Undecidable Problems About Timed Automata

Undecidable problems ab out timed auto mata Olivier Finkel Equip e de Logique Math´ ematique U.F.R. de Math´ ematiques, Univ ersit ´ e Paris 7 2 Place Jussi eu 75251 P aris cedex 05, F rance finkel@log ique.juss ieu.fr . Abstract. W e solve some decision problems fo r timed aut o mata which w ere raised by S. T ri pakis in [T ri04] and by E . Asarin in [Asa04]. In particular, w e sho w th at one cannot d ecide whether a given timed au- tomaton is determinizable or w hether the complemen t of a timed regular language is timed regular. W e sho w that the problem of th e minimization of the num ber of clocks of a timed automato n is undecidable. It is also undecidable whether th e shuffle of t w o timed regula r languages is timed regular. W e show that in t h e case of timed B¨ uchi automata accepting infinite timed w ords some of these problems are Π 1 1 -hard, hence highly undecidable (located b eyo nd the arithmetical hierarc h y). 1 Keywords: Timed automata; timed B ¨ uc hi automata; timed regular ( ω )-languages; decision problems; universalit y problem; determinizabilit y; complementa bilit y; shuffle operation; minimization of the num ber of cloc ks. 1 In tro duction R. Alur and D. Dill in troduced in [AD94] the notion of timed automata reading timed words. Since then the theo ry of timed automata has b een muc h studied and used for s p ecification and v erification of timed sy stems. In a recent pap er, E. Asarin raised a series of questio ns ab out the theoretical foundations of timed automata and timed languages which were s till ope n and wrote: “I believe that getting answ ers to them would substan tially impro ve our understanding of the area” of timed systems, [Asa04]. Some o f these q ue s tions concern decision pro blems “` a la [T ri04]”. F or instance : “Is it p ossible, given a timed automaton A , to decide whether it is equiv alen t to a deterministic one ?” . S. tripak is show ed in [T ri04] that there is no algorithm which , given a timed au- tomaton A , de c ides whether it is equiv alen t to a deterministic one, and if this is the cas e giv es an equiv alen t deterministic a utomaton B . But the ab o v e ques tion 1 P art of the results stated in th is p aper were presented very recen tly in the Bulletin of th e EA TCS [Fin05 ,Fin06]. of the decidability of the determinizability alone (where we do not require the construction of the witnes s B ) was still op e n. W e give in th is pap er the answer to this q uestion and to several other ones of [T ri04,Asa04]. In particular, w e sho w that one cannot decide whether a giv en timed automa ton is determinizable or whether the complement of a timed reg- ular languag e is timed re gular. W e study a lso the corresp onding problems but with “bo unded re s ources” stated in [T ri04]. F or that purp ose w e use a metho d which is very similar to that one used in [Fin03b] to pro v e undecidability results about infinitary rational r elations, re- ducing the universality problem, whic h is undecidable, to some other decision problems. W e study also the problem of the minimization of the num ber of clo c ks of a timed automaton, showing that one canno t decide, for a given timed a ut omaton A with n clo c ks, n ≥ 2, whether there is an equiv a len t timed automaton B with at most n − 1 clo c ks. The questio n of the closure of the class of timed regula r langua ges under sh uffle was also r aised by E. Asarin in [Asa04]. C. Dima proved in [Dim05] that timed regular express ions with shu ffle c haracter iz e timed langua ges a c cepted b y stop- watc h automata. This implies that the class o f timed reg ular la nguages is no t closed under shuffle. W e pr o v ed this result indep enden tly in [Fin06]. W e rec a ll the pro of here, g iving a simple example of tw o timed reg ular langua ges who se shuffle is not timed r egular. Next w e use this example to prov e that one ca n not decide whether the shuffle of t w o given timed regular lang uages is timed regular . W e extend also the