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Determinization of transducers over finite and infinite
words
Marie-Pierre Béal, Olivier Carton
To cite this version:
Marie-Pierre Béal, Olivier Carton. Determinization of transducers over finite and infinite words.
Theoretical Computer Science, Elsevier, 2002, 289 (1), pp.225-251. �hal-00619203�
over nite and in nite words
Marie-Pierre B
eal InstitutGaspardMonge, UniversitedeMarne-la-Vallee http://www-igm.univ-mlv.fr/~bea l/
Olivier Carton InstitutGaspardMongeandCNRS
UniversitedeMarne-la-Vallee http://www-igm.univ-mlv.fr/ ~ art on/
Mar h 27,2001
Abstra t
We studythe determinization of transdu ers over niteand in nite words. The rstpartofthepaperisdevotedto nitewords. Were allthe hara terizationofsubsequentialfun tionsduetoCho rut. Wedes ribe hereaknownalgorithmtodeterminizeatransdu er.
In the ase of in nite words, we onsider transdu ers with all their states nal. Wegiveane e tive hara terization ofsequentialfun tions overin nitewords. Wedes ribeanalgorithmtodeterminizetransdu ers overin nitewords. Thispart ontainsthemainnovelresultsofthepaper.
1 Introdu tion
The aim of this paper is the study of determinization of transdu ers, that is ofma hines realizingrational transdu tions. A transdu eris a nitestate au-tomaton(or a nitestate ma hine) whoseedges arelabeledby pairsofwords takenin nite alphabets. The rst omponent ofea hpair is alled theinput label. These ondonetheoutputlabel. Thetransdu ersthatwe onsiderhave a epting(or nal)states. Su htransdu ersaresometimes alleda-transdu ers (a for a epting). The rational relation de ned by a transdu er is the set of pairs of words whi h are labels of an a epting path in the transdu er. We assumethat the relationsde ned by ourtransdu ers are fun tions. This is a de idableproperty.
The study of transdu ers has many appli ations. Transdu ers are used to model oding s hemes ( ompression s hemes, onvolutional oding s hemes,
omputerarithmeti [17℄and innaturallanguagepro essing[25℄. Transdu ers arealsousedin programsanalysis [14℄. Thedeterminizationofatransdu eris the onstru tionofanothertransdu erwhi hde nesthesamefun tionandhas adeterministi (orrightresolving)inputautomaton. Su htransdu ersallowa sequentialen odingandthusare alledsequentialtransdu ers.
Inthe rstpartofthepaper,wepresentashortsurveyofthe determiniza-tionoftransdu ers realizingfun tions over nite words. Ourtransdu ers may have nalstates. Wepresentsomeknownresultsaboutsubsequentialfun tions, thatisfun tions that anberealizedbytransdu erswithadeterministi input butthat mayhaveanoutputfun tion de ned onstates. Thenotionof subse-quential fun tions has beenintrodu ed by S hutzenberger [28℄. We re all the hara terizationofsubsequentialfun tions obtainedbyCho rut [11,12℄. This hara terizationgivesade isionpro edureforthesubsequentialityoffun tions de nedbyatransdu er. Ithasbeenprovedin[30,31℄thatthis anbede ided inpolynomialtime. Wegiveanotherproofofthisresultwhi hisa onsequen e ofthede idability in polynomialtime of fun tionality overin nitewords[10℄. Anotherproofofthesameresultisgivenin[4℄. Thede idabilityoffun tionality wasalreadyprovedbyGire[18℄. Wealsodes ribethealgorithmtodeterminize atransdu er. Thisalgorithmtakesareal-timetransdu erwhi hrealizesa sub-sequentialfun tion and outputsasubsequentialtransdu er. This algorithmis a tually ontainedin the proof of Cho rut [11, 12℄ (see also [5, p. 109{110℄). This algorithm has also beendes ribed byMohri [22℄ and Ro he and Shabes [25,p.223{233℄.
The determinization of a transdu er realizing a subsequential fun tion f provides a subsequentialtransdu er realizing f. If the fun tion is sequential, this subsequentialtransdu er anbe transformedinto asequentialone. This anbeobtained by thenormalization of atransdu er introdu ed by Cho rut [12, 13℄. EÆ ient algorithmsthat ompute thenormalizationhavebeengiven in[21,23℄, [8,9℄and[2℄.
Inthese ondpartofthepaper,we onsidertransdu ersandfun tionsover in nitewordsandourtransdu ershavealltheirstates nal. Thereasonwhywe assumethatallstatesare nal isthat the aseof transdu erswith nal states seemstobemu hmore omplex. Indeed,thedeterminizationofautomataover in nitewordsisalreadyverydiÆ ult[26℄. Inparti ular,itisnottruethatany rationalset of in nitewords is re ognized by adeterministi automatonwith nal states. Other a epting onditions, asthe Muller onditionfor instan e, mustbeused.
We rstgiveane e tive hara terizationofsequentialfun tionsoverin nite words. This hara terization extends to in nite words the twinning property introdu ed by Cho rut [11℄. We prove that a fun tion is sequential if it is a ontinuous map whose domain an be re ognized by a deterministi Bu hi automaton,andsu hthatthetransdu erobtainedafterremovingsomespe ial states has the twinning property. These onditions an be simpli ed in the asewherethetransdu erhasno y lingpathwithanemptyoutputlabel. We usethis hara terization todes ribean algorithm he kingwhether afun tion
thetransdu erhasno y lingpathwithanemptyoutputlabel. Finally,wegive analgorithmtodeterminizeareal-timetransdu er. Thealgorithm anbeeasily adaptedtothe asewhenthetransdu erisnotreal-time. Thealgorithmismu h more omplexthaninthe aseof nitewords. Itisthemainresultofthepaper. Thesedeterminizationsdonotpreservethedynami propertiesofthe trans-du ersas the lo ality of itsoutput. We mentionthat in [19℄, an algorithm is giventodeterminizetransdu ersoverbi-in nitewordsthathavearight losing input (or that are n-deterministi or deterministi with a nite delay in the input)and a lo al output (see also[20, p. 143℄ and [1, p. 110{115℄). This al-gorithmpreservesthelo ality oftheoutput. These features areimportantfor odingappli ations.
Thepaperis organizedasfollows. Se tion2isdevotedto transdu ersover nite words. Basi notions of transdu ers of rationalfun tions are de ned at thebeginning of this se tion. The hara terizationof subsequential fun tions isre alledinSe tion2.1whilethealgorithmfordeterminizationoftransdu ers isdes ribedinSe tion 2.2. The hara terizationofsequentialfun tionsamong subsequentialonesisre alledinSe tion2.3. Se tion3isdevotedtotransdu ers over in nite words. We give in Se tion 3.1 a hara terization of sequential fun tions while the algorithm for determinization of transdu ers is des ribed in Se tion 3.2. In both asesof nite and in nite words, wegiveexamplesof determinizationoftransdu ers.
Part of the results of the present paper was presented at the onferen e ICALP'2000[3℄.
2 Transdu ers over nite words
Inthesequel,Aand B denote nitealphabets. ThefreemonoidA
is theset of nite wordsor sequen es ofletters of A. Theempty word isdenoted by". Wedenotethefa t thata niteworduis apre xof a nite wordv byu v. Therelation isapartial order. Ifuis apre x ofv, wedenote by u
1 v the uniquewordwsu hthatv=uw.
A transdu er overthemonoid A
B
is omposed of asetQ ofstates, a set E QA
B
Qof edges andtwosets I;F Q of initial and nal states. An edgee=(p;u;v;q)from pto q isdenoted by p
ujv
!q. Thestatep istheorigin, uistheinputlabel,v is theoutputlabel,and qisthe end. Thus, atransdu eris thesameobje tasanautomaton,ex eptthatthelabelsof the edgesarepairsofwordsinsteadofletters.
Atransdu erisoftendenotedbyA=(Q;E;I;F),oralsoby(Q;E;I)ifall statesare nal,i.e., Q=F.
