Myhill-nerode, Theory of Computation

Assignment Help:

Theorem (Myhill-Nerode) A language L ⊆ Σ is recognizable iff ≡L partitions Σ* into ?nitely many Nerode equivalence classes.

Proof: For the "only if" direction (that every recognizable language has ?nitely many Nerode equivalence classes) observe that L ∈ Recog iff L = L(A) for some DFA A and that if δ(q0,w) = δ(q0, u) (i.e., if the path from the start state labeled w and that labeled u end up at the same state) then w ≡L u. This is a consequence of the fact that the state ˆ δ(q0,w) encodes all the information the automaton remembers about the string w. If v extends w to wv ∈ L(A) then v is the label of a path to an accepting state from δ(q0,w). Since this is the same state as δ(q0, u) the same path witnesses that uv ∈ L. Similarly, if the path leads one to a non-accepting state then it must necessarily lead the other to the same state. The automaton has no way of distinguishing two strings that lead to the same state and, consequently, the language it recognizes cannot distinguish them. Since A is deterministic, every string in Σ* labels a path leading to some state, hence the equivalence classes corresponding to the states partition Σ*. Since the automaton has ?nitely many states, it distinguishes ?nitely many equivalence classes.


Related Discussions:- Myhill-nerode

Exhaustive search, A problem is said to be unsolvable if no algorithm can s...

A problem is said to be unsolvable if no algorithm can solve it. The problem is said to be undecidable if it is a decision problem and no algorithm can decide it. It should be note

Strictly local languages, We have now de?ned classes of k-local languages f...

We have now de?ned classes of k-local languages for all k ≥ 2. Together, these classes form the Strictly Local Languages in general. De?nition (Strictly Local Languages) A langu

D c o, Prove xy+yz+ýz=xy+z

Prove xy+yz+ýz=xy+z

Turing, turing machine for prime numbers

turing machine for prime numbers

Two-tape turing machine, Let there L1 and L2 . We show that L1 ∩ L2 is CFG ...

Let there L1 and L2 . We show that L1 ∩ L2 is CFG . Let M1 be a decider for L1 and M2 be a decider for L2 . Consider a 2-tape TM M: "On input x: 1. copy x on the second

Positiveness problem - decision problems, For example, the question of whet...

For example, the question of whether a given regular language is positive (does not include the empty string) is algorithmically decidable. "Positiveness Problem". Note that

Normal forms, how to convert a grammar into GNF

how to convert a grammar into GNF

Transition and path functions, When an FSA is deterministic the set of trip...

When an FSA is deterministic the set of triples encoding its edges represents a relation that is functional in its ?rst and third components: for every q and σ there is exactly one

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

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