I would like some help for the computation theory.

There is a PDA that accepts the language $\{0^n 1^a 2^b \mid n = a+b\}$, so how can I express it into context free grammar? Any help would be appreciated!

That's what I was trying, I am not sure if it is correct or not

S -> 0SA | 0SB | empty
A -> 1
B -> 2

Unfortunately your grammar generates mixed $1$s and $2$s.

You can try something like this:

$S\rightarrow 0SB | S'$
$S' \rightarrow 0S'A | \varepsilon$
$A\rightarrow 1$
$B\rightarrow 2$

Notice that I don't know if in your problem $n$ can be $0$ or not (and similarly for $a$ and $b$), so maybe you have to fix something.

Anyway, here the idea is starting generating first some $0$s an all the $2$s, and then add the remaining $0$s and the $1$s: it is similar to the "standard" grammar used to generate $0^n 1^n$.

  • $\begingroup$ thank you for your help! $\endgroup$
    – randomguy
    Dec 14 '20 at 23:40
  • $\begingroup$ one more thing, can this context free grammar generate any regular lanuage ? and how should I determine if a context free grammar can generate a regular lanuage? Thanks for your help $\endgroup$
    – randomguy
    Dec 14 '20 at 23:44
  • $\begingroup$ Any regular language is generated by a CFG, but this specific grammar generates a single language, that is not regular (you van prove it using Pumping Lemma). There's no general algorithm to decide if a CFG actually generates a regular language, i.e., it's an undecidable problem. Anyway, once you have a specific CFG, there's some standard strategies (again, Pumping Lemma) to prove if the generated language is regular. Anyway, this question is interesting, and maybe deserves an other post. $\endgroup$
    – user6530
    Dec 14 '20 at 23:54
  • $\begingroup$ oh, thank you so much! basically, by using the pumping lemma on the lanuage generated by CFG. Then, i will be able to see whether this particular CFG could generate a regular lanuage? thats a good tactic i have not thought about it. thanks for that $\endgroup$
    – randomguy
    Dec 15 '20 at 0:11
  • $\begingroup$ and i just want to clarify one thing: some of the CFG cannot generate the regular lanuage even CFG being more powerful than regular lanuage. $\endgroup$
    – randomguy
    Dec 15 '20 at 0:13

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