Apath inthetransdu erT isasequen e
p 0 u 0 jv 0 !p 1 u 1 jv 1 ! u n jv n !p n of onse utiveedges. Its input label is the word u = u
1 u 2 u n whereas its output label is theword v = v
1 v
2 v
n
. The path leaves p 0
and ends in p n
p 0 ujv !p n :
Apathissu essful ifitleavesaninitialstateandendsina nalstate. Theset re ognized bythetransdu eristhesetoflabelsofitssu essfulpaths,whi his a tuallyarelationRA
B
. Thetransdu er omputesafun tionifforany wordu2A
, there existsat mostone wordv 2B
su h that (u;v)2 R . We allit the fun tion realized by the transdu er. A transdu er whi h realizes a fun tion issometimes alledsingle-valued intheliterature. Thusatransdu er anbe seen asa ma hine omputing nondeterministi allyoutput words from inputwords. Wedenotebydom(f)thedomainof thefun tion f.
A transdu eris nite ifitsset ofstatesanditsset oftransitions are nite. Itis a onsequen eof Kleene's theoremthat asubsetof A
B
is arational relationifandonlyifitistheset re ognizedbya nitetransdu er.
0 1 2 3 4 ajb aj ajb ajb aj aj
Figure1: Atransdu erfortherelation(a 2 ;b 2 ) [(a 2 ; 2 ) (a; ).
Example1 (from[5℄)TheautomatonofFigure1re ognizestherelation(a 2 ;b 2 ) [ (a 2 ; 2 )
(a; )overthealphabetsA=fagandB=fb; g. Thisrelationis a tu-allythefun tion whi hmapsa
n tob n ifnisevenandto n ifnisodd.
Let T bea transdu er. Theunderlying inputautomaton (respe tively un-derlying output automaton) ofT is obtainedbyomittingtheoutputlabel (re-spe tivelyinputlabel)ofea hedge.
Atransdu erissaidtobereal-timeifitislabeledinAB
. It anbeproved that anyrational fun tion anberealizedbya real-timetransdu er. Further-more,fromanytransdu errealizingafun tion anbe omputedinpolynomial timeanequivalentreal-timetransdu er(seeforinstan e[31,Prop.1.1℄). Wesay thatatransdu erT issequential ifitisreal-timeandifthefollowing onditions aresatis ed.
These onditionsensurethat forea hwordu2A
, thereis atmostoneword v 2 B
su h that (u;v) is re ognized by T. Thus, the relation omputed by T is a partial fun tion from A
into B
. A fun tion is sequential if it anbe realizedbyasequentialtransdu er.
Remark2 In[16,p.299℄,[5℄and[6℄,itisassumedthatallstatesofa sequen-tial transdu er are nal. We follow the de nition of Cho rut [11, 12℄ where sequentialtransdu ermayhave nalstates. Thus,some hara terizationsthat wegivebelowdi erfromthosepresentedin[5℄forthisreason. Whenallstates are nal,thedomainofasequentialfun tionispre x losed,i.e., ifuv belongs tothedomainthenualsobelongstothedomain. Asourde nitionallows nal states,thedomainofasequentialfun tionisnotne essarilypre x losed.
0 1
aja aja
bjb
bja
Figure2: A sequentialtransdu er.
Example3 LetA=B =fa;bgbethe inputandthe outputalphabets. The transdu er ofFigure 2, whose initial stateis 0, issequential. It repla es bya thoseb's whi h appear after an oddnumberof b. Onthe ontrary,the trans-du erofExample1is notsequential. A tually, thefun tion omputedbythis transdu erisnotsequential. Indeed,onemayverifythatiff issequential,and ifuandv aretwowordsofdom(f)su hthat u v,thenf(u) f(v).
Remark4 If f isa sequentialfun tion and if f(")is de ned,then f(")=". Toremovethisrestri tion,itispossibletoadd aninitial word asso iatedwith theinitialstate. Thiswordisoutputbeforeany omputation. Thisinitialword isne essarytogettheuni ityofaminimalsequentialtransdu er[28,12℄.
A subsequential transdu er (A;) over A
B
is a pair omposed of a sequential transdu er A over A
B
with F as set of nal states, and of a fun tion:F !B
. Thefun tionf omputed by(A;)isde nedasfollows. LetubeawordinA
. Thevaluef(u)isde ned ifandonlyifthereis apath i
ujv
!q in A with inputlabel u, from theinitial state i to a nal stateq. In this ase,onehasf(u)=v(q). Thus,thefun tionisusedtoappendaword to theoutput atthe end ofthe omputation. Afun tion issubsequential ifit anberealizedbyasubsequentialtransdu er.
fun tionsu hthat(q)="forany nal stateq. 0 1 a b aja aja bjb bjb
Figure3: A subsequentialtransdu er.
Example6 Thefun tion f realizedbythe subsequentialtransdu erpi tured in Figure3appends to ea h wordits lastletter. Theworduis mappedto ua ifit endswithanaanditismappedtoubifitendswithab. This fun tionis subsequentialbutit is notsequential. Indeed,for any wordw, f(wa)is nota pre xoff(wab).
2.1 Subsequential fun tions
Inthis se tion, wepresentsomeknownresults aboutsubsequential fun tions. Were allthe hara terizationofsubsequentialfun tions obtainedbyCho rut [11, 12℄. It is known that it is de idable whether a fun tion realized by a transdu er is subsequential. It has been proved in [30, 31℄ that this an be de ided in polynomial time. We givehere another proof of this result whi h is a onsequen e of the de idability in polynomial time of fun tionality over in nitewords[10℄. Wealso des ribe thealgorithm to determinize areal-time transdu er. This algorithm takes atransdu er whi h realizes a subsequential fun tion and outputs a subsequential transdu er. This algorithm is a tually ontainedintheproofofCho rut[11,12℄and[5,p.109{110℄. Ithasalsobeen des ribedbyMohri[21,23℄andRo heandShabes[25,p.223{233℄.
Ifthefun tion isa tuallysequential,thissubsequentialtransdu erisagain transformedinasequentialtransdu erbythealgorithmdes ribedinSe tion2.3. We give below two hara terizations of subsequential fun tions that have beenobtainedbyCho rut(see[11,12℄and[5,p. 105℄). The rst hara teriza-tionisintrinsi tothefun tion. Itisbasedonmetri propertiesofthefun tion. These ond hara terizationis e e tive. It isbasedonaproperty alled twin-ning propertyof atransdu er realizing thefun tion. As ithas been shown in [30,31℄,thisproperty anbede idedin polynomialtime.
Somenotationisneededtostatethe hara terizationofsubsequential fun -tions. We rst introdu e adistan e d on nite words. Let u;v be two nite words,wedenotebydthedistan esu hthat
Apartialfun tion f :A
!B
hasboundedvariation ifandonlyif:
8k09K08u;v2dom(f) d(u;v)k)d(f(u);f(v))K :
Thede idabilityofthesubsequentialityisessentiallybasedonthefollowing notionintrodu edby Cho rut [12, p. 133℄(see also[5, p. 128℄). Twostatesq andq
0
ofatransdu eraresaidto betwinned i foranypairofpaths
i uju 0 !q vjv 0 !q i 0 uju 0 0 !q 0 vjv 0 0 !q 0 where i and i 0
are two initial states, the output labels satisfy the following property. Either v
0 =v
00
= " orthere exists a nite word w su h that either u 00 = u 0 w and wv 0 0 = v 0 w, or u 0 = u 00 w and wv 0 = v 00
w. The latter ase is equivalenttothefollowingtwo onditions:
(i) jv 0 j=jv 00 j, (ii) u 0 v 0 ! =u 00 v 0 0 !
Atransdu erhasthetwinning property ifanytwostatesaretwinned.
Proposition 7 (Choffrut) Let f : A
! B
be a partial fun tion realized byatransdu erT. The followingthreepropositions areequivalent.
The fun tionf issubsequential. The fun tionf hasboundedvariation. The transdu erT has thetwinning property.
The equivalen ebetweenthe rst twostatements is an intrinsi hara teriza-tionofsubsequentialfun tionsamongrationalfun tions. Ita tuallysuÆ esto supposethattheinverseimagebyf ofanyrationalsetisstillrationalandthat f hasboundedvariationto insurethat f issubsequential. However,weare in hispaperinterestedine e tivemattersandwealwayssupposethatafun tion onwordsis given bya transdu er whi h realizes it. The equivalen ebetween thelast twostatements allowsus tode ide thesubsequentiality. The proofof thisequivalen eisessentiallytheproofofLemma16below.
We mention here another hara terization of the subsequentiality. For a partialfun tion f :A
!B
,de netheright ongruen eonA
byuu 0
i therearetwowords nitewordsvandv
0
su hthatthefollowingtwoproperties holdforany nite wordw. First,theworduw is in thedomainof f i u
0 wis inthedomainoff. Se ond,ifuwandu
0
wareinthedomain,thenv 1
f(uw)= v
0 1
f(vw). Thefun tion f isthensubsequentiali theright ongruen ehas niteindex. Inthat ase,the ongruen eallowsoneto onstru tdire tly a subsequentialtransdu errealizingf. Furthermore,thissequentialtransdu eris minimalinthesensethatanyothersubsequentialtransdu errealizingf anbe proje tedontothis one. The algorithm presentedin Se tion 2.2 allowsoneto omputee e tivelytheright ongruen e.
Example8 WehavealreadymentionedinExample3thatthefun tion(a ;b ) [ (a 2 ; 2 )
(a; )ofExample1isnotsequential. A tually,thisrelationisnot sub-sequentialas it anbeeasily shown with Proposition 7. Indeed, the fun tion doesnothaveboundedvariation. Foranyintegern, onehas
d(a 2n ;a 2n+1 )=1 while d(b 2n ; 2n+1 )=4n+1:
Wenowgivetwode idabilityresultsaboutrationalrelations. The rstone isdue toS hutzenberger[27℄ (seealso[7℄). These ond oneis dueto Cho rut [11,12℄(seealso[5,p.128℄).
Proposition 9 (S h
utzenberger) LetT be atransdu er over A
B
. It isde idable whether the relationde nedbyT is afun tion.
Cho rutalsoprovedthede idabilityofthesubsequentiality. Heshowedthat itsuÆ esto he kthetwinningpropertywhenthelengthsofthewordsuandv areboundedbythesquareof thenumberof states[12,p. 133℄and [5,p.128℄. However,thisalgorithmdoesnotseemtobepolynomial.
Proposition 10(Choffrut) LetT beatransdu erlabeledinA
B
whi h realizes afun tion f,then thesubsequentialityof f isde idable.
Thefollowingresultisdue toWeberandKlemm[30, 31℄.
Proposition 11 Letf bethe fun tionrealizedbyatransdu er labeledinA
B
. Itisde idable inpolynomialtimewhether f issubsequential.
The proof of the proposition followsdire tly from Proposition 7 and from thefollowinglemma. Wegivebelowanotherproofbasedonthede idabilityin polynomialtimeofthefun tionalityoverin nitewords. A thirdproofisgiven in[4℄.
Lemma12 The twinning property of a transdu er is de idable in polynomial time.
Proof LetT =(Q;E;I;F)beatransdu er. Wede idethetwinningproperty ofT intwosteps. We rstde ideinpolynomialtimethe ondition(i)andthen the ondition(ii).
We de ne an automaton A whose statesare the pairs of states of T and whoseedgesarelabeledbyintegers. Thereisanedge (p;p
0 ) n !(q;q 0 )i there aretwoedgesp
aju !q andp 0 aju 0 !q 0 in Asu hthat n=ju 0
j juj. Thelabel ofapathin Aisthesumofthelabelsoftheedgesofthepath. We laimthat thetransdu erT satis es ondition(i) i thelabel ofany y learound apair (q;q
0
)a essiblefrom somepair(i;i 0
)fortwoinitialstatesiandi 0
,isequalto zero. This anbedonebyadepth- rstsear h.
Weassumethatthetransdu eralreadysatis es ondition(i). This rst on-ditioninsuresthattheoutputlabelv
0
isemptyi v 0 0
isempty. The ondition(ii) isthen equivalentto thefun tionalityofthe relationonin nitewordsde ned
relationde nedbyT isafun tion,thenanytwostatesaretwinned. Conversely, ifthis relation is nota fun tion, there exist twoin nite paths labeled by xjy andxjy 0 withy6=y 0 . Letp 0 p 1 p 2 ::: andp 0 0 p 0 1 p 0 2
::: bethestatesvisitedbythe twopaths. Letk anindex su h that y
k 6=y
0 k
. Thereexist indi es m>nsu h that (p m ;p 0 m )=(p n ;p 0 n
). Moreover,nmay be hosengreat enoughsu h that theoutputs alongthe pathsfrom the initial stateto p
n and p
0 n
havea length greaterthank. Thenthestatesp
m andp
0 m
arenottwinned.
Itisde idableinpolynomialtimewhether arelationonin nitewords real-izedbyatransdu erisafun tion [10℄.
2.2 Determinization of transdu ers over nite words
Inthis se tion,wedes ribeanalgorithm whi hdeterminizesareal-time trans-du erwhi h hasthetwinning property. Thisalgorithm provesthat the ondi-tionsofProposition7aresuÆ ient.
Let T =(Q;E;I;F) be areal-time transdu er, that islabeledin AB
, realizingafun tion whi hissubsequential. Wegivebelowanalgorithm to de-terminizethetransdu erT,that is,whi hprodu esasubsequentialtransdu er realizing f. The algorithm is exponentialin the numberof states of T. The determinizationofanautomatonisalreadyexponential.
Wede ne asubsequentialtransdu er Dasfollows. A stateP ofD isaset ofpairs(q;w)whereqisastateofT andwisawordoverB. Wenowdes ribe thetransitions ofT. LetP bestateofDandletabealetter. Thepair(P;a) determinesasetRde ned by
R=f(q 0
;wu)jthereexist(q;w)2P andq aju
!q 0
2Eg:
IfR isempty, thereisno transitionfrom P input labeledbya. Otherwise, let vbethelongest ommonpre xofthewordswufor(q
0 ;wu)2Rand P 0 =f(q 0 ;w 0 )j(q 0 ;vw 0 )2R g:
ThereisthenatransitionP ajv
!P 0
. TheinitialstateofDisthesetJ =f(i;")j i 2 Ig where I is the set of initial statesof T. It follows from the de nition ofthetransitionsof Dthat ifP isastatea essiblefrom theinitialstate, the longest ommon pre x of thewordsw for (q;w)2 P is theempty word. We onlykeepin Dthea essiblepartfromtheinitialstate. Thetransdu erDhas adeterministi inputautomaton.
Thefollowinglemmastatesthemain propertyofthetransitionsofD.
Lemma13 Let ube a nite word. LetJ ujv
!P be the unique path inD with inputlabelufromthe initial state. Then, thestateP isequalto
P =f(q;w)jthere existsapath i ujvw
onsiderthefollowingpathinD J ujv !P ajt !P 0
where a is a letter. Let (q 0
;w 0
) be a pair in P 0
. By the de nition of the transitionsofD,thereisapair(q;w)inP andatransitionq
ajt 0 !q 0 in T su h that tw 0 =wt 0
. Bythe indu tion hypothesis, there is apath i ujvw
!q in T. Finally,onehasvtw
0 =vwt
0 .
The pre eding lemma has the following onsequen e. If both pairs (q;w) and(q
0 ;w
0
)belong toastateP whi h isa essiblefrom theinitialstateand if bothqand q
0
are nal statesin T,thentheequalityw=w 0
ne essarilyholds. Otherwise,therelationrealizedbyT isnotafun tion. This remarkallowsus tode ne thesetof nal statesofDandthefun tion . A stateP is nalifit ontainsasleast onepair (q;w) where q is a nal state of T. Thefun tion mapssu ha nalstateP to thewordw.
0 1 2 3 4 5 ajb ajba ajba aj" aj" aj" aj" bja aja
Figure4: Transdu erofExample14
Example14 Consider the transdu er pi tured in Figure 4. If the algorithm for determinization is applied to this transdu er, one gets the subsequential transdu erpi tured in Figure5. This subsequentialtransdu eris transformed intoasequentialtransdu erin Examples19.
Thisde nesasubsequentialtransdu erwhi hmayhaveanin nitenumber ofstates. However,we laimthattheboundedvariationpropertyofT implies thatthelengthsofthewordsinstatesofDare bounded. Thus thenumberof statesofDisa tually nite.
Lemma15 Let v 1 , v 2 ,v 0 1 andv 0 2
befour nite wordssu hthat jv 2 j=jv 0 2 jand v 1 v ! 2 =v 0 1 v 0 2 !
