[PDF] [PDF] COMP481 Review Problems Turing Machines and (Un)Decidability





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    The halting problem and the acceptance problem are “equivalent”. Theorem: the halting problem HP is semi-decidable. Proof: we construct Turing machine M which takes (?M?,w) and simulates the execution of M on input w. If (the simulation of) M halts, M accepts its input.

  • Is semi-decidable decidable?

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  • What is a semi-decidable problem?

    Definition. A problem (D,Q) is semi-decidable if there is a TM/RM that returns “yes” for any d ? Q, but may return “no” or loop forever when d /? Q. Semi-decidable problems are sometimes called recognisable.
  • The halting problem is partially computable
    (p, x) ? HALT.

COMP481ReviewProblems

TuringMachinesand(Un)Decidability

LuayK.Nakhleh

NOTES:

HP=fhM;wijMisaTMandithaltsonstringwg.

HP,anddenedas

L 2".

5.ShowingL0·mLisequivalenttoshowing

L0·mL.

6.Rice'sTheorem(theversionIuse):

L

C=fhMi:L(M)2Cg

isnotrecursive." isnotrecursive.

THEPROBLEMS:

steps. 1

²L2=fhMijMisaTMandjL(M)j·3g.

-NotRE.Weprovethisbyareductionfrom

HP.¿(hM;xi)=hM0i.M0oninputw:it

Wenowprovethevalidityofreduction:

3)M02L2.

²L3=fhMijMisaTMandjL(M)j¸3g.

onx.

Wenowprovethevalidityofthereduction:

3)M0=2L3.

-NotRE.Weprovethisbyareductionfrom

HP.¿(hM;xi)=hM0i.M0oninputw:it

Wenowprovethevalidityofthereduction:

evennumbers)M02L4. M

0=2L4.

swers.

²L5=fhMijMisaTMandL(M)isniteg.

-NotRE.Weprovethisbyareductionfrom

HP.¿(hM;xi)=hM0i.M0oninputw:it

runsMonx.

Wenowprovethevalidityofthereduction:

nite)M02L5. 2 M

0=2L5.

²L6=fhMijMisaTMandL(M)isinniteg.

-NotRE.Weprovethisbyareductionfrom

HP.¿(hM;xi)=hM0i.M0oninputw:it

Wenowprovethevalidityofthereduction:

)M02L6.

²L7=fhMijMisaTMandL(M)iscountableg.

alphabetsandnite-lengthstrings).

²L8=fhMijMisaTMandL(M)isuncountableg.

nite-lengthstrings). runsMonxandacceptsifMhaltsonx.

Wenowprovethevalidityofthereduction:

)"2L(M0)[L(M0))hM0;M0i2L9. runsMonxandacceptsifMhaltsonx.

Wenowprovethevalidityofthereduction:

)"2L(M0)\L(M0))hM0;M0i2L10. -NotRE.Weprovethisbyareductionfrom

HP.¿(hM;xi)=hM1;M2i.M1isaTMthat

acceptsifMhaltsonw.

Wenowprovethevalidityofthereduction:

3 "=2L(M1)nL(M2))hM1;M2i=2L11.

CµRE:trivial.

C6=RE:e.g.,;2REnC.

L

12isnotrecursive.

input,mightnothalt!

L(M0)g.

TM's. -R.TheproofisverysimilartoL1. haltsg. Monx. usedbefore. k,andatleastk=2TM'sofM1;:::;Mkhaltonxg. 4 L 18.

²L19=fhMijMisaTM,andjMj<1000g.

TM.So,L19isnite,andhencerecursive.

²L20=fhMij9x;jxj´51,andx2L(M)g.

fhMij9x;jxj´51,andx2L(M)g62R. inRE!Wereduce M steps,reject,otherwiseaccept.

Provethevalidityofthereduction.

Theproofisbyareductionfrom

M

Provethevalidityofthereduction.

-RE.Easy;proveityourself. -NotR.Alsoeasy;proveityourself. L 0g. language. answer.

²L26=fhMijMisaTMsuchthatbothL(M)and

L(M)areinniteg.

5 M ifxisofoddlength,accept.

Let'sprovethevalidityofthisone.

hM;wi2 L(M) M=2 -R.Easy.

²L28=fhMijMisaTM,andjL(M)jisprimeg.

-NotRE.Reduce reject. numberofstrings.Formalizethis!) -NotRE.Reduce

IfhM;wi2

IfhM;wi=2

²L32=fhM1;M2ijL(M1)·mL(M2)g.

-NotRE.Reductionfrom

HP.¿(hM;wi)=hM1;M2i.

M

2oninputx:reject.(i.e.,L(M2)=;).

