Finite Nondeterministic Automata: Simulation and Minimality

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1 Finite Nondeterministic Automt: Simultion nd Minimlity Cristin S. Clude, Elen Clude, Bkhdyr Khoussinov Abstrct Motivted by recent pplictions of finite utomt to theoreticl physics, we study the minimiztion problem for nondeterministic utomt (with outputs, but no initil sttes). We use Ehrenfeucht Frïsse-like gmes to model utomt responses nd simultions. The miniml utomton is constructed nd, in contrst with the clssicl cse, proved to be unique up to n isomorphism. Finlly, we investigte the prtil ordering induced by utomt simultions. For exmple, we prove tht, with respect to this ordering, the clss of deterministic utomt forms n idel in the clss of ll utomt. 1 Introduction Automt hve been used s toy models for physicl prticles for mny yers (see [12, 6, 7, 9]). Recent ppers (see [1, 14, 17, 15, 8, 16, 3, 2, 5, 4]) hve imposed the notion of finite utomton with outputs nd no initil sttes s bsic model. In this context the interest is not directed to the lnguges ccepted by vrious utomt but to utomt behviour nd simultions. The behviour of n utomton is described by its responses to vrious experiments (expressed s sequences of input symbols). An utomton A simultes the behviour of n utomton B in cse A cn perform ny computtion B cn perform nd the outputs produced will be the sme. The cse of deterministic utomt (both complete nd incomplete) being disposed in [2, 5], we concentrte our ttention on nondeterministic utomt. Vrious models of simultions will be considered nd investigted. In constructing the miniml nondeterministic utomton we will rely on the notion of indistinguishble sttes 1 which will The first nd third uthors hve been prtilly supported by AURC A18/XXXXX/629/F341456, 1996; the second uthor hs been prtilly supported by AURC A18/XXXXX/629/F34147, Computer Science Deprtment, The University of Aucklnd, Privte Bg 9219, Aucklnd, New Zelnd, e-mil: cristin@cs.ucklnd.c.nz. Computer Science Deprtment, The University of Aucklnd, Privte Bg 9219, Aucklnd, New Zelnd, e-mil: elen@cs.ucklnd.c.nz. Computer Science Deprtment, The University of Aucklnd, Privte Bg 9219, Aucklnd, New Zelnd nd Mthemtics Deprtment, Cornell University, Ithc, USA, e-mil: bmk@cs.ucklnd.c.nz. 1 Informlly, two sttes p, q of A re distinguishble if there is n experiment which mkes A rect (respond) differently on p nd q.

2 be described by n equivlence induced by suitble clss of Ehrenfeucht Frïsse-like gmes. Miniml utomt will be proven to be unique up to n isomorphism; this sitution differs from the clssicl theory of nondeterministic utomt (see for instnce, [1, 11, 13, 18]) but it prllels nd extends the theory of deterministic utomt developed in [2, 5]. While for the deterministic cse ll constructions mde use of utomt responses only, i.e., no informtion bout the internl mchinery ws necessry, for nondeterministic utomt we need the full internl mchinery. Here is brief review of the pper. Section 2 is devoted to bsic notions nd nottions. Section 3 introduces utomt trjectories nd responses. Section 4 introduces nd briefly discusses clss of Ehrenfeucht Frïsse-like gmes useful in modeling the nondeterministic utomton behviour. In Section 5 we review five unsuccessful ttempts in modeling the notion of stte indistinguishbility ; this discussion motivtes the introduction, in Section 6, of well-behved equivlence reltion which will be essentil for defining the notion of simultion nd for constructing the miniml nondeterministic utomton in Section 7. Finlly, we investigte the prtil ordering induced by utomt simultions. 2 Nottions We begin by introducing some nottions nd bsic definitions. If S is finite set, then S denotes the crdinlity of S. Let Σ be finite set (sometimes clled lphbet); the set Σ stnds for the set of ll finite words over Σ with the empty word denoted by λ. The length of string x is denoted by x. We fix two finite lphbets Σ nd O: Σ contins input symbols, nd O contins output symbols. A nondeterministic finite utomton over the lphbet Σ nd O is triple A =(S A, A,F A ), where S A is finite nonempty set of sttes, A is function from S A Σ to the set 2 S A of ll subsets of S A, clled the trnsition tble, F A is mpping from the set of sttes S A into the output lphbet O, clled output function. The bove definition does not include the so clled initil sttes which mkes our definition different from the clssicl one. In drwing grph representtions of utomt, we denote sttes by nd lbel them with symbols from the output lphbet. 2 The picture q ν σ p µ mens tht there is trnsition σ from q to p, tht is p A (q, σ), nd F A (q) = ν, F A (p) =µ. 2 Sometimes, we omit the nme of the stte. 2

3 In contrst with the fct tht miniml deterministic utomt (with initil sttes) ccepting the sme lnguge re isomorphic, for nondeterministic utomt (with initil sttes) there exist miniml non-isomorphic nondeterministic utomt A nd B which ccept the sme lnguge (for the clssicl theory of utomt see [1, 11, 13, 18]). We give n exmple. The grph representtion of A is in Figure 1; the output function is given by F A (s )=F A (s 1 ) = 1 nd the initil stte is s. The utomton ccepts the lnguge { n b m n, m }. b s s 1 Figure 1. In Figure 2 we hve n utomton B whose initil stte is p nd F B (p )=F B (p 1 )= 1; B ccepts the sme lnguge s A. b p b p 1 Figure 2. Both nondeterministic utomt re miniml but they re not isomorphic. Informlly, one cn sy even more: neither A nor B simulte ech other; they ccept the sme lnguge just by chnce. This type of negtive phenomenon does not occur under n pproprite definition of simultion for nondeterministic utomt with no initil sttes. 3 Trjectories nd Responses Let A = (S A, A,F A ) be nondeterministic utomton. There re severl wys to introduce the notion of response of A to n input sequence of signls. Tke w = σ 1...σ n Σ nd s S A.Atrjectory of A on s nd w is sequence s,s 1,...,s n of sttes such tht s i+1 A (s i,σ +1 ) for ll i n 1. A trjectory s,s 1,...,s n emits the output F A (s )F A (s 1 ) F A (s n ). The totl response, denoted by R A, is function which to ny (s, w) S A Σ ssigns the set R A (s, w) of ll outputs emitted by ll trjectories of A on s nd w. The finl response of A is function f A which to ny pir (s, w) S A Σ ssigns the subset of ll lst symbols occurring in words in R A (s, w). 3

4 These functions permit the identifiction of those sttes of A which give the sme response to the sme inputs. Indeed, we cn consider two equivlence reltions nd 1 defined s follows. We sy tht two sttes p nd q of S A re 1 equivlent if for ll w Σ, R A (p, w) =R A (q, w). Similrly, we sy tht two sttes p nd q of S A re equivlent if for ll w Σ, f A (p, w) =f A (q, w). It is cler tht if p 1 q, then p q. The exmple below shows tht in contrst to the deterministic cse (see [2]), 1 is not the sme s. Exmple 3.1 Consider the utomton A whose stte digrm is given in Figure 3. We hve p q nd p 1 q. p 1 1 q 1 1 Figure 3. Exmple 3.2 The utomton A in Figure 4 hs the following property: there exist two sttes p, q S A such tht p 1 q, but for ll p A (p, σ) nd q A (q, σ), we hve p 1 q. Indeed, it is not hrd to see tht for ll w Σ R A (p, w) =R A (q, w). It follows tht p 1 q. However, no p A (p, ) is 1 equivlent to ny q A (q, ). 4

5 p 1 q 1 Figure 4. Motivted by the phenomenon described in Exmple we will be interested in those equivlence reltions on S A which re well behved with respect to the trnsition tble A. Here is the pproprite definition. An equivlence reltion on S A is well behved if for ll p q (p, q S A ) nd for every σ Σ the following properties hold: 1. For every p A (p, σ) there is stte q A (q, σ) such tht p q. 2. For every q A (q, σ) there is stte p A (p, σ) such tht q p. A well-behved equivlence reltion should gurntee tht ny two equivlent sttes simulte ech other. Hving well behved equivlence reltion, one cn consider the fctor utomton A/ nd prove tht it is miniml. 4 4 Gme Responses The bove nlysis of totl nd finl responses suggests gme theoretic pproch in formlizing the notion of response. Informlly, the behviour of nondeterministic utomton A receiving n input w cn be thought s gme with two plyers: Plyer nd Plyer 1. A move of ny plyer consists of picking up stte of A. Plyer picks stte p. Plyer 1 tries to pick up stte q such tht the observer cnnot distinguish p nd q using responses coming from p nd q; Plyer tries to prove the opposite. For the ske of completeness we include some simple fcts bout finite gmes. Let T be finite tree, nd W be set of some pths from T. Nodes on even positions re positions of Plyer ; the remining nodes re positions of Plyer 1. A ply is 3 See lso Lemm 2.2 in [2]. 4 Exmple 3.2 shows tht the equivlence reltion 1 is not well behved, so 1 cnnot be used for constructing the miniml utomton. 5

6 finite sequence of nodes x y...x k y k such tht x is the root of T nd the sequence x y...x k y k ispthint. A gme is the pir (T,W). A strtegy for Plyer (Plyer 1) is function which mps every position x of Plyer (Plyer 1) to child (i.e., n immedite successor) of x. For instnce, Plyer cn follow strtegy g nd n initil ply ccording to this strtegy cn be: g(x )y g(y )y 1 g(y 1 )y 2 g(y 2 ), where x is the root of T. We sy tht Plyer 1 wins the gme (T,W) if there is strtegy g for Plyer 1 such tht every ply plyed following g belongs to W ; otherwise Plyer 1 looses. Fct 4.1 In the gme (T,W) one of the plyers wins. If Plyer 1 does not win this gme, then there is strtegy g for Plyer such tht every ply plyed following g does not belong to W. Proof. Let C W be the set of ll nodes in T which re the lst elements of the pths in W. We mrk elements of T s follows: Stge. Every element in C W is mrked. Stge i + 1. Consider node x. If x is position of Plyer, then x is mrked t this stge if ll children of x re mrked. Otherwise we do not mrk x t this stge (x my be mrked t lter stges). If x is position of Plyer 1, then x is mrked if some child of x is mrked. Otherwise x is not mrked t this stge. Clerly there is stge fter which no node will be mrked. Thus, there re two cses: Cse 1. If the root is mrked, then Plyer 1 wins. The winning strtegy for Plyer 1 is the following: if x is mrked nd is position for Plyer 1, then tke mrked child of x. Cse 2. If the root is not mrked, then Plyer wins. The winning strtegy for Plyer is the following: if x is n unmrked position for Plyer, then tke n unmrked child of x. From the proof of this fct we get the following: Corollry 4.2 Consider the gme (T,W). A strtegy g is winning strtegy for Plyer 1 if nd only if every ply ccording to g goes through mrked nodes. s 6

7 5 Unsuccessful Models Fix nondeterministic utomton A, two sttes p, q S A nd string w = σ 1...σ n Σ. We define finite gme G w (p, q), clled w response gme, with two plyers: Plyer nd Plyer 1. Plyer lwys moves first, nd Plyer 1 responds to ech move. A ply is sequence p 1 q 1 p 2 q 2...p k q k such tht the following conditions hold: 1. p = p 1, q = q 1, 2. q i+1 A (q i,σ i ), for ech 1 i k 1, p i+1 A (p i,σ i ). Thus every ply is sequence of sttes. The letters on even positions re clled positions of Plyer ; the others re positions of Plyer 1. Since w is finite, every ply in w response gme is finite. A strtegy for Plyer (Plyer 1) is function which mps the set of ll finite words of even (odd) length from S A to S A. Note tht since G w (p, q) is finite, every strtegy of this gme is function with finite domin, hence the number of strtegies is finite. If g is strtegy for Plyer 1, nd p 1 q 1 p 2 q 2...p k is ply plyed by Plyer 1 following g, then the next move of Plyer 1 is g(p 1 q 1 p 2 q 2...p k ). For exmple, the following is n initil segment of ply ccording to g: pqp 1 g(pqp 1 ) p 2 g(pqp 1 g(pqp 1 )p 2 ) p 3 g(pqp 1 g(pqp 1 )p 2 g(pqp 1 g(pqp 1 )p 2 )p 3 ). Similrly, Plyer cn follow strtegy in the gme G w (p, q). We sy tht Plyer 1 wins the ply p 1 q 1 p 2 q 2...p k q k if R A (p i,σ i...σ n )=R A (q i,σ i...σ n ), for ll 1 i k. Otherwise, Plyer wins. A plyer wins the gme if it hs strtegy g such tht the plyer wins every ply following g. Since G w (p, q) is finite gme, one of the plyers wins the gme, by Fct 4.1. We sy tht the sttes p nd q re 2 equivlent if for every w Σ, Plyer 1 wins the gmes G w (p, q) ndg w (q, p). The next result follows from the definition. Lemm 5.1 For ll sttes p, q, ifp 2 q, then p 1 q. Lemm 5.2 The reltion 2 is n equivlence reltion. Proof. It is cler tht the reltion is symmetric nd reflexive. Suppose tht p 2 q nd q 2 s. We need to show tht p 2 s, tht is Plyer 1 wins both gmes G w (p, s) nd G w (s, p). We explin how Plyer 1 wins the gme G w (p, s); by symmetry, one cn then see how Plyer 1 wins the other gme G w (s, p). Let g 1 nd g 2 be winning strtegies of Plyer 1 in gmes G w (p, q) ndg w (q, s), respectively. Then the winning strtegy g for Plyer 1 in the gme G w (p, s) cn be described by the following instructions: 7

8 First, think of ny move of Plyer s move in the gme G w (p, q). Secondly, using the strtegy g 1, respond to the move s you were in the gme G w (p, q). Thirdly, consider the response of Plyer 1 s move of Plyer in the gme G w (q, s). Finlly respond, using the strtegy g 2, to the move s you were in the gme G w (s, q). It is not hrd to see tht this strtegy g is winning strtegy for Plyer 1. Unfortuntely the equivlence reltion 2 is not well-behved. Exmple 5.3 The utomton A in Figure 5 hs the following property: there exist two sttes p 2 q, but for ll p A (p, σ) nd q A (q, σ), we hve p 2 q. Indeed, p 2 q, but for ll p A (p, σ) ndq A (q, σ), p 2 q. p 1 b q b 1 Figure 5. The bove exmple suggests modifiction of the gme G w (p, q). In the new gme, clled G(p, q, w), every ply is the sme s in G w (p, q), but we sy tht Plyer 1 strongly wins the ply p 1 q 1 p 2 q 2...p k q k if p i 1 q i, for ll 1 i k. Agin, since G(p, q, w) is finite gme, one of the plyers wins the gme. We sy tht the sttes p nd q re strongly 3 equivlent, ndwe denote this by 3, if for every w Σ, Plyer 1 strongly wins the gmes G(p, q, w) nd G(q, p, w). Lemm 5.4 For ll sttes p, q, ifp 3 q, then p 2 q, nd hence p 1 q. Agin, however, the negtive phenomenon occurs: Exmple 5.5 There is n utomton A such tht p 3 q for some p, q S A, but for ll p A (p, σ) nd ll q A (q, σ), p 3 q. 8

9 The sttes of A ccessible from p, respectively, q re given in Figure 6. 1 b 1 b b b 3 b p 1 b 1 b b b 1 1 b b b b q 1 b b 1 b b b Figure 6. 9

10 The bove nlysis shows tht we need to further refine the equivlence reltion 3. To this im we define two new equivlence reltions. For p, q S A nd w Σ, consider gin the gme G(p, q, w). A continution of this gme is ny gme G(p, q, wu), where u Σ. Clerly, if Plyer 1 wins G(p, q, wu), then he wins G(p, q, w) too. One of the min resons tht the equivlence reltions 3 nd 2 re not well behved is hidden in the following fct: In the gme G(p, q, w) Plyer 1 cn not predict future ctions of Plyer when new input u is inserted into A fter w. In other words, winning strtegy for Plyer 1 in the gme G(p, q, w) cn not lwys be extended to winning strtegy in ny continution of the gme. Thus we re led to sy tht Plyer 1 strtegiclly wins the gme G(p, q, w) if there is strtegy h for Plyer 1 in the gme G(p, q, w) such tht for ll u Σ the strtegy h cn be extended to winning strtegy of the gme G(p, q, wu). Clerly, if Plyer 1 strtegiclly wins the gme G(p, q, w), then he wins the gme G(p, q, w) itself. Now this definition llows us to consider n equivlence reltion 4 finer thn 3. We sy tht p nd q re 4 equivlent if for every w, Plyer 1 strtegiclly wins both gmes G(p, q, w) ndg(q, p, w). Thus, if p 4 q, then for every w there is winning strtegy g (g ) for Plyer 1 in the gme G(p, q, w) ((G(q, p, w)) such tht for ll u Σ the strtegy g (g ) cn be extended to winning strtegy in the gme G(p, q, wu) ((G(q, p, wu)). There is nother possibility to refine 3 by defining new gme, denoted by G(p, q, n), s follows: A ply is sequence p q p 1 q 1...p k q k of sttes such tht p = p, q = q, nd for every 1 i k 1 there re σ 1,σ 2 Σsuch tht p i+1 A (p i,σ 1 )ndq i+1 A (q i,σ 2 ), 1 k n. Thus, in this ply Plyer chooses p = p, Plyer 1 chooses q = q, Plyer responds by tking ny stte p 1,etc. We sy tht Plyer 1 wins the ply if for ll 1 i n nd σ Σ, p i+1 A (p i,σ)if nd only if q i+1 A (q i,σ), nd F A (p i )=F A (q i ). We sy tht Plyer 1 wins the gme G(p, q, n) if there is winning strtegy h for Plyer 1 in the gme G(p, q, n). If Plyer 1 wins the gme G(p, q, n), then clerly he wins the gme G(p, q, w), for ll w Σ, w n. Note tht if Plyer 1 wins the gme G(p, q, n + 1), then he wins the gme G(p, q, n) s well. Two sttes p nd q of A re 5 equivlent if for every n, Plyer 1 wins the gmes G(p, q, n) ndg(q, p, n). Now the following lemm is consequence of definitions nd Lemm 5.2. Lemm The reltions 4 nd 5 re equivlence reltions. 2. For ll sttes p, q nd i =4,5,ifp i q, then p 3 q. Agin, it turns out tht neither 4 nor 5 re well behved. We stte this fct without giving ny exmples nd turn our interest to the construction of well behved equivlence reltion. Fct 5.7 The equivlence reltions 4 nd 5 re not well behved. 1

