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Fall 2005 Costas Busch - RPI 1

Recursively Enumerable

andRecursive Languages

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Definition:

A language is recursively enumerableif there is a Turing machine that accepts it

Also known as: Turing Recognizable languages

orTuring Acceptable languages

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For any string :

Let be a recursively enumerable languageL

and the Turing Machine that accepts itM

Lw

w

if then halts in an accept stateM

wif then halts in a non-accept stateM

or loops forever

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Definition:

A language is recursiveif there is a Turing machine accepts it

and the machine halts on every input string

Also known as decidable languages

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For any string :

Let be a recursive languageL

that accepts such that:M

Lw

w

if then halts in an accept stateM

wif then halts in a non-accept stateM

Then there is a Turing machineL

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We will prove:

1. There is a specific languagewhich is not recursively enumerable

(not accepted by any Turing Machine)

2. There is a specific language

which is recursively enumerable

but not recursive

L

L

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Recursive


Non Recursively Enumerable

L

L

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A Language which

is notRecursively Enumerable

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Consider alphabet }{a

Strings: -,,,, aaaaaaaaaa

1a

2a

3a

4a .

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Consider Turing Machines

that accept languages over alphabet }{a

They are countable:

-,,,, 4321 MMMM

(There is an enumeration procedure that

generates them)

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-

,,,, 4321MMMM

Each machine accepts some language over }{a

-),(),(),(),( 4321 MLMLMLML

Note that it is possible to have

)()( ji MLML ! ji {

Since, a language could be accepted by more than one

Turing machine

for

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Example language accepted by

},,{)( aaaaaaaaaaaaML i !

},,{)( 642 aaaML i !

iM

Binary representation

1a 2a 3a 4a 5a 6a 7a

)( iML

.

0 1 1 10 0 0 .

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1a

2a

3a

4a

)( 1ML

.

0 1 10 .

)( 2ML

)( 3ML

01 0 1 .

0 1 11 .

)( 4ML 0 10 .0

Example of binary representations

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Consider the language

)}(:{i

iiMLaaL !

L consists of the 1s in the diagonal

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1a

2a 3a

4a

)( 1ML

.

0 1 10 .

)( 2ML

)( 3ML

01 0 1 .

0 1 11 .

)( 4ML 0 10 .0

},,{ 43 -aaL !

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Consider the language

)}(:{ iii

MLaaL !

L consists of the 0s in the diagonal

)}(:{ iii

MLaaL !

L

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1a 2a

3a

4a

)( 1ML

.

0 1 10 .

)( 2ML

)( 3ML

01 0 1 .

0 1 11 .

)( 4ML 0 10 .0

},,{ 21 -aaL !

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Proof:

is recursively enumerableL

Assume for contradiction that

There must exist some machine

that accepts :kM

L LML k !)(

Theorem:

Language is not recursively enumerableL

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1a 2a

3a

4a

)( 1ML

.

0 1 10 .

)( 2ML

)( 3ML

01 0 1 .

0 1 11 .

)( 4ML 0 10 .0

Question: ?1MMk ! LML k !)(

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1a 2a

3a

4a

)( 1ML

.

0 1 10 .

)( 2ML

)( 3ML

01 0 1 .

0 1 11 .

)( 4ML 0 10 .0

Answer: 1MMk {

)(

)(

11

1

MLa

MLa k

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1a 2a

3a

4a

)( 1ML

.

0 1 10 .

)( 2ML

)( 3ML

01 0 1 .

0 1 11 .

)( 4ML 0 10 .0


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1a 2a

3a

4a

)( 1ML

.

0 1 10 .

)( 2ML

)( 3ML

01 0 1 .

0 1 11 .

)( 4ML 0 10 .0

Answer: 2MMk {

)(

)(

22

2

MLa

MLa k

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1a 2a

3a

4a

)( 1ML

.

0 1 10 .

)( 2ML

)( 3ML

01 0 1 .

0 1 11 .

)( 4ML 0 10 .0


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1a 2a

3a

4a

)( 1ML

.

0 1 10 .

)( 2ML

)( 3ML

01 0 1 .

0 1 11 .

)( 4ML 0 10 .0

Answer: 3MMk {

)(

)(

33

3

MLa

MLa k

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Similarly: ik MM {

)(

)(

i

i

ki

MLa

MLa

)(

)(

i

i

ki

MLa

MLa

for any i

Because either:

or

the machine cannot existkM

is not recursively enumerableL

End of Proof

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Recursive



L

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A Language which is

Recursively Enumerableand not Recursive

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There is a

Turing Machine

that accepts

Every machine

that accepts

doesnt halt

on some input string

Is recursively

enumerable

But not

recursive

)}(:{ iii

MLaaL !

We will prove that the language

L

L

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1a

2a 3a

4a

)( 1ML

.

0 1 10 .

)( 2ML

)( 3ML

01 0 1 .

0 1 11 .

)( 4ML 0 10 .0

},,{ 43 -aaL !

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The languageTheorem:

)}(:{ iii

MLaaL !is recursively enumerable

Proof: We will give a Turing Machine that

accepts L

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Turing Machine that accepts L

For any input string w

Compute , for which iaw !

Find Turing machine iM

(using the enumeration procedure

for Turing Machines)

Simulate on inputiMi

a If accepts, then acceptiM w

i

End of Proof

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Observation:

)}(:{ iii

MLaaL !

)}(:{ iii MLaaL !

Recursively enumerable

Not recursively enumerable

(Thus, is not recursive either)L

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Theorem:

If a language is recursive,

then its complement is recursive tooX

X

Proof:

Build a Turing machine thataccepts and halts on every inputX

Md

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(accepts and halts on every input string)

1. Let be the Turing machine that acceptsM X

and halts on every input string

If accepts then reject

If rejects then acceptM

M

X

MdTuring Machine

On any input string do:w

2. Run with input stringM w

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Md

iq iq

jq jq

Rxx ,p

For all tape symbols

not read in the othertransitions of

x

iq

M

Transform toM Md

ENDOF PROOF

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Theorem: )}(:{ iii

MLaaL !

is not a recursive language

Proof: If is recursive

Then is recursive

L

L

The complementof a recursive language

is recursive

Contradiction!!!!

However, is not Turing acceptable!L

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L

Recursive



L

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Turing acceptable languagesand

Enumeration Procedures

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We will prove:

If a language is recursive thenthere is an enumeration procedure for it

A language is recursively enumerable

if and only ifthere is an enumeration procedure for it

(weak result)

(strong result)

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Theorem:

if a language is recursive then

there is an enumeration procedure for it

L

Proof:

Let be the Turing machine that accepts

and halts on every input

M L

Use to build the enumeration procedure

for

M

L

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Example:

alphabet is },{ ba

ab

aaabba

bbaaaaab......

(proper order)

Let be an enumerator that prints

all strings from input alphabet in proper order

M~

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Enumeration procedure for

Repeat:

M~ generates a string

M checks if Lw

YES: print to outputw

NO: ignore w

L

This part terminates,

because is recursiveL

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MM

~

Enumeration Machine for

string

Give menext stringEnumerates all

strings of

input alphabet

Generates allStrings in alphabet

iw

If accepts

then print to

output

M iw

iw

Tests each stringif it is accepted by

L

M

output

All strings

of L

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Example:

M~

a

b

aaabba

bbaaaaab

M

,....},,,{ aaabbabbL !Enumeration

Output

b

ab

bbaaa

reject

reject

reject

accept

accept

accept

accept

reject

ENDOF PROOF

1w

2w

3w

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Theorem:

if language is recursively enumerable thenthere is an enumeration procedure for it

L

Proof:

Let be the Turing machine that acceptsM L

Use to build the enumeration procedure

for

M

L

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MM~

Enumeration Machine for

Accepts LEnumerates allstrings of input alphabet

in proper order

L

NAIVE APPROACH

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Enumeration procedure for

Repeat: M~ generates a string

M

w

checks if Lw

YES: print to outputw

NO: ignore w

NAIVE APPROACH

Problem: If

machine may loop forever

Lw

M

L

BE E PP

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M

~

M

1w

executes first step on

BETTER APPROACH

1w

M~ Generates second string 2w

M executes first step on 2w

second step on 1w

Generates first string

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M~ Generates third string

3w

M executes first step on 3w

second step on 2w

third step on 1w

And so on............

St i

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1w 2w 3w 4w ..

1

Step in

computationof string

1 1 1

2 2 2 2

3 3 3 3

String:

4 4 4 4

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If for any string

machine halts in an accepting state

then it prints on the output

M

iw

iw

End of Proof

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Theorem:

If for language

there is an enumeration procedure

then is recursively enumerable

L

L

Proof:

Using the enumerator for

we will build a Turing machine

that accepts

L

L

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w

Input Tape

Enumerator

for L

Compare

Turing Machine that acceptsL

w

iw

If same,Accept and HaltGive me thenext string

in the

enumeration

sequence

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Turing machine that accepts L

Loop:

U

sing the enumerator of ,generate the next string of L

For any input string w

Compare generated string with w

If same, accept and exit loop

End of Proof

L

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By Combining the last two theorems,

we have proven:

A language is recursively enumerable

if and only if

there is an enumeration procedure for it

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