Forum:On oracles and admissibles

For countless months (but countable, there's about 15 of them) we all have been considering higher order busy beaver functions \(\Sigma_k(n)\) to be on par with FGH extended, in some uniform way no one ever cared about, to admissible ordinal \(\omega^\text{CK}_k\). I believe I have actually been the first of us who started stating that. Argument I gave was that with halting oracle we can compute ordinal \(\omega^\text{CK}_1\), by trying out all machines, checking if they code ordinals and then adding up all the ordinals, and then we can sort of recurse through this ordinal and reach anywhere below \(\omega^\text{CK}_2\) (for higher orders it's just induction).

However, some time ago, somewhere on this wiki, I was told by Deedlit that checking if machine computes ordinal is actually a hard problem. A really hard problem. To be exact how hard it is, it's \(\Pi^1_1\)-complete problem, so it's at least as hard as any other \(\Pi^1_1\) problem, so it's far more complicated than any hyperarithmetical or arithmetical problem, including halting problem for many oracles. This means that it cannot be solved on halting oracle TM.

After a lot of time to rethink this, I came to a conclusion that it's actually impossible to, using any reasonable oracle machine, to compute CK ordinal, by following argument - we all know Kleene's O. Kleene's O can be interpreted as tree of height \(\omega^\text{CK}_1\), so we probably can, given some way of working with \(\omega^\text{CK}_1\), decide if number is part of Kleene's O, which is again \(\Pi^1_1\)-complete  problem, so that's impossible.

I'm still not sure about this, but if my rethought intuition is correct, not even \(\Sigma_k(n)\) reaches \(f_{omega^\text{CK}_1}(n)\). I wanted to ask you guys for your thoughts after reading this wall of text. LittlePeng9 (talk) 20:58, July 6, 2014 (UTC)