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163692498 grammar
- 1. 1 Grammars
- 2. 2 Grammars Grammars express languages Example: the English language verbpredicate nounarticlephrasenoun predicatephrasenounsentence → → → _ _
- 4. 4 A derivation of “the boy walks”: walksboythe verbboythe verbnounthe verbnounarticle verbphrasenoun predicatephrasenounsentence ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ _ _
- 5. 5 A derivation of “a dog runs”: runsdoga verbdoga verbnouna verbnounarticle verbphrasenoun predicatephrasenounsentence ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ _ _
- 6. 6 Language of the grammar: L = { “a boy runs”, “a boy walks”, “the boy runs”, “the boy walks”, “a dog runs”, “a dog walks”, “the dog runs”,
- 8. 8 Another Example Grammar: Derivation of sentence : λ→ → S aSbS abaSbS ⇒⇒ ab aSbS → λ→S
- 9. 9 aabbaaSbbaSbS ⇒⇒⇒ aSbS → λ→S aabb λ→ → S aSbS Grammar: Derivation of sentence :
- 11. 11 Language of the grammar: L = { “a boy runs”, “a boy walks”, “the boy runs”, “the boy walks”, “a dog runs”, “a dog walks”, “the dog runs”, “the dog walks” }
- 12. 12 Language of the grammar λ→ → S aSbS }0:{ ≥= nbaL nn
- 13. 13 More Notation Grammar ( )SPTVG ,,,= :V :T :S :P A finiteSet of variables A finiteSet of terminal symbols Start variable Set of Production rules V and T are assumed to be disjoint
- 14. 14 Example Grammar : λ→ → S aSbSG ( ), , ,G V T P S= }{SV = },{ baT = },{ λ→→= SaSbSP
- 15. 15 More Notation Sentential Form: A sentence that contains variables and terminals Example: aaabbbaaaSbbbaaSbbaSbS ⇒⇒⇒⇒ Sentential Forms sentence
- 17. 17 In general we write: If: nww * 1 ⇒ nwwww ⇒⇒⇒⇒ 321
- 21. 21 Another Grammar Example Grammar : λ→ → → A aAbA AbS Derivations: G aabbbaaAbbbaAbbAbS abbaAbbAbS →→→→ →→→
- 24. 24 Language of a Grammar For a grammar with start variable : G S }:{)( wSwGL ∗ ⇒= String of terminals
- 25. 25 Example For grammar : λ→ → → A aAbA AbS }0:{)( ≥= nbbaGL nn Since: bbaS nn ∗ ⇒ G
- 26. 26 A Convenient Notation λ→ → A aAbA λ|aAbA → thearticle aarticle → → theaarticle |→
- 28. 28 Linear Grammars Grammars with at most one variable at the right side of a production Examples: λ→ → → A aAbA AbS λ→ → S aSbS
- 29. 29 A Non-Linear Grammar bSaS aSbS S SSS → → → → λ Grammar :G )}()(:{)( wnwnwGL ba == Number of in stringa w Why S SS What will happen if S S instead
- 30. 30 Another Linear Grammar Grammar : AbB aBA AS → → → λ| }0:{)( ≥= nbaGL nn G
- 31. 31 Right-Linear Grammars All productions have form: Example: xBA → xA →or aS abSS → → string of terminals
- 32. 32 Left-Linear Grammars All productions have form: Example: BxA → aB BAabA AabS → → → | xA → or string of terminals
- 34. 34 Regular Grammars A regular grammar is any right-linear or left-linear grammar Examples: aS abSS → → aB BAabA AabS → → → | 1G 2G
- 35. 35 Observation Regular grammars generate regular languages Examples: aS abSS → → aabGL *)()( 1 = aB BAabA AabS → → → | *)()( 2 abaabGL = 1G 2G
- 38. 38 Theorem - Part 1 Languages Generated by Regular Grammars Regular Languages ⊆ Any regular grammar generates a regular language
- 39. 39 Theorem - Part 2 Languages Generated by Regular Grammars Regular Languages ⊇ Any regular language is generated by a regular grammar
- 40. 40 Proof – Part 1 Languages Generated by Regular Grammars Regular Languages ⊆ The language generated by any regular grammar is regular )(GL G
- 41. 41 The case of Right-Linear Grammars Let be a right-linear grammar We will prove: is regular Proof idea: We will construct NFA with L(M)=L(G) G )(GL M )()( GLML =
- 43. 43 Construct NFA such that every state is a grammar variable: M aBbB BaaA BaAS | | → → → S FV A B special final state
- 44. 44 Add edges for each production: S FV A B a aAS →
- 51. 51 In General A right-linear grammar has variables: and productions: G ,,, 210 VVV jmi VaaaV 21→ mi aaaV 21→ or
- 52. 52 We construct the NFA such that: each variable corresponds to a node: M iV 0V FV 1V 2V 3V 4V special final state
- 53. 53 For each production: we add transitions and intermediate nodes jmi VaaaV 21→ iV jV………1a 2a ma
- 54. 54 For each production: we add transitions and intermediate nodes mi aaaV 21→ iV FV………1a 2a ma
- 55. 55 Resulting NFA looks like this:M 0V FV 1V 2V 3V 4V 1a 3a 3a 4a 8a 2a 4a 5a 9a 5a 9a )()( MLGL =It holds that:
- 56. 56 The case of Left-Linear Grammars Let be a left-linear grammar We will prove: is regular Proof idea: We will construct a right-linear grammar with G )(GL G′ R GLGL )()( ′=
- 57. 57 Since is left-linear grammar the productions look like: G kaaBaA 21→ kaaaA 21→
- 58. 58 Construct right-linear grammar G′ Left linear G kaaBaA 21→ G′ BaaaA k 12→ BvA → BvA R → Right linear
- 59. 59 Construct right-linear grammar G′ G kaaaA 21→ G′ 12aaaA k → vA → R vA → Left linear Right linear
- 60. 60 It is easy to see that: Since is right-linear, we have: R GLGL )()( ′= )(GL ′ R GL )( ′ G′ )(GL Regular Language Regular Language Regular Language
- 61. 61 Proof - Part 2 Languages Generated by Regular Grammars Regular Languages ⊇ Any regular language is generated by some regular grammar L G
- 62. 62 Proof idea: Let be the NFA with . Construct from a regular grammar such that Any regular language is generated by some regular grammar L G M )(MLL = M G )()( GLML =
- 63. 63 Since is regular there is an NFA such that L M )(MLL = Example: a b a bλ *)*(* abbababL = )(MLL = M 1q 2q 3q 0q
- 64. 64 Convert to a right-linear grammarM a b a bλ M 0q 1q 2q 3q 10 aqq →
- 68. 68 In General For any transition: a q p Add production: apq → variable terminal variable
- 69. 69 For any final state: fq Add production: λ→fq
- 70. 70 Since is right-linear grammar is also a regular grammar with G G LMLGL == )()(
- 71. 71 For any regular language one can construct left linear as well as right linear grammar.
- 72. 72 S FV A B a λ abBB BaaA BaAS | | → → → a a b a F → Ba B → Bb here F is start symbol B → Aaa B → S A → Sa S → λ this is new rule as S is now a final state
- 73. 73 A → aB B → bB B → aC C → b D D →λ B D → λ D → Cb D → Bλ C → B a B → B b Now D is the start variable B → Aa A → λ
- 74. 74 Language generated by both aaab* a +b* a
- 75. 75
- 76. 76
- 77. 77
- 79. 79 Regular Languages }{ nn ba }{ R ww Context-Free Languages
- 82. 82 Example A context-free grammar : λ→ → S aSbS aabbaaSbbaSbS ⇒⇒⇒ G A derivation:
- 84. 84 λ→ → → S bSbS aSaS abaabaabaSabaabSbaaSaS ⇒⇒⇒⇒ A context-free grammar :G Another derivation:
- 86. 86 λ→ → → S SSS aSbS ababSaSbSSSS ⇒⇒⇒⇒ A context-free grammar :G A derivation: Example
- 87. 87 λ→ → → S SSS aSbS abababaSbabSaSbSSSS ⇒⇒⇒⇒⇒ A context-free grammar :G A derivation:
- 89. 89 Definition: Context-Free Grammars Grammar Productions of the form: xA → x is string of variables and terminals ),,,( SPTVG = Variables Terminal symbols Start variable
- 90. 90 Definition: Context-Free Languages A language is context-free if and only if there is a C.F.grammar with L G )(GLL =
- 91. 91 Derivation Order ABS →.1 λ→ → A aaAA .3 .2 λ→ → B BbB .5 .4 aabaaBbaaBaaABABS 54321 ⇒⇒⇒⇒⇒ Leftmost derivation: aabaaAbAbABbABS 32541 ⇒⇒⇒⇒⇒ Rightmost derivation:
- 94. 94 Def: G =(V,T,P,S) An ordered tree is a derivation tree for G iff 1 The root is labeled S 2 every leaf has a label from T∪{ λ} 3 Every interior vertex has a label from V 4 If a vertex has label A(variable) and its children are labeled from (L to R) a1,a2,…an then P must contain a production A a1 a2 …an 5 A leaf labeled λ has no sibling.
- 95. 95 Yield: The string of terminals obtained by reading the leaves of the tree from left to right omitting any λ’s encountered is called yield of the tree.
