FiniteStateMachines1

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SomeResultsFirstComputingModel

FiniteAutomata

PushdownAutomata

LinearBoundedAutomata

TuringMachines

LanguageClass RegularLanguages

Context-FreeLanguages

Context-SensitiveLanguages

RecursivelyEnumerableLanguages

Non-determinism

Makesnodifference

Makesadifference

Nooneknows Makesnodifference

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Alphabets

• S (sigma):afinite(non-empty)setofsymbolscalledthealphabet.

• EachsymbolinS isaletter.

• Lettersinthealphabetareusuallydenotedbylowercaseletters:a,b,c,…

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Strings• Aword wisastringoflettersfromS inalinearsequence.

• Weareinterestedonlyinfinitewords(boundedlength).

• |w| denotesthelength ofwordw.

• Theemptystring containsnolettersandis

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Languages

• Alanguage Lisaset(finiteorinfinite)ofwordsfromagivenS.

• ThesetofallstringsoversomefixedalphabetS isdenotedbyS*.

• Forexample,ifS ={a},thenS*={e,a,aa,aaa,…}.

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Languages

• Thesetofallstringsoflengthi oversomefixedalphabetS isdenotedbySi.

• Forexample,letS ={a,b}.• ThenL=S2 ={aa,ab,ba,bb}isthesetofwordswsuchthat|w| =2.

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OperationsonWordsandLanguages

• Concatenation: puttingtwostringstogetherx=aa;y=bb;x.y=xy=aabb

• Power: concatenatingmultiplecopiesofaletterorwordan =a.an-1;a1 =a;a2 =a.a;etc.x=ab;x3 =ababab

• KleeneStar: zeroormorecopiesofaletterorworda*={e,a,aa,aaa,…}x=ab;x*={e,ab,abab,ababab,…}

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Deterministic,FiniteStateAutomata

• Afinite-stateautomaton comprisesthefollowingelements:– Asequenceofinputsymbols (theinput“tape”).– Thecurrentlocationintheinput,whichindicatesthecurrentinputsymbol(theread“head”).

– Thecurrentstate ofthemachine(denotedq0,q1,…,qn).

– Atransitionfunction whichinputsthecurrentstateandthecurrentinput,andoutputsanew(next)state.

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Duringcomputation,v TheFSAbeginsinthestartstate(usually,q0).v Ateachstep,thetransitionfunctioniscalledonthecurrentinputsymbolandthecurrentstate,thestateisupdatedtothenewstate,andthereadheadmovesonesymboltotheright.

v TheendofcomputationisreachedwhentheFSAreachestheendoftheinput.

Oneormorestatesmaybemarkedasfinalstates,suchthatthecomputationisconsideredsuccessfulifandonlyifcomputationendsinafinalstate.

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AnExample

• AnFSAcanberepresentedgraphicallyasadirectedgraph,wherethenodesinthegraphdenotestates andtheedgesinthegraphdenotetransitions.Finalstatesaredenotedbyadoublecircle.

• Forexample,hereisagraphicalrepresentationofaDFSAthatacceptsthelanguageL={a2n :n³ 1}:

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• Input:aaaStates:q0,q1,q2,q1 (notaccepted)

• Input:aaaaStates:q0,q1,q2,q1,q2 (accepted)

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DFSADefinition

ADFSAcanbeformallydefinedasA=(Q,S,¶,q0,F):

– Q,afinitesetofstates– S,afinitealphabetofinputsymbols– q0 Î Q,aninitialstartstate– FÍ Q,asetoffinalstates

– ¶ (delta):QxS® Q,atransitionfunction

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Transitionfunction- ¶

• Wecanexpandthenotionof¶ onlettersto¶onwords,¶w,byusingarecursivedefinition:

• ¶w :QxS*® Q- (afunctionof(state,word)toastate)• ¶w(q,e)=q - (instateq,outputstateqifwordise)• ¶w(q,xa)=¶(¶w(q,x),a)- (otherwise,use¶ foronestep

andrecurse)

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LanguageRecognition• ForanautomatonA,wecandefinethelanguageofA:- L(A)={wÎ S*:¶w(q0,w)Î F}- L(A)isasubsetofallwordswoffinitelengthoverS,suchthatthetransitionfunction¶w(q0,w)producesastateinthesetoffinalstates(F).- Intuitively,ifwethinkoftheautomatonasagraphstructure,thenthewordsinL(A)representthe“paths”whichendinafinalstate.Ifweconcatenatethelabelsfromtheedgesineachsuchpath,wederiveastringwÎ L(A).

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ImplementingDFSAinJava(afirstattempt)

• ImplementingaDFSAinJavahassomesimilaritiestoimplementingagraphstructure.Asmentionedearlier,thestatesinaDFSAcorrespondtonodesinagraph,andthetransitionscorrespondtoedgesinagraph.

• First,let’sconsiderhowtoimplementthetransitionfunction,¶.Recallthat¶(<state>,<letter>)=<state>.So,foranygivenstateq,weneedtoknowifthereisatransitiontotheotherstates,andifso,whatletterofthealphabetmustbereadforthetransitiontooccur.

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• Ifeachvertexinadirectedgraphhasatmostoneedgeleadingtoanothervertex(possiblythevertexitself)thenwecanmodelthegraphusingatwo-dimensionalarray.

• SupposewewanttomodelaDFSAinthismanner.ForeachstateintheDFSA,thereisarowinthearray;eachcolumninthatrowcorrespondstoa(possible)transitiontoanotherstate.Eachcellisempty(null)ifthereisnosuchtransition;otherwise,acellcontainstheletterofthealphabetwhichmustbereadforthetransitiontobevalid.

• Assumingwenamethearraydelta,thendelta(m,n)==nullifthereisnotransitionfromQm toQn,anddelta(m,n)==<letter>if¶(Qm,<letter>)=Qn.

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Whatiswrongwiththisapproach?

TheDFSAonthepreviouspagehas3states,soitcanbemodeledbya3x3array:

Q0 Q1 Q2 array transitionfunction

Q0 null a null delta(0,1)=a ¶(Q0,a)=Q1

Q1 null null a delta(1,2)=a ¶(Q1,a)=Q2

Q2 null a null delta(2,1)=a ¶(Q2,a)=Q1

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• Notethatthisworksonlywhenthereisasingletransitionfromonestatetoanother.IfS wereexpandedtocontaintwocharactersratherthanone,andifthereweretwotransitionsleadingfromQ0tosay,Q1,thenwecouldnotusethisimplementation.

• BeforewewriteaJavaclassforthisDFSA,wehaveonemorethingtoconsider.– Howcanwerepresentthesetoffinalstates?

• Astraightforwardsolutionistouseaone-dimensionalbooleanarray,final,oflengthn(assumingwehavenstates).Thenfinal(i)=trueifqi isafinalstate,andfinal(i)=falseotherwise.

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publicclassDFSA{

privateboolean[]finalStates;privatechar[][]delta;privateintstartState;privateintcurrentState;privateinttotalStates;

publicDFSA(intn,intstart){totalStates=n;finalStates=newboolean[totalStates];delta=newchar[totalStates][totalStates];startState=start;

}

privatebooleanisFinal(){returnfinalStates[currentState];

}

publicvoidaddTransition(intfromState,inttoState,charletter){delta[fromState][toState]=letter;

}publicvoidaddFinalState(intq){finalStates[q]=true;

}

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publicbooleanisAccepted(Strings){currentState=startState;readingSymbols:for(inti=0;i<s.length();i++){System.out.println("currentstate:"+currentState);System.out.println("nextsymbol:"+s.charAt(i));for(intj=0;j<totalStates;j++){

if(delta[currentState][j]==s.charAt(i)){currentState=j;continuereadingSymbols;

}}System.out.println("d("+currentState+","+s.charAt(i)+")isundefined");returnfalse;

}System.out.println("finalstate:"+currentState);returnisFinal();

}

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publicstaticvoidmain(Stringargs[]){

//Definethethree-stateDFSAfromthehandout.

