Java Reference
In-Depth Information
1 S
→
ABc
2 A
→
a
3
| λ
4 B
→
b
5
| λ
Level
First
ans
Marker
Done?
Comment
X β
(
=
Ye s )
F
irst
(B)
0
B
⊥{ }
12
1
b
⊥{
b
}
B
→
b
11
1
⊥⊥ { }
B
→λ
10
0
{
b
}
Final answer
15
F
irst
(A)
0
A
⊥{ }
12
1
a
⊥{
a
}
A
→
a
11
1
⊥⊥ { }
A
→λ
10
0
{
a
}
Final answer
15
F
irst
(S)
0
S
⊥{ }
12
1 ABc
{
a
}
Computation shown
above
16
⇒
λ
2 Bc
{
b
}
Because A
;
computation shown
above
16
⇒
λ
3
c
⊥{
c
}
Because B
11
2
{
b,c
}
15
1
{
a,b,c
}
15
0
{
a,b,c
}
15
Figure 4.10: A grammar and its First sets.
