Turing Machine

Source of Slides: Introduction to Automata Theory, Languages, and Computation
By John E. Hopcroft, Rajeev Motwani and Jeffrey D. Ullman

Dept. of Computer Science & IT, FUUAST
Theory of Computation
2
Turing MachineTuring Machine
Church-Turing’s ThesisChurch-Turing’s Thesis
Any mathematical problem
solving that can be described
by an algorithm can be modeled
by a Turing Machine.

3
1)1) Multiple trackMultiple track
2) Shift over Turing Machine2) Shift over Turing Machine
3) Nondeterministic3) Nondeterministic
4) Two way Turing Machine4) Two way Turing Machine
5) Multitape Turing Machine5) Multitape Turing Machine
6) Multidimensional Turing Machine6) Multidimensional Turing Machine
7) Composite Turing Machine7) Composite Turing Machine
8) Universal Turing Machine8) Universal Turing Machine
Types of Turing Machine

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Formal Definition

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1. Start in state q0
2. Read symbol under head
3. Write new symbol
4. Shift left/right
5. Enter new state qj
Steps

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Notational Conventions For Turing Machines

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A Turing Machine M that accepts the language { 0n
1n
| n ≥0 }

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Moves for input 0011:
Moves for input 0010:

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Transition Diagram for 0011 input

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Pushdown AutomataPushdown Automata
A Turing Machine M computes a function ( proper subtraction) for
0m
10n
on the tape.
means if m ≥ n then m - n else if m < n then 0

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Evaluating function

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Evaluating function

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Transition Table for the function

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Transition Diagram for

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Transition Table for the function
Transition Diagram for

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0 0 1 1 B B
q0
1n
| n ≥0 }

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X 0 1 1 B B
q1
1n
| n ≥0 }

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X 0 1 1 B B
q1
1n
| n ≥0 }

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X 0 Y 1 B B
q2
1n
| n ≥0 }

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X 0 Y 1 B B
q2
1n
| n ≥0 }

22
X 0 Y 1 B B
q0
1n
| n ≥0 }

23
X 0 Y 1 B B
q0
1n
| n ≥0 }

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X X Y 1 B B
q1
1n
| n ≥0 }

25
X X Y 1 B B
q1
1n
| n ≥0 }

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X X Y Y B B
q2
1n
| n ≥0 }

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X X Y Y B B
q2
1n
| n ≥0 }

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X X Y Y B B
q0
1n
| n ≥0 }

29
X X Y Y B B
q0
1n
| n ≥0 }

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X X Y Y B B
q3
1n
| n ≥0 }

31
X X Y Y B B
q3
1n
| n ≥0 }

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X X Y Y B B
q4
1n
| n ≥0 }

End of Simulation
33

34
Evaluating function
0 0 0 0 0 1 0 0 0 B B
q0

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Evaluating function
B 0 0 0 0 1 0 0 0 B B
q1

36
Evaluating function
B 0 0 0 0 1 0 0 0 B B
q1

37
Evaluating function
B 0 0 0 0 1 0 0 0 B B
q1

38
Evaluating function
B 0 0 0 0 1 0 0 0 B B
q1

39
Evaluating function
B 0 0 0 0 1 0 0 0 B B
q1

40
Evaluating function
B 0 0 0 0 1 0 0 0 B B
q2

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Evaluating function
B 0 0 0 0 1 1 0 0 B B
q3

42
Evaluating function
B 0 0 0 0 1 1 0 0 B B
q3

43
Evaluating function
B 0 0 0 0 1 1 0 0 B B
q3

44
Evaluating function
B 0 0 0 0 1 1 0 0 B B
q3

45
Evaluating function
B 0 0 0 0 1 1 0 0 B B
q3

46
Evaluating function
B 0 0 0 0 1 1 0 0 B B
q3

47
Evaluating function
B 0 0 0 0 1 1 0 0 B B
q0

48
Evaluating function
B B 0 0 0 1 1 0 0 B B
q1

49
Evaluating function
B B 0 0 0 1 1 0 0 B B
q1

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Evaluating function
B B 0 0 0 1 1 0 0 B B
q1

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Evaluating function
B B 0 0 0 1 1 0 0 B B
q1

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Evaluating function
B B 0 0 0 1 1 0 0 B B
q2

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Evaluating function
B B 0 0 0 1 1 0 0 B B
q2

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Evaluating function
B B 0 0 0 1 1 0 0 B B
q2

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Evaluating function
B B 0 0 0 1 1 1 0 B B
q3

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Evaluating function
B B 0 0 0 1 1 1 0 B B
q3

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Evaluating function
B B 0 0 0 1 1 1 0 B B
q3

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Evaluating function
B B 0 0 0 1 1 1 0 B B
q3

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Evaluating function
B B 0 0 0 1 1 1 0 B B
q3

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Evaluating function
B B 0 0 0 1 1 1 0 B B
q0

61
Evaluating function
B B B 0 0 1 1 1 0 B B
q1

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Evaluating function
B B B 0 0 1 1 1 0 B B
q1

63
Evaluating function
B B B 0 0 1 1 1 0 B B
q1

64
Evaluating function
B B B 0 0 1 1 1 0 B B
q2

65
Evaluating function
B B B 0 0 1 1 1 0 B B
q2

66
Evaluating function
B B B 0 0 1 1 1 0 B B
q2

67
Evaluating function
B B B 0 0 1 1 1 1 B B
q3

68
Evaluating function
B B B 0 0 1 1 1 1 B B
q3

69
Evaluating function
B B B 0 0 1 1 1 1 B B
q3

70
Evaluating function
B B B 0 0 1 1 1 1 B B
q3

71
Evaluating function
B B B 0 0 1 1 1 1 B B
q3

72
Evaluating function
B B B 0 0 1 1 1 1 B B
q3

73
Evaluating function
B B B 0 0 1 1 1 1 B B
q0

74
Evaluating function
B B B B 0 1 1 1 1 B B
q1

75
Evaluating function
B B B B 0 1 1 1 1 B B
q1

76
Evaluating function
B B B B 0 1 1 1 1 B B
q2

77
Evaluating function
B B B B 0 1 1 1 1 B B
q2

78
Evaluating function
B B B B 0 1 1 1 1 B B
q2

79
Evaluating function
B B B B 0 1 1 1 1 B B
q2

80
Evaluating function
B B B B 0 1 1 1 1 B B
q4

81
Evaluating function
B B B B 0 1 1 1 B B B
q4

82
Evaluating function
B B B B 0 1 1 B B B B
q4

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Evaluating function
B B B B 0 1 B B B B B
q4

84
Evaluating function
B B B B 0 B B B B B B
q4

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Evaluating function
B B B B 0 B B B B B B
q4

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Evaluating function
B B B 0 0 B B B B B B
q6
Final State

87
End of Simulation

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0m
10n
1
Multiplication

89
0m
10n
1
Multiplication

90

91

92

93

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Variants of Turing Machine
A normal TM is a 7-tuple
(Q, Σ, Γ, δ, q0, qaccept, qreject) where:
everything is the same as a TM except the
transition function:
δ: Q x Γk
→ Q x Γk
x {L, R}k
δ(qi, a1,a2,…,ak) = (qj, b1,b2,…,bk, L, R,…, L) =
“in state qi, reading a1,a2,…,ak on k tapes, move to
state qj, write b1,b2,…,bk on k tapes, move L, R on
k tapes as specified.”
o Multitape Turing Machine:

95
k-tape TM
0 1 1 0 0 1 1 1 0 1 0 0
q0
input tape
finite control
…
k read/write
heads
0 1 1 0 0 1
“work tapes”
…
0 1 1 0 0 1 1 1 0 1 0 0 …
0 …
…

96
Informal description of k-tape TM:
 Input written on left-most squares of tape #1
 Rest of squares are blank on all tapes
 At each point, take a step determined by
• current k symbols being read on k tapes
• current state of finite control
 A step consists of
• writing k new symbols on k tapes
• moving each of k read/write heads left or right
• changing state

97
Theorem: Every k-tape TM has an equivalent single-tape TM.
Proof: Simulate k-tape TM on a 1-tape TM.
. . .a b a b
a a
b b c d
. . .
. . .
(input tape)
# a b a b # a a # b b c d # . . .
• add new symbol x for
each old x
•marks location of “virtual
heads”

98
Theorem:
The time taken by the one-tape TM to
simulate n moves of the k-tape TM is
O(n2
).

99
o Nondeterministic Turing Machine (NTM):
A nondeterministic Turing Machine (NTM) differs from the
deterministic variety by having a transition function δ such that
for each state q and tape symbol X, δ(q, X) is a set of triples
{(q1,Y1,D1), (q2,Y2,D2), ……….. (qk,Yk,Dk)}
Where k is any finite integer. The NTM can choose, at each step,
any of the triples to be the next move. It cannot, however, pick a
state from one, a tape symbol from another, a the direction from
yet another.

100
Theorem: Every NTM has an equivalent (deterministic) TM.
Proof: Simulate NTM with a deterministic TM
Cstart
• computations of M are a tree
• nodes are configurations
• fanout is b = maximum number of
choices in transition function
• leaves are accept/reject
configurations.
accrej

101
Simulating NTM M with a deterministic TM:
Breadth-first search of tree
• if M accepts: we will encounter accepting leaf and
accept
• if M rejects: we will encounter all rejecting leaves,
finish traversal of tree, and reject
• if M does not halt on some branch: we will not halt
as that branch is infinite…

102
Simulating NTM M with a deterministic TM:
o use a 3 tape TM:
• tape 1: input tape (read-only)
• tape 2: simulation tape (copy of M’s tape at point
corresponding to some node in the tree)
• tape 3: which node of the tree we are exploring (string
in {1,2,…b}*)
o Initially, tape 1 has input, others blank

103

104
Turing Machine and Computer
o A computer can simulate Turing Machine.
o A Turing Machine can simulate a computer.

105
Simulating a Turing Machine by Computer

106
Simulating a Computer by Turing Machine
Both addresses and
contents are written in
binary. The marker
symbol * and # are
used to find the ends
of addresses and
contents. Another
marker,$, indicates
the beginning of the
sequence of
addresses and
contents
Both addresses and
contents are written in
binary. The marker
symbol * and # are
used to find the ends
of addresses and
contents. Another
marker,$, indicates
the beginning of the
sequence of
addresses and
contents

107
A universal Turing machine is a Turing machine Tu that
works as follows. It is assumed to receive an input string
of the form e(T )e(z), where T is an arbitrary TM, z is a
string over the input alphabet of T , and e is an encoding
function whose values are strings in {0, 1} . The∗
computation performed by Tu on this input string satisfies
these two properties:
1. Tu accepts the string e(T )e(z) if and only if T accepts z.
2. If T accepts z and produces output y, then Tu produces
output e(y).
Universal Turing Machine

108
Universal Turing Machine

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END

Turing Machine

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