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!
This close relationship between the SL2 languages and the recognizable languages lets us use some of what we know about SL2 to discover properties of the recognizable languages. Because they are SL2 languages, the runs of an automaton A (and, equivalently, the strings of pairs licensed by G2A) will satisfy the 2-suffix substitution closure property. This means that every recognizable language L is a homomorphic image of some language L′ (over an alphabet Σ′ , say) for which
u′1σ′v′1 ∈ L′ and u′2 σ′v′2 ∈ L′⇒ u′1σ′v′2( and u′2σ′v′1) ∈ L′.
Moreover, u′1σ′v′1 ∈ L′ and u′1σ′v′2 ∈ L′⇒ u′2σ′v′2 ∈ L′
The hypothetical u′1σ′ and u′2σ′ are indistinguishable by the language. Any continuation that extends one to a string in L′ will also extend the other to a string in L′ ; any continuation that extends one to a string not in L′ will extend the other to a string not in L′.
For the SL2 language L′ the strings that are indistinguishable in this way are marked by their ?nal symbol. Things are not as clear for the recognizable language L because the homomorphism may map many symbols of Σ′ to the same symbol of Σ. So it will not generally be the case that we can easily identify the sets of strings that are indistinguishable in this way. But they will, nevertheless, exist. There will be pairs of strings u1 and u2 - namely the homomorphic images of the pairs u′1σ′ and u′2σ′-for which any continuation v, it will be the case that u1v ∈ L iff u2v ∈ L.
This equivalence between strings (in the sense of being indistinguishable by the language in this way) is the key to characterizing the recognizable languages purely in terms of the strings they contain in a way analogous to the way suffix substitution closure characterizes the SL2.
20*2
Exercise: Give a construction that converts a strictly 2-local automaton for a language L into one that recognizes the language L r . Justify the correctness of your construction.
Differentiate between DFA and NFA. Convert the following Regular Expression into DFA. (0+1)*(01*+10*)*(0+1)*. Also write a regular grammar for this DFA.
The initial ID of the automaton given in Figure 3, running on input ‘aabbba' is (A, aabbba) The ID after the ?rst three transitions of the computation is (F, bba) The p
A problem is said to be unsolvable if no algorithm can solve it. The problem is said to be undecidable if it is a decision problem and no algorithm can decide it. It should be note
how to understand DFA ?
A.(A+C)=A
Applying the pumping lemma is not fundamentally di?erent than applying (general) su?x substitution closure or the non-counting property. The pumping lemma is a little more complica
proof ogdens lemma .with example i am not able to undestand the meaning of distinguished position .
As we are primarily concerned with questions of what is and what is not computable relative to some particular model of computation, we will usually base our explorations of langua
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