meljun cortes -- automata theory lecture - 18

7/31/2019 MELJUN CORTES -- AUTOMATA THEORY LECTURE - 18

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CSC 3130: Automata theory and formal languages

Andrej Bogdanov

http://www.cse.cuhk.edu.hk/~andrejb/csc3130

More undecidable languages

Fall 2008


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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}

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

decidable recognizable but

undecidable

unrecognizable


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Example 1

Step 1: You gotta believe it

To know ifMaccepts e, it looks like we have to

simulate it

But then we might end up in a loop

Step 2: Use what you know


ATM is undecidable


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Proof by reduction

Show that ifL1 can be decided,

... so canATM


Areject if not

accept ifMaccepts eM

?M, w

reject if not

accept ifMaccepts w


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Proof by reduction

M, w

reject

if not

accept

ifMaccepts w

Areject if not

accept ifMaccepts eM

AM

M is a Turing Machine such that:

IfMaccepts w, then Maccepts e

IfMdoes not accept w, then Mdoes not accept e


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Proof by reduction

M, w

reject

if not

acceptifMaccepts w

AM

M: On input z,

Ifz = e, then simulateMon w

Otherwise, reject

constructM

M

q0

qaccqrej

/L

q1

M


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Recognizable or not?



undecidable

unrecognizable

Algorithm forL1

On input M:SimulateMon input e


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Example 2



To know ifMaccepts, it looks like we have to simulate

it


Step 2: Use what you knowATM is undecidable L1 is undecidable


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Example 2

M

reject

if not

accept

ifMaccepts e

Areject if not

accept ifMaccepts some inputM

AM


IfMaccepts e, then Maccepts some input

IfMdoes not accept e, then Mdoes not accept anything


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Example 2

M

reject

if not

acceptifMaccepts w

AMconstruct

M

M: On input w,

Ifw = e, then simulateMon e

Otherwise, reject


undecidable

unrecognizable


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Is it recognizable?

Attempt to recognize L2:


SimulateMon input e

SimulateMon input 0

SimulateMon input 1

SimulateMon input00

...

Accept if one of them accepts

... but there are infinitely many!


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Is it recognizable?

Attempt to recognize L2:


For all possible strings x(in lexicographic order):

SimulateMon input x

If it accepts, accept.

If it rejects, reject.

lexicographic order: e, 0, 1, 00, 01, 10, 11, 000, 001, ...

what ifMloops on e

butMaccepts, say, 11?


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Is it recognizable?

Description of Turing Machine that recognizes L2:


For all possible strings x(in lexicographic order):

For all stringsythat come before x

If it accepts, accept.

If it rejects, reject.

k := 1

k := k + 1

SimulateMonyfork steps


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Explanation of algorithm

Execution of algorithm

inputs e 0 1 00 01 ...

SimulateMon e for1 step

SimulateMon e for2 stepsSimulateMon 0for2 steps


SimulateMon 1for3 steps..

.

IfMaccepts some w,algorithm will see this

in some stage of the

simulation


undecidable

unrecognizable


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Example 3


To know ifMaccepts, it looks like we have to simulateit


Step 2: Use what you know... by yourself this

time!



undecidable

unrecognizable


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Is it recognizable?

Lets try...





SimulateMon 1for3 steps..

.

and then what?

accept if all of them accept

but there are infinitely many!


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Is it recognizable?

To check that Mis in L3, it looks like we have to wait

for an infinite numberof simulations to finish

But we dont have that much time!


... but I dont!



undecidable

unrecognizable


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How to argue unrecognizability

First attempt: Remember that

... so we can try to argue that L3is recognizable


IfLand Lare both recognizable, then Lis

decidable.


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First attempt


L3 = {M:Mis a TM that does not accept all inputs}

Lets try...



SimulateMon e for3 stepsSimulateMon 0for3 stepsSimulateMon 1for3 steps

.

.

.

accept ifMrejects or loops

but how to know if it loops?


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How to argue unrecognizability

Second attempt: Use what you know

We can try to argue that


ATM is not recognizable

If we can recognize L3then we can also

recognizeATM


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Proof by reduction

Arej/loop if not

accept ifMaccepts all inputsM

M, w

rej/loop

ifMaccepts w

accept

ifMdoes not accept w

AMconstruct

M


IfMdoes not accept w, thenMaccepts all inputs

IfMaccepts w, thenM rejects or loops on some input


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Constructing M

To do this, we want to simulateMon w... but what if simulation

loops?

Idea:M keeps accepting more and more of its

inputs as long asMhas not halted on w

IfMneverhalts on w, Mwill accept all inputs

Mis a Turing Machine such that:




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Constructing M

Description ofM:

On input z,

SimulateMon input wfor|z| steps

If simulation ofMreaches state qacc, reject.If simulation ofMreaches state qrej, accept.

If neither state is reached, accept.




length ofz


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Correctness of construction




On input z,


If simulation ofMreaches state qacc, reject.

If simulation ofMreaches state qrej, accept.If neither state is reached, accept.

Description ofM:


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Correctness of construction


If simulation ofMreaches state qacc, reject.

If simulation ofMreaches state qrej, accept.If neither state is reached, accept.




On input z,

Description ofM: In k steps z= 0k

M rejects input 0k


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Example 3 recap


We showed that

... but we knowATM is not recognizableso

If we can recognize L3then we can alsorecognizeATM


undecidable

unrecognizable


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Example 4


Step 2: Use what you know

decidable recognizable butundecidable

unrecognizable

ATM is recognizable

but undecidable

L1 is recognizable

but undecidable

L2 is recognizable

but undecidable

L3 is unrecognizableATM is unrecognizable

L4 = {(M1, M2):M1andM2 accept the same inputs


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Example 4

L4 = {(M1, M2):M1andM2 accept the same inputs}


?M

rej/loop if not

accept ifMaccepts all inputs

Arej/loop if not

accept ifM1, M2 accept same inputsM1, M2


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Example 4

Arej/loop if not

accept ifM1, M2 accept same inputsM1, M2

M

rej/loop

if not

accept

ifMaccepts all inputs

AM1

M2

IfMaccepts all inputs, then M1, M2 accepts same inputsIfMdoes not, then M1, M2 do not accept same inputs

M1 =M M2 = TM that always accepts

qacc

meljun cortes -- automata theory lecture - 18

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