1 the chomsky hierarchy. 2 unrestricted grammars: rules have form string of variables and terminals...

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The Chomsky Hierarchy

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Unrestricted Grammars:

Rules have form vu

String of variablesand terminals


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Example grammar:

dAc

cAaB

aBcS

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A language is generated by anunrestricted grammarif and only if is recursively enumerableL

L

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Context-Sensitive Grammars:

Rules have form vu



And: |||| vu

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The language }{ nnn cba

is context-sensitive:

aaAaaaB

BbbB

BbccAc

bAAb

aAbcabcS

|

|

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A language is context sensistiveif and only if is accepted by a Linear-Boundedautomaton

L

L

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There is a language which is context-sensitivebut not recursive

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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 ?M

• Is string a binary number? w

• Does DFA accept any input? M

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A problem is decidable

if some Turing machine solves the problem

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Some decidable problems:

• Does Machine have three states ?M

• Is string a binary number? w

• Does DFA accept any input? M

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The machine that decides the 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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Difference between

Recursive Languages andDecidable problems

The YES halting states may not be final states

For decidable problems:

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There are some problemswhich are undecidable:

There is no Turing Machine thatsolves all instances of the problem

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A famous undecidable problem:

The halting problem

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The Halting Problem

Inputs: •Turing MachineM

•String w

Question: Does halt on ? M w

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THEOREM

The halting problem is undecidable

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THEOREM


PROOF

Assume for contradiction thatthe halting problem is decidable

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There exists Turing Machinethat solves the halting problem

H

HM

wYES or NO

YES: M accepts w

NO: M Doesn’t accept w

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H

wwM 0q

yq

nq

Initial tapecontents

Encodingof M w

String

YES

NO

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H Construct machine :

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

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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 wwrepeat

MwH

H

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Run machine H with input itself

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HRun machine

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


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


PROOF

Assume for contradiction thatthe halting problem is decidable

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Turing Machine solves the halting problemH

For inputs: machine M and string w

H Returns:

M halts on input wYES: if

M doesn’t halt on inputwNO: if

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Let be a recursively enumerable language L

Let be the Machine that acceptsM L

We will prove that is recursive L

We will describe a membership algorithm

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Run with inputs: M wandH

HIf Returns:

M doesn’t halt on inputwNO:

Then M doesn’t accept w

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HIf Returns:

M halts on input wYES:

Run M wwith input

M will halt, and either accept or reject w

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Therefore L is recursive

But there are recursively enumerablelanguages which are not recursive

Contradiction!!!!

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Therefore, the halting problemis undecidable

END OF PROOF

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Reducibility

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Problem is reduced to problemA B

BIf is decidable then is decidableA

means

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Problem is reduced to problemA B

AIf is undecidable then is undecidableB

also means

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Example:

The halting problem

is reduced to

The state-entry problem

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The state-entry problem

Inputs: M•Turing Machine

•State q

Question:

Does M

•String w

enter state q

on input ?w

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THEOREM

The state-entry problem is undecidable

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THEOREM

The state-entry problem is undecidable

PROOF

Reduce the halting problem to

the state-entry problem

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Suppose we have an algorithmfor the state-entry problem

We will construct an algorithmfor the halting problem

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Inputs for the halting problem

•A machine M

•A string w

Algorithm for halting problem must determineif halts on inputM w

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Modify machine :M

•Add new state q

•From any halting state add transitions to q

M q

halting statesSinglehalt state

M

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M halts if and only if

M halts on state q

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Algorithm for halting problem:

Inputs: M wand

1. Construct with state M q

2. Run algorithm for state-entry problem with inputs: M wq, ,

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Generate

M M

w

M qw

State-entry

machine

Halting problem machine

YES

NO

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Since the halting problem is undecidable,

the state-entry problem is also undecidable

END OF PROOF

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Another example:

The halting problem

is reduced to

The blank-tape halting problem

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The blank-tape halting problem

Input: MTuring Machine

Question:

Does M halt when started with

a blank tape?

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THEOREM

The blank-tape halting problem is undecidable

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THEOREM

PROOF

Reduce halting problem to

Blank-tape halting problem

The blank-tape halting problem is undecidable

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Suppose we have an algorithmfor the blank-tape halting problem

We will construct an algorithmfor the halting problem

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Inputs for the halting problem

•A machine M

•A string w

Algorithm for halting problem must determineif halts on inputM w

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wMConstruct a new machine

• On blank tape writes w

• Then continues execution like M

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M halts if and only if

wM halts when started with blank tape

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Algorithm for halting problem:

Inputs: M wand

1. Construct wM

2. Run algorithm for blank-tape halting problem with input

, ,

wM

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Generate

wM

M

w

blank-tape halting machine

Halting problem machine

YES

NO

wM

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Since the halting problem is undecidable,

the blank-tape halting problem is also undecidable

END OF PROOF

1 the chomsky hierarchy. 2 unrestricted grammars: rules have form string of variables and terminals...

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