Turingmachines
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The document discusses Turing machines, which are an abstract mathematical model of computation that is capable of simulating any computer algorithm. Some key points: - Turing machines were invented by Alan Turing in 1936 and use an infinite tape divided into cells where a read/write head can access one cell at a time. - Turing machines can model any calculation that can be performed by a real-world computer and are used to formally define what is a computable function. - The Chomsky hierarchy shows the relationships between formal language classes based on the type of grammar that can generate them, with Turing machines being able to recognize recursively enumerable languages. - While all recursive languages are recursively enumerable,
Turingmachines
- 2. Invented by Alan Turing in 1936. A simple mathematical model of a general purpose computer. It is capable of performing any calculation which can be performed by any computing machine.
- 3. The Language Hierarchy *a Regular Languages Context-Free Languages nn ba R ww nnn cba ww? **ba ?
- 4. *a Regular Languages Context-Free Languages nn ba R ww nnn cba ww **ba Languages accepted by Turing Machines Finite Automata NDPA
- 5. A Turing Machine ............ Tape Read-Write head Control Unit
- 6. The Tape ............ Read-Write head No boundaries -- infinite length The head moves Left or Right
- 7. ............ Read-Write head The head at each time step: 1. Reads a symbol 2. Writes a symbol 3. Moves Left or Right
- 8. ............ Example: Time 0 ............ Time 1 1. Reads 2. Writes a a cb a b k c a k 3. Moves Left
- 9. ............ Time 1 a b k c ............ Time 2 a k cf 1. Reads 2. Writes b f 3. Moves Right
- 10. The Input String ............ Blank symbol head a b ca Head starts at the leftmost position of the input string Input string Are treated as left and right brackets for the input written on the tape.
- 11. States & Transitions 1q 2qLba , Read Write Move Left 1q 2qRba , Move Right
- 12. Example: 1q 2qRba , ............ a b ca Time 1 1q current state
- 13. ............ a b ca Time 1 1q 2qRba , ............ a b cb Time 2 1q 2q
- 14. ............ a b ca Time 1 1q 2qLba , ............ a b cb Time 2 1q 2q Example:
- 15. ............ a b ca Time 1 1q 2qRg, ............ ga b cb Time 2 1q 2q Example:
- 16. Determinism 1q 2qRba , Allowed Not Allowed 3qLdb , 1q 2qRba , 3qLda , No lambda transitions allowed Turing Machines are deterministic
- 17. Partial Transition Function 1q 2qRba , 3qLdb , ............ a b ca 1q Example: No transition for input symbol c Allowed:
- 18. Halting The machine halts if there are no possible transitions to follow
- 19. Example: ............ a b ca 1q 1q 2qRba , 3qLdb , No possible transition HALT!!!
- 20. Final States 1q 2q Allowed 1q 2q Not Allowed • Final states have no outgoing transitions • In a final state the machine halts
- 21. Acceptance Accept Input If machine halts in a final state Reject Input If machine halts in a non-final state or If machine enters an infinite loop
- 22. Turing Machine Example A Turing machine that accepts the language: *aa 0q Raa , L, 1q
- 27. aaTime 4 1q a 0q Raa , L, 1q Halt & Accept
- 28. Rejection Example 0q Raa , L, 1q baTime 0 0q a
- 29. 0q Raa , L, 1q baTime 1 0q a No possible Transition Halt & Reject
- 30. Infinite Loop Example 0q Raa , L, 1q Lbb ,
- 34. baTime 2 0q a baTime 3 0q a baTime 4 0q a baTime 5 0q a ... Infinite Loop
- 35. Because of the infinite loop: •The final state cannot be reached •The machine never halts •The input is not accepted
- 36. Another Turing Machine Example Turing machine for the language }{ nn ba 0q 1q 2q3q Rxa , Raa , Ryy , Lyb , Laa , Lyy , Rxx , Ryy , Ryy , 4q L,
- 37. 0q 1q 2q3q Rxa , Raa , Ryy , Lyb , Laa , Lyy , Rxx , Ryy , Ryy , 4q L, ba 0q a bTime 0
- 38. 0q 1q 2q3q Rxa , Raa , Ryy , Lyb , Laa , Lyy , Rxx , Ryy , Ryy , 4q L, bx 1q a bTime 1
- 39. 0q 1q 2q3q Rxa , Raa , Ryy , Lyb , Laa , Lyy , Rxx , Ryy , Ryy , 4q L, bx 1q a bTime 2
