Solution for Midterm
1.a) (8 Points)
First(Expr) =
-, (, id
; Follow(Expr) =
$, )
, First(Var) =
id
; Follow(Var) =
$, ), -
First(ExprTail) =
-,
; Follow(ExprTail) =
$, )
; First(VarTail) =
),
Follow(VarTail) =
$, ), -
1.b (11 Points) The predictive parser table for the grammar:
id
$
Expr Var ExprTail - Expr ( Expr)
ExprTail - Expr
Var id VarTail
VarTail
(Expr)
1.c (3+5 Points) The grammar is not LL(1).
and
both contain symbol
. Hence, on input
, B can either be
expanded by nCfgRuleBb, or by
.
1.d (3+5 Points) The grammar is LL(1) because the rules of the nonterminals can be distinguished by their different starting
terminal symbols. Hence, it is possible to look at an input symbol and decide which rule to use for each nonterminal.
2.a (13 Points) Start by adding a new nonterminal S’ and rule
! " $#
. The states are:
s0:
[S’ -> .SL]
[SL -> .SL; S]
[SL -> .] goto(s0, SL) = s1;
s1:
[S’ -> SL.]
[SL -> SL.;S] goto(s1, ;) = s2;
s2:
[SL -> SL;.S]
[S -> .stmt] goto(s2, S) = s3;
s3:
[SL-> SL; S.] goto(s2, stmt) = s4
s4:
[S -> stmt.]
2.b (12 Points) First we calculate the follow for nonterminals:
Follow(S’) = { $ }, Follow(SL) = {; $}, Follow(S) = { ;, $}
| ; | stmt | $ | SL | S |
----- ------------------------------------------------
s0 | r2 | | r2 | 1 | |
------------------------------------------------------
s1 | s2 | | accept | | |
------------------------------------------------------
s2 | | s4 | | | 3 |
------------------------------------------------------
s3 | r1 | | r1 | | |
------------------------------------------------------
s4 | r3 | | r3 | | |
------------------------------------------------------
2.c (3+7 Points) The grammar is LR(1) (In fact, it is SLR(1)). There are no shift/reduce or reduce/reduce conflicts. This can be
discerned by looking at the grammar carefully and examining rules that can cause conflicts. The first possibility appears to be in
the following rules:
id
%'&)(+*
(
id
%,&-(
ECS 142 –1– Spring 1997
