[Haskell-beginners] Simplified Luhn Algorithm

[John Lusk]

Looks fine to me. Maybe drop the if-then, and simply return the result of
the == ? (Maybe not possible in Haskell (I'm just a duffer myself) but
extraneous trues and falses always drive me nuts.)

--
Sent from my tablet,  which has a funny keyboard. Makes me sound more curt
and muted than normal.

On Dec 30, 2017 11:03 PM, "trent shipley" <trent.shipley at gmail.com> wrote:

> I have the following, and it works, but I am trying teach myself Haskell,
> and I have the suspicion that my solutions is both inefficient and
> graceless. Any feedback would be appreciated.
>
> Trent.
>
> ------------------------------------
>
> {-
> 8.The Luhn algorithm is used to check bank card numbers for simple errors
> such as mistyping a digit, and proceeds as follows:
>
> * consider each digit as a separate number;
> * moving left, double every other number from the second last;
> * subtract 9 from each number that is now greater than 9;
> * add all the resulting numbers together;
> * if the total is divisible by 10, the card number is valid.
>
> Define a function luhnDouble :: Int -> Int that doubles a digit
> and subtracts 9 if the result is greater than 9.
>
> For example:
>
> > luhnDouble 3
> 6
>
> > luhnDouble 6
> 3
>
> Using luhnDouble and the integer remainder function mod, define a function
> luhn :: Int -> Int -> Int -> Int -> Bool
> that decides if a four-digit bank card number is valid.
>
> For example:
> > luhn 1 7 8 4
> True
>
> > luhn 4 7 8 3
> False
>
> In the exercises for chapter 7 we will consider a more general version of
> this function that accepts card numbers of any length.
>
> Hutton, Graham. Programming in Haskell (pp. 45-46). Cambridge University
> Press. Kindle Edition.
> -}
>
> luhnDouble :: Int -> Int
> luhnDouble x = if (2 * x) > 9
>     then (2 * x) - 9
>     else 2 * x
>
>
> luhn :: Int -> Int -> Int -> Int -> Bool
> luhn x1 x2 x3 x4 = if 0 == sum[luhnDouble x1, x2, luhnDouble x3, x4] `mod`
> 10
>     then True
>     else False
>
>
>
>
>
>
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