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Calculating Luhn-algorithm checksum digit

Submitted by: @import:stackexchange-codereview··
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checksumalgorithmcalculatingdigitluhn

Problem

Today I decided to learn some basic Haskell, and for starters I made a program for calculating the checksum of a Swedish personal identification number. It uses the Luhn-algorithm, aka. IBM MOD-10.

Explanation of this code can be found on Swedish Wikipedia and English Wikipedia

Here's a description of how the algorithm works:

Given a string of 9 digits, abcdefghi compute:

array = [a*2, b, c*2, d, e*2, f, g*2, h, i*2]


Then you computer the sum of the digits in this array, so for example if a*2 is 16, then that part counts as 1 + 6.

Then the result is how much more you have to add to make it evenly divisible by 10. For example, if sum == 54 then the result is 6 as 60 - 54 = 6.

import Data.Char

sumOfChars :: String -> Int
sumOfChars "" = 0
sumOfChars str = digitToInt(str !! 0) + sumOfChars(tail str)

twoMult :: Char -> String
twoMult c = show (digitToInt(c) * 2)

identificationSum :: String -> Int
identificationSum "" = 0
identificationSum str = if length str `mod` 2 == 1 then sumOfChars(twoMult(str !! 0)) + identificationSum(tail str)
  else digitToInt(str !! 0) + identificationSum(tail str)

remainingToTens :: Int -> Int
remainingToTens x = ceiling(fromIntegral x / 10) * 10 - x

determineLastDigit :: String -> Int
determineLastDigit str = remainingToTens(identificationSum(str))


To test the code:

determineLastDigit("811228987")


Prints:

4


As this is the first time ever I managed to make something in Haskell, any comments are welcome. Even though it is the first time though, feel free to rip my code apart as much as you would like and suggest any advanced things, I am always eager to learn.

Solution

This is a good start! I'll do two passes over your code, one to address issues of style, and another to leverage functions from the Haskell Prelude to decrease code length and bring it more in line with typical Haskell usage and idioms.

Style

Your sumOfChars function is a good example of tail-recursion, but in Haskell we would take that String apart through pattern matching. A Haskell String is really a list of characters, or [Char], so we'll decompose it as a list.

sumOfChars :: [Char] -> Int -- String is a synonym for [Char]
sumOfChars [] = 0 -- The empty list is the same value as the empty string, i.e. [] == ""
sumOfChars (c:cs) = digitToInt c + sumOfChars cs


The : is the list cons operator, c is the head of the list and cs is the tail. So, c == str !! 0 and cs == tail str. (And head is the function we usually use for getting the element at index 0 in a list.)

twoMult isn't wrong, but function application in Haskell doesn't require any parentheses. Parentheses are for grouping since function application has the highest precedence of any operation.

twoMult :: Char -> String
twoMult c = show (digitToInt c * 2)


Using an if statement at the top level of a function is usually an indication that you can use a guard. Guards are a bit like a multi-way if.

identificationSum :: [Char] -> Int
identificationSum [] = 0
identificationSum s@(c:cs) -- The 's@' part is an as-pattern, explained below
    | length s `mod` 2 == 1 = sumOfChars (twoMult c) + identificationSum cs
    | otherwise             = digitToInt c + identificationSum cs


Each guard begins with a pipe (|) and evaluates to a Bool value. First pattern matching takes place, then guards are evaluated in order. otherwise is just a synonym for True, i.e., a guard that always succeeds when evaluation reaches it.

I used an as-pattern there to bind the value of c:cs to an identifier. This is a handy shortcut and much preferable to writing c:cs all over the place if you need both the original value and its decomposition.

remainingToTens is a little odd, and I'd guess that you got to that answer when the compiler complained about there not being an instance for Fractional Int, yeah? Integer division uses a function called div, / is for Fractional values (like Float or Double) (yeah it's a little strange the first time you come across this).

remainingToTens :: Int -> Int
remainingToTens x = (x `div` 10 + 1) * 10 - x -- Edit: This is incorrect, use version below


And determineLastDigit just has a few extra parentheses.

determineLastDigit :: String -> Int
determineLastDigit s = remainingToTens (identificationSum s)


Idiomatic Haskell

Many of the operations you implemented can be expressed using some of the higher-order functions we have in the Prelude and coding in a functional style.

Instead of using primitive recursion on the elements of a list in sumOfChars, typical Haskell usage would see us using functions like sum and map to manipulate the entire list without getting into the weeds.

sumOfChars :: [Char] -> Int
sumOfChars cs = sum (map digitToInt cs)


maps type is (a -> b) -> [a] -> [b], that is, it takes a function - that itself takes an element of type a and returns something of type b - and a list and returns a list where that function has been applied to every element of the first list. sum adds all the values in a list of numbers.

One further thing we could do to that function is to write it in pointfree style. Writing functions pointfree is definitely an aspect of Haskell style, but don't concentrate on it overmuch until you have a solid grasp on the basics of the language. I've included an Appendix A at the bottom of this post where functions have been written pointfree for your reference.

identificationSum could be written many, many ways (some as in the appendices below) but given the context of sumOfChars, twoMult, and the specification of the algorithm I would favor separating out the functionality into two phases. First, constructing a new string based on doubling every other digit, and secondly summing all of the digits. To do this I'll introduce a new function.

doubleAlternating :: [Char] -> [Char]
doubleAlternating []       = []
doubleAlternating (c:[])   = twoMult c
doubleAlternating (c:d:cs) = twoMult c ++ [d] ++ doubleAlternating cs


And then identificationSum is a composition of doubleAlternating and sumOfChars.

identificationSum :: String -> Int
identificationSum s = sumOfChars (doubleAlternating s)


remainingToTens could really just benefit from some modular arithmetic. Use the rem function.

remainingToTens :: Int -> Int
remainingToTens x = negate x `mod` 10


I'll take it a little further with inlining in the appendices, but as it stands this is a very readable translation of the problem as it was stated. Having a close correspondence to the problem domain can be muc

Code Snippets

sumOfChars :: [Char] -> Int -- String is a synonym for [Char]
sumOfChars [] = 0 -- The empty list is the same value as the empty string, i.e. [] == ""
sumOfChars (c:cs) = digitToInt c + sumOfChars cs
twoMult :: Char -> String
twoMult c = show (digitToInt c * 2)
identificationSum :: [Char] -> Int
identificationSum [] = 0
identificationSum s@(c:cs) -- The 's@' part is an as-pattern, explained below
    | length s `mod` 2 == 1 = sumOfChars (twoMult c) + identificationSum cs
    | otherwise             = digitToInt c + identificationSum cs
remainingToTens :: Int -> Int
remainingToTens x = (x `div` 10 + 1) * 10 - x -- Edit: This is incorrect, use version below
determineLastDigit :: String -> Int
determineLastDigit s = remainingToTens (identificationSum s)

Context

StackExchange Code Review Q#57846, answer score: 17

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