ไวยกรณ์ไม่พึงบริบท CONTEXT-FREE GRAMMARS

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.

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.

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.

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).

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

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.

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.

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

? 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.

? โจทย์ น่าคิด กำหนดให้ ภาษาบรรยายโดย จงเขียน ไวยกรณ์ไม่พึงบริบท โจทย์ น่าคิด กำหนดให้ ภาษาบรรยายโดย * จงเขียน ไวยกรณ์ไม่พึงบริบท ของภาษานี้ ?

? โจทย์ น่าคิด กำหนดให้ ภาษา 0n1n เมื่อ n  0 โจทย์ น่าคิด กำหนดให้ ภาษา 0n1n เมื่อ n  0 { , 01, 0011, 000111, … } จงเขียน ไวยกรณ์ไม่พึงบริบท ของภาษานี้ ?

? โจทย์ น่าคิด กำหนดให้ ภาษาบรรยายโดย จงเขียน ไวยกรณ์ไม่พึงบริบท โจทย์ น่าคิด กำหนดให้ ภาษาบรรยายโดย (0+001)*(10+111)*1 จงเขียน ไวยกรณ์ไม่พึงบริบท ของภาษานี้ ?

(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, 01110100 or 10010110 These strings can be obtained by the concatenation of two strings in L. Rule4: S  SS. Example, 01110100 can be obtained from 01 and 110100.

(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.

(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  .

(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.

(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.

(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.

(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 SS1S. ( d(wn)>0 and d(zn)>0, wn and zn L0).

(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 SS1S. ( 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 SS1S.

(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

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 | SS | SS | (S) | a where a is a terminal symbol.

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.

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.

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

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.

ภาษาไม่พึงบริบท ทฤษฎีบท Case: L1L2 V = V1  V2  { S } CONTEXT-FREE GRAMMARS ภาษาไม่พึงบริบท CONTEXT-FREE LANGUAGES ทฤษฎีบท Case: L1L2 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 |  }

? ภาษาไม่พึงบริบท Is every regular language a context-free language ? CONTEXT-FREE GRAMMARS ภาษาไม่พึงบริบท CONTEXT-FREE LANGUAGES ? Is every regular language a context-free language ?

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.

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: 01110101 can be derived from CFG.

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 | 

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

ภาษาไม่พึงบริบท ต้นไม้ DERIVATION TREE How the sentence CONTEXT-FREE GRAMMARS ภาษาไม่พึงบริบท CONTEXT-FREE LANGUAGES ต้นไม้ DERIVATION TREE How the sentence 0111011011010101 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

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.

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. SS+Sa+Sa+S+Sa+a+Sa+a+a SS+SS+S+Sa+S+Sa+a+Sa+a+a This language is ambiguous.

? 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 (1ik) = n. ?

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

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.

NORMAL FORMS

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.

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.

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.

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

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 

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 }.

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.

-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 | 

-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 | 

-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

-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

-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

-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

-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

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) – {  }.

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

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.

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.

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).

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

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.

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.

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

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 = 

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 = 

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

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

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

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) – {  }.