previous undecidability res ults to the ca se of timed B ¨ uchi au- tomata a ccepting infinite timed w ords. In this case ma n y problems are Π 1 1 -hard, hence highly undecidable (lo cated b eyond the arithmetical hiera rc h y), be cause the universality pr oblem for timed B¨ uc hi automata, whic h is itse lf Π 1 1 -hard, [AD94], can be reduced to these other dec ision pro blems. W e men tion that part of the results stated in this paper were presen ted very recently in the Bulletin of the E A TCS [Fin05,Fin06]. The pap er is orga nized as follows. W e r ecall usual n otations in Sectio n 2 . The undecidability of determinizability or r egular c omplemen tabilit y for timed regu- lar lang ua ges is prov ed in Section 3. The problem of minimization o f the n um be r of clo cks is studied in Sectio n 4. Res ults ab out the sh uffle op eration are stated in Section 5. Finally we extend in Section 6 so me undecidability r esults to the case of timed B¨ uc hi auto ma ta. 2 Notations W e assume the reader to b e familiar with the basic theor y of timed languages and timed automata (T A) [AD94]. The s e t of p ositiv e reals will be denoted R . A (finite length) timed w ord over a finite alphab et Σ is of the for m t 1 .a 1 .t 2 .a 2 . . . t n .a n , where, for all integers i ∈ [1 , n ], t i ∈ R and a i ∈ Σ . It ma y b e s een a s a time-event se quenc e , where the t i ∈ R repre sen t time lapses betw een events a nd the letters a i ∈ Σ r epresen t even ts. The set of all (finite length) timed words over a finite a lpha bet Σ is the set ( R × Σ ) ⋆ . A timed language is a subset o f ( R × Σ ) ⋆ . The complemen t ( in ( R × Σ ) ⋆ ) of a timed lang ua ge L ⊆ ( R × Σ ) ⋆ is ( R × Σ ) ⋆ − L denoted L c . W e consider a basic mo del of timed a ut omaton, as introduced in [AD94]. A timed automato n A has a finite set of states and a finite set of transitio ns. Each transition is lab elled with a letter of a finite input alphab et Σ . W e assume that each tra ns ition of A has a set of clocks to res e t to zer o and only di agonal-fr e e clo c k gua rd [AD94]. A timed automa to n A is said to b e deterministic iff it satisfies the tw o fo llo wing requirements: (a) A has o nly one start state, and (b) if there are multiple transitions starting at the same state with the sa me lab el, then their clo c k co nstrain ts are mutually exclusiv e. Then a deterministic timed automaton A has at most one run o n a g iv en timed word [AD94]. As usual, we denote by L ( A ) the timed langua ge accepted (b y final states) by the timed automaton A . A timed language L ⊆ ( R × Σ ) ⋆ is said to b e timed regular iff there is a timed automaton A suc h tha t L = L ( A ). An infinite timed word over a finite alphab et Σ is of the form t 1 .a 1 .t 2 .a 2 .t 3 .a 3 . . . , where, for all integers i ≥ 1, t i ∈ R a nd a i ∈ Σ . It ma y b e seen as an infinite time-event se quenc e . The s et of all infinite timed words ov e r Σ is the set ( R× Σ ) ω . A timed ω -languag e is a subs e t of ( R × Σ ) ω . The complement ( in ( R × Σ ) ω ) of a timed ω -lang uage L ⊆ ( R × Σ ) ω is ( R × Σ ) ω − L denoted L c . W e cons ider a basic mode l of timed B¨ uch i automaton, (TBA), as introduced in [AD94]. W e a ssume, as in the case o f T A accepting finite timed w o rds, that eac h transition of A has a set of clo c ks to re s et to zer o and only diagonal-fr e e clock guard [AD94]. The timed ω -languag e accepted b y the timed B ¨ uc hi automa to n A is denoted L ω ( A ). A timed languag e L ⊆ ( R × Σ ) ω is sa id to b e timed ω -regular iff there is a timed B ¨ uc hi automaton A suc h that L = L ω ( A ). 