class06_001 pumping lemma

7/31/2019 Class06_001 Pumping Lemma

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Formal Languages

Non-Regular LanguagesSlides based on RPI CSCI 2400

Thanks to Petros Drineas


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Evert: Session 3Posix notation


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Evert: Session 1Vollstaendige Induktion


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Regular languagesa* cb*

.et

(acb

Non-regular languages:{ nbann

,{:{

avvvR


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How can we prove that a language

is not regular?L

Prove that there is no DFA that acceptsL

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 mustcontain at least two pigeons


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...........

...........

pigeons

pigeonholes

n

m n


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...........

pigeonspigeonholes

nm

n

There is a pigeonhole

with at least 2 pigeons


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and

DFAs


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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

aa

a

In walks of strings:

aab

aa

a no state

is repeated


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In walks of strings:

1q 2q 3qa

b

4q

b

b

b

aa

a

...abbbabbabb

abbabb

bbaa

aabb a state

is repeated


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If string has length :

1q 2q 3qa

b

4q

b

b

b

aa

a

w ||w

Thus, a state must be repeated

Then the transitions of stringare more than the states of the DFAw

l


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In general, for any DFA:

String has length number of statesw

A state must be repeated in the walk ofw

q

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

walk of w

Repeated state


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In other words for a string :

transitions are pigeons

states are pigeonholesq

a

w

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

walk of w

Repeated state


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The Pumping Lemma


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Take an infinite regular language L

There exists a DFA that accepts L

mstates


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Take string withw w

There is a walk with label :w

.........

walk w

If t i h l th ||


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If string has lengthw w|| (numberof states

of DFA)

then, from the pigeonhole principle:

a state is repeated in the walk w

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

walk w


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q

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

walk w

Let be the first state repeated in the

walk of w


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Write xw

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

x

y

z


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

x

y

z

Observations: yx||length number

of states

of DFA||ylength


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The string

is accepted

xObservation:

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

x

y

z


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The string

is accepted

yxObservation:

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

x

y

z


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The string

is accepted

yyObservation:

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

x

y

z

i


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The string

is accepted

yxi

In General:

,,1,0i

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

x

y

z


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zyxiIn General: ,,1,0

i

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

x

y

z

Language accepted by the DFA


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In other words, we described:

The Pumping Lemma !!!


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The Pumping Lemma:

Given an infinite regular languageL

there exists an integer m

for any string with lengthw w||

we can write xw

with andyx|| ||y

such that:zyxi ,,1,0i


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Applications

of

the Pumping Lemma


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Theorem: The language :{ nbaLnn

is not regular

Proof: Use the Pumping Lemma

{bLnn


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Assume for contradiction

that is a regular languageL

Since is infinitewe can apply the Pumping Lemma

L

:{ nbaLnn

{bLnn


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Let be the integer in the Pumping Lemma

Pick a string such that:w w

w||length

awWe pick

m

:{ nbaLnn

bmm


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it must be that length

From the Pumping Lemma||,||yx

aabayzmm......

, kayk

x y z

m m

Write: xbamm

Thus:


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kk


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From the Pumping Lemma:

aaaazy ...........2

x y

m m

Thus:

zyx2

azy , kayk

y

ba

km

b

km k


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bakm

:{ nbaLnn

BUT:

bamkm

CONTRADICTION!!!

k


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Our assumption that

is a regular language is not true

L

Conclusion: L is not a regular language

Therefore:


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Regular languages

Non-regular languages :{ nbann

class06_001 pumping lemma

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