SLIDE 1 1
CS 301 - Lecture 7 More Pumping Lemma
Fall 2008
Review
– Alphabets, strings, languages
– Deterministic Finite Automata – Nondeterministic Finite Automata – Equivalence of NFA and DFA – Minimizing a DFA – Regular Expressions – Regular Grammars – Properties of Regular Languages – Pumping lemma for Regular Languages
– More pumping lemma for regular languages
The Pumping Lemma:
- Given a infinite regular language L
- there exists an integer
m
- for any string with length
L w∈ m w ≥ | |
z y x w =
m y x ≤ | | 1 | | ≥ y
L z y x
i
∈ ... , 2 , 1 , = i
Theorem: The language
} : { ≥ = n b a L
n n is not regular
Proof:
Use the Pumping Lemma
SLIDE 2
2
Assume for contradiction that is a regular language
L
Since is infinite we can apply the Pumping Lemma
L } : { ≥ = n b a L
n n Let be the integer in the Pumping Lemma Pick a string such that:
w L w ∈ m w ≥ | |
length m mb
a w =
We pick
m } : { ≥ = n b a L
n n it must be that length From the Pumping Lemma
1 | | , | | ≥ ≤ y m y x b ab aa aa a b a xyz
m m
... ... ... ... = = 1 , ≥ = k a y
k
x y z m m
Write:
z y x b a
m m
=
Thus: From the Pumping Lemma:
L z y x
i
∈ ... , 2 , 1 , = i
Thus: m mb
a z y x = L z y x ∈
2
1 , ≥ = k a y
k
SLIDE 3
3
From the Pumping Lemma:
L b ab aa aa aa a z xy ∈ = ... ... ... ... ...
2
x y z k m + m
Thus:
L z y x ∈
2 m mb
a z y x = 1 , ≥ = k a y
k
y L b a
m k m
∈
+
L b a
m k m
∈
+
} : { ≥ = n b a L
n n
BUT:
L b a
m k m
∉
+
CONTRADICTION!!!
Our assumption that is a regular language is not true
L Conclusion: L is not a regular language
Therefore: Regular languages Non-regular languages
} : { ≥ n b a
n n
SLIDE 4 4
More Applications
the Pumping Lemma
Regular languages Non-regular languages
*} : { Σ ∈ = v vv L
R
Theorem: The language
is not regular
Proof:
Use the Pumping Lemma
*} : { Σ ∈ = v vv L
R
} , { b a = Σ
Assume for contradiction that is a regular language
L
Since is infinite we can apply the Pumping Lemma
L
*} : { Σ ∈ = v vv L
R
SLIDE 5
5
m m m m
a b b a w =
We pick Let be the integer in the Pumping Lemma Pick a string such that:
w L w ∈ m w ≥ | |
length
m
and
*} : { Σ ∈ = v vv L
R Write
z y x a b b a
m m m m
=
it must be that length From the Pumping Lemma
a ba bb ab a aa a xyz ... ... ... ... ... ... = x y z m m m m 1 | | , | | ≥ ≤ y m y x 1 , ≥ = k a y
k Thus: From the Pumping Lemma:
L z y x
i
∈ ... , 2 , 1 , = i
Thus:
L z y x ∈
2
1 , ≥ = k a y
k m m m m
a b b a z y x =
From the Pumping Lemma:
L a ba bb ab a aa aa a z xy ?¸ ... ... ... ... ... ... ... =
2
x y z
k m +
m m m 1 , ≥ = k a y
k
y L z y x ∈
2 Thus: m m m m
a b b a z y x = L a b b a
m m m k m
∈
+
SLIDE 6
6 L a b b a
m m m k m
∈
+
L a b b a
m m m k m
∉
+
BUT: CONTRADICTION!!!
