FRESH

Hacker News

Fizz Buzz without conditionals or booleans

55 points by ingve

by susam

1 subcomments

Late to the party but here's a solution I worked out using roots of unity and cosines:

  from math import cos, pi
  for n in range(1, 101):
      print([n, 'Fizz', 'Buzz', 'FizzBuzz'][round((1 + 2 * cos(2 * pi * n / 3)) / 3 + 2 * (1 + 2 * cos(2 * pi * n / 5) + 2 * cos(4 * pi * n / 5)) / 5)])

by userbinator

1 subcomments

challenged listeners to “write Fizz Buzz with no booleans, no conditionals, no pattern matching, or other things that are like disguised booleans.”
Without any other constraints, this is not an interesting challenge.
```
    print("<precalculated output goes here>")
```

by jswelker

1 subcomments

3 lines of python, no imports.

  fizzbuzz = [None, None, "Fizz", None, "Buzz", "Fizz", None, None, "Fizz", "Buzz", None, "Fizz", None, None, "FizzBuzz"]

  for i in range(1,100):
    print(fizzbuzz[i % 15] or i)

Edit: I see I was a few hours late, and someone posted nearly the exact same solution. :(

by ramenprofitable

0 subcomment

The suckless approach

  #include <stdio.h>
  #include <stdint.h>
  #include <stdlib.h>

  void fizzbuzz(int i) {
      uint32_t r3 = i % 3;
      uint32_t r5 = i % 5;
      uint32_t is_nonzero_3 = (r3 | -r3) >> 31;
      uint32_t is_nonzero_5 = (r5 | -r5) >> 31;
      uint32_t is_zero_3 = is_nonzero_3 ^ 1;
      uint32_t is_zero_5 = is_nonzero_5 ^ 1;
      uint32_t idx = (is_zero_5 << 1) | is_zero_3;

      const char *fmt[] = {
          "%d\n",
          "Fizz\n",
          "Buzz\n",
          "FizzBuzz\n"
      };

      printf(fmt[idx], i);
  }

  typedef void (*func_t)(int);

  void run_loop(int i);
  void stop_loop(int i) { exit(0); }

  func_t actions[] = { stop_loop, run_loop };

  void run_loop(int i) {
      fizzbuzz(i);
      int d = 99 - i;
      uint32_t is_neg = ((uint32_t)d) >> 31;
  }

  func_t actions_table[] = { run_loop, stop_loop };

  void run_loop_wrapper(int i) {
      fizzbuzz(i);
      int d = 99 - i;
      uint32_t is_neg = ((uint32_t)d) >> 31;
      actions_table[is_neg](i + 1);
  }

  int main() {
      actions_table[0] = run_loop_wrapper;
      run_loop_wrapper(1);
      return 0;
  }

by brudgers

1 subcomments

Obviously FizzBuzz is a property of integers

  Integer extend [
    fizzbuzz [ 
        (self \\ 15 = 0)
        ifTrue: ['fizzbuzz' printNl]
        ifFalse: [
             (self \\ 3 = 0)
             ifTrue: ['fizz' printNl]
             ifFalse: [
                  (self \\ 5 = 0)
                  ifTrue: ['buzz' printNl]
                  ifFalse: [self printNl]
             ]
        ]
    ]
  ]

  1 to: 100 by: 1 do: [:i | i fizzbuzz]

by kiratp

1 subcomments

A for loop has a conditional in it.
Unless by conditionals we mean “no if/else” and not “no branch instructions”.

by andriamanitra

0 subcomment

I gave it a go in C (I wanted to do assembly but couldn't be arsed to write string to int and int to string conversions). [1] The trickiest part was figuring out how to terminate the program. My first attempt invoked nasal demons through dividing by zero, but I then realized I could intentionally cause a segfault with high probability (which is much better, right?). One could argue that my `fizz` and `buzz` variables are still "disguised booleans", but at least the generated assembly contains no branching or cmov instructions (aside from the ones inside libc functions like atoi and sprintf).
[1] https://gist.github.com/Andriamanitra/5c20f367dc4570dd5c8068...

by HWR_14

4 subcomments

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

0 subcomment

by brudgers

1 subcomments

No matter how you calculate FizzBuzz, it is bad engineering.
```
  const Fizzbuzz = "1 2. Fizz, 4 ... "
  Print Fizzbuzz
```

by unsnap_biceps

2 subcomments

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]}")

by seanhunter

2 subcomments

   from itertools import cycle
   
   fizz = cycle(["","","fizz"])
   buzz = cycle(["","","","","buzz"])
   for z in zip(fizz,buzz):
      print(f"{z[0]}{z[1]}")

by sovnwnt

0 subcomment

> No disguised booleans

Do indexes count? If not, there's a simple one liner

    def fizzbuzz(n):
        return [str(n), "fizz", "buzz", "fizzbuzz"][1-min(1, n%3) + (1-min(1,n%5))*2]

by dullcrisp

0 subcomment

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\+$/\1'
Now I see it's the same solution as in the post.
Edit: Actually all you need is vim -es +'exec "norm! i\r\rFizz\r\rBuzz\rFizz\r\r\rFizz\rBuzz\r\rFizz\r\r\rFizzBuzz\<esc>Vggy7P101GdG"|%s/^$/\=line(".")/|%p|q!'

by MathMonkeyMan

0 subcomment

Here's my attempt:

    # Multi-pass FizzBuzz
    n = 100 
    # [['1'], ['2'], ['3'], ['4'], ['5'], ...]
    seq = [[str(i)] for i in range(1, n + 1)] 
    # [['1'], ['2'], ['3', 'Fizz'], ['4'], ['5'], ...]
    for i in range(3, n +  1, 3): 
        seq[i-1].append('Fizz')
    # [['1'], ['2'], ['3', 'Fizz'], ['4'], ['5', 'Buzz'], ..., ['15', ''Fizz', 'Buzz'], ...]
    for i in range(5, n + 1, 5): 
        seq[i-1].append('Buzz')
    # Arithmetic equivalent to:
    # len=1 -> the whole thing (from zero to end, because zero = -zero)
    # len=2 -> the length-1 suffix (just Fizz or Buzz)
    # len=3 -> the length-2 suffix (Fizz and Buzz)
    # The branch is hidden in the slice syntax:
    # Python has to check whether `x` is negative in `terms[x:]`. 
    for terms in seq:
        print(''.join(terms[-(len(terms) - 1):]))

Here's a version that uses generators instead of multiple passes over a list:

    # Single-pass FizzBuzz
    n = 100

    def numbers():
        for i in range(1, n+1):
            yield [str(i)]

    def fizzies():
        nums = numbers()
        try:
            while True:
                yield next(nums)
                yield next(nums)
                yield [*next(nums), 'Fizz']
        except StopIteration:
            pass

    def buzzies():
        fzs = fizzies()
        try:
            while True:
                yield next(fzs)
                yield next(fzs)
                yield next(fzs)
                yield next(fzs)
                yield [*next(fzs), 'Buzz']
        except StopIteration:
            pass

    for terms in buzzies():
        print(''.join(terms[-(len(terms) - 1):]))

Edit: Can't resist -- unbounded without loops, but recursion blows the call stack (granted, well after 100):

    def numbers(i=1):
        yield [str(i)]
        yield from numbers(i+1)
    
    def fizzies(source=numbers()):
        yield next(source)
        yield next(source)
        yield [*next(source), 'Fizz']
        yield from fizzies(source)
    
    def buzzies(source=fizzies()):
        yield next(source)
        yield next(source)
        yield next(source)
        yield next(source)
        yield [*next(source), 'Buzz']
        yield from buzzies(source)
    
    def main(source=buzzies()):
        terms = next(source)
        print(''.join(terms[1-len(terms):]))
        main(source)
    
    main()

by somat

3 subcomments

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 nether of which is present in python. perhaps a recursive solution(throws up a little in mouth)
baring that I too got nerd sniped by this and unsatisfied 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 elegant duel state engine was wasted) it turns out the problem solution could be as simple as
```
    f_out = ['', '', 'fizz']
    b_out = ['', '', '', '', 'buzz']
    
    def fizz_buz(n):
        return(f_out[n % 3] +  b_out[n % 5])
```
anyhow the rest of my clever but unneeded and useless enumeration system, remember to read the spec first.
```
    f_state = {
        0:1,
        1:2,
        2:0,
        }

    b_state = {
        0:1,
        1:2,
        2:3,
        3:4,
        4:0,
        }

   def fizz_buzz_all():
        f_index = 0
        b_index = 0
        while 1: #how to loop with no end check?
            print(f_out([f_index] + b_out[b_index] )
            f_index = f_state[f_index]
            b_index = b_state[b_index]
```
and the recursive solution:
```
    def fizz_buzz_recurse(n):
        print(fizz_buzz(n))
        fizz_buzz_recurse(n + 1)
```

by kstrauser

0 subcomment

How about:

  print(filter(None, [f + b, str(n)])[0])

Would that be not-Boolean enough?

by bluGill

0 subcomment

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.

by stop50

0 subcomment

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

by LanceH

1 subcomments

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])
   }
  }

by cestith

0 subcomment

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)
```

by qrian

0 subcomment

itertools.count is probably what OP is looking for

by Mikhail_Edoshin

1 subcomments

What is the canonical formulation of FizzBuzz?

by swisniewski

2 subcomments

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”.

by mapehe

0 subcomment

This is pretty cool, actually

by pbiggar

0 subcomment

Simple solution using modulo arithmetic and arrays. Relies on python shorthands that hides implicit branches for the number case though.

    def main():
      fizz_array = ["Fizz", "", ""]
      buzz_array = ["Buzz", "", "", "", ""]

      for n in range(1, 101):
        # Use modulo to index into arrays
        f = fizz_array[n % 3]
        b = buzz_array[n % 5]

        # Combine fizz and buzz
        result = f + b
    
        output = result + str(n)[len(result):]
    
        print(output)
    
    main()

I couldn't figure out this line and had to rely on AI for it.

    output = result + str(n)[len(result):]