Equivalence of nfas, Theory of Computation

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 a path that includes some number of ε-transitions (before the σ-transition, after it or both), we can get the same effect by extending the transition relation to include a σ-transition directly from q to v. So, in the example we could add ‘a' edges from 0 to 1 (accounting for the path 0 2072_Equivalence of NFAs.png 3) and from 1 to 3 (accounting for the path 1 27_Equivalence of NFAs1.png 3) and ‘b' edges from 1 to 3 (accounting for the path 1 1649_Equivalence of NFAs2.png  3), from 3 to 2 (accounting for the path 3 1088_Equivalence of NFAs3.png2), and from 1 to 2 (accounting for the  path 1 2144_Equivalence of NFAs4.png2), Note that in each of these cases this corresponds to extending δ(q, σ) to include all states in ˆ δ(q, σ). The remaining effect of the ε-transition from 0 to 2 is the fact that the automaton accepts ‘ε'. This can be obtained, of course, by simply adding 0 to F. Formalizing this  we get a lemma.

Posted Date: 3/21/2013 2:54:59 AM | Location : United States







Related Discussions:- Equivalence of nfas, Assignment Help, Ask Question on Equivalence of nfas, Get Answer, Expert's Help, Equivalence of nfas Discussions

Write discussion on Equivalence of nfas
Your posts are moderated
Related Questions
We developed the idea of FSA by generalizing LTk transition graphs. Not surprisingly, then, every LTk transition graph is also the transition graph of a FSA (in fact a DFA)-the one

Application of the general suffix substitution closure theorem is slightly more complicated than application of the specific k-local versions. In the specific versions, all we had

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


Construct a PDA that accepts { x#y | x, y in {a, b}* such that x ? y and xi = yi for some i, 1 = i = min(|x|, |y|) }. For your PDA to work correctly it will need to be non-determin

1. Simulate a TM with infinite tape on both ends using a two-track TM with finite storage 2. Prove the following language is non-Turing recognizable using the diagnolization

The Equivalence Problem is the question of whether two languages are equal (in the sense of being the same set of strings). An instance is a pair of ?nite speci?cations of regular

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

And what this money. Invovle who it involves and the fact of,how we got itself identified candidate and not withstanding time date location. That shouts me media And answers who''v