Comment by qrendel
11 years ago
1) My submission: modifying Ackermann's function to describe a sequence of Busy Beaver numbers of sequential hypercomputation systems. E.g. ABB(n+1) = BB_n(n) Of course it's the same problem, you could just as easily do AABB(n+1) = ABB_n(n), but still grows faster than a Busy Beaver sequence for any single level of the hypercomputation hierarchy.
2) Question: I understand the argument for noncomputability of Busy Beavers given, but couldn't you just argue that BB(n) is finite, therefore BB(n) is a computable sum of 1+1+...+1, therefore BB(n) is computable given a finite amount of time, so any given number in the sequence is itself computable, therefore by mathematical induction all elements in the sequence are computable? Clearly not, but I don't understand why this doesn't work.
1) If you're aware of the BB_n hierarchy, you've outgrown the need for recursion tricks. Instead, talk about halting times of programs that can make calls to the oracle F(n,m)=BB_n(m). That leaves any Ackermann-like approach in the dust. I leave it to you to come up with an even better approach (yes, it exists, and then there's one or two levels after that). Please don't say "I'll just use Ackermann on top of your idea", at that level it's the moral equivalent of "your number plus one".
2) For each individual BB number, there's a program that prints it. But there's no program that prints the whole infinite sequence of BB numbers one by one. All finite sequences are computable, but not all infinite sequences are.
To say that the sequence is computable means that you must present a Turing machine, call it T, that will produce BB(n) in finitely many steps after taking input n. Our specific Turing machine T will have a fixed number of rules call it N rules, then by the definition of the Busy Beaver T(N+1) <= BB(N) = T(N) but BB(N+1) > BB(N) so T(N+1) does not compute BB(N+1). The flaw in your inductive proof is that you have shown there is a Turing machine that can compute BB(N) for each N, but you haven't shown that the same Turing machine can compute all numbers in the sequence.
And just to be clear here, this isn't just because it can't recognize which numbers are the Busy Beavers as it arrives at them while summing ones indefinitely? If T(N) tries to compute BB(N) by summing ones then, since BB(N) is finite, then it just fails to halt? But then there would be finite numbers that can't be described as the sum of one from 0 to BB(N)? It's just hard to reconcile the concept of indefinite-but-finite running time, all numbers being able to be expressed as a sum of ones, and the BB sequence still not being computable.
Edit: Ah wait, I get it now. Misunderstanding was in confusing "Turing machine" with "universal Turing machine." Any given T(N) will have a finite number of steps it can complete before halting, and since a universal Turing machine or equivalent (e.g. rule 110) can emulate any arbitrary Turing machine, I was modeling it as just letting an arbitrary UTM run forever. But the UTM is just emulating a specific T(N) each time, and you'd have to keep going up emulating different machines for each N to get to each next BB without running out of steps. So you can write programs to calculate arbitrary BBs indefinitely, but no single program to calculate them all. Hence the infinite sequence is noncomputable despite any finite length of it being computable. Thanks for the point in the right direction.