CS1010: Theory of Computation · – If you had to list the elements of A DFA what would they be? • All the pairs (Automata, string) such that “string” is accepted by “automata”

CS1010:TheoryofComputation

LorenzoDeStefaniFall2019

Lecture11:DecidabilityandLanguages

LimitationsofAlgorithmicSolvability

•  Inthispartoftheclassweinvestigatethepowerofalgorithmstosolveproblems–  Somecanbesolvedalgorithmicallyandsomecannot

•  Whydowestudyunsolvability?–  Importantguideline

•  Ifaproblemisunsolvablesearchingforanalgorithmicsolutionisawasteoftime

•  Perhapstheproblemcanbesimplified–  Gainunderstandingoncomputabilityanditslimits–  Firststeptowardscomplexitycharacterization

•  Firstweascertainwhetherthereisanalgorithmicsolutionandthenwestudywhetherthereisan“efficient”(polynomial-time)one

10/10/19 Theory of Computation - Fall'19

Lorenzo De Stefani 1

Outline

•  Decidablelanguages•  Regularlanguagesaredecidable•  Context-freelanguagesaredecidable•  Languageshierarchy



FromSipserChapter4.1

DecidableLanguages

•  Westartwithproblemsthataredecidable– Wefirstlookatproblemsconcerningregularlanguagesandthenthoseforcontext-freelanguages

•  …eventuallywewillmovetoproblemsconcerningTuringMachinesandshowthatsomeproblemsarenotdecidable!



DecidableProblemsforRegularLanguages

•  Weconsideralgorithmsfortesting–  whetherafiniteautomatonacceptsastring–  whetherthelanguageofafiniteautomatonisempty–  whethertwofiniteautomataareequivalent

•  Werepresenttheproblemsaslanguages(notFAs)–  LetADFA={(B,w)|BisaDFAthatacceptsstringw}–  TheproblemoftestingwhetheraDFABacceptsaspecificinputwisthesameastestingwhether(B,w)isamemberofthelanguageADFA.

–  IfyouhadtolisttheelementsofADFAwhatwouldtheybe?•  Allthepairs(Automata,string)suchthat“string”isacceptedby“automata”


Lorenzo De Stefani

Showingthatthelanguageisdecidableisthesamethingasshowingthatthecomputationalproblemoftestingacceptanceisdecidable

4

ADFAisaDecidableLanguage

ProofIdea:ConstructaTuringMachineMthatdecidesADFAM=Oninput(<B>,w),whereBisaDFAandwisastring:

1.  SimulateBoninputw2.  Ifthesimulationendsinanacceptstate,thenaccept;else

reject


Lorenzo De Stefani

Theorem:ADFAisadecidablelanguage

5

OutlineofProof•  MmusttakeBasinput,encodedasastring,andthensimulateit

–  ThealgorithmforsimulatinganyDFAmustbeembodiedintheTM’sstatetransitions

–  Givenacurrentstateandinputsymbol,scanthetapefortheencodedtransitionfunctionandthenusethatinfotodeterminenewstate

•  TheactualproofdescribeshowaTMsimulatesaDFA–  CanassumeBisrepresentedbyits5componentsandthenwehavew

•  NotethattheTMmustbeabletohandleanyDFA•  Keeptrackofcurrentstateandpositioninwbywritingonthetape•  Initiallycurrentstateisq0andcurrentpositionisleftmostsymbolofw

–  Thestatesandpositionareupdatedusingthetransitionfunctionδ•  TMM’sδnotthesameasDFAB’sδàthelatterisencodedastheinput

–  WhenMfinishesprocessing,acceptifinanacceptstate;elsereject.Theimplementationwillmakeitclearthatwillcompleteinfinitetime.



ANFAisaDecidableLanguage

ProofIdea:–  BecausewehaveprovendecidabilityforDFAs,allweneedto

doisconverttheNFAtoaDFA.Oninput(B,w)whereBisanNFAandwisastring

1.  ConvertNFABtoanequivalentDFAC,usingtheprocedureforconversiongiveninTheorem1.39

2.  RunTMMoninput(C,w)usingthetheoremwejustproved3.  IfMaccepts,thenaccept;elsereject

–  RunningTMMinstep2meansincorporatingMintothedesignofNasasubroutine



ComputingwhetheraDFAacceptsanyString

•  EDFA={<A>|AisaDFAandL(A)=∅)isadecidablelanguage•  Proof:

–  ADFAacceptssomestringifandonlyifitispossibletoreachtheacceptstatefromthestartstate.Howcanwecheckthis?

–  Wecanuseamarkingalgorithmsimilartotheoneusedtodecidethelanguageofconnectedgraphs.

–  T=Oninput(A)whereAisaDFA:1.  MarkthestartstateofA2.  Repeatuntilnonewstatesgetmarked:

3.  Markanystatethathasatransitioncomingintoitfromanystatealreadymarked

4.  Ifnoacceptstateismarked,accept;otherwisereject



EQDFAisaDecidableLanguage•  EQDFA={(A,B)|AandBareDFAsandL(A)=L(B)}•  Proofidea

–  ConstructaDFACfromAandB,whereCacceptsonlythosestringsacceptedbyeitherAorBbutnotboth(symmetricdifference)•  IfAandBacceptthesamelanguage,thenCwillacceptnothingandwe

canusethepreviousproof(forEDFA)tocheckforthis.•  So,theproofis:

–  F=Oninput(A,B)whereAandBareDFAs:1.  ConstructDFACthatisthesymmetricdifferenceofAandB

(detailsonhowtodothisonnextslide)2.  RunTMTfromtheprooffromlastslideoninput(C)3.  IfTaccepts(sym.diff=∅)thenaccept.IfTrejectsthenreject



HowtoConstructC

L(C)=(L(A)∩L(B))∪(L(A)∩L(B))– Weusedproofsbyconstructionthatregularlanguagesareclosedunder∪,∩,andcomplement

– WecanusethoseconstructionstoconstructaFAthatacceptsL(C)• Weneverprovedregularlanguagesareclosedunder∩!

L(A) L(B)



RegularLanguagesClosedunderIntersection

IfLandMareregularlanguages,thensoisL∩M• Proof:LetAandBbeDFAswhoseregularlanguagesareLandM,respectively

–  ConstructC,the“productautomation”ofAandB–  CtracksthestatesinAandB(justlikewhenwedidtheproofofunionwithoutusingNFAs)

– MakethefinalstatesofCbethepairsconsistingoffinalstatesofbothAandB•  IntheunioncasewethefinalstateanystatewithafinalstateinAorB





ACFGisaDecidableLanguageProofIdea:–  ForCFGGandstringwwewanttodecidewhetherGgeneratesw.

–  OneideaistouseGtogothroughallderivations.Thiswillnotwork,why?•  ThismethodabestwillyieldaTMthatisarecognizer,notadecider.•  Cangenerateinfinitestringsandifnotinthelanguage,willneverknowit.

–  Canweusetheinformationonthelengthofw?–  Astringwoflengthnwillhaveaderivationthatuses2n-1stepsiftheCFGisin

Chomsky-NormalForm.•  FirstconverttoChomsky-NormalForm•  Thenlistallderivationsoflength2n-1steps.Ifanygeneratesw,thenaccept,elsereject.

•  Thisisavariantofbreadthfirstsearch,butinsteadofextendedthedepth1atatimeweallowittogo2n-1atatime.Aslongasfinitedepthextension,weareokay



ECFGisaDecidableLanguage

•  ECFG={<G>|GisaCFGandL(G)=∅)isadecidablelanguage

•  Whatisthe“bruteforceapproach”?–  Tryallpossiblestringsw.Willthiswork?

•  Thenumberisnotbounded,sothiswouldnotbedecidable•  Instead,thinkofthisasagraphproblemwhereyouwanttoknowifyoucanreachastringofterminalsfromthestartstate•  Doyouthinkitiseasiertoworkforwardorbackwards?•  Answer:backwards



ECFGisaDecidableLanguage(cont)

ProofIdea:–  Canthestartvariablegenerateastringofterminals?– Determineforeachvariableifitcangenerateanystringofterminalsandifso,markit

–  Keepworkingbackwardssothatiftheright-sideofanyrulehasonlymarkeditems,thenmarktheLHS•  Forexample,ifXàYZandYandZaremarked,thenmarkX•  IfyoumarkS,thendone;ifnothingelsetomarkandSnotmarked,thenreject

•  Youstartbymarkingallterminalsymbols



EQCFGisnotaDecidableLanguage

•  WecannotreusethesamereasoningusedtoshowthatEQDFAisadecidablelanguagesinceCFGsarenotclosedundercomplementandintersection

•  Asitturnsout,EQCFGisnotdecidable!



EveryContext-FreeLanguageisDecidable

•  WepreviouslyshowedACFGisdecidable.•  WewanttoknowifA,whichisaCFL,isdecidable.–  AwillhavesomeCFGGthatgeneratesit– WhenweprovedthatACFGisdecidable,weconstructedaTMSthatwouldtellusifanyCFGacceptsaparticularinputw.

–  NowweusethisTMandrunitoninput<G,w>andifitaccepts,weaccept,andifitrejects,wereject.



HierarchyofClassesofLanguages

Regular

Context-Free

Decidable

Turing-recognizable

•  WeprovedRegular⊆Context-freesincewecanconvertaFAintoaCFG

•  WejustprovedthateveryContext-freelanguageisdecidable

•  FromthedefinitionsinChapter3itisclearthateveryDecidablelanguageistriviallyTuring-recognizable.

•  NoteveryTuring-recognizablelanguageisDecidable!



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CS1010: Theory of Computation · – If you had to list the elements of A DFA what would they be? • All the pairs (Automata, string) such that “string” is accepted by “automata”