Credit Card Validation

April 8, 2011

Most credit card numbers, and many other identification numbers including the Canadian Social Insurance Number, can be validated by an algorithm developed by Hans Peter Luhn of IBM, described in U. S. Patent 2950048 in 1954 (software patents are nothing new!), and now in the public domain. The Luhn algorithm will detect almost any single-digit error, almost all transpositions of adjacent digits except 09 and 90, and many other errors.

The Luhn algorithm works from right-to-left, with the right-most digit being the check digit. Alternate digits, starting with the first digit to left of the check digit, are doubled. Then the digit-sums of all the numbers, both undoubled and doubled, are added. The number is valid if the sum is divisible by ten.

For example, the number 49927398716 is valid according to the Luhn algorithm. Starting from the right, the sum is 6 + (2) + 7 + (1 + 6) + 9 + (6) + 7 + (4) + 9 + (1 + 8) + 4 = 70, which is divisible by 10; the digit-sums of the doubled digits have been shown in parentheses.

Your task is to write two functions, one that adds a check digit to a identifying number and one that tests if an identifying number is valid. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.

About these ads

Pages: 1 2

14 Responses to “Credit Card Validation”

  1. My Haskell solution (see http://bonsaicode.wordpress.com/2011/04/08/programming-praxis-credit-card-validation/ for a version with comments):

    import Data.Char
    import Data.List.HT
    
    luhnSum :: Integral a => a -> a
    luhnSum n = digitSum a + digitSum (show . (* 2) . digitToInt =<< b)
        where [a,b] = sliceHorizontal 2 . reverse $ show n
              digitSum = sum . map (fromIntegral . digitToInt)
    
    isValid :: Integral a => a -> Bool
    isValid n = mod (luhnSum n) 10 == 0
    
    makeValid :: Integral a => a -> a
    makeValid n = 10*n + mod (10 - mod (luhnSum $ 10*n) 10) 10
    
  2. […] today’s Programming Praxis exercise our goal is to write to functions related to the Luhn credit card […]

  3. Graham said

    I came up with two versions of my intermediate Luhn sum function: one that
    creates a list and modifies it in-place, and another that uses iterators
    instead. My solution is available on github.

  4. Jussi Piitulainen said

    Just numbers; just one recursive branch / two vectors.
    First I missed that (- 10 (modulo 30 10)) is 10, not 0.
    Then Ikarus thought that (modulo -30 10) => 10. Sigh.

    (define (luhn-digit digits)
      (define (modulo n m) (- n (* m (floor (/ n m))))) ;...                                                                     
      (let loop ((digits digits) (neg-sum 0)
                 (vig '#(0 2 4 6 8 1 3 5 7 9))
                 (vug '#(0 1 2 3 4 5 6 7 8 9)))
        (if (zero? digits)
            (modulo neg-sum 10)
            (loop (quotient digits 10)
                  (- neg-sum (vector-ref vig (modulo digits 10)))
                  vug vig))))
    
    (define (luhn? digits) (= digits (luhn (quotient digits 10))))
    
    (define (luhn digits) (+ (* digits 10) (luhn-digit digits)))
    
    (write (luhn? 49927398716))
    (write (luhn? 3020))
    (write (luhn? 3129))
    (newline)
    
  5. arturasl said

    Wanted to see if I still remember Linux assembly so I wrote my answer in it: github. It might be possible to make the answer more readable and shorter, but that would take me an additional hour…

  6. Mike said
    dd = ((0,0), (1,2), (2,4), (3,6), (4,8),
          (5,1), (6,3), (7,5), (8,7), (9,9))
    
    def luhn(ds):
        '''
        >>> n = "49927398716"
        >>> luhn(n[:-1])
        6
        '''
    
        return -sum(dd[int(d)][~n&1] for n,d in enumerate(ds[::-1])) % 10
    
    def is_valid(ds):
        '''
        n = "49927398716"
        >>> is_valid(n)
        True
        '''
    
        return luhn(ds[:-1]) == int(ds[-1])
    
    def add_check_digit(ds):
        '''
        n = "4992739871"
        >>> add_check_digit(n)
        '49927398716'
        '''
    
        return ds + str(luhn(ds))
    
  7. David said

    Factor Language solution.

    USING: kernel math math.parser math.ranges sequences ;
    IN: luhn
    
    : reversed-digits ( n -- list )
        { } swap
        [ dup 0 > ]
            [ 10 /mod  swapd suffix  swap ]
        while drop ;
    
    : luhn-sum  ( n -- n )
        reversed-digits dup length iota [
            2dup swap nth
            swap odd? [ 2 *  10 /mod + ] when
        ] map sum
        nip ;
    
    : luhn? ( n -- ? )
        luhn-sