On counting functions and slenderness of languages

TL;DR

This paper investigates counting-regular languages, showing certain automata accept such languages, explores their closure and decidability properties, and studies the concept of $k$-slender languages within formal language theory.

Contribution

It establishes the class of languages accepted by specific automata as counting-regular, analyzes their closure and decidability properties, and examines $k$-slender languages in formal language theory.

Findings

Languages accepted by unambiguous one-way Turing machines with reversal-bounded worktapes are counting-regular.

Closure properties of counting-regular languages are characterized, with some undecidability results.

Decidability of $k$-slenderness for languages in semilinear full trios is proven.

Abstract

We study counting-regular languages -- these are languages $L$ for which there is a regular language $L'$ such that the number of strings of length $n$ in $L$ and $L'$ are the same for all $n$. We show that the languages accepted by unambiguous nondeterministic Turing machines with a one-way read-only input tape and a reversal-bounded worktape are counting-regular. Many one-way acceptors are a special case of this model, such as reversal-bounded deterministic pushdown automata, reversal-bounded deterministic queue automata, and many others, and therefore all languages accepted by these models are counting-regular. This result is the best possible in the sense that the claim does not hold for either $2$-ambiguous PDA's, unambiguous PDA's with no reversal-bound, and other models. We also study closure properties of counting-regular languages, and we study decidability problems in regards to counting-regularity. For example, it is shown that the counting-regularity of even some restricted subclasses of PDA's is undecidable. Lastly, $k$-slender languages -- where there are at most $k$ words of any length -- are also studied. Amongst other results, it is shown that it is decidable whether a language in any semilinear full trio is $k$-slender.

Equations

$$f_{L_{6}}(n)=\begin{cases}{2n\choose n}&\mbox{~{}if $n\equiv 0$ (mod 3)}\\ 0&\mbox{else}.\end{cases}$$

$$f_{L_{2}}(n)=\begin{cases}2^{n-2}+{n-1\choose{n-1\over 2}}&\mbox{~{}if $n$ is odd and~{}}n\geq 3,\\ 2^{n-2}+{{n-2}\choose{(n-2)\over 2}}&\mbox{~{}if $n$ is even (and $n\geq 2$)}.\end{cases}$$

$$L_{\text{RCM}}=\{w\in \{a,b{\}}^{\ast }|w[|w{|}_a]=b\}.$$

$$|u|=|\zeta (u)|,$$

$$|\chi (x)|=\sum _{1\le i\le k}|x{|}_{a_i}|\chi (a_i)|=\sum _{1\le i\le k}{l}_i|\chi (a_i)|=\sum _{1\le i\le k}{l}_i|u_i|=|u|.$$

$$\sum _{m=0}^{\lfloor n/2\rfloor }\left(\genfrac{}{}{0ex}{}{n}{2m}\right){\left(\frac{a}{b}\right)}^{n-2m}{\left({\left(\frac{a}{b}\right)}^2-1\right)}^m.$$

$$L_{1}^{\prime}=\{\begin{array}[t]{l}ID_{1}\#ID_{3}\#\cdots\#ID_{2k-1}\$ID_{2k}^{R}\#\cdots\#ID_{4}^{R}\#ID_{2}^{R}\mid k\geq 2,\mbox{~{}each~{}}ID_{i}\\ \mbox{is a configuration of~{}}Z,ID_{1}\mbox{~{}is initial~{}}ID\mbox{~{}of~{}}Z\mbox{~{}on blank tape,}\\ ID_{2k}\mbox{~{}is a unique halting~{}}ID,ID_{i}\Rightarrow ID_{i+1}\mbox{~{}for~{}}i=1,3,\ldots,2k-1\},\end{array}$$

$$L_{2}^{\prime}=\begin{array}[t]{l}\{ID_{1}\#ID_{3}\#\cdots\#ID_{2k-1}\$ID_{2k}^{R}\#\cdots\#ID_{4}^{R}\#ID_{2}^{R}\mid k\geq 2,\mbox{~{}each~{}}ID_{i}\\ \mbox{is a configuration of~{}}Z,ID_{1}\mbox{~{}is initial~{}}ID\mbox{~{}of~{}}Z\mbox{~{}on blank tape},\\ ID_{2k}\mbox{~{}is a unique halting~{}}ID,ID_{i}\Rightarrow ID_{i+1}\mbox{~{}for~{}}i=2,4,\ldots,2k-2\},\end{array}$$

$$L_1$$

$$L_2$$

$$L_1\cap L_2=\{xw{1}^n|x\in (a+b{)}^{+},|x{|}_a=|x{|}_b=n\}.$$

$$f(z)=\frac{1}{3}\left(\frac{1}{\sqrt{1-4z^3}}+\frac{1}{\sqrt{1-4\omega z^3}}+\frac{1}{\sqrt{1-4{\omega }^2{z}^3}}\right)z^t.$$

$$L_{1}=\begin{array}[t]{l}\{ID_{1}\#ID_{3}\#\cdots\#ID_{2k-1}\$ID_{2k}^{R}\#\cdots\#ID_{4}^{R}\#ID_{2}^{R}\mid k\geq 2,\mbox{~{}each~{}}ID_{i}\\ \mbox{is a configuration of~{}}Z,ID_{1}\mbox{~{}is initial~{}}ID\mbox{~{}of~{}}Z\mbox{~{}on blank tape,~{}}ID_{2k}\mbox{~{}is the}\\ \mbox{halting~{}}ID,ID_{i}\Rightarrow ID_{i+1}\mbox{~{}for~{}}i=1,3,\ldots,2k-1,|ID_{1}|=\cdots=|ID_{2k}|\},\end{array}$$

$$L_{2}=\begin{array}[t]{l}\{ID_{1}\#ID_{3}\#\cdots\#ID_{2k-1}\$ID_{2k}^{R}\#\cdots\#ID_{4}^{R}\#ID_{2}^{R}\mid k\geq 2,\mbox{~{}each~{}}ID_{i}\mbox{~{}is}\\ \mbox{~{}a configuration of~{}}Z,ID_{1}\mbox{~{}is initial~{}}ID\mbox{~{}of~{}}Z\mbox{~{}on blank tape},ID_{2k}\mbox{~{}is the}\\ \mbox{halting~{}}ID,ID_{i}\Rightarrow ID_{i+1}\mbox{~{}for~{}}i=2,4,\ldots,2k-2,|ID_{1}|=\cdots=|ID_{2k}|\},\end{array}$$

$$w=I{D}_1\#I{D}_3\#\cdots \#I{D}_{2k-1}\$I{D}_{2k}^R\#\cdots \#I{D}_4^R\#I{D}_2^R,$$

$$L_1$$

$$L_2$$

$$L=\{a^i{b}^i\mid i\ge 1\}\cup \{c^i{a}^i{d}^i\mid i\ge 1\}.$$

$$f_{L}(n)=\begin{cases}0&\mbox{if $n=0$ or $n=1$ or not divisible by $2$ and $3$,}\\ 1&\mbox{if $n\geq 2$, is divisible by $2$, and not divisible by $3$,}\\ 1&\mbox{if $n\geq 3$, is divisible by $3$, and not divisible by $2$,}\\ 2&\mbox{if $n\geq 2$, is divisible by $2$ and divisible by $3$}.\end{cases}$$

$$w=x\$y_1\#\cdots \#y_r,$$

Full text

We also study closure properties of counting-regular languages, and we study decidability problems in regards to counting-regularity. For example, it is shown that the counting-regularity of even some restricted subclasses of $\PDA$’s is undecidable. Lastly, $k$-slender languages — where there are at most $k$ words of any length — are also studied. Amongst other results, it is shown that it is decidable whether a language in any semilinear full trio is $k$-slender.

††journal: Theoretical Computer Science

\newclass\DCM

DCM \newclass\RCMRCM \newclass\eDCMeDCM \newclass\eNPDAeNPDA \newclass\DPDADPDA \newclass\RDPDARDPDA \newclass\PDANPDA \newclass\MAJMAJ \newclass\DCMNEDCM_NE \newclass\TwoDCM2DCM \newclass\NCMNCM \newclass\NCPMNCPM \newclass\eNCMeNCM \newclass\eNQAeNQA \newclass\eNSAeNSA \newclass\eNPCMeNPCM \newclass\eNQCMeNQCM \newclass\eNSCMeNSCM \newclass\DPCMDPCM \newclass\NPCMNPCM \newclass\NQCMNQCM \newclass\NSCMNSCM \newclass\NPDANPDA \newclass\TRETRE \newclass\NFANFA \newclass\DFADFA \newclass\NCANCA \newclass\DCADCA \newclass\DTMDTM \newclass\NTMNTM \newclass\DLOGDLOG \newclass\CFGCFG \newclass\ETOLET0L \newclass\EDTOLEDT0L \newclass\CFPCFP \newclass\ORDERO \newclass\MATRIXM \newclass\BDBD \newclass\LBLB \newclass\ALLALL \newclass\decLBDdecLBD \newclass\StLBStLB \newclass\SBDSBD \newclass\TCATCA \newclass\RNCSARNCSA \newclass\RDCSARDCSA \newclass\ctCOUNT \newclass\DCSADCSA \newclass\NCSANCSA \newclass\DCSACMDCSACM \newclass\NCSACMNCSACM \newclass\NTMCMNTMCM \newclass\codecode \newclass\UFIN\LL(IND_UFIN) \newclass\UFINONE\LL(IND_UFIN_1) \newclass\FIN\LL(IND_FIN) \newclass\ILIN\LL(IND_LIN) \newclass\ETOLfin\LL(ET0L_FIN)

label1]ibarra@cs.ucsb.edu

label2]mcquillan@cs.usask.ca

label3]ravikuma@sonoma.edu

1 Introduction

This work is concerned with the counting function $f_{L}(n)$ of a language $L$, equal to the number of strings of length $n$ in $L$. We say that a language $L$ is counting-regular if there is a regular language $L^{\prime}$ such that $f_{L}(n)$ = $f_{L^{\prime}}(n)$. If we can effectively find a deterministic finite automaton ($\DFA$) for $L^{\prime}$ from a representation of $L$ using standard techniques, then we can efficiently compute $f_{L}(n)$ even if membership in $L$ can be difficult to answer. We say that $L$ is strongly counting-regular if for any regular language $L_{1}$, $L\cap L_{1}$ is counting-regular. Being strongly counting-regular will allow us to calculate not only the number of strings of length $n$ in the language, but also the number of strings of length $n$ in $L$ that satisfy additional conditions, such as strings in which some symbol occurs an odd number of times, or strings that avoid some substring etc. A language family $\LL$ is said to be (strongly) counting-regular if all $L\in\LL$ are (strongly) counting regular.

As a simple motivating example, consider the language $L$ = $\{x\#y\ |\ x,y\in\{a,b\}^{*},|x|_{a}=|y|_{b}\}$. The number of strings of length $n$ in $L$ is $2^{n-1}$ (for $n\geq 1$) which follows after a little bit of work from the series sum $\sum_{i=0}^{n-1}\sum_{k=0}^{i}{i\choose k}{n-i-1\choose k}$ for $f_{L}(n)$. (This result can also be established by exhibiting a bijective mapping between the set of strings of length $n$ in $L$ and the set of binary strings of length $n-1$.) Therefore this language is a counting-regular language while a simpler looking, and closely related language $L_{eq}$ = $\{x\ |\ x\in\{a,b\}^{*},|x_{a}|=|x_{b}|\}$ is not counting-regular. Is there an ‘explanation’ for the disparate behaviors of these two languages? In [1], it was determined that all languages accepted by one-way deterministic counter machines where the counter is reversal-bounded, are counting-regular. Indeed, $L$ above is in this family. Of interest is the problem of determining other language families that are (strongly) counting-regular. It can be observed that there are arbitrarily complex languages (e.g., languages that are not recursively enumerable) that are counting-regular. Therefore, to get some insight into counting-regularity, we should restrict the class of languages, such as to the class of context-free languages $\CFL$’s and some subclasses of $\CFL$’s.

Slender languages (those that are $k$-slender, for some $k$, i.e. $f_{L}(n)\leq k$ for all $n$) were studied in [2, 3] in order to better understand the structure of slender context-free languages ($\CFL$’s) as well as to present decision algorithms for problems such as equivalence and containment when $\CFL$’s are restricted to slender languages. Similar results were shown for the more general class of languages generated by matrix grammars [4]. It is natural to determine larger families of slender languages for which equivalence, containment etc. are decidable.

We briefly state some of the motivations for the study presented in this paper. The counting functions of regular languages are well understood [5] and there are very efficient algorithms to compute $f_{L}(n)$ when $L$ is regular. But for other classes, such algorithms are not known although there are a few exceptions such as linearly constrained languages [6]. Hence, as outlined above, a strongly counting-regular language admits an efficient algorithm for counting the number of strings of length $n$ in it. In some applications, we are not necessarily interested in computing exactly the number of strings of length $n$ for a specific $n$, but in asymptotic growth rate of $f_{L}(n)$ as $n\rightarrow\infty$. In the works [7, 8, 9, 1] etc., the issue of how well a non-regular language $L$ can be approximated by a $\DFA$ $M$ asymptotically was studied. The asymptotic approximation bound is the density of the language $L(M)\oplus L$, and so if $L$ is strongly counting-regular, we can compute the approximation bound. Another area of application is the static analysis of programs that involves computing the execution information rate [10]. The information rate can be measured by modeling a control-flow of a program using a $\DFA$ or a $\DFA$ augmented with counters. Since the information rate of a regular language is efficiently computable, it is also efficiently computable for languages that are (effectively) counting-regular.

A main result shown here is that the languages accepted by unambiguous nondeterministic Turing machines ($\NTM$s) with a one-way read-only input and a reversal-bounded worktape are strongly counting-regular. Many different families of languages accepted by one-way unambiguous and nondeterministic (and one-way deterministic) machine models are contained in this family, and are therefore counting-regular.

Counting-regularity is a special case of the notion of a language being commutatively equivalent to a regular language, which requires not only a length preserving bijection to the regular language, but a bijection that preserves the Parikh map. It has been shown that every bounded semilinear language is commutatively equivalent to a regular language (proof split across three papers [11, 12, 13]). Here, we provide an alternate significantly shorter proof that they are all counting-regular. This is then used to show that in every semilinear trio (a family closed under $\epsilon$-free homomorphism, inverse homomorphism, intersection with regular languages, and every language has the same Parikh map as some regular language), all bounded languages are counting-regular. More generally, the counting functions of the bounded languages in every semilinear trio are exactly the same, no matter the family. There are many families for which this holds in the literature, such as Turing machines with a one-way read-only input tape plus a finite-crossing worktape [14], and one-way machines with $k$ pushdown stacks where the machines can write to any of the stacks but only pop from the first non-empty pushdown [15], uncontrolled finite-index indexed grammars [16], amongst many others discussed in [17, 14].

The rest of the paper is organized as follows: in Section 2, basic definitions, terminology and notation are presented. In Section 3, the languages accepted by the Turing machine model above are shown to be strongly counting-regular. We also present some specific language families that are not all counting-regular. Next, in Section 4, we consider bounded languages and show that all bounded languages in any semilinear trio are counting-regular. In Section 5, closure properties of the class of counting-regular languages (and context-free counting-regular languages) are studied. Then, in Section 6, we present some undecidability results regarding counting-regular languages. The main result is that it is undecidable, given a machine of any of the following types: (i) $2$-ambiguous $\PDA$’s which make only one reversal on the stack, (ii) nondeterministic one-counter machines that make only one reversal on the counter, and (iii) $2$-ambiguous nondeterministic one counter machines, whether the language accepted is counting-regular. Then, in Sections 7 and 8, decidability properties are discussed for slender languages. It is shown that for any semilinear full trio (where certain closure properties are constructive) $\LL$, it is decidable whether a language $L\in\LL$ is $k$-slender, and containment and equality of two $k$-slender languages in $\LL$ are decidable. It is also shown that in every such family, every $k$-slender language in $\LL$ is counting-regular.

2 Basic Terminology and Notation

We assume an introductory knowledge in formal language and automata theory [18], including the definitions of finite automata, context-free languages, Turing machines etc. Next, some notations are given. An alphabet is a finite set of symbols. A word $w$ over $\Sigma$ is any finite sequence of symbols from $\Sigma$. Given an alphabet $\Sigma$, then $\Sigma^{*}$ is the set of all words over $\Sigma$, including the empty word $\epsilon$, and $\Sigma^{+}$ is the set of non-empty words over $\Sigma$. A language $L$ is any subset of $\Sigma^{*}$. The complement of a language $L\subseteq\Sigma^{*}$ with respect to $\Sigma$ is $\overline{L}=\Sigma^{*}-L$. Given a word $w\in\Sigma^{*}$, $w^{R}$ is the reverse of $w$, $w[i]$ is the $i$’th character of $w$, and $|w|$ is the length of $w$. Given an alphabet $\Sigma=\{a_{1},\ldots,a_{m}\}$ and $a\in\Sigma$, $|w|_{a}$ is the number of $a$’s in $w$. The Parikh map of $w$ is $\psi(w)=(|w|_{a_{1}},\ldots,|w|_{a_{m}})$, which is extended to the Parikh map of a language $L$, $\psi(L)=\{\psi(w)\mid w\in L\}$. Also, $\operatorname{alph}(w)=\{a\in\Sigma\mid|w|_{a}>0\}$. Given languages $L_{1},L_{2}$, the left quotient of $L_{2}$ by $L_{1}$, $L_{1}^{-1}L_{2}=\{y\mid xy\in L_{2},x\in L_{1}\}$, and the right quotient of $L_{1}$ by $L_{2}$ is $L_{1}L_{2}^{-1}=\{x\mid xy\in L_{1},y\in L_{2}\}$.

For a language $L$, let $f_{L}(n)$ be the number of strings of length $n$ in $L$. A language $L$ is called counting-regular if there exists a regular language $L^{\prime}$ such that for all integers $n\geq 0$, $f_{L}(n)$ = $f_{L^{\prime}}(n)$. Furthermore, $L$ is called strongly counting-regular if, for any regular language $L_{1}$, $L\cap L_{1}$ is counting-regular. Let $k\geq 1$. A language $L$ is $k$-slender if $f_{L}(n)\leq k$ for all $n$, and $L$ is thin if is is $1$-slender. Furthermore, $L$ is slender if it is $k$-slender for some $k$.

A language $L\subseteq\Sigma^{*}$ is bounded if there exist (not necessarily distinct) words $w_{1},\ldots,w_{k}\in\Sigma^{+}$ such that $L\subseteq w_{1}^{*}\cdots w_{k}^{*}$; it is also letter-bounded if each of $w_{1},\ldots,w_{k}$ are letters.

Let $\mathbb{N}$ be the set of positive integers and $\mathbb{N}_{0}=\mathbb{N}\cup\{0\}$. A linear set is a set $Q\subseteq\mathbb{N}_{0}^{m}$ if there exist $\vec{v_{0}},\vec{v_{1}},\ldots,\vec{v_{n}}$ such that $Q=\{\vec{v_{0}}+i_{1}\vec{v_{1}}+\cdots+i_{n}\vec{v_{n}}\mid i_{1},\ldots,i_{n}\in\mathbb{N}_{0}\}$. The vector $\vec{v_{0}}$ is called the constant, and $\vec{v_{1}},\ldots,\vec{v_{n}}$ are called the periods. We also say that $Q$ is the linear set generated by constant $\vec{v_{0}}$ and periods $\vec{v_{1}},\ldots,\vec{v_{n}}$. A linear set is called simple if the periods form a basis. A semilinear set is a finite union of linear sets. And, a semilinear set is semi-simple if it is the finite disjoint union of simple sets [19, 20].

A language $L\subseteq\Sigma^{*}$ is semilinear if $\psi(L)$ is a semilinear set. Equivalently, a language $L$ is semilinear if and only if there is a regular language $L^{\prime}$ with the same Parikh map (they have the same commutative closure) [21]. The length set of a language $L$ is the set $\{n\mid w\in L,n=|w|\}$. A language $L$ is length-semilinear if the length set is a semilinear set; i.e. after mapping all letters of $L$ onto one letter, the Parikh map is semilinear, which is equivalent to it being regular.

A language $L\subseteq\Sigma^{+}$ is a code if $x_{1}\cdots x_{n}=y_{1}\cdots y_{m},x_{i},y_{j}\in L$, implies $n=m$ and $x_{i}=y_{j}$, for $i$, $1\leq i\leq n$. Also, $L$ is a prefix code if $L\cap L\Sigma^{+}=\emptyset$, and a suffix code if $L\cap\Sigma^{+}L=\emptyset$. See [22] for background on coding theory.

A language family $\LL$ is said to be semilinear if all $L\in\LL$ are semilinear. It is said that language family $\LL$ is a trio if $\LL$ is closed under inverse homomorphism, $\epsilon$-free homomorphism, and intersection with regular languages. In addition, $\LL$ is a full trio if it is a trio closed under homomorphism; and a full AFL is a full trio closed under union, concatenation, and Kleene-*. Many well-known families form trios, such as each family of the Chomsky hierarchy [18]. The theory of these types of families is explored in [23]. When discussing a language family that has certain properties, such as a semilinear trio, we say that the family has all properties effective if all these properties provide effective constructions. For semilinearity, this means that there is an effective construction to construct the constant and periods from each linear set making up the semilinear set.

A pushdown automaton $M$ is $t$-reversal-bounded if $M$ makes at most $t$ changes between non-decreasing and non-increasing the size of its pushdown on every input, and it is reversal-bounded if it is $t$-reversal-bounded for some $t$. A pushdown automaton $M$ is unambiguous if, for all $w\in\Sigma^{*}$, there is at most one accepting computation of $w$ by $M$. More generally, $M$ is $k$-ambiguous if there are at most $k$ accepting computations of $w$.

Let $\DPDA$ ($\PDA$) denote the class of deterministic (nondeterministic) pushdown automata (and languages). We also use $\CFL=\PDA$, the family of context-free languages.

We make use of one particular family of languages, which we will only describe intuitively (see [24] for formal details). Consider a nondeterministic machine with a one-way input and $k$ pushdowns, where each pushdown only has a single symbol plus a bottom-of-stack marker. Essentially, each pushdown operates like a counter, where each counter contains some non-negative integer, and machines can add or subtract one, and test for emptiness or non-emptiness of each counter. When $k=1$, we call these nondeterministic (and deterministic) one counter machines. Although such a machine with two counters has the same power as a Turing machine [18], if the counters are restricted, then the machine can have positive decidability properties. Let $\NCM(k,t)$ be the family of $k$ counter machines where the counters are $t$-reversal-bounded counters, and let $\DCM(k,t)$ be the deterministic subset of these machines. Also, let $\NCM$ be $\bigcup_{k,t\geq 1}\NCM(k,t)$ and $\DCM$ be $\bigcup_{k,t\geq 1}\DCM(k,t)$. The class of $\PDA$’s or $\DPDA$’s augmented with reversal-bounded counter machines is denoted by $\NPCM$ or $\DPCM$ [24]. It is known that $\NCM$ has a decidable emptiness problem, and $\DCM$ also has a decidable containment problem, with both being closed under intersection [24]. Furthermore, $\NCM$ and $\NPCM$ are semilinear trios.

3 Counting-Regular Languages

Obviously every regular language is counting-regular, and so are many non-regular languages. For example, $L_{Sq}$ = $\{ww\mid w\in\{a,b\}^{*}\}$ is counting-regular since $L^{\prime}$ = $(a(a+b))^{*}$ has the same number of strings of length $n$ as $L_{Sq}$ for all $n$. It is a simple exercise to show that $L_{Sq}$ is actually strongly counting-regular. It is also easy to exhibit languages that are not counting-regular, e.g., $L_{bal}$ = $\{w\mid w$ is a balanced parentheses string$\}$ over alphabet $\{[,]\}$. The reason is that there is no regular language $L$ such that the number of strings of length $2n$ in $L$ is the $n$’th Catalan number, as will be seen from the characterization theorem stated below.

Our goal in this section is to explore general families of languages that are counting-regular. We briefly note the following:

Theorem 1

Any family $\LL$ that contains some non-length-semilinear language $L$ is not counting regular.

Proof 1

Given such an $L$, then examine the set of all $n$ with $f_{L}(n)>0$. But, as every regular language $R$ is length-semilinear, the set of all $n$ with $f_{R}(n)>0$ must be different. \qed

Thus, it is immediate that e.g. any family that contains some non-semilinear unary language must not be counting-regular. This includes such families as those accepted by checking stack automata [25], and many others. Of interest are exactly what families of length-semilinear languages are counting-regular. We will investigate these questions here.

The following result is due to Berstel [5].

Theorem 2

Let $L$ be a regular language and let $f_{L}(n)$ denote the number of strings of length $n$. Then, one of the following holds:

1. (i)

$f_{L}(n)$* is bounded by a constant $c$.* 2. (ii)

There is an integer $k>0$ and a rational $c>0$ such that $limsup_{n\rightarrow\infty}{{f_{L}(n)}\over{n^{k}}}=c$. 3. (iii)

*There exists an integer $k\geq 0$ and an algebraic number $\alpha$ such that *

$limsup_{n\rightarrow\infty}{{f_{L}(n)}\over{\alpha^{n}n^{k}}}=c$* (where $c\neq 0$ is rational).*

We also need the following theorem due to Soittola [26]. We begin with the following definitions. A sequence $s$ = $\{s_{n}\}$, $n\geq 0$, is said to be the merge of the sequences $\{s^{(0)}\},\ldots,\{s^{(p-1)}\}$, where $p$ is a positive integer, if $s_{n}^{(i)}$ = $s_{i+np}$ for $0\leq i\leq p-1$. A sequence $\{s_{n}\}$ is said to be regular if there exists a regular language $L$ such that $f_{L}(n)$ = $s_{n}$ for all $n$.

Next we define a $\mathbb{Z}$-rational sequence as follows: A sequence $\{s_{n}\}$, $n\geq 0$, is $\mathbb{Z}$-rational if there is a matrix $M$ of order $d\times d$, a row vector $u$ of order $1\times d$, and a column vector $v$ of order $d\times 1$ such that $s_{n}$ = $u\ M^{n}v$. All the entries in $M$, $u$ and $v$ are over $\mathbb{Z}$. A $\mathbb{Z}$-rational sequence is said to have a dominating pole if its generating function $s(z)$ = $\sum_{n=0}^{\infty}s_{n}z^{n}$ can be written as a rational function $s(z)$ = $p(z)/q(z)$ where $p$ and $q$ are relatively prime polynomials, and $q$ has a simple root $r$ such that $r^{\prime}>r$ for any other root $r^{\prime}$.

Soittola’s theorem [26] can be stated as follows.

Theorem 3

A $\mathbb{Z}$-rational sequence with non-negative terms is regular if and only if it is the merge of $\mathbb{Z}$-rational sequences with a dominating pole.

We will also need the following theorem due to Béal and Perrin [27].

Theorem 4

A sequence $s$ is the generating sequence of a regular language over a $k$-letter alphabet if and only if both the sequences $s$ = $\{s_{n}\}$, $n\geq 0$ and $t$ = $\{(k^{n}-s_{n})\}$, $n\geq 0$ are regular.

We now show the main result of this section. This result can be viewed as a strengthening of a result of Baron and Kuich [28] that $L(M)$ has a rational generating function if $M$ is an unambiguous finite-turn $\PDA$.

Here, an $\NTM$ is considered to have a one-way read-only input tape, plus a two-way read/write worktape that uses blank symbol ␣ . Such a machine is said to be reversal-bounded if there is a bound on the number of changes in direction between moving left and right and vice versa on the worktape. An $\NTM$ is said to be in normal form if, whenever the worktape head moves left or right (or at the beginning of a computation) to a cell $c$, then the next transition can change the worktape letter in cell $c$, but then the cell does not change again until after the worktape head moves. This essentially means that if there is a sequence of ‘stay’ transitions (that do not move the read/write head) followed by a transition that moves, then only the first such transition can change the tape contents.

Lemma 5

Given an unambiguous reversal-bounded $\NTM$ $M$, there exists an unambiguous reversal-bounded $\NTM$ $M^{\prime}$ in normal form such that $L(M)=L(M^{\prime})$.

Proof 2

Given $M$, an $\NTM$ $M^{\prime}$ is constructed as follows: After a transition that moves the read/write head (or at the first move of the computation), instead of simulating a ‘stay’ transition directly, $M^{\prime}$ guesses the final value to be written on the cell before the head moves, and writes it during the first ‘stay’ transition. $M^{\prime}$ then continues to simulate the sequence of stay transitions without changing the value in the cell but remembering it in the finite control. Then, during the last transition of this sequence that moves the tape head, $M^{\prime}$ verifies that it guessed the final value correctly. Certainly, $M^{\prime}$ is reversal-bounded if and only if $M$ is reversal-bounded. Further, as $M$ is unambiguous, there is only one computation that is accepting on every word in $L(M)$. And, in $M^{\prime}$ therefore, there can only be one value guessed on each sequence of ‘stay’ transitions that leads to acceptance. Thus, $M^{\prime}$ is unambiguous. \qed

It will be shown that every such $\NTM$ is counting regular.

Theorem 6

Let $M$ be an unambiguous reversal-bounded $\NTM$ over a $k$ letter alphabet. Then $L(M)$ is strongly counting regular, where the regular language is over a $k+1$ letter alphabet.

Proof 3

First we note that it is enough to show that $L(M)$ is counting regular, as the class of languages accepted by unambiguous reversal-bounded $\NTM$s are closed under intersection with regular languages.

Let $M=(Q,\Sigma,\Gamma,\delta,q_{0},F)$ be an unambiguous $\NTM$ that is $t$-reversal-bounded such that $M$ is in normal form. Also, assume without loss of generality that $t$ is odd.

Intuitively, the construction resembles the construction that the store languages (the language of all contents of the worktape that can appear in an accepting computation) of every such $\NTM$ is a regular language [29]. A $2\NFA\ M^{\prime}$ is constructed that has $t+1$ “tracks”, and it uses the first track for simulating $M$ before the first reversal, the second track for simulation of $M$ between the first and the second reversal, etc. Thus the input to $M^{\prime}$ is the set of strings in which the first track contains an input string $x$ of $M$, and the other tracks are annotated with the contents of the read-write tape during moves of $M$ between successive reversals. Formal details are given below.

A $2\NFA$ $M^{\prime}=(Q^{\prime},\Sigma^{\prime},\delta^{\prime},q_{0}^{\prime},F^{\prime})$ is constructed as follows: Let $C=[(Q\times(\Sigma\cup\{\epsilon\})\times Q\times\Gamma)\cup\{\leavevmode\hbox{\ignorespaces\verb*!␣! }\}]^{t+1}$ (the $t+1$ tracks; each track is either a blank, or some tuple in $Q\times(\Sigma\cup\{\epsilon\})\times Q\times\Gamma$). Also, let $C_{i}$ have $t+1$ tracks where the $i$’th track, for $1\leq i\leq t+1$, contains an element from $Q\times(\Sigma\cup\{\epsilon\})\times Q\times\Gamma$, and all other tracks contain new symbol $\#$. Let $\Sigma^{\prime}=C\cup C_{1}\cup\cdots\cup C_{t+1}$. To simulate moves between the $(i-1)$st reversal and the $i$’th reversal, $M^{\prime}$ will examine track $i$ of a letter of $C$ to simulate the first transition after the tape head moves to a different cell (or at the first step of a computation), and track $i$ of letters of $C_{i}$ to simulate any stay transitions that occur before the tape head moves again, followed by the transition that moves.

Let $X=C_{t+1}^{*}\cdots C_{4}^{*}C_{2}^{*}CC_{1}^{*}C_{3}^{*}\cdots C_{t}^{*}$. Let $h_{i}$ be a homomorphism that maps each string in $(\Sigma^{\prime})^{*}$ to the $i$’th track for symbols in $C\cup C_{i}$, and erases all symbols of $C_{j},j\neq i$. Also, $\bar{h_{i}}$ is a homomorphism that maps each string in $(\Sigma^{\prime})^{*}$ to the $i$’th track if it is not ␣ , and $\epsilon$ otherwise, for symbols in $C\cup C_{i}$, and erases all symbols of $C_{j}$, $j\neq i$. Then $M^{\prime}$ does the following:

$M^{\prime}$* verifies that the input $w$ is in $X^{*}$, and that no letter of $[\leavevmode\hbox{\ignorespaces\verb*!␣! }]^{t+1}$ is used in $w$.* 2. 2.

$M^{\prime}$* verifies that for each $i$, $h_{i}(w)\in\leavevmode\hbox{\ignorespaces\verb*!␣! }^{*}(Q\times(\Sigma\cup\{\epsilon\})\times Q\times\Gamma)^{*}\leavevmode\hbox{\ignorespaces\verb*!␣! }^{*}$, so blanks can only occur at the ends.* 3. 3.

$M^{\prime}$* verifies that $w$ represents an accepting computation of $M$ as follows: $M^{\prime}$ goes to the first symbol of $C$ with a non-blank in the first track. Say $\bar{h_{1}}(w)=(p_{1},a_{1},p_{1}^{\prime},d_{1})\cdots(p_{m},a_{m},p_{m}^{\prime},d_{m}),m\geq 1$. For $j$ from $1$ to $m$, we say that $j$ is from $C$ if $(p_{j},a_{j},p_{j}^{\prime},d_{j})$ is from a symbol of $C$ and not $C_{1}$, and we say $j$ is from $C_{1}$ otherwise. It verifies from left-to-right on $w$ that for each $j$ from $1$ to $m$, there is a transition of $M$ that switches from $p_{j}$ to $p_{j}^{\prime}$ while reading $a_{j}\in\Sigma\cup\{\epsilon\}$ as input on worktape letter *␣

• if $j$ is in $C$, and $d_{j-1}$ otherwise, replacing it with $d_{j}$, that:*

• •

moves right on the worktape, if $j<m$ and $j+1$ is from $C$,

• •

‘stay’s on the worktape, if $j<m$ and $j+1$ is from $C_{1}$,

• •

moves left on the worktape, if $j=m$.

$M^{\prime}$* also verifies that $p_{1}=q_{0}$, and that for each $1\leq j<m$, $p_{j}^{\prime}=p_{j+1}$, At the symbol of $w$ where $(p_{m},a_{m},p_{m}^{\prime},d_{m})$ occurs (at a point of reversal), $M^{\prime}$ verifies that all symbols of $C$ to the right in track 1 and track 2 are blanks as well as there being no symbols from $C_{1}\cup C_{2}$. Then $M^{\prime}$ returns to the point of reversal, remembers $p_{m}^{\prime}$ in the finite control, and then returns to the rightmost symbol of $C$ in $h_{1}(w)$. From this symbol to the right, all values in $C$ have blanks in the second track. It is verified that the second track has a non-blank and $p_{m}^{\prime}$ for the first state. Then on $\bar{h_{2}}(w)$ from right-to-left (using symbols of $C_{2}$ instead of $C_{1}$ that occur to the left, read after a symbol from $C$), $M^{\prime}$ continues the simulation in a similar fashion until the second reversal, but $M^{\prime}$ instead verifies that for all symbols of $C$ read, it represents a transition that rewrites the worktape symbol on the first track with the worktape symbol on the second track. Since $M$ only changes values in tape cells the first move after arriving at a cell (corresponding to letters of $C$), after each reversal, $M^{\prime}$ can “lookup the value in the cell” by using the previous track. $M^{\prime}$ continues this process until up to after reversal $t$. Then $M^{\prime}$ accepts if this process results in a final state.*

It is evident that $w$ describes both the tape contents and also encodes the input for every accepting computation. Since each $x\in L(M)$ has exactly one accepting computation, and by the ordering of $X$, there must be a unique $w\in L(M^{\prime})$ with $x$ as the input word. Call this unique word $w$, $\code(x)$. Furthermore, consider homomorphisms $g_{1},\ldots,g_{t+1}$ such that $g_{i}$ maps each symbol of $C\cup C_{i}$ to the letter of $\Sigma\cup\{\epsilon\}$ in track $i$, and erases all other symbols. Given every word $w\in L(M^{\prime})$, the word $g_{1}(w)g_{2}(w)^{R}\cdots g_{t}(w)g_{t+1}(w)^{R}$ gives this word $x$. Hence, there is a bijection between $x\in L(M)$ and $\code(x)\in L(M^{\prime})$.

However, the length of $\code(x)$ can be different than that of $x$. Every letter in $C_{1}\cup\cdots\cup C_{t+1}$ encodes either one or zero letters of $\Sigma$ (depending on whether it contains a letter from $\Sigma$ or $\epsilon$). For every letter of $C$, it encodes between [math] and $t+1$ letters of $\Sigma$ (depending on the number of tracks encoding letters or $\epsilon$). Let $h$ be a homomorphism from $(\Sigma^{\prime})^{*}$ to (\Sigma^{\prime}\cup\{\})^{}$(where$$$is a new symbol) that acts as follows:$h$fixes all letters of$C_{1}\cup\cdots\cup C_{t+1}$that have letters of$\Sigma$, and erases those in$\epsilon$; it erases all symbols of$C$where every non-blank track encodes a transition on$\epsilon$(call these letters$C_{\epsilon}$), and it maps all other letters of$C$(call the set of these$\bar{C}$)$c\in\bar{C}$to$c$^{l-1}$, where$c$has$l$tracks encoding transitions on$\Sigma$($l\geq 1$, otherwise$c$would be in$C_{\epsilon}$). Define$\overline{\code}(x)=h(\code(x))$.*

To see that there is a bijection from $\code(x)$ to $\overline{\code}(x)$, first notice that $h$ is a function. In the other direction, the number of \$$’s is determined by the preceding letter of C$; and even though letters of$\Sigma^{\prime}$are erased, as$\overline{\code}(x)$still encodes all letters of$\Sigma$read (i.e.$x=g_{1}(\overline{\code}(x))g_{2}(\overline{\code}(x))^{R}\cdots g_{t+1}(\overline{\code}(x))^{R}$, where each$g_{i}$is extended to erase$$$), and therefore this coding is still unique.

In addition, let $R=\{\overline{\code}(x)\mid x\in L(M)\}$ which is regular since languages accepted by $2\NFA$s are regular, and regular languages are closed under homomorphism. Lastly, for each $x\in L(M)$, $|\overline{\code}(x)|=|x|$ by the reasoning above.

Surprisingly, we can reduce the size of the alphabet to $s+1$. More precisely, let $M$ be an unambiguous $\NTM$ as above over $\Sigma$ with $s=|\Sigma|$. We will show that there is a regular language $L_{1}$ over an alphabet of size $s+1$ such that $f_{L(M)}(n)$ = $f_{L_{1}}(n)$ for all $n$. As we have shown above, there is a regular language $L(M_{1})$ such that $f_{L(M_{1})}(n)$ = $f_{L(M)}(n)$ for all $n$. Thus, by Theorem 3, the generating function $a(z)$ associated with the sequence $(f_{L(M)}(n))$, $n\geq 0$, is a rational function $p(z)/q(z)$ that satisfies the conditions of Theorem 3. Let $b_{n}$ be defined as $b_{n}$ = $(k+1)^{n}-a_{n}$. The generating function $b(z)$ for the sequence $(b_{n})$, $n\geq 0$, is ${p(z)}\over{(1-(k+1)z)\ q(z)}$. This is a rational function with dominating pole $1\over{k+1}$ and hence it satisfies Theorem 3. Since both $(a_{n})$ and $(b_{n})$ = $((k+1)^{n}-a_{n})$ are regular sequences, by Theorem 4, there is a regular language $L_{1}$ over a $k+1$ letter alphabet such that $f_{L_{1}}(n)$ = $f_{L(M)}(n)$ for all $n$. \qed

We conjecture that the result also holds for $\NTM$s with finite-crossing (where there is a bound on the number of times the boundary between any two cells are crossed) worktapes.

Many different machine models can be simulated by unambiguous $\NTM$s with a one-way read-only input tape and a reversal-bounded worktape. These include unambiguous reversal-bounded queue automata where the store is a queue with a bound on the number of switches between enqueueing an dequeueing, and also unambiguous reversal-bounded $k$-flip $\NPDA$s, which are like $\NPDA$s with the additional ability to flip the stores up to $k$ times. As deterministic models are all unambiguous, this applies to deterministic models as well. We leave the details of the simulation to the reader.

Corollary 7

Let $M$ be a machine accepted by any of the machine models below, with a one-way input:

• •

*unambiguous nondeterministic reversal-bounded *$\NPDA$s,

• •

*reversal-bounded *$\DPDA$s,

• •

unambiguous nondeterministic reversal-bounded queue automata,

• •

deterministic reversal-bounded queue automata,

• •

*unambiguous nondeterministic reversal-bounded $k$-flip *$\NPDA$s,

• •

*reversal-bounded $k$-flip *$\DPDA$s,

• •

$\DTM$s with a reversal-bounded worktape.

Then the language $L(M)$ is strongly counting-regular.

The next theorem shows that some natural extensions of the model in the above theorem accept some non-counting-regular languages. In the proofs below, we use the well-known fact [2] that if $L$ is regular, then $f_{L}(n)$ is rational.

Theorem 8

The following families of languages are not counting-regular:

languages accepted by deterministic $1$-counter machines (no reversal-bound), 2. 2.

$\DCM(2,1)$* (deterministic $1$-reversal-bounded $2$-counter languages),* 3. 3.

$\NCM(1,1)$* (nondeterministic $1$-reversal-bounded $1$-counter languages),* 4. 4.

$2$-ambiguous $\NCM(1,1)$ languages.

Proof 4

Consider the language $L_{Eq}$ over $\{a,b\}$ that accepts strings with an equal number of $a$’s and $b$’s. Clearly $L_{Eq}$ can be accepted by a deterministic $1$-counter machine (without restriction on the number of reversals). We will now show that there is no regular language $L$ such that $f_{L}(n)$ = $f_{L_{Eq}}(n)$ for all $n$. It is easy to see that $f_{L_{Eq}}(n)$ is 0 if $n$ is odd, and and is ${2n}\choose n$ if $n$ is even. But ${{2n}\choose n}\approx{{c2^{n}}\over{\sqrt{n}}}$ for a constant $c$. Since the exponent of $n$ in the asymptotic expression for $f_{L_{Eq}}(n)$ is $-{1\over 2}$ which is not a positive integer, the conclusion follows from Theorem 2.

The same language is a witness for the second class: it is easy to construct a $\DCM(2,1)$ machine that accepts $L_{Eq}$: increment counter one (two) for each $a$ ($b$) read and verify the counter values are the same at the end by decrementing simultaneously and accept if the counters reach the value 0 at the same time.

To show 3, we use a language from [30]: Let $S$ = $\{w\ |\ w=a^{n}bv_{1}a^{n}v_{2}$ for some $v_{1}$ and $v_{2}$ in $\{a,b\}^{*}\}$. It is easy to see that $S$ can be accepted by an $\NCM(1,1)$. It is shown in [30] that the generating function $S(z)$ for $S$ is $S(z)$ = ${z(1-z)}\over{1-2z}$ $\sum_{n\geq 1}{z^{2n}\over{1-2z+z^{n+1}}}$. Since $S(z)$ has countably many poles (one for each $n$ in the infinite sum), it follows that $S(z)$ is not algebraic. Since the generating function for any regular language is algebraic, it follows that $S$ is not counting-regular.

Finally, we will show 4. Note that the claim of 4 is stronger than 3, but the counter-examples we offer for 3 and 4 exhibit an interesting contrast. The former example has the property that its generating function has an infinite number of poles. The example we present now has only a finite number of poles but is not algebraic. Consider the languages $L_{3}$ = $\{x1^{n}\ |$ $x\in\{a,b\}^{*},\ |x|_{a}=n\}$ and $L_{4}$ = $\{x1^{n}\ |$ $x\in\{a,b\}^{*},\ |x|_{b}=n\}$. Clearly $L_{3}$ and $L_{4}$ can be accepted by a $\DCM(1,1)$ and hence are strongly counting-regular by Corollary 7. (For $L_{3}$, the counter machine pushes a 1 on the counter on each $a$, skipping over $b$’s. When a first 1 is reached, it starts popping and makes sure that no $a$ or $b$ is seen again, and accepts when the counter reaches 0.) We will now show that $L_{5}$ = $L_{3}\cup L_{4}$ is not counting-regular, by showing that its generating function is not rational.

Let $L_{6}$ = $L_{3}\cap L_{4}$. It is easy to see that $L_{6}$ = $\{x1^{n}\ |$ $x\in\{a,b\}^{*},\ |x|_{a}=|x|_{b}=n\}$. It can be checked that

[TABLE]

So the generating function of $L_{6}$ is $f_{L_{6}}(z)$ = $1\over{\sqrt{1-4z^{3}}}$. Clearly $f_{L_{6}}(z)$ is not rational. Suppose $L_{5}$ is counting-regular. Then, $f_{L_{5}}(z)$ is rational, and $f_{L_{6}}(z)$ = $f_{L_{3}}(z)+f_{L_{4}}(z)-f_{L_{5}}(z)$ is rational, a contradiction.

Since $L_{5}$ can be accepted by a $2$-ambiguous $\NCM(1,1)$, the claim follows. \qed

Are there counting-regular languages that are not strongly counting-regular? Specifically, since we showed that all unambiguous reversal-bounded $\PDA$’s are strongly counting-regular, is there an unambiguous $\PDA$ $L$ (that is not reversal-bounded) such that $L$ is counting-regular, but not strongly counting-regular? We discuss this issue next. In fact, we show that there is a deterministic $1$-counter language (also a $\DCM(2,1)$ language) that is counting-regular, but not strongly counting-regular, namely: $L_{\MAJ}$ = \{x\in(0+1)^{*}\ |\ $x$ has more 1’s than 0’s or has equal number of 0’s and 1’s and starts with 1$\}$.

Theorem 9

$L_{\MAJ}$* is counting-regular but is not strongly counting-regular.*

Proof 5

Since the number of strings of length $n$ in $L_{\MAJ}$ is exactly $2^{n-1}$ for all $n\geq 1$, it is counting-regular.

Since $L_{\MAJ}$ is counting-regular, the smallest $\DFA$ $M$ such that $L_{\MAJ}\cap L(M)$ does not have a regular counting function must have at least two states. Somewhat surprisingly, such a $2$-state $\DFA$ exists. Consider the $\DFA$ $M_{2}$ in Figure 1.

Let $L_{2}$ = $L_{\MAJ}\cap L(M_{2})$. To show our claim, we need to establish that there is no regular language $L$ such that $f_{L_{2}}(n)$ = $f_{L}(n)$.

We will obtain a closed-form expression for $f_{L_{2}}(n)$ and from it, we will obtain the generating function for the sequence $\{f_{L_{2}}(n)\}$, $n$ = $0,\ 1,\ldots$ as follows. Note that the $\DFA$ $M$ has the property that all the transitions into a state are labeled by the same input. This means that there is a bijective mapping between an input string $w$ and the sequence of states visited on input $w$. Specifically, if we omit the first state (which is $q_{0}$, the start state), every input string of length $n$ can be bijectively mapped to the sequence of length $n$ of states visited. Thus, when $n$ is odd, the number of strings of length $n$ in $L_{2}$ is equal to the number of sequence of states of length $n$ that have more $q_{0}$ than $q_{1}$ and ending with $q_{0}$. (The requirement of ending with $q_{0}$ ensures that the string is in $L(M_{2})$.) When $n$ is even, the number of strings of length $n$ in $L_{2}$ is equal to the number of sequences of states of length $n$ that have more $q_{0}$ than $q_{1}$ and ending with $q_{0}$, plus the number of sequences of states of length $n$ with equal number of $q_{0}$ and $q_{1}$ that begins and ends with $q_{0}$. Thus:

[TABLE]

From this, we can explicitly obtain the generating function of $L_{2}$ as: $1+{z^{2}\over{1-2z}}+{{z^{2}+z}\over{\sqrt{1-4z^{2}}}}$. This function is not a rational function and the conclusion follows. \qed

Finally, we will show an interesting language that is both context-free and strongly counting-regular. This language was presented in [31] as a member of the class $\RCM$ — a class defined as follows: $\RCM$ is the class of the languages given by $<R,C,\mu>$ where $R$ is a regular language, $C$ a system of linear constraints and $\mu$, a length-preserving morphism that is injective on $R\cap[C]$. The specific language we denote by $L_{\RCM}$ is defined as follows:

[TABLE]

(From the definition, it is clear that $a^{n}$ is not in $L_{\RCM}$. But the definition does not specify the status of the string $b^{n}$ since there is no position 0 in the string. We resolve this by explicitly declaring the string $b^{n}$ to be in $L_{\RCM}$ for all $n$.)

We first observe that $L_{\RCM}$ is in $\NCM(1,1)$ and hence is context-free: We informally describe a nondeterministic 1-reversal-bounded 1-counter machine $M$ for $L_{\RCM}$. Let $w$ = $w_{1}\cdots w_{n}\in L_{\RCM}$ and suppose $|w|_{a}$ = $i$ and $|w_{1}w_{2}\cdots w_{i-1}|_{a}$ = $j$. Then, it is clear that $w_{i}$ = $b$, $j<i$ and $|w_{i+1}\cdots w_{n}|_{a}$ = $i-j$. Now we describe the operation of $M$ on input string $w$ (not necessarily in $L_{\RCM}$) as follows. $M$ increments the counter for every $b$ until it reaches position $i$. $M$ guesses that it has reached position $i$, and checks that the symbol currently scanned is $b$, then switches to decrementing phase in which it decrements the counter for each $a$ seen. When the input head falls off the input tape, if the counter becomes 0 it accepts the string. If $w\in L_{\RCM}$, it is clear that when $M$ switches from incrementing to decrementing phase, the counter value is $i-j$ and hence when it finishes reading the input, the counter will become 0. The converse is also true. Thus it is clear that $L(M)$ = $L_{\RCM}$.

An interesting aspect of the next result is that it applies the simulation technique of Theorem 6 twice — by first bijectively (and in length-preserving way) mapping $L_{\RCM}$ to a language accepted by $\DCM(1,1)$, then applying Theorem 6 to map it to a regular language.

Theorem 10

$L_{\RCM}$* is strongly counting-regular.*

Proof 6

Consider $L^{\prime}$ = $\{[a_{1},b_{1}][a_{2},b_{2}]\cdots[a_{n},b_{n}]\ |\ a_{1}a_{2}\cdots a_{n}\in L_{\RCM}$, and $b_{i}$ = $c$ for all $1\leq i\leq k$, and $b_{i}$ = $d$ for all $k+1\leq i\leq n$ where $k$ is the number of $a$’s in $a_{1}a_{2}\cdots a_{n}\}$. Thus, $L^{\prime}$ is defined over $\Sigma_{1}$ = $\{[a,c],[b,c],[a,d],[b,d]\}$. As an example, the string $[a,c][b,c][a,d]$ is in $L^{\prime}$ since $aba\in L_{\RCM}$. It is easy to see that there is a bijective mapping between strings in $L_{\RCM}$ and strings in $L^{\prime}$. Given a string $w\in L_{\RCM}$, there is a unique string $w^{\prime}$ in $L^{\prime}$ whose upper-track consists of $w$ and the lower-track consists of $c^{k}d^{n-k}$ where $|w|$ = $n$ and $|w|_{a}$ = $k$. Since $k$ and $n$ are uniquely specified for a given string $w$, the string $w^{\prime}$ is uniquely defined for a given $w$. Conversely, the projection of a $w^{\prime}\in L^{\prime}$ to its upper-track string uniquely defines a string in $L_{\RCM}$.

Let $R$ be a regular language over $\{a,b\}$. Define a language $R^{\prime}=\{[a_{1},b_{1}][a_{2},b_{2}]\cdots[a_{n},b_{n}]\ |\ a_{1}\cdots a_{n}\in L^{\prime}\cap R\}$. It is clear that there is a 1-1 correspondence between strings in $R^{\prime}$ and $L_{RCM}\cap R$. Finally, we note that there is a $\DCM(1,1)$ $M$ that accepts $R^{\prime}$: $M$ simulates the $\DFA$ for $R$ on the upper-track input and at the same time, performs the following operation. For every input $[b,c]$, the counter is incremented, and for every $[a,d]$ the counter is decremented. It also remembers the previous symbol scanned, and checks that the previous symbol scanned before the first occurrence of $[b,d]$ is $[a,c]$. Finally, when the counter value becomes 0, the string is accepted. Note that since all $d$’s in the second track occur after the $c$’s, the counter reverses at most once.

Since all $\DCM(1,1)$’s are $1$-reversal-bounded $\DPDA$’s, then by Corollary 7, $R^{\prime}$ is counting-regular. Hence, $L_{\RCM}\cap R$ is counting-regular. \qed

In this section, we have shown that the languages accepted by unambiguous nondeterministic Turing machines with a one-way read-only input tape and a reversal-bounded worktape are strongly counting-regular. We also showed that some natural extensions of this class fail to be counting-regular. We presented some relationships between counting-regular languages and the class $\RCM$. However, our understanding of which languages in $\DCM$, $\NCM$, or $\CFL$ are (strongly) counting-regular is quite limited at this time.

4 Bounded Semilinear Trio Languages are Counting-Regular

In this section, we will show that all bounded languages in any semilinear trio are counting-regular.

First, the following is known [17] (follows from results in [24], and the fact that all bounded $\NCM$ languages are in $\DCM$ [32]):

Lemma 11

Let $u_{1},\ldots,u_{k}\in\Sigma^{+}$, and let $\phi$ be a function from $\mathbb{N}_{0}^{k}$ to $u_{1}^{*}\cdots u_{k}^{*}$ which associates to every vector $(l_{1},\ldots,l_{k})$, the word $\phi((l_{1},\ldots,l_{k}))=u_{1}^{l_{1}}\cdots u_{k}^{l_{k}}$. Then the following are true:

• •

given a semilinear set $Q$, then $\phi(Q)\in\DCM$,

• •

given an $\NCM$ (or $\DCM$) language $L\subseteq u_{1}^{*}\cdots u_{k}^{*}$, then $IND(L)=\{(l_{1},\ldots,l_{k})\mid\phi((l_{1},\ldots,l_{k}))\in L\}$ is a semilinear set.

Moreover, both are effective.

Recall that two languages $L_{1}$ and $L_{2}$ are called commutatively equivalent if there is a Parikh-map preserving bijection between them. Therefore, a language $L$ being commutatively equivalent to some regular language is stronger than saying it is counting-regular. Split across three papers in [11, 12, 13], it was shown that all bounded semilinear languages — which are all bounded languages where $IND(L)$ is a semilinear set — are commutatively equivalent to some regular language, and are therefore counting-regular. Recently, it was shown that all bounded languages from any semilinear trio are in $\DCM$ [17] (and are therefore bounded semilinear by Lemma 11). This enables us to conclude that all bounded languages in any semilinear trio are commutatively equivalent to some regular language, and thus counting-regular. However, as the proof that all bounded semilinear languages are commutatively equivalent to some regular language is quite lengthy, we provide a simple alternate proof that all bounded languages in any semilinear trio (all bounded semilinear languages) are counting-regular. The class $\DCM$ plays a key role in this proof.

Lemma 12

Let $u_{1},\ldots,u_{k}\in\Sigma^{+}$, $L\subseteq u_{1}^{*}\cdots u_{k}^{*}$ be a bounded $\DCM$ language, and let $\phi$ be a function from Lemma 11. There exists a semilinear set $B$ such that $\phi(B)=L$ and $\phi$ is injective on $B$. Also, the construction of $B$ is effective.

Proof 7

Let $A=\{a_{1},\ldots,a_{k}\}$, and consider the homomorphism $h$ that maps $a_{i}$ to $u_{i}$, for $1\leq i\leq k$. It is known that there exists a regular subset $R$ of $a_{1}^{*}\cdots a_{k}^{*}$ that $h$ maps bijectively from $R$ onto $u_{1}^{*}\cdots u_{k}^{*}$ (the Cross-Section Theorem of Eilenberg [19]). Let $L^{\prime}=h^{-1}(L)\cap R$. Then $L^{\prime}$ is in $\DCM$ since $\DCM$ is closed under inverse homomorphism and intersection with regular languages [24]. Hence, there is a semilinear set $B=IND(L^{\prime})$ from Lemma 11(2). Then $\phi(B)=L$ since, given $(l_{1},\ldots,l_{k})\in B$, then $u_{1}^{l_{1}}\cdots u_{k}^{l_{k}}\in L$, and given $w\in L$, by the bijection $h$, there exists a string $a_{1}^{l_{1}}\cdots a_{k}^{l_{k}}$ of $R$ such that $a_{1}^{l_{1}}\cdots a_{k}^{l_{k}}=h^{-1}(w)$, and so $w\in\phi(B)$. Also, $\phi$ is injective on $B$, as given two distinct elements $(l_{1},\ldots,l_{k})$ and $(j_{1},\ldots,j_{k})$ in $B$, then both $a_{1}^{l_{1}}\cdots a_{k}^{l_{k}}$ and $a_{1}^{j_{1}}\cdots a_{k}^{j_{k}}$ are in $R$, which means that $h$ maps them onto different words in $u_{1}^{*}\cdots u_{k}^{*}$ since $h$ is a bijection. \qed

The proof of the next result uses similar techniques as the proof that all bounded context-free languages are counting-regular from [19]. But because there are key differences, we include a full proof for completeness.

Lemma 13

Let $L\subseteq u_{1}^{*}\cdots u_{k}^{*}$ be a bounded $\DCM$ language for given words $u_{1},\ldots,u_{k}$. Then there exists an effectively constructible bounded regular language $L^{\prime}$ such that, for every $n\geq 0$, $f_{L}(n)=f_{L^{\prime}}(n)$.

Proof 8

Let $\phi$ be a function from $\mathbb{N}_{0}^{k}$ to $u_{1}^{*}\cdots u_{k}^{*}$ such that $\phi((l_{1},\ldots,l_{k}))=u_{1}^{l_{1}}\cdots u_{k}^{l_{k}}$. By Lemma 11, there exists a semilinear set $B$ of $\mathbb{N}_{0}^{k}$ such that $\phi(B)=L$ [24]. Let $B=B_{1}\cup\cdots\cup B_{m}$, where $B_{i}$, $1\leq i\leq m$ are linear sets. Let $L_{1}=\phi(B_{1})\in\DCM$ (by Lemma 11), $L_{2}=\phi(B_{2})-L_{1}\in\DCM$ (by Lemma 11 and since $\DCM$ is closed under intersection and complement [24]), etc. until $L_{m}=\phi(B_{m})-(L_{1}\cup\cdots\cup L_{m-1})\in\DCM$ (inductively, by Lemma 11, by closure of $\DCM$ under intersection, complement, and union). Then $L_{1}\cup\cdots\cup L_{m}=L$, and also $L_{1},\ldots,L_{m}$ are pairwise disjoint, and therefore by Lemma 12, there is a semilinear set $B_{i}^{\prime}$ such that $\phi(B_{i}^{\prime})=L_{i}$, and $\phi$ is injective on $B_{i}^{\prime}$. It is known that, given any set of constants and periods generating a semilinear set $Q$, there is a procedure to effectively construct another set of constants and periods that forms a semi-simple set, also generating $Q$ [19, 20] (this is phrased more generally in both works, to say that the rational sets of a commutative monoid are semi-simple; but in our special case it amounts to constructing a new set of constants and periods generating the same semilinear set such that the linear sets are disjoint, and the periods generating each linear set form a basis). Hence, each $B_{i}^{\prime}$ must also be semi-simple as well (generated by a possibly different set of constants and periods). Let $B^{\prime}=B_{1}^{\prime}\cup\cdots\cup B_{m}^{\prime}$. Since each word in $L$ is only in exactly one language of $L_{1},\ldots,L_{m}$, it follows that for each $(l_{1},\ldots,l_{k})$, $\phi((l_{1},\ldots,l_{k}))$ is in at most one language of $L_{1},\ldots,L_{m}$. And, since $\phi$ is injective on each $B_{i}^{\prime}$, it therefore follows that $\phi$ is injective on $B^{\prime}$. Also, $\phi(B^{\prime})=L=\phi(B)$.

The rest of the proof then continues just as Theorem 10 of [19] (starting at the second paragraph) which we describe. That is, define an alphabet $A=\{a_{1},\ldots,a_{k}\}$, and let $\psi$ be the Parikh map of $A^{*}$ to $\mathbb{N}_{0}^{k}$. For every linear set $B^{\prime\prime}$ making up any of the semilinear sets of some $B_{i}^{\prime}$, let $B^{\prime\prime}$ have constant $b_{0}$ and periods $b_{1},\ldots,b_{t}$. Define the regular language $R_{B^{\prime\prime}}=v_{0}v_{1}^{*}\cdots v_{t}^{*}$ where $v_{0},\ldots,v_{t}$ are any fixed words of $A^{*}$ such that, for every $i$, $\psi(v_{i})=b_{i}$. Thus, $\psi(R_{B^{\prime\prime}})=B^{\prime\prime}$, for each $B^{\prime\prime}$. Let $R$ be the union of all $C_{B^{\prime\prime}}$ over all the linear sets making up $B_{1}^{\prime},\ldots,B_{m}^{\prime}$. Certainly $R$ is a regular language.

It is required to show that $\psi$ is injective on $R$. Indeed, consider $x,y$ be two distinct elements in $R$. If $x,y$ are constructed from two different linear sets from distinct semilinear sets $B_{i}^{\prime},B_{j}^{\prime},i\neq j$, then $\psi(x)\neq\psi(y)$ since the semilinear sets $B_{i}^{\prime}$ and $B_{j}^{\prime}$ are disjoint. If $x,y$ are constructed from two different linear sets making up the same semilinear set $B_{i}^{\prime}$, then since $B_{i}^{\prime}$ is semi-simple, the linear sets must be disjoint, and hence $\psi(x)\neq\psi(y)$. If $x,y$ are in the same linear set, then $\psi(x)\neq\psi(y)$ since the linear set must be simple, its periods form a basis, and therefore, there is only one linear combination giving each. Hence, $\psi$ is injective on $R$.

Consider the map such that, for every $i$, $1\leq i\leq k$, $a_{i}$ maps to $a_{i}^{|u_{i}|}$, and extend this to a homomorphism $\chi$ from $A^{*}$ to $A^{*}$. Since $\chi(A)$ is a code, $\chi$ is an injective homomorphism of $A^{*}$ to itself. Let $L^{\prime}=\chi(R)$. Then $L^{\prime}$ is a regular language.

Next, it will be shown that $f_{L}(n)=f_{L^{\prime}}(n)$. Consider the relation $\zeta=\phi^{-1}\psi^{-1}\chi$. Then, when restricting $\zeta$ to $L$, this is a bijection between $L$ and $L^{\prime}$, since $\phi$ is a bijection from $B^{\prime}$ to $L$, $\psi$ is a bijection of $R$ to $B^{\prime}$, and $\chi$ is a bijection of $R$ to $L^{\prime}$. It only remains to show that, for each $u\in L$,

[TABLE]

which therefore would imply $f_{L}(n)=f_{L^{\prime}}(n)$. For each $u\in L$, then $u=u_{1}^{l_{1}}\cdots u_{k}^{l_{k}}=\phi((l_{1},\ldots,l_{k}))=\phi(\psi(x))$, where $x$ is in $R$ and $\psi^{-1}((l_{1},\ldots,l_{k}))$. Since $|x|=\sum_{1\leq i\leq k}|x|_{a_{i}}=\sum_{1\leq i\leq k}l_{i}$, then

[TABLE]

Thus, 1 is true, and the theorem follows. \qed

We should note that if the bounded language $L\subseteq a_{1}^{*}\cdots a_{k}^{*}$, where $a_{1},\ldots,a_{k}$ are distinct symbols, then there is a simpler proof of the theorem above as follows: Let $L\subseteq a_{1}^{*}\cdots a_{k}^{*}$ where the symbols are distinct. Then the Parikh map of $L$ is semilinear, and therefore a regular language $L^{\prime}$ can be built with the same Parikh map, and in this language $f_{L}(n)=f_{L^{\prime}}(n)$. But when the bounded language $L$ is not of this form, this simpler proof does not work.

Our next result is a generalization of the previous result.

Theorem 14

Let $L\subseteq u_{1}^{*}\cdots u_{k}^{*}$, for words $u_{1},\ldots,u_{k}$ where $L$ is in any semilinear trio $\LL$. There exists a bounded regular language $L^{\prime}$ such that, for every $n\geq 0$, $f_{L}(n)=f_{L^{\prime}}(n)$. Moreover, $L$ is strongly counting-regular. Furthermore, if $u_{1},\ldots,u_{k}$ are given, and all closure properties are effective in $\LL$, then $L^{\prime}$ is effectively constructible.

Again, this follows from [17] since it is known that every bounded language from any such semilinear trio where the closure properties are effective can be effectively converted into a $\DCM$ language. Strong counting-regularity follows since intersecting a bounded language in a trio with a regular language produces another bounded language that is in $\LL$, since trios are closed under intersection with regular languages.

Also, since the family of regular languages is the smallest semilinear trio [23], it follows that the counting functions for the bounded languages in every semilinear trio are identical.

Corollary 15

Let $\LL$ be any semilinear trio. The counting functions for the bounded languages in $\LL$ are identical to the counting functions for the bounded regular languages.

This works for many semilinear full trios. We will briefly discuss some in the next example.

Example 1

The families accepted/generated from the following grammar/machine models form semilinear full trios (the closure properties and semilinearity are effective):

*the context-free languages, *$\CFL$s, 2. 2.

*one-way nondeterministic reversal-bounded multicounter machines, $\NCM$s, **[24]**, * 3. 3.

finite-index $\ETOL$ systems ($\ETOL$* systems where the number of non-active symbols in each derivation is bounded by a constant) [33].* 4. 4.

$k$-flip $\NPDA$s ($\NPDA$s with the ability to “flip” their pushdown up to $k$ times) **[34]**. 5. 5.

one-way reversal-bounded queue automata (queue automata with a bound on the number of switches between enqueueing and dequeueing) **[14]**. 6. 6.

$\NTM$s with a one-way read-only input tape and a finite-crossing worktape **[14]**, 7. 7.

uncontrolled finite-index indexed grammars (a restricted version of indexed grammars, where every accepting derivation has a bounded number of nonterminals), **[16]**. 8. 8.

multi-push-down machines (a machine with multiple pushdowns where the machine can simultaneously push to all pushdowns, but can only pop from the first non-empty pushdown) **[15]**.

Moreover, all of these machine models can be augmented by reversal-bounded counters and the resulting machines are semilinear full trios [14, 35].

Corollary 16

Let $L\subseteq u_{1}^{*}\cdots u_{k}^{*}$, be a bounded language for given words $u_{1},\ldots,u_{k}$, such that $L$ is from any of the families listed in Example 1. Then there exists an effectively constructible bounded regular language $L^{\prime}$ such that, for every $n\geq 0$, $f_{L}(n)=f_{L^{\prime}}(n)$.

Note that it is not assumed for these models that the machines are unambiguous, like in Theorem 6.

The results in this section assumed that the words $u_{1},\ldots,u_{k}$ such that $L\subseteq u_{1}^{*}\cdots u_{k}^{*}$ are given. However, it is an open problem whether, given a language $L$ in an arbitrary semilinear trio $\LL$, it is possible to determine whether $L\subseteq u_{1}^{*}\cdots u_{k}^{*}$ for some words $u_{1},\ldots,u_{k}$.

5 Closure Properties for Counting-Regular Languages

In this section, we will address the closure properties of counting-regular languages, and also counting-regular $\CFL$’s.

First, it is immediate that counting-regular languages are closed under reversal (and since the $\CFL$s are closed under reversal, so are the counting-regular $\CFL$s). Next Kleene-* will be addressed.

Theorem 17

If $L$ is counting-regular and $L$ is a code, then $L^{*}$ is counting-regular.

Proof 9

Since $L$ is a code, for each word $w\in L^{*}$, there is a unique decomposition of $w=u_{1}\cdots u_{k}$, where each $u_{i}\in L$. Since $L$ is counting-regular, there is some regular language $R$ with the same counting function. From $R$, make $R^{\prime}$ where the first letter of each word is tagged with a prime, and all other letters are unmarked. Now, $R^{\prime}$ is a code because of the tagged letters, and $R^{\prime}$ has the same counting function as $R$.

Moreover, $(R^{\prime})^{*}$ has the same counting function as $L^{*}$. Indeed, let $n\geq 0$. Consider all sequences $u_{1},\ldots,u_{k}$ such that $n=|u_{1}|+\cdots+|u_{k}|$. Then for each $u_{i}$, $L$ has the same number of words of length $|u_{i}|$ as does $R^{\prime}$. Since $R^{\prime}$ is a code, it follows that there are the same number of such sequences using elements from $R^{\prime}$. \qed

A similar relationship to codes exists for concatenation.

Theorem 18

If $L_{1},L_{2}$ are counting-regular and either $L_{1}$ is a prefix code or $L_{2}$ is a suffix code, then $L_{1}L_{2}$ is counting-regular.

Proof 10

Assume first that $L_{1}$ is a prefix code, so that $L\cap L\Sigma^{+}=\emptyset$. Let $w=uv,u\in L_{1},v\in L_{2}$. Then this decomposition is unique since $L_{1}$ is a prefix code. Let $R_{1},R_{2}$ be regular languages with the same counting functions as $L_{1},L_{2}$ respectively. Let $R_{1}^{\prime}$ be obtained from $R_{1}$ by tagging the last letter with a prime. Then $R_{1}^{\prime}$ is also a prefix code. Further, the counting function for $R_{1}^{\prime}R_{2}$ is equal to that of $L_{1}L_{2}$.

The case is similar for $L_{2}$ being a suffix code. \qed

Corollary 19

If $L_{1},L_{2}\subseteq\Sigma^{*}$ are counting-regular, and \,#$are new symbols, then$L^{R},L_{1}$L_{2},$L_{1}\cup#L_{2}$, and$(L$)^{}$ are counting-regular.*

Proof 11

The first was discussed above. The second follows from Theorem 18 and since L_{1}\$$ is a prefix code. The fourth follows from Theorem [17](#Thmtheorem17) and since L$$is a code. For the third, since$L_{1},L_{2}$are counting-regular, this implies there exist regular languages$R_{1},R_{2}$with the same counting functions, and$$L_{1}\cup#L_{2}$has the same counting function as$$R_{1}\cup#R_{2}$. \qed

This means that even though e.g. non-reversal-bounded $\DPDA$s can accept non-counting-regular languages but reversal-bounded $\DPDA$s cannot, if a $\DPDA$ was reversal-bounded but reading a \$$ caused a “reset” where the pushdown emptied, and another reversal-bounded computation was then possible, then this model would only accept counting-regular languages. This is also the case with say \DTM$s where the worktape was reversal-bounded, but reading a$$$ caused a reset, where more reversal-bounded computations were again possible. This is quite a general model for which this property holds.

The next questions addressed are whether these are true when removing the \$$ and #$ (or removing or weakening the coding properties).

Theorem 20

The counting-regular languages (and the counting-regular $\CFL$’s) are not closed under union or intersection with regular languages.

Proof 12

Recall the $\DFA$ $M_{2}$ presented in Figure 1. Since $L_{\MAJ}$ is a counting-regular $\CFL$, and $L_{MAJ}\cap L(M_{2})$ is not, the non-closure under intersection with regular sets follows.

For non-closure under union with regular sets, we show that $L_{\MAJ}\cup L(M_{2})$ is not counting-regular by explicitly computing the generating function for this language using the fact that there is a 1-1 mapping between strings of $L_{\MAJ}$ and $\overline{L_{MAJ}}$ and between strings of $L(M_{2})$ and $\overline{L(M_{2})}$. Thus the generating functions for both $f_{L_{MAJ}}(n)$ and $f_{L(M_{2})}(n)$ are ${1-z}\over{1-2z}$, from which it follows that the generating function for $L_{MAJ}\cup L(M_{2})$ is ${{1-2z-z^{2}}\over{1-2z}}-$ ${{z^{2}+z}\over{\sqrt{1-4z^{2}}}}$. Since this is not a rational function, the claim follows. \qed

Thus, Corollary 19 cannot be weakened to remove the marking from the marked union.

It is an open question as to whether Theorem 17 can be weakened to remove the code assumption, but we conjecture that it cannot. However, for concatenation, we are able to show the following:

Theorem 21

The counting-regular languages (and counting-regular $\CFL$’s) are not closed under concatenation with regular languages.

Proof 13

Let $S_{1}$ = $\{w\ |\ w=a^{n}bva^{n}$ for some $v\in\{a,b\}^{*}\}$ and let $S$ = $\{w\ |\ w=a^{n}bv_{1}a^{n}v_{2}$ for some $v_{1},v_{2}\in\{a,b\}^{*}\}$ (as in Theorem 8). It is easy to see that $S$ = $S_{1}(a+b)^{*}$. We already showed that $S$ is not counting-regular. It is easy to show that $S_{1}$ is a counting-regular $\CFL$. In fact, we can explicitly exhibit $f_{S_{1}}(n)$ as follows: $f_{S_{1}}(0)$ = 0 and, for $n\geq 1$, $f_{S_{1}}(n)$ = $\sum_{j=0}^{\lfloor(n-1)/2\rfloor}2^{n-2i-1}$. From this, it is easy to see that $f_{S_{1}}(n)$ = $f_{L}(n)$ for the regular language $L$ with regular expression $(aa)^{*}b(a+b)^{*}$. \qed

Hence, Corollary 19 and Theorem 18 cannot be weakened to remove the marking with marked concatenation or the coding properties. It is an open problem as to whether Theorem 18 is true when $L_{1}$ or $L_{2}$ are codes (the set of suffix codes together with the set of prefix codes is a strict subset of the set of codes [22]).

Theorem 22

The counting-regular languages (and counting-regular $\CFL$’s) are not closed under right quotient with a single symbol, and are not closed under left quotient with a symbol.

Proof 14

(sketch) First, it will be shown for right quotient. The language $L_{\MAJ}$ used in Theorem 9 is counting-regular. In fact, it has exactly $2^{n-1}$ strings of length $n$ for all $n\geq 1$. We will outline an argument that $L$ = $L_{\MAJ}\{0\}^{-1}$ is not counting-regular. The number of strings of length $n$ in $L$ is $2^{n-1}-{n\choose{n-1\over 2}}$ (for odd $n$), $2^{n-1}-{n-2\choose{n\over 2}}$ (for even $n$). Using a technique similar to the proof of Theorem 9, we can show that the generating function for $L$ is not rational.

Next, it will be shown for left quotient. Since $L_{\MAJ}$ is counting regular, and the reversal of every counting regular language has the same counting function, then $L_{\MAJ}^{R}$ is also counting-regular. But, as in in proof for right quotient, $\{0\}^{-1}L_{\MAJ}^{R}$ is not counting-regular. \qed

The language $L_{MAJ}$ used is a deterministic one counter language with no reversal bound. The following however provides a contrast, as it follows from closure properties of deterministic machines. When the languages are accepted by reversal-bounded machines, we have:

Theorem 23

If $L$ is accepted by a reversal-bounded $\DTM$, and $R$ is a regular language, then $LR^{-1}$ is counting-regular. 2. 2.

If $L$ is accepted by an unambiguous reversal-bounded $\NTM$, and $x$ is a string, then $L\{x\}^{-1}$ is also counting-regular.

Proof 15

Part 1 follows from the fact that the languages accepted by reversal-bounded $\DTM$’s are closed under right-quotient with regular languages [29].

For Part 2, clearly, if $L$ is accepted by an unambiguous reversal-bounded $\NTM$ $M$, we can construct an unambiguous reversal-bounded $\NTM$ $M$ accepting $L\{x\}^{-1}$. \qed

This result also holds for all machine models in Corollary 7 (that is, all deterministic models listed there work with right quotient with regular languages [29], and the unambiguous nondeterministic models there work with right quotient with a word). It is an open question as to whether unambiguous nondeterministic $\NTM$s with a reversal-bounded worktape are closed under right quotient with regular languages, which would allow part 2 to be strengthened.

Part 1 of of the next theorem contrasts Part 1 of the previous theorem.

Theorem 24

There is a counting-regular language $L$ accepted by a $\DCM(1,1)$ and distinct symbols \$$* and *#$* such that *${$,#}^{-1}L$ is not counting-regular. 2. 2.

If $L$ is accepted by a reversal-bounded $\DPDA$ (resp., reversal-bounded unambiguous $\NPDA$, reversal-bounded unambiguous $\NTM$), and $x$ is a string, then $\{x\}^{-1}L$ is also counting-regular.

Proof 16

For Part 1, let $L_{1}=\{x1^{n}\mid x\in(a+b)^{+},n=|x|_{a}\}$ and $L_{2}=\{x1^{n}\mid x\in(a+b)^{+},n=|x|_{b}\}$. Then $L_{1}$ and $L_{2}$ can each be accepted by a $\DCM(1,1)$. Let $L=L_{1}\cup L_{2}$, shown in Theorem 8 to not be counting-regular. Let L^{\prime}=\L_{1}\cup#L_{2}$, which can also be accepted by a$\DCM(1,1)$, hence it is counting-regular. However,${$,#}^{-1}L^{\prime}=L$ is not counting-regular. Part 2 is obvious. \qed

It may seem obvious that for any counting-regular language $L$, $\overline{L}$ is counting-regular because of the following putative reasoning: If there is a regular language $L^{\prime}$ whose counting function equals that of the counting function of $L$, the complement of $L^{\prime}$ (which is regular) has the same counting function as that of $\overline{L}$. The fallacy in this argument is as follows. Suppose that the size of the alphabet over which $L$ is defined is $k$. The size of alphabet $k_{1}$ over which $L^{\prime}$ is defined may be larger, i.e., $k_{1}>k$. Thus, the complement of $L^{\prime}$ has a counting function ${k_{1}}^{n}-f_{L^{\prime}}(n)$ which is not the same as the counting function $k^{n}-f_{L}(n)$ of $\overline{L}$.

In fact, the following result shows that the exact opposite of the fallacy is actually true.

Theorem 25

There is a counting-regular language $L$ (that is in $\P$, i.e., $L$ is deterministic polynomial time computable) such that $\overline{L}$ is not counting-regular.

Proof 17

The proof relies on a result presented in Béal and Perrin [27]. Béal and Perrin [27] provide an example of a sequence $\{r_{n}\}$, $n\geq 0$ and an integer $k$ such that $\{r_{n}\}$ is not a counting function of any regular language, but $\{k^{n}-r_{n}\}$ is the counting function of a regular language. Specifically, it is shown in [27] that the sequence $r_{n}$ = $b^{2n}\cos^{2}(n\theta)$, with $cos\ \theta$ = $a\over b$, where the integers $a$, $b$ are such that $b\neq 2a$ and $0<a<b$, and $k$ such that $b^{2}<k$, satisfies the properties stated above.

Define a language $L$ over an alphabet of size $k$ as follows: arrange the strings of length $n$ over the alphabet $\{0,1,\ldots,k-1\}$ lexicographically. A string $w$ of length $n$ is defined to be in $L$ if and only if the rank of $w$ (in the lexicographic order) is greater than $r_{n}$. (Rank count starts at 1.) Clearly, the number of strings of length $n$ in $L$ is exactly $s_{n}$ = $k^{n}-r_{n}$. As shown in [27], $L$ is counting-regular and $\overline{L}$ is not counting-regular.

Finally, we provide a sketch of the proof that there is a deterministic polynomial time algorithm for $L$. Given a string $w$ of length $n$, and an integer $T$ (in binary) where the number of bits in $T$ is $O(n)$, it is easy to see that there is a deterministic algorithm that determines in time polynomial in $n$ if the rank of $w$ is greater than $T$. (This algorithm simply converts $T$ from binary to base $k$ and compares the resulting string to $w$ lexicographically. Base conversion can be shown to have complexity no more than that of integer multiplication.) To complete the algorithm, we need to show how to compute $r_{n}$ (in binary), given $n$, in time polynomial in $n$. Note that $r_{n}$ is given by $b^{2n}\cos^{2}(n\theta)$. Clearly, $b^{2n}$ can be computed in polynomial time by repeated multiplication by $b$ . We don’t even need repeated squaring to achieve a polynomial bound since the time complexity is measured in terms of $n$, not $log\ n$. Also, $\cos\ (n\ arccos(a/b))$ can be computed as follows: $\cos\ (n\ arccos(a/b))$ is the well-known Tchebychev polynomial $T_{n}(a/b)$ which is explicitly given by the series [36]:

[TABLE]

Since each of the terms in the above series can be computed in time polynomial in $n$, and since there are $\lfloor{n\over 2}\rfloor$ terms in the series, it is clear that $r_{n}$ can be computed in time polynomial in $n$. \qed

It is evident from Corollary 7 and closure of reversal-bounded $\DPDA$’s under complement that counting-regular reversal-bounded $\DPDA$’s are closed under complement. For $\DPDA$’s generally, it is open whether counting-regular $\DPDA$’s are closed under complement. At the end of the proof of Theorem 6, we observed that the mapping that we used to map the strings from a language $L$ accepted by an unambiguous reversal-bounded $\NTM$ to a regular language increased the size of the alphabet from $k$ to $k+1$. If for every counting-regular $\DPDA$, the size of the alphabet of the simulating $\NFA$ is $k$, then it will follow that counting-regular $\DCFL$’s are closed under complement.

