Concatenation, Theory of Computation

Assignment Help:

We saw earlier that LT is not closed under concatenation. If we think in terms of the LT graphs, recognizing the concatenation of LT languages would seem to require knowing, while scanning a string in L1 . L2, for instance, when to switch from keeping track of factors for L1 to keeping track of factors from L2.

Assuming that the alphabets were not disjoint, there is (evidently, since LT is not closed under concatenation) no way, in general, to know that. For the recognizable languages, on the other hand, we have the convenience of being able to work with non-determinism. We don't actually have to know when to switch from one automaton to the next. Whenever we get to a point in the string that could possibly be the end of the pre?x that is in L1 we can just allow for a non-deterministic choice of whether to continue scanning for A1 (the machine recognizing L1) or to switch to scanning for A2. Since whenever the string is in L1 .  L2 there will be some correct place to switch and since acceptance by a NFA requires only that there some accepting computation, the combined automaton will accept every string in L1 . L2. Moreover, the combined automaton will accept a string iff there is some point at which it can be split into a string accepted by A1 followed by one accepted by A2: it accepts all and only the strings in L1 . L2.


Related Discussions:- Concatenation

Algorithm for the universal recognition problem, Sketch an algorithm for th...

Sketch an algorithm for the universal recognition problem for SL 2 . This takes an automaton and a string and returns TRUE if the string is accepted by the automaton, FALSE otherwi

Decision problems, In Exercise 9 you showed that the recognition problem an...

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

Powerset construction, As de?ned the powerset construction builds a DFA wit...

As de?ned the powerset construction builds a DFA with many states that can never be reached from Q′ 0 . Since they cannot be reached from Q′ 0 there is no path from Q′ 0 to a sta

Strictly local generation automaton, Another way of interpreting a strictly...

Another way of interpreting a strictly local automaton is as a generator: a mechanism for building strings which is restricted to building all and only the automaton as an inexh

#turing machine, #can you solve a problem of palindrome using turing machin...

#can you solve a problem of palindrome using turing machine with explanation and diagrams?

Deterministic finite automata, conversion from nfa to dfa 0 | 1 ____...

conversion from nfa to dfa 0 | 1 ___________________ p |{q,s}|{q} *q|{r} |{q,r} r |(s) |{p} *s|null |{p}

Myhill-nerode, Theorem (Myhill-Nerode) A language L ⊆ Σ is recognizable iff...

Theorem (Myhill-Nerode) A language L ⊆ Σ is recognizable iff ≡L partitions Σ* into ?nitely many Nerode equivalence classes. Proof: For the "only if" direction (that every recogn

Tuning machine, design a tuning machine for penidrome

design a tuning machine for penidrome

Kleene Closure, 1. Does above all''s properties can be used to prove a lang...

1. Does above all''s properties can be used to prove a language regular? 2..which of the properties can be used to prove a language regular and which of these not? 3..Identify one

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