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The Church-Turing Thesis

Chapter 18

Are We Done?

FSM  PDA  Turing machine Is this the end of the line?

There are still problems we cannot solve: ● There is a countably infinite number of Turing machines since we can lexicographically enumerate all the strings that correspond to syntactically legal Turing machines.

● There is an uncountably infinite number of languages over any nonempty alphabet. ● So there are more languages than there are Turing machines.

What Can Algorithms Do?

Is it possible to axiomatize all of the mathematical structures of interest in such a way that every true statement become a theorem?

Does there exist an algorithm that can examine a string and determine whether or not it is an axiom?

Can we decide whether a statement is a theorem?

Does there exist an algorithm that always halts and that returns

True

when given a theorem and

False

otherwise?

G ödel’s Incompleteness Theorem

Kurt Gödel showed, in the proof of his Incompleteness Theorem [Gödel 1931], that the answer to question 1 is no. In particular, he showed that there exists no decidable axiomatization of Peano arithmetic that is both consistent and complete. Axiom: Proposition regarded as self-evidently true without proof Peano Arithmetic: Theory of natural numbers defined by the five Peano's axioms First order logic: Propositional logic + quantification + predicate 4

The Entscheidungsproblem

Entscheidungsproblem (German for “decision problem”) Three equivalent ways to state the problem: 1. Does there exist an algorithm to decide, given an arbitrary sentence

in first order logic, whether

is valid?

2. Given a set of axioms

and a sentence

, does there exist an algorithm to decide whether

is entailed by

3. Given a set of axioms,

, and a sentence,

, does there exist an algorithm to decide whether

can be proved from

The Entscheidungsproblem

To answer the question, in any of these forms, requires formalizing the definition of an algorithm: ● Turing: Turing machines.

● Church: lambda calculus. Turing proved that Turing machines and the lambda calculus are equivalent.

The Entscheidungsproblem can be solved in neither!

Church's Thesis (Church-Turing Thesis)

All formalisms powerful enough to describe everything we think of as a computational algorithm are equivalent. This isn’t a formal statement, so we can’t prove it. But many different computational models have been proposed and they all turn out to be equivalent.

The Church-Turing Thesis

Examples of equivalent formalisms: ● Modern computers (with unbounded memory) ● Lambda calculus ● Partial recursive functions ● Tag systems (FSM plus FIFO queue) ● Unrestricted grammars: a

a 

● Post production systems ● Markov algorithms ● Conway’s Game of Life ● One dimensional cellular automata ● DNA-based computing ● Lindenmayer systems 8

The Unsolvability of the Halting Problem

Chapter 19

Languages and Machines

SD D Context-Free Languages Regular Languages

reg exps

FSMs

cfgs

PDAs

unrestricted grammars

Turing Machines

D and SD

● A TM

for any string

● ● if if

w w

with input alphabet   

L L

  then then *,

M M

accepts rejects

decides

, and a language

  * iff, A language

machine

decidable

that decides it. (in D) iff there is a Turing ● A TM

M w

  *, ● if

● if

with input alphabet 

semidecides L

 

L L

then then

M M

accepts

does not accept

iff for any string may reject or loop.

A language

semidecidable

machine that semidecides it. (in SD) iff there is a Turing

Defining the Universe

What is the complement of: •A n B n = { a

 0} •{<

> : TM

halts on input string

Defining the Universe

3 = {<

> : TM

= {<

> :

halts on input string halts on nothing}.

= {<

a ,

b > :

a and

b halt on the same strings}.

For a string

to be in

● encode a machine 1 , it must: ● be syntactically well-formed.

and a string

when started on

such that

halts Define the universe from which we are drawing strings to contain only those strings that meet the syntactic requirements of the language definition. This convention has no impact on the decidability of any of these languages since the set of syntactically valid strings is in D. (A “checker” can be built to check the correct syntax.)

A Different Definition of Complement

Our earlier definition: 

1 = {

is not a syntactically well formed <

> pair}  {<

> : TM

does not halt on input string

We will use a different definition: Define the

complement

of any language

whose member strings include at least one Turing machine description to be with respect to a universe of strings that are of the same syntactic form as

. Now we have: 

1 = {<

> : TM

does not halt on input string

The Language H

H = {<

> : TM

halts on input string

}

Theorem:

The language: H = {<

> : TM

halts on input string

} ● is semidecidable, but ● is not decidable.