3 Complemen tabilit y and det er minizability W e first state the undecida bilit y of determiniza bilit y or regular co mplemen tabil- it y for timed regular la nguages. Theorem 1. It is unde cidable to determine, for a given T A A , whether 1. L ( A ) is ac c epte d by a deterministic T A. 2. L ( A ) c is ac c epte d by a T A. Pro of. It is well known that the clas s of timed regular la nguages is not clo sed under co mplemen tation. Let Σ b e a finite alpha bet and let a ∈ Σ . Let A b e the set of timed w or ds of the form t 1 .a.t 2 .a . . . t n .a , where, for all in tegers i ∈ [1 , n ], t i ∈ R and there is a pair o f integers ( i, j ) suc h that i, j ∈ [1 , n ], i < j , and t i +1 + t i +2 + . . . + t j = 1. The timed languag e A is formed by timed words containing only letters a a nd such that there is a pa ir o f a ’s which are separated by a time distance 1. The timed language A is r egular but its complemen t can not b e accepted b y a ny timed automaton b ecause suc h an automaton should hav e an unbo unded n umber o f clo cks to c heck that no pair of a ’s is separated by a time distance 1, [AD94 ]. W e shall use the undecidability of the universality pr oblem for timed reg ula r languages : one cannot dec ide, for a given timed automa to n A with input alpha bet Σ , whether L ( A ) = ( R × Σ ) ⋆ , [AD94]. Let c b e an additiona l letter not in Σ . F or a given timed regula r language L ⊆ ( R × Σ ) ⋆ , w e are going to construct a nother timed language L o ver t he alphab et Γ = Σ ∪ { c } defined as the union of the follo wing three languages . – L 1 = L. ( R × { c } ) . ( R × Σ ) ⋆ – L 2 is the set of timed words ov er Γ ha ving no c ’s or having a t least t wo c ’s. – L 3 = ( R × Σ ) ⋆ . ( R × { c } ) .A , wher e A is th e ab o ve defined timed regular language ov er the a lphabet Σ . The timed language L is regular b ecause L a nd A a re re gular timed lang ua ges. There are no w tw o cases. (1) First case. L = ( R × Σ ) ⋆ . Then L = ( R × ( Σ ∪ { c } )) ⋆ . Ther efore L has the minim um p ossible complexity . L is of course ac cepted by a deterministic timed automato n (without any clo ck). Mo reo ver its co mplement L c is empty th us it is also accepted b y a det erministic timed automaton (without an y clo c k). (2) Second case. L is strictly included in to ( R× Σ ) ⋆ . Then there is a timed w ord u = t 1 .a 1 .t 2 .a 2 . . . t n .a n ∈ ( R × Σ ) ⋆ which do es not b elong to L . Co nsider now a timed w o rd x ∈ ( R × Σ ) ⋆ . It holds that u. 1 .c .x ∈ L iff x ∈ A . Then we hav e also : u. 1 .c.x ∈ L c iff x ∈ A c . W e are going to show that L c is not timed regular. Assume on the con trar y that there is a timed automato n A such that L c = L ( A ). There a re only finitely man y p ossible global states (including the clock v alues ) of A after the reading of the initial segment u . 1 .c . It is clearly not p ossible that the timed automa ton A , from these glo bal s tates, accept all timed words in A c and only these ones, for the sa me reasons whic h imply that A c is not timed regular . Th us L c is not timed r egular. This implies that L is not ac c epted by any determinis tic timed automaton b ecause the class of determinis tic reg ular timed languages is closed under complement. In the first case L is accepted b y a deterministic timed automaton and L c is timed regular . In the second case L is not accepted by a n y deterministic timed automaton and L c is not timed reg ular. But o ne cannot decide which case holds bec ause of the undecidability of the univ ersality pr oblem for timed r egular lan- guages.  Below T A ( n, K ) denotes the c lass o f timed automata having at most n clocks and wher e constants are at most K . In [T ri04], T ripak is stated the following problems whic h are similar to the ab o ve ones but with “b ounded res o urces”. Problem 10 of [T ri0 4 ]. Given a T A A and non-negative integers n, K , do es there exist a T A B ∈ T A ( n, K ) s uc h that L ( B ) c = L ( A ) ? If so, construct suc h a B . Problem 11 of [T ri0 4 ]. Given a T A A and non-negative integers n, K , do es there exist a deterministic T A B ∈ T A ( n, K ) such that L ( B ) = L ( A ) ? If s o , c onstruct such a B . T r ipakis s ho wed that these pr oblems are no t a lg orithmically solv able. He asked also whether thes e b ounded-resource v er sions of previous problems remain un- decidable if w e do not require the construction of the witness B , i.e. if w e omit the sentence “If so construct such a B ” in the statement of Problems 1 0 and 11. It is easy to see, from the pro of of preceding Theore m, that this is ac tua lly the case bec ause we hav e seen that, in the firs t case, L and L c are accepted by deterministic timed automata without any clo ck . 4 Minimization of t he n um b er of clo c ks The following problem w a s shown to b e undecidable b y S. T ripa kis in [T r i04]. Problem 5 of [T r i04]. Giv en a T A A with n clo c ks, do es there exists a T A B with n − 1 clo c ks, suc h that L ( B ) = L ( A ) ? If so, construct suc h a B . The co rrespo nding decision problem, where w e require o nly a Y es / No ans wer but no witness in the case of a positive answ e r , was left op en in [T ri04]. Using a v ery similar reasoning a s in the preceding section, w e can pro ve that this problem is also undecidable. Theorem 2. L et n ≥ 2 b e a p ositive inte ger. It is u n d e cidable to de termine, for a given T A A with n clo cks, whether ther e exists a T A B with n − 1 clo cks, such that L ( B ) = L ( A ) . Pro of. Le t Σ b e a finite alpha bet and let a ∈ Σ . Le t n ≥ 2 be a p ositive int e g er, a nd A n be the set o f timed w or ds of the form t 1 .a.t 2 .a . . . t k .a , where, for all integers i ∈ [1 , k ], t i ∈ R and there a re n pairs o f integers ( i, j ) such that i, j ∈ [1 , k ], i < j , a nd t i +1 + t i +2 + . . . + t j = 1. The timed langua ge A n is formed by timed words c o n taining o nly letters a and such that there are n pa irs of a ’s whic h are separated by a time distance 1. A n is a timed reg ula r languag e but it can not be accepted by any timed automaton with less than n clo c k s , see [HKW95]. Let c b e an additiona l letter not in Σ . F or a given timed regula r language L ⊆ ( R × Σ ) ⋆ accepted by a T A with at mo st n clocks, w e construct another timed languag e V n ov er the alphab et Γ = Σ ∪ { c } defined as the union of the following three languag es. – V n, 1 = L . ( R × { c } ) . ( R × Σ ) ⋆ – V n, 2 is the set of timed w ords over Γ having no c ’s or having at leas t tw o c ’s. – V n, 3 = ( R × Σ ) ⋆ . ( R × { c } ) .A n . The timed language V n is regular b ecause L and A n are regular timed languag es. Moreov er it is ea sy to see that V n is accepted by a T A with at most n c lo cks, bec ause L a nd A n are acc e pted by timed automata with at most n clo cks. There are now tw o cases. (1) First case. L = ( R × Σ ) ⋆ . Then V n = ( R × ( Σ ∪ { c } )) ⋆ , th us V n is ac c e pt ed by a (deterministic) timed automaton without any clo ck . (2) Second case. L is strictly included in to ( R× Σ ) ⋆ . Then there is a timed w ord u = t 1 .a 1 .t 2 .a 2 . . . t k .a k ∈ ( R × Σ ) ⋆ which do es not belong to L . Co ns ider now a timed word x ∈ ( R × Σ ) ⋆ . It holds tha t u. 1 .c.x ∈ V n iff x ∈ A n . T owards a contradiction, assume that V n is a c cepted b y a timed automaton B with at most n − 1 clo cks. There are only finitely man y possible global states (including the clo ck v alues) of B a fter the reading of the initial se gmen t u. 1 .c . It is clea rly no t p ossible that the timed automa ton B , from these globa l states, accept all timed w ords in A n and only these ones, b ecause it has less than n clocks. But one cannot decide whic h ca se holds b ecause o f the undecidability o f the universalit y problem for tim ed regular languag es acce pt ed by timed automata with n clocks, wher e n ≥ 2.  Remark 3. F or time d automata with only one clo ck, the inclusion pr oblem, henc e also the un i versality pr oblem, have r e c ently b e en shown to b e de cidable by J. Ouaknine and J. Worr el l [OW04]. Then the ab ove metho d c an not b e applie d. It is e asy t o se e that it is de cidable whether a time d r e gular language ac c epte d by a time d automaton with only one clo ck is also ac c epte d by a time d automaton without any clo ck. 5 Sh uffle op eration It is well known that the clas s of timed regular language s is closed under union, int e r section, but not under co mplemen tation. Another usual op eration is the shuffle o