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CSCI 2670 Introduction to Theory of Computing

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Assume L is decidable. Let N be a TM that decides the language ... Proof: Assume ETM is decidable with decider TM R. Use R to decide ATM. ... – PowerPoint PPT presentation

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Title: CSCI 2670 Introduction to Theory of Computing


1
CSCI 2670 Introduction to Theory of Computing
October 27, 2004
2
Agenda
  • Yesterday
  • Revisit the ATM proof
  • Finish Chapter 4
  • Problems
  • Today
  • Begin Chapter 5 (pp. 171 176)

3
Announcements
Announcements
  • No tutorials this week -- extra office hours
    instead
  • Tuesday Wednesday 300 500
  • Or by appointment

4
Some decidable languages
  • FDFA ltAgt A is a DFA and L(A) is finite
  • PRIME n n is a prime number
  • CONN ltGgt G is a connected graph
  • L10DFA D D is a DFA that accepts every
    string w with w 10
  • INTCFG ltG1, G2, wgt G1 and G2 are CFGs and w
    is accepted by both
  • INTLCFG L(G1 ? G2), where G1 and G2 are CFGs

5
Classes of languages
  • We have shown some language falls within each of
    the following classes
  • Regular
  • Context-free
  • Decidable
  • Non-Turing recognizable (ATM)
  • Today we will show some languages are undecidable

6
Undecidable languages
  • We can prove a problem is undecidable by
    contradiction
  • Assume the problem is decidable
  • Show that this implies something impossible

7
The halting problem HALTTM
  • HALTTM ltM,wgt 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
  • If M does not halt on w, then reject
  • If M halts on w, then run M on w and return result

8
Proof (cont.)
  • Assume R decides HALTTM
  • Let S be the following TM
  • S on input ltM,wgt
  • Run R on ltM,wgt
  • If R rejects, reject
  • If R accepts, simulate M on w until it halts
  • If M accepts, accept if M rejects, reject

9
Recap
  • If HALTTM is decidable then ATM is decidable
  • Since ATM is not decidable, HALTTM cannot be
    decidable

10
Proving language L is undecidable
  • Assume L is decidable
  • Let N be a TM that decides the language
  • 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
  • The language L is not decidable
  • Trick Figuring out which undecidable language
    to use

11
Undecidability of ETM
  • ETM ltMgt 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 ltM,wgt M is a TM that accepts
    w.
  • How can we use R (which takes ltMgt as input) to
    determine if M accepts w?
  • Hint make new TM

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

13
Use R and M1 to decide ATM
  • Consider the following TM
  • S On input ltM,wgt
  • Construct M1 that rejects all but w and simulates
    M on w
  • Run R on ltM1gt that accepts iff L(M1)?
  • If R accepts, reject if R rejects, accept
  • S decides ATM a contradiction
  • Therefore, ETM is not decidable

14
Have a fantastic fall break!
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