Closure properties of recognizable languages, Theory of Computation

Assignment Help:

We got the class LT by taking the class SL and closing it under Boolean operations. We have observed that LT ⊆ Recog, so certainly any Boolean combination of LT languages will also be recognizable. But what about the class of recognizable languages as a whole? Are Boolean combinations of recognizable (not just LT) languages also recognizable. In answering we can use the same methodology we use to show that any language is recognizable: consider what we need to keep track of in scanning a string in order to determine if it belongs to the language or not and then use that information to build our state set.

Suppose, then, that L = L1 ∩ L2, where L1 and L2 are both recognizable. A string w will be in L iff it is in both L1 and L2. Since they are recognizable there exist DFAs A1 and A2 for which L1 = L(A1) and L2 = L(A2). We can tell if the string is in L1 or L2 simply by keeping track of the state of the corresponding automaton. We can tell if it is in both by keeping track of both states simultaneously.


Related Discussions:- Closure properties of recognizable languages

Ogdens lemma, proof ogdens lemma .with example i am not able to undestand ...

proof ogdens lemma .with example i am not able to undestand the meaning of distinguished position .

Synthesis theorem, Kleene called this the Synthesis theorem because his (an...

Kleene called this the Synthesis theorem because his (and your) proof gives an effective procedure for synthesizing an automaton that recognizes the language denoted by any given r

Prism algorithm, what exactly is this and how is it implemented and how to ...

what exactly is this and how is it implemented and how to prove its correctness, completeness...

Decidability, examples of decidable problems

examples of decidable problems

Pendulum Swings, how many pendulum swings will it take to walk across the c...

how many pendulum swings will it take to walk across the classroom?

Construct a recognizer, Let L1 and L2 be CGF. We show that L1 ∩ L2 is CFG t...

Let L1 and L2 be CGF. We show that L1 ∩ L2 is CFG too. Let M1 be a decider for L1 and M2 be a decider for L2 . Consider a 2-tape TM M: "On input x: 1. copy x on the sec

Write Your Message!

Captcha
Free Assignment Quote

Assured A++ Grade

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!

All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd