I need a Context Free Grammar for this language. I could come up with this solution:

S -> AB
A -> aA | ε
B -> bbbB | ε

But, this grammar is clearly wrong, since the number of a's can still exceed the number of b's. How do I draw a link between the two variables A and B such that $n\leq m+3$?

P.S. It is a humble request to anyone who downvotes this question to please specify the reason in the comments below

  • 2
    $\begingroup$ Create up to 3 a’s. Then repeatedly either create ab or b. $\endgroup$ – gnasher729 Apr 28 at 15:14
  • $\begingroup$ @gnasher729 sorry I couldn't get it, could you write an answer please? $\endgroup$ – Infinity Apr 28 at 15:30

Based on gnasher729's comment, I came up with the following CFG that I believe meet the requirements:

$$ S \to AAAB \\ A \to a \;|\; \varepsilon\\ B \to aBb \;|\; Bb \;|\; \varepsilon $$

  • $\begingroup$ looks fine to me too $\endgroup$ – Infinity Apr 28 at 16:42

It is hard to come up with a CFG directly. A better approach can be, making a PDA and converting the PDA to CFG. Refer to this answer.

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    $\begingroup$ Huh? Took me 10 seconds to write down how to do it, and Acid probably not more than a minute to do it (and most of that time figuring out how to type an epsilon). $\endgroup$ – gnasher729 Apr 28 at 23:18
  • $\begingroup$ @gnasher729 maybe because you are more proficient at it? $\endgroup$ – Infinity Apr 29 at 3:33
  • $\begingroup$ @gnasher729 in case anyone of you has downvoted this question, please read the post again and do the needful $\endgroup$ – Infinity Apr 29 at 3:37
  • $\begingroup$ @gnasher729 I just pointed out a procedure to go about it. I personally had a hard time coming up with the grammars directly in the past. $\endgroup$ – SiluPanda Apr 29 at 4:26
  • $\begingroup$ @SiluPanda hello sir! you are from iit b? $\endgroup$ – Infinity Apr 29 at 8:04

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