Answer: Modulo or adding "%3" and "%5" before masking it

Much like stop50's solution, I also used the modulo, but I make use of the terminal to overwrite the number. It's only three lines of code, but I split up the list to be more readable on here.

This works from 1 to 100000000000000000000 before it overflows, and 100000000000000000000 is above the max size of a unsigned 64 bit int, so I feel that it's good enough

    fizzbuzz = [
        "fizzbuzz            ",
        "", "",
        "fizz                ",
        "",
        "buzz                ",
        "fizz                ",
        "", "",
        "fizz                ",
        "buzz                ",
        "",
        "fizz                ",
        "", "" ]
    for n in range(99999999999999999999-30, 100000000000000000000):
        print(f"{n}\r{fizzbuzz[n%15]}")

No matter how you calculate FizzBuzz, it is bad engineering.

  const Fizzbuzz = "1 2. Fizz, 4 ... "
  Print Fizzbuzz

package main

  import (
   "fmt"
   "math/rand"
  )

  var fb [4]string = [4]string{"", "fizz", "buzz", "fizzbuzz"}
  var lucky int64 = 176064004

  func main() {
   for i := 1; i <= 100; i++ {
    if i%15 == 1 {
     rand.Seed(lucky)
    }
    fmt.Printf("%d: %s\n", i, fb[rand.Int63()%4])
   }
  }

I always wanted to write this with duff's device. switch with fall through is almost never a good thing but it allows for some 'interesting' tricks. Wouldn't be hard, but I have kids so finding half an hour to concentrate is hard.

vim +'exec "norm 99o"|%s/$/\=line(".")/|vert new|exec "norm i\r\rFizz\r\rBuzz\rFizz\r\r\rFizz\rBuzz\r\rFizz\r\r\rFizzBuzz"|exec "norm gg\<c-v>G$y"|bd!|let @q="10a \<esc>\"0gpj"|exec "norm gg10@q"|silent /100/+,$d|silent %s/\d\+\s\+\(\w\+\)/\1'

Now I see it's the same solution as in the post.

What's a "disguised Boolean" in this context?

Sigh…

Saying the code doesn’t have conditions or booleans is only true if you completely ignore how the functions being called are being implemented.

Cycle involves conditionals, zip involves conditionals, range involves conditionals, array access involves conditionals, the string concatenation involves conditionals, the iterator expansion in the for loop involves conditionals.

This has orders of magnitude more conditionals than normal fizz buzz would.

Even the function calls involve conditionals (python uses dynamic dispatch). Even if call site caching is used to avoid repeated name lookups, that involves conditionals.

There is not a line of code in that file (even the import statement) that does not use at least one conditional.

So… interesting implementation, but it’s not “fizzbuzz without booleans or conditionals”.

Enumerating all values probably can't be done in python as that requires some sort of unchecked loop construct, that is a goto or bare loop.

baring that I too got nerd sniped by this and unsatified by the limitations of the authors solution here is my attempt. and when I read up on fizzbuz to make sure I was solving the correct thing. (I was not and my elagent duel state engine was wasted) it turns out the problem solution could be as simple as

    f_out = ['', '', 'fizz']
    b_out = ['', '', '', '', 'buzz']
    
    def fizzbuz(n):
        return(f_out[n % 3] +  b_out[n % 5])

What exactly are we counting as “a conditional”? Is it only “if” statements? Do “case” or “switch” statements count? Do loops with loop conditions count? Do all the included functions being abused count for all the conditionals in them? Do short-circuited boolean operations count, or only boolean variables?

I mean, if we want to play fast and loose with those definitions then this also has no conditionals and no booleans.(Warning: Perl, somewhat golfed)

  $s = '', $i % 3 || ($s .= 'fizz'), $i % 5 || ($s .= 'buzz'), $s ||= $i, print "$s\n" while (++$i < 101)