Fizz Buzz without conditionals or booleans

https://evanhahn.com/fizz-buzz-without-conditionals-or-booleans/

by ingve • 8 months ago

58 68 8 months ago

stop50• 8 months ago

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]}")

floxy• 8 months ago

Very nice.

kiratp3• 8 months ago

A for loop has an implicit conditional in its stop condition check.

kstrauser3• 8 months ago

I could see that both ways. Python’s for loops are different than, say, C’s, in that they always consume an iterator. The implementation is that it calls next(iter) until it raises a StopIteration exception, but you could argue that’s just an implementation detail and not cheating.

If you wanted to be more general, you could use map() to apply the function to every member of the iterator, and implementation details aside, that feels solidly in the spirit of the challenge.

brudgers2• 8 months ago

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

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

hyperhello• 8 months ago

What’s the point of the game without booleans?

LanceH2• 8 months ago

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])
   }
  }

frosting1337• 8 months ago

There's a conditional, though?

albedoa• 8 months ago

Yeah this is weird. It's against the rules to speculate about whether a commenter read the article, but what about the title of the article?

bluGill• 8 months ago

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.

dullcrisp• 8 months ago

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!'

HWR_14• 8 months ago

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

gridspy• 8 months ago

I'd imagine that as any variable that only has two states.

nucleogenesis• 8 months ago

I think it was more about doing it without a Boolean-based branch construct like a ternary or switch or whatever flavor of thing that abstracts away the explicit checks for true/false by other means. Idk though for sure

kqr2• 8 months ago

Something like this: https://entropicthoughts.com/fizzbuzz-through-monoids

swisniewski2• 8 months ago

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”.

swisniewski2• 8 months ago

Not sure why this got downvoted.

The technique could be implemented without conditionals, but not in python, and not using iterators.

You could do it in C, and use & and ~ to make the cyclic counters work.

But, like I mentioned, the code in the article is very far from being free of conditionals.

abustamam• 8 months ago

I didn't down vote, but it does seem like unnecessary pedantry. Maybe it could be better phrased as "without writing any conditionals"

kstrauser3• 8 months ago

I think that’s kind of vacuously true. Like, good luck writing this in any language where the resulting assembler all the way at the bottom of the runtime has zero branch operations. And I bet even then that most CPUs’ microcode or superscalar engine would have conditionals underlying the opcodes.

I’d settle for just not writing conditionals in the user’s own code. Range doesn’t have to be implemented with branches. Hypothetically, Python could prefill a long list of ints, and range could return the appropriate slice of it. That’d be goofy, of course, but the main idea is that the user doesn’t know or really care exactly how range() was written and optimized.

somat2• 8 months ago

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)

soxfox42• 8 months ago

That solution fails for any value that is a multiple of neither 3 nor 5. In those cases, the result should be the original number.

somat2• 8 months ago

Sigh, Even after I reread the spec... I did not in fact read the spec. complete failure on my part.

drdeca• 8 months ago

You can replace `while 1:` with `for x in iter(int, 1):` .

jeremysalwen• 8 months ago

Make it throw an exception with an index out of bounds to terminate the loop.

cestith• 8 months ago

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)

userbinator• 8 months ago

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>")

moritzwarhier• 8 months ago

  prompt("please enter the expected output of a solution to the fizz buzz problem, without any further enclosing text or explanations; especially: do not enter program code")

should work in a browser JS environment (with layer 8) and with modern AI

kiratp3• 8 months ago

A for loop has a conditional in it.

Unless by conditionals we mean “no if/else” and not “no branch instructions”.

WhyNotHugo• 8 months ago

The conditional here only makes it stop when it reaches 100. The solution can be adapted to use a while loop if you’re okay with it running indefinitely.

kiratp3• 8 months ago

A loop either never halts or has a conditional. I guess a compiler could elide a “while True:” to a branch-less jump instruction.

One hack would be to use recursion and let stack exhaustion stop you.

brudgers2• 8 months ago

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]

kqr2• 8 months ago

How is this without conditionals?

carlmr• 8 months ago

Bin ifTrue

kstrauser3• 8 months ago

How about:

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

Would that be not-Boolean enough?

MathMonkeyMan• 8 months ago

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()

andriamanitra• 8 months ago

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...

mapehe• 8 months ago

This is pretty cool, actually

sovnwnt• 8 months ago

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]

Mikhail_Edoshin• 8 months ago

What is the canonical formulation of FizzBuzz?

genewitch3• 8 months ago

i've seen worse than this sort of thing:

https://github.com/EnterpriseQualityCoding/FizzBuzzEnterpris...

ramenprofitable• 8 months ago

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;
  }

jswelker4• 8 months ago

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. :(

genewitch3• 8 months ago

does not "or" imply a boolean

jswelker4• 8 months ago

It does.Will you accept max?

  fizzbuzz = ["", "", "Fizz", "", "Buzz", "Fizz", "", "", "Fizz", "Buzz", "", "Fizz", "", "", "FizzBuzz"]

  for i in range(1,101):
    print(max((fizzbuzz[(i-1)% 15],str(i),)))

Probably still bools under the hood.

susam• 8 months ago

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)])

nik9000• 8 months ago

As it should be.

pbiggar• 8 months ago

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):]

seanhunter3• 8 months ago

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

bjoli• 8 months ago

When I find a problem I can solve with ciclical structures I always feel like Da Vinci.

Fizzbuzz is probably the perfect fit, which is sad since it is a synthetic problem. I did find one problem once where I could save about 30s of user waiting time by generating (had to be done dynamically) wheels in the 3-10k element range.

throwaway1503• 8 months ago

Doesn't look right. The numbers are missing from the output. If you don't print the numbers, sure it's trivial.

seanhunter3• 8 months ago

I didn't know that was a requirement. OK then

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

Happy now?

qrian• 8 months ago

itertools.count is probably what OP is looking for

Fizz Buzz without conditionals or booleans

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