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Determinization of transducers over infinite words: the
general case
Marie-Pierre Béal, Olivier Carton
To cite this version:
Marie-Pierre Béal, Olivier Carton. Determinization of transducers over infinite words: the general
case. Theory of Computing Systems, Springer Verlag, 2004, 37 (4), pp.483-502. �hal-00619205�
words: the general ase
Marie-Pierre B
eal InstitutGaspardMonge, UniversitedeMarne-la-Vallee http://www-igm.univ-mlv.fr/~bea l/
Olivier Carton LIAFA, UniversiteParis7
http://www.liafa.jussieu.fr/~ arto n/
June 17,2003
Abstra t
We onsidertransdu ersoverin nitewords withaBu hior aMuller a eptan e ondition. We give hara terizations of fun tions that an be realized by Bu hi and Muller sequential transdu ers. We des ribe analgorithm to determinize transdu ers de ningfun tions over in nite words.
1 Introdu tion
Theaim of this paperis the study of thedeterminization of transdu ers over in nitewords,that isofma hines realizingrationaltransdu tionsoverin nite words. Transdu ersare nite stateautomatawithedges labeledbypairsof -nitewords(aninputandanoutputlabel). Theyareveryusefulinalotofareas like oding[10℄, omputerarithmeti [11℄,languagepro essing(seeforinstan e [16℄and[13℄)or inprogramanalysis[8℄. Transdu ersthathaveadeterministi inputautomatonare alledsequentialtransdu ers[19℄andfun tionalrelations that anberealizedbyasequentialtransdu er are alledsequentialfun tions. Theyplayanimportantrole sin ethey allowsequentialen oding. The deter-minizationofatransdu eristhe onstru tionofasequentialtransdu erwhi h de nes the same fun tion. We refer the reader to [4℄ and [18℄ for omplete introdu tionstotransdu ers.
The determinization of an automaton over nite words is easily solvedby a subset onstru tion. The determinization of a transdu er is more omplex thanthedeterminizationofanautomatonsin eit involvesboththeinputand
whogivesin [6, 7℄ a hara terization of subsequentialfun tions and an algo-rithmthattransformsatransdu erwhi hrealizesasubsequentialfun tioninto asubsequentialtransdu er (see also [4, p. 109{110℄, [16, p. 223{233℄ and[2℄). Cho rut provedthat the subsequentialityof fun tions realized bytransdu ers over nite wordsisde idable. A polynomialtimede isionpro edure hasbeen obtained by Weber and Klemm in [22℄, see also [3℄. The determinization of transdu ersover nitewordsisthe rststepbeforeaminimization pro ess in-trodu edbyCho rutin[6℄and[7℄. EÆ ientalgorithmstominimizesequential transdu ershavebeendes ribedlaterin [13℄,[14℄, [5℄and[1℄.
We onsider here transdu ers that de ne fun tional relations over in nite words. The determinization of automata overin nite words is already mu h morediÆ ultthanover nite words. First,noteveryBu hi automaton anbe determinized. Muller automata whi h have a more powerful a eptan e on-dition must be used [12℄. Se ond, all determinization algorithms of automata over in nite words that have been given so far are omplex [17℄. In [2℄, we have oped withthis diÆ ulty by onsidering transdu ers without a eptan e ondition, that is, alltheir states are nal. This ase is indeed mu h simpler be ause the determinization of anautomaton overin nite wordswithout any a eptan e ondition an be a hieved by a simple subset onstru tion. How-ever,inthis ase,thedeterminizationofatransdu erisalreadynon-trivialand needsnew te hniques like thenotionof onstantstates. In[2℄, wehavegiven a hara terizationof sequentialfun tions, andadeterminization algorithm,in the asewhereallstatesof thetransdu erare nal.
Inthispaper,wesolvethegeneral ase,thatis,where thetransdu ersover in nitewordshaveBu hi orMullera eptan e onditions. Wegive hara ter-izationsof fun tions that anberealized byBu hi orMuller sequential trans-du ers. Inthe asewherethefun tionisBu hiorMullersequential,wegivean e e tivealgorithmto onstru tasequentialtransdu er(i.e.,atransdu erwith adeterministi inputautomaton) whi h realizes thesame fun tion. However, this result does not ompletely over those in [2℄. Indeed, this general algo-rithmapplied toatransdu erwithouta eptan e onditionyieldsasequential transdu erwithana eptan e onditionalthoughwehaveprovedin [2℄that a sequentialfun tionrealizedbyanon-sequentialtransdu erwithouta eptan e ondition ana tuallyberealizedbyasequentialtransdu erwithouta eptan e ondition.
Thepaperusesnotionsalready onsideredin[2℄likethenotionofa onstant statein atransdu er (astatesu hthat allpathsgoingoutofithavethesame in niteoutputlabel)butitalsointrodu esnewmethods. The hara terizations arebased onthe ontinuityof the fun tion realized by the transdu er andon anewnotionwhi his avariantof thetwinning propertyintrodu ed by Chof-frut[6,7℄ thatwe allweaktwinningproperty. Thedeterminizationalgorithm is performed in two main steps. The rst step onstru ts a sequential trans-du erwithouta eptan e onditionwhi hrealizesanextensionofthefun tion f realized bythe initial transdu er. These ond step ombines, with aneasy produ t onstru tion,thetransdu erobtainedatthe rststepwitha
determin-(orMuller)sequentialtransdu erthatrealizesexa tlyf. Theproblemthat the determinizationoftransdu ersin ludesthedeterminizationofautomataisthus avoidedbythisse ondstep. Roughlyspeaking,the rststepmainlydealswith the outputs of the transdu er whereas the se ond oneignores ompletely the outputsanddealsonlywiththeinputs.
Wemention that the ontinuityof fun tions realized byBu hi transdu ers is de idable in polynomial time [15℄. The de idability of the weak twinning property that we introdu e is notdis ussed in thepaper. Seethe Con lusion forafurtherdis ussion.
A onsequen e of our hara terizations is that any fun tion realized by a Mullersequentialtransdu eristherestri tionofafun tionrealizedbyaBu hi sequentialtransdu er. Thismeansthatthedi eren ebetweenfun tionsrealized byBu hi and Muller sequentialtransdu ers is entirelydue to the domains of thefun tions andnottotheoutputs.
The paper is organized as follows. Basi notions about transdu ers and a eptan e onditions over in nite words are de ned in Se tion 2. The two main results (Theorem 3 and Theorem 4) that state the hara terizations of Bu hiandMullersequentialfun tionsaregiveninSe tion3. Se tion4 ontains thedeterminizationalgorithmandanexampleofthe onstru tionofasequential transdu er.
2 Transdu ers
Inthe sequel, A and B denote nite alphabets. The set of nite and in nite words over A are denoted by A
and A
!
, respe tively. The empty word is denotedby".
