courtesy costas busch - rpi1 linear bounded automata lbas
TRANSCRIPT
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Courtesy Costas Busch - RPI 1
Linear Bounded AutomataLBAs
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Courtesy Costas Busch - RPI 2
Linear Bounded Automata (LBAs)are the same as Turing Machineswith one difference:
The input string tape spaceis the only tape space allowed to use
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Courtesy Costas Busch - RPI 3
[ ]a b c d e
Left-endmarker
Input string
Right-endmarker
Working space in tape
All computation is done between end markers
Linear Bounded Automaton (LBA)
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We define LBA’s as NonDeterministic
Open Problem:
NonDeterministic LBA’shave same power withDeterministic LBA’s ?
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Example languages accepted by LBAs:
}{ nnn cbaL
}{ !naL
LBA’s have more power than NPDA’s
LBA’s have also less power than Turing Machines
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The Chomsky Hierarchy
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Unrestricted Grammars:
Productionsvu
String of variablesand terminals
String of variablesand terminals
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Example unrestricted grammar:
dAc
cAaB
aBcS
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A language is recursively enumerableif and only if is generated by anunrestricted grammar
LL
Theorem:
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Context-Sensitive Grammars:
and: |||| vu
Productionsvu
String of variablesand terminals
String of variablesand terminals
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The language }{ nnn cba
is context-sensitive:
aaAaaaB
BbbB
BbccAc
bAAb
aAbcabcS
|
|
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A language is context sensistive if and only if is accepted by a Linear-Bounded automatonL
LTheorem:
There is a language which is recursivebut not context-sensitive
Observation:
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Non-recursively enumerable
Recursively-enumerable
Recursive
Context-sensitive
Context-free
Regular
The Chomsky Hierarchy
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Decidability
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Consider problems with answer YES or NO
Examples:
• Does Machine have three states ?M
• Is string a binary number? w
• Does DFA accept any input? M
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A problem is decidable if some Turing machinedecides (solves) the problem
Decidable problems:
• Does Machine have three states ?M
• Is string a binary number? w
• Does DFA accept any input? M
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Turing MachineInputprobleminstance
YES
NO
The Turing machine that decides (solves) a problem answers YES or NO for each instance of the problem
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The machine that decides (solves) a problem:
• If the answer is YES then halts in a yes state
• If the answer is NO then halts in a no state
These states may not be final states
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YES states
NO states
Turing Machine that decides a problem
YES and NO states are halting states
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Difference between
Recursive Languages and Decidable problems
The YES states may not be final states
For decidable problems:
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Some problems are undecidable:
which means:there is no Turing Machine thatsolves all instances of the problem
A simple undecidable problem:
The membership problem
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The Membership Problem
Input: •Turing MachineM
•String w
Question: Does accept ? M w
?)(MLw
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Theorem:
The membership problem is undecidable
Proof: Assume for contradiction thatthe membership problem is decidable
(there are and for which we cannotdecide whether )
M w)(MLw
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Thus, there exists a Turing Machinethat solves the membership problem
H
HM
w
YES M accepts w
NO M rejects w
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Let be a recursively enumerable language L
Let be the Turing Machine that acceptsM L
We will prove that is also recursive: L
we will describe a Turing machine thataccepts and halts on any inputL
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M accepts ?wNO
YESM
w
Hacceptw
Turing Machine that acceptsand halts on any input
L
reject w
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Therefore, L is recursive
But there are recursively enumerablelanguages which are not recursive
Contradiction!!!!
Since is chosen arbitrarily, every recursively enumerable language is also recursive
L
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Therefore, the membership problem is undecidable
END OF PROOF
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Another famous undecidable problem:
The halting problem
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The Halting Problem
Input: •Turing MachineM
•String w
Question: Does halt on input ? M w
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Theorem:
The halting problem is undecidable
Proof: Assume for contradiction thatthe halting problem is decidable
(there are and for which we cannotdecide whether halts on input )
M wM w
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Thus, there exists Turing Machinethat solves the halting problem
H
HM
w
YES M halts on w
Mdoesn’t halt on
wNO
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H
wwM 0q
yq
nq
Input:initial tape contents
Encodingof M w
String
YES
NO
Construction of H
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Construct machine :H
If returns YES then loop forever H
If returns NO then haltH
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H
wwM 0q
yq
nq NO
aq bq
H
Loop forever
YES
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HConstruct machine :
Input:
If M halts on input Mw
Then loop forever
Else halt
Mw (machine )M
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MwMM wwcopy
MwH
H
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HRun machine with input itself:
Input:
If halts on input
Then loop forever
Else halt
Hw ˆ (machine )H
H Hw ˆ
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on input H Hw ˆ
If halts then loops forever
If doesn’t halt then it halts
:
H
H
NONSENSE !!!!!
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Therefore, we have contradiction
The halting problem is undecidable
END OF PROOF
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Another proof of the same theorem:
If the halting problem was decidable thenevery recursively enumerable languagewould be recursive
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Theorem:
The halting problem is undecidable
Proof: Assume for contradiction thatthe halting problem is decidable
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There exists Turing Machinethat solves the halting problem
H
HM
w
YES M halts on w
Mdoesn’t halt on
wNO
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Let be a recursively enumerable language L
Let be the Turing Machine that acceptsM L
We will prove that is also recursive: L
we will describe a Turing machine thataccepts and halts on any inputL
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M halts on ?wYES
NOM
w
Run with input
Mw
Hreject w
acceptw
rejectw
Turing Machine that acceptsand halts on any input
L
Halts on final state
Halts on non-final state
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Therefore L is recursive
But there are recursively enumerablelanguages which are not recursive
Contradiction!!!!
Since is chosen arbitrarily, every recursively enumerable language is also recursive
L
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Therefore, the halting problem is undecidable
END OF PROOF