1. Reducibility The Chinese University of Hong Kong Fall 2010
CSCI 3130: Automata theory and formal languages
Reducibility
Andrej Bogdanov

2. Summary of last lecture
program 〈M〉
accept, if M accepts w
reject, if M rejects w
input w
loops, if M loops on w
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!

3. Summary of last lecture
qacc
qrej
accept
reject
loop
halt
recognizable:
undecidable:
By universal TM U
Turing’s Theorem
ATM = {〈M, w〉: M is a TM that accepts input w}

4. Another undecidable language
HALTTM = {〈M, w〉: M is a TM that halts on input w}
HALTTM is an undecidable language
We will argue that … so by Turing’s Theorem it is undecidable.
If HALTTM is decidable, so is ATM.

5. Undecidability of halting
If HALTTM can be decided, so can ATM.
Suppose H decides HALTTM
accept if M halts on w
〈M, w〉 R
reject if M loops on w
We want to use R to design TM S that decides ATM
〈M, w〉 ?
accept if M accepts w
reject if M rejects
or loops on w

6. Undecidability of halting
accept if M halts on w
〈M, w〉 R
reject if M loops on w S U R
rej
acc
simulate
M on w
rej
acc
accept if M accepts w
〈M, w〉
reject if M rejects
or loops on w

7. Undecidability of halting
HALTTM = {〈M, w〉: M is a TM that halts on input w}
ATM = {〈M, w〉: M is a TM that accepts input w}
Suppose that HALTTM is decidable. Let H be a TM that decides HALTTM. Then this TM decides ATM:
S :=
On input 〈M, w〉,
Run R on input 〈M, w〉.
If R rejects, reject.
If R accepts, run U on input 〈M, w〉
If U accepts, accept. If U rejects, reject.
Impossible, because ATM is undecidable.

8. Reducibility Suppose you think L is undecidable.
How do you know for sure?
Show:
If some TM R decides L
then using R, I can build another TM S
so that S decides ATM
...but this is impossible, because ATM is undecidable.

9. Example 1 AEPSTM = {〈M〉: M is a TM that accepts input e}
decidable
undecidable
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: Prove it!

10. Example 1: Figuring out the reduction
reject if not
accept if M’ accepts e
〈M’〉
〈M, w〉
reject
if not
accept
if M accepts w S R
〈M’〉
M’ should be a Turing Machine such that:
M’ on input e = M on input w
If M accepts w, then M’ accepts e
If M does not accept w, then M’ does not accept e

11. Example 1: Implementing the reduction
〈M, w〉
construct M’
〈M’〉
M’ should be a Turing Machine such that:
M’ on input e = M on input w
M’ := On input z,
Simulate M on input w
If M accepts, accept
Otherwise, reject

12. Example 1: Impementing the reduction
accept
if M accepts w
construct M’ R
〈M’〉
〈M, w〉
reject
if not
S := On input 〈M, w〉:
Construct the following TM M’:
M’ := On input z,
Simulate M on input w and output answer
Run R on input 〈M’〉 and output its answer.

13. Example 1: Writing it up undecidable
AEPSTM = {〈M〉: M is a TM that accepts input e}
ATM = {〈M, w〉: M is a TM that accepts input w}
Suppose AEPSTM is decidable and R decides it. Let
S := On input 〈M, w〉 where M is a TM and w is a string
M’ := On input z,
Simulate M on input w and output answer
Run R on input 〈M’〉 and output its answer.
Construct the following TM M’:
Then S accepts 〈M, w〉 if and only if M accepts w
So S decides ATM, which is impossible.

14. Example 2 SOMETM = {〈M〉: M is a TM that accepts some input}
decidable
undecidable
Step 1: You gotta believe it
To know if M is in SOMETM, looks like we have to simulate
But then we might end up in a loop
Step 2: Prove it!

15. Example 2: Setting up the reduction
SOMETM = {〈M〉: M is a TM that accepts some input}
Show: If SOMETM can be decided by some TM R, R
reject if M’ accepts no inputs
accept if M’ accepts some input
〈M’〉
... then ATM can be decided by some other TM S S
〈M, w〉
reject if not
accept if M accepts w

16. Example 2: Setting up the reduction
accept
if M accepts w
construct M’ R
〈M’〉
〈M, w〉
reject
if not
Task
Given 〈M, w〉, construct M’ so that:
If M accepts w,
then M’ accepts some input
If M does not accept w,
then M’ accepts no inputs

17. Example 2: Implementing the reduction
Task
Given 〈M, w〉, construct M’ so that:
If M accepts w,
then M’ accepts some input
If M does not accept w,
then M’ accepts no inputs
M’ := On input z,
Simulate M on input w
If M accepts, accept
Otherwise, reject

18. Example 2: Writing it up undecidable
SOMETM = {〈M〉: M is a TM that accepts some input}
ATM = {〈M, w〉: M is a TM that accepts input w}
Suppose SOMETM is decidable and R decides it. Let
S := On input 〈M, w〉 where M is a TM and w is a string
M’ := On input z,
Simulate M on input w and output answer
Run R on input 〈M’〉 and output its answer.
Construct the following TM M’:
Then S accepts 〈M, w〉 if and only if M accepts w
So S decides ATM, which is impossible.

19. Example 3 ETM = {〈M〉: M is a TM that accepts no input}
decidable
undecidable
Show: If ETM can be decided by some TM R,
... then SOMETM can be decided by another TM S
SOMETM = {〈M〉: M is a TM that accepts some input}

20. Example 3 undecidable Suppose ETM can be decided by some TM R.
ETM = {〈M〉: M is a TM that accepts no input}
SOMETM = {〈M〉: M is a TM that accepts some input}
Suppose ETM can be decided by some TM R.
Consider the following TM S:
S := On input 〈M〉 where M is a TM
If R accepts, reject. If R rejects, accept.
Run R on input 〈M〉.
Then S decides SOMETM, a contradiction.

21. Example 4 EQTM = {〈M1, M2〉: M1 and M2 accept the same inputs}
decidable
undecidable
Show: If EQTM can be decided by some TM R,
... then ETM can be decided by another TM S

22. Example 4: Setting up the reduction
EQTM = {〈M1, M2〉: M1 and M2 accept the same inputs}
ETM = {〈M〉: M is a TM that accepts no input}
Task
Given 〈M〉, construct 〈M1, M2〉 so that:
If M accepts no input,
then M1, M2 accept same inputs
If M accepts some input,
then M1, M2 do not accept same inputs
Idea
Make M1 = M
Make M2 accept nothing

23. Example 2: Writing it up undecidable
EQTM = {〈M1, M2〉: M1 and M2 accept the same inputs}
ETM = {〈M〉: M is a TM that accepts no input}
Suppose EQTM is decidable and R decides it. Let
S := On input 〈M〉 where M is a TM
M2 := On input z, reject.
Run R on input 〈M, M2〉 and output its answer.
Construct the following TM M2:
Then S accepts 〈M〉 if and only if M accepts no input
So S decides ETM, which is impossible.