Which one of these DFAs is thebest one One answer to this question is to try to build a minimal DFA one with the smallest possible number of states But building a minimal DFA from scratch can be dicult Instead we will show a procedure that converts ID: 79228
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21.q1=qandq01=q0,2.qi+1=(qi;wi)andq0i+1=(q0i;wi),fori=1;:::;m�1,and3.Exactlyoneofthestatesqmandq0misinF.2CharacterizingminimalDFAsWenowdescribethemainfeatureofaminimalDFA.Theorem4.LetM=(Q;;;q0;F)beaDFA.ThenMisminimalifandonlyifeverystateofMisreachableandeverypairofstatesofMisdistinguishable.Toprovethistheorem,wehavetodotwothings:First,wehavetoshowthatifeverystateisreachableandeverypairofstatesofMisdistinguishable,thenMisminimal.Then,wehavetoshowthatifMisminimal,theneverystateisreachableandeverypairofstatesisdistinguishable.Lemma5.IfeverystateofMisreachableandeverypairofstatesofMisdistinguishable,thenMisminimal.Proofsketch.WeneedtoarguethatifeverystateofMisreachableandeverypairofstatesofMisdistinguishable,thenMisminimal,thatistherecanbenosmallerM0=(Q0;;0;q00;F0)(i.e.,onewithjQ0jjQj)forthesamelanguage.WewillassumesuchanM0existsandarguethatsomethingmustbewrongwithit.BecauseeverystateofMisreachable,foreverystateqi2Q,thereisastringwithatreachesqiinM.Bythepigeonholeprinciple,thereexistsareachablestateq0ofQ0andapairofstringswi;wj(wherewi6=wj)suchthatwireachesq0inM0andwjreachesq0inM0.Byassumption,qiandqjaredistinguishableinM,sothereexistsastringwthatdistinguishesthem.NowthinkwhathappenswhenwerunMonthestringswiwandwjw.Sinceqiandqjaredistinguishable,exactlyoneofwiwandwjwreachesanacceptingstateofM.Ontheotherhand,bothwiwandwjwreachthesamestateofM0.ThereforethelanguagesofMandM0mustbedi erent. Fortheotherdirection,weneedtoarguethatifMisminimal,theneverystateofMisreachableandeverypairofstatesofMisdistinguishable.Wewillreasonbycontradiction:IfeitherMhasastatethatisnotreachable,orMhasapairofstatesthatareindistinguishable,thenwewillshowthatMisnotminimal{itcanbemadesmaller.LetusstartbyseeingwhathappenswhenMhasanunreachablestate.Lemma6.IfMhasanunreachablestate,thenMisnotminimal.Proof.SupposeMhasastateqthatisnotreachable.LetM0betheDFAobtainedbyremovingallunreachablestatesfromM,togetherwiththeiroutgoingtransitions.ThatisM0=(Q0;;0;q0;F0)where1.ThestatesofM0are:Q0=fq:q2QandqisreachableinMg, 4classA,andona1-transition,everystatefromclassAmovestoclassB.Thesamethinghappenswiththeotherclasses.Sowecantaketogetherallthestatesfromthesameclass,mergethemtogetherintonew\megastates",andobtainasmallerDFA: Infact,thismergingofindistinguishablestatesinto\megastates"canalwaysbedoneinsuchaway.Thereasonisthatifapairofstatesqiandqjareindistinguishable,thenforeverya2,ri=(qi;a)andrj=(qj;a)mustalsobeindistinguishable:Forifriandrjcanbedistinguishedbyw,thenqiandqjcanbedistinguishedbyaw.Sotransitionsoutofindistinguishablestatespointintoindistinguishablestates.Withthisinmind,wecanproveLemma7.ProofofLemma7.AssumeMhasapairofdistinguishablestates.WewillshowhowtoconstructanequivalentDFAM0=(Q0;;0;q00;F0)withfewerstates.Todoso,wedividethestatesQofMintoindistinguishabilityclassesq00;q01;:::;q0m0.EachoftheseclassesrepresentsasubsetofstatesofMwhicharemutuallyindistinguishable.Wewillsaythatastateqisrepresentedbyitsindistinguishabilityclassq0.Wecanassumethatq0isrepresentedbyq00.SinceMhasatleasttwoindistinguishablestates,itfollowsthatm0jQj.WecannowdescribeM0:1.States:Q0=fq00;q01;:::;q0m0g,2.Transitions:Foreveryq02Q0anda2,ifq0representsq,then0(q0;a)represents(q;a).3.Acceptingstates:F0=fq0:Thestatesinclassq0areallinFg.Forthisdenitiontomakesense,wehavetoarguethatthetransitionsofM0arewelldened:Ifqiandqjarebothrepresentedbythesamestate,weneedtoknowthatri=(qi;a)andrj=(qj;a)arebothrepresentedbythesamestate.Indeed,thismustbethecase:Ifriandrjarerepresentedbydi erentstates,thentheymustbedistinguishable|saybyw.Butthenqiandqjaredistinguishablebyaw,andthiscontradictsthefactthattheybelongtothesameindistinguishabilityclass.NowwecanarguethatL(M)=L(M0).Letw=w1:::wkbeanystring.SupposethatwhenwerunMoninputw,Mgoesthroughthestatesq0;q1;:::;qk,thatis,(qi;wi+1)=qi+1fori=0;:::;k�1.Thenonthesameinputw,M0willgothroughsomesequenceofstatesq00;q01;:::;q0k,whereq0iistherepresentativeofqi. 5Thelaststateq0kwillbeacceptingifandonlyifqkisaccepting:Ifqkisrejecting,thenq0kcontainsarejectingstatesoitisalsorejecting.Ontheotherhand,ifqkisaccepting,thensomustbealltheotherstatesrepresentedbyq0k(becausetheyareindistinguishable),soq0kisalsoaccepting.ItfollowsthatMacceptswifandonlyifM0acceptsw{soMandM0areequivalent.ButM0hasfewerstatesthanM. 4TheDFAminimizationalgorithmNowweunderstandwhatminimalDFAsmustlooklike:Alltheirstatesarereachable,andalltheirpairsofstatesaredistinguishable.Moreover,wesawthatifMisnotminimal,wecanmakeitsmallerbygroupingtogetherstatesintoindistinguishabilityclasses.Itremainstoseehowwecanndtheseindistinguishabilityclassessystematically.Todoso,wemustrstgureoutwhichpairsofstatesofMareindistinguishable.Aneasiertaskistondthepairsthataredistinguishable.Todoso,wewilliterativelyupdatea\table"Xof(unordered)pairsofdistinguishablestates(q;q0)usingthefollowingrules:Initialization:RemoveallunreachablestatesofM.SetXtobeempty.Rule1:Ifqisacceptingandq0isrejecting,addthepair(q;q0)toX.Rule2:If(q;q0)isalreadyinXandr;r0isapairsuchthatq=(r;a)andq0=(r0;a)forsomea2,thenaddthepair(r;r0)toX.Weapplyrules1and2aslongasnewpairscanbeaddedtoXusingtheserules.Oncewearenished,Xwillcontainallpairsofdistinguishablestates.Allpairsofunmarkedstateswillbeindistinguishableandtheycanbemergedtogetherintoindis-tinguishableclasses.TheresultingDFAwillbeminimal.Letusexplainhowthisworksontheaboveexample.WestartwithanemptyX.ApplyingRule1,wecanaddallthepairs(q";q11);(q0;q11);(q1;q11);(q00;q11);(q01;q11),and(q10;q11)toX.Afterthisstep,X=f(q";q11);(q0;q11);(q1;q11);(q00;q11);(q01;q11);(q10;q11)g:NowwecanstartapplyingRule2.Forexample,(q1;q11)isalreadyinX,andweseethat(q";1)=q1,and(q1;1)=q11,sowealsoaddthepair(q";q1)toX.Similarly,wecanalsoincludethepairs(q0;q1),(q00;q1),(q10;q1),(q";q01),(q0;q01),(q00;q01),and(q10;q01).NowX=f(q";q11);(q0;q11);(q1;q11);(q00;q11);(q01;q11);(q10;q11);(q";q1);(q0;q1);(q00;q1);(q10;q1);(q";q01);(q0;q01);(q00;q01);(q10;q01)g:Atthispoint,thereisnothingmoretoaddusingRules1and2.TheonlypairsofstatesnotincludedinXarenowthoseinthelist(1).Thosearethepairsofindistinguishablestates.Aftermergingtogethertheindistinguishablestates,weobtaintheminimizedDFA.