User blog:LittlePeng9/Functions between computability and busy beaver

Warning: I'm going to use concepts from computability theory which I'm not going to describe in great detail (mainly Turing degrees), so if interested, I recommend first of all checking out what the idea behind Turing degree is, and for other things feel free to ask in comments.

So, I have recently been wondering, is it possible to have a function \(F\) which is, in fundamental way, inbetween all computable functions and the busy beaver function (for simplicity, let it be \(S(n)\)).

Turns out, as expected, that the answer depends on what we precisely mean with "fundamentally between". There is quite intuitive notion for this, but I'd rather go about making it more formal.

First, I want my function to be above all computable functions. Two most obvious formalizations for this intuitive notion are: Of course, first one implies the second.
 * \(F\) dominates all computable functions, which means that for any computable function \(f\) and sufficiently large \(n\) we have \(F(n)>f(n)\).
 * \(F\) isn't bounded by any any computable function, i.e. no computable function \(f\) satisfies, for all \(n\), \(F(n)<f(n)\). Equivalently, \(F\) isn't dominated by any computable function.

Next we want \(F\) to be below \(S\). But thing like \(F(n)=S(n)-1\) isn't at all satisfactory, because we can still reach \(S\) with \(F\). We actually want \(S\) to not only be above \(F\), we would like it beyond all recursive extensions of \(F\). More formally, again I had two ideas:


 * \(S\) dominates all functions computable with \(F\).


 * \(S\) is not bounded by any function computable with \(F\).

Again, first implies second.

Optimally, we want \(F\) to dominate computable functions and \(S\) to dominate all \(F\)-computable functions. This, however, is impossible. There is a theorem, called Martin's domination theorem, which says that, if we have Turing degrees \(A,B\) then there is a \(A\)-computable function which dominates all \(B\)-computable functions if and only if \(A\) is high above \(B\), i.e. \(A'\geq_T B\) (the proof uses the fact that \(B\) is Turing-equivalent to set of machines which \(B\)-compute total functions).

So if the optimal scenario held, we would need to have that \(F'\geq_T 0\) and \(S'\geq_T F\geq_T 0'''\). But \(S\equiv_T 0'\) and \(0\not\geq_T 0'\). So it's impossible.

Other cases, however, can hold. First I'll construct a function \(F\) which isn't bounded by computable function and \(S\) dominates all \(F\)-computable functions. For this, let's take \(A\) to be any undecidable recursively enumerable set which is low (i.e. \(A'=0'\)) (such set exists, but construction uses priority methods which I don't exactly understand). Because \(A\) is r.e. there exists a total computable function \(a\) such that elements of \(A\) form the range of \(a\). Now I define \(F(n)\) as: the least number such that, for every \(m\leq n\), if \(m\) is in \(A\), then there is \(k\leq F(n)\) such that \(a(k)=m\). Less formally, if \(a\) sees that \(m\in A\), then it sees so with at most \(F(n)\) entries.

Obviously, we can compute \(F\) given \(A\) as oracle. We'll now show that every recursive extension \(f\) of \(F\) is dominated by \(S(n)\). (to be continued)