3 ms·

More generally, only if your input is an integer multiple of your alphabet. However, there are methods of converting uniform distributions that will preserve u

More generally, only if your input is an integer multiple of your alphabet. However, there are methods of converting uniform distributions that will preserve uniformity. I'll have to find a good reference.

heavenlyblue 9y ago
Well, it's obvious for me. Issue is that the root reply seemed to have taken it as a JS implementation issue (due to math.random) as it's quite obvious why you should not do the modulus (or when you can), if you know the underlying maths. So did I, as a matter of fact. You - on the other hand had given the answer to the wrong confusion.
- AndrewCHM 9y ago
  they could of responded, "no, random is uniform" (the answer to the written question) and that wouldn't of answered anything!
- falsedan 9y ago
  > Issue is that the root reply seemed to have taken it as a JS implementation issue I'm the root reply. jimktrains2's explanation exactly addressed my confusion. To get a uniform distribution over an alphabet from random, I would do: var bucket_size = (random.MAX_VALUE - random.MIN_VALUE) / count(alphabet) return alphabet[random/bucket_size]
  - jimktrains2 9y ago
    Actually, that doesn't work either max = 4, min = 0, count = 3 bucket_size = 4 / 3 0,1,2,3,4 0 / (4/3) = 0 1 / (4/3) = 3/4 = 0 2 / (4/3) = 6/3 = 2 3 / (4/3) = 9/4 = 2 4 / (4/3) = 12/4 = 3 = invalid index but even if you adjust this to count - 1 bucket_size = 4 / 2 = 2 0 = 0/2 = 0 1 = 1/2 = 0 2 = 2/2 = 1 3 = 3/2 = 1 4 = 4/2 = 2 So, 0s and 1s are still more likely than 2 to appear. One of the easiest, and slowest, ways is just to discard any trial in which your uniform RNG generates a number above the largest index and try again. For a larger alphabet, you essentially just collect multiple trials, treat them as a single trial, and do similar tactics.
    - falsedan 9y ago
      > bucket_size = 4 / 3 I have an off-by-one error… your random function has a range of 5, so it should be max - min + 1: bucket_size = 5 / 3 then 0 / (5/3) = 0 1 / (5/3) = 3/5 = 0 2 / (5/3) = 6/5 = 1 3 / (5/3) = 9/5 = 1 4 / (5/3) = 12/5 = 2 > So, 0s and 1s are still more likely than 2 to appear. Yeah, I planned for 65535 randoms and 52 alphabet. It would still mildly bias to the first range mod alphabet_size elements.

heavenlyblue 9y ago

Well, it's obvious for me. Issue is that the root reply seemed to have taken it as a JS implementation issue (due to math.random) as it's quite obvious why you should not do the modulus (or when you can), if you know the underlying maths. So did I, as a matter of fact. You - on the other hand had given the answer to the wrong confusion.

AndrewCHM 9y ago
they could of responded, "no, random is uniform" (the answer to the written question) and that wouldn't of answered anything!
falsedan 9y ago
> Issue is that the root reply seemed to have taken it as a JS implementation issue I'm the root reply. jimktrains2's explanation exactly addressed my confusion. To get a uniform distribution over an alphabet from random, I would do: var bucket_size = (random.MAX_VALUE - random.MIN_VALUE) / count(alphabet) return alphabet[random/bucket_size]
- jimktrains2 9y ago
  Actually, that doesn't work either max = 4, min = 0, count = 3 bucket_size = 4 / 3 0,1,2,3,4 0 / (4/3) = 0 1 / (4/3) = 3/4 = 0 2 / (4/3) = 6/3 = 2 3 / (4/3) = 9/4 = 2 4 / (4/3) = 12/4 = 3 = invalid index but even if you adjust this to count - 1 bucket_size = 4 / 2 = 2 0 = 0/2 = 0 1 = 1/2 = 0 2 = 2/2 = 1 3 = 3/2 = 1 4 = 4/2 = 2 So, 0s and 1s are still more likely than 2 to appear. One of the easiest, and slowest, ways is just to discard any trial in which your uniform RNG generates a number above the largest index and try again. For a larger alphabet, you essentially just collect multiple trials, treat them as a single trial, and do similar tactics.
  - falsedan 9y ago
    > bucket_size = 4 / 3 I have an off-by-one error… your random function has a range of 5, so it should be max - min + 1: bucket_size = 5 / 3 then 0 / (5/3) = 0 1 / (5/3) = 3/5 = 0 2 / (5/3) = 6/5 = 1 3 / (5/3) = 9/5 = 1 4 / (5/3) = 12/5 = 2 > So, 0s and 1s are still more likely than 2 to appear. Yeah, I planned for 65535 randoms and 52 alphabet. It would still mildly bias to the first range mod alphabet_size elements.