1. More undecidable languages
The Chinese University of Hong Kong
Fall 2008
CSC 3130: Automata theory and formal languages
More undecidable languages
Andrej Bogdanov

2. More undecidable problems
L1 = {〈M〉: M is a TM that accepts input e}
L2 = {〈M〉: M is a TM that accepts some input}
L3 = {〈M〉: M is a TM that accepts all inputs}
L4 = {〈M, M’〉: M and M’ accept the same inputs}
decidable
recognizable but
undecidable
unrecognizable

3. Example 1 L1 = {〈M〉: M is a TM that accepts input e}
Step 1: You gotta believe it
To know if M accepts 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

4. Proof by “reduction” ? L1 = {〈M〉: M is a TM that accepts input e}
Show that if L1 can be decided, so can ATM
〈M〉 A
accept if M accepts e
reject if not ?
accept if M accepts w
〈M〉, w
reject if not

5. Proof by “reduction” A A M’ is a Turing Machine such that:
accept if M accepts e
reject if not
accept
if M accepts w A
〈M’〉
〈M〉, w
reject
if not
M’ is a Turing Machine such that:
If M accepts w, then M’ accepts e
If M does not accept w, then M’ does not accept e

6. Proof by “reduction” A M’ M’: On input z,
accept
if M accepts w
construct M’ A
〈M’〉
〈M〉, w
reject
if not M’
M’: On input z, q1
☐/☐R
If z = e, then simulate M on w
Otherwise, reject
☐/☐L q0 M
qrej
qacc

7. Recognizable or not? L1 = {〈M〉: M is a TM that accepts input e}
decidable
recognizable but
undecidable
unrecognizable
Algorithm for L1
On input 〈M〉:
Simulate M on input e

8. 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

9. Example 2 A A M’ is a Turing Machine such that:
accept if M accepts some input
reject if not
accept
if M accepts e A
〈M’〉
〈M〉
reject
if not
M’ is a Turing Machine such that:
If M accepts e, then M’ accepts some input
If M does not accept e, then M’ does not accept anything

10. Example 2 A M’: On input w, If w = e, then simulate M on e
accept
if M accepts w
construct M’ A
〈M’〉
〈M〉
reject
if not
M’: On input w,
If w = e, then simulate M on e
Otherwise, reject
decidable
recognizable but
undecidable
unrecognizable

11. ... Is it recognizable? L2 = {〈M〉: M is a TM that accepts some input}
Attempt to recognize L2:
Simulate M on input e
Simulate M on input 0
Simulate M on input 1
Accept if one of them accepts
Simulate M on input 00
...
... but there are infinitely many!

12. Is it recognizable? L2 = {〈M〉: M is a TM that accepts some input}
Attempt to recognize L2:
For all possible strings x (in lexicographic order):
Simulate M on input x
what if M loops on e but M accepts, say, 11?
If it accepts, accept.
If it rejects, reject.
lexicographic order: e, 0, 1, 00, 01, 10, 11, 000, 001, ...

13. Is it recognizable? L2 = {〈M〉: M is a TM that accepts some input}
Description of Turing Machine that recognizes L2:
k := 1
For all possible strings x (in lexicographic order):
For all strings y that come before x
Simulate M on y for k steps
If it accepts, accept.
If it rejects, reject.
k := k + 1

14. Explanation of algorithm
...
inputs e 1 00 01
Execution of algorithm
Simulate M on e for 1 step
If M accepts some w,
algorithm will see this
in some stage of the
simulation
Simulate M on e for 2 steps
Simulate M on 0 for 2 steps
Simulate M on e for 3 steps
Simulate M on 0 for 3 steps
Simulate M on 1 for 3 steps
...
decidable
recognizable but
undecidable
unrecognizable

15. Example 3 L3 = {〈M〉: M is a TM that accepts all inputs}
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!
decidable
recognizable but
undecidable
unrecognizable

16. Is it recognizable? and then what? ...
L3 = {〈M〉: M is a TM that accepts all inputs}
Let’s try...
Simulate M on e for 1 step
and then what?
Simulate M on e for 2 steps
Simulate M on 0 for 2 steps
accept if all of them accept
Simulate M on e for 3 steps
Simulate M on 0 for 3 steps
but there are infinitely many!
Simulate M on 1 for 3 steps
...

17. Is it recognizable? L3 = {〈M〉: M is a TM that accepts all inputs}
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!
decidable
recognizable but
undecidable
unrecognizable

18. How to argue unrecognizability
L3 = {〈M〉: M is a TM that accepts all inputs}
First attempt: Remember that so we can try to argue that L3 is recognizable
If L and L are both recognizable, then L is decidable.

19. First attempt ... Let’s try...
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 e for 1 step
Simulate M on e for 2 steps
Simulate M on 0 for 2 steps
Simulate M on e for 3 steps
Simulate M on 0 for 3 steps
Simulate M on 1 for 3 steps
...
accept if M rejects or loops
but how to know if it loops?

20. How to argue unrecognizability
L3 = {〈M〉: M is a TM that accepts all inputs}
Second attempt: Use what you know
We can try to argue that
ATM is not recognizable
If we can recognize L3 then we can also recognize ATM

21. Proof by “reduction” A A M’ is a Turing Machine such that:
accept if M accepts all inputs
rej/loop if not
accept
if M does not accept w
construct M’ A
〈M’〉
〈M〉, w
rej/loop
if M accepts w
M’ is a Turing Machine such that:
If M does not accept w, then M’ accepts all inputs
If M accepts w, then M’ rejects or loops on some input

22. Constructing M’
M’ is a Turing Machine such that:
If M does not accept w, then M’ accepts all inputs
If M accepts w, then M’ rejects or loops on some input
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

23. Constructing M’ Description of M’: M’ is a Turing Machine such that:
If M does not accept w, then M’ accepts all inputs
If M accepts w, then M’ rejects or loops on some input
Description of M’:
length of z
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.

24. Correctness of construction
M’ is a Turing Machine such that:
If M does not accept w, then M’ accepts all inputs
If M accepts w, then M’ rejects or loops on some input
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.

25. Correctness of construction
M’ is a Turing Machine such that:
If M does not accept w, then M’ accepts all inputs
If M accepts w, then M’ rejects or loops on some input
In k steps
z = 0k
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’ rejects input 0k

26. 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 but
undecidable
unrecognizable

27. Example 4 L4 = {(〈M1〉, 〈M2〉): M1 and M2 accept the same inputs}
Step 1: You gotta believe it
Step 2: Use what you know
decidable
recognizable but
undecidable
unrecognizable
ATM is recognizable but undecidable
L1 is recognizable but undecidable
L2 is recognizable but undecidable
L3 is unrecognizable
ATM is unrecognizable

28. Example 4 ? A L4 = {(〈M1〉, 〈M2〉): M1 and M2 accept the same inputs}
accept if M1, M2 accept same inputs
rej/loop if not
L3 = {〈M〉: M is a TM that accepts all inputs} ?
accept if M accepts all inputs
〈M〉
rej/loop if not

29. Example 4 A A If M accepts all inputs, then M1, M2 accepts same inputs
accept if M1, M2 accept same inputs
rej/loop if not
accept
if M accepts all inputs
〈M〉 A
〈M1〉
〈M2〉
qacc
rej/loop
if not
If M accepts all inputs, then M1, M2 accepts same inputs
If M does not, then M1, M2 do not accept same inputs
M1 = M
M2 = TM that always accepts