Transcript enumeration procedure

CDT314
FABER
Formal Languages, Automata
and Models of Computation
Lecture 15-1
Mälardalen University
2012
1
1
Countable Sets
Based on course materials by Prof. Costas Busch
2
Infinite sets are either:
• Countable
• Uncountable
3
Countable set:
There is a one to one correspondence
between
elements of the set
and
positive integers
4
Example:
The set of even integers
is countable
Even integers:
0, 2, 4, 6, 
Correspondence:
Positive integers:
2n
1, 2, 3, 4, 
corresponds to
n 1
5
Example:
The set of rational numbers
is countable
Rational numbers:
1 3 7
, , ,
2 4 8
6
Naive Approach
Rational numbers:
1 1 1
, , ,
1 2 3
Correspondence:
Positive integers:
1, 2, 3, 
Doesn’t work:
we will never count numbers with nominator 2
2 2 2
, , ,
1 2 3
7
Better Approach
1
1
1
2
1
3
2
1
2
2
2
3
3
1
3
2
4
1
1
4




8
1
1
1
2
1
3
2
1
2
2
2
3
3
1
3
2
4
1
1
4




9
1
1
1
2
1
3
2
1
2
2
2
3
3
1
3
2
4
1
1
4




10
1
1
1
2
1
3
2
1
2
2
2
3
3
1
3
2
4
1
1
4




11
1
1
1
2
1
3
2
1
2
2
2
3
3
1
3
2
4
1
1
4




12
1
1
1
2
1
3
2
1
2
2
2
3
3
1
3
2
4
1
1
4




13
Rational Numbers:
1 1 2 1 2
, , , , ,
1 2 1 3 2
Correspondence:
Positive Integers:
1, 2, 3, 4, 5, 
14
We proved:
the set of rational numbers is countable
by giving
an enumeration procedure
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Definition
Let S be a set of strings
An enumeration procedure for S is a
Turing machine that generates
any string of S in finite number of steps.
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strings
s1, s2 , s3 ,  S
Enumeration
Machine for S
output
Finite time:
s1, s2 , s3 , 
t1, t2 , t3 ,
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Enumeration Machine
Configuration
Time 0
# #
q0
Time t1
x1  s1
qs
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Time t 2
x2  s2
qs
Time t3
x3  s3
qs
19
A set is countable if there is an
enumeration procedure for it
20
Example:
The set of all strings
is countable.

{a, b, c}
We will describe the enumeration procedure.
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Naive procedure:
Produce the strings in lexicographic order:
a
aa
aaa
...
Doesn’t work!
Strings starting with b will never be produced.
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Better procedure: Proper Order
Produce all strings of length 1
Produce all strings of length 2
Produce all strings of length 2
..........
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Produce strings:
Proper Order
a, b, c
aa
ab
ac
ba
bb
bc
ca
cb
cc
aaa
aab
aac
......
Length 1
Length 2
Length 3
......
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Theorem:
The set of all Turing Machines
is countable.
Proof:
Any Turing machine is a finite string
Encoded with a sequence of 0’s and 1’s.
Find an enumeration procedure
for the set of Turing Machine strings.
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Enumeration Procedure:
Repeat
1. Generate the next string of 0’s and 1’s
in proper order
2. Check if the string defines a
Turing Machine
if YES: print string on output
if NO: ignore string
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Uncountable Sets
27
Definition:
A set is uncountable if it is not countable
28
Theorem:
Let S be an infinite countable set.
The powerset 2
S
of S is uncountable.
The power set of natural numbers has the same cardinality as the set of real numbers.
(Using the Cantor–Bernstein–Schröder theorem, it is easy to prove
that there exists a bijection between the set of reals and the power set of the natural
numbers).
29
Proof:
Since S is countable, we can write
S  {s1, s2 , s3 ,}
Element of S
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Elements of the powerset have the form:
{s1, s3}
{s5 , s7 , s9 , s10 }
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We encode each element of the power set
with a string of 0’s and 1’s *
Encoding
Powerset
s
s
s
s
3
1
2
4 
element
{s1}
1
0
0
0

