. See Figure 5.

Note that all the states $s_{\_}1,\ldots,s_{\_}n$ are $\oplus$-states and $p_{\_}0,\ldots,p_{\_}m$ are $\infty$-states.

In order to reduce the clutter in the notation, we treat the whole sequence $p_{\_}0,\ldots,p_{\_}{m-1},r_{\_}0,\ldots,s_{\_}1,s$ and $s^{\prime}$ as a single indexed sequence $p_{\_}0,\ldots,p_{\_}{t-1},p_{\_}{t,1}$ and $p_{\_}{t,2}$ (see Figure 6). Observe that each $p_{\_}i$ (but possibly $p_{\_}{t,1}$ and $p_{\_}{t,2}$) is either an $\infty$- or a $\oplus$-state, and each of them is a zig-zag state. Moreover, all these states are pairwise different (thus inequivalent).

In the first step we show that if there is a specific pattern (which is a bit more general than a cycle of prime length) appears in a reversible automaton $N^{\prime}$, then every factor automaton of $N^{\prime}$ is “large”.

Lemma 2**.**

Assume $N^{\prime}$ is a reversible automaton, $k\geq 1$ is a prime number, $t\geq 0$ and $i<k$ are integers, $q_{\_}0,\ldots,q_{\_}{k-1}$, $p^{\prime}_{\_}1,\ldots,p^{\prime}_{\_}t$, $p^{\prime\prime}_{\_}1,\ldots,p^{\prime\prime}_{\_}t$ are states, $a_{\_}1,\ldots,a_{\_}t\in\Sigma$ are letters, $e_{\_}1,\ldots,e_{\_}t\in\{+,-\}$ are directions and $w\in\Sigma^{+}$ is a word satisfying the following conditions:

• •

$p^{\prime}_{\_}t$* is not equivalent to $p^{\prime\prime}_{\_}t$,*

• •

$q_{\_}jw=q_{\_}{j+1}$* for each $0\leq j<k$ with the convention that $q_{\_}{k}=q_{\_}0$, that is, indices of the *$q$s are taken modulo $k$,

• •

for each $1\leq j\leq t$ with $e_{\_}j=+$ we have $p_{\_}{j-1}^{\prime}a_{\_}j=p_{\_}j^{\prime}$ and $p_{\_}{j-1}^{\prime\prime}a_{\_}j=p_{\_}j^{\prime\prime}$,

• •

and for each $1\leq j\leq t$ with $e_{\_}j=-$ we have $p_{\_}{j}^{\prime}a_{\_}j=p_{\_}{j-1}^{\prime}$ and $p_{\_}{j}^{\prime\prime}a_{\_}j=p_{\_}{j-1}^{\prime\prime}$

with setting $p_{\_}0^{\prime}:=q_{\_}0$ and $p_{\_}0^{\prime\prime}:=q_{\_}i$ (See Figure 7).

Then whenever $\Theta$ is a reversible congruence on $N^{\prime}$, the states $q_{\_}j$ belong to pairwise different $\Theta$-classes. (In particular, $N^{\prime}/\Theta$ has at least $k$ states.)

Before proceeding with the proof, the reader is encouraged to check that the above described pattern appears in the automaton $N^{\prime}$ of Figure 4 with the choice of $k=5$, $q_{\_}0=q_{\_}{51}$, $q_{\_}1=q_{\_}{52}$,…,$q_{\_}4=q_{\_}{55}$, $i=1$, $w=b$, $t=3$, $p_{\_}1^{\prime}=q_{\_}2$, $p_{\_}1^{\prime\prime}=q_{\_}2^{\prime}$, $p_{\_}2^{\prime}=q_{\_}4$, $p_{\_}2^{\prime\prime}=q_{\_}4^{\prime}$, $p_{\_}3^{\prime}=q_{\_}1$, $p_{\_}3^{\prime\prime}=q_{\_}3$, $a_{\_}1=a$, $a_{\_}2=b$, $a_{\_}3=a$, $e_{\_}1=-$, $e_{\_}2=+$ and $e_{\_}3=-$, with states appearing on the left-hand side of these equations are the states from the pattern of Lemma 2 while states on the right-hand side are states of $N^{\prime}$.

