2 comments

  • jdb1729 2 minutes ago
    > The new result does not directly imply new complexity bounds for polynomial multiplication over a fixed finite field, say multiplication in F2[x]. The state of the art for this problem is an algorithm that multiplies polynomials of degree n in time O(n log n 4^log*(n)) [HvdH19].

    @platypii and I obtained a Lean proof of the same improvement for DFT over prime-order finite fields (yielding polynomial multiplication via the convolution theorem): https://github.com/platypii/integer-mult-bounds-lean/blob/ma...

  • flebron 16 minutes ago
    Note in this case the title says "O(n log n)", but the important part is this is "o(n log n)", not just "O(n log n)'. The former is a strict subset of the latter. We already knew how to multiply two n-bit integers in O(n log n) time, via the FFT.
    • onionisafruit 11 minutes ago
      For those of us uninformed, what’s the difference between O and o?
      • theokrueger 7 minutes ago
        Big O basically drops constants

        O(2n+1) = O(2n) = O(n)

        o(2n+1) > o(2n) > o(n)

        Both define upper boundary, but big O can read like "the function is generally proportional to ___" and little o can read like "the function is never more than _____"

      • boothby 7 minutes ago
        If f is O(g) and o(h) then

          Lim f/g -> C a constant
        
        And

          Lim f/h -> 0
        • appplication 3 minutes ago
          Lmao! If someone is asking the difference between big and little o notation this is absolutely not the right way to clarify it.

          If you were joking then maybe you got me

      • Ygg2 4 minutes ago
        Both are upper bounds. But small o is strictly smaller than upper bound. Big O is not greater than upper bound.

        https://stackoverflow.com/questions/1364444/difference-betwe...

    • craigcitro 8 minutes ago
      ty, fixed. (ahh, copy paste.)