# Why is $FOLLOW$ not necessary for $LL(1)$ grammars with no $\epsilon$ transitions?

I'm aware of how $$FIRST$$ and $$FOLLOW$$ sets are used to construct a parsing table for $$LL(1)$$ grammars.

However, I've encountered this statement from my notes:

With $$\epsilon$$ productions in the grammar, we may have to look beyond the current non-terminal to what can come after it

In my opinion, this suggests that $$FOLLOW$$ is not necessary for $$LL(1)$$ grammars that have no $$\epsilon$$ transition. Am I wrong? And if I'm not, why is this the case?

Thanks

That's correct for $$LL(1)$$: if there are no $$\epsilon$$ productions, the $$LL$$ parser-generation algorithm will never consult $$FOLLOW$$, because it only does that if it finds $$\epsilon$$ in the $$FIRST$$ set for the first non-terminal in the right-hand side of a production. (So it might not need the $$FOLLOW$$ sets even if there are some $$\epsilon$$ productions, provided that none of those productions occur at the beginning of a right-hand side.)
The observation doesn't generalise well to other values of $$k$$. You'll need $$FOLLOW_k(\alpha)$$ if any production can derive a string whose length is less than $$k$$.