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@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 

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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