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The Chinese University of Hong Kong

CSCI 3130: Formal languages and automata theory

Decidable andundecidable languages

Andrej Bogdanov

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

Problems about automata

- We can formulate this question as a language:

b

a

a

Does

accept input abb?

q0

q1

b

ADFA = {〈D, w〉: Dis a DFA that accepts input w}

Is ADFA decidable?

((q0,q1)(a,b)((q0,a,q0)(q0,b,q1)(q1,a,q0)(q1,b,q1)(q0)(q1))(abb)

w

D = (Q, S, d, q0, F)

Problems about automata

ADFA = {〈D, w〉: Dis a DFA that accepts input w}

pseudocode:

TM description:

On input 〈D, w〉,where D = (Q, S, d, q0, F):

On input 〈D, w〉,where D is a DFA, w is a string

Set q := q0

SimulateD on input w

For i := 1 to length(w):

If simulation ends in acc state,

accept. Otherwise, reject.

q := d(q, wi)

If q∈F accept, else reject

Problems about automata

ADFA = {〈D, w〉: Dis a DFA that accepts input w}

Turing Machine details:

Check input is in correct format.

(Transition function is complete, no duplicate transitions)

Perform simulation:

state

input symbol

.

.

.

.

.

.

((q0,q1)(a,b)((q0,a,q0)(q0,b,q1)(q1,a,q0)(q1,b,q1)(q0)(q1))(abb)

qacc

Problems about automata

decidable

ADFA = {〈D, w〉: Dis a DFA that accepts input w}

Turing Machine details:

Check input is in correct format.

(Transition function is complete, no duplicate transitions)

Perform simulation:

Put markers on start state of D and first symbol of w

Until marker for w reaches last symbol:

Update both markers

If state marker is on accepting state, accept. Else reject.

Acceptance problems about automata

decidable

ADFA = {〈D, w〉: Dis a DFA that accepts input w}

ANFA = {〈N, w〉: Nis an NFA that accepts w}

AREX = {〈R, w〉: Ris a regular expression that generates w}

Which of these is decidable?

Acceptance problems about automata

- The following TM decides ADFA:

decidable

ADFA = {〈D, w〉: Dis a DFA that accepts input w}

M :=

On input 〈D, w〉, where D is a DFA and w is a string

Simulate D on input w

If the simulation ends in acc state of D, accept. If it doesn’t, reject.

Acceptance problems about automata

- The following TM decides ANFA:

decidable

ANFA = {〈N, w〉: Nis an NFA that accepts input w}

N :=

On input 〈N, w〉, where N is an NFA and w is a string

Convert N to a DFA D using the conversion procedure from Lecture 2

Run the TM M for ADFA on input 〈D, w〉

If M accepts, accept. Otherwise, reject.

Acceptance problems about automata

- The following TM decides AREX:

decidable

AREX = {〈R, w〉: Ris a regular expression that generates w}

P :=

On input 〈R, w〉, where R is a reg exp and w is a string

Convert R to an NFA N using the conversion procedure from Lecture 4

Run the TM N for ANFA on input 〈N, w〉

If N accepts, accept. Otherwise, reject.

Other problems about automata

- The following TM decides MINDFA:

decidable

MINDFA = {〈D〉: Dis a minimal DFA}

R :=

On input 〈D〉, where D is a DFA

Run the distinguishable states algorithm

from Lecture 7 on D

If every pair of states is distinguishable, accept.

Otherwise, reject.

Other problems about automata

- The following TM decides EQDFA:

decidable

EQDFA = {〈D1, D2〉: D1, D2are DFAs and L(D1) = L(D2)}

S :=

On input 〈D1, D2〉, where D1 and D2 are DFAs

Run the DFA minimization algorithm

from Lecture 7 on D1 to obtain a DFA D1’

Run the DFA minimization algorithm

from Lecture 7 on D2 to obtain a DFA D2’

If D1’= D2’accept, otherwise reject.

Other problems about automata

- The following TM decides EDFA:

decidable

EDFA = {〈D〉: D is a DFAs and L(D) is empty}

T :=

On input 〈D〉, where D is a DFA

Run the TM S for EQDFA on input 〈D,〉

If S accepts accept, otherwise reject.

Problems about context-free grammars

decidable

ACFG = {〈G, w〉: Gis a CFG that generates w}

V :=

On input 〈G, w〉, where G is a CFG and w is a string

Eliminate the nullable and unit productions from G

Convert G to Chomsky Normal Form

Run the Cocke-Younger-Kasami algorithm on 〈G, w〉

If the CYK algorithm produces a parse tree, accept.

Otherwise, reject.

Decidability of context-free languages

Every context-free language is decidable.

Let L be a context-free language.

There is a CFG G for L.

The following TM decides L:

MG :=

On input w,

Run TM V on input 〈G, w〉.

If Vaccepts accept, otherwise reject.

Are all languages about CFGs decidable?

EQCFG = {〈G1, G2〉: G1 and G2are context-free grammars that generate the same strings}

undecidable

What’s the difference between EQDFA and EQCFG?

To decide EQDFA, we minimized both DFAs

But there is no method that, given a CFG or PDA,

produces a unique equivalent minimal CFG or PDA

Turing Machines versus computers

computer

program

output

data

A computer is a machine that manipulates data according to a list of instructions.

How does a Turing Machine take a program

as part of its input?

The Universal Turing Machine

U

program 〈M〉

whatever

M does on x

input x for M

The universal TMU takes as inputs a program Mand a string x and simulatesM on x

The program M itself is specified as a TM!

Turing Machines as strings

- This Turing Machine can be described by the string

M

☐/☐R

qa

0/0R

- A Turing Machine is (Q, ,, , q0, qacc, qrej)

q

1/1R

qr

〈M〉 = (q,qa,qr)(0,1)(0,1,☐) ((q,q,☐/☐R) (q,qa,0/0R) (q,qr,1/1R)) (q)(qa)(qr)

The universal Turing Machine

U

(q,qa,qr)(0,1)(0,1,

001

input w for M

program 〈M〉

U := On input 〈M, w〉,

SimulateM on input w

If M enters accept state, accept.

If M enters reject state, reject.

Acceptance of Turing Machines

ATM = {〈M, w〉: Mis a TM that accepts w}

U := On input 〈M, w〉,

SimulateM on input w

M accepts w

M rejects w

M loops on w

U accepts 〈M, w〉

U rejects 〈M, w〉

U loops on 〈M, w〉

TM Urecognizes but does not decideATM

Recognizing versus deciding

qacc

qrej

accept

reject

loop

halt

The language recognized by a TM is the set ofall inputs that make it accept

A TM decides language Lif it recognizes L andhalts (does not loop) on every input

Undecidability

- Turing’s Theorem:
- Before we show this, let’s observe one thing:

The language ATMis undecidable.

A Turing Machine M can be given its own

description〈M〉 as an input!

Proof of Turing’s Theorem

- Proof by contradiction:

Suppose ATM is decidable.

Then there is a TM H that decides ATM:

accept ifM accepts w

〈M, w〉

H

reject ifM rejects wor M loops on w

accept ifM accepts 〈M〉

〈M,〈M〉〉

What happens whenw = 〈M〉?

reject ifM rejects 〈M〉or M loops on 〈M〉

Proof of undecidability

accept ifM accepts 〈M〉

H

〈M,〈M〉〉

reject ifM rejects 〈M〉or M loops on 〈M〉

Let H’ be a TM that does the opposite of H

acc

acc

H

H’

To go from H to H’,

just switch its accept

and reject states

rej

rej

Proof of undecidability

accept ifM accepts 〈M〉

H

〈M,〈M〉〉

reject ifM rejects 〈M〉or M loops on 〈M〉

ifM rejects 〈M〉

or M loops on 〈M〉

accept

H’

〈M,〈M〉〉

reject ifM accepts 〈M〉

Proof of undecidability

ifM rejects 〈M〉

or M loops on 〈M〉

accept

H’

〈M,〈M〉〉

reject ifM accepts 〈M〉

Let D be the following TM:

〈M〉

H’

〈M,〈M〉〉

copy

Proof of undecidability

ifM rejects 〈M〉

or M loops on 〈M〉

accept

D

〈M〉

reject ifM accepts 〈M〉

What happens when M = D?

H never loops, soD never loops either

If D accepts 〈D〉

ifD rejects 〈D〉

or D loops on 〈D〉

〈D〉

then D rejects 〈D〉

so D does not exist!

reject ifD accepts 〈D〉

If D rejects 〈D〉

then D accepts 〈D〉

Proof of undecidability: conclusion

- Proof by contradiction
- We assumed ATM was decidable
- Then we built Turing Machines H, H’, D
- But D does not exist!
- Conclusion

The language ATMis undecidable.

What happened?

all possible inputs w

…

00

e

0

1

M1

acc

rej

rej

acc

M2

…

rej

acc

loop

rej

all possible

Turing Machines

M3

rej

loop

rej

rej

…

…

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

What happened?

…

〈M4〉

〈M1〉

〈M2〉

〈M3〉

M1

acc

loop

rej

acc

M2

…

rej

acc

rej

acc

M3

loop

rej

loop

rej

…

…

Now let’s look only at those wthat describe a TM M

What happened?

…

〈M4〉

〈M1〉

〈M2〉

〈M3〉

M1

acc

loop

rej

acc

M2

…

rej

acc

rej

acc

…

…

D

rej

rej

acc

rej

…

…

If ATM is decidable, then TM D is in the table

What happened?

…

〈M4〉

〈M1〉

〈M2〉

〈M3〉

M1

acc

loop

rej

acc

M2

…

rej

acc

rej

acc

…

…

D

rej

rej

acc

rej

D does the opposite of the diagonal entries

〈M〉

accept ifM rejects or loops on 〈M〉

D

reject ifM accepts 〈M〉

What happened?

…

〈M4〉

〈M1〉

〈M2〉

〈M3〉

〈D〉

M1

acc

loop

rej

acc

loop

M2

…

rej

acc

rej

acc

acc

…

…

?

…

D

rej

rej

acc

rej

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

Unrecognizable languages

- How about languages that are not recognizable?

The language ATMis recognizable but not decidable.

ATM = {〈M, w〉: Mis a TM that does not accept w}

= {〈M, w〉: Mrejects or loops on input w}

The language ATMis not recognizable.

Unrecognizable languages

- Theorem
- We know ATM is recognizable, so if ATM were also,then ATM would be decidable
- But Turing’s Theorem says ATM is not decidable

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

Unrecognizable languages

- Proof idea

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

accept ifw ∈ L

M

w

accept

rej/loop ifw ∉ L

w

accept ifw ∉ L

M’

w

reject

rej/loop ifw ∈ L

Unrecognizable languages

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

Problem: If M loops on w, we will never get to step 2!

Let M = TM for L, M’ = TM for L

On input w,

Simulate M on input w. If it accepts, accept.

①

Simulate M’ on input w. If it accepts, reject.

②

Bounded simulation

- Turing Machine that decides L:

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

Let M= TM for L, M’= TM for L

On input w,

For t := 0 to infinity

Do ttransitions of M on w. If it accepts, accept.

Do ttransitions of M’ on w. If it accepts, reject.

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