ไวยกรณ์ไม่พึงบริบท context-free grammars

ไวยกรณ์�ไม่พึ�งไวยกรณ์�ไม่พึ�งบร�บทบร�บท

CONTEXT-FREECONTEXT-FREEGRAMMARSGRAMMARS

ว�ตถุ�ประสงค์�ว�ตถุ�ประสงค์�A context-free grammar (CFG) is aA context-free grammar (CFG) is asimple recursive method of specifyingsimple recursive method of specifyinggrammar rules by which strings ingrammar rules by which strings ina language can be generated.a language can be generated.

To a large extent, they are capableTo a large extent, they are capableof specifyingof specifying

the syntax of high-level programming languages

and other formal languages.and other formal languages.

CONTEXT-FREE GRAMMARS

DESCRIBE THE LANGUAGESการนิ�ยาม่ภาษาการนิ�ยาม่ภาษาUsing recursive definition to describeUsing recursive definition to describeany languages (regular or nonregular).any languages (regular or nonregular).

Example, palindrome language.

Alphabet = { a, b}Rule1: , a, b PALINDROMERule2: For any S PALINDROME,

then aSa, bSb are in PALINDROME.Note: S can be considered as a variable,representing an arbitrary element of PALINDROME.


Alphabet = { a, b}Rule1: , a, b PALINDROMERule2: For any S PALINDROME,

then aSa, bSb are in PALINDROME.

S can be considered as a variable,representing an arbitrary element of PALINDROME.

Rule can be considered as a relation,representing a transformation, denoted by “”.

Rule1: S | a | bRule2: S aSa | bSb

Note: The vertical bar “|” means “or”.

, a and b are terminals.

S is a nonterminal or variable.

DESCRIBE THE LANGUAGESการนิ�ยาม่ภาษาการนิ�ยาม่ภาษา


ต�วแปต�วแปรรThe production of the grammar

S Start variable.

others represent some auxiliaryclass of strings.(class of recursive definition).



Alphabet Alphabet = {a, b} = {a, b}L is defined byL is defined bythe following rules:the following rules:

S S aSa | bSb | V aSa | bSb | VV V aTb | bTa aTb | bTaT T aT | bT | aT | bT |

Some words generated by these rules,S aSa aVa abTaa abaa.S aSa abSba abVba abaTbba abaabba.

NONPALINDROME

ต�วอยต�วอยางาง



A context-free grammar (CFG) is a4-tiple G = (V, , S, P) where

V and are disjoint finite sets,S is an element of V andP is a finite set of formulasof the form A ,

where A V and (V )*.S: start symbolV: set of variables or nonterminals: alphabet or set of terminals.P: set of grammar rules or productions.

นิ�ยาม่นิ�ยาม่DESCRIBE THE LANGUAGES

การนิ�ยาม่ภาษาการนิ�ยาม่ภาษาCONTEXT-FREE GRAMMARS

G = (V, , S, P)Alphabet = {a, b}L is defined by the following rules:

S aSa | bSb | VV aTb | bTaT aT | bT |

aVa G abTaa denotes abTaa can be obtainedfrom aVa by one of the productionsin G ( V bTa ).

S *G aaaba denotes aaaba is derived from Sin zero or more steps in G.




Let G = (V, , S, P) be a CFG.The language generated by G is

L(G) = { x * | S *G x }.A language L is a

Context-free language (CFL)

if there is a CFG G so that L = L(G).

A string x (V )* is a sentential form of G

if there is a derivation S *G x in G.

A string x * is a sentence of G

if there is a derivation S *G x in G.

ค์�าค์�าศั�พึธ์�ศั�พึธ์�



?To demonstrate that a CFG generatesTo demonstrate that a CFG generatesa language, we must show two things:a language, we must show two things:

1.1. every string in the language can beevery string in the language can bederived from the grammarderived from the grammar

2.2. no other string can be.no other string can be.

ป!ญหป!ญหาา



โจทย์� น่�าคิ�ดน่�าคิ�ด

?ก�าหนิดให& ภาษาบรรยายโดยก�าหนิดให& ภาษาบรรยายโดย

**

จงเขี,ยนิ จงเขี,ยนิ ไวยกรณ์�ไม่พึ�งไวยกรณ์�ไม่พึ�งบร�บทบร�บท

ขีองภาษานิ,-ขีองภาษานิ,-


?ก�าหนิดให& ภาษาก�าหนิดให& ภาษา 0n1n เม่./อ เม่./อ n n 00

{ { , 01, 0011, 000111, … }, 01, 0011, 000111, … }




?ก�าหนิดให& ภาษาบรรยายโดยก�าหนิดให& ภาษาบรรยายโดย

(0+001)*(10+111)*1(0+001)*(10+111)*1



L = { x {0,1}* | n0(x) = n1(x) }.

(ni(x) = number of i’s of x.)

Clearly that is in this language. Rule1: S One way to produce other strings in L, add one 0 and one 1.

Rule2: S 0S1Rule3: S 1S0

Consider strings which do not start and end with the same symbol.Example, 01110100 or 10010110These strings can be obtained by the concatenation of two strings in L.

Rule4: S SS.Example, 01110100 can be obtained from 01 and 110100.




ConclusionConclusionLet G = (V, Let G = (V, , S, P) with, S, P) with

S S 0S1 | 1S0 | SS | 0S1 | 1S0 | SS | ..

We have that L(G) We have that L(G) L. L.

L = { x {0,1}* | n0(x) = n1(x) }.





Now we will show that L L(G)where G = (V, , S, P) with

S 0S1 | 1S0 | SS | .By mathematical induction on |x|.Let d(x) = n0(x) – n1(x).

Basic step: if |x| = 0 and x L then d(x) = 0, x = .Of course, x L(G) by S .

L = { x {0,1}* | n0(x) = n1(x) }.

(ni(x) = number of i’s of x.)พึ�ส0จนิ�พึ�ส0จนิ�



Inductive step: for |x|≤ k, d(x) = 0, x L.Let x L and |x| = k+1.Case1: x begins with 0, ends with 1, then x = 0y1 and y L(G).

Then S *G y. Since S 0S1, we also have x L(G).Case2: x begins with 1, ends with 0, similar to the case 1.Case3: x begins with 1, ends with 1, then x = 1y1 and d(x) = 0.

d(1) = -1 and d(1y) = 1, then u such that u is a prefix of 1y,d(u) = 0. x = uw, u and w L and L(G). Since S SS, x L(G).

Case4: x begins with 0, ends with 0, similar to the case 3. Q.E.D.

L = { x {0,1}* | n0(x) = n1(x) }.




L = { x {0,1}* | n0(x) n1(x) }.





Let L0 be the language of stringswith more 0’s than 1’s.

Let L1 be the language of strings

with more 1’s than 0’s.

Then L = L0 L1.

Consider L0.Clearly that 0 L0, and for any x in L0, 0x and x0 are in L0.This suggests the production S 0 | 0S | S0.1 should be introduced in L0.Consider two strings x and y in L0,xy is at least two more 0’s than 1’s.Therefore adding a single 1 will still yield an element of L0.Then G0 = (V, , S, P)

S 0 | 0S | S0 | 1SS | S1S | SS1.It is clear that L(G0) L0.

L = { x {0,1}* | n0(x) n1(x) }.




Now prove that L0 L(G0).S 0 | 0S | S0 | 1SS | S1S | SS1.

By mathematical induction on |x| in L0.Basic step: |x| = 1 and x L0, x = 0.Since S 0, then x L(G0).Inductive step: |x|≤ k, d(x) > 0 then x L(G0).For x L0 and |x| = k+1 and d(x) > 0.Case x = 0y0. Rewrite x = wi1zi, for 1 i n; n = 1’s.

if d(wn)>0 then x is obtained by starting SS1S.( d(wn)>0 and d(zn)>0, wn and zn L0).

L = { x {0,1}* | n0(x) n1(x) }.




Inductive step: |x|≤ k, d(x) > 0 then x L(G0).For x L0 and |x| = k+1 and d(x) > 0.Case x = 0y0. Rewrite x = wi1zi, for 1 i n; n = 1’s.

if d(wn)>0 then x is obtained by starting SS1S.( d(wn)>0 and d(zn)>0, wn and zn L0).

if d(wn)0 then m that d(wm)0 (d(w1)>0)and also d(wm-1) = 1.

Then d(wm-1)> 0 and d(zm-1)>0.x is obtained by starting SS1S.

L = { x {0,1}* | n0(x) n1(x) }.




Context-free grammar of this languageis the following:

S A | BA 0 | 0A | 1AA | A1A | AA1B 1 | 1B | 0BB | B0B | BB0

L = { x {0,1}* | n0(x) n1(x) }.