11 6 A Well Behved Equivlence Reltion Let A nd B be two, not necessrily distinct, nondeterministic utomt. Tke sttes p S A nd q S B, nd fix positive integer n 1. We define gme Γ(p, q, n) between two plyers: Plyer nd Plyer 1. Agin, Plyer tries to prove tht outputs emitted by trjectories which begin in p re different from outputs emitted by trjectories originted in q. Plyer 1 tries to show the opposite. The difference from the previous gmes is tht Plyer (Plyer 1) is not restricted to consider computtions which begin from p (q) only. Plyer (Plyer 1) is llowed to pick up ny instnce of computtion which begins from q (p) swell. Here is description of ply. Every ply hs t most n stges. Ech stge begins with move of Plyer nd ends with response of Plyer 1. Stge. Plyer picks up either p or q. Plyer 1 responds by picking up the other stte. Stge k + 1 n.attheendofstgekwe hve two sequences p p 1...p k nd q q 1...q k where p = p nd q = q. Now Plyer chooses stte either from σ Σ A(p k,σ) or from σ Σ B(q k,σ). If Plyer chooses p k+1 from σ Σ A(p k,σ), then Plyer 1 responds by choosing stte q k+1 from σ Σ B(q k,σ). If Plyer chooses q k+1 from σ Σ A(q k,σ), then Plyer 1 responds by choosing stte p k+1 from σ Σ B(p k,σ). This ends description of stge k + 1 of ply. Let p p 1...p t nd q q 1...q t be sequences produced during ply. We sy tht Plyer 1 wins the ply if for ll <i t,σ Σ, we hve p i A (p i 1,σ)iffq i B (q i 1,σ)ndF A (p i )=F B (q i ). From the definition of the gme Γ(p, q, n) we hve the following lemm. Lemm 6.1 If plyer wins the gme Γ(p, q, n) then he wins the gme Γ(q, p, n). To formulte the next theorem we suppose tht in the gme Γ(p, q, n) the utomt A nd B coincide. We sy tht tht p is equivlent to q if Plyer 1 wins the gme Γ(p, q, n), for ll positive integers n. Theorem 6.2 The reltion is well-behved equivlence reltion on S A. Proof. The first prt of the theorem follows, with slight modifiction, from the proof of Lemm 5.2. Suppose tht p q nd q s. We need to show tht p s, tht is Plyer 1 wins the gme Γ(p, s, n) for every n. Let g 1 nd g 2 be winning strtegies for Plyer 1 in gmes Γ(p, q, n) ndγ(q, s, n), respectively. Then winning strtegy g for Plyer 1 in the gme Γ(p, s, n) cn be described s follows. Suppose tht t the end of stge k (k <n) of ply the plyers hve produced two sequences p p 1...p k 11

12 nd s s 1...s k where p = p nd s = s. If t stge k + 1 Plyer chooses stte p k+1 from σ Σ A(p k,σ), then Plyer 1 follows the instructions bellow: First, think of this move of Plyer s move in the gme Γ(p, q, n). Secondly, using the strtegy g 1, respond to the move s you were in the gme Γ(p, q, n). Thirdly, consider this response of Plyer 1 s move of Plyer in the gme Γ(q, s, n). Finlly respond, using the strtegy g 2, to the move s you were in the gme Γ(q, s, n). On the other hnd, if Plyer chooses stte s k+1 from σ Σ A(s k,σ), then Plyer 1 follows the instructions: First, think of this move of Plyer s move in the gme Γ(q, s, n). Secondly, using the strtegy g 2, respond to the move s you were in the gme Γ(q, s, n). Thirdly, consider this response of Plyer 1 s move of Plyer in the gme Γ(p, q, n). Finlly respond, using the strtegy g 1, to the move s you were in the gme Γ(p, q, n). In both cses the strtegy is clerly winning strtegy for Plyer 1. We prove the second prt. Suppose tht p is equivlent to q. We need to show tht for every σ Σ nd every p A (p, σ) there is q A (q, σ) such tht p is equivlent to q. Let q 1,...,q s be ll sttes belonging to A (q, σ). Suppose tht none of q i is equivlent to p. Then for every q i there is n n i such tht Plyer wins the gme Γ(p,q i,n i ). Let h i be strtegy for Plyer to win the gme Γ(p,q i,n i ). Then Plyer wins lso ny continution of the gme, Γ(p,q,n i +t),for every nturl number t. Let n be the mximl number mong ll n 1,..., n s nd consider the gme Γ(p, q, n). Suppose tht in this gme the first move of Plyer is p. If Plyer 1 responses by not tking stte from {q 1,...,q s }, then clerly Plyer 1 looses the gme. On the other hnd, if Plyer 1 chooses stte q i, then Plyer simply follows the strtegy h i. It is cler tht in this cse Plyer wins the gme Γ(p, q, n) which contrdicts the fct tht p q. 7 Simultions nd Minimlity Let A n B be nondeterministic utomt. We sy tht A is simulted by B, or equivlently, B simultes A, if there is mpping h : S A S B such tht for ll s S A, the sttes s nd h(s) re equivlent. We denote this fct by A B. 5 Thus, the function h in this definition mens tht Plyer 1 wins the gme Γ(p, h(p), n), for every n. Let A be nondeterministic utomton. We define the utomton M(A) sfollows: 1. The set of sttes S M(A) of M(A) is{[s] s S A }, where [s] ={q S A s q}. 2. For ll [q], [s] S M(A) nd σ Σ, [q] M(A) ([s],σ) if nd only if q A (s, σ). 5 Note tht the simultion reltion defined bove coincides with the simultions of deterministic utomt, in cse A nd B re deterministic; see [2]. 12

13 3. F M(A) ([s]) = F A (s). The next lemm, concerning the reltionship between A nd M(A), is n exct nlogue of the cse for deterministic utomt (see [2]). Lemm 7.1 The utomt A nd M(A) simulte ech other. Proof. We prove tht utomton A is simulted by M(A) vi the mpping s [s], for ll s S A. We need to show tht Plyer 1 hs strtegy to win the gme Γ(s, [s],n), for ech n. Suppose tht t the end of stge k (k <n) of ply the plyers hve produced two sequences s s 1...s k nd [p ][p 1 ]...[p k ] where s = s nd [p ]=[s]. By induction, we cn ssume tht p k s k. Suppose tht t stge k + 1 Plyer chooses s k+1 from σ Σ A(s k,σ). Since s k p k, by Theorem 6.2, there exists p k+1 σ Σ A(p k,σ) such tht s k+1 p k+1. Hence Plyer 1 picks up this p k+1. Suppose tht t stge k +1Plyerchooses[p k+1 ] from σ Σ M(A)([p k ],σ). Agin by the sme theorem Plyer 1 cn choose s k+1 such tht s k+1 [p k+1 ]. Similrly, one cn prove tht the utomton M(A) is simulted by A vi the mpping [s] min[s], where min[s] is the miniml element in [s] under some fixed liner ordering in S A. Sy tht two utomt A nd B re equivlent (nd denote this by A B) ifa B nd B A. Clerly, the reltion is n equivlence reltion. A nondeterministic utomton A is miniml if for every nondeterministic utomton B such tht A B one hs S A S B. Our gol is to prove tht ech clss [A] ={B A B}contins miniml utomton which is unique up to n isomorphism. We recll tht two utomt A nd B re isomorphic if there is bijective mpping h : S A S B such tht for ll s, p S A,σ Σ, p A (s, σ) if nd only if h(p) B (h(s),σ)ndf A (s)=f B (h(s)). Lemm 7.2 The utomton M(A) is miniml. Proof. The proof is similr to the deterministic cse. Suppose tht B is miniml. Let h : S M(A) S B be mpping such tht M(A) is simulted by B vi h. Thenhis one to one. Otherwise, there exist two sttes [p] [q]ins M(A) such tht h([p]) = h([q]). Hence p h(p), h(p) =h(q), nd h(q) q. It follows tht [p] [q], nd consequently, p q, i.e., [p] =[q]. This is contrdiction. Thus, S M(A) S B. In the lst step we show the unicity up to n isomorphism of the miniml utomton. Lemm 7.3 If B is miniml nd A B, then B is isomorphic to M(A). Proof. Suppose tht B is miniml. There exists mpping h : S M(A) S B such tht M(A) is simulted by B vi h. From the proof of Lemm 7.2 we see tht h must 13

14 be one to one mpping. Since the utomton B is miniml, h must be onto. Indeed, ssume by contrdiction tht there is mpping g : S B S M(A) such tht B is simulted by M(A) vi g nd g(p) =g(q), for some p, q S B. Hence p q. Since M(B) B, B cnnot miniml, contrdiction. Consequently, h is bijection from S M(A) to S B. We need to prove tht h is n isomorphism. It is cler tht F M(A) ([s]) = F B (h([s])), for ll s S A. Suppose tht [s] M(A) ([p],σ). We need to show tht h[s] B (h([p]),σ). Since [p] h([p]), there exists q B (h([p]),σ) such tht q [s]. Hence q h([s]) since h estblishes simultion. If q h([s]), then since q h([s]), we hve S M(B) < S B. This is gin contrdiction with the ssumption tht B is miniml. Hence q = h([s]) nd h([s]) B (h([p]),σ). The bove lemms prove the min theorem of this section. Theorem 7.4 For every nondeterministic utomton A, the utomton M(A) stisfies the following properties: 1) The utomt A nd M(A) simulte ech other. 2) The utomton M(A) is miniml. 3) The utomton M(A) is unique up to isomorphism. 8 Simultion s Prtil Ordering The gol of this section is to investigte the prtil ordering induced by, the simultion of nondeterministic utomt. Recll tht [A] ={B B A}. We sy tht[a]is simulted by B, nd denote this by [A] [B], if A B. In other words, the reltion nturlly induces prtil ordering in the clss K of ll equivlences clsses [A]. We dd to K the empty utomton E with mening tht E [A], for every utomton A. Thus, we hve prtilly ordered set K =(K, ) with the lest element E. In this section we investigte this prtilly ordered set nd give chrcteriztion of in terms of embeddings of miniml utomt. A morphism from n utomton A to n utomton B is mpping h : S A S B hving the following properties: 1. F A (s) =F B (h(s)), for ll s S A, 2. p A (s, σ) if nd only if h(p) B (h(s),σ), for ll p, s S A nd σ Σ, 3. for ll q B (h(s),σ), there is p A (s, σ) such tht q = h(p). If h is one to one, then A is embedded into B. The following lemm follows from the bove definition. Lemm 8.1 If there is morphism from A to B, then A B. 14

15 Proof. Indeed, suppose tht h estblishes morphism from A to B. We need to show tht Plyer 1 wins the gme Γ(p, h(p),n) for ech p S A nd positive integer n. Suppose thtttheendofstgek(k<n) of ply the plyers hve produced two sequences p p 1...p k nd s s 1...s k where p = p nd s = h(p). Suppose tht t stge k + 1 Plyer chooses stte p k+1 from σ Σ A(p k,σ). Then Plyer 1 chooses h(p k+1 ). Suppose tht t stge k + 1 Plyer chooses s k+1 from σ Σ A(s k,σ). Then since h is morphism there is p k+1 such tht h(p k+1 )=s k+1. Hence the response of Plyer 1 is simply p k+1. One cn see tht this is indeed winning strtegy for Plyer 1. The following result connects the ordering with the notion of embedding. Theorem 8.2 For ll [A], [B] K, M(B). [A] [B] if nd only if M(A) is embedded into Proof. In view of the previous Lemm 8.1, it is not hrd to check tht if M(A) is embedded into M(B), then [A] [B]. Suppose tht [A] [B]. Consider the miniml utomt M(A) ndm(b). There is mpping h : S M(A) S M(B) such tht M(A) is simulted by M(B) vi h. The function h must be injection. Otherwise, using stndrd resoning from the previous section we cn prove tht M(A) is not miniml. Similrly, one cn see tht h is n embedding. In fct the bove proof gives stronger result. Corollry 8.3 For ll A nd B if [A] [B], there is unique embedding of M(A) into M(B). Now we show some other lgebric properties of the prtilly ordered set K. Alower (upper) lttice is prtil ordered set in which every two elements hve supremum (infimum). A lttice is prtil ordered set which is both n upper nd lower lttice. Lemm 8.4 (K, ) is n upper lttice. Proof. Tke two clsses [A] nd[b] nd ssume tht S A nd S B re disjoint. Therefore we cn consider new utomton, denoted by A B, which is obtined by tking the union of the set of sttes, trnsition digrms, nd output functions of the utomt A nd B. Itisclertht[A] [A B]swells[B] [A B]. We wnt to show tht for ny [C] if[a] [C]nd[B] [C], then [A B] [C]. Indeed, suppose tht [A] [C] vi h 1 : S A S C nd [B] [C] vi h 2 : S B S C. The function h = h 1 h 2 is clerly well-defined nd one cn esily see tht [A B] [C] vi h. Lemm 8.5 (K, ) is lower lttice. 15

