CS1010:TheoryofComputation

LorenzoDeStefaniFall2020

Lecture1:DeterministicFiniteStateAutomata(DFA)

Outline• WhatisaFiniteStateAutomaton• DFAdefinition• ExampleDFAconstruction• ThelanguageofaDFA• RegularOperations• Closureunderunion• Closureunderconcatenation

1

FromSipser Chapter1.1

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WhatisaComputer?

• Westartwithasimplecomputationalmodel– Differentmodelswillhavedifferentfeaturesandmaymatcharealcomputerbetterinsomewaysandworseinothers

– Tradeoffbetweengeneralityandprecision• Ourfirstmodelisthefinitestatemachineorfiniteautomaton– Modelformachineswithfinitememory

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FiniteAutomataModelsofcomputerswithextremelylimitedmemory

– Manysimplecomputershaveextremelylimitedmemories andare,infact,finitestatemachines

– Canyounameany?• Vendingmachine• Elevator• Thermostat• Automaticdooratsupermarket

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AutomaticDoor

• Whatisthedesiredbehavior?Describetheactionsandthenlistthestates.– Personapproaches,doorshould open– Doorshould stayopenwhilepersongoingthrough– Doorshould shutifnoonenearthedoorway– Statesareopen andclosed

• Moredetailsaboutautomaticdoor– FrontpadDoorRearPad– Describebehaviornow:

• Hint:actiondependsnotjustonwhathappens,butwhatstate youarecurrentlyin

• Ifyouwalkthrough,thedoorshouldstayopenwhileyouareonrearpad• Butifdoorisclosedandsomeonestepsonrearpad,doordoesnotopen

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Input“signals/actions”

AutomaticDoor

NEITHER FRONT REAR BOTH

CLOSED CLOSED OPEN CLOSED CLOSED

OPEN CLOSED OPEN OPEN OPEN

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CLOSED OPEN

FRONT

NEITHER

REAR,BOTH,NEITHER FRONT,REAR,BOTH

9/15/20

MoreonFiniteAutomata

• HowmanybitsofdatadoesthisFiniteStatesMachine(FSM)store?– 1bitforstates:openorclosed

• Whataboutstateinformationforelevators,thermostats,vendingmachines,etc?

• FSMusedinspeechprocessing,opticalcharacterrecognition,etc.

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Afiniteautomaton

AfiniteautomatonM1 with3states– Statediagram

• Startstateq0 (enteringarrowfromnostate),• Acceptstateq1 (doublecircle),• Transitions(arrows)

– M1 willaccept(orrecognize)astringlike“1101”ifitendsinacceptstate.Otherwisethestringisrejected!

– Canyoudescribeallstrings(i.e.,apropertysharedbyallandonlythestrings)thatthisautomatonwillaccept?

• Itwillacceptallstringsendingina1andanystringwithanevennumberof0’sfollowingthelast1

00

1

1

0,1

q0 q2q1

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FormalDefinitionofFiniteAutomata

Afiniteautomatonisa5-tuple(Q,S,δ,q0,F)– Qisafinitesetcalledstates– S isafinitesetcalledthealphabet– δ:QxS® Qisthetransitionfunction– q0Î Qisthestartstate– FÍ Qisthesetofacceptstates

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FormalDefinition

• M1 =(Q,S,δ,q0,F)– Q=– S =– q0 isthestartstate– F=

0 1q0 q0 q1q1 q2 q1q2 q1 q1

0 0

1

1

0,1

q0 q2q1

Transitionfunctionδ

{q0,q1,q2}{0,1}

{q1}

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TheLanguageofanautomaton• ThelanguageofDFAMisthesetAofallstringsacceptedbytheDFAM– L(M)=A– WealsosaythatMrecognizesAorMacceptsA

• Convention:Macceptsstringsandrecognizesalanguage

• Attentiontoquantifiers:a machinemayacceptmanystrings,butonlyonelanguage

• Multiple automatacanrecognizethesame language,butanautomatononly recognizesasingle language

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WhatistheLanguageofM1?

• L(M1)=AorM1 recognizesA• WhatisA?

– A={w|…….}– A={w|wcontainsatleastone1andanevennumberof0’sfollowsthelast1}

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WhatistheLanguageofM2?

• M2 ={{q0,q1},{0,1},δ,q0,{q1}}– Ileaveδ asanexercise– WhatisthelanguageofM2?

• L(M2)={w|?}• L(M2)={w|wendsina1}

0

1

q0

1

0

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q1

WhatistheLanguageofM3?

• M3 isM2 withadifferentacceptstate• WhatisthelanguageofM3?

– L(M3)={w|?}– L(M3)={w|wendsin0}[Notquiteright!Why?]– L(M3)={w|wistheemptystringε orendsin0}

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1

0

q1q0

0 1

WhatistheLanguageofM4?WhatlanguagedoesM4 recognize?

s

r1

r2

q1

q2

Figure1.12onpage38

a b b

bb

b

a

a a

a

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WhatistheLanguageofM4?

WhatlanguagedoesM4 recognize?– M4 acceptsallstringsthatstartandendwith“a”orstartandendwith“b”

– Moresimply,languageL(M4)isthesetofallstringstartingandendingwiththesamesymbol

• Notethatlength1isallowed!

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ConstructM5 todoModulo-3Arithmetic

• Alphabetå ={RESET,0,1,2}• ConstructM5 toacceptastringonlyifthesumofeachinputsymbolmodulo3is0andRESETsetsthesumbackto0(1.13,page39)– ({q0,q1,q2},{0,1,2,RESET},δi,q0,{q0})

• δi(qj,0)=qj• δi(qj,1)=qk wherek=j+1modulo3• δi(qj,2)=qk wherek=j+2modulo3• δi(qj,RESET)=qo

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NowGeneralizeM5• GeneralizeM5 toacceptifsumofsymbolsisamultipleofi insteadof3– ({q0,q1,q2,q3,…,qi-1},{0,1,2,RESET},δi,q0,{q0})

• δi(qj,0)=qj• δi(qj,1)=qk wherek=j+1moduloi• δi(qj,2)=qk wherek=j+2moduloi• δi(qj,RESET)=qo

• Note:aslongasi isfinite,weonlyneedfinitememory (log#ofstates)

• Couldyougeneralizeonå ={1,2,3,…k}?– Setofstateswouldnotchange– Moretransitions!

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FormalDefinitionofAccept

• LetM=(Q,S,δ,q0,F)beafiniteautomatonandletw=w1w2 …wn beastringwherewi Î å

• ThenMacceptswifthereexistsasequenceofstatesr0,r1,…,rn inQsuchthat:– r0=q0– δ(ri,wi+1)=ri+1,fori =0,1,…,n-1– rn Î F

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RegularLanguages

Definition: Alanguageiscalledaregularlanguage ifthereexistsafiniteautomatonwhichrecognizesit• Thatis,ifthereexistsafiniteautomatonthatacceptsallandonly thestringsinthelanguage

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DesigningFiniteAutomata• YouwillneedtodesignFAwhichacceptagivenlanguage

• Strategy:– Determinewhatyouneedtoremember(thestates)

• E.g.,Howmanystatestodetermineeven/oddnumberof1’sinaninput?

• Whatdoeseachstaterepresent– Setthestartandacceptstatesbasedonwhateachstaterepresents

– Assignthetransitions– Checkyoursolution:itshouldacceptwÎ LandnotacceptwnotinL

– Becarefulabouttheemptystring

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DesigningFAs

• DesignaFAtoacceptthelanguageofbinarystringswherethenumberof1’sisodd,zerocountsaseven(page43)

• DesignaFAtoacceptallstringwith001asasubstring(page44)

• DesignaFAtoacceptastringwithsubstringabab

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ALWAYSstartbyasking:“WhatdoIneedtoremember?”

RegularOperations

LetAandBbelanguages.Wedefine3regularoperations:• Union:AÈ B={x|xÎAorxÎB}• Concatenation:A× Bwhere{xy|xÎAandyÎB}• Star:A* ={x1x2 ….xk|k≥0andeachxi Î A}

– Starisaunaryoperatoronasinglelanguage– Starrepeatsastring0ormoretimes

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ExamplesofRegularOperations

• LetA={0,1}andB={c,d}• Then:

– AU B=– A× B=– A*=

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{0,1,c,d}{0c,0d,1c,1d}

{ε,0,1,00,01,10,11,000,…}

Closure

• Acollectionofobjectsisclosedunderanoperation ifapplyingthatoperationtomembersofthecollectionreturnsanobjectinthecollection

• E.g.,Thesetofnaturalnumbersisclosedunderadditionandmultiplication(butnotdivisionandsubtraction)

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ClosureforRegularLanguages

• Regularlanguagesareclosedundertheregularoperatorsunion,concatenationandstar

• Whywecare?– Iftheseoperatorsareclosed,then,ifwecanimplementeachoperatorusingaFA,wecanalsobuildaFAtorecognizeanyoftheircombinations

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ClosureofUnionTheorem1.25: Theclassofregularlanguagesisclosedundertheunionoperation• IfA1 andA2 areregularlanguagesthensoisA1È A2• Howcanweprovethis?Useproofbyconstruction:

– AssumeM1 acceptsA1 andM2 acceptsA2– ConstructM3 usingM1 andM2 toacceptA1È A2– Weneedtosimulate M1 andM2 runninginparallel andstopifeitherreachesanacceptstate

• Thislastpartisfeasiblesincewecanhavemultipleacceptstates• Youneedtoremember whereyouwouldbeinbothmachines

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ClosureofUnion• YouneedtogenerateastatetorepresentthestateyouwouldbeinforbothM1 andM2

• LetM1=(Q1,S,δ1,q1,F1)andM2=(Q2,S,δ2,q2,F2)• BuildM3=(Q,S, δ,q0,F)asfollows:

– Q={(r1,r2)|r1 Î Q1 andr2 Î Q2 }(CartesianproductQ=Q1xQ2)• Careful:orderof(r1,r2)doesnotmatter!• Morecorrect:QisthesetofunorderedpairssuchthatonestateisinQ1 andtheotherisinQ2

– S =S1È S2– q0 isthepair(q1,q2)– F={(r1,r2)|r1 Î F1 orr2 Î F2}– δ((r1,r2),a)=(δ(r1,a),δ2(r2,a))

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ClosureofConcatenationTheorem1.26:Theclassofregularlanguagesisclosedundertheconcatenationoperator

– IfA1 andA2 areregularlanguagesthensoisA1 × A2– CanweconstructaDFAwhichrecognizesA1 × A2?– Canyouseehowtodothissimply?

• SinceA1 andA2 areregular,thereexistsaDFAM1 whichrecognizesA1andM2 whichrecognizesA2

• CombiningM1 andM2 isnottrivial:– WecannotjustconcatenateM1 andM2,wherestartstateofM2becomethefinishstatesofM1

– Becausewedonotacceptastringassoonasitentersthefinishstate(waituntilstringisdone)itcanleaveandcomeback

– ThuswedonotknowwhentostartusingM2;ifwemakethewrongchoicewillnotacceptastringthatcanbeaccepted

– ThisproofiseasyifwehaveNondeterministicFA(NFA)

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Concatenation:SimpleExample

Concatenationofthefollowing:– L(M1)=A,whereS ={0,1}andA=binarystringwithexactly2“1”’s

– L(M2)=B,whereS ={0,1}andB=binarystringwithexactly3“1”’s

• M1 willenteracceptstateassoonassees2“1”’s.• Itcanthenloopbackonany“0”’sormove toM2withoutissue.

• ItcanmoveimmediatelytoM2 ona1,andnothaveanissuesinceitcannotloopback

• SinceAincludesonlystringswithexactly 2“1”’s.OnceinM2 everythingwillworkokay.

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Concatenation:HardExample• L(M1)=A,whereS ={0,1}andA=binarystringwithatleast2“1”’s• L(M2)=B,whereS ={0,1}andB=binarystringwithexactly 2“1”’s• Previousconstructiondoesnotwork(buteasywithNFAormore

complicatedDFA)– IfwhileinM1 andsee2“1”’s,enteracceptstate.Whenseeanother1,

havechoicetoloopbackintoacceptstateinM1,orstartmovingintoM2,tothestatethatrepresentssawfirst1forstringinB.

• Iftheconcatenatedstringhasexactly4“1”’stotal,thenwillonlyacceptifmoveintoM2 asearlyaspossible(afterseeingthefirst21’s)

• Iftheconcatenatedstringhasmorethan4“1”’s,thenwillonlyacceptifloopinM1 acceptstateuntilonly2“1”’sleft.

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