Turing machines

1. Costas Busch - LSU 1
Turing Machines

2. Costas Busch - LSU 2
The Language Hierarchy
*a
Regular Languages
Context-Free Languages
nn
ba R
ww
nnn
cba ww?
**ba
?

3. Costas Busch - LSU 3
*a
Regular Languages
Context-Free Languages
nn
ba R
ww
nnn
cba ww
**ba
Languages accepted by
Turing Machines

4. Costas Busch - LSU 4
A Turing Machine
............
Tape
Read-Write head
Control Unit

5. Costas Busch - LSU 5
The Tape
............
Read-Write head
No boundaries -- infinite length
The head moves Left or Right

6. Costas Busch - LSU 6
............
Read-Write head
The head at each transition (time step):
1. Reads a symbol
2. Writes a symbol
3. Moves Left or Right

7. Costas Busch - LSU 7
............
Example:
Time 0
............
Time 1
1. Reads
2. Writes
a a cb
a b k c
a
k
3. Moves Left

8. Costas Busch - LSU 8
............
Time 1
a b k c
............
Time 2
a k cf
1. Reads
2. Writes
b
f
3. Moves Right

9. Costas Busch - LSU 9
The Input String
............    
Blank symbol
head
a b ca
Head starts at the leftmost position
of the input string
Input string

10. Costas Busch - LSU 10
States & Transitions
1q 2qLba ,
Read Write
Move Left
1q 2qRba ,
Move Right

11. Costas Busch - LSU 11
Example:
1q 2qRba ,
............    a b ca
Time 1
1q
current state

12. Costas Busch - LSU 12
............    a b ca
Time 1
1q 2qRba ,
............    a b cb
Time 2
1q
2q

13. Costas Busch - LSU 13
............    a b ca
Time 1
1q 2qLba ,
............    a b cb
Time 2
1q
2q
Example:

14. Costas Busch - LSU 14
............    a b ca
Time 1
1q 2qRg,
............    ga b cb
Time 2
1q
2q
Example:

15. Costas Busch - LSU 15
Determinism
1q
2qRba ,
Allowed Not Allowed
3qLdb ,
1q
2qRba ,
3qLda ,
No lambda transitions allowed
Turing Machines are deterministic

16. Costas Busch - LSU 16
Partial Transition Function
1q
2qRba ,
3qLdb ,
............    a b ca
1q
Example:
No transition
for input symbol c
Allowed:

17. Costas Busch - LSU 17
Observation:
In order to accept an input string,
it is not necessary to scan all the
symbols in the string

18. Costas Busch - LSU 18
Halting
The machine halts in a state if there is
no transition to follow

19. Costas Busch - LSU 19
Halting Example 1:
............    a b ca
1q
1q No transition from
HALT!!!
1q

20. Costas Busch - LSU 20
Halting Example 2:
............    a b ca
1q
1q
2qRba ,
3qLdb ,
No possible transition
from and symbol
HALT!!!
1q c

21. Costas Busch - LSU 21
Accepting States
1q 2q Allowed
1q 2q Not Allowed
•Accepting states have no outgoing transitions
•The machine halts and accepts

22. Costas Busch - LSU 22
Acceptance
Accept Input
If machine halts
in an accept state
Reject Input
If machine halts
in a non-accept state
or
If machine enters
an infinite loop
string
string

23. Costas Busch - LSU 23
Turing Machine Example
Accepts the language: *a
0q
Raa ,
L,
1q
Input alphabet },{ ba

24. Costas Busch - LSU 24
   aaTime 0
0q
a
0q
Raa ,
L,
1q

25. Costas Busch - LSU 25
   aaTime 1
0q
a
0q
Raa ,
L,
1q

26. Costas Busch - LSU 26
   aaTime 2
0q
a
0q
Raa ,
L,
1q

27. Costas Busch - LSU 27
   aaTime 3
0q
a
0q
Raa ,
L,
1q

28. Costas Busch - LSU 28
   aaTime 4
1q
a
0q
Raa ,
L,
1q
Halt & Accept

29. Costas Busch - LSU 29
Rejection Example
0q
Raa ,
L,
1q
   baTime 0
0q
a

30. Costas Busch - LSU 30
0q
Raa ,
L,
1q
   baTime 1
0q
a
No possible Transition
Halt & Reject

31. Costas Busch - LSU 31
Accepts the language: *a
0q
but for input alphabet }{a
A simpler machine for same language

32. Costas Busch - LSU 32
   aaTime 0
0q
a
0q
Halt & Accept
Not necessary to scan input

33. Costas Busch - LSU 33
Infinite Loop Example
0q
Raa ,
L,
1q
Lbb ,
A Turing machine
for language *)(* baba 

34. Costas Busch - LSU 34
   baTime 0
0q
a
0q
Raa ,
L,
1q
Lbb ,

35. Costas Busch - LSU 35
   baTime 1
0q
a
0q
Raa ,
L,
1q
Lbb ,

36. Costas Busch - LSU 36
   baTime 2
0q
a
0q
Raa ,
L,
1q
Lbb ,

37. Costas Busch - LSU 37
   baTime 2
0q
a
   baTime 3
0q
a
   baTime 4
0q
a
   baTime 5
0q
a
Infiniteloop

38. Costas Busch - LSU 38
Because of the infinite loop:
•The accepting state cannot be reached
•The machine never halts
•The input string is rejected

39. Costas Busch - LSU 39
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,
1n

40. Costas Busch - LSU 40
Match a’s with b’s:
Repeat:
replace leftmost a with x
find leftmost b and replace it with y
Until there are no more a’s or b’s
If there is a remaining a or b reject
Basic Idea:

