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Introduction to Computability Theory

Lecture12: ReductionsProf. Amos Israeli

The rest of the course deals with an important tool in Computability and Complexity theories, namely: Reductions.

The reduction technique enables us to use the undecidability of to prove many other languages undecidable.

Introduction

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TMA

A reduction always involves two computational problems. Generally speaking, the idea is to show that a solution for some problem A induces a solution for problem B. If we know that B does not have a solution, we may deduce that A is also insolvable. In this case we say that B is reducible to A.

Introduction

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In the context of undecidability: If we want to prove that a certain language L is undecidable. We assume by way of contradiction that L is decidable, and show that a decider for L, can be used to devise a decider for . Since is undecidable, so is the language L.

Introduction

4

TMATMA

Using a decider for L to construct a decider for , is called reducing L to .

Note: Once we prove that a certain language L is undecidable, we can prove that some other language, say L’ , is undecidable, by reducing L’ to L.

Introduction

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TMA TMA

1. We know that A is undecidable.

2. We want to prove B is undecidable.

3. We assume that B is decidable and use this assumption to prove that A is decidable.

4. We conclude that B is undecidable.

Note: The reduction is from A to B.

Schematic of a Reduction

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1. We know that A is undecidable. The only undecidable language we know, so far, is whose undecidability was proven directly. (In the discussion you also proved directly that is undecidable). So we pick to play the role of A.

2. We want to prove B is undecidable.

Demonstration

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TMHALT

TMA

TMA

2. We want to prove B is undecidable. We pick to play the role of B that is: We want to prove that is undecidable.

3. We assume that B is decidable and use this assumption to prove that A is decidable.

Demonstration

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TMHALT

TMHALT

3. We assume that B is decidable and use this assumption to prove that A is decidable.In the following slides we assume (towards a contradiction) that is decidable and use this assumption to prove that is decidable.

4. We conclude that B is undecidable.

Demonstration

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TMA

TMHALT

Consider

Theorem

is undecidable.

Proof

By reducing to .

The “Real” Halting Problem

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wMwMHALTTM on haltsTM that a is ,

TMHALT

TMATMHALT

Assume by way of contradiction that is decidable.

Recall that a decidable set has a decider R: A TM that halts on every input and either accepts or rejects, but never loops!.

We will use the assumed decider of to devise a decider for .

Discussion

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TMHALT

TMHALT

TMA

Recall the definition of :

Why is it impossible to decide ?

Because as long as M runs, we cannot determine whether it will eventually halt.

Well, now we can, using the decider R for .

Discussion

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TMA

TMA

wMwMATM acceptsTM that a is ,

TMHALT

Assume by way of contradiction that is decidable and let R be a TM deciding it. In the next slide we present TM S that uses R as a subroutine and decides . Since is undecidable this constitutes a contradiction, so R does not exist.

Proof

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TMHALT

TMA TMA

S=“On input where M is a TM:

1. Run R on input until it halts.

2. If R rejects, (i.e. M loops on w ) - reject.

(At this stage we know that R accepts, and we conclude that M halts on input w.)

3. Simulate M on w until it halts.

4. If M accepts - accept, otherwise - reject. “

Proof (cont.)

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wM ,

wM ,

In the discussion, you saw how Diagonalization can be used to prove that is not decidable.

We can use this result to prove by reduction that is not decidable.

Another Example

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TMA

TMHALT

Note: Since we already know that both and are undecidable, this new proof does not add any new information. We bring it here only for the the sake of demonstration.

Another Example

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TMA

TMHALT

1. We know that A is undecidable. Now we pick to play the role of A.

2. We want to prove B is undecidable. We pick to play the role of B, that is: We want to prove that is undecidable.

3. We assume that B is decidable and use this assumption to prove that A is decidable.

Demonstration

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TMHALT

TMA

TMA

3. We assume that B is decidable and use this assumption to prove that A is decidable.In the following slides we assume that is decidable and use this assumption to prove that is decidable.

4. We conclude that B is undecidable.

Demonstration

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TMA

TMHALT

Let R be a decider for . Given an input for , R can be run with this input :If R accepts, it means that .This means that M accepts on input w. In particular, M stops on input w. Therefore, a decider for must accept too.wM ,

Discussion

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wM ,

TMHALT

TMA

TMAwM ,

If however R rejects on input , a decider for cannot safely reject: M may be halting on w to reject it. So if M rejects w, a decider for must accept .

Discussion

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TMHALT

wM ,

wM ,

TMHALT

How can we use our decider for ?The answer here is more difficult. The new decider should first modify the input TM, M, so the modified TM, , accepts, whenever TM M halts.

Since M is a part of the input, the modification must be a part of the computation.

Discussion

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TMA

1M

Faithful to our principal “ If it ain’t broken don’t

fix it”, the modified TM keeps M as a subroutine, and the idea is quite simple:Let and be the accepting and rejecting states of TM M, respectively. In the modified TM, , and are kept as ordinary states.

Discussion

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acceptq

1M

rejectq

acceptq rejectq

We continue the modification of M by adding a new accepting sate . Then we add two new transitions: A transition from to , and another transition from to .

This completes the description of . It is not hard to verify that accepts iff M halts.

acceptnq

Discussion

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acceptq

rejectq

acceptnq

acceptnq

1M

1M

Discussion

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acceptnq

1M

acceptq

rejectq

M

The final description of a decider S for is:

S=“On input where M is a TM:

1. Modify M as described to get .

2. Run R, the decider of with input .

3. If R accepts - accept, otherwise - reject. ”

Discussion

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TMHALT

TMA

1M

wM ,1

wM ,

It should be noted that modifying TM M to get , is part of TM S, the new decider for , and can be carried out by it.

