CSC 3130: Automata theory and formal languages

More undecidable languages

Fall 2008MELJUN P. CORTES, MBA,MPA,BSCS,ACSMELJUN P. CORTES, MBA,MPA,BSCS,ACS

MELJUN CORTESMELJUN CORTES

More undecidable problems

L1 = { 〈 M 〉 : M is a TM that accepts input }

L2 = { 〈 M 〉 : M is a TM that accepts some input}

L4 = { 〈 M, M’ 〉 : M and M’ accept the same inputs}

L3 = { 〈 M 〉 : M is a TM that accepts all inputs}

decidable recognizable butundecidable

unrecognizable

Example 1

• Step 1: You gotta believe it– To know if M accepts , it looks like we have to

simulate it– But then we might end up in a loop

• Step 2: Use what you know

L1 = { 〈 M 〉 : M is a TM that accepts input }

ATM is undecidable

Proof by “reduction”

• Show that if L1 can be decided,

... so can ATM

L1 = { 〈 M 〉 : M is a TM that accepts input }

Areject if not

accept if M accepts 〈M〉

?〈 M 〉 , w

reject if not

accept if M accepts w

Proof by “reduction”

〈 M 〉 , w

reject if not

accept if M accepts w

Areject if not

accept if M accepts 〈 M〉

A〈 M’ 〉

M’ is a Turing Machine such that:

If M accepts w, then M’ accepts If M does not accept w, then M’ does not accept

Proof by “reduction”

〈 M 〉 , w

reject if not

accept if M accepts w

A〈 M’ 〉

M’: On input z,

If z = , then simulate M on wOtherwise, reject

construct

M’

M’

q0

qaccqrej

☐/☐R

☐/☐

L

q1

M

Recognizable or not?

L1 = { 〈 M 〉 : M is a TM that accepts input }

decidable recognizable butundecidable

unrecognizable

Algorithm for L1

On input 〈 M 〉 : Simulate M on input

Example 2

L2 = { 〈 M 〉 : M is a TM that accepts some input}

• Step 1: You gotta believe it– To know if M accepts, 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 L1 is undecidable

Example 2

〈 M 〉

reject if not

accept if M accepts

Areject if not

accept if M accepts some input〈 M〉

A〈 M’ 〉

M’ is a Turing Machine such that:

If M accepts , then M’ accepts some inputIf M does not accept , then M’ does not accept anything

Example 2

〈 M 〉

reject if not

accept if M accepts w

A〈 M’〉

construct

M’

M’: On input w,

If w = , then simulate M on Otherwise, reject

decidable recognizable butundecidable

unrecognizable

Is it recognizable?

• Attempt to recognize L2:

L2 = { 〈 M 〉 : M is a TM that accepts some input}

Simulate M on input Simulate M on input Simulate M on input Simulate M on input ...

Accept if one of them accepts

... but there are infinitely many!

Is it recognizable?

• Attempt to recognize L2:

L2 = { 〈 M 〉 : M is a TM that accepts some input}

For all possible strings x (in lexicographic order):

Simulate M on input x

If it accepts, accept.If it rejects, reject.

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

what if M loops on but M accepts, say, 11?

Is it recognizable?

• Description of Turing Machine that recognizes L2:

L2 = { 〈 M 〉 : M is a TM that accepts some input}

For all possible strings x (in lexicographic order):

For all strings y that come before x

If it accepts, accept.If it rejects, reject.

k := 1

k := k + 1

Simulate M on y for k steps

Explanation of algorithm

• Execution of algorithm

inputs 0 1 00 01 ...

Simulate M on for 1 step

Simulate M on for 2 stepsSimulate M on 0for 2 steps

Simulate M on for 3 stepsSimulate M on 0for 3 stepsSimulate M on 1for 3 steps...

If M accepts some w,algorithm will see thisin some stage of thesimulation

decidable recognizable butundecidable

unrecognizable

Example 3

• Step 1: You gotta believe it– To know if M accepts, it looks like we have to simulate

it– But then we might end up in a loop

• Step 2: Use what you know... by yourself this time!

L3 = { 〈 M 〉 : M is a TM that accepts all inputs}

decidable recognizable butundecidable

unrecognizable

Is it recognizable?

• Let’s try...

L3 = { 〈 M 〉 : M is a TM that accepts all inputs}

Simulate M on for 1 step

Simulate M on for 2 stepsSimulate M on 0for 2 steps

Simulate M on for 3 stepsSimulate M on 0for 3 stepsSimulate M on 1for 3 steps...

and then what?

accept if all of them accept

but there are infinitely many!