DFSAa=newDFSA(3,0);a.addTransition(0,1,'a');a.addTransition(1,2,'a');a.addTransition(2,1,'a');a.addFinalState(2);

//Foreachinputstring,checkvalueofisAccepted()

for(inti=0;i<args.length;i++){System.out.println("\nInputString:"+args[i]);if(a.isAccepted(args[i])){System.out.println("Accepted");

}else{System.out.println("Rejected");

}}

}

}

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>javaDFSAaaaaaaaaabaa

InputString:aaacurrentstate:0nextsymbol:acurrentstate:1nextsymbol:acurrentstate:2nextsymbol:afinalstate:1Rejected

InputString:aaaacurrentstate:0nextsymbol:acurrentstate:1nextsymbol:acurrentstate:2nextsymbol:acurrentstate:1nextsymbol:afinalstate:2Accepted

InputString:aabaacurrentstate:0nextsymbol:acurrentstate:1nextsymbol:acurrentstate:2nextsymbol:bd(2,b)isundefinedRejected

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Let’sconsidermakingcertainmodificationstotheprogramabovesothatitallowsformultipletransitionsfromonestatetoanotherbasedonaS withmorethanonecharacter.

Thereareseveralwaysthiscouldbedone.- Onemightusealargetwo-dimensionalarraywhoserowsareindexedwithstatenumbersandwhosecolumnsareindexedwithallthecharactersfoundinS.ThiswouldgiveO(1)transitionlookupsbutwouldbeinefficientwithrespecttospace.- Anotherapproachwouldbetouseanadjacenysetrepresentation.Asingledimensionalarray,saystates,couldbeindexedbasedonthestatenumber.Theelementatstates[i]wouldcontainasetof<character><nextstate>pairs.Eachsetwouldallowfast<nextstate>lookupsgiventhecharactersymbolfromS.Thesetcouldbeimplementedwithalinkedlistor,perhaps,withabalancedtree.

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• Considermodifyingtheprogramabovesothatitusesanadjacenysetrepresentationtocompute¶(Qm,<letter>).Thesetcouldbeimplementedasadigitalsearchtree.TheDSTclasswouldallowfortheinsertionandsearchof(<letter>,<next-state>)pairs.The<letter>partofthepairwouldbethesearchkey.Wheninsertingorsearching,thedecisiontomoveleftorrightinthesearchtreewouldbebasedonthebitsofthekey.Theimplementationeffortforthistypeofdatastructureisaboutthesameasbinarysearchtreesbutusuallyhasbetterperformance.

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AnotherExampleConsiderrunningsuchaprogramthatwouldmodeltheDFSAbelow.Suchaprogrammightreadaseriesofstringsfromthecommandline(justliketheprogramabove)andtelltheuserwhetherthemachinebelowacceptsorrejectseachstring.

Q={q0,q1,q2 }S ={R,0,1,2}q0:thestartstateF={q0 }¶ (delta):QxS® Q

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Thisautomatonkeepsarunningcountofthesumofthenumericalinputsymbolsitreads,modulo3.EverytimeitreceivestheR(reset)symbolitresetsthecountto0.Itacceptstheifthesumis0,modulo3,orinotherwords,ifthesumisamultipleof3.Thisautomatonisfrom“IntroductiontotheTheoryofComputation”byMichaelSipser

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HomeworkQuestions(nottobeturnedinbuttoprepareforexam)

GivestatediagramsforDFA’srecognizingthefollowinglanguages.S ={0,1}.

1.{w|wbeginswitha1andendswitha0}2.{w|wcontainsatleastthree1’s}3.{w|thelengthofwisatmost5}4.{w|wcontainsatleasttwo0’sandatmostone1.}5.{w|wcontainsanevennumberof0’s,orexactlytwo1’s}

6.{w|wisnote }

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