Path function of a nfa, Theory of Computation

Assignment Help:

The path function δ : Q × Σ* → P(Q) is the extension of δ to strings:

969_Path Function of a NFA.png

This just says that the path labeled ε from any given state q goes only to q itself (or rather never leaves q) and that to ?nd the set of states reached by paths labeled wσ from q one ?rst ?nds all the states q′ reached by paths labeled w from q and then takes the set of all the states reached by an edge labeled σ from any of those q′.

We will still accept a string w i? there is a path labeled w leading from the initial state to a ?nal state, but now there may be many paths labeled w from the initial state, some of which reach ?nal states and some of which do not. When thinking in terms of the path function, we need to modify the de?nition of the language accepted by A so it includes every string for which at least one path ends at a ?nal state.


Related Discussions:- Path function of a nfa

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

Third model of computation, Computer has a single LIFO stack containing ?xe...

Computer has a single LIFO stack containing ?xed precision unsigned integers (so each integer is subject to over?ow problems) but which has unbounded depth (so the stack itself nev

Answer, And what this money. Invovle who it involves and the fact of,how we...

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

Xx, Ask queyystion #Minimum 100 words accepted#

Ask queyystion #Minimum 100 words accepted#

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

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 .

Decidability, examples of decidable problems

examples of decidable problems

D c o, Prove xy+yz+ýz=xy+z

Prove xy+yz+ýz=xy+z

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