41. Costas Busch - LSU 41
0q 1q 2q3q
Rxa ,
Raa ,
Ryy ,
Lyb ,
Laa ,
Lyy ,
Rxx ,
Ryy ,
Ryy ,
4q
L,
 ba
0q
a bTime 0 

42. Costas Busch - LSU 42
0q 1q 2q3q
Rxa ,
Raa ,
Ryy ,
Lyb ,
Laa ,
Lyy ,
Rxx ,
Ryy ,
Ryy ,
4q
L,
 bx
1q
a b Time 1

43. Costas Busch - LSU 43
0q 1q 2q3q
Rxa ,
Raa ,
Ryy ,
Lyb ,
Laa ,
Lyy ,
Rxx ,
Ryy ,
Ryy ,
4q
L,
 bx
1q
a b Time 2

44. Costas Busch - LSU 44
0q 1q 2q3q
Rxa ,
Raa ,
Ryy ,
Lyb ,
Laa ,
Lyy ,
Rxx ,
Ryy ,
Ryy ,
4q
L,
 yx
2q
a b Time 3

45. Costas Busch - LSU 45
0q 1q 2q3q
Rxa ,
Raa ,
Ryy ,
Lyb ,
Laa ,
Lyy ,
Rxx ,
Ryy ,
Ryy ,
4q
L,
 yx
2q
a b Time 4

46. Costas Busch - LSU 46
0q 1q 2q3q
Rxa ,
Raa ,
Ryy ,
Lyb ,
Laa ,
Lyy ,
Rxx ,
Ryy ,
Ryy ,
4q
L,
 yx
0q
a b Time 5

47. Costas Busch - LSU 47
0q 1q 2q3q
Rxa ,
Raa ,
Ryy ,
Lyb ,
Laa ,
Lyy ,
Rxx ,
Ryy ,
Ryy ,
4q
L,
 yx
1q
x b Time 6

48. Costas Busch - LSU 48
0q 1q 2q3q
Rxa ,
Raa ,
Ryy ,
Lyb ,
Laa ,
Lyy ,
Rxx ,
Ryy ,
Ryy ,
4q
L,
 yx
1q
x b Time 7

49. Costas Busch - LSU 49
0q 1q 2q3q
Rxa ,
Raa ,
Ryy ,
Lyb ,
Laa ,
Lyy ,
Rxx ,
Ryy ,
Ryy ,
4q
L,
 yx x y
2q
Time 8

50. Costas Busch - LSU 50
0q 1q 2q3q
Rxa ,
Raa ,
Ryy ,
Lyb ,
Laa ,
Lyy ,
Rxx ,
Ryy ,
Ryy ,
4q
L,
 yx x y
2q
Time 9

51. Costas Busch - LSU 51
0q 1q 2q3q
Rxa ,
Raa ,
Ryy ,
Lyb ,
Laa ,
Lyy ,
Rxx ,
Ryy ,
Ryy ,
4q
L,
 yx
0q
x y Time 10

52. Costas Busch - LSU 52
0q 1q 2q3q
Rxa ,
Raa ,
Ryy ,
Lyb ,
Laa ,
Lyy ,
Rxx ,
Ryy ,
Ryy ,
4q
L,
 yx
3q
x y Time 11

53. Costas Busch - LSU 53
0q 1q 2q3q
Rxa ,
Raa ,
Ryy ,
Lyb ,
Laa ,
Lyy ,
Rxx ,
Ryy ,
Ryy ,
4q
L,
 yx
3q
x y Time 12

54. Costas Busch - LSU 54
0q 1q 2q3q
Rxa ,
Raa ,
Ryy ,
Lyb ,
Laa ,
Lyy ,
Rxx ,
Ryy ,
Ryy ,
4q
L,
 yx
4q
x y 
Halt & Accept
Time 13

55. Costas Busch - LSU 55
If we modify the
machine for the language }{ nn
ba
we can easily construct
a machine for the language }{ nnn
cba
Observation:

56. Costas Busch - LSU 56
Formal Definitions
for
Turing Machines

57. Costas Busch - LSU 57
Transition Function
1q 2qRba ,
),,(),( 21 Rbqaq 

58. Costas Busch - LSU 58
1q 2qLdc ,
),,(),( 21 Ldqcq 
Transition Function

59. Costas Busch - LSU 59
Turing Machine:
),,,,,,( 0 FqQM  
States
Input
alphabet
Tape
alphabet
Transition
function
Initial
state
blank
Accept
states

60. Costas Busch - LSU 60
Configuration
  ba
1q
a
Instantaneous description:
c
baqca 1

61. Costas Busch - LSU 61
 yx
2q
a b
Time 4
 yx
0q
a b
Time 5
 
A Move: aybqxxaybq 02 

62. Costas Busch - LSU 62
 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
 
A computation

63. Costas Busch - LSU 63
bqxxyybqxxaybqxxaybq 1102 
bqxxyxaybq 12

Equivalent notation:

64. Costas Busch - LSU 64
Initial configuration: wq0
 ba
0q
a b 
w
Input string

65. Costas Busch - LSU 65
The Accepted Language
For any Turing Machine M
}:{)( 210 xqxwqwML f

 
Initial state Accept state

66. Costas Busch - LSU 66
If a language is accepted
by a Turing machine
then we say that is:
•Turing Acceptable
•Recursively Enumerable
M
L
L
•Turing Recognizable
Other names used: