Download - The Halting Problem - Brown University

TheHaltingProblemThereisaspecificproblemthatisalgorithmicallyunsolvable!–  Oneofthemostphilosophicallyimportanttheoremsinthetheoryofcomputation

–  Infact,ordinary/practicalproblemsmaybeunsolvable–  Softwareverification:Givenacomputerprogramandaprecisespecificationofwhattheprogramissupposedtodo(e.g.,sortalistofnumbers).Comeupwithanalgorithmtoprovetheprogramworksasrequired

–  Thiscannotbedone!–  Butwait,can’tweproveanaddition,multiplication,sortingalgorithmworks?

–  Note:Theproofisnotonlytoproveitworksforaspecifictask,likesortingnumbersbutthatitsbehavioralwaysfollowsthespecification!

10/17/19 Theory of Computation - Fall'19

Lorenzo De Stefani 1

Thefirstundecidableproblem

DoesaTMacceptagiveninputstring?– WehaveshownthataCFLisdecidableandaCFGcanbesimulatedbyaTM.

– Thisdoesnotyieldacontradiction.TMsaremoreexpressivethanCFGs.



Why“Halting”problem?

•  ATM={(M,w)|MisaTMandMacceptsw}•  ATMisundecidable–  ItcanonlybeundecidableduetoaloopofMonw.–  Ifwecoulddetermineifitwillloopforever,thencouldreject.HenceATMisoftencalledthehaltingproblem.•  AsitisimpossibletodetermineifaTMwillalwayshaltoneverypossibleinput

–  NotethatthisisTuringrecognizable!WecansimulateMoninputw•  IfMacceptswthenaccept(M,w)•  IfMrejectswthenreject(M,w)



ComparisonofinfinitesetsIn1873mathematicianCantorwasconcernedwiththeproblemofmeasuringthesizesofinfinitesets–  Howcanwetellifoneinfinitesetisbiggerthananotheroriftheyarethesamesize?•  Wecannotusethecountingmethodthatwewoulduseforfinitesets.Example:howmanyevenintegersarethere?

–  Cantorobservedthattwofinitesetshavethesamesizeifeachelementinonesetcanbepairedwiththeelementintheother



FunctionPropertyDefinitionsGivenasetAandBandafunctionffromAtoB–  fisone-to-oneifitnevermapstwoelementsinAtothesameelementinB•  Thefunctionadd-twoisone-to-onewhereasabsolute-valueisnot

–  fisontoifeveryiteminBisreachedfromsomevalueina(i.e.,f(a)=bforeveryb∈B).•  Forexample,ifAandBarethesetofintegers,thenadd-twoisontobutifAandBarethepositiveintegers,thenitisnotontosinceb=1isneverhit.

– Afunctionthatisone-to-oneandontohasa(one-to-one)correspondence•  ThisallowsallitemsinAandBtobepaired



AnExampleofPairingSetItems•  LetNbethesetofnaturalnumbers{1,2,3,…}andletEbethesetofevennaturalnumbers{2,4,6,…}.

•  UsingCantor’sdefinitionofsizewehavethatNandEhavethesamesize.–  ThecorrespondenceffromNtoEisf(n)=2n.

•  ThisissomehowcounterintuitivesinceEisapropersubsetofN!!

•  Focusononthedefinition:sincef(n)isa1:1correspondence,sowesaytheyarethesamesize.

•  Definition:AsetiscountableifeitheritisfiniteorithasthesamesizeasN,thesetofnaturalnumbers(infinitelycountable)



Example:RationalNumbers

•  Q={m/n:m,n∈N},thesetofpositiveRationalNumbers

•  QseemsmuchlargerthanN,butaccordingtoourdefinition,theyarethesamesize.– Hereisthe1:1correspondencebetweenQandN– WeneedtolistalloftheelementsofQandthenlabelthefirstwith1,thesecondwith2,etc.• WeneedtomakesureeachelementinQislistedonlyonce



CorrespondencebetweenNandQ

–  Writingthelistbygoingrow-to-

roworcolumnbycolumnisabadidea!•  Since1strowisinfinite,wouldnevergettothesecondrow

–  Weusediagonals,notaddingthevaluesthatareequivalent•  Sotheorderis1/1,2/1,½,3/1,1/3,…

–  ThisyieldsacorrespondencebetweenQandN•  Thatis,N=1correspondsto1/1,N=2correspondsto2/1,N=3correspondsto½etc.

1/1 1/2 1/3 1/4 1/5 2/1 2/2 2/3 2/4 2/5 3/1 3/2 3/3 3/4 3/5 4/1 4/2 4/3 4/4 4/5 5/1 5/2 5/3 5/4 5/5


Lorenzo De Stefani

Tobuildourcorrespondence,webuildaninfinitematrixcontainingallthepositiverationalnumbers

8

RisUncountable

•  ArealnumberisonethathasadecimalrepresentationandRissetofRealNumbers–  Includesthosethatcannotberepresentedwithafinitenumberofdigits(e.g.,)

•  Willshowthattherecanbenopairing-nopossibleone-to-onecorrespondence-ofelementsbetweenRandN– Proofbycontradiction:Givenanypossibleparingwewillfindsomevaluexthatnotinthepairing



⇡,p2, 3.3̄

<latexit sha1_base64="(null)">(null)</latexit><latexit sha1_base64="(null)">(null)</latexit><latexit sha1_base64="(null)">(null)</latexit><latexit sha1_base64="(null)">(null)</latexit>

•  Assumethatonecompletemappingexits•  Wenowdescribearecipetoobtainavaluex

between0and1whichisnotintheinfinitelist–  Toensurethatx≠f(1),pickadigitnotequaltothe

firstdigitafterthedecimalpoint.Anyvaluenotequalto1willwork.Pick4sowehave.4

–  Tox≠f(2),pickadigitnotequaltotheseconddigit.Anyvaluenotequalto5willwork.Pick6.Wehave.46

–  Continue,choosingvaluesalongthe“diagonal”ofdigits(i.e.,ifwetookthef(n)columnandputonedigitineachcolumnofanewtable).

•  Theselectedvaluexisguaranteedtonotalreadybeinthelistsinceitdiffersinatleastonepositionwitheveryothernumberinthelist.

n f(n)

1 3.14159…

2 55.5555…

3 0.12345…

4 0.500000

. .


Lorenzo De Stefani

RisUncountable

10

ImplicationsRbeinguncountablehasanimportantapplicationinthetheoryofcomputation•  TherearecountablymanyTuringMachines•  Thereareuncountablymanylanguages•  EachTMrecognizesonesinglelanguageàsomelanguagesarenotrecognizedbyanyTuringmachine.– Corollary:somelanguagesarenotTuring-recognizable



SomeLanguagesareNotTuring-recognizableProof:

–  Theset∑*iscountable:thereareonlyafinitenumberofstringsofeachlength,wemayformalistof∑*bywritingdownallstringsoflength0,length1,length2,etc.

–  ThesetofallTuringMachinesMiscountablesinceeachTMMhasanencodingintoastring<M>•  ThesetofvalidTM’sisasubsetofthesetofpossiblestrings.•  Asthelatteriscountable,soistheformer.

–  ThesetofalllanguagesLover∑isuncountable•  thesetofallinfinitebinarysequencesBisuncountable(eachsequenceisinfinitelylong)–  ThesamediagonalizationproofweusedtoproveRisuncountable

•  LisuncountablebecauseithasacorrespondencewithB–  Assume∑*={s1,s2,s3…}.Wecanencodeanylanguageasacharacteristicbinary

sequence,wherethebitindicateswhetherthecorrespondingsiisamemberofthelanguage.Thus,thereisa1:1mapping.

•  SinceBisuncountableandLandBareofequalsize,Lisuncountable–  SincethesetofTMsiscountableandthesetoflanguagesisnot,we

cannotputthesetoflanguagesintoacorrespondencewiththesetofTuringMachines.ThusthereexistssomelanguageswithoutacorrespondingTuringmachine



HaltingProblemisUndecidable

Provethathaltingproblemisundecidable• LetATM={<M,w>|MisaTMandacceptsw}• ProofTechnique:– AssumeATMisdecidableandobtainacontradiction– Adiagonalizationargument



Proof:HaltingProblemisUndecidable

•  AssumeATMisdecidable•  LetHbeadeciderforATM

–  Oninput<M,w>,whereMisaTMandwisastring,HhaltsandacceptsifMacceptsw;otherwiseitrejects

•  ConstructaTMDusingHasasubroutine–  DcallsH(M,<M>)todeterminewhatMdoeswhentheinputstringisits

owndescription<M>.–  DthenoutputstheoppositeofH’sanswer–  D(<M>)acceptsifMdoesnotaccept<M>andrejectsifMaccepts<M>

•  AssumewerunDonitsowndescriptionD(<D>)–  DinvokesH(D,<D>)whichacceptsifDaccepts<D>;otherwiseitrejects–  D(<D>)=acceptifDdoesnotaccept<D>andrejectifDaccepts<D>–  AcontradictionsoHcannotbeadeciderforATM



ConstructingDbydiagonalization

<M1>

<M2> <M3> <M4> … <D>

M1 Accept Reject Accept Reject … Accept

M2 Accept Accept Accept Accept … Accept

M3 Reject Reject Reject Reject … Reject

M4 Accept Accept Reject Reject … Accept

.

D Reject Reject Accept Accept … Contradiction

.



Softwarechecking•  Youwriteaprogram,halts(P,X)thattakesasinputanyprogram,P,andtheinputtothatprogram,X–  Yourprogramhalts(P,X)analyzesPandreturns“yes”ifPwillhaltonXand“no”ifPwillnothaltonX

•  Younowwriteaprocedurelul(X)withassingleinstruction:lul(X){a:ifhalts(X,X)thengotoaelsehalt}ThisprogramhaltsifPdoesnothaltonX;otherwiseitdoeshalt

•  Doeslul(lul)halt?–  Ithaltsifandonlyifhalts(lul,lul)returnsno

•  Ithaltsifandonlyifitdoesnothalt.Contradiction.•  Thuswehaveproventhatyoucannotwriteaprogramtodetermineifanarbitraryprogramwillhaltorloop



Whatdoesthismean?•  Thehaltingproblemaskswhetherwecantellif

someTMMwillacceptaninputstring•  Weareaskingifthelanguagebelowisdecidable–  ATM={(M,w)|MisaTMandMacceptsw}

•  Itisnotdecidable–  Misainputvariabletoo!–  Somealgorithmsaredecidable,likesorting

algorithms•  ItisTuring-recognizable–  SimulatetheTMonwandifitaccepts/rejects,

thenaccept/reject.10/17/19

Theory of Computation - Fall'19 Lorenzo De Stefani 17

Co-TuringRecognizable•  Alanguageisco-TuringrecognizableifitisthecomplementofaTuring-recognizablelanguage

•  Theorem:AlanguageisdecidableifandonlyifitisTuring-recognizableandco-Turing-recognizable–  IfalanguageLisTuring-recognizablethenthereexistsaTM1whichacceptsitsstringsinfinitetime•  IftheTM1doesnotaccept,itmayrejectorloop(itwhichcaseitmaybenotdecidable).

–  IfLisco-Turing-recognizablethenitscomplementLisTuring-recognizable•  HencethereexistTM2whichacceptsstringsfromLinfinitetime•  IfastringisacceptedbyTM2thenwehavethatitisnotinL

– Lisdecidable10/17/19

Theory of Computation - Fall'19 Lorenzo De Stefani 18

ComplementofATMisnotTuring-recognizable

•  Ifalanguageisundecidable,theneitherthelanguageoritscomplementisnotTuring-recognizable!

•  ATMisnotTuring-recognizableProof:– WeknowthatATMisTuring-recognizablebutnotdecidable

–  IfATMwerealsoTuring-recognizable,thenATMwouldbedecidable,whichitisnot

– ThusATMisnotTuring-recognizable



Download - The Halting Problem - Brown University

Top Related