1. Linear Bounded Automata LBAs
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Linear Bounded Automata (LBAs)
are the same as Turing Machines
with one difference:
The input string tape space
is the only tape space allowed to use
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Linear Bounded Automaton (LBA)
Input string
Working space
in tape
Left-end
marker
Right-end
marker
All computation is done between end markers
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We define LBA’s as NonDeterministic
Open Problem:
NonDeterministic LBA’s
have same power with
Deterministic LBA’s ?
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Example languages accepted by LBAs:
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:
Productions
String of variables
and terminals
String of variables
and terminals
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Example unrestricted grammar:
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Theorem:
A language is recursively enumerable
if and only if is generated by an
unrestricted grammar
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Context-Sensitive Grammars:
Productions
String of variables
and terminals
String of variables
and terminals
and:
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The language
is context-sensitive:
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Theorem:
A language is context sensistive
if and only if
is accepted by a Linear-Bounded automaton
Observation:
There is a language which is recursive
but not context-sensitive
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The Chomsky Hierarchy
Non-recursively enumerable
Recursively-enumerable
Recursive
Context-sensitive
Context-free
Regular
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Decidability
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Consider problems with answer YES or NO
Examples:
Does Machine have three states ?
Is string a binary number?
Does DFA accept any input?
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A problem is decidable if some Turing machine
decides (solves) the problem
Decidable problems:
Does Machine have three states ?
Is string a binary number?
Does DFA accept any input?
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The Turing machine that decides (solves)
a problem answers YES or NO
for each instance of the problem
Input
problem
instance
YES
Turing Machine NO
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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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Turing Machine that decides a problem
YES states
NO states
YES and NO states are halting states
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Difference between
Recursive Languages and Decidable problems
For decidable problems:
The YES states may not be final states
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Some problems are undecidable:
which means:
there is no Turing Machine that
solves all instances of the problem
A simple undecidable problem:
The membership problem
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The Membership Problem
Input:
Turing Machine
String
Question:
Does accept ?
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Theorem:
The membership problem is undecidable
(there are and for which we cannot
decide whether )
Proof:
Assume for contradiction that
the membership problem is decidable
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Thus, there exists a Turing Machine
that solves the membership problem
accepts
YES NO
rejects
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Let be a recursively enumerable language
Let be the Turing Machine that accepts
We will prove that is also recursive:
we will describe a Turing machine that
accepts and halts on any input
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Turing Machine that accepts
and halts on any input
YES
accept
accepts ? NO
reject
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Therefore,
is recursive
Since is chosen arbitrarily, every recursively enumerable language is also recursive
But there are recursively enumerable
languages which are not recursive
Contradiction!!!!
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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 Machine
String
Question:
Does halt on input ?
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Theorem:
The halting problem is undecidable
(there are and for which we cannot
decide whether halts on input )
Proof:
Assume for contradiction that
the halting problem is decidable
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Thus, there exists Turing Machine
that solves the halting problem
YES
halts on NO
doesn’t
halt on
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Construction of
Input:
initial tape contents
YES NO
Encoding of
String
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Construct machine :
If returns YES then loop forever
If returns NO then halt
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Loop forever
YES NO
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Construct machine :
Input:
(machine ) If
halts on input
Then loop forever
Else halt
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copy
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Run machine with input itself:
Input:
(machine ) If
halts on input
Then loop forever
Else halt
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on input :
If halts then loops forever
If doesn’t halt then it halts
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 then
every recursively enumerable language
would be recursive
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Theorem:
The halting problem is undecidable
Proof:
Assume for contradiction that
the halting problem is decidable
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There exists Turing Machine
that solves the halting problem
YES
halts on NO
doesn’t
halt on
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Let be a recursively enumerable language
Let be the Turing Machine that accepts
We will prove that is also recursive:
we will describe a Turing machine that
accepts and halts on any input
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Turing Machine that accepts
and halts on any input NO
reject
halts on ?
YES
accept
Halts on final state
Run
with input
reject
Halts on non-final
state
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Therefore
is recursive
Since is chosen arbitrarily, every
recursively enumerable language
is also recursive
But there are recursively enumerable
languages which are not recursive
Contradiction!!!!
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Therefore, the halting problem is undecidable
END OF PROOF
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