• VicFic!@iusearchlinux.fyi
    link
    fedilink
    English
    arrow-up
    18
    ·
    1 year ago

    Wait till you include floating numbers. “There are an infinite numbe of numbers between any two natural numbers” So technically you could increase that percentage to 99.9999…%

  • xthexder@l.sw0.com
    link
    fedilink
    English
    arrow-up
    9
    ·
    edit-2
    1 year ago

    A few calculations I did last time I saw this meme (over at !programmer_humor@programming.dev):

    • There are 9592 prime numbers less than 100,000. Assuming the test suite only tests numbers 1-99999, the accuracy should actually be only 90.408%, not 95.121%
    • The 1 trillionth prime number is 29,996,224,275,833. This would mean even the first 29 trillion primes would only get you to 96.667% accuracy.

    In response to the question of how long it would take to round up to 100%:

    • The density of primes can be approximated using the Prime Number Theorem: 1/ln(x). Solving 99.9995 = 100 - 100 / ln(x) for x gives e^200000 or 7.88 × 10^86858. In other words, the universe will end before any current computer could check that many numbers.

    Edit: Fixed community link

    • smitten@lemmy.blahaj.zoneOP
      link
      fedilink
      English
      arrow-up
      2
      ·
      1 year ago

      I think a more concise answer to the second one would be; it depends on where you decide to round, but as you run it, it approaches 100%, or 99.99 repeating (which is 100%)

      • xthexder@l.sw0.com
        link
        fedilink
        English
        arrow-up
        3
        ·
        1 year ago

        The screenshot displays 3 decimal places, which is the the precision I used. As it turns out, even just rounding to the nearest integer still requires checking more numbers than we even have the primes enumerated for (e^200 or 7x10^86)

  • NotAUser@lemmy.blahaj.zoneM
    link
    fedilink
    English
    arrow-up
    5
    ·
    1 year ago

    By the prime number theorem, if the tests go from 1 to N, the accuracy will be 1 - 1 / ln(N). They should have kept going for better accuracy.