如何显示语法不是LL（1）并将语法转换为LL（1）

Question

I'm trying to find the ambiguity in this grammar so I can remove it and convert it to LL(1), however for the life of me I can't find the ambiguity. Any help will be much appreciated.

D -> if (C) {S} | if (C) {S} else {S}
S -> D | SA | A
A -> V = T;
V -> x | y
T -> 1 | 2
C -> true | false

Answer 1

The grammar is not ambiguous. Nonetheless, it is not LL(1) because when the lookahead token is if , it is not possible to know which of the two productions for D will be used.

To make it LL(1), you will need to left-factor D .