. Forany wordsv 3 andv 0 3 , 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 ):
0;" A B 1;" 2;a 3;a C 1;a 2;" 4;a a 5;" D ajb aj" aj" aja bjaa bja
Figure5: Determinizationofthetransdu erofFigure4
Proof Bysymmetry, we may suppose that jv 1
j jv 0 1
j. There is then a nite wordwsu h thatv
0 1 =v 1 w andwv 0 2 =v 2
w. Thusthewordv 0 1 v 0 2 v 0 3 isequalto v 1 v 2 wv 0 3 . It followsthat 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 ):
Thefollowinglemmastatesthatifatransdu erT hasthetwinningproperty, then the outputs labels of two paths with the same input label have a long ommonpre x. Itprovesthat iftherelationrealizedbyT isafun tion,ithas boundedvariation. Theproofisvery losetotheproofofProposition6.4in[5℄ but we do not assume that the relation realized by T is a fun tion. This is usefulwhentransdu ersrealizingrelationsonin nitewordsare onsidered.
Lemma16 LetT beatransdu erwhi h has the twinningproperty. Thereis a onstantK su hthat the outputsof twopaths i
ujv !q andi 0 ujv 0 !q 0 from two initial statesiandi
0 satisfy
d(v;v 0
)K :
ProofLetKbeequalto2n 2
Mwherenisthenumberofstatesofthetransdu er and M is the maximal length of the output label of a transition. We prove d(v;v
0
)K by indu tion on thelength ofu. If jujn 2
, theresult holdsby de nitionofK. Otherwise,bothpaths anbefa torized
i u1jv1 !p u2jv2 !p u3jv3 !q i 0 u1jv 0 1 !p 0 u2jv 0 2 !p 0 u3jv 0 3 !q 0 where u 1 u 2 u 3 = u, v 1 v 2 v 3 = v, v 0 1 v 0 2 v 0 3 = v 0 and ju 2 j > 0. By the twinning property,onehasd(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
) andtheresultfollowsfrom theindu tion hypothesis.
(q;w) in thestatesof D arebounded. This impliesthat thenumberof states ofDis nite.
Lemma17 There is a onstant K su h that for any pair (q;w) in a state P ofD,onehas jwjK.
Proof Let J ujv
! P be a path in D. Let (q;w) be a pair in some state P. Byde nition of the transitions of D, there is another pair (q
0 ;w 0 ) in D su h that w and w 0
have no ommon pre x. By Lemma 13, there are twopaths, i ujvw !qandi 0 ujvw 0 !q 0
in T. ByLemma16,there isa onstantK su hthat d(vw;vw
0
)KandthusjwjK.
Thefollowingproposition nallystatesthatthesubsequentialtransdu erD isequivalentto the transdu erT. It followsdire tly from Lemma 13 and the de nitionofthefun tion.
Proposition18 The sequential transdu er D realizes the same fun tion f as the transdu erT.
We have already mentioned in Proposition 11 that it an be de ided in polynomial timewhether afun tion realized by atransdu er is subsequential. Thealgorithm des ribedaboveis exponentialbut it providesanother de ision pro edure.Indeed,Lemma17givesaupperboundofthelengthsofwordswhi h anappearinstatesofD. ByLemma16,thisupperboundis2n
2
M wherenis thenumberofstatesofT andM isthemaximallengthoftheoutputlabelofa transitionofT. LetT beatransdu er realizingafun tion f. Ifthealgorithm is applied to T, either it stops and gives a subsequential transdu er D or it reatesa stateP ontaining apair(q;w) su h that the lengthof w isgreater than 2n
2
M. In the former ase, the subsequentialtransdu er D is equivalent to T and thefun tion f issubsequential. Inthelatter ase,the fun tion f is notsubsequential.
2.3 Sequential fun tions
Thedeterminization of atransdu er realizing a subsequentialfun tion f pro-videsasubsequentialtransdu errealizingf. Evenifthefun tionfissequential, thealgorithmdoesnotgiveasequentialtransdu erbutthissubsequential trans-du er an betransformedintoasequentialone.
Thistransformationis basedonanormalizationofsubsequential transdu -ersintrodu ed byCho rut [12, 13℄. This normalization onsists in pushingas mu h aspossiblethe output labels from nal states towardsthe initial state. Algorithms omputingthenormalizedtransdu eraregivenin[21,23℄,[8,9℄and [2℄. Thealgorithmsgivenin[21,23℄and[2℄runintimeO(jEjP)whereEisthe setof transitionsof thetransdu er, andwhere P isthemaximal lengthof the greatest ommonpre x oftheoutput labelsof pathsleaving ea h stateof the transdu er. If the normalizationis applied to asubsequentialtransdu er, the
ization an beperformedin polynomialtime, it anbe he kedin polynomial timewhetherafun tionrealizedbyasubsequentialtransdu erissequential. It anbeshownthatafun tionrealizedbyasubsequentialtransdu erissequential i itpreservespre xes. Thiswasalready provedin [30,31℄ that thisproperty anbe he kedin polynomialtime.
Inordertotransformasubsequentialtransdu erintoasequentialone,itis notne essaryto push asmu h aspossiblethe output labels from nal states towards the initial state, as the normalization does. It suÆ es to push these outputlabelsuntiltheoutputofallstatesareempty. Therefore,thealgorithm givenin [2℄ anbeadapted to meet thisrequirements. Thisgivesatime om-plexity of O(jEjL) instead of O(jEjP) where L is the maximal length of the outputwords. A B C D ajba aj" aj" aja bja bj
Figure6: Sequentialtransdu erofExample19
Example19 Consider the transdu er pi tured in Figure 5 where the states havebeenrenamedA,B,CandD. Ifthenormalizationisappliedtothis sub-sequentialtransdu er,onegetsthesequentialtransdu er pi turedin Figure6.
3 Transdu ers over in nite words
Inthisse tion,we onsidertransdu ersoverin nitewordswithallstatesbeing nal. We rstgiveane e tive hara terizationofsequentialfun tions over in- nitewords. This hara terizationextendstoin nitewordsthetwinning prop-ertyintrodu ed byCho rut [11, 12℄. We usethis hara terizationto des ribe analgorithmto he kwhetherafun tionrealizedbyatransdu erissequential. Finally,wegiveanalgorithmtodeterminizeatransdu er.
In this se tion, we denote by A !
the set of all (right-)in nite words over thealphabet A. We onsidertransdu ersoverin nitewords. Theedgesof the transdu ersare stilllabeledin A
B
. Thetransdu er hasinitial statesbut wesupposethat all statesare nal. Thus weomitthe set F of nal statesin thenotation. Anin nitepathisthensu essful ifitleavesaninitialstate. The relationoverin nitewordsde ned by thetransdu er istheset R A
! B
in nitewordsx su h that there issomein nite wordy su h that (x;y)labels asu essful pathin thetransdu er. Whenthe transdu er realizesafun tion, its domain is also the domain of the fun tion. A fun tion from A
! to B
! is sequential if it is realized by asequential transdu er. We point out that the notionofsubsequentialfun tion isirrelevantinthe aseofin nitewords.
3.1 Chara terization of sequential fun tions
Inthisse tion,we hara terizefun tionsrealizedbytransdu erswithallstates nalthat an berealizedbysequentialtransdu ers. This hara terizationuses topologi alpropertiesofthefun tionandsometwinningpropertyofthe trans-du er. Inthisse tion,weassumethatallstatesoftransdu ersare nal.
We rst introdu e a de nition. We de ne a subset of states whi h play aparti ular role in the sequel. We say that a state q of atransdu er is non onstant iftherearetwopathsleavingqlabelledbytwopairs(x;y)and(x
0 ;y
0 ) ofin nitewordssu hthaty6=y
0
. Ifastateqis onstant,eitherthereisnopath leavingqlabelledbyapairofin nitewordsorthereisanin nitewordy
q alled the onstantofqsu h thatforanypair(x;y)ofin nitewordslabellingapath leavingq,then y=y
q
. Intheformer ase,thestateq anberemovedsin eit annoto urin ana eptingpath labelled bya pairof in nitewords. In the sequel,wealwaysassumethatsu hstateshavebeenremoved. The onstanty
q is anultimately periodi word. It should be noti ed that anystate a essible from a onstant state is also onstant. We now state the hara terization of sequentialfun tions.
Proposition 20 Let f beafun tion realizedby atransdu er T withall states nal. LetT
0
bethetransdu erobtainedbyremovingfromT all onstantstates. Then thefun tion f issequentiali the following threepropertieshold:
the domainof f an bere ognizedbyadeterministi B u hiautomaton, the fun tion f is ontinuous,
the transdu er T 0
has the twinningproperty.
Sin ethefun tionf isrealizedbyatransdu er,thedomainoff isrational. However,itis nottruethat anyrational set ofin nite wordsis re ognizedby a deterministi Bu hi automaton. Landweber's theorem states that a set of in nitewordsisre ognizedbyadeterministi Bu hiautomatoni itisrational and G
Æ
[29℄. Re all that a set is said to be G Æ
is it is equal to a ountable interse tionofopensetsfortheusualtopologyofA
! .
Itisworthpointingoutthatthedomainofafun tionrealizedbyatransdu er maybeanyrationalsetalthoughitissupposedthatallstatesofthetransdu er are nal. The nalstatesofaBu hiautomaton anbeen odedintheoutputsof atransdu erinthefollowingway. LetA=(Q;E;I;F)beaBu hi automaton. We onstru t atransdu er T by adding an output to anytransition of A. A transitionp
a
!qof Abe omesp ajv
in nitei thepathgoesin nitelyoftenthrougha nalstate. Thusthedomain of the transdu er T is the set re ognized by A. For instan e, the domain of atransdu er maybe notre ognizable by adeterministi Bu hi automaton as in the following example. It is howevertrue that the domain is losed if the transdu erhasno y lingpathwithanemptyoutput.
0 1
aj"
bj"
bjb
bjb
Figure7: Transdu erofExample21
Example21 Thedomain ofthefun tion f realizedbythetransdu erof Fig-ure7istheset(a+b)
b
!
ofwordshavinga nitenumberofa. Thefun tionf annotberealizedbyasequentialtransdu ersin eitsdomain isnotaG
Æ set.
Itmustbealsopointedoutthatafun tionrealizedbyatransdu ermaybe not ontinuousalthoughitissupposedthatallstatesofthetransdu erare nal asitisshownin thefollowingexample.
0 1
aja
bj"
bjb
bjb
Figure8: Transdu erofExample22
Example22 The image of an in nite word x by the fun tion f realized by the transdu er of Figure 8is f(x) =a
!
f(x)=a b ifthenumberofainxisn. Thefun tionf isnot ontinuous. For instan e,thesequen ex
n =b n ab ! onvergestob ! whilef(x n )=ab ! doesnot onvergetof(b ! )=b ! .
Proof We rst explainwhythe abovethree onditionsof theproposition are ne essary. Thefa tthatthe onditionsaresuÆ ientfollowsfromthealgorithm thatwedes ribeinSe tion 3.2.
If the fun tion f is realized by a sequential transdu er D, adeterministi Bu hi automatonre ognizing thedomain of f an bededu ed from theinput automatonofDinthefollowingway. Ea hstateqis rstsplitintwostatesq
1 andq
2
. Wedistributethentheedgesarrivinginqbetweenq 1
andq 2
a ording totheemptinessoftheiroutput. Edgeswithanemptyoutputarriveinq
1 while edgeswithanonemptyoutputarriveinq
2
. Thestateq 2
isthen nal andq 1
is not. Ifqwasinitial,exa tlyoneamongq
1 andq
2
istheninitial. Alledgesgoing outof qaredupli atedin edgesgoingoutofq
1 andq
2
. Insymboli dynami s, su h a transformation is alled an input state splitting. It is lear that this deterministi Bu hiautomatonre ognizesthedomainoff. Itisalso learthat anysequentialfun tionis ontinuous.
We now prove that the third ondition is ne essary. We suppose that we havethefollowingpi turerepresentingpathsinT.
0 1 2 3 uju 0 uju 00 vjv 0 vjv 0 0
where 0and 1are initial states, u, u 0 , u 00 , v, v 0 and v 0 0
are nite words. Let Dbeasequentialtransdu errealizingthesamefun tion asT. There areinD paths 4 5 uv l jw v k jw 0
where0istheinitial state,wand w 0
are nite words. Byprolongingthepath inT from0to2(respe tivelyfrom1to3)withliterationsofthepatharound2 (respe tively around 3), we anassume withoutloss of generality that l = 0. Byrepla ing the y ling patharound2(respe tivelyaround3)byk iterations ofthispath,we anassumethatk=1.
We laim that if the state 2 is not onstant, then the equality jwj = jv 0
j holds. Sin estates2and3arenot onstant,thenifv="thenv
0 =v
00
thetwinningpropertyissatis ed. Wenowassumethatvisnotempty. Letxjx andyjy
0
bethein nitelabelsoftwoin nitepathsleaving2su h that x 0
6=y 0
. Thereare inD twoin nitepaths labeledbyxjx
0 0
and yjy 00
leavingthestate5 su hthat u 0 v 0 n x 0 =ww 0 n x 0 0 u 0 v 0 n y 0 =ww 0 n y 00 : Ifjv 0 j <jw 0 j, thewordsx 0 andy 0
havea ommonpre x oflength jwj ju 0 j+ n(jw 0 j jv 0
j)foranylargen. Thisleadstothe ontradi tionthatx 0 =y 0 . Ifjv 0 j> jw 0 j,thewordsx 00 andy 00
havea ommonpre xoflengthjuj jwj+n(jv 0
j jw 0
j) foranylargen. Thisleadstothe ontradi tionthatx
00 =y 00 andx 0 =y 0 . By symmetry, if the state3 is not either onstant, then the equality also jwj=jv
0 0
jholdsandthereforejv 0 j=jv 0 0 j. Ifbothwordsv 0 andv 0 0
arenonempty,thenf(uv ! )=u 0 v 0 ! =u 00 v 0 0 ! .
Beforedes ribingthealgorithmfordeterminization,we rststudya parti -ular ase. Itturnsoutthatthe rsttwo onditionsofthepropositionaredueto thefa t thatthe transdu erT mayhave y lingpathswith anemptyoutput. If the transdu er T has no y ling path with an empty output, the previous proposition an bestatedinthefollowingway.
Proposition23 Let f beafun tion realizedby atransdu er T withall states nal. Suppose also that T has no y ling path with an empty output. Let T
0 be the transdu er obtained by removing from T all onstant states. Then the fun tionf issequential i thetransdu erT
0
has thetwinning property.
Ifthetransdu erT hasno y lingpathwithanemptyoutput,anyin nite pathhasanin niteoutput. Thus,anin nitewordxbelongstothedomainoff i itistheinputlabelofanin nitepathinT. Thedomainoff isthena losed set. Itis thenre ognizedby adeterministi Bu hiautomatonwhose allstates are nal. This automaton an be obtained by the usual subset onstru tion ontheinputautomatonofT. Furthermore,ifthetransdu erT hasno y ling pathwithanemptyoutput,thefun tionf isne essarily ontinuous. This ould beproveddire tlybutitfollowsfrom Lemma31.
Wenowstudy thede idabilityofthe onditionsofPropositions 20and23. Wehavethefollowingresults.
Proposition24 Itis de idable if afun tion f given by a transdu er with all states nal is sequential. Furthermore, if the transdu er has no y ling path withan emptyoutput,this anbede idedin polynomial time.
Notethat theresultdoesnotholdifitisnotsupposed thatthetransdu er hasno y ling path with an empty output. Inthe general ase, the problem is NP-hard. For any Bu hi automaton, onsider the transdu er obtained by repla ingea htransitionp
a
!qoftheBu hiautomatonbyatransitionp aj"
!q ifp is not nal and by p
ajb
mapsanyin nitewordtob anditsdomainisexa ltythesetofin nitewords re ognizedby theBu hi automaton. This fun tion is sequentiali itsdomain isdeterministi . Sin etestingwhetherthesetofin nitewordsre ognizedbya givennondeterministi Bu hi automatonisdeterministi is anNP-hard prob-lem,testingwhetherafun tionis sequentialisalsoNP-hard.
ProofAsexplainedintheproofofProposition20,aBu hiautomaton re ogniz-ingthedomainofthefun tion anbeeasilydedu edfromthetransdu er. Itis thende idableifthisset anbere ognizedbyadeterministi Bu hiautomaton [29,Thm5.3℄.
It isde idable in polynomialtime ifafun tion givenbya transdu erwith nalstatesis ontinuous[24℄.
Wenowshowthat the third ondition ofProposition20 anbede ided in polynomialtime. Sin ewehavealreadyprovedinLemma12thatthetwinning property an be de ided in a polynomial time, it suÆ es to prove that the transdu er T
0
anbe omputed in polynomial time. We laimthat it an be de idedinpolynomialtimewhether agivenstateis onstant.
