1. CSCI 2670 Introduction to Theory of Computing
October 27, 2004

2. Agenda Yesterday Today Revisit the ATM proof Finish Chapter 4 Problems
Begin Chapter 5 (pp. 171 – 176)
October 27, 2004

3. Announcements Announcements
No tutorials this week -- extra office hours instead
Tuesday & Wednesday 3:00 – 5:00
Or by appointment
October 27, 2004

4. Some decidable languages
FDFA = {<A> | A is a DFA and L(A) is finite}
PRIME = { n | n is a prime number}
CONN = {<G> | G is a connected graph}
L10DFA = {D | D is a DFA that accepts every string w with |w| = 10}
INTCFG = {<G1, G2, w> | G1 and G2 are CFG’s and w is accepted by both}
INTLCFG = L(G1  G2), where G1 and G2 are CFG’s
October 27, 2004

5. Classes of languages
We have shown some language falls within each of the following classes
Regular
Context-free
Decidable
Non-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 27, 2004

6. Undecidable languages
We can prove a problem is undecidable by contradiction
Assume the problem is decidable
Show that this implies something impossible
October 27, 2004

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
If M does not halt on w, then reject
If M halts on w, then run M on w and return result
HALTTM is undecidable because there’s no single strategy that will work for all TM’s
October 27, 2004

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 27, 2004

9. Recap If HALTTM is decidable then ATM is decidable
Since ATM is not decidable, HALTTM cannot be decidable
October 27, 2004

10. Proving language L is undecidable
Assume L is decidable
Let N be a TM that decides the language
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
The language L is not decidable
Trick: Figuring out which undecidable language to use
October 27, 2004

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?
Hint: make new TM
October 27, 2004

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 27, 2004

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> that accepts iff L(M1)=
If R accepts, reject; if R rejects, accept”
S decides ATM – a contradiction
Therefore, ETM is not decidable
October 27, 2004

14. Have a fantastic fall break!
October 27, 2004