8 comments

  • Dylan16807 1 hour ago
    > I find it interesting to consider that if you pick a value at random, it will usually fail! That is, most 64-bit integers cannot be written as the product of two 32-bit integers.

    While I find the 17% number interesting to think about, "most" is far less interesting. Multiplication doesn't care about order so you're instantly cutting 2^64 possibilities down to about 2^63. That's a hair's breadth away from "most" already, and considering even a tiny amount of overlapping results gets you there.

    What gets interesting is actually trying to quantify the overlapping results.

    • thaumasiotes 13 minutes ago
      > While I find the 17% number interesting to think about, "most" is far less interesting. Multiplication doesn't care about order so you're instantly cutting 2^64 possibilities down to about 2^63. That's a hair's breadth away from "most" already

      It's much worse than that. It's difficult for a 64-bit product to have the high bit set if the multiplicands are both no larger than 32 bits.

    • danbruc 1 hour ago
      All the primes above 2^32 are out, but that accounts for only two point something percent.
    • PaulHoule 1 hour ago
      ... or just considering the even numbers almost all of them are 2 x N where N>2^32 and that gets you to within a hair of "most" and if you add in the odd thirds for which the same is true you get a bound of 2/3 - epsilon.
    • adgjlsfhk1 1 hour ago
      A lot of the remaining is multiples of 4, which you can either get from having a 2 in both factors or a 4 in one (multiples of 9 are similar).
  • crest 2 minutes ago
    So you're better of using a 8x8->16 widening multiplication SIMD instruction or even just a multi register TBL/TBX instruction?
  • henry2023 55 minutes ago
    There are about 4 billion more 64 bit integers than 32 bit integers.

    The chance of a random 64 bit integer being a 32 bit integer is 0.0000000233 %

    The chance of a random 64 bit integer being a product of two 32 bit integers is 17%

    Nice

    • HWR_14 50 minutes ago
      There are about 18.446 quintillion more 64-bit integers than 32-bit integers.
      • adrian_b 4 minutes ago
        True, but there are as many 64-bit integers as pairs of 32-bit integers.

        Therefore the fact that relatively few 64-bit numbers are products of 32-bit integers means that a lot of pairs of 32-bit integers give by multiplication the same product.

      • moefh 16 minutes ago
        I think they meant to write "There are about 4 billion TIMES more 64 bit integers than 32 bit integers".
    • layer8 37 minutes ago
      The chance of a random 64-bit integer matching some pair of 32-bit integers is a 100%, though.
    • rao-v 27 minutes ago
      Wonder what the limit is as you add more 32 bit integers to the product. Just the primes over 32 bit?
      • thaumasiotes 10 minutes ago
        If you're allowed to multiply as many 32-bit numbers as you want, the only numbers you won't be able to achieve by so doing are those whose prime factors are all larger than 2^32.

        If you're only considering numbers that will fit into 64 bits, all of the ones that satisfy that requirement will themselves be prime. (Because the product of two numbers both greater than 2^32 is greater than 2^64.)

    • brookst 33 minutes ago
      Or, the odds of a random 64-bit integer being a 32-bit integer are the same as you or me guessing a random 32 bit integer.
  • pants2 1 hour ago
    I dream of a future where all 64-bit integers are products of 32-bit integers. Together, we can change math for the better.
    • jerf 2 minutes ago
      Indeed, but justice requires that we recursively continue all the way to the base case, until all 32-bit integers are products of 16-bit integers, all 16-bit integers are products of 8-bit integers, all 8-bit integers are products of 4-bit integers, all 4-bit integers are products of 2-bit integers, and all 2-bit integers are products of 1-bit integers. Only when we have reach all the way down that list to the very, very smallest of the numbers around us and brought justice to them will the future be able to arrive. I literally can not wait for that day.
    • jihadjihad 51 minutes ago
      Why stop there? We can dream of a future where math is bent to our will [0] for the betterment of all mankind!

      0: https://en.wikipedia.org/wiki/Indiana_pi_bill

    • dvh 44 minutes ago
      1 + 1 = 3 (for sufficiently large values of 1)
      • sshine 29 minutes ago
        It helps if you take the limit of 1 going towards 1.5.

        Most 1s won't go towards 1.5, but sometimes you're lucky.

    • brookst 33 minutes ago
      There should be a law!
    • kleiba2 59 minutes ago
      I upvoted you, not because I think your joke is particularly great, but I hate that HN has this tendency to downvote comments that are clearly meant as a humorous contribution. And I get it, no-one wants HN to turn into Reddit. I also understand that not every joke lands. But I just think it's unnecessary to downvote, you could simply ignore.
      • zamadatix 54 minutes ago
        "Ignore" is one of those things that sounds like it's a neutral choice but really isn't in practice - it's still just saying "can only ever be positively pressured". IMO people shouldn't go as far as flag though, at the very least, and if it's already at the bottom of the sort there is no sense dumping on it further.

        My current comment itself, for instance, also doesn't really add anything to the discussion about the article and I'd have no expectation people leave it from going negative. Maybe the will, maybe they won't, but there is no reason to expect they should in principle of me loving tangents :D.

  • da_chicken 43 minutes ago
    This feels like a underlying property that contributes to of Benford's Law[0]. That is, most numbers we measure and record are the results of various independent (addition) and dependent (multiplication) factors stacking together, and we observe this property in the distribution of them.

    [0]: https://en.wikipedia.org/wiki/Benford%27s_law

  • MarkusQ 52 minutes ago
    If this seems counterintuitive, consider that only about a third of the two-digit numbers ({0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 24, 25, 27, 28, 30, 32, 35, 36, 40, 42, 45, 48, 49, 54, 56, 63, 64, 72, 81}) can be written as the product of two one-digit numbers.
    • childintime 7 minutes ago
      where is the graph and the theorem for integers of n bits, with n going to infinity?
  • nicechianti 28 minutes ago
    [dead]