I have the following grammar, which I know it is regular because it can be represented by a finite state automata:

\begin{array}{l} \mathrm{S} \rightarrow \mathrm{X} \mid \mathrm{Y} \\ \mathrm{X} \rightarrow \mathrm{a} \mathrm{Y} \mathrm{b} \mid \mathrm{ab} \\ \mathrm{Y} \rightarrow \mathrm{b} \mathrm{X} \mathrm{a} \mid \mathrm{ba} \end{array}

According to Chomsky hierarchy:

Type-3 grammars generate the regular languages. Such a grammar restricts its rules to a single nonterminal on the left-hand side and a right-hand side consisting of a single terminal, possibly followed by a single nonterminal (right regular). Alternatively, the right-hand side of the grammar can consist of a single terminal, possibly preceded by a single nonterminal (left regular).

However, I do not find the hierarchy applied to the grammar. Am I missing something?


Every regular grammar produces a regular language.


Not every grammar for a regular language is regular.

In other words, a regular language can be described by any of an infinitude of grammars, some of which are regular and some not.

In this case, you can easily create either a left- or a right-regular grammar for your language. For example:

\begin{array}{l} \mathrm{S} \rightarrow \mathrm{X} \mid \mathrm{Y} \\ \mathrm{X} \rightarrow \mathrm{X} \mathrm{a}\mathrm{b} \mid \mathrm{ab} \\ \mathrm{Y} \rightarrow \mathrm{Y} \mathrm{b} \mathrm{a} \mid \mathrm{ba} \end{array}

Of course, that transformation only preserves the language parsed, not the induced parse trees.

  • $\begingroup$ Thank you very much! :) $\endgroup$ – Kevin Dec 4 '20 at 17:51

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