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Unconditional Challenge: FizzBuzz without `if`

Nathan Kallman on June 15, 2020

As the title says, make a classic FizzBuzz function or method without using if/else (or equivalents like ternaries, ?:, sneaky). Specifically: T...

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Heiker • Jun 16 '20 • Edited

I'm sure I could make a fancy function composition but my brain is satisfied with this.

const Troo = (iff, elz) => iff;
const Falz = (iff, elz) => elz;

const If  = (boolz, iff, elz) => boolz(iff, elz);

const Boolz = {
  from_bool: bool => [Falz, Troo][Number(bool)],
};

const is_fizz = n => Boolz.from_bool(n % 3 === 0);
const is_buzz = n => Boolz.from_bool(n % 5 === 0);

const fizzbuzz = n => If(
  is_fizz(n),
  If(is_buzz(n), "FizzBuzz", "Fizz"),
  If(is_buzz(n), "Buzz", n)
);

const range = (end) => Array.from({ length: end }, (val, index) => index + 1);

console.log(
  range(15)
    .map(fizzbuzz)
    .join(", ")
);

Nathan Kallman • Jun 16 '20

Awesome! I think you're really getting at the heart of the prompt by defining true and false as functions.

Bashu Naimi-Roy • Jun 21 '20

this blew me away, thank you for posting it

Heiker • Jun 21 '20

Glad you like it. I got the idea from this talk by Anjana Vakil.

Bashu Naimi-Roy • Jun 21 '20

Aha! She has several functional programming in JavaScript videos I've been putting off for a while, I think this is my cue to watch one. I'd like to better understand your solution. Cheers!

JP Antunes • Jun 15 '20

Does this one count for the hard mode?

const fizzBuzz = n => {
    const mapper = (arr, modulo, txt) => arr
                                    .filter(e => e % modulo == 0)
                                    .forEach(e => arr[arr.indexOf(e)] = txt);
    let x = 1;
    const range = [...Array(n)].map(_ => x++)

    mapper(range, 15, 'FizzBuzz');
    mapper(range, 5, 'Buzz');
    mapper(range, 3, 'Fizz');

    return range.toString();
}

Nathan Kallman • Jun 15 '20

Nice solution! For hard mode, I think either do it without .filter or show how .filter could be implemented according to the rules.

JP Antunes • Jun 15 '20

How about with for loops?

const fizzBuzz = n => {
    let x = 1;
    const range = [...Array(n)].map(_ => x++);
    for (let i = 2; i <= n; i += 3) range[i] = 'Fizz';
    for (let i = 4; i <= n; i += 5) range[i] = 'Buzz';
    for (let i = 14; i <= n; i += 15) range[i] = 'FizzBuzz';
    return range.toString();
}

Nathan Kallman • Jun 15 '20

Nice! I'll accept it even though I said no for loops because I like that you used the stepping statement to simply go through the multiples of 3,5,15. That's creative!

I was more expecting someone to abuse the i <= n into the n % 3 type of check in a traditional fizz buzz. (especially abusing while(n % 3) into a more convoluted if)

JP Antunes • Jun 15 '20

I missed that! Sorry, but I was running out of ideas :-) Looking forward to seeing your solution(s)!

Nathan Kallman • Jun 19 '20

Happy Juneteenth! I love everyone's approaches here, it's fun to see the way each person thinks through the problem.

I'll be posting my answers on Monday (along with the next installment of With Only CSS) so encourage your friends to give this a shot over the weekend (or take another go yourself)!

Follow me and react with a 🔖 so you remember to come back and see what I post here.

Drop a 🦄 or ❤️ if you'd be interested in some follow up posts going more in depth into each (OO/FP) solution: how they work; how they are similar; and how they are different.

Thanks everyone who took the time to post a solution to my little challenge; I'll see you all on Monday!

Nathan Kallman • Jun 22 '20 • Edited

As promised, here is my functional approach to solving this (deeper article on it to come soon)

const functionalTrue = (onTrue, onFalse) => onTrue;
const functionalFalse = (onTrue, onFalse) => onFalse;
const isDivisible = (dividend, divisor) => [functionalTrue, ...Array(divisor).fill(functionalFalse)][dividend % divisor];

const functionalFizzBuzz = (n) => {
  const divisible_by_three = isDivisible(n, 3);
  const divisible_by_five = isDivisible(n, 5);
  return divisible_by_three(divisible_by_five("FizzBuzz", "Fizz"), divisible_by_five("Buzz", n));
};

Similar to Heiker's solution above; the main part of this solution is defining "true" and "false" as functions taking the same two parameters but returning one or the other.

