CMSC 430 Midterm Answers March 18, 2004
1
Answer all questions in the exam book. You may keep the
exam questions after you are done.
1. [56] Consider the following grammar G:
1. S X a ⊥
2. X Y b
3. X Y Z c
4. Y d
5. Z b
For each of the following grammar classes, if G is of that class,
give the appropriate parsing table. If it is not of that class, fully
explain why it isn’t.
(a) LL(0)
(b) LL(1)
(c) LR(0)
(d) LR(1)
(e) SLR(1)
(f) LALR(1)
(g) General precedence
(a) An LL(0) grammar generates a single string. This
grammar generates 2 strings so can’t be LL(0).
(b) First(Yb) = First(Yzc) = {d} so can’t be LL(1).
(c) Try to generate LR(0) table first:
S0 [S.Xa]
[X.Yb]
[X.YZc]
[Y.d]
X 1
Y 2
d3
S1 SX.a] a4
S2 [XY.b]
[XY.Zc]
[Z.b]
b5
Z6
S3 [Yd.] Reduce 4
S4 SXa.] Reduce 1
S5 [XYb.]
[Zb.]
Reduce – reduce conflict so not LR(0).
(d) LR(1)
S0 [S.Xa,]
[X.Yb,a]
[X.YZc,a]
[Y.d,b]
X 1
Y 2
d3
S1 SX.a,] a4
S2 [XY.b,a]
[XY.Zc,a]
[Z.b,c]
b5
Z6
S3 [Yd.,b] Reduce 4
S4 SXa.,] Reduce 1
S5 [XYb.,a]
[Zb.,c]
a Reduce 2
c Reduce 5
S6 [XYZ.c,a] c7
S7 [XYZc.,a] Reduce 3
No conflicts so LR(1).
(e) SLR(1)
Copy over table from (c) -- LR(0)
S0 [S.Xa]
[X.Yb]
[X.YZc]
[Y.d]
X 1
Y 2
d3
S1 SX.a] a4
S2 [XY.b]
[XY.Zc]
[Z.b]
b5
Z6
S3 [Yd.] Reduce 4
S4 SXa.] Reduce 1
S5 [XYb.]
[Zb.]
aReduce 2
cReduce 5
In S5: Follow(X) ={a}
Follow(Z) = {c}
Disjoint so fix up above table and continue
S6 [XYZ.c] c7
S7 [XYZc.] Reduce 3
No further conflicts so SLR(1)
(f) LALR(1) – ALL item lists in (e) (LR(1)) are disjoint,
so grammar is LALR(1) by default.
(g) Generate precedence relations:
From rule 2:
Y = b
From rule 3:
Y = Z
Z = c
Y < First(Z) = b
Thus Y=b and Y<b. Precedence relations are not unique,
so not precedence grammar.
2. [8] Is the language given by the grammar in problem
above LL(1)? Prove your answer.
Yes LL(1). Grammar only generates 2 strings: dba⊥ and
dbca⊥. Write grammar as follows:
1. S dbX ⊥
2. Xa
3. X ca
Language generated by this grammar is same as original
grammar and it is LL(1) since
First(a) = {a}, First(ca)={c} and
First(a) First(ca) = Ø
3. [12] For the regular expression
(00)* (0 | 1)* (11)*
(a) Give the non-deterministic FSA that accepts the same set.
