Given language $L$, why is it not necessarily true that the number of equivalence classes of $L$ is equal to the number of equivalence classes of $L^R$?

And for the private case that $L$ is regular, is that true?


1 Answer 1


Assuming you refer to Myhill-Nerode equivalence classes:

First notice, that for a non-regular language $L$, the number of classes is necessarily infinite. This is one direction of the Myhill-Nerode theorem.

Since regular languages are closed under reversal, it follows that for a non-regular language $L$, both $L$ and $L^R$ have the same ``number'' (namely, infinitely) of equivalence classes.

For the case of regular languages, the Myhill-Nerode relation is very ``right-biased'', since it relates to the existence of a separating suffix. This is why reversing the language may change the number of classes. For example, consider the language

$$L_k=\{w\in \{a,b\}^*: w_k=a\}$$ i.e. the language of words whose $k$-th letter is $a$.

It's easy to show that the number of classes for $L_k$ is roughly $k$ (maybe $\pm 1$).

However, the language $L_k^R$ of words whose $k$-before last letter is $a$, is known not to have a DFA with less than $2^{k-1}$ states (and hence it has $\Omega(2^k)$ classes).

Interestingly, reversal and congruence classes do have an interplay: if you take a DFA, reverse it, determinize it, reverse it again, determinize again, and remove all unreachable states, you will obtain a minimal DFA for the original language, which is exactly what the equivalence classes characterize. This is Brzozowski's algorithm.

  • $\begingroup$ Thank you! got it. $\endgroup$
    – Ella
    Jun 21, 2020 at 7:33

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