L and L’ are Turing-recognizable, prove L is Turing-decidable

1 / 3

# L and L’ are Turing-recognizable, prove L is Turing-decidable - PowerPoint PPT Presentation

L and L’ are Turing-recognizable, prove L is Turing-decidable. M TR. <w>. accept. w. B. accept. w. accept. reject. A. B checks if string w is in L, A checks if w is in L’ M TR halts because w is in either L or L’; B and A are run once.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about 'L and L’ are Turing-recognizable, prove L is Turing-decidable' - kioshi

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.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
L and L’ are Turing-recognizable, prove L is Turing-decidable

MTR

accept

w

B

accept

w

accept

reject

A

• B checks if string w is in L, A checks if w is in L’
• MTR halts because w is in either L or L’; B and A are run once
Exercise 4.3MS* = { | Z is a DFA, L(Z) = S*}

MS*

accept

reject

MEQdfa

reject

accept

F

• F creates DFA E such that L(E) = S*
• MEQdfa accepts if L(Z) = L(E), rejects otherwise.
• MS* accepts if and only if L(Z) = L(E) if and only if L(Z) = S*
• MS* halts because F, MEQdfa are decidable and run only once
Exercise 4.2: MT = { | B is a DFA, E is a regular expression and B = E }

MT

accept

accept

MEQdfa

reject

reject

D

• D converts regular expression E into equivalent DFA F
• MEQdfa accepts if L(B) = L(F), rejects otherwise
• MT accepts if and only if L(B) = L(F) if and only if L(B) = L(E)
• MT halts since D and MEQdfaare deciable and run once