It is not hard to see that S decides . Since

is undecidable, we conclude that is undecidable too.

Discussion

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1M

TMHALT

TMHALT

TMHALTTMA

We continue to demonstrate reductions by showing that the language , defined by

is undecidable.

Theorem

is undecidable.

The TM Emptiness Problem

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M AndTM a is LMMETM

TME

TME

The proof is by reduction from :

1. We know that is undecidable.

2. We want to prove is undecidable.

3. We assume toward a contradiction that is decidable and devise a decider for .

4. We conclude that is undecidable.

Proof Outline

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TME

TMA

TMA

TMATME

TME

Assume by way of contradiction that is decidable and let R be a TM deciding it. In the next slides we devise TM S that uses R as a subroutine and decides .

Proof

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TME

TMA

Given an instance for , , we may try to run R on this instance. If R accepts, we know that . In particular, M does not accept w so a decider for must reject .

Proof

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TMA

ML

wM ,

wM ,

TMA

What happens if R rejects? The only conclusion we can draw is that . What we need to know though is whether .

In order to use our decider R for , we once again modify the input machine M to obtain TM :

Proof

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ML

MLw

TME

1M

We start with a TM satisfying .

Description of___

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1M

MLML 1

1Macceptq

rejectq

M

startq

acceptnq

rejectnq

startnq

wM

wMwML

rejects if

accepts if 1

Now we add a filter to divert all inputs but w.

Description of___

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1M

wx

wx

1Macceptq

rejectq

M

startq

acceptnq

rejectnq

startnq

wx filter

no

yes

TM has a filter that rejects all inputs excepts w, so the only input reaching M, is w.

Therefore, satisfies:

wM

wMwML

rejects if

accepts if 1

Proof

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1M

1M

Here is a formal description of :

“On input x :

1. If - reject . 2. If - run M on w and accept if M accepts. ”

Note: M accepts w if and only if .

Proof

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wx 1M

wx

1ML

1M

S=“On input where M is a TM:

1. Compute an encoding of TM . 2. Run R on input .

3. If R rejects - accept, otherwise - reject.

Proof

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wM ,

1M 1M

1M

Recall that R is a decider for . If R rejects the modified machine , , hence by the specification of , , and a decider for must accept .If however R accepts, it means that , hence , and S must reject . QED

Proof

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1M

TME

1ML

MLw1M

TMA

1ML

wM ,

MLw wM ,

We continue to demonstrate reductions by showing that the language :

is undecidable.

Theorem

is undecidable.

_______ is undecidable

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TMREGULAR

TMREGULAR

TMREGULAR

RegularIS M AndTM a is LMMREGULARTM

The proof is by reduction from :

1. We know that is undecidable.

2. We want to prove is undecidable.

3. We assume that is decidable and devise a decider for .

4. We conclude that is undecidable.

Proof Outline

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TMA

TMA

TMA

TMREGULAR

TMREGULAR

TMREGULAR

Consider an instance of ATM . Once again, the

idea is to transform TM M to another TM, , such that is regular if and only if .

Once we have such a machine we can run a decider for with as input, and accept , if the decider accepts.

Proof

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TMAwM ,

wM ,

2M 2ML

TMREGULAR 2M

wM ,

Here the idea is to devise a TM that satisfy:

.

Description of___

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2M

wMn

wMML

nn rejects if 0|10

accepts if *

2

TM M2 is obtained as follows:

1. Start with M.

2. Add a filter in front of M such that for every input x:

2.1 If x is of the form 0n1n - accept. 2.2 Send w to M – If M accepts - accept.

Description of___

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2M

We start with a TM satisfying .

Description of___

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2M

MLML 2

2M

M

rejectq

acceptq

startq

acceptnq

rejectnq

startnq

Add a filter to accept all inputs of form .

Description of___

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1M

nnx 102M

M

rejectq

acceptq

startq

acceptnq

rejectnq

startnq

nn10filter

no

yes

w

nn10

“On input x :

1. If x has the form , accept . 2. If x does not have this form, run M on w and accept if M accepts w . ”

Note: If M accept w then . If M does not accepts w then .

Proof

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nn10

2M

0|102 nML nn

*2 ML

Now consider S:

“On input :1. Construct as described using .

2. Run R on .

3. If R accepts (Meaning ) - accept , otherwise ( ) - reject .”

Proof

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S wM ,

2M wM ,

2M

*2 ML

0|102 nML nn

Consider

Theorem

is undecidable.

TM Equivalence is Undecidable

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2121, MLMLMMEQTM

TMEQ

The proof is by reduction from . We have to show that if is decidable, so is .The idea is very intuitive: In order to check that a language of TM M is empty, as required by , we will check whether M is equivalent to a TM that rejects all its inputs.

Proof

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TME

ML

TMEQ TME

TME

S=“On input , where M is a TM:

1. Run R on input where is a TM that rejects all its inputs.

2. If R accepts – accept, otherwise – reject .

If R decides , S decides . Since is undecidable, so is . QED

Proof

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M

1, MM 1M

TMEQ TME

TMEQ

TME

Mapping Reductions are the most common reductions in Computer Science. In this lecture we define mapping reductions and demonstrate the way in which they are used.

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

Let A and B be two languages over . 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 both 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 two 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

Let A and B be two languages over . A computable function is a mapping reduction from A to B if for every it holds that follows: For each it holds that 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 .

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:

Mapping Reductions

57

BA m

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

61

At the beginning of this lecture we looked at

and proved that it is undecidable.

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

The Halting Problem Revisited

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wMwMHALTTM on haltsTM that a is ,

TMHALTTMA

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 . “

The Halting Problem Revisited

63

wM ,

wM ,'

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