CSCI 2670 Introduction to Theory of Computing

CSCI 2670Introduction to Theory of Computing October 26, 2005

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

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

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 October 26, 2005

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

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

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 October 26, 2005

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

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

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

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

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

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

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

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

Proof (cont.) 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 October 26, 2005

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 October 26, 2005

Have a fantastic fall break!

CSCI 2670 Introduction to Theory of Computing