6 ms·

If it’s round robin then it should be an even load, how does the modulo change that exactly? Also what number are they using to modulo and where is that happen

If it’s round robin then it should be an even load, how does the modulo change that exactly? Also what number are they using to modulo and where is that happening? Because at that point don’t they already have an incrementing ID before generating another one?

andreareina 4y ago
Take a 3-bit counter: 0->A 1->B 2->C 3->A 4->B 5->C 6->A 7->B A and B get hit three times while C only twice, so it will see 66% utilization compared to A and B EDITED s/once/twice/ thanks CyberDildonics
- CyberDildonics 4y ago
  while C only once You listed C twice
  - woodruffw 4y ago
    That's a typo. You can still see they're correct about the ratio (two "C"s for every three "A"s and "B"s).
    - manigandham 4y ago
      There's no ratio. It's even across all of them, as long as the integer keeps incrementing. One more number (9) instead of stopping at 8 and there would be an even spread.
      - ahmedtd 4y ago
        Not when the counter overflows back to 0. If it's a 3 bit counter, 0 is A again, not C.
        manigandham 4y ago
        The comment says a 32-bit signed int. Where is the 3-bit assumption coming from?
        iforgotpassword 4y ago
        The comment starts with that assumption for the sake of a concise example. do you expect them to write the whole sequence out using a 32bit counter? :-D
        manigandham 4y ago
        Then it's irrelevant. The point is that the remainders are evenly distributed as the integer is incremented, with at most a single round being short. That's just how the math works.
        iainmerrick 4y ago
        "A single round being short" is exactly what they're talking about. Edit to add: whether that's a big enough effect for the use case they're talking about, I don't know. This sort of thing is definitely significant in cryptographic code, though.
        Thorrez 4y ago
        When you divide 2^32 across 3 machines, you get 1431655766, 1431655765, 1431655765. They're nearly identical, just an itsy bitsy bit different. This isn't going to cause one machine to have noticeably less load than the others.
        iforgotpassword 4y ago
        That's not what's being discussed here though.
        Thorrez 4y ago
  - andreareina 4y ago
    Edited, thanks
- manigandham 4y ago
  That doesn't change anything... it's still round robin. You just stopped at an arbitrary number of 8 integers instead of 9.
  - lofatdairy 4y ago
    It's not arbitrary, GP stated it was a 3 bit counter. In GGP (or something, not sure how far down in the thread we are), they were referring to a 32 bit int counter until overflow. If you bin each number from 0 to 2^32 - 1 by mod 3, you don't get 3 bins of equal sizes, 1 bin always comes out smaller.
    - manigandham 4y ago
      Smaller by a single number at most, it's effectively insignificant. Not sure where the 3-bit counter assumption came from as the original post said 32bit signed int.

andreareina 4y ago

Take a 3-bit counter: 0->A 1->B 2->C 3->A 4->B 5->C 6->A 7->B A and B get hit three times while C only twice, so it will see 66% utilization compared to A and B EDITED s/once/twice/ thanks CyberDildonics

CyberDildonics 4y ago
while C only once You listed C twice
- woodruffw 4y ago
  That's a typo. You can still see they're correct about the ratio (two "C"s for every three "A"s and "B"s).
  - manigandham 4y ago
    There's no ratio. It's even across all of them, as long as the integer keeps incrementing. One more number (9) instead of stopping at 8 and there would be an even spread.
    - ahmedtd 4y ago
      Not when the counter overflows back to 0. If it's a 3 bit counter, 0 is A again, not C.
      - manigandham 4y ago
        The comment says a 32-bit signed int. Where is the 3-bit assumption coming from?
        iforgotpassword 4y ago
        The comment starts with that assumption for the sake of a concise example. do you expect them to write the whole sequence out using a 32bit counter? :-D
        manigandham 4y ago
        Then it's irrelevant. The point is that the remainders are evenly distributed as the integer is incremented, with at most a single round being short. That's just how the math works.
        iainmerrick 4y ago
        "A single round being short" is exactly what they're talking about. Edit to add: whether that's a big enough effect for the use case they're talking about, I don't know. This sort of thing is definitely significant in cryptographic code, though.
        Thorrez 4y ago
        When you divide 2^32 across 3 machines, you get 1431655766, 1431655765, 1431655765. They're nearly identical, just an itsy bitsy bit different. This isn't going to cause one machine to have noticeably less load than the others.
        iforgotpassword 4y ago
        That's not what's being discussed here though.
        Thorrez 4y ago
- andreareina 4y ago
  Edited, thanks
manigandham 4y ago
That doesn't change anything... it's still round robin. You just stopped at an arbitrary number of 8 integers instead of 9.
- lofatdairy 4y ago
  It's not arbitrary, GP stated it was a 3 bit counter. In GGP (or something, not sure how far down in the thread we are), they were referring to a 32 bit int counter until overflow. If you bin each number from 0 to 2^32 - 1 by mod 3, you don't get 3 bins of equal sizes, 1 bin always comes out smaller.
  - manigandham 4y ago
    Smaller by a single number at most, it's effectively insignificant. Not sure where the 3-bit counter assumption came from as the original post said 32bit signed int.