16 Proof. Tke two clsses [A] nd[b] nd ssume tht S A nd S B re disjoint. Consider the utomt M(A) ndm(b) s well s ll utomt C such tht C cn be embedded into M(A) ndm(b). The number of ll nonisomorphic utomt which cn be embedded into A nd B is finite. If this number is, then clerly E =[A] [B]. Let A 1,...A n be ll utomt embedded into A s well s into B with pirwise disjoint domins. Then it is not hrd to see tht the utomton A 1... A n, denoted by A B, hs the property tht for ll A i, A i A B. Moreover [A B] [A] nd [A B] [B]. It follows tht K is lower lttice. A covering of clss [A] is clss [B] such tht [A] [B], [A] [B], nd for ll [C] with [A] [C] [B], either [A] =[C]or[B]=[C]. Lemm 8.6 Suppose tht O > 1. Then every element of K hs infinitely mny coverings. Proof. Let σ Σ nd suppose tht, 1 O. For ech prime number p consider the utomton A p with the following properties: 1. S A hs exctly p number of sttes s 1,...,s p, 2. Ap (s i,σ)={s i+1 }, for ll i p 1, nd Ap (s p,σ)={s 1 }, 3. F Ap (s 2 )=...=F Ap (s p )=ndf Ap (s 1 )=1. It is not hrd to see tht if p p, then neither A p nor A p simulte ech other. It cn lso be checked tht A p = M(A p ), for ll p. Finlly, tke ny utomton A nd suppose tht S M(A) = n. Then for ll p>n,[a] [A p ] is covering of [A]. An idel of K is subset I Ksuch tht for ll [A], [B] Ithe following properties hold: 1. If [A] I nd [B] [A], then [B] I. 2. If [A], [B] I, then [A] [B] I. Lemm 8.7 The set K d = {[A] A is deterministic} is n idel of K. Proof. If[B] [A]nd[A] K d, then M(B) isembeddedintom(a). Hence M(B) is deterministic utomton, so [B] K d. If [A],[B] I, then clerly the disjoint union of A nd B is deterministic utomton. Hence [A] [B] belongs to K d. From the bove lemms we get the following: Theorem 8.8 The prtilly ordered set K is lttice ech element of which hs infinitely mny coverings. Moreover, the set K d is n idel of K. 16

17 Acknowledgment We thnk Dr. Mrjo Lipponen for her criticl comments nd suggestions. References [1] Bruer, W. Automtentheorie. Teubner, Stuttgrt, [2] Clude, C., Clude, E., nd Khoussinov, B. Deterministic utomt: simultion, universlity nd minimlity. Annls of Pure nd Applied Logic 9, 1-3 (1997), [3] Clude, C., Clude, E., Svozil, K., nd Yu, S. Physicl versus computtionl complementrity I. Interntionl Journl of Theoreticl Physics 36 (1997), [4] Clude, C. S., nd Lipponen, M. Computtionl complementrity nd sofic shifts. in X. Lin (ed.). Theory of Computing 98, Proceedings of the 4th Austrlsin Theory Symposium, CATS 98, Springer-Verlg, Singpore, 1998, 1998, [5] Clude, E., nd Lipponen, M. Deterministic incomplete utomt: simultion, universlity nd complementrity. Journl of Universl Computer Science 11 (1997), [6] Chitin, G. J. An improvement on theorem by E. F. Moore. IEEE Trnsctions on Electronic Computers EC-14 (1965), [7] Conwy, J. H. Regulr Algebr nd Finite Mchines. Chpmn nd Hll Ltd., London, [8] Dvurečenskij, A., Pulmnnová, S., nd Svozil, K. Prtition logics, ortholgebrs nd utomt. Helvetic Physic Act 68 (1995), [9] Finkelstein, D., nd Finkelstein, S. R. Computtionl complementrity. Interntionl Journl of Theoreticl Physics 22, 8 (1983), [1] Grib, A. A., nd Zptrin, R. R. Automt simulting quntum logics. Interntionl Journl of Theoreticl Physics 29, 2 (199), [11] Hopcroft, J. E., nd Ullmn, J. D. Introduction to Automt Theory, Lnguges, nd Computtion. Addison-Wesley, Reding, MA, [12] Moore, E. F. Gednken-experiments on sequentil mchines. In Automt Studies, C. E. Shnnon nd J. McCrthy, Eds. Princeton University Press, Princeton, [13] Slom, A. Computtion nd Automt. Cmbridge University Press, Cmbridge, [14] Schller, M., nd Svozil, K. Prtition logics of utomt. Il Nuovo Cimento 19B (1994),

18 [15] Schller, M., nd Svozil, K. Automton prtition logic versus quntum logic. Interntionl Journl of Theoreticl Physics 34, 8 (August 1995), [16] Schller, M., nd Svozil, K. Automton logic. Interntionl Journl of Theoreticl Physics 35, 5 (My 1996), [17] Svozil, K. Rndomness & Undecidbility in Physics. World Scientific, Singpore, [18] Yu, S. Regulr lnguges. In Hndbook of Forml Lnguges, Vol. 1-3, G. Rozenberg nd A. Slom, Eds. Springer-Verlg, Heidelberg, 1997,