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TK PrasadPumping Lemma

2

Nonregularity Proofs

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Regular Languages: Grand UnificationGrand Unification


3

)()( RELRGL

)(

)()(

DFAsL

NFAsLsNFAL

)()( RELFAL

(Parallel Simulation) (Rabin and Scott’s work)

(Collapsing graphs; Structural Induction)(S. Kleene’s work)

)()( RGLFAL (Construction)(Solving linear equations)


Role of various representations for Regular Languages


4Closure under complemention. (DFAs)Closure under union, concatenation, and

Kleene star. (NFA-s, Regular expression.) Consequence:

Closure under intersection by De Morgan’s Laws.

Relationship to context-free languages. (Regular Grammars.)

Ease of specification. (Regular expression.)

Building tokenizers/lexical analyzers. (DFAs)



5

regular.not is }0|{Show ibaL ii

Consider pairs of strings:

......:'

......:'n

si

nsi

bbbbv

aaaau

jiFbaq

Fbaqji

M

iiM

if ),(

),(

0*

0*

If L were regular, then there exists a DFA MM accepting L with the following property:

jibaqbaq jiM

iiM if ),(),( 0

*0

* admission.edhole.com

TK PrasadPumping Lemma 6

jiaqaq jM

iM if ),(),( 0

*0

*

jibaqbaq ijM

iiM for ),(),( 0

*0

*

JUSTIFICATION: Otherwise, from the definition of DFA,

CLAIM:

which contradicts the earlier conclusion.

In order to satisfy

jiaqaq jM

iM if ),(),( 0

*0

*

the machine M must have a unique state for every i.Thus, M must have infiniteinfinite number of states, if Lis regular. This violates the definition of DFA.So, L must be non-regular.


Using Closure Properties


7 Regular languages are closed under set-intersection.

Note that regularity is a property of a collection, and not a property of an individual string in the collection.

21

21

21

LL

LL

LLL

L1=bit strings with even parityL2=bit strings with number of 1’s divisible by 3L=bit strings with number of 1’s a multiple of 6



8

If R is a regular language and C is context-free, then may not be regular.

Proof:

• Show that

is not regular. • Proof: If L were regular, ought to be

regular. However, is known to be non-regular. Hence, L cannot be regular.

CR

CCR

ibaC

baRii

}0|{

**

} in s'# in '# | *},{{ bsabaL

RL CRL


Prelude to Pumping Lemma


9

Is 46551 divisible by 46?



Necessary vs sufficient condition


Pumping Lemma for Regular Languages


10

It is a necessary condition. Every regular language satisfies it. If a language violates it, it is not regular.

RL => PL not PL => not RL

It is not a sufficient condition. Not every non-regular language violates it.

not RL =>? PL or not PL (no conclusion)


Basic Idea:


11

q0

b a

a

a,b

bb

a

q2 q3

q1

)(MLababbaaab

3102312310 qqqqqqqqqqbaaabbaba


TK PrasadPumping Lemma 12

3102312310 qqqqqqqqqqbaaabbaba

Note,

)(MLababbSo,

)(MLabaaab3102312310 qqqqqqqqqq

baaabbaba

)()()( :, MLaaababbabji ji admission.edhole.com

Fundamental Observation


13

Given a “sufficiently” long string, the states of a DFA must repeat in an accepting computation. These cycles can then be used to predict (generate) infinitely many other strings in (of) the language.

Pigeon-Hole Principle


Pumping Lemma


14Let L be a regular language that is accepted by a DFA M with k states. Let z be any string in L with . Then z can be decomposed as uvw with

Lwuvi

vlength

kuvlength

i

:0

and ,0)(

,)(

kzlength )(



15

For all sufficiently long strings (z) There exists non-null prefix (uv) and substring (v) For all repetitions of the

substring (v), we get strings in the

language.

)0 :(

0) || ( ) || (

)( :

|| :

Lwuvii

vkuv

suvwu,v,w

ksLs

i


Proving non-regularity


16If there exists an arbitrarily long string

s L, and for each decomposition s = uvw, there exists an i such that , then L is non-regular.

)0 :(

0) || ( ) || (

)( :

|| :

Lwuvii

vkuv

suvwu,v,w

ksLs

i

Negation of the necessary condition:

Lwuvi


APPLYING PUMPING LEMMA


17

Examples



18

Proof by contradiction:Let be accepted by a k-state DFA.ChooseFor all prefixes of length show there exists such thati.e.,

regular.not is

}number primea is |{ paL pp

pLk nas n primea is where,

,kj ,ji p

jnij Laa j )(

number. compositea is )( jnij j



19

Choose (For this specific problem happens to be

independent of j, but that need not always be the case.)

is non-regular because it violates the

necessary condition.

1ni j

ji

number! composite

)1(*

)1(

jn

nn*jjnnj*

pL

,...)( , 21211p

nnp

nn LaaLaa



20

Proof : (For this example, choice of initial string is crucial.)

For this choice of s, the pumping lemma cannot generate a contradiction!

However, let instead.

}|{ mnbaL mnp

DFAof states ofnumber where nas n

1 nn bas

:String Pumped

:String Original1*

1

njnji

njnj

baa

baas



21For

Thus, by pumping the substring containing a’s 0 times (effectively deleting it), the number of a’s can be made smaller than the number of b’s.

So, by pumping lemma, L is non-regular.

njn

jni

1

1 0



22

Proof by contradiction: If is regular, then so is , the

complement of But which is known to be

non-regular. So, cannot be regular.

regular.not is

}number compositea is |{ caL cc

pc LLa cLa

cL .cL

cL


Summary: Proof Techniques


23

Counter ExamplesConstructions/Simulations

Induction ProofsImpossibility Proofs

Proofs by ContradictionReduction Proofs : Closure Properties


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