M

Youcannowprovethat:(1)IfhM;wi2

HPthenL(M1)=;(inthiscaseL(M1)·m

HPthenL(M1)=

Therefore,

HP·mL32andhenceL32isnotinRE.

6

Provethevalidityofthereduction.

-NotRE.ProofsimilartoL33.

²L35=fhM;xijxisprexofhMig.

isaprexofthestring hMi.

²L36=fhM1;M2;M3ijL(M1)=L(M2)[L(M3)g.

-NotRE.Reductionfrom

HP.¿(hM;wi)=hM1;M2;M3i,where

M

1onx:runsMonw;ifMhalts,M1accepts.

M

2ony:reject.(i.e.,L(M2)=;).

M

3onz:reject.(i.e.,L(M3)=;).

Provethevalidityofthereduction.

-NotRE.Reductionfrom

HP.¿(hM;wi)=hM1;M2;M3i,where

M

1onx:runsMonw;ifMhalts,M1accepts.

M M

3onz:reject.

Provethevalidityofthereduction.

thatacceptseverything,forexample). M

J=fwjw=0xforsomex2ATMorw=1yforsomey2

ATMg. 7

UsingDeMorgan'slaw,

(a)ShowthatJisnotinRE.

Reduce

(b)Showthat

JisnotinRE.

Reduce

(c)ShowthatJ·m J.

¿(w)=(1xifw=0x

0xifw=1x

Provethevalidityofthereduction.

4.ShowthatifalanguageAisinREandA·m

A,thenAisrecursive.

SinceA·m

bothAand

5.AlanguageLisRE-Completeif:

²L2RE,and

²L0·mLforallL02RE.

Recallthefollowinglanguages:

L

HP=fhM;wijMhaltsonwg

HP—similartothereductionusedwithL6).

(b)IsHPRE-Completeornot?Proveyouranswer. toHP. follows: decidableornot?Proveyouranswer. 8

Fori=1to“1"

ProvethatM0indeedsemi-decidesL0.

beabletoprove)thesefacts.

9.PROBLEMFORMULATION.

thatitisundecidable.

Thelanguageis:

TheTMM0oninputhM;wi:

leftmosttapecellwith]. ii.TMM00runsMonw. simulatesMreachingtheleftmosttapecell. leftmosttapecell

Therfore,thelanguageLisnotdecidable.

thatitisdecidable.

Thelanguageis:

L=fhM;wijMmovesitsheadleftoninputwg.

numberofstatesofthemachineM.Proveit! 9 Bµ

AisinRE,andM1semi-decidesit.

BisinRE,andM2semi-decidesit.

Moninputw:

runM1andM2onw(ininterleavedmode).

NoticethatMalwayshalts;wetakeC=L(M).

²ThereisareductionfromAtoB.

²ThereisareductionfromBtoC.

²ThereisareductionfromDtoC.

Please,justifyyouranswer!

MAYBETRUE.

inR. numbersasfollows: 10

²f1(n)=3n+5.

Solution:S1!111S,SS!11111.

²f2(n)=8

:1ifn´0(mod3)

11ifn´1(mod3)

111ifn´2(mod3)

Solution:111!",SS!1,S1S!11,S11S!111.

²f3(n)=n¡1.

S1!",1S!1.

²f4(n)=n=2.Assumeniseven.

S11!1S,SS!".

Thesolutionisinthecoursepacket.

a's.Forexample,f6(aaba)=bbab).

Sa!bS,Sb!aS,SS!".

f

7(aaba)=aaaabbaa.

Sa!aaS,Sb!bbS,SS!".

²f8(w)=(f

6(w)iftherightmostsymbolofwisa

f aS!Tb,bS!Ubb, aT!Tb,bT!Ta,ST!" aU!Uaa,bU!Ubb,SU!".

TMsthatdecidethelanguages!

²L40=fhMijMisaDFAandL(M)isniteg.

symbols.

²L42=fhM;xijMisaDFAandMacceptsxg.

RunMonx.

²L43=fhM;xijMisaDFAandMhaltsonxg.

DFA'salwayshalt!!!

²L44=fhGijGisaCFGandL(G)=;g.

aDFA). 11 generatethem.

S!S1XT

S

1!aS1ajbS1bj]TY

T!aTjbTj"

Y!YA

Aaa!aAa

Abb!bAb

Aba!aAb

Aab!bAa

AaX!Xa

AbX!Xb

YX!".

S!ABSCjABSjT

T!ATj"

AB!BA BA!AB BC!CB CB!BC AC!CA CA!AC A!aquotesdbs_dbs21.pdfusesText_27
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