FizzBuzz

So, I ran across the "FizzBuzz" interview question in a geek article at Los Techies:

http://lostechies.com/jimmybogard/2013/01/29/fizzbuzz-is-dead-long-live-fizzbuzz

By the time I got tired of obsessing over it, I had eight distinct JavaScript solutions (see http://pastebin.com/bsKvNzMD ). Of course, this was all hacked out in the comfort of my favorite overstuffed armchair at home. Here's my pick of the litter.

	function FizzBuzz(fizz, buzz, fizzNum, buzzNum, last) {
		var msgs = [0, fizz, buzz, fizz + buzz];
		var rpt = new Array();
		for (var i = 1; i <= last; i++) {
			msgs[0] = "" + i;
			rpt.push(msgs[((0 == i%fizzNum)?1:0) + ((0 == i%buzzNum)?2:0)]);
		}
		return rpt;
	};

The "standard answer" is given by the call:

	FizzBuzz("Fizz", "Buzz", 3, 5, 100)

The expression

  ((0 == i%fizzNum)?1:0) + ((0 == i%buzzNum)?2:0)

implements the following partition of the decision universe and the corresponding numerical results:

      Fizz          Buzz
   i divisible   i divisible
   by fizzNum    by buzzNum          Fizz + Buzz
   -----------   -----------   ------------------------
     F => 0        F => 0       0 + 0 = 0  =>   "" + i
     T => 1        F => 0       1 + 0 = 1  =>    fizz
     F => 0        T => 2       0 + 2 = 2  =>    buzz
     T => 1        T => 2       1 + 2 = 3  => fizz + buzz


Prob'ly not the answer I'd have produced during a job interview, since this version is a refinement of the fifth solution generated during my geek sprint.

Ah, OCD! Okay, one more generalization of the FizzBuzz notion...

	function Fuzz(fuzzes, fuzzNums, last) {
		var msg;
		var rpt = new Array();
		var lastFuzz = fuzzes.length;
		for (var i = 1; i <= last; i++) {
			msg = "";
			for (var j = 0; j < lastFuzz; j++) {
				if (0 == i%fuzzNums[j]) msg += fuzzes[j];
			}
			rpt.push(("" == msg)?"" + i:msg);
		}
		return rpt;
	};

Classical FizzBuzz is:

	Fuzz( ["Fizz", "Buzz"], [3, 5], 100 )

For wackier fun, try

	Fuzz( ["Fizz", "Buzz", "Your", "Mama"], [3, 5, 7, 11], 3*5*7*11 )