Transcript Turing Machines Fall 2006 Costas Busch - RPI The Language Hierarchy ww ? n n n ? a b c Context-Free Languages n n R a b ww Regular Languages a* Fall 2006 a.

Turing Machines
Fall 2006
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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*
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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*
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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
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a  b, R
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q2
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Example:
Time 1
......
  a b a c
  
......
q1
current state
q1
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a  b, R
q2
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......
Time 1
  a b a c
  
......
  
......
q1
......
Time 2
  a b b c
q2
q1
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a  b, R
q2
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Example:
......
Time 1
  a b a c
  
......
  
......
q1
......
Time 2
  a b b c
q2
q1
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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
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  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
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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
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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
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  , L
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q1
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Time 0
  a a a  
q0
a  a, R
q0
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  , L
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q1
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Time 1
  a a a  
q0
a  a, R
q0
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  , L
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q1
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Time 2
  a a a  
q0
a  a, R
q0
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  , L
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q1
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Time 3
  a a a  
q0
a  a, R
q0
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  , L
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q1
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Time 4
  a a a  
q1
a  a, R
q0
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Halt & Accept
  , L
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q1
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Rejection Example
Time 0
  a b a  
q0
a  a, R
q0
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  , 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
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  , 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
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  , L
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q1
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Time 0
  a b a  
q0
b  b, L
a  a, R
q0
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  , L
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q1
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Time 1
  a b a  
q0
b  b, L
a  a, R
q0
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  , L
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q1
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Time 2
  a b a  
q0
b  b, L
a  a, R
q0
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  , 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
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  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
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Initial
blank
stateCostas Busch - RPI
Accept
states
59
Configuration
  c a b a  
q1
Instantaneous description:
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ca q1 ba
60
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:
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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 :

q0 w  x1 q f x2 }
Initial state
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M
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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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Computing Functions
with
Turing Machines
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A function
Domain:
has:
f (w)
Result Region:
D
f (w)
w D
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S
f ( w)  S
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A function may have many parameters:
Example:
Addition function
f ( x, y )  x  y
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Integer Domain
Decimal:
5
Binary:
101
Unary:
11111
We prefer unary representation:
easier to manipulate with Turing machines
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Definition:
f
A function
is computable if
there is a Turing Machine M such that:
Initial configuration

w
Final configuration


qf
q0
initial state
For all
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f (w) 
accept state
w D Domain
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In other words:
f
A function
is computable if
there is a Turing Machine M such that:

q0 w  q f f ( w)
Initial
Configuration
For all
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Final
Configuration
w D Domain
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Example
The function
f ( x, y )  x  y is computable
x, y
are integers
Turing Machine:
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Input string:
x0 y
unary
Output string:
xy 0
unary
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x
Start
 1 1

y
1 0 1  1 
q0
initial state
The 0 is the delimiter that
separates the two numbers
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y
x
Start
 1 1
1 0 1  1 

q0 initial state
x y
Finish
 1 1

1 1 0 
q f final state
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The 0 here helps when we use
the result for other operations
x y
Finish
 1 1

1 1 0 
q f final state
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Turing machine for function
1 1, R
f ( x, y )  x  y
1 1, R
1 1, L



,
L
0

1
,
R
1
0
,
L
q
q0
q3
q1
2
  , R
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q4
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Execution Example:
x  11 (=2)
Time 0
y
x
 1 1 0 1 1 
y  11 (=2)
q0
Final Result
x y
 1 1 1 1 0 
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q4
78
Time 0
 1 1 0 1 1 
q0
1 1, R
1 1, R
1 1, L



,
L
0

1
,
R
1
0
,
L
q
q0
q3
q1
2
  , R
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q4
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Time 1
 1 1 0 1 1 
q0
1 1, R
1 1, R
1 1, L



,
L
0

1
,
R
1
0
,
L
q
q0
q3
q1
2
  , R
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q4
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Time 2
 1 1 0 1 1 
q0
1 1, R
1 1, R
1 1, L



,
L
0

1
,
R
1
0
,
L
q
q0
q3
q1
2
  , R
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q4
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Time 3
 1 1 1 1 1 
q1
1 1, R
1 1, R
1 1, L



