1. CSCI 2670 Introduction to Theory of Computing
November 1, 2005

2. Agenda Last week Today Test & undecidability
Review one undecidability proof
Rice’s Theorem
Reductions (Section 5.3)
November 1, 2005

3. Announcements Announcements
Homework due next Tuesday (11/8)
5.7, 5.12 (do not use Rice’s Theorem), 5.20, 5.22, 5.30 b
Old text 5.7
Get handout for 5.12, 5.20, 5.22 and 5.30 b
November 1, 2005

4. An 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.
November 1, 2005

5. Proof (cont.) Consider the following TM S = “On input <M,w>
Construct the following TM M2
M2 = “On input x
If x = 0n1n for some n, accept
Otherwise, run M on w. If M accepts w, accept”
Run R on M2 (accepts iff L(M2) is a RL)
If R accepts, accept; if R rejects, reject”
S decides ATM if R decides REGULARTM
L(M2) = 0n1n if M does not accept w
L(M2) = all strings if M accepts w
November 1, 2005

6. Insight
The TM M2 is specially designed to be regular if and only if M accepts w
Then call TM that decides REGULARTM on M2
November 1, 2005

7. Rice’s theorem
Determining whether a TM satisfies any non-trivial property is undecidable
A property is non-trivial if:
It depends only on the language of M, and
Some, but not all, Turing machines have the property
Examples: Is L(M) regular? A CFG? Finite?
November 1, 2005

8. Proof of Rice’s theorem
Assume there is some decidable non-trivial property P for Turing machines
Assume TM’s that accept  do not satisfy P
If they do, just consider P
November 1, 2005

9. Proof of Rice’s theorem
Let TM B decide P
On input <M>, B accepts iff TM M has property P
Let MP be a TM that satisfies P
Since P is non-trivial, there is some MP satisfying P
Use B and MP to decide ATM
Create a new TM S that decides ATM using B and MP
November 1, 2005

10. ATM decider using MP S = “On input <M,w>
Create the following TM N
N = “On input x
Run M on w until it accepts
If M accepts w, run MP on x
If MP accepts x, accept; if MP rejects x, reject”
Run B on <N(<M,w>)>
If B accepts, accept; if B rejects, reject”
L(N) = L(MP) if M accepts w; otherwise L(N) = 
S decides ATM if B decides TM’s satisfying P
Therefore, B cannot exist
November 1, 2005