On Bounded Rational Trace Languages

Transcription

1 On Bounded Rational Trace Languages Christian Choffrut Laboratoire LIAFA, Université de Paris 7 2, pl. Jussieu, Paris Cedex 05 cc@liafa.jussieu.fr, cc Flavio D Alessandro Dipartimento di Matematica, Università di Roma La Sapienza Piazzale Aldo Moro 2, Roma, Italy. dalessan@mat.uniroma1.it, Stefano Varricchio Dipartimento di Matematica, Università di Roma Tor Vergata, via della Ricerca Scientifica, Roma, Italy. varricch@mat.uniroma2.it, February 24, 2008 Abstract ** In this paper, for a finitely generated monoid M, we tackle the following three questions: Is it possible to give a characterization of rational subsets of M which have polynomial growth? Which is the structure of the counting function of rational sets which have polynomial growth? Is it true that every rational subset of M has either exponential growth or it has polynomial growth? Can one decide for a given rational set which of the two options holds? We give a positive answer to all the previous questions in the case that M is a direct product of free monoids. Some of the proved results also extend to trace monoid. ** 1

2 1 Introduction Right from the beginning, the notion of automaton has been extended, in at least two main directions. Indeed, a finite automaton can be accommodated to accept tuples of words leading to the notion of rational relations, as done by Elgot and Mezei. Transitions can also be provided with numerical values, called multiplicities by Eilenberg, though these values can be taken from different semirings: the nonnegative integers, the integers, different fields, more exotic semirings such as the tropical semiring,... This second direction leads to the theory of rational series which is rich on instances of gap theorems. E. g., the growth of the coefficients in a rational series over the semirings N or Z is either polynomial or exponential and in the former case the degree of the polynomial can be effectively calculated. It covers in particular the case where the function computes the number of words of a given length of a rational subset of the free monoid. The present work mixes the two approaches. Indeed, we show that the same gap theorem holds for the family of rational relations over a direct product of finitely generated monoids and more generally for the family of rational languages in trace monoids. We not only prove that it is decidable whether or not the growth is polynomial or exponential but in the former case we can compute exactly the number of elements of a given length. We can be more precise. As in the case of the rational sets of the free monoid, we cannot avoid periodicity (think of a rational subsets whose words have even length). The same applies for relations. We show that given an automaton accepting a k-ary relation R having polynomial growth, one can compute a family of m polynomials p 0,..., p m 1 such that the number of elements of R of a given length N is equal to p i (N) where i = N mod m. A few words on previous and connected works are in order. Let us mention Schützenberger s result whose discussion we postpone to paragraph 3.2 as it would be too technical at this point. Along with the notion of length, a natural notion for trace monoids is that of height. The term stems from the fact that elements of trace monoids can be represented by heaps of pieces popular in the community of Petri nets and scheduling. The growth function can be considered relative to the height instead of the length. As far as we know, this problem is solved in special cases only. Another reference is [3] where the generating function of rational relations and rational subsets of general trace monoids is studied. It is shown that such functions may not be algebraic for the direct product of free monoids. Our result can be interpreted as asserting that the generating function is rational when the relation is bounded. The paper is organized as follows. In Section 2 we introduce the definition of counting function for an arbitrary finitely generated monoid, along with the notion of growth function which is its cumulative counterpart. We also adapt the classical concept of boundedness to general finitely generated monoids. A few elementary properties concerning these first definitions are given. In Section 3 we recall the notion of multitape automaton and we discuss how 2

3 Schützenberger s result relates to ours. Section 4 starts with a reformulation of the notion of boundedness for k-ary relations, which allows us to prove the Gap Theorem stating that the growth function is either bounded from above by a polynomial or that it is bounded from below by an exponential. Section 5 focusses on effectiveness issues. We give a structural characterization of the bounded relations which is the main tool for proving the recursive decidability of the property of boundedness and for obtaining an explicit expression for the counting function. The last section is concerned with an extension of the Gap Theorem to trace monoids. 2 Preliminaries The free monoid generated by the set A is denoted by A. Its elements are words and the empty word is denoted by 1. Here we are mainly concerned with finite direct products of finitely generated free monoids and more precisely with rational subsets of such monoids. We take for granted that the reader is familiar with these notions. Otherwise we recommend the various handbooks on the subject, cf. [2, 12, 18]. The notion of growth function of a subset can be defined on an arbitrary finitely generated monoid, denoted by M throughout this section. Let ψ : A M, be a morphism of the free monoid A onto M. For all m M, the length of m is defined as m = min{n 0 ψ 1 (m) A n }. With every subset L M, we associate its counting function f L : N N defined as f L (n) = Card({m L, m = n}). The growth function of L g L : N N can be seen as the cumulative counting function. It is defined as g L (n) = Card({m L, m n}) = f L (i). (1) 0 i n Definition 1 Let L be a subset of M. We say that L has exponential growth if there exist an integer n 0 0 and a real number k > 1 such that, for every n n 0, g L (n) k n. We say that L has polynomial growth if g L is upper bounded by a polynomial. A set whose growth is polynomial is sparse. It is well-known that the notion of polynomial growth is closely related to the notion of boundedness introduced by Ginsburg in the early sixties. Here is the formal definition. In order to simplify notations we adopt the convention of identifying a singleton with its element, i.e., we write x instead of the more correct {x}. 3

4 Definition 2 Let L be a subset of M and n a positive integer. Then L is n- bounded in M, or simply n-bounded when the monoid is understood, if there exist elements u 1,..., u n M such that L u 1 u n. Moreover, L is bounded if L is n-bounded for some integer n 1. The following lemmas give some useful properties of the sets of polynomial growth. The first one shows that the notion is meaningful since it is in a sense intrinsic. Its proof easily follows from that one of [10], Lemma Lemma 1 Let L be a subset of a finitely generated monoid. The property of having polynomial growth is independent of the choice of the system of generators of M. In the particular case where the monoid is free we have that boundedness implies sparsity. Lemma 2 ([10], Prop ) Let L A be a bounded language over the alphabet A. Then the growth function of L is polynomially upper bounded. For more general monoids, we have Lemma 3 Let L be a subset of M. The following conditions hold: 1. L has polynomial growth if and only if the counting map f L of L is upper bounded by a polynomial. 2. Let N be a finitely generated submonoid of M and let L be a subset of N. If L has polynomial growth in M then it has polynomial growth in N. Moreover, assume that for some choice of the generating sets for M and N and for some integer K the following holds x N K x M, (2) where x N and x M are the lengths of x in the monoids N and M respectively. Then if L has polynomial growth in N then it has polynomial growth in M. Proof. Condition (1) derives easily from Eq. 1. Let us prove Condition (2). Denote by g (M) L and g (N) L the growth maps of L with respect to the monoids M and N, respectively. Suppose that there exists a polynomial p where, for every n 0, g (M) L (n) p(n). Since the submonoid N is finitely generated, there exists an integer H such that x M H x N holds for all x N, thus x N n implies x M Hn, i.e., {x L x N n} {x L x M Hn} which yields (n) g(m) (Hn) p(hn). g (N) L L Conversely, suppose that there exists a polynomial p where, for every n 0, g (N) L (n) p(n). Because of Eq. 2, for every x L, x M n implies x N Kn. Therefore we have {x L x M n} {x L x N Kn} and thus (n) g(n) (Kn) = p(kn) which completes the proof. g (M) L L 4

5 3 Multitape automata We essentially follow [2, section 6 of Chapter 3] and [18, section 1.1. of Chapter II] for the main definitions and preliminary results. From now on, unless otherwise stated, M is the direct product A 1 A k of the free monoids generated by the alphabets A 1,..., A k. It is sometimes convenient to view its elements as vectors with k components. In particular if u A 1 A k, then u i represents the i-th component of u, so that u = (u 1,..., u k ). We use the term relation as a synonym for a subset of A 1 A k. 3.1 Finite automata A finite multitape automaton (simplified hereafter as finite automaton) is a construct A = (Q, M, E, I, T ) where Q is a finite set of states, I, T Q are the sets of initial and terminal states respectively and E Q M Q is a finite set of transitions. We assume the reader familiar with the basic notions such as path, label of a path etc.... In particular we shall use the standard notation q m p to represent a path from state q to state p labeled by m M. A state q Q is accessible if there exists a path that starts from an initial state and ends in q. A state q of A is said to be co-accessible if there exists a path that starts from q and ends in a terminal state of A. A state q is useful if it is accessible and co-accessible. The automaton A is in normal form if all its states are useful. The following fundamental result was established by Elgot and Mezei in the mid sixties (cf [18], Thm 1.1). Theorem 1 A subset R A 1 A k is rational if and only if there exists a finite automaton, in normal form, that accepts it. 3.2 Schützenberger s result We discuss how our contribution relates and differs from the previous work of Schützenbeger, [19]. A binary rational relation R A B can be viewed as a partial function f R from A into the semiring of rational sets of B which to each u A assigns the set of v B such that (u, v) R holds. Intuitively, this amounts to viewing the elements of A as inputs and those of B as outputs. From the onset, Schützenbeger gives a very simple characterization for the images of each u A to have finite cardinality, denote it f R (u), and he works under this hypothesis. He shows that two possibilities hold. Either for an infinite sequence of inputs of increasing lengths, u 1 < u 2 < < u n the cardinality of their images grows at least as fast as an exponential in the length of the inputs, i.e., f R (u n ) = Ω(K un ) for some K > 1 or else for all inputs, the cardinality of the images is bounded by some polynomial in the length of the input. Next we reformulate our problem in the same spirit. We are given a k-ary rational relation R A 1 A k and a finite automaton that accepts it. 5

6 We transform it into an automaton over the direct product B A 1 A k where B = {b} is a unary alphabet by associating with every transition (q, x 1,, x k, p) E the transition (q, b n, x 1,, x k, p) where n is the sum of the lengths of x 1,..., x k. View the relation accepted by this automaton as a partial function of B into the power set of A 1 A k as observed above. Then the cardinality of the image of the input b n is precisely the number of elements of R having length equal to n. 4 Rational relations 4.1 Reformulating the boundedness We take advantage of the commutativity among the different components of a direct product of free monoids in order to reformulate Definition 2. The following lemma provides an alternative definition for boundedness. Proposition 1 A relation R A 1 A k is bounded if and only if it is included in R 1 R k where each subset R i is bounded in A i. Proof. Indeed, assume R i = u i,1 u i,n i for some u i,1,..., u i,ni A i. For all i and l n i, define v i,l as the vector having all components equal to the empty word except the i-th component equal to u i,l. Because v i,l and v i,l commute whenever i i, we have R vi,1 vi,n i. 1 i k ** Conversely, if a subset is bounded as specified in Definition 2, then let R i be the projection of R to the i th component. One easily verifies that R i is bounded and R R 1 R k. ** We now return to the notion of polynomial growth for subsets of the monoid M = A 1 A k and refine it in the following way. With R M we associate a map γ R : N k N, defined as γ R (n 1,..., n k ) = Card({(u 1,..., u k ) R u 1 = n 1,..., u k = n k }). We start with an elementary property of polynomials with several variables. Lemma 4 Let f : N N and g : N k N be two functions satisfying the condition f(n) = g(n 1,..., n k ). Then f is bounded by some polynomial n 1+ +n k =n if and only if g is bounded by some polynomial in k variables. Proof. Suppose g is bounded by a polynomial q. We may assume that all the coefficients of q are non negative. Then each term g(n 1,..., n k ) is bounded 6

7 by q(n,..., n) where n = n n k. The number of such terms is upper bounded by (n + 1) k so that f(n) is upper bounded by (n + 1) k q(n,..., n). Conversely, if f is upper bounded by polynomial p then we consider the function q(n 1,..., n k ) = p(n n k ) which, by composition of polynomial functions is again polynomial. Then we obtain g(n 1,..., n k ) f(n n k ) p(n n k ) = q(n 1,..., n k ). Consequently we get Corollary 1 Consider a subset R of the direct product M = A 1 A k. Then R has polynomial growth if and only if the function γ R is upper bounded by a polynomial in k variables. 4.2 The Gap Theorem The following notion is instrumental in characterizing sparse relations. Indeed, we shall see that sparsity of a rational relation is reflected in all finite automata accepting a relation, provided this automaton be in normal form. Let M = A 1 A k. It is useful to remark that, assuming the minimal set of generators for the monoid M, for every m = (u 1,..., u k ) M, we have that m = u u k. Definition 3 Let A = (Q, M, E, I, T ) be a finite automaton over M. Let q Q and let i [1, k]. We set and C q = {m A 1 A k there exists a path q m q}, C i q = {u A i there exists a path q m q, m A 1 A k, m i = u}. The set C q is the language of cycles associated with q and the set C i q is the language of cycles associated with the pair (q, i). Clearly, C q is a submonoid of A 1 A k and, for every i = 1,..., k, Ci q is a submonoid of A i. With these notations we have Lemma 5 If there exists a positive integer i such that C i q contains two words which are not powers of a common word, then R has exponential growth. Proof. Let i be a fixed positive integer such that u, v are words of C i q which are not powers of a common word. By the Defect Theorem [17], the submonoid {u, v} generated by the set {u, v} is a free submonoid of C i q. Let m u, m v be two elements of C q such that: m u = (u 1,..., u k ), m v = (v 1,..., v k ), 7

8 where u i = u, v i = v. Set α = max{max{ u j, j = 1,..., k}, max{ v j, j = 1,..., k}}. Moreover, consider two paths q 0 x q and q y q1 where q 0 I and q 1 T. Since the monoid {u, v} is free, for every n 0, the subset {m u, m v } n of M has at least 2 n elements. On the other hand, for every w {m u, m v } n, xwy R and xwy x + y + nαk. These facts imply that the growth of R is exponential Now we turn to the main results of this section. We start by giving a structural characterization of the finite automata accepting rational sets of A 1 A k of polynomial growth. The intuition is to think of the graph underlying the finite automaton and to consider its decomposition into maximal strongly connected components. The boundedness of the relation is reflected in a simple property of the restriction of the automaton to its maximal strongly connected components. The technique is reminiscent of that employed in [19, Section 3]. Theorem 2 Let A = (Q, A 1 A k, E, I, T ) be a finite automaton and let R be the rational set it accepts. Then the following conditions are equivalent: 1. R has polynomial growth. 2. For every q Q, for every i = 1,..., k, there exist effectively computable words v i A i such that Ci q v i. 3. R is bounded in A 1 A k. Proof. (1 2) Indeed, Lemma 5 asserts that all elements in C i q are powers of the same word. It suffices to set v i equal to their common root. The fact that the v i s can be effectively computed is proved in Proposition 3. (2 3) For every p, q Q, define the set Because of equality R pq = {m A 1 A k there exists a path p m q}. R = p I,q F R pq, in order to prove the claim, it suffices to prove that, for all p, q Q, R pq is bounded in A 1 A k. Let s = (q 1,..., q n ) be a repetition-free sequence of states of A, ie, for every positive integers i, j, if i j, then q i q j. Clearly, n is less than or equal to the number of states of A and therefore the set S of all repetition-free sequences is finite. For every p, q Q, define the set S pq = {s S s = (q 1,..., q n ), q 1 = p, q n = q}. 8

9 and the set L pq as: L pq = {m A 1 A k For every s = (q 1,..., q n ) S, define the set (p, m, q) E}. R s = C q1 L q1q 2 C q2 C qn 1 L qn 1q n C qn A 1 A k. (3) All C qi s are bounded by hypothesis. Because the family of bounded sets is closed under union, in order to prove that R pq is bounded, it suffices to show the equality R pq = R s. (4) s S pq The inclusion R pq s S pq R s is obvious so we only have to prove the opposite inclusion. We start with a simple, though useful observation. Consider a repetition-free sequence s = (q 1,..., q n ). Then C q1 R s = R s holds because C q1 is a submonoid. If furthermore q 0 does not occur in s, then L q0q 1 R s R s where s is the repetition-free sequence obtained by adjoining q 0 to the left of s. Let m R pq be the label of a path from p to q. This path is a sequence of transitions m p = q 1 m 1 2 m q2 n qn 1 qn = q, with m = m 1 m 2 m n. We proceed by induction on n and observe that the case n = 1 is trivial. If q 1 has a single occurrence in the sequence, then m 1 L q1q 2 and m 2 m n R q2q n. By induction we have m 2 m n R s where s = (p 1,..., p l ) is in S and q 2 = p 1. This implies m L q1q 2 R s R s where s is the repetition-free sequence obtained by adjoining q 1 to the left of s. If q 1 occurs more than once, consider its rightmost occurrence, say q i = q 1 and q j q 1 for all j > i. Then we have m 1 m i 1 C q1q 1. By induction we have m i m n R s where s = (p 1,..., p l ) is in S with p 1 = q 1. This implies m C q1 R s = R s. (3 1) Suppose that R is bounded in A 1 A k. According to Proposition 1, we have the inclusion R R 1 R k, where, for every i = 1,..., k, we have R i = u i,1 u i,n i, and u i,1,..., u i,ni A i. According to Lemma 2, for every i = 1,..., k, there exists a polynomial p i such that: g Ri (n) p i (n), (5) holds, where g Ri denotes the growth function of the set R i. We have γ R (n 1,..., n k ) p i (n i ). Apply Lemma 4. The proof is complete. As a consequence we have i=1,...,k 9

10 Theorem 3 The growth of a rational relation in R A 1 A k is either exponential or polynomial. Proof. Suppose that for every q Q and for every i [1, k], C i q is 1-bounded. Then, by Theorem 2 R has polynomial growth. Otherwise, if there exists q Q and i [1, k] such that C i q is not 1-bounded, by Lemma 5, R has exponential growth. 5 Effectiveness issues In this section we first give a structural description of bounded rational relations. We prove that the Gap Theorem is effective: it is recursively decidable whether or not a rational relation has polynomial growth. Then we show how the growth function of bounded rational relations can be effectively computed. 5.1 A structural characterization In view of Theorem 2, it is interesting to investigate the structure of bounded rational relations. In order to prove the main result of this section we need the following proposition. Proposition 2 Let A 1 A k be a direct product of free monoids. Let v 1, v 2,..., v k be words, with v i A i. Let R Rat(A 1 A k ). If R v1 vk, then R Rat(v 1 vk ). Proof. Let us consider the commutative monoid a 1 a k, where the a i s are new letters. Consider the morphism ψ : a 1 a k A 1 A k, defined by ψ(a x1 1,..., ax k k ) = (vx1 1,..., vx k k ). One can prove that ψ 1 (R) Rat(a 1 a k )(cf [6, Proposition 2]). Moreover, since R v 1 vk, one has ψ(ψ 1 (R)) = R. Because ψ is a morphism from a 1 a k to v1 vk, and because rationality is preserved under morphisms [2], one has R Rat(v1 vk ). Theorem 4 Let R A 1 A k be a rational relation. Then R is bounded if and only if it is a finite union of relations of the form: where m i, a i A 1 A k. m 1a 1 m 2a 2 a n 1 m n, (6) Proof. The sufficiency immediately follows from Eq. 6 and the fact that the family of bounded sets is closed under product and set union. 10

