The Accepting Power of Finite. Faculty of Mathematics, University of Bucharest. Str. Academiei 14, 70109, Bucharest, ROMANIA

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1 The Accepting Power of Finite Automata Over Groups Victor Mitrana Faculty of Mathematics, University of Bucharest Str. Academiei 14, 70109, Bucharest, ROMANIA Ralf STIEBE Faculty of Computer Science, University of Magdeburg P.O.Box 4120, D-39016, Magdeburg, GERMANY Turku Centre for Computer Science TUCS Technical Report No 70 November 1996 ISBN ISSN

2 Abstract Some results from [2], [5], [6] are generalized for nite automata over arbitrary groups. The accepting power is smaller when abelian groups are considered, in comparison with the non-abelian groups. We prove that this is due to the commutativity. Each language accepted by a nite automaton over an abelian group is actually a unordered vector language. Finally, deterministic nite automata over groups are investigated. TUCS Research Group Mathematical Structures of Computer Science

3 1 Introduction One of the oldest and most investigated machine in the automata theory is the nite automaton. Many fundamental properties have been established and many problems are still open. Unfortunately, the nite automata without any external control have a very limited accepting power. Dierent directions of research have been considered for overcoming this limitation. The most known extension added to a nite automata is the pushdown memory. In this way, a considerable increasing of the accepting capacity has been achieved. The pushdown automata are able to recognize all context-free languages. Another simple and natural extension, related somehow to the pushdown memory, was considered in a series of papers [2,5,6], namely to associate to each conguration an element of a given group, but no information regarding the associated element is allowed. This value is stored in a counter. An input string is accepted if and only if the automaton reaches a designated nal state with its counter containing the neutral element of the group. Thus, new characterizations of unordered vector languages [6] or contextfree languages [2] have been reported. These results are, in a certain sense, unexpected since in such an automaton the same choice is available regardless the content of its counter. More precisely, the next action is determined just by the input symbol currently scanned and the state of the machine. In this paper, we shall consider only acceptors with a one-way input tape read from left to right and a counter able to store elements from a given group. The aforementioned papers deal with nite automata over very well dened groups e.g. the additive group of integers, the multiplicative group of non-null rational numbers or the free group. The aim of this paper is to provide some general results regardless the associated group. We shall prove that the addition of an abelian group to a nite automaton is less powerful than the addition of the multiplicative group of rational numbers. An interchange lemma points out the main reason of the power decrease of nite automata over abelian groups. Characterizations of the context-free and recursively enumerable languages classes are set up in the case of non-abelian groups. As far as the deterministic variants of nite automata over groups are concerned we shall show their considerable lack of accepting power. 1

4 2 Preliminaries We assume the reader familiar with the basic concepts in automata and formal language theory and in the group theory. For further details, we refer to [4] and [8], respectively. For an alphabet, we denote by the free monoid generated by under the operation of concatenation; the empty string is denoted by " and the semigroup? f"g is denoted by +. The length of x 2 is denoted by jxj. Let K = (M; ; e) be a group under the operation denoted by with the neutral element denoted by e. An extended nite automaton (EFA shortly) over the group K is a construct A = (Z; ; K; q 0 ; F; ) where Z; ; q 0 ; F have the same meaning as for a usual nite automaton [4], namely the set of states, the input alphabet, the initial state and the set of nal states, respectively, and : Z [ f"g?! P f (Z M) This sort of automaton can be viewed as a nite automaton having a counter in which any element of M can be stored. The relation (q; m) 2 (s; a); q; s 2 Z; a 2 [ f"g; m 2 M means that the automaton A changes its current state s into q, by reading the symbol a on the input tape, and writes in the register x m, where x is the old content of the register. The initial value registered is e. We shall use the notation (q; aw; m) j= A (s; w; m r) i (s; r) 2 (q; a) for all s; q 2 Z a 2 [ f"g; m; r 2 M. The reexive and transitive closure of the relation j= A is denoted by j= A. Sometimes, the subscript identifying the automaton will be omitted when it is self-understood. The word x 2 is accepted by the automaton A if and only if there is a nal state q such that (q 0 ; x; e) j= (q; "; e). In other words, a string is accepted if the automaton completely reads it and reaches a nal state when the content of the register is the neutral element of M. The language accepted by an extended nite automaton over a group A as above is L(A) = fx 2 j(q 0 ; x; e) j= A (q; "; e); for some q 2 F g 2

5 For two groups K 1 = (M 1 ; 1 ; e 1 ) and K 2 = (M 2 ; 2 ; e 2 ), we dene the triple K 1 K 2 = (M 1 M 2 ; ; (e 1 ; e 2 )) with (m 1 ; m 2 )(n 1 ; n 2 ) = (m 1 1 n 1 ; m 2 2 n 2 ). It is well-known that K 1 K 2 is also a group. We are going to provide some results that will be useful in what follows. The notation L(REG) identies the class of regular languages. Theorem 1 For any group K, L(EF A(K)) = L(REG) i all nitely generated subgroups of K are nite. Proof. Let K be a group such that any nitely generated subgroup of K is nite. Let A = (Z; ; K; z 0 ; F; ) be an EFA over K = (M; ; e). We denote by X the nite subset of M X = fm 2 M j (z 0 ; m) 2 (z; a) for some z; z 0 2 Z; a 2 [ f"gg: Let H = (hxi; ; e) be the subgroup generated by X. We construct the nite automaton with "-moves B = (Z hxi; ; (z 0 ; e); F feg; ') with '((z; m); a) = f(z 0 ; m n) j (z 0 ; n) 2 (z; a), for all z 2 Z, m 2 M, a 2 [ f"g. One can easily prove that (z 0 ; w; e) j= A (z; "; m) i ((z 0 ; e); w) j= B ((z; m); "), which implies L(A) = L(B). It remains to prove that for any innite group K, nitely generated, exists an EFA over K accepting a non-regular language. Let K = (hxi; ; e) be such a group with the nite set of generators X. Consider the (deterministic) EFA A = (fzg; Y; K; z; fzg; ), with Y = X [ fx?1 j x 2 Xg and (z; a) = (z; a), for all a 2 Y. The following facts about L(A) are obvious: 1. For any m 2 hxi, exist a word v 2 Y such that (z; v; e) j= A (z; "; m). 2. For any v 2 Y, exists a word w 2 Y such that vw 2 L(A). 3. For any k 0, the set is nite. X k = fm 2 hxi j 9v 2 Y ; jvj k : (z; v; e) j= A (z; "; m)g As a consequence of these facts and of the inniteness of hxi, we obtain: For all k 0, there is a word v k such that v k v =2 L(A), for all v 2 Y, jvj k. But, one can easily prove that for any regular language L Y, exists k 0 such that for all vw 2 L, there is w 0 2 Y, jw 0 j k, with vw 0 2 L. Hence, L(A) cannot be regular. 2 A nitely generated abelian group is nite if all its elements are of nite order. Hence, for an abelian group K, L(EF A(K)) = L(REG) i all elements of K have nite order. This is not necessarily true for non-abelian groups. We can, however, prove a pumping lemma which is very similar to the pumping lemma for regular languages. 3

6 Lemma 1 Let K be some group without elements of innite order. For any language L 2 L(EF A(K)), there is a constant n 1 such that, for all x 2 L, jxj n, there exist a decomposition x = uvw and a natural number q 1 with juvj n, jvj 1, uv iq+1 w 2 L, for all i 0. Moreover, if K has the nite exponent p then q can uniformly be chosen as q = p. Proof. Let A = (Z; ; K; z 0 ; F; ) be an EFA over K. We choose n = jzj + 1. Now consider a word x 2 with jxj n. Similar to the proof of the pumping lemma for regular languages, it can be shown that there is a decomposition x = uvw, juvj n, jvj 1, such that (z 0 ; uvw; e) j= A (z; vw; m 1 ) j= A (z; w; m 1 m 2 ) j= A (f; "; e); z 2 Z; f 2 F: Now choose q such that m q 2 = e. Obviously, any word uv iq+1 w is accepted by A. 2 As a consequence of the above pumping lemma, we obtain: Theorem 2 For any group K, L(EF A(K)) contains the language L = fa n b n j n 1g i at least one element of K has an innite order. Proof. Let K = (M; ; e). If M contains an element m of innite order then the nitely generated subgroup (hmi; ; e) is isomorphic to ( Z; +; 0), hence L is in L(EF A(K)). If all elements of M have nite order, a simple application of the above pumping lemma yields L =2 L(EF A(K)): 2 For a group K, let F(K) denote the family of all nitely generated subgroups of K. Theorem 3 For any group K, L(EF A(K)) = [ H2F(K) L(EF A(H)) Proof. Let K = (M; ; e) be a group. The inclusion L(EF A(K)) [ L(EF A(H)) H2F(K) holds, since L(EF A(K)) L(EF A(H)), for any subgroup H of K. On the other hand, let A = (Z; ; K; z 0 ; F; ) be an EFA over K. The group H = (hxi; ; e), where X = fm 2 Mj(q; m) 2 (z; a) for some q; z 2 Z; a 2 [ f"gg is a nitely generated subgroup of K. Obviously, during any computation in the counter of A appear only elements of hxi. Therefore, the automaton A can be viewed as an automaton over H. More precisely, A 0 = (Z; ; H; z 0 ; F; ) accepts the same language as A does. This proves the second inclusion and thus the theorem. 2 4

7 3 EFA over abelian groups Valence grammars and EFA have initially been introduced for the groups Z k = ( Z k ; +; 0), k 1 and Q = (Q? f0g; ; 1). In what follows we shall show that the accepting capacity of EFA does not increase if we consider arbitrary abelian groups instead of Q. Thus, every language accepted by an EFA over an abelian group is a (unordered) vector language [1]. The deeper reason of this fact is the following fundamental result in the group theory. Theorem 4 A nitely generated abelian group is the direct product of a nite number of cyclic groups. As a consequence, a nitely generated abelian group is either nite or isomorphic to a group Z k H, where k is a positive integer and H is a nite abelian group. Theorem 5 For a group K and a nite group H, L(EF A(K H)) = L(EF A(K)) Proof. Let K and H be given by K = (M 1 ; 1 ; e 1 ) and H = (M 2 ; 2 ; e 2 ), and let A = (Z; ; K H; z 0 ; F; ) be an EFA over K H. We construct the EFA over K, A 0 = (Z 0 ; ; K; z 0 0 ; F 0 ; 0 ) with Z 0 = Z 0 M 2, z 0 0 = (z 0; e 2 ), F 0 = F fe 2 g and 0 ((z; n 2 ); a) = f((z 0 ; n 2 2 m 2 ); m 1 ) j (z 0 ; (m 1 ; m 2 )) 2 (z; a)g; z; z 0 2 Z; a 2 [ f"g; m 1 2 M 1 ; m 2 ; n 2 2 M 2 : By induction on the number of steps, one can show that ((z 0 ; e 2 ); w; e 1 ) j= A 0 ((z; m 2 ); w 0 ; m 1 ) i (z 0 ; w; (e 1 ; e 2 )) j= A (z; w 0 ; (m 1 ; m 2 )), hence L(A) = L(A 0 ). 2 We are now ready to prove the main result of this section. Theorem 6 For an abelian group K, one of the following relations hold. L(EF A(K)) = L(REG); L(EF A(K)) = L(EF A(Z k )); for some k; L(EF A(K)) = L(EF A(Q)): Proof. As it was shown in Theorem 3, L(EF A(K)) = S H2F(K) L(EF A(H)). Every H 2 F(K) is either nite or isomorphic to a group Z k H 0 where k 1 and H 0 is a nite group. Hence for all H 2 F(K), either L(EF A(H)) = 5

