definition: A set, S, of strings from some symbol alphabet is decidable ( or recursively decidable or recursive) if there is Turing machine , M. which , given any string in the alphabet, always halts in some special state,H,after printing Y (for yes) if the string is in S and N (for no) if the string is not in S.

M is called the decision machine for S. So, one says that a set if decidable iff it has a decision machine.

Church's Thesis: Any intuitively computable process can be simulated by some Turing machine( or any of several models of computation that have been shown to be equivalent to Turing machines. This statement is a thesis ( not a theorem) because it can never be proved. It is a statement of belief about what it means to be intuitively computable. However, there is strong evidence that it is true. The evidence is that all known models of computation either compute the same functions and accept the same languages as do Turing machines or else a subset thereof., being less powerful than Turing machines. No one has every conceived of any computable process that can not be done on a Turing machine, subject to suitable encoding.

The intuitive notion of decidable appeared before the notion of Turing machines. In fact, Turing introduced the notion of Turing machine in order to give a precise mathematical characterization of decidability,

definition: A set, S, of strings from some symbol alphabet is (intuitively) decidable ( ( or recursively decidable or recursive) if there is some computable process, or program,. which , given any string in the alphabet, always halts , after returning Y if the string is in s and halts after returning N , if not.

If one believes Church's Thesis, the two definitions of decidable are equivalent.

Because of the wide acceptance of Church's Thesis, one can prove something is computable merely by exhibiting a program or intuitively computable process for it, instead of producing an actual Turing machine.

Church's Thesis also provides a

new intuitive definition of Turing machine language: A set L is a Turing machine language iff

there is some program p which given input x, halts if x Î L and which loops forever if x Ï L.

Turing machine machine language might well be called semi-decidable, because there is a program that

halts when the answer is "yes", but unlike a decision program, fails to halt when the answer is "no".

The famous halting problem is: Is the set K, just defined, decidable?

Or to put it another way: Is there a computable process for determining whether a Turing machine will halt , given itself as input?

The answer is no, and the result follows from what we already know about K together with the following: theorem.

Theorem: A set S is decidable iff both S and ~S are Turing languages.

Proof ( if) Assume S is decidable. Prove that both S and ~S are Turing machine languages.

Let M be the decision machine for S. We first modify M to build a TM that accepts S.

To do this, add code to M so that instead on halting after it prints N, it jumps into an infinite loop.

To find a machine that accepts ~M, modify it so that after printing Y , it goes into an infinite loop , instead of halting, and also change it so that it prints Y when it used to print N.

Part 2 (if) This part is hard to do on a Turing machine ,so let us give an intuitive approach first.

Assume S and ~S are both TM languages. ( We must prove that S is decidable.)

Therefore, there are Turing machines, say M and ~M that accept S and ~S respectively.

Given some string , w, the idea is to run M and ~M concurrently on w. Since w has to be in either S or ~S,

one of the 2 machines must halt. If M halts, output Y; if ~M halts, output N. This is a decision process.

To carry this process out on a Turing machine, one must combine the two machines M and ~M into on machine that alternates moves between the 2 machines. They both work on the same tape, but a marker can separate the parts of the tape used by the same machines, so that they do not interfere with each other.

The read head can alternately move between the two areas. The current states and the current positions of the read-head can be written on the tapes in the same way that this is done with the universal machine. The location of a special encoding of the current state is placed on each tape at the left of the symbol the

original machine would be on.

The undecidability of K now follows from our theorem since ~K is not a Turing machine language.

The result on the previous page can also be stated as : A set S is decidable iff both S and ~S are r.e.

Consequently, all decidable sets are r.e.

But K is an example of an r.e. set that is not decidable.

All of the sets that you are likely to think of, are decidable:

Try to find decision processes for these sets of natural numbers:

(evens}

{odds}

( multiples of any given m}

{primes}

{squares}

{cubes}

{all p powers for any fixed p}

{perfect numbers} ( A perfect number is one that is equal to the sum of its proper divisors, such as 6 = 1+2+3 and 28 = 1 + 2 + 4 + 7 + 14 = 28.

{abundant numbers} sum of the proper divisors exceed the number.

{deficient numbers}

{numbers that can be expressed as the sum of two primes}

Goldbach's famous unproven conjecture is that the set above is the set of even numbers.

{x | x is the (code number) of a person who will die today}.

About Godel numberings

The listing W₀ ,W_1, W₂, ..... is a called a Godel numbering , this time of the the r.e. sets.

j is called the index of W_j .

Earlier, we defined the subscripts to be (encoded) Turing machine and made a special definition:

in the case that v is not the number for any Turing machine , namely , W_v = { 0,1,2,3,...} ( one can think of the numbers are being codes for string.). To avoid this, one can remove the invalid Turing machine coded

from the list and renumber. Thus W_j now means the language of the jth Turing machine code, in

numerical order. Most importantly, since a Turing machine can determine whether a number is a valid TM code, a program can still find the actual TM from the renumbered index. Further, since one can uniformly

(computably) find , given any Turing machine, an enumeration machine, for the language of the Turing machine, one can consider j to be a program for enumerating the member of W_j . There are many ways

to number the r.e. sets: the idea of a Godel numbering is that the index is a program for enumerating the set W_j . Further this program can be translated uniformly into an executable program: there is a computable function, U(j,x) which returns the output of program j, running on input x, if there is output .

Also worth mentioning: we can define a Godel numbering of function by letting, f_j º the function computed by Turing machine , number j. The listing f_0, f_2, f_3, ... is thus a Godel numbering of all the computable functions. The index, j, is a program for the function f_j .

Until otherwise stated, we will think of strings as being natural numbers and think about r.e. and decidable sets of natural numbers.`

Prop: All finite sets are decidable.

Proof: ?

Prop: If D is decidable, ~D is decidable.

Prop: If C and D are both decidable, then C È D is decidable

Prop If C and D are both decidable, then C.Ç D is decidable.

Prop. The union and interesection of r.e. sets is r.e. but the complement need not be.

K is r.e., but ~K is not r.e.

Prop: An infinite set S, is decidable iff it can be enumerated in natural order.

proof (if) A decision process for S is:

given x, enumerate S in order until either x is found or something bigger than x is found.

Output "yes" in the first case and "no" in the second case.

Q: why must the enumeration be in order for this to work?

(only if) given a decision process for S, test the numbers 0,1,2,3,... in order. Whenever the process returns "yes", output the number. The result is an enumeration in order.

Corollary: Every infinite r.e. set contains an infinite decidable subset.

Corollay: Every infinitie decidable set can be split into 2 disjoint infinitie decidable subsets.

Cor.: Every infinitie decidable set can be split into any given number of infinite, pairwise disjoint decidable subsets.

The next result shows that there are infinitely many r.e. sets that are not decidable.

Prop: Every infinite r,e, set ,S, contains a subset which is r.e. but not decidable.

Choose an enumeratiion method for S and let a_0, a_1, a_2, a_{3, ...}

be this enumeration.

Define B = { a_j | j Î K } where K is any undecidable, r.e. set.

B is r.e. since K and S are both r.e.

Enumeration method for B.

Loop

Enumerate next member of K, say j

Enumerate S until a_j appears, the print a_j

End Loop

B is not decidable.

proof :

suppose we did have a decision procedure, d(x) for B,

We could then use it to create a decision procedure, k(x) for K as follows:

Given x, to compute k(x),

enumerate S to find a_x .

call d(a_x )

If "yes" is returned then output "yes"

If "no" is returned then output "no"

Therefore, since no decision procedure for K can exist, we conclude that d(x) can not exist.