Separability by Piecewise Testable Languages is PTime-Complete

TL;DR

This paper investigates the computational complexity of determining whether two regular languages can be separated by piecewise testable languages, establishing PTime-completeness for DFAs and NFAs.

Contribution

It proves that the separability problem by piecewise testable languages is PTime-complete for both minimal DFAs and NFAs, filling a gap in complexity understanding.

Findings

Separability by piecewise testable languages is PTime-hard for DFAs.

The problem remains PTime-hard even for minimal DFAs.

Membership testing for NFAs is PSpace-complete.

Abstract

Piecewise testable languages form the first level of the Straubing-Th\'erien hierarchy. The membership problem for this level is decidable and testing if the language of a DFA is piecewise testable is NL-complete. The question has not yet been addressed for NFAs. We fill in this gap by showing that it is PSpace-complete. The main result is then the lower-bound complexity of separability of regular languages by piecewise testable languages. Two regular languages are separable by a piecewise testable language if the piecewise testable language includes one of them and is disjoint from the other. For languages represented by NFAs, separability by piecewise testable languages is known to be decidable in PTime. We show that it is PTime-hard and that it remains PTime-hard even for minimal DFAs.

Full text

1 Introduction

A regular language over $\Sigma$ is piecewise testable if it is a finite boolean combination of languages of the form $\Sigma^{*}a_{1}\Sigma^{*}a_{2}\Sigma^{*}\cdots\Sigma^{*}a_{n}\Sigma^{*}$, where $a_{i}\in\Sigma$ and $n\geq 0$. If $n$ is bounded by a constant $k$, then the language is called $k$-piecewise testable. Piecewise testable languages are exactly those regular languages whose syntactic monoid is $\mathcal{J}$-trivial [36]. Simon [37] provided various characterizations of piecewise testable languages, e.g., in terms of monoids or automata. These languages are of interest in many disciplines of mathematics, such as semigroup theory [2, 3, 28] for their relation to Green’s relations or in logic on words [10] for their relation to first-order logic $\textrm{FO}[<]$ and the Straubing-Thérien hierarchy [40, 43].

For an alphabet $\Sigma$, level 0 of the Straubing-Thérien hierarchy is defined as $\mathscr{L}(0)=\{\emptyset,\Sigma^{*}\}$. For integers $n\geq 0$, level $\mathscr{L}(n+\frac{1}{2})$ consists of all finite unions of languages $L_{0}a_{1}L_{1}a_{2}\ldots a_{k}L_{k}$ with $k\geq 0$, $L_{0},\ldots,L_{k}\in\mathscr{L}(n)$, and $a_{1},\ldots,a_{k}\in\Sigma$, and level $\mathscr{L}(n+1)$ consists of all finite Boolean combinations of languages from level $\mathscr{L}(n+\frac{1}{2})$. The levels of the hierarchy contain only star-free languages [27]. Piecewise testable languages form the first level of the hierarchy. The hierarchy does not collapse on any level [5]. In spite of a recent development [1, 29, 32], deciding whether a language belongs to level $\ell$ of the hierarchy is open for $\ell>\frac{7}{2}$. The Straubing-Thérien hierarchy is further closely related to the dot-depth hierarchy [5, 7, 23, 41] and to complexity theory [45].

The fundamental question is how to efficiently recognize whether a given regular language is piecewise testable. Stern [39] provided a solution that was later improved by Trahtman [44] and Klíma and Polák [21]. Stern presented an algorithm deciding piecewise testability of a regular language represented by a DFA in time $O(n^{5})$, where $n$ is the number of states of the DFA. Trahtman improved Stern’s algorithm to time quadratic with respect to the number of states and linear with respect to the size of the alphabet, and Klíma and Polák found an algorithm for DFAs that is quadratic with respect to the size of the alphabet and linear with respect to the number of states. Cho and Huynh [6] proved that deciding piecewise testability for DFAs is NL-complete. Event though the complexity for DFAs has intensively been investigated, a study for NFAs is missing in the literature. We fill in this gap by showing that deciding piecewise testability for NFAs is PSpace-complete (Theorem 2).

The knowledge of the minimal $k$ or a reasonable bound on $k$ for which a piecewise testable language is $k$-piecewise testable is of interest in several applications [24, 16]. The complexity of finding the minimal $k$ has been investigated in the literature [16, 20, 21, 26]. Testing whether a piecewise testable language is $k$-piecewise testable is coNP-complete for $k\geq 4$ if the language is represented as a DFA [20] and PSpace-complete if the language is represented as an NFA [26]. The complexity for DFAs and $k<4$ has also been discussed in detail [26]. Klíma and Polák [21] further showed that the upper bound on $k$ is given by the depth of the minimal DFA. This result has recently been generalized to NFAs [25].

The recent interest in piecewise testable languages is mainly because of the applications of separability of regular languages by piecewise testable languages in logic on words [31] and in XML schema languages [8, 16, 24]. Given two languages $K$ and $L$ and a family of languages $\mathcal{F}$, the separability problem asks whether there exists a language $S$ in $\mathcal{F}$ such that $S$ includes one of the languages $K$ and $L$ and is disjoint from the other. Place and Zeitoun [31] used separability to obtain new decidability results of the membership problem for some levels of the Straubing-Thérien hierarchy. The separability problem for regular languages represented by NFAs and the family of piecewise testable languages is decidable in polynomial time with respect to both the number of states and the size of the alphabetby [8, 30]. Separability by piecewise testable languages is of interest also outside regular languages. Although separability of context-free languages by regular languages is undecidable [17], separability by piecewise testable languages is decidable (even for some non-context-free languages) [9]. Piecewise testable languages are further investigated in natural language processing [11, 33], cognitive and sub-regular complexity [34], and learning theory [12, 22]. They have been extended from word languages to tree languages [4, 13, 14].

In this paper, we show that separability of regular languages represented as NFAs by piecewise testable languages is a PTime-complete problem (Theorem 3) and that it remains PTime-hard even for minimal DFAs. Consequently, the separability problem is unlikely to be solvable in logarithmic space or effectively parallelizable.

2 Preliminaries

We assume that the reader is familiar with automata theory [38]. The cardinality of a set $A$ is denoted by $|A|$ and the power set of $A$ by $2^{A}$. The free monoid generated by an alphabet $\Sigma$ is denoted by $\Sigma^{*}$. A word over $\Sigma$ is any element of $\Sigma^{*}$; the empty word is denoted by $\varepsilon$. For a word $w\in\Sigma^{*}$, $\operatorname{\rm alph}(w)\subseteq\Sigma$ denotes the set of all symbols occurring in $w$.

A nondeterministic finite automaton (NFA) is a quintuple $\mathcal{A}=(Q,\Sigma,\delta,Q_{0},F)$, where $Q$ is the finite nonempty set of states, $\Sigma$ is the input alphabet, $Q_{0}\subseteq Q$ is the set of initial states, $F\subseteq Q$ is the set of accepting states, and $\delta\colon Q\times\Sigma\to 2^{Q}$ is the transition function extended to the domain $2^{Q}\times\Sigma^{*}$ in the usual way. The language accepted by $\mathcal{A}$ is the set $L(\mathcal{A})=\{w\in\Sigma^{*}\mid\delta(Q_{0},w)\cap F\neq\emptyset\}$.

A path $\pi$ from a state $q_{0}$ to a state $q_{n}$ under a word $a_{1}a_{2}\cdots a_{n}$, for some $n\geq 0$, is a sequence of states and input symbols $q_{0},a_{1},q_{1},a_{2},\ldots,q_{n-1},a_{n},q_{n}$ such that $q_{i+1}\in\delta(q_{i},a_{i+1})$, for all $i=0,1,\ldots,n-1$. Path $\pi$ is accepting if $q_{0}\in Q_{0}$ and $q_{n}\in F$. We write $q_{0}\xrightarrow{a_{1}a_{2}\cdots a_{n}}q_{n}$ to denote that there is a path from $q_{0}$ to $q_{n}$ under the word $a_{1}a_{2}\cdots a_{n}$.

We says that $\mathcal{A}$ has a cycle over an alphabet $\Gamma\subseteq\Sigma$ if there is a state $q$ in $\mathcal{A}$ and a word $w$ over $\Sigma$ such that $q\xrightarrow{\,w\,}q$ and $\operatorname{\rm alph}(w)=\Gamma$.

The NFA $\mathcal{A}$ is deterministic (DFA) if $|Q_{0}|=1$ and $|\delta(q,a)|=1$ for every $q\in Q$ and $a\in\Sigma$. Although we define DFAs as complete, we mostly depict only the most important transitions in our illustrations. The reader can easily complete such an incomplete DFA.

