3 ms·

I believe fastest in this context means that it has been proven that there can be no faster algorithm.

• MojoJolo 13y ago
  Is it really possible? I believe the "fastness" of something is infinite because of numbers being endless.
  • olalonde 13y ago
    The "fast" here refers to the complexity of the algorithm. http://en.wikipedia.org/wiki/Computational_complexity_theory http://en.wikipedia.org/wiki/Computational_complexity_theory
• wlievens 13y ago
  Indeed, in terms of complexity theory I would guess, not in terms of milliseconds.
  • jd007 13y ago
    Read http://en.wikipedia.org/wiki/Big_O_notation http://en.wikipedia.org/wiki/Big_O_notation Basically it's talking about time complexity, or how fast does the time it takes to compute the solution grow as the input grows in size.
  • ginko 13y ago
    An Algorithm with a lower time complexity will always take a shorter amount of time once the problem size grows large enough. You could run one algorithm on a modern cpu, utilizing Simd and tune for cache coherency and the other in QBASIC on an old 386 and the O(n log n) algorithm would still terminate earlier than the O(n^2) one past a certain n.
    • ufo 13y ago
      the problem is that while dropping from n^2 to n log n is a big deal, dropping from (n log n) to n won't help much if you end up increasing the constant costs at the same time.
      • tel 13y ago
        It will eventually. Practical problems are pretty much solved.