1 Uncountable Sets continued.... 2 Theorem: Let be an infinite countable set. The powerset of is...
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Transcript of 1 Uncountable Sets continued.... 2 Theorem: Let be an infinite countable set. The powerset of is...
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Application: Languages
Example Alphabet : },{ ba
Set of Strings:
},,,,,,,,,{},{ * aabaaabbbaabaababaS infinite and countable
Powerset: all languages
}},,,}{,{},{},{{2 aababaabaaS 1L 2L 3L 4L
uncountable
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Languages: Uncountable
Turing machines: Countable
1L 2L 3L kL
1M 2M 3M
?
There are infinitely many more languagesthan Turing Machines
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There are some languages not acceptedby Turing Machines
These languages cannot be describedby algorithms
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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
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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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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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Enumeration procedure
Repeat:
M~ generates a string
M
w
checks if Lw
YES: print to output w
NO: ignore w
End of proof
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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
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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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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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We want to find a language thatis not Recursively Enumerable
This language is not accepted by anyTuring Machine
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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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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
Answer: 1MMk )(
)(
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1
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
Answer: 2MMk )(
)(
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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
Answer: 3MMk )(
)(
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3
MLa
MLa k
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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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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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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
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Observation:
)}(:{ iii MLaaL
)}(:{ iii MLaaL
Recursively enumerable
Not recursively enumerable
(Thus, 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!!!!