Laboratoire d Informatique Fondamentale de Lille

99{02 Jan. 99 LIFL Laboratoire d Informatique Fondamentale de Lille Publication 99{02 Synchronized Shue and Regular Languages Michel Latteux Yves Roos Janvier 1999 c LIFL USTL UNIVERSITE DES SCIENCES ET TECHNOLOGIES DE LILLE U.F.R. d'i.e.e.a. B^at. M3 { 59655 VILLENEUVE D'ASCQ CEDEX Tel. 03 20 43 47 24 Telecopie 03 20 43 65 66

Synchronized Shue and Regular Languages M.Latteux Y.Roos CNRS URA 369, L.I.F.L. Universite des Sciences et Technologies de Lille U.F.R. I.E.E.A. Informatique F59655 Villeneuve d'ascq cedex latteux@li.fr yroos@li.fr Summary. New representation results for three families of regular languages are stated, using a special kind of shue operation, namely the synchronized shue. First, it is proved that the family of regular star languages is the smallest family containing the language (a + bc) and closed under synchronized shue and length preserving morphism. The second representation result states that the family of "- free regular languages is the smallest family containing the language (a + bc) d and closed under synchronized shue, union and length preserving morphism. At last, it is proved that Reg is the smallest family containing the two languages (a+bb) and a+(ab) +, closed under synchronized shue, union and length preserving morphism. 1 Introduction Finite automata are very popular objects in Computer Science and are now used in several other scientic domains. The family of regular languages, called Reg, is a basic class in Chomsky hierarchy. The signicance of this family is strengthened by a great number of its nice characterizations. Some of these characterizations state that Reg is equal to the closure of a small family under a given set of operations. Besides the famous Kleene's theorem, one can recall the morphic compositional representation of regular languages establishing that Reg is the closure of the family reduced to the single language a b under morphism and inverse morphism. This nice result was proved in 1982 by Culik, Fich and Salomaa [1]. After the existence of such a morphic representation had been proved there followed a sequence of papers improving and generalizing this theorem (see [4], [6] and [9]). We shall prove here similar characterizations of Reg involving a special kind of shue. The shue operation appears as a fundamental operation in the theory of concurrency and admits several variations such as literal shue, insertion, etc.. Recently, Mateescu, Rozenberg and Salomaa introduced the notion of shue on trajectories [7] that provides a way to nd again the most of these variations. Synchronized shue was introduced in [2] by De Simone under the name of produit de mixage(see also [5], where a closely related operation was introduced). This special shue can bee seen as a shue with rendez-vous and is very useful in modelling the behaviours of

2 Michel Latteux, Yves Roos parallel processes (see [3], [8]). This powerful operation is quite amazing. For example, contrary to what was stated in [2], it is not associative. The study of synchronized shue in connection with union and length preserving morphism provides us to obtain new representation results for three families of regular languages. First, we prove that the family of regular star languages is the smallest family containing the language (a+bc) and closed under synchronized shue and length preserving morphism. Our second representation result states that the family of "-free regular languages is the smallest family containing the language (a + bc) d and closed under synchronized shue, union and length preserving morphism. At last, we prove that Reg is the smallest family containing the two languages (a + bb) and a + (ab) +, closed under synchronized shue, union and length preserving morphism. 2 Preliminaries 2.1 Notations Let X be an alphabet. For any word w 2 X, we shall denote by jwj the length of the word w and for any letter a 2 X, we shall denote by jwj a the number of occurrences of the letter a that appear in the word w. For any language L X, we denote by (L), the alphabet of L that is : (L) = fx 2 X j 9u 2 L : juj x > 0g For example we get : (a + ab) = (a 2 + b 3 ) = fa; bg, (;) = (") = ;. We denote by X;Y the projection over the sub-alphabet Y, i.e. the homomorphism from X to Y dened by: 8x 2 X; if x 2 Y then X;Y (x) = x ; else X;Y (x) = " When there is no ambiguity, we shall use notation Y instead of X;Y. u tt v is the shue of the two words u and v that is u tt v = fu 1 v 1 u 2 v 2 :::u n v n j u i 2 ; v i 2 ; u = u 1 u 2 :::u n ; v = v 1 v 2 :::v n g This denition is extended over languages by : 8L 1 ; L 2 X ; L 1 tt L 2 = S w 1 tt w 2 w 1 L 1 w 2 2 L 2 When there is no ambiguity, for a language reduced to a single word, we shall write this language u rather than fug.

Synchronized Shue and Regular Languages 3 2.2 Synchronized shue Denition 1. [2] Let X be an alphabet and L 1, L 2 be two languages included in X. The synchronized shue of L 1 and L 2, denoted by L 1 uu L 2 is dened by : L 1 uu L 2 = fw 2 ((L 1 ) [ (L2)) j (Li)(w) 2 L i ; i 2 f1; 2gg Examples : 1. bab uu cac = fbcabc; bcacb; cbabc; cbacbg 2. bac uu cab = ; 3. fa; abbg uu ab = ; and a uu ab = ab 4. (a uu fa; bg) uu ab = a uu ab = ab 5. a uu (fa; bg uu ab) = a uu ; = ; Example 3 shows that it generally does not hold that S L 1 uu L 2 = w 1 uu w 2 w 1 L 1 w 2 2 L 2 Examples 4 and 5 show that synchronized shue is not an associative operation. The following properties and proposition easily come from denition : the synchronized shue is commutative :L 1 uu L 2 = L 2 uu L 1 L uu " = L L uu ; = ; if (L 1 ) = (L 2 ) then L 1 uu L 2 = L 1 \ L 2 if (L 1 ) \ (L 2 ) = ; then L 1 uu L 2 = L 1 tt L 2 (L 1 uu L 2 ) = (L 1 uu L 2 ) (" 2 L 1 uu L 2 ) () (" 2 L 1 \ L 2 ) if L 1 and L 2 are "-free languages then for any letter a :(a 2 L 1 uu L 2 ) () (a 2 L 1 \ L 2 ) Proposition 1. Let L 1 and L 2 be languages with X i = (L i ) for i 2 f1; 2g. Then the following equality holds : L 1 uu L 2 = (L 1 tt (X 2 n X 1 ) ) \ (L 2 tt (X 1 n X 2 ) ) 2.3 synchronized shue and associativity As we have seen in examples, the synchronized shue is not associative. We shall give now a sucient condition for the synchronized shue to be associative over a (uu-closed) family of language. For every family of language L, we denote by (L) uu the least family of languages containing L and closed under uu. First we can state :

4 Michel Latteux, Yves Roos Proposition 2. Let L be a family of languages. If for every languages L 1, L 2 2 (L) uu, the equality (L 1 uu L 2 ) = (L 1 ) [ (L 2 ) holds then uu is associative over (L) uu. Proof. Let L 1 ; L 2 ; L 3 2 (L) uu then, from denition, we have u 2 (L 1 uu L 2 ) uu L 3 () ( (L1 uu L 2) (u) 2 L 1 uu L 2 ) ^ ( (L3) (u) 2 L 3 ) Since (L 1 uu L 2 ) = (L 1 ) [ (L 2 ), we get : u 2 (L 1 uu L 2 ) uu L 3 () ( (L1[L 2) (u) 2 L 1 uu L 2 ) ^ ( (L3) (u) 2 L 3 ) Moreover (Li)(u) = (Li)( (L1)[(L2) (u)) for i 2 f1; 2g, hence we have : u 2 (L 1 uu L 2 ) uu L 3 m ( (L1) (u) 2 L 1 ) ^ ( (L2) (u) 2 L 2 ) ^ ( (L3) (u) 2 L 3 ) and uu is associative over (L) uu. ut Denition 2. Let L be a language. A language R is said to be L-compatible if the two following conditions are satised : 1. (R) (L) R 2. (R) (L) A family of languages L is said to be L-compatible if every language R in L is L-compatible. Lemma 1. Let L be a language and R 1 ; R 2 be two languages which are L- compatible. Then : 1. (R 1 uu R 2 ) = (R 1 ) [ (R 2 ). 2. R 1 uu R 2 is L-compatible. Proof. From 1 of denition 2, (R1)[(R2) (L) R 1 uu R 2 and from 2 of denition 2, we also have ( (R1)[(R2) (L)) = (R 1 ) [ (R2) thus (R 1 ) [ (R 2 ) (R 1 uu R 2 ). From denition of synchronized shue, (R 1 uu R 2 ) (R 1 ) [ (R 2 ) then (R 1 uu R 2 ) = (R 1 ) [ (R 2 ) (L). It follows that (R1 uu R 2) (L) = (R1)[(R2) (L) R 1 uu R 2 hence R 1 uu R 2 is L-compatible. ut Then, by induction and from proposition 2 it directly follows : Proposition 3. Let L be a family of languages. If there exists a language L such that L is L-compatible, then (L) uu is L-compatible and uu is associative over (L) uu In the following, we shall omit use of parenthesis in expressions involving synchronized shue over such families of languages.

3 A new characterization of Reg Synchronized Shue and Regular Languages 5 Here, we shall prove characterization results involving synchronized shue. These results concern several families of regular languages. First, let us consider Fin ", the family of "-free nite languages. Proposition 4. The family Fin " is the smallest family containing the language a + ab and closed under union, length preserving morphisms and synchronized shue. Proof. Let L be the smallest family containing the language a+ab and closed under the three above operations. Clearly, L Fin ". Conversely, since union is allowed it is sucient to prove that : 1. ; is in L 2. for every alphabet X = fx 1 ; x 2 ; : : : ; x n g; n > 0, the language reduced to the single word x 1 x 2 : : : x n is in L. For 1, we get ; = (a + ab) uu (b + ba). For 2, we make an induction on n. If n < 3, we get a = [(a+ab)+(b+bc)] uu [(a+ac)+(c+cb)] and ab = (a+ab) uu b. If n 3, it is easily seen that x 1 x 2 : : : x n = x 1 x 2 : : : x n?1 uu x 2 : : : x n. ut We shall prove a similar result for Reg, the family of regular star languages. The starting language is (a + bc) and we use only length preserving morphisms and synchronized shue. First, we consider monoids generated by special nite languages. Denition 3. A marked language F is a nite language such that : 8u; v 2 F; 8x 2 (F ); (juj x > 0 ^ jvj x > 0) =) (u = v ^ juj x = 1) Next, we show that if F is a marked language, then F can be obtained from languages of the type (a + bc) using synchronized shue. Denition 4. Two languages L 1 and L 2 are said equivalent if there is a length preserving morphism which maps bijectively L 1 onto L 2. Lemma 2. For every marked language F, F can be obtained from languages equivalent to (a + bc) using synchronized shue. Proof. Let L be the family of languages which can be obtained by synchronized shue from languages equivalent to (a + bc). The languages a = (a + bc) uu (a + cb) and (ab + cd) = (a + cd) uu (c + ab) belong to L. Hence, (a + b) = a uu b, (ab) = (ab + cd) uu (ab + dc), " = (ab) uu (ba) are in L. Let us consider now a marked language F = u+v with jvj juj 0. From the above equalities, if jvj 2 then F 2 L. Assume that v = x 1 x 2 : : : x k with k 3. The languages L 1 = (u + x 1 : : : x k?1 ), L 2 = (u + x 2 : : : x k ) and L 3 =

6 Michel Latteux, Yves Roos (u + x 1 x k ) are F -compatible. From the equality (u + v) = L 1 uu L 2 uu L 3, we get, by induction, that (u + v) 2 L. At last, if F = u 1 + u 2 + : : : + u n is a marked language with n 3, the languages R 1 = (u 1 + : : : + u n?1 ), R 2 = (u 2 + : : : + u n ) and R 3 = (u 1 + u n ) are F -compatible. Once again, the equality F = R 1 uu R 2 uu R 3 implies by induction that F 2 L. ut We can now get easily our characterization result for the family Reg. Proposition 5. The family of regular star languages is the smallest family containing the language (a + bc) and closed under length preserving morphisms and synchronized shue. Proof. Clearly, Reg is closed under the above operations. For the reverse inclusion, let R be a regular language. It is known (see [4], [6]) that there exist two nite languages F 1 and F 2 and a length preserving morphism such that R = g(f 1 \ F 2 ). One can assume that (F 1 ) = (F 2 ). Hence, R = g(f 1 uu F ). Moreover, 2 F 1 and F 2 are the image by a length preserving morphism of marked languages. Thus, lemma 2 implies the result. ut We are now able to state our rst characterization of Reg. Proposition 6. The family of regular languages Reg is the smallest family containing the languages (a+bc) and a, closed under union, length preserving morphisms and synchronized shue. Proof. Let L be the smallest family containing the languages (a + bc) and a, closed under the three above operations. Clearly L is included in Reg. For the reverse inclusion, let us consider a language R 2 Reg. From proposition 5, we know that " is in L, so we may suppose that " 62 R since union is allowed. Hence one may assume, without loss of generality, that R A A 0 where A and A 0 are disjoint alphabets. Then R = R uu A 0. From proposition 5, R can be obtained from (a + bc) using length preserving morphisms and synchronized shue. On the other hand, A 0 can be obtained from a using union and length preserving morphisms. ut We can observe that it is not possible to enunciate a similar result with a single generator. Indeed, since " 2 L 1 uu L 2 if and only if " 2 L 1 \ L 2, we need to start from a family containing at least two languages L 1 and L 2 with " 2 L 1 and " 62 L 2. The following result concerns the family of "-free regular languages, that is regular languages which do not contain the empty word. For this family, it is possible to start from a single generator. Proposition 7. The family of regular "-free languages Reg " is the smallest family containing the language (a + bc) d and closed under union, length preserving morphisms and synchronized shue.

Synchronized Shue and Regular Languages 7 Proof. Let L be the smallest family containing the language (a + bc) d and closed under the three above operations. Clearly L is included in Reg ". For the reverse inclusion, let us consider a language R 2 Reg ". Without loss of generality, one may assume that R is a nite union of languages in the form Kd with K 2 Reg and d a letter not in (K). It remains to prove that such a language Kd is in L. This can be done by induction over the construction of K with respect to proposition 6. If K = (a + bc) then Kd is in L. If K = a, observe rst that d = (a + ab) d uu (b + ba) d is in L. Moreover a d, which can be obtained from (a + bc) d using a length preserving morphism, is in L. Then we get ad = a uu a d belongs to L. Now, if K = h(k 0 ) for some K 0 2 L and some length preserving morphism h, we may suppose that d 62 (K 0 ) then K 0 d 2 L which implies Kd 2 L. if K = K 1 + K 2 with K i 2 L for i = 1; 2 then Kd = K 1 d + K 2 d 2 L. if K = K 1 uu K 2 with K i 2 L for i = 1; 2. We may suppose that d 62 (K i ) for i = 1; 2 then Kd = K 1 d uu K 2 d 2 L. ut 4 Binary generators The single generator for the family of "-free regular languages Reg " used in proposition 7 is built over a four letter alphabet. The following proposition shows that it is possible to start from a language dened over a three letter alphabet : Proposition 8. The family of regular "-free languages Reg " is the smallest family containing the language (a + bc) b and closed under union, length preserving morphisms and synchronized shue. Proof. Let L be the smallest family containing the language (a + bc) b and closed under union, length preserving morphisms and synchronized shue. From proposition 7, we have to prove that (a + bc) d is in L. Observe rst that b = (a + bc) b uu (c + ba) b is in L. Then a b = (a + bc) b uu b and (a + bc) bd = (a + bc) b uu b d are in L. It follows that (a + bc) bc is also in L. Let us consider now the length preserving morphism h dened by : h(a) = a; h(b) = h(d) = b and h(c) = c. We get (a + bc) bc uu h((a + cb) c uu d) = a (bc) + 2 L then a (bb) + 2 L and a b + = a b + a (bb) + + h(a (bb) + uu a d) is in L. Moreover a + b = a + + (a b + uu a + ) is also in L since, clearly, a + 2 L. Now, with the length preserving morphism g dened by : g(a) = g(e) = a; g(b) = b; g(c) = c and g(d) = d, we obtain (a + bc) bca d = g((a + bc) bc uu c + e uu e d uu c d) 2 L

