decid ability

7/30/2019 Decid Ability

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Recursively Enumerable

andRecursive Languages


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Definition:

A language is recursively enumerableif some Turing machine accepts it


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For string :

Let be a recursively enumerable languageL

and the Turing Machine that accepts it

Lww

if then halts in a final state

Lwif then halts in a non-final state

or loops forever


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Definition:

A language is recursive

if some Turing machine accepts it

and halts on any input string

In other words:A language is recursive if there is

a membership algorithm for it


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For string :

Let be a recursive languageL

and the Turing Machine that accepts it

Lw

w

if then halts in a final state

Lwif then halts in a non-final state


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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


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Recursive


Non Recursively Enumerable


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A Language whichis not



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We want to find a language that

is not Recursively Enumerable

This language is not accepted by any

Turing Machine


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Consider alphabet }{a

Strings: ,,,, aaaaaaaaaa

1

a2

a3

a4

a


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Consider Turing Machines

that accept languages over alphabet }{a

They are countable:

,,,, 4321 MMM


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Example language accepted by

},,{)( aaaaaaaaaaaaML i

},,{)( 642 aaaML i

i

Alternative 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


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Consider the language

)}(:{ iii MLaaL

L consists from 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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Theorem:

Language is not recursively enumerableL


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Proof:

is recursively enumerableLAssume for contradiction that

There must exist some machine

that acceptsk

L

LMLk)(

1 2 3 4


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1a 2a 3a 4a

)(1

ML

0 1 10

)( 2ML

)( 3ML

01 0 1

0 1 11

)( 4ML 0 10 0

Question: ?1Mk

1 2 3 4


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1a 2a 3a 4a

)(1

ML

0 1 10

)( 2ML

)( 3ML

01 0 1

0 1 11

)( 4ML 0 10 0

Answer: 1Mk

)(

)(

11

1

MLa

MLa k

1 2 3 4


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1a 2a 3a 4a

)(1

ML

0 1 10

)( 2ML

)( 3ML

01 0 1

0 1 11

)( 4ML 0 10 0

Question: ?2Mk

1 2 3 4


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1a 2a 3a 4a

)(1

ML

0 1 10

)( 2ML

)( 3ML

01 0 1

0 1 11

)( 4ML 0 10 0

Answer: 2Mk

)(

)(

22

2

MLa

MLa k

1 2 3 4


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1a 2a 3a 4a

)(1

ML

0 1 10

)( 2ML

)( 3ML

01 0 1

0 1 11

)( 4ML 0 10 0

Question: ?3Mk

1 2 3 4


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1a 2a 3a 4a

)(1

ML

0 1 10

)( 2ML

)( 3ML

01 0 1

0 1 11

)( 4ML 0 10 0

Answer: 3Mk

)(

)(

33

3

MLa

MLa k


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Similarly: ik M

)(

)(

ii

ki

MLa

MLa

)(

)(

ii

ki

MLa

MLa

for any i

Because either:

or


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Therefore, the machine cannot existk

Therefore, the languageis not recursively enumerable

L

End of Proof


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Observation:

There is no algorithm that describes L

(otherwise would be accepted by

some Turing Machine)

L


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Recursive



L


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A Language which isRecursively Enumerable

and not Recursive


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We want to find a language which

There is a

Turing Machinethat accepts

the language

The machine

doesnt halt

on some input

Is recursively

enumerable

But not

recursive


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We will prove that the language

)}(:{ iii MLaaL

Is recursively enumerable

but not recursive


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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 language

Theorem:

)}(:{ iii MLaaL

is recursively enumerable


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Proof:

We will give a Turing Machine thataccepts L

Turing Machine that accepts L


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For any input string w

Compute , for which iaw

Find Turing machine i

(using an enumeration procedurefor Turing Machines)

Simulate on inputi

i

a If accepts, then accepti w

i

End of Proof


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Observation:

)}(:{ iii MLaaL

)}(:{ iii MLaaL

Recursively enumerable

Not recursively enumerable

(Thus, also not recursive)


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Theorem:

The language )}(:{ iii MLaaL

is not recursive


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Proof:

Assume for contradiction that is recursive

Then is recursive:

L

L

Take the Turing Machine that accepts L

halts on any input:

If accepts then reject

If rejects then accept


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Therefore:

L is recursive

But we know:

L is not recursively enumerable

thus, not recursive

CONTRADICTION!!!!


