Merging nodes, Theory of Computation

Assignment Help:

Another striking aspect of LTk transition graphs is that they are generally extremely ine?cient. All we really care about is whether a path through the graph leads to an accepting node or not. From that perspective, there is surely no reason to distinguish the nodes in the region marked H in Figure 2. Every one of these is an accepting node and every path from any one of them leads only to others in the same region. Every string with an initial segment which reaches one of these nodes will be accepted regardless of what the rest of the string looks like.

With a little more thought, it should become clear that the nodes in each of the other regions marked out in the ?gure are equivalent in a similar way. Any path which, when appended to a path leading to any one of the nodes, extends it to a path leading to an accepting state will do the same for paths leading to any node in the same region.

We can characterize the paths leading to the nodes in each region in terms of the components of aa ∧ (¬bb ∨ ba) they satisfy. Paths leading to region H satisfy aa ∧ ba. Strings starting this way will be accepting no matter what occurs in the remainder of the string. Regions D, F and G all satisfy aa. D and F also satisfy ¬bb and, so, are accepting. Paths reaching region G have seen bb and no longer accept until they have been extended with an a, thus satisfying aa ∧ ba and entering region H. We need to distinguish the nodes inregions D and F because paths leading to D end in a and, therefore, can be extended with b harmlessly, while if a path leading to F is extended with b we will no longer accept it.


Related Discussions:- Merging nodes

Defining strictly local automata, One of the first issues to resolve, when ...

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

Two-tape turing machine, Let there L1 and L2 . We show that L1 ∩ L2 is CFG ...

Let there L1 and L2 . We show that L1 ∩ L2 is CFG . 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 second

Formal languages and grammar, The universe of strings is a very useful medi...

The universe of strings is a very useful medium for the representation of information as long as there exists a function that provides the interpretation for the information carrie

Automaton theory, let G=(V,T,S,P) where V={a,b,A,B,S}, T={a,b},S the start ...

let G=(V,T,S,P) where V={a,b,A,B,S}, T={a,b},S the start symbol and P={S->Aba, A->BB, B->ab,AB->b} 1.show the derivation sentence for the string ababba 2. find a sentential form

Finiteness of languages is decidable, To see this, note that if there are a...

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

Automata, automata of atm machine

automata of atm machine

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

Automata answer, build a TM that enumerate even set of even length string o...

build a TM that enumerate even set of even length string over a

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