In fact, we can show the following stronger claim.

Theorem 26

The following statements are equivalent:

For every counting-regular $\DCFL$ over a $k$-letter alphabet, there is a regular language $L^{\prime}$ over a $k$-letter alphabet such that $f_{L}(n)$ = $f_{L^{\prime}}(n)$ for all $n$. 2. 2.

Counting-regular $\DCFL$’s are closed under complement.

Proof 18

(1) $\Rightarrow$ (2) is immediate from the above discussion.

To show that (2) $\Rightarrow$ (1), let $L$ be a counting-regular $\DCFL$ over a $k$-letter alphabet. This means $f_{L}(n)$ is a regular sequence. By (2), the complement of $L$ is counting-regular, so $k^{n}-f_{L}(n)$ is also a regular sequence. From Theorem 4, it follows that there is a regular language $L^{\prime}$ over a $k$-letter alphabet such that $f_{L}(n)$ = $f_{L^{\prime}}(n)$ for all $n$. \qed

Our conclusion is that the class of counting-regular languages (or counting-regular $\CFL$s) are very fragile in that it is not closed under basic operations such as union or intersection with regular languages. We conjecture that the counting-regular $\CFL$s are also not closed under Kleene star.

6 Some Decision Problems Related to Counting-Regularity

In this section, some decision problems in regards to counting-regularity are addressed. In particular, we will show the following: (1) It is undecidable, given a real-time $1$-reversal-bounded $2$-ambiguous $\PDA$ $M$, whether $L(M)$ (resp. $\overline{L(M)}$) is counting-regular; (2) It is undecidable, given a real-time $\NCM(1,1)$ $M$, whether $L(M)$ (resp. $\overline{L(M)}$) is counting-regular.

We begin with the following result:

Theorem 27

*It is undecidable, given two real-time $1$-reversal-bounded $\DPDA$’s $M_{1}$ and $M_{2}$ accepting strongly counting-regular languages, whether $L(M_{1})\cap L(M_{2})$ is counting-regular. *

Proof 19

Let $Z$ be a single-tape $\DTM$ working on an initially blank tape. We assume that if $Z$ halts on blank tape, it makes $2k$ steps for some $k\geq 2$. We assume that $Z$ has one-way infinite tape and does not write blank. Let

[TABLE]

[TABLE]

with $L_{1}^{\prime},L_{2}^{\prime}\subseteq\Sigma^{*}$. Clearly, $L_{1}^{\prime}$ and $L_{2}^{\prime}$ can be accepted by real-time $1$-reversal-bounded $\DPDA$’s $M_{1}^{\prime}$ and $M_{2}^{\prime}$. Moreover, $L_{1}^{\prime}\cap L_{2}^{\prime}$ is empty or a singleton (if and only if $Z$ accepts, which is undecidable). Let $a$, $b$, $1$ be three new symbols. Let

[TABLE]

We can construct a real-time $1$-reversal-bounded $\DPDA$ $M_{1}$ accepting $L_{1}$ as follows: $M_{1}$ reads $x$ and stores the number of $a$’s in the stack. Then it simulates the $1$-reversal-bounded $\DPDA$ $M_{1}^{\prime}$ on $w$, and finally checks (by continuing to pop the stack) that the number of $a$’s stored in the stack is $n$. Similarly, we can construct a real-time $1$-reversal-bounded $\DPDA$ $M_{2}$ accepting $L_{2}$. By Corollary 7, $L_{1}$ and $L_{2}$ are strongly counting-regular.

Clearly, if $L(M_{1}^{\prime})\cap L(M_{2}^{\prime})$ = $\emptyset$, then $L_{1}\cap L_{2}$ = $\emptyset$, hence $L_{1}\cap L_{2}$ is counting-regular.

On the other hand, if $L(M_{1}^{\prime})\cap L(M_{2}^{\prime})$ is not empty, the intersection is a singleton $w$ and thus $L_{1}\cap L_{2}$ is given by:

[TABLE]

As in the proof of Theorem 8(ii), we can show that $L_{1}\cap L_{2}$ is not counting-regular. In fact, if $|w|$ = $t$, then the number of strings of length $3n+t$ is ${2n}\choose{n}$ from which we can explicitly construct the generating function $f(z)$ for $L_{1}\cap L_{2}$ as:

[TABLE]

Clearly $f(z)$ is not rational and the claim follows. \qed

This result is used within the next proof:

Theorem 28

It is undecidable, given a $\PDA$ $M$ that is real-time $2$-ambiguous and $1$-reversal-bounded, whether $\overline{L(M)}$ is counting-regular. Also, it is undecidable, given such a machine $M$, whether $L(M)$ is counting-regular.

Proof 20

Consider the languages $L_{1}$ and $L_{2}$ in the proof of Theorem 27. Since $L_{1}$ and $L_{2}$ can be accepted by real-time $1$-reversal-bounded $\DPDA$’s, $\overline{L_{1}}$ and $\overline{L_{2}}$ can also be accepted by real-time $1$-reversal-bounded $\DPDA$’s. Hence $L=\overline{L_{1}}\cup\overline{L_{2}}$ can be accepted by a real-time $1$-reversal-bounded $2$-ambiguous $\PDA$ $M$. It follows that $\overline{L(M)}$ is counting-regular if and only if $L_{1}\cap L_{2}$ is counting-regular, which is undecidable from Theorem 27.

Now consider $L(M)$. If $Z$ does not halt on blank tape, then $L(M)=\Sigma^{*}$, which is counting-regular. If $Z$ halts on blank tape, then we will show that $L(M)$ is not counting-regular as follows: the generating function $g(z)$ of $L(M)$ is given by ${1\over{1-sz}}-f(z)$ where $f(z)$ is the generating function of $\overline{L(M)}$, and $s$ is the size of the alphabet over which $M$ is defined. If $L(M)$ is counting-regular, then $g(z)$ is rational, then so is $f(z)$ = ${1\over{1-kz}}-g(z)$, contradicting the fact that $f(z)$ is not rational. Hence, $L(M)$ is counting-regular if and only if $Z$ does not halt. This completes the proof. \qed

Next, we will consider $\NCM(1,1)$ ($1$-reversal-bounded $1$-counter machines).

Theorem 29

It is undecidable, given a real-time $\NCM$(1,1) $M$, whether $\overline{L(M)}$ is counting-regular. Also, it is undecidable, given such a machine $M$, whether $L(M)$ is counting-regular.

Proof 21

Again, we will use the undecidability of the halting problem for a $\DTM$ $Z$ on an initially blank tape. As before, assume that $Z$ has a one-way infinite tape and does not write blank symbols. Represent each $ID$ (configuration) of $Z$ with blank symbols filled to its right, since for the languages we will define below, we require that all $ID_{i}$’s have the same length. So, e.g, $ID_{1}=q_{0}B\cdots B$ (where $B$ is the blank symbol). We also require that the halting $ID$ have all non-blanks in state $f$, which is unique. Clearly since $Z$ does not write any blank symbols, if $Z$ halts on the blank tape, the lengths of the $ID$’s in the halting sequence of $ID$’s do not decrease in length. Assume that if $Z$ halts, it halts after $k$ steps for some $k\geq 2$.

Let

[TABLE]

[TABLE]

with $L_{1},L_{2}\subseteq\Sigma^{*}$. Note that $ID_{2k}$ must have all non-blanks in state $f$.

Let $a,b,1$ be new symbols. We can construct a real-time $\NCM$(1,1) $M_{1}$ which operates as follows. When given input string $z$, $M_{1}$ nondeterministically selects one of the following tasks to execute:

$M_{1}$* checks and accepts if $z$ is not a string of the form $xw1^{n}\in(a+b)^{+}\Sigma^{+}1^{+}$. (This does not require the use of the counter.)* 2. 2.

$M_{1}$* checks and accepts if $z$ is of the form $xw1^{n}\in(a+b)^{+}\Sigma^{+}1^{+}$ but $|x|_{a}\neq n$. (This requires only one counter reversal.)* 3. 3.

$M_{1}$* checks that $z$ is of the form $xw1^{n}\in(a+b)^{+}\Sigma^{+}1^{+}$, but $w$ is not a string of the form in $L_{1}$. $M_{1}$ does not check the lengths of the $ID$’s and whether $ID_{i+1}$ is a successor of $ID_{i}$. (This does not require a counter.)* 4. 4.

$M_{1}$* checks that $z$ is of the form $xw1^{n}\in(a+b)^{+}\Sigma^{+}1^{+}$ and*

[TABLE]

for some $k\geq 2$ but $|ID_{i}|\neq|ID_{j}|$ for some $i\neq j$, or $ID_{1}$ is not the initial $ID$, or $ID_{2k}$ is not the halting $ID$. (This requires one counter reversal.) 5. 5.

$M_{1}$* assumes that $z$ is of the form $xw1^{n}\in(a+b)^{+}\Sigma^{+}1^{+}$ and*

[TABLE]

for some $k\geq 2$ , $|ID_{1}|=\cdots=|ID_{2k}|$, $ID_{1}$ is the initial $ID$, $ID_{2k}$ is the halting $ID$, and accepts if $ID_{i+1}$ is not the successor of $ID_{i}$ for some $i$ = $1,3,\ldots,2k-1$. Since all the $ID$’s are assumed to have the same length, $M_{1}$ needs to only use a counter that reverses once to check one of the conditions.

Similarly, we can construct a real-time $\NCM$(1,1) $M_{2}$ as above using $L_{2}$. Let $M$ be a real-time $\NCM(1,1)$ accepting $L(M_{1})\cup L(M_{2})$ and consider $\overline{L(M)}$.

If $Z$ does not halt on blank tape, then $\overline{L(M)}=\overline{L(M_{1})\cup L(M_{2})}=\overline{L(M_{1})}\cap\overline{L(M_{2})}=\emptyset$, which is counting-regular.

If $Z$ halts on blank tape, then $\overline{L(M)}$ = $\{xw1^{n}\ |\ x\in(a+b)^{+},\ |x|_{a}=|x|_{b}=n\}$ for some $w$ which is not counting-regular (as shown in Theorem 27).

Now consider $L(M)$. If $Z$ does not halt on blank tape, then $L(M)$ = $\Sigma^{*}$, hence is counting-regular. If $Z$ halts on blank tape, then $L(M)$ = $L(M_{1})\cup L(M_{2})$, which we will show to be not counting-regular as follows: the generating function $g(z)$ of $L(M)$ is given by ${1\over{1-sz}}-f(z)$ where $f(z)$ is the generating function of $\overline{L(M)}$, and $s$ is the size of the alphabet over which $M$ is defined. If $L(M)$ is counting-regular, then $g(z)$ is rational, then so is $f(z)$ = ${1\over{1-kz}}-g(z)$, contradicting the fact that $f(z)$ is not rational. \qed

Note that the machine $M$ constructed in the proof above is 1-reversal-bounded but not finitely-ambiguous. Next, it is shown to be undecidable for $2$-ambiguous machines but without the reversal-bound.

Theorem 30

It is undecidable, given a one-way 2-ambiguous nondeterministic one counter machine $M$, whether $\overline{L(M)}$ is counting-regular. Also, it is undecidable, given such a machine $M$, whether $L(M)$ is counting-regular.

Proof 22

It is known that it is undecidable, given two deterministic one counter machines (with no restriction on counter reversals) $M_{1}$ and $M_{2}$, whether $L(M_{1})\cap L(M_{2})=\emptyset$ (shown implicitly in [37]). Moreover, if the intersection is not empty, it is a singleton. Let $\Sigma$ be the input alphabet of $M_{1}$ and $M_{2}$. Let $a,b,1$ be three new symbols. Let

[TABLE]

Clearly, we can construct deterministic one counter machines accepting $L_{1}$ and $L_{2}$. Hence, $\overline{L_{1}}$ and $\overline{L_{2}}$ can be accepted by deterministic one counter machines. It follows that $\overline{L_{1}}\cup\overline{L_{2}}$ can be accepted by a $2$-ambiguous nondeterministic one counter machine $M$. Then, as in the proof of Theorem 28, $\overline{L(M)}$ (resp., $L(M)$) is counting-regular if and only if $L(M_{1})\cap L(M_{2})=\emptyset$, which is undecidable. \qed

In view of the above theorems, it is an interesting open question whether the undecidability holds for reversal-bounded finitely-ambiguous $\NCM$ machines.

7 Slender Semilinear and Length-Semilinear Languages and Decidability Problems

A topic closely related to counting functions of formal languages is that of slenderness. Decidability and closure properties of context-free languages ($\CFL$s) have been investigated in [3, 4, 2, 38, 39]. For example, [3] shows that it is decidable whether a $\CFL$ is slender, and in [4], it is shown that for a given $k\geq 1$, it is decidable whether a language generated by a matrix grammar is $k$-slender (although here, the $k$ needs to be provided as input in contrast to the $\CFL$ result).

In this section, we generalize these results to arbitrary language families that satisfy certain closure properties. These generalizations would then imply the known results for context-free languages and matrix languages, and other families where the problem was open.

First, we discuss the bounded language case. The constructive result of Theorem 14 above, plus decidability of slenderness for regular languages implies the following:

Corollary 31

Let $\LL$ be a semilinear trio (with all properties effective). Then, it is decidable, given $L$ bounded in $\LL$ and words $u_{1},\ldots,u_{k}$ such that $L\subseteq u_{1}^{*}\cdots u_{k}^{*}$, whether $L$ is slender.

In [39], the similar result was shown that it is decidable whether or not a given bounded semilinear language $L$ is slender.

In our definition of a semilinear family of languages $\LL$, we only require that every language in $\LL$ has a semilinear Parikh map. However, it is known that in every semilinear trio $\LL$, all bounded languages are bounded semilinear [17], and therefore the result of [39] also implies Corollary 31. Conversely, all bounded semilinear languages are in the semilinear trio $\NCM$ [24, 17]; hence, given any bounded semilinear language (in any semilinear family so long as we can construct the semilinear set), slenderness is decidable. This method therefore also provides an alternate proof to the result in [39].

Next, we will examine the case where $L$ is not necessarily bounded. One recent result is quite helpful in studying $k$-slender languages. In [35], the following was shown.

Theorem 32

[35]** Let $\LL$ be any semilinear full trio where the semilinearity and intersection with regular languages properties are effective. Then the smallest full AFL containing intersections of languages in $\LL$ with $\NCM$, denoted by $\hat{{\cal F}}(\LL\wedge\NCM)$, is effectively semilinear. Hence, the emptiness problem for $\hat{{\cal F}}(\LL\wedge\NCM)$ is decidable.

We make frequent use of this throughout the proofs of the next two sections.

First, decidability of $k$-slenderness is addressed.

Theorem 33

Let $\LL$ be a full trio which is either:

• •

semilinear, or

• •

is length-semilinear, closed under concatenation, and intersection with $\NCM$,

with all properties effective. It is decidable, given $k$ and $L\in\LL$, whether $L$ is a k-slender language.

