[Haskell-beginners] any feedback on this solution

[Joel Williamson]

Roelof,

It doesn't look like you ever call toDigitsRev. To keep the solution
clearer, you should remove any unneeded definitions.

Joel

On Sat, 7 Feb 2015 07:50 Kees Bleijenberg <K.Bleijenberg at lijbrandt.nl>
wrote:

> 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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> >
> >
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