Reputation: 159
S -> 1S2 | S0 | epsilon
I thought it would be LL(1) because it is possible to determine.
For example if the next input symbol was 0 i'd know it was S -> S0
Does the epsilon mean that it cannot be LL(1)?
Upvotes: 1
Views: 1008
Reputation: 641
This grammar is not LL(1) just because it has a left recursion on it. When you will try to build the parser table, you will came up with a conflict.
http://en.wikipedia.org/wiki/LL_parser
Upvotes: 1