Exercise 4.4.1 d)

[huangxueqin]

The production S -> S + S | S S | (S) | S * | a, after eliminate left recursion you got

S -> TB
B -> AB | ε
A -> +S | TB | *          <--- TB should replace with S
T -> (S) | a  

so it turns to be

S -> TB
B -> AB | ε
A -> +S | S | *
T -> (S) | a  

And the follow set should be

FOLLOW(S) = {),$}
FOLLOW(B) = {),$}
FOLLOW(A) = {+,*,(,a,),$}
FOLLOW(T) = {+,*,(,a,),$}

[Jin Jianian]

Because A -> +S and Grammer Rule No.3, all elements in FOLLOW(A) should in FOLLOW(S), which means they are all {+, *, (, a, ), $}.

[Φ²]

Well, I agree with @NoteBookie

[Jason Chen]

Yes, I think the FOLLOW sets are all the same too.

[DiuR21]

So. The origin answer should be changed as soon as possible.

[Yu_dezhi]

yeah,I get the same answer as @NoteBookie ,it should be changed.

[theo-one]

i have a question，why $ is in FOLLOW(A) and FOLLOW(T)?

[Aditya Soni]

For the Follow of T, we need first of B, now first of B(remember second grammar has a epsilon, so we can substitute it) means first of A, which means +,* and first of S, which implies first of T, which has ( and a, now come back to the epsilon, using the 1st grammar, we reach at the end of it since we can substitute that B with € and thus we hit the end of it(1st production) and thus we include follow of (left side)S which has $ and )
Hope it helps @theo-one.