1. Decidable and undecidable languages
The Chinese University of Hong Kong
Fall 2010
CSCI 3130: Formal languages and automata theory
Decidable and undecidable languages
Andrej Bogdanov

2. Problems about automata
Does q0 q1
accept input abb? b
We can formulate this question as a language:
ADFA = {〈D, w〉: D is 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)
D = (Q, S, d, q0, F) w

3. Problems about automata
ADFA = {〈D, w〉: D is 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
Simulate D 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

4. Problems about automata
ADFA = {〈D, w〉: D is 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

5. Problems about automata
decidable
ADFA = {〈D, w〉: D is 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 symol of w
Until w marker reaches last symbol:
Update both markers
If state marker is on accepting state, accept. Else reject.

6. Acceptance problems about automata
decidable
ADFA = {〈D, w〉: D is a DFA that accepts input w}
ANFA = {〈N, w〉: N is an NFA that accepts w}
AREX = {〈R, w〉: R is a regular expression that generates w}
Which of these is decidable?

7. Acceptance problems about automata
decidable
ADFA = {〈D, w〉: D is a DFA that accepts input w}
The following TM decides ADFA:
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.

8. Acceptance problems about automata
decidable
ANFA = {〈N, w〉: N is an NFA that accepts input w}
The following TM decides ANFA:
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.

9. Acceptance problems about automata
decidable
AREX = {R is a regular expression that generates w}
The following TM decides AREX:
P :=
On input 〈R, w〉, where R is a reg exp and w is a string
Convert R to an NFA A using the conversion procedure from Lecture 4
Run the TM N for ANFA on input 〈A, w〉
If N accepts, accept. Otherwise, reject.

10. Other problems about automata
decidable
MINDFA = {〈D〉: D is a minimal DFA}
The following TM decides MINDFA:
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.

11. Other problems about automata
decidable
EQDFA = {〈D1, D2〉: D1, D2 are DFAs and L(D1) = L(D2)}
The following TM decides EQDFA:
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.

12. Other problems about automata
decidable
EDFA = {〈D〉: D is a DFAs and L(D) is empty}
The following TM decides EDFA:
T :=
On input 〈D〉, where D is a DFA
Run the TM S for EQDFA on input 〈D, 〉
If S accepts accept, otherwise reject.

13. Problems about context-free grammars
decidable
ACFG = {〈G, w〉: G is 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.

14. 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 V accepts accept, otherwise reject.

15. Are all languages about CFGs decidable?
EQCFG = {〈G1, G2〉: G1 and G2 are 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

16. The universal Turing Machine and undecidability

17. Turing Machines versus computers
program
output
input
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?

18. The Universal Turing Machine
program 〈M〉
M on input x
input x
The universal TM U takes as inputs a program M and a string x and simulates M on x
The program M itself is specified as a TM!

19. Turing Machines as stringsqa
0/0R
A Turing Machine is (Q, , , , q0, qacc, qrej) q
1/1R qr
This Turing Machine can be described by the string
〈M〉 = (q,qa,qr)(0,1)(0,1,☐) ((q,q,☐/☐R) (q,qa,0/0R) (q,qr,1/1R)) (q)(qa)(qr)

20. The universal Turing Machine
(q,qa,qr)(0,1)(0,1,
001
program 〈M〉
input w for M
U := On input 〈M, w〉,
Simulate M on input w
If M enters accept state, accept.
If M enters reject state, reject.

21. Acceptance of Turing Machines
ATM = {〈M, w〉: M is a TM that accepts w}
U := On input 〈M, w〉,
Simulate M 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 U recognizes but does not decide ATM

22. Recognizing versus deciding
qacc
qrej
accept
reject
loop
halt
The language recognized by a TM is the set of all inputs that make it accept
A TM decides language L if it recognizes L and halts (does not loop) on every input

23. Undecidability Turing’s Theorem: The language ATM is undecidable.
Before we show this, let’s observe one thing:
The language ATM is undecidable.
A Turing Machine M can be given its own
description 〈M〉 as an input!

