Simplified Luhn Algorithm

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Simplified Luhn Algorithm

trent shipley
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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Re: Simplified Luhn Algorithm

Patrik Iselind
I guess your issue is how to represent the card numbers of arbitrary length, or? Wouldn't a list work?

Patrik

Den 31 dec 2017 05:03 skrev "trent shipley" <[hidden email]>:
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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http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners



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Re: Simplified Luhn Algorithm

John Lusk-2
In reply to this post by trent shipley
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" <[hidden email]> 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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http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners


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Re: Simplified Luhn Algorithm

Alex Rozenshteyn
Haskell can totally return the result of the "==", and that would be one of my suggestions as well. The other suggestion is for "luhnDouble": I would just compute `rem (2 * x) 9`, but if you need to explicitly subtract, you can do `let d = 2 * x in if d > 9 then d - 9 else d`, which does the computation just once.

On Sun, Dec 31, 2017 at 9:12 AM John Lusk <[hidden email]> wrote:
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" <[hidden email]> 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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Beginners mailing list
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[hidden email]
http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners

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Re: Simplified Luhn Algorithm

trent shipley

luhnDouble = (2*x) `mod` 9   almost works, but produces 0 for 2*9 when the answer should be 9.

So.

luhnDouble :: Int -> Int
luhnDouble x = let y = (2 * x) in if y >  9
    then y - 9
    else y

(Note, at this point in the book, the "let" trick has not been introduced.)

luhn :: Int -> Int -> Int -> Int -> Bool
luhn x1 x2 x3 x4 = 0 == sum[luhnDouble x1, x2, luhnDouble x3, x4] `mod` 10

(This suggested modification works wonderfully.)

A solution for a list of arbitrary length awaits chapter 7, this question was from chapter 4.  Note, Wikipedia.en implies that originally the digits of the multiplication result were to be added, as in 8 * 2 = 16 = 1 + 6 = 7  OR 16 - 9 = 7. 

On Sun, Dec 31, 2017 at 8:09 AM Alex Rozenshteyn <[hidden email]> wrote:
Haskell can totally return the result of the "==", and that would be one of my suggestions as well. The other suggestion is for "luhnDouble": I would just compute `rem (2 * x) 9`, but if you need to explicitly subtract, you can do `let d = 2 * x in if d > 9 then d - 9 else d`, which does the computation just once.


On Sun, Dec 31, 2017 at 9:12 AM John Lusk <[hidden email]> wrote:
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" <[hidden email]> 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



      


_______________________________________________
Beginners mailing list
[hidden email]
http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners

_______________________________________________
Beginners mailing list
[hidden email]
http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
_______________________________________________
Beginners mailing list
[hidden email]
http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners

_______________________________________________
Beginners mailing list
[hidden email]
http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners