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

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

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

Mealy machine, Construct a Mealy machine that can output EVEN or ODD Accord...

Construct a Mealy machine that can output EVEN or ODD According to the total no. of 1''s encountered is even or odd.

Class of local languages is not closed under union, Both L 1 and L 2 are ...

Both L 1 and L 2 are SL 2 . (You should verify this by thinking about what the automata look like.) We claim that L 1 ∪ L 2 ∈ SL 2 . To see this, suppose, by way of con

Instantaneous description of an fsa, De?nition Instantaneous Description of...

De?nition Instantaneous Description of an FSA: An instantaneous description (ID) of a FSA A = (Q,Σ, T, q 0 , F) is a pair (q,w) ∈ Q×Σ* , where q the current state and w is the p

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

Agents architecture, Describe the architecture of interface agency

Describe the architecture of interface agency

Moore machine, Construct a Moore machine to convert a binary string of radi...

Construct a Moore machine to convert a binary string of radix 4.

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