Class8

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Class8

  1. 1. Pushdown Automata (PDA) Pushdown Automata PDAs Homework (p.183) 2, 3abc, 4a~h, 5, 6, 9~11 1
  2. 2. Pushdown Automata (PDA) Pushdown Automaton -- PDA Input String* Stack with symbol from  States  Gamma  Sigma 2
  3. 3. Pushdown Automata (PDA) Initial Stack Symbol Stack z bottom special symbol 3
  4. 4. Pushdown Automata The States (PDA) Symbol on Input symbol top of the The string may be  stack that replaces q1 a,b, x q 2 a   {} b x * Input alphabet Stack alphabet b & x are all for stack processing 4
  5. 5. Pushdown Automata (PDA) q1 a,b, ? q2 ?   (q1, a, b) input … a … stack b top h For the execution of the stack, there e are 4 kinds of operations: replace, z push, pop, and no change. 5
  6. 6. Pushdown Automata (PDA) q1 a,b, ? q2 (q2, c)   (q1, a, b): when it is on the state q1 and input a is read, then b is replace by c and moves to q2 … a … … a … stack b top c h Replace h e e z z 6
  7. 7. Pushdown Automata (PDA) q1 a,b, c q2 (q2, c)   (q1, a, b): when it is on the state q1 and input a is read, then b is replace by c and moves to q2 … a … … a … stack b top c h Replace h e e z z 7
  8. 8. Pushdown Automata (PDA) q1 a,b, ? q2 (q2, cb)   (q1, a, b): when it is on the state q1 and a is read, input then c is pushed, i.e. b is replace by cb, and moves to q2 … a … … a … stack c b top b h Push h e e z In Def 7.1 (p.177), it is written as (q2, cb)  (q1, a, b) for pushing. z For the execution of the stack, there are 4 kinds of operations: replace, push, pop, and no change. 8
  9. 9. Pushdown Automata (PDA) q1 a,b, cb q2 (q2, cb)   (q1, a, b): when it is on the state q1 and a is read, input then c is pushed, i.e. b is replace by cb, and moves to q2 … a … … a … stack c b top b h Push h e e z z 9
  10. 10. Pushdown Automata (PDA) q1 a,b, ? q2 (q2, )   (q1, a, b): when it is on the state q1 and a is read, input then b is popped, i.e. b is replace by , and moves to q2 … a … … a … stack b top Pop h h e e Beware, when we push a symbol into stack—we do z not really care about what is the top symbol of the z stack. But if we want to pop a symbol out of a stack, we need to know what is the top stack symbol (will be popped up). 10
  11. 11. Pushdown Automata a,b,  (PDA) q1 q2 (q2, )   (q1, a, b): when it is on the state q1 and a is read, input then b is popped, i.e. b is replace by , and moves to q2 … a … … a … stack b top h Pop h e e z z 11
  12. 12. Pushdown Automata (PDA) q1 a,b, ? q2 (q2, b)   (q1, a, b): when it is on the state q1 and a is read, input nothing will be changed on the stack and moves to q2 … a … … a … stack b top b h No Change h e e z z 12
  13. 13. Pushdown Automata (PDA) q1 a,b, b q2 (q2, b)   (q1, a, b): when it is on the state q1 and a is read, input nothing will be changed on the stack and moves to q2 … a … … a … stack b top b h No Change h e e z z 13
  14. 14. Pushdown Automata (PDA) Every FA is a PDA Consider regular language L={ a2nbm : n, m  0} with FA as: b , z, z a , z, z a , z, z  , z, z It becomes a PDA ! 14
