FizzBuzz in One Line

2013.02.03

Over the past month, I’ve relocated to San Francisco and started as part of the spring cohort at DevBootcamp. As a result, I have had less time to add new posts here.

Among the many highlights of last week at DBC, I learned a one-line implementation of FizzBuzz in Ruby. It’s a clever piece of code and worth taking apart.

First, to review, the requirements of the classic FizzBuzz program are as follows: print out the numbers one to one hundred. If a number is divisible by three, print out “Fizz” instead of the number. Likewise, if a number is divisible by five, print out “Buzz.” Finally, if a number is divisible by both three and five, print out “FizzBuzz.”

Here’s how to do all that using just one line in Ruby.

puts (1..100).map { |i| (fb = [["Fizz"][i % 3], ["Buzz"][i % 5]].compact.join).empty? ? i : fb }

Naturally, the code above is far from transparent. Let’s take it apart piece by piece.

We start by calling puts on an array, which will be the return value of the map function. The nice thing about puts is that it will iterate over an array, printing an element and then a new line each time.

Next comes the hairy part. We call map on one through one hundred. For each number, we assign a particular value to the fb local variable. Understanding that particular value is at the heart of this one-liner.

Let’s start by assuming that we’ve passed three into the block, represented by i. The expression i % 3 becomes 3 % 3 which evaulates to zero. Therefore, we’ll have ["Fizz"][0]. The zeroeth index of a one member array is the first member, and so the result will be "Fizz". Second, the expression i % 5 will become 3 % 5 which will evaluate to 3. And this is the most important trick. In Ruby, when we ask for an index on an array which is larger than the array, e.g., ["a", "b", "c"][10], we won’t get an error. Instead, we’ll simply get nil. As a result, the expression ["Buzz"][3 % 5] will become ["Buzz"][5] which will finally become simply nil.

So, when i is three, the resulting array will be:

# step one
[["Fizz"][3 % 3], ["Buzz"][3 % 5]]
# step two
[["Fizz"][0], ["Buzz"][5]]
# step three
["Fizz", nil]

In other words, every time we have a number which is evenly divisible by three, i.e., i % 3 == 0, we’ll get a "Fizz" in our array. And every time we get a number which is evenly divisible by five, i.e., i % 5 == 0, we’ll get a "Buzz". If both conditions are satisfied, e.g., as with 15, we’ll get an array with both "Fizz" and "Buzz", e.g., ["Fizz", "Buzz"].

With that down, we simply call compact on the resulting array, removing any nil members, and join the two members together into a string. If we have a "Fizz" and a nil, the join method is smart enough to drop the nil, and just leave "Fizz". A "Fizz" and a "Buzz" will become "FizzBuzz. If we have two nil members, join will return an empty string.

We then assign the result of compact and join to a variable local to the block, i.e., fb.

The rest of the line is simply a ternary expression. If our fb string is an empty string, we return i, otherwise we return whatever has been assigned to fb.

And that is the FizzBuzz problem in one line. Just please don’t write code like this in production.