- 40. 0q 1q 2q3q Rxa , Raa , Ryy , Lyb , Laa , Lyy , Rxx , Ryy , Ryy , 4q L, yx 2q a bTime 3
- 41. 0q 1q 2q3q Rxa , Raa , Ryy , Lyb , Laa , Lyy , Rxx , Ryy , Ryy , 4q L, yx 2q a bTime 4
- 42. 0q 1q 2q3q Rxa , Raa , Ryy , Lyb , Laa , Lyy , Rxx , Ryy , Ryy , 4q L, yx 0q a bTime 5
- 43. 0q 1q 2q3q Rxa , Raa , Ryy , Lyb , Laa , Lyy , Rxx , Ryy , Ryy , 4q L, yx 1q x bTime 6
- 44. 0q 1q 2q3q Rxa , Raa , Ryy , Lyb , Laa , Lyy , Rxx , Ryy , Ryy , 4q L, yx 1q x bTime 7
- 45. 0q 1q 2q3q Rxa , Raa , Ryy , Lyb , Laa , Lyy , Rxx , Ryy , Ryy , 4q L, yx x y 2q Time 8
- 46. 0q 1q 2q3q Rxa , Raa , Ryy , Lyb , Laa , Lyy , Rxx , Ryy , Ryy , 4q L, yx x y 2q Time 9
- 47. 0q 1q 2q3q Rxa , Raa , Ryy , Lyb , Laa , Lyy , Rxx , Ryy , Ryy , 4q L, yx 0q x yTime 10
- 48. 0q 1q 2q3q Rxa , Raa , Ryy , Lyb , Laa , Lyy , Rxx , Ryy , Ryy , 4q L, yx 3q x yTime 11
- 49. 0q 1q 2q3q Rxa , Raa , Ryy , Lyb , Laa , Lyy , Rxx , Ryy , Ryy , 4q L, yx 3q x yTime 12
- 50. 0q 1q 2q3q Rxa , Raa , Ryy , Lyb , Laa , Lyy , Rxx , Ryy , Ryy , 4q L, yx 4q x y Halt & Accept Time 13
- 51. If we modify the machine for the language }{ nn ba we can easily construct a machine for the language }{ nnn cba Observation:
- 53. 1q 2qLdc , ),,(),( 21 Ldqcq Transition Function
- 54. Turing Machine: ),,,,,,( 0 FqQM States Input alphabet Tape alphabet A partial Transition function Initial state Blank : a special symbol Of Final states
- 56. yx 2q a b Time 4 yx 0q a b Time 5 A Move: aybqxxaybq 02
- 57. yx 2q a b Time 4 yx 0q a b Time 5 bqxxyybqxxaybqxxaybq 1102 yx 1q x b Time 6 yx 1q x b Time 7
- 58. bqxxyybqxxaybqxxaybq 1102 bqxxyxaybq 12 Equivalent notation:
- 59. Initial configuration: wq0 ba 0q a b w Input string
- 60. The Accepted Language For any Turing Machine M }:{)( 210 xqxwqwML f Initial state Final state
- 61. Standard Turing Machine • Deterministic • Infinite tape in both directions •Tape is the input/output file The machine we described is the standard:
- 62. Design a Turing machine to recognize all strings in which 010 is present as a substring. q0 q1 q2 H 0,0,R 1,1,R 0,0,R 0,0,R 1,1, R 1,1,R
- 63. DFA for the previous language q0 q1 q2 0 1 0 0 1 1 0 , 1
- 64. Turing machine for odd no of 1’s 1, 1 , R 1, b , R 1, b , R
- 66. Definition: A language is recursively enumerable if some Turing machine accepts it
- 67. For string : Let be a recursively enumerable languageL and the Turing Machine that accepts itM Lw w if then halts in a final stateM Lwif then halts in a non-final stateM or loops forever
- 68. Definition: A language is recursive if some Turing machine accepts it and halts on any input string In other words: A language is recursive if there is a membership algorithm for it
- 69. For string : Let be a recursive languageL and the Turing Machine that accepts itM Lw w if then halts in a final stateM Lwif then halts in a non-final stateM
- 70. We will prove: 1. There is a specific language which is not recursively enumerable (not accepted by any Turing Machine) 2. There is a specific language which is recursively enumerable but not recursive
- 71. Recursive Recursively Enumerable Non Recursively Enumerable
- 72. We will first prove: • If a language is recursive then there is an enumeration procedure for it • A language is recursively enumerable if and only if there is an enumeration procedure for it
- 74. Unrestricted Grammars: Productions vu String of variables and terminals String of variables and terminals
- 76. A language is recursively enumerable if and only if is generated by an unrestricted grammar L L Theorem:
- 77. Context-Sensitive Grammars: and: |||| vu Productions vu String of variables and terminals String of variables and terminals
- 78. The language }{ nnn cba is context-sensitive: aaAaaaB BbbB BbccAc bAAb aAbcabcS | |
- 79. The language }{ nnn cba is context-sensitive: aaAaaaB BbbB BbccAc bAAb aAbcabcS | |
- 80. A language is context sensistive if and only if is accepted by a Linear-Bounded automatonL L Theorem:
- 81. There is a language which is context-sensitive but not recursive Observation: