fall 2006costas busch - rpi1 undecidable problems (unsolvable problems)
TRANSCRIPT
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Fall 2006 Costas Busch - RPI 1
Undecidable Problems(unsolvable problems)
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Fall 2006 Costas Busch - RPI 2
Recall that: A language is decidable, if there is a Turing machine (decider) that accepts the language and halts on every input string
AM
A
Turing Machine
Inputstring
Accept
Reject
M
Decider for A
DecisionOn Halt:
Decidable Languages
YES
NO
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Fall 2006 Costas Busch - RPI 3
A computational problem is decidableif the corresponding language is decidable
We also say that the problem is solvable
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Fall 2006 Costas Busch - RPI 4
Problem:Does DFA acceptthe empty language ?
M
Corresponding Language:
} language empty accepts that DFA a is :{
MM
EMPTYDFA
X
Description of DFA as a stringM
)(ML
(For example, we can represent as a binary string, as we did for Turing machines)
M
(Decidable)
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Fall 2006 Costas Busch - RPI 5
Determine whether there is a path from the initial state to any accepting state
Decider for :
On input : M
DFA
)(ML
M DFA M
)(ML
Reject MDecision: Accept M
DFAEMPTY
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Fall 2006 Costas Busch - RPI 6
Problem: Does DFA accept a finite language?
M
Corresponding Language:
}language finite a accepts that DFA a is :{ MM
FINITE DFA X (Decidable)
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Fall 2006 Costas Busch - RPI 7
Decider for :
On input : M
DFA M DFA M
Reject MDecision: Accept M
Check if there is a walk with a cyclefrom the initial state to an accepting state
infinite finite
DFAFINITE
(NO) (YES)
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Fall 2006 Costas Busch - RPI 8
Problem: Does DFA accept string ? M
Corresponding Language:
} string accepts that DFA a is :,{ wMwM
ADFA X
w
(Decidable)
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Fall 2006 Costas Busch - RPI 9
Decider for :
On input string :
DFAA
Run DFA on input string M w
If accepts Then accept (and halt)Else reject (and halt)
M wwM,
wM,
wM,
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Fall 2006 Costas Busch - RPI 10
Problem:Do DFAs and accept the same language?
1M
Corresponding Language:
}languages same the
accept that DFAs are and :,{ 2121 MMMM
EQUALDFA X
2M
(Decidable)
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Fall 2006 Costas Busch - RPI 11
Let be the language of DFA Let be the language of DFA
1L
2L
)()()( 2121 LLLLML
Decider for :
On input : 21,MM
1M
2M
Construct DFA such that:M
(combination of DFAs)
DFAEQUAL
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Fall 2006 Costas Busch - RPI 12
)()( 2121 LLLL
21 LL 21 LLand
21 LL
1L 2L 1L2L
21 LL 12 LL 2L 1L
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Fall 2006 Costas Busch - RPI 13
)()( 2121 LLLL
21 LL 21 LLor
1L 2L 1L2L
21 LL 12 LL
21 LL
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Fall 2006 Costas Busch - RPI 14
Therefore, we only need to determine whether
)()()( 2121 LLLLML
which is a solvable problem for DFAs:
DFAEMPTY
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Fall 2006 Costas Busch - RPI 15
Undecidable Languages
there is no Turing Machine which accepts the language and makes a decision (halts)for every input string
undecidable language = not decidable language
There is no decider:
(machine may make decision for some input strings)
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Fall 2006 Costas Busch - RPI 16
For an undecidable language,the corresponding problem is undecidable (unsolvable):
there is no Turing Machine (Algorithm)that gives an answer (yes or no) for every input instance
(answer may be given for some input instances)
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Fall 2006 Costas Busch - RPI 17
We have shown before that there are undecidable languages:
Decidable
Turing-Acceptable L
L
Lis Turing-Acceptable and undecidable
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Fall 2006 Costas Busch - RPI 18
We will prove that two particular problemsare unsolvable:
Membership problem
Halting problem
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Fall 2006 Costas Busch - RPI 19
Membership Problem
Input: •Turing MachineM•String w
Question: Does accept ? M w?)(MLw
Corresponding language:
} string accepts
that machine Turing a is :,{
w
MwMATM
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Fall 2006 Costas Busch - RPI 20
Theorem:
(The membership problem is unsolvable)
Proof:
We will assume that is decidable;We will then prove that every decidable languageis Turing-Acceptable
TMA is undecidable
TMABasic idea:
A contradiction!
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Fall 2006 Costas Busch - RPI 21
Suppose that is decidable
HM
w
YES M accepts w
NO M rejects w
TMA
Deciderfor TMAwM,
Input string
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Fall 2006 Costas Busch - RPI 22
Let be a Turing recognizable language L
Let be the Turing Machine that acceptsLM L
We will prove that is also decidable: L
we will build a decider for L
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Fall 2006 Costas Busch - RPI 23
NO
YESLM
s
accept
L
reject
Decider for
(and halt)
(and halt)
Decider for TMA
Inputstring
s
sLM accepts ?s
String description of LM
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Fall 2006 Costas Busch - RPI 24
Therefore, L is decidable
But there is a Turing-Acceptable languagewhich is undecidable
Contradiction!!!!
Since is chosen arbitrarily, every Turing-Acceptable language is decidable
L
END OF PROOF
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Fall 2006 Costas Busch - RPI 25
We have shown:
Decidable
TMAUndecidable
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Fall 2006 Costas Busch - RPI 26
We can actually show:
Decidable
TMATuring-Acceptable
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Fall 2006 Costas Busch - RPI 27
Turing machine that accepts :TMA
wM,
1. Run on input
2. If accepts then accept
M w
M wwM,
TMA is Turing-Acceptable
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Fall 2006 Costas Busch - RPI 28
Halting Problem
Input: •Turing MachineM•String w
Question: Does halt whileprocessing input string ?
M
w
Corresponding language:
} string input on halts
that machine Turing a is :,{
w
MwMHALTTM
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Fall 2006 Costas Busch - RPI 29
Theorem:
(The halting problem is unsolvable)
Proof:
Suppose that is decidable;we will prove that every decidable languageis also Turing-Acceptable
TMHALT is undecidable
Basic idea:
A contradiction!
TMHALT
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Fall 2006 Costas Busch - RPI 30
M
w
YES Mhalts oninput
w
Mdoesn’t halt on input wNO
Suppose that is decidableTMHALT
Decider for
TMHALT
wM,
Input string
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Fall 2006 Costas Busch - RPI 31
Let be a Turing-Acceptable language L
Let be the Turing Machine that acceptsLM L
We will prove that is also decidable: L
Lwe will build a decider for
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Fall 2006 Costas Busch - RPI 32
haltsand accepts
LM halts on ?sYES
NOLM
Run with input
LM
rejects
accept
reject
LDecider for
s
s
s
Inputstring
Decider forTMHALT
s
LM
and halt
haltsand rejects
LM
and halt
and halt
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Fall 2006 Costas Busch - RPI 33
Contradiction!!!!END OF PROOF
Therefore, L is decidable
But there is a Turing-Acceptable languagewhich is undecidable
Since is chosen arbitrarily, every Turing-Acceptable language is decidable
L
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Fall 2006 Costas Busch - RPI 34
Theorem:
Proof:
Assume for contradiction thatthe halting problem is decidable;
(The halting problem is unsolvable)
TMHALT is undecidable
we will obtain a contradictionusing a diagonilization technique
An alternative proof
Basic idea:
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Fall 2006 Costas Busch - RPI 35
HM
w
YES M halts on w
Mdoesn’t halt on
wNO
Suppose that is decidableTMHALT
Deciderfor TMHALTwM,
Input string
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Fall 2006 Costas Busch - RPI 36
H
wM, 0q
acceptq
rejectq
Input string:YES
NO
Looking inside
H
Decider for TMHALT
M halts on ?w
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Fall 2006 Costas Busch - RPI 37
H
0q
NO
aq bq
H Loop forever
YES
wM,
Construct machine :H
If halts on input Then Loop Forever Else Halt
M w
M halts on ?w
acceptq
rejectq
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Fall 2006 Costas Busch - RPI 38
Construct machine :
MM,Copyon tape
H F
If M halts on input M
Then loop forever
Else halt
MM
F
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Fall 2006 Costas Busch - RPI 39
Run with input itself
FF ,Copyon tape
H
If F halts on input F
Then loops forever on input
Else halts on input
FF
F
F
FF
FF
END OF PROOFCONTRADICTION!!!
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Fall 2006 Costas Busch - RPI 40
We have shown:
Decidable
Undecidable TMHALT
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Fall 2006 Costas Busch - RPI 41
We can actually show:
Decidable
Turing-AcceptableTMHALT
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Fall 2006 Costas Busch - RPI 42
Turing machine that accepts :
wM,
1. Run on input
2. If halts on then accept
M w
M wwM,
TMHALT
is Turing-AcceptableTMHALT