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CS103
HO#43
SlidesContextFree Languages
5/10/10
1
The Pumping Lemma
If
A is a regular language, then
there is a number p
(the pumping length)
such that, if s is
any string in A of length at least p
, then s may be divided
into three pieces,
s = xyz
satisfying the following three conditions:
1.
for each i
0, xy
i
z
A,
2. y > 0, and
3. xy
p.
FOR ANY
regular language A,
THERE EXISTS
a number p such that
FOR ANY
string s in A with s
p
THERE EXIST
strings x, y, z,
such that s = xyz and …
The Pumping Lemma
If
A is a regular language, then
there is a number p
(the pumping length)
such that, if s is
any string in A of length at least p
, then s may be divided
into three pieces,
s = xyz
satisfying the following three conditions:
1.
for each i
0, xy
i
z
A,
2. y > 0, and
3. xy
p.
To show that language A is
NOT REGULAR
, assume it is and get a contradiction
from this
FOR ANY
regular language A,
THERE EXISTS
a number p such that
FOR ANY
string s in A with s
p
THERE EXIST
strings x, y, z,
such that s = xyz and .
..
The Pumping Lemma
If
A is a regular language, then
there is a number p
(the pumping length)
such that, if s is
any string in A of length at least p
, then s may be divided
into three pieces,
s = xyz
satisfying the following three conditions:
1.
for each i
0, xy
i
z
A,
2. y > 0, and
3. xy
p.
To show that language A is
NOT REGULAR
, assume it is and get a contradiction.
If A is regular, there is a pumping length p that works for any s where s
p.
Given p,
show how to find an s
of sufficient length that cannot be pumped.
That is,
no matter how we divide s in to xyz
there is an i such that
xy
i
z
A.
(This may involve analyzing several cases.)
FOR ANY
regular language A,
THERE EXISTS
a number p such that
FOR SOME
string s in A with s
p
FOR ALL
strings x, y, z,
such that s = xyz
and .
..
Show that L = { w  w
{
)
,
(
}* and w has balanced parentheses }
is not regular.
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This document was uploaded on 02/08/2011.
 Fall '09

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