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Exploring Undecidable Languages in Theory of Computing

Uncover the intricacies of undecidable languages in theory of computing, including proofs and examples of HALTTM, ETM, REGULARTM, and more. Discover how these languages challenge the limits of computation.

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Exploring Undecidable Languages in Theory of Computing

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  1. CSCI 2670Introduction to Theory of Computing October 26, 2005

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

  3. 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

  4. 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

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

  6. 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

  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 October 26, 2005

  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 26, 2005

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

  10. 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

  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? Make new TM, M1, with L(M1) =  iff M accepts w October 26, 2005

  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 26, 2005

  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> • If R accepts, reject; if R rejects, accept” S decides ATM – a contradiction Therefore, ETM is not decidable October 26, 2005

  14. 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

  15. 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

  16. 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

  17. 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

  18. Have a fantastic fall break!

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