Introduction Turing machine and short overview Taken from internet

1. Turing Machines

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

23. aaTime 0
0q
a
0q
Raa ,
L,
1q

24. aaTime 1
0q
a
0q
Raa ,
L,
1q

25. aaTime 2
0q
a
0q
Raa ,
L,
1q

26. aaTime 3
0q
a
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 ,

31. baTime 0
0q
a
0q
Raa ,
L,
1q
Lbb ,

32. baTime 1
0q
a
0q
Raa ,
L,
1q
Lbb ,

33. baTime 2
0q
a
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:

52. Formal Definitions
for
Turing Machines

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

55. Configuration
ba
1q
a
Instantaneous description:
c
baqca 1

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

65. Recursively Enumerable
and
Recursive
Languages

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

73. The Chomsky Hierarchy

74. Unrestricted Grammars:
Productions
vu
String of variables
and terminals
String of variables
and terminals

75. Example unrestricted grammar:
dAc
cAaB
aBcS

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:

82. Non-recursively enumerable
Recursively-enumerable
Recursive
Context-sensitive
Context-free
Regular
The Chomsky Hierarchy