,
L
0

1
,
R
1
0
,
L
q
q0
q3
q1
2
  , R
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q4
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Time 4
 1 1 1 1 1 
q1
1 1, R
1 1, R
1 1, L



,
L
0

1
,
R
1
0
,
L
q
q0
q3
q1
2
  , R
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q4
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Time 5
 1 1 1 1 1 
q1
1 1, R
1 1, R
1 1, L



,
L
0

1
,
R
1
0
,
L
q
q0
q3
q1
2
  , R
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q4
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Time 6
 1 1 1 1 1 
q2
1 1, R
1 1, R
1 1, L



,
L
0

1
,
R
1
0
,
L
q
q0
q3
q1
2
  , R
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Costas Busch - RPI
q4
85
Time 7
 1 1 1 1 0 
q3
1 1, R
1 1, R
1 1, L



,
L
0

1
,
R
1
0
,
L
q
q0
q3
q1
2
  , R
Fall 2006
Costas Busch - RPI
q4
86
Time 8
 1 1 1 1 0 
q3
1 1, R
1 1, R
1 1, L



,
L
0

1
,
R
1
0
,
L
q
q0
q3
q1
2
  , R
Fall 2006
Costas Busch - RPI
q4
87
Time 9
 1 1 1 1 0 
q3
1 1, R
1 1, R
1 1, L



,
L
0

1
,
R
1
0
,
L
q
q0
q3
q1
2
  , R
Fall 2006
Costas Busch - RPI
q4
88
Time 10
 1 1 1 1 0 
q3
1 1, R
1 1, R
1 1, L



,
L
0

1
,
R
1
0
,
L
q
q0
q3
q1
2
  , R
Fall 2006
Costas Busch - RPI
q4
89
Time 11
 1 1 1 1 0 
q3
1 1, R
1 1, R
1 1, L



,
L
0

1
,
R
1
0
,
L
q
q0
q3
q1
2
  , R
Fall 2006
Costas Busch - RPI
q4
90
Time 12
 1 1 1 1 0 
q4
1 1, R
1 1, R
1 1, L



,
L
0

1
,
R
1
0
,
L
q
q0
q3
q1
2
HALT & accept
Fall 2006
Costas Busch - RPI
  , R
q4
91
Another Example
The function
f ( x)  2 x
x
is computable
is integer
Turing Machine:
Input string:
Output string:
Fall 2006
x
unary
xx
unary
Costas Busch - RPI
92
x
Start
 1 1

1 
q0 initial state
2x
Finish
 1 1

1 1 1 
q f accept state
Fall 2006
Costas Busch - RPI
93
Turing Machine Pseudocode for
f ( x)  2 x
• Replace every 1 with $
• Repeat:
• Find rightmost $, replace it with 1
• Go to right end, insert 1
Until no more $ remain
Fall 2006
Costas Busch - RPI
94
Turing Machine for
1  $, R
f ( x)  2 x
1 1, L
1 1, R
q0   , L q1 $  1, R
  , R
q3
Fall 2006
q2
  1, L
Costas Busch - RPI
95
Start
Example
 1 1 
Finish
 1 1 1 1 
q0
q3
1  $, R
1 1, L
1 1, R
q0   , L q1 $  1, R
Fall 2006
  , R
q3
q2
  1, L
Costas Busch - RPI
96
Another Example
The function
f ( x, y ) 
is computable
Input:
Output:
Fall 2006
1
if
x y
0
if
x y
x0 y
1
or
0
Costas Busch - RPI
97
Turing Machine Pseudocode:
• Repeat
Match a 1 from
Until all of
x with a 1 from y
x or y is matched
• If a 1 from x is not matched
erase tape, write 1
else
erase tape, write 0
Fall 2006
Costas Busch - RPI
( x  y)
( x  y)
98
Combining Turing Machines
Fall 2006
Costas Busch - RPI
99
Block Diagram
input
Fall 2006
Turing
Machine
Costas Busch - RPI
output
100
Example:
x  y if x  y
f ( x, y) 
0
x, y
x, y
Comparator
Adder
x y
Eraser
0
x y
x y
Fall 2006
if x  y
Costas Busch - RPI
101