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Today's Topics:
1. Re: Simplified Luhn Algorithm (trent shipley)
----------------------------------------------------------------------
Message: 1
Date: Tue, 02 Jan 2018 08:48:34 +0000
From: trent shipley <trent.ship...@gmail.com>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <beginners@haskell.org>
Subject: Re: [Haskell-beginners] Simplified Luhn Algorithm
Message-ID:
<CAEFLyb+TR88kmoKsc=y8w4fyprocuuik8-v1ql94yeo4d9z...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
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 <rpglove...@gmail.com>
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 <johnlu...@gmail.com> 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" <trent.ship...@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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