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A CFG for Strings with Equal Numbers of
a
s and
b
s
•
G
= (
{
S
}
,
{
a, b
}
, R, S
) where
R
has rules:
S
→
ε
S
→
SS
S
→
aSb
S
→
bSa
•
Claim
:
L
(
G
) =
{
x
:
x
has the same # of
a
’s and
b
’s
}
(1)
x
∈
L
(
G
)
⇒
x
has the same # of
a
’s and
b
’s
Pf:
Easy, every RHS has the same number of
a
’s and
b
’s. Formal proof by
induction on length
k
of the derivation.
(2)
x
has the same # of
a
’s and
b
’s
⇒
x
∈
L
(
G
)
Proof:
by induction on

x

(a)

x

= 0: then
x
=
ε
and is generated by a single rule
(b)

x

=
k
+ 2
4 subcases depending on ﬁrst and last symbols of
x
(i)
x
=
ayb
for some
y
∈
Σ
*
:
So

y

=
k
, and
y
has the same # of
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This document was uploaded on 12/24/2011.
 Fall '09

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