11 In order to prove that the condition is necessary, we return to the combinatorial argument used in the proof of the implication (2) (3) of Theorem 2 of which we reuse the notations. By hypothesis, R is a bounded rational relation. Hence, by Theorem 2, for every i = 1,..., r and for every j = 1,..., k, there exists a word v jqi A j such that C j q i v jq i. This implies that, for every i = 1,..., r, C qi is a submonoid of the commutative monoid Cq 1 i Cq k i (v 1qi ) (v kqi ), We know that C qi Rat(A 1 A k ). By Proposition 2, we can say that C qi Rat(v1q i vkq i ). Because of [13], every rational subset of a free finitely commutative monoid is a finite union of linear sets. Thus, C qi is a finite union of sets of the form m 0 q i ( m 1 qi ) ( m l i q i ) for some m 0 q i, m 1 q i,..., m li q i (v 1qi ) (v kqi ). By substituting this finite union in Eq. 3 shows that each R s is a finite union of subsets of the form 6, proving thus the claim. The above result can be slightly generalized. We recall that a submonoid N of a monoid M is said to be subtractive if it satisfies the following property: for any x, y M, x, xy N implies y N. Theorem 5 Let R A 1 A k be a relation. Assume that R Rat(N), where N is a subtractive submonoid of A 1 A k. Then R is bounded if and only if it is a finite union of relations of the form: where m i, a i N. m 1a 1 m 2a 2 a n 1 m n, (7) Proof. The proof is based upon the same argument used to prove the previous theorem. We consider in more detail the sole point that needs care. By hypothesis, R Rat(N) so that there exists a finite N-automaton that accepts R. As done in the proof of the previous theorem, one proves that, for any state q of this automaton, the set C q is a finite union of linear sets of the form such that X = m 0 q(m 1 q) (m l q), (8) X N, m 0 q,..., m l q A 1 A k. To get the claim, we need to show that, for every i = 0,..., l, we have m i q N. Since X N, (8) implies that m 0 q N. Moreover, since, for every i = 1,..., l, the condition m 0 qm i q N holds, the subtractivity of N implies m i q N. The final part of the proof runs as for the previous theorem. 11

12 5.2 Computing the counting function We show that Theorems 3 and 4 are effective. proposition. First we need the following Proposition 3 The condition 2 of Theorem 2 can be effectively tested and the v i s can be effectively computed. Proof. The subset C i q consists of the labels of all paths taking the state q to itself. We may thus restrict our attention to the maximal connected component of the graph underlying the automaton and containing the state q. This leads us to investigating the following decision problem. Instance: a k-tape finite automaton whose underlying graph is connected, a specific state q 0 Q and an integer 1 i k. Question: does there exist v A i, such that for all paths, the following holds q 0 (z 1,...,z k ) q 0 z i v? (9) Call cyclic an automaton satisfying the above condition for i = 1,..., k. We proceed as follows. We assume without loss of generality that k = 1 which allows us to drop the indices for z i. We first verify whether or not there exists a nonempty label of a path from q 0 to q 0. Since the underlying graph is connected, this is equivalent to saying that there exists a transition labeled by some nonempty word, which can be done by inspecting all transitions. If all labels are empty, we are done otherwise there exists a transition with a nonempty label v r where r > 0 and v is primitive (i.e., v v1 for some v 1 implies v 1 = v). Consider an arbitrary path from q to q 0 and let u be its label. If the automaton w is cyclic, for all paths q 0 q we have wu v which shows that there exists an assignment of a prefix of v, different from v, to each state q, say v q such that all labels of a path from q 0 to q belong to v v q. In particular we have v q0 = 1. Observe that the assignment q v q can be achieved by a depth-first search, which is linear in the dimension of the underlying graph. Furthermore, for all z transitions q p we have. z (v q ) 1 v v p (10) Conversely, if such an assignment exists, then the automaton is cyclic. Theorem 6 It is recursively decidable whether a rational subset of A 1 A k has polynomial (resp. exponential) growth. If it has polynomial growth an expression such as 6 can be effectively computed. 12

13 k Proof. Let A = (Q, A 1 A k, E, I, T ) be an automaton accepting R. By Theorem 3, either R has exponential growth or it has polynomial growth. Thus the problem amounts to deciding whether or not R has polynomial growth. By Theorem 2, the latter test is equivalent to checking whether, for every q Q and for every i = 1,..., k, the language Cq i is 1-bounded or not. Compute all the maximal connected components of the graph underlying the automaton, i.e., the graph obtained by ripping off the labels of all transitions. For each of them, verify whether or not it is cyclic in the sense given in the proof of Proposition 3. If some of them is noncyclic, then the relation has nonpolynomial growth function else we proceed and show how to obtain a bounded expression for a given connected component over the set P of states. So far we have computed words v 1 A 1,..., v k A k such that for all transitions p (z1,...,z k) p with p, p P the words z 1,..., z k are factors of words in v1,..., vk, see expression 10. We set l 1 = v 1,..., l k = v k. Let {a 1,..., a k } be a new set of symbols. Transform the k-automaton restricted to P, into a k-tape automaton on the direct product a 1 a k by replacing each transition q (z1,...,z k) p by the p-tuple (a z1 1,..., a z k ). A rational expression, and therefore a semilinear expression, can be obtained from this new automaton. Because of the cyclicity of the component, we know furthermore that the expression is actually semilinear in the free commutative monoid generated by the elements {a l1 1,..., al k k }. A semilinear expression on a free commutative monoid is clearly a bounded expression. It then suffices to substitute u i for each occurrence of a li i obtain a bounded expression for the component. in order to Based on Theorem 4 we are able to design an algorithm which allows us to compute the counting function of a bounded rational relation. Theorem 7 Let R A 1 A k be a bounded rational relation. Then there exist c > 0 effectively computable polynomials with rational coefficients p 0,..., p c 1 and a positive integer n 0 such that, for any n n 0, the following holds f R (n) = p l (n) where the positive integer l is chosen so that l n mod c. To keep our notation simple, we will do the proof of Theorem 7 in the case of k = 2, that is for bounded rational relations of the product monoid A B. The general case adds only technicalities. The proof is a natural adaptation of the scheme used in [8] to prove the very same result for bounded context-free languages. Proof of Theorem 7: 13

