Fall 2006Costas Busch - RPI1 Reductions. Fall 2006Costas Busch - RPI2 Problem is reduced to problem If we can solve problem then we can solve problem.

Fall 2006 Costas Busch - RPI 1

Reductions


Problem is reduced to problemX Y

If we can solve problem then we can solve problemX

Y


Language is reduced tolanguage

There is a computable function (reduction) such that:f

BwfAw )(

A

B

Definition: A B

w )(wf


Computable function : f

which for any string computes )(wfwThere is a deterministic Turing machineM

Recall:


If: a: Language is reduced to b: Language is decidableThen: is decidable

Theorem:

Proof:

A BB

A

Basic idea:Build the decider for using the decider for

A

B


Decider for B

Decider for A

compute

)(wf

)(wfw

accept

reject

accept

reject

(halt)

(halt)(halt)

(halt)

Inputstring

BwfAw )(

END OF PROOF

Reduction YES YES

NO NO


Example:

}languages same the accept that

DFAs are and :,{ 2121 MMMMEQUALDFA

} language empty the

accepts that DFA a is :{

MMEMPTYDFA

is reduced to:


Turing Machinefor reduction

DFADFA EMPTYMEQUALMM 21,

f21,MM M

MMf

21,

DFA

We only need to construct:


21,MM M

MMf

21,

Let be the language of DFA Let be the language of DFA

1L

2L1M

2M

)()()( 2121 LLLLML

construct DFA by combining and so that:

M

DFA

1M 2M

Turing Machinefor reduction f


)(21 MLLL

)()()( 2121 LLLLML

DFADFA EMPTYMEQUALMM 21,


Decider

Decider for

compute M

Inputstring

DFAEQUAL

21,MM 21,MMf DFAEMPTY

YESYES

NONO

Reduction


If: a: Language is reduced to b: Language is undecidableThen: is undecidable

Theorem (version 1):

A B

BA

(this is the negation of the previous theorem)

Proof:

Using the decider for build the decider forA

BSuppose is decidable B

Contradiction!


Decider for B

Decider for A

compute

)(wf

)(wfw

accept

reject

accept

reject

(halt)

(halt)(halt)

(halt)

Inputstring

BwfAw )(

Reduction

END OF PROOF

If is decidable then we can build:B

CONTRADICTION!

YES YES

NO NO


Observation:

In order to prove that some language is undecidablewe only need to reduce a known undecidable language to

B

BA


State-entry problem

Input: M•Turing Machine•State q

Question: Does M

•Stringw

enter state qwhile processing input string ?w

Corresponding language:

} string input on state enters

that machine Turing a is :,,{

wq

MqwMSTATE TM


Theorem:

(state-entry problem is unsolvable)

Proof: Reduce (halting problem) to (state-entry problem)

TMSTATE is undecidable

TMHALT

TMSTATE


Decider for

YES

NO

wM,

state-entry problem decider

TMSTATEDeciderCompute

Reduction

wMf ,

wqM ,,ˆ

TMHALT

YES

NO

Given the reduction,if is decidable,then is decidable

TMSTATE

TMHALT

A contradiction!sinceis undecidable

Halting Problem Decider

TMHALT


wM,Compute

Reduction

wMf ,wqM ,,ˆ

We only need to build the reduction:

TMHALTwM , TMSTATEqwM ,,ˆ

wMf ,

So that:


Mqhalting

states

specialhalt state

M̂

Rxx ,

Construct from :M̂ M

A transition for every unused tape symbol of x

iq

iq


M̂ halts on state qM halts

Mqhalting

states

specialhalt state

M̂

iq


M̂ halts on state on inputq

M halts on input w

w

Therefore:

Equivalently:

END OF PROOF

TMHALTwM , TMSTATEqwM ,,ˆ


Blank-tape halting problem

Input: MTuring Machine

Question: Does M halt when started with

a blank tape?


}tape blank on started when halts

that machin aTuring is :{ eMMBLANKTM


Theorem:

(blank-tape halting problem is unsolvable)

Proof: Reduce (halting problem) to (blank-tape problem)

TMBLANK is undecidable

TMHALT

TMBLANK


Decider for

YES

NO

wM,

blank-tape problem decider

DeciderCompute

Reduction

wMf ,

M̂

TMHALT

YES

NO

Given the reduction,If is decidable,then is decidableTMHALT


Halting Problem Decider

TMHALT

TMBLANK

TMBLANK


wM,Compute

Reduction

wMf ,M̂


TMHALTwM , TMBLANKM ˆ

wMf ,

So that:


no

yes

M̂

Write on tape w

Tape is blank?

Run

with input

Construct from :M̂ wM,

If halts then halt

M

w

M


M̂ halts when started on blank tape

M halts on input

no

yesM

Write on tape w

Tape is blank?

Run

with inputw

M̂

w


END OF PROOF

M̂ halts when started on blank tape

M halts on inputw

TMHALTwM , TMBLANKM ˆ

Equivalently:


If: a: Language is reduced to b: Language is undecidableThen: is undecidable

Theorem (version 2):

A B

BA

Proof:

Using the decider for build the decider for A

B

Suppose is decidable B

Contradiction!

