LL(1) grammar and first & follow sets

Question

I think the below is actually LL(1) but I am not 100% sure. are we able to prove that this iss LL(1) grammar and then also, is the first and follow sets given correct? I dont really understand for sure how to actually get the follow sets.

Grammar

expr ::= term {addop term}
term ::= factor {mulop factor}
factor ::= variable| unsigned-number| ( expr)
variable ::= identifier
addop ::= + | -
mulop ::= * | /

First Sets

First(expr) = identifier, unsigned number, (
First(Term) = identifier, unsigned number, (
First(Factor) = identifier, unsigned number, (
First(Variable) = Identifier
First(addop) = +, -
First(mulop) = *, /

Follow Sets

Follow(Expr) = First(term)
Follow(Term) = First(Factor)

Do not understand how to do the Follow sets of the below sets
Follow(Factor) = First( ")" ) = )
Follow(Variable) Follow(addop) Follow(mulop)

Venkatesh Nandigama Venkatesh Nandigama · Accepted Answer · 2019-11-14T03:34:38

FIRST sets are correct.

FOLLOW sets:

FOLLOW(A) of non-terminal A is the set of terminal symbols that can follow in the derivation sequence

FOLLOW(expr): check where it appeared in the right-hand side of production. It is there is factor:=(expr), when we take this production in the derivation what follows expr is ) and expr is a start symbol.

   FOLLOW(expr)={),$}

similarly,

   FOLLOW(term) = FIRST(addop) U FOLLOW(expr) = {+,-,),$}
   FOLLOW(factor) = FIRST(mulop) U FOLLOW(trem) = {*,/,+,-,),$}
   FOLLOW(addop) = FIRST(term) = {identifier,unsigned number,(}
   FOLLOW(multop) = FIRST(factor) = {identifier,unsigned number,(}
   FOLLOW(Variable)= FOLLOW(factor) = {*,/,+,-,),$}

where $ is the end of the input.

Given grammar is LL(1). FIRST sets of factor are disjoint and we don't need check the FOLLOW sets as there are no epsilon productions.

LL(1) grammar and first & follow sets

1 Answers