Is it recognizable?

– To check that 〈 M 〉 is in L3, it looks like we have to wait for an infinite number of simulations to finish

– But we don’t have that much time!

• Step 1: You gotta believe it... but I don’t!

L3 = { 〈 M 〉 : M is a TM that accepts all inputs}

decidable recognizable butundecidable

unrecognizable

How to argue unrecognizability

• First attempt: Remember that

... so we can try to argue that L3 is recognizable

L3 = { 〈 M 〉 : M is a TM that accepts all inputs}

If L and L are both recognizable, then L isdecidable.

First attempt

L3 = { 〈 M 〉 : M is a TM that accepts all inputs}

L3 = { 〈 M 〉 : M is a TM that does not accept all inputs}

• Let’s try...Simulate M on for 1 step

Simulate M on for 2 stepsSimulate M on 0for 2 steps

Simulate M on for 3 stepsSimulate M on 0for 3 stepsSimulate M on 1for 3 steps...

accept if M rejects or loops

but how to know if it loops?

How to argue unrecognizability

• Second attempt: Use what you know

• We can try to argue that

L3 = { 〈 M 〉 : M is a TM that accepts all inputs}

ATM is not recognizable

If we can recognize L3 then we can also recognize ATM

Proof by “reduction”

Arej/loop if not

accept if M accepts all inputs〈 M〉

〈 M 〉, w

rej/loop if M accepts w

accept if M does not accept w

A〈 M’ 〉

construct M’

M’ is a Turing Machine such that:

If M does not accept w, then M’ accepts all inputsIf M accepts w, then M’ rejects or loops on some input

Constructing M’

• To do this, we want to simulate M on w... but what if simulation loops?

• Idea: M’ keeps accepting more and more of its inputs as long as M has not halted on w– If M never halts on w, M’ will accept all inputs

M’ is a Turing Machine such that:

If M does not accept w, then M’ accepts all inputsIf M accepts w, then M’ rejects or loops on some input

Constructing M’

• Description of M’:

On input z,

Simulate M on input w for |z| steps

If simulation of M reaches state qacc, reject.If simulation of M reaches state qrej, accept.If neither state is reached, accept.

M’ is a Turing Machine such that:

If M does not accept w, then M’ accepts all inputsIf M accepts w, then M’ rejects or loops on some input

length of z

Correctness of construction

M’ is a Turing Machine such that:

If M does not accept w, then M’ accepts all inputsIf M accepts w, then M’ rejects or loops on some input

On input z,

Simulate M on input w for |z| steps

If simulation of M reaches state qacc, reject.If simulation of M reaches state qrej, accept.If neither state is reached, accept.

Description of M’:

Correctness of construction

Simulate M on input w for |z| steps

If simulation of M reaches state qacc, reject.If simulation of M reaches state qrej, accept.If neither state is reached, accept.

M’ is a Turing Machine such that:

If M does not accept w, then M’ accepts all inputsIf M accepts w, then M’ rejects or loops on some input

On input z,

Description of M’: In k steps

z = 0k

M’ rejects input 0k

Example 3 recap

L3 = { 〈 M 〉 : M is a TM that accepts all inputs}

• We showed that

... but we know ATM is not recognizable so

If we can recognize L3 then we can also recognize ATM

decidable recognizable butundecidable

unrecognizable

Example 4

• Step 1: You gotta believe it

• Step 2: Use what you know

decidable recognizable butundecidable

unrecognizable

ATM is recognizablebut undecidable

L1 is recognizablebut undecidable

L2 is recognizablebut undecidable

L3 is unrecognizableATM is unrecognizable

L4 = {( 〈 M1 〉 , 〈 M2 〉 ): M1 and M2 accept the same inputs}

Example 4

L4 = {( 〈 M1 〉 , 〈 M2 〉 ): M1 and M2 accept the same inputs}

L3 = { 〈 M 〉 : M is a TM that accepts all inputs}

?〈 M 〉

rej/loop if not

accept if M accepts all inputs

Arej/loop if not

accept if M1, M2 accept same inputs〈 M1 〉 , 〈 M2 〉

Example 4

Arej/loop if not

accept if M1, M2 accept same inputs〈 M1 〉 , 〈 M2 〉

〈M〉

rej/loop if not

accept if M accepts all inputs

A

〈 M1 〉〈 M2 〉

If M accepts all inputs, then M1, M2 accepts same inputsIf M does not, then M1, M2 do not accept same inputs

M1 = M M2 = TM that always accepts

qacc