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

Graph Connectivity, Let G be a graph with n > 2 vertices with (n2 - 3n + 4)...

Let G be a graph with n > 2 vertices with (n2 - 3n + 4)/2 edges. Prove that G is connected.

CNF, S-->AAA|B A-->aA|B B-->epsilon

S-->AAA|B A-->aA|B B-->epsilon

Closure properties of recognizable languages, We got the class LT by taking...

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

Equivalence of nfas, It is not hard to see that ε-transitions do not add to...

It is not hard to see that ε-transitions do not add to the accepting power of the model. The underlying idea is that whenever an ID (q, σ  v) directly computes another (p, v) via

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

Create a general algorithm from a checking algorithm, Claim Under the assum...

Claim Under the assumptions above, if there is an algorithm for checking a problem then there is an algorithm for solving the problem. Before going on, you should think a bit about

Designing finite automata, a finite automata accepting strings over {a,b} e...

a finite automata accepting strings over {a,b} ending in abbbba

Dfa to re, c program to convert dfa to re

c program to convert dfa to re

Prove the arden''s theorem, State and Prove the Arden's theorem for Regular...

State and Prove the Arden's theorem for Regular Expression

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