LL(1) grammar verification - parsing

Question

Let G be a grammar such that:
S -> aBa
B -> bB | \epsilon
where \epsilon represents the empty string.
After computing FIRST and FOLLOW, is there a way to tell if G is LL(1) without resorting to the parsing table?

Answer 1

After computing the FIRST and FOLLOW sets for the variables of G, you can compute the length 1 lookahead sets LA(1) for the variables and rules of G. Then G is strong LL(1) iff the following condition holds:
LA(1)(A -> wi) partition LA(1)(A) for each variable A such that A -> wi is a rule.
Alternatively, you can prove that G is strong LL(1) from the definition of a strong LL(k) grammar without computing the FIRST and FOLLOW sets. This is oftentimes easier and less tedious than computing FIRST and FOLLOW for small grammars like G.
I don't have a book handy, so there might be an error in some of these definitions or computations. But this is how I would approach the problem. Computing the FIRST and FOLLOW sets gives:
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}
Computing the length 1 lookahead sets for the variables and rules gives:
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}
Since LA(1)(B -> ε) and LA(1)(B -> bB) partition LA(1)(B) and LA(1)(S -> aBa) trivially partitions LA(1)(S), G is strong LL(1).