计算语法的 FIRST 和 FOLLOW 集？

Question

I have to compute the FIRST and FOLLOW sets of the following grammar:

S -> ABC
A -> a | Cb | 1
B -> c | dA | 1
C -> e | f

However, I am not entirely sure if I understand how to do it. According to my understanding I get the following computation:

FIRST(S)= {a, e, f, b, c, d}
FIRST(A)= {a, e, f, 1}
FIRST(B)= {c, d, a, e, f, b, 1}
FIRST(C)= {e, f}

FOLLOW(S)= {$}
FOLLOW(A)= {c, d, a, e, f, b}
FOLLOW(B)= {e, f}
FOLLOW(C)= {$}

Is this correct? Did I miss something? If so, could someone please explain what I missed and how I would go about computing the correct FIRST and FOLLOW sets? Also, by looking at the sets produced, how can I tell if its an LL(1) grammar?

Answer 1

When S (the goal nonterminal symbol) expands, the first terminal in its derivation must come from A only, since that is exactly what the grammar indicates in S->ABC. (Unless "1" in your grammar means epsilon, an empty right-hand side; in this case, we move on to B and find B's FIRST set.) A can expand to a and 1 directly, and via C to e and f. So the FIRST set of S is a,1,e,f (the union of all these terminals). The grammar looks to be LL(1), since each actual token appears to determine which right-hand-side to use at each step in the derivation. Calculate all the FIRST sets to confirm this guess. If the same terminal is in the FIRST set for two RHSs of the same nonterminal LHS, then the grammar is not LL(1) since that terminal would cause an ambiguity in the choice of which production to use in the derivation. LL(1) grammars must be unambiguous and have a nonempty FIRST entry for each nonterminal, meaning that the grammar has no left-recursion.

Note that the FOLLOW set is not needed since there are no productions with empty right-hand-sides (such productions can be used to model LL(2) languages in LL(1) grammars at the cost of an increase in parser complexity).