0
$\begingroup$

From Wikipedia, Regular Language

All finite languages are regular.

and Also Regular Grammar, is a way to describe the Regular Language

Right regular grammar (also called right linear grammar).

Left regular grammar (also called left linear grammar).

From Wikipedia its Example : a* b c* can be described as Regular Grammar..

but it generate infinite number of 'a's and infinite number of 'c's.

Recall Regular Language can be described by Regular Expression and Also Finite Language ..

$\endgroup$

closed as unclear what you're asking by D.W., Juho, Luke Mathieson, David Richerby, Nicholas Mancuso Jan 23 '15 at 7:45

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ All finite languages are regular but not all regular languages are finite. $\endgroup$ – jmite Jan 17 '15 at 10:03
  • 1
    $\begingroup$ So what's the question? I don't see a question in there.... And what reading and self-study have you done? These topics are covered in standard resources (e.g., textbooks). $\endgroup$ – D.W. Jan 17 '15 at 11:50
  • $\begingroup$ @D.W. my question is that how is regular and generate infinite number of symbols .. the second question is that why a^n b^n where n>=0, this is not a Regular language its context-free, so why a^n b^m is regular .. am confused about differentiate regular and it language, and context-free, $\endgroup$ – Yassine Jan 17 '15 at 11:52
0
$\begingroup$

This might be a language issue. In (mathematical) English, the statement all finite languages are regular means: if $L$ is a finite language then $L$ is regular. No implication in the other direction ("if a language is regular then it is finite") is implied.

$\endgroup$
  • $\begingroup$ Okey, what about a^n b^n where n>=0, this is not a Regular language its context-free, so why a^n b^m is regular ?? am a little confused .. $\endgroup$ – Yassine Jan 17 '15 at 11:45
  • $\begingroup$ Some infinite languages are regular, some aren't. Beyond that, it depends on the particular language. $\endgroup$ – Yuval Filmus Jan 17 '15 at 13:46

Not the answer you're looking for? Browse other questions tagged or ask your own question.