outline...turing machine scheme control a b a b – – – … • turing machine include an...

Outline

•  WhatareTuringMachines•  TuringMachineScheme•  FormalDefinition•  LanguagesofaTM•  Decidability

10/1/19 TheoryofComputation-Fall'19LorenzoDeStefani 1

FromSipserChapter3.1

FA,PDAandTuringMachines•  FiniteAutomata:–  Modelsfordeviceswithfinitememory

•  PushdownAutomata:– Modelsfordeviceswithunlimitedmemory(stack)thatisaccessibleonlyinLast-In-First-Outorder

•  TuringMachines(Turing1936)– Usesunlimitedmemoryasaninfinitetapewhichcanberead/writtenandmovedtoleftorright

– Onlymodelthusfarthatcanmodelgeneralpurposecomputers– Church-Turingthesis

– Still,TMcannotsolveallproblems10/1/19 TheoryofComputation-Fall'19LorenzoDeStefani 2

TuringMachineScheme

Control

a b a b – – – …

•  Turingmachineincludeaninfinitetape– TapeusesitsownalphabetΓ,with–  Initiallycontainstheinputstringandblankseverywhereelse

– Machinecanreadandwritefromtapeandmoveleftandrightaftereachaction

– MuchmorepowerfulthanFIFOstackofPDAs

⌃ ⇢ �(null)(null)(null)(null)

Tape

–

Head


TuringMachineSchemeControl

a b a b – – – …

•  Thecontroloperatesasastatemachine– Startsinaninitialstate– Proceedsbyaseriesoftransitionbasedonthevalueonthetape

– Themachinecontinuesuntilitentersanacceptorrejectstateatwhichpointitimmediatelyhaltsandoutputsacceptorreject

– NotethisisverydifferentfromFAsandPDAs


TuringMachineSchemeControl

a b a b – – – …

•  Themachinecanloopforever!–  InthiscasewesaythattheTMdoesnothaltforagiveninput

•  CanaFAoraPDAloopforever?– NO!itwillterminatewheninputstringisfullyprocessedandwillonlytakeone“action”foreachinputsymbol


DesigningTuringMachines

DesignaTMtorecognizethelanguage:B={w#w|w∈{0,1}*}–  Thinkofaninformaldescription– NOTPALINDROMES–  ImaginethatyouarestandingonaninfinitetapewithsymbolsonitandwanttochecktoseeifthestringbelongstoB?•  Whatprocedurewouldyouusegiventhatyoucanread/writeandmovethetapeinbothdirections?

•  Youhaveafinitecontrolsocannotremembermuchandthusmustrelyontheinformationonthetape


ATuringMachineforB ={w#w|w ∈{0,1}*}

•  M1loopsandineachiterationitmatchessymbolsoneachsideofthe#–  Itreadstheleftmostsymbolremainingandreplacesitwith“x”–  Scanstotherightuntilthe“#”andproceedstothefirstnon-xsymbol–  Isitthesame?

•  Yes!Wehaveamatch!Wegobacktotheleftmostremainingsymbolandrepeat

•  No!Wehaveamismatch=(theTMtransitiontoa“rejectstate”andhalts

–  Ifboththesymbolstotherightandtotheleftofthe“#”arexthenwehaveacompletematch!Alsochecksamelength!!!

–  TheTMhaltsandthestringisaccepted•  Isloopingpossible?

–  NO!Guaranteedtoterminate/haltsincemakesprogresseachiteration.


•  M1loopsandineachiterationitmatchessymbolsoneachsideofthe#–  Itreadstheleftmostsymbolremainingandreplacesitwith“blank”–  Scanstotherightuntilthe“#”andproceedstothefirstnon-blanked

symbol–  Isitthesame?

•  Yes!Wehaveamatch!Wegobacktotheleftmostremainingsymbolandrepeat

•  No!Wehaveamismatch=(theTMtransitiontoa“rejectstate”andhalts

–  Ifthesymboltotherightofthe“#”isblankthewehaveacompletematch!TheTMhaltsandthestringisaccepted

•  Isloopingpossible?–  NO!Guaranteedtoterminate/haltsincemakesprogresseach

iteration.




qstart

q0

q1

q#1

q#0

qrevqrejectqrev2

qalmost

qbegin

0àx,R

#à#,R

0à0,R1à1,R xàx,L

0àx,L

1à1,R

xàx,L

#à#,L

0à0,L

1à1,L

1àx,R

#à#,R

1àx,L 0à0,R

_à_,R

_à_,R

xàx,L

qaccept

0à0,L1à1,L

xàx,R

0àx,R

1àx,R

#à#,R

_à_,R

xàx,R

0à0,R1à1,R


Conventionsofrepresentation

•  Read“aàb,R”as:ifsymbol“a”isreadonthetapethenreplaceitwith“b”andmovetotheright

•  Weassumemissingtransitionsleadtorejectstate


ExecutionExample

Inputstring011000#011000Thetapeheadisatrightofthestate qstart011000#011000-- xq111000#011000-- … X11000#qrevX11000-- qrev2X11000#X11000-- xxq11000#X11000-- … XXXXXX#XXXXXX--


FormalDefinitionofaTuringMachine

ATuringMachineisa7-tuple{Q,Σ,Γ,δ,q0,qaccept,qreject},where:•  Qsetofstates•  Σistheinputalphabetnotcontainingtheblank•  Γisthetapealphabet,whereblank_∈ΓandΣ⊆Γ•  δ:QxΓ→QxΓx{L,R}isthetransitionfunction•  q0,qaccept,andqrejectarethestart,accept,andrejectstates–  Doweneedmorethanonerejectoracceptstate?–  No:sinceonceentersuchastateyouterminate


TransitionsintheTM

Thetransitionfunctionδiskey:– QxΓ→QxΓx{L,R}

•  Amachineisinastateqandtheheadisoverthetapeatsymbola,thenafterthemoveweareinastaterwithbreplacingtheaonthetapeandtheheadhasmovedeitherleft(L)orright(R)


Configurationsandtransitions•  AtanystepaTMisinacertainconfigurationwhichisspecifiedby:–  thestate–  thecurrentheadlocation–  thesymbolatthecurrentheadlocation

•  WesaythataconfigurationC1yieldsC2ifthereisatransitionwhichallowsgofromC1toC2


Typesofconfigurations•  Startingconfiguration:instartingstate,headpositionatthebeginningoftheinput– Leftmostpositionofthetapeoccupiedbytheinput

•  Acceptingconfiguration:inacceptingstate•  Rejectingconfiguration:inrejectingstate•  Haltingconfiguration:eitheracceptingorrejectingconfigurations


TuringRecognizable&DecidableLanguages

•  ThesetofstringsthataTuringMachineMacceptsisthelanguageofM,orthelanguagerecognizedbyM,L(M)–  AlanguageisTuring-recognizableifsomeTuringmachinerecognizesit•  Sometimesreferredas“recursivelyenumerable”

–  ATuringmachinethathaltsonallinputsisadecider.Adeciderthatrecognizesalanguagedecidesit.

–  AlanguageisTuring-decidableorsimplydecidableifsomeTuringmachinedecidesit.•  Sometimesreferredas“recursive”

•  Remarks:–  DecidableifTuring-recognizableandalwayshalts–  EverydecidablelanguageisTuring-recognizable–  ItispossibleforaTMtohaltonlyonthosestringsitaccepts


LimitsofTuringMachines

•  Church-Turingthesis:AnythingthatcanbeprogrammedcanbeprogrammedonaTM

•  NotalllanguagesareTuringDecidable–  ATM={,MisadescriptionofaTuringMachineTM,wisadescriptionofaninputandTMacceptsw}• WeshallseethisinChapter4•  ATMisnotevenTuring-recognizable!


TuringMachineExampleDesignaTMM2thatdecidesA={02n|n≥0},thelanguageofallstringsof0swithlength2n.•  Withoutdesigningit,doyouthinkthiscanbedone?Why?

