제 4 장. Regular Language의특성

학습목표

정규언어의일반적특성에대해이해하고, 주어진언어가 regular인지판별하는방법학습

Closure Property & Pumping Lemma

개요

정의→표현→유용성→특성?

Every formal language is regular?accepts by some complex finite automaton?

① operations on RL : set operations, changing operations, closure question

② ability to decide on certain properties : finite or not?

③ regular language or not?DFA, RE, RG

다양한 language 집합의분별도구

no

general properties를만족하나?

할수있는일, 없는일

Closure Properties : Set Operations (1)

① closure under simple set operationsRL is closed under union, intersection, concatenation, complementation, star-closure

Thm

pf:

RL:,,,,RL:, 1121212121

∗∩∪→ LLLLLLLLLL

⎩⎨⎧

∈−∈∈∈∴

−Σ=→Σ=

⋅+==∃

∗

∗∗

∗

LwifFQwqLwifFwq

FQqQMFqQM

rrrrrrLLrLLrr

LL

),(),(fn total:

),,,,(ˆ),,,,(

ationcomplement,, definitionby ),(),(such that,

0

0

00

12121221121

11

δδδ

δδ

예) RL is closed under union

한두개의 instance가아닌모든원소의특성을설명하기위한방법

Intersection은? 좀더복잡함!

MwiffLLw

FpFqpqF-

papqaq

pqapq-pMqMpqPQQ-

FpqQM

FpPMMLLFqQMMLL

jiji

lj

ki

lkji

j

iji

ˆ by accepted

,),(:ˆ

),(),(

whenever),()),,((ˆ in

,in),,(:ˆ

where)ˆ),,(,ˆ,,ˆ(ˆcombined aconstruct

),,,,(where)(),,,,(where)(

onintersecti

21

21

2

1

2

1

00

202222

101111

∩∈∴

∈∈→

⎜⎜⎝

⎛==

=

×=

Σ=⋅

⎜⎜⎝

⎛Σ==Σ==

⋅

δδ

δ

δ

δδ

Closure Properties : Set Operations (2)

Constructive proof의또다른등장, 여러곳에서유용!

Closure Properties : Reversal

2121

2121

differenceunder closed )

law sDeMorgan':cf)

LLLL

LLLL

∩=−

∪=∩

예

⎪⎩

⎪⎨

⎧→→

→→

directionreverseinitialfinal

finalinitial accepts nfa accepts NFA'NFA' NFA RL :L

wiffwR

Thm RL : closed under reversal

단순히증명만을원한다면이걸로…

단순한 set operation이외에좀더복잡한연산에대한 closure 특성이해필요

Closure Properties : Homomorphism (1)

② Closure under other operations : homomorphism, right quotientDef 1:

}:)({)(image chomomorphi,on language:

)()()()(:

smhomomorphi alphabets,:,

21

21

LwwhLhL

ahahahwhaaawh

n

n

∈=Σ

⎜⎜⎝

⎛==Γ→Σ

ΓΣ∗

L

L

abbbcababahbbcbhabahhcbaba

====Γ=Σ

)()(,)(:},,,{},,{)1 예

∗∗∗∗ +=→+= ))()((:)())((: 1 dbccdbccbdcdbccrLhaabarL

bdcbhdbccahhdcbba ===Γ=Σ )(,)(:},,,{},,{)2 예

},{)(},,{ abbbcabababLhabaaaL ==

Closure Properties : Homomorphism (2)

Pf:

regular:)(RL:sm,homomorphi: LhLh →

)( such that ))(())(()()(

of for )(by )(,with RL:.

whvLwrhLvrhLwhrLw

raahrhrLer

=∈∃→∈∀∈→∈∀

Σ∈

Thmsmshomomorphiarbitrary under closed : RL

Closure Properties: Right Quotient (1)

⎜⎜⎝

⎛

≥=∪≥≥=

}1:{}{}0,1:{)

2

1

mbLbamnbaL

m

mn예

q0

a,ba,b

a

bba

a

bb

a

q1

q3 q5

q2

q4

)/(),( such that state final a walk to,

210

2

LLFqxqxLvvQq

∈=∀∈→∃∈∀

δ

}0,1:{/ 21 ≥≥= mnbaLL mn

} somefor ,:{/ with ofquotient right alphabet, sameon language ,

2121

2121

LyLxyxLLLLLL

∈∈=

Def 2:

① take all strings in L1 having a suffix belonging to L2② for the string, removing this suffix belongs to L1/L2

기본정의가좀이해하기어려우려나?

Closure Properties: Right Quotient (2)

Thm

Pf:

⎜⎜⎝

⎛→≠∩

Σ=

∈=∈∃∈∀

Σ=

Σ==

∗

ˆ to add)(),,,,(

),( such that determine ,

)ˆ,,,,(ˆ),,,,( where),(

2

2

0

01

FqMLLFqQM

FqyqLyQq

FqQM

FqQMMLL

ii

ii

fii

φδ

δ

δ

δ

regular :/ regular :,quotientright under closed : RL

2121 LLLL →

증명도좀해보고…

211

2

0

),(such that on,constructiby

ˆ),(,ˆby accepted ii)

/LLx,LxyFyqLy

FqxqMx

∈∈∴∈∈∃

∈=∀∗

∗

δ

δ

FxqxMFq

Fyq

qxq

FxyqLxyLyLLx

LLML

ˆ),( accepts ˆ,ˆon constructiby

),(

,),(

),(such that ,/

:/)ˆ( i)

0

0

01221

21

∈∈

∈

=∃

∈⇒∈∈∃∈

=

∗

∗

∗

∗

δ

δ

δ

δ

Q

Closure Properties: Right Quotient (3)증명의마무리!

따라서, 는 regular!21 / LL

Closure Properties: Right Quotient (4)

)(

)(/

2

121∗

∗∗

=

=

abLL

baaaLLLL for

φφφ

φ

=∩≠=∩≠=∩

=∩

23

22

21

20

)(}{)(}{)(

)(

LMLaLMLaLML

LML

∗∗∗∗∗ ⎯→⎯+ baabaaaba

3

0 1 2

a,b

b

aab

b

a

예)

예를가지고이해해봅시다

Closure Properties: Exercises 4.1

- 2 : constructive proof를이용한 NFA의구축

- 6 : 집합의특성을이용해볼것

- 10 : Right quotient를제대로이해하고있는지체크하는쉬운문제

- 15 : Right quotient의응용능력 (left quotient)…그리쉽지만은않을걸…

- 20 : 이런류의문제를접해본다는의미이상은없음

Elementary Questions about RL

Thm 1

Thm 2

Thm 3

.acceptance dfa.in not or whether

determinigfor algorithman , ,on RLGiven Lw

wL ∃Σ∈Σ ∗

infinite :ⓕ ⓘ Cycle emptynot :ⓕ ⓘpath simple :dfa ofgraph transition

infinite finite, empty, is RL awhether gdetermininfor algorithman

→

→

∃

213

21213

2121

:)()(

.,,

LL iffLLLLLL

LLLL

==∩∪∩=

=∃

closureby regular

whether determine to algorithm anGiven

φ

Elementary Questions : Exercises 4.2

- 5 : palindrome이 regular인가? 일단 DFA를만들어보고…

- 9 : 좀어려운문제지만, 이것도일단 DFA를만들어보고…

- 12 : 약간의트릭이필요하긴한데, 뒤에답안이있군!

Identifying Nonregular Languages

language is regular only if the information that has to be remembered at any stage is strictly limited

Pigeonhole Principle– If put n objects into m boxes, n>m

then at least one box must have more than one item in it

상자의수보다더많은공을넣어야한다면?어딘가 1개이상의공이들어가는상자가있을수밖에!

Nonregular Languages: 예

예)

regular becannot ifonly accepts that scontradict

),()),,((),(

),( accepts Since

),(&),(with such that state some principle, pigeonholeby

),, },,{ ,(DFA regular. is Suppose

00

00

0

LmnbaM

qbqbaqbaq

FqbqbaM

qaqqaqmnq

FqbaQmL

nm

fnnmnm

fn

nn

mn

∴=

===∴

∈=

==

≠∃=∃

∗∗∗∗

∗

∗∗

δδδδ

δ

δδ

δ

regular?:}0:{ ≥= nbaL nn

pf)

대강은이해할수있겠는데, 이를좀더정형화한틀이필요하지않을까?

Pumping Lemma (1)

Idea : in a transition graph with n vertices, any walk of length n or longer must repeat some vertex (contain a cycle)

L,2,1,0 in also is such that 1||,||

with as decomposedbecan |w|with such thatinteger positive

RL infinite:

=∀=

≥≤=

≥∈∃

iLzxywymxyxyzw

mLwm

L

ii

Thm

그것이바로펌핑렘마!!!

Finite한상태로 infinite한스트링을표현하자니어쩔수없이어떤상태는한번이상사용될수밖에없겠죠!

Pumping Lemma (2)

Pf

L

LLL

L

Q

L

,),(,),(1||,1 with

),(,),(,),(such that of,, substrings and

,,,,,,,,move. n later tha no repeated bemust state oneleast at

entries 1|| of ,,,,,for states ofset infinite is 1||such thatin

,,, states with recognizesDFA regular :

200

0

0

0

10

ff

frrrr

frrji

th

fji

n

qzxyqqxzqymn|xy|

qzqqyqqxqwzyx

qqqqqqn

wqqqqwLnmwLw

qqqLL

==∴

≥=+≤

===

∃

∴

++=≥

∃→

∗∗

∗∗∗

δδ

δδδ

그걸한번증명해봅시다

Pumping Lemma: 예

예1)

• pumping lemma holds for every w∈L & every i→ if it is violated even for one w or i, then language cannot be regular

regularnot is }0:{ ≥= nbaL nn

false bemust regular is lemma pumping contadicts

0 usingby obtained string then , Supposes' ofentirely consist must substring Choose

hold.must lemma pumping regular : that Assume

0

LLbaw

ik |y|aynm

L

mkm

→∉=

==→=

→

−

증명도증명이지만이를다양한문제에적용할수있어야함

Pumping Lemma: 예

예2)

예3)

예4)

예5)

regularnot is }:{ *∑∈= wwwL R

regularnot is )}()(:{ * wnwnwL ba <∑∈=

regularnot is }0,:){( ≥>= kknaabL kn

regularnot is }0:{ ! ≥= naL n

Pumping Lemma: 예예

• closure property to show that L is not regular

예2)

• 그밖의여러문제들… 진득한연습, 또연습

is}0,0:{ ≥≥= + kncbaL knkn예1)

arnot regul then

Takesm.homomorphiunder closurearnot regul

:}0:{}0:{)(

)(,)(,)(

≥=≥+=

===++ icakncaLh

cchabhaahiiknkn

regularnot is }:{ lnbaL ln ≠=

Pumping Lemma : Exercises 4.3

- 4 (b), (d), (f) : pumping lemma의적용능력함양문제

- 14 (8) : pumping lemma의직접적인적용은좀어려울듯. 답은… false

- 15 (a), (c), (e) (9) : 정규언어에대한직관을키울수있는문제들.

- 21 (15) : finite 언어와 infinite 언어의차이…