[Haskell-beginners] any feedback on this solution

[Kees Bleijenberg]

Roelof,

Say, you find after a few months a bug or a better algorithm for toDigits,
then you have to remember that sumDigits uses this algorithm (in a bit
different form) too. In this case the functions are so small that it is not
a problem.

If you apply toDigits to all elements of a list of int's and  you get almost
the solution
apply toDigits to alle elements of [4,12,3] and you get [[4],[1,2],[3]],
which is a small step to [4,1,2,3]

Kees

-----Oorspronkelijk bericht-----
Van: Beginners [mailto:beginners-bounces at haskell.org] Namens Roelof Wobben
Verzonden: zaterdag 7 februari 2015 11:02
Aan: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell
Onderwerp: Re: [Haskell-beginners] any feedback on this solution

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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