Nathan Kallman • Jun 22 '20

And here is my object-oriented approach (deeper article on it to come soon)

const baseBoolean = {
  setThen: function(then) { return { ...this, then }; },
  setOtherwise: function(otherwise) { return { ...this, otherwise }; },
};

const objectOrientedTrue = {
  ...baseBoolean,
  evaluate: function() { return this.then; },
};

const objectOrientedFalse = {
  ...baseBoolean,
  evaluate: function() { return this.otherwise; },
};

const objectOrientedNumber = {
  value: 0,
  isaMultipleCache: [objectOrientedFalse],
  setValue: function(n) { return { ...this, value: n, isaMultipleCache: [objectOrientedTrue, ...Array(n).fill(objectOrientedFalse)] }; },
  isaMultipleOf: function(dividend) { return this.isaMultipleCache[dividend.value % this.value]; }
};

const objectOrientedFizzBuzz = {
  for: function(n) {
    const number = objectOrientedNumber.setValue(n);
    return this.three
               .isaMultipleOf(number)
               .setThen(
                 this.five
                     .isaMultipleOf(number)
                     .setThen("FizzBuzz")
                     .setOtherwise("Fizz")
                     .evaluate()
               )
               .setOtherwise(
                 this.five
                     .isaMultipleOf(number)
                     .setThen("Buzz")
                     .setOtherwise(number.value)
                     .evaluate()
               )
               .evaluate();
  },
  three: objectOrientedNumber.setValue(3),
  five: objectOrientedNumber.setValue(5),
};

Similarly to the functional approach; the main part of this solution is defining "true" and "false" as objects. But now instead of parameters, the objects have the same interface and return one or the other of the attributes set on the object.

Nathan Kallman • Jun 22 '20 • Edited

You can see both solutions in action on this codepen (planning one final post doing a compare/contrast between the two approaches)

Nathan Kallman • Jun 17 '20

I think your .get is a secret conditional. (It will return the value at the given key or the default IF the value is undefined)

If you can show how to implement .get without an if then I think this is a great answer!

Aldwin Vlasblom • Jun 15 '20

Do you consider something like this cheating, because it builds on filter, a built-in, to handle the conditional logic?

const isFizz = n => n % 3 === 0
const isBuzz = n => n % 5 === 0
const complement = f => x => !(f (x))

const fizzbuzz = n => {
  const fizz = [n].filter(isFizz).map(_ => 'Fizz')
  const buzz = [n].filter(isBuzz).map(_ => 'Buzz')
  const num = [n].filter(complement(isFizz)).filter(complement(isBuzz)).map(String)
  return `${fizz.join('')}${buzz.join('')}${num.join('')}`
}

Nathan Kallman • Jun 15 '20

Clever! I'll accept it for the normal challenge :)

If you want hard mode though, I think you'll have to show an implementation of .filter that doesn't use if, ?:, ||, &&, ?., or ??. (I'll allow .reduce to be used though, as well as .map)

Well done, thanks for commenting!

Aldwin Vlasblom • Jun 15 '20 • Edited

Okay, well, my next solution is probably not what you were expecting. It builds on the idea that a boolean can be used as a number. I think it meets the hardcore requirements.

const isFizz = n => n % 3 === 0
const isBuzz = n => n % 5 === 0
const isNeither = n => !isFizz(n) && !isBuzz(n)

// the magic:
const optional = (f, x) => Array(Number(f(x))).fill(x)

const fizzbuzz = n => {
  const fizz = optional(isFizz, n).map(_ => 'Fizz')
  const buzz = optional(isBuzz, n).map(_ => 'Buzz')
  const num = optional(isNeither, n).map(String)
  return `${fizz.join('')}${buzz.join('')}${num.join('')}`
}

Nathan Kallman • Jun 15 '20

Very nice! I think you have met the hardcore requirements.

And I think slightly less code than my functional solution will be, well done!

