1 uncountable sets continued.... 2 theorem: let be an infinite countable set. the powerset of is...
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1
Uncountable Sets
continued...
2
Theorem:
Let be an infinite countable set.
The powerset of is uncountable S2 S
S
3
Application: Languages
Example Alphabet : },{ ba
Set of Strings:
},,,,,,,,,{},{ * aabaaabbbaabaababaS infinite and countable
Powerset: all languages
}},,,}{,{},{},{{2 aababaabaaS 1L 2L 3L 4L
uncountable
4
Languages: Uncountable
Turing machines: Countable
1L 2L 3L kL
1M 2M 3M
?
There are infinitely many more languagesthan Turing Machines
5
There are some languages not acceptedby Turing Machines
These languages cannot be describedby algorithms
6
Recursively Enumerable Languages
and Recursive Languages
7
Definition:
A language is recursively enumerableif some Turing machine accepts it
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For string :
Let be a recursively enumerable languageL
and be the Turing Machine that accepts itM
Lw
w
if then halts in a final state M
Lwif then halts in some stateM
or loops forever
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Definition:
A language is recursiveif some Turing machine accepts itand halts on any input string
In other words: A language is recursive if there is a membership algorithm for it
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For string :
Let be a recursive languageL
and be the Turing Machine that accepts itM
Lw
w
if then halts in a final state M
Lwif then halts in a non-final stateM
11
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
12
Recursive
Recursively Enumerable
Non Recursively Enumerable
13
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 is recursive then there is an enumeration procedure for it
L
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Proof:
MM~
Enumeration Machine
Accepts LEnumerates allstrings of input alphabet
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Enumeration procedure
Repeat:
M~ generates a string
M
w
checks if Lw
YES: print to output w
NO: ignore w
End of proof
17
Theorem:
if language is recursively enumerable thenthere is an enumeration procedure for it
L
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Proof:
MM~
Enumeration Machine
Accepts LEnumerates allstrings of input alphabet
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Enumeration procedure
Repeat: M~ generates a string
M
w
checks if Lw
YES: print to output w
NO: ignore w
NAIVE APPROACH
Problem:If machine may loop forever
LwM
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M~
M
1w
executes first step on
BETTER APPROACH
1w
M~ Generates second string 2w
M executes first step on 2w
second step on 1w
Generates first string
21
M~ Generates third string 3w
M executes first step on 3w
second step on 2w
third step on 1w
And so on............
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1w 2w 3w 4w
1
Move
2
3
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If for string machine halts in a final statethen it prints on the output
Mw
w
End of proof
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Theorem:
If for language there is an enumeration procedurethen is recursively enumerable
L
L
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Proof:
w
Input Tape
Enumeratorfor L
Compare
Machine that accepts L
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Turing machine that accepts L
Repeat:
• Using the enumerator, generate the next string ofL
For input string w
• Compare generated string with w
If same, accept and exit loop
End of proof
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This is not a membership algorithm.Why?
Question:
Answer:
The enumeration procedure may not produce strings in proper order
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We have proven:
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 want to find a language thatis not Recursively Enumerable
This language is not accepted by anyTuring Machine
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Consider alphabet }{a
Strings: ,,,, aaaaaaaaaa
1a 2a 3a 4a
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Consider Turing Machinesthat accept languages over alphabet }{a
They are countable:
,,,, 4321 MMMM
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Example language accepted by
},,{)( aaaaaaaaaaaaML i
},,{)( 642 aaaML i
iM
Alternative representation
1a 2a 3a 4a 5a 6a 7a
)( iML
0 1 1 10 0 0
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1a 2a 3a 4a
)( 1ML
0 1 10
)( 2ML
)( 3ML
01 0 1
0 1 11
)( 4ML 0 10 0
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Consider the language
)}(:{ iii MLaaL
L consists from the 1’s in the diagonal
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1a 2a 3a 4a
)( 1ML
0 1 10
)( 2ML
)( 3ML
01 0 1
0 1 11
)( 4ML 0 10 0
},,{ 43 aaL
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Consider the language
)}(:{ iii MLaaL
L consists from the 0’s in the diagonal
)}(:{ iii MLaaL
L
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1a 2a 3a 4a
)( 1ML
0 1 10
)( 2ML
)( 3ML
01 0 1
0 1 11
)( 4ML 0 10 0
},,{ 21 aaL
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Theorem:
Language is not recursively enumerableL
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Proof:
is recursively enumerableL
Assume for contradiction that
There must exist some machinethat accepts
kM
L
LML k )(
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1a 2a 3a 4a
)( 1ML
0 1 10
)( 2ML
)( 3ML
01 0 1
0 1 11
)( 4ML 0 10 0
Question: ?1MMk
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1a 2a 3a 4a
)( 1ML
0 1 10
)( 2ML
)( 3ML
01 0 1
0 1 11
)( 4ML 0 10 0
Answer: 1MMk )(
)(
11
1
MLa
MLa k
43
1a 2a 3a 4a
)( 1ML
0 1 10
)( 2ML
)( 3ML
01 0 1
0 1 11
)( 4ML 0 10 0
Question: ?2MMk
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1a 2a 3a 4a
)( 1ML
0 1 10
)( 2ML
)( 3ML
01 0 1
0 1 11
)( 4ML 0 10 0
Answer: 2MMk )(
)(
22
2
MLa
MLa k
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1a 2a 3a 4a
)( 1ML
0 1 10
)( 2ML
)( 3ML
01 0 1
0 1 11
)( 4ML 0 10 0
Question: ?3MMk
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1a 2a 3a 4a
)( 1ML
0 1 10
)( 2ML
)( 3ML
01 0 1
0 1 11
)( 4ML 0 10 0
Answer: 3MMk )(
)(
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3
MLa
MLa k
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Similarly: ik MM
)(
)(
ii
ki
MLa
MLa
)(
)(
ii
ki
MLa
MLa
for any i
Because either:
or
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Therefore the machine cannot exist kM
CONTRADICTION!!!
Therefore the languageis not recursively enumerable
L
End of proof
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Observation:
There is no algorithm thatdescribes L
(otherwise it would be accepted by a Turing Machine)
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A Language which is Recursively Enumerable
and not Recursive
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We want to find a language which
There is a Turing Machine that accepts the language
The machinedoesn’tnecessarily halt on any input
Is recursively enumerable
But notrecursive
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We will prove that the language
)}(:{ iii MLaaL
Is recursively enumerablebut not recursive
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The language
Theorem:
)}(:{ iii MLaaL
is recursively enumerable
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Proof:
We will give a Turing Machine thataccepts L
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Turing Machine that accepts L
For any input string w
• Write iaw• Find Turing machine iM
(using the enumeration procedure for Turing Machines)
• Simulate on inputiMia
• If accepts, then accept iM w
End of proof
56
Observation:
)}(:{ iii MLaaL
)}(:{ iii MLaaL
Recursively enumerable
Not recursively enumerable
(Thus, not recursive)
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Theorem:
The language )}(:{ iii MLaaL
is not recursive
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Proof:
Assume for contradiction that is recursive
Then is recursive:
L
L
Take the Turing Machine that accepts M L
halts on any inputM
If accepts then rejectIf rejects then accept
MM
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Therefore:
L recursive
But we know:
Lnot recursively enumerable
thus, not recursive
CONTRADICTION!!!!
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Therefore, is not recursiveL
End of proof
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Recursive
Recursively Enumerable
Non Recursively EnumerableL
L