LL（1）语法验证
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令“ 0”是一个语法，使得：
S -> aBa
B -> bB | \\epsilon
\\epsilon
FIRST
FOLLOW
G
LL(1)
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伞腹
变量的和集后，就可以计算出length0ѭ的变量和规则的长度1个超前集。如果满足以下条件，则就是strong6ѭ：或者，您可以从强ѭ17语法的定义中证明是强strong6ѭ，而无需计算和集。这通常比计算像这样的小语法的和更容易，也没有那么繁琐。 我没有一本方便的书，因此其中某些定义或计算可能有误。但这就是我要解决的问题。计算和集可得出：FIRST(1)(S) = trunc(1)({x : S =>* x AND x IN Σ*}) = trunc(1)({ab^na : n >= 0}) = {a} FIRST(1)(B) = trunc(1)({x : B =>* x AND x IN Σ*}) = trunc(1)({b^n : n >= 0}) = {ε,b} FOLLOW(1)(S) = trunc(1)({x : S =>* uSv AND x IN FIRST(1)(v)}) = trunc(1)({x : x IN FIRST(1)(ε)}) = trunc(1)(FIRST(1)(ε)) = {ε} FOLLOW(1)(B) = trunc(1)({x : S =>* uBv AND x IN FIRST(1)(v)}) = trunc(1)({x : x IN FIRST(1)(a)}) = trunc(1)(FIRST(1)(a)) = {a}LA(1)(S) = trunc(1)(FIRST(1)(S)FOLLOW(1)(S)) = trunc(1)({a}{ε}) = trunc(1){a} = {a} LA(1)(B) = trunc(1)(FIRST(1)(B)FOLLOW(1)(B)) = trunc(1)({ε,b}{a}) = trunc(1){a,b} = {a,b} LA(1)(S -> aBa) = trunc(1)(FIRST(1)(a)FIRST(1)(B)FIRST(1)(a)FOLLOW(1)(S)) = trunc(1){a} = {a} LA(1)(B -> bB) = trunc(1)(FIRST(1)(b)FIRST(1)(B)) = trunc(1){b} = {b} LA(1)(B -> ε) = trunc(1)(FIRST(1)(ε)FOLLOW(1)(b)) = trunc(1)({ε}{a}) = {a}和分隔和琐碎地分隔，所以是强。