{s2 , s3 }
0
1
1
0

{s1,s3, s4}

1
0
1

1

*Cantor’s diagonal argument
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Let’s assume the contrary, that the
powerset is countable.
We can enumerate the elements of the
powerset.
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Powerset
element
Encoding
t1
1
0
0
0
0 
t2
1
1
0
0
0

t3
1
1
0
1
0

t4
1
1
0
0
1 


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Take the powerset element
whose bits are the complements
of the diagonal.
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t1
1
0
0
0
0 
t2
1
1
0
0
0

t3
1
1
0
1
0

t4
1
1
0
0
1 
New element:
0011
(Diagonal complement)
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The new element must be some ti
This is impossible:
The i-th bit must be the complement
of itself.
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We have contradiction!
Therefore the powerset is uncountable.
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Theorem:
Let S be an infinite countable set.
S
The powerset 2 of S is uncountable.
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Application: Languages
Alphabet: {a, b}
Set of Strings:
S  {a, b}  { , a, b, aa, ab, ba, bb, aaa, aab,}
*
infinite and countable
Powerset: all languages
2  {{},{a},{a, b}{aa, ab, aab},}
L1 L2 L3
L4

S
uncountable
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Languages: uncountable
L1
L2
L3
M1
M2
M3

Lk

?
Turing machines: countable
There are infinitely many more languages
than Turing machines!
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There are some languages not accepted
by Turing Machines.
These languages cannot be described
by algorithms.
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Recursively Enumerable Languages
and
Recursive Languages
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Definition:
A language is recursively enumerable
if some Turing machine accepts it.
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Let L be a recursively enumerable language
and M be the Turing Machine that accepts it.
For a string w :
if
w L
then M halts in a final state
if
w L
then M halts in some state
or loops forever
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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
46
Let L be a recursive language
and M be the Turing Machine that accepts it.
For a string w :
if w L then M halts in a final state.
if w L then M halts in a non-final state.
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We will prove:
1. There is a specific language
which is not recursively enumerable.
2. There is a specific language
which is recursively enumerable
but not recursive.
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Non Recursively Enumerable
Recursively Enumerable
Recursive
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First we 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.
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Theorem:
if a language L is recursive then
there is an enumeration procedure for it.
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Proof:
Enumeration Machine
~
M
Enumerates all
strings of input alphabet
M
Accepts L
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Enumeration procedure
Repeat:
~ generates a string w
M
M
checks if
w L
YES: print w to output
NO:
ignore w
End of proof
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Theorem:
if language L is recursively enumerable
then there is
an enumeration procedure for it.
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Proof:
Enumeration Machine
~
M
Enumerates all
strings of input alphabet
M
Accepts L
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NAIVE APPROACH
Enumeration procedure
~
Repeat: M generates a string w
M checks if
w L
YES: print w to output
NO:
ignore w
w L
Problem: If
machine M may loop forever
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BETTER APPROACH
~ generates first string w
M
1
M executes first step on w1
~ generates second string w
M
2
M executes first step on w2
second step on w1
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~ Generates third string
M
M
w3
executes first step on
w3
second step on w2
third step on
w1
And so on............
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w1
w2
w3
w4
1




2




3





Move

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If for string w
machine M halts in a final state
then it prints w on the output.
End of proof
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Theorem:
If for language L
there is an enumeration procedure
then L is recursively enumerable.
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Proof:
Input Tape
w
Machine that
accepts L
Enumerator
for L
Compare
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Turing machine that accepts L
For input string w
Repeat:
• Using the enumerator,
generate the next string of L
• Compare generated string with w
If same, accept and exit loop
End of proof
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Question:
This is not a membership algorithm.
Why?
Answer:
The enumeration procedure
may not produce strings in proper order
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We have shown:
A language is recursively enumerable
if and only if
there is an enumeration procedure for it.
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A Language which
is not
Recursively Enumerable
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We search for a language that
is not Recursively Enumerable.
This language is not accepted by any
Turing Machine.
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Consider alphabet {a}
Strings:
a, aa, aaa, aaaa, 
1
a a
2
a
3
a
4

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Consider Turing Machines
that accept languages over alphabet {a}
They are countable:
M1, M 2 , M 3 , M 4 , 
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Example language accepted by M i
L( M i )  {aa, aaaa, aaaaaa}
2
4
6
L( M i )  {a , a , a }
Alternative representation
1
2
a
a
L( M i ) 0
1
a
3
0
4
a
1
0
a
5
a
6
1
a
7
0


70
1
a
a
2
3
a
L ( M1 )
0
1
0
1

L( M 2 )
1
0
0
1

L( M 3 )
0
1
1
1

L( M 4 )
0
0
0
1

a
4

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Consider the language
i
i
L  {a : a  L( M i )}
L consists of the 1’s on the diagonal
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1
2
3
4

a
a
L ( M1 )
0
1
0
1

L( M 2 )
1
0
0
1

L( M 3 )
0
1
1
1

L( M 4 )
0
0
0
1

a
3
a
4
L  {a , a ,}
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Consider the language
i
i
i
i
L
L  {a : a  L( M i )}
L  {a : a  L( M i )}
L
consists from of 0’s on the diagonal
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1
a
a
2
a
L ( M1 )
0
1
0
1

L( M 2 )
1
0
0
1

L( M 3 )
0
1
1
1

L( M 4 )
0
0
0
1

1
3
a
4

2
L  {a , a ,}
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Theorem:
Language L is not recursively enumerable.
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Proof:
Assume on the contrary that
L is recursively enumerable
There must exist some machine M k
that accepts L
L( M k )  L
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1
a
a
2
a
L ( M1 )
0
1
0
1

L( M 2 )
1
0
0
1

L( M 3 )
0
1
1
1

L( M 4 )
0
0
0
1

3
a
4

Question: M k  M1 ?
78
1
a
a
2
a
L ( M1 )
0
1
0
1

L( M 2 )
1
0
0
1

L( M 3 )
0
1
1
1

L( M 4 )
0
0
0
1

3
a
4

1
Answer:
M k  M1
a  L( M k )
1
a  L ( M1 )
79
1
a
a
2
a
L ( M1 )
0
1
0
1

L( M 2 )
1
0
0
1

L( M 3 )
0
1
1
1

L( M 4 )
0
0
0
1

3
a
4

Question: M k  M 2 ?
80
1
a
a
2
a
L ( M1 )
0
1
0
1

L( M 2 )
1
0
0
1

L( M 3 )
0
1
1
1

L( M 4 )
0
0
0
1

3
a
4

2
Answer:
Mk  M2
a  L( M k )
2
a  L( M 2 )
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1
a
a
2
a
L ( M1 )
0
1
0
1

L( M 2 )
1
0
0
1

L( M 3 )
0
1
1
1

L( M 4 )
0
0
0
1

3
a
4

Question: M k  M 3 ?
82
1
a
a
2
a
L ( M1 )
0
1
0
1

L( M 2 )
1
0
0
1

L( M 3 )
0
1
1
1

L( M 4 )
0
0
0
1

3
a
4

3
Answer:
M k  M3
a  L( M k )
3
a  L( M 3 )
83
Similarly:
M k  Mi
for any
i
Because either:
i
a  L( M k )
i
a  L( M i )
i
or
a  L( M k )
i
a  L( M i )
84
Therefore the machine
M kcannot exist
CONTRADICTION!!!
The language L
is not recursively enumerable.
End of proof
85
Observation:
There is no algorithm that
describes L
(otherwise it would be accepted by
a Turing Machine)
86
A Language
which is Recursively Enumerable
and not Recursive
87
We want to find a language which
Is recursively
enumerable
There is a
Turing Machine
that accepts
the language
But not
recursive
The machine
doesn’t
necessarily halt
on any input
88
We will prove that the language
i
i
L  {a : a  L( M i )}
Is recursively enumerable
but not recursive.
89
Theorem:
The language
i
i
L  {a : a  L( M i )}
is recursively enumerable
90
Proof:
We will give a Turing Machine that
accepts L
91
Turing Machine that accepts L
For any input string w
• Write
wa
i
• Find Turing machine M i
(using the enumeration procedure
for Turing Machines)
• Simulate M i on input ai
• If M i accepts, then accept w
End of proof
92
Observation:
Recursively enumerable
i
i
L  {a : a  L( M i )}
Not recursively enumerable
i
i
L  {a : a  L( M i )}
(Thus, not recursive)
93
Theorem:
The language
i
i
L  {a : a  L( M i )}
is not recursive.
94
Proof:
Assume on the contrary that L is recursive.
Then L is recursive:
Take the Turing Machine M that accepts L
M halts on any input
If M accepts then reject
If M rejects then accept
95
Therefore:
L recursive
But we know:
L not recursively enumerable
thus, not recursive
CONTRADICTION!
Therefore, L is not recursive
End of proof
96
Non Recursively Enumerable
L
Recursively Enumerable
L
Recursive
97