Mike Talbot ⭐ • Jun 19 '20 • Edited

function map(compare, say, next = v => v) {
    return function(value) {
        return [() => say, next][Math.sign(compare(value))](value)
    }
}
const process = map(
    v => (v % 3) + (v % 5),
    "fizzbuzz",
    map(v => v % 5, "buzz", map(v => v % 3, "fizz"))
)

for (let i = 1; i < 31; i++) {
    console.log(process(i))
}

Nathan Kallman • Jun 19 '20

Well done! Clever use of Math.sign to select either the first or second element from an array

Mike Talbot ⭐ • Jun 19 '20 • Edited

Slightly shorter without bothering to call out for variable potential conditions and using anons:


let map = (compare, say, next = v => v) => v => [() => say, next][Math.sign(v % compare)](v)
const process = map(15, "FizzBuzz", map(5, "Buzz", map(3, "Fizz")))

JP Antunes • Jun 17 '20

Still not perfect, but this one comes with a special thanks to Mr. Kevlin Henney

const fizzBuzz = n => {
    const isFizzBuzz = n => ( {false: '', true: 'Fizz'}[n % 3 == 0] 
                            + {false: '', true: 'Buzz'}[n % 5 == 0] 
                            || n.toString() );
    let x = 1;
    return [...Array(n)].map(_ => isFizzBuzz(x++)).toString();                             
}

Nathan Kallman • Jun 17 '20

Nice! I like how it uses false and true as keys on an object to select the resulting string.

If you can get rid of the || usage, then this will meet the hard mode requirements...

JP Antunes • Jun 17 '20 • Edited

Well Mr. Henney has a great answer:

const fizzBuzz = n => {
  const test = (d, s, x) => n % d == 0 ? _ => s + x('') : x;
  const fizz = x => test(3, 'Fizz', x);
  const buzz = x => test(5, 'Buzz', x);

  return fizz(buzz(x => x))(n.toString());
}

edit: but it still has a conditional...

Vidit Sarkar • Jun 15 '20

Thanks for the problem. I guess this will not satisfy your conditions, but here is my solution,

const fizzbuzz = n => (!(n%15) && "FizzBuzz") ||
                      (!(n%5) && "Buzz") ||
                      (!(n%3) && "Fizz") ||
                      n

Nathan Kallman • Jun 15 '20

It's a nice answer! It satisfies the normal rules (no if or ?:)

Thanks for taking the time to submit a solution!

Vidit Sarkar • Jun 17 '20

Thanks for the reply. I think re-submission is also allowed. So, here is a C++ solution, which again, may not satisfy all your conditions.

string fizzbuzz(int n){
    string num_string = to_string(n);

    string all_options[5][3] = {
        {"FizzBuzz" ,"Buzz" , "Buzz"},
        {"Fizz" , num_string, num_string},
        {"Fizz" , num_string, num_string},
        {"Fizz" , num_string, num_string},
        {"Fizz" , num_string, num_string}
    };

    return all_options[n%5][n%3];
}

Nathan Kallman • Jun 17 '20

Well done! I think you've got a solution

Eric Nijman • Jun 23 '20

This could probably get simplified (I've seen the one-liner of this one), but I still wanted to post mine

const fbMap = [
  () => 'FizzBuzz',
  String,
  String,
  () => 'Fizz',
  String,
  () => 'Buzz',
  () => 'Fizz',
  String,
  String,
  () => 'Fizz',
  () => 'Buzz',
  String,
  () => 'Fizz',
  String,
  String,
];
const fizzBuzz = x => fbMap[x % fbMap.length](x);

Bashu Naimi-Roy • Jun 21 '20

holy COW did I find out a lot of things about object destructuring while making this. Thanks for the challenge, Nathan. It was so great to notice how instinctively I reach for a conditional.

Nathan Kallman • Jun 21 '20

I'm glad you liked it! I think its good to do a little self reflection on the way we do things every once in a while.

And WOW is that an impressive use of destructuring in JS. I knew about each of those things independently; I never have thought about putting them all together like that!

Danny Engelman • Mar 27 '22

A bit late to the party...

Array(100)
  .fill((x, div, label) => x % div ? "" : label) //store Function in every index
  .map((func, idx) =>
    func(++idx, 3, "Fizz") + func(idx, 5, "Buzz") || idx
  );

Nathan Kallman • Mar 27 '22

Welcome to the party!

Bashu Naimi-Roy • Jun 23 '20 • Edited

I am posting for a friend who found some absolutely mind blowing ideas.

First a one-liner that extends my destructuring idea.

const fn = n => ({ ['2xx0x10xx01x0'[n % 15]]: result = n } = ['Fizz', 'Buzz', 'FizzBuzz'], result);

secondly a solution built around a regex from hell/heaven

let fn = n => String(n).replace(/^(?=.*[05]$)(?:[0369]|[147](?:[0369]|[147][0369]*(?:[258]|[147][0369]*[147]))*(?:[258]|[147][0369]*[147])|[258](?:[258]|[258][0369]*(?:[147]|[258][0369]*[258]))*(?:[147]|[258][0369]*[258]))*$/, 'FizzBuzz').replace(/^(?:[0369]|[147](?:[0369]|[147][0369]*(?:[258]|[147][0369]*[147]))*(?:[258]|[147][0369]*[147])|[258](?:[258]|[258][0369]*(?:[147]|[258][0369]*[258]))*(?:[147]|[258][0369]*[258]))*$/, 'Fizz').replace(/^\d*[05]$/, 'Buzz');

Aldwin Vlasblom • Jun 17 '20

You are referring to the following clause:

Do this without "secret" conditionals like || and && (in loosely typed languages like JavaScript)

My understanding is that this applies to the usage of such operators as conditionals when they are applied to non-boolean types. For example:

const fizz = n % 3 === 0 && 'Fizz' || String(n)

In my code, it is used strictly on Boolean values, which I don't think was against the rules.

Nathan Kallman • Jun 17 '20

It uses it in a strictly boolean sense, which I'll allow. What isn't allowed is the loose/early-return way JS can use &&; a contrived example that would not pass hard mode:

const fizzbuzz = n => (n % 3 && (n % 5 && n || "Buzz")) || (n % 5 && "Fizz" || "FizzBuzz")

(an easy litmus test: does changing the order of the && expression change the result?)

Jeremy Jacob Anderson • Oct 16 '20 • Edited

Has someone done this yet? It's really very fast.
There's a ternary there, I admit.

function fizzbuzz(i) {
  let test = (d, s, x) => i % d == 0 ? _ => s + x('') : x
  let fizz = x => test(3, 'Fizz', x)
  let buzz = x => test(5, 'Buzz', x)
  return fizz(buzz(x=>x))(i.toString())
}

Array.apply(0, Array(100)).map(function (_, i) {
  i += 1
  return fizzbuzz(i)
}).join("\n")

Nathan Kallman • Oct 16 '20

Nice! But test uses a ternary ?: (which is just a different way to spell if in my opinion)

If you can get rid of that I think you'll have a solution!

Jeremy Jacob Anderson • Oct 16 '20 • Edited

I can't bring myself to be terribly interested in re-implementing language features like boolean or ternary. I loved seeing some of the really outlandish solutions like that un-sane RegEx that was posted. I enjoyed seeing the contortions done to avoid conditionals, too, but that's not something I'm into for it's own sake. I'm happy to fail at it.

There is a meaningful distinction between ternary and if.

if is a wilderness with few boundaries, as a form of Many-Valued Logic with an arbitrary number of values. Ternary is a specific subset of Many-Valued Logic; Three-Valued Logic. The Three-Valued Logic of ternary operators provides specific limitations that are not necessarily implied by the N-Valued Logic of the if statement. In most programming languages ternary is an operator, and if is a control structure block. Operators can be used in expressions, which is generally not true of if statements. Ternary is usually slightly computationally more expensive than if but has some advantages in terms of limiting complexity.

In real-world code I'm going to write an if (without else if or else if I can help it) about 95% of the time. It is both more readable and more performant in most cases. But when I need an expression that can make a decision between boolean and a maybe, I reach for ternary.

If you wanted to argue that switch is basically goto but potentially worse due to unintended fallthrough, then we'd have something to agree on.

The magic of the fizzbuzz problem is, of course, that it helps us understand in more granular detail that our design decisions, and design non-decisions, come with unavoidable trade-offs. It is a meditation on (a horrible term) non-functional requirements.

Nathan Kallman • Aug 13 '21

Very nice! I feel like this is actually pretty similar to my own Object-oriented approach (if you wanted to read):

Unconditional Fizzbuzz: an Object-oriented approach

Nathan Kallman ・ Jul 6 '20 ・ 3 min read

#oop #computerscience #challengeanswer

Alfrederson • Jul 23 '20

How about this:

Nathan Kallman • Jul 23 '20

Nice!

Nathan Kallman • Jun 17 '20

Excellent! I think you've checked off all the boxes. Thanks for submitting!

Bashu Naimi-Roy • Jun 21 '20

I'm thinking of an absolutely egregious version using try and catch, but I'm going to go ahead and assume that try/catch would be a "surrogate conditional"

Nathan Kallman • Jun 21 '20

😂 it probably is, but now I want to see it implemented with just try/catch!