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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
• 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
• 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
• 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!