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.
Why is this not LL(1) grammar?
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07-07-2023 - |
Question
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)?
Solution
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