LANGUAGE CLASSES ASSOCIATED WITH AUTOMATA OVER MATRIX GROUPS

Consider the language $\mathtt{L}=\{a^{n}b^{n}|n\geq 0\}$ which is a context-free language. Since context-free languages are closed under star, $\mathtt{L}^{*}$ is a context-free language whereas it cannot be recognized by any $\mathbb{Z}^{k}$-automaton for all $k\geq 1$ by [11]. On the other hand, the non-context-free language $\mathtt{L}^{\prime}=\{a^{n}b^{n}c^{n}|n\geq 0\}$ can be recognized by a $\mathbb{Z}^{2}$-automaton. ∎

We are going to use Fact 3.2 to prove the result. Since $SL(2,\mathbb{Z})$ has index 2 in $GL(2,\mathbb{Z})$ and $GL(2,\mathbb{Z})$ is finitely generated, $\mathfrak{L}(GL(2,\mathbb{Z}))=\mathfrak{L}(SL(2,\mathbb{Z}))$. Since $\mathbf{F}_{2}$ has index 12 in $SL(2,\mathbb{Z})$ [2] and $SL(2,\mathbb{Z})$ is finitely generated, $\mathfrak{L}(SL(2,\mathbb{Z}))=\mathfrak{L}(\mathbf{F}_{2})$ which is equal to the family of context-free languages by Fact 3.1. ∎

It is obvious that an $SL(2,\mathbb{Z})$-automaton can be simulated by an $SL(3,\mathbb{Z})$-automaton. Note that $\mathfrak{L}(SL(2,\mathbb{Z}))$ is the family of context-free languages by Theorem 3.2. Since $\mathfrak{L}(\mathbf{H})\subsetneq\mathfrak{L}(SL(3,\mathbb{Z}))$ and the non-context-free language $\mathtt{MULT}=\{x^{p}y^{q}z^{pq}|p,q\geq 0\}$ can be recognized by an $\mathbf{H}$-automaton [19], the result follows. ∎

Since $GL(3,\mathbb{Z})$ is a finitely generated group and $SL(3,\mathbb{Z})$ has finite index in $GL(3,\mathbb{Z})$, the result follows by Fact 3.2. ∎

Since $\mathbb{Z}^{2}$ is a subgroup of $\mathbf{H}$, $\mathfrak{L}(\mathbb{Z}^{2})\subseteq\mathfrak{L}(\mathbf{H})$ follows. The inclusion is proper since $\mathbf{H}$ can recognize the language $\mathtt{MULT}=\{x^{p}y^{q}z^{pq}|p,q\geq 0\}$ [19], whereas any bounded language in $\mathfrak{L}(\mathbb{Q}^{+})$ is semilinear [13]. ∎

It is obvious that $\mathfrak{L}(SL(2,\mathbb{Z}))\subseteq\mathfrak{L}(SL(2,\mathbb{Q}))$. We will prove that the inclusion is proper.

Let us construct an $SL(2,\mathbb{Q})$-automaton $\mathcal{G}$ recognizing the language $\mathtt{L}=\{a^{2^{2n+1}}|$ $n\geq 0\}$. The state diagram of $\mathcal{G}$ and the matrices are given in Figure 1. Without scanning any input symbol, $\mathcal{G}$ first multiplies its register with the matrix $A_{1}$. $\mathcal{G}$ then multiplies its register with the matrix $A_{2}$ successively until nondeterministically moving to the next state. After that point, $\mathcal{G}$ starts reading the string and multiplies its register with the matrix $A_{3}$ for each scanned $a$. At some point, $\mathcal{G}$ nondeterministically stops reading the rest of the string and multiplies its register with the matrix $A_{4}$. After successive multiplications with $A_{4}$, $\mathcal{G}$ nondeterministically decides moving to an accept state.

Let us trace the value of the register at different stages of the computation. Before reading the first input symbol, the register has the value

For the rest of the computation, $\mathcal{G}$ will multiply its register with $A_{4}$ until nondeterministically moving to the final state. As a result of $z$ multiplications with $A_{4}$, the register will have the value

The final value of the register is equal to the identity matrix when $y=2^{2x+1}$ and $z=x+1$, which is possible only when the length of the input string is $2^{2x+1}$ for some $x\geq 0$. In the successful branch, the register will be equal to the identity matrix and $\mathcal{G}$ will end up in the final state having successfully read the input string. For input strings which are not members of $\mathtt{L}$, either the computation will end before reading the whole input string, or the final state will be reached with the register value not equaling the identity matrix.

Since the matrices used during the computation are 2 by 2 rational matrices with determinant 1, $\mathtt{L}\in\mathfrak{L}(SL(2,\mathbb{Q}))$. $\mathfrak{L}(SL(2,\mathbb{Q}))$ contains a unary nonregular language, which is not true for $\mathfrak{L}(SL(2,\mathbb{Z}))$ by Theorem 3.2 and we conclude the result. ∎

Let $\mathtt{L}\in\mathfrak{L}(\mathbb{Q}^{+})$ and let $\mathcal{G}$ be a $\mathbb{Q}^{+}$-automaton recognizing $\mathtt{L}$. We will construct an $SL(2,\mathbb{Q})$-automaton $\mathcal{G}^{\prime}$ recognizing $\mathtt{L}$. Let $S=\{s_{1},\dots,s_{n}\}$ be the set of elements multiplied with the register during the computation of $\mathcal{G}$. We define the mapping $\varphi$ as follows.

The elements $\varphi(s_{i})$ are $2\times 2$ rational matrices with determinant 1. Let $\delta$ and $\delta^{\prime}$ be the transition functions of $\mathcal{G}$ and $\mathcal{G}^{\prime}$ respectively. We let $(q^{\prime},s_{i})\in\delta(q,\sigma)\iff(q^{\prime},\varphi(s_{i}))\in\delta^{\prime}(q,\sigma)$ for every $q,q^{\prime}\in Q$, $\sigma\in\Sigma$ and $s_{i}\in S$. The resulting $\mathcal{G}^{\prime}$ recognizes $\mathtt{L}$.

The inclusion is proper since $\mathtt{L}=\{a^{2^{2n+1}}|n\geq 0\}\in\mathfrak{L}(SL(2,\mathbb{Q}))$ by Theorem 3.2, and $\mathfrak{L}(\mathbb{Q}^{+})$ does not contain any unary nonregular languages by Fact 3.1, noting that $\mathbb{Q}^{+}$-automata are equivalent to 1NFAMW’s. ∎

Let us construct a $BS(1,2)$-automaton $\mathcal{G}$ recognizing the language $\mathtt{UPOW}=\{a^{2^{n}}|n\geq 0\}$. The state diagram of $\mathcal{G}$ and the matrices are given in Figure 2. Without scanning any input symbol, $\mathcal{G}$ multiplies its register with the matrix $A_{1}$ successively. $\mathcal{G}$ nondeterministically moves to the next state reading the first input symbol without modifying the register. After that point, $\mathcal{G}$ starts reading the string and multiplies its register with the matrix $A_{2}$ for each scanned $a$. At some point, $\mathcal{G}$ nondeterministically stops reading the rest of the string and multiplies its register with the element $A_{3}$. After successive multiplications with $A_{3}$, $\mathcal{G}$ nondeterministically decides to move to an accept state.

As a result of $i$ multiplications with $A_{1}$, the register has the value

before reading the first input symbol. Multiplication with each $A_{2}$ leaves $2^{i}$ unchanged while subtracting 1 from $2^{i}-1$ for each scanned $a$. The register will have the value

as a result of $j$ multiplications with the matrix $A_{2}$.

For the rest of the computation, $\mathcal{G}$ will multiply its register with $A_{3}$ resulting in the register value

since each multiplication with $A_{3}$ divides $2^{i}$ by 2.

The register contains the identity matrix at the end of the computation if $i=k$ and $j=2^{i}-1$ which is possible if the input string is of the form $a^{1+2^{i}-1}=a^{2^{i}}$. In the successful branch, the register will be equal to the identity matrix and $\mathcal{G}$ will end up in the final state having successfully read the input string.

For input strings which are not members of $\mathtt{UPOW}$, either the computation will end before reading the whole input string or the final state will be reached with the register value being different from the identity matrix. Note that $A_{1}=B^{-1}A^{-1}$, $A_{2}=A$ and $A_{3}=B$, where $A$ and $B$ are the generators of the group $G_{BS}$ and recall that $G_{BS}$ is isomorphic to $BS(1,2)$. Since $\mathtt{UPOW}$ is a unary nonregular language, it is not context-free and we conclude the result. ∎

The first equality is Fact 3.3. Recall from Section 3.2 that $\varphi$ is an isomorphism from $\mathbf{F}_{2}$ onto $\mathbf{G}$, the matrix group generated by the matrices $M_{a}$ and $M_{b}$. Let $\mathbf{G}^{\prime}$ be the following group of matrices

We will define the mapping $\psi:\mathbf{F}_{2}\times\mathbf{F}_{2}\rightarrow\mathbf{G}^{\prime}$ as $\psi(g_{1},g_{2})=(\varphi(g_{1}),\varphi(g_{2}))$ for all $(g_{1},g_{2})\in\mathbf{F}_{2}\times\mathbf{F}_{2}$ which is an isomorphism from $\mathbf{F}_{2}\times\mathbf{F}_{2}$ onto $\mathbf{G}^{\prime}$.

This proves that $\mathbf{F}_{2}\times\mathbf{F}_{2}$ is isomorphic to a subgroup of $SL(4,\mathbb{Z})$. The fact that $\mathfrak{L}(\mathbf{F}_{2}\times\mathbf{F}_{2})$ is the set of recursively enumerable languages helps us to conclude that $\mathfrak{L}(SL(n,\mathbb{Z}))$ is the set of recursively enumerable languages for $n\geq 4$. ∎

Note that any finite automaton over a matrix group can be simulated by a nondeterministic Turing machine which keeps track of the register simply by multiplying the matrices and checking whether the identity matrix is reached at the end of the computation, provided that the matrix entries are computable numbers. Since $\mathsf{RE}=\mathfrak{L}(SL(4,\mathbb{Z}))$ and $G$ contains $SL(4,\mathbb{Z})$ as a subgroup, $\mathfrak{L}(G)$ is the set of recursively enumerable languages. ∎

Suppose for a contradiction that there exists a weakly $t(n)$ time-bounded $G$-automaton $\mathcal{G}$ recognizing $\mathtt{L}$ in time $t(n)$. For a sufficiently large $n$, let $S$ be the set of uniformly $n$-dissimilar strings such that $|S|=U_{\mathtt{L}}(n)$. For every string $w_{i}\in S$, there exists a string $v_{i}$ such that $w_{i}v_{i}\in\mathtt{L}$ and $w_{j}v_{i}\notin\mathtt{L}$ for all $w_{j}\in S$ with $i\neq j$ .

Let $S_{acc}$ be the set of accepted extended strings of the form $w_{i}v_{i}\in\mathtt{L}$ with $|w_{i}v_{i}|\leq n$ where $w_{i}\in S$ and $w_{j}v_{i}\notin\mathtt{L}$ for all $w_{j}\in S$ with $i\neq j$ and $|w_{j}v_{i}|\leq n$. Let $C$ be the set of $t(n)$ time bounded accepting computation paths for the strings in $S_{acc}$. The computation $c_{w_{i}v_{i}}\in C$ on the string $w_{i}v_{i}$ can be written as

where $c_{w_{i}v_{i}}^{w_{i}}$ represents the computation up to the end of the prefix $w_{i}$ and $c_{w_{i}v_{i}}^{v_{i}}$ represents the rest of the computation on the string $v_{i}$.

A configuration of a group automaton is a pair consisting of a state and a group element. Let us count the number of configurations that can be reached at the end of the computation $c_{w_{i}v_{i}}^{w_{i}}$. Since the number of states is constant, the number of configurations that can be reached is dependent on the number of different group elements that can appear in the register. After reading a prefix $w_{i}$ with $|w_{i}|=m\leq n$, the product of the labels on the edges can be given by $l=g_{i_{1}}g_{i_{2}}\dots g_{i_{k}}$ for some $k\leq t(m)$, since the computation in consideration is time bounded. $l$ can be expressed as a product of $\kappa$ generators, where $\kappa$ is at most $C\cdot k$ for some constant $C$, since each group element labeling a transition in $\mathcal{G}$ is composed of at most some constant number of generators, which is independent of the length of the string. The number of elements in $G$ which can be represented as a product of at most $\kappa$ generators is given by $g_{G}(\kappa)$ by the definition of the growth function of $G$. Hence, the number of different values that can appear in the register after reading a string of length exactly $m$ is less than or equal to $g_{G}(\kappa)$. Since $\kappa\leq C\cdot k$ and $k\leq t(m)$ and $g_{G}(t(n))\in o(U_{\mathtt{L}}(n))$, we can conclude that

Now it is easy to see that the number of different configurations that can be reached at the end of a computation $c_{w_{i}v_{i}}^{w_{i}}$ is $o(U_{\mathtt{L}}(n))$. Note that the cardinality of the set $S$, and thus that of $S_{acc}$, is equal to $U_{L}(n)$. Due to the pigeonhole principle, the same configuration must be reached at the end of two computations $c_{w_{i}v_{i}}^{w_{i}}$ and $c_{w_{j}v_{j}}^{w_{j}}$ for some $i\neq j$. This will result in the acceptance of the strings $w_{i}v_{j}$ and $w_{j}v_{i}$, which are not members of $\mathtt{L}$. We arrive at a contradiction and conclude that $\mathtt{L}$ cannot be recognized by any weakly $t(n)$ time-bounded $G$-automaton. ∎

Let $X$ be the generator set of $G$. The number of distinct elements $g$ in $G$ which can be represented by a word of length less than or equal to $\lfloor\frac{n}{2}\rfloor$ is $g_{G}(\lfloor\frac{n}{2}\rfloor)$, which is the cardinality of the set $B_{G}^{X}(\lfloor\frac{n}{2}\rfloor)=\{g\in G,|g|_{X}\leq\lfloor\frac{n}{2}\rfloor\}$. Let $S$ be the set containing the string representations of the elements in $B_{G}^{X}(\lfloor\frac{n}{2}\rfloor)$. Every $w_{i}\in S$ can be extended with $w_{i}^{-1}$ so that the extended string represents the identity element of $G$ and has length less than or equal to $n$. Since the strings in $W(G)$ are those which belong to $(X\cup X^{-1})^{*}$ and represent the identity element of $G$, the extended string $w_{i}w^{-1}_{i}\in W(G)$. For every string $w_{j}\in S$ such that $i\neq j$, $w_{j}w_{i}^{-1}\notin W(G)$ since it is not possible for $w_{j}w_{i}^{-1}$ to represent the identity element of $G$. We conclude that the set $S$ is uniformly $n$-dissimilar. Since $|S|=|B_{G}^{X}(\lfloor\frac{n}{2}\rfloor)|=g_{G}(\lfloor\frac{n}{2}\rfloor)$, it follows that $U_{W(G)}(n)\geq g_{G}(\lfloor\frac{n}{2}\rfloor)$. ∎

Since $U_{W(H)}(n)\geq g_{H}(\lfloor\frac{n}{2}\rfloor)$ by Lemma 4.2, and $g_{H}(n)$ is an exponential function, $U_{W(H)}(n)$ is also at least exponential. $g_{G}(t(n))$ is a polynomial function, since both $g_{G}(n)$ and $t(n)$ are polynomial. Hence, $W(H)\notin\mathfrak{L}(G)_{t(n)}^{w}$ by Theorem 4.2, and the result follows since $W(H)$ is trivially in $\mathfrak{L}(H)$. ∎

It is known that the word problem of the free group of rank 2, $W(\mathbf{F}_{2})$, has an exponential growth function [12]. Assuming that $G$ is a group with polynomial growth function, $W(\mathbf{F}_{2})$ cannot be recognized by any weakly $t(n)$ time-bounded $G$-automaton by Theorem 4.2. Since $W(\mathbf{F}_{2})$ is a context-free language, the proof is complete. ∎

We are going to use the construction of Kambites [14] to prove that any context-free language can be recognized by a weakly linear-time bounded $\mathbf{F}_{2}$-automaton.

Let $\mathtt{L}$ be a context-free language and let $\mathcal{M}=\{Q,\Sigma,K_{X},\delta,q_{0},Q_{a}\}$ be a polycyclic monoid automaton recognizing $\mathtt{L}$. $K_{X}$ is the polycyclic monoid on $X$ where the cardinality of the set $X$ is $n$ for some $n\geq 2$. Let $e$ be the identity element of $K_{x}$. The construction of Kambites provides an $\mathbf{F}_{n+1}$-automaton $\mathcal{G}=\{Q^{\prime},\Sigma,\mathbf{F}_{n+1},\delta^{\prime},q_{0}^{\prime},Q_{a}^{\prime}\}$ recognizing the language $\mathtt{L}$. The generator set for $\mathbf{F}_{n+1}$ is $X^{\prime}$, where $X^{\prime}=X\cup\#$.

Let us analyze the construction in more detail.

• •

  $Q^{\prime}=Q_{-}\cup Q_{+}$ where $Q_{-}=\{q_{-}|q\in Q\}$ and $Q_{+}=\{q_{+}|q\in Q\}$

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  $q_{0}^{\prime}$=$q_{+}$ where $q=q_{0}$.

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  $Q_{a}^{\prime}=\{q_{-}|q\in Q_{a}\}$.

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  $\delta^{\prime}(p_{+},\sigma)=(q_{+},x\#)$ if $\delta(p,\sigma)=(q,x\#)$ where $x$ is a positive generator for all $\sigma\in\Sigma$.

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  $\delta^{\prime}(p_{-},\sigma)=(q_{+},x^{\prime}\#)$ if $\delta(p,\sigma)=(q,x^{\prime}\#)$ where $x^{\prime}$ is a negative generator for all $\sigma\in\Sigma$.

• •

  $\delta^{\prime}(p_{+},\sigma)=(q_{+},e)$ if $\delta(p,\sigma)=(q,e)$ for all $\sigma\in\Sigma$.

• •

  $\delta^{\prime}(q_{+},\epsilon)=(q_{-},e)$ for each $q\in Q$.

• •

  $\delta^{\prime}(q_{-},\epsilon)=(q_{-},\#^{-1})$ for each $q\in Q$.

We will prove that $\mathcal{G}$ actually runs in linear time. There are two transitions where the automaton is allowed to move without consuming any input symbols.

For each state $q\in Q$, there are two states $q_{+}$ and $q_{-}$ in $\mathcal{G}$ which are connected with an edge labeled $(\epsilon,e)$. These transitions do not change the register value, and cannot contribute more than half of the runtime of the machine, since at least one input symbol has to be consumed between any two executions of such transitions.

$\epsilon$-loops exist in the machine $\mathcal{G}$ for each state $q_{-}$ where the loop is labeled by $(\epsilon,\#^{-1})$. Although this looks worrisome at first for the purpose of bounding the runtime, the number of times these loops are traversed is actually bounded, as the following argument suggests. Suppose that the register is multiplied with $l_{1}$, $l_{2}$, $\cdots$, $l_{m}$ while reading some input string $w$ of length $n$, resulting in the register value $l=l_{1}l_{2}\cdots l_{m}(\#^{-1})^{k}$, where $k\in\mathbb{N}$, at the end of the computation. If $w$ is accepted by the machine, $l$ should satisfy the following, as well as being equal to the identity element:

This is called a permissible padding in [14]. By looking at the transition function of $\mathcal{G}$, one can see that the register is multiplied by a $\#$ only when an input symbol is consumed. Hence, the number of $\#$’s that occur in $l$ is less than or equal to the length of the string. The register is multiplied with $\#^{-1}$ without consuming any input symbol. In order for the $\#$’s and $\#^{-1}$’s to cancel each other, they should be equal in number. Therefore, it can be concluded that the $\epsilon$-loops are traversed at most $n$ times.