Does This Program Halt?

times3

(

: positive integer) = While

 1 do: If

is even then

/2.

Else

= 3

+ 1 25

Does This Program Halt?

times3

(

: positive integer) = While

 1 do: If

is even then

/2.

Else

= 3

+ 1 25 76 38 19 58

Does This Program Halt?

times3

(

: positive integer) = While

 1 do: If

is even then

/2.

Else

= 3

+ 1 25 76 38 19 58 29 88 44 22 11

Does This Program Halt?

times3

(

: positive integer) = While

 1 do: If

is even then

/2.

Else

= 3

+ 1 25 76 38 19 58 29 88 44 22 11 34 17 52 26 13

Does This Program Halt?

times3

(

: positive integer) = While

 1 do: If

is even then

/2.

Else

= 3

+ 1 25 76 38 19 58 29 88 44 22 11 34 17 52 26 13 40 20 10 5 16

Does This Program Halt?

times3

(

: positive integer) = While

 1 do: If

is even then

/2.

Else

= 3

+ 1 25 76 38 19 58 29 88 44 22 11 34 17 52 26 13 http://www.numbertheory.org/php/collatz.html

40 20 10 5 16 8 4 2 1

H is Semidecidable

Lemma:

The language: H = {<

> : TM

halts on input string

} is semidecidable.

Proof:

H is Semidecidable

Lemma:

The language: H = {<

: TM

halts on input string

} is semidecidable.

Proof:

The TM

M H

semidecides H:

M H

>) = 1. Run

2. Accept

M H

halts iff

halts on

. Thus

M H

semidecides H.

The Unsolvability of the Halting Problem

Lemma:

The language: H = {<

> : TM

halts on input string

} is not decidable.

Proof:

If H were decidable, then some TM

M H

decide it.

M H

would would implement the specification:

halts

: string,

: string>) = If <

> is a Turing machine description and

halts on input

then accept else reject

Trouble

Trouble

(

: string) = if

halts

(

) then loop forever, else halt.

If there exists an

H that computes the function

halts

Trouble

exists:

C #

is a Turing machine that writes

on the tape given

(i.e., input) on the tape What is

Trouble

>)? What is

halts

Trouble

>)? ● If

halts

reports that

Trouble

>) halts,

Trouble

● But if

halts

reports that

Trouble

loops.

>) does not halt, then

Trouble

halts.

Viewing the Halting Problem as Diagonalization

● ● Lexicographically enumerate Turing machines.

Let 1 mean halting, blank mean non halting.

1 1

…

Trouble

machine

… Trouble

… … 1 1 1 1 1 1 1 1 But

Trouble

behaves as:

Trouble

1 1 … 1 1 Or maybe

halts

said that

trouble

>) would halt. But then

trouble

would loop.

If H were in D

H = {<

>: TM

halts on input string

}

Theorem:

If H were in D then every SD language would be in D.

Proof:

Let

be any SD language. There exists a TM

M L

semidecides it. that If H were also in D, then there would exist a Turing machine

that decides it.

If H were in D

To decide whether

is in

(

M L

(

: string) = 1. Run 2. If

O O

on <

M L

>. accepts (i.e.,

M L

will halt), then: 2.1. Run

M L

2.2. If it accepts, accept. Else reject.

3. Else reject.

So, if H were in D, all SD languages would be.

Back to the Entscheidungsproblem

Theorem:

The Entscheidungsproblem is unsolvable.

Proof:

(Due to Turing) 1. If we could solve the problem of determining whether a given Turing machine ever prints the symbol 0 , then we could solve the problem of determining whether a given Turing machine halts. 2. But we can’t solve the problem of determining whether a given Turing machine halts, so neither can we solve the problem of determining whether it ever prints 0 .

3. Given a Turing machine

, we can construct a logical formula

that is true iff

ever prints the symbol 0 . 4. If there were a solution to the Entscheidungsproblem, then we would be able to determine the truth of any logical sentence, including

and thus be able to decide whether

ever prints the symbol 0 . 5. But we know that there is no procedure for determining whether

ever prints 0 .

6. So there is no solution to the Entscheidungsproblem.