CSCI 2670Introduction to Theory of

Computing

November 1, 2005

November 1, 2005

Agenda

• Last week– Test & undecidability

• Today– Review one undecidability proof– Rice’s Theorem– Reductions (Section 5.3)

November 1, 2005

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

Announcements

November 1, 2005

An undecidable languageLet 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

Proof (cont.)

Consider the following TMS = “On input <M,w>

1. Construct the following TM M2

M2 = “On input x1. If x = 0n1n for some n, accept2. Otherwise, run M on w. If M accepts w,

accept”

2. Run R on M2 (accepts iff L(M2) is a RL)

3. If R accepts, accept; if R rejects, reject”

S decides ATM if R decides REGULARTM

November 1, 2005

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

Rice’s theorem

• Determining whether a TM satisfies any non-trivial property is undecidable

– A property is non-trivial if:1. It depends only on the language of M,

and 2. Some, but not all, Turing machines have

the property

– Examples: Is L(M) regular? A CFG? Finite?

November 1, 2005

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

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

ATM decider using MP

S = “On input <M,w>1. Create the following TM N

N = “On input x1. Run M on w until it accepts2. If M accepts w, run MP on x3. If MP accepts x, accept; if MP rejects x,

reject”

2. 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