1 Introduction to Computability Theory Lecture13: Mapping Reductions Prof. Amos Israeli

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Introduction to Computability Theory

Lecture13: Mapping Reductions

Prof. Amos Israeli

So far, we presented several reductions:

From to , from to , from to , from to , and several other.

Each of these reductions had some special form not shared with the others. Can we generalize?

Mapping Reductions

2

TMHALTTMA

TMREGULARTMA TMEQTME

TMHALT TMA

Mapping reductions constitute the generalization we are looking for. They are always in a specified form which:

1. Preserves the intrinsic difficulty of finding a reduction.

2. Ease the “red tape” requirements.

Mapping Reductions

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In other words: Assume we want to prove that B is undecidable by reduction from A that is known to be undecidable.

If our reduction has the form of a Mapping reduction then the proof is complete. Nothing else is required.

Mapping Reductions

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The idea of a mapping reduction is very simple: If the instances (candidate elements) of one language, say A, are mapped to the instances of another language, say B, by a computable mapping M in a way that iff , then a decider for B can be used to devise a decider for A.

Mapping Reductions

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AI BIM

The sought for decider for A works as follows:

Upon input I do:

1. Compute .

2. Use the decider for B to determine whether .

3. If accept - otherwise - reject.

Mapping Reductions

6

IM

BIM

BIM

This scheme works but there is one delicate point here. Can you see where it is?

The mapping must be computable.

Given the input I, your ordinary Turing machine must be able to compute .

For this we need the notion of Computable Functions.

Mapping Reductions

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IM

IM

Let be an alphabet: A function is a computable function if there exists a TM, M such that for every , if M computes with input w, it halts with on its tape.

Computable Functions

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**: f

*w

wf

1. Let m and n be natural numbers, let be a string encoding of m and n, and let be the string representing their sum. The function , is computable.

2. The function is a computable function.

Can you devise TM-s to compute f and g?

Examples of Computable Functions

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nmnmf ,

rwwg

nm,

nm

3. Let be an encoding of TM M and let be an encoding of another TM M’ satisfying:1. .2. TM M’ never makes four consecutive steps in the same head direction.The function t defined below is computable:

Examples of Computable Functions

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M

'MLML

TM any of ecodingnot is if

TM of ecoding an is if '

M

MMMMt

'M

In the reduction from to - TM M from the input to was modified by TM S, the decider to TM s.t. all ending computations on M would become accepting computations on .

Since was carried out by TM S, it is obviously a computable function.

Use of Comp. Func. for Reductions

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1M

wM ,

TMHALT TMA

TMHALT

1M

1MMf

Another example is the reduction from to - TM M from the input to , was modified by TM S, to TM s.t. is regular iff . Once again, this is a computable function.

Now we turn to define mapping reductions:

Use of Comp. Func. for Reductions

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2M

TMAwM ,

TMA

TMREGULAR

TMA

2ML

wM ,

Let A and B be two languages over . A computable function is a mapping reduction from A to B if the following holds: For each , iff . The function f is called reduction of A to B. The arguments of the reduction are often called instances.

Mapping Reductions

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**: f

*

AI *I BIf

If there exists a mapping reduction from A to B, We say that A is mapping reducible to B and denote it by .

A pictorial illustration of a mapping reduction appears in the next slide.

Mapping Reductions

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BA m

.

A Pictorial Illustration of_______

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BA f

A

A Af

Af

f

Assume that .

Can we say something about the relationship between and ?

Answer: Yes

An Observation about__and_____

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A

A Af

Af

f

A B

A B

BA m

BA m

If language A is mapping reducible to language B, then a solution for B, can be used to derive a solution for A. This fact is made formal in the following theorem:

Using Mapping Reductions

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If and B is decidable, then A is decidable.

Theorem

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BA m

Let M be a decider for B and let f be the reduction from A to B. Consider TM N:

N=“On input w :

1. Compute .

2. Run M on and output whatever M outputs. “

Clearly N accepts iff . QED

Proof

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wf

wf

Bw

If and A is undecidable, then B is undecidable.

Corollary

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BA m

The previous corollary is our main tool for proving undecidability.

Notice that in order to prove B undecidable we reduce from A which is known to be undecidable to B. The reduction direction is often a source of errors.

A similar tool is used in Complexity theory.

Discussion

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On the previous lecture we proved that

is undecidable.

Now, we prove this theorem once more by demonstrating a mapping reduction from to .

Revisited

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wMwMHALTTM on haltsTM that a is ,

TMHALTTMA

TMHALT

Theorem

The language

Is undecidable.

Revisited

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TMHALT

wMwMHALTTM on haltsTM that a is ,

Proof

The proof is by reduction from . In the next slide we show that . By the previous corollary it follows that is undecidable.

Revisited

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TMmTM HALTA TMA

TMHALT

TMHALT

The mapping reduction is presented by TM F :

F=“On input :

1. Construct TM M’ .

M’=“On input x:

1. Run M on x.

2. if M accepts accept.

3. If M rejects, enter a loop.”

2. Output .”

Mapping Reduction from to__

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wM ,

wM ,'

TMHALTTMA

It is not hard to see that iff . In other words: . Since is undecidable, the previous corollary implies that is undecidable too.

Proof (cont.)

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TMAwM ,

TMHALTwM ,'

TMmTM HALTA TMA

TMHALT

What happens if the input does not contain a valid description of a TM?

By the specification of we know that in this case . Therefore in this case TM F should output any string s satisfying .

The Halting Problem Revisited

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wM ,

TMA

TMAwM ,

TMHALTs

Theorem

The language

Is undecidable.

Revisited

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TMEQ

NLMLNMNMEQTM and s-TM are and ,

Proof

The proof is by reduction from which is known to be undecidable. In the next slide we show that . By the previous corollary it follows that is undecidable.

Revisited

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TMmTM EQE

TME

TMEQ

TMHALT

Let be an encoding of a TM that rejects all its inputs. The mapping f is defined as follows:

For each instance of , , .

It is not hard to verify that

Meaning: . QED

Mapping Reduction from to__

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M

rMLMLML

TMEQTME

rM

TME rMMMf ,

TMrTM EQMMEM ,

If and B is Turing recognizable, then A is Turing recognizable.

Theorem

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BA m

Let M be a decider for B and let f be the reduction from A to B. Consider TM N:

N=“On input w :

1. Compute .

2. Run M on and output whatever M outputs.”

Clearly N accepts iff . QED

Proof

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wf

wf

Bw

Turing Recognizer

N recognizes B.

- if M halts,

If and A is not Turing recognizable, then B is not Turing recognizable.

Corollary

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BA m

Theorem

is neither Turing recognizable nor co-Turing recognizable.

Reminder

A language L is co-Turing recognizable if its complement, , is Turing recognizable.

The Status of______

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TMEQ

TMEQ

L

If by a mapping f, then by the same mapping it holds that . Since we know that is not Turing recognizable, we can prove that language A is not Turing recognizable by reducing to .

Proof

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TMA A

BA m

BA m

TMA

Now we use the previous remark and prove that is not Turing recognizable, by showing that .

Since , we conclude that and since we know that is not Turing recognizable we conclude that so is .

Proof

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TMEQ

TMMTM EQA

TMMTM EQA

TMMTM EQA TMA

TMEQ

F=“ on input , where M is a TM:

1. Construct that accepts on any input and

that accepts if M accepts w.

2. Output .

It is not hard to see that F computes a reduction from to .

Proof

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1M

TMA TMEQ

wM ,

2M

21, MM

G=“ on input , where M is a TM:

1. Construct that rejects on any input and that accepts if M accepts w.

2. Output .

It is not hard to see that G computes a reduction from to .

Proof

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1M

TMA TMEQ

wM ,

2M

21, MM

Since we conclude that is not Turing recognizable.

Since we conclude that is not co-Turing recognizable, hence is not Turing recognizable.

QED

Proof

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TMEQTMMTM EQA

TMMTM EQA TMEQ

TMTM EQEQ