Prof. Busch - LSU1 Reductions. Prof. Busch - LSU2 Problem is reduced to problem If we can solve problem then we can solve problem.

Prof. Busch - LSU 1

Reductions

Prof. Busch - LSU 2

Problem is reduced to problemX Y

If we can solve problem then we can solve problemX

Y

Prof. Busch - LSU 3

Language is reduced tolanguage

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

BwfAw )(

A

B

Definition: A B

w )(wf

Prof. Busch - LSU 4

Computable function : f

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

Recall:

Prof. Busch - LSU 5

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

Prof. Busch - LSU 6

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

Prof. Busch - LSU 7

Example:

}languages same the accept that

DFAs are and :,{ 2121 MMMMEQUALDFA

} language empty the

accepts that DFA a is :{

MMEMPTYDFA

is reduced to:

Prof. Busch - LSU 8

Turing Machinefor reduction

DFADFA EMPTYMEQUALMM 21,

f21,MM M

MMf

21,

DFA

We only need to construct:

Prof. Busch - LSU 9

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

Prof. Busch - LSU 10

)(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

Accept and halt


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

Accept and halt


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:


•Write on tape, and•Simulate on input

wM w M w

M̂

sTape of M̂

input string

accepts ?

Louisiana?s


yes

Turing Machine

Accepts

yes


The only possible accepted string

Louisiana


s



wM w M waccepts

?

Louisiana?s

yes

Turing MachineM̂

Accepts

yes


accepts }Louisiana{)ˆ(MLM w

does notaccept

M w )ˆ(ML

Prof. Busch - LSU 43Prof. Busch - LSU 43Prof. Busch - LSU 43


wM w M waccepts

?

Louisiana?s

yes

Turing MachineM̂

Accepts

yes


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:


sTape of M̂

input string


Prof. Busch - LSU 50Prof. Busch - LSU 50Prof. Busch - LSU 50Prof. Busch - LSU 50


wM w M w

Accepts

accepts ?

?kkbas

yes yes

Turing MachineM̂)0 some (for k


accepts }0:{)ˆ( nbaML nnM w

does notaccept

M w )ˆ(ML

not regular

regular

Prof. Busch - LSU 51Prof. Busch - LSU 51Prof. Busch - LSU 51Prof. Busch - LSU 51Prof. Busch - LSU 51


wM w M w

Accepts

accepts ?

?kkbas

yes yes

Turing Machine)0 some (for k

M̂


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 (two strings)?

Size2 language problem


Question: )(ML


strings} two exactly accepts

that machine aTuring is :{2 MMSIZE TM

?2|)(| ML


Theorem:

(size2 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:


sTape of M̂

input string


Prof. Busch - LSU 58Prof. Busch - LSU 58Prof. Busch - LSU 58Prof. Busch - LSU 58Prof. Busch - LSU 58


wM w M w

Accepts

accepts ?

?}Rouge,Baton{s

yes yes

Turing MachineM̂


accepts }Rouge,Baton{)ˆ( MLM w

does notaccept

M w )ˆ(ML

2 strings

0 strings

M̂

Prof. Busch - LSU 59Prof. Busch - LSU 59Prof. Busch - LSU 59Prof. Busch - LSU 59Prof. Busch - LSU 59Prof. Busch - LSU 59


wM w M w

Accepts

accepts ?

?}Rouge,Baton{s

yes yes

Turing Machine


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

}Louisiana{L

YES

NO

}Rouge,Baton{LNO

P

1P


: is regular?

More examples of non-trivial properties:

L

}0:{ nbaL nn

YES

NO

L2P

}0:{ naL nYES

: has size 2?L3PLNO

YES

NO }Louisiana{L}Rouge,Baton{L


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:

1LYES

NO

NO

P

}{ 1LP

If language has property then PL PL

}Louisiana{2 L

}Rouge,Baton{3 L


: 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:


sTape of M̂

input string


?Xs


wM w M w

Accepts

accepts ?

yes yes

Turing MachineM̂


For this we can run machine ,that accepts language , with input string

XMX

s

?Xs


wM w M w

Accepts

accepts ?

yes yes

Turing MachineM̂


accepts XML )ˆ(M w

does notaccept

M w )ˆ(ML

P

P


?Xs


wM w M w

Accepts

accepts ?

yes yes

Turing MachineM̂


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:


sTape of M̂

input string


?Xs


wM w M w

Accepts

accepts ?

yes yes

Turing MachineM̂


accepts XML )ˆ(M w

does notaccept

M w )ˆ(ML

P

PM̂

?Xs


wM w M w

Accepts

accepts ?

yes yes

Turing Machine


Therefore:

acceptsM w PML )ˆ(

Equivalently:


END OF PROOF

Prof. Busch - LSU1 Reductions. Prof. Busch - LSU2 Problem is reduced to problem If we can solve problem then we can solve problem.

Documents

problem slide

reduction slide

busch lsu10 slide

undecidable slide

busch lsu11 decider

busch lsu2 problem

busch lsu24 decider

busch lsu6 decider