Pda accept context free
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The document discusses how to convert between context-free grammars and pushdown automata (PDAs). It presents proofs that: 1) Any context-free grammar can be converted to an equivalent PDA by constructing transition rules based on the grammar productions. 2) Any PDA can be converted to an equivalent context-free grammar by defining variables for each PDA state and production rules for state transitions. The document outlines the specific procedures for performing these conversions in both directions.
Pda accept context free
- 1. © 2020 Wael Badawy 1 PDAs Accept Context-Free Languages 1 DISCLAIMER: This video is optimized for HD large display using patented and patent-pending “Nile Codec”, the first Egyptian Video Codec for more information, PLEASE check https://NileCodec.com Also available as a PodCast
- 2. © 2020 Wael Badawy Copyright © 2020 Wael Badawy. All rights reserved n This video is subject to copyright owned by Wael Badawy “WB”. Any reproduction or republication of all or part of this video is expressly prohibited unless WB has explicitly granted its prior written consent. All other rights reserved. n This video is intended for education and information only and is offered AS IS, without any warranty of the accuracy or the quality of the content. Any other use is strictly prohibited. The viewer is fully responsible to verify the accuracy of the contents received without any claims of costs or liability arising . n The names, trademarks service marked as logos of WB or the sponsors appearing in this video may not be used in any any product or service, without prior express written permission from WB and the video sponsors n Neither WB nor any party involved in creating, producing or delivering information and material via this video shall be liable for any direction, incidental, consequential, indirect of punitive damages arising out of access to, use or inability to use this content or any errors or omissions in the content thereof. n If you will continue to watch this video, you agree to the terms above and other terms that may be available on http://nu.edu.eg & https://caiwave.net 2
- 3. © 2020 Wael Badawy
- 4. © 2020 Wael Badawy 4 Context-Free Languages (Grammars) Languages Accepted by PDAs = Theorem:
- 5. © 2020 Wael Badawy 5 Context-Free Languages (Grammars) Languages Accepted by PDAs Í Proof - Step 1: Convert any context-free grammar to a PDA with: G M )()( MLGL =
- 6. © 2020 Wael Badawy 6 Context-Free Languages (Grammars) Languages Accepted by PDAs Ê Proof - Step 2: Convert any PDA to a context-free grammar with:G M )()( MLGL =
- 8. © 2020 Wael Badawy 8 MWe will convert to a PDA such that:G Take an arbitrary context-free grammar G )()( MLGL =
- 9. © 2020 Wael Badawy 9 q0 q1 2qS®ee, $$, ®e wA ®,e e®aa, production in wA ® terminal in a G G Conversion Procedure: For each For each Add transitions
- 10. © 2020 Wael Badawy 10 q0 q1 2q Grammar PDA ee e e e ® ® ® ® T TaT bS aSTbS , , , , e® ® ® ® T TaT bS aSTbS e e ® ® bb aa , , Example S®ee, $$, ®e
- 11. © 2020 Wael Badawy 11 nkk mk XX S ssss ss !! ! !! ! 11 11 +Þ Þ Þ Þ ,$),( $),,( $),,( ,$),( 2 111 111 110 e ss ssss ssss q XXq Sq q mnk nkk nkk ! "! ""! "! ""! "" + + + PDA simulates leftmost derivations Grammar Leftmost Derivation PDA Computation Scanned symbols Stack contents