Proof.

Assume for the sake of contradiction that $\Theta$ is a reversible congruence on $N^{\prime}$ collapsing the states $q_{\_}\ell$ and $q_{\_}j$ for some $1\leq\ell<j\leq k$, that is, $q_{\_}\ell~{}\Theta~{}q_{\_}{j}$.

We claim that $q_{\_}{\ell+d}~{}\Theta~{}q_{\_}{j+d}$ for each $d\geq 0$. This holds by assumption for $d=0$. Using induction on $d$, assuming $q_{\_}{\ell+d}~{}\Theta~{}q_{\_}{j+d}$ we get by applying $w$ that $q_{\_}{\ell+d+1}=q_{\_}{\ell+d}w~{}\Theta~{}q_{\_}{j+d}w=q_{\_}{j+d+1}$ as $\Theta$ is a congruence. Hence, writing $j=\ell+\delta$ we get that $q_{\_}{\ell+d}~{}\Theta~{}q_{\_}{\ell+\delta+d}$ for each $d\geq 0$, thus in particular for multiples of $\delta$: $q_{\_}{\ell+d\cdot\delta}~{}\Theta~{}q_{\_}{\ell+(d+1)\cdot\delta}$.

Hence we have that $q_{\_}\ell~{}\Theta~{}q_{\_}{\ell+\delta}~{}\Theta~{}q_{\_}{\ell+2\delta}~{}\Theta\ldots$. As $k$ is assumed to be a prime number, there are integers $d_{\_}1$ and $d_{\_}2$ with $\ell+d_{\_}1\delta\equiv 0~{}\mathrm{mod}~{}k$ and $\ell+d_{\_}2\delta\equiv i~{}\mathrm{mod}~{}k$, thus $q_{\_}0~{}\Theta~{}q_{\_}i$. As $p_{\_}0^{\prime}$ is defined as $q_{\_}0$ and $p_{\_}0^{\prime\prime}$ is defined as $q_{\_}i$, we have $p_{\_}0^{\prime}~{}\Theta~{}p_{\_}0^{\prime\prime}$.

Now for any integer $d\geq 0$, the relation $p^{\prime}_{\_}{d}~{}\Theta~{}p^{\prime\prime}_{\_}d$ implies $p^{\prime}_{\_}{d+1}~{}\Theta~{}p^{\prime\prime}_{\_}{d+1}$: if $e_{\_}{d+1}=+$, then by applying $a_{\_}{d+1}$ (since $\Theta$ is a congruence), while if $e_{\_}{d+1}=-$, then be reversibly applying $a_{\_}{d+1}$ (since $\Theta$ is a reversible congruence). Hence, it has to be the case $p^{\prime}_{\_}t~{}\Theta~{}p^{\prime\prime}_{\_}t$ which is nonsense since these two states are assumed to be inequivalent and $\Theta$ is a congruence. ∎

Observe that if some reversible automaton $N^{\prime}$ recognizing $L$ admits the pattern of Lemma 2 for some prime number $k$, then there exists a reduced reversible automaton of the form $N^{\prime}/\Theta$ (also recognizing $L$) which then has at least $k$ states.

In the rest of this part we show that if there exists an irreversible zig-zag state in $M$, then we can construct such an automaton $N^{\prime}$ for arbitrarily large primes $k$, given a reversible automaton $N$ recognizing $L$. Thus in that case it is clear that there exists an infinite number of reduced reversible automata (up to isomorphism) recognizing $L$.

First we show that even a weaker condition suffices.

Lemma 3**.**

Suppose $N$ is a reversible automaton recognizing $L$ such that for each zig-zag state $p$ of $M$ there exist at least two states $p^{\prime}$ and $p^{\prime\prime}$ of $N$ with $p\equiv p^{\prime}\equiv p^{\prime\prime}$ and to the $\infty$-state $p_{\_}0$ of $M$ (of Figure 5), there exist at least $k$ different states $q_{\_}0,\ldots,q_{\_}{k-1}$ in $N$, each being equivalent to $p_{\_}0$, with $q_{\_}jw=q_{\_}{j+1}$ for each $0\leq j<k$ (again, with $q_{\_}k=q_{\_}0$).

Then there exists a reversible automaton $N^{\prime}$ also recognizing $L$ which admits the zig-zag pattern.

For an example reversible automaton $N$ recognizing $L$ but not admitting the zig-zag pattern the reader is referred to Figure 8. We not prove that in these cases the transitions can be “rewired”.

Proof.

We will construct a sequence $N=N_{\_}0$, $N_{\_}1$,…, $N_{\_}t=N^{\prime}$ of reversible automata, each recognizing $L$ (having the same set of states, and even $(N_{\_}i,p)\equiv(N_{\_}j,p)$ for each $i,j$ and $p$, that is, we do not change the languages recognized by any of the states of $N$) and sequences $p_{\_}0^{\prime},\ldots,p_{\_}t^{\prime}$ and $p_{\_}0^{\prime\prime},\ldots,p_{\_}t^{\prime\prime}$ of states such that for each $0\leq j\leq t$ the following all hold:

• •

$p_{\_}0^{\prime}=q_{\_}0$, $p_{\_}0^{\prime\prime}=q_{\_}1$,

• •

if $e_{\_}j=+$, then $p^{\prime}_{\_}{j-1}b_{\_}j=p^{\prime}_{\_}j$ and $p^{\prime\prime}_{\_}{j-1}b_{\_}j=p^{\prime\prime}_{\_}j$ for each automaton $N_{\_}\ell$ with $\ell\leq j$,

• •

if $e_{\_}j=-$, then $p^{\prime}_{\_}{j}b_{\_}j=p^{\prime}_{\_}{j-1}$ and $p^{\prime\prime}_{\_}{j}b_{\_}j=p^{\prime\prime}_{\_}{j-1}$ for each automaton $N_{\_}\ell$ with $\ell\leq j$,

• •

if $j<t$ and $\ell\leq j$, then $p_{\_}j$ is equivalent to both $p^{\prime}_{\_}j$ and $p^{\prime\prime}_{\_}j$ in $N_{\_}\ell$,

• •

$p^{\prime}_{\_}t\equiv p_{\_}{t,1}$ and $p^{\prime\prime}_{\_}t\equiv p_{\_}{t,2}$ in $N_{\_}t$.

If we manage to achieve this, then the automaton $N^{\prime}:=N_{\_}t$ indeed admits the zig-zag pattern and still recognizes $L$.

We construct the above sequence $N_{\_}j$ by induction on $j$. For $j=0$, choosing $N_{\_}0=N$ and $p_{\_}0^{\prime}:=q_{\_}0$, $p_{\_}0^{\prime\prime}:=q_{\_}1$ suffices. Having constructed $N_{\_}j$, we construct $N_{\_}{j+1}$ based on whether the direction $e_{\_}{j+1}$ is $+$ or $-$, the latter one having several subcases.

If $e_{\_}{j+1}=+$, then let us set $N_{\_}{j+1}:=N_{\_}j$, $p_{\_}{j+1}^{\prime}:=p_{\_}j^{\prime}b_{\_}{j+1}$ (in $N_{\_}j$) and $p_{\_}{j+1}^{\prime\prime}:=p^{\prime\prime}_{\_}jb_{\_}{j+1}$ (also in $N_{\_}j$). This choice satisfies the conditions. 2. 2.

If $e_{\_}{j+1}=-$, then we have three subcases, based on whether there exist states in $N_{\_}j$ equivalent to $p_{\_}{j+1}$ from which $b_{\_}{j+1}$ leads to either $p_{\_}j^{\prime}$ or $p_{\_}j^{\prime\prime}$.

1. (a)

If there exist such states $r_{\_}1$ and $r_{\_}2$ with $r_{\_}1b_{\_}{j+1}=p_{\_}j^{\prime}$ and $r_{\_}2b_{\_}{j+1}=p_{\_}j^{\prime\prime}$, then again, setting $N_{\_}{j+1}:=N_{\_}j$, $p^{\prime}_{\_}{j+1}:=r_{\_}1$ and $p^{\prime\prime}_{\_}{j+1}:=r_{\_}2$ suffices. 2. (b)

Assume there is no such $r_{\_}1$ nor $r_{\_}2$. Then, as $p_{\_}{j+1}$ is a zig-zag state, there exist, by the assumption on $N$, two different states $r_{\_}1$ and $r_{\_}2$, each being equivalent to $p_{\_}{j+1}$. Moreover, as $p_{\_}{j+1}b_{\_}{j+1}=p_{\_}j$ holds, we have that $r_{\_}1b_{\_}{j+1}$ and $r_{\_}2b_{\_}{j+1}$ (in $N_{\_}j$) are equivalent to $p_{\_}j$. So let us define $N_{\_}{j+1}$ as follows: $r_{\_}1b_{\_}{j+1}:=p^{\prime}_{\_}j$, $r_{\_}2b_{\_}{j+1}:=p^{\prime\prime}_{\_}j$ and for all the other pairs $(r,b)$ let us leave the transitions of $N_{\_}j$ unchanged. Then, setting $p^{\prime}_{\_}{j+1}:=r_{\_}1$ and $p^{\prime\prime}_{\_}{j+2}:=r_{\_}2$ suffices. 3. (c)

Finally, assume that exactly one of these predecessor states exists. By symmetry, we can assume that it is $r_{\_}1$, that is, $r_{\_}1b_{\_}{j+1}=p^{\prime}_{\_}j$ in $N_{\_}j$ but there is no state $r$ equivalent to $p_{\_}{j+1}$ with $rb_{\_}{j+1}=p^{\prime\prime}_{\_}j$. Since $p_{\_}{j+1}$ is a zig-zag state, there exists some $r_{\_}2\neq r_{\_}1$ in $N_{\_}j$, still equivalent to $p_{\_}{j+1}$. In this case we set the transitions in $N_{\_}{j+1}$ as $r_{\_}2b_{\_}{j+1}p^{\prime\prime}_{\_}j$, leaving the other transitions unchanged suffices with $p^{\prime}_{\_}{j+1}:=r_{\_}1$ and $p^{\prime\prime}_{\_}{j+1}:=r_{\_}2$.

∎

Hence, given $N$, it suffices to construct a reversible automaton satisfying the conditions of Lemma 3. First we construct a reversible automaton $N^{\prime}$ in which for each $\oplus$-state $p$ of $M$ there exist at least two different equivalent states. Let $U\subseteq\Sigma^{*}$ be the set of words $u$ such that $u=\varepsilon$ or $q_{\_}0u$ is not a $\infty$-state. Note that $U$ is a nonempty finite set. Now let us define the state set of $N^{\prime}$ as the finite set $Q^{\prime}=Q\times U$, equipped with the following transition function

[TABLE]

For an illustration of the construction starting from $N$ of Figure 4 (a), consult Figure 9.

It is clear that $(q_{\_}0,\varepsilon)u=(q_{\_}0u,v)$ for some prefix $v$ of $u$, moreover, if $q_{\_}0u$ is a $\oplus$-state, then $v=u$. Also, states of the form $(q,u)$ in $N^{\prime}$ are equivalent to $q$ (if we set $F\times U$ as the accepting set). Hence, whenever $q$ is a $\oplus$-state which can be reached by the words $u_{\_}1,\ldots,u_{\_}n$, $n\geq 2$, then $(q,u_{\_}1),\ldots,(q,u_{\_}n)$ are pairwise different states in $N^{\prime}$, reachable from $(q_{\_}0,\varepsilon)$ and thus to each $\oplus$-state $p$ of $M$ there exist at least two equivalent states in $N^{\prime}$. Now starting from $N^{\prime}$ we will construct an automaton $N^{\prime\prime}$ satisfying the conditions of the Lemma.

