• aio@awful.systems
    link
    fedilink
    English
    arrow-up
    0
    ·
    3 months ago

    the computational cost of operating over a matrix is always going to be convex relative to its size

    This makes no sense - “convex” doesn’t mean fast-growing. For instance a constant function is convex.

    • bitofhope@awful.systems
      link
      fedilink
      English
      arrow-up
      0
      ·
      3 months ago

      Hell, so is 1/x for positive values of x. Or any linear function, including those with negative slope.

    • dorian@awful.systems
      link
      fedilink
      English
      arrow-up
      0
      ·
      3 months ago

      you will be pleased to know that the original text said “superlinear”; i just couldn’t remember if the lower bound of multiplying a sufficiently sparse matrix was actually lower than O(n²) (because you could conceivably skip over big chunks of it) and didn’t feel like going and digging that fact out. i briefly felt “superlinear” was too clunky though and switched it to “convex” and that is when you saw it.