Everything can be done in constant time, at least during runtime, with a sufficiently large look-up table. It’s easy! If you want to simulate the universe exactly, you just need a table with nxm entries, where n is the number of plank volumes in the universe, and m is the number of quantum fields. Then, you just need to compute all of them at compile time, and you have O(1) time complexity during runtime.
I know I shouldn’t do it here, but let me ask a serious question : does the square in O(n!²) really matter ? I have a confused intuition that the factorial grows so much faster than the square that it kind of disapears assymptoticaly.
no it’s the joke. In o-notation you always use the highest approximation, so o(n!²) does not exist, it’s only o(n!)
Otherwise there would never be o(1) or o(n), because o(1) would imply that the algorithm only has a single line of instructions, same for o(n)
T = O(n) means that there exists a single constant k such that T < kn for all sufficiently large n. Therefore O(n!^2) is not the the same as O(n!), but for example both 10n!, 10000n!, n! + n^2 (note the plus) are O(n!).
Another way to think about this: suppose you believe that O(n) and O(n^2) are distinct. Now plug in only numbers that are factorials (2, 6, 24, …).
O((n!)!)
