[Haskell-beginners] any feedback on this solution
Roelof Wobben
r.wobben at home.nl
Sat Feb 7 10:01:55 UTC 2015
Why,
They have opposite terms.
SumDigits takes a array and has a integer as output.
ToDigits takes a integer and puts out a array.
Roelof
Kees Bleijenberg schreef op 7-2-2015 om 10:56:
> Roelof,
>
> Maybe it's better to reuse toDigits in sumDigits.
>
> Kees
>
> -----Oorspronkelijk bericht-----
> Van: Beginners [mailto:beginners-bounces at haskell.org] Namens Roelof Wobben
> Verzonden: zaterdag 7 februari 2015 10:37
> Aan: The Haskell-Beginners Mailing List - Discussion of primarily
> beginner-level topics related to Haskell
> Onderwerp: [Haskell-beginners] any feedback on this solution
>
> Hello,
>
> I finally solved exercise 1 where I had to write a programm which checks
> if a creditcard number is valid.
>
> I solved it this way :
>
> toDigits :: Integer -> [Integer]
> toDigits n
> | n < 0 = []
> | n < 10 = [n]
> | otherwise = toDigits (n `div` 10) ++ [n `mod` 10]
>
> -- | convert a number to a reversed array where a negative number will
> be a empty array
> toDigitsRev :: Integer -> [Integer]
> toDigitsRev 0 = [0]
> toDigitsRev n
> | n < 0 = []
> | n < 10 = [n]
> | otherwise = n `mod` 10 : toDigitsRev (n `div` 10)
>
> -- | Doubles every second number from the right.
> doubleEveryOther :: [Integer] -> [Integer]
> doubleEveryOther [] = []
> doubleEveryOther (x:[]) = [x]
> doubleEveryOther (x:(y:zs))
> | length (x:(y:zs)) `mod` 2 /= 0 = [x] ++ (y * 2) : doubleEveryOther zs
> | otherwise = [x *2] ++ y : doubleEveryOther zs
>
>
> -- | sum all the digits of a array
> sumDigits :: [Integer] -> Integer
> sumDigits [] = 0
> sumDigits (x:zs)
> | x < 10 = x + sumDigits zs
> | otherwise = x `mod` 10 + x `div` 10 + sumDigits zs
>
>
> -- | validate a number by looking if a number can be divided by 10
> validate :: Integer -> Bool
> validate n = sumDigits(doubleEveryOther(toDigits(n))) `mod` 10 == 0
>
>
> -- | The main entry point.
> main :: IO ()
> main = do
> print $ validate 4012888888881881
>
> Any remarks about this solution.
>
> I know there are higher function solutions but that part is not
> descrived in chapter 1 .
>
> Roelof
>
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