Finite-state automaton, Theory of Computation

Assignment Help:

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 not (there is only one such path). Those leading to region C end in b. Note that once we are in region C the question of whether we have seen bb or not is no longer relevant; in order to accept we must see aa and, since the path has ended with b, we cannot reach aa without ?rst seeing ba (hence, passing through region E). Finally, in region A we have not looked at anything yet. This where the empty string ends up.

331_Finite-state automaton.png

Putting this all together, there is no reason to distinguish any of the nodes that share the same region. We could replace them all with a single node. What matters is the information that is relevant to determining if a string should be accepted or can be extended to one that should be. In keeping with this insight, we will generalize our notion of transition graphs to graphs with an arbitrary, ?nite, set of nodes distinguishing the signi?cant states of the computation and edges that represent the transitions the automaton makes from one state to another as it scans the input. Figure 3 represents such a graph for the minimal equivalent of the automaton of Figure 1.


Related Discussions:- Finite-state automaton

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

Abstract model for an algorithm solving a problem, These assumptions hold f...

These assumptions hold for addition, for instance. Every instance of addition has a unique solution. Each instance is a pair of numbers and the possible solutions include any third

Non-determinism - recognizable language, Our DFAs are required to have exac...

Our DFAs are required to have exactly one edge incident from each state for each input symbol so there is a unique next state for every current state and input symbol. Thus, the ne

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

Qbasic, Ask question #Minimum 100 words accepte

Ask question #Minimum 100 words accepte

Ogdens lemma, proof ogdens lemma .with example i am not able to undestand ...

proof ogdens lemma .with example i am not able to undestand the meaning of distinguished position .

Can you help me in automata questions, i have some questions in automata, c...

i have some questions in automata, can you please help me in solving in these questions?

Xx, Ask queyystion #Minimum 100 words accepted#

Ask queyystion #Minimum 100 words accepted#

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

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