Already have an account? Get multiple benefits of using own account!
Login in your account..!
Remember me
Don't have an account? Create your account in less than a minutes,
Forgot password? how can I recover my password now!
Enter right registered email to receive password!
The upper string r ∈ Q+ is the sequence of states visited by the automaton as it scans the lower string w ∈ Σ*. We will refer to this string over Q as the run of A on w. The automaton A accepts w iff the run of A on w ends in an accepting state. (If A is non-deterministic there will potentially be many runs with the automaton accepting if any one of them ends in an accepting state.) Note that the set of runs of an automaton is an SL2 language, recognized by the SL2 automaton (over Q) one gets by projecting away the third component of the triples of GA. Thus there is some kind of close relationship between the strictly local languages and the recognizable languages.
To get at this we will start by working in the other direction, extending our tiles to hold four symbols. The idea is to include, for each tile (q, p, σ) ∈ GA, a tile extended with σ′ for each σ′ ∈ Σ. (We don't actually need tiles for all such σ′ , only for those that occur on tiles (x, q, σ′) which might precede this one in a tiling, but including all of them will be harmless-the ones that do not occur on such tiles will just be useless.)
(c) Can you say that B is decidable? (d) If you somehow know that A is decidable, what can you say about B?
Claim Under the assumptions above, if there is an algorithm for checking a problem then there is an algorithm for solving the problem. Before going on, you should think a bit about
To see this, note that if there are any cycles in the Myhill graph of A then L(A) will be infinite, since any such cycle can be repeated arbitrarily many times. Conversely, if the
wht is pumping lema
proof of arden''s theoram
In Exercise 9 you showed that the recognition problem and universal recognition problem for SL2 are decidable. We can use the structure of Myhill graphs to show that other problems
One of the first issues to resolve, when exploring any mechanism for defining languages is the question of how to go about constructing instances of the mechanism which define part
One might assume that non-closure under concatenation would imply non closure under both Kleene- and positive closure, since the concatenation of a language with itself is included
The Recognition Problem for a class of languages is the question of whether a given string is a member of a given language. An instance consists of a string and a (?nite) speci?cat
Ask question #Minimum 100 words accepte
Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!
whatsapp: +91-977-207-8620
Phone: +91-977-207-8620
Email: [email protected]
All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd