UNIT II: Regular Expressions

THEORY OF COMPUTATION

1

Jagadish Kashinath Kamble (M.Tech, IIT Kharagpur)

Assistant Professor,Dept. of Information Technology, Pune Institute of Computer Technology, Pune.

Regular ExpressionsUNIT II:

Regular Language Acceptor Finite

Automata

GeneratorRegular

Grammer

Right Linear Grammer

Left Linear Grammer

FA with Output

FA without Output

Representator

Regular

Expression

Regular Expressions

Regular expressions

describe regular languages

Example:

describes the language

*)( cba +

,...,,,,,*, bcaabcaabcabca =

Recursive Definition

,,

( )1

1

21

21

*

r

r

rr

rr

+

Are regular expressions

Primitive regular expressions:

2r1r

Given regular expressions

and

1) + : Union 2) . : Concatenation 3) * : Kleene Closure

{ }

{ }

{ a }

Examples

( ) )(* ++ ccbaA regular expression:

( )++ baNot a regular expression:

Languages of Regular Expressions

: language of regular expression

Example

( )rL r

( ) ,...,,,,,*)( bcaabcaabcacbaL =+

Definition

For primitive regular expressions:

( )

( )

( ) aaL

L

L

=

=

=

Definition (continued)

For regular expressions and1r 2r

( ) ( ) ( )2121 rLrLrrL =+

( ) ( ) ( )2121 rLrLrrL =

( ) ( )( )** 11 rLrL =

( )( ) ( )11 rLrL =

Example

Regular expression: ( ) *aba +

( )( )*abaL + ( )( ) ( )*aLbaL +=

( ) ( )*aLbaL +=

( ) ( )( ) ( )( )*aLbLaL =

( ) ( )*aba =

,...,,,, aaaaaaba =

,...,,,...,,, baababaaaaaa=

Example

Regular expression ( ) ( )bbabar ++= *

( ) ,...,,,,, bbbbaabbaabbarL =

Example

Regular expression ( ) ( ) bbbaar **=

( ) }0,:{ 22 = mnbbarL mn

Equivalent Regular Expressions

Definition:

Regular expressions and

are equivalent if

1r 2r

)()( 21 rLrL =

Regular Expressionsand

Regular Languages

• Regular Language

• ∅

• {ε}

• {a,b}*

• {aab}*{a,ab}

• {aa,bb}U {ab,ba}{aa,bb}*

• Regular Expression

• ∅

• {}

• (a+b)*

• (aab)*(a+ab)

• (aa+bb+(ab+ba)(aa+bb)*)

Theorem

Languages

Generated by

Regular Expressions

Regular

Languages=

Find RE for the Language of strings of Length 2 over the alphabet {a,b}

Find RE for the Language of strings of Length at least 2 over the alphabet {a,b}

Find RE for the Language of strings of Length at most 2 over the alphabet {a,b}

Find RE for the Language of strings of even Length over the alphabet {a,b}

Find RE for the Language of strings of odd Length over the alphabet {a,b}

Find RE for the Language of strings which is divisible by 3 over the alphabet {a,b}

Find RE for the Language of strings = 2 mod 3 over the alphabet {a,b}

~

Find RE for the Language of strings of a’s exactly 2 over the alphabet {a,b}

• Find RE for the Language of strings of a’s atleast 2 over the alphabet {a,b}

• Find RE for the Language of strings of a’s atmost 2 over the alphabet {a,b}

• Find RE for the Language of strings of even length a’s over the alphabet {a,b}

• Find RE for the Language of strings which starts with a over the alphabet {a,b}

• Find RE for the Language of strings which ends with a over the alphabet {a,b}

• Find RE for the Language of strings which contains a over the alphabet {a,b}

• Find RE for the Language of strings which starts and ends with different symbol over the alphabet {a,b}

• Find RE for the Language of strings which starts and ends with Same symbol over the alphabet {a,b}

• Give RE for Following over the alphabet {0,1} (Aug-2015 6 Marks) – All binary strings with at least one 0

– All binary strings with at Most one 0

• Language of all strings containing substring 00

• The Language of Strings in {a, b}∗ Ending with b and Not Containing aa

• Find RE for the Language of strings which does not contains two a’s together over the alphabet {a,b}

Review


Identities Related to Regular Expressions

• Given R, P, L, Q as regular expressions, the followingidentities hold −

• ∅* = ε

• ε* = ε

• RR* = R*R

• R*R* = R*

• (R*)* = R*

• RR* = R*R

• (PQ)*P =P(QP)*

• (a+b)* = (a*b*)* = (a*+b*)* = (a+b*)* = a*(ba*)*

• R + ∅ = ∅ + R = R (The identity for union)

• R ε = ε R = R (The identity for concatenation)

• ∅ L = L ∅ = ∅ (The annihilator for concatenation)

• R + R = R (Idempotent law)

• L (M + N) = LM + LN (Left distributive law)

• (M + N) L = ML + NL (Right distributive law)

• ε + RR* = ε + R*R = R*

Theorem

Languages

Generated by

Regular Expressions

Regular

Languages=

Languages

Generated by

Regular Expressions

Regular

Languages

Languages

Generated by

Regular Expressions

Regular

Languages

Proof:

Proof - Part 1

r

)(rL

For any regular expression

the language is regular

Languages

Generated by

Regular Expressions

Regular

Languages

Proof by induction on the size of r

Induction Basis

Primitive Regular Expressions: ,,

Corresponding

NFAs

)()( 1 == LML

)(}{)( 2 LML ==

)(}{)( 3 aLaML ==

regular

languages

a

Inductive Hypothesis

Suppose

that for regular expressions and ,

and are regular languages1r 2r

)( 1rL )( 2rL

Inductive Step

We will prove:

( )

( )

( )

( )( )1

1

21

21

*

rL

rL

rrL

rrL

+

Are regular

Languages

Using the regular closure of these operations,

we can construct recursively the NFA

that accepts

M)()( rLML =

Example:21 rrr +=

)()( 11 rLML =

)()( 22 rLML =

)()( rLML =

Regular Expression r1

1M

Single accepting state

NFA 2M


Regular Expression r2

NFA

Union

NFA for r1+r2

1M

2M

Concatenation

NFA for r1tr2

1M 2M

Star Operation

NFA for r1*

1M

)( *

1 bar =

a

b

1M

)(2 bar = ab

2M

Example

a

b

ab

}{1 baL n=

}{2 baL =

)()( *

21 babarr +=+NFA for

Example

)(2 bar =

)( *

1 bar =

NFA for

a

b ab

}{1 baL n=}{2 baL =

)())(( **

21 bbaababarr ==

Example

)(2 bar =)( *

1 bar =

• An NFA Corresponding to ((aa +b)*(aba)*bab)*

a ε a

b

ε

ε

ε

ε

a ε b ε aε

ε

b ε a ε b

ε

ε

ε

ε

• Construct an NFA with null-moves, which accepts the language defined by:

• ((0 + 1)* 10 + (00)* (11)*)*

• 01[((10)+ + 111)* + 0]* 1

• (a / b)* ab.

Problems

Problems

• Construct DFA for the R.E l0 + (0 + ll)

• Construct DFA for regular expression (0 + l)*, (00 + ll)

Review


• Construction of RE from Language

• RE and DFA

• Conversion of RE to DFA

For any regular language there is

a regular expression with

Proof - Part 2

Languages

Generated by

Regular Expressions

Regular

Languages

L

r LrL =)(

We will convert an NFA that accepts

to a regular expression

L

Arden’s Theorem

In order to find out a regular expression of aFinite Automaton, we use Arden’s Theoremalong with the properties of regularexpressions.

Statement −

Let P, Q and R be the regular expressions.

If P does not contain null string, then

R = Q + RP has a unique solution that is

R = QP*

Proof −

R = Q + (Q + RP)P [After putting the value R = Q + RP]

= Q + QP + RPP

When we put the value of R recursively again and again, we get the following equation −

R = Q + QP + QP2 + QP3…..

R = Q (ε + P + P2 + P3 + …. )

R = QP* [As P* represents (ε + P + P2 + P3 + ….) ]

Hence, proved.

Assumptions for Applying Arden’s Theorem

• The transition diagram must not have NULL transitions

MethodStep 1 − Create equations as the following form for all the

states of the DFA having n states with initial state q1 (on incoming edges only and add ε to the initial state ).

q1 = q1R11 + q2R21 + … + qnRn1 + ε

q2 = q1R12 + q2R22 + … + qnRn2

…………………………

qn = q1R1n + q2R2n + … + qnRnn

Rij represents the set of labels of edges from qi to qj, if no such edge exists, then Rij = ∅

Step 2 − Solve these equations to get the equation for the final state in terms of Rij

Problem

Construct a regular expression correspondingto the automata given below:

Solution −

Here the initial state is q2 and the final state is q1.

The equations for the three states q1, q2, and q3 are as follows −

q1 = q1a + q3a + ε (ε move is because q1 is the initial state0

q2 = q1b + q2b + q3b

q3 = q2a

Now, we will solve these three equations −

q2 = q1b + q2b + q3b

= q1b + q2b + (q2a)b (Substituting value of q3)

= q1b + q2(b + ab)

= q1b (b + ab)* (Applying Arden’s Theorem)

q1 = q1a + q3a + ε

= q1a + q2aa + ε (Substituting value of q3)

= q1a + q1b(b + ab*)aa + ε (Substituting value of q2)

= q1(a + b(b + ab)*aa) + ε

= ε (a+ b(b + ab)*aa)*

= (a + b(b + ab)*aa)*

Hence, the regular expression is

(a + b(b + ab)*aa)*.

Problem


Solution −

Here the initial state is q1 and the final state is q2

Now we write down the equations −

q1 = q10 + ε

q2 = q11 + q20

q3 = q21 + q30 + q31

Now, we will solve these three equations −

q1 = ε0* [As, εR = R]

So, q1 = 0*

q2 = 0*1 + q20

So, q2 = 0*1(0)* [By Arden’s theorem]


0*10*.

• Construct the regular expressions for the following DFAs:

Find the regular expression for the following .(Nov-2017 4 Marks)

Review


• Construction of RE from Language

• RE and DFA

• Conversion of RE to DFA

• Conversion of DFA to RE

Construct Regular Expression for the following transitiondiagram using Arden’s theorem.(Nov 2014 4 Marks)

• From construct the equivalent

• Generalized Transition Graph

• in which transition labels are regular expressions

M

Example:

a

ba,

c

Ma

ba +

c

Corresponding

Generalized transition graph

State Elimination Method

Problem


q1 = q1a + q3a + ε

= q1a + q2aa + ε (Substituting value of q3)

= q1a + q1b(b + ab*)aa + ε (Substituting value of q2)

= q1(a + b(b + ab)*aa) + ε

= ε (a+ b(b + ab)*aa)*

= (a + b(b + ab)*aa)*


(a + b(b + ab)*aa)*.

• Another Example:

ba +a

b

b

0q 1q 2q

ba,a

b

b

0q 1q 2q

b

bTransition labels

are regular

expressions

In General

• Removing a state:

iq q jq

a b

cd

e

0q fq

1r

2r

3r4r

*)*(* 213421 rrrrrrr +=

LMLrL == )()(

The resulting regular expression:

By repeating the process until

two states are left, the resulting graph is

Initial graph Resulting graph

End of Proof-Part 2

Standard Representations of Regular Languages

Regular Languages

DFAs

NFAsRegular

Expressions

When we say: We are given

a Regular Language

We mean:

L

Language is in a standard

representation

L

(DFA, NFA, or Regular Expression)

Non-regular languages

(Pumping Lemma)

Regular languages

ba* acb +*

...etc

*)( bacb ++

Non-regular languages}0:{ nba nn

}*},{:{ bavvvR

How can we prove that a language

is not regular?

L

Prove that there is no DFA or NFA or RE

that accepts L

Difficulty: this is not easy to prove

(since there is an infinite number of them)

Solution: use the Pumping Lemma !!!

The Pigeonhole Principle

pigeons

pigeonholes

4

3

A pigeonhole must

contain at least two pigeons

...........

...........

pigeons

pigeonholes

n

m mn


...........

pigeons

pigeonholes

n

m

mn There is a pigeonhole

with at least 2 pigeons


and

DFAs

Consider a DFA with states 4

1q 2q 3qa

b

4q

b

a b

b

a a

Consider the walk of a “long’’ string:

1q 2q 3qa

b

4q

b

b

b

a a

a

aaaab

1q 2q 3q 2q 3q 4qa a a a b

A state is repeated in the walk of

(length at least 4)

aaaab

aaaab

1q 2q 3q 2q 3q 4qa a a a b

1q 2q 3q 4q

Pigeons:

Nests:(Automaton states)

Are more than

Walk of

The state is repeated as a result of

the pigeonhole principle

(walk states)

Repeated

state

Consider the walk of a “long’’ string:

1q 2q 3qa

b

4q

b

b

b

a a

a

aabb

1q 2q 3q 4q 4qa a b b

A state is repeated in the walk of

(length at least 4)

Due to the pigeonhole principle:

aabb

aabb

1q 2q 3q 4q

Automaton States

Pigeons:


Are more than

Walk of

The state is repeated as a result of

the pigeonhole principle

(walk states)1q 2q 3q 4q 4qa a b b

Repeated

state

iq...... ......

Repeated state

kw 21=

1 2 k

Walk of

iq.... iq.... ....1 2 ki j1+i 1+j

Arbitrary DFA

DFA of states#|| wIf ,

by the pigeonhole principle,

a state is repeated in the walk

In General:

1qzq

zq1q

w

mDFA of states#|| =w

1q 2q 1−mq mq

Walk of wPigeons:


Are

more

than

(walk states)

iq....iq.... ....

1q

iq.... ....

zq

A state is

repeated

The Pumping Lemma

Take an infinite regular language L

There exists a DFA that accepts L

mstates

(contains an infinite number of strings)

mw ||(number of

states of DFA)

then, at least one state is repeated

in the walk of w

q...... ......1 2 k

Take string with Lw

kw 21=

Walk in DFA of

Repeated state in DFA

Take to be the first state repeatedq

q....

w

There could be many states repeated

q.... ....

Second

occurrence

First

occurrence

Unique states

One dimensional projection of walk :

1 2 ki j1+i 1+j

Review

• DFA to RE using State Elimination Method


Regular Languages: Grand Unification

)(

)()(

DFAsL

NFAsLsNFAL

=

=−

)()( RELFAL =

(Parallel Simulation)

(Rabin and Scott’s work)

(Collapsing graphs;

Structural Induction)

(S. Kleene’s work)

)()( RGLFAL = (Construction)

(Solving linear

equations) study later)()( RELRGL =

Pumping Lemma for Regular Languages

• 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)

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

Second

occurrence

First

occurrence

1 2 ki j1+i 1+j

wOne dimensional projection of walk :

ix 1= jiy 1+= kjz 1+=

xyzw =We can write

zyxw =

q... ...

x

y

z

In DFA:

...

...

1 ki

1+ij

1+j

contains only

first occurrence of q

Observation: myx ||length number

of states

of DFA

Since, in no

state is repeated

(except q)

xy

Unique States

q...

x

y

...

1 i

1+ij

Observation: 1|| ylength

Since there is at least one transition in loop

q

y

...

1+ij

We do not care about the form of string z

q...

x

y

z

...

z may actually overlap with the paths of and x y

The string

is accepted

zxAdditional string:

q... ...

x z

...

Do not follow loopy

...

1 ki

1+ij

1+j

The string

is accepted

zyyx

q... ... ...

x z

Follow loop

2 times

Additional string:

y

...

1 ki

1+ij

1+j

The string

is accepted

zyyyx

q... ... ...

x z

Follow loop

3 times

Additional string:

y

...

1 ki

1+ij

1+j

The string

is accepted

zyx iIn General:

...,2,1,0=i

q... ... ...

x z

Follow loop

timesiy

...

1 ki

1+ij

1+j

Lzyx i Therefore: ...,2,1,0=i

Language accepted by the DFA

q... ... ...

x z

y

...

1 ki

1+ij

1+j

In other words, we described:

The Pumping Lemma !!!

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 ...,2,1,0=i

(critical length)

)0 :(

0) || ( ) || (

)( :,,

|| :

Lzxyii

ymxy

wxyzzyx

mwLw

i

=

For all sufficiently long strings (w)

There exists non-null prefix (xy)

and substring (y)

For all repetitions of the substring (y),

we get strings in the language.

In the book:

Critical length = Pumping lengthm p

Applications

of

the Pumping Lemma

Observation:

Every language of finite size has to be regular

Therefore, every non-regular language

has to be of infinite size (contains an infinite number of strings)

(we can easily construct an NFA

that accepts every string in the language)

Suppose you want to prove that

An infinite language is not regular

1. Assume the opposite: is regular

2. The pumping lemma should hold for

3. Use the pumping lemma to obtain a

contradiction

L

L

L

4. Therefore, is not regular L

Explanation of Step 3: How to get a contradiction

2. Choose a particular string which satisfies

the length conditionLw

3. Write xyzw =

4. Show that Lzxyw i = for some 1i

5. This gives a contradiction, since from

pumping lemma Lzxyw i =

mw ||

1. Let be the critical length form L

Note: It suffices to show that

only one string

gives a contradiction

Lw

You don’t need to obtain

contradiction for every Lw

Theorem: The language }0:{ = nbaL nn

is not regular

Proof: Use the Pumping Lemma

Example of Pumping Lemma application

Assume for contradiction

that is a regular languageL

Since is infinite

we can apply the Pumping Lemma

L

}0:{ = nbaL nn

Let be the critical length for

Pick a string such that: w Lw

mw ||and length

mmbaw =We pick

m

}0:{ = nbaL nn

L

with lengths

From the Pumping Lemma:

1||,|| ymyx

babaaaaabaxyz mm ............==

mkay k = 1,

x y z

m m

we can write zyxbaw mm ==

Thus:

=w

From the Pumping Lemma: Lzyx i

...,2,1,0=i

Thus:

mmbazyx =

Lzyx 2

mkay k = 1,


Lbabaaaaaaazxy = ...............2

x y z

km+ m

Thus:

Lzyx 2

mmbazyx =

y

Lba mkm +

mkay k = 1,

Lba mkm +

}0:{ = nbaL nnBUT:

Lba mkm +

CONTRADICTION!!!

1≥k

Our assumption that

is a regular language is not true

L

Conclusion: L is not a regular language

Therefore:

END OF PROOF

Regular languages

Non-regular language }0:{ nba nn

)( **baL

More Applications

of

the Pumping Lemma

The Pumping Lemma:

• Given a infinite regular language L

• there exists an integer (critical length)m

• for any string with length Lw mw ||

• we can write zyxw =

• with andmyx || 1|| y

• such that: Lzyx i ...,2,1,0=i

Regular languages

Non-regular languages *}:{ = vvvL R

Theorem: The language

is not regular


*}:{ = vvvL R },{ ba=



Since is infinite


L

*}:{ = vvvL R

mmmm abbaw =We pick

Let be the critical length for


mw ||length

m

and

*}:{ = vvvL R

L

we can write: zyxabbaw mmmm ==

with lengths:


ababbabaaaaxyz ..................=

x y z

m m m m

1||,|| ymyx

mkay k = 1,Thus:

=w


...,2,1,0=i

Thus: Lzyx 2

mmmm abbazyx = mkay k = 1,


Lababbabaaaaaazxy ∈.....................=2

x y z

km + m m m

y

Lzyx 2

Thus:

mmmm abbazyx =

Labba mmmkm +

mkay k = 1,

Labba mmmkm +

Labba mmmkm +

BUT:

CONTRADICTION!!!

1k

*}:{ = vvvL R

Our assumption that


L


Therefore:

END OF PROOF

Regular languages


}0,:{ = + lncbaL lnln


is not regular





Since is infinite


L


mmm cbaw 2=We pick

Let be the critical length of


mw ||length

m


and

L

We can write zyxcbaw mmm == 2

With lengths


cccbcabaaaaaxyz ..................=

x y z

m m m2

1||,|| ymyx

Thus:

=w

mkay k = 1,


...,2,1,0=i

Thus:

mmm cbazyx 2=

Lxzzyx ∈=0

mkay k = 1,


Lcccbcabaaaxz = ...............

x z

km− m m2

mmm cbazyx 2=

Lxz

Thus: Lcba mmkm − 2

mkay k = 1,

Lcba mmkm − 2

Lcba mmkm − 2

BUT:

CONTRADICTION!!!


1k

Our assumption that


L


Therefore:

END OF PROOF

Regular languages

Non-regular languages }0:{ ! = naL n


is not regular


}0:{ ! = naL n

nnn −= )1(21!



Since is infinite


L

}0:{ ! = naL n

!maw =We pick

Let be the critical length of


mw ||length

m

}0:{ ! = naL n

L

We can write zyxaw m == !

With lengths


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

x y z

m mm −!

1||,|| ymyx

mkay k = 1,Thus:

=w


...,2,1,0=i

Thus:

!mazyx =

Lzyx 2

mkay k = 1,


Laaaaaaaaaaaazxy = ..................2

x y z

km+ mm −!

Thus:

!mazyx = mkay k = 1,

Lzyx 2

y

La km +!

La km +!

!! pkm =+

}0:{ ! = naL nSince:

mk 1

There must exist such that: p

La km +!

La km +!

BUT:

CONTRADICTION!!!

}0:{ ! = naL n

mk 1

Our assumption that


L


Therefore:

END OF PROOF

Review

• Pumping Lemma

Closure Properties of Regular Languages

Regular Languages

• If ∑ is an alphabet, the set R of regular languages over ∑ is defined as follows.

• 1. The language ∅ is an element of R, and for every a ∈ , the language {a} is in R.

• 2. For any two languages L1 and L2 in R, the three languages L1 ∪ L2, L1L2, and L1* are elements of R.

1L 2L

21LLConcatenation:

*1LStar:

21 LL Union:

Are regularLanguages

For regular languages and we will prove that:

1L

21 LL

Complement:

Intersection:

RL1

Reversal:

We say: Regular languages are closed under

21LLConcatenation:

*1LStar:

21 LL Union:

1L

21 LL

Complement:

Intersection:

RL1

Reversal:

a

b

b

aNFA

Equivalent NFA

a

b

b

a

A useful transformation: use one accept state

2 accept states

1 accept state

NFA

Equivalent NFA

Singleacceptingstate

In General

NFA without accepting state

Add an accepting statewithout transitions

Extreme case

1LRegular language

( ) 11 LML =

1M


NFA 2M

2L


( ) 22 LML =

Regular language

NFA

Take two languages

}{1 baL n=

a

b

1M

baL =2ab

2M

0n

Example

Union

NFA for

1M

2M

21 LL

a

b

ab

}{1 baL n=

}{2 baL =

}{}{21 babaLL n =NFA for

Example

Concatenation

NFA for 21LL

1M 2M

NFA for

a

b ab

}{1 baL n=}{2 baL =

}{}}{{21 bbaababaLL nn ==

Example

Star Operation

NFA for *1L

1M

*1L

1

21

Lw

wwww

i

k

=

NFA for *}{*1 baL n=

a

b

}{1 baL n=

Example

Reverse

RL1

1M

NFA for

1M

1. Reverse all transitions

2. Make initial state accepting state and vice versa

1L

}{1 baL n=

a

b

1M

}{1nR

baL =

a

b

1M

Example

Complement

1. Take the DFA that accepts 1L

1M1L 1M1L

2. Make accepting states non-final, and vice-versa

}{1 baL n=

a

b

1M

ba,

ba,

}{*},{1 babaL n−= a

b

1M

ba,

ba,

Example

Intersection

1L regular

2L regular

We show 21 LL

regular

DeMorgan’s Law: 2121 LLLL =

21 , LL regular

21 , LL regular

21 LL regular

21 LL regular

21 LL regular

Example

}{1 baL n=

},{2 baabL =

regular

regular

}{21 abLL =

regular

1Lfor for 2LDFA

1M

DFA

2M

Construct a new DFA that accepts

Machine Machine

M 21 LL

M simulates in parallel and 1M 2M

Another Proof for Intersection Closure

States in M

ji pq ,

1M 2MState in State in

1M 2M

1q 2qa

transition

1p 2pa

transition

DFA DFA

11, pq a

New transition

MDFA

22, pq

0q

initial state

0p

initial state

New initial state

00, pq

1M 2MDFA DFA

MDFA

iq

accept state

jp

accept states

New accept states

ji pq ,

kp

ki pq ,

1M 2MDFA DFA

MDFA

Both constituents must be accepting states

Example:

}{1 baL n=

a

b

1M

0n

}{2mabL =

b

b

2M

0q 1q0p 1p

0m

2q2p

a

a

ba, ba,

ba,

00, pq

Automaton for intersection

}{}{}{ ababbaL nn ==

10, pqa

21, pq

b

ab11, pq

20, pq

a

12, pq

22, pq

b

ba,

a

b

ba,

b

a

M simulates in parallel and 1M 2M

M accepts string w if and only if:

accepts string w1M

and accepts string w2M

)()()( 21 MLMLML =

Applications of Regular Expression

• Regular Expressions in Lexical Analysis

• Regular Expressions in Web Search Engines

• Regular Expressions in Software Engineering

•Thank You….

Documents

UNIT II: Regular Expressions