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Therefore, is not recursiveL

End of Proof


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L

Recursive



L


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Turing acceptable languagesand

Enumeration Procedures

W ill


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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 thenthere is an enumeration procedure for it

L


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Proof:

~

Enumeration Machine

Accepts LEnumerates all

strings of input alphabet


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If the alphabet is then

can enumerate strings as follows:

},{ ba~

ab

aaabba

bbaaaaab......

E ti d


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Enumeration procedure

Repeat:~ generates a string w

checks if Lw

YES: print to outputw

NO: ignore w

End of Proof

E l }{ bbbbL


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Example:

~

a

baaabba

bbaaaaab

......

)(ML

b

ab

bbaaa

......

,....},,,{ aaabbabbL

Enumeration

Output

b

ab

bbaaa

...... Theorem:


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Theorem:

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

Proof:


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Proof:

~

Enumeration Machine

Accepts LEnumerates all

strings of input alphabet


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If the alphabet is then

can enumerate strings as follows:

},{ ba~

ab

aaabba

bbaaaaab

NAIVE APPROACH


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Enumeration procedure

Repeat: ~ generates a string w

checks if Lw

YES: print to outputw

NO: ignore w

NAIVE APPROACH

Problem: If

machine may loop forever

Lw

BETTER APPROACH


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~

1

w

executes first step on

BETTER APPROACH

1w

~ Generates second string 2w

executes first step on 2w

second step on 1w

Generates first string


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~ Generates third string 3w

executes first step on 3w

second step on 2wthird step on 1w

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


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1w 2w 3w 4w

1

Stepin

string

1 1 1

2 2 2 2

3 3 3 3


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If for any string

machine halts in a final state

then it prints on the output

iw

iw

End of Proof

Theorem:


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Theorem:

If for languagethere is an enumeration procedure

then is recursively enumerable

L

L

Proof:I t T


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w

Input Tape

Enumerator

for L Compare

Machine that

accepts L

Turing machine that accepts L


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Turing machine that accepts L

Repeat:

Using the enumerator,generate the next string of L

For input string w

Compare generated string with wIf same, accept and exit loop

End of Proof


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We have proven:

A language is recursively enumerable

if and only if

there is an enumeration procedure for it


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The Chomsky Hierarchy


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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?w

Does DFA accept any input?

A problem is decidable if some Turing machine


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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?w

Does DFA accept any input?

The Turing machine that decides (solves)


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Turing MachineInput

problem

instance

YES

NO

g

a problem answers YES or NO

for each instance of the problem

Th hi th t d id ( l ) bl


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The machine that decides (solves) a problem:

If the answer is YESthen halts in ayes state

If the answer is NO

then halts in a no state

These states may not be final states

Turing Machine that decides a problem


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YES states

NO states

YES and NO states are halting states

Difference between


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Difference between

Recursive Languages and Decidable problems

The YES states may not be final states

For decidable problems:

Some problems are undecidable:


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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

The Membership Problem


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The Membership Problem

Input: Turing Machine

String w

Question: Does accept ?w

?)(MLw

Theorem:


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Theorem:

The membership problem is undecidable

Proof: Assume for contradiction thatthe membership problem is decidable

(there are and for which we cannot

decide whether )

w

)(MLw

Th s th ists T i M hi H


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Thus, there exists a Turing Machine

that solves the membership problem

H

Hw

YES accepts w

NO rejects w



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Let be the Turing Machine that accepts L

We will prove that is also recursive:L

we will describe a Turing machine that

accepts and halts on any inputL



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accepts ?wNO

YES

w

Hacceptw

Turing Machine that accepts

and halts on any inputL

reject w

Therefore L is recursive


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Therefore, L is recursive

But there are recursively enumerable

languages which are not recursive

Contradiction!!!!

Since is chosen arbitrarily, everyrecursively enumerable language is alsorecursive

L


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Therefore, the membership problem

is undecidable

END OF PROOF

A th f d id bl bl


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Another famous undecidable problem:

The halting problem

The Halting Problem


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The Halting Problem


String w

Question: Does halt on input ?w

Theorem:


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Theorem:

The halting problem is undecidable

Proof: Assume for contradiction thatthe halting problem is decidable

(there are and for which we cannot

decide whether halts on input )

w

w

Thus there exists Turing Machine H


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Thus, there exists Turing Machine

that solves the halting problem

H

Hw

YES halts on w

doesnt

halt onwNO

Construction of H


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H

wwM 0q

yq

nq

Input:

initial tape contents

Encoding

of w

String

YES

NO


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Construct machine :H

If returns YES then loop foreverH

If returns NO then haltH


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H

wwM 0q

yq

nq NO

aq bq

H

Loop forever

YES

HConstruct machine :


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HConstruct machine :

Input:

If halts on input Mw

Thenloop forever

Elsehalt

Mw (machine )


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Mw MMwwcopy

Mw H

H

HRun machine with input itself:


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HRun machine with input itself:

Input:

If halts on input

Thenloop forever

Elsehalt

Hw (machine )H

H Hw


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on inputH Hw

If halts then loops forever

If doesnt halt then it halts

:

H

H

NONSENSE !!!!!

Therefore we have contradiction


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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 language

would be recursive

Theorem:


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Theorem:


Proof: Assume for contradiction that

the halting problem is decidable

There exists Turing Machine H


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There exists Turing Machine

that solves the halting problem

H

Hw

YES halts on w

doesnt

halt onwNO



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Let be a recursively enumerable language

Let be the Turing Machine that accepts L

We will prove that is also recursive:L

we will describe a Turing machine that

accepts and halts on any inputL

Turing Machine that acceptsd h lt i t

L


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halts on ?w

YES

NO

w

Run

with input w

H reject w

accept w

reject w

and halts on any input

Halts on final state

Halts on non-final

state

Therefore L is recursive


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But there are recursively enumerable

languages which are not recursive

Contradiction!!!!

Since is chosen arbitrarily, everyrecursively enumerable language

is also recursive

L


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Therefore, the halting problem is undecidable

END OF PROOF


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Reducibility

Problem is reduced to problemA B


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If we can solve problem then

we can solve problem

B

A

B

A



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If is undecidable then is undecidable

If is decidable then is decidableB A

A B



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Example: the halting problem

is reduced to

the state-entry problem

The state-entry problem


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y p

Inputs: Turing Machine

State q

Question: Does

String w

enter state q

on input ?w

Theorem:


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The state-entry problem is undecidable

Proof: Reduce the halting problem to

the state-entry problem


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We want to build a decider


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Halting problem

deciderw

YES

NO

halts on w

doesnt

halt onw

for the halting problem:


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We need to convert one problem instance

to the other problem instance


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qw

Halting problem decider

YES

NO

YES

NOw

Convert

Inputs

?


State-entryproblem

decider

Convert to :


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Add new state q

From any halting state of add transitions toq

q

halting statesSingle

halt state


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halts on input

halts on state on inputq

if and

only if

w

w


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Generate

w

qw


YES

NO

YES

NOState-entryproblem

decider

We reduced the halting problem


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Since the halting problem is undecidable,the state-entry problem is undecidable

END OF PROOF

g p

to the state-entry problem

Another example:


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the halting problem

is reduced to

the blank-tape halting problem

The blank-tape halting problem


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Question: Does halt when started with

a blank tape?

Theorem:


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Proof: Reduce the halting problem to the

blank-tape halting problem

The blank-tape halting problem is undecidable

Suppose we have a decider for theblank-tape halting problem:


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blank-tapehalting problem

decider

YES

NO

halts on

blank tape

doesnt halt

on blank tape

blank tape halting problem:

We want to build a decider


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halting problem

deciderw

YES

NO

halts on w

doesnt

halt onw


We want to reduce the halting problem to

the blank-tape halting problem:


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w

Blank-tapehalting problem

decider


YES

NOw

YES

NO

the blank tape halting problem:




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w

Blank-tapehalting problem

decider


YES

NOw

YES

NO


ConvertInputs

?

wConstruct a new machine


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When started on blank tape, writes w

Then continues execution like

w

then write w

step 1 step2

if blank tape executewith input w


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halts on input string

w halts when started with blank tape

if and

only if

w


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We reduced the halting problem


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Since the halting problem is undecidable,the blank-tape halting problem is undecidable

END OF PROOF

to the blank-tape halting problem

Summary of Undecidable Problems


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Does machine halt on input ?

Halting Problem:

w

Membership problem:

Does machine accept string ?w

Blank-tape halting problem:


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Does machine enter stateon input ?

Does machine halt when starting

on blank tape?

State-entry Problem:

wq


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Uncomputable Functions

Uncomputable Functions


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A function is uncomputable if it cannotbe computed for all of its domain

Domain Values

region

f


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An uncomputable function:

)(nf

maximum number of moves until

any Turing machine with states

halts when started with the blank tape

n

Theorem: Function is uncomputable)(nf


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Proof:

Then the blank-tape halting problem

is decidable

Assume for contradiction thatis computable)(nf


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Therefore, the blank-tape halting

l l


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problem is decidable

However, the blank-tape haltingproblem is undecidable

Contradiction!!!


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132/177

Take a recursively enumerable language L


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is empty?L

L is finite?L contains two different strings

of the same length?