14 Let R be a rational relation of the monoid A B. Suppose that R is bounded so that via Proposition 1 there exist nonempty words u 1,..., u m A, v 1,..., v n B such that: R u 1 u m v 1 v n. Let us now consider two new alphabets X = {a 1,..., a m }, Y = {b 1,..., b n }. Consider the morphisms φ 1 : X A and φ 2 : Y B respectively defined as: φ 1 (a i ) = u i and φ 2 (b i ) = v i. Finally consider the morphism φ : X Y A B defined as: φ(x, y) = (φ 1 (x), φ 2 (y)). One can prove that φ 1 (R) Rat(X Y ) (cf [6, Proposition 2]). By using the Cross Section Theorem, [12, Theorem IX.7.1], we can find rational languages R 1 Rat(X ) and R 2 Rat(Y ) where and such that the restrictions R 1 a 1 a m, R 2 b 1 b n, φ 1 : R 1 u 1 u m, φ 2 : R 2 v 1 v n of the morphisms φ 1 and φ 2 to the languages R 1 and R 2 respectively, are bijections. This implies that the restriction φ : R 1 R 2 u 1 u m v 1 v n of the morphism φ to the set R 1 R 2 is a bijection. Moreover, since R 1 Rec(X ) and R 2 Rec(Y ), by Elgot and Mezei s Theorem, one has that R 1 R 2 Rec(X Y ) so that the set R = φ 1 (R) (R 1 R 2 ) is a rational set of X Y contained in the set a 1 a m b 1 b n. By construction, the restriction φ : R R, of the morphism φ to the set R is a bijection. Let us now consider the map defined as ι : N m N n a 1 a m b 1 b n, ι(x 1,..., x m, y 1,..., y n ) = (a x1 1 axm m, b y1 1 byn n ). 14

15 Set B = ι 1 (R ). By Proposition 3 (point 1) of [6], we have B Rat(N m N n ). Now let ψ be the map ψ = φ ι : N m N n A B, such that, for every x N m N n, ψ(x) = φ(ι(x)). Since φ is a bijection from R to R and ι is a bijection from B to R, the restriction ψ : B R, (11) of the map ψ from B to R is a bijection. Moreover, by applying the well-known characterization of rational sets of N m N n by Eilenberg and Schützenberger, [13], we may suppose that B is a semi-linear set of N m N n of the form B = B 1 B 2 B r, (12) where the sets B i (i = 1,..., r) form a finite family of pairwise disjoint simple sets. ** We recall that a subset of N k is simple if it can be represented unambiguously as b 0 + Nb Nb s, i.e. the vectors b 1,..., b s are linearly independent. ** By Eq. 11 and Eq. 12, we have R = i=1,...,r where, for every i = 1,..., r, R i is defined as: R i = ψ(b i ). Since the map ψ is a bijection and since the sets B i are pairwise disjoint, so are the sets R i. Therefore the counting function of R satisfies the equality f R (n) = f Ri (n). (13) i=1,...,r In order to prove the claim, by Eq. (13), we first consider the computation of the counting function f Ri, where i = 1,..., r. Let B i be a simple set of the union (12) and assume R i B i = b 0 + Nb Nb s, where b 0,..., b s are the vectors of the representation of B i and b 1,..., b s are linearly independent. Thus, for every integer n 0, the number of distinct elements of length n of R i equals the number s(n) of positive integer solutions of the equation E i (n): where, for every l = 1,..., s, λ 0 + λ 1 x λ s x s = n, (14) λ l = ψ(b l ). 15

