Not sure if that is counter to the spirit of the post, because while the answer is either −1 or 0, it involves an intermediate calculation which vastly exceeds 10100.
Feynman once challenged people to come up with a problem that could be stated quickly but he couldn’t solve to within 10% in a minute, and a colleague stumped him with finding tan(10100).
If you mean something like cos(3^^^3 * pi), then I think that one should be solveable. Maybe try induction.
EDIT: Whoops, I didn’t see the floor symbols. Nevermind me. With not-quite-so-large numbers it’s an interesting problem that might be solveable by finding digits of pi in linear time, but linear time is of limited help here.
n=⌊sin(3↑↑↑↑3)⌋
Not sure if that is counter to the spirit of the post, because while the answer is either −1 or 0, it involves an intermediate calculation which vastly exceeds 10100.
Feynman once challenged people to come up with a problem that could be stated quickly but he couldn’t solve to within 10% in a minute, and a colleague stumped him with finding tan(10100).
If you mean something like cos(3^^^3 * pi), then I think that one should be solveable. Maybe try induction.EDIT: Whoops, I didn’t see the floor symbols. Nevermind me. With not-quite-so-large numbers it’s an interesting problem that might be solveable by finding digits of pi in linear time, but linear time is of limited help here.
It’s easy to estimate cos(floor(pi * 3^^^3)) to within 30%. Try it!
↑ is not ^
https://en.wikipedia.org/wiki/Knuth%27s_up-arrow_notation
I was too lazy to find the right character. Whoops, looks like I misread the comment though.