More Applications of the Pumping Lemma
description
Transcript of More Applications of the Pumping Lemma
![Page 1: More Applications of the Pumping Lemma](https://reader035.fdocuments.in/reader035/viewer/2022070500/56816810550346895ddd9d6e/html5/thumbnails/1.jpg)
1
More Applications
of
the Pumping Lemma
![Page 2: More Applications of the Pumping Lemma](https://reader035.fdocuments.in/reader035/viewer/2022070500/56816810550346895ddd9d6e/html5/thumbnails/2.jpg)
2
The Pumping Lemma:• Given a infinite regular language L
• there exists an integer m
• for any string with length Lw mw ||
• we can write zyxw
• with andmyx || 1|| y
• such that: Lzyx i ...,2,1,0i
![Page 3: More Applications of the Pumping Lemma](https://reader035.fdocuments.in/reader035/viewer/2022070500/56816810550346895ddd9d6e/html5/thumbnails/3.jpg)
3
Regular languages
Non-regular languages *}:{ vvvL R
![Page 4: More Applications of the Pumping Lemma](https://reader035.fdocuments.in/reader035/viewer/2022070500/56816810550346895ddd9d6e/html5/thumbnails/4.jpg)
4
Theorem:The language
is not regular
Proof: Use the Pumping Lemma
*}:{ vvvL R },{ ba
![Page 5: More Applications of the Pumping Lemma](https://reader035.fdocuments.in/reader035/viewer/2022070500/56816810550346895ddd9d6e/html5/thumbnails/5.jpg)
5
Assume for contradictionthat is a regular languageL
Since is infinitewe can apply the Pumping Lemma
L
*}:{ vvvL R
![Page 6: More Applications of the Pumping Lemma](https://reader035.fdocuments.in/reader035/viewer/2022070500/56816810550346895ddd9d6e/html5/thumbnails/6.jpg)
6
mmmm abbaw We pick
Let be the integer in the Pumping Lemma
Pick a string such that: w Lw
mw ||length
m
and
*}:{ vvvL R
![Page 7: More Applications of the Pumping Lemma](https://reader035.fdocuments.in/reader035/viewer/2022070500/56816810550346895ddd9d6e/html5/thumbnails/7.jpg)
7
Write zyxabba mmmm
it must be that lengthFrom the Pumping Lemma
ababbabaaaaxyz ..................
x y z
m m m m
1||,|| ymyx
1, kay kThus:
![Page 8: More Applications of the Pumping Lemma](https://reader035.fdocuments.in/reader035/viewer/2022070500/56816810550346895ddd9d6e/html5/thumbnails/8.jpg)
8
From the Pumping Lemma: Lzyx i
...,2,1,0i
Thus: Lzyx 2
1, kay kmmmm abbazyx
![Page 9: More Applications of the Pumping Lemma](https://reader035.fdocuments.in/reader035/viewer/2022070500/56816810550346895ddd9d6e/html5/thumbnails/9.jpg)
9
From the Pumping Lemma:
Lababbabaaaaaazxy ∈.....................=2
x y z
km + m m m
1, kay k
y
Lzyx 2
Thus:
mmmm abbazyx
Labba mmmkm
![Page 10: More Applications of the Pumping Lemma](https://reader035.fdocuments.in/reader035/viewer/2022070500/56816810550346895ddd9d6e/html5/thumbnails/10.jpg)
10
Labba mmmkm
Labba mmmkm
BUT:
CONTRADICTION!!!
1k
*}:{ vvvL R
![Page 11: More Applications of the Pumping Lemma](https://reader035.fdocuments.in/reader035/viewer/2022070500/56816810550346895ddd9d6e/html5/thumbnails/11.jpg)
11
Our assumption thatis a regular language is not true
L
Conclusion: L is not a regular language
Therefore:
![Page 12: More Applications of the Pumping Lemma](https://reader035.fdocuments.in/reader035/viewer/2022070500/56816810550346895ddd9d6e/html5/thumbnails/12.jpg)
12
Regular languages
Non-regular languages}0,:{ lncbaL lnln
![Page 13: More Applications of the Pumping Lemma](https://reader035.fdocuments.in/reader035/viewer/2022070500/56816810550346895ddd9d6e/html5/thumbnails/13.jpg)
13
Theorem:The language
is not regular
Proof: Use the Pumping Lemma
}0,:{ lncbaL lnln
![Page 14: More Applications of the Pumping Lemma](https://reader035.fdocuments.in/reader035/viewer/2022070500/56816810550346895ddd9d6e/html5/thumbnails/14.jpg)
14
Assume for contradictionthat is a regular languageL
Since is infinitewe can apply the Pumping Lemma
L
}0,:{ lncbaL lnln
![Page 15: More Applications of the Pumping Lemma](https://reader035.fdocuments.in/reader035/viewer/2022070500/56816810550346895ddd9d6e/html5/thumbnails/15.jpg)
15
mmm cbaw 2We pick
Let be the integer in the Pumping Lemma
Pick a string such that: w Lw
mw ||length
m
}0,:{ lncbaL lnln
and
![Page 16: More Applications of the Pumping Lemma](https://reader035.fdocuments.in/reader035/viewer/2022070500/56816810550346895ddd9d6e/html5/thumbnails/16.jpg)
16
Write zyxcba mmm 2
it must be that lengthFrom the Pumping Lemma
cccbcabaaaaaxyz ..................
x y z
m m m2
1||,|| ymyx
1, kay kThus:
![Page 17: More Applications of the Pumping Lemma](https://reader035.fdocuments.in/reader035/viewer/2022070500/56816810550346895ddd9d6e/html5/thumbnails/17.jpg)
17
From the Pumping Lemma: Lzyx i
...,2,1,0i
Thus:
mmm cbazyx 2
Lxzzyx ∈=0
1, kay k
![Page 18: More Applications of the Pumping Lemma](https://reader035.fdocuments.in/reader035/viewer/2022070500/56816810550346895ddd9d6e/html5/thumbnails/18.jpg)
18
From the Pumping Lemma:
Lcccbcabaaaxz ...............
x z
km m m2
mmm cbazyx 2 1, kay k
Lxz
Thus: Lcba mmkm 2
![Page 19: More Applications of the Pumping Lemma](https://reader035.fdocuments.in/reader035/viewer/2022070500/56816810550346895ddd9d6e/html5/thumbnails/19.jpg)
19
Lcba mmkm 2
Lcba mmkm 2
BUT:
CONTRADICTION!!!
}0,:{ lncbaL lnln
1k
![Page 20: More Applications of the Pumping Lemma](https://reader035.fdocuments.in/reader035/viewer/2022070500/56816810550346895ddd9d6e/html5/thumbnails/20.jpg)
20
Our assumption thatis a regular language is not true
L
Conclusion: L is not a regular language
Therefore:
![Page 21: More Applications of the Pumping Lemma](https://reader035.fdocuments.in/reader035/viewer/2022070500/56816810550346895ddd9d6e/html5/thumbnails/21.jpg)
21
Regular languages
Non-regular languages }0:{ ! naL n
![Page 22: More Applications of the Pumping Lemma](https://reader035.fdocuments.in/reader035/viewer/2022070500/56816810550346895ddd9d6e/html5/thumbnails/22.jpg)
22
Theorem:The language
is not regular
Proof: Use the Pumping Lemma
}0:{ ! naL n
nnn )1(21!
![Page 23: More Applications of the Pumping Lemma](https://reader035.fdocuments.in/reader035/viewer/2022070500/56816810550346895ddd9d6e/html5/thumbnails/23.jpg)
23
Assume for contradictionthat is a regular languageL
Since is infinitewe can apply the Pumping Lemma
L
}0:{ ! naL n
![Page 24: More Applications of the Pumping Lemma](https://reader035.fdocuments.in/reader035/viewer/2022070500/56816810550346895ddd9d6e/html5/thumbnails/24.jpg)
24
!mawWe pick
Let be the integer in the Pumping Lemma
Pick a string such that: w Lw
mw ||length
m
}0:{ ! naL n
![Page 25: More Applications of the Pumping Lemma](https://reader035.fdocuments.in/reader035/viewer/2022070500/56816810550346895ddd9d6e/html5/thumbnails/25.jpg)
25
Write zyxam !
it must be that lengthFrom the Pumping Lemma
aaaaaaaaaaaxyz m ...............!
x y z
m mm !
1||,|| ymyx
mkay k 1,Thus:
![Page 26: More Applications of the Pumping Lemma](https://reader035.fdocuments.in/reader035/viewer/2022070500/56816810550346895ddd9d6e/html5/thumbnails/26.jpg)
26
From the Pumping Lemma: Lzyx i
...,2,1,0i
Thus:
!mazyx
Lzyx 2
mkay k 1,
![Page 27: More Applications of the Pumping Lemma](https://reader035.fdocuments.in/reader035/viewer/2022070500/56816810550346895ddd9d6e/html5/thumbnails/27.jpg)
27
From the Pumping Lemma:
Laaaaaaaaaaaazxy ..................2
x y z
km mm !
Thus:
!mazyx mkay k 1,
Lzyx 2
y
La km !
![Page 28: More Applications of the Pumping Lemma](https://reader035.fdocuments.in/reader035/viewer/2022070500/56816810550346895ddd9d6e/html5/thumbnails/28.jpg)
28
La km !
!! pkm
}0:{ ! naL nSince:
mk 1
There must exist such that: p
![Page 29: More Applications of the Pumping Lemma](https://reader035.fdocuments.in/reader035/viewer/2022070500/56816810550346895ddd9d6e/html5/thumbnails/29.jpg)
29
However:
)!1()1(!!!
!!!
mmmmmm
mmmmkm ! for 1m
)!1(! mkm
!! pkm for any p
![Page 30: More Applications of the Pumping Lemma](https://reader035.fdocuments.in/reader035/viewer/2022070500/56816810550346895ddd9d6e/html5/thumbnails/30.jpg)
30
La km !
La km !
BUT:
CONTRADICTION!!!
}0:{ ! naL n
mk 1
![Page 31: More Applications of the Pumping Lemma](https://reader035.fdocuments.in/reader035/viewer/2022070500/56816810550346895ddd9d6e/html5/thumbnails/31.jpg)
31
Our assumption thatis a regular language is not true
L
Conclusion: L is not a regular language
Therefore: