ดาวน์โหลดงานนำเสนอ
งานนำเสนอกำลังจะดาวน์โหลด โปรดรอ
1
ไวยกรณ์ไม่พึงบริบท CONTEXT-FREE GRAMMARS
2
the syntax of high-level programming languages
CONTEXT-FREE GRAMMARS วัตถุประสงค์ A context-free grammar (CFG) is a simple recursive method of specifying grammar rules by which strings in a language can be generated. To a large extent, they are capable of specifying the syntax of high-level programming languages and other formal languages.
3
การนิยามภาษา Alphabet = { a, b}
CONTEXT-FREE GRAMMARS การนิยามภาษา DESCRIBE THE LANGUAGES Using recursive definition to describe any languages (regular or nonregular). Example, palindrome language. Alphabet = { a, b} Rule1: , a, b PALINDROME Rule2: For any S PALINDROME, then aSa, bSb are in PALINDROME. Note: S can be considered as a variable, representing an arbitrary element of PALINDROME.
4
การนิยามภาษา Alphabet = { a, b} Rule1: S | a | b
CONTEXT-FREE GRAMMARS การนิยามภาษา DESCRIBE THE LANGUAGES Alphabet = { a, b} Rule1: , a, b PALINDROME Rule2: 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 | b Rule2: S aSa | bSb Note: The vertical bar “|” means “or”. , a and b are terminals. S is a nonterminal or variable.
5
การนิยามภาษา ตัวแปร The production of the grammar S Start variable.
CONTEXT-FREE GRAMMARS การนิยามภาษา DESCRIBE THE LANGUAGES ตัวแปร The production of the grammar S Start variable. others represent some auxiliary class of strings. (class of recursive definition).
6
การนิยามภาษา ตัวอย่าง NONPALINDROME Alphabet = {a, b}
CONTEXT-FREE GRAMMARS การนิยามภาษา DESCRIBE THE LANGUAGES ตัวอย่าง Alphabet = {a, b} L is defined by the following rules: S aSa | bSb | V V aTb | bTa T aT | bT | Some words generated by these rules, S aSa aVa abTaa abaa. S aSa abSba abVba abaTbba abaabba. NONPALINDROME
7
การนิยามภาษา นิยาม A context-free grammar (CFG) is a
CONTEXT-FREE GRAMMARS การนิยามภาษา DESCRIBE THE LANGUAGES นิยาม A context-free grammar (CFG) is a 4-tiple G = (V, , S, P) where V and are disjoint finite sets, S is an element of V and P is a finite set of formulas of the form A , where A V and (V )*. S: start symbol V: set of variables or nonterminals : alphabet or set of terminals. P: set of grammar rules or productions.
8
การนิยามภาษา ตัวอย่าง G = (V, , S, P) Alphabet = {a, b}
CONTEXT-FREE GRAMMARS การนิยามภาษา DESCRIBE THE LANGUAGES ตัวอย่าง G = (V, , S, P) Alphabet = {a, b} L is defined by the following rules: S aSa | bSb | V V aTb | bTa T aT | bT | aVa G abTaa denotes abTaa can be obtained from aVa by one of the productions in G ( V bTa ). S *G aaaba denotes aaaba is derived from S in zero or more steps in G.
9
การนิยามภาษา คำศัพธ์ L(G) = { x * | S *G x }.
CONTEXT-FREE GRAMMARS การนิยามภาษา DESCRIBE THE LANGUAGES คำศัพธ์ 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
10
? การนิยามภาษา ปัญหา To demonstrate that a CFG generates
CONTEXT-FREE GRAMMARS การนิยามภาษา DESCRIBE THE LANGUAGES ปัญหา ? To demonstrate that a CFG generates a language, we must show two things: every string in the language can be derived from the grammar no other string can be.
11
? โจทย์ น่าคิด กำหนดให้ ภาษาบรรยายโดย จงเขียน ไวยกรณ์ไม่พึงบริบท
โจทย์ น่าคิด กำหนดให้ ภาษาบรรยายโดย * จงเขียน ไวยกรณ์ไม่พึงบริบท ของภาษานี้ ?
12
? โจทย์ น่าคิด กำหนดให้ ภาษา 0n1n เมื่อ n 0
โจทย์ น่าคิด กำหนดให้ ภาษา 0n1n เมื่อ n 0 { , 01, 0011, , … } จงเขียน ไวยกรณ์ไม่พึงบริบท ของภาษานี้ ?
13
? โจทย์ น่าคิด กำหนดให้ ภาษาบรรยายโดย จงเขียน ไวยกรณ์ไม่พึงบริบท
โจทย์ น่าคิด กำหนดให้ ภาษาบรรยายโดย (0+001)*(10+111)*1 จงเขียน ไวยกรณ์ไม่พึงบริบท ของภาษานี้ ?
14
(ni(x) = number of i’s of x.)
CONTEXT-FREE GRAMMARS การนิยามภาษา DESCRIBE THE LANGUAGES ตัวอย่าง 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 0S1 Rule3: S 1S0 Consider strings which do not start and end with the same symbol. Example, or These strings can be obtained by the concatenation of two strings in L. Rule4: S SS. Example, can be obtained from 01 and
15
(ni(x) = number of i’s of x.)
CONTEXT-FREE GRAMMARS การนิยามภาษา DESCRIBE THE LANGUAGES ตัวอย่าง L = { x {0,1}* | n0(x) = n1(x) }. (ni(x) = number of i’s of x.) Conclusion Let G = (V, , S, P) with S 0S1 | 1S0 | SS | . We have that L(G) L.
16
(ni(x) = number of i’s of x.)
CONTEXT-FREE GRAMMARS การนิยามภาษา DESCRIBE THE LANGUAGES พิสูจน์ L = { x {0,1}* | n0(x) = n1(x) }. (ni(x) = number of i’s of 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 .
17
(ni(x) = number of i’s of x.)
CONTEXT-FREE GRAMMARS การนิยามภาษา DESCRIBE THE LANGUAGES พิสูจน์ L = { x {0,1}* | n0(x) = n1(x) }. (ni(x) = number of i’s of x.) 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.
18
(ni(x) = number of i’s of x.)
CONTEXT-FREE GRAMMARS การนิยามภาษา DESCRIBE THE LANGUAGES ตัวอย่าง L = { x {0,1}* | n0(x) n1(x) }. (ni(x) = number of i’s of x.) Let L0 be the language of strings with more 0’s than 1’s. Let L1 be the language of strings with more 1’s than 0’s. Then L = L0 L1.
19
(ni(x) = number of i’s of x.)
CONTEXT-FREE GRAMMARS การนิยามภาษา DESCRIBE THE LANGUAGES พิสูจน์ L = { x {0,1}* | n0(x) n1(x) }. (ni(x) = number of i’s of x.) 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.
20
(ni(x) = number of i’s of x.)
CONTEXT-FREE GRAMMARS การนิยามภาษา DESCRIBE THE LANGUAGES พิสูจน์ L = { x {0,1}* | n0(x) n1(x) }. (ni(x) = number of i’s of 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).
21
(ni(x) = number of i’s of x.)
CONTEXT-FREE GRAMMARS การนิยามภาษา DESCRIBE THE LANGUAGES พิสูจน์ L = { x {0,1}* | n0(x) n1(x) }. (ni(x) = number of i’s of 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.
22
(ni(x) = number of i’s of x.)
CONTEXT-FREE GRAMMARS การนิยามภาษา DESCRIBE THE LANGUAGES พิสูจน์ L = { x {0,1}* | n0(x) n1(x) }. (ni(x) = number of i’s of x.) Context-free grammar of this language is the following: S A | B A 0 | 0A | 1AA | A1A | AA1 B 1 | 1B | 0BB | B0B | BB0
23
การนิยามภาษา ตัวอย่าง ALGEBRAIC EXPRESSION
CONTEXT-FREE GRAMMARS การนิยามภาษา DESCRIBE THE LANGUAGES ตัวอย่าง ALGEBRAIC EXPRESSION Context-free grammar of this language is the following: S S+S | S-S | SS | SS | (S) | a where a is a terminal symbol.
24
ภาษาไม่พึงบริบท ทฤษฎีบท For any context-free languages L1 and L2,
CONTEXT-FREE GRAMMARS ภาษาไม่พึงบริบท CONTEXT-FREE LANGUAGES ทฤษฎีบท For any context-free languages L1 and L2, the languages L1 L2, L1L2 and L1* are also context-free languages.
25
ภาษาไม่พึงบริบท แนวคิด L(G1) = { 00 } and L(G2) = { 11 }. S S1 | S2
CONTEXT-FREE GRAMMARS ภาษาไม่พึงบริบท CONTEXT-FREE LANGUAGES แนวคิด 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 XB X 0 X 1 A 0 B 1 01 L(G1) L(G2) But S * 01.
26
ภาษาไม่พึงบริบท แนวคิด L(G1) = { 00 } and L(G2) = { 11 }. S S1 | S2
CONTEXT-FREE GRAMMARS ภาษาไม่พึงบริบท CONTEXT-FREE LANGUAGES แนวคิด 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 YB X 0 Y 1 A 0 B 1
27
ภาษาไม่พึงบริบท ทฤษฎีบท For any context-free languages L1 and L2,
CONTEXT-FREE GRAMMARS ภาษาไม่พึงบริบท CONTEXT-FREE LANGUAGES ทฤษฎีบท For any context-free languages L1 and L2, the languages L1 L2, L1L2 and L1* are also context-free languages.
28
ภาษาไม่พึงบริบท ทฤษฎีบท Case: L1L2 V = V1 V2 { S }
CONTEXT-FREE GRAMMARS ภาษาไม่พึงบริบท CONTEXT-FREE LANGUAGES ทฤษฎีบท 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 | }
29
? ภาษาไม่พึงบริบท Is every regular language a context-free language ?
CONTEXT-FREE GRAMMARS ภาษาไม่พึงบริบท CONTEXT-FREE LANGUAGES ? Is every regular language a context-free language ?
30
ภาษาไม่พึงบริบท ทฤษฎีบท
CONTEXT-FREE GRAMMARS ภาษาไม่พึงบริบท CONTEXT-FREE LANGUAGES ทฤษฎีบท Every regular language is a context-free language. By definition of regular language, , { }and { a } are regular languages, and regular is closed under union, concatenation and Kleene’s star. To prove this theorem, we have to show that any regular can be produced by some context-free grammars.
31
ภาษาไม่พึงบริบท ตัวอย่าง Regular language (011+1)*(01)*
CONTEXT-FREE GRAMMARS ภาษาไม่พึงบริบท CONTEXT-FREE LANGUAGES ตัวอย่าง Regular language (011+1)*(01)* From the regular expression, we have S XY X represents (011+1)* Y represents (01)* X 011X | 1X | Y 01Y | Example: can be derived from CFG.
32
ภาษาไม่พึงบริบท ตัวอย่าง
CONTEXT-FREE GRAMMARS ภาษาไม่พึงบริบท CONTEXT-FREE LANGUAGES ตัวอย่าง Let L be the language { 0i1j0k | j > i + k }. Find the context-free grammar for L. 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 | 1 Z 1Z0 |
33
ภาษาไม่พึงบริบท ความเข้าใจ DERIVATION TREE
CONTEXT-FREE GRAMMARS ภาษาไม่พึงบริบท CONTEXT-FREE LANGUAGES ความเข้าใจ Understanding the sentence begins with understanding its grammatical structure ( i.e., knowing how it is derived from the grammar rules ). DERIVATION TREE
34
ภาษาไม่พึงบริบท ต้นไม้ DERIVATION TREE How the sentence
CONTEXT-FREE GRAMMARS ภาษาไม่พึงบริบท CONTEXT-FREE LANGUAGES ต้นไม้ DERIVATION TREE How the sentence can be generated? S X Y 011 X 01 Y 1 X 01 Y Given a CFG S XY X 011X | 1X | Y 01Y | X 011 01 011
35
ภาษาไม่พึงบริบท ต้นไม้ ความกำกวม AMBIGUOUS DERIVATION TREE
CONTEXT-FREE GRAMMARS ภาษาไม่พึงบริบท CONTEXT-FREE LANGUAGES ต้นไม้ DERIVATION TREE ความกำกวม AMBIGUOUS A context-free grammar G is ambiguous if there is at least one string in L(G) having two or more distinct derivation trees.
36
ภาษาไม่พึงบริบท ต้นไม้ ความกำกวม AMBIGUOUS
CONTEXT-FREE GRAMMARS ภาษาไม่พึงบริบท CONTEXT-FREE LANGUAGES ต้นไม้ ความกำกวม AMBIGUOUS Given a context-free grammar G with = { a, + } and S S + S | a The sentence a+a+a is in L(G) which has two derivations. SS+Sa+Sa+S+Sa+a+Sa+a+a SS+SS+S+Sa+S+Sa+a+Sa+a+a This language is ambiguous.
37
? โจทย์ น่าคิด Let G be a context-free and x nG y be
โจทย์ น่าคิด Let G be a context-free and x nG y be a derivation in G where x can be written x = y1A1y2A2…ykAkyk+1 with yi *. Show that there are strings pi ( V)* that satisfy 1. Ai ti pi. 2. y = y1p1y2p2…ykpkyk+1 3. Sum of all ti (1ik) = n. ?
38
ภาษาไม่พึงบริบท ตัวอย่าง if exp1 if exp2 statement1 else statement2
CONTEXT-FREE GRAMMARS ภาษาไม่พึงบริบท CONTEXT-FREE LANGUAGES ตัวอย่าง Given a context-free grammar G with S if E S | if E S else S | S This language is ambiguous. if exp1 if exp2 statement1 else statement2 S S if E S if E S else S PARENTHESES CAN BE USED TO REMOVE THE AMBIGUITY exp1 exp1 if E S if E S else S statement2 exp2 statement2 exp2 statement1 statement1
39
ภาษาไม่พึงบริบท ตัวอย่าง Given a context-free grammar G with
CONTEXT-FREE GRAMMARS ภาษาไม่พึงบริบท CONTEXT-FREE LANGUAGES ตัวอย่าง Given a context-free grammar G with S S + S| S S | S | a This language is ambiguous. Theorem The context-free grammar G1 with productions S S + T | T T T F | F F (S) | a is unambiguous.
40
NORMAL FORMS
41
วัตถุประสงค์ Ambiguity is one undesirable property of a
NORMAL FORMS วัตถุประสงค์ Ambiguity is one undesirable property of a context-free grammar that we might wish to Eliminate. In this section we discuss some more straightforward ways of “improving” a grammar without changing the resulting language.
42
วัตถุประสงค์ We try to eliminate “-productions”, of the form
NORMAL FORMS วัตถุประสงค์ We try to eliminate “-productions”, of the form A , and unit productions,” in which one variable is simply replaced by another. REMARK : production G If there are no -productions, || ||. If there are no unit productions, if || = || then this step consists of replacing a variable by a single terminal.
43
วัตถุประสงค์ We try to eliminate “-productions”, of the form
NORMAL FORMS วัตถุประสงค์ We try to eliminate “-productions”, of the form A , and unit productions,” in which one variable is simply replaced by another. 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 is the number of steps S G x is at most 2k-1 steps.
44
ตัวอย่าง -Production ELIMINATING THE -PRODUCTION
NORMAL FORMS -Production CONTEXT-FREE LANGUAGES ตัวอย่าง ELIMINATING THE -PRODUCTION Given a context-free grammar G with productions S ABCBCDA A CD B Cb C a | D bD | 1. Replacing ABCBCDA by A1BC1BC2DA2 2. Create 32 productions S A1BC1BC2DA2 and the others with or without A1, C1, C2, D and A2. 3. Adding A CD | C | D B Cb | b C a D bD | b
45
นิยาม -Production A is nullable if and only if A *G NULLABLE
NORMAL FORMS -Production CONTEXT-FREE LANGUAGES นิยาม NULLABLE A nullable variable in a context-free grammar G = ( V, , S, P ) is defined as follows: 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
46
From the previous example, N0 = { C, D } and N1 = { A, C, D }.
NORMAL FORMS -Production CONTEXT-FREE LANGUAGES อัลกอริทึม Algorithm: FindNull 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 }.
47
อัลกอริทึม -Production Algorithm: RemoveNull CONTEXT-FREE LANGUAGES
NORMAL FORMS NORMAL FORMS -Production CONTEXT-FREE LANGUAGES อัลกอริทึม Algorithm: RemoveNull 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 obtained form this one by deleting from or more the occurrences in of nullable variables. Delete all -production from P1. Delete any duplicates. Delete productions of the from A A.
48
-Production ตัวอย่าง Given a context-free grammar G with productions
NORMAL FORMS NORMAL FORMS -Production CONTEXT-FREE LANGUAGES ตัวอย่าง Given a context-free grammar G with productions S ABC | ACB A ACD | CD B Cb C a | D bD |
49
-Production ตัวอย่าง Given a context-free grammar G with productions
NORMAL FORMS NORMAL FORMS -Production CONTEXT-FREE LANGUAGES ตัวอย่าง Given a context-free grammar G with productions S ABC | ACB A ACD | CD B Cb C a | D bD |
50
-Production ตัวอย่าง Given a context-free grammar G with productions
NORMAL FORMS NORMAL FORMS -Production CONTEXT-FREE LANGUAGES ตัวอย่าง Given a context-free grammar G with productions S ABC | ACB A ACD | CD | AC | C B Cb C a | D bD | b
51
-Production ตัวอย่าง Given a context-free grammar G with productions
NORMAL FORMS NORMAL FORMS -Production CONTEXT-FREE LANGUAGES ตัวอย่าง Given a context-free grammar G with productions S ABC | ACB A ACD | CD | AC | C B Cb C a | D bD | b
52
-Production ตัวอย่าง Given a context-free grammar G with productions
NORMAL FORMS NORMAL FORMS -Production CONTEXT-FREE LANGUAGES ตัวอย่าง Given a context-free grammar G with productions S ABC | ACB | AB A ACD | CD | AC | C | AD | D | A | B Cb | b C a D bD | b
53
-Production ตัวอย่าง Given a context-free grammar G with productions
NORMAL FORMS NORMAL FORMS -Production CONTEXT-FREE LANGUAGES ตัวอย่าง Given a context-free grammar G with productions S ABC | ACB | AB A ACD | CD | AC | C | AD | D | A | B Cb | b C a D bD | b
54
-Production ตัวอย่าง Given a context-free grammar G with productions
NORMAL FORMS NORMAL FORMS -Production CONTEXT-FREE LANGUAGES ตัวอย่าง Given a context-free grammar G with productions S ABC | ACB | AB | BC | CB | B A ACD | CD | AC | C | AD | D | A B Cb | b C a D bD | b
55
ทฤษฎีบท -Production Context-free grammar
NORMAL FORMS -Production CONTEXT-FREE LANGUAGES ทฤษฎีบท Context-free grammar 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, and L(G1) = L(G) – { }.
56
Suppose that the context-free grammar G has no -productions.
NORMAL FORMS Unit-Production CONTEXT-FREE LANGUAGES นิยาม 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. Unit productions
57
นิยาม Unit-Production B is A-derivable if and only if A *G B.
NORMAL FORMS Unit-Production CONTEXT-FREE LANGUAGES นิยาม An A-derivable variable in a context-free grammar G = ( V, , S, P ) is defined as follows: 1. Any variable B for which P contains the production A B, then B is A- derivable. 2. If C is A-derivable, if C B is in P, and if 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.
58
Algorithm RemoveUnit REMOVING THE UNIT PRODUCTIONS FROM A CFG.
NORMAL FORMS Algorithm RemoveUnit 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.
59
THEOREM Let G = ( V, , S, P ) be a context-free
NORMAL FORMS THEOREM Let G = ( V, , S, P ) be a context-free grammar with no -productions, and let G1 = ( V, , S, P ) be a context-free grammar obtained from G by Algorithm RemoveUnit. Then G1 contains no unit productions, and L(G1) = L(G).
60
EXAMPLE Given a context-free grammar G without
NORMAL FORMS EXAMPLE Given a context-free grammar G without -productions, and with productions S S + T | T T T F | F F (S) | a S-derivable = { T, F } T-derivable = { F } Productions in P1 S S + T | T | T F | (S) | a T T F | F | (S) | a F (S) | a
61
CHOMSKY NORMAL FORM Eliminating specific types of productions,
NORMAL FORMS CHOMSKY NORMAL FORM Eliminating specific types of productions, such as -productions and unit productions, it may also be useful to impose restrictions on the form of the remaining productions. Several types of “Normal forms” have been introduced; we shall present one of them, the Chomsky normal form.
62
CNF CHOMSKY NORMAL FORM
NORMAL FORMS CNF CHOMSKY NORMAL FORM Definition A context-free grammar G = ( V, , S, P ) is in Chomsky normal form if every production is of one of these two types: A BC A a where A, B and C are variables, and a is a terminal symbol.
63
CNF CHOMSKY NORMAL FORM
NORMAL FORMS CNF CHOMSKY 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 2 A a Replace A B1B2…Bk by A B1X1, X1 B2X2, …, Xk-1 Bk-1Bk
64
EXAMPLE Given a context-free grammar G with productions
NORMAL FORMS EXAMPLE Given a context-free grammar G with productions S ABC | ACB | AB | BC | CB | B A ACD | CD | AC | C | AD | D | A B Cb | b C a D bD | b S-derivable = { B } A-derivable = { C, D } B-derivable = C-derivable = D-derivable =
65
EXAMPLE Given a context-free grammar G with productions
NORMAL FORMS EXAMPLE Given a context-free grammar G with productions S ABC | ACB | AB | BC | CB | Cb | b A ACD | CD | AC | AD | a | bD | b B Cb | b C a D bD | b S-derivable = { B } A-derivable = { C, D } B-derivable = C-derivable = D-derivable =
66
EXAMPLE Given a context-free grammar G with productions
NORMAL FORMS EXAMPLE Given a context-free grammar G with productions S ABC | ACB | AB | BC | CB | Cb | b A ACD | CD | AC | AD | a | bD | b B Cb | b C a D bD | b Xb b Xbc BC Xcb CB Xcd CD
67
EXAMPLE Given a context-free grammar G with productions
NORMAL FORMS EXAMPLE Given a context-free grammar G with productions S ABC | ACB | AB | BC | CB | CXb | b A ACD | CD | AC | AD | a | XbD | b B CXb | b C a D XbD | b Xb b Xbc BC Xcb CB Xcd CD
68
EXAMPLE Given a context-free grammar G with productions
NORMAL FORMS EXAMPLE Given a context-free grammar G with productions S AXbc | AXcb | AB | BC | CB | CXb | b A AXcd | CD | AC | AD | a | XbD | b B CXb | b C a D XbD | b Xb b Xbc BC Xcb CB Xcd CD
69
THEOREM For any G = ( V, , S, P ) be a context-free grammar,
NORMAL FORMS THEOREM For any G = ( V, , S, P ) be a context-free grammar, there is a G1 = ( V1, , S, P1 ) be a context-free grammar in CNF so that L(G1) = L(G) – { }.
งานนำเสนอที่คล้ายกัน