24. A TM that decides ATM is so potent that it will destroy itself.
Turing’s theorem
A TM that
decides ATM is so
potent that it will
destroy itself.

25. Proof of Turing’s Theorem
Proof by contradiction:
Suppose ATM is decidable.
Then there is a TM H that decides ATM:
〈M, 〈M〉〉
accept if M accepts 〈M〉
reject if M rejects 〈M〉 or M loops on 〈M〉 H
〈M, w〉
accept if M accepts w
reject if M rejects w or M loops on w
What happens when w = 〈M〉?

26. Proof of undecidability
〈M, 〈M〉〉 H
accept if M accepts 〈M〉
reject if M rejects 〈M〉 or M loops on 〈M〉
Let H’ be a TM that does the opposite of H
H = (Q, , , , q0, qacc, qrej) H’ H
acc
rej
acc
rej
H’ = (Q, , , , q0, qrej, qacc)

27. Proof of undecidability
〈M, 〈M〉〉 H
accept if M accepts 〈M〉
reject if M rejects 〈M〉 or M loops on 〈M〉
if M rejects 〈M〉
or M loops on 〈M〉
accept
〈M, 〈M〉〉 H’
reject if M accepts 〈M〉

28. Proof of undecidability
if M rejects 〈M〉
or M loops on 〈M〉 H’
accept
〈M, 〈M〉〉
reject if M accepts 〈M〉
Let D be the following TM:
〈M〉
〈M, 〈M〉〉
copy H’

29. Proof of undecidability
reject if D accepts 〈D〉
if D rejects 〈D〉
or D loops on 〈D〉
if M rejects 〈M〉
or M loops on 〈M〉
accept
〈M〉 D
reject if M accepts 〈M〉
What happens when M = D?
If D accepts 〈D〉
then D rejects 〈D〉
H never loops, so H’ and D never loop either
so D does not exist!
If D rejects 〈D〉
then D accepts 〈D〉

30. Proof of undecidability: conclusion
Proof by contradiction
We assumed ATM was decidable
Then we had Turing Machines H, H’, D
But D does not exist!
Conclusion
The language ATM is undecidable.

31. What happened? We can write an infinite table for every pair (M, w)
all possible inputs w … e 1 00 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)

32. What happened? Now let’s look only at those w that describe a TM M …
acc
loop
rej
acc M2
rej
acc
rej
acc … M3
loop
rej
loop
rej … …
Now let’s look only at those w that describe a TM M

33. What happened? If ATM is decidable, then TM D is in the table … 〈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

34. What happened? D does the opposite of the diagonal entries D … 〈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 if M rejects or loops on 〈M〉 D
reject if M accepts 〈M〉

35. What happened? ? We run into trouble when we look at (D, 〈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〉)!

36. Unrecognizable languages
The language ATM is recognizable but not decidable.
How about languages that are not recognizable?
ATM = {〈M, w〉: M is a TM that does not accept w}
= {〈M, w〉: M rejects or loops on input w}
The language ATM is not recognizable.

37. Unrecognizable languages
Theorem
We know ATM is recognizable, so if ATM were also, then ATM would be decidable
Impossible by Turing’s Theorem
If L and L are both recognizable, then L is decidable.

38. Unrecognizable languages
If L and L are both recognizable, then L is decidable.
Proof idea w
accept
reject M
rej/loop if w ∉ L
accept if w ∈ L w M’
rej/loop if w ∈ L
accept if w ∉ L

39. Unrecognizable languages
If L and L are both recognizable, then L is decidable.
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, ①
Simulate M on input w. If it accepts, accept. ②
Simulate M’ on input w. If it accepts, reject.

40. Bounded simulation
If L and L are both recognizable, then L is decidable.
Decider for L:
Let M = TM for L, M’ = TM for L
On input w,
For k = 0 to infinity
Simulate M on input w for k steps. If it accepts, accept.
Simulate M’ on input w for k steps. If it accepts, reject.