10 Jul
2023
10 Jul
'23
1:03 p.m.
if we consider making one f() that can resolve whether or not any g() halts, and we simply consider that f() must not halt, but can run *must halt, but can run arbitrarily long forever, i think the cryptographic solution might work. the user must be able to provide unlimited cryptographic confirmation that they are not evaluating f() for the purposes of influencing another evaluation of f(). i feel like, if that works, there are other approaches too.
it's notable these means that g() cannot use f() simply because of the reason of use, or access to the cryptographic material, and i'm guessing but not sure that these means that f() does not meet the requirements of correctly evaluating all data on all functions