Comment by lupire

8 months ago

It has to come from a finite value (specifically, the amount of complexity that can be enocoded in 745 pieces of information https://turingmachinesimulator.com/shared/vgimygpuwi), because the finite size 745 with infinite tape leads to uncomputability, but the size 5 does not.

In a very real sense, a deep kind of infinite complexity can be generated from 745 objects of certain kind, but not from 5 objects of that kind..

Turing machines have infinite tape, not infinite state. The entire set of all halting machines of a given size collectively only use finite tape. Totally finite. Only (some of) the non-halting machines use infinite tape.

The problem is that we don't know in advance how large the (definitely finite) upper bound on the amount of tape all the size-N halting machines use, until after enough of them (one per known equivalence class) halt. And we don't know (in general) how to run all the halting ones until they halt, without also running a non-halting program for an unbounded amount of time.

TL:DR: unbounded is not infinite, but big enough to be a problem.

I am aware it's an infinite tape and finite state (maybe I misspoke somewhere), as well as the halting machines using finite tape (because of course they do).

But the overall 'complexity' (at a timestep, say) is going to be due to the states and the tape together. The BB(5) example that was analyzed, iirc, was a Collatz-like problem (Aaronson describes it here: https://scottaaronson.blog/?p=8088 ). My interpretation of this is that:

1. collatz-like functions have a lot of complexity just due to math alone 2. 5 states turned out to be enough to "reach" that one that 3. more states means you're going to reach more possible Collatz-like functions (they don't have to be Collatz-like; it's just easier to think about them like that) 4. eventually you reach ones that ZFC cannot show to halt, because there is effectively no way to prove it other than running them, and then you would have to solve the halting problem.

The part that was helpful for me to be less unsettle by BB(745) being independent of the ZFC was the notion that it eventually boils down to a halting problem, and asking ZFC to "solve" it... which is more agreeable than the idea that "ZFC cannot compute a function that seems to be solvable by brute force".