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Class5
- 1. 1 Reverse of a Regular Language
- 2. 2 Theorem: The reverse of a regular language is a regular language R L L Proof idea: Construct NFA that accepts : R L invert the transitions of the NFA that accepts L
- 3. 3 Proof Since is regular, there is NFA that accepts L Example: baabL += * a b b a L
- 5. 5 Make old initial state a final state a b b a
- 6. 6 Add a new initial state a b b a λ λ
- 7. 7 a b b a λ λ Resulting machine accepts R L baabL += * ababLR += * R L is regular
- 8. 8 Grammars
- 9. 9 Grammars Grammars express languages Example: the English language verbpredicate nounarticlephrasenoun predicatephrasenounsentence → → → _ _
- 11. 11 A derivation of “the boy walks”: walksboythe verbboythe verbnounthe verbnounarticle verbphrasenoun predicatephrasenounsentence ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ _ _
- 12. 12 A derivation of “a dog runs”: runsdoga verbdoga verbnouna verbnounarticle verbphrasenoun predicatephrasenounsentence ⇒ ⇒ ⇒ ⇒ ⇒ ⇒ _ _
- 13. 13 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” }
- 15. 15 Another Example Grammar: Derivation of sentence : λ→ → S aSbS abaSbS ⇒⇒ ab aSbS → λ→S
- 16. 16 aabbaaSbbaSbS ⇒⇒⇒ aSbS → λ→S aabb λ→ → S aSbSGrammar: Derivation of sentence :
- 18. 18 Language of the grammar λ→ → S aSbS }0:{ ≥= nbaL nn
- 19. 19 More Notation Grammar ( )PSTVG ,,,= :V :T :S :P Set of variables Set of terminal symbols Start variable Set of Production rules
- 20. 20 Example Grammar : λ→ → S aSbSG ( )PSTVG ,,,= }{SV = },{ baT = },{ λ→→= SaSbSP
- 21. 21 More Notation Sentential Form: A sentence that contains variables and terminals Example: aaabbbaaaSbbbaaSbbaSbS ⇒⇒⇒⇒ Sentential Forms sentence
- 23. 23 In general we write: If: nww * 1 ⇒ nwwww ⇒⇒⇒⇒ 321
- 27. 27 Another Grammar Example Grammar : λ→ → → A aAbA AbS Derivations: aabbbaaAbbbaAbbS abbaAbbAbS bAbS ⇒⇒⇒ ⇒⇒⇒ ⇒⇒ G
- 29. 29 Language of a Grammar For a grammar with start variable : G S }:{)( wSwGL ∗ ⇒= String of terminals
- 30. 30 Example For grammar : λ→ → → A aAbA AbS }0:{)( ≥= nbbaGL nn Since: bbaS nn ∗ ⇒ G
- 31. 31 A Convenient Notation λ→ → A aAbA λ|aAbA → thearticle aarticle → → theaarticle |→
- 33. 33 Linear Grammars Grammars with at most one variable at the right side of a production Examples: λ→ → → A aAbA AbS λ→ → S aSbS
- 34. 34 A Non-Linear Grammar bSaS aSbS S SSS → → → → λ Grammar :G )}()(:{)( wnwnwGL ba ==
- 35. 35 Another Linear Grammar Grammar : AbB aBA AS → → → λ| }0:{)( ≥= nbaGL nn G
- 36. 36 Right-Linear Grammars All productions have form: Example: xBA → xA → or aS abSS → →
- 37. 37 Left-Linear Grammars All productions have form: Example: BxA → aB BAabA AabS → → → | xA → or
- 39. 39 Regular Grammars A regular grammar is any right-linear or left-linear grammar Examples: aS abSS → → aB BAabA AabS → → → | 1G 2G
- 40. 40 Observation Regular grammars generate regular languages Examples: aS abSS → → aabGL *)()( 1 = aB BAabA AabS → → → | *)()( 2 abaabGL = 1G 2G
- 43. 43 Theorem - Part 1 Languages Generated by Regular Grammars Regular Languages ⊆ Any regular grammar generates a regular language
- 44. 44 Theorem - Part 2 Languages Generated by Regular Grammars Regular Languages ⊇ Any regular language is generated by a regular grammar
- 45. 45 Proof – Part 1 Languages Generated by Regular Grammars Regular Languages ⊆ The language generated by any regular grammar is regular )(GL G
- 46. 46 The case of Right-Linear Grammars Let be a right-linear grammar We will prove: is regular Proof idea: We will construct NFA with G )(GL M )()( GLML =
- 48. 48 Construct NFA such that every state is a grammar variable: M aBbB BaaA BaAS | | → → → S FV A B special final state
- 49. 49 Add edges for each production: S FV A B a aAS →
- 56. 56 In General A right-linear grammar has variables: and productions: G ,,, 210 VVV jmi VaaaV 21→ mi aaaV 21→ or
- 57. 57 We construct the NFA such that: each variable corresponds to a node: M iV 0V FV 1V 2V 3V 4V special final state
- 58. 58 For each production: we add transitions and intermediate nodes jmi VaaaV 21→ iV jV………1a 2a ma
- 59. 59 For each production: we add transitions and intermediate nodes mi aaaV 21→ iV FV………1a 2a ma
- 60. 60 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:
- 61. 61 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 )()( ′=
- 62. 62 Since is left-linear grammar the productions look like: G kaaBaA 21→ kaaaA 21→
- 63. 63 Construct right-linear grammar G′ In :G kaaBaA 21→ In :G′ BaaaA k 12→ vBA → BvA R →
- 64. 64 Construct right-linear grammar G′ In :G kaaaA 21→ In :G′ 12aaaA k → vA → R vA →
- 65. 65 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
- 66. 66 Proof - Part 2 Languages Generated by Regular Grammars Regular Languages ⊇ Any regular language is generated by some regular grammar L G
- 67. 67 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 =
- 68. 68 Since is regular there is an NFA such that L M )(MLL = Example: a b a bλ *)*(* abbababL = )(MLL = M 1q 2q 3q 0q
- 69. 69 Convert to a right-linear grammarM a b a bλ M 0q 1q 2q 3q 10 aqq →
- 73. 73 In General For any transition: a q p Add production: apq → variable terminal variable
- 74. 74 For any final state: fq Add production: λ→fq
- 75. 75 Since is right-linear grammar is also a regular grammar with G G LMLGL == )()(