16 By applying a result of [1] (cf also [8], Thm 1) to the equation E i (n), one can compute polynomials with rational coefficients p 0,..., p c 1 and a positive integer n 0 such that, for any n n 0, one has that: where the positive integer l is chosen so that f Ri (n) = s(n) = p l (n), (15) l n mod c. Now, starting from Eq. 15, we construct a set of polynomials useful to compute f R. For the sake of simplicity, assume that in Eq. 12, r = 2, that is: R = R 1 R 2, the proof in the general case being completely similar. By Eq. 13, we have that, for every n 0: f R (n) = f R1 (n) + f R2 (n) = s 1 (n) + s 2 (n), (16) where s 1 (n) and s 2 (n) are the numbers of positive integer solutions of the equations E 1 (n) and E 2 (n) associated with R 1 and R 2 respectively. Now there exist two families of polynomials and integers n 1, n 2 0 such that: {p 0,..., p h1 1}, {q 0,..., q h2 1}, n n 1, s 1 (n) = p l (n), (resp. n n 2, s 2 (n) = q l (n)) where n l mod h 1 (resp. n l mod h 2 ). For every i = 0,..., h 1 1 and for every j = 0,..., h 2 1, let us define a new family of h 1 h 2 polynomials as: Then, for any n n 1 n 2, we have r (i,j) = p i + q j. f R (n) = s 1 (n) + s 2 (n) = p i (n) + q j (n) = r (i,j) (n), where i, j are the minimal non negative integers such that The proof is thus complete. n i mod h 1, n j mod h 2. Counting arguments usually require the set to be counted given by an unambiguous expression. It is worthwhile noticing that we could prove Theorem 7 though ambiguous rational relations exist. Let us study the following example which clarifies the apparent paradox. 16

17 Example 1 Let us consider the relation R in the monoid {a, b} {c} defined as: R = {(a n b m, c q ) n = q or m = q} = (a, c) (b, 1) (a, 1) (b, c). This relation is ambiguous (cf [2]). However it is the image under the map φ of the nonambiguous relation R N 3 defined as: with R = R 1 R 2 R 3 R 4 R 5, R 1 = (1, 1, 1), (describing the elements where q = n = m), R 2 = (1, 1, 1) + (0, 1, 0) +, (describing the elements where q = n < m), R 3 = (1, 1, 1) + (0, 0, 1) +, (describing the elements where m = n < q), R 4 = (1, 1, 1) + (1, 0, 1) +, (describing the elements where m < q = n), R 5 = (1, 1, 1) + (1, 1, 0) +, (describing the elements where q < m = n). 6 The case of trace monoids The aim of this section is to extend the previous theorem to rational subsets of trace monoids. The reader is referred to [11] for the main definitions and results of this subject. Let us recall only few elements. Given a finite set of generators A and a symmetric reflexive relation D on A, the trace monoid generated by A with dependence relation D, denoted by M(A, D), is defined by the monoid presentation A ab = ba, (a, b) / D The trace monoid can be embedded into a direct product of free monoids as follows. Consider the graph (technically known as the graph of dependence) whose vertex set is A and whose edges are all the pairs (a, b) D. Next consider a collection of cliques of this graph covering A: A 1,..., A p A with A = p i=1 A i. Then M(A, D) maps injectively into the product M = A 1 A p under the morphism ϕ(a) = (v 1,..., v p ) where v i = { a if a Ai 1 otherwise Let x M(A, D) and let x M(A,D) be the length of x in M(A, D) with respect to the system of generators of M(A, D). Moreover, let x M be the length of ϕ(x) in the direct product M. The following should be clear from the definition. Property 1 From the definition of the morphism ϕ, one has that: x M(A,D) x M p x M(A,D) 17

18 By Lemma 3 and by Property 1, we also have. Lemma 6 Let L M(A, D) be a rational subset. Then L has polynomial (resp. exponential) growth in M(A, D) if and only if ϕ(l) has polynomial (resp. exponential) growth in A 1 A p. Proposition 4 Let ϕ : M(A, D) A 1 A p be the natural embedding as defined above. Then ϕ(m(a, D)) is a subtractive submonoid of A 1 A p. Proof. The set T = ϕ(a) is the minimal set of generators of the submonoid ϕ(m(a, D)). Therefore, it suffices to prove that for any t T and m A 1 A p, the condition tm ϕ(m(a, D)) implies m ϕ(m(a, D)). We prove this by induction on the length l of m. If l = 0, the implication trivially holds. Suppose now l > 0. If tm ϕ(m(a, D)), then there exist t 1, t 2,... t l T, such that tm = t 1 t 2 t l. If t = t 1, then m = t 2 t l, since the monoid A 1 A p is cancellative so m M(A, D). Assume now that t t 1. One has t = ϕ(a) and t 1 = ϕ(b), with a b. It is easily checked that the equality tm = t 1 t 2 t l implies that in the vectors t = ϕ(a) and t 1 = ϕ(b) the sets of positions containing a letter are disjoint, so t and t 1 commute. Moreover, one can write m = t 1 m, with m A 1 A p and m < m. One has tm = tt 1 m = t 1 tm = t 1 t 2 t l. By cancellation one derives tm = t 2 t l, and, by the inductive hypothesis m ϕ(m(a, D)). Therefore, m = t 1 m ϕ(m(a, D)). The results proved in the previous sections may be extended to the family of rational sets of a trace monoid. In particular, we obtain the following. Theorem 8 Let L M(A, D) be a rational set. The following conditions hold: 1. L has polynomial growth if and only if it is bounded. 2. L has either polynomial or exponential growth. 3. It is decidable whether L has polynomial (resp. exponential) growth. Proof. We prove the sufficiency of Condition (1). Suppose that L Rat(M(A, D)) has polynomial growth. Then it is accepted by some finite automaton A over M(A, D). An automaton B on A 1 A p accepting the image ϕ(l) is defined on the same set of states and the same sets of initial and terminal states. Its x set of transitions is obtained by transforming each transition q p of A into the transition q ϕ(x) p. By Proposition 4, ϕ(m(a, D)) is a subtractive submonoid of A 1 A p. By theorems 2 and 5, the subset ϕ(l) is equal to a finite union of linear subsets of the form (7): m 1a 1 a n 1 m n, where a i, m i ϕ(m(a, D)). 18

