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I'm curious about the right way to characterize symbol $A$ in a CFG like this one:

$$ \begin{align*} A &\to A B\\ A &\to x\\ B &\to y\\ B &\to \varepsilon \end{align*} $$

$B$ is certainly nullable. However, should $A$ be considered nullable? It feels like the answer is probably "no" (and most first-follow implementations I've seen either agree or crash on this). However, you can derive an infinitely large parse tree for the null symbol sequence like $A \to A(A(A(...) B()) B()$.

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The infinitely large parse tree you propose does not represent the parse of any sentence, because its leaves are not all terminals.

A more precise statement would be that $A$ cannot derive the empty sentence. You can produce an arbitrarily long derivation of $A$ by repeatedly using the rules $A \to A B$ and $B \to \epsilon$ but none of those derivations is empty. Since the precise definition of "nullable" is "a non-terminal which can derive the empty sentence", $A$ is not nullable.

If a FIRST/FOLLOW implementation crashes on that input, then the implementation is buggy. The correct computations are:

$$\begin{align} FIRST(A) &= \{x\} \\ FIRST(B) &= \{y, \epsilon\} \\ FOLLOW(A) &= \{y, \epsilon\} \\ FOLLOW(B) &= \{y, \epsilon\} \end{align}$$

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  • $\begingroup$ Why is FOLLOW(B) just ϵ? I would have thought that since A(A(B(y))B(y)) is a valid parse tree that FOLLOW(B) should include y. $\endgroup$ – ChaseMedallion Jan 12 at 22:47
  • $\begingroup$ @chase: quite right, fixed. $\endgroup$ – rici Jan 12 at 23:08
  • $\begingroup$ @chase: by the way, it's usually easier to think in terms of derivations (which is how all the concepts are defined) than parse trees. The derivation I missed was $A\to AB\to ABB\to ABy$. $\endgroup$ – rici Jan 13 at 0:09
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A variable is nullable if you can derive the empty word from it. Hence, in your example, $A$ is not nullable. However, if you add a rule as, for example,

$$A \to B$$ then it is, as $A\Rightarrow B\Rightarrow \varepsilon$.

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