courtesy costas busch - rpi1 non-regular languages

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Courtesy Costas Busch - RPI 1 Non-regular languages

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Page 1: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 1

Non-regular languages

Page 2: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 2

Regular languages

ba* acb *

...etc

*)( bacb

Non-regular languages}0:{ nba nn

}*},{:{ bavvvR

Page 3: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 3

How can we prove that a languageis not regular?

L

Prove that there is no DFA that accepts L

Problem: this is not easy to prove

Solution: the Pumping Lemma !!!

Page 4: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 4

The Pigeonhole Principle

Page 5: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 5

pigeons

pigeonholes

4

3

Page 6: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 6

A pigeonhole mustcontain at least two pigeons

Page 7: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 7

...........

...........

pigeons

pigeonholes

n

m mn

Page 8: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 8

The Pigeonhole Principle

...........

pigeons

pigeonholes

n

m

mn There is a pigeonhole with at least 2 pigeons

Page 9: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 9

The Pigeonhole Principle

and

DFAs

Page 10: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 10

DFA with states 4

1q 2q 3qa

b

4q

b

b b

b

a a

Page 11: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 11

1q 2q 3qa

b

4q

b

b

b

a a

a

In walks of strings:

aab

aa

a no stateis repeated

Page 12: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 12

In walks of strings:

1q 2q 3qa

b

4q

b

b

b

a a

a

...abbbabbabb

abbabb

bbaa

aabb a stateis repeated

Page 13: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 13

If string has length :

1q 2q 3qa

b

4q

b

b

b

a a

a

w 4|| w

Thus, a state must be repeated

Then the transitions of string are more than the states of the DFA

w

Page 14: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 14

In general, for any DFA:

String has length number of states w

A state must be repeated in the walk of wq

q...... ......

walk of w

Repeated state

Page 15: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 15

In other words for a string : transitions are pigeons

states are pigeonholesq

a

w

q...... ......

walk of w

Repeated state

Page 16: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 16

The Pumping Lemma

Page 17: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 17

Take an infinite regular languageL

There exists a DFA that accepts L

mstates

Page 18: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 18

Take string with w Lw

There is a walk with label :w

.........

walk w

Page 19: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 19

If string has length w mw || (number of statesof DFA)

then, from the pigeonhole principle: a state is repeated in the walkw

q...... ......

walk w

Page 20: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 20

q

q...... ......

walk w

Let be the first state repeated in thewalk of w

Page 21: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 21

Write zyxw

q...... ......

x

y

z

Page 22: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 22

q...... ......

x

y

z

Observations: myx ||length numberof statesof DFA1|| ylength

Page 23: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 23

The string is accepted

zxObservation:

q...... ......

x

y

z

Page 24: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 24

The string is accepted

zyyxObservation:

q...... ......

x

y

z

Page 25: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 25

The string is accepted

zyyyxObservation:

q...... ......

x

y

z

Page 26: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 26

The string is accepted

zyx iIn General:

...,2,1,0i

q...... ......

x

y

z

Page 27: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 27

Lzyx i ∈In General: ...,2,1,0i

q...... ......

x

y

z

Language accepted by the DFA

Page 28: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 28

In other words, we described:

The Pumping Lemma !!!

Page 29: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 29

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 30: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 30

Applications

of

the Pumping Lemma

Page 31: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 31

Theorem:The language }0:{ nbaL nn

is not regular

Proof: Use the Pumping Lemma

Page 32: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 32

Assume for contradictionthat is a regular languageL

Since is infinitewe can apply the Pumping Lemma

L

}0:{ nbaL nn

Page 33: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 33

Let be the integer in the Pumping Lemma

Pick a string such that: w Lw

mw ||length

mmbawWe pick

m

}0:{ nbaL nn

Page 34: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 34

it must be that length

From the Pumping Lemma 1||,|| ymyx

babaaaaabaxyz mm ............

1, kay k

x y z

m m

Write: zyxba mm

Thus:

Page 35: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 35

From the Pumping Lemma: Lzyx i

...,2,1,0i

Thus:

mmbazyx

Lzyx 2

1, kay k

Page 36: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 36

From the Pumping Lemma:

Lbabaaaaaaazxy ...............2

x y z

km m

Thus:

Lzyx 2

mmbazyx 1, kay k

y

Lba mkm

Page 37: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 37

Lba mkm

}0:{ nbaL nnBUT:

Lba mkm

CONTRADICTION!!!

1≥k

Page 38: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 38

Our assumption thatis a regular language is not true

L

Conclusion: L is not a regular language

Therefore:

Page 39: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 39

Regular languages

Non-regular languages }0:{ nba nn

Page 40: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 40

Regular languages

Non-regular languages *}:{ vvvL R

Page 41: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 41

Theorem:The language

is not regular

Proof: Use the Pumping Lemma

*}:{ vvvL R },{ ba

Page 42: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 42

Assume for contradictionthat is a regular languageL

Since is infinitewe can apply the Pumping Lemma

L

*}:{ vvvL R

Page 43: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 43

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 44: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 44

Write zyxabba mmmm

it must be that length

From the Pumping Lemma

ababbabaaaaxyz ..................

x y z

m m m m

1||,|| ymyx

1, kay kThus:

Page 45: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 45

From the Pumping Lemma: Lzyx i

...,2,1,0i

Thus: Lzyx 2

1, kay kmmmm abbazyx

Page 46: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 46

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 47: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 47

Labba mmmkm

Labba mmmkm

BUT:

CONTRADICTION!!!

1k

*}:{ vvvL R

Page 48: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 48

Our assumption thatis a regular language is not true

L

Conclusion: L is not a regular language

Therefore:

Page 49: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 49

Regular languages

Non-regular languages

}0,:{ lncbaL lnln

Page 50: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 50

Theorem:The language

is not regular

Proof: Use the Pumping Lemma

}0,:{ lncbaL lnln

Page 51: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 51

Assume for contradictionthat is a regular languageL

Since is infinitewe can apply the Pumping Lemma

L

}0,:{ lncbaL lnln

Page 52: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 52

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 53: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 53

Write zyxcba mmm 2

it must be that length

From the Pumping Lemma

cccbcabaaaaaxyz ..................

x y z

m m m2

1||,|| ymyx

1, kay kThus:

Page 54: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 54

From the Pumping Lemma: Lzyx i

...,2,1,0i

Thus:

mmm cbazyx 2

Lxzzyx ∈=0

1, kay k

Page 55: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 55

From the Pumping Lemma:

Lcccbcabaaaxz ...............

x z

km m m2

mmm cbazyx 2 1, kay k

Lxz

Thus: Lcba mmkm 2

Page 56: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 56

Lcba mmkm 2

Lcba mmkm 2

BUT:

CONTRADICTION!!!

}0,:{ lncbaL lnln

1k

Page 57: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 57

Our assumption thatis a regular language is not true

L

Conclusion: L is not a regular language

Therefore:

Page 58: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 58

Regular languages

Non-regular languages }0:{ ! naL n

Page 59: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 59

Theorem:The language

is not regular

Proof: Use the Pumping Lemma

}0:{ ! naL n

nnn )1(21!

Page 60: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 60

Assume for contradictionthat is a regular languageL

Since is infinitewe can apply the Pumping Lemma

L

}0:{ ! naL n

Page 61: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 61

!mawWe pick

Let be the integer in the Pumping Lemma

Pick a string such that: w Lw

mw ||length

m

}0:{ ! naL n

Page 62: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 62

Write zyxam !

it must be that length

From the Pumping Lemma

aaaaaaaaaaaxyz m ...............!

x y z

m mm !

1||,|| ymyx

mkay k 1,Thus:

Page 63: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 63

From the Pumping Lemma: Lzyx i

...,2,1,0i

Thus:

!mazyx

Lzyx 2

mkay k 1,

Page 64: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 64

From the Pumping Lemma:

Laaaaaaaaaaaazxy ..................2

x y z

km mm !

Thus:

!mazyx mkay k 1,

Lzyx 2

y

La km !

Page 65: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 65

La km !

!! pkm

}0:{ ! naL nSince:

mk 1

There must exist such that: p

Page 66: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 66

However:

)!1(

)1(!

!!

!!

!

m

mm

mmm

mm

mmkm ! for 1m

)!1(! mkm

!! pkm for any p

Page 67: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 67

La km !

La km !

BUT:

CONTRADICTION!!!

}0:{ ! naL n

mk 1

Page 68: Courtesy Costas Busch - RPI1 Non-regular languages

Courtesy Costas Busch - RPI 68

Our assumption thatis a regular language is not true

L

Conclusion: L is not a regular language

Therefore:

Page 69: Courtesy Costas Busch - RPI1 Non-regular languages

69

SummaryShowing regularconstruct DFA, NFAconstruct regular expressionshow L is the union, concatenation, intersection (regular operations) of regular languages.

Showing non-regularpumping lemmaassume regular, apply closure properties of regular languages and obtain a known non-regular language.