ALGEBRAIC EXPRESSIONALGEBRAIC EXPRESSION

Context-free grammar of this languageis the following:

S S+S | S-S | SS | SS | (S) | a

where a is a terminal symbol.




For any context-free languages L1 and L2,the languages L1 L2, L1L2 and L1* arealso context-free languages.

ทฤษฎี,ทฤษฎี,บทบท

CONTEXT-FREE LANGUAGESภาษาไม่พึ�งบร�บทภาษาไม่พึ�งบร�บท


Let = { 0, 1 }.Consider two context-free grammars,G1 = (V1, , S1, P1) and G2 = (V2, , S2, P2).P1 = { S1 XA, X 0, A 0 }P2 = ( S2 XB, X 1, B 1 }.Find G = G1 G2.

L(G1) = { 00 } and L(G2) = { 11 }.S S1 | S2

S1 XA S2 XBX 0 X 1A 0 B 1

01 L(G1) L(G2)But S * 01.



แนิวค์�แนิวค์�ดด

L(G1) = { 00 } and L(G2) = { 11 }.S S1 | S2

S1 XA S2 YBX 0 Y 1A 0 B 1

Let = { 0, 1 }.Consider two context-free grammars,G1 = (V1, , S1, P1) and G2 = (V2, , S2, P2).P1 = { S1 XA, X 0, A 0 }P2 = ( S2 XB, X 1, B 1 }.Find G = G1 G2.



แนิวค์�แนิวค์�ดด

For any context-free languages L1 and L2,the languages L1 L2, L1L2 and L1* arealso context-free languages.




Case: L1L2 V = V1 V2 { S }P = P1 P2 { S S1 | S2 }

Case: L1L2 V = V1 V2 { S }P = P1 P2 { S S1S2 }

Case: L1* V = V1 { S }P = P1 { S S1S | }




??Is every regular language Is every regular language a context-free language ?a context-free language ?



By definition of regular language, , { }and { a } are regular languages, and regular is closed underunion, concatenation and Kleene’s star.

To prove this theorem, we have to show that anyregular can be produced by some context-free grammars.




Every regular language is a context-

free language.

From the regular expression, we haveS XY X represents (011+1)*

Y represents (01)*

X 011X | 1X |

Y 01Y |

Example: 01110101 can be derived from CFG.

Regular language (011+1)*(01)*Regular language (011+1)*(01)*ต�วอยต�วอยางาง



This language can be considered as 0i1i1m1k0k

for m > 0. S XYZ X represents 0i1i

Y represents 1m

Z represents 1k0k

X 0X1 | Y 1Y | 1Z 1Z0 |


Let L be the language { 0 i1 j0k | j > i + k }.

Find the context-free grammar for L.



Understanding the sentence begins with understanding its grammaticalstructure ( i.e., knowing how it isderived from the grammar rules ).

DERIVATION TREE

ค์วาม่เขี&าใจ



How the sentence0111011011010101can be generated?

Given a CFGS XYX 011X | 1X | Y 01Y |

S

X Y

011 X 01 Y

1 X 01 Y

01 011 X

011

ต&นิไม่&CONTEXT-FREE LANGUAGES

ภาษาไม่พึ�งบร�บทภาษาไม่พึ�งบร�บทCONTEXT-FREE GRAMMARS

DERIVATION TREE

ค์วาม่ก�ากวม่

AMBIGUOUSA context-free grammar G is ambiguousif there is at least one string in L(G) havingtwo or more distinct derivation trees.



DERIVATION TREE

Given a context-free grammar G with = { a, + } and S S + S | aThe sentence a+a+a is in L(G) which has two derivations.

SS+Sa+Sa+S+Sa+a+Sa+a+aSS+SS+S+Sa+S+Sa+a+Sa+a+a

This language is ambiguous.

ค์วาม่ก�ากวม่

AMBIGUOUS




?Let G be a context-free and x n

G y bea derivation in G where x can be written

x = y1A1y2A2…ykAkyk+1

with yi *.Show thatthere are strings pi ( V)* that satisfy

1. Ai ti pi.2. y = y1p1y2p2…ykpkyk+1

3. Sum of all ti (1ik) = n.

Given a context-free grammar G withS if E S | if E S else S | SThis language is ambiguous.

S

if E S

if E S else S

statement2

exp1

statement1exp2

S

if E S

if E S else S

statement2

exp1

statement1exp2

PARENTHESES CAN

BE USED TO REMOVE

THE AMBIGUITY




if exp1 if exp2 statement1 else statement2

Given a context-free grammar G withS S + S| S S | S | aThis language is ambiguous.

TheoremThe context-free grammar G1 with productions

S S + T | TT T F | FF (S) | a

is unambiguous.




NORMALNORMALFORMSFORMS

Ambiguity is one undesirable property of acontext-free grammar that we might wish toEliminate. In this section we discuss somemore straightforward ways of “improving”a grammar without changing the resultinglanguage.

NORMAL FORMS

ว�ตถุ�ประสงค์�ว�ตถุ�ประสงค์�

We try to eliminate “-productions”, of the formA , and unit productions,” in which onevariable is simply replaced by another.

REMARK : production G

If there are no -productions, || ||.If there are no unit productions,

if || = || then this step consists ofreplacing a variable by a single terminal.

NORMAL FORMS

ว�ตถุ�ประสงค์�ว�ตถุ�ประสงค์�

Let L be the length of the current string.Let T be the number of terminals of the current string.For any string x Language, of length k, then L+T = 2k.

That isthe number of steps S G x is at most 2k-1 steps.

NORMAL FORMS

ว�ตถุ�ประสงค์�ว�ตถุ�ประสงค์�We try to eliminate “-productions”, of the formA , and unit productions,” in which onevariable is simply replaced by another.

Given a context-free grammar G with productionsS ABCBCDAA CDB CbC a | D bD |

NORMAL FORMS


CONTEXT-FREE LANGUAGES -Production-Production

ELIMINATING THE -PRODUCTION

1. Replacing ABCBCDA by A1BC1BC2DA2

2. Create 32 productionsS A1BC1BC2DA2 and the otherswith or without A1, C1, C2, D and A2.

3. Adding A CD | C | DB Cb | bC aD bD | b

CONTEXT-FREE LANGUAGES

NORMAL FORMS

นิ�ยาม่นิ�ยาม่ -Production-Production

NULLABLE

A nullable variable in a context-freegrammar G = ( V, , S, P ) is defined

asfollows:

1. Any variable A for which P contains

the production A is nullable.2. If A B1B2…Bn and B1, B2, …, Bn

are nullables, then A is also nullable.3. No other variables in V are

nullable.

A is nullable if and only if A *G

N0 = { A V | P contains the production A };j = 0;Do j = j + 1;

Nj = Nj-1 { A | P contains A B that B Nj-1* }While Nj Nj-1;

Nj is the set of all nullable variables.

From the previous example, N0 = { C, D } and N1 = { A, C, D }.

NORMAL FORMS


อ�ลกอร�อ�ลกอร�ท�ม่ท�ม่

-Production-ProductionAlgorithm: FindNull

Given a CFG G=(V,,S,P), construct a CFG G1=(V,,S,P1)without -productions as follows:Initialize P1 to be P.Find all nullable variables in V, using FindNull.For every A in P,

add to P1 every production that can be obtainedform this one by deleting from or morethe occurrences in of nullable variables.

Delete all -production from P1.Delete any duplicates.Delete productions of the from A A.

NORMAL FORMSNORMAL FORMS


อ�ลกอร�อ�ลกอร�ท�ม่ท�ม่

-Production-ProductionAlgorithm: RemoveNull

ตั�วอย์�างGiven a context-free grammar G with productions

S ABC | ACBA ACD | CDB CbC a | D bD |



ตั�วอย์�าง



Given a context-free grammar G with productions

S ABC | ACBA ACD | CDB CbC a | D bD |

ตั�วอย์�างGiven a context-free grammar G with productions

S ABC | ACBA ACD | CD | AC | CB CbC a | D bD | b



ตั�วอย์�างGiven a context-free grammar G withproductions

S ABC | ACB | ABA ACD | CD | AC | C | AD | D | A | B Cb | bC a D bD | b



ตั�วอย์�างGiven a context-free grammar G withproductions

S ABC | ACB | AB | BC | CB | BA ACD | CD | AC | C | AD | D | AB Cb | bC a D bD | b



Let G = ( V, , S, P ) be a context-free

grammar, and let G1 = ( V, , S, P ) be

a context-free grammar obtained from

G by Algorithm RemoveNull.

Then G1 has no -productions, andL(G1) = L(G) – { }.

NORMAL FORMS


ทฤษฎี,บทฤษฎี,บทท

-Production-ProductionContext-free grammar

Unit productionsUnit productions

We try to eliminate “unit productions”, of the form A B, and A G* B.

Suppose that the context-free grammar G has no -productions.


NORMAL FORMS

นิ�ยาม่นิ�ยาม่ UnitUnit-Production-Production

An A-derivable variable in a context-free

grammar G = ( V, , S, P ) is defined as

follows:

1. Any variable B for which P containsthe production A B, then B is A-

derivable.2. If C is A-derivable, if C B is in P,

andif B A, then B is A-derivable.3. No other variables in V are A-

derivable.

B is A-derivable if and only if A *G B.


NORMAL FORMS

นิ�ยาม่นิ�ยาม่ UnitUnit-Production-Production

Algorithm RemoveUnitAlgorithm RemoveUnitREMOVING THE UNIT PRODUCTIONS FROM A CFG.REMOVING THE UNIT PRODUCTIONS FROM A CFG.

Given a CFG G=(V,,S,P) without -productions,construct a CFG G1=(V,,S,P1) having no unit productions as follows:

Initialize P1 to be P.For each A in V, find the set of all A-derivable variables.For every pair (A, B) such that B is A-derivable,

and every non-unit production B ,add the production A to P1.

Delete all unit productions from P1.

NORMAL FORMS

THEOREMTHEOREMLet G = ( V, , S, P ) be a context-

freegrammar with no -productions,and let G1 = ( V, , S, P ) be a

context-freegrammar obtained fromG by Algorithm RemoveUnit.

Then G1 contains no unit productions, and

L(G1) = L(G).

NORMAL FORMS

EXAMPLEEXAMPLEGiven a context-free grammar G without-productions, and with productions

S S + T | TT T F | FF (S) | a

S-derivable = { T, F }T-derivable = { F }

Productions in P1

S S + T | T | T F | (S) | aT T F | F | (S) | aF (S) | a

NORMAL FORMS

CHOMSKY NORMAL FORMCHOMSKY NORMAL FORMEliminating specific types of

productions,such as -productions and unit

productions,it may also be useful to impose

restrictionson the form of the remaining

productions.

Several types of “Normal forms” havebeen introduced; we shall present one

ofthem, the Chomsky normal form.

NORMAL FORMS

CNFCNFCHOMSKY NORMAL FORMCHOMSKY NORMAL FORM

Definition

A context-free grammar G = ( V, , S, P )

is in Chomsky normal form if everyproduction is of one of these two

types:

A BCA a

where A, B and C are variables, and a is

a terminal symbol.

NORMAL FORMS

CNFCNFCHOMSKY NORMAL FORMCHOMSKY NORMAL FORM

Construct a CFG in CNF

Given context-free grammar G = ( V, , S, P )

Construct G1 = ( V, , S, P1 ) with no-productions and with no unit

productions.

Construct G2 = ( V2, , S, P2 ) with productions

A B1B2…Bk , for k 2A a

Replace A B1B2…Bk byA B1X1, X1 B2X2, …, Xk-1 Bk-1Bk

NORMAL FORMS

EXAMPLEEXAMPLEGiven a context-free grammar G withproductions

S ABC | ACB | AB | BC | CB | BA ACD | CD | AC | C | AD | D | AB Cb | bC a D bD | b

S-derivable = { B }A-derivable = { C, D }B-derivable = C-derivable = D-derivable =

NORMAL FORMS


S ABC | ACB | AB | BC | CB | Cb | bA ACD | CD | AC | AD | a | bD | bB Cb | bC a D bD | b

S-derivable = { B }A-derivable = { C, D }B-derivable = C-derivable = D-derivable =

NORMAL FORMS


S ABC | ACB | AB | BC | CB | Cb | bA ACD | CD | AC | AD | a | bD | bB Cb | bC a D bD | b

Xb bXbc BCXcb CBXcd CD

NORMAL FORMS


S ABC | ACB | AB | BC | CB | CXb | bA ACD | CD | AC | AD | a | XbD | bB CXb | bC a D XbD | b


NORMAL FORMS


S AXbc | AXcb | AB | BC | CB | CXb | bA AXcd | CD | AC | AD | a | XbD | bB CXb | bC a D XbD | b


NORMAL FORMS

THEOREMTHEOREMFor any G = ( V, , S, P ) be a

context-freegrammar,there is a G1 = ( V1, , S, P1 ) bea context-free grammar in CNFso that

L(G1) = L(G) – { }.

NORMAL FORMS

ไวยกรณ์ไม่พึงบริบท context-free grammars

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