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제 3 장 정규 언어

컴파일러 입문컴파일러 입문

Regular Language Page 2

목 차목 차

3.1 정규 문법과 정규 언어

3.2 정규 표현

3.3 유한 오토마타

3.4 정규 언어의 속성

정규 문법과 정규 언어정규 문법과 정규 언어 A study of the theory of regular languages is often justified by

the fact that they model the lexical analysis stage of a compiler.

Type 3 Grammar(N. Chomsky)

RLG : A → tB, A → t

LLG : A → Bt, A → t

where, A,B V∈ N and t V∈ T*.

It is important to note that grammars in which left-linear productions are intermixed with right-linear productions are not regular.

For example,

G : S → aR S → c R → Sb

L(G) = {ancbn | n 0} is a cfl.

Definition

(1) A grammar is regular if each rule isi) A aB, A a, where a VT, A, B VN.

ii) if S ε P, then S doesn't appear in RHS.

우선형 문법 A tB, A t 의 형태에서 t 가 하나의 terminal 로 이루어진 경우로 정규 문법에 관한 속성을 체계적으로 전개하기 위하여 바람직한 형태이다 .

(2) A language is said to be a regular language(rl) if it can be generated by a regular grammar.

ex) L = { anbm| n, m ≥1 } is rl.

S aS | aA

A bA | b

[Theorem] The production forms of regular grammar can be derived from those of RLG.(RLGRLG => => RGRG) (Text p.69)

(proof)

A tB, where t VT*.Let t = a1a2... an, ai VT.

A a1A1

A1 a2A2 . . .An-1 anB.

If t = , then A B (single production) or A (epsilon production).

⇒ These forms of productions can be easily removed.

(Text pp.175-181)

ex) S abcA ⇒ S aS1, S1 bS2 S2 cA

A bcA ⇒ A bA1, A1 cA

A cd ⇒ A cA1', A1' d

Right-linear grammar : A → tB or A → t,

where A, B ∈ VN and t ∈ VT*.

EquivalenceEquivalence1. 언어 L 은 우선형 문법에 의해 생성된다 .

2. 언어 L 은 좌선형 문법에 의해 생성된다 .

3. 언어 L 은 정규 문법에 의해 생성된다 .

정규 언어

[ 예 ] L = {anbm | n,m ≥ 1} : rl

S aS | aA

A bA | b

Text p. 70

토큰의 구조를 정의하는데 정규 언어를 사용하는 이유(1) 토큰의 구조는 간단하기 때문에 정규 문법으로 표현할 수 있다 .

(2) context-free 문법보다는 정규 문법으로부터 효율적인 인식기를 구현할 수 있다 .

(3) 컴파일러의 전반부를 모듈러하게 나누어 구성할 수 있다 .

(Scanner + Parser)

문법의 형태가 정규 문법이면 그 문법이 나타내는 언어의 형태를 체계적으로 구하여 정규 표현으로 나타낼 수 있다 .

if G = rg, L: re.

G Lderivation

A notation that allows us to describe the structures of sentences in regular language.

The methods for specifying the regular languages(1) regular grammar(rg)(2) regular expression(re)(3) finite automata(fa)

정규 표현정규 표현

Text p. 71 Definition :

A regular expression over the alphabet T and the language denoted by that expression are defined recursively as follows :

I. Basis : , , a T.(1) is a regular expression denoting the empty set.(2) is a regular expression denoting {}.(3) a where a T is a regular expression denoting {a}.

II. Recurse : + , • , *

If P and Q are regular expressions denoting Lp and Lq respectively, then

(1) (P + Q) is a regular expression denoting Lp U Lq. (union)

(2) (P • Q) is a regular expression denoting Lp Lq. (concatenation)

(3) (P*) is a regular expression denoting (closure)

{{ee} U } U LLpp U U LLpp22 U ... U U ... U LLpp

nn ... ...

Note : precedence : + < • < *

II. Nothing else is a regular expression.

ex) (0+1)* denotes {0,1}*. (0+1)*011 denotes the set of all strings of 0s and 1s

ending in 011.

Definition : if α is α regular expression, L(α) denotes the language associated with α. (Text p.72) Let a and b be regular expressions. Then, (1) L(α+ β) = L(α) L(β)

(2) L(α β) = L(α) L(β) (3) L(α*) = L(α)*

examples :

(1) L(a*) = {, a, aa, aaa, … } = {an | n 0}

(2) L((aa)*(bb)*b) = {a2nb2m+1| n,m 0}

(3) L((a+b)*b(a+ab)*) --- 연습문제 3.2 (3) - text p.115

= { b, ba, bab, ab, bb, aab, bbb, … }

Definition : Two regular expressions are equal if and only if they denote the same language. α= β if L(α) = L(β).

Axioms : Some algebraic properties of regular expressions. Let a, b and g be regular expressions. Then, (Text p.73)

A1. α+β = β+α A2. (α+β) +γ = α+ (β+γ)

A3. (αβ) γ = α (βγ) A4. α(β+γ) = αβ +αγ

A5. (β + γ) α = βα + γα A6. α+α=α

A7. α + = α A8. α = = α

A9. α = α = α A10. α* = +α•α*

A11. α* = ( + α)* A12. (α* )* = α*

A13. α* + α = α * A14. α* + α+ = α*

A15. (α + β)* = (α* β *) *

All of these identities(=Axioms) are easily proved by the definition of regular expression.

A8. α = = α

(proof) α = { xy | x Lα and y L }

Since y L is false, (x Lα and y L) is false.

Thus α = .

Definitions : regular expression equations.::= the set of equations whose coefficient

are regular expressions.

ex) α,β 가 정규 표현이면 , X = αX+β 가 정규 표현식이다 . 이때 , X 의 의미는 nonterminal 심볼이며 우측의 식이 그 nonterminal 이 생성하는 언어의 형태이다 .

▶ The solution of the regular expression equation

X = αX + β

When we substitute X = α*β in both side of the equation, each side of the equation represents the same language.

X = αX + β = α(α*β) + β = αα*β + β = (αα* + ε)β = α*β.

fixed point iteration

X = αX + β = α(αX + β) + β

= α2X + αβ + β = α2X + (ε + α)β...

= αk+1X + (ε + α + α2 + ... αk )β= (ε + α + α2 + ... + αk + ...)β = α*β.

Not all regular expression equations have unique solution. X = αX + β

(a) If ε is not in α, then X = α*β is the unique solution.

(b) If ε is in α, then X = α*(β + L) for some language L.

So it has an infinity of solutions.

⇒ Smallest solution : X = α*β.

ex) X = X + a : not unique solution

⇒ X = a + b or X = b*a or X = (a + b)* etc.

X = X + a X = X + a

= a + b + a = b*a + a

= a + a + b = (b* + ε) a

= a + b. = b*a

Finding a regular expression denoting L(G) for a given rg G.

L(A) where A VN denotes the language generated by A.

By definition, if S is a start symbol, then L(G)= L(S).

Two steps :1. Construct a set of simultaneous equations from G.

A aB, A a

L(A) = {a}·L(B) U {a} A = aB + a

In general, X α |β| γ ⇒ X = α + β + γ.

2. Solve these equations.X = αX + β X = α*β.

if G = rg, L: re.

G Lderivation

ex1) S aS S bR S ε R aS

L(S) = {a}L(S) U {b}L(R) U{ε} L(R) = {a}L(S)

ree: S = aS + bR + ε R = aS S = aS + baS + ε = (a + ba)S + ε = (a + ba)* ε = (a + ba)*

ex2) S aA | bB | b A bA | ε B bS

ree: S = aA + bB + b A = bA + ε ⇒ A = b*ε = b*

B = bS S = ab* + bbS + b = bbS + ab* + b = (bb)*(ab*+b)

Text p.1163.5(5)

풀이

ex5) S 0A | 1B | 0 ex6) X1 = 0X2 + 1X1 + ε A 0A | 0S | 1B X2 = 0X3 + 1X2

B 1B | 1 | 0 X3 = 0X1 + 1X3

ex7) A1 = (01* + 1) A1 + A2 ex8) A aB | bA

A2 = 11 + 1A1 + 00A3 B aB | bC

A3 = A1 + A2 + ε C bD | aB D bA | aB |ε

ex9) X α1X + α2Y + α3 ex10) PR b DL SL e

Y β1X + β2Y + β3 DL d ; DL | ε SL SL ; s | s

인식기인식기 (Recognizer)(Recognizer)☞ A recognizer for a language L is a program that takes

as input string x and answers “yes ” if x is a sentence of L and “no ” otherwise.

• Turing Machine• Linear Bounded Automata• Pushdown Automata• Finite Automata

a0a1a2 … aiai+1ai+2 … an

Finite State Control

input head

Auxiliary Storage

Definition : fa

A finite automaton M over an alphabet is a system (Q, , , q0, F)

where, Q : finite, non-empty set of states.

: finite input alphabet.

: mapping function.

q0 Q : start(or initial) state.

F ⊆ Q : set of final states.

mapping : Q x 2Q.

i,e. (q,a) = {p1, p2, ... , pn}

DFA , NFA.Regular Language Page 19

G = (VN, VT, P, S)

re : , , a, + , •, *

M = (Q, , , q0, F)

Text p. 78

유한 오토마타유한 오토마타

목차 목차 - FA- FA

1. DFA

2. NFA

3. Converting NFA into DFA

4. Minimization of FA

5. Closure Properties of FA

1. Deterministic Finite 1. Deterministic Finite Automata(DFA)Automata(DFA)

deterministic if (q,a) consists of one state.

We shall write "(q,a) = p " instead of (q,a) = {p} if deterministic.

If δ(q,a) always has exactly one number,

We say that M is completely specified.

extension of : Q x Q x ⇒ *

(q, ) = q

(q,xa) = ((q,x),a), where x * and a .

A sentence x is said to be accepted by M

if (q0, x) = p , for some p F.

The language accepted by M :

L(M) = { x | (q0,x) F }

ex) M = ( {p, q, r}, {0, 1}, , p, {r} )

: (p,0) = q (p,1) = p

(q,0) = r (q,1) = p

(r,0) = r δ(r,1) = r

1001 L(M) ?

(p,1001) = (p,001) = (q,01) = (r,1) = r F ． ∴ 1001 L(M).

1010 L(M) ?

(p,1010) = (p,010) = (q,10) = (p,0) = q F.

∴ 1010 L(M).

: matrix 형태로 transition table. ex)

rrrprq

10Input symbols

Definition : State (or Transition) diagram for automaton.

The state diagram consists of a node for every state and a directed arc from state q to state p with label a if (q,a) = p.

Final states are indicated by a double circle and the initial state is marked by an arrow labeled start.

p rstart

(1+01)*00(0+1)*

Astart

letter, digit

Sletter

Identifier :

Text p. 82Algorithm : w L(M).assume M = (Q, , , q0, F);

currentstate := q0; (* start state *)

get(nextsymbol);

while not eof do

begin currentstate := (currentstate, nextsymbol);

get(nextsymbol)

if currentstate in F then write(‘Valid String’)

else write(‘Invalid String’);

2. Nondeterministic Finite 2. Nondeterministic Finite Automata(NFA)Automata(NFA)

nondeterministic if (q,a) = {p1, p2, ..., pn}

In state q, scanning input data a, moves input head one symbol right and chooses any one of p1, p2, ..., pn as the next state.

ex) NFA (Nondeterministic Finite Automata) M = ( {q0,q1,q2,q3,qf}, {0,1}, , q0, {qf} )

if (q,a) = , then (q,a) is undefined.

δ 0 1 q0 {q1, q2} {q1, q3} q1 {q1, q2} {q1, q3} q2 {qf} q3 {qf} qf {qf} {qf}

To define the language recognized by NFA, we must extend .(i) : Q x * → 2Q

( q, ε ) = { q }

( q, xa ) = U (p,a), where a VT and x VT*.

p ( q, x )

(ii) : 2Q x * → 2Q

({p1, p2, ..., pk}, x) =

Definition : A sentence x is accepted by M if there is a state p in both F and (q0, x).

ex) 1011 L(M) ?

(q0, 1011) = ({q1,q3}, 011) = ({q1,q2},11)

= ({q1,q3},1) = {q1,q3,qf}

1011 L(M) ( {q∵ 1,q3,qf} ∩ {qf} Φ)

ex) 0100 L(M) ?

i=1 (pi,x)

Nondeterministic behavior

q1 q3 qf

If the number of states |Q| = m and input length |x| = n, then there are mn nodes.

In general, NFA can not be easily simulated by a simple program, but DFA can be simulated easily.

And so we shall see DFA is constructible from the NFA.

Text p. 863. Converting NFA into DFA3. Converting NFA into DFA

NFA : easily describe the real world.DFA : easily simulated by a simple program. ===> Fortunately, for each NFA we can find a DFA

accepting the same language.

Accepting Sequence(NFA)(q0, a1a2 ... an) = ({q1,q2, … ,qi}, a2a3 ... an)

... ...

= ({p1,p2, … ,pj}, ai ... an) ... ...

= {r1,r2, ... ,rk} Since the states of the DFA represent subsets of the

set of all states of the NFA, this algorithm is often called the subset construction.

[Theorem] Let L be a language accepted by NFA. Then there exists DFA which accepts L. Text p.86

(proof) Let M = (Q, , , q0, F) be a NFA accepting L.

Define DFA M' = (Q', , ', q0', F') such that

(1) Q' = 2Q, {q1, q2, ..., qi} ∈ Q', where qi ∈ Q.

denote a set of Q' as [q1, q2, ..., qi].

(2) q0' = {q0} = [q0]

(3) F' = {[r1, r2, ..., rk] | ri ∈ F}

(4) ' : ' ([q1, q2, ...,qi], a) = [p1, p2, ..., pj]

if ({q1, q2, ..., qj}, a) = {p1, p2, ..., pj}.

Now we must prove that L(M) = L(M’) i.e,

' (q0',x) F' (q0, x) ∩ F . we can easily show that by inductive hypothesis on the

length of the input string x.

ex1) M = ({q0,q1}, {0,1}, , q0, {q1}),

dfa M' = (Q', , ', q0', F'),

where Q' = 2Q = {[q0], [q1], [q0,q1]}

q0' = [q0]

F' = {[q1], [q0,q1]}

δ' :δ'([q0],0) = δ({q0},0) = {q0,q1} = [q0,q1]

δ'([q0],1) = {q0} = [q0]

δ' ([q1],0) = δ(q1,0) =

δ' ([q1],1) = δ(q1,1) = {q0,q1} = [q0,q1]

δ' ([q0,q1],0) = δ({q0,q1},0) = {q0,q1} = [q0,q1]

δ' ([q0,q1],1) = δ({q0,q1},1) = {q0,q1} = [q0,q1]

q0 {q0 , q1} {q0}

q1 {q0 , q1}

State renaming : [q0] = A, [q1] = B, [q0,q1] = C.

Since B is an inaccessible state, it can be removed.

’ 0 1A C AB CC C C

A Cstart

Definition : we call a state p accessible if there is w such that (q0, w) (p, ε) , where q0 is the initial state.

ex2) NFA DFA NFA : 0 1

q0 {q1,q2} {q1,q3}

q1 {q1,q2} {q1,q3}

q2 {qf}

q3 {qf}

qf {qf} {qf}

DFA : ’ 0 1q0 q1q2 q1q3

q1q2 q1q2qf q1q3

q1q3 q1q2 q1q3qf

q1q2qf q1q2qf q1q3qf

q1q3qf q1q2qf q1q3qf

Definition : - NFA M = (Q, , , q0, F) : Q ( {} ) 2Q

- CLOSURE : 을 보고 갈 수 있는 상태들의 집합 s 가 하나의 상태

-CLOSURE(s) = {s}{q|(p, )=q, p -CLOSURE(s)}

T 가 하나 이상의 상태 집합인 경우

-CLOSURE(T) =

ex) - NFA 에서 CLOSURE 를 구하기

CLOSURE (A) = {A, B, D}CLOSURE({A,C}) = CLOSURE(A) CLOSURE(C) = {A, B, C, D}

A Dstarta

-CLOSURE(q) ∪q∈T

Ex) - NFA DFA

A = [1,3,4], B = [2], C = [3,4], D = [4]

Dstarta b

1start

ε ε3

CLOSURE(1) = {1,3,4} [1,3,4]

CLOSURE(2) = {2} [2]

CLOSURE(3) = {3,4} [3,4]

CLOSURE(4) = {4} [4]

CLOSURE(3) = {3,4} [3,4]

Text p. 954. Minimization of FA4. Minimization of FA State minimization => state merge

Definition : ω * distinguishes q1 from q2 if (q1,ω) = q3, (q2,ω) = q4 and exactly one of q3, q4 is in F.

Algorithm : equivalence relation() ⇒ partition.

(1) : final state 인가 아닌 가로 partition.(2) : input symbol 에 따라 다른 equivalence class 로 가는가 ? 그 symbol 로 distinguish 된다고 함 . :(3) : 더 이상 partition 이 일어나지 않을 때까지 .

The states that can not be distinguished are merged into a

single state.

Text p. 119 3.11Ex)

: {A,F}, {B, C, D, E} : 처음에 final, nonfinal 로 분할한다 . : {A,F}, {B,E}, {C,D} : {B, C, D, E} 가 input symbol b 에

의해 partition 됨 : {A,F}, {B,E}, {C,D}.

[AF] [BE]

[BE] [CD]

[CD] [AF]

How to minimize the number of states in a fa.

<step 1> Delete all inaccessible states;<step 2> Construct the equivalence relations;<step 3> Construct fa M’ = (Q’, , ’, q0’, F’),

(a) Q’ : set of equivalence classes under Let [p] be the equivalence class of state p under .

(b) ’([p],a) = [q] if (p,a) = q. (c) q0’ is [q0].

(d) F' = {[q] | q F}.

Definition : M is said to be reduced. if (1) no state in Q is inaccessible and

(2) no two distinct states of Q are indistinguishable

ex) Find the minimum state finite automaton for the language specified by the finite automaton M = ({A,B,C,D,E,F}, {0,1}, , A, {E,F}),

where is given by

: {A, B, C, D}, {E, F} : {A}, {C}, {B, D}, {E, F}

BEAFDD

CFAEFE

0 1Text p. 119 3.11(2)

[A]=p[C]=q

[B,D]=r s s

[E, F]=s r s

ProgrammingProgramming<< 연습문제 연습문제 3.20>3.20> --- 교과서 121 쪽

Input Design Data Structure

Minimization of DFA

NFA to DFA DFANFAReduced

5. Closure properties of FA5. Closure properties of FA

[Theorem] If L1 and L2 are finite automaton languages (FAL),

then so are (i) L1 U L2 (ii) L1 • L2 (iii) L1*.

(proof) M1 = (Q1, , 1, q1, F1)

M2 = (Q2, , 2, q2, F2), Q1 Q2 = (∵ renaming)

(i) M = (Q1 U Q2 U {q0}, , , q0, F)

where, (1) q0 is a new state.

(2) F = F1 U F2 if L1 U L2.

F1 U F2 U {q0} if L1 U L2.

(3) (a) (q0,a) = (q1,a) U (q2,a) for all a .

(b) (q,a) = 1(q,a) for all q Q1, a .

(c) (q,a) = 2(q,a) for all q Q2, a .

새로운 시작 상태를 만들어 각각의 fa 에 마치 각 fa 의 시작 상태에서 온 것처럼 연결한다 . 그리고 를 인식하면 새로 만든 시작 상태도 종결 상태로 만든다 .

ex) p.98 [ 예 28]

(ii) M = (Q1 U Q2, , , q0, F)

(1) F = F2 if q2 F2

F1 U F2 if q2 F2

(2) (a) (q,a) = 1(q,a) for all q Q1 - F1.

(b) (q,a) = 1(q,a) U 2(q2,a) for all q F1.

(c) (q,a) = 2(q,a) for all q Q2.

M1 의 종결 상태에서 M2 의 시작 상태에서 온 것처럼 연결한다 . 그리고 M1 의 시작 상태가 접속한 오토마타의 시작 상태가 된다 .

A Bstart

0 M1 : => 01*

X Ystart

0 M2 : => 01*

A Ystart

0 M1 •M2 : => 01*01*B

정규 언어의 속성정규 언어의 속성

Regular grammar (rg)

Finite automata (fa) Regular expression (re)

※ re ===> fa : scanner generator

목 차 목 차

1. RG & FA

2. FA & RE

3. Closure Properties of Regular

Language

4. The Pumping Lemma for Regular

Language

1. RG & FA1. RG & FA Given rg, there exists a fa that accepts the same

language generated by rg and vice versa.

Given rg, G = (VN, VT, P, S) , construct M = (Q, , , q0, F).

(1) Q = VN U {f}, where f is a new final state.

(2) = VT.

(3) q0 = S.

(4) F = {f} if L(G)

= {S, f} otherwise.

(5) : if A aB P then (A,a) B.

if A a P then (A,a) f.

(proof)

If is accepted by fa then it is accepted in some sequence of

moves through states, ending in f.

But if (A,a) = B and B f , then A aB is a productions.

Also if (A,a) = f then A a is a production.

So we can use the same series of productions to generate

Thus S => .

ex) p.101 [ 예 29]

fa rgGiven M = (Q, , , q0, F), construct G = (VN, VT, P, S).

(1) VN = Q

(2) VT =

(3) S = q0

(4) P : if (q,a) = r then q ar.

if p F then p .

ex) p rstart

L(P)=(1+01)*00(0+1)*

p 1p | 0q q 1p | 0r r 0r | 1r | ε

2. FA & RE2. FA & RE fa rg re ex) p.118 3.10 (1)

A Dstart

A = bA + aBB = aB + bCC = aB + bDD = aB + bA + = A +

A = (a+b)*abb

re fa (※ scanner generator)For each component, we construct a fa inductively :

1. basis

2. induction - combine the components.

i f ε :ε

i fa : a

(1) N1 + N2

(3) N*

i f ε ε

(2) N1 •N2

N1i N2 f

Definition : The size of a regular expression is the number of operations and operands in the expression.

ex) size(ab + c*) = 6

decomposition:

The number of state is at most twice the size of the expression.

(∵ each operand introduces two states and each operator introduces at

most two states.)

The number of arcs is at most four times the size of the expression.

Simplifications : p.106

※ -arc 로 연결된 두 상태는 소스 상태에서 나가는 다른 arc 가

없으면 같은 상태로 취급될 수 있다 .

ex) p.105 [ 예 31]

re -NFA ( 간단화 ) DFA ex) p.109 [ 예 33]

The following statements are equivalent :

1. L is generated by some regular grammar.2. L is recognized by some finite automata.3. L is described by some regular expression.

A B ε

p.120 3.14

(1) (b + a(aa* b)*b)*

(2) (b + aa + ac + aaa + aac)*

(3) a(a+b)*b(a+b)*a(a+b)*b(a+b)*

Za a b

a, b a, b a, b

3. Closure Properties of Regular 3. Closure Properties of Regular LanguageLanguage

[Theorem] If L1 and L2 are regular languages,

then so are (i) L1 U L2 , (ii) L1L2, and (iii) L1*.