Let $K$ and $L$ be languages. A language $S$ separates $K$ from $L$ if $S$ contains $K$ and does not intersect $L$. Languages $K$ and $L$ are separable by a family of languages $\mathcal{F}$ if there exists a language $S$ in $\mathcal{F}$ that separates $K$ from $L$ or $L$ from $K$.

3 Piecewise Testability for NFAs

Given an NFA $\mathcal{A}$ over an alphabet $\Sigma$, the piecewise-testability problem asks whether the language $L(\mathcal{A})$ is piecewise testable. Although the membership in PSpace follows basically from the result by Cho and Huynh [6], we prefer to provide the proof here for two reasons: (i) we would like to provide unfamiliar readers with a method to recognize whether a regular language is piecewise testable, (ii) Cho and Huynh assume that the input is a minimal DFA, hence it is necessary to extend their algorithm with a non-equivalence check. We use the following characterization in our proof.

Proposition 1** (Simon [37], Cho and Huynh [6, Proposition 2.3(b)]).**

A regular language $L$ is not piecewise testable if and only if the minimal DFA for $L$ either

contains a nontrivial (non-self-loop) cycle or 2. 2.

there are three distinct states $p$, $q$, $q^{\prime}$ such that $q$ and $q^{\prime}$ are reachable from $p$ by words over the symbols that form self-loops on both $q$ and $q^{\prime}$; formally, there are paths $p\xrightarrow{w}q$ and $p\xrightarrow{w^{\prime}}q^{\prime}$ in the DFA with $w,w^{\prime}\in\Sigma(q)\cap\Sigma(q^{\prime})$, where $\Sigma(q)=\{a\in\Sigma\mid q\xrightarrow{a}q\}$.

We now prove the first result of this paper.

Theorem 2**.**

The piecewise-testability problem for NFAs is PSpace-complete.

Proof.

To prove that piecewise testability is in PSpace, let $\mathcal{A}=(Q,\Sigma,\delta,Q_{0},F)$ be an NFA. Since $\mathcal{A}$ is nondeterministic, we cannot directly use the algorithm of Cho and Huynh [6]. Instead, we consider the DFA $\mathcal{A}^{\prime}$ obtained from $\mathcal{A}$ by the standard subset construction where the states of $\mathcal{A}^{\prime}$ are subsets of states of $\mathcal{A}$. We now need to modify Cho and Huynh’s algorithm to check whether the guessed states are distinguishable.

For a set of states $X\subseteq Q$, let $\Sigma(X)=\{a\in\Sigma\mid X\xrightarrow{a}X\}$. The entire algorithm is presented as Algorithm 1.

In line 1 the algorithm guesses two states, $X$ and $Y$, of $\mathcal{A}^{\prime}$ that are verified to be reachable and in a cycle in line 2. If so, it is verified in lines 10–15 that the states $X$ and $Y$ are not equivalent in $\mathcal{A}^{\prime}$. If there is no nontrivial cycle in $\mathcal{A}^{\prime}$ or the guess in line 1 fails, property (2) of Proposition 1 is verified in lines 3–9, and the guessed states $X$ and $Y$ are checked to be non-equivalent in lines 10–15. Notice that in lines 6–9, the algorithm verifies that the states $X$ and $Y$ are reachable from a state $P$ by paths of the same length rather than by paths of different lengths. This is not a problem because line 7 considers only symbols from $\Sigma(X)\cap\Sigma(Y)$. If $\mathcal{A}^{\prime}$ reaches $X$ under $\Sigma(X)\cap\Sigma(Y)$, it stays in $X$ under those symbols (and analogously for $Y$). Thus, under $\Sigma(X)\cap\Sigma(Y)$, the states $X$ and $Y$ are reachable from state $P$ by paths of different lengths if and only if they are reachable by paths of the same length. The algorithm is in NPSpace $=$ PSpace [35] and returns a positive answer if and only if $\mathcal{A}$ does not accept a piecewise testable language. Since PSpace is closed under complement [19, 42], piecewise testability is in PSpace.

