$S \rightarrow Sa \mid S \mid \epsilon$
This is a weird case where I have only one non-terminal. I'm trying to apply the algorithm 4.19 in the dragon book. It don't think it should be applicable but I think it gives something like that :
$S \rightarrow S'$
$S' \rightarrow aS \mid S' \mid \epsilon $
but this is equivalent to say
$S' \rightarrow aS' \mid S' \mid \epsilon $
Is this the right way to think about this?