| ▲ | jiggawatts 3 hours ago | |
> the algorithm requires you to multiply complex numbers whose magnitudes grow superpolynomially with n. Err... doesn't that completely invalidate the big-O analysis!? I suspected that these Open AI "proofs" of sub-nlogn times were cheating somehow, by sweeping something under a rug. This seems to be it! | ||
| ▲ | LegionMammal978 3 hours ago | parent | next [-] | |
I can't speak for the FFT claim, but least the sub-O(n log n) multiplication claim is for the multitape Turing machine model with a fixed number of tapes, which if anything is more restrictive than actual random-access memory. Of course, it's still galactic just like the O(n log n) algorithm, but it definitely seems like there's something there. | ||
| ▲ | LPisGood 3 hours ago | parent | prev [-] | |
You can be linear in the bit length of the input number, but exponential in the magnitude. This is a common thing in complexity, and I think even the abstract says it is talking about “length n” and not “magnitude n” | ||