1. CSCI 2670 Introduction to Theory of Computing
October 26, 2005

2. Agenda Yesterday Today Test Begin Chapter 5 Pages 187 – 192
Old text pages 171 – 176)
October 26, 2005

3. Announcement In Chapter 5 we will skip from page 192 to page 206
Old text skip from 176 to 183
Homework due next Wednesday (11/3)
Show the following languages are not decidable
T = {<M> | M is a TM with L(M)R = L(M)}
i.e., wRL(M) whenever wL(M) (prob. 5.9)
G = {<M> | M is a TM s.t. L(M) is context-free}
F = {<M> | M is a TM with |L(M)| = 5}
October 26, 2005

4. Classes of languages
We have shown some language falls within each of the following classes
Regular
Context-free
Decidable
Turing recognizable
Co-Turing recognizable (ATM)
Today we will show some languages are undecidable
The fact that ATM is not Turing recognizable means there is no computer algorithm that recognizes it!
October 26, 2005

5. Undecidable languages
Turing recognizable
Co-Turing recognizable
Decidable
October 26, 2005

6. Undecidable languages
We can prove a problem is undecidable by contradiction
Assume the problem is decidable
Show that this implies something impossible
October 26, 2005

7. The halting problem HALTTM
HALTTM = {<M,w> | M is a TM and M halts on input w}
Theorem: HALTTM is undecidable
Proof: (by contradiction)
Show that if HALTTM is decidable then ATM is also decidable
HALTTM is undecidable because there’s no single strategy that will work for all TM’s
October 26, 2005

8. Proof (cont.) Assume R decides HALTTM Let S be the following TM
S = “on input <M,w>
Run R on <M,w>
If R rejects, reject
If R accepts, simulate M on w until it halts
If M accepts, accept; if M rejects, reject”
October 26, 2005

9. Proof (cont.) If HALTTM is decidable then S decides ATM
Since ATM is not decidable, HALTTM cannot be decidable
October 26, 2005

10. Proving language L is undecidable
Assume L is decidable
Let N be a TM that decides L
Show that a known undecidable language L’ will be decidable if it can use N to make decisions
This is called reducing problem L’ to problem L
Conclude N cannot exist
I.e., the language L is not decidable
Trick: Figuring out which undecidable language to use
October 26, 2005

11. Undecidability of ETM ETM = {<M> | M is a TM and L(M) = }
Theorem: ETM is undecidable
Proof: Assume ETM is decidable with decider TM R. Use R to decide ATM.
Recall ATM = {<M,w> | M is a TM that accepts w}.
How can we use R (which takes <M> as input) to determine if M accepts w?
Make new TM, M1, with L(M1) =  iff M accepts w
October 26, 2005

12. Proof (cont.)
New TM: Reject everything other than w, do whatever M does on input w.
M1 = “On input x
If x ≠ w, reject
If x = w, run M on input w
Accept if M accepts”
L(M1) ≠  if and only if M accepts w
October 26, 2005

13. Use R and M1 to decide ATM Consider the following TM
S = “On input <M,w>
Construct M1 that rejects all but w and simulates M on w
Run R on <M1>
If R accepts, reject; if R rejects, accept”
S decides ATM – a contradiction
Therefore, ETM is not decidable
October 26, 2005

14. Recap Assume R decides ETM
Create Turing machine M1 such that L(M1) ≠  iff M accepts w
Create Turing machine S that decides ATM by running R on input M1
Conclude R cannot exist
ETM cannot be decidable
October 26, 2005

15. Another undecidable language
Let REGULARTM = {<M> | M is a
TM and L(M) is a regular language}
Theorem: REGULARTM is undecidable
Proof: Assume R decides REGULARTM and use R to decide ATM (reduce the ATM problem to the REGULARTM problem).
As before, make a new TM, M2, that accepts a regular language iff M accepts w.
October 26, 2005

16. Proof (cont.) If M accepts w, then L(M2) = Σ* Otherwise, L(M2) = 0n1n
M2 = “On input x
If x = 0n1n for some n, accept
Otherwise, run M on w. If M accepts w, accept”
If M accepts w, then L(M2) = Σ*
A regular language
Otherwise, L(M2) = 0n1n
Not a regular language
L(M2) = 0n1n if M does not accept w
L(M2) = all strings if M accepts w
October 26, 2005

17. Proof (cont.) Assuming R decides REGULARTM consider the following TM
S = “On input <M,w>
Construct M2 s.t. L(M2) is regular iff M accepts w
Run R on M2
If R accepts, accept; if R rejects, reject”
S decides ATM iff R decides REGULARTM
L(M2) = 0n1n if M does not accept w
L(M2) = all strings if M accepts w
October 26, 2005

18. Have a fantastic fall break!