2 Cyclic languages

Strongly cyclic languages

Marie-Pierre Beal¹, Olivier Carton² and Christophe Reutenauer³

1 LITP - Institut Blaise Pascal, Universite Denis Diderot 2 Place Jussieu 75251 Paris cedex 05.[email protected]

2 Institut Gaspard Monge, Universite de Marne-la-Vallee 93166 Noisy le Grand cedex.[email protected]

3 Mathematiques, Universite du Quebec a Montreal C.P. 8888, succ. Centre-Ville, Montreal Canada H3C 3P8.

[email protected]

Abstract. We prove that cyclic languages are the boolean closure of languages called strongly cyclic languages. The result is used to give another proof of the rationality of the zeta function of rational cyclic languages.

1 Introduction

Cyclic languages and strongly cyclic languages are two classes of languages of nite words over a nite alphabet. A cyclic language is conjugation-closed and for any two words having a power in common, if one of them is in the language, then so is the other. A strongly cyclic language is the set of words stabilizing a subset of the set of states of a nite deterministic automaton, the subset stabilized depending on the word stabilizing it. One says that the language stabilizes the automaton. A strongly cyclic language is rational.

We prove that a rational cyclic language is a boolean combination of strongly cyclic languages. More precisely, each rational cyclic language can be written as a chain of strongly cyclic languages. This result allows us to extend the com- putation of the zeta function (and generalized zeta function) of strongly cyclic languages done in [Bea95] to the class of rational cyclic languages. The zeta function of a formal languageLis(L) = exp(^Pa_{n zn}ⁿ), wherea_n is the number of words of lengthninL. The motivations of this denition and the connections with algebraic geometry and symbolic dynamics are discussed in [BR90]. The zeta function of a strongly cyclic languageL is equal to the zeta function of the soc system dened by the nite automaton stabilized byL. The rationality and computability of the zeta function of a soc system have been established in [Bow78] and [Man71]. The formula of computation given in [Bow78] are proved in [Bea95] by the use of a construction on nite automata called external power.

The rationality of the zeta function of a rational cyclic language has been es- tablished in [BR90]. The result we give here leads to another proof and to a dierent computation.

We assume that the reader knows the basics of formal languages (see [Eil72]).

We also assume that the reader is familiar with the elementary notions of semi- group theory. For example, notions like syntactic monoid, Green relations, regu- lar^D-classes, minimal ideal and 0-minimal ideal are supposed to be known. We refer to [Lal79] and [Pin86] for a presentation of this subject.

The paper is organized as follows. Section 2 and 3 give the basic properties of cyclic languages and strongly cyclic languages. The chain-decomposition of a rational cyclic language in strongly cyclic languages is established in section 4.

The computation of the zeta function and the generalized zeta function is done in the last section.

2 Cyclic languages

In this section, we introduce cyclic languages and give some basic properties. In the following, we denote byAa nite alphabet. In the sequel,Mwill always de- note a nite monoid. Every elementsofMhas a power which is an idempotent.

We denote bys^!this idempotent.

Denition1.

A languageLofA is said to be cyclic if it satises

8u²A; ⁸n >0 u²L ^,uⁿ²L

8u;v²A uv²L^,vu²L

A language is cyclic if it is closed under conjugation, power and root. IfLis a submonoid ofA which is cyclic, it is then pure [BP84].

Example 1. IfA =^fa;b^g, the languageL =AaA=A b of words having at least oneais cyclic.

Example 2. The language

L=^fa^pbⁿ¹aⁿ²bⁿ²:::a^nkb^nka^{q j}ni0 andp+q=n^1g[

fb^paⁿ¹bⁿ¹aⁿ²bⁿ²:::a^nkb^{q j}n_i0 andp+q=n_k^g is cyclic but not rational.

We will now only consider rational cyclic languages. Rational cyclic languages have the following straightforward characterization in terms of nite monoids.

Proposition2.

Let LA be a rational language. Let ':A^!!M be a mor- phism from A onto a monoidM such that L = ' ¹(P). The languageL is cyclic if and only if

8s²M; ⁸n >0 s²P ^,s^{n 2}P

8s;t²M st²P ^, ts²P

From the previous characterization, we deduce some useful facts about the structure of the image of a rational cyclic language onto a nite monoid recog- nizing this language. We also deduce a property of the syntactic monoid of a cyclic language.

Corollary3.

^{Let ': A!!M} be a morphism from A onto a monoidM such thatL=' ¹(P). LetH a regular^H-class ofM. One hasH P orH^\P =^;. Proof. Let us suppose h^{1 2} H belongs to P. For anyh^{2 2} H, we have h^!1 = h^!2 =ewhereeis the idempotent ofH. We then haveh^!1 =h^{!2 2}P andh²²P.

Corollary4.

Let ': A^!!M be a morphism from A onto a monoidM such thatL=' ¹(P). LetH¹ andH² be two regular^H-classes of a regular^D-class.

If one hasH¹P, one also has H²P.

Proof. Lete¹ande²be the respective idempotents ofH¹andH². As two idem- potents of a same^D-class are conjugated (see [Pin86]), there are two elements x¹ andx² ofM such that x¹x² = e¹ and x²x¹ = e². If H¹ P, we have e¹=x¹x²²P and thene²=x²x¹²P and the^H-classH² satisesH²P by the previous corollary.

Corollary5.

The syntactic monoid of a cyclic language has a zero.

Proof. LetMbe the syntactic monoid of a rational cyclic language and letDthe minimal ideal ofM. Then D is a completely regular^D-class. By the previous corollary, we have D P or D^\P = ^;. Let s be an element ofD. For any x;y ² M, we havexsy ²D because D is the minimal ideal ofM. If we have D P, the contexts ofsareMM. If we haveD^\P =^;, the contexts ofs are ^;;. In both cases, all the elements ofD are equivalent for the Nerode congruence, andDhas only one element. The syntactic monoid ofLhas then a zero.

3 Strongly cyclic languages

We now dene the notion of a strongly cyclic language.

Denition6.

^{LetA= (Q;A;E}) be a deterministic nite automaton whereQ is the set of states andE the set of transitions. We say that a wordwstabilizes a subsetP Qof states if we haveP:w=P. This means

8p²P p:w²P

8p⁰²P ⁹p²P p:w=p⁰

We denote by Stab(^A) the set of the wordsw such thatw stabilizes a sub- setP of states in the automaton^A. It should be noticed that in this denition the subsetP of states stabilized byw may depend onw. We point out that the empty word"stabilizes the setQand therefore belongs to Stab(^A) for any automaton^A. We say that a languageLis strongly cyclic if there is an automa- ton^Asuch thatL= Stab(^A). In this case, we say that the languageLstabilizes the automaton^A. The terminology is justied since strongly cyclic languages are cyclic. It could be proved directly but we shall obtain this fact as a consequence of the characterization of strongly cyclic languages.

1

1 2

a b

A

1

A

2 b

Fig.1.Automata^A1 and^A2

Example 3. The languages (b+aa)+ (aba)+a andb are respectively the strongly cyclic languages associated with the automata^A1and^A2 of Figure 1.

The following result gives a characterization of the wordsw stabilizing a subset of states in an automaton.

Proposition7.

^{LetA= (Q;A;E}) be a deterministic nite automaton. A wordw stabilizes a subsetP of states in^Aif and only if there is some stateqof^Asuch that for any integern, the transitionq:wⁿ exists.

Proof. Suppose rst that the wordw stabilizes the subsetP of states. By de- nition, for any statep²P, the transitionp:wⁿ exists.

Conversely, suppose all the transitionsq:wⁿexist for some stateqof^A. Since the automaton is nite, there are two integersl < msuch thatq:w^l=q:w^m. Let P be the set^fq:w^{i j}lim^g. It is straightforward that the wordwstabilizes the subsetP.

The following theorem gives a characterization of the strongly cyclic lan- guages.

Theorem8.

LetLbe a rational language dierent fromA. The following con- ditions are equivalent.

1. The languageL is strongly cyclic.

2. There is a morphism ' from A onto a monoidM having a zero such that L=' ¹(^fs²M ^js^!6= 0^g).

3. The syntactic monoidM(L) ofL has a zero and the image ofLin M(L) is

fs²M^js^!6= 0^g.

Proof. Suppose rst that the language stabilizes the automaton^A. Let M the transition monoid of^A and ' the canonical morphism from A ontoM. We show then that this monoid has a zero and that the image ofL is equal to

fs² M ^j s^{! 6}= 0^g. Letw be a word not belonging toL. By Proposition 7, for eachq²Q, there is an integernq such that the transitionq:w^nq does not exist.

For n greater than everynq, the transitions q:wⁿ do not exist for anyq. The transition induced bywⁿ is then the empty transition and the element '(wⁿ)

is a zero of the monoidM. This means that '(w)^! = 0. On the contrary, for any wordw²L, the transition induced bywⁿ is not the empty transition since there is a state q ² Q such that the transition q:wⁿ exists. This proves that '(w)^!6= 0. We have then proved thatLis equal to' ¹(^fs²M^js^!6= 0^g)

Suppose now there is a morphism'fromAonto a monoidN having a zero such thatL=' ¹(^ft²N ^jt^!6= 0^g). Since the morphism is onto, the syntactic monoidMofLis a quotient ofN: there is morphism :N^!!MfromNontoM. The image (0) of the zero ofNis then a zero ofM. Since the zero ofNdoes not belong to the image ofL, the zero ofMdoes not belong to the image ofL. Lett a element ofN such thatt^!= 0. We have (t)^!= (t^!) = 0. On the contrary, if t^{! 6}= 0, the element t^! belongs to the image ofL, and so does (t^!). This implies (t^!)⁶= 0. Finally, we have^ft²N ^jt^!6= 0^g= ¹(^fs²M^js^!6= 0^g).

Suppose that the syntactic monoid M ofL has a zero and that the image ofL in M is^fs²M ^j s^!6= 0^g. We denote by'the canonical morphism from AontoM. We build the following deterministic automaton^A= (Q;A;E). The set of states of^A is the setQ=M ^f0^gof the non-zero elements ofM. The transitionq:aisq:a=q'(a) ifq'(a)⁶= 0 and does not exist otherwise. It can be easily checked that a transitionq:w is q:w=q'(w) if q'(w)⁶= 0 and does not exist otherwise. We show now that the languageLstabilizes the automaton^A. Letw a word inL. For q ='(w), the transitionq:wⁿ is q:wⁿ ='(w)ⁿ⁺¹ and exists since'(w)^{n+1 6}= 0. On the contrary, for a wordw not inL, there is an integerm such that '(w)ⁿ = 0. The transition q:wⁿ does not exist for anyq. By the previous Proposition, the languageL stabilizes the automatonL. This nishes the proof.

Corollary9.

The previous characterization shows that strongly cyclic languages are cyclic.

1

P

*

*b

Fig.2.Structure of the syntactic monoid of^L.

The previous theorem can be used to prove that a given language is not strongly cyclic. Without this characterisation, such results are sometimes hard to obtain.

Example 4. The syntactic monoid of the languageL=AaAis the two elements monoidM =^fb= 1;a= 0^g. The ^D-class structure ofM is shown in Figure 2.

The image ofLinM is the singleton^f0^gand the previous theorem states that the languageLis not strongly cyclic.

4 Decomposition of cyclic languages

In this section, we prove the main result.

By the denition of cyclic languages, a boolean combination of cyclic lan- guages is still a cyclic language. In particular, a boolean combination of strongly cyclic languages is a rational cyclic language. The following result gives some converse.

Theorem10.

Any rational cyclic language is a boolean combination of strongly cyclic languages.

The proof of the theorem is based on the following lemma. By a strict quotient ofM, we mean a quotient which is strictly smaller thanM.

Lemma11.

Let L be a rational cyclic language and ' : A^!!M a morphism fromAonto a nite monoidMsuch thatL=' ¹(P). Let suppose furthermore that the monoidM has a zero and that this zero does not belong toP.

Then, eitherLis recognized by a strict quotient ofMor there exists a strongly cyclic languageL⁰ such thatLL⁰ and such that the languageL⁰ Lis recog- nized by a strict quotient ofM.

In both cases, the zero of the quotient does not belong to the image of the language (L in the rst case andL⁰ Lin the second case).

Proof. Let P the image ofL inM. SupposeD¹;:::;Dn are the 0-minimal ^D- classes ofM. For anys²Diand x;y²M, we havexsy= 0 orxsy²Di.

SupposeD¹satisesD^1\P =^;. For anyx;y²M, we havexsy⁶²P. All the elements ofD¹are equivalent to 0 by the Nerode congruence. The languageLis then recognized by the Rees quotientM=I whereI is the idealI =D^1[f0^g.

We can suppose that everyDi satises Di^\P ⁶=^;. There exists then si ²

Di^\P. Since si ² P, the idempotent s_!i is not zero and belongs then toDi

because this^D-class is 0-minimal. The elementsi belongs then to a regular^H- classHi ofDi. Let s be an element ofDi. If s^! = 0, the elements does not belong toP because 0 does not belong toP. On the contrary, If s^{! 6}= 0, the elements belongs to a regular ^H-classH_i⁰ ofDi. Since Hi ^\P ⁶= ^;, we have HiP andH_i⁰ P by Corollaries 3 and 4. Finally, we have proved that

Di^\P=^fs²Di^js^!6= 0^g

LetP⁰ be dened byP⁰ =^fs²M ^j s^!6= 0^g. By Theorem 8, the language L⁰ = ' ¹(P⁰) is strongly cyclic. It is straightforward that P P⁰. Since we haveDi^\(P⁰ P) =^;, the languageL⁰ Lis recognized by the Rees quotient M=I where the ideal is equal to^f0^g[Sn_i⁼¹Di. This quotient is strictly smaller thanM and this nishes the proof of the lemma.

The following lemma states that the class of strongly cyclic languages is closed under union and intersection.

Lemma12.

Let L¹ and L² be two strongly cyclic languages. Both languages L^1[L² andL^1\L² are then strongly cyclic.

Proof. We suppose that L¹ andL² respectively stabilizes the automata ^A1 = (Q¹;A;E¹) and ^A2 = (Q²;A;E²). We can suppose that Q^{1 \}Q² = ^;. The languageL^1[L²stabilizes then the automaton^A1[A2. The languageL^1\L² stabilizes the automaton^A1A2= (Q¹Q²;A;E³) where the transitionE³is dened by (p¹;p²):a= (q¹;q²) if the transitionsp¹:a=q¹ andp²:a=q²exist.

We can now complete the proof of the theorem.

Proof. We prove that every cyclic languageL can be written as a chain of strongly cyclic languages. This means that there are strongly cyclic languages L¹;:::;Ln satisfyingL¹L²Ln such that

L=L¹ L²+L³ Ln

We prove the result by induction on the size of a nite monoidM having a zero and recognizing the languageL. We suppose that there is a morphism': A^!!MfromAonto a nite monoidMsuch thatL=' ¹(P). We also suppose that the monoidM has a zero. By Corollary 5, the syntactic monoid ofL has this property. If the monoidM has only one element, the languageLis either^; orAwhich are both strongly cyclic. The empty language^;stabilizes the empty automaton. The full language stabilizes the automaton having one state and a transition for each letter from this state to this state.

If the zero ofM belongs to the image ofL inM, we replace L by A L which is also cyclic. It is then sucient to prove the result forA L. Indeed, if the complementA LofLcan be written as a chainA L=L¹ L²+Ln, the languageLcan be writtenL=A L¹+L² Lnwhich is also a chain of strongly cyclic languages.

We can now suppose that the zero ofMdoes not belong to the imageP ofL inM. By Lemma 11, either L is recognized by a strict quotient of M and the induction hypothesis immediately applies or there is a strongly cyclic languageL⁰ such that L⁰ L is recognized by a strict quotient ofM. By the induction hypothesis, the language L⁰ L can be written as a chain of strongly cyclic languages, i.e.,L⁰ L=L² L³+Ln. The languageLcan be then written as the chainL=L¹ L²+Ln where the languageL¹is equal toL^0[L² which is strongly cyclic by Lemma 12.

We point out that the zero of the smaller monoid recognizing L⁰ L does not belong to the image ofL⁰ L. It is not necessary to replace this language by its complement any more.

We give here an example.

*

*

* a

*

*

*1

ab

bab = 0

P

P’

aab aa b

aba ba

Fig.3.Structure of the syntactic monoid of^L.

Example 5. LetLbe the language (b+aa)+ (aba)+a b. The structure of the syntactic monoid ofLis given on gure 3. The imageP ofLin M(L) is equal toP =^fa;aa;aba;aab^g.

The subsetP⁰ dened in the proof is equal toP⁰ =^f1;a;aa;b;aba;aab^gand the languageL⁰ is (b+aa)+ (aba)+a. The languageL⁰ L is then equal tobwhich is strongly cyclic. The languagesL⁰andL⁰ Lstabilize respectively the automata^A1 and^A2 (see Figure 1 p. 4).

We have directly proved that every rational cyclic language can be written as a chain of strongly cyclic languages. In fact, it is just necessary to prove that rational cyclic languages are boolean combination of strongly cyclic languages.

A general result states that if a class^F of sets is closed under union and in- tersection, every set belonging to the boolean closure of^F can be written as a chain of sets of^F. For further details see [Car93] (chapter 3).

5 Zeta function of a cyclic language

We rst give the denitions of generalized zeta function and zeta function of a language of nite words over a nite alphabetA.

If A is a nite alphabet, we note

Z

^{hhAii (resp.}

Z

^[[A]]) the algebra of non commutative (resp. commutative) formal series with coecients in

Z

^{over the}

alphabetA. The subset of non commutative polynomials is denoted by

Z

^{hAi. We}

note' the natural algebra homomorphism from

Z

^{hhAii to}

Z

^[[A]] which makes the letters commute. For example,'(2ab 3ba) = ab. We notethe morphism from

Z

[[A]] to

Z

[[z]] dened by(a) =z for each letteraofA.

LetLbe a language of nite words overA, we noteLthe characteristic series ofL. This series belongs to

Z

^hhAiiand admits the decomposition:

L= ^X

w²Lw=^X

n⁰L_n; whereL_n is the homogeneous part of degreenofL.

Denition13.

The generalized zeta function of a languageLover the alphabet Ais now the following commutative series:

Z(L) = exp(^X

n¹

'(L_n) n ⁾:

Denition14.

The zeta function of a languageL over the alphabet A is the following series :

(L) =(Z(L)) = exp(^X

n¹

anzⁿ n ⁾; wherean is the number of words ofLof lengthn.

It is shown for example in [Bea95] that the generalized zeta function of a strongly cyclic languageLof an automaton^Ais the generalized zeta function of the soc system dened by^A, that is the set of bi-innite words that are labels of bi-innite paths of ^A. The zeta function of a soc system counts periodic orbits of the symbolic dynamic system. The zeta function of a strongly cyclic language is a rational series. The computation of the generalized zeta function and the zeta function of a strongly cyclic language was done in [Bea95] by using a construction on nite automata called external power.

If ^A= (Q=^f1;2;:::;n^g;E;T) is a deterministic automaton, the external power of order k, where 1 k ^jQ^j, of the automaton ^A is the automaton (Q⁰;E⁰) labelled in ^f 1;1^gA, where Q⁰ = ^f(i¹;i²;:::;ik)¹i¹<i²<<i^kn^g. There exists an edge labelled()afrom (i¹;i²;:::;ik) to (j¹;j²;:::;jk) if for each lwith 1 l k, there exists one edge in E labelled a going out from il and (j¹;j²;:::;jk) is the image of (i¹a;:::;ika) by a permutation of signature (). The automaton^Ais equal to its external power of order 1 by identication of +atoa. The matrix associated to an automaton labelled on^f 1;1^gAis the square matrix (xij)¹i;jn wherexpqis the sum (in

Z

^hAi) of the labels of edges fromptoq. The commutative matrix associated is the matrix ('(xij))¹i;jn.

We denote byQi the commutative matrix associated to the external power of orderiof the automaton^A. We then have (see [Bea95])

Z(L) =^Yn

i⁼¹(det(I Qi))^{( 1)i} whereI is the identity matrix of the same size asQi.

Computation of the zeta function of a cyclic language

The result of section 4 can be used to extend the previous computation of zeta functions of strongly cyclic languages to all cyclic languages. This gives an other proof of the rationality of the zeta function of a cyclic language established in [BR90]. The computation is the following:

LetL be a cyclic language. By section 4 it can be written as a chain L¹ L²++ ( 1)^{r 1}Lr, where Lj⁺¹ Lj for 1 j (r 1) and where all Lj

are strongly cyclic languages. By denition of the generalized zeta function we have:

Z(L) = exp(^X

n¹

'(L_n) n ⁾

= exp(^X

n¹ r

X

j⁼¹( 1)^{j 1}'(L_jn) n ⁾

= exp(^Xr

j⁼¹( 1)^{j 1X}

n¹

'(L_jn) n ⁾

=^Yr

j⁼¹exp(^X

n¹

'(L_jn) n ^{)( 1)j 1}

=^Yr

j⁼¹(Z(Lj))^{( 1)j 1}:

Example 6. We compute the generalized zeta functionZ(L) and the zeta function (L) of the cyclic language L = (b+aa)+ (aba)+a b introduced in example 5. The languageLadmits the chain-decompositionL=L¹ L²where L¹= Stab(^A1) (see gure 1) andL²= Stab(^A2) (see gure 1). We get:

Z(L¹) = ^j1 +a^j

1 b a

a 1

Z(L²) = 1

j1 b^j Z(L) = Z(L¹)

Z(L²) =(1 +a)(1 b)

1 b aa

(L) = 1 z² 1 z z²:

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