Myhill-Nerode theorem
Let be a language on the finite alphabet and let be its Nerode equivalence. The following are equivalent.
-
1.
is recognized by a deterministic finite automaton.
-
2.
is finite.
Moreover, the number of classes of is the smallest number of states of a DFA recognizing .
Proof. First, suppose where is the empty word on . Construct a DFA (called the Nerode automaton for ) with defined by
(1) |
and
(2) |
Then is well defined because implies . It is also straightforward that recognizes .
On the other hand, let be a DFA that recognizes . Extend to by putting and for every , , . Define as
(3) |
Then is well defined. In fact, suppose : then either , or there are such that . But in the latter case, for any , hence since recognizes . Finally, for any we have so every class of has a preimage according to ; consequently, .
Title | Myhill-Nerode theorem |
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Canonical name | MyhillNerodeTheorem |
Date of creation | 2013-03-22 18:52:13 |
Last modified on | 2013-03-22 18:52:13 |
Owner | Ziosilvio (18733) |
Last modified by | Ziosilvio (18733) |
Numerical id | 14 |
Author | Ziosilvio (18733) |
Entry type | Theorem |
Classification | msc 68Q70 |
Classification | msc 20M35 |
Synonym | Myhill-Nerode theorem for languages |