19 Since a i, m i ϕ(m(a, D)), there exist α i, µ i M(A, D) such that ϕ(α i ) = a i, ϕ(µ i ) = m i. Because ϕ is a morphism, we have: m 1a 1 a n 1 m n = ϕ(µ 1 ) ϕ(α 1 ) ϕ(α n 1 )ϕ(µ n ) = ϕ(µ 1α 1 α n 1 µ n). Because of the injectivity of ϕ, the previous equality implies that L is a finite union of sets of the form: µ 1α 1 α n 1 µ n, where α i, µ i M(A, D) and this shows that L is bounded. We prove the necessity of Condition (1). Suppose that L is bounded. Hence ϕ(l) A 1 A p is also bounded, so that, by Theorem 2, ϕ(l) has polynomial growth in A 1 A p. By Lemma 6, the latter implies that L has polynomial growth. Condition (1) is thus proved. We prove Condition (2). Suppose that L does not have polynomial growth. By Lemma 6, ϕ(l) does not have polynomial growth either. This implies that ϕ(l) has exponential growth. Therefore, for sufficiently large positive integers n, ϕ(l) has at least K n elements of length n, for some K > 1. Since ϕ is injective and all these elements are images of elements of length at most n in L, we have g L (n) g ϕ(l) (n) K n. Let us finally prove Condition (3). Let L Rat(M(A, D)). By the previous Condition (2), it suffices to decide whether L has polynomial growth or not. Since L Rat(M(A, D)), we have ϕ(l) Rat(A 1 A p) and by Lemma 6, condition (2) follows by applying the procedure of Theorem 6 to ϕ(l). References [1] E. T. Bell, Interpolated denumerants and Lambert series, Amer. J. Math., 65, , [2] J. Berstel, Transductions and Context-Free Languages. Teubner, Stuttgart, [3] Alberto Bertoni and Nicoletta Sabadini. Generating functions of trace languages. Bulletin of the EATCS, 35, , [4] A. Bouillard, J. Mairesse, Generating series of the trace group, LNCS vol. 2710, pp , 2003, [5] C. Choffrut, S. Grigorieff, Separability of rational subsets by recognizable subsets is decidable in Σ N m, Information Processing Letters 99 (1), (2006). [6] C. Choffrut, F. D Alessandro, S. Varricchio, On the separability of bounded context-free languages and bounded rational relations, Theoret. Comput. Sci. 381, (2007). 19

20 [7] Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. Introduction to Algorithms, Second Edition. The MIT Press and McGraw-Hill Book Company, [8] F. D Alessandro, B. Intrigila, S. Varricchio, On the structure of the counting function of context-free languages, Theoret. Comput. Sci. 356, (2006). [9] F. D Alessandro, S. Varricchio, On the growth of context-free languages, Technical report, Università di Roma La Sapienza, (2006). [10] A. de Luca, S. Varricchio, Finiteness and Regularity in Semigroups and Formal Languages Springer-Verlag, [11] V. Diekert, G. Rozenberg Editors, The book of traces, World scientific Pu. Co., [12] S. Eilenberg. Automata, Languages and Machines, volume A. Academic Press, [13] S. Eilenberg and M.-P. Schützenbeger. Rational sets in commutative monoids. Journal of Algebra, 13(2), , [14] S. Ginsburg, The mathematical theory of context-free languages, McGraw- Hill Book Co., [15] J. Honkala, Decision problems concerning thinness and slenderness of formal languages, Acta Informatica 35, (1998). [16] J. Hopcroft and J. Ullman, Introduction to Automata Theory, Languages and Computation, Addison-Wesley Pub. Co., [17] Lothaire. Combinatorics on Words, volume 17 of Encyclopedia of Mathematics and its Applications. Addison-Wesley, Gian-Carlo Rota edition, [18] J. Sakarovitch, Éléments de théorie des automates. Vuibert, Paris, [19] M. P. Schützenberger, Sur les relations rationnelles entre monoides libres, Theoret. Comput. Sci. 3, (1976). 20