PSpace-hardness follows from a result by Hunt III and Rosenkrantz [18], who have shown that a property $\mathbb{P}$ of languages over the alphabet $\{0,1\}$ such that (i) $\mathbb{P}(\{0,1\}^{*})$ is true and (ii) there exists a regular language that is not expressible as a quotient $x\backslash L=\{w\mid xw\in L\}$, for some $L$ for which $\mathbb{P}(L)$ is true, is as hard as to decide “$=\{0,1\}^{*}$”. Since piecewise testability is such a property (piecewise testable languages are closed under quotient) and universality is PSpace-hard for NFAs, the result implies that piecewise testability for NFAs is PSpace-hard. ∎

4 Separability of Regular Languages by Piecewise Testable Languages

We now show that separability of regular languages by piecewise testable languages is PTime-complete. Since the membership in PTime is known [8, 30], we prove PTime-hardness by constructing a log-space reduction from the PTime-complete monotone circuit value problem [15].

The monotone circuit value problem consists of a set of boolean variables $g_{1}$, $g_{2}$, …, $g_{n}$ called gates, whose values are defined recursively by equalities of the forms $g_{i}=\operatorname{\bf 0}$ (then $g_{i}$ is called a $\operatorname{\bf 0}$-gate), $g_{i}=\operatorname{\bf 1}$ ($\operatorname{\bf 1}$-gate), $g_{i}=g_{j}\wedge g_{k}$ ($\wedge$-gate), or $g_{i}=g_{j}\vee g_{k}$ ($\vee$-gate), where $j,k<i$. Here $\operatorname{\bf 0}$ and $\operatorname{\bf 1}$ are symbols representing the boolean values. The aim is to compute the value of $g_{n}$.

A word $a_{1}a_{2}\cdots a_{n}$ with $a_{i}\in\Sigma$ is a subsequence of a word $w$ if $w\in\Sigma^{*}a_{1}\Sigma^{*}a_{2}\Sigma^{*}\cdots\Sigma^{*}a_{n}\Sigma^{*}$. For languages $K$ and $L$, a sequence $(w_{i})_{i=1}^{r}$ of words is a tower between $K$ and $L$ if $w_{1}\in K\cup L$ and, for all $i=1,2,\ldots,r-1$, $w_{i}$ is a subsequence of $w_{i+1}$, $w_{i}\in K$ implies $w_{i+1}\in L$, and $w_{i}\in L$ implies $w_{i+1}\in K$. The number of words in the sequence is the height of the tower; the height may be infinite. Languages $K$ and $L$ are not required to be disjoint, but a $w\in K\cap L$ implies an infinite tower $w,w,\ldots$ between $K$ and $L$.

Our proof is based on the fact that non-separability of languages $K$ and $L$ by a piecewise testable language is equivalent to the existence of an infinite tower between the languages $K$ and $L$ [8].

Theorem 3**.**

Deciding separability of regular languages represented as NFAs by piecewise testable languages is PTime-complete. It remains PTime-hard even for minimal DFAs.

Proof.

The membership in PTime was independently shown by Czerwiński et al. [8] and Place et al. [30].

We prove PTime-hardness by reduction from the monotone circuit value problem (MCVP). Given an instance $g_{1},g_{2},\ldots,g_{n}$ of MCVP, we construct two minimal DFAs $\mathcal{A}$ and $\mathcal{B}$ using a log-space reduction and prove that there exists an infinite tower between their languages if and only if the circuit evaluates gate $g_{n}$ to $\operatorname{\bf 1}$. The theorem then follows from the fact that non-separability of two regular languages by a piecewise testable language is equivalent to the existence of an infinite tower [8].

Let $f(i)$ be the element of $\{\wedge,\vee,\operatorname{\bf 0},\operatorname{\bf 1}\}$ such that $g_{i}$ is an $f(i)$-gate. For every $\wedge$-gate and $\vee$-gate, we set $\ell(i)$ and $r(i)$ to be the indices such that $g_{i}=g_{\ell(i)}f(i)g_{r(i)}$ is the defining equality of $g_{i}$. If $g_{i}$ is a $\operatorname{\bf 0}$-gate, we set $f(i)=\ell(i)=r(i)=\operatorname{\bf 0}$, and if $g_{i}$ is a $\operatorname{\bf 1}$-gate, we set $f(i)=\ell(i)=r(i)=\operatorname{\bf 1}$.

We first construct an automaton $\mathcal{A}^{\prime}=(Q_{\mathcal{A}^{\prime}},\Sigma,\delta_{\mathcal{A}^{\prime}},s,F_{\mathcal{A}^{\prime}})$ with states $Q_{\mathcal{A}^{\prime}}=\{s,\operatorname{\bf 0},\operatorname{\bf 1},1,2,\dots,n\}$, the input alphabet $\Sigma=\{x,y\}\cup\{a_{i},b_{i}\mid i=1,\ldots,n\}$, and accepting states $F_{\mathcal{A}^{\prime}}=\{\operatorname{\bf 0},\operatorname{\bf 1}\}$. The initial state of $\mathcal{A}^{\prime}$ is $s$ and the transition function $\delta_{\mathcal{A}^{\prime}}$ is defined by $\delta_{\mathcal{A}^{\prime}}(i,a_{i})=\ell(i)$ and $\delta_{\mathcal{A}^{\prime}}(i,b_{i})=r(i)$. In addition, there are two special transitions $\delta_{\mathcal{A}^{\prime}}(s,x)=n$ and $\delta_{\mathcal{A}^{\prime}}(\operatorname{\bf 1},y)=s$.

To construct automaton $\mathcal{B}=(Q_{\mathcal{B}},\Sigma,\delta_{\mathcal{B}},q,F_{\mathcal{B}})$, let $Q_{\mathcal{B}}=\{q,t\}\cup\{i\mid f(i)=\wedge\}$ and $F_{\mathcal{B}}=\{q\}$, where $q$ is also the initial state of $\mathcal{B}$. If $f(i)=\vee$ or $f(i)=\operatorname{\bf 1}$, we define $\delta_{\mathcal{B}}(t,a_{i})=\delta_{\mathcal{B}}(t,b_{i})=t$. If $f(i)=\wedge$, we define $\delta_{\mathcal{B}}(t,a_{i})=i$ and $\delta_{\mathcal{B}}(i,b_{i})=t$. Finally, we define $\delta_{\mathcal{B}}(q,x)=t$ and $\delta_{\mathcal{B}}(t,y)=q$.

All undefined transitions go to the unique sink states of the respective automata. The automata $\mathcal{A}^{\prime}$ and $\mathcal{B}$ can be constructed from $g_{1},\ldots,g_{n}$ in logarithmic space. An example of the construction for the circuit $g_{1}=\operatorname{\bf 0}$, $g_{2}=\operatorname{\bf 1}$, $g_{3}=g_{1}\wedge g_{2}$, $g_{4}=g_{3}\vee g_{3}$ is illustrated in Figure 1.

The languages $L(\mathcal{A}^{\prime})$ and $L(\mathcal{B})$ are disjoint, the automata $\mathcal{A}^{\prime}$ and $\mathcal{B}$ are deterministic, and $\mathcal{B}$ is minimal. However, automaton $\mathcal{A}^{\prime}$ need not be minimal because the circuit may contain gates that do not contribute to the definition of the value of $g_{n}$. We therefore define a minimal deterministic automaton $\mathcal{A}$ by adding new transitions into $\mathcal{A}^{\prime}$, each under a fresh symbol, from state $s$ to each of the states $1,2,\dots,n-1$, from each of the states $1,2,\dots,n$ to state $\operatorname{\bf 0}$, and from state $\operatorname{\bf 0}$ to state $\operatorname{\bf 1}$. This can again be done in logarithmic space. No new transition is defined in $\mathcal{B}$.

Since the language of $\mathcal{B}$ is over $\Sigma$, the symbols of $\mathcal{A}$ not belonging to $\Sigma$ have no effect on the existence of an infinite tower between $L(\mathcal{A})$ and $L(\mathcal{B})$. Namely, there exists an infinite tower between the languages $L(\mathcal{A})$ and $L(\mathcal{B})$ if and only if there exists an infinite tower between $L(\mathcal{A}^{\prime})$ and $L(\mathcal{B})$. It is therefore sufficient to prove that the circuit evaluates gate $g_{n}$ to $\operatorname{\bf 1}$ if and only if there is an infinite tower between the languages $L(\mathcal{A}^{\prime})$ and $L(\mathcal{B})$.

The intuition behind the construction is that the symbols of an infinite tower with unbounded number of occurrences correspond to gates that evaluate to $\operatorname{\bf 1}$ to satisfy $g_{n}$, and that the non-existence of an infinite tower implies the existence of a symbol with bounded number of occurrences in $\mathcal{A}^{\prime}$ that appears in a non-trivial cycle of the form $a_{j}b_{j}$ in $\mathcal{B}$. Such a state corresponds to an $\land$-gate, $g_{j}$, which cannot be satisfied and causes that $g_{n}$ evaluates to $\operatorname{\bf 0}$ (cf. symbol $a_{3}$ in Figure 1).

If there are no $\land$-gates, $g_{n}$ is satisfied if and only if state $\operatorname{\bf 1}$ is reachable from state $n$ in $\mathcal{A}^{\prime}$. Let $w$ be a word under which state $\operatorname{\bf 1}$ is reachable from state $n$. Then $xw\in L(\mathcal{A}^{\prime})$, $xwy\in L(\mathcal{B})$, $xwyxw\in L(\mathcal{A}^{\prime}),\ldots$ is an infinite tower between $L(\mathcal{A}^{\prime})$ and $L(\mathcal{B})$. If state $\operatorname{\bf 1}$ is not reachable from state $n$ in $\mathcal{A}^{\prime}$, then the language $L(\mathcal{A}^{\prime})$ is finite and there is indeed no infinite tower between $L(\mathcal{A}^{\prime})$ and $L(\mathcal{B})$.

The problem with $\land$-gates is how to ensure that both children of an $\land$-gate $g_{j}$ are satisfied. To this aim, we use the nontrivial cycle under $a_{j}b_{j}$ in $\mathcal{B}$, which enforces that both $a_{j}$ and $b_{j}$ appear in the words of an infinite tower. Speaking intuitively, automata $\mathcal{A}^{\prime}$ and $\mathcal{B}$ encode the satisfiability check of $g_{j}$ (see state $g_{3}$ in Figure 1) in the following way. Automaton $\mathcal{A}^{\prime}$ checks reachability of state $\operatorname{\bf 1}$ from state $j$ under a word in $a_{j}\Sigma^{*}\cup b_{j}\Sigma^{*}$ and automaton $\mathcal{B}$ ensures that $a_{j}$ appears in a word in $L(\mathcal{B})$ if and only if $b_{j}$ does. The main idea now is that if there is an infinite tower $(w_{i})_{i=1}^{\infty}$ and $a_{j}$ appears in a word $w_{i}\in L(\mathcal{A}^{\prime})$, then both $a_{j}$ and $b_{j}$ appear in $w_{i+1}\in L(\mathcal{B})$. By the construction of $\mathcal{A}^{\prime}$, symbol $x$ appears between any two occurrences of $a_{j}$ and $b_{j}$, hence $\mathcal{B}$ increases the number of occurrences of $a_{j}$ and $b_{j}$ in the words of the tower as the height grows. Since the tower is infinite, the number of their occurrences is unbounded. However, to read an unbounded number of $a_{j}$ and $b_{j}$ in $\mathcal{A}^{\prime}$ requires that there is a path from state $j$ to state $\operatorname{\bf 1}$ under a word in $a_{j}\Sigma^{*}$ as well as under a word in $b_{j}\Sigma^{*}$, which (using inductively the same argument for other $\land$-gates) is possible only if $g_{j}$ is satisfied. In Figure 1, the words of $L(\mathcal{A}^{\prime})$ contain at most one occurrence of $a_{3}$, whereas those of $L(\mathcal{B})$ require unbounded number of occurrences of $a_{3}$. Thus, there is no infinite tower between the languages of Figure 1.

We now formally prove that the circuit evaluates gate $g_{n}$ to $\operatorname{\bf 1}$ if and only if there is an infinite tower between the languages $L(\mathcal{A}^{\prime})$ and $L(\mathcal{B})$. The dependence between the gates $g_{1},g_{2},\ldots,g_{n}$ can be depicted as a directed acyclic graph $G=(\{1,2,\ldots,n\},E)$, where $E$ is defined as $\delta_{\mathcal{A}^{\prime}}$ without the labels, multiplicities and states $s,\operatorname{\bf 0},\operatorname{\bf 1}$. We say that $i$ is accessible from $j$ if there is a path from $j$ to $i$ in $G$.

