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CSCI 2670 Introduction to Theory of Computing. October 26, 2004. Agenda. Last week Decidability of languages involving DFA’s & CFG’s Non-decidability of the Turing machine acceptance problem A TM Today Revisit the A TM proof Finish Chapter 4 Problems Tomorrow

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Presentation Transcript
agenda
Agenda
  • Last week
    • Decidability of languages involving DFA’s & CFG’s
    • Non-decidability of the Turing machine acceptance problem ATM
  • Today
    • Revisit the ATM proof
    • Finish Chapter 4
    • Problems
  • Tomorrow
    • Begin Chapter 5 (pp. 171 – 176)

October 26, 2004

announcements
Announcements

Announcements

  • Homework due next Tuesday (11/2)
    • 4.4, 4.8, 5.1, 5.2, 5.13
  • No quiz this week
    • Quiz next Tuesday
  • No tutorials this week -- extra office hours instead
    • Tuesday & Wednesday 3:00 – 5:00
    • Or by appointment

October 26, 2004

undecidability of a tm
Undecidability of ATM

Theorem: ATM is undecidable

Proof: (By contradiction) Assume ATM is decidable and let H be a decider for ATM

H(<M,w>) = {

accept if M accepts w

reject if M does not accept w

October 26, 2004

proof cont
Proof (cont.)
  • Consider the TM D that submits the string <M> as input to the TM M

D = “On input <M>, where M is a TM:

    • Run H on input <M,<M>>
    • If H accepts <M,<M>>, reject
    • If H rejects <M,<M>>, accept”
      • Since H is a decider, it must accept or reject
      • Therefore, D is a decider as well
      • D is a diagonalizing TM

October 26, 2004

proof cont1
Proof (cont.)
  • What happens if D’s input is <D>?

D(<D>) = {

  • D cannot exist!
  • Therefore, H cannot exist – i.e., ATM is undecidable

reject if D accepts <D>

accept if D does not accept <D>

October 26, 2004

recap
Recap
  • Assume H decides ATM
    • H(<M,w>) = accept if TM M accepts w, reject otherwise
  • Define D using H
    • D(<M>) returns opposite of H(<M,<M>>)
  • Consider D(<D>)
    • D accepts <D> if and only if D rejects <D>

October 26, 2004

apply this method to a dfa
Apply this method to ADFA
  • Assume H decides ADFA
    • H(<A,w>) = accept if DFA A accepts w, reject otherwise
  • Define D using H
    • D(<A>) returns opposite of H(<A,<A>>)
  • Consider D(<D>)
    • D only accepts DFA’s as input … D is a TM
    • Method does not apply

October 26, 2004

apply this method to a tm d
Apply this method to ATM-D
  • Assume H decides ATM-D
    • H(<M,w>) = accept if decider TM M accepts w, reject otherwise
  • Problem
    • What if M is not a decider and just a regular TM?
    • How can we test our input?
  • ATM-D is not decidable

October 26, 2004

co turing recognizable
Co-Turing recognizable

Definition: A language A is co-Turing-recognizable if Ā is Turing-recognizable.

Example:

EDFA = {<A> | A is a DFA and L(A) = }

EDFA = {S | S does not describe a DFA or S describes a DFA with a non-empty language}

  • How would we decide this language?

October 26, 2004

decidability and recognizability
Decidability and recognizability

Theorem: The language A is decidable if and only if it is both Turing-recognizable and co-Turing-recognizable

Proof: In two parts

  • If A is decidable then A and Ā are Turing-recognizable
    • Follows from definitions
  • If A and Ā are Turing-recognizable then A is decidable

October 26, 2004

decider for a
Decider for A
  • Assume M1 recognizes A and M2 recognizes Ā
  • Consider the following TM

M = “On input w:

    • Run M1 and M2 on input w in parallel
    • If M1 accepts, accept; if M2 accepts, reject”
  • M must halt on w because at least on of the recognizers must halt on w
    • If w  A, M1 must halt; otherwise, M2 must halt

October 26, 2004

corollary
Corollary
  • ATM is not Turing-recognizable

Proof: ATM is Turing-recognizable, but not decidable. By previous theorem, ATM cannot be co-Turing-recognizable.

October 26, 2004

proving dedicability of l
Proving dedicability of L
  • Create a TM that decides L
  • Your TM may invoke another TM that decides a language you know is decidable

October 26, 2004

slide16

Some decidable languages

  • FDFA = {<A> | A is a DFA and L(A) is finite}
  • PRIME = { n | n is a prime number}
  • CONN = {<G> | G is a connected graph}
  • L10DFA = {D | D is a DFA that accepts every string w with |w| = 10}
  • INTCFG = {<G1, G2, w> | G1 and G2 are CFG’s and w is accepted by both}
  • INTLCFG = L(G1 G2), where G1 and G2 are CFG’s

October 26, 2004

group project 1
Group project 1

FDFA = {<A> | A is a DFA and L(A) is finite}

October 26, 2004

group project 2
Group project 2

PRIME = { n | n is a prime number}

October 26, 2004

group project 3
Group project 3

CONN = {<G> | G is a connected graph}

October 26, 2004

group project 4
Group project 4

L10DFA = {D | D is a DFA that accepts every string w with |w| = 10}

October 26, 2004

group project 5
Group project 5

INTCFG = {<G1, G2, w> | G1 and G2 are CFG’s and w is accepted by both}

October 26, 2004

group project 6
Group project 6

INTLCFG = L(G1 G2), where G1 and G2 are CFG’s

October 26, 2004