Using any of the techniques (substitution, factoring, left-recursion removal), construct an LL(1) grammar accepting the same language as G.
G over Σ = {0, 1, 2}:
S → Y | 1X
X → 1X | 0
Y → Y0 | 1X1 | 2X2
I did this so far:
X is left recursive so:
X -> 1F | 0F
F -> 1F | e
What else would i need to do to construct an LL(1), could i factor Y?
First, X is not left recursive, as its RHS doesn't begin with X. Rather it's tail-recursive and it's all right. However Y -> Y0 tells you that Y is left-recursive. In this case you do the following:
S -> Y | 1X
X -> 1X | 0
Y -> 1X1F | 2X2F
F -> 0F | e
You may also want to add and epsilon rule to X as well making it
X -> 1X | 0 | e
just to make sure you'll never end up with infinite sentences.
