Myhill-nerode Theorem-1
Myhill-Nerode Relations
Isomorphism: Two DFAs given by M QM m sm Fm = ( , S,d , , ) and N QN n sn Fn = ( , S,d , , ) are said to be “isomorphic” if there is a one-to-one and onto mapping
f : QM ®QN such that
- f (sM ) = sN ,
- (d M ( p, a)) = d N ( f ( p), a) for all P QM a Î , ÎS,
- p FM if f p FN Î ( )Î .
Isomorphic automata accept same set.
Myhill-Nerode Relations: Let R Í S* be a regular set, and let M = (Q, S,d, s, F ) be a DFA for R with no inaccessible states.
The automaton M induces an equivalence relation º M on S* defined by
It is easy to show that the relation º M is an equivalence relation, meaning it is reflexive, symmetric and transitive.
A few properties satisfied by º M are as follows:
(a) It is a right congruence: for any x, yÎS* and a ÎS,
Proof: Assume x º _{M }y.
Therefore we have
(b) It refines R : for any x, yÎS* ,
Proof: Since d$ (s, x) = d$ (s, y), which is either an accept state or a reject state, so either both x and y are accepted or both are rejected.
(c) It is of “Finite index”: i.e., it has only finitely much equivalence class.
This is because there is exactly one equivalence class
{x ÎS* | d$ (s, x) = q}
Corresponding to each state q of M.
Hence the equivalence relation º on S* is a “Myhill-Herode relation” for R if it satisfies properties (a), (b) and (c). i.e., if it is a right congruence of finite index refining R.
Myhill-Nerode Theorem: Let R Í S* . The following statements are equivalent.
- R is regular
- There exists a Myhill-Nerode relation for R
- The relation º R is of finite index.
Example 1: Using Myhill-Nerode Theorem verify whether L = {a nbn : n ³ 0} is regular or not.
Solu tion: This is done by determining the º R -classes. If k ¹ m, then since
a k bk ÎL but a mbk ÏL. Hence there are infinitely many º L -classes, at least one for each a k , k ³ 0. Hence by Myhill-Nerode Theorem L is not regular. (The application of Myhill-Nerode theorem has been illustrated above).