4 ms·

This is a cute solution but I feel like the exposition makes it more mysterious seeming then necessary. If you have n%15, that's enough to do FizzBuzz on it's o

by tgb 6y ago

This is a cute solution but I feel like the exposition makes it more mysterious seeming then necessary. If you have n%15, that's enough to do FizzBuzz on it's own, without having to take a fourth power. It's just then you need to map n%15=3,6,9,12 to "Fizz" instead of just 6. Taking the fourth power simplifies things because 3^4, 6^4, 9^4 and 12^4 all equal 6 (mod 15). And similarly for the multiples of 5. Though this explanation doesn't generalize as well.

kingaillas 6y ago
His point was writing a function without conditional logic, and then of course generalizing to any "fizzbuzz category" problem. I thought it was a fantastic post on a extremely well trodden subject, up there with solving fizzbuzz in Tensorflow post. (https://joelgrus.com/2016/05/23/fizz-buzz-in-tensorflow/ https://joelgrus.com/2016/05/23/fizz-buzz-in-tensorflow/)
- LanceH 6y ago
  My entry to use random-isn't-random: https://github.com/LanceH/fizzbuzz/blob/master/fb.go https://github.com/LanceH/fizzbuzz/blob/master/fb.go I feel like I need to revisit after seeing tensorflow and Euler. I need to find some way of making the lucky number using things around a room, like a mentalist or something. Some other subversive ones I've seen of are the java enterprise version, and several that import a fizzbuzz library and just run.
  - isaacimagine 6y ago
    I find that random-not-random implementation pretty incredible.
    - LanceH 6y ago
      Thanks. It seems kind of obvious compared to the Euler method, though. Once you realize it's 4 bits across a cycle of 15, it's just just a matter of finding the right seed. It should be on the order of 1 in a billion, which is no big deal. In hindsight, if I had combined my number theory knowledge I might have been able to accomplish the Euler solution, but I didn't, so hats off to them. The presentation of the tensorflow provides more a much higher contempt for the question than I've achieved as well. The fact it isn't 100% accurate, but can be trained to be accurate to a certain level, just makes it better. I will redouble my research efforts on this or some other problem which doesn't need solving.
- uuidgen 6y ago
  But there is conditional logic "hidden" in lambda mapping. You want it without ANY conditional logic try this (python3): [str(n)*(n%3!=0)*(n%5!=0) + 'Fizz'*(n%3==0) + 'Buzz'*(n%5==0) for n in range(1,101)]
  - chongli 6y ago
    There’s tons of hidden conditional logic in the Python string multiplication operator. It’s implemented via the C function unicode_repeat which handles all the conditional logic.
    - uuidgen 6y ago
      If you're talking about assembly then every single statement in python will involve hidden conditions. I was talking about the idea behind the code - that map is a pythonic shorthand for a bunch of if-then statements, it still expresses conditionals. While using string multiplication doesn't.
chronolitus 6y ago
Right, but your comment doesn't do justice to the beauty of the solution being so concise. Is it a coincidence that the same exponent 4 works for both multiples of 3 and 5? What if instead of 3 and 5, we did FizzBuzz with 7 and 11? Let's do n%77, now we need to map 7, 14, 21, 28, 35, 42, 49, 56, 63, 70 to Fizz, and 11, 22, 33, 44, 55, 66 to Buzz, unless we can find some way to have them all reduce to the same number. Is that really so straightforward? Now that we know the trick with the exponents, we could try a few to start. For multiples of 11 it seems the smallest one which works is 6! Great, let's try it on multiples of 7.. Nope. Ok, for multiples of 7, I see that ^10 works.. but that doesn't work for multiples of 11. * the article addresses this at the end, and tells us that the correct exponent is ^30, the lowest common denominator of 6 and 10. But if I'd been given this problem, I'd have remained stuck on the second paragraph of this comment, with no idea where to start when seeking to reduce the divisors to a single number.
- namelessUser 6y ago
  If a^n = 0 (or 1) mod b, then a^(kn) = (a^n)^k = 0 (or 1) mod b for any positive integer k. Therefore you need to pick a common multiple of 6 and 10.

by tgb 6y ago

kingaillas 6y ago

His point was writing a function without conditional logic, and then of course generalizing to any "fizzbuzz category" problem. I thought it was a fantastic post on a extremely well trodden subject, up there with solving fizzbuzz in Tensorflow post. (https://joelgrus.com/2016/05/23/fizz-buzz-in-tensorflow/ https://joelgrus.com/2016/05/23/fizz-buzz-in-tensorflow/)

LanceH 6y ago
My entry to use random-isn't-random: https://github.com/LanceH/fizzbuzz/blob/master/fb.go https://github.com/LanceH/fizzbuzz/blob/master/fb.go I feel like I need to revisit after seeing tensorflow and Euler. I need to find some way of making the lucky number using things around a room, like a mentalist or something. Some other subversive ones I've seen of are the java enterprise version, and several that import a fizzbuzz library and just run.
- isaacimagine 6y ago
  I find that random-not-random implementation pretty incredible.
  - LanceH 6y ago
    Thanks. It seems kind of obvious compared to the Euler method, though. Once you realize it's 4 bits across a cycle of 15, it's just just a matter of finding the right seed. It should be on the order of 1 in a billion, which is no big deal. In hindsight, if I had combined my number theory knowledge I might have been able to accomplish the Euler solution, but I didn't, so hats off to them. The presentation of the tensorflow provides more a much higher contempt for the question than I've achieved as well. The fact it isn't 100% accurate, but can be trained to be accurate to a certain level, just makes it better. I will redouble my research efforts on this or some other problem which doesn't need solving.
uuidgen 6y ago
But there is conditional logic "hidden" in lambda mapping. You want it without ANY conditional logic try this (python3): [str(n)*(n%3!=0)*(n%5!=0) + 'Fizz'*(n%3==0) + 'Buzz'*(n%5==0) for n in range(1,101)]
- chongli 6y ago
  There’s tons of hidden conditional logic in the Python string multiplication operator. It’s implemented via the C function unicode_repeat which handles all the conditional logic.
  - uuidgen 6y ago
    If you're talking about assembly then every single statement in python will involve hidden conditions. I was talking about the idea behind the code - that map is a pythonic shorthand for a bunch of if-then statements, it still expresses conditionals. While using string multiplication doesn't.

chronolitus 6y ago

Right, but your comment doesn't do justice to the beauty of the solution being so concise. Is it a coincidence that the same exponent 4 works for both multiples of 3 and 5? What if instead of 3 and 5, we did FizzBuzz with 7 and 11? Let's do n%77, now we need to map 7, 14, 21, 28, 35, 42, 49, 56, 63, 70 to Fizz, and 11, 22, 33, 44, 55, 66 to Buzz, unless we can find some way to have them all reduce to the same number. Is that really so straightforward? Now that we know the trick with the exponents, we could try a few to start. For multiples of 11 it seems the smallest one which works is 6! Great, let's try it on multiples of 7.. Nope. Ok, for multiples of 7, I see that ^10 works.. but that doesn't work for multiples of 11. * the article addresses this at the end, and tells us that the correct exponent is ^30, the lowest common denominator of 6 and 10. But if I'd been given this problem, I'd have remained stuck on the second paragraph of this comment, with no idea where to start when seeking to reduce the divisors to a single number.

namelessUser 6y ago
If a^n = 0 (or 1) mod b, then a^(kn) = (a^n)^k = 0 (or 1) mod b for any positive integer k. Therefore you need to pick a common multiple of 6 and 10.