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1
Module 26
•
Pumping Lemma
–
A technique for proving a language L is NOT
regular
–
What does the Pumping Lemma mean?
–
Proof of Pumping Lemma
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Pumping Lemma
How do we use it?
3
Pumping Condition
•
A language L satisfies the pumping
condition if:
–
there exists an integer n > 0 such that
–
for all strings x in L of length at least n
–
there exist strings u, v, w such that
•
x = uvw and
•
uv ≤ n and
•
v ≥ 1 and
•
For all k ≥ 0, uv
k
w is in L
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Pumping Lemma
•
All regular languages satisfy the pumping
condition
All languages over {a,b}
Regular languages
“Pumping Languages”
5
Implications
•
We can use the pumping lemma to prove a
language L is not regular
–
How?
•
We cannot use the pumping lemma to prove
a language is regular
–
How might we try to use the pumping lemma to
prove that a language L is regular and why does
it fail?
Regular
Pumping
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Pumping Lemma
What does it mean?
7
Pumping Condition
•
A language L satisfies the pumping condition if:
–
there exists an integer n > 0 such that
–
for all strings
x in L
of length at least n
–
there exist strings u, v, w such that
•
x = uvw
and
•
uv
≤
n and
•
v ≥ 1 and
•
For all k
≥
0, uv
k
w is in L
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v can be
pumped
•
Let x = abcdefg be in L
•
Then there exists a substring
v
in x such that
v
can be
repeated (pumped) in place any number of times and the
resulting string is still in L
–
u
v
k
w is in L for all k ≥ 0
•
For example
–
v =
cde
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This note was uploaded on 07/25/2008 for the course CSE 460 taught by Professor Torng during the Fall '07 term at Michigan State University.
 Fall '07
 TORNG

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