Transcript Turing Machines

Turing Machines
Prof. Busch - LSU
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The Language Hierarchy
ww ?
n n n ?
a b c
Context-Free Languages
n n
R
a b
ww
Regular Languages
a*
a *b *
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Languages accepted by
Turing Machines
ww
n n n
a b c
Context-Free Languages
n n
R
a b
ww
Regular Languages
a*
a *b *
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Tape
......
A Turing Machine
......
Read-Write head
Control Unit
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The Tape
No boundaries -- infinite length
......
......
Read-Write head
The head moves Left or Right
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......
......
Read-Write head
The head at each transition (time step):
1. Reads a symbol
2. Writes a symbol
3. Moves Left or Right
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Example:
Time 0
......
a b a c
Time 1
......
1. Reads
2. Writes
a b k c
......
......
a
k
3. Moves Left
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Time 1
......
a b k c
Time 2
......
1. Reads
2. Writes
a f
k c
......
......
b
f
3. Moves Right
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The Input String
Input string
......
  a b a c
Blank symbol
  
......
head
Head starts at the leftmost position
of the input string
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States & Transitions
Read
q1
Write
a  b, L
Move Left
q2
Move Right
q1
a  b, R
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q2
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Example:
Time 1
......
  a b a c
  
......
q1
current state
q1
a  b, R
q2
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......
Time 1
  a b a c
  
......
  
......
q1
......
Time 2
  a b b c
q2
q1
a  b, R
q2
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Example:
......
Time 1
  a b a c
  
......
  
......
q1
......
Time 2
  a b b c
q2
q1
a  b, L
q2
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Example:
......
Time 1
  a b a c
  
......
q1
......
Time 2
  a b b c
g  
......
q2
q1
  g, R
q2
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Determinism
Turing Machines are deterministic
Not Allowed
Allowed
a  b, R
q2
a  b, R
q2
a  d, L
q3
q1
q1
b  d, L
q3
No lambda transitions allowed
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Partial Transition Function
Example:
......
  a b a c
  
......
q1
a  b, R
q2
q1
b  d, L
q3
Allowed:
No transition
for input symbol
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c
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Halting
The machine halts in a state if there is
no transition to follow
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Halting Example 1:
......
  a b a c
  
......
q1
q1
No transition from q1
HALT!!!
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Halting Example 2:
......
  a b a c
  
......
q1
a  b, R
q2
b  d, L
q3
q1
No possible transition
from q1 and symbol c
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HALT!!!
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Accepting States
q1
q2
Allowed
q1
q2
Not Allowed
•Accepting states have no outgoing transitions
•The machine halts and accepts
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Acceptance
Accept Input
string
If machine halts
in an accept state
Reject Input
string
If machine halts
in a non-accept state
or
If machine enters
an infinite loop
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Observation:
In order to accept an input string,
it is not necessary to scan all the
symbols in the string
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Turing Machine Example
Input alphabet
  {a , b }
Accepts the language:
a*
a  a, R
q0
  , L
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q1
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Time 0
  a a a  
q0
a  a, R
q0
  , L
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q1
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Time 1
  a a a  
q0
a  a, R
q0
  , L
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q1
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Time 2
  a a a  
q0
a  a, R
q0
  , L
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q1
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Time 3
  a a a  
q0
a  a, R
q0
  , L
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q1
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Time 4
  a a a  
q1
a  a, R
q0
Halt & Accept
  , L
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q1
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Rejection Example
Time 0
  a b a  
q0
a  a, R
q0
  , L
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q1
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Time 1
  a b a  
q0
No possible Transition
Halt & Reject
a  a, R
q0
  , L
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q1
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A simpler machine for same language
but for input alphabet
  {a }
Accepts the language:
a*
q0
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Time 0
  a a a  
q0
Halt & Accept
q0
Not necessary to scan input
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Infinite Loop Example
A Turing machine
for language a * b(a  b) *
b  b, L
a  a, R
q0
  , L
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q1
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Time 0
  a b a  
q0
b  b, L
a  a, R
q0
  , L
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q1
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Time 1
  a b a  
q0
b  b, L
a  a, R
q0
  , L
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q1
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Time 2
  a b a  
q0
b  b, L
a  a, R
q0
  , L
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q1
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Time 2
  a b a  
q0
  a b a  
q0
Time 4
  a b a  
q0
Time 5
  a b a  
q0
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Infinite loop
Time 3
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Because of the infinite loop:
•The accepting state cannot be reached
•The machine never halts
•The input string is rejected
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Another Turing Machine Example
Turing machine for the language
n n
{a b }
n 1
y  y, R
q3
q4
  , L
y  y, R
q0
y  y, R
a  a, R
a  x, R
q1
y  y, L
a  a, L
b  y, L
q2
x  x, R
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Basic Idea:
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
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 a a b b  
Time 0
q0
y  y, R
q3
q4
  , L
y  y, R
q0
y  y, R
a  a, R
a  x, R
q1
y  y, L
a  a, L
b  y, L
q2
x  x, R
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 x a b b  
Time 1
q1
y  y, R
q3
q4
  , L
y  y, R
q0
y  y, R
a  a, R
a  x, R
q1
y  y, L
a  a, L
b  y, L
q2
x  x, R
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 x a b b  
Time 2
q1
y  y, R
q3
q4
  , L
y  y, R
q0
y  y, R
a  a, R
a  x, R
q1
y  y, L
a  a, L
b  y, L
q2
x  x, R
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 x a y b  
Time 3
q2
y  y, R
q3
q4
  , L
y  y, R
q0
y  y, R
a  a, R
a  x, R
q1
y  y, L
a  a, L
b  y, L
q2
x  x, R
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 x a y b  
Time 4
q2
y  y, R
q3
q4
  , L
y  y, R
q0
y  y, R
a  a, R
a  x, R
q1
y  y, L
a  a, L
b  y, L
q2
x  x, R
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 x a y b  
Time 5
q0
y  y, R
q3
q4
  , L
y  y, R
q0
y  y, R
a  a, R
a  x, R
q1
y  y, L
a  a, L
b  y, L
q2
x  x, R
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 x x y b  
Time 6
q1
y  y, R
q3
q4
  , L
y  y, R
q0
y  y, R
a  a, R
a  x, R
q1
y  y, L
a  a, L
b  y, L
q2
x  x, R
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 x x y b  
Time 7
q1
y  y, R
q3
q4
  , L
y  y, R
q0
y  y, R
a  a, R
a  x, R
q1
y  y, L
a  a, L
b  y, L
q2
x  x, R
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 x x y y  
Time 8
q2
y  y, R
q3
q4
  , L
y  y, R
q0
y  y, R
a  a, R
a  x, R
q1
y  y, L
a  a, L
b  y, L
q2
x  x, R
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 x x y y  
Time 9
q2
y  y, R
q3
q4
  , L
y  y, R
q0
y  y, R
a  a, R
a  x, R
q1
y  y, L
a  a, L
b  y, L
q2
x  x, R
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 x x y y  
Time 10
q0
y  y, R
q3
q4
  , L
y  y, R
q0
y  y, R
a  a, R
a  x, R
q1
y  y, L
a  a, L
b  y, L
q2
x  x, R
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 x x y y  
Time 11
q3
y  y, R
q3
q4
  , L
y  y, R
q0
y  y, R
a  a, R
a  x, R
q1
y  y, L
a  a, L
b  y, L
q2
x  x, R
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 x x y y  
Time 12
q3
y  y, R
q3
q4
  , L
y  y, R
q0
y  y, R
a  a, R
a  x, R
q1
y  y, L
a  a, L
b  y, L
q2
x  x, R
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 x x y y  
Time 13
q4
Halt & Accept
y  y, R
q3
q4
  , L
y  y, R
q0
y  y, R
a  a, R
a  x, R
q1
y  y, L
a  a, L
b  y, L
q2
x  x, R
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Observation:
If we modify the
machine for the language
we can easily construct
a machine for the language
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n n
{a b }
n n n
{a b c }
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Formal Definitions
for
Turing Machines
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Transition Function
q1
a  b, R
q2
 (q1, a)  (q2 , b, R)
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Transition Function
q1
c  d, L
q2
 (q1, c)  (q2 , d , L)
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Turing Machine:
States
Input
alphabet
Tape
alphabet
M  (Q, , ,  , q0 , , F )
Transition
function
Initial
blank
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Accept
states
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Configuration
  c a b a  
q1
Instantaneous description:
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ca q1 ba
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Time 4
Time 5
 x a y b  
q2
A Move:
 x a y b  
q0
q2 xayb  x q0 ayb
(yields in one mode)
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Time 4
Time 5
 x a y b  
 x a y b  
q2
q0
Time 6
Time 7
 x x y b  
 x x y b  
q1
q1
A computation
q2 xayb  x q0 ayb  xx q1 yb  xxy q1 b
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q2 xayb  x q0 ayb  xx q1 yb  xxy q1 b

Equivalent notation:
q2 xayb  xxy q1 b
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Initial configuration:
q0 w
Input string
w
 a a b b  
q0
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The Accepted Language
For any Turing Machine
L( M )  {w :
M

q0 w  x1 q f x2 }
Initial state
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Accept state
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If a language L is accepted
by a Turing machine M
then we say that L is:
•Turing Recognizable
Other names used:
•Turing Acceptable
•Recursively Enumerable
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