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CSCI 2670 Introduction to Theory of Computing. November 1, 2005. Agenda. Last week Test & undecidability Today Review one undecidability proof Rice’s Theorem Reductions (Section 5.3). Announcements. Announcements. Homework due next Tuesday (11/8)
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CSCI 2670Introduction to Theory of Computing November 1, 2005
Agenda • Last week • Test & undecidability • Today • Review one undecidability proof • Rice’s Theorem • Reductions (Section 5.3) November 1, 2005
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
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
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 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: • 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
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> • 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