8 Michel Latteux, Yves Roos At last, we get (a + bc) d = a d + (a + bc) bca d 2 L. ut We have seen that the family of ("-free) regular languages can be obtained from languages whose cardinality alphabet is less or equal to three. A natural question is whether it is possible to start from binary languages which are languages built over two letter alphabets. In a rst time, we shall establish a negative answer for the family of star regular languages. Denition 5. A star language L satises the (P ) property if there exist three distinct letters a; b; c such that the words a and bc are in L, but the word bac is not in L. Lemma 3. Let S be a set of star languages and L be the smallest family of languages, containing S and closed under length preserving morphism and synchronized shue. If L contains a language which satises the (P ) property then so do S. Proof. First, it is clear that L contains only star languages. We shall prove this lemma by induction. Let L be a language in L with a, bc 2 L and bac 62 L. If there exist some language L 0 2 L and some length preserving morphism h such that L = h(l 0 ) then it is obvious that L 0 satises the (P ) property. Let us suppose now that L = L 1 uu L 2 such that neither L 1 nor L 2 satises (P ). We shall show that it will lead to the following contradiction bac 2 L. For i = 1; 2, we denote i = (L i ) \ fa; b; cg. Observe that bac 2 L if and only if i (bac) 2 L i for i = 1; 2. Moreover, since L is a star language, the words abc and bca are in L. Now, for i = 1; 2 : if i = fa; b; cg, we get i (bac) 2 i (abc + bca) L i if i = fa; b; cg, then a 2 L i and bc 2 L i. Since L i does not satisfy (P ), it follows that i (bac) = bac 2 L i. The contradiction bac 2 L implies that L 1 satises (P ) or L 2 satises (P ). ut Since the language (a + bc) satises the (P ) property, we can state, as a corollary of the above lemma : Proposition 9. The family of regular star languages can not be obtained as the closure of a set of binary languages under length preserving morphism and synchronized shue. For the family of ("-free) regular languages, the answer is positive. In order to establish this result, we shall rst prove the following lemma : Lemma 4. Let L be a family of languages containing the language a b and closed under length preserving morphism and synchronized shue, then L is closed under product.

Synchronized Shue and Regular Languages 9 Proof. Let L 1 and L 2 be two languages of L. We may suppose that 1 = (L 1 ) and 2 = (L 2 ) are disjoint. The family fx y j x 2 1 ; y 2 2 g is clearly L 1 L 2 -compatible, then synchronized shue is associative over this family and uu x y = 1 2 L 2 x 2 1 y 2 2 Then L 1 L 2 = (L 1 uu L 2 ) uu 1 2 is in L. We are now able to enunciate our last proposition which states that Reg, the family of regular languages can be obtained from binary generators. Proposition 10. The family of regular languages Reg is the smallest family containing the languages a + (ab) + and (a + bb), and closed under union, length preserving morphism and synchronized shue. Proof. Let L be the smallest family containing the languages a + (ab) + and (a + bb), closed under the three above operations. Clearly L is included in Reg. For the reverse inclusion, observe rst that (a + bb) uu (a + (ab) + ) = a 2 L It remains to prove that (a + bc) 2 L. Let h be the length preserving morphism dened by h(a) = h(b) = a, we get h((a + bb) ) = a 2 L and h(a uu b) = a + 2 L. It follows that (a + (ab) + ) uu b = ab 2 L and (a + (ab) + ) uu b + = (ab) + 2 L. Now, if g is the length preserving morphism dened from by g(a) = g(d) = g(e) = a, g(b) = b and g(c) = c, we get g((ac) + uu bd) uu g((ca) + uu be) = b(ac) + a 2 L Then b(aa) + a in L, g(b(aa) + a uu bd) = b(aa) + aa 2 L and g(ba uu ad) = baa 2 L so : ba = b + ba + baa + b(aa) + a + b(aa) + aa 2 L In a same way we can prove a b 2 L. It follows that a b uu bc = a bc is in L then a b + is in L. Now, since a b + + a = a b 2 L and from lemma 4, we get that L is closed under product. We are now able to obtain (a + bc) : (((a + bb) uu (a + cc) ) uu (bc) + ) + a = (a + bcbc) 2 L The family L is closed under product then we also have bc(a + bcbc) bc 2 L. Then L 1 = (a + bcbc) uu bc(d + bcbc) bc = a bc(d bca bc) d bca 2 L ut

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