Credit Card Validation

creditcard.hs

toDigits :: Integer -> [Integer]
toDigits 0 = []
toDigits x = toDigits (div x 10) ++ [(mod x 10)]

toDigitsRev :: Integer -> [Integer]
toDigitsRev 0 = []
toDigitsRev x = (mod x 10) : toDigitsRev (div x 10)

doubleEveryOther :: [Integer] -> [Integer]
doubleEveryOther =  (reverse . (zipWith ($) (cycle [id, (*2)])) . reverse)

sumDigits :: [Integer] -> Integer
sumDigits xs = sum (concatMap toDigits xs)

validate :: Integer -> Bool
validate x = (mod ((sumDigits . doubleEveryOther . toDigits) x) 10) == 0

My doubleEveryOther is pretty naive with the double reverse but I figured that since we know n is always small it doesn't really matter.

I also had the doubled up reverse, I wonder what the trick might be to avoid that.

This is one way to avoid double-reverse; due to the right-associative nature of foldr, the initial argument will pair with the rightmost element of the list. We can pair each argument with an extra value (isomorphic to Boolean, if you don't want to create a new type for it) to keep track of whether we should double or not. At the end of the recursion, simply discard the parity value.

data Parity = Even | Odd
            deriving (Eq, Show)

doubleEveryOther = snd . foldr place init
  where
    init = (Even, [])
    place x (Even, xs) = (Odd , x:xs)
    place x (Odd,  xs) = (Even, (x*2):xs)

You can write it a bit more succinctly, but I prefer clarity over succinctness when they conflict.