Let $q$ be a state of $N^{\prime}$, equivalent to $p_{\_}0$. Since $p_{\_}0w=p_{\_}0$ in $M$ for the nonempty word $w$, the sequence $q$, $qw$, $qw^{2}$,…contains some repetition. Let $i$ be the least integer with $qw^{i}=qw^{j}$ for some $j>i$. Then if $i>0$, then we have $qw^{i-1}\cdot w=qw^{i}=qw^{j}=qw^{j-1}w$, thus $qw^{i-1}=qw^{j-1}$ since $N^{\prime}$ is reversible. Hence $q=qw^{j}$ for some integer $j$. In particular, $q$ belongs to a nontrivial SCC of $N^{\prime}$. Let $av$ be a shortest nonempty word with $qav=q$ (such a word exists since $q$ is in a nontrivial SCC). For the fixed integer $k\geq 1$, let us define $N^{\prime\prime}$ as the automaton over the state set $Q^{\prime}\times\{0,\ldots,k-1\}$, with transition function

[TABLE]

That is, we increase the index $i$ (modulo $k$) if we get the input $a$ in the state $q$, and in all the other cases the index remains untouched. For an example with $k=5$, see Figure 10.

As $av$ is a shortest word with $qav=q$, it is clear that $q$ does not occur on the $v$-path from $qa$ to $qav=q$. Hence, in $N^{\prime\prime}$ we have $(q,i)av=(q,(i+1)~{}\mathrm{mod}~{}k)$. Moreover, if $u$ is a shortest word in $N^{\prime}$ leading into $q$, then it leads into $(q,0)$ in $N^{\prime\prime}$. Thus, in $N^{\prime\prime}$ we have the reachable states $q_{\_}0=(q,0)$, $q_{\_}1=(q,1)$, …, $q_{\_}{k-1}=(q,k-1)$ with $q_{\_}iav=q_{\_}{(i+1)~{}\textrm{mod}~{}k}$ for each $i$, and still, for each $\oplus$-state $p$ there are at least two different states in $N^{\prime\prime}$ equivalent to $p$. Hence, applying Lemma 3 we get that there exists some reversible automaton $N^{\prime\prime\prime}$ (note that the automaton $N^{\prime\prime}$ we constructed is also reversible) admitting the zig-zag pattern.

Figure 11 shows the result of this last step: first, since $(q_{\_}5,b,0)$ has an incoming $a$-edge from $(q_{\_}2,b)$ but $(q_{\_}5,b,1)$ has no such edge, we search for another state equivalent to $q_{\_}2$, that’s $(q_{\_}2,d)$. Then we set $(q_{\_}2,d)a$ to $(q_{\_}5,b,1)$. Then, we can follow the $b$-transitions into $(q_{\_}4,bb)$ and $(q_{\_}4,db)$ respectively. After that, we should follow $a$-edges backwards into $q_{\_}1$ and $q_{\_}3$. Hence we rewire the outgoing transitions as $(q_{\_}1,a)a=(q_{\_}4,bb)$ and $(q_{\_}3,c)a=(q_{\_}4,db)$ and all is set.

Thus, by Lemma 2 we get the main result of the subsection:

Theorem 3**.**

If there is an irreversible zig-zag state in $M$, then for an arbitrarily large $k$ one can effectively construct a reduced reversible automaton equivalent to $M$, having at least $k$ states.

Now Theorem 1 is the conjunction of Theorems 2 and 3.

5 Conclusion and acknowledgements

We extended the current knowledge on the reversible regular languages by further analyzing the structure of the minimal automaton of the language in question. In particular, we gave a forbidden pattern characterization of those reversible languages having only a finite number of reduced reversible automata. As the characterization relies on the existence of a “forbidden pattern” (that of Figure 6), it gives an efficient decision procedure, namely an $\mathbf{NL}$ (nondeterministic logspace) algorithm: one has to guess a state $p_{\_}0$, then guess some loop from $p_{\_}0$ to itself, then following some back-and-forth walk in the graph of the automaton to two distinct states $p_{\_}{t,1}$ and $p_{\_}{t,2}$. In the process we also have to check that no $1$-state is encountered during this walk (which can clearly also be done in $\mathbf{NL}$). It can be an interesting question to study the notion of reduced reversible automata in other reversibility settings, as e.g. in the case of [18]. The authors wish to thank Giovanni Pighizzini, Giovanna Lavado and Luca Prigioniero for their useful comments on a much earlier version of this paper.