Proof 23

Let $L\subseteq\Sigma^{*}$, and let $\#$ be a new symbol. First construct an $\NCM$ $M_{1}$ which when given $x_{1}\#x_{2}\#\ldots\#x_{k+1}$, $x_{i}\in\Sigma^{*},1\leq i\leq k+1$, accepts if $|x_{1}|=\cdots=|x_{k+1}|$, and $x_{i}\neq x_{j}$ are different, for all $i\neq j$. To do this, $M_{1}$ uses (many) counters to verify the lengths, and it uses counters to guess positions and verify the discrepancies between $x_{i}$ and $x_{j}$, for each $i\neq j$. Let $\LL^{\prime}=\hat{{\cal F}}(\LL\wedge\NCM)$ (or just $\LL$ in the second case), which is semilinear by Theorem 32, (or length-semilinear in the second case, by assumption). Construct $L_{2}\in\LL^{\prime}$ which consists of all words of the form $x_{1}\#x_{2}\#\ldots\#x_{k+1}$, where $x_{i}\in L$ (every full AFL is closed under concatenation). Then $L_{3}=L(M_{1})\cap L_{2}\in\LL^{\prime}$. Clearly, $L$ is not $k$-slender if and only if $L_{3}$ is not empty, which is decidable by Theorem 32. \qed

There are many known semilinear full trios listed in Example 1. Plus, it is known that languages generated by matrix grammars form a length-semilinear (but not semilinear, in general) full trio closed under concatenation and intersection with $\NCM$ ([4], where is it is shown that the languages are closed under intersection with BLIND multicounter languages, known to be equivalent to $\NCM$ [40]). Therefore, the result is implied for matrix grammars as well, although this is already known.

Corollary 34

Let $\LL$ be any of the families listed in Example 1. Then, the problem, “for $k\geq 1$ and $L\in\LL$, is $L$ a $k$-slender language?” is decidable.

All the machine models used in Example 1 have one-way inputs, however with a two-way input, the problem is more complicated. Now let $2\DCM(k)$ (resp., $2\NCM(k)$) be a two-way $\DFA$ (resp., two-way $\NFA$) with end-markers on both sides of the input, augmented with $k$ reversal-bounded counters.

Theorem 35

It is decidable, given $k$ and $2\DCM(1)$ $M$, whether $M$ accepts a $k$-slender language.

Proof 24

We may assume that $M$ always halts [41]. Given $M$, construct another $2\DCM(1)$ $M^{\prime}$ with a $(k+1)$-track tape. First for each $1\leq i<k+1$, $M^{\prime}$ checks that the string in track $i$ is different from the strings in tracks $i+1,\ldots,k+1$. Thus, $M$ needs to make multiple sweeps of the $(k+1)$-track input.

Then $M^{\prime}$ checks that the string in each track is accepted. Clearly, $L(M)$ is not $k$-slender if and only if $L(M^{\prime})$ is not empty. The result follows, since emptiness is decidable for $2\DCM(1)$ [41]. \qed

The above result does not generalize for $2\DCM(2)$:

Theorem 36

The following are true:

It is undecidable, given $k$ and a $2\DCM(2)$ $M$, whether $M$ accepts a $k$-slender language, even when $M$ accepts a letter-bounded language that is a subset of $a_{1}^{*}\cdots a_{r}^{*}$ for given $a_{1},\ldots,a_{r}$. 2. 2.

It is undecidable, given $k$ and a $2\DCM(2)$ $M$, whether $M$ accepts a slender language, even when $M$ accepts a letter-bounded language.

Proof 25

It is known [24] that it is undecidable, given a $2\DCM(2)$ $M$ accepting a language that is in $b_{1}^{*}\cdots b_{r}^{*}$ for given $b_{1},\ldots,b_{r}$, whether $L(M)=\emptyset$. Let $c$ and $d$ be new symbols. Construct another $2\DCM(2)$ $M^{\prime}$ which when given a string $w=b_{1}^{i_{1}}\cdots b_{k}^{i_{r}}c^{i}d^{j}$, simulates $M$ on $b_{1}^{i_{1}}\cdots b_{k}^{i_{r}}$, and when $M$ accepts, $M^{\prime}$ accepts $w$. Then $L(M^{\prime})$ is not $k$-slender for any given $k$ (resp., not slender) if and only if $L(M)$ is not empty, which is undecidable. \qed

Whether or not Theorem 33 holds for $2\NCM(1)$ $M$ is open. However, we can prove a weaker version using the fact that it is decidable, given a $2\NCM(1)$ $M$ accepting a bounded language over $w_{1}^{*}\cdots w_{r}^{*}$ for given $w_{1},\ldots,w_{r}$, whether $L(M)=\emptyset$ [42].

Theorem 37

It is decidable, given $k$ and $2\NCM(1)$ $M$ that accepts a language over $w_{1}^{*}\cdots w_{r}^{*}$ for given $w_{1},\ldots,w_{r}$, whether $M$ accepts a $k$-slender language.

Proof 26

We construct from $M$ another $2\NCM(1)$ $M^{\prime}$ which, when given $x_{1}\#x_{2}\#\cdots\#x_{k+1}$, where each $x_{i}$ is in $w_{1}^{*}\cdots w_{r}^{*}$ first checks that all $x_{i}$’s are different. For each $i$, and $j=i+1,\ldots,k+1$, $M^{\prime}$ guesses the position of discrepancy between $x_{i}$ and $x_{j}$ and records this position in the counter so that it can check the discrepancy. (Note that only one reversal-bounded counter is needed for this.) Then $M^{\prime}$ checks that each $x_{i}$ is accepted. Clearly, $M$ is not $k$ slender if and only if $L(M^{\prime})$ is not empty, which is decidable, since the language accepted by $M^{\prime}$ is bounded. \qed

We can also prove some closure properties. Here is an example:

Theorem 38

The following are true:

Slender (resp., thin) $\NCM$ languages are closed under intersection. 2. 2.

Slender (resp., thin) $2\DCM(1)$ languages ($2\NCM(1)$* languages) are closed under intersection.*

Proof 27

Straightforward since the families of languages above are closed under intersection. \qed

Deciding if an $\NCM$ language (or anything more general than $\CFL$s) is $k$-slender for some $k$ (where $k$ is not part of the input) is open, although as we showed in Theorem 35, for a given $k$, we can decide if an $\NCM$ or $\NPCM$ accepts a $k$-slender language. We conjecture that every $k$-slender $\NPCM$ language is bounded. If this can be proven, we will also need an algorithm to determine words $w_{1},\ldots,w_{r}$ such that the language is a subset of $w_{1}^{*}\cdots w_{r}^{*}$, which we also do not yet know how to do.

Let $c\geq 1$. A $2\DCM(k)$ ($2\NCM(k))$ $M$ is $c$-crossing if the number of times the input head crosses the boundary of any two adjacent cells of the input is at most $c$. Then $M$ is finite-crossing if it is $c$-crossing for some $c$. It is known that a finite-crossing $2\NCM(k)$ can can be converted to an $\NCM(k^{\prime})$ for some $k^{\prime}$ [43]. Hence every bounded language accepted by any finite-crossing $2\NCM(k)$ is counting-regular. The next result shows that this is not true if the two-way input is unrestricted:

Theorem 39

The letter-bounded language $L=\{a^{i}b^{ij}~{}|~{}i,j\geq 1\}$ is accepted by a $2\DCM(1)$ whose counter makes only $1$ reversal, but $L$ is not counting-regular.

Proof 28

A $2\DCM(1)$ $M$ accepting $L$ operates as follows, given input $a^{i}b^{k}$ ($i,k\geq 1$): $M$ reads and stores $k$ in the counter. Then it makes multiple sweeps on $a^{i}$ while decrementing the counter to check that $k$ is divisible by $i$.

To see that $L$ is not counting-regular, we note that, for any $n\geq 2$, the number of strings of length $n$ in $L$ is $\phi(n)$, Euler’s totient function is equal to the number of divisors of $n$. In fact, let the divisors of $n$ be $d_{1},d_{2},\ldots,d_{m}$. Then, there are exactly $m$ strings of length $n$, namely: $a^{d_{1}}b^{(n/d_{1}-1)d_{1}},a^{d_{2}}b^{(n/d_{2}-1)d_{2}},\ldots,\\ a^{d_{m}}b^{(n/d_{m}-1)d_{m}}$. Conversely, for each string $w$ of length $n$ in $L$, there is a unique divisor of $n$ (namely the number of $a$’s in $w$) associated with the string. This means the generating function of $L$ is $f(z)$ = $\sum_{n\geq 2}\phi(n)z^{n}$.

It can be shown that $f(z)$ = $\sum_{n\geq 2}a_{n}{z^{n}\over{1-z^{n}}}$ where $a_{n}$ = $\sum_{d|n}\phi(d)\mu(n/d)$ where $\mu$ is the Mobius function. From this expression, it is clear that every solution to $z^{n}=1$ is a pole of $f(z)$ and so the $n$’th root of unity is a pole of $f(z)$ for each positive integer $n\geq 2$. Since a rational function can have only a finite number of poles, it follows that $f(z)$ is not rational and hence $L$ is not counting-regular. \qed

However, it is known that unary languages accepted by $2\NCM(k)$s are regular [41]; hence, such languages are counting-regular.

This is interesting, as $2\DCM(1)$ has a decidable emptiness and slenderness problem, yet there is a letter-bounded language from the theorem above accepted by a $2\DCM(1)$ that is not counting-regular.

The following result provides an interesting contrast, as it involves a model with an undecidable membership (and emptiness) problem, but provides an example of a (non-recursively enumerable) language that is all of slender, thin, bounded, semilinear, but also counting-regular.

Theorem 40

There exists a language $L$ that is letter-bounded and semilinear and thin and counting regular but is not recursively enumerable. Moreover, we can effectively construct a $\DFA$ $M$ such that $f_{L(M)}(n)=f_{L}(n)$.

Proof 29

Let $L\subseteq a^{*}$ be a unary language that is not recursively enumerable, which is known to exist [44]. Assume without loss of generality that the empty word is not in $L$, but the letter $a$ itself is in $L$.

Let $L^{\prime}$ be the language consisting of, for each $n\geq 1$, the single word with all $a$’s except for one $b$ in the position of the largest $m\leq n$ such that $a^{m}\in L$.

Then $L^{\prime}$ has one word of every length, and is therefore thin. Also, it has one word of every length and exactly one $b$ in every word, so it has the same Parikh map as $a^{*}b$, so is semilinear. Also, it is clearly bounded in $a^{*}ba^{*}$. It is also not recursively enumerable, otherwise if it were, then make a gsm [18] that outputs $a$’s for every $a$ until a $b$, then it outputs $a$. Then, for every remaining $a$, it outputs the empty word. Applying this to $L^{\prime}$ gives $L$ (the position of the $b$ lets the gsm recover the words of $L$). But the recursively enumerable languages are closed under gsm mappings, a contradiction. \qed

8 Characterization of $k$-Slender Semilinear and Length-Semilinear Languages

This section discusses decidability properties (such as the problem of testing whether two languages are equal, or one language is contained in another) for $k$-slender languages in arbitrary families of languages satisfying certain closure properties. It is known that the equivalence problem for $\NPCM$ languages that are subsets of $w_{1}^{*}\cdots w_{r}^{*}$ for given $w_{1},\ldots,w_{r}$ is decidable. However, as mentioned above, we do not know yet if $k$-slender languages are bounded and even if they are, we do not know yet how to the determine the associated words $w_{1},\ldots,w_{r}$. Hence, the definition and results below are of interest.

The following notion is useful for studying decidability properties of slender languages. Let $k\geq 1$ be given. A language $L$ is $k$-slender effective if we can effectively construct a $\DFA$ over a unary alphabet $\{1\}$ with $k+1$ distinguished states $s_{0},s_{1},\ldots,s_{k}$ (other states can exist) which, when given an input $1^{n}$ where $n\geq 0$, halts in state $s_{i}$ if $f_{L}(n)=i$, where $0\leq i\leq k$. (Hence, the $\DFA$ can determine the number of strings in $L$ of length $n$, for every $n$.)

For example, consider the language

[TABLE]

Then,

[TABLE]

Clearly, $L$ is $2$-slender effective.

Next, we will discuss which $k$-slender languages are $k$-slender effective. First, we say that the length-semilinear property is effective if it is possible to effectively construct, for $L\in\LL$, a $\DFA$ accepting $\{1^{n}\mid f_{L}(n)\geq 1\}$.

Theorem 41

Let $\LL$ be an effective length-semilinear trio. A finite union of thin languages in $\LL$ is $k$-slender effective.

Proof 30

Let $L$ is the finite union of $L_{1},\ldots,L_{k}\in\LL$, each thin. Then construct $h(L_{i})$, where $h$ maps $L_{i}$ onto the single letter $1$. Then $h(L_{i})=\{1^{n}\mid f_{L_{i}}(n)=1\}$. Then since $\LL$ is length-semilinear, each of $h(L_{i})$ are regular, and can be accepted by a $\DFA$ $M_{i}$. Thus, make another $\DFA$ $M^{\prime}$ such that, on input $1^{n}$, $M^{\prime}$ runs each $M_{i}$ in parallel on $1^{n}$, and then switches to distinguished state $s_{j}$ if there are $j$ of the $\DFA$s $M_{1},\ldots,M_{k}$ that are accepting. \qed

It will be seen next that for all $k$-slender languages in “well-behaved” families, they are $k$-slender effective. First, the following lemma is needed.

Lemma 42

Let $\LL$ be a full trio which is either:

• •

semilinear, or

• •

is length-semilinear, closed under concatenation, and intersection with $\NCM$,

with all properties effective. Let $k\geq 1$ and $L\in\LL$, a $k$-slender language such that $f_{L}(n)$ is equal to [math] or $k$ for every $n$. There is a $\DFA$ that can determine $f_{L}(n)$. Hence $L$ is a $k$-slender effective language.

Proof 31

Let $\LL^{\prime}=\hat{{\cal F}}(\LL\wedge\NCM)$ (or just $\LL$ in the second case), which is semilinear by Theorem 32, (or length-semilinear by assumption).

Consider $L^{\prime}=\{x_{1}\#\cdots\#x_{k}\mid x_{1},\ldots,x_{k}\in L\}\in\LL^{\prime}$. Create $L^{\prime\prime}$ by intersecting $L^{\prime}$ with an $\NCM$ language that enforces that all words of the form $x_{1}\#\cdots\#x_{k}$ have $|x_{1}|=\cdots=|x_{k}|$, and $x_{i}\neq x_{j}$ for each $i\neq j$. Thus $L^{\prime\prime}=\{x_{1}\#\cdots\#x_{k}\mid x_{1},\ldots,x_{k}\in L,|x_{1}|=\cdots=|x_{k}|,x_{i}\neq x_{j}\mbox{~{}for each~{}}i\neq j\}$. Hence, $L^{\prime\prime}\in\LL^{\prime}$. Let $L^{\prime\prime\prime}$ be the language obtained from $L^{\prime\prime}$ by homomorphism that projects onto the single letter $1$. Since $L^{\prime\prime\prime}$ is length-semilinear, it can be accepted by a $\DFA$. Moreover, the length $n$ of a word $x_{1}\#\cdots\#x_{k}\in L^{\prime\prime}$ can be transformed into $|x_{1}|$ via $\frac{n-(k-1)}{k}$. Given the $\DFA$, then another $\DFA$ can be built that can determine $f_{L}(n)$. \qed

Then, the following is true.

Theorem 43

Let $\LL$ be a full trio which is either:

• •

semilinear, or

• •

is length-semilinear, closed under concatenation, union, and intersection with $\NCM$,

with all properties effective. If $L\in\LL$ be a $k$-slender language $L$, then $L$ is $k$-slender effective.

Proof 32

The case $k=1$ is true by Theorem 41. Assume by induction that the theorem is true for $k\geq 1$.

Now consider an $L\in\LL$ that is a $(k+1)$-slender language, $k\geq 1$. Hence $f_{L}(n)\leq(k+1)$ for each $n\geq 0$.

Let $\LL^{\prime}=\hat{{\cal F}}(\LL\wedge\NCM)$ (or just $\LL$ in the second case), which is semilinear by Theorem 32, (or length-semilinear by assumption). Let $A=\{x_{1}\#\cdots\#x_{k+1}\mid x_{1},\ldots,x_{k+1}\in L\}\in\LL^{\prime}.$ Then intersect $A$ with an $\NCM$ that enforces that all words $x_{1}\#\cdots\#x_{k+1}$ have $|x_{1}|=\cdots=|x_{k+1}|$, and $x_{i}\neq x_{j}$ for each $i\neq j$. Let $A^{\prime}$ be the resulting language. Then $A^{\prime}=\{x_{1}\#\cdots\#x_{k+1}\mid x_{1},\ldots,x_{k+1}\in L\mbox{~{}such that~{}}|x_{1}|=\cdots=|x_{k+1}|,x_{i}\neq x_{j}\mbox{~{}for each~{}}i\neq j\}.$ By Lemma 42, a $\DFA$ accepting $\{1^{n}\mid f_{L}(n)=k+1\}$ can be effectively constructed. Thus, a $\DFA$ accepting $\{1^{n}\mid f_{L}(n)\neq k+1\}$ can also be constructed. Furthermore, $B=\{w\mid w\in L,f_{L}(|w|)\neq k+1\}\in\LL^{\prime}$. Then, $B$ is $k$-slender and, hence $k$-slender effective by induction hypothesis. Hence, $L$ is $(k+1)$-slender effective. \qed

The proof of Theorem 43 actually shows the following:

Corollary 44

Let $\LL$ be a full trio closed under concatenation, union, and intersection with $\NCM$, and is length-semilinear with all properties effective. Let $k\geq 1$. A language $L\in\LL$ is a $k$-slender language if and only if $L=L_{1}\cup\cdots\cup L_{k}$, where for $1\leq i\leq k$, $L_{i}$ is an $i$-slender effective language such that $f_{L_{i}}(n)$ is equal to [math] or $i$ for each $n$.

