Then the left side is clearly \an y sequence of 0's and - - PDF document

SLIDE 1 Extended RE's UNIX pioneered the use

• f

additional

• p

erators and notation for RE's:

• E

? =

• r

1

• ccurrences
• f

E =

• +

E .

• E

+ = 1

• r

more

• ccurrences
• f

E = E E

• .
• Char

acter classes [a

• z

GX ] = the union

• f

all (ASCI I) c haracters from a to z , plus the c haracters G and X , for example. Algebraic La ws for RE's If t w

• expressions

E and F ha v e no v ariables, then E = F means that L(E ) = L(F ) (not that E and F are iden tical expressions).

• Example:

1 + = 11

• .

If E and F are RE's with v ariables, then E = F (E is e quivalent to F ) means that whatev er languages w e substitute for the v ariables (pro vided w e substitute the same language ev erywhere the same v ariable app ears), the resulting expressions denote the same language.

• Example:

R + = RR

• .

With t w

• notable

exceptions, w e can think

• f

union (+) as if it w ere addition with ; in place

• f

the iden tit y 0, and concatenation, with

• in

place

• f

the iden tit y 1, as m ultiplication.

• +

and concatenation are b

• th

asso ciativ e.

• +

is comm utativ e.

• La

ws

• f

the iden tities hold for b

• th.
• ;

is the annihilator for concatenation.

• The

exceptions: 1. Concatenation is not comm utativ e: ab 6= ba. 2. + is idemp

• tent

: E + E = E for an y expression E . Chec king a La w Supp

• se

w e are told that the la w (R + S )

• =

(R

• S
• )
• holds

for RE's. Ho w w

• uld

w e c hec k that this claim is true?

• Think
• f

R and S as if they w ere single sym b

• ls,

rather than placeholders for languages, i.e., R = f0g and S = f1g.

✦

Then the left side is clearly \an y sequence

• f

0's and 1's. 1

SLIDE 2

✦

The righ t side also denotes an y string

• f

0's and 1's, since and 1 are eac h in L(0

• 1
• ).
• That

test is ne c essary (i.e., if the test fails, then the la w do es not hold.

✦

W e ha v e particular languages that serv e as a coun terexample.

• But

is it sucient (if the test succeeds, the la w holds)? Pro

• f
• f

Suciency The b

• k

has a fairly simple argumen t for wh y , when the \concretized" expressions denote the same language, then the languages w e get b y substituting an y languages for the v ariables are also the same.

• But

if y

• u

think that's

• b

vious, the b

• k

also has an example

• f

\RE's with in tersection" where the same statemen t is false.

• Also

| is it clear that w e can tell whether t w

• RE's

without v ariables denote the same language?

✦

Algorithm to do so will b e co v ered. Closure Prop erties

• Not

ev ery language is a regular language.

• Ho

w ev er, there are some rules that sa y \if these languages are regular, so is this

• ne

deriv ed from them.

• There

is also a p

• w

erful tec hnique | the pumping lemma | that helps us pro v e a language not to b e regular.

• Key

to

• l:

Since w e kno w RE's, DF A's, NF A's,

• NF

A's all dene exactly the regular languages, w e can use whic hev er represen tation suits us when pro ving something ab

• ut

a regular language. Pumping Lemma If L is a regular language, then there exists a constan t n suc h that ev ery string w in L,

• f

length n

• r

more, can w e written as w = xy z , where: 1. < jy j. 2. jxy j

• n.

2

SLIDE 3 3. F

• r

all i

• 0,

w y i z is also in L.

✦

Note y i = y rep eated i times; y = .

• The

alternating quan tiers in the logical statemen t

• f

the PL mak es it v ery complex: (8L)(9n)(8w )(9x; y ; z )(8i). Pro

• f
• f

Pumping Lemma

• Since

w e claim L is regular, there m ust b e a DF A A suc h that L = L(A).

• Let

A ha v e n states; c ho

• se

this n for the pumping lemma.

• Let

w b e a string

• f

length

• n

in L, sa y w = a 1 a 2

• a

m , where m

• n.
• Let

q i b e the state A is in after reading the rst i sym b

• ls
• f

w .

✦

q = start state, q 1 =

• (q

; a 1 ), q 2 = ^

• (q

; a 1 a 2 ), etc.

• Since

there are

• nly

n dieren t states, t w

• f

q ; q 1 ; : : : ; q n m ust b e the same; sa y q i = q j , where

• i

< j

• n.
• Let

x = a 1

• a

i ; y = a i+1

• a

j ; z = a j +1

• a

m .

• Then

b y rep eating the lo

• p

from q i to q i with lab el a i+1

• a

j zero times

• nce,
• r

more, w e can sho w that xy i z is accepted b y A. PL Use W e use the PL to sho w a language L is not regular.

• Start

b y assuming L is regular.

• Then

there m ust b e some n that serv es as the PL constan t.

✦

W e ma y no kno w what n is, but w e can w

• rk

the rest

• f

the \game" with n as a parameter.

• W

e c ho

• se

some w that is kno wn to b e in L.

✦

T ypically , w dep ends

• n

n.

• Applying

the PL, w e kno w w can b e brok en in to xy z , satisfying the PL prop erties.

✦

Again, w e ma y not kno w ho w to break w , so w e use x; y ; z as parameters.

• W

e deriv e a con tradiction b y pic king i (whic h migh t dep end

• n

n, x, y , and/or z ) suc h that xy i z is not in L. 3

SLIDE 4 Example Consider the set

• f

strings

• f

0's whose length is a p erfect square; formally L = f0 i j i is a squareg.

• W

e claim L is not regular.

• Supp
• se

L is regular. Then there is a constan t n satisfying the PL conditions.

• Consider

w = n 2 , whic h is surely in L.

• Then

w = xy z , where jxy j

• n

and y 6= .

• By

PL, xy y z is in L. But the length

• f

xy y z is greater than n 2 and no greater than n 2 + n.

• Ho

w ev er, the next p erfect square after n 2 is (n + 1) 2 = n 2 + 2n + 1.

• Th

us, xy y z is not

• f

square length and is not in L.

• Since

w e ha v e deriv ed a con tradiction, the

• nly

unpro v ed assumption | that L is regular | m ust b e at fault, and w e ha v e a \pro

• f

b y con tradiction" that L is not regular. Closure Prop erties Certain

• p

erations

• n

regular languages are guaran teed to pro duce regular languages.

• Example:

the union

• f

regular languages is regular; start with RE's, and apply + to get an RE for the union. Substitu ti

• n
• T

ak e a regular language L

• v

er some alphab et .

• F
• r

eac h a in , let L a b e a regular language.

• Let

s b e the substitution dened b y s(a) = L a for eac h a.

✦

Extend s to strings b y s(a 1 a 2

• a

n ) = s(a 1 )s(a 2 )

• s(a

n ); i.e., concatenate the languages L a 1 L a 2

• L

a n .

✦

Extend s to languages b y s(M ) =[ w in M s(w ).

• Then

s(L) is regular. Pro

• f

That Substitu ti

• n
• f

Regular Languages In to a Regular Language is Regular

• Let

R b e a regular expression for language L. 4

SLIDE 5

• Let

R a b e a regular expression for language s(a) = L a , for all sym b

• ls

a in .

• Construct

a RE E for s(L) b y starting with R and replacing eac h sym b

• l

a b y the RE L a .

• Pro
• f

that L(E ) = s(L) is an induction

• n

the heigh t

• f

(the expression tree for) RE R. Basis : R is a single sym b

• l,

a. Then E = R a , L = fag, and s(L) = s(fag) = L(R a ).

• Cases

where R is

• r

; easy . Induction : There are three cases, dep ending

• n

whether R = R 1 + R 2 , R = R 1 R 2 ,

• r

R = R

• 1

. W e'll do

• nly

R = R 1 R 2 .

• L

= L 1 L 2 , where L 1 = L(R 1 ) and L 2 = L(R 2 ).

• Let

E 1 b e R 1 , with eac h a replaced b y R a , and E 2 similarly .

• By

the IH, L(E 1 ) = s(L 1 ) and L(E 2 ) = s(L 2 ).

• Th

us, L(E ) = s(L 1 )s(L 2 ) = s(L). Application s

• f

the Substitu ti

• n

Theorem

• If

L 1 and L 2 are regular, so is L 1 L 2 .

✦

Let s(a) = L 1 and s(b) = L 2 . Substitute in to the regular language fabg.

• So

is L 1 [ L 2 .

✦

Substitute in to fa; bg.

• Ditto

L

• 1

.

✦

Substitute in to L(a

• ).
• Closure

under homomorphism = substitution

• f
• ne

string for eac h sym b

• l.

✦

Sp ecial case

• f

a substitution. Example: Homomorphism Let L = L(0

• 1
• ),

and let h b e a homomorphism dened b y h(0) = aa and h(1) = .

• Then

h(L) = L

• aa)
• =

all strings

• f

an ev en n um b er

• f

a's. Closure Under In v erse Homomorphism

• h

1 (L) = fw j h(w ) is in Lg. 5

SLIDE 6

• See

argumen t in course reader. Briey:

✦

Giv en homomorphism h and regular language L, start with a DF A A for L.

✦

Construct DF A B for h 1 (L), b y ha ving B go from state q to state p

• n

input a if ^

• q

; h(a)

• =

p. Closure Under Rev ersal

• The

r everse

• f

a string w = a 1 a 2

• a

n is a n

• a

2 a 1 .

✦

Denoted w R .

✦

Note

• R

= .

• The

rev erse

• f

a language L is the set con taining the rev erse

• f

eac h string in L.

• If

L is regular, so is L R .

✦

Pro

• f:

use RE's, recursiv e rev ersal as in course reader. 6