1 / 6

The universal Turing machine

CSC 4170 Theory of Computation. The universal Turing machine. Section 4.2. 4.2.a. The universal Turing machine. Let A TM = {<M,w> | M is a TM and M accepts string w} A TM is misleadingly called the “halting problem” in the textbook. . Theorem: A TM is Turing-recognizable. .

chaela
Download Presentation

The universal Turing machine

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. CSC 4170 Theory of Computation The universal Turing machine Section 4.2

  2. 4.2.a The universal Turing machine Let ATM = {<M,w> | M is a TM and M accepts string w} ATM is misleadingly called the “halting problem” in the textbook. Theorem:ATM is Turing-recognizable. Proof idea. The following TM U, called the universal TM, recognizes ATM: U = “On input <M,w>, where M is a TM and w is a string: 1. Simulate M on input w. 2. If M ever enters its accept state, accept; if M ever enters its reject state, reject.” Does U also decide ATM?

  3. 4.2.b.1 ATM is undecidable Theorem 4.11:ATM is undecidable. Proof idea. Suppose, for a contradiction, that ATM is decidable. That is, there is a TM H that decides ATM. Thus, that machine H behaves as follows: accept if M accepts w reject if M does not accept w H(<M,w>) = Using H as a subroutine, we can construct the following TM D: D = “On input <M>, where M is a TM: 1. Run H on input <M,<M>>. 2. Do the opposite of what H does. That is, if H accepts, reject, and if H rejects, accept.” accept if M does not accept <M> reject if M accepts <M> Thus, D(<M>) =

  4. 4.2.b.2 ATM is undecidable accept if D does not accept <D> reject if D accepts <D> But then D(<D>) = Contradiction! • To summarize: • H accepts <M,w> exactly when M accepts w. • D rejects <M> exactly when M accepts M. • D rejects <D> exactly when D accepts <D>.

  5. 4.2.c Theorem 4.23: A is Turing-unrecognizable. TM A Turing-unrecognizable language ATM = {<M,w> | M is a TM and M accepts string w} ATM = {<M,w> | M is a TM and M does not accept string w} Proof idea. Suppose, for a contradiction, that ATM is Turing-recognizable. That is, there is a TM U that recognizes ATM. • U recognizes ATM (slide 4.2.a) • U recognizes ATM Thus, Let M = “On input w: 1. Run both U and U on input w in parallel; 2. If U accepts, accept; if U accepts, reject.” It can bee seen that M decides ATM, which contradicts Theorem 4.11.

  6. 4.2.d The language hierarchy summary All languages Turing-recognizable languages Turing-decidable languages Context-free languages Regular languages {anbn | n0} ATM ATM {anbncn | n0}

More Related