Next, decidability of containment is addressed.

Theorem 45

Let $\LL$ be a full trio which is either:

• •

semilinear, or

• •

is length-semilinear, closed under concatenation, and intersection with $\NCM$,

with all properties effective. It is decidable, given $L_{1},L_{2}\in\LL$ with $L_{2}$ being a $k$-slender language, whether $L_{1}\subseteq L_{2}$.

Proof 33

Then $L_{2}$ is $k$-slender effective by Theorem 43. Without loss of generality, assume that the input alphabet of both $L_{1}$ and $L_{2}$ is $\Sigma$. Let $1,\#$, and \$$ be new symbols. Let \LL^{\prime}=\hat{{\cal F}}(\LL\wedge\NCM)$(or just$\LL$ in the second case), which is semilinear by Theorem 32, (or length-semilinear by assumption). We will construct a sequence of machines and languages below.

First, let $M_{1}^{\prime}$ (resp. $M_{2}^{\prime}$) be the unary $\DFA$ accepting all words $1^{n}$ where a word of length $n$ is in $L_{1}$ (resp. in $L_{2}$). Let $A_{1}=L(M_{1})-L(M_{2})$. (This is empty if and only if all lengths of words in $L_{1}$ are lengths of words in $L_{2}$. This language is regular. 2. 2.

Construct $A_{2}\in\LL^{\prime}$ consisting of all words w=1^{n}\x$y_{1}#\cdots y_{r}$$*, where *$x\in L_{1}$* and each *$y_{j}\in L_{2}$. 3. 3.

Construct an $\NCM$ $A_{3}$ which, when given w=1^{n}\x$y_{1}#\cdots y_{r}$$*, accepts *$w$ if the following is true:

1. (a)

$r=f_{L_{2}}(n)$* (which can be tested since $L_{2}$ is $k$-slender effective).* 2. (b)

$|x|=|y_{1}|=\cdots=|y_{r}|=n$. 3. (c)

$y_{i}\neq y_{j}$* for each $i\neq j$.* 4. (d)

$x\neq y_{i}$* for each $i$.*

Note that $A_{3}$ needs multiple reversal-bounded counters to carry out the four tasks in parallel. 4. 4.

Construct $A_{4}=A_{2}\cap A_{3}\in\LL^{\prime}$. 5. 5.

Finally construct an $A_{5}=A_{4}\cup A_{1}\in\LL^{\prime}$ (full trios are closed under union with regular languages **[23]**).

It is easy to verify that $L_{1}\not\subseteq L_{2}$ if and only if $A_{5}$ is not empty, which is decidable, since emptiness is decidable. \qed

Corollary 46

Let $\LL$ be a full trio which is either:

• •

semilinear, or

• •

is length-semilinear, closed under concatenation, and intersection with $\NCM$,

with all properties effective. It is decidable, given $L_{1},L_{2}\in\LL$ that are $k$-slender languages, whether $L_{1}=L_{2}$.

There are many semilinear full trios in the literature for which the properties in this section hold.

Corollary 47

Let $\LL$ be any of the families from Example 1. The following are decidable:

• •

For $L_{1},L_{2}$ with $L_{2}$ being $k$-slender, is $L_{1}\subseteq L_{2}$?

• •

For $L_{1},L_{2}$ being $k$-slender languages, is $L_{1}=L_{2}$?

Furthermore, matrix grammars are an example of a length-semilinear [45] full trio closed under concatenation, union, and intersection with $\NCM$ (although they do accept non-semilinear languages). We therefore get all these properties for matrix grammars as a consequence of these proofs. However, this result is already known [4].

Using the ideas in the constructive proof of the theorem above, we can also show:

Theorem 48

Let $\LL$ be a union and concatenation closed length-semilinear full trio with all properties effective that is closed under intersection with $\NCM$. Let $L_{1},L_{2}\in\LL$ with $L_{2}$ a $k$-slender language. Then $L_{1}-L_{2}\in\LL$. Hence, the complement of any $k$-slender language in $\LL$ is again in $\LL$.

Proof 34

Let $\Sigma$ be the (without loss of generality) joint alphabet of $L_{1}$ and $L_{2}$, and let $\Sigma^{\prime}$ be the set of the primed versions of the symbols in $\Sigma$. Let $\#$, and $$$ be new symbols. Consider input

[TABLE]

where $x$ is in $(\Sigma^{\prime})^{*}$ and $y_{1},\ldots,y_{r}$ are in $\Sigma^{*}$, for some $0\leq r\leq k$. By Theorem 43, $L_{2}$ is $k$-slender effective. Let $M^{\prime}$ be this unary $\DFA$ accepting all words of lengths in $L_{2}$. Build an $\NCM$ $M^{\prime\prime}$ that on input $w$, verifies:

$r=f(n)$. 2. 2.

$|x|=|y_{1}|=\cdots=|y_{r}|$. 3. 3.

$y_{i}\neq y_{j}$* for each $i\neq j$.* 4. 4.

$h(x)\neq y_{i}$* for each $i$, where $h(a^{\prime})=a$ for each $a^{\prime}\in\Sigma^{\prime}$.*

Consider $L^{\prime\prime\prime}\in\LL$ consisting of all words of the form of $w$ in Equation 2, where $x\in L_{1}$, and each $y_{i}\in L_{2}$. This is in $\LL$ since $\LL$ is closed under concatenation.

Now define a homomorphism $h_{1}$ which maps \#,\$$, and symbols in \Sigma$to$\epsilon$and fixes letters in$\Sigma^{\prime}$. Clearly,$h_{1}(L^{\prime\prime\prime}\cap L(M^{\prime\prime}))$is$L_{1}-L_{2}$, and it is in$\LL$. \qed

This holds for not only the matrix languages, but also concatenation and union-closed semilinear full trios closed under intersection with $\NCM$. Some examples are:

Corollary 49

Let $\LL$ be any family of languages that are accepted by a machine model in Example 1 that are augmented by reversal-bounded counters. Given $L_{1},L_{2}\in\LL$ with $L_{2}$ being $k$-slender, then $L_{1}-L_{2}\in\LL$. Furthermore, the complement of any $k$-slender language in $\LL$ is again in $\LL$.

Next, decidability of disjointness for $k$-slender languages will be addressed.

Theorem 50

Let $\LL$ be a full trio which is either:

• •

semilinear, or

• •

is length-semilinear, closed under concatenation, union, and intersection with $\NCM$,

with all properties effective. Given $L_{1},L_{2}\in\LL$ being $k$-slender languages, it is decidable whether $L_{1}\cap L_{2}=\emptyset$.

Proof 35

Let $\LL^{\prime}=\hat{{\cal F}}(\LL\wedge\NCM)$ (or just $\LL$ in the second case), which is semilinear by Theorem 32, (or length-semilinear by assumption).

Notice that $L_{1}\cap L_{2}=(L_{1}\cup L_{2})-((L_{1}-L_{2})\cup(L_{2}-L_{1}))$. By Theorem 48, $L_{1}-L_{2}\in\LL$ and $L_{2}-L_{1}\in\LL$, and both must be $k$-slender since $L_{1}$ and $L_{2}$ are both $k$-slender. Certainly $(L_{1}-L_{2})\cup(L_{2}-L_{1})\in\LL$, and is also $2k$-slender. Also, $L_{1}\cup L_{2}\in\LL$. Hence, by another application of Theorem 48, $(L_{1}\cup L_{2})-((L_{1}-L_{2})\cup(L_{2}-L_{1}))\in\LL$. Since emptiness is decidable in $\LL$, the theorem follows. \qed

This again holds for all the families in Example 1 plus the languages accepted by matrix grammars.

An interesting open question is whether every $k$-slender $\NCM$ language (or other more general families) can be decomposed into a finite disjoint union of thin $\NCM$ languages. Although we have not been able to show this, we can give a related result. To recall, in [14], the model $\TCA$ is introduced consisting of a nondeterministic Turing machine with a one-way read-only input tape, a finite-crossing read/write tape, and reversal-bounded counters. It is shown that this model only accepts semilinear languages, and indeed, it is a full trio. Clearly, the model is closed under intersection with $\NCM$ by adding more counters. Although we do not know whether it is possible to decompose $\NCM$ slender languages into thin $\NCM$ languages, we can decompose them into thin $\TCA$ languages.

Theorem 51

Every $k$-slender $\NCM$ language $L$ is a finite union of thin $\TCA$ languages.

Proof 36

Let $M$ be an $\NCM$ accepting $L$. Since $L$ is $k$-slender, for each $n$, there are either exactly $k$ words of length $n$, or $k-1$ words of length $n$, etc. or [math] words of length $n$. Let $A_{k}=\{x_{1}\#\cdots\#x_{k}\mid x_{1},\ldots,x_{k}\in L(M),|x_{1}|=\cdots=|x_{k}|$, $x_{1}<\cdots<x_{k}\}$ (the $<$ relation uses lexicographic ordering). For all such words $x_{1}\#\cdots\#x_{k}$, then the lengths of the first parts, $|x_{1}|$, are exactly those lengths $n$ such that $f_{L}(n)=k$. Then build a $\TCA$ $M^{\prime}$ accepting $A_{k}$ as follows: $M^{\prime}$ reads $x_{1}\#\cdots\#x_{k}$, and verifies each $x_{i}\in L(M)$ using a set of counters, while in parallel verifying their lengths are the same. In parallel, $M^{\prime}$ writes $x_{1}$ on the worktape; then when reading $x_{2}$ letter-by-letter, it scans $x_{1}$ on the worktape also letter-by-letter, and in the first position where they differ, it verifies that $x_{2}>x_{1}$. From that point on, it replaces $x_{1}$ on the tape with $x_{2}$. It then repeats up to $x_{k}$.

Let $G_{i}$ be a gsm that extracts the $i$’th “component” of $A_{k}$. Then $G_{1}(A_{k}),\ldots,G_{k}(A_{k})$ are all thin languages. As they are thin, there is a $\DFA$ $M_{k}$ accepting all these lengths of words. Next, let $A_{k-1}=\{x_{1}\#\cdots\#x_{k-1}\mid x_{1},\ldots,x_{k-1}\in L(M),|x_{1}|=\cdots=|x_{k-1}|$, $x_{1}<\cdots<x_{k-1},1^{|x_{1}|}\notin L(M_{k})\}$. These must be those words $x_{1}\#\cdots\#x_{k-1}$ where there cannot be a $k$’th such word (since $1^{|x_{1}|}\notin L(M_{k})$). And again, we can use $C_{1}(A_{k-1}),\ldots,C_{k-1}(A_{k-1})$ to separate based on lexicographic order, and each such language is thin.

Continuing in this fashion down to $1$, we see that $L$ is a finite union of thin languages in $\TCA$ (for all $C_{i}(A_{j}),1\leq i\leq k$). \qed

A language $L$ is $k$-counting-regular if there exists a regular language $L^{\prime}$ such that $f_{L^{\prime}}(n)=f_{L}(n)\leq k$ for $n\geq 0$ (this is equivalent to $L$ being $k$-slender and counting-regular). $L$ is finite-counting regular if it is $k$-counting regular for some $k$. The next result shows that all $k$-slender languages in “well-behaved” language families are $k$-counting-regular.

Theorem 52

Let $\LL$ be a full trio which is either:

• •

semilinear, or

• •

is length-semilinear, closed under concatenation, and intersection with $\NCM$,

with all properties effective. Let $L$ be a $k$-slender language in $\LL$. Then $L$ is $k$-counting-regular (and thus counting-regular) and we can effectively construct a $\DFA$ $M$ such that $f_{L}(n)=f_{L(M)}(n)\leq k$ for $n\geq 0$. Moreover, $L(M)$ is bounded with $L(M)\subseteq 1^{*}\{\#_{1},\ldots,\#_{k}\}$ for some distinct symbols $\#_{1},\ldots,\#_{k}$.

Proof 37

From Lemma 42, $L$ is $k$-slender effective, and we can construct a $\DFA$ $M^{\prime}$ such that when given input $1^{n}$ , halts in state $s_{i}$ $(0\leq i\leq k)$ if $f_{L}(n)=i$.

Let $\#_{1},\ldots,\#_{k}$ be new symbols. Construct a $\DFA$ $M$ which, on input $w=1^{n-1}\#_{s}$ (for some $1\leq s\leq k$) simulates $M^{\prime}$ on $w$. ($M$ pretends that $\#_{s}$ is $1$ in the simulation). If $M^{\prime}$ lands in state $s_{t}$ ($1\leq t\leq k$), $M$ accepts $w$ if and only of $s=1,\ldots,t$. (Of course, if none of $M^{\prime}$ lands in state $s_{0}$, $M$ does not accept $w$).

It is easy to verify that $f_{L}(n)=L_{L(M)}(n)\leq k$ for all $n\geq 0$. Hence, $L$ is $k$-counting-regular. \qed

Corollary 53

Let $\LL$ be a full trio which is either:

• •

semilinear, or

• •

is length-semilinear, closed under concatenation, and intersection with $\NCM$,

with all properties effective. Then

$L$* is $k$-slender if and only if $L$ is $k$-counting regular.* 2. 2.

$L$* is slender if and only if $L$ is finite-counting regular.*

From Corollary 31 and the above corollary:

Corollary 54

Let $\LL$ be a semilinear trio (with all properties effective). Then, it is decidable, given $L$ bounded in $\LL$ and words $u_{1},\ldots,u_{k}$ such that $L\subseteq u_{1}^{*}\cdots u_{k}^{*}$, whether $L$ is finite-counting-regular.

As in Section 4, we assumed in this section that the closure properties are effective, since we wanted the results to be effective. However, we can remove this assumption and many of the results would still hold existentially. For example, the closure properties would still hold.

9 Conclusions

In this work, we attempted to understand languages with simple counting functions: those that have counting functions that belong to the class of counting functions of regular languages (known as counting-regular languages), and those for which the counting function is bounded by a constant $k$, (known as $k$-slender languages). First, it is shown that all unambiguous $\NTM$s with a one-way input and a reversal-bounded worktape are counting-regular. Then, certain “well-behaved” language families are considered, that form semilinear full trios $\LL$. It is shown that the counting functions for the bounded languages in $\LL$ coincide with the counting functions for the bounded regular languages. Also, all $k$-slender languages in $\LL$ have the same counting function as some bounded regular language. The containment, equality, and disjointment problems are shown to be decidable for $k$-slender languages in $\LL$. Most results are general enough to cover more general families that are not semilinear should they satisfy certain other closure properties (such families include the languages generated by matrix grammars).

We conclude with some open problems arising from this study. It is open whether for every counting-regular $\CFL$ $L$ over a $k$-letter alphabet, there is a regular language $L^{\prime}$ over a $k$-letter alphabet such that $f_{L}$ = $f_{L^{\prime}}$. Regarding closure properties, we conjecture that the counting-regular $\CFL$s are not closed under Kleene star. It is unknown whether every $k$-slender language is bounded within every semilinear full trio (this is true within the context-free languages). Also, the decidability status of whether a language $L$ in an arbitrary full trio is slender ($k$ is not given) is open (this is again decidable for the context-free languages). Similarly, it is unknown whether there is a procedure to determine words $w_{1},\ldots,w_{n}$, if they exist, such that $L\subseteq w_{1}^{*}\cdots w_{n}^{*}$, for $L$ in such full trios (this is again possible for context-free languages).

Acknowledgements

We thank Flavio D’Alessandro for pointing out to us that Theorem 14 is also implied by the main result in [13].