형식 언어

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Transcript 형식 언어

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Basic definitions
(1) alphabet
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a finite set of symbols.
ex) T1 = {ㄱ,ㄴ,ㄷ,...,ㅎ,ㅏ,ㅑ, … ,ㅡ,ㅣ}
T2 = {A,B,C, … ,Z}
T3 = {auto, break, case, … , while}
(2) string(or sentence, word)
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a sequence of symbols from some alphabet T.
(3) length
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the number of symbols in the string.
denoted by |ω|
꼭 기억해야 할 세 가지 개념
1. 언어의 정의
2. 문법의 정의 및 개념
3. 인식기의 의미
(4) empty string
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a string consisting of no symbols.
denoted by ε or λ.
(5) T* denotes the set of all strings of symbols over the
alphabet T, including the empty string.
T+ = T* - {ε}
T* : T star
T+ : T dagger
(6) Language is any set of strings over an alphabet.(Text p.40)
(or A Language L over the alphabet T is a subset of T*.)
L ⊆ T*
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Text p. 44
More definitions
(1) concatenation
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u = a1a2a3...an, v = b1b2b3...bm , u • v = a1a2a3...anb1b2b3...bm
u • v를 보통 uv로 표기.
uε= u = εu
∀u,v ∈ T*, uv ∈ T*.
|uv| = |u| + |v|
(2) an represents n a's.
a0 = ε
(3) the reversal of a string ω, denoted ωR is the string ω
written in reverse order:
i.e., if ω = a1a2...an then ωR = anan-1...a1.
(ωR)R=ω
(4) language product
LL' = {xy| x ∈ L and y ∈ L'}
(5) The powers of a language L are defined recursively by:
L0 = {ε}
Ln = LLn-1 for n  1.
(6) L* : reflexive transitive closure
= L0 ∪ L1 ∪ L2 ∪ ...∪ Ln ∪… =
(7) L+ : transitive closure
= L1 ∪ L2 ∪... ∪ Ln ∪ ...
= L* - L0
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Language
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문장(sentence)들을 원소로 갖는 집합
언어를 어떻게 표현할 것인가 ?
Grammar
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terminal : 정의된 언어의 알파벳
nonterminal :
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스트링을 생성하는 데 사용되는 중간 과정의 심볼
언어의 구조를 정의하는데 사용
grammar symbol (V)
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G = (VN, VT, P, S)
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VN : a finite set of nonterminal symbols
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VT : a finite set of terminal symbols
VN ∩ VT =  , VN∪ VT = V
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P : a finite set of production rules
α → β, α∈ V+, β∈ V*
lhs rhs
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S : start symbol(sentence symbol)
Text p. 46
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[예] G = ( {S, A}, {a, b}, P, S )
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P:
S → aAS
A → SbA
S→a
A → ba
⇒ S → aAS | a
A → SbA | ba | SS
Text p.47 [예제 2.8]
A → SS
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Derivation
1. ⇒ : "directly produce" or "directly derive"
if α → β∈ P and  , δ∈ V* then
 αδ ⇒  βδ
* : Suppose α1,α2,...,αn ∈ V* and α1 ⇒α2 ⇒ … ⇒αn,
2. ⇒
*
then α1 ⇒ αn
(zero or more derivations)
+
3. ⇒
: one or more derivations.
cf)
→ : production rule에서 사용.
“may be replaced by”
⇒ : derivation할 때 사용한다.
Inroduction to FL theory
[8/25]
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L(G) : Language generated by grammar G
L(G) = {ω | S ⇒*ω, ω ∈ VT*}
☞ ω is a sentential form of G if S ⇒* ω and ω ∈ V*.
* ω and ω V *.
ω is a sentence
of G if S ⇒
∈ T
P:
S → aA | bB | ε
A → bS
B → aS
*
S⇒
abba 유도 과정
S ⇒ aA (생성규칙 S → aA)
⇒ abS (생성규칙 A → bS)
⇒ abbB (생성규칙 S → bB)
⇒ abbaS (생성규칙 B → aS)
⇒ abba (생성규칙 S → )
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G1 = ( {S}, {a}, P, S ) 을 이용하여 L(G1)
P : S → a | aS
L (G1) = { an | n  1 }
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Language design
Text p. 46
generation
Grammar
design
Language
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G = ( {A, B, C}, {a, b, c}, P, A)
P : A → abc
Bb → bB
bC → Cb
aC → aa
L(G) = { anbncn | n  1 }
A → aBbc
Bc → Cbcc
aC → aaB
(===>) ex1) S → 0S1 | 01
ex2) S → aSb | c
ex3) A → aB
B → bB | b
ex4) A → abc
Bb → bB
bC → Cb
aC → aa
A → aBbc
Bc → Cbcc
aC → aaB
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Grammar Design
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L = { an | n  0 } 일 때 문법 :
A → aA | ε
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L = { an | n  1 } 일 때 문법 :
A → aA | a
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Embedded production
A → aAb
ex1) L1 = { anbn | n  0 }
ex2) L2 = { 0i1j | i  j, i,j  1 }
ex3) Constructs of Conventional PL
1) 파스칼 언어의 상수 정의 부분 :
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상수정의 부분은 CONST라는 예약어로 시작하며 하나의 상수 정
의는 a=b의 형태를 갖는다. 여기서, a는 identifier를 b는 상수를 나
타내는 terminal 심벌이다. 상수정의부분은 선택적이며 각각의 상
수정의는 ; 으로 구분한다.
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다음은 상수정의 부분의 예이다.
CONST ON = TRUE;
OFF = FALSE;
EPSILON = 1.0E-10;
2) C 언어의 정수 선언 부분 :
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정수선언 부분은 여러 개의 정수선언으로 구성되며 하나의 선언은
int a,a,a;와 같은 형태를 갖는다. 여기서 a는 임의의 identifier를 나
타낸다.
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그리고 ; 으로 각각의 선언을 구분한다. 예를 들어, int i,j; int sum;
과 같다.
※ 문법을 고안할 때, nonterminal의 이름은 구문 구조를
대변할 수 있는 명칭으로 쓰는 것이 바람직하다.
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In order to prove that a grammar generates a language L
i) Every sentence generated by the grammar is in L.
ii) Every string in L can be generated by the grammar.
교과서 55쪽
[예제 2.16]
G = ( { S }, { ( , ) }, {S → (S)S |ε}, S )
⇔ All strings of balanced parentheses.
proof) (=>) Every sentence derivable from S is balanced.
(<=) Every balanced string is derivable from S.
(=>) Every sentence derivable from S is balanced.
* ω: balanced)
(i.e., S ⇒ ω,
By induction on the number of steps in a derivation.
i) n = 1 일 때, S ⇒ ε, ε is surely balanced.
ii) Suppose that all derivations of fewer than n steps
produce balanced sentences.
iii) Consider a leftmost derivation of exactly n steps.
* (x)S ⇒* (x)y
S ⇒ (S)S ⇒
By the hypothesis x, y : balanced.
Thus (x)y balanced.
(<=) Every balanced string is derivable from S.
By induction on the length of a string.
i) |ω| = 0, S ⇒ ε
(the empty string is derivable from S.)
ii) Assume that every balanced string of length less than 2n is derived from S.
iii) Consider a balanced string ω of length 2n.
Let (x) : shortest prefix of ω being balanced.
Thus ω = (x)y, where x, y : balanced.
Since |x|, |y | < 2n, they are derivable from S by inductive hypothesis.
* (x)S ⇒
* (x)y = ω
Thus S ⇒ (S)S ⇒
Therefore, (x)y is also derivable from S.
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Noam Chomsky
According to the form of the productions.
α→β∈P
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Type 0 : No restrictions(unrestricted grammar)
Type 1 : Context-sensitive grammar(CSG).
 → β, |  |  | β|
Type 2 : Context-free grammar(CFG).
A → , where A : nonterminal,  ∈ V*.
Type 3 : Regular grammar(RG).
A → tB or A → t, (right-linear)
A → Bt or A → t, (left-linear)
where, A, B : nonterminal, t ∈ VT*.
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REL (Recursively Enumerable Language)
CSL (Context Sensitive Language)
CFL (Context Free Language)
RL (Regular Language)
Examples of Formal Language
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simple matching language :
double matching language :
mirror image language
:
palindrome language
:
parenthesis language
:
Lm = {anbn | n ≥ 0}
Ldm = {anbncn | n ≥ 1}
Lmi = {ωωR | ω ∈ VT*}
Lr = {ω | ω = ωR }
Lp = {ω | ω: balanced parenthesis}
CFL
CSL
CFL
CFL
CFL
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The Chomsky Hierarchy of Languages
regular language
context-free language
context-sensitive language
unrestricted language
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Languages & Recognizers
Grammar
Language
Recognizer
type 0
(unrestricted)
recursively
enumerable set
Turing Machine
type 1
(context-sensitive)
context-sensitive
language
Linear Bounded
Automata
type 2
(context-free)
context-free
language
Pushdown
Automata
type 3
(regular)
regular language
Finite Automata