8 L(REG), or L(EF A(H)) = L(EF A(Z k H 0 )) = L(EF A(Z k )), for some k 1. If all nitely generated subgroups of K are nite, then L(EF A(K)) = L(REG) holds, due to Theorem 1. Otherwise, let N(K) be the set of all k such that L(EF A(H)) = L(EF A(Z k )), for some H 2 F(K). If N(K) is nite then L(EF A(K)) = L(EF A(Z k )), where k = max(n(k)). If N(K) is innite then L(EF A(K)) = L(EF A(Q)). 2 It is known that languages as L 1 = fa n b n j n 1g or L 2 = fwcw R j w 2 fa; bg + g are not in L(EF A(Q)). Therefore, L 1 ; L 2 =2 L(EF A(K)), for any abelian group K. In [2] it was conjectured that the commutativity of the multiplication of rational numbers is responsible for this fact. We shall formally prove this conjecture by help of the following "interchange lemma". Lemma 2 Let K = (M; ; e) be some abelian group, and let L be a language in L(EF A(K)). There is a constant k such that, for any x 2 L, jxj k, and any decomposition x = v 1 w 1 v 2 w 2 : : : v k w k v k+1, jw i j 1, exist two integers 1 r < s k such that the word x 0 = v 1 w 0 1v 2 w 0 2 : : : v k w 0 kv k+1 with w 0 r = w s, w 0 s = w r, w 0 i = w i, for i =2 fr; sg, is in L. Proof. Let A = (Z; ; K; z 0 ; F; ) be an EFA over K = (M; ; e). We choose k = jzj For a word x 2 L(A), jxj k, let be given a decomposition x = v 1 w 1 v 2 w 2 : : : v k w k v k+1, jw i j 1. There are the states y i ; z i 2 Z, 1 i k, q 2 F, with (z i?1 ; v i ; e) j= (y i ; "; m i ); 1 i k; (y i ; w i ; e) j= (z i ; "; n i ); 1 i k; (z k ; v k+1 ; e) j= (q; "; m k+1 ); and m 1 n 1 m 2 n 2 : : : m k n k m k+1 = e. By the pigeon-hole principle, there are two numbers 1 r < s k with (y r ; z r ) = (y s ; z s ). Now consider x 0 = v 1 w 0 1 v 2w 0 2 : : : v kw 0 k v k+1 with w 0 r = w s, w 0 s = w r, w 0 i = w i, for i =2 fr; sg. For the words v i, 1 i k + 1, and w 0 i, 1 i k + 1 the relations (z i?1 ; v i ; e) j= (y i ; "; m i ); 1 i k; (y i ; w 0 i ; e) j= (z i ; "; n 0 i ); 1 i k; (z k ; v k+1 ; e) j= (q; "; m k+1 ); 6

9 hold, with n 0 r = n s, n 0 s = n r, n 0 i = n i, for i =2 fr; sg. By the commutativity of K it follows m 1 n 0 1 m 2 n 0 2 : : : m k n 0 k m k+1 = m 1 n 1 m 2 n 2 : : : m k n k m k+1 = e; implying that x 0 is accepted by A. 2 Theorem 7 For any abelian group K, the languages L 1 = fa n b n j n 1g and L 2 = fwcw R j w 2 fa; bg + g are not in L(EF A(K)). Proof. Assume that L 1 2 L(EF A(K), for some K, and let k be the constant from the interchange lemma. Now consider the word x = aba 2 b 2 : : : a k b k and the decomposition v i = a i ; w i = b i, 1 i k, v k+1 = ". There are 1 r < s k such that x 0 = aw 0 1 a2 w 0 : : : 2 ak wk 0 with = w0 r bs, ws 0 = br, wi 0 = b i, for i =2 fr; sg; is in L 1, contradiction. A similar reasoning for the relation L 2 =2 L(EF A(K)) is left to the reader. 2 4 EFA over non-abelian groups In this section, we restrict our investigation to the free groups, since for any (non-abelian) group K there is a homomorphism from a free group to K [8]. In this way, we get a characterization of the context-free languages class in terms of languages accepted by extended nite automata over the free group with just two generators [2]. The free group with n generators is denoted by F n. Recall from [2] Theorem 8 The family of context-free languages equals L(EF A(F 2 )). Lemma 3 Let K 1 and K 2 be two groups. For two languages L i 2 L(EF A(K i )), i = 1; 2, the languages L 1 \ L 2 and L 1 L 2 are in L(EF A(K 1 K 2 )). Proof. Let A i = (Z i ; i ; K i ; z 0 i ; F i; i ), i = 1; 2; be two EFA over K i, respectively. We assume that Z 1 \ Z 2 is empty. We have L(B) = L 1 \ L 2 and L(C) = L 1 L 2, where B = (Z 1 Z 2 ; 1 \ 2 ; K 1 K 2 ; (z 0 1 ; z0 2 ); F 1 F 2 ; ) with ((z 1 ; z 2 ); a) = f((z 0 1; z 0 2); (m 1 ; m 2 )) j(z 0 1; m 1 ) 2 1 (z 1 ; a); (z 0 2; m 2 ) 2 2 (z 2 ; a)g ; 7

