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

Kleene closure, One might assume that non-closure under concatenation would...

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

Gastric juice, what are composition and its function of gastric juice

what are composition and its function of gastric juice

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

Transition graphs, We represented SLk automata as Myhill graphs, directed g...

We represented SLk automata as Myhill graphs, directed graphs in which the nodes were labeled with (k-1)-factors of alphabet symbols (along with a node labeled ‘?' and one labeled

Design and implementation of the state machine, You are required to design ...

You are required to design a system that controls the speed of a fan's rotation. The speed at which the fan rotates is determined by the ambient temperature, i.e. as the temperatur

Strictly local languages, While the SL 2 languages include some surprising...

While the SL 2 languages include some surprisingly complex languages, the strictly 2-local automata are, nevertheless, quite limited. In a strong sense, they are almost memoryless

Context free grammar, A context free grammar G = (N, Σ, P, S)  is in binary...

A context free grammar G = (N, Σ, P, S)  is in binary form if for all productions A we have |α| ≤ 2. In addition we say that G is in Chomsky Normaml Form (CNF) if it is in bi

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

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

Finite-state automaton, Paths leading to regions B, C and E are paths which...

Paths leading to regions B, C and E are paths which have not yet seen aa. Those leading to region B and E end in a, with those leading to E having seen ba and those leading to B no

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