peration. Recall that the shuffle x ⋊ ⋉ y of tw o elemen ts x and y of a monoid M is the set o f all pro ducts of the form x 1 · y 1 · x 2 · y 2 · · · x n · y n where x = x 1 · x 2 · · · x n and y = y 1 · y 2 · · · y n . This op eration ca n naturally b e extended to subsets o f M by setting, for R 1 , R 2 ⊆ M , R 1 ⋊ ⋉ R 2 = { x ⋊ ⋉ y | x ∈ R 1 and y ∈ R 2 } . W e kno w that the class of reg ular (un timed) languages is clo sed under shuffle. The questio n of the closure of the class of timed regula r langua ges under sh uffle was raise d by E. Asarin in [Asa 04 ]. C. Dima pr o ved in [Dim05] that timed reg ular expressions w ith shuffle characterize timed languag es accepted b y stopw a tc h automata. This implies that the class of timed regular languages is not closed under sh uffle. W e prov ed this result indep enden tly in [Fin06]. W e are g oing to reprov e this here, giving a simple ex ample of tw o timed regula r languages who se sh uffle is not timed r egular. Next we shall use this example to prove that one canno t decide whether the s h uffle of t wo given timed regula r languages is timed r egular. Theorem 4. The shuffle of time d r e gu la r language s is not alw ays time d r e gular. Pro of. Let a, b b e tw o different letters a nd Σ = { a, b } . Let R 1 be the langua ge of timed w ords ov er Σ of the form t 1 · a · 1 · a · t 2 · a for some positive r eals t 1 and t 2 such that t 1 + 1 + t 2 = 2 , i.e. t 1 + t 2 = 1. It is c lear that R 1 is a timed r egular lang uage of finite timed words. Remark. As remarked in [AD94, page 217], a timed automaton can compare delays with co nstan ts, but it cannot remember delays. If w e would lik e a timed automaton to be able to co mpare delays, we should add clock constra in ts of the form x + y ≤ x ′ + y ′ for some clock v alues x, y , x ′ , y ′ . But t his w ould greatly increase the expressive p ow er of automata: the languages accepted b y such au- tomata are not alwa ys timed reg ular, and if we allow the addition primitiv e in the syntax of clo c k constraints, then the emptiness problem fo r timed automata would b e undecidable [AD 94, page 217]. Notice that the abov e languag e R 1 is timed regular because a timed automaton B reading a word o f the form t 1 · a · 1 · a · t 2 · a , for some p ositive reals t 1 and t 2 , ca n co mpa re the delays t 1 and t 2 in or der to c he ck that t 1 + t 2 = 1. This is due to the fact that the delay b et ween the t wo first o ccurrences of the event a is c onstant equal to 1. Using the shuffle oper ation we sha ll co nstruct a languag e R 1 ⋊ ⋉ R 2 , for a reg ular timed languag e R 2 . Informally sp eaking, this will “inser t a v a riable delay” b e- t ween the tw o first o ccurrences of the even t a a nd the re s ulting languag e R 1 ⋊ ⋉ R 2 will not be timed regular . W e now giv e the details of this constr uction. Let R 2 be the langua ge of timed w ords ov er Σ of the form 1 · b · s · b for some positive r eal s . The language R 2 is of course als o a timed regular language. W e ar e going to pro ve that R 1 ⋊ ⋉ R 2 is not timed r egular. T owards a con tra diction, a ssume that R 1 ⋊ ⋉ R 2 is timed reg ular. Let R 3 be the set of timed words ov er Σ of the for m t 1 · a · 1 · b · s · b · 1 · a · t 2 · a for some pos itive reals t 1 , s, t 2 . It is clear that R 3 is timed regular. On the other hand the class of timed re gular language s is closed under in ter section thus the timed language ( R 1 ⋊ ⋉ R 2 ) ∩ R 3 would b e also timed regular . But this languag e is simply the set of timed w or ds of the form t 1 · a · 1 · b · s · b · 1 · a · t 2 · a , for some po sitiv e reals t 1 , s, t 2 such that t 1 + t 2 = 1. Assume that this timed la nguage is accepted b y a timed automaton A . Consider now the reading by A of a w or d of the fo rm t 1 · a · 1 · b · s · b · 1 · a · t 2 · a , for some positive r eals t 1 , s, t 2 . After reading the initial segment t 1 · a · 1 · b · s · b · 1 · a the v alue of an y clo c k of A can only b e t 1 + s + 2, 2 + s , 1 + s , or 1. If the clo ck v alue o f a clock C has b een at so me time r eset to zero, its v alue may be 2 + s , 1 + s , or 1 . So the v alue t 1 is no t sto red in the clock v alue and this clo c k can not be used to compare t 1 and t 2 in order to c heck that t 1 + t 2 = 1 . On the other ha nd if the clo c k v alue of a clo c k C has not b een at some time r eset to ze r o, then, a fter reading t 1 · a · 1 · b · s · b · 1 · a , its v alue will b e t 1 + s + 2 . This must hold for uncoun tably many v alues of the real s , and aga in the v alue t 1 + s + 2 can not b e used to accept, from the globa l state of A a fter reading the initial segment t 1 · a · 1 · b · s · b · 1 · a , only the word t 2 · a for t 2 = 1 − t 1 . This implies that ( R 1 ⋊ ⋉ R 2 ) ∩ R 3 hence a lso ( R 1 ⋊ ⋉ R 2 ) ar e no t timed reg ula r.  W e can now state the following res ult: Theorem 5. It is unde cidable t o determine whether the shuffle of two given time d r e gu la r languages is time d r e gular. Pro of. W e shall use ag ain the undecidability of the universality pr oblem for timed reg- ular la ng uages: one ca nnot de c ide , for a giv e n timed automaton A with input alphab et Σ , whether L ( A ) = ( R × Σ ) ⋆ . Let Σ = { a, b } , and c b e a n additional letter not in Σ . F or a g iv en timed regular language L ⊆ ( R × Σ ) ⋆ , we a re going firstly to construct a nother timed language L o ver the alphab et Γ = Σ ∪ { c } . The language L is defined as the union of the following three la nguages. – L 1 = L. ( R × { c } ) . ( R × Σ ) ⋆ – L 2 is the set of timed words ov er Γ ha ving no c ’s or having a t least t wo c ’s. – L 3 = ( R × Σ ) ⋆ . 1 .c.R 1 , where R 1 is the ab o ve defined timed regula r lang uage ov er the alphabet Σ . The timed lang uage L is regular be cause L and R 1 are re gular timed la nguages. Consider now the languag e L ⋊ ⋉ R 2 , where R 2 is the ab o ve defined regular timed language. There are no w tw o cases. (1) First case. L = ( R × Σ ) ⋆ . Then L = ( R × ( Σ ∪ { c } )) ⋆ and L ⋊ ⋉ R 2 = ( R × ( Σ ∪ { c } )) ⋆ . Th us L ⋊ ⋉ R 2 is timed regular. (2) Second case. L is strictly included in to ( R × Σ ) ⋆ . T owards a contradiction, a ssume that L ⋊ ⋉ R 2 is timed regular. Then the timed language L 4 = ( L ⋊ ⋉ R 2 ) ∩ [( R × Σ ) ⋆ . 1 .c.R 3 ], wher e R 3 is the ab o ve defined timed reg ular lang uage, would b e also timed regular b ecause it would be the intersection of tw o timed regula r languages. On the o ther hand L is strictly included into ( R × Σ ) ⋆ th us there is a timed word u = t 1 .a 1 .t 2 .a 2 . . . t n .a n ∈ ( R × Σ ) ⋆ which do es not b elong to L . Consider now a timed w or d x ∈ ( R × Σ ) ⋆ . It holds that u . 1 .c.x ∈ L 4 iff x ∈ ( R 1 ⋊ ⋉ R 2 ) ∩ R 3 . W e are going to show now that L 4 is not timed regular. Assume on the contrary that there is a timed automaton A such that L 4 = L ( A ) . There are only finitely many p ossible glo bal states (including the clo ck v a lues) of A a ft er the rea ding of the initial seg men t u. 1 .c . It is clearly not p ossible tha t the timed automaton A , from these global states, a ccept all timed w or ds in ( R 1 ⋊ ⋉ R 2 ) ∩ R 3 and o nly these ones, for the same reaso ns which imply that ( R 1 ⋊ ⋉ R 2 ) ∩ R 3 is not timed r egular. Thus L 4 is not timed r egular and this implies that L ⋊ ⋉ R 2 is not timed r egular. In the first case L ⋊ ⋉ R 2 is timed re gular. In the seco nd ca se L ⋊ ⋉ R 2 is not timed regular . But one cannot decide whic h ca se holds b ecause of t he undecidability of the univ er salit y problem for timed regula r language s .  W e can also study the corresp onding problems with “b ounded r esources”: Problem 1. Given tw o timed automata A and B and non-negative integers n, K , do es there exist a T A C ∈ T A ( n, K ) such that L ( C ) = L ( A ) ⋊ ⋉ L ( B ) ? Problem 2 . Giv en tw o timed automata A and B and an integer n ≥ 1, do es there exist a T A C with less than n clocks suc h that L ( C ) = L ( A ) ⋊ ⋉ L ( B ) ? Problem 3. Given tw o timed automata A and B , do es ther e exist a de ter ministic T A C suc h that L ( C ) = L ( A ) ⋊ ⋉ L ( B ) ? F r om the pro of of ab ove Theorem 5, it is e asy to see that these problems are a ls o undecidable. Indeed in the first case L ⋊ ⋉ R 2 was accepted b y a deterministic timed automaton without any clo cks. And in the second case L ⋊ ⋉ R 2 was not accepted b y any timed a uto maton. E. Asar in, P . Ca rpi, and O. Maler hav e pr o ved in [A CM02] that the formalism of timed regular expressions (with in ter section and re na ming) has the same ex- pressive pow er than timed automata. C. Dima prov e d in [Dim05] that timed regular express ions with shu ffle c haracter iz e timed langua ges a c cepted b y s top- watc h automata. W e refer the reader to [Dim05] for the definition of s top watc h automata. Dima show ed that, from tw o timed a uto mata A and B , one can construct a stopw atch automaton C suc h that L ( C ) = L ( A ) ⋊ ⋉ L ( B ). Th us we can infer the following co r ollaries from the above results. Notice that in [ACM02,Dim05] the a uthors consider automa ta w ith epsilon- transitions while in this pa per we have only consider e d timed automata without epsilon-trans itio ns, althoug h we think that many results co uld be extended to the cas e of automata with e psilon-transitions. So in the statemen t of the follow- ing co rollaries we c onsider s top watc h automata with epsilon-tra nsitions but only timed automata without epsilon-transitions. Corollary 6. On e c ann o t de cide, for a given st op watch automaton A , whether ther e exists a time d automaton B (r esp e ctively, a deterministic time d automaton B ) such t h at L ( A ) = L ( B ) . Corollary 7. On e c annot de cide, for a given s top watch automaton A and non- ne gative inte gers n, K , whe t he r ther e exists a time d automaton B ∈ T A ( n, K ) such that L ( A ) = L ( B ) . Corollary 8. On e c annot de cide, for a given stopwatch automaton A and an inte ger n ≥ 1 , whether ther e exists a t i me d aut omaton B with less than n clo cks such that L ( A ) = L ( B ) . 6 Timed B ¨ uc hi automata The previous undecidabilit y res ults can be extended to the case of timed B ¨ uchi automata accepting infinite timed words. Moreov er in this case many pr oblems are hig hly undecidable ( Π 1 1 -hard) b ecause the universality problem for timed B ¨ uc hi automata, which is its e lf Π 1 1 -hard, [AD94], can b e reduced to these prob- lems. F o r mor e informa tion ab out the a nalytical hie r arc hy (co n taining in particular the class Π 1 1 ) see the textb o o k [Rog6 7]. W e now consider first the problem of determiniza bility or regular co mplementabil- it y for timed regular ω -la ng uages. Theorem 9. The fol lo wing p ro blems ar e Π 1 1 -har d. F or a given TBA A , determine whether : 1. L ω ( A ) is ac c epte d by a deterministic TBA. 2. L ω ( A ) c is ac c epte d by a TBA. Pro of. Let Σ b e a finite alphab et and let a ∈ Σ . Let, as in Section 3, A be the set of timed words con taining o nly letter s a a nd suc h that there is a pair o f a ’s which ar e s eparated by a time distance 1. The timed languag e A is reg ula r but its complement is not timed regular [AD94]. W e sha ll use the Π 1 1 -hardness of the universalit y problem fo r tim e d regula r ω - languages : Let c b e an additional letter not in Σ . F or a given timed regular ω -language L ⊆ ( R × Σ ) ω , we can construct ano ther timed la nguage L o ver the alphabet Γ = Σ ∪ { c } defined as the union of the following three lang uages. – L 1 = A. ( R × { c } ) . ( R × Σ ) ω , where A is t he above defined timed regular language ov er the a lphabet Σ . – L 2 is the set of infinite timed w o r ds ov er Γ having no c ’s or ha v ing at least t wo c ’s. – L 3 = ( R × Σ ) ⋆ . ( R × { c } ) .L . The timed ω - language L is regula r b ecause L is a reg ula r timed ω - la nguage and A is a r egular timed langua g e. There are now t wo ca ses. (1) First case. L = ( R × Σ ) ω . Then L = ( R × ( Σ ∪ { c } )) ω . Therefore L has the minimum po s sible complexity and it is accepted by a deterministic TBA (without any clo ck). Moreo ver its complement L c is empt