10 z 1 2 Z 1 ; z 2 2 Z 2 ; a 2 1 \ 2 ; ((z 1 ; z 2 ); ") = f((z 0 1; z 2 ); (m 1 ; e 2 )) j (z 0 1; m 1 ) 2 1 (z 1 ; ")g [ f((z 1 ; z 0 2); (e 1 ; m 2 )) j (z 0 2; m 2 ) 2 2 (z 2 ; ")g ; z 1 2 Z 1 ; z 2 2 Z 2 : and C = (Z 1 [ Z 2 ; 1 [ 2 ; K 1 K 2 ; z 0 1; F 2 ; ) with (z; a) = f(z 0 ; (m; e 2 )) j (z 0 ; m) 2 1 (z; a)g; z 2 Z 1 ; a 2 1 ; ( f(z0 ; (m; e (z; ") = 2 )) j (z 0 ; m) 2 1 (z; ")g; z 2 Z 1 n F 1 ; f(z 0 ; (m; e 2 )) j (z 0 ; m) 2 1 (z; ")g [ f(z2; 0 (e 1 ; e 2 ))g; z 2 F; (z; a) = f(z 0 ; (e 1 ; m)) j (z 0 ; m) 2 2 (z; a)g; z 2 Z 2 ; a 2 2 [ f"g: which concludes the proof. 2 It is well-known that every recursively enumerable language can be expressed as the homomorphical image of the intersection of two linear languages. It is obvious that each family L(EF A(K) is closed under homomorphims. In conclusion, due to the previous lemma as well as to Theorem 8, we have just proved: Theorem 9 L(EF A(F 2 F 2 )) equals the family of recursively enumerable languages. 5 Deterministic EFA In the case of EFA, the determinism signicantly decreases the accepting capacity. Denote by L(DEF A(K)) the family of languages recognized by deterministic EFA over the group K. Lemma 4 For any group K, the languages L 1 = fa n j n 1g [ fa n b n j n 1g, L 2 = fa m b n j m ng and L 3 = fa; bg n fa n b n j n 1g are not in L(DEF A(K)). Proof. Let K = (M; ; e), and let A = (Z; fa; bg; K; z 0 ; F; ) be a DEFA over K such that L 1 = L(A). Since a n 2 L(A), for all n 1, there are the integers 1 r < s jf j + 1 such that (z 0 ; a r ; e) j= A (q; "; e) and (z 0 ; a s ; e) j= A (q; "; e), for some q 2 Z. Since a n b n 2 L(A), for all n 1, we have also (z 0 ; a r b r ; e) j= A (q; br ; e) j= A ; "; e), for some q (q0 0 2 F. Hence, (z 0 ; a s b r ; e) j= A (q; br ; e) j= A ; "; e), implying a s b r 2 L(A), contradiction. (q0 In conclusion, L 1 6= L(A). By similar arguments it can be shown that L 2 and L 3 are not in L(DEF A(K)). 2 8