  15. 15. Pushdown Automata Every FA is a PDA (PDA) LM   L(a * b) a, z, z a , z, z a b b,,z, z a, z, z b, z, z b, , b z a z, q0 q1 q2 It becomes a PDA ! Do Hw#3 on p. 183 15
  16. 16. Pushdown Automata (PDA) While it is on state q1 and reads a Non-Determinism If top of stack symbol is b: (q2, c)   (q1, a, b) q2   transition a,b, c q1  ,b, c q2 q1 If top of stack symbol is b: moves to q2 without consuming any input symbol, i.e. (q2,c) (q1,, b) a,d , ed q3 If top of stack symbol is d: (q3, ed)   (q1, a, d) These are allowed transitions in a Non-deterministic PDA (NPDA) 16
  17. 17. Formal Definition Pushdown Automata (PDA) Non-Deterministic Pushdown Automaton NPDA M  (Q, Σ, Γ, δ, q0 , z, F ) Final States states Input Stack alphabet Transition Initial start Stack function state symbol alphabet 17
  18. 18. NPDA: Non-Deterministic Pushdown Automata (PDA) PDA Example: a, z, az a, a, aa b, a,  q0  , z, z q1 b,a,  q2  , z, z q 3 According to the def. 7.1, , should be replaced by ,$$ transitions on q1 should be { a,$a$, a, aaa} 18
  19. 19. Execution Example: Time 0 (initial configuration) Pushdown Automata (PDA) a a a b b b z Stack a, z, az current a, a, aa b, a,  state q0  , z, z q1 b,a,  q2  , z, z q 3 19
  20. 20. Execution Example: Time 1 Pushdown Automata (PDA) a a a b b b z Stack a, z, az a, a, aa b, a,  q0  , z, z q1 b,a,  q  ,z, z q 2 3 20
  21. 21. Execution Example: Time 1 Pushdown Automata (PDA) a a a b b b z Stack a, z, az a, a, aa b, a,  q0  , z, z q1 b,a,  q2  ,z, z q 3 21
  22. 22. Execution Example: Time 2 Pushdown Automata (PDA) a a a b b b a z z Stack a, Z , aZ a, a, aa b, a,  q0  ,Z, Z q b,a,  q2  ,z, z q 1 3 演變過程請看投影片動畫 22
  23. 23. Execution Example: Time 3 Pushdown Automata (PDA) a a a a b b b a a z z Stack a, Z , aZ a, a, aa b, a,  q0  , z, z q1 b,a,  q2  ,z, z q 3 演變過程請看投影片動畫 23
  24. 24. Execution Example: Time 4 Pushdown Automata (PDA) a a a a b b b a a a a z z Stack a, z, az a, a, aa b, a,  q0  , z, z q1 b,a,  q2  ,z, z q 3 演變過程請看投影片動畫 24
  25. 25. Execution Example: Time 5 Pushdown Automata (PDA) a a a a b b b a a z Stack a, z, az a, a, aa b, a,  q0  , z, z q1 b,a,  q2  ,z, z q 3 演變過程請看投影片動畫 25
  26. 26. Execution Example: Time 6 Pushdown Automata (PDA) a a a a b b b a a z Stack a, z, az a, a, aa b, a,  q0  , z, z q1 b,a,  q2  ,z, z q 3 演變過程請看投影片動畫 26
  27. 27. Execution Example: Time 7 Pushdown Automata (PDA) a a a a b b b a a z Stack a, z, az a, a, aa b, a,  q0  , z, z q1 b,a,  q2  ,z, z q 3 演變過程請看投影片動畫 27
  28. 28. Execution Example: Time 8 Pushdown Automata (PDA) a a a a b b b a a z Stack a, z, az a, a, aa b, a,  q0  , z, z q1 b,a,  q2  ,z, z q 3 演變過程請看投影片動畫 28
  29. 29. Execution Example: End of input tape Pushdown Automata (PDA) a a a a b b b a a z Stack a, z, az a, a, aa b, a,  accept q0  , z, z q1 b,a,  q2  ,z, z q 3 演變過程請看投影片動畫 29
  30. 30. Pushdown Automata (PDA) A string is accepted if there is a computation such that: • All the input is consumed and • The last state is a final state At the end of the computation, we do not care about the stack contents 30
  31. 31. Pushdown Automata (PDA) Therefore: The input string aaabbb is accepted by the NPDA: a, z, az a, a, aa b, a,  q0  , z, z q1 b,a,  q2  , z, z q 3 31
  32. 32. Pushdown Automata (PDA) L( M ) = ? { anbn : n  0 } is the language accepted by the NPDA: a, z, az a, a, aa b, a,  q0  , z, z q1 b,a,  q2  , z, z q 3 32
  33. 33. Pushdown Automata (PDA) L( M ) = { anbn : n  0 } Can we find a machine with 3 states? Can we find a machine with 2 states? What is L(M) if both q0 & q1 are final states? a, z, az a, a, aa b, a,  q0  , z, z q1 b,a,  q2  , z, z q 3 33
  34. 34. Pushdown Automata (PDA) Pushing Strings Input Top stack Push symbol symbol string q1 a, b, w q2 The top stack symbol b is replaced by the string w 34
  35. 35. Example: Pushdown Automata (PDA) q1 a, b, cdf q2 input top   a     a   c pushed stack d top string b f h Push h e e Z Z 35
  36. 36. Example: Pushdown Automata (PDA) input Beware: we can push a string into a stack but we do not pop a string from a stack--- only a      pop one symbol or nothing at a a   time!  c pushed stack d top string b f h Push h e e Z Z In fact, to read from input tape or pop up from stack, there is at most one symbol at a time. The main reason is we can only see one symbol at a time. As to push into stack is like to record an information so we can write a string, or push a string into a stack. 36
  37. 37. Pushdown Automata Examples for NPDA (PDA) Construct an NPDA for LM   {a b n 2n : n  0} LM   {a b : n  0} 2n n LM   {a b c n m nm : n, m  0} 37
  38. 38. Formal Def. for NPDA M Pushdown Automata (PDA) L ={ anbn : n  0 } M  ({q0 , q1 , q2 , q3},   {a, b},   {a, b, z},  , q0 , z, F  {q0 , q3}) a, z, az a, a, aa b, a,  q0  , z, z q1 b,a,  q2  , z, z q 3 38
  39. 39. Pushdown Automata Transition function: (PDA) q2 a, b, w q1 a, b, w q3  (q1, a, b)  {(q2 , w), (q3 , w)}  : Q  ({}  )    finite subsets of Q   * 39
  40. 40. M  ({q0 , q1 , q2 , q3},   {a, b},   {a, b, z}, Pushdown Automata (PDA)  , q0 , z, F  {q0 , q3}) Transition Functions  ? a, z, az a, a, aa b, a,  q0  , z, z q1 b,a,  q2  , z, z q 3 40
  41. 41. Pushdown Automata (PDA) Transition Functions  (q1, a, z) = {(q1, az)} (q1, a, a) = {(q1, aa)} (q2, b, a) = {(q2, )} a, z, az a, a, aa b, a,  q0  , z, z q1 b,a,  q2  , z, z q 3 (q0, , z) = {(q1, z)} (q1, b, a) = {(q2, )} (q2, , z) = {(q3, z)} 41
  42. 42. Pushdown Automata (PDA) Instantaneous Description ( q, u , s ) Current Current Remaining state stack input contents 42
  43. 43. Pushdown Automata (PDA) Time 4 a a a a b b b a a z Instantaneous ( q1, bbb, aaaz) Description Stack a, z, az a, a, aa b, a,  q0  , z, z q1 b,a,  q2  ,z, z q 3 43
  44. 44. Execution Example: Time 5 Pushdown Automata (PDA) a a a a b b b a a z Instantaneous ( q2, bb, aaz) Description Stack a, z, az a, a, aa b, a,  q0  , z, z q1 b,a,  q2  ,z, z q 3 44
  45. 45. Pushdown Automata (PDA) We write: (q1 , bbb, aaaZ ) ├ (q2 , bb, aaZ ) Time 4 Time 5 45
  46. 46. Pushdown Automata (PDA) A computation for w= aaabbb: (q0 , aaabbb, z )├ (q1 , aaabbb, z ) ├ (q1 , aabbb, az)├ (q1 , abbb, aaz) ├ (q1 , bbb, aaaz) ├ (q2 , bb, aaz)├ (q2 , b, az) ├ (q2 ,  , z ) ├(q3 ,  , z ) a, z, az a, a, aa b, a,  q0  , z, z q1 b,a,  q2  , z, z q 3 46
  47. 47. Pushdown Automata (PDA) (q0 , aaabbb, z ) (q1 , aaabbb, z )├ ├ (q1 , aabbb, az)├ (q1 , abbb, aaz) ├ (q1 , bbb, aaaz) ├ (q2 , bb, aaz) (q2 , b, az)├ (q2 ,  , z )├ (q3 ,  , z ) ├ For convenience we write: (q0 , aaabbb, z ) ├ (q3 ,  , z ) * 47
  48. 48. Pushdown Automata (PDA) Formal Definition Language of NPDA M : *  L( M )  {w : (q0 , w, z )├ (q f ,  , u )} Initial state Final state empty 48
  49. 49. Example: Pushdown Automata (PDA)  (q0 , aaabbb, z ) ├ (q3 ,  , z ) aaabbb L(M ) NPDA M : a, z, az a, a, aa b, a,  q0  , z, z q1 b,a,  q2  , z, z q 3 49
  50. 50. Pushdown Automata (PDA)  (q0 , a b , z ) ├ (q3 ,  , z ) n n a b  L(M ) n n NPDA M : a, z, az a, a, aa b, a,  q0  , z, z q1 b,a,  q2  , z, z q 3 50
  51. 51. Pushdown Automata (PDA) Therefore: L( M )  {a b : n  0} n n NPDA M : a, z, az a, a, aa b, a,  q0  , z, z q1 b,a,  q2  , z, z q 3 51
  52. 52. Pushdown Automata (PDA) Another NPDA example NPDA M for L( M )  {ww : w {a, b}*} R 52
  53. 53. Pushdown Automata NPDA M for (PDA) L( M )  {ww : w {a, b}*} R a, Z , aZ b, Z , bZ a, a,  a, a, aa b, a, ba b, b,  a, b, ab b, b, bb , Z , Z  , a, a  ,Z, Z q2 q0  , b, b q1 Move to Move to Push w next state Pop wR final state 53
  54. 54. NPDA M for L( M )  {ww : w  {a, b}*} R Pushdown Automata (PDA) Try strings like: , abba, baa. If it is accepted, write the computation. a , z , az b, z , bz a , a , aa b, a, ba a, a,  a , b, ab b, b, bb b, b,   , z, z  , a, a  ,z, z q0 q1 q2  , b, b 54
  55. 55. NPDA M for L( M )  {ww : w  {a, b}*} R Pushdown Automata (PDA) M  ({q0 , q1 , q2 },   {a, b},   {a, b, z},  , q0 , z, F  {q2 }) How about for L( M )  {wcw : w  {a, b}*} R a , z , az b, z , bz a , a , aa b, a, ba a, a,  a , b, ab b, b, bb b, b,   , z, z  , a, a  ,z, z q0 q1 q2  , b, b 55
  56. 56. Pushdown Automata (PDA) Another NPDA example NPDA M for L( M )  {w : na  nb } How to use the stack to recognize the strings like : abbbaa babbaaba  56
  57. 57. M for L( M )  {w : na  nb } Pushdown Automata NPDA (PDA) M  ({q1 , q2},   {a, b},   {0,1, z},  , q1 , Z , F  {q2}) Write a computation for abbbaa. a,z,0 z b,z,1z a,0,00 b,1,11 a,1,  b,0,  q1  , z, z q2 57
  58. 58. M for L( M )  {w : na  nb } Pushdown Automata NPDA (PDA) How about NPDA for {w: na = nb+1 } ? How about NPDA for {w: na = 2nb } ? a,z,0 z b,z,1z a,0,00 b,1,11 a,1,  b,0,  q1  , z, z q2 {w: na = 2nb }: one b and push 2 bs, or push one b and pop up one a, or pop up two as, i.e., b, z, bbz; b, b, bbb in q1, but b, a, (pop up and go to another state, say q’, such that ,a,  (pop up one more time); or , z, bz; or , b, bb; and all go back to state q1. 58
  59. 59. M for L( M )  {w : na  2nb } Pushdown Automata NPDA (PDA) a,Z ,0Z b,Z ,1Z b,Z ,11Z a,0,00 b,1,11 b,1,111 a,1,  b,0,  q3  ,Z ,1Z  ,0,  q1 q2  , Z, Z 59
  60. 60. Pushdown Automata More Example on designing: (PDA) LM   {a b : n  m  2n} n m How about NPDA for {anbm: n < m } ? How about NPDA for {anbm: n > m } ? How about NPDA for {anbm: n  m } ? 60

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