(proof) (ii) Since L1 and L2 are rl, rg G1 = (VN1, VT1, P1, S1) and rg G2 = (VN2,VT2, P2, S2), such that L(G1) = L1 and L(G2) = L2.

Construct G=(VN1 U VN2,VT1 U VT2, P, S1) in which P is defined as follows :

(1) If A aB P1, A aB P.

(2) If A a P1, A aS2 P.

(3) All productions in P2 are in P.

We must prove that L(G) = L(G1) . L(G2).

Since G is rg, L(G) is rl. Therefore L(G1) . L(G2) is rl.

ex) P1 : S aS | bA A aA | a

P2 : X 0X | 1Y Y 0Y | 1

P : S aS | bA A aA | aX X 0X | 1Y Y 0Y | 1

(iii) L : rl, rg G = (VN, VT, P, S) such that L(G) = L.

Let G' = (VN U {S'}, VT, P', S')

P' : (1) If A aB P, then A aB P'.

(2) If A a P, then A a, A aS' P'.

(3) S' S ┃ε P'.

We must prove that L(G') = (L(G))*.

L(G), S => . S' => S => wS' => w*S' => w*.

∴ (L(G))* = L(G').

ex) P : S aS, S bP' : S aS, S b, S bS', S' S, S'

.notenote P : S = aS + b = a*b

P' : S = aS + b + bS' = a*(b+bS') = a*b + a*bS'

∴ S' = S + = a*bS' + a*b + = (a*b)*(a*b + ) = (a*b)*(a*b) + (a*b)* = (a*b)*

4. The Pumping Lemma for Regular 4. The Pumping Lemma for Regular LanguageLanguage

It is useful in proving certain languages not to be regular.

[Theorem] Let L be a regular language. There exists a constant p such that if a string w is in L and |ω| p, then w can be written as xyz, where 0 < |y| ≤p and xyiz L for all i 0. (proof) Let M = (Q, , , q0, F) be a fa with n states such that L(M) = L. Let p = n. If L and |ω| n, then consider the sequence of configurations entered by M in accepting w. Since there are at least n+1 configurations in the sequence, there must be two with the same state among the first n+1 configurations.

Thus we have a sequence of moves such that (q0,xyz) = (q1,yz) = δ(q1,z) = qf F for some q1.

But then, (q0,xyiz) = (q1,yiz) = (q1,yi-1z) = ... = (q1,z) = qf F. Since w = xyz L, xyiz≤ L for all i 0.

Consequently, we say that “finite automata can not count”,meaning they can not accept a language which requires

thatthey count the number exactly.

ex) L = {0n1n | n ≥1} is not type 3.

(Proof) Suppose that L is regular.

Then for a sufficiently large n, 0n1n can be written as xyz

such that | y| 0 and xyiz L for all i 0.

If y 0+ or y 1+ , then xz = xy0z L. If y 0+1+, then xyz L.

We have a contradiction, so L can not be regular.

ancbn not rl

ancbm rl

연습문제 연습문제 3.5 3.5 풀이풀이교과서 교과서 116116 쪽쪽

A = aB + bA ……………………… (1)B = aB + bC ……………………… (2)C = bD + aB ……………………… (3)D = bA + aB + ……………………… (4)

식 (4) 에서 bA + aB = aB + bA = A 이므로D = A + ……………………… (5)

식 (3) 에 식 (5) 를 대입C = b(A + ) + aB = bA + aB + b

= A + b ……………………… (6)식 (2) 에 식 (6) 을 대입

B = aB + b(A + b) = aB + bA + bb = A + bb ……………………… (7)식 (1) 에 식 (7) 을 대입

A = aB + bA = a(A + bb) + bA = aA + abb + bA = (a + b)A + abb = (a+b)*abb

L(G) = (a+b)*abb

http://plac.dongguk.ac.kr 제 3 장 정규 언어 컴파일러 입문. regular language page 2 목...

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