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Introduction to Computability Theory

Lecture11: The Halting Problem

Prof. Amos Israeli

In this lecture we present an undecidable language.

The language that we prove to be undecidable is a very natural language namely the language consisting of pairs of the form where Mis a TM accepting string w:

The Halting Problem

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wMwMATM acceptingTM a is ,

wM ,

Since this language requires to decide whether the computation of TM M halts on input w, it is often called The Halting Problem.

Theorem

The halting problem is Turing Recognizable.

The Halting Problem

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Consider a TM U that gets a pair as input and simulates the run of M on input w. If M accepts or rejects so does U. Otherwise, U loops.

Note: U recognizes ATM ,since it accepts any pair

, that is: any pair in which M accepts input w.

Proof

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wM ,

LwM ,

On the previous lecture, we detailed the simulation of a DFA by a TM.

Simulating one TM by another, using the encoding of the first TM is a very similar process. In the next slide we review the main characteristics of TM N simulating TM M, using M’s encoding <M>.

Simulating an Input TM

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TM N works as follows:

1. Mark M’s initial state and w’s initial symbol as the “current state” and “current head location”.

2. Look for M’s next transition on the description of its transition function.

3. Execute M’s transition.

Simulating an Input TM

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4. Move M’s “current state” and “current head location” to their new places.

5. If M’s new state is a deciding state decide according to the state, otherwise – repeat stages 2-5.

Simulating an Input TM

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Our goal in this lecture is to prove that ATM is

not decidable. The proof uses a common mathematical technique known as Diagonalization.

Diagonalization was first used by Cantor when he found a way to distinguish between several types of infinite sets.

Diagonalization

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Cantor dealt with questions like:

How many natural numbers are there? Infinity!

How many real numbers are there? Infinity!

Does the amount of natural numbers equal to the amount of real numbers?

How is the size of infinite sets measured?

Cardinality

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Cantor’s answer to these question was the notion of Cardinality.

The cardinality of a set is a property marking its size.

Two sets has the same cardinality if there is a correspondence between their elements

Cardinality

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At this point of the lecture, think about a correspondence between sets A and B as 2 lists: A list of A’s elements and in parallel a list of B’s elements. These 2 lists are juxtaposed so that each element of A corresponds to a unique element of B.

Intuitive Notion of Correspondence

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Clearly, the cardinality of A is equal to the cardinality of B.

How about the cardinality of infinite sets?

Example

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4,3,2,1A 8,6,4,2B86424321

BA

How about the cardinality of infinite sets?

Is the cardinality of natural numbers larger than the cardinality of even natural numbers?

Intuitively, the cardinality of any set should be larger that the cardinality of any of its proper subsets. Alas, our intuition of sets is driven by our daily experience with finite sets.

Example

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So let us try to create a correspondence between the natural numbers the even natural numbers?

Indeed defines the wanted correspondence between the 2 sets.

Example

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,...,...,4,3,2,1 nN

,...2,...,8,6,4,2 nEN ...2...642......321

nENnN

nnf 2

So the cardinality of N is equal to the cardinality of EN.

Example

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,...,...,4,3,2,1 nN

,...2,...,8,6,4,2 nEN ...2...642......321

nENnN

This last example suggests the notion of Countable Sets:

A set A is countable if it is either finite or its cardinality is equal to the cardinality of N.

A cool way of looking at countable sets is:“A set is countable if a list of its elements can be created”.

Countable Sets

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“A set is countable if a list of its elements can be created”.

Note: This list does not have to be finite, but for each natural number i, one should be able to specify the i-th element on the list.

For example, for the set EN the i-th element on the list is 2i .

Countable Sets

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We just proved that EN, the set of even natural numbers is countable. What about the set of rational numbers?

Is the set Q of rational numbers countable?

Can its elements be listed?

Countable Sets

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Theorem

The set of rational numbers is countable.

Proof

In order to prove this theorem we have to show how a complete list of the rational numbers can be formed.

The set of Rationals is Countable

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Recall that each natural number is defined by a pair of natural numbers.

One way to look

at the Rationals

is by listing them

in an infinite Rectangle.

The set of Rationals is Countable

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5/54/53/52/51/55/45/35/25/1

4/43/42/41/44/33/32/31/34/23/22/21/24/13/12/11/1 ………

………………………

………

………………………………………………...

How can we form a list including all these numbers?

If we first list

The first row –

We will never

reach the second.

The set of Rationals is Countable

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5/54/53/52/51/55/45/35/25/1

4/43/42/41/44/33/32/31/34/23/22/21/24/13/12/11/1 ………

………………………

………

………………………………………………...

5/54/53/52/51/55/45/35/25/1

4/43/42/41/44/33/32/31/34/23/22/21/24/13/12/11/1

The set of Rationals is Countable

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One way to do it is to start from the upper left corner,and continue in this fashion

Note that some rational numbers appear more than once. For example: all numbers on the main diagonal are equal to 1, so this list is not final.

In order to compute the actual place of a given rational, we need to erase all duplicates, but this is a technicality…

The set of Rationals is Countable

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Can you think of any infinite set whose elements cannot be listed in one after the other?

Well, there are many:

Theorem

The set of infinite binary sequences is not countable.

So perhaps all sets are countable

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Assume that there exists a list of all binary sequences. Such a list may look like this:

Uncountable Sets

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100011110

0111000000111101 ………

………………………

………

…………………….................

But can you be sure that all sequences are in this list?

In fact, There exist many sequences that are not on the list:

Uncountable Sets

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100011110

0111000000111101 ………

………………………

………

…………………….................

Consider for example S=0,0,1,1,0,… . The sequence S is formed so that 1st elt. Of 1st seq.

2nd elt. Of 2nd seq.

3rd elt. Of 3rd seq.

And so on …

Uncountable Sets

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100011110

0111000000111101 ………

………………………

………

……………………...........

1S

2S

3S

In general: The i-th element of S, differs from the i-th element of the i-th sequence in the list. Note: the i-th element of the i-th sequence in the list is always the element on the diagonal.

Can the sequence S appear on the list?

Uncountable Sets

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iS

?...01100..............................10001......10111......10000......10011......01101

5

4

3

2

1

S

SSSSS

In this case,

Uncountable Sets

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Assume there exists an index j such that jSS

jSjS j

jSjS jBut by definition:

Contradiction!!

For any sequence on the list, , the sequence differs from by its element on the diagonal, that is: , so cannot be on the list.

For obvious reasons, this technique is called Diagonalization.

Uncountable Sets

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jSS jS

jSjS j S

We just used diagonalization to prove that the set of infinite binary sequences is uncountable.

Can a a similar proof for the set of real numbers?

Uncountable Sets

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Corollary

Some Languages are not Turing-recognizable.

Proof

For any (finite) alphabet, , the set of (finite) strings , is countable. A list of all elements in is obtained by first listing strings of length 1, then 2, …, then n…

Turing Unrecognizable Languages

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*

*

The set of all TM-s is also countable because every TM, , can be described by its encoding , which is a string over . So the set of TM-s corresponds to a subset of .

Note: Here we use the (unproven but correct) fact that the cardinality of a set is always not greater then the cardinality of any of its supersets.

Proof (cont.)

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MM

*

Since each TM recognizes exactly a single language, a list of all TM-s can be used as a list of all recognizable languages.

If we show that the set of languages over is uncountable, we can deduce that at least a single language is not on the list, that is: it is not recognized by any TM.

Proof (cont.)

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We have already seen that the set of infinite binary sequences is uncountable. If we form a correspondence between the set of languages and the set of infinite binary sequences we will show that the set of languages is uncountable.

Proof (cont.)

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Consider a fixed list l of all words . The correspondence is formed as follows: For every infinite binary sequence S, corresponds the language:

QED

Proof (cont.)

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*

1 where| isilSL

So far we proved the existence of a language which is not Turing recognizable. Now we continue our quest to prove:

Theorem

The language

is undecidable.

The Language ___ Is Undecidable

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TMA

wMwMATM acceptingTM a is ,

Before we start the proof let us consider two ancient questions:

Question1:

Can god create a boulder so heavy that god cannot lift?

The Language ___ Is Undecidable

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TMA

Question2:

In the small town of L.J. there is a single barber:

Over the barber’s chair there is a note saying:“I will shave you on one condition: Thaw shall never shave thyself!!!”

Who Shaves the Barber?

The Language ___ Is Undecidable

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TMA

Assume, by way of contradiction, that is decidable and let H be a TM deciding .

That is

Define now another TM new D that uses H as a subroutine as follows:

Proof

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TMA

wMrejectwMaccept

wMH rejects if accepts if

,

TMA

Define now another TM new D that uses H as a subroutine as follows:

D=“On input where N is a TM:

1. Run H on input .

2. Output the opposite of H’s output namely: If H accepts reject, otherwise accept.“

Proof

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MMM ,

Note: What we do here is taking advantage of the two facts:

Fact1: TM M should be able to compute with any string as input.

Fact2: The encoding of M, , is a string.

Proof

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M

Running a machine on its encoding is analogous to using a compiler for the computer language Pascal to, that is written in Pascal, to compile itself.

As we recall from the two questions self-reference is sometimes means trouble (god forbid…)

Proof

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What we got now is:

Consider now the result of running D with input . What we get is:

Proof

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M accepts if M ejects if

MrejectMaccept

MDr

D

D accepts if D ejects if

DrejectDaccept

DDr

So if D accepts, it rejects wand if it rejects it accepts. Double Trouble.

And it all caused by our assumption that TM H exists!!!

Proof

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D accepts if D ejects if

DrejectDaccept

DDr

1. Define .

2. Assume that id decidable and let H be a TM deciding it.

3. Use H to build TM D that gets a TM encoding and behaves exactly opposite to H’s behavior, namely:

Proof Review

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wMwMATM acceptingTM a is ,

TMA

M

M accepts if M ejects if

MrejectMaccept

MDr

4. Run TM D on its encoding and conclude:

Contradiction.

Proof Review

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D accepts if D ejects if

DrejectDaccept

DDr

D

The following table describes the behavior of each machine on its encoding:

So Where is the Diagonalization?

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acceptacceptMM

acceptacceptacceptacceptMacceptacceptM

MMMM

4

3

2

1

4321

This table describes the behavior of TM H. Note: TM H rejects where loops.

So Where is the Diagonalization?

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rejectrejectacceptacceptMrejectrejectrejectrejectMacceptacceptacceptacceptMrejectacceptrejectacceptMMMMM

4

3

2

1

4321

iM

Now TM D is added to the table…

Proof Review

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???

4

3

2

1

4321

acceptacceptrejectrejectD

acceptrejectrejectacceptacceptMrejectrejectrejectrejectrejectMacceptacceptacceptacceptacceptMacceptrejectacceptrejectacceptM

DMMMM