11 On the other hand, L(DEF A(K)) generally contains languages with undecidable membership problem. Theorem 10 There is a nitely generated group K such that the following question is undecidable. Given a DEFA A over K and a word w over the input alphabet of A, is w 2 L(A)? Proof. For a nitely generated group K = (M; ; e) with the set of generators X, the word problem is the following question. Is a given term x 1 x 2 : : : x n, x i 2 X [ X?1, 1 i n, X?1 = fx?1 j x 2 Xg equal to the neutral element e? It is a well-known result from group theory that there is a nitely generated group K with undecidable word problem. Let K = (M; ; e) with the nite set of generators X be such a group. We construct the DEFA A = (fzg; X [ X?1 ; K; z; fzg; ) with the transition function (z; x) = (z; x), for all x 2 X [ X?1. Obviously, A accepts a word x 1 x 2 : : : x n i x 1 x 2 : : : x n = e. The undecidability of the membership problem for A follows directly from the undecidability of the word problem for K. 2 Theorem 11 For every group K having at least one element of innite order, we have L(DEF A(K)) L(EF A(K)), strict inclusion. Proof. Let K be an arbitrary given group and A = (Z; ; K; z 0 ; F; ) be a deterministic EFA over K. If there is an element of M of innite order, then the language fa n jn 1g [ fa n b n jn 1g 2 L(EF A(K)), since L(EF A(K)) is closed under union. In conclusion, L(DEF A(K)) L(F EA(K)). 2 6 Deterministic EFA over abelian groups The next result is a consequence of Theorem 11. Theorem 12 For every abelian group K, we have either L(DEF A(K)) = L(REG) or L(DEF A(K)) L(EF A(K)), strict inclusion. Obviously, the statement of Theorem 6 is also valid for the deterministic EFA over abelian groups. As we have seen, neither L(EF A((Q)) [6] nor L(DEF A((Q)) are closed under complement. In the end of this section we will show that the complement of a language from L(DEF A((Q)) is in L(EF A((Q)). In order to handle the diculties owing to the existence of "-steps, we introduce some notations. 9

12 Let A = (S; ; Z k ; s 0 ; F; ) be a DEFA over (Z k ), for some k 1. For all s 2 S, we dene the sets N s = fr 2 Z k j (s; "; 0) j= (q; "; r); for some q 2 F g Lemma 5 Let A be a DEFA as above. For all s 2 S, there is an EFA A s = (Y s ; ; Z k ; s; fq s g; s ) such that s (s; a) = ;, for all a 2, and (s; "; 0) j= (q s ; "; r) i r 2 Z k n N s. Proof. In what follows let e i, 1 i k, denote the i-th unit vector of Z k. For any s 2 S, we construct the EFA A 1 (s) = (S [ fqg; T; Z k ; s; fqg; 1 ) with q =2 S, T = fa 1 ; : : : ; a k g [ fb 1 ; : : : ; b k g, and the transition relation 1 dened as 1 (p; a) = ;; for p 2 S; a 2 T; 1 (p; ") = (p; "); for p 2 S n F; 1 (p; ") = (p; ") [ f(q; 0)g; for p 2 F; 1 (q; a i ) = f(q;?e i ); for 1 i k; 1 (q; b i ) = f(q; e i ); for 1 i k; 1 (q; ") = ;: Obviously, the language accepted by A 1 (s) is L s = fw 2 T j (jwj a1? jwj b1 ; : : : ; jwj ak? jwj bk ) 2 N s g: Let : T! IN 2k be the Parikh mapping with (w) = (jwj a1 ; jwj b1 ; : : : ; jwj ak ; jwj bk ), for all w 2 T. The Parikh set (L s ) is semilinear, and its complement (L s ) is semilinear, too. Therefore, there is a nite automaton A 2 (s) = (Y s ; T; s; fq s g; 0 s) such that the Parikh set of L(A 2 (s)) is (L s ). From A 2 (s) we can construct A s as follows: A s = (Y s ; ; Z k ; s; fq s g; s ) with s (p; a) = ;; for p 2 Y s ; a 2 ; s (p; ") = f(p 0 ; e i ) j p s(p; a i ); 1 i kg [ f(p 0 ;?e i ) j p s (p; b i); 1 i kg; for p 2 Y s : Obviously, (s; "; 0) j= As (q s; "; r) i L(A 2 (s)) contains a word w with r = (jwj a1? jwj b1 ; : : : ; jwj ak? jwj bk ), i.e. i L s contains no word v with r = (jvj a1? jvj b1 ; : : : ; jvj ak? jvj bk ), hence i r =2 N z. 2 Theorem 13 For all L 2 L(DEF A(Q)), the complement of L is contained in L(EF A(Q)). 10