Atransdu er overAB is omposedofa nitesetQofstates,a niteset EQA
B
Qofedges,asetI Qofinitial statesandana eptan e ondition. An edge e =(p;u;v;q)from pto qis denoted by p
ujv
!q. The wordsuandv are alledtheinputlabel andtheoutputlabel oftheedge. Thus, atransdu eris thesameobje tasanautomaton,ex eptthatthelabelsof the edgesarepairsofwordsinsteadofletters(as usual)orwords.
A nite path inatransdu er isa nitesequen e
q 0 u1jv1 !q 1 u2jv2 ! unjvn !q n
of onse utive edges. Its input label is the word u = u 1 u 2 :::u n , its output label is the word v = v
1 v
2 :::u
n
and its label is the pair (u;v) (also denoted ujv) of nite words. Su h a path is sometimes denoted by q
0 ujv
! q n
like a transition. Wesayitstarts at q
0
andends atq n
. Similarly, anin nite path inatransdu erisanin nitesequen e
q 0 u 0 jv 0 !q 1 u 1 jv 1 !q 2 u 2 jv 2 !q 3
0 1 2 isthe wordy =v 0 v 1 v 2
::: and itslabel isthe pair(x;y) (alsodenoted xjy)of words. Notethattheinputlabelortheoutputlabelofanin nitepathmaybe a nitewordbe ausetheinputlabelortheoutputlabelofatransitionmaybe theemptyword. Wesaythat the pathstarts at q
0
. Wedenote bylim ( ) the setofstatesthatappearin nitelyoftenalong . Sin ethenumberofstatesof thetransdu eris nite,lim( )isalwaysnonempty.
Thea eptan e onditiondeterminesafamilyof nalpathsasfollows. A pathis nal ifitsatis esandifbothitsinputandoutputlabelsarein nite words. A path is su essful ifit is nal and ifit starts at an initial state. In thispaper,we onsider twotypesof a eptan e ondition: Bu hi andMuller a eptan e onditions. In aB u hi transdu er the a eptan e ondition is a set F ofstates, alled nal states,and apath satis es ifit goesin nitely oftenthrough a nal state, i.e., lim( )\F 6= ?. Ina Muller transdu er the a eptan e ondition is a family F of sets of states, and apath satis es iflim ( ) 2 F. Observe that whether or not satis es depends onlyon theset lim ( )ofstatesthat o urin nitelyoftenalongthepath . Therefore, removing a nite pre x of a nal path or pre xing a nal path with a nite pathalwaysyieldsa nalpath.
Inthesequelwesaythata nite y lingpatharoundastateq(i.e.,starting and ending at q), also alled aloop, is a epting ifthe in nite path madeby loopingin nitelyoftenalongthisloopis nal. ForaBu hia eptan e ondition, aloopisa eptingifit ontainsa nalstate. ForaMullera eptan e ondition, aloopisa eptingifthesetofstatesthatareen ounteredalongthepathbelongs tothefamilyF.
A pair (x;y) of in nitewordsis re ognized ifit is thelabelof asu essful path. Theset ofallre ognizedpairsis therelationrealized bythetransdu er. This relationR isof ourse afun tion f if foranywordx 2 A
!
, there exists atmostonewordy 2B
!
su h that(x;y)2R . Inthat ase, atransdu er an beseenasama hine omputingnondeterministi allytheoutputwordy=f(x) fromtheinputwordx. Wedenotebydom(f)thedomainofthefun tion f.
Asinthe aseofautomata,nondeterministi Bu hiandMullertransdu ers havethesamepower. First,anyBu hi transdu er withaset F of nal states anbeviewedasMullertransdu erwhosea eptan e onditionisgivenbythe family F = fP Q j P \F 6= ?g. Conversely, any Muller transdu er an be simulatedbyaBu hi transdu er. This equivalentBu hi transdu er anbe obtainedbythesame onstru tionasforautomata[21,p. 417℄.
A transdu er is trim if ea h stateis a essible from an initial state and if thereisatleastone nalpathstartingatea hstate. Stateswhi hdonotsatisfy these onditions anberemoved. Therefore, we assumein the sequelthat all transdu ers aretrim. Note that it anbee e tively he ked whether agiven stateisa essiblefromaninitialstate. It analsobee e tively he kedwhether itisthe rststateofa nalpath. Indeedastateisthe rststateofa nalpath ifana eptingloop isa essiblefrom that state. Therefore,atransdu er an be e e tively made trim. This a tion an be seen as a prepro essing of the transdu er.
A transdu er is saidto bereal-time if it islabeledin AB , that is, the inputlabelofea htransitionisaletter. Wesaythatatransdu erT issequential ifthefollowing onditionsaresatis ed:
itisreal-time,
ithasauniqueinitialstate,
foranystateq andanylettera,thereisatmostonetransitiongoingout ofqandinputlabeledbya.
These onditionsensurethat forea h wordx2A !
,there isat mostone word y2B
!
su h that(x;y)isre ognizedbyT. Thus, therelationrealizedbyT is afun tion fromA
! intoB
!
. A fun tionissaidtobeB u hisequential (respe -tivelyMullersequential)ifit anberealizedbyasequentialBu hi(respe tively Muller)transdu er.
Inthe aseof nitewords,oneoftendistinguishessequentialand subsequen-tialfun tions. Inasubsequentialtransdu er,anadditional niteword depend-ingontheendingstateis appended totheoutput labelofthepath. However, thenotionofsubsequentialtransdu erisirrelevantinthe aseofin nitewords.
0 1 2 0j0 1j0 1j1 0j0 0j1 1j1
Figure1: SequentialBu hitransdu erofExample1
0 0 0 1 2 0j0 1j0 0j0 1j0 1j1 0j0 0j1 1j1 F=ff0g;f0;0 0 ;1g;f0 0 ;1;2g;f0;0 0 ;1;2g;f1;2gg
Figure 2: SequentialMullertransdu erofExample1
Example1 Let A = f0;1g be the binary alphabet. Considerthe sequential transdu erT pi turedinFigure1. Ifthein nitewordxisthebinaryexpansion of areal number 2 [0;1), the output orresponding to x in T is thebinary expansionof =3. If allstates of this transdu er are nal, it a epts both as inputandasoutputlabel binaryexpansionswhi hare notnormalized,thatis oftheform(0+1)
1
!
asshown in Figure 1. Inorder to reje tthese expansionsalso asinput label, thestate 0must be split and thetransdu er must be equippedwith aMuller a eptan e onditionasshowninFigure2.
Thefollowingpropositionallowsusin thesequelto only onsiderreal-time transdu ers. Thisresultis dueto Gire [9℄in themoregeneral aseofrational relationsofin nitewords. Wegivebelowasimplerproofforrationalfun tions.
Proposition2 ForanyB u hitransdu errealizingafun tionofin nitewords, one an ompute areal-time B u hi transdu er realizing thesame fun tion.
Proof LetT beaBu hi transdu er realizingafun tion. We anassumethat ea htransitionislabeledbyapair"jaoraj"whereaisaletteror". Otherwise, ea h transitionp ujv !q where u=a 1 :::a m andv =b 1 :::b m anberepla ed byn+m onse utivetransitions p a1j" !q 1 a2j" !q 2 q n 1 anj" !q n "jb1 !q n+1 q m+n 1 "jbm !q; whereq 1 ;:::;q m+n 1
arenewstates.
LetQbethesetofstatesofT andletF beitssetof nalstates. Wede ne areal-timetransdu erT
0
asfollows.
Letabealetterof theinputalphabet,letpand qbetwostatesofT,and lete be 0 or1. If e = 0, let V
a;e p;q
be the set of words v su h that there is a pathp
ajv
!q from p toq with input label aand output labelv. If e=1, let V
a;e p;q
be theset ofwordsv su hthat there is apathp ajv
!q fromp toq with input label a and output label v and whi h goesthrough a nal state. Note that V
a;e p;q
is always arationalsubset of B and that V a;1 p;q is asubsetof V a;0 p;q . Suppose that two nonempty words v and v
0 belong to aset V a;e p;q . Whenever the path p ajv
! q o urs in a su essful path of T, it an berepla ed by the pathp
ajv 0
! q. Indeed, sin ethe transdu er T realizesafun tion, theoutput wordofthe su essfulpathremains un hanged. Thismeans thatit suÆ esto keeponenonemptywordinea hsetV
a;e p;q . FromV a;e p;q ,wepi kasubsetW a;e p;q of ardinalityat most2asfollows.
IfV a;e p;q
ontainstheemptyword,theemptywordisalsoputin W a;e p;q . IfV a;e p;q
ontainsatleastonenonemptyword,oneofthemisputin W a;e p;q
.
ThesetofstatesofT 0
istheset Q 0
=Qf0;1g. Thesetofinitialstatesis I
0
=f(q;0)jq2Igandtheset of nalstatesisF 0
=f(q;1)jq2Qg. Theset oftransitionsofT
0
isde nedasfollows. Letabealetteroftheinputalphabet andlet(q;e)and(q
0 ;e
0
)betwostatesofT 0
. Thereisatransitionfrom(q;e)to (q 0 ;e 0 )labeledbyajvifv2W a;e 0 p;q . Thetransdu erT 0
realizesthesamefun tion asT. Thisisindependentof the hoi eofthe nite subsetsW
a;e p;q
.
The domain of a fun tion realized by a Bu hi or Muller transdu er is a rationalset of in nitewords. Re all that a set of in nite words is said to be
theedgesarelabeledbylettersinsteadofpairsofwords. Thelabelofapathin anautomatonis thus aword. A Bu hi (respe tivelyMuller)automaton isan automatonequipped witha Bu hi (respe tivelyMuller)a eptan e ondition. Wereferthereaderto [20℄or[21℄ fora ompleteintrodu tionto automataon in nitewords.
It is nottrue that anyrational set ofin nite words is re ognized by a de-terministi Bu hi automaton. However, any rational set of in nite words is re ognized by a deterministi Muller automaton [21, Thm 5.1℄. Furthermore anequivalentdeterministi Mullerautomaton anbe omputed from a Bu hi automaton. Sets of in nite words that an be re ognized by a deterministi Bu hiautomatonare alleddeterministi . It anbee e tively he kedwhether the set of words re ognized by a given Bu hi automaton is deterministi [20, Thm5.3 ℄. Furthermore,ifthatsetisdeterministi ,anequivalentdeterministi Bu hiautomaton ane e tivelybe omputed[20,Lem 5.4℄.
ABu hiautomatonre ognizingthedomainofafun tion anbee e tively omputedfromatransdu errealizingthefun tion. Theroughideaistoremove the output labels of the edges. We refer the reader to the proof of the main resultin [2℄.
3 Chara terization of sequential fun tions
The hara terizationsofBu hiandMullersequentialfun tionsneedthenotion of ontinuity of a fun tion. First re all that the set A
!
is endowed with the usualtopology. Thistopology anbede ned bythedistan edgivenby
d(x;y)= ( 0 ifx=y 2 n wheren=minfkjx k 6=y k gotherwise:
Intuitively two in nite words are lose if they share a long ommon pre x. Therefore, a sequen e of in nite words (x
n )
n0
onverges to a word x if for any integerk, there is an integern
k
su hthat any word x n
for nn k
has a ommonpre x with x oflength greaterthan k. We re allnow ade nition of the ontinuity that we uselater. A fun tion f is said to be ontinuous iffor anysequen e (x
n )
n0
ofelementsofitsdomain onvergingtoan elementx of itsdomain,thesequen e(f(x
n ))
n0
onvergestof(x).
The hara terizationsofBu hiandMullersequentialfun tionsalsoneedthe notionofa onstantstateinatransdu er. Wesaythatastateqofatransdu er is onstant ifall nalpaths startingat this state havethesameoutput label. Theterminology omesfrom the fa t that the transdu er in whi h q is initial realizesa onstantfun tion. Fora onstant stateq, the ommon outputlabel ofall nalpathsstartingatqisdenotedbyy
q
. Thisin nitewordalwaysexists sin ethetransdu erisassumedtobetrim.
Inordertoilluminatethenotionofa onstantstate,wemakesomeremarks andweprovesomeeasyproperties. Note rstthatinthede nitionofa onstant
with either a nite output label oran in nite output label whi h is di erent fromtheoutputofa nalpath.
Notethatifqisa onstantstateandifthestateq 0
isa essiblefromq,q 0
is alsoa onstantstate. Indeed,supposethat thereisa nitepath fromq toq
0 whoseoutputlabelisv. If
1 and
2
aretwo nalpathsstartingatq 0 ,thetwo paths 1 and 2
aretwo nalpathsstartingat q. Itfollowsthattheoutput labelsof
1 and
2
mustbeequalandq 0
isa onstantstate. Furthermore,the outputlabelsy q and y q 0 satisfyy q =vy q 0 . Note also that the ommonoutput labely
q
of a onstant stateq is an ul-timately periodi word, that is anin nitewordofthe form uv
!
fortwo nite words uand v. If there is a nal path starting at q, then there is alwaysan ultimatelyperiodi nalpathstartingatqsin ethenumberofstatesis nite.
Note nally that if q is a onstant state and there is a nite path from q toq(aloop)withanonemptyoutputlabelv,thentheoutputlabely
q
isequal tov
!
. Thisistrueevenifthelooparoundqisnota epting. Let bethe nite pathfromqtoqwiththeoutputlabelvandlet
1
bea nalpathstartingatq. Byde nition, the output label of
1 is y
q
. Sin e the path 1
is also a nal pathstartingatq,theequalityvy
q =y
q
holds. Sin evisnonempty,y q
isequal tov
! .
The hara terization of sequentiality is essentially based on the following notionwhi h is avariant of thetwinning property introdu ed byCho rut [7, p. 133℄ (see also [4, p. 128℄). This property is a kind of ompatibility of the outputs of paths with the same inputs. A transdu er has the weak twinning property ifforanypairofpaths
i tju !q vjw !q i 0 tju 0 !q 0 vjw 0 !q 0 ; whereiandi 0
areinitial states,thefollowingtwopropertieshold.
Ifbothq andq 0
arenot onstant, theneither w=w 0
=" orthere exists a nite word s su h that either u
0 =us and sw 0 = ws, or u=u 0 s and sw=w 0
s. Thelatter aseisequivalenttothefollowingtwo onditions:
(i) jwj=jw 0 j, (ii) uw ! =u 0 w 0 ! Ifq isnot onstant,q 0
is onstant,andw isnonempty, thentheequality u 0 y q 0 =uw !
holds. Notethatifw 0 isnonempty,theny q 0 =w 0 ! .
No property is required when both q and q 0
are onstant states. In that ase, the ompatibility of the outputs is already ensured by the fun tionality ofthetransdu er. Thepropertyrequiredwhenbothq andq
0
arenot onstant is exa tlythe twinning property asde ned by Cho rut [7℄ (requiredfor all q andq
0
). The weaktwinning propertyand thetwinning propertyonly di erin theway onstantstatesaretreated.
tions.
Theorem 3 Let f be a fun tion realized by a real-time B u hi transdu er T. Then thefun tion f is Muller sequential i the following twoproperties hold:
the fun tion f is ontinuous,
the transdu er T has the weaktwinning property.
Theorem 4 Let f be a fun tion realized by a real-time B u hi transdu er T. Then thefun tion f isB u hi sequentiali the following twopropertieshold:
the domainof f an bere ognizedbyadeterministi B u hiautomaton, the fun tion f isMullersequential.
Beforepro eeding totheproofsof thetheoremsweprovidesomeexamples showingthatthe onditionsareindependent.
0 1
aja
bj" bjb
bjb
Figure3: Transdu erofExample5
Example5 TheBu hitransdu erpi turedinFigure3isequippedwithaBu hi a eptan e ondition. Itrealizesanon ontinuousfun tionf. Indeed,theimage ofanin nitewordxisf(x)=a
!
ifxhasin nitelymanyo urren esofaand itisf(x)=a
n b
!
ifxhasno urren esofa. Althoughthesequen ex n =b n ab ! onvergestox=b ! ,thesequen ef(x n )=ab !
doesnot onvergetof(x)=b !
. State1is onstantbut state0is not. This transdu er hastheweaktwinning property. Notealsothatitdoesnothavethetwinning propertysin ethereare paths0 bj" ! 0 bj" ! 0and 0 bjb ! 1 bjb
! 1. This shows that the weak twinning propertyisreallyweaker.
Example6 TheBu hitransdu erpi turedin Figure4realizesthe ontinuous fun tion de ned by f(a ! ) = a ! , f(a n bx) = a n bx and f(a n x) = a 2n x for anyn0andx2fa;b; g
!
. However,thistransdu erdoesnothavetheweak twinningproperty. Thestates1and2arenot onstantbutonehasthefollowing paths0 aja !1 aja !1and0 ajaa !2 ajaa !2.
0 1 2 3 aja ajaa j bjb bjb ajaa j aja;bjb; j
Figure4: Transdu erofExample6
Example7 LetAbethealphabetfa;bgandletX =A
b !
bethesetofin nite wordshaving nitely manya. Let f betheidentityfun tion restri ted to the setX. Thisfun tionisMullersequentialbutitisnotBu hisequentialsin eits domainisnotdeterministi .
The proofs of Theorems 3 and 4 are given in the remainder of the paper. We provebelow that the onditions in Theorems 3and 4 are ne essary. The onversefollowsfrom thealgorithmthat wedes ribeinthefollowingse tion.
We rstprovethatafun tionf realizedbyaMullersequentialtransdu erS mustbe ontinuous. Supposethat thesequen e(x
n )
n0
of in nitewords on-vergesto xandthatallx
n
andxareinthedomainoff. Sin eS issequential, ea h wordofthe domainisthe inputlabelof exa tlyonepath. Let
n be the pathlabeledbyx
n
andlet bethepathlabeledbyx. Sin ex n
onvergestox, the ommonpre x ofx
n
andx be omeslonger and longer. It followsthat n onvergesto and hen ef(x
n
) onvergestof(x).
ItisalmoststraightforwardthatthedomainofaBu hisequentialfun tionf isre ognizedbyadeterministi Bu hiautomaton. Anin nitewordbelongsto the domain of f if it is the input label of a path whi h goes in nitely often througha nal state and throughatransition with a nonempty output label. A Bu hi automatonre ognizing the domain an beeasily onstru tedfrom a sequentialBu hitransdu errealizingf.
Itremainstoprovethatatransdu erT realizingaMullersequentialfun tion hastheweaktwinning property. Wesuppose thatwehavethefollowingpaths inT. i tju !q vjw !q i 0 tju 0 !q 0 vjw 0 !q 0 ; whereiandi 0
areinitialstates. LetSbeasequentialMullertransdu errealizing thesamefun tionf asT. Letxjybethelabelofa nalpathinT startingatq.
Foranyintegern,theequalityf(tv x)=uw y holds. Sin eS realizesf,there mustbeasu essfulpath in S with labeltv
n xjuw
n
y for any n. Forn greater than the number of states of S, thesame state appears twi e. Then there is inS apath i 00 tv l ju 00 !q 00 v k jw 00 !q 00 wherel0,k1,andi 00
istheinitialstateofS. ByprolongingthepathinT from i to q (respe tively from i
0 to q
0
) with l iterations of thepath around q (respe tively aroundq
0
), we an assumewithoutloss ofgeneralitythat l= 0. Byrepla ingthe y lingpatharoundq(respe tivelyaroundq
0
)bykiterations ofthispath,we analsoassumewithoutlossofgeneralitythatk=1.
We laim that if the state q is not onstant, then the equality jwj =jw 00
j holds. Indeed, let xjy and x
0 jy
0
be the labels of two nal paths starting at q su hthat y6=y
0
. ThereareinS twopathslabeledbyxjzandx 0 jz 0 startingat thestateq 00
su h thatforanyn0
f(tv n x)=uw n y =u 00 w 00 n z f(tv n x 0 )=uw n y 0 =u 00 w 00 n z 0 : Ifjwj <jw 00
j, thewords y and y 0
havea ommon pre x oflengthju 00
j juj+ n(jw
00
j jwj) for any largen. This leadsto the ontradi tion that y = y 0
. If jw
00
j<jwj,thewordszandz 0
havea ommonpre xoflengthjuj ju 00
j+n(jwj jw
00
j)for any largen. This leadsto the ontradi tionthat z =z 0
and y =y 0
. Thisprovesthatjwj=jw
00
jand,iftheyarenonempty,thatuw ! =u 00 w 00 ! . We rst suppose that q 0
is alsonot onstant. Bysymmetryone hasjwj = jw
00 j=jw
0
j. Furthermore,iftheyarenonempty,onehasuw ! =u 00 w 00 ! =u 0 w 0 ! . We now suppose that q
0
is onstant and that w is nonempty. This last assumption implies that w
00
is also nonempty and that the equality uw ! = u 00 w 00 ! holds. Letx 0 jy q
0 bethelabelof a nalpathstarting atq 0
. Thenthere isapathinS withlabelx
0 jz 00 startingatq 00 su hthat f(tv n x 0 )=u 0 w 0 n y q 0 =u 00 w 00 n z 00
foranyn0. Sin eq 0
is onstant,thewordw 0 n y q 0 is equalto y q 0 . Therefore, thewordu 0 y q 0 isequaltou 00 w 00 n z 00
foranyintegern. Thus,itisequaltou 00 w 00 ! sin ew 00
isnonempty. Thisendstheproofofthene essityofthe onditionsin Theorems3and4.
4 Determinization algorithm
In this se tion, we des ribe an algorithm to determinize a Bu hi transdu er whi h satis esthe onditionsofTheorem3or4. Wedes ribethe onstru tion of a sequential transdu er S from a Bu hi transdu er T. The transdu er S has a trivial a eptan e ondition. This means that any in nite path in S whi hhasin niteinputandoutputlabelsis nal. Ifthetransdu erT satis es the onditionsofTheorem 3,thefun tion realized byS isanextension of the
Mullerautomaton re ognizing the domain of T to obtain a Mullersequential transdu er whi h realizes thesame fun tion asT. If furthermorethe domain ofT isre ognized byadeterministi Bu hi automatonA, thetransdu erS is ombined with A to obtain a Bu hi sequential transdu er whi h realizes the samefun tion asT.
The sequential transdu er S is obtained from T by performing a kind of subset onstru tion. Fora xed nite wordu,allstateswhi h anbea essed fromtheinitialstatesbysomepathwhoseinputlabelisu,aregroupedtogether intoastateofS. Toea hofthesestatesisasso iatedaword. This wordgives whatremainsto beoutput. Foranon onstantstate,thiswordis nite andit isthesuÆxoftheoutputobtainedbydeletingtotheleftthemaximal ommon pre x of theoutputs labelling these paths. For a onstantstate, this wordis in nite and it it equals vw where v is as in the previous ase and w is the unique ultimately periodi output the state an produ e. The onstru tion yields potentially in nitely many omposite states onsisting of pairs (state, output word). It just happens that under the assumptions of Theorem 3 it leadstoa niteobje t.
We now des ribe the sequential transdu er S. By Proposition 2, we an supposethat the transdu erT isreal-time. This meansthat thelabelsof the edgesbelongtoAB
. The onstru tion ana tuallybeadaptedtodealwith transdu erswith edges labeled byA
B
but this isa bit te hni al. Let us denoteby Q, E,I, and C the set ofstates, edges, initial states,and onstant statesofT respe tively. A stateof S isa nite set P ontainingtwokindsof pairs. The rstkindarepairs(q;z)whereq belongs toQnC andz isa nite word overB. The se ond kindare pairs(q;z)where q belongs to C and z is anultimately periodi in nitewordoverB. Wenowdes ribethe transitions ofS. LetP beastateofSandletabealetterinA. LetRbeequaltotheset de nedasfollows R=f(q 0 ;zv 0 )jq 0 = 2C and9(q;z)2P; q2=C andq ajv 0 !q 0 2Eg [f(q 0 ;zv 0 y q 0 )jq 0 2C and9(q;z)2P; q2=Candq ajv 0 !q 0 2Eg [f(q 0 ;z)jq 0 2C and9(q;z)2P; q2Candq ajv 0 !q 0 2Eg: There areonly three asesin thede nition of R be ause q
0
is onstant ifq is already onstant. Wenowde nethetransitionfrom thestateP withinput la-bela. IfRisempty,thereisnotransitionfromPwithinputlabela. Otherwise, theoutputlabelofthistransitionisthewordvde ned asfollows. Wede nev asthe rstletterofthewordzifR only ontainspairs(q
0
;z)withq 0
2C and allthein nitewordszareequal. Otherwise,wede nevasthelongest ommon pre xofallthe niteorin nitewordszfor(q
0 ;z)2R . ThestateP 0 isde ned asfollows P 0 =f(q 0 ;z 0 )j(q 0 ;vz 0 )2R g: Then there is atransition P
ajv ! P
0
in S. The initial stateof S is the set J whereJ =f(i;")ji2I andi2=Cg[f(i;y
i
Itturns out that thetransdu er S hasa nite numberof states. This will be provedin Lemma14.
Somede nitionsareneededtoprovethe orre tnessofthe onstru tion. We introdu e rstadistan edon nitewords. Thisdistan eshould notbemixed upwith thedistan e thatwehaveusedat thebeginningof Se tion3tode ne thetopologyonA
!
. For nitewordsuandv,wedenotebydthedistan esu h that
d(u;v)=juj+jvj 2ju^vj;
whereu^visthelongest ommonpre xofuandv(see[4,p.104℄).Weextend thisdistan ewhenv isrepla edbyanin niteword. Letubea nitewordand letx beanin niteword. Wede ne
d(u;x)=juj ju^xj;
whereu^xisthelongest ommonpre xofuandx. Inthat ase,thefun tiond isnotadistan ebutitmeasureshowfaruisfrombeingapre xofx. Notethat ifuandwaretwo nite wordsand ifz isa niteorin niteword,theequality d(wu;wz)=d(u;z)holds. The following lemma statessomerelationbetween the distan e d and the weak twinning property. This is an easy property of ombinatori sofwords. Lemma8 Letv 1 ,v 2 ,v 0 1 andv 0 2
be nitewordssu hthatjv 2 j=jv 0 2 jandv 1 v ! 2 = v 0 1 v 0 2 !
. Forany nite wordv 3
andfor any niteor in nitewordv 0 3 ,onehas d(v 1 v 2 v 3 ;v 0 1 v 0 2 v 0 3 )=d(v 1 v 3 ;v 0 1 v 0 3 ):
Proof We rst suppose that jv 1
j jv 0 1
j. Then there is a nite word w su h thatv 0 1 =v 1 w andwv 0 2 =v 2
w. Thusthe wordv 0 1 v 0 2 v 0 3 isequalto v 1 v 2 wv 0 3 and itfollowsthat d(v 1 v 2 v 3 ;v 0 1 v 0 2 v 0 3 )=d(v 3 ;wv 0 3 )=d(v 1 v 3 ;v 0 1 v 0 3 ):
The asewherejv 1
jjv 0 1
j anbehandled similarly.
The transdu er S is sequential but it may notbe omplete. Forastate q andalettera,theremaybenotransitiongoingoutofqandinputlabeledbya. Foranynonempty niteworduandanystatesP andP
0 ofS,thereisatmost onepathP ujv !P 0 fromP toP 0
. Thefollowinglemmaandits orollary state themain property ofthe transitions of S. This property omes dire tly from thede nition ofthetransitionsofS. NopropertyofT isassumed.
Lemma9 Letubeanonempty niteword.
(a) LetP ujv !P 0 beapathfromP toP 0
in S with inputlabelu. If (q 0 ;z 0 )2 P 0
, then there is a pair (q;z)2 P and a path q ujv 0 ! q 0 in T su h that zv 0 =vz 0 if q;q 0 = 2C, zv 0 y q 0 =vz 0 if q2= C and q 0 2 C, andz =vz 0 if q;q 0 2C.
(b) LetP beastateofS. If(q;z)2P andq ujv
!q 0
isapathinT,thenthere isapath P ujv !P 0 inS andawordz 0 su hthat (q 0 ;z 0 )2P 0 ,zv 0 =vz 0 if q;q 0 = 2C,zv 0 y q 0 =vz 0 ifq2=C andq 0 2C,andz=vz 0 ifq;q 0 2C.
Proof We rst prove the statement (a). The proof is an easy indu tion on thelength of the word u. If u is aletter, the result follows dire tly from the de nitionofthetransitionsofS. Otherwise,theworduisequaltou
0 u 1 where u 0 andu 1
aretwononemptywords. Thepathfrom P to P 0 anbefa torized P u 0 jv 0 !P 00 u 1 jv 1 !P 0 where v = v 0 v 1
. For ea h pair (q 0
;z 0
) of P 0
, there are from the indu tion hypothesis twopairs(q;z)and(q
00 ;z
00
)inP andP 00
andtwopathsq u 0 jv 0 0 !q 00 andq 00 u 1 jv 0 1 !q 0
in T. Wedis ussonthemembership ofq,q 00 andq 0 toC. If q2=C, q 00 = 2C and q 0 =
2C,the indu tion hypothesis giveszv 0 0 =v 0 z 00 andz 00 v 0 1 =v 1 z 0 . Thisimplieszv 0 0 v 0 1 =v 0 z 00 v 0 1 =v 0 v 1 z 0 ,thatiszv 0 =vz 0 . If q2=C, q 00 = 2C and q 0
2C,the indu tion hypothesis giveszv 0 0 =v 0 z 00 and z 00 v 0 1 y q 0 =v 1 z 0 . This implieszv 0 0 v 0 1 y q 0 =v 0 z 00 v 0 1 y q 0 =v 0 v 1 z 0 , that is zv 0 y q 0 =vz 0 . Ifq2=C,q 00 2Candq 0
2C,theindu tionhypothesisgiveszv 0 0 y q 00 =v 0 z 00 and z 00 =v 1 z 0 . Sin ey q 00 =v 0 1 y q
0,this impliesthat zv 0 0 v 0 1 y q 0 =zv 0 0 y q 00 = v 0 z 00 =v 0 v 1 z 0 ,thatiszv 0 y q 0 =vz 0 . Ifq2C,q 00 2C andq 0
2C,theindu tionhypothesisgivesz=v 0 z 00 and z 00 =v 1 z 0 . Thisimpliesz=v 0 v 1 z 0 ,thatisz=vz 0 .
Theproofofthestatement(b) anbehandledsimilarly.
The following orollary just states the result of the previous lemma when thestateP istheinitial stateJ ofS.
Corollary10 Letubeanonempty nite word.
(a) LetJ ujv
!P beapathfromtheinitialstateJ toP inS withinputlabelu. If(q;z)2P,thenthereisapathi
ujv 0 !qinT su hthatv 0 =vzifq2=C, andv 0 y q =vz ifq2C. (b) Ifi ujv 0
!qisapathinT,thenthereisapathJ ujv !P inS andawordz su hthat (q;z)2P,v 0 =vz if q2=C,andv 0 y q =vz if q2C.
Proof These ond omponentz ofapair(i;z)inJ iseithertheemptywordif iisnot onstantorthewordy
i
ifiis onstant. Thentheresultfollowsdire tly fromthepreviouslemma.
The followingfour lemmas aredevoted to theproof that thetransdu er S has nitelymanystates. Itis rstprovedinthenextlemmathatinea hstateP ofS there isatmostoneo urren eofea h stateq. Therefore, thenumberof
is proved in the next two lemmas that the lengths of the nite words whi h appearin thepairsare bounded. It is nally provedin thefourthlemma that thenumberofin nitewordswhi h anappearinthepairsisbounded.
Lemma11 LetT beatransdu errealizinga fun tion f. Letq bea stateofT andletP be astateof S. There isatmostone wordz su hthat (q;z)belongs toP.
Proof LetJ ujv
!P beapathinS andlet(q;z)and(q;z 0
)betwopairsin P. We rst suppose that q is not onstant and thus that z and z
0
are nite. Letxjy andx
0 jy
0
bethelabelsoftwo nalpathsstartingatqsu hthaty6=y 0
. Bythe previous orollary, there are twopaths i
ujvz ! q and i 0 ujvz 0 ! q in T. Onehasf(ux)=vzy =vz
0 y and f(ux 0 )=vzy 0 =vz 0 y 0 . If z 6=z 0 , it may be assumedbysymmetrythatjz
0
j>jzjand thatz 0
=zw forsome nite wordw. Thisleadsto the ontradi tiony=y
0 =w
! .
Wenowsupposethat qis onstantandthusthat z andz 0
arein nite. Let xjy
q
bethelabelofa nalpath startingatq. Bytheprevious orollary,there aretwopathsi
ujw !qand i 0 ujw 0 !q in T su h thatwy q =vz andw 0 y q =vz 0 . Furthermore,onehasf(ux)=wy
q =w 0 y q andthusz=z 0 .
Fromnowon,wealwaysassumethatthetransdu erT realizesafun tionf.
Lemma12 LetT beatransdu erwhi hhasthe weaktwinningproperty. There isa onstantKsu hthatforanytwopathsi
ujv !qandi 0 ujv 0 !q 0 whereiandi 0 areinitial statesandq2=C,onehas
d(v;v 0 )K ifq 0 = 2C d(v;v 0 y q 0 )K ifq 0 2C
Proof Let K beequalto 2n 2
M wheren isthenumberofstatesof the trans-du erT and M is themaximal lengthoftheoutput label ofatransition. We provetheinequalitiesbyindu tiononthelengthofu.Ifjujn
2
,thentheresult followseasilyfromjvj;jv
0 jn
2
M. Otherwise,bothpaths anbefa torized
i u 1 jv 1 !p u 2 jv 2 !p u 3 jv 3 !q i 0 u1jv 0 1 !p 0 u2jv 0 2 !p 0 u3jv 0 3 !q 0 whereu 1 u 2 u 3 =u,v 1 v 2 v 3 =v, v 0 1 v 0 2 v 0 3 =v 0 , ju 2 j>0and ju 3 jn 2 . Sin e qis not onstant,pisalsonot onstant.
We rst suppose that p 0
is not onstant. By the weak twinning property and by Lemma 8, one haseither d(v
1 v 2 v 3 ;v 0 1 v 0 2 v 0 3 ) =d(v 1 v 3 ;v 0 1 v 0 3 ) if q 0 is not onstantord(v 1 v 2 v 3 ;v 0 1 v 0 2 v 0 3 y q 0)=d(v 1 v 3 ;v 0 1 v 0 3 y q
0)otherwise. Theresultfollows fromtheindu tionhypothesis.
We now suppose that p 0
is onstant. Therefore, q 0
is also onstant and y p 0 = v 0 3 y q 0 and y p 0 = v 0 2 y p 0 . If v 2
is empty, one has d(v 1 v 2 v 3 ;v 0 1 v 0 2 v 0 3 y q 0 ) =
d(v 1 v 3 ;v 1 v 3 y q 0 ) sin e v 2 v 3 y q 0 = v 3 y q 0.
The result follows from the indu tion hypothesis. Ifv
2
isnonempty,theweaktwinningpropertyimpliesthatv 0 1 y p 0 = v 1 v ! 2 . Therefore,d(v 1 v 2 v 3 ;v 0 1 v 0 2 v 0 3 y q 0 )jv 3 jK.
The following lemma states that the lengths of the nite words z of the pairs(q;z)in thestatesofS arebounded. Itisessentiallydueto thetwinning propertyofT.
Lemma13 LetT beatransdu erwhi hhasthe weaktwinningproperty. There isa onstant K su hthat for any pair (q;z) in astate P of S, z is in nite if q2C,andjzjK ifq2=C.
Proof LetKbethe onstantgivenbythepreviouslemma. Let(q;z)beapair inastateP su hthatthestateqofT isnot onstant. If(q;z)istheonlypair inthestateP,thewordzmustbeemptyandtheresultholds. Otherwise,there is anotherpair (q
0 ;z
0
)in P su h that z and z 0
do not havea ommon pre x. Onehasjzjd(z;z
0
)K.
It is now possible to prove that the transdu er S has a nite number of states. However,thenumberofstatesofS anbeexponentialasinthe aseof nitewords.
Lemma14 Let T be atransdu er whi h has the weak twinning property. The numberofstates ofS is nite.
Proof We haveproved in the pre eding lemma that thelengths of the nite words z are bounded. It remains to show that there is a nite number of di erentin nitewordsz whi h anappearinsomepair(q;z). Byde nitionof thetransitions,anyin nitewordz ofapairisthesuÆxofz
0 wy p where(p 0 ;z 0 ) isa pairsu hthat p
0 =
2 C and z 0
is nite and where p2 C and p 0
ajw !pis a transition of T. Sin e the lengthof z
0
is bounded, the numberof su h words z
0 wy
p
is niteandtheyareultimatelyperiodi . Thentherearea nitenumber ofsuÆxesofsu hwords.
ThefollowinglemmastatesthekeypropertyofS. Itspurposeistoguarantee thatthetransdu erS hasthesameoutputasT uptoaboundedsuÆx.
Lemma15 Let T be a transdu er satisfying the onditions of Theorem 3and letS bethe orresponding sequential transdu er. Letq
ujv
!q andP ujv
0 !P be y ling paths in T and S where the stateP ontains apair (q;z). If the path q
ujv
!q ontainsa nal stateandifv isnonempty,thenv 0
isalso nonempty.
Proof ByLemma 11, there is onlyonewordz su h that (q;z)belongsto P. Sin ethestateP isa essible,there isapathJ
tjw 0
!P inS. ByCorollary10, thereisapathi
tjw
i tjw ! q ujv ! q J tjw 0 !P ujv 0 !P
Weassumethattheloopq ujv
!q aroundq ontainsa nalstate. Sin ev 6=", thewordtu
!
belongstothedomainofthefun tionandf(tu !
)=wv !
. Wedistinguish two main ases depending onwhether q is a onstant state ornot. Inthe asethat qisnot onstant,thehypothesisthatthepathq
ujv !q goesthrough a nal stateis notneeded. This fa t is used in the proof of the other ase.
We rst suppose that q is not onstant. The word z is thus nite. By Corollary10applied to thepaths J
tjw 0 ! P and J tujw 0 v 0 !P, bothequalities w=w 0 z andwv =w 0 v 0
z hold. This impliesthat jv 0
j=jvjand thewordv 0
is nonempty.
We now suppose that q is onstant. The word z is thus in nite. We dis-tinguishagain two asesdependingonwhether thestateP ontainsatleast a non onstantstateor not. Inboth ases,weuse the following laim. Forany pair(q 0 ;z 0 ) in P, there is a pair (q 00 ;z 00
) in P su h that there are paths in T andS asshowninthefollowingdiagram
i 0 tjw 00 !q 00 u k jv 00 ! q 00 u l jv 000 !q 0 J tjw 0 ! P u k jv 0k ! P u k jv 0l !P
where k is a positive integer and l is a nonnegative integer. Let (q 0
;z 0
) be any pair in P. De ne byindu tion the sequen e (q
n ;z n ) n0 of pairs in P as follows. Let(q 0 ;z 0 )bethepair(q 0 ;z 0
). Supposethatthepair(q n
;z n
)isalready de ned. ByLemma9,thereisapair(q
n+1 ;z
n+1
)inP su hthatthereisapath q n+1 ujw n !q n
in T. Sin etheset P is nite, there aretwointegersk1and l0su h thatq k +l =q l andthusz k +l =z l byLemma11. Let(q 00 ;z 00 )denote thepair(q l ;z l ). By onstru tionofq 00
,thereisinT a y lingpathq 00 u k jv 00 !q 00 andthereisalsoapathq
00 u l jv 000 !q 0
. Sin ethepair(q 00
;z 00
)belongstoP,there is,byCorollary10,apathi
0 tjw 00 !q 00 in T,withi 0
2I. Thisprovesthe laim. We rstsupposethatP ontainsapair(q
0 ;z 0 )su hthat q 0 isnot onstant. Let(q 00 ;z 00
)bethepairgivenbytheprevious laim. Sin ethereisapathfromq 00 to q
0
, thestate q 00
is alsonot onstant. We proveby ontradi tionthat v 00
is nonempty. Letus assumethatv
00
=". Sin e q 00
isnot onstant,there aretwo nalpathsstartingatq
00
withdi erentoutputlabels. Letxjy andx 0
jy 0
bethe labelsofthesetwo nalpathswithy6=y
0
. Theimagesf(tu k n x)andf(tu k n x 0 ) areequalto w 00 y andw 00 y 0
foranyintegern. Both sequen es(tu k n x) n0 and (tu k n x 0 ) n0 onverge to tu !
. Sin e the fun tion f is ontinuous, both words w 00 y and w 00 y 0
areequaltof(tu !
)=wv !
. Thisis a ontradi tionsin ey6=y 0
. Thisprovesthatv
00
6=". Sin eq 00
isnot onstant,theproofofthe rst ase an beappliedto thepathsq
00 u k jv 00 !q 00 andP u k jv 0k
!P. Thisprovesthat v 0k
and thusv
0
We nallysupposethatforeverypair(q;z)inP,thestateq is onstant. Let(q
0 ;z
0
)beanypairinP andlet(q 00
;z 00
)bethepairgivenbythe laimabove. Weprovethatz=z
00
. Byhypothesis,thestateq 00
is onstant. Letxjy q
00 bethe labelofa nalpathstartingatq
00
. Sin eq 00
is onstant,theequalityv 00 y q 00 =y q 00 holds. The image f(tu
k n x) is equalto w 00 v 00n y q 00 = w 00 y q 00
for any integern. Thesequen e(tu
k n x)
n0
onvergestotu !
. Sin ethefun tionf is ontinuous, thewordw 00 y q 00 isequaltof(tu ! )=wv ! =wy q
. ByCorollary10appliedtothe pathJ tjw 0 !P, bothequalitieswy q =w 0 z andw 00 y q 00 =w 0 z 00 hold. Combined withtheequalityw
00 y q 00 =wy q ,onegetsz=z 00 . ByLemma9applied tothepathP
u l jv 0l !P, theequalityz 00 =v 0l z 0 holds. Ifv 0 =",then z 00 =z=z 0
. Sin e thisequalityholds foranypair(q 0 ;z 0 )ofP, allwordsz 0 of thepairs(q 0 ;z 0
)in P areequal. This ontradi ts thede nition ofthetransitionsofSsin etheoutputv
0
alongthepathP ujv
0
!P isnonempty inthis ase. Thisimpliesthatv
0
6=".
Thefollowingpropositionstatesthatthefun tionrealizedbythesequential transdu erS isanextensionofthefun tionrealizedbythetransdu erT.
Proposition16 LetT beatransdu ersatisfying the onditionsof Theorem 3 andletS bethe orresponding sequential transdu er. Letf andf
0
bethe fun -tionsrealizedbythetransdu ersT andS. Thenthein lusiondom(f)dom(f
0 ) holdsandfor anyx in dom(f),the equality f(x)=f
0
(x)holds.
Proof Weprovethatifthein nitewordxbelongsto thedomainoff,italso belongsto thedomainoff
0
and itsimagesbyf andf 0
areequal.
Let x bean in nite word whi h belongs to the domain of f and let be asu essfulpath in T withinput label x. Therefore, this pathgoesin nitely oftenthrougha nalstateanditsoutputlabelisanin niteword. Considerthe uniquepath in S withinputlabelx.
We laimthattheoutputlabelalong isnonemptyandthatitisequalto theoutputlabelalong . Sin ebothtransdu ersT andS (byLemma14)have a nitenumberofstates,bothpaths and anbefa torized
=i u 0 jv 0 !q u 1 jv 1 !q u 2 jv 2 !q =J u0jv 0 0 !P u1jv 0 1 !P u2jv 0 2 !P
Sin e the output along the path is in nite, it an be assumed that ea h wordv
n
is nonemptyandsin ethepath goesin nitelyoftenthrougha nal state,it anbealso assumed that ea hpath q
unjvn
!q ontainsa nal state. ByCorollary10,thestateP ofS ontainsapair(q;z)forsome niteorin nite wordz. ByLemma15,ea h wordv
0 n
isnonempty. By Corollary10, onehas for ea h n, v
0 :::v n = v 0 0 :::v 0 n z ifq is not on-stantand onehasv
0 :::v n y q =v 0 0 :::v 0 n
z otherwise. Thisimplies theequality v 0 v 1 v 2 :::=v 0 0 v 0 1 v 0 2
:::ofthetwooutputs.
Bythelastproposition,thefun tionrealizedbythesequentialtransdu erS extendsthefun tionrealizedbythegiventransdu erT. Toobtainasequential
This is a hieved by onstru ting anew sequentialtransdu er S 0
whi h is the syn hronizedprodu tofS andofanautomatonforthedomainofT.
Re allthatthetransdu erS hasnoa eptan e ondition. Thismeansthat anin nitepathis nali bothitsinputandoutputlabelsarein nitewords.
LetX bethedomainofthefun tionrealizedbyT. LetAbeadeterministi Bu hiautomatonre ognizingX ifX isdeterministi orletAbeadeterministi Mullerautomatonre ognizingX otherwise. Intheformer ase,its a eptan e ondition is a set F of nal states and, in the latter ase, its a eptan e onditionisafamilyF ofsetsofstates. AsexplainedattheendofSe tion2, theautomatonA anbe omputedfrom thetransdu erT.
Wenowdes ribethetransdu erS 0
. Thestateset ofS 0 isQQ 0 whereQ andQ 0
arethestate setsof S and A. Theinitial stateis (i;i 0
)where iand i 0 are the initial states of S and A. There is a transition (p;p
0 ) aju ! (q;q 0 ) i p aju !qand p 0 a !q 0
are transitionsof S and A. Thea eptan e ondition 0 ofS
0
mimi sthatofA. Moreformally,ifAisaBu hi automaton,thenS 0
isa Bu hi transdu erand itsset of nal statesisF
0 =f(q;q 0 )jq 0 2Fg. If Ais a Mullerautomaton, then S
0
is aMuller transdu er andits familyF 0
of sets of statesisde ned asfollows.
F 0 =ff(q 1 ;q 0 1 );:::;(q k ;q 0 k )gjfq 0 1 ;:::;q 0 k g2Fg:
Itispureroutineto he kthatS 0 isequivalenttoT. 0 1 aja bjb j" aja j" ajaa j"
Figure 5: Transdu erT ofExample17
0;" A 0;" 1;a ! B bjb aja j" bjb aja j"
x y z a b a a b F =ffxg;fx;zgg
Figure7: AMullerautomatonforthedomainofT
Ax Bx Ay Bz aja bjb j" aja bjb j" bjb aja j" j" aja bjb F 0 =ffBxg;fBx;Bzgg Figure8: Transdu erS 0 ofExample17
Weillustratethe onstru tionofS andS 0
bythefollowingexample.
Example17 Let A be the alphabet fa;b; g and onsider the transdu er T pi tured in Figure 5. Note that the state 1 is onstant whereas the state 0 is not. Applying the onstru tiondes ribed above, onegets thetransdu er S pi tured in Figure 6. The domain of T is A
(
a)
!
but the domain of S is A ( (a+b)) !
. The Mullerautomaton A for thedomain of T is pi tured in Figure 7. The transdu er S
0
obtained by ombining S and A is pi tured in Figure8.
5 Con lusion
Inthispaper,wehaveprovided hara terizationsofsequentialfun tionsof in -nitewordsrealizedbyMullerandBu hitransdu ers. Whenatransdu errealizes asequentialfun tion,wehavegivenanalgorithmto omputeanequivalent se-quentialtransdu er. Sin ethisdeterminization in ludesthedeterminizationof anautomatonforthe domainof thefun tion,the omplexityisat least expo-nential.
In the ase of nite words, the determinization is also exponential but it anbe he ked in polynomial time whether a fun tion givenby a transdu er is sequential. The ontinuity an be he ked in polynomial time [15℄. The
inthepaper. Wedonotknowwhetherthis anbe he kedinpolynomialtime. However,sin ethis notionis lose tothe twinning propertyofCho rut [6,7℄, wethinkthatthemethodsusedin[22℄or[3℄ anbeusedtoobtainapolynomial timealgorithmto he kthisproperty.
A knowledgments
The authors would like to thank Joost Engelfriet and the anonymous refer-ees fortheirhighly relevant omments andveryhelpful suggestionsabout the preliminaryversionofthispaper.
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