meljun cortes automata theory 17

8/13/2019 MELJUN CORTES Automata Theory 17

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CSC 3130: Automata theory and formal languages

Undecidable Languages

MELJUN P. CORTES MBA MPA BSCS ACS

MELJUN CORTES


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The Universal Turing Machine

U

input x

program M

Mon input x

The universal TMUtakes as inputs a programM

and a string xand simulatesMon x

The programMitself is specified as a TM!


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Turing Machines as strings

This Turing Machine can be described by the

string

M= ((q0, qacc, qrej),((q0, q0, /R),(q0, qacc, 0/0R),

(q0, qrej, 1/1R)))

M

q0

qacc

qrej

/R


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Languages about automata

A TM can take another TM (or DFA, CFG) asinput

ADFA= {(D, w): Dis a DFA that accepts input w}

APDA= {(P, w): Pis a PDA that accepts w}

ATM= {(M, w):Mis a TM that accepts w}

Which of these is decidable?


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Languages about automata

Idea: SimulateMon input w

Simulation is correct: IfMaccepts w, we accept

IfMrejects w, we reject

IfMloops on w, we loop

This TM recognizesbut does not decideATM

ATM= {(M, w):Mis a TM that accepts w}


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Recognizing versus deciding

This kind of TM is called a decider

The textbook uses different terminology:

The language recognized by a TM is theset ofall inputs that make it reach qacc

A TM decides language Lif it recognizes L

anddoes not loop on any input

recognizable = recursively enumerable (r. e.)

decidable = recursive


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Undecidability

Turings Theorem:

Before we prove this, lets observe one thing:

The languageATMis undecidable.

A Turing MachineMcan be given its owndescriptionMas an input!


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Proof of undecidability

Proof by contradiction:

SupposeATMis decidable.

Then there is a TM Hthat decidesATM

:

HM, w acceptifMaccepts w

rejectifMrejects w

orMloops onw

M, M

acceptifMaccepts M

rejectifMrejects M

orMloops on M

What happens whenw= M?


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HM, M

acceptifMaccepts M

rejectifMrejects M

orMloops on M

Let Hbe a TM that does the oppositeof H

H

acc

rej

acc

rej

H


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HM, M acceptifMaccepts M

rejectifMrejects M

orMloops on M

HM, M

rejectifMaccepts M

ifMrejects MorMloops on M

accept


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HM, M

rejectifMaccepts M


accept

Let Dbe the following TM:

copy

M M, M

H


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DM

rejectifMaccepts

M


accept

What happens whenM= D?

D

rejectifDacceptsD

ifDrejectsD

or Dloops onD

If Daccepts Dthen Drejects D

If Drejects Dthen Daccepts Dso Ddoes not exist!


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Proof of undecidability: conclusion

Proof by contradiction We assumedATMwas decidable

Then we had Turing Machines H, H, D

But Ddoes not exist!

Conclusion

The languageATMis undecidable.


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What happened?

M1

M2M3

e 0 1 00

acc rej rej acc

rej acc loop rejrej loop rej rej

TuringMachine

input w

We can write an infinite table for every pair (M, w)


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What happened?

M1

M2M3

M1 M2 M4

acc loop rej acc

rej acc rej accloop rej loop rej

Now lets look only at those wthat describe a TMM

M3


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What happened?

M1

M2

M1 M2 M4

acc loop rej acc

rej acc rej acc

IfATMis decidable, then TM Dis in the table

M3

D rej rej acc rej


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What happened?

M1

M2

M1 M2 M4

acc loop rej acc

rej acc rej acc

Ddoes the opposite of the diagonal entries

M3

D rej rej acc rej

DM

reject ifMaccepts M

accept ifMrejects or loops on M


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What happened?

M1

M2

M1 M2 M4

acc loop rej acc

rej acc rej acc

We run into trouble when we look at (D, D)!

M3

D rej rej acc rej

D

loop

acc

?


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Unrecognizable languages

General Theorem

We knowATMis recognizable, so ifATMwere

also,

thenATMwould be decidable

Impossible by Turings Theorem

If Land Lare both recognizable, then Lis

decidable.


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Proof idea

If Land Lare both recognizable, then Lisdecidable.

M rej/loop ifwLaccept ifwLw

M rej/loop ifwLaccept ifwLw

w

accept

reject


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Proof attempt

LetM= TM for L, M= TM for L

On input w,

SimulateMon input w. If it accepts, accept.

Simulate Mon input w. If it accepts, reject.


Problem: IfMloops on w, we

will never get to step 2!


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Bounded simulation

ProofLetM= TM for L, M= TM for L

On input w,

SimulateMon input w for ksteps. If it accepts, accept.

Simulate Mon input w for ksteps. If it accepts, reject.


For k= 0 to infinity


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Another undecidable language

To prove this, we will show that

so by Turings Theorem it is undecidable.

HALTTM= {(M, w):Mis a TM that halts on input w}

If HALTTM

can be decided, so canATM

.

HALTTMis an undecidable language


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Undecidability of halting

Suppose Hdecides HALTTM

H reject ifMloops onw

accept ifMhalts on wM, w

If HALTTMcan be decided, so canATM.

We want to use Hto design DforATM

? reject ifMrejectsorloops on w

accept ifMaccepts wM, w


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Undecidability of halting

H reject ifMloops onw

accept ifMhalts on wM, w

D

reject ifMrejects

orloops on w

accept ifMaccepts wM, w

Hrej

accsimulateMon w rej

acc


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More undecidable problems

L1= {M:Mis a TM that accepts input e}

L2= {M:Mis a TM that accepts some input}

L4= {M,M:Mand M accept the same inputs}

Which of these are recognizable?

L3= {M:Mis a TM that accepts all inputs}


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Gdels incompleteness theorem

Proof due to Turing:

Some mathematical

statements

are truebut not provable.

If every true statement is provable, thenATMcan be decided.


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Proof sketch of Gdels theorem

Suppose every true statement has a proof

For every TMMand input w, one of the

statements

is true, so it must have a proof

Maccepts w Mdoes not accept wor


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Proof sketch of Gdels theorem

Consider this algorithm forATM:

On input M, w:

Fork= 0 to infinity

For all possible proofspof length kIfpis a proof of Maccepts w, accept.

Ifpis a proof of Mdoes not accept w, reject.

One of them is true, so it has a proofp

Eventually, the algorithm will find this proof


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meljun cortes automata theory 17

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