LetAbetheoutputautomatonofthetransdu er. Byadepth rstsear h,it anbefoundtwo nitewordsuandvsu hthatjuj+jvjnandsu hthatuv
! labelsapathleavingq. One onstru tsa ompletedeterministi automatonB re ognizinguv
!
with asinkstate0whi h isthe onlynona eptingstate. We then onsider the syn hronized produ t automaton of A and B. There is a transitionfrom (p;r)to(p
0 ;r
0
)labelled by a nite wordw(perhapsempty)i thereisatransitionfromptop
0
inAandapathfromrtor 0
inB. Thein nite worduv
!
isthelabelofallpathsleavingqi nostate(q 0 ;0)isa essiblefrom (q;i B )wherei B
istheinitialofB. Thisnaivealgorithmrunsinquadrati time forea hstateq. Thereforethe onstantstatesofatransdu er anbe omputed in ubi time. Itturnsoutthatthey anbe omputedin lineartime[10℄.
3.2 Determinization of transdu ers over in nite words
Inthis se tion,wedes ribeanalgorithmto determinizeareal-timetransdu er whi h satis es the properties of Proposition 20. This algorithm an easily be adaptedtothe asewhenthetransdu erisnotreal-time. Thisalgorithmproves thatthe onditionsoftheproposition aresuÆ ient.
Let T = (Q;E;I) be a transdu er and let T 0
bethe transdu er obtained byremovingfrom T all onstantstates. Weassumethat T
0
hasthetwinning property. WedenotebySthesetof onstantstates. ForastateqofS,wedenote byy
q
,thesingleoutputofqwhi hisanultimatelyperiodi word. Wesuppose that the domain of f is re ognized by the deterministi Bu hi automaton A. Thisautomatonisusedinthe onstru tedtransdu ertoinsurethattheoutput isin niteonlywhentheinputbelongsto thedomain ofthefun tion.
Wedes ribethedeterministi transdu erDrealizingthefun tionf. Astate ofDisapair(p;P)wherepisastateofAandP isaset ontainingtwokinds ofpairs. The rstkindarepairs(q;z)whereqbelongto QnS andzisa nite word over B. The se ond kind are pairs(q;z) where q belongs to S and z is anultimately periodi in nitewordoverB. Wenowdes ribethe transitions
de nedas follows R=f(q 0 ;zw)jq 0 = 2S and9(q;z)2P; q2=S andq ajw !q 0 2Eg [f(q 0 ;zwy q 0 )jq 0 2S and9(q;z)2P; q2=S andq ajw !q 0 2Eg [f(q 0 ;z)jq 0 2S and 9(q;z)2P; q2S andq ajw !q 0 2Eg:
Wenow de nethe transitionfrom the state(p;P) inputlabeledbya. IfR is empty, there is no transition from (p;P) input labeled by a. Otherwise, the output of this transition is the word v de ned as follows. Letp
a !p
0 be the transitioninAfromplabeledbya. Ifp
0
isnota nalstateofA,wede nevas theemptyword. Ifp
0
isa nalstate,wede nev asthe rstletterofthewords z ifR only ontainspairs(q
0
;z)withq 0
2S andifall thein nitewordsz are equal. Otherwise,wede nev asthelongest ommonpre x ofallthe nite or in nitewordsz for(q
0
;z)2R . ThestateP 0
isthende nedasfollows
P 0 =f(q 0 ;z)j(q 0 ;vz)2R g:
There is then a transition (p;P) ajv
! (p 0
;P 0
) in D. The initial state of D is thepair (i
A
;J) where i A
is the initial state of A and where J = f(i;")j i 2 I andi 2= Sg[f(i;y
i
)ji2 I andi 2Sg. Ifthe statep 0
is not nal in A, the outputofthetransition from(p;P)to (p
0 ;P
0
)is emptyandthewordszof the pairs(q;z)inP,mayhaveanonempty ommonpre x. Weonlykeepin Dthe a essiblepartfromtheinitialstate. Thetransdu erDhasadeterministi input automaton. It turns outthat the transdu er Dhas a nite number ofstates. ThiswillbeprovedinLemma33. Itwillbealsoprovedin Proposition34that thetransdu erDrealizesthesamefun tion asT.
0 1 aja bjb aj" aj" jaa
Figure9: Transdu erofExample25
Example25 Consider the transdu er pi tured in Figure 9. A deterministi Bu hi automatonre ognizing thedomainispi tured inFigure 10. Ifthe algo-rithmfordeterminizationisapplied tothistransdu er,onegetsthetransdu er pi turedinFigure11.
A B C D a b a
Figure10: A deterministi Bu hiautomatonforthedomain
A 0;" B 0;" 1;a ! C 1;a ! D 1;a ! aja aja aj" aj" bjb bjb ja ja ja
Figure11: Determinizationofthetransdu erofFigure9
Lemma26 Let u be a nite word. Let (i A
;J) ujv
! (p;P) be the unique path in D with input label u from the initial state. Then, the statep is the unique stateofAsu hthat i
A u
!pisapathin Aandthe setP isequal to
P =f(q;z)j9i ujv 0 !qin T su hthatv 0 =vz if q2=S v 0 y q =vz if q2Sg:
Proof The proof of the lemma is by indu tion on the length of u. Let us onsiderthefollowingpathinD
(i A ;J) ujv !(p;P) ajt !(p 0 ;P 0 )
wherea isaletter. Let (q 0 ;z 0 )beapairin P 0 . If q 0 = 2S, there isapair(q;z) in P and a transition q ajt 0 ! q 0
in T. If both states q and q 0
do not belong toS, the proofis similar tothe proofof Lemma 13. If q2=S andq
0 2S, one hastz 0 =zt 0 y q 0
. By theindu tion hypothesis, there is apath i ujvz
! q in T. One nally gets vtz
0 =vzt 0 y q 0 . If q2 S and q 0 2S, one hastz 0 =z. By the indu tionhypothesis, thereis apath i
ujv 0 !qin T su hthat v 0 y q =vz. Sin e y q =t 0 y q 0
,one nallygetsvtz 0 =v 0 t 0 y q 0 .
Thepreviouslemma hasthe orollarywhi h statesthat ea h stateqis the rst omponentofatmostonepair(q;z)inthese ond omponentP ofastate (p;P)ofD.
Corollary27 LetqbeastateofT andlet(p;P)beastateofD. ThesubsetP ontainsatmostonepair (q;z).
Proof Let(i A
;J) ! (p;P)bea pathin D and let(q;z)and (q;z 0
) betwo pairsin P.
We rst suppose that q is not onstant. Let xjy and x 0
jy 0
be two pairs of in nite words whi h label two paths leaving q su h that y 6= y
0
. By the previouslemma, there are two pathsi
ujvz ! q and i 0 ujvz 0 ! q in T. One has f(ux)=vzy=vz 0 yandf(ux 0 )=vzy 0 =vz 0 y 0 . Ifz6=z 0 ,itmaybeassumedby symmetrythat jz 0 j>jzjand that z 0
=zw forsome nite word w. This leads tothe ontradi tiony=y
0 =w
! .
Wenowsupposethatqis onstant. Letxjybeapairofin nitewordswhi h labelsa pathleaving q. Bythe previouslemma, there are twopathsi
ujw !q and i 0 ujw 0
! q in T su h that wy =vz and w 0
y =vz 0
. Furthermore,one has f(ux)=wy=w
0
y andthusz=z 0
.
Wenowintrodu esomete hni alpropertyofthepathsofatransdu er. This propertyis akindof twinning propertywhen theoutput ofone of the y ling pathsisempty. Itsturns outthat thispropertyisequivalentto the ontinuity ofthefun tion realizedbythetransdu erwhenitisalreadysupposedthat the transdu erhasthetwinningproperty. LetT beatransdu erandletSbeitsset of onstantstates. Thetransdu erT issaidtohavethe"- ompatibilityproperty i foranypairofpaths
i uju 0 !q vjv 0 !q i 0 uju 00 !q 0 vj" !q 0
su hthat iandi 0
aretwoinitialstatesandv 0
isanonemptyword,thestateq 0 is onstantandits onstanty
q 0 satis esu 00 y q 0 =u 0 v 0 !
. Ifthestatesqandq 0
are twinned,there annotbeapairof su hpaths. If theoutputalong these ond y ling path is empty, the output along the rst y ling path should also be empty. The above onditions add some ompatibility of the outputs when q andq
0
arenottwinned.