- 96. 96 Partial derivation tree A tree that has properties 3,4,5 but 1 need not And 2 is replaced by V ∪ T ∪ {λ }
- 97. 97 ABS ⇒ ABS → λ|aaAA → λ|BbB → S BA
- 98. 98 ABS → λ|aaAA → λ|BbB → aaABABS ⇒⇒ a a A S BA
- 99. 99 ABS → λ|aaAA → λ|BbB → aaABbaaABABS ⇒⇒⇒ S BA a a A B b
- 100. 100 ABS → λ|aaAA → λ|BbB → aaBbaaABbaaABABS ⇒⇒⇒⇒ S BA a a A B b λ
- 101. 101 ABS → λ|aaAA → λ|BbB → aabaaBbaaABbaaABABS ⇒⇒⇒⇒⇒ S BA a a A B b λ λ Derivation Tree
- 102. 102 aabaaBbaaABbaaABABS ⇒⇒⇒⇒⇒ yield aab baa = λλ S BA a a A B b λ λ Derivation Tree ABS → λ|aaAA → λ|BbB →
- 103. 103 Partial Derivation Trees ABS ⇒ S BA Partial derivation tree ABS → λ|aaAA → λ|BbB →
- 104. 104 aaABABS ⇒⇒ S BA a a A Partial derivation tree
- 105. 105 aaABABS ⇒⇒ S BA a a A Partial derivation tree sentential form yield aaAB
- 106. 106 aabaaBbaaBaaABABS ⇒⇒⇒⇒⇒ aabaaAbAbABbABS ⇒⇒⇒⇒⇒ S BA a a A B b λ λ Same derivation tree Sometimes, derivation order doesn’t matter Leftmost: Rightmost:
- 107. 107 Ambiguity
- 108. 108 aEEEEEE |)(|| ∗+→ aaa ∗+ E EE EE + a a a ∗ aaaEaa EEaEaEEE *+⇒∗+⇒ ∗+⇒+⇒+⇒ leftmost derivation
- 109. 109 aEEEEEE |)(|| ∗+→ aaa ∗+ E EE + a a ∗ EE a aaaEaa EEaEEEEEE ∗+⇒∗+⇒ ∗+⇒∗+⇒∗⇒ leftmost derivation
- 110. 110 aEEEEEE |)(|| ∗+→ aaa ∗+ E EE + a a ∗ EE a E EE EE + a a a ∗ Two derivation trees
- 111. 111 The grammar aEEEEEE |)(|| ∗+→ is ambiguous: E EE + a a ∗ EE a E EE EE + a a a ∗ string aaa ∗+ has two derivation trees
- 112. 112 string aaa ∗+ has two leftmost derivations aaaEaa EEaEEEEEE ∗+⇒∗+⇒ ∗+⇒∗+⇒∗⇒ aaaEaa EEaEaEEE *+⇒∗+⇒ ∗+⇒+⇒+⇒ The grammar aEEEEEE |)(|| ∗+→ is ambiguous:
- 113. 113 Definition: A context-free grammar is ambiguous if some string has: two or more derivation trees G )(GLw∈
- 114. 114 In other words: A context-free grammar is ambiguous if some string has: two or more leftmost derivations G )(GLw∈ (or rightmost)
- 115. 115 Why do we care about ambiguity? E EE + a a ∗ EE a E EE EE + a a a ∗ aaa ∗+ take 2=a
- 116. 116 E EE + ∗ EE E EE EE + ∗ 222 ∗+ 2 2 2 2 2 2
- 117. 117 E EE + ∗ EE E EE EE + ∗ 6222 =∗+ 2 2 2 2 2 2 8222 =∗+ 4 2 2 2 6 2 2 24 8
- 118. 118 E EE EE + ∗ 6222 =∗+ 2 2 2 4 2 2 2 6 Correct result:
- 119. 119 • We want to remove ambiguity • Ambiguity is bad for programming languages
- 120. 120 We fix the ambiguous grammar: aEEEEEE |)(|| ∗+→ New non-ambiguous grammar: aF EF FT FTT TE TEE → → → ∗→ → +→ )(
- 121. 121 aF EF FT FTT TE TEE → → → ∗→ → +→ )( aaaFaaFFa FTaTaTFTTTEE ∗+⇒∗+⇒∗+⇒ ∗+⇒+⇒+⇒+⇒+⇒ E E T+ T ∗ F F a T F a a aaa ∗+
- 122. 122 E E T+ T ∗ F F a T F a a aaa ∗+ Unique derivation tree
- 123. 123 The grammar : aF EF FT FTT TE TEE → → → ∗→ → +→ )( is non-ambiguous: Every string has a unique derivation tree G )(GLw∈
- 124. 124 Inherent Ambiguity Some context free languages have only ambiguous grammars Example: }{}{ mmnmnn cbacbaL ∪= λ| |11 aAbA AcSS → → λ| |22 bBcB BaSS → →21 | SSS →
- 125. 125 The string nnn cba has two derivation trees S 1S S 2S 1S c 2Sa
- 127. 127 A parser knows the grammar of the programming language
- 128. 128 The parser finds the derivation of a particular input 10 + 2 * 5 Parser E -> E + E | E * E | INT E => E + E => E + E * E => 10 + E*E => 10 + 2 * E => 10 + 2 * 5 input derivation
- 129. 129 10 E 2 5 E => E + E => E + E * E => 10 + E*E => 10 + 2 * E => 10 + 2 * 5 derivation derivation tree E E E E + *