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I went through a question asking me to categorize the following grammar.

$$S → AA, S → AB, A → a, A→BB, B → b, B → e$$

From the production rules, clearly it is Context-Free. But it accepts a finite set of strings. $\{e, a, aa, ab, abb, ba, bba, b, bb, bbb, bbbb\}$ which is regular language.

So, is the above grammar regular? Though it does not follow from the rules.

Basically my question is: Is the grammar $\{S → AA, A → a\}$ regular?.

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The answer is no. Whilst the language is finite and therefore regular the grammar itself is not regular. I think you already knew this as you said that this grammar's production rules are not of the form of a regular grammar.

The point here is that just because a grammar isn't regular doesn't mean the language it generates must not be regular.

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  • $\begingroup$ I got that. But according to [wikipedia][en.wikipedia.org/wiki/Regular_grammar], a regular grammar is a formal grammar that describes a regular language. Now, my grammar is describing a regular language. The set of rules are mentioned later, not as definition. $\endgroup$ – Shashwat Dec 18 '12 at 11:56
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    $\begingroup$ @Shashwat: You referring to a property of a regular grammar (it accepts a regular language), but this is not the definition of a regular language. The definition goes over the restricted left hand sides of the rules. $\endgroup$ – A.Schulz Dec 18 '12 at 13:29
  • $\begingroup$ As the above comment says, the statement "a regular grammar is a formal grammar that describes a regular language" is not the definition of a regular grammar. That wikipedia page doesn't explicitly use the word definition at all. A better place to learn the basics and formal definitions would be a textbook. My favourite is en.wikipedia.org/wiki/… $\endgroup$ – Sam Jones Dec 18 '12 at 14:13
  • $\begingroup$ Ok. If restriction of rules is the definition, then my doubt is cleared. The on wikipedia, that was the first line; so I assumed it to be the definition. Thank you. $\endgroup$ – Shashwat Dec 18 '12 at 17:05
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    $\begingroup$ I just fixed the wording in Wikipedia. $\endgroup$ – reinierpost Jan 19 '15 at 7:46
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This a type-2 grammar (CFG) not type-3 (regular garmmar), because it consists of both left hand derivation and right hand derivation. There are CFGs which generate regular languages. That's a property.

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