y thus it is also accepted b y a deterministic TBA (without an y clock). (2) Second ca s e. L is str ictly included into ( R × Σ ) ω , i.e. L c is non-empty . It is then easy to see that : L c = A c . ( R × { c } ) .L c where L c = ( R × Γ ) ω − L , A c = ( R × Σ ) ⋆ − A , and L c = ( R × Σ ) ω − L . W e a re going to show that L c is no t timed ω -regula r . Assume on the contrary that there is a TB A A such that L c = L ω ( A ). Consider the reading of a timed ω -w o rd of the form x. 1 .c.u , where x ∈ ( R × Σ ) ⋆ and u ∈ ( R × Σ ) ω , by the TBA A . W he n reading the init ial seg ment x. 1 .c , the TBA A has t o chec k that x ∈ A c , i.e. that no pair of a ’s in x is separated b y a time distance 1; t his is clear ly not p ossible for the same reasons which imply that A c is not timed r egular (see ab o ve Section 3). Th us L c is not timed ω -regular. This implies that L is not accepted by a n y deterministic TBA be cause the class of deter ministic regula r timed ω -langua ges is clo sed under co mplemen t, [AD94]. In the first case L is accepted b y a deterministic TBA and L c is timed ω -regular. In the second case L is not accepted by any de ter ministic TBA and L c is not timed ω -regular. This ends the pro of b ecause the univ ers a lit y problem for timed B¨ uchi automa ta is Π 1 1 -hard, [AD94].  As in the case of T A rea ding finite leng th timed words, we can consider the corres p onding pr oblems with “b ounded r esources”. Below T B A ( n, K ) deno tes the class o f timed B¨ uchi automata having a t most n clo c ks, where constants are at most K . Problem A. Given a TBA A and non-neg ativ e in tegers n, K , does there exist a TBA B ∈ T B A ( n, K ) such that L ω ( B ) c = L ω ( A ) ? Problem B. Given a TBA A a nd non-negative integers n, K , do es there exist a deterministic TBA B ∈ T B A ( n, K ) such that L ω ( B ) = L ω ( A ) ? W e can infer from the pro of o f pr eceding Theorem, that these pr oblems are also Π 1 1 -hard, b ecause we have seen that, in the first ca s e, L and L c are ac c epted b y deterministic timed B¨ uc hi automata without any clo ck . In a v ery similar manner, u sing the same ideas a s in the proof of Theorems 2 and 9, we can study the problem of minimization of the num b er o f clo c k s for timed B ¨ uc hi automata. W e can then show that it is Π 1 1 -hard, by reducing to it the univ er salit y problem for timed B ¨ uchi automata with n clo cks, where n ≥ 2, which is Π 1 1 -hard. So w e get the following result. Theorem 10. L et n ≥ 2 b e a p ositive inte ger. It is Π 1 1 -har d to determine, for a given TBA A with n clo cks, whether t her e exists a TBA B wi t h n − 1 clo cks, such that L ω ( B ) = L ω ( A ) . Remark 11. We have alr e ady ment io n e d that, for time d aut omata with only one clo ck, the universality pr oblem is de cidable [OW04]. On the other hand, for time d B¨ uchi aut oma t a with only one clo ck, the universality pr oblem has b e en r e c ently shown to b e unde cidable by P. A. Ab dul la, J. Deneux, J. Ouaknine, and J. Worr el l in [ADO W05]. Howeve r it se ems to u s that, in the p ap er [ADOW05] , this pr oblem is just pr ove d to b e unde cidable and not Π 1 1 -har d. Then we c an just infer t ha t the ab ove the or em is stil l true for n = 1 if we r eplac e “ Π 1 1 -har d” by “unde cidable”. Ac kno w ledgemen ts. Tha nks to the a non ymous referees for useful co mmen ts on a preliminary version of this paper. References AD94. R . Alur and D. Dill, A Theory of Timed Automata, Theoretical Computer Science, V olume 126, p. 183 - 23 5, 1994 . AM04. R . Alur and P . Ma d h usud an , Decis ion Pro blems for Timed Automata: A Sur- vey , in F ormal Metho ds for th e Design of Real-Time Sy stems, International School on F ormal Metho ds fo r the Design of Co mputer, Comm u nica tion and Softw are Systems, SFM-R T 2004, Revised Lectures. Lecture N otes in Computer S ci ence, V olume 3185, Springer, 2004, p. 1-24. ADOW05. P . A. Ab dulla, J. Den eux, J. Ouaknine, and J. 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