I’ve been playing around with problem 31 of Project Euler which is defined as follows:

In England the currency is made up of pound, £, and pence, p, and there are eight coins in general circulation: 1p, 2p, 5p, 10p, 20p, 50p, £1 (100p) and £2 (200p). It is possible to make £2 in the following way: 1 £1 + 150p + 220p + 15p + 12p + 31p How many different ways can £2 be made using any number of coins?

Having coded way too much in Java my first thought was that the coins could be represented as an Enum but I wasn’t sure how to do that in Haskell.

As it turns out the question has been asked on Stack Overflow and there’s an Enum type class we can implement to achieve the same thing.

In this case:

``````
data UKCoin = OnePence | TwoPence | FivePence | TenPence | TwentyPence | FiftyPence | OnePound | TwoPounds
deriving (Eq, Show)
``````
``````
instance Enum UKCoin where
fromEnum = fromJust . flip lookup table
toEnum = fromJust . flip lookup (map swap table)
table = [(OnePence, 1), (TwoPence, 2), (FivePence, 5), (TenPence, 10), (TwentyPence, 20),
(FiftyPence, 50), (OnePound, 100), (TwoPounds, 200)]
``````

If we want to get the monetary value of a coin we use ‘fromEnum’:

``````
200
``````

And if we wanted to go back to the enum value we’d do this:

``````
> (toEnum 200) :: UKCoin
TwoPounds
``````

If we want to get a list of all the instances of ‘UKCoin’ we can just do a map over ‘table’:

``````
> let ukCoins = map fst table
> ukCoins
[OnePence,TwoPence,FivePence,TenPence,TwentyPence,FiftyPence,OnePound,TwoPounds]
``````