bark­madley

thinker, hacker and other things

fun with fizzbuzz and haskells monoids

2009-01-22

In this short article I will be exploring FizzBuzz, and showing how the monoidal approach helps combine the logic of the problem in a reasonable way when compared to the naive solution.

The problem of FizzBuzz

The FizzBuzz problem has been written about extensively many times before, most commonly as a way to exspanss the failure of the majority of programmers applying for jobs to grok it.

The problem itself is as follows

Write a program that prints the numbers from 1 to 100.
But for multiples of three print "Fizz" instead of the
number and for the multiples of five print "Buzz".
For numbers which are multiples of both three and five
print "FizzBuzz".

Naive Haskell implementation of FizzBuzz

To attack the problem using Haskell, I split the problem into a purely function one, with a main that does the heavy IO lifting. This took me all of 2 seconds to write down, especially the if exspanssions.

fizz, buzz :: Num a => a -> Bool
fizz n = n `mod` 3 == 0
buzz n = n `mod` 5 == 0
fizzbuzz :: Num a => a -> String
fizzbuzz n = if fizz n && buzz n then "FizzBuzz" else
    if fizz n then "Fizz" else
        if buzz n then "Buzz" else show n
main :: IO ()
main = mapM (putStrLn . fizzbuzz) [1..100] >> return ()

The main problem I have with this implementation is that it doesn't compose the fizz and buzz functions very well into a coherent fizzbuzz. It is very imperative, in that it checks a, then b, then c etc. If you think that this approach will suffice then you don't have to continue reading, but I do have more to say.

Monoids

Monoids have been getting a lot of blog attention very recently because of an article Brian Hurt wrote about his thoughts on Haskell, which included a contraversial statement about monoids, so everyone and their mum decided to write about them. I am adding to the hysteria even though it is a dead horse, except that my goal is merely to solidify their usefulness in my mind by applying monoids to a not so practical application.

So what is a monoid? Briefly a monoid is a way to accumulate results together into a whole result. Any data type that wants to become a monoid must support two operations, empty and append, and follow 3 laws (not strictly speaking but it is prefered), these are written about here.

The reason I thought of fizzbuzz and monoids in the same thought was the realisation that the fizzbuzz function should be a composition of the fizz and the buzz function, and theoretically you should be able to compose many other modulous computations that result in strings.

Monoidal FizzBuzz

So how would you go about composing the fizzbuzz problem? First we need to identify the sub problems. One of the things that makes the naive solution both uglier and less efficient is the fact that it potentially invokes 4 condition functions, where we have only defined 2. i.e. fizz and buzz could be called twice.

What I would like to say is that the functions fizz and buzz return something significant, and that we want the composition of those two results to be our final result. In this case significance is denoted by whether or not a number is divisible by 3 or 5, and the words Fizz and Buzz which sometimes need to be composed to be FizzBuzz. So there are 4 cases then, FizzBuzz, Fizz, Buzz and Nothing. Haskell already has a data type that encapsulates Nothing so lets look at the Maybe data type.

data Maybe a = Just a | Nothing

What I got to thinking was that we can use the Maybe type to hold either Fizz, Buzz, or FizzBuzz, and if the value is Nothing, we should show the number.

My first stab at the new fizzbuzz function might look like this

fizz n = if n`mod`3 == 0 then Just "Fizz" else Nothing
buzz n = if n`mod`5 == 0 then Just "Buzz" else Nothing
fizzbuzz n = fizz n `mappend` buzz n

This is made possible by the monoidal instance of the Maybe and String (list) types.

instance Monoid a => Monoid (Maybe a) where
    mempty = Nothing
    mappend (Just x) (Just y) = Just (x `mappend` y)
    mappend Nothing x = x
    mappend x Nothing = x
instance Monoid [a] where
    mempty = []
    mappend = (++)

Note that the Maybe instance requires that the data that Maybe encapsulates needs to also be an instance of monoid, which is why the list instance is required.

Finally I needed some way to use the number instead of the result from fizzbuzz, and the maybe function provided the answer.

maybe :: b -> (a -> Maybe b) -> a -> b

The maybe function encodes the idea of using a function that returns a maybe type, with a default value incase it happens to return the value Nothing.

The Final Solution

So the final code looks like this

fizz, buzz :: Num a => a -> Maybe String
fizz n = if n `mod` 3 == 0 then Just "Fizz" else Nothing
buzz n = if n `mod` 5 == 0 then Just "Buzz" else Nothing

fizzbuzz :: Num a => a -> Maybe String
fizzbuzz n = mconcat [fizz n, buzz n]

applyfizzbuzz :: Num a => a -> String
applyfizzbuzz n = maybe (show n) id (fizzbuzz n)

main :: IO ()
main = mapM (putStrLn . applyfizzbuzz) [1..100] >> return ()

Where mconcat is defined as a sequential application of mappend and the next element in the list. Personally I think that this code is easier to read than the original version, while at the same time making it easy to add new conditions, for example if the problem was extended to include a check for numbers that were divisible by 7 (and print "Elfs"), the code would only been to be changed on two lines like so:

elfs n = if n `mod` 7 == 0 then Just "Elfs" else Nothing
fizzbuzz n = mconcat [fizz n, buzz n, elfs n]

Finally a note about point free style. I recently noticed that functions of type a to b are instances of monoids if b is an instance of monoid. Since out functions fizz, buzz and elfs all return Maybe Monoids, then we can simply mconcat the functions together and leave out the parameters as follows.

fizzbuzz' = mconcat [fizz, buzz, elfs]