Regularity of languages generated by non context-free grammars over a singleton terminal alphabet

TL;DR

This paper proves that any language over a singleton alphabet satisfying the Pumping Lemma is regular, extending known results about the regularity of certain non-context-free languages.

Contribution

It demonstrates that all languages in the Pumping Lemma superclass over a singleton alphabet are regular, without using Parikh's Theorem, broadening previous understanding.

Findings

Languages satisfying the Pumping Lemma over a singleton alphabet are regular

The proof is based on a transformational approach, not Parikh's Theorem

Extends known results to languages not necessarily context-free or satisfying Parikh's Theorem

Abstract

It is well-known that: (i) every context-free language over a singleton terminal alphabet is regular, and (ii) the class of languages that satisfy the Pumping Lemma is a proper super-class of the context-free languages. We show that any language in this superclass over a singleton terminal alphabet is regular. Our proof is based on a transformational approach and does not rely on Parikh's Theorem. Our result extends previously known results because there are languages that are not context-free, do satisfy the Pumping Lemma, and do not satisfy the hypotheses of Parikh's Theorem.

Full text

††journal: Information Processing Letters

Let us begin by introducing our terminology and notations.

The set of the natural numbers is denoted by $N$. The set of the $n$-tuples of natural numbers is denoted by $N^{n}$. We say that a language $L$ is over the terminal alphabet $\Sigma$ iff $L\!\subseteq\!\Sigma^{*}$. Given a word $w\!\in\!\Sigma^{*}$, $w^{0}$ is the empty word $\varepsilon$, and, for any $i\!\geq\!0$, $w^{i+1}$ is $w^{i}\,w$, that is, the concatenation of $w^{i}$ and $w$. The length of a word $w$ is denoted by $|w|$. Given a symbol $a\!\in\!\Sigma$, the number of occurrences of $a$ in $w$ is denoted by $|w|_{\displaystyle a}$. The cardinality of a set $A$ is denoted by $|A|$.

Given an alphabet $\mathrm{\Sigma}$ such that $|\Sigma|\!=\!1$, the concatenation of any two words $w_{1},w_{2}$ in $\mathrm{\Sigma}^{*}$ is commutative, that is, $w_{1}\,w_{2}=w_{2}\,w_{1}$.

In Theorem 2 below we extend the well known result stating that any context-free language over a singleton terminal alphabet is a regular language [5]. An early proof of this result appears in a paper by Ginsburg and Rice [4]. That proof is based on Tarski’s fixpoint theorem and it is not based on the Pumping Lemma (contrary to what has been stated in a paper by Andrei et al. [2]). Our extension is due to the facts that: (i) our proof does not rely on Parikh’s Theorem [7], like the proof in Harrison’s book [5], and (ii) there are non context-free languages that do satisfy the Pumping Lemma (see Definition 1) and do not satisfy Parikh’s Condition (see Definition 2) (and thus Parikh’s Theorem cannot be applied) [8]. Our proof is very much related to one presented in a book by Shallit [9], but we believe that ours is more elementary.

$\exists\,n\!>\!0$, $\forall\,z\in L$, if $|z|\geq n$, then $\exists\,u,v,w,x,y\in\mathrm{\Sigma}^{*}$, such that

(1)  $z=u\,v\,w\,x\,y$,

(2)  $v\,x\not=\varepsilon$,

(3)  $|v\,w\,x|\leq n$,  and

(4)  $\forall\,i\geq 0$, $u\,v^{i}\,w\,x^{i}\,y\in L$.

• $\Box$*

Let us first state and prove the following lemma whose proof is by transformation from Definition 1.

$\exists\,n\!>\!0$, $\forall\,z\!\in\!L$, if $|z|\!\geq\!n$, then $\exists\,p\!\geq\!0,$ $q\!\geq\!0,$ $m\!\geq\!0$, such that

${\rm{(1.1)}}~{}$ $|z|=p+q$,

${\rm{(2.1)}}~{}$ $0\!<\!q\!\leq\!n$,

${\rm{(3.1)}}~{}$ $0\!<\!m\!+\!q\!\leq\!n$,  and

${\rm{(4.1)}}~{}$ $\forall\,s\in\mathrm{\Sigma}^{*},$ $\forall\,i\!\geq\!0$, if $|s|=p+i\,q$, then $s\in L$.**

${\rm{(1^{\prime})}}~{}$* $z=\widetilde{u}\,w\,\widetilde{v}$*,**

${\rm{(2^{\prime})}}~{}$* $\widetilde{v}\not=\varepsilon$*,**

${\rm{(3^{\prime})}}~{}$* $|w\,\widetilde{v}|\leq n$*, and**

${\rm{(4^{\prime})}}~{}$* $\forall\,i\!\geq\!0$*, $\widetilde{u}\,w\,\widetilde{v}\hskip 0.7pt^{i}\in L$.**

Now if we take the lengths of the words and we denote $|\widetilde{u}\,w|$ by $p$, $|\widetilde{v}|$ by $q$, and $|w|$ by $m$, we get:

$\exists\,n\!>\!0$, $\forall\,z\in L$, if $|z|\!\geq\!n$, then $\exists\,p\!\geq\!0,$ $q\!\geq\!0,$ $m\!\geq\!0$, such that

${\rm{(1^{\prime\prime})}}~{}$* $|z|=p+q$*,**

${\rm{(2^{\prime\prime})}}~{}$* $q\!>\!0$*,**

${\rm{(3^{\prime\prime})}}~{}$* $m\!+\!q\leq n$*,  and**

${\rm{(4^{\prime\prime})}}~{}$* $\forall\,s\!\in\!\mathrm{\Sigma}^{*},$* $\forall\,i\!\geq\!0$, if $|s|=p+i\,q$, then $s\in L$.**

For all $n\!>\!0$, $q\!>\!0$, and $m\!\geq\!0$, we have that $(q\!>\!0\wedge m\!+\!q\!\leq\!n)$ iff $(0\!<\!q\!\leq\!n~{}\wedge~{}0\!<\!m\!+\!q\!\leq\!n)$. Thus, we get $\mathit{PL}1(L)$. $\Box$

We say that $\mathit{PL}1(L)$ holds for $b$ if $b$ is a witness of the quantification ‘$\exists\,n\!>\!0$’ in $\mathit{PL}1(L)$. The following theorem states our main result.

(i) $L_{\displaystyle{<\!b}}=\{a^{k}\mid a^{k}\!\in\!L\wedge k\!<\!b\}$   and   (ii) $L_{\displaystyle{\geq\!b}}=\{a^{k}\mid a^{k}\!\in\!L\wedge k\!\geq\!b\}$.

Now, $L_{\displaystyle{<\!b}}$ is a regular language, because it is finite.* Since regular languages are closed** under finite union and intersection [6], in order to prove that $L$ is regular, it is enough to prove, as we now do, that*

$L_{\displaystyle\geq\!b}=\bigcup\mathcal{S}~{}\cap~{}\{a^{i}\mid i\!\geq\!b\}$* $(\dagger 1)$*

where: (i) $\mathcal{S}$ is a set of languages which is* a subset of the following finite set $\mathcal{L}$ of languages ($k,p_{h},q_{0},\ldots,q_{k}$ are integers):*

$\mathcal{L}=\{L^{\displaystyle{\langle p_{h},q_{0},\ldots,q_{k}\rangle}}\mid~{}(0\!\leq\!k\!<\!b)~{}\wedge~{}(0\!\leq\!p_{h}\!<\!b)~{}\wedge~{}$* $(\dagger 2)$*

$(0\!<\!q_{0}\!\leq\!b)~{}\wedge\ldots\wedge~{}(0\!<\!q_{k}\!\leq\!b)~{}\wedge~{}$

$q_{0},\ldots,q_{k}$* are all distinct $\}$*

and (ii) for all $k,p_{h},q_{0},\ldots,q_{k}$, the language:

$L^{\displaystyle{\langle p_{h},q_{0},\ldots,q_{k}\rangle}}\!=\!\{a^{\displaystyle{\hskip 1.42262ptp_{h}+i_{0}\,q_{0}+\ldots+i_{k}\,q_{k}}}\mid i_{0}\!>\!0~{}\wedge\ldots\wedge~{}i_{k}\!>\!0\}$* $(\dagger 3)$*

is regular.

Indeed, (i) $\{a^{i}\mid i\!\geq\!b\}$ is regular, (ii) $\mathcal{L}$ is a finite set of languages because, for any $b$, there exists only a finite number of tuples $\langle p_{h},q_{0},\ldots,q_{k}\rangle$ satisfying all the conditions stated inside the set expression $(\dagger 2)$, and (iii) the language $L^{\displaystyle{\langle p_{h},q_{0},\ldots,q_{k}\rangle}}$ is regular because it is recognized by the following nondeterministic finite automaton with initial state $A$ and final state $B$:

\VCDraw{VCPicture}

(0,-1.5)(10,2.5) \FixStateDiameter12mm \FixStateLineDouble0.41.3 \ChgStateLineWidth1.2 \SetEdgeArrowWidth6pt \SetEdgeArrowLengthCoef1.8

\State

A1 \FinalStateB2 \SetStateLineColorwhite \FixStateDiameter1mm \State.dot1 \State.dot2 \State.dot3

\Initial

1 \Edge12\LabelL*[0.47]a^ ph​+​q0​+​…​+​qk \LoopN2*\LabelL*[0.74] a^ q0 \LoopE2*\LabelL*[0.5]a^ qk*

In order to prove Equality $(\dagger 1)$ it remains to prove that, for any $z\!\in\!L_{\displaystyle{\geq\!b}}$, there exists a tuple of the form $\langle p_{h},q_{0},q_{1},\ldots,q_{k}\rangle$ such* that $z\!\in\!L^{\displaystyle{\langle p_{h},q_{0},q_{1},\ldots,q_{k}\rangle}}$.*

Given any word $z\!\in\!L_{\displaystyle{\geq\!b}}$, the following algorithm constructs a tuple of the form $\langle p_{h},q_{0},q_{1},\ldots,q_{h}\rangle$, for some $h\!\geq\!0$.

$\{\,z\!\in\!L_{\displaystyle{\geq\!b}}\,\}$* Tuple Generation Algorithm*

$\ell\!:=\!|z|;~{}~{}~{}i\!:=\!0;~{}~{}~{}\langle p_{0},q_{0}\rangle\!:=\!\pi(\ell);$**

$\{\,|z|=p_{i}+\sum_{j=0}^{i}q_{j}~{}~{}\wedge~{}~{}\bigwedge_{j=0}^{i}~{}0\!<\!q_{j}\!\leq\!b~{}~{}\wedge~{}~{}0\!\leq\!p_{i}\,\}$**

while*  $p_{i}\!\geq\!b$  do  $\ell\!:=\!p_{i};~{}~{}i\!:=\!i\!+\!1;~{}~{}\langle p_{i},q_{i}\rangle\!:=\!\pi(\ell)$  od;*

$h\!:=\!i;$**

$\{\,|z|=p_{h}+\sum_{j=0}^{h}q_{j}~{}~{}\wedge~{}~{}\bigwedge_{j=0}^{h}~{}0\!<\!q_{j}\!\leq\!b~{}~{}\wedge~{}~{}0\!\leq\!p_{h}\!<\!b\,\}\vspace*{-2mm}$**

In this algorithm $\pi$ is a function from $N$ to $N\!\times\!N$, whose existence follows from the validity of $\mathit{PL}1(L)$, satisfying the following condition: for every $\ell\!\geq\!b$, $\pi(\ell)\!=\!\langle p,q\rangle$ such that $\ell\!=\!p\!+\!q$ and $0\!<\!q\!\leq\!b$ (take $i\!=\!1$ in Condition (4.1) of $\mathit{PL}1(L)$ in Lemma 1). The termination of the Tuple Generation Algorithm is a consequence of the fact that, for every $z\!\in\!L_{\displaystyle{\geq\!b}}$, for every $i\!\geq\!0$, $p_{i}\!=\!p_{i+1}+q_{i+1}$ and $q_{i}\!>\!0$. This implies that $p_{0},p_{1},\ldots$ is a* strictly decreasing sequence of integers, and eventually in that sequence we will get an element smaller than $b$, and the while-loop terminates.*

Thus, for every $z\!\in\!L_{\displaystyle{\geq\!b}}$, there exist $h\!\geq\!0,$ $p_{0},$ $q_{0},$ $p_{1},$ $q_{1},$ $p_{2},$ $q_{2},$ $\ldots,$ $p_{h},q_{h}$ such that:

$z$$=a^{\displaystyle{\,(p_{h}+q_{h})+q_{h-1}+\ldots+q_{2}+q_{1}+q_{0}}}$* $(\dagger 4)$*

where:* $0\!\leq\!p_{h}\!<\!b$  and  for every $i$, if $0\!\leq\!i\!<\!h$, then ($p_{i}\!\geq\!b$ and $0\!<\!q_{i}\!\leq\!b$).*

In general, in Equality $(\dagger 4)$ the $q_{i}$’s are not all distinct. Thus, by rearranging the summands, and writing $i_{j}\,q_{j}$, instead of $({q_{j}+\ldots+q_{j}})$ with $i_{j}$ occurrences of $q_{j}$, we have that, for every word $z\!\in\!L_{\displaystyle{\geq\!b}}$, there exist some integers $k,p_{h},i_{0},q_{0},\ldots,i_{k},q_{k}$ such that

$z=a^{\displaystyle{\,p_{h}+i_{0}\,q_{0}+\ldots+i_{k}\,q_{k}}}$,     where:

($\ell$ 0)$0\!\leq\!k$,               **($\ell$ 1)$0\!\leq\!p_{h}\!<\!b$,               ($\ell$ 2)$0\!<q_{0}\!\leq\!b,\ldots,0\!<q_{k}\!\leq\!b$,**

($\ell$ 3)$q_{0},\ldots,q_{k}$* are all distinct,   and   *($\ell$ 4)$i_{0}\!>\!0,\ldots,i_{k}\!>\!0$.