(Only if) Assume that $g_{n}$ is evaluated to $\operatorname{\bf 1}$. We construct an alphabet $\Gamma$, $\{x,y\}\subseteq\Gamma\subseteq\Sigma$, under which both automata $\mathcal{A}^{\prime}$ and $\mathcal{B}$ have a cycle containing the initial and an accepting state. These cycles then imply the existence of an infinite tower between the languages $L(\mathcal{A}^{\prime})$ and $L(\mathcal{B})$. Symbol $a_{i}$ belongs to $\Gamma$ if and only if $g_{i}$ is evaluated to $\operatorname{\bf 1}$, $i$ is accessible from $n$, and either $\ell(i)=\operatorname{\bf 1}$ or $g_{\ell(i)}$ is evaluated to $\operatorname{\bf 1}$. Similarly, $b_{i}$ belongs to $\Gamma$ if and only if $i$ is accessible from $n$, $g_{i}$ is evaluated to $\operatorname{\bf 1}$, and either $r(i)=\operatorname{\bf 1}$ or $g_{r(i)}$ is evaluated to $\operatorname{\bf 1}$. It is not hard to observe that each transition labeled by a symbol $a_{i}$ or $b_{i}$ from $\Gamma$ is part of a path from $n$ to $\operatorname{\bf 1}$ in $\mathcal{A}^{\prime}$, hence it appears on a cycle in $\mathcal{A}^{\prime}$ from the initial state $s$ back to state $s$ through the accepting state $\operatorname{\bf 1}$. Moreover, the definition of $\wedge$ implies that $a_{i}\in\Gamma$ if and only if $b_{i}\in\Gamma$ for each $i=1,2,\dots,n$ such that $f(i)=\wedge$. Notice that $\mathcal{B}$ has a cycle from $q$ to $q$ labeled by $xa_{i}b_{i}y$ for each $i=1,2,\dots,n$ with $f(i)\neq\operatorname{\bf 0}$. Therefore, both automata $\mathcal{A}^{\prime}$ and $\mathcal{B}$ have a cycle over the alphabet $\Gamma$ containing the initial and accepting states. The existence of an infinite tower follows.

(If) Assume that there exists an infinite tower $(w_{i})_{i=1}^{\infty}$ between $L(\mathcal{A}^{\prime})$ and $L(\mathcal{B})$, and, for the sake of contradiction, assume that $g_{n}$ is evaluated to $\operatorname{\bf 0}$. Note that any path from $i$ to $\operatorname{\bf 1}$ in $\mathcal{A}^{\prime}$, where $g_{i}$ is evaluated to $\operatorname{\bf 0}$, must contain a state corresponding to an $\wedge$-gate that is evaluated to $\operatorname{\bf 0}$. In particular, this applies to any path in $\mathcal{A}^{\prime}$ accepting a word of the infinite tower of length at least $n+2$, since such a path contains a subpath from $n$ to $\operatorname{\bf 1}$. Let $j$ denote the smallest positive integer such that $f(j)=\wedge$, gate $g_{j}$ is evaluated to $\operatorname{\bf 0}$, and $a_{j}$ or $b_{j}$ is in $\cup_{i=1}^{\infty}\operatorname{\rm alph}(w_{i})$. The construction of $\mathcal{B}$ implies that both $a_{j}$ and $b_{j}$ are in $\cup_{i=1}^{\infty}\operatorname{\rm alph}(w_{i})$ because of the nontrivial cycle $a_{j}b_{j}$. Since $g_{j}$ is evaluated to $\operatorname{\bf 0}$, there exists $c\in\{a,b\}$ such that the transition from $j$ under $c_{j}$ leads to a state $\sigma$, where either $\sigma=\operatorname{\bf 0}$ or $\sigma<j$ and $g_{\sigma}$ is evaluated to $\operatorname{\bf 0}$. Consider a word $w_{i}\in L(\mathcal{A}^{\prime})$ of the infinite tower containing $c_{j}$. If $w_{i}$ is accepted in $\operatorname{\bf 1}$, then the accepting path contains a subpath from $\sigma$ to $\operatorname{\bf 1}$, which yields a contradiction with the minimality of $j$. Therefore, $w_{i}$ is accepted in $\operatorname{\bf 0}$. However, no symbol of a transition to state $\operatorname{\bf 0}$ appears in a word accepted by $\mathcal{B}$ (cf. the symbols $a_{1}$ and $b_{1}$ in Figure 1), a contradiction again. ∎

Acknowledgements

The author is very grateful to Štěpán Holub for his comments on the preliminary version of this work, and to an anonymous reviewer for valuable comments. The research was supported by the German Research Foundation (DFG) in Emmy Noether grant KR 4381/1-1 (DIAMOND).