1 ≥ k
*} : { Σ ∈ = v vv L
R Our assumption that is a regular language is not true
L Conclusion: L is not a regular language
Therefore: Regular languages Non-regular languages
} , : { ≥ =
+
l n c b a L
l n l n
Theorem: The language
is not regular
Proof:
Use the Pumping Lemma
} , : { ≥ =
+
l n c b a L
l n l n
SLIDE 7
7
Assume for contradiction that is a regular language
L
Since is infinite we can apply the Pumping Lemma
L } , : { ≥ =
+
l n c b a L
l n l n m m m
c b a w
2
=
We pick Let be the integer in the Pumping Lemma Pick a string such that:
w L w ∈ m w ≥ | |
length
m } , : { ≥ =
+
l n c b a L
l n l n and Write
z y x c b a
m m m
=
2 it must be that length From the Pumping Lemma
c cc bc ab aa aa a xyz ... ... ... ... ... ... = x y z m m m 2 1 | | , | | ≥ ≤ y m y x 1 , ≥ = k a y
k Thus: From the Pumping Lemma:
L z y x
i
∈ ... , 2 , 1 , = i
Thus: m m m
c b a z y x
2
= 1 , ≥ = k a y
k
SLIDE 8
8
From the Pumping Lemma:
L c cc bc ab aa a xz ∈ = ... ... ... ... ... x z k m − m m 2
m m m
c b a z y x
2
= 1 , ≥ = k a y
k
L xz ∈
Thus:
L c b a
m m k m
∈
− 2
L c b a
m m k m
∈
− 2
L c b a
m m k m
∉
− 2
BUT: CONTRADICTION!!!
} , : { ≥ =
+
l n c b a L
l n l n
1 ≥ k
Our assumption that is a regular language is not true
L Conclusion: L is not a regular language
Therefore: Regular languages Non-regular languages
} : {
!
≥ = n a L
n
SLIDE 9
9
Theorem: The language
is not regular
Proof:
Use the Pumping Lemma
} : {
!
≥ = n a L
n
n n n ⋅ − ⋅ = ) 1 ( 2 1 !
Assume for contradiction that is a regular language
L
Since is infinite we can apply the Pumping Lemma
L } : {
!
≥ = n a L
n ! m
a w =
We pick Let be the integer in the Pumping Lemma Pick a string such that:
w L w ∈ m w ≥ | |
length
m } : {
!
≥ = n a L
n Write
z y x am =
! it must be that length From the Pumping Lemma
a aa aa aa aa a a xyz
m
... ... ... ... ...
! =
= x y z m m m − ! 1 | | , | | ≥ ≤ y m y x m k a y
k
≤ ≤ = 1 ,
Thus:
SLIDE 10
10
From the Pumping Lemma:
L z y x
i
∈ ... , 2 , 1 , = i
Thus: ! m
a z y x = L z y x ∈
2
m k a y
k
≤ ≤ = 1 ,
From the Pumping Lemma:
L a aa aa aa aa aa a z xy ∈ = ... ... ... ... ... ...
2
x y z k m + m m − !
Thus: ! m
a z y x = m k a y
k
≤ ≤ = 1 , L z y x ∈
2
y L a
k m
∈
+ !
L a
k m
∈
+ !
! ! p k m = + } : {
!
≥ = n a L
n Since:
m k ≤ ≤ 1
There must exist such that:
p
However:
)! 1 ( ) 1 ( ! ! ! ! ! ! + = + = + < + ≤ + ≤ m m m m m m m m m m k m + !
for
1 > m )! 1 ( ! + < + m k m ! ! p k m ≠ +
for any p
SLIDE 11 11 L a
k m
∈
+ !
L a
k m
∉
+ !
BUT: CONTRADICTION!!!
} : {
!
≥ = n a L
n
m k ≤ ≤ 1
Our assumption that is a regular language is not true
L Conclusion: L is not a regular language
Therefore:
What’s Next
– Linz Chapter 1, 2.1, 2.2, 2.3, (skip 2.4) 3, 4, 5.1-5.3 – JFLAP Startup, Chapter 1, 2.1, 3, 4, 6.1
- Next Lecture Topics from Chapter 5.1-5.3
– Context Free Grammars
- Quiz 1 in Recitation on Wednesday 9/17
– Covers Linz 1.1, 1.2, 2.1, 2.2, 2.3 and JFLAP 1, 2.1 – Closed book, but you may bring one sheet of 8.5 x 11 inch paper with any notes you like. – Quiz will take the full hour
– Homework Due Thursday