- 12. © 2020 Wael Badawy 12 Bzyx xAy ji ss ! ! Þ Þ Þ Grammar Leftmost Derivation BzA ji ss !® Production applied Terminals Leftmost variable Variables or terminals Terminals Variable Variables or terminals
- 13. © 2020 Wael Badawy 13 Bzyx xAy ji ss ! ! Þ Þ Þ $),,( $),,( 1 1 Bzyq Ayq jini ni ssss ss !!" !" !" Grammar Leftmost Derivation PDA Computation BzA ji ss !® Production applied q0 q1 2qS®ee, $$, ®e BzA ji sse !®, Transition applied
- 14. © 2020 Wael Badawy 14 Bzyx xAy ji ss ! ! Þ Þ Þ $),,( $),,( $),,( 111 1 1 Bzyq Bzyq Ayq jini jini ni ssss ssss ss !!" !!" !" !" ++ Grammar Leftmost Derivation PDA Computation q0 q1 2q ess ®ii , Transition applied Read from input and remove it from stack is S®ee, $$, ®e
- 15. © 2020 Wael Badawy 15 Bzyx xAy ji ss ! ! Þ Þ Þ $),,( $),,( $),,( $),,( 11 111 1 1 Bzyq Bzyq Bzyq Ayq nj jini jini ni ss ssss ssss ss !" !" !!" !!" !" !" + ++ Grammar Leftmost Derivation PDA Computation q0 q1 2q ess ®jj , Last Transition applied All symbols have been removed from top of stack ji ss ! S®ee, $$, ®e
- 16. © 2020 Wael Badawy 16 Cpzyx Bzyx xAy kjji ji ssss ss !! ! ! 1+Þ Þ Þ Þ $),,( $),,( $),,( 11 111 11 Cpzyq Cpzyq Bzyq nk kjnj nj ss ssss ss !" !" !!" !" !" + ++ + CpB kj ss !1+® Production applied The process repeats with the next leftmost variable And so on……
- 17. © 2020 Wael Badawy 17 0q q1 2q Input Stack $ a ab Time 0 b Example: ee e e e ® ® ® ® T TaT bS aSTbS , , , , e e ® ® bb aa , , S®ee, $$, ®e
- 18. © 2020 Wael Badawy 18 q0 q1 2q Input Stack $ a ab b S Time 1 Derivation: S ee e e e ® ® ® ® T TaT bS aSTbS , , , , e e ® ® bb aa , , S®ee, $$, ®e
- 19. © 2020 Wael Badawy 19 q0 2q Input Stack $ a ab b a b S T q1 Time 2 Derivation: aSTbS Þ ee e e e ® ® ® ® T TaT bS aSTbS , , , , e e ® ® bb aa , , S®ee, $$, ®e
- 20. © 2020 Wael Badawy 20 q0 2q Input Stack $ a ab b a b S T q1 Time 3 aSTbS ÞDerivation: ee e e e ® ® ® ® T TaT bS aSTbS , , , , e e ® ® bb aa , , S®ee, $$, ®e
- 21. © 2020 Wael Badawy 21 q0 2q Input Stack $ a ab b b T b q1 Time 4 Derivation: abTbaSTbS ÞÞ ee e e e ® ® ® ® T TaT bS aSTbS , , , , e e ® ® bb aa , , S®ee, $$, ®e
- 22. © 2020 Wael Badawy 22 q0 2q Input Stack $ a ab b b T b q1 Time 5 Derivation: abTbaSTbS ÞÞ ee e e e ® ® ® ® T TaT bS aSTbS , , , , e e ® ® bb aa , , S®ee, $$, ®e
- 23. © 2020 Wael Badawy 23 q0 2q Input Stack $ a ab b b T a q1 Time 6 Derivation: abTababTbaSTbS ÞÞÞ ee e e e ® ® ® ® T TaT bS aSTbS , , , , e e ® ® bb aa , , S®ee, $$, ®e
- 24. © 2020 Wael Badawy 24 q0 2q Input Stack $ a ab b b T a q1 Time 7 Derivation: abababTababTbaSTbS ÞÞÞÞ ee e e e ® ® ® ® T TaT bS aSTbS , , , , e e ® ® bb aa , , S®ee, $$, ®e
- 25. © 2020 Wael Badawy 25 q0 2q Input Stack $ a ab b b a q1 Time 8 Derivation: abababTababTbaSTbS ÞÞÞÞ ee e e e ® ® ® ® T TaT bS aSTbS , , , , e e ® ® bb aa , , S®ee, $$, ®e
- 26. © 2020 Wael Badawy 26 q0 2q Input Stack $ a ab b b q1 Time 9 Derivation: abababTababTbaSTbS ÞÞÞÞ ee e e e ® ® ® ® T TaT bS aSTbS , , , , e e ® ® bb aa , , S®ee, $$, ®e
- 27. © 2020 Wael Badawy 27 q0 q1 2q Input Stack $ a ab b accept Time 10 Derivation: abababTababTbaSTbS ÞÞÞÞ ee e e e ® ® ® ® T TaT bS aSTbS , , , , e e ® ® bb aa , , S®ee, $$, ®e
- 28. © 2020 Wael Badawy 28 abab abTab abTb aSTb S Þ Þ Þ Þ ,$),( ,$),( $),,( $),,( $),,( $),,( $),,( $),,( $),,( ,$),( 2 1 1 1 1 1 1 1 1 0 e e q q bbq ababq Tababq Tbabq bTbbabq STbbabq Sababq ababq ! ! ! ! ! ! ! ! ! Grammar Leftmost Derivation PDA Computation
- 29. © 2020 Wael Badawy 29 Grammar generates string G wS * Þ PDA accepts M ,$),(,$),( 2 * 0 eqwq ! In general, it can be shown that: w wIf and Only if Therefore )()( MLGL =
- 31. © 2020 Wael Badawy 31 We will convert to a context-free grammar such that:G Take an arbitrary PDA )()( GLML = M M
- 32. © 2020 Wael Badawy 32 First modify PDA so that:M 1. The PDA has a single accept state 4. Each transition either pushes a symbol or pops a symbol but not both together 3. On acceptance the stack contains only stack symbol # (this symbol is not used in any transition) 2. Use new initial stack symbol #
- 33. © 2020 Wael Badawy 33 PDA fq eee ®, eee ®, e ee ®, Old accept states New accept state M 1. The PDA has a single accept state PDA 1M
- 34. © 2020 Wael Badawy 34 Z®ee, PDA 1M 2. Use new initial stack symbol # new initial stack symbol initial stack symbol of M # Z Top of stack still thinks that Z is the initial stack PDA 2M 1M @, ®ee @ auxiliary stack symbol
- 35. © 2020 Wael Badawy 35 fq ee ®x, Empty stack }#{@,-GÎ"x ee ®@, New accept state PDA 2M eee ®, Old accept state PDA 3M 3. On acceptance the stack contains only stack symbol # (this symbol is not used in any transition)
- 36. © 2020 Wael Badawy 36 iq jq ba ®,s iq jq es ®a, b®ee, 4. Each transition either pushes a symbol or pops a symbol but not both together PDA 3M PDA 4M
- 37. © 2020 Wael Badawy 37 iq jq ees ®, iq jq des ®, ede ®, PDA 3M PDA 4M Where is a symbol of the stack alphabetd
- 38. © 2020 Wael Badawy 38 4MPDA is the final modified PDA Note that the new initial stack symbol # is never used in any transition
- 39. © 2020 Wael Badawy 39 Example: aa ®e, e®ab, M aa ®e, e®ab, a®ee, ee ®a, ee ®a, ee ®b, ee ®Z, ee ®@, 4M q 0q 2q 3q 4q 5q 1q @, ®ee Z®ee,
- 40. © 2020 Wael Badawy Grammar Construction 40 ji qqA , Variables: States of PDA
- 41. © 2020 Wael Badawy 41 q PDA Grammar e®qqA Kind 1: for each state
- 42. © 2020 Wael Badawy 42 p q r PDA Grammar rqprpq AAA ® Kind 2: for every three states
- 43. © 2020 Wael Badawy 43 p qr PDA Grammar baAA rspq ® sta ®e, e®tb, Kind 3: for every pair of such transitions
- 44. © 2020 Wael Badawy 44 0q PDA Grammar fqqA 0 fq Initial state Accept state Start variable
- 45. © 2020 Wael Badawy 45 Example: PDA aa ®e, e®ab, a®ee, ee ®a, ee ®a, ee ®b, ee ®Z, ee ®@, 4M 0q 2q 3q 4q 5q 1q @, ®ee Z®ee,
- 46. © 2020 Wael Badawy 46 Grammar e e e e e e ® ® ® ® ® ® 55 44 33 22 11 00 qq qq qq qq qq qq A A A A A A Kind 1: from single states
- 47. © 2020 Wael Badawy 47 50qqAStart variable 55555445533552255115500555 55505440533052205110500050 15501440133012201110100010 05500440033002200110000000 ||||| ||||| ||||| ||||| qqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqqqqqqqqqq AAAAAAAAAAAAA AAAAAAAAAAAAA AAAAAAAAAAAAA AAAAAAAAAAAAA ® ® ® ® ! ! Kind 2: from triplets of states
- 48. © 2020 Wael Badawy 48 Kind 3: from pairs of transitions 4150 qqqq AA ® baAA qqqq 2222 ® 3242 qqqq aAA ® 4242 qqqq aAA ® bAA qqqq 2322 ® 4241 qqqq AA ® 3342 qqqq AA ® 4342 qqqq AA ® aa ®e, e®ab, a®ee, ee ®a, ee ®a, ee ®b, ee ®Z, ee ®@, 4M 0q 2q 3q 4q 5q 1q @, ®ee Z®ee,
- 49. © 2020 Wael Badawy 49 We need to prove that Suppose that a PDA is converted to a context-free grammar G M )()( MLGL = )()( MLGL Í )()( MLGL Ê or equivalently
- 50. © 2020 Wael Badawy 50 )()( MLGL Í We need to show that if has derivation: wA fqq * Þ0 Then there is an accepting computation in :M )#,,()#,,( 0 efqwq * ! G (string of terminals) with input string w
- 51. © 2020 Wael Badawy 51 We will actually show that if has derivation: wApq * Þ Then there is a computation in :M ),,(),,( eee qwp * ! G
- 52. © 2020 Wael Badawy 52 wA fqq * Þ0 )#,,()#,,( 0 efqwq * ! ),,(),,( 0 eee fqwq * ! Therefore: Since there is no transition with the # symbol
- 53. © 2020 Wael Badawy 53 Lemma: wApq * ÞIf (string of terminals) then there is a computation from state to state on string which leaves the stack empty: ),,(),,( eee qwp * ! p q w
- 54. © 2020 Wael Badawy 54 Proof Intuition: wApq ÞÞ ! wbaAA rspq ÞÞÞ ! wAAA rqprpq ÞÞÞ !Case 1: Case 2: Type 2 Type 3
- 55. © 2020 Wael Badawy 55 wAAA rqprpq ÞÞÞ !Case 1: Type 2 Stack height Input string Generated by prA p r q Generated by rqA
- 56. © 2020 Wael Badawy 56 Stack height Input string Generated by rsA p q wbaAA rspq ÞÞÞ !Case 2: Type 3 a b r s
- 57. © 2020 Wael Badawy 57 We formally prove this claim by induction on the number of steps in derivation: wApq ÞÞ ! Formal Proof: number of steps
- 58. © 2020 Wael Badawy 58 Induction Basis: wApq Þ (one derivation step) A Kind 1 production must have been used: e®ppA Therefore, and e=wqp = ),,(),,( eeee pp * ! This computation of PDA trivially exists:
- 59. © 2020 Wael Badawy 59 Induction Hypothesis: wApq ÞÞ ! k derivation steps ),,(),,( eee qwp * ! suppose it holds:
- 60. © 2020 Wael Badawy 60 Induction Step: wApq ÞÞ ! 1+k derivation steps ),,(),,( eee qwp * ! We have to show:
- 61. © 2020 Wael Badawy 61 wApq ÞÞ ! 1+k derivation steps wbaAA rspq ÞÞÞ ! wAAA rqprpq ÞÞÞ !Case 1: Case 2: Type 2 Type 3
- 62. © 2020 Wael Badawy 62 wAAA rqprpq ÞÞÞ !Case 1: Type 2 yzw =We can write yApr ÞÞ ! kAt most steps zArq ÞÞ ! kAt most steps 1+k steps
- 63. © 2020 Wael Badawy 63 yApr ÞÞ ! kAt most steps zArq ÞÞ ! kAt most steps From induction hypothesis, in PDA: ),,(),,( eee ryp * ! ),,(),,( eee qzr * ! From induction hypothesis, in PDA:
- 64. © 2020 Wael Badawy 64 ),,(),,( eee ryp * ! ),,(),,( eee qzr * ! ),,(),,(),,( eeee qzryzp ** !! ),,(),,( eee qwp * ! yzw =since
- 65. © 2020 Wael Badawy 65 wbaAA rspq ÞÞÞ !Case 2: Type 3 aybw =We can write yArs ÞÞ ! kAt most steps 1+k steps
- 66. © 2020 Wael Badawy 66 yArs ÞÞ ! kAt most steps From induction hypothesis, the PDA has computation: ),,(),,( eee syr * !
- 67. © 2020 Wael Badawy 67 p qr sta ®e, e®tb, And PDA Contains transitions Type 3 baAA rspq ® Grammar contains production wbaAA rspq ÞÞÞ !
- 68. © 2020 Wael Badawy 68 p qr sta ®e, e®tb, ),,(),,( tybraybp !e ),,(),,( eeqtbs !
- 69. © 2020 Wael Badawy 69 ),,(),,( tybraybp !e ),,(),,( eee syr * ! ),,(),,( eeqtbs ! ),,(),,( tbstybr * ! We know We also know Therefore: ),,(),,(),,(),,( eee qtbstybraybp !!! *
- 70. © 2020 Wael Badawy 70 ),,(),,(),,(),,( eee qtbstybraybp !!! * ),,(),,( eee qwp * ! aybw =since END OF PROOF
- 71. © 2020 Wael Badawy 71 So far we have shown: )()( MLGL Í With a similar proof we can show )()( MLGL Ê Therefore: )()( MLGL =