Thefollowinglemmastatesthatifthefun tionrealizedbythetransdu eris ontinuous,thenthetransdu erhasthe"- ompatibilityproperty. The onverse isestablishedin Lemma31.
Lemma28 LetT betransdu er realizing afun tion f onin nite words. If the fun tionf is ontinuous,thenthetransdu erT hasthe"- ompatibilityproperty.
Proof Letxjy be apair of in nitewordswhi h labelsapath leaving q 0
. For any integern, onehas f(uv
n x) = u 00 y and f(uv ! )= u 00 y by ontinuity of f. Sin e f(uv ! ) = u 0 v 0 ! , the state q 0
is onstant and its onstant y q 0 satis es u 00 y q 0 =u 0 v 0 ! .
Fora nitewordwandanin nitewordx,wedenotebyd(w;x)theinteger jwj jw^xjwherew^xisthelongest ommonpre xofwandx. Remarkthat disnotadistan e butLemma15stillholds whenv
0 3
ertyandthatT 0
has thetwinningproperty. Thereisa onstantKsu hthatfor anytwopaths i
ujv !qandi 0 ujv 0 !q 0 whereiandi 0
areinitial states,q2=Sand q 0 2S,onehas d(v;v 0 y q 0 )K :
ProofLetKbeequalton 2
Mwherenisthenumberofstatesofthetransdu er and M is the maximal length of the output label of a transition. We prove d(v;v
0 y
q
0)K byindu tiononthelengthofu. Ifjujn 2
,theresultholdsby de nitionofK. Otherwise,bothpaths anbefa torized
i u 1 jv 1 !p u 2 jv 2 !p u 3 jv 3 !q i 0 u 1 jv 0 1 !p 0 u 2 jv 0 2 !p 0 u 3 jv 0 3 !q 0 : where ju 2 j > 0and ju 3 j n 2 . If both wordsv 2 and v 0 2
are empty, the result followsdire tlyfromtheindu tion hypothesis. Thus, wemayassumethat one thewordsv
2 orv
0 2
isnotempty. Sin eqdoesnotbelongtoS,pdoesnotbelong toSeither. The"- ompatibilitypropertyimpliesthenthatv
2
annotbeempty. We rst suppose that p
0 =
2 S. Bythe twinning property, Lemma 15 and theaboveremark,onehasd(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
)andtheresult followsfromtheindu tionhypothesis.
Wenowsupposethat p 0
2S andwe laimthat v 0 1 v 0 2 v 0 3 y q 0 =v 1 v ! 2 . Sin e p 0 is onstant,y p 0 =v 0 3 y q 0 . If theword v 0 2
is empty, the"- ompatibilityproperty impliesthatv 0 1 y p 0 =v 1 v ! 2 . Ifv 0 2 isnonempty,y p 0 =v 0 2 ! . Sin ef(u 1 u ! 2 )=v 1 v ! 2 = v 0 1 v 0 2 !
,the laimedequalityholds. Inboth ases,onehasd(v 1 v 2 v 3 ;v 0 1 v 0 2 v 0 3 y q 0 )= d(v 1 v 2 v 3 ;v 1 v ! 2 )jv 3 jK.
Thefollowinglemmastatessomete hni al onsequen eofthe"- ompatibility property.
Lemma30 LetT beatransdu erwhi hhasthe "- ompatibilitypropertyandlet f thefun tionrealizedbyT. Then ifxisinthedomainof f andx isthe input label of a path entirely out of S,the output of this path is in nite and isthus equal tothe image ofx by f.
Proof Suppose that x is theinputlabelof twopaths and 0
. Suppose also that all states of do not belong to S and the output along
0
is an in nite word. Sin ethenumberofstatesis nite,bothpaths and
0 anbefa torized =i u0jv0 !q u1jv1 !q u2jv2 !q 0 =i 0 u0jv 0 0 !q 0 u1jv 0 1 !q 0 u2jv 0 2 !q 0
Furthermore,it anbeassumedthatea hv 0 k isnonemptysin ev 0 0 v 0 1 v 0 2 ::: isan in niteword. Byhypothesis,thisimpliesthat ea hv
k
parti ularthatifatransdu erT hasno y lingpathwithanemptyoutputand ifT
0
hasthetwinning property,thenthefun tion realizedbyT is ontinuous. Ifx andy aretwoin nitewords,d(x;y) denotestheusualdistan e betweenx andy whi hmakesthesetA
!
ofallin nitewordsa ompa tspa e.
Lemma31 Let T be atransdu er whi h has the "- ompatibility andsu h that T
0
has the twinningproperty. Then thefun tion realizedby T is ontinuous.
Proof Let f be the fun tion realized by the transdu er T and let x be an in nitewordin the domainof f. We laimthat for any integerm there isan integerksu hthatforanyin nitewordx
0
alsointhedomainf,theinequality d(x;x
0 )2
k
impliestheinequalityd(f(x);f(x 0
))2 m
. Lety=f(x)bethe imageof x. Let beapath labeled byxjy and leti bethe initial stateof . Let 0 beapathlabelledby(x 0 ;y 0 )wherey 0 =f(x 0
). A ordingtotheprevious lemma,it anbeassumedthat eitherthere isapathentirelyoutofS whi his labeledbyxjyorthat xisnottheinputlabelof apathentirelyoutofS.
We rstsupposethat thepath isentirelyoutofS. ByLemma15, there is a onstant K su h that ifi
ujv ! q andi 0 ujv 0 ! q 0
are twopathswith q 2= S andq
0 =
2S, thenonehasd(v;v 0
)K. ByLemmas28and29,thereisanother onstant K 0 su h that ifi ujv !q andi 0 ujv 0 !q 0
aretwopaths withq2= S and q
0
2 S, then onehas d(v;v 0
y q
0
) K. Let k be hosen su h that the output along the rst k transitions of has a length greater then m+max(K ;K
0 ). Letqbethestateof rea hedafter ktransitionsandletv betheoutputof
0 alongthe rstk transitions. Supposethat x
0 satis es d(x;x 0 )k and that 0 is apath labeled byx 0 jy 0 where y 0 = f(x 0 ). Let i 0
the initial state of 0 and letq 0 bethestateof 0
rea hed afterk transitions. Ifq 0
doesnotbelongto S, onehasd(v;v 0 )K wherev 0 isthe outputof 0
alongthe rstk transitions. Sin ejvjm+K, onehasjv^v
0
jmand thusd(y;y 0 )2 m . Ifq 0 belongs toS,onehasd(v;y 0 )K 0 . Sin ejvjm+K 0 , onehasjv^y 0 jmandthus d(y;y 0 )2 m .
We now suppose that x is notthe input label of apath entirely outof S. ThereisthenanintegerK su hthatanypathinputlabeledbyapre xofxof lengthgreaterthanKendsinastateofS. LetkbeequaltoK+K
0
whereK 0 isthelengthofpartof inside S whi h ontainsat leastn
2 transitions witha nonemptyoutput. Ifd(x;x 0 )2 k
, bothpaths and 0 anbefa torized =i u 0 jv 0 !q u 1 jv 1 !q u 2 jv 2 ! 0 =i 0 u 0 jv 0 0 !q 0 u 1 jv 0 1 !q 0 u 0 2 jv 0 2 ! whereu 0 u 1 u 2 =x,u 0 u 1 u 0 2 =x 0 ,v 1
is nonemptyandqandq 0 belong toS. We laim that y = y 0 . One has y = v 0 y q and y 0 = v 0 0 y q 0 . Sin e v 1 is nonempty, onealsohasy
q =v ! 1 . Ifv 0 1
isalso nonempty,one hasy q 0 =v 0 1 ! andf(u 0 u ! 1 )= v 0 v ! 1 = v 0 0 v 0 1 ! and thus y = y 0 . If the word v 0 1
is empty, the "- ompatibility propertyimpliesv 0 y q =v 0 0 y q 0 andy=y 0 .
Thefun tion f isthen ontinuous.
Thefollowinglemmastatesthatthelengthsofthewordszofthepairs(q;z) in thestatesof D arebounded. It is essentiallydue to the twinning property ofT
0 .
Lemma32 Thereis a onstant K su hthat for any pair (q;z)in P ofa state (p;P) ofDwhere q2=S andz isa nite word, onehas jzjK.
ProofLetmandnbetherespe tivenumbersofstatesofAandT. ByLemma 16and 29, there isa onstant K
0 su h that if i ujv !q and i 0 ujv 0 ! q 0 are two pathssu h that q2=S, thenonehasd(v;v
0 )K 0 ifq 0 = 2S ord(v;v 0 y q 0)K 0 ifq 0 2 S. LetK =K 0
+mnM where M is themaximal lengthof theoutput labelofatransitionin T. Let(p;P)beastateofDand onsiderapath
(i A ;J) u 0 jv 0 !(p 0 ;P 0 ) ujv !(p;P) wherep 0
is a nal stateofA. If there isno pathfrom (i A
;J)to (p;P)whi h goes througha state(p
0 ;P
0
) with p 0
nal, weassume that (p 0 ;P 0 ) is a tually (i A
;J). The proof is by indu tion on thelength of u. If juj = 0,the state p isa tuallya nalstate ofA. Inthe asewherepis nal, thelongest ommon pre xof thewordsz ofthe pairs(q;z) in P isempty. Lemmas 16, 26 and29 imply that jzj K
0
. We now suppose that p is not nal. If juj mn, the resultfollowsfrom thede nition ofthetransitionsofD. Wenowsupposethat juj>mnandthat(p
0 ;P
0
)isthelaststatealongthepathfrom(i A
;J)to(p;P) su h thatp
0
isa nal stateof A. Let(q;z)beapairinP su h thatq2=S and z is a nite word. Byde nition of thetransitions of D, there isapair (q
0 ;z 0 ) inP 0 andapathq 0 ujw !qinT su hthatz 0
w=vz. Thereisalsoapathp 0
u !p inA. Sin ejuj>mn,bothpaths anbefa torized
p 0 u1 ! p 00 u2 ! p 00 u3 ! p q 0 u1jw1 !q 00 u2jw2 !q 00 u3jw3 !q where u 1 u 2 u 3 =u and w 1 w 2 w 3
= w. Sin e the y ling path p 00
u2 ! p
00 in A does not ontain any nal state, the in nite word u
0 u 1 u ! 2
doesnot belong to thedomainoff. Thisimpliesthatthewordw
2
isempty. Wethen onsiderthe path (p 0 ;P 0 ) u1u3jv 00 !(p;P 00 ) inD. ThesubsetP 00 ontainsapair(q;z 00
)forsome nitewordz 00
. We laim that z
00
= z. Indeed, one has z 0 w 1 w 2 w 3 = z 0 w 1 w 3 = vz = v 00 z 00 . As both pathsp 0 u 1 u 2 u 3 !pandp 0 u 1 u 3
!pinA ontainnoother nalstatethanp,both outputsv andv
0 0
alongthe orrespondingpathsinDareempty. Thusonegets z=z
00
. Bytheindu tion hypothesis, onehasjzj=jz 00
states. However,thenumberofstatesofD anbeexponentialasinthe aseof nitewords.
Lemma33 The numberof statesofD 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 q where(q 0 ;z 0 ) is apairsu h that q
0 =
2 S and z 0
is nite and where q 2S and q 0
ajw ! q is a transition of T. Sin e the lengthof z
0
is bounded, the numberof su h words z
0 wy
q
is niteandtheyareultimatelyperiodi . Therearethena nitenumber ofsuÆxesofsu hwords.
Thefollowingproposition nallystatesthat thesequentialtransdu er Dis equivalenttothetransdu erT. Bothtransdu ersrealizethesamefun tionover in nitewords.
Proposition34 The sequential transdu er D realizes the same fun tion f as the transdu erT.
Proof Werespe tivelydenotebyf andf 0
thefun tions realizedbythe trans-du erT andD. We rstprovethatifanin nitewordxbelongstothedomain off,italsobelongsto thedomainoff
0 andf(x)=f 0 (x). Letx =a 0 a 1 a 2
::: be anin nitewordwhi his in thedomain of f. Let beapath =i a 0 jv 0 !q 1 a 1 jv 1 !q 2 a 2 jv 2 ! (1)
be apath in T input labeled by x and whose output v 0 v 1 v 2 ::: is an in nite word. Considertheuniquepath in Dinputlabeledbyx
=(i A ;J) a0jv 0 0 !(p 1 ;P 1 ) a1jv 0 1 !(p 2 ;P 2 ) a2jv 0 2 ! (2)
ByLemma26, ea hstateP n ontainsapair(q n ;z n ).
We rst suppose that x input labels a path in T entirely out of S. By Lemma30,it anbeassumedthatea hstateq
n
doesnotbelongtoS andthat ea h z
n
is nite. By Lemma 26, the equality v 0 :::v n =v 0 0 :::v 0 n z n holds for any integern. ByLemma 32, thelengths ofthe words z
n
are bounded. This impliestheequalityv
0 v 1 v 2 :::=v 0 0 v 0 1 v 0 2
::: ofthetwooutputs.
We now suppose that x is notthe input label of apath entirely outof S. There is then an integern su h that for any m n, P
m
only ontains pairs (q;z)withq2S andz in nite. Both path and anbefa torized
=i u 0 jv 0 !q u 1 jv 1 !q u 2 jv 2 !q =(i A ;J) u0jv 0 0 !(p;P) u1jv 0 1 !(p;P) u2jv 0 2 !(p;P)
n p
u k
!pinA ontainsa nalstateofA. Thesingleoutputofthestateq isv ! 1 . ByLemma26, thesubsetP ontainsapair(q;z)andv
0 y q =v 0 v ! 1 =v 0 0 z. Let(q 0 ;z 0
)beanotherpairinP. Byde nitionofthetransitionsofD,there isasequen e(q
n )
n0
ofstatessu hthatthepairs(q n ;v 0 1 n z 0 )belongtoP. Sin e there is a nite number of states, there are n < m su h that q
n = q
m . This impliesthatthereisinT a y lingpatharoundq
n inputlabeledbyu m n 1 . Let q 00 =q n =q m . We rst laimthat v 0 1 n z 0 =z 0 . Ifthewordv 0 1 isempty, thisis obvious. Otherwise, Corollary27 implies that v
0 1 n z 0 =v 0 1 m z 0 . Thus z 0 = v 0 1 ! and theequality v
0 1 n z 0 =z 0
also holds. The subsetP ontainsa pair(q 0 0
;z 0
). ByLemma 26, there is apath i
u 0 jv 00 ! q 00 in T su h that v 0 0 z 0 = v 00 y q 00. By onstru tion,thereisalsoa y lingpatharoundq
0 0
inputlabeledbyu m n 1
. We supposethattheoutputlabelofthis y lingpathisthewordw. Ifthewordw isempty, Lemma 28,statesthat v
0 0 y q 0 0 =v 0 v ! 1
. Thus,onehasv 0 0 z 0 =v 0 0 y q 00 = v 0 v ! 1 = v 0 0 z and z = z 0
. If the word w is nonempty, one has y q 0 0 = w ! and f(u 0 u ! 1 )=v 00 w ! =v 0 v ! 1 . Thisimpliesz=z 0 .
Sin ewehaveprovedthatallpairs(q;z)inP sharethesamein nitewordz andsin eea hpathp
ui
!p ontainsa nalstate,ea hwordv 0 i
isnonemptyby de nition of the transitions of D and the equality v
0 1 v 0 2 v 0 3 ::: =z holds. This lastequalityimpliesthatv
0 y q =v 0 0 z=v 0 0 v 0 1 v 0 2 ::: and thatf(x)=f 0 (x). Conversely, the de nition of the transitions of D implies that the domain off
0
is ontainedin thedomain of f. Thus bothfun tions f and f 0
have the samedomainandf =f
0 .
WehavealreadymentionedinProposition24thatit anbede idedwhether afun tion overin nite words realized by atransdu er with all states nal is sequential. Asinthe aseof nitewords,thealgorithmdes ribedaboveprovides anotherde isionpro edure. Indeed,Lemma32givesanupperbound Kof the lengthsof nite wordswhi h anappearin statesof D. LetT beatransdu er with all states nal realizing afun tion f. If the algorithm is applied to T, either it stops and givesasequentialtransdu er D orit reates astate (p;P) ontaining a pair (q;z) su h that the length of z is greater than K. In the former ase,thesequentialtransdu erDisequivalentto T and thefun tionf issequential. Inthelatter ase,thefun tionf isnotsequential.
A knowledgments
Theauthorswouldliketo thankJean Berstelforveryhelpfulsuggestionsand ChristianCho rut for his relevant omments on apreliminary versionof this paper.
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