–  Yes:wecouldwriteaprogramtodoitandthereforeweknowaTMcoulddoitsincewesaidaTMcandoanythingacomputercando

•  Howwouldyoudesignit?•  Solution:

Idea:divideby2eachtimeandseeifresultisaone1.  Sweeplefttorightacrossthetape,crossingoffeveryother0.2.  Ifinstep1:

–  thetapecontainsexactlyone0,thenaccept–  thetapecontainsanoddnumberof0’s,rejectimmediately–  Onlyalternativeiseven0’s.Inthiscasereturnheadtostartandloop

backtostep1.


SampleExecutionofTMM20000-- Numberis4,whichis22

x000--x0x0-- Nowwehave2,or21

x0x0--x0x0-- xxx0--

xxx0-- Nowwehave1,or20xxx0-- Seekbacktostartxxx0-- Scanright;one0,soaccept


TuringMachineExampleII

DesignTMM3todecidethelanguage:C={aibjck|ixj=kandi,j,k≥1}– WhatisthistestingaboutthecapabilityofaTM?

•  Thatitcando(oratleastcheck)multiplication•  Aswehaveseenbefore,weoftenuseunary

– Howwouldyouapproachthis?•  Imaginethatweweretrying2x3=6


TuringMachineExampleII

Solution:1.  Firstscanthestringfromlefttorighttoverifythatitisof

forma+b+c+;ifitisscantostartoftapeandifnot,reject.2.  Crossoffthefirstaandscanuntilthefirstboccurs.

Shuttlebetweenb’sandc’scrossingoffoneofeachuntilallb’saregone.Ifallc’shavebeencrossedoffandsomeb’sremain,reject.

3.  Restorethecrossedoffb’sandrepeatstep2iftherearea’sremaining.Ifalla’sgone,checkifallc’sarecrossedoff;ifso,accept;elsereject.

*Usedifferentsymbolsforcrossingouttheb’ssothatiseasiertorestorethem=)


Transducers

•  Sofarwehavealwaystalkedaboutrecognizingalanguage,notgeneratingalanguage.Thisiscommoninlanguagetheory.

•  Whentalkingaboutcomputationthisseemsstrangeandlimiting.–  Computerstypicallytransforminputintooutput–  Forexample,wearemorelikelytohaveacomputerperformmultiplicationthancheckthattheequationiscorrect.

–  TuringMachinescanalsogenerate/transduce–  Howwouldyoucomputeckgivenaibjandixj=k

•  Inasimilarmanner.Foreverya,youscanthroughtheb’sandforeachyougototheendofthestringandaddac.Thusbyzig-zaggingatimes,youcangeneratetheappropriatenumberofc’s.


TuringMachineExampleIV

Solvetheelementdistinctnessproblem:Givenalistofstringsover{0,1}eachseparatedbya#,acceptifallstringsaredifferent.E={#x1#x2#…#xn|eachxi∈{0,1}*andxi≠xjforeachi≠j}


TuringMachineExampleIV•  Solution:

1.  Placeamarkontopoftheleft-mostsymbol.Ifitwasablank,accept.Ifitwasa#continue;elsereject

2.  Scanrighttonext#andplaceamarkonit.Ifno#isencountered,weonlyhadx1soaccept.

3.  Byzig-zagging,comparethetwostringtotherightofthetwomarked#s.Iftheyareequal,reject.

4.  Movetherightmostofthetwomarkstothenext#symboltotheright.Ifno#symbolisencounteredbeforeablank,movetheleftmostmarktothenext#toitsrightandtherightmostmarktothe#afterthat.Thistime,ifno#isavailablefortherightmostmark,allthestringshavebeencompared,soaccept.

5.  Gobacktostep3


Decidability•  Alloftheseexampleshavebeendecidable,andhenceTuring-recognizable.

•  Howdoweknowthattheseexamplesaredecidable?–  Ateachiterationprogressismadetowardthegoal,sothegoalitselfisreachable

–  Nothardtoproveformally.Forexample,ifthetheinputiscomposedbynsymbolsandasymboliserasedateachiteration,thealgorithmwillfinishafterniterations

•  ShowingthatalanguageisTuringrecognizablebutnotdecidableischallenging


outline...turing machine scheme control a b a b – – – … • turing machine include an...

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