13 Proof. Let A = (S; ; Z k ; s 0 ; F; ) be a DEFA over Z k. The set of states can be partitioned into the set R consisting of all states s 2 S, such that A can perform only "-steps if s is reached, and its complement S n R. The complement of L(A) consists of two sets; rst, the set of all words w = w 1 w with w 2 6= " and (s 0 ; w; 0) j= A (p; w 2; r) for some r 2 Z k and some p 2 R; second, the set of all words ww 1 a 2 + with a 2 and (s 0 ; w; 0) j= A (p; a; r 0 ) j= A (s; "; r) for some p; s 2 S, and r + t 6= 0, for all t 2 N s. The last condition is equivalent to r + t = 0, for some t 2 Z k n N s. Now let for all s 2 S, A s = (Y s ; ; Z k ; s; fq s g; s ) be the DEFA constructed in the last lemma. Without loss of generality we may assume that Y s and Y s 0 are disjoint for s 6= s 0 and that Y s and S are disjoint for all s 2 S. Now we construct B = (S 0 ; ; Z k ; s 0 ; F 0 ; 0 ) with the set of states S 0 = S [ S s2s Y s [ffg, the set of nal states F 0 = fq s j s 2 Sg[ffg, and the following transition mapping 0 (s; ") = 0 (s; a) = 8 >< 0 (f; a) = f(f; 0)g; (s; ") if s 2 S? fs 0 g (s; ") [ f(f; 0)g if s = s 0 x (s; ") if s 2 Y x ; for some x 2 S >: ( (s; a) [ f(ys 0; r)j(s 0 ; r) 2 (s; a)g if s 2 S? R (s; a) [ f(y s 0; r)j(s 0 ; r) 2 (s; a)g [ f(f; 0)g if s 2 R 0 (f; ") = f(f; m e i )g; m 2 f?1; 0; 1g; 1 i k: for all a 2. It is easy to see that B accepts all words not in L(A). On the other hand, no word from L(A) is accepted by B. 2 An important consequence of the last theorem is the decidability of the inclusion problem (and of the equivalence problem, too) for DEFA over Q, which is an interesting contrast to the undecidability of the universe problem for nondeterministic one-turn counter automata, a proper subclass of EF A over ( Z; +; 0). Theorem 14 Let A and B be DEFA over Q. It is decidable whether or not L(A) L(B). Proof. Given B, one can construct a (nondeterministic) EF A B over Q with L(B) = L(B). Clearly, L(A) L(B) i L(B) \ L(A) is empty. Now an EFA C over Q can be constructed such that L(C) = L(B) \ L(A). Hence, the inclusion problem for DEFA over Q is reduced to the emptiness problem for EFA over the same group, which is decidable [1]. 2 11

14 References [1] J. Dassow, Gh. Paun, Regulated rewriting in formal language theory, Akademie-Verlag, Berlin, [2] J. Dassow, V. Mitrana, Finite automata over free generated groups, submitted, [3] S. Ginsburg, S. A. Greibach, Abstract families of languages. Memoirs Amer. Math. Soc. 87(1969), 1{32. [4] S. Ginsburg, Algebraic and automata-theoretic properties of formal languages, North-Holl., Amsterdam, Oxford, [5] S. A. Greibach, Remarks on blind and partially blind one-way multicounter machines. Th. Comp. Sci. 7(1978), 311{324. [6] O. H. Ibarra, S. K. Sahni, C. E. Kim, Finite automata with multiplication. Th. Comp. Sci., 2(1976), 271{294. [7] Gh. Paun, A new generative device: valence grammars, Rev. Roum. Math. Pures et Appl., 25, 6(1980), 911{924. [8] J. J. Rotman, An Introduction to the Theory of Groups, Springer-Verlag,

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16 Turku Centre for Computer Science Lemminkaisenkatu 14 FIN Turku Finland University of Turku Department of Mathematical Sciences Abo Akademi University Department of Computer Science Institute for Advanced Management Systems Research Turku School of Economics and Business Administration Institute of Information Systems Science