Courtesy Costas Busch

Courtesy Costas Busch - RPI 1

Linear Bounded AutomataLBAs


Linear Bounded Automata (LBAs)are the same as Turing Machineswith one difference:

The input string tape spaceis the only tape space allowed to use


[ ]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)


We define LBA’s as NonDeterministic

Open Problem:

NonDeterministic LBA’shave same power withDeterministic LBA’s ?


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


The Chomsky Hierarchy


Unrestricted Grammars:

Productionsvu

String of variablesand terminals



Example unrestricted grammar:

dAc

cAaB

aBcS


A language is recursively enumerableif and only if is generated by anunrestricted grammar

LL

Theorem:


Context-Sensitive Grammars:

and: |||| vu

Productionsvu




The language }{ nnn cba

is context-sensitive:

aaAaaaB

BbbB

BbccAc

bAAb

aAbcabcS

|

|


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:


Non-recursively enumerable

Recursively-enumerable

Recursive

Context-sensitive

Context-free

Regular

The Chomsky Hierarchy


Decidability


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


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


Turing MachineInputprobleminstance

YES

NO

The Turing machine that decides (solves) a problem answers YES or NO for each instance of the problem


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


YES states

NO states

Turing Machine that decides a problem

YES and NO states are halting states


Difference between

Recursive Languages and Decidable problems

The YES states may not be final states

For decidable problems:


Some problems are undecidable:

which means:there is no Turing Machine thatsolves all instances of the problem

A simple undecidable problem:

The membership problem


The Membership Problem

Input: •Turing MachineM

•String w

Question: Does accept ? M w

?)(MLw


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


Thus, there exists a Turing Machinethat solves the membership problem

H

HM

w

YES M accepts w

NO M rejects w


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


M accepts ?wNO

YESM

w

Hacceptw

Turing Machine that acceptsand halts on any input

L

reject w


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


Therefore, the membership problem is undecidable

END OF PROOF


Another famous undecidable problem:

The halting problem


The Halting Problem

Input: •Turing MachineM

•String w

Question: Does halt on input ? M w


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


Thus, there exists Turing Machinethat solves the halting problem

H

HM

w

YES M halts on w

Mdoesn’t halt on

wNO


H

wwM 0q

yq

nq

Input:initial tape contents

Encodingof M w

String

YES

NO

Construction of H


Construct machine :H

If returns YES then loop forever H

If returns NO then haltH


H

wwM 0q

yq

nq NO

aq bq

H

Loop forever

YES


HConstruct machine :

Input:

If M halts on input Mw

Then loop forever

Else halt

Mw (machine )M


MwMM wwcopy

MwH

H


HRun machine with input itself:

Input:

If halts on input

Then loop forever

Else halt

Hw ˆ (machine )H

H Hw ˆ


on input H Hw ˆ

If halts then loops forever

If doesn’t halt then it halts

:

H

H

NONSENSE !!!!!


Therefore, we have contradiction


END OF PROOF


Another proof of the same theorem:

If the halting problem was decidable thenevery recursively enumerable languagewould be recursive


Theorem:


Proof: Assume for contradiction thatthe halting problem is decidable


There exists Turing Machinethat solves the halting problem

H

HM

w

YES M halts on w

Mdoesn’t halt on

wNO


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


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


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


Therefore, the halting problem is undecidable

END OF PROOF

Courtesy Costas Busch - RPI1 Linear Bounded Automata LBAs.

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