Supposewehavecomputedn1Nowwehavethatpnqifandonlyifpn1qandforeverya2wehavepan1qaThesecondpartrequiressomejusti cationandweproveitasalemmabelowLemma10LetMQq0FbeaDFAForan ID: 296019
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Forexampleonecanverifythatf0;00;000garedistinguishablewithrespecttoEQandthatf00;01;10;11garedistinguishablewithrespecttoA.Wenowprovethemainresultofthissection.Lemma5LetLbealanguage,andsupposethereisasetofkdistinguishablestringswithrespecttoL.TheneveryDFAforLhasatleastkstates.Proof:IfLisnotregular,thenthereisnoDFAforL,muchlessaDFAwithlessthankstates.IfLisregular,letMbeaDFAforL,letx1;:::;xkbethedistinguishablestrings,andletqibethestatereachedbyMafterreadingxi.Foreveryi6=j,wehavethatxiandxjaredistinguishable,andsoqi6=qjbecauseofLemma3.Sowehavekdi erentstatesq1;:::;qkinM,andsoMhasatleastkstates.UsingLemma5andthefactthatthestringsf00;01;10;11garedistinguishablewithrespecttoAweconcludethateveryDFAforAhasatleast4states.Foreveryk1,considerthesetf0;00;:::;0kgofstringsmadeofkorfewerzeroes.ItiseasytoseethatthisisasetofdistinguishablestringswithrespecttoEQ.ThismeansthattherecannotbeaDFAforEQ,because,iftherewereone,itwouldhavetohaveatleastkstatesforeveryk,whichisclearlyimpossible.2StateMinimizationLetLbealanguageoveranalphabet.WehaveseenintheprevioussectionthedenitionofdistinguishablestringswithrespecttoL.Wesaythattwostringsxandyareindistinguishable,andwewriteitxLyiftheyarenotdistinguishable.Thatis,xLymeansthat,foreverystringz,thestringxzbelongstoLifandonlyifthestringyzdoes.Bydenition,xLyifandonlyifyLx,andwealwayshavexLx.ItisalsoeasytoseethatifxLyandyLwthenwemusthavexLw.Inotherwords:Fact6TherelationLisanequivalencerelationoverthestringsin.AsyoumayrememberfromsuchclassesasMath55orCS70,whenyoudeneanequivalencerelationoverasetyoualsodeneawaytopartitionthesetintoacollectionofsubsets,calledequivalenceclass.AnequivalenceclassinwithrespecttoLisasetofstringsthatareallindistinguishablefromoneanother,andthatarealldistinguishablefromalltheothersnotintheset.Wedenoteby[x]theequivalenceclassthatcontainsthestringx.AfancywayofstatingLemma5istosaythateveryDFAforLmusthaveatleastasmanystatesasthenumberofequivalenceclassinwithrespecttoL.Perhapssurprisingly,theconverseisalsotrue:thereisalwaysaDFAthathaspreciselyasmanystatesasthenumberofequivalenceclasses.Theorem7(Myhill-Nerode)LetLbealanguageover.IfhasinnitelymanyequivalenceclasseswithrespecttoL,thenLisnotregular.Otherwise,LcanbedecidedbyaDFAwhosenumberofstatesisequaltothenumberofequivalenceclassesinwithrespecttoL.Proof:Ifthereareinnitelymanyequivalenceclasses,thenitfollowsfromLemma5thatnoDFAcandecideL,andsoLisnotregular.2 Supposewehavecomputedn�1.Nowwehavethatpnqifandonlyif{pn�1qand{foreverya2wehave(p;a)n�1(q;a).Thesecondpartrequiressomejustication,andweproveitasalemmabelow.Lemma10LetM=(Q;;;q0;F)beaDFA.Foranytwostatesp;qandintegern1,wehavethatp6nqifandonlyifp6n�1qorthereisana2suchthat(p;a)6n�1(q;a).Proof:Ifp6nq,thenthereisastringxsuchthatMacceptswhenstartingfrompandgivenx,butitrejectswhenstartingfromqandgivenx.Ifthelengthofxisn�1,thenwehavep6n�1q.Otherwise,letuswritex=ax0,whereaistherstcharacterofx,andletuscallp0=(p;a)andq0=(q;a).Thenthestringx0haslengthn�1anditshowsthatp06n�1q0.Fortheotherdirection,ifp6n�1qthenclearlyalsop6nq.Otherwise,ifthereisanasuchthat(p;a)6n�1(q;a),thenletx0bethestringoflengthn�1thatshowsthat(p;a)6n�1(q;a).Thenthestringax0oflengthnshowsthatp6nq.ThisgivesanO(jjjQj)timealgorithmtocomputengivenn�1,andsoanO(njjjQj)timealgorithmtocomputenfromscratch.Nowwecometoanimportantobservation:If,forsomek,wehavethattherelationkisthesameastherelationk+1,thenalsokisthesameasnforalln�k.Toseethatitistrue,notethattheprocessthatweusetogofromktok+1isindependentofk,andsoifitdoesnotchangetherelationwhenappliedonce,itwillnotchangetherelationifappliedanarbitrarynumberoftimes.ThisshowsthatthealgorithmconvergestoaxedpartitioninjQj�1steps,becauseitstartswithtwoequivalenceclasses,itcannotcreatemorethanjQjequivalenceclasses,andateachstepitmusteitherincreasethenumberofequivalenceclassesorreachthenalpartition.Inparticular,ifweletk=jQj�1,thenwehavethatkisthesameasnforeverynk,anditcanbecomputedintimeatmostO(jQj2jj).Wecanalsoseethatkisthesamerelationas.Thisistruebecauseifpqthencertainlypkq.Butalso,ifp6q,thenthereisastringxthatshowsthisisthecase,andifwecallnthelengthofxwehavethatp6nq.Ifnk,thencertainlythisalsomeansthatp6kq;ifn�kwecanstillsaythatp6kqbecausewehaveobservedabovethatnandkarethesameforn�k.Wecannownallydescribeourstateminimizationalgorithm.GivenM=(Q;;;q0;F).Letk=jQj�1andcomputetheequivalenceclassesofQwithrespecttok.DeneanewautomatonM0=(Q0;;0;q00;F0)asfollows.ThereisastateinQ0foreachequivalenceclass.Theinitialstateq00istheequivalenceclass[q0].ThesetF0containalltheequivalenceclassesthatcontainnalstatesinF.1Dene0([q];a)=[(q;a)].RemovefromQ0allthestatesthatarenotreachablefromq00.Suchstatescanbeeasilyfounddoingadepth-rstsearchofthegraphoftheautomatonstartingfromq00.Theremovalofthesestatesdoesnotchangethelanguageacceptedbyautomatonbecausetheyneveroccurinacomputation.LetM00=(Q00;;00;q00;F00)bethenewautomaton. 1NotethatanequivalenceclasscontainseitheronlystatesinForonlystatesnotinF.4