From ($\ell$* 2) and ($\ell$ 3), we have that $k\!<\!b$. Hence, Condition ($\ell$ 0) can be strengthened to: *($\ell\hskip 0.7pt{0}^{*})$$0\!\leq\!k\!<\!b$. We also have that $k\!\leq\!h$, and $k\!=\!h$ when in Equality $(\dagger 4)$ the values of $q_{0},\ldots,q_{h}$ are all distinct.

Since Conditions $(\ell\hskip 0.7pt{0}^{*}$), $(\ell\hskip 0.7pt{1})$, $(\ell\hskip 0.7pt{2})$, and $(\ell\hskip 0.7pt{3})$ are those occurring in the set expressions $(\dagger 2)$, and Condition $(\ell\hskip 0.7pt{4})$ is the one occurring in the set expressions $(\dagger 3)$, we have concluded the proof of Equality $(\dagger 1)$ and that of Theorem 2.  $\Box$

Let us make a few of remarks on the proof of Theorem 2.

(i) The validity of $\mathit{PL}1(L)$ tells us that the function $\pi$ exists, but it does not tell us how to compute $\pi(\ell)$, for any given $\ell\!\geq\!b$.

(ii) Since summation is commutative, it may be the case that a language in $\mathcal{L}$ corresponds to more than one tuple $\langle p_{h},q_{0},\ldots,q_{k}\rangle$. In particular, we have that $L^{\displaystyle{\langle p_{h},q_{0},\ldots,q_{k}\rangle}}$ $=$ $L^{\displaystyle{\langle p_{h},q^{\prime}_{0},\ldots,q^{\prime}_{k}\rangle}}$, whenever $\langle q_{0},\ldots,q_{k}\rangle$ is a permutation of $\langle q^{\prime}_{0},\ldots,q^{\prime}_{k}\rangle$.

(iii) If $b\!=\!1$, then $k\!=\!h\!=\!0$. Thus, from Conditions ($\ell$ 1) and ($\ell$ 3) we have: $\langle p_{h},q_{h}\rangle\!=\!\langle p_{0},q_{0}\rangle\!=\!\langle 0,1\rangle$. We also have that $\mathcal{L}$ is the singleton $\{L^{\textstyle{\langle 0,1\rangle}}\}$, where $L^{\textstyle{\langle 0,1\rangle}}$ is the language $\{a^{i}\mid i\!>\!0\}$.

(iv) In Equality $(\dagger 1)$ the set $\mathcal{S}$ of languages may be a proper subset of $\mathcal{L}$. Indeed, let us consider the language $L\!=\!\{a\,(a\,a)^{n}\mid n\!\geq\!0\}$ generated by the context-free grammar $S\rightarrow a\,S\,a\mid a$. Since $\mathit{PL}1(L)$ holds for $3$, we can take the constant $b$ occurring in Equality $(\dagger 1)$ to be $3$. If we consider the word $z\!=\!a\,a\,a$, then the set $\mathcal{L}$ of languages includes, among others, the languages $L^{\langle 0,3\rangle}\!=\!\{a^{0\,+\,i\cdot 3}\mid i\!>\!0\}$, $L^{\langle 1,2\rangle}\!=\!\{a^{1\,+\,i\cdot 2}\mid i\!>\!0\}$, and $L^{\langle 2,1\rangle}\!=\!\{a^{2\,+\,i\cdot 1}\mid i\!>\!0\}$ (these three languages are obtained for $k\!=\!h\!=\!0$). Now, $L_{\displaystyle\geq 3}=L^{\langle 1,2\rangle}\cap\{a^{i}\mid i\!\geq\!3\}=L^{\langle 1,2\rangle}$, while $L^{\langle 0,3\rangle}\!\not\subseteq\!L_{\displaystyle\geq 3}$ and $L^{\langle 2,1\rangle}\!\not\subseteq\!L_{\displaystyle\geq 3}$.

(v) It may be the case that the length ${p_{h}\!+\!q_{0}\!+\!\ldots\!+\!q_{k}}$ of the word labeling the arc from state $A$ to state $B$ of the finite automaton depicted above, is smaller than $b$. Thus, in the definition of $L_{\displaystyle{\geq\!b}}$ the intersection of $\bigcup\mathcal{S}$ with $\{a^{i}\mid i\!\geq\!b\}$ ensures that only words whose length is at least $b$ are considered.

Acknowledgements

This work has been partially funded by INdAM-GNCS (Italy).