Non Regular Languages 2

8/8/2019 Non Regular Languages 2

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NON-REGULARLANGUAGES

Junaid Khan

Department of Computer Science University of

Peshawar Pakistan

[email protected]

Presented By


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Standard Representations

ofRegular Languages


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Standard Representations

of Regular Languages

Regular Languages

DFAs

NFAsRegular

Expressions

Regular

Grammars


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Non-Regular languages

Languages that cannot be represented by

Finite Automation

Languages that cannot be defined by a

regular expression and TG


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Regular languagesba *

acb *

...etc

*)( bacb

Non-regular languages

}0:{ unba nn

}*},{:{ bawwwR


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How can we prove that a language

is not regular?

L

Prove that there is no DFA that accepts L

Problem: this is not easy to prove

Solution: the Pumping Lemma !!!


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The Pigeonhole

Principle


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pigeons

pigeonholes

4

3


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A pigeonhole must

contain at least two pigeons


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...........

...........

pigeons

pigeonholes

n

m mn "


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The Pigeonhole Principle

...........

pigeons

pigeonholes

n

m

mn "

There is a pigeonhole

with at least 2 pigeons


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The Pigeonhole

Principle

and

DFAs


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Conti.

DFA with states4

1q 2q 3qa

b

4q

b

b b

b

a a


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1q 2q 3qa

b

4q

b

b

b

a a

a

In walks of strings:

aabaa

a no state

is repeated


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In walks of strings:

1q 2q 3qa

b

4q

b

b

b

a a

a

...abbbabbabb

abbabb

bbaa

aabb a state

is repeated


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If the walk of string has length

1q 2q 3qa

b

4q

b

b

b

a a

a

w 4|| uw

then a state is repeated


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Pumping Lemma

Pumping: pump more stuff into the middleof the world.

Without changing front and the back part ofthe string (swelling)..

Lemma: Help in proving otherresults E.g. help us certain specific languages are

nonregular.


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Pumping lemma

String w divided into three parts x, y, z

X=all state of w, at the beginning to the first

state that is revisited, (it may be null)

Y=substring travels around the circuit (cannot

be null)

Z=rest strings of w (may be null)


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Theorem: The language }0:{ u! nbaL nn

is not regular

Proof: Use the Pumping Lemma


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Assume for contradiction

that is a regular languageL

Since is infinitewe can apply the Pumping Lemma

L

}0:{ u! nbaL nn


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Let be the integer in the Pumping Lemma

Pick a string such that:wL

w

mw u||length

Example: mmbaw !pick

}0:{ u! nbaL nn


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Write: zyxbamm!

it must be that: length

From the Pumping Lemma

1||,|| ue ymyx

Therefore: babaaaabamm

............!

1,u! k

ay

k

x y z

m m


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From the Pumping Lemma: Lzyxi

...,2,1,0!i

Thus:

mmbazyx !

Lbazyyxzyxmkm

!!2

Lzyx 2

We have: 1, u! kayk


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Lbamkm

Therefore:

}0:{ u! nbaLnn

BUT:

Lba

mkm

CONTRADICTION!!!


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Our assumption that

is a regular language is not true

L

Conclusion:

Lis not a regular language

Therefore:

Non Regular Languages 2

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