Decision problems:

All these problems are undecidable

Theorem:For any recursively enumerable language L


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y y g g

it is undecidable to determine whether

is emptyL

Proof:

We will reduce the membership problemto this problem

Let be the TM with LML )(


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empty language

problem

decider

YES

NO

Suppose we have a decider for theempty language problem:

)(ML

)(ML

empty

not empty


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We want to reduce the membership problem to

the empty language problem:


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YES

NOw

NO

YES

Membership problem decider

w empty language

problem

decider




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YES

NOw

NO

YES

Membership problem decider

w empty language

problem

decider

Convert

inputs

?


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Undecidable problemsfor

Recursively enumerable languages

continued

Take a recursively enumerable language L


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is empty?L

L is finite?L contains two different strings

of the same length?

Decision problems:

All these problems are undecidable


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finite language

problem

decider

YES

NO

Suppose we have a deciderfor the finite language problem:

)(ML

)(ML

finite

not finite

We will build a decider


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Halting problem

deciderw

YES

NO

halts on w

We will build a decider


doesnthalt on

w

We want to reduce the halting problem tothe finite language problem


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YES

NOw

NO

YES


wfinite language

problem

decider

We need to convert one problem instanceto the other problem instance


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YES

NOw

NO

YES


wfinite language

problem

decider

convert

input

?

Construct machine :w

On arbitrary input string s


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If enters a halt state,accept ( inifinite language)

Initially, simulates on input w

Otherwise, reject ( finite language)

y p g

s

s

*


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halts on

)( wML is infinite

if andonly if

w

*)( wML


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construct

w

YES

NOw

NO

YES

halting problemdecider

finite language

problemdecider

w


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Theorem:

For a recursively enumerable language L


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For a recursively enumerable language L

it is undecidable to determine whethercontains two different strings of

same length

L

Proof: We will reduce the halting problemto this problem



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Two-strings

problem

decider

YES

NO

Suppose we have the deciderfor the two-strings problem:

)(ML

)(ML

contains

Doesntcontain

two equal length strings

We will build a decider for


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Halting problem

deciderw

YES

NO

halts on w

the halting problem:

doesnthalt on

w

We want to reduce the halting problem tothe empty language problem


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YES

NOw

YES

NO


Two-strings

problemdecider

w

We need to convert one problem instanceto the other problem instance


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YES

NOw

YES

NO


Two-strings

problemdecider

wconvert

inputs?

Construct machine :w

O bit i t st i s


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When enters a halt state,accept if or

Initially, simulate on input w

as bs (two equal length strings )

On arbitrary input string s

},{)( baMLw

Otherwise, reject ( )s )( wML


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halts on

w

if andonly if

w

accepts two equal length strings

w accepts anda b


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construct

w

YES

NOw

YES

NO


Two-strings

problemdecider

w


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Definition:


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Non-trivial properties ofrecursively enumerable languages:

any property possessed by some (not all)

recursively enumerable languages


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164/177


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The Post Correspondence

Problem

Some undecidable problems for

context-free languages:


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context free languages:

Is context-free grammar ambiguous?G

Is ? )()( 21 GLGL

21,GG are context-free grammars

We need a tool to prove that the previous

problems for context-free languages


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are undecidable:

The Post Correspondence Problem

The Post Correspondence Problem

I t T f i


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Input: Two sequences of strings

nwwwA ,,, 21

nvvvB ,,, 21

n


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Example:

11100 1111w 2w 3w

:A


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001 111 11

1v 2v 3v

:B

PC-solution: 3,1,2 312312 vvvwww

1110011

Example:

00100 10001w 2w 3w

:A


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0 11 011

1v 2v 3v

:B

There is no solution

Because total length of strings from

is smaller than total length of strings from

B

A

We will show:


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1. The MPC problem is undecidable

2. The PC problem is undecidable

(by reducing MPC to PC)

(by reducing the membership to MPC)


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Some undecidable problems forcontext-free languages:


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Is context-free grammar

ambiguous?

G

Is ? )()( 21 GLGL

21,GG are context-free grammars

We reduce the PC problem to these problems

Theorem: Let be context-free

grammars. It is undecidable21,GG


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Proof:

to determine if )()(

21GLGL

Rdeduce the PC problem to this

problem

Suppose we have a decider for the

empty-intersection problem


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Empty-interection

problem

decider

?)()( 21 GLGL

1G

2G

YES

NO

Context-free

grammars

For a context-free grammar ,GTheorem:

it i d id bl t d t i


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it is undecidable to determine

if G is ambiguous

Proof: Reduce the PC problem

to this problem

decid ability

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