Then is decidableB


Suppose is decidableB

Decider for B

s

accept

reject

(halt)

(halt)


Suppose is decidableB

Decider for B

s

accept

reject

(halt)

(halt)

Then is decidableB(we have proven this in previous class)

reject

accept(halt)

(halt)

Decider for BNO YES

YES NO


Decider for B

Decider for A

compute

)(wf

)(wfw

accept

reject

accept

reject

(halt)

(halt)(halt)

(halt)

Inputstring

BwfAw )(

Reduction

If is decidable then we can build:B

CONTRADICTION!

YES YES

NO NO


Decider for B

Decider for A

compute

)(wf

)(wfw

accept

reject accept

reject

(halt)

(halt)(halt)

(halt)

Inputstring

BwfAw )(

Reduction

END OF PROOFCONTRADICTION!

Alternatively:

NO YES

YES NO


Observation:

In order to prove that some language is undecidablewe only need to reduce some known undecidable languagetoor to B

(theorem version 1)

(theorem version 2)

B

AB


Undecidable Problems for Turing Recognizable languages

• is empty?L

L• is regular?

L• has size 2?

Let be a Turing-acceptable language L

All these are undecidable problems


• is empty?L

L• is regular?

L• has size 2?



Empty language problem


Question: Is )(ML empty?


} language empty the accepts

that machine aTuring is :{

MMEMPTYTM

?)( ML


Theorem:

(empty-language problem is unsolvable)

is undecidable

Proof: Reduce (membership problem) to

(empty language problem)

TMA

TMEMPTY

TMEMPTY


Decider for

YES

NO

wM,

empty problem decider

DeciderCompute

Reduction

wMf ,

M̂YES

NO


membership problem decider

TMA

TMATMEMPTY

TMEMPTY


TMA


wM,Compute

Reduction

wMf ,M̂


TMATwM , TMEMPTYM ˆ

wMf ,

So that:


skip inputstring s

write on tape

run

on input

w Mw

M w

M̂

yes

then accept

If

s

sTape of M̂

input string

accepts ?

troys



skip inputstring s

write on tape

run

on input

w Mw

M w

M̂

yes

then accept

If

s

s

accepts ?

troys

Tape of M̂


skip inputstring s

write on tape

run

on input

w Mw

M w

M̂

yes

then accept

If

s

s

accepts ?

w

troys

Tape of M̂


skip inputstring s

write on tape

run

on input

w Mw

M w

M̂

yes

then accept

If

s

s

accepts ?

During this phase this area is not touched

working area of

wM

troys


skip inputstring s

write on tape

run

on input

w Mw

M w

M̂

yes

then accept

If

s

s

accepts ?

waltered

Simply check if entered an accept stateM

troys


skip inputstring s

write on tape

run

on input

w Mw

M w

M̂

yes

then accept

If

s

s

accepts ?

Now check input string

troys


skip inputstring s

write on tape

run

on input

w Mw

M w

M̂

yes

then accept

If

saccepts ?

The only possible accepted string

troys

t r o y


skip inputstring s

write on tape

run

on input

w Mw

M w

M̂

yes

then accept

If

saccepts ?

troys

accepts }{)ˆ( troyMLM w

does notaccept

M w )ˆ(ML


Therefore:

acceptsM w )ˆ(ML

Equivalently:

TMATwM , TMEMPTYM ˆ

END OF PROOF


• is empty?L

L• is regular?

L• has size 2?



Regular language problem


Question: Is )(ML a regular language?


language} regular a accepts

that machine aTuring is :{ MMREGULARTM


Theorem:

(regular language problem is unsolvable)

is undecidable

Proof: Reduce (membership problem) to (regular language problem)

TMA

TMREGULAR

TMREGULAR


Decider for

YES

NO

wM,

regular problem decider

DeciderCompute

Reduction

wMf ,

M̂YES

NO

Given the reduction,If is decidable,then is decidable


TMA

TMA

TMREGULAR


TMATMREGULAR


wM,Compute

Reduction

wMf ,M̂


TMATwM , TMREGULARM ˆ

wMf ,

So that:


skip inputstring s

write on tape

run

on input

w Mw

M w

M̂

yes

then accept

If has the form

s

sTape of M̂

input string

accepts ?

nnba


s


skip inputstring s

write on tape

run

on input

w Mw

M w

M̂

yesaccepts ?

accepts }0:{)ˆ( nbaML nnM w

does notaccept

M w )ˆ(ML

then accept

If has the form

s

nnbas

not regular

regular


Therefore:

acceptsM w )ˆ(ML

Equivalently:

TMATwM , TMREGULARM ˆ

END OF PROOF

is not regular


• is empty?L

L• is regular?

L• has size 2?



Does have size 2?

Size2 language problem


Question: )(ML


strings} two exactly accepts

that machine aTuring is :{2 MMSIZE TM

?2|)(| ML

(accepts exactly two strings?)


Theorem:

(regular language problem is unsolvable)

is undecidable

Proof: Reduce (membership problem) to (size 2 language problem)

TMA

TMSIZE 2

TMSIZE 2


Decider for

YES

NO

wM,

size2 problem decider

DeciderCompute

Reduction

wMf ,

M̂YES

NO

Given the reduction,If is decidable,then is decidable


TMA

TMA


TMA

TMSIZE 2

TMSIZE 2


wM,Compute

Reduction

wMf ,M̂


TMATwM , TMSIZEM 2ˆ

wMf ,

So that:


skip inputstring s

write on tape

run

on input

w Mw

M w

M̂

yes

then accept

If

s

sTape of M̂

input string

accepts ?


},{ albanytroys


skip inputstring s

write on tape

run

on input

w Mw

M w

M̂

yesaccepts ?

accepts },{)ˆ( albanytroyML M w

does notaccept

M w )ˆ(ML

2 strings

0 strings

then accept

If

s

},{ albanytroys


Therefore:

acceptsM w )ˆ(ML

Equivalently:

TMATwM , TMSIZEM 2ˆ

END OF PROOF

has size 2


RICE’s Theorem

• is empty?L

L• is regular?

L• has size 2?

Undecidable problems:

This can be generalized to all non-trivialproperties of Turing-acceptable languages


Non-trivial property:

A property possessed by some Turing-acceptable languages but not all

: is empty?LExample:

L

}{troyL YES

NO

},{ albanytroyL NO

P

1P


: is regular?

More examples of non-trivial properties:

L

}0:{ nbaL nn

YES

NO

L2P

}0:{ naL nYES

: has size 2?L3PL

}{troyL },{ albanytroyL

NO

YES

NO


Trivial property:

A property possessed by ALL Turing-acceptable languages

P

: has size at least 0?LExamples: 4PTrue for all languages

: is accepted by some Turing machine?

L5P

True for allTuring-acceptable languages


We can describe a property as the setof languages that possess the property

P

: is empty?LExample:

L

}{troyL YES

NO

},{ albanytroyL NO

P

}{P

If language has property then PL PL


: has size 1?LP

}{a

},{ a

NO

NO

YES

Example: Suppose alphabet is }{a

}{aa},{ aa

}{aaa}{},{ aaa

}},{},{},{},{},{{ aaaaaaaaaaP

},,{ aaaNO },,{ aaaaaaaaa


Non-trivial property problem

Does have the non-trivial property ?


Question: )(ML


})( is, that , property

trivial-non the has )( that such

machine aTuring is :{

PMLP

ML

MMPROPERTYTM

?)( PML P


Rice’s Theorem: TMPROPERTY is undecidable

(the non-trivial property problem is unsolvable)

Proof: Reduce (membership problem)

to

TMA

TMPROPERTY TMPROPERTYor


We examine two cases:

Case 1:

Case 2:

P

P

Examples: : is empty?)(MLP

: is regular?)(MLP

: has size 2?)(MLPExample:


Let be the Turing machine thataccepts

Case 1: P

Since is non-trivial, there is a Turing-acceptable languagesuch that:

PX

PX

XM

X


Reduce (membership problem) to

TMA

TMPROPERTY


Decider for

YES

NO

wM,

Non-trivial property problem decider

DeciderCompute

Reduction

wMf ,

M̂YES

NO



TMA

TMA


TMA

TMPROPERTY

TMPROPERTY


wM,Compute

Reduction

wMf ,M̂


TMATwM , TMPROPERTYM ˆ

wMf ,

So that:


skip inputstring s

write on tape

run

on input

w Mw

M w

M̂

yes

then accept

If

s

sTape of M̂

input string

accepts ?


Xs


skip inputstring s

write on tape

run

on input

w Mw

M w

M̂

yes

then accept

If

saccepts ?

Xs

For this phase we can run machinethat accepts , with input string

XMX s


skip inputstring s

write on tape

run

on input

w Mw

M w

M̂

yesaccepts ?

accepts XML )ˆ(M w

does notaccept

M w )ˆ(ML

P

P

then accept

If

s

Xs


Therefore:

acceptsM w PML )ˆ(

Equivalently:



Let be the Turing machine thataccepts

Case 2: P

Since is non-trivial, there is a Turing-acceptable languagesuch that:

PX

PX

XM

X


Reduce (membership problem) to

TMA

TMPROPERTY


Decider for

YES

NO

wM,

Non-trivial property problem decider

DeciderCompute

Reduction

wMf ,

M̂YES

NO



TMA

TMA


TMA

TMPROPERTY

TMPROPERTY


wM,Compute

Reduction

wMf ,M̂



wMf ,

So that:


skip inputstring s

write on tape

run

on input

w Mw

M w

M̂

yes

then accept

If

s

sTape of M̂

input string

accepts ?


Xs


skip inputstring s

write on tape

run

on input

w Mw

M w

M̂

yesaccepts ?

accepts XML )ˆ(M w

does notaccept

M w )ˆ(ML

P

P

then accept

If

s

Xs


Therefore:

acceptsM w PML )ˆ(

Equivalently:


END OF PROOF

Fall 2006Costas Busch - RPI1 Reductions. Fall 2006Costas Busch - RPI2 Problem is reduced to problem If we can solve problem then we can solve problem.

Documents