prof. busch - lsu1 decidable languages. prof. busch - lsu2 recall that: a language is...

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Decidable Languages

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Recall that:

A language is Turing-Acceptableif there is a Turing machine that accepts

Also known as: Turing-Recognizable or Recursively-enumerable languages

LM

L

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For any string :

Lw

w

halts in an accept state M

Lw halts in a non-accept stateM

or loops forever

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Definition:

A language is decidableif there is a Turing machine (decider) which accepts and halts on every input string

Also known as recursive languages

L

LM

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For any string :

Lw

w

halts in an accept state M

Lw halts in a non-accept stateM

Every decidable language is Turing-Acceptable

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Sometimes, it is convenient to have Turing machines with single accept and reject states

acceptqrejectq

These are the only halting states

That result to possible halting configurations

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We can convert any Turing machine to have single accept and reject states

acceptq

Old machine New machine

Rxx ,

Rxx ,

Rxx ,

Multiple accept states

For each tape symbol x

One accept state

Lxx ,

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iq

iq Rxx ,

For all tape symbols not used for read in the other transitions of

x

iq

Old machine

New machine

rejectq

Multiple reject states One reject state

Do the following for each possible halting state:

For each

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Inputstring

Accept

Reject

Decider for LDecisionOn Halt:

acceptq

rejectq

For a decidable language : L

For each input string, the computationhalts in the accept or reject state

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Inputstring

Turing Machine forLacceptq

rejectq

For a Turing-Acceptable language : L

It is possible that for some input string the machine enters an infinite loop


} 7, 5, 3, 2, {1, PRIMES

Is number prime? x

Corresponding language:

Problem:

We will show it is decidable


On input number :x

Divide with all possible numbersbetween and

If any of them divides Then rejectElse accept

x2 x

Decider for : PRIMES

x


(Input string)is prime?

Decider for PRIMES

Input numberxYES

NO

the decider for the languagesolves the corresponding problem

(Accept)

(Reject)

x

acceptq

rejectq


Theorem:

If a language is decidable,then its complement is decidable too

LL

Proof:

Build a Turing machine that accepts and halts on every input string

M L

( is decider for )LM


M acceptq

M

rejectqacceptq

rejectq

aq

rq

aq

rq

Transform accept state to rejectand vice-versa


Accepts and halts on every input string

1. Let be the decider forM

If accepts then rejectIf rejects then accept

MM

M Turing Machine

On each input string do: w

2. Run with input stringM w

L

L

END OF PROOF


Undecidable Languages

An undecidable language has no decider:

There is a language which isTuring-Acceptable and undecidable

We will show that:

each Turing machine that acceptsdoes not halt on some input string

L


We will prove that there is a language :

• is not Turing-acceptable

(not accepted by any Turing Machine)

• is Turing-acceptable

LL

L

Therefore, is undecidableL

the complement of a decidable language is decidable


Decidable

Turing-Acceptable

Non Turing-Acceptable L

L


A Language which is not

Turing Acceptable


Consider alphabet }{a

,,,, aaaaaaaaaa

1a 2a 3a 4a

}{aStrings of :


Consider Turing Machinesthat accept languages over alphabet }{a

They are countable:

,,,, 4321 MMMM

(There is an enumerator that generates them)


,,,, 4321 MMMM

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 oneTuring machine

for


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


1a 2a 3a 4a

)( 1ML

0 1 10

)( 2ML

)( 3ML

01 0 1

0 1 11

)( 4ML 0 10 0

Example of binary representations


Consider the language

)}(:{ iii MLaaL

L consists of the 1’s in the diagonal


1a 2a 3a 4a

)( 1ML

0 1 10

)( 2ML

)( 3ML

01 0 1

0 1 11

)( 4ML 0 10 0

},,{ 43 aaL


Consider the language

)}(:{ iii MLaaL

L consists of the 0’s in the diagonal

)}(:{ iii MLaaL

L


1a 2a 3a 4a

)( 1ML

0 1 10

)( 2ML

)( 3ML

01 0 1

0 1 11

)( 4ML 0 10 0

},,{ 21 aaL


Proof:

is Turing-AcceptableL

Assume for contradiction that

There must exist some machinethat accepts :

kML LML k )(

Theorem:

Language is not Turing-AcceptableL


1a 2a 3a 4a

)( 1ML

0 1 10

)( 2ML

)( 3ML

01 0 1

0 1 11

)( 4ML 0 10 0

Question: ?1MMk LML k )(


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


1a 2a 3a 4a

)( 1ML

0 1 10

)( 2ML

)( 3ML

01 0 1

0 1 11

)( 4ML 0 10 0



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


1a 2a 3a 4a

)( 1ML

0 1 10

)( 2ML

)( 3ML

01 0 1

0 1 11

)( 4ML 0 10 0



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


Similarly: ik MM

)(

)(

ii

ki

MLa

MLa

)(

)(

ii

ki

MLa

MLa

for any i

Because either:

or

the machine cannot exist kM

is not Turing-AcceptableLEnd of Proof


Decidable

Turing-Acceptable

Non Turing-Acceptable

L


A Language which is Turing-Acceptableand Undecidable


There is a Turing machine that accepts

Each machinethat accepts doesn’t halt on some input string

Is Turing-Acceptable

Undecidable

)}(:{ iii MLaaL

We will prove that the language

LL


1a 2a 3a 4a

)( 1ML

0 1 10

)( 2ML

)( 3ML

01 0 1

0 1 11

)( 4ML 0 10 0

},,{ 43 aaL


The languageTheorem:

)}(:{ iii MLaaL

Is Turing-Acceptable

Proof:We will give a Turing Machine thataccepts L


Turing Machine that accepts LFor any input string w

• Compute , for which iaw

• Find Turing machine iM(using the enumerator for Turing Machines)

• Simulate on inputiMia

• If accepts, then accept iM w

i

End of Proof


Observation:

)}(:{ iii MLaaL

)}(:{ iii MLaaL

Turing-Acceptable

Not Turing-acceptable

(Thus, is undecidable)L


Decidable

Turing-Acceptable

Non Turing-Acceptable L

L

L ?


Theorem: )}(:{ iii MLaaL

is undecidable

Proof: If is decidable

Then is decidable

L

L

Contradiction!!!!

However, is not Turing-Acceptable!L

the complement of a decidable language is decidable


Decidable

Turing-Acceptable

Not Turing-Acceptable L

L


Turing acceptable languages and

Enumerators


We will prove:

• If a language is decidable then there is an enumerator for it

• A language is Turing-acceptable if and only if there is an enumerator for it

(weak result)

(strong result)


Theorem:

if a language is decidable then there is an enumerator for it

L

Proof:

Let be the decider for M L

Use to build the enumerator for M L


Example: alphabet is },{ ba

abaaabbabbaaaaab......

(proper order)

Let be an enumerator that printsall strings from input alphabet in proper order

M~


Enumerator for

Repeat:

M~1. generates a string

M

w

2. checks if Lw

YES: print to output w

NO: ignore w

L

This part terminates,because is decidable L


MM~Enumerator for

string

Give me next stringEnumerates all

strings of

input alphabet

Generates allStrings in alphabet

iwIf acceptsthen print tooutput

M iw

iw

Tests each stringif it is accepted by

L

M

output

All strings

of L


Example:

M~

abaaabbabbaaaaab

M

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

Output

b

ab

bbaaa

reject

reject

reject

accept

accept

accept

accept

reject

END OF PROOF

1w

2w3w


Theorem:

if language is Turing-Acceptable thenthere is an enumerator for it

L

Proof:

Let be the Turing machine that accepts M L

Use to build the enumerator for M L


MM~

Enumerator for

Accepts LEnumerates allstrings of input alphabetin proper order

L


Enumerator for

Repeat: M~ generates a string

M

w

checks if Lw

YES: print to output w

NO: ignore w

NAIVE APPROACH

Problem:If machine may loop forever

LwM

L


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


M~ Generates third string 3w

M executes first step on 3w

second step on 2w

third step on 1w

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


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


If for any string machine halts in an accepting statethen print on the output

Miw

iw

End of Proof


Theorem:

If for language there is an enumeratorthen is Turing-Acceptable

L

L

Proof:

Using the enumerator forwe will build a Turing machine that accepts

L

L


wInput Tape

Enumeratorfor L

Compare

Turing Machine that accepts Lw

iwIf same, Accept and Halt

Give me thenext stringin the enumeration sequence


Turing machine that accepts L

Loop:• Using 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


By combining the last two theorems,we have proven:

A language is Turing-Acceptableif and only if there is an enumerator for it

prof. busch - lsu1 decidable languages. prof. busch - lsu2 recall that: a language is...

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