Which symbols in the following grammar $G$ are generating?

Question

I have been given the following set of production rules $P$ for a grammar $G(A) = \newcommand{\perm}[1]{\left\langle #1 \right\rangle}\perm{\Sigma, V, P, S}$, generated from a pushdown automaton $A = \perm{Q, \Sigma, \Gamma, \delta, q_0, F}$, that recognizes the language $\newcommand{\lang}{\mathcal L}\lang = \newcommand{\set}[1]{\left\{ #1 \right\}}\newcommand{\Nset}{\mathbb{N}}\set{0^n1^n \mid n \in \Nset \cup \set{0}}$:

\begin{align*} \newcommand{\rewrite}{\longrightarrow} \perm{q, Sp} \in \delta(s,\epsilon, \epsilon) &: A(s, \epsilon, f) \rewrite A(q, S, q) A( q, p, f ) \\ \perm{q, \epsilon} \in \delta(q, 0, 0) &: A(q, 0, q) \rewrite 0 \\ \perm{q, \epsilon} \in \delta(q, 1, 1) &: A(q, 1, q) \rewrite 1 \\ \perm{q, 0S1} \in \delta(q, \epsilon, S) &: A(q, S, q) \rewrite A(q, 0, q) A(q, S, q) A(q, 1, q) \\ \perm{q, \epsilon} \in \delta(q, \epsilon, S) &: A(q, S, q) \rewrite \epsilon \\ \perm{f, \epsilon} \in \delta(q, \epsilon, S) &: A(q, p, f) \rewrite \epsilon, \end{align*} In addition to these, we have the following productions related to the initial and accepting states $s$ and $f$ of the automaton $A$: $$ S \rewrite A(s, \epsilon, s), \quad A(s, \epsilon, s) \rewrite \epsilon \quad\text{and}\quad S \rewrite A(s, \epsilon, f)\,. $$

I'm supposed to transform this grammar into Chomsky normal form, but to this end I would need to first find out which of the variables or non-terminals $v \in V$ are generating, as in a word of the language $\lang$ can be derived from them: $\newcommand{\derive}{\Longrightarrow} v \derive_G^\ast x \in \lang$.

The following algorithm to find out whether a variable $V_i$ is generating can be used:

1. Mark all alphabet of the language as generating (which makes sense as there is no non-empty language without the alphabet).
2. Mark all variables $V_i$ in the rules $V_i\rewrite v$, where the string $v$ only contains generating symbols as generating.
3. Repeat step 2 until all symbols have all either been marked as generating or not.

Now obviously the variables $A(q, 0, q)$ and $A(q, 1, q)$ are generating, as the alphabet $0$ and $1$ are generating. But what about the variables $S$, $A(s,\epsilon,s)$, $A(q, S, q)$ and $A(q, p, f)$? The empty string $\epsilon$ is obviously a part of the language $\lang$, but can it be considered as a part of the alphabet $\Sigma$?

If it can, then the symbols $A(s, \epsilon, s)$, $S$ and $A(q, p, f)$ are also generating. But what about $A(q, S, q)$? It is used in its own definition, so can it be considered generating? If this is the case, then there would be no simplifying the production rules.

But is this the case?

Answer 1

The empty string $\epsilon$ is obviously a part of the language $\mathcal{L}$, but can it be considered as a part of the alphabet $\Sigma$?

$\epsilon$ is clearly not an element of the alphabet, since it is not a symbol. $\epsilon$ is a string of length 0. So it is certainly the case that every symbol contained in $\epsilon$ is a generating symbol. (Every symbol in $\epsilon$ is also a unicorn. The statement is trivially true.)

[W]hat about $A(q, S, q)$? It is used in its own definition, so can it be considered generating?

If every production for $A(q, S, q)$ were recursive, then it would not be generating. That's demonstrated by the algorithm you quote: Initially, like all other variables, it is marked non-generating, and thus step 2 would never be executed for it.

But if there is another rule for $A(q, S, q)$, that other rule could trigger step 2. And note that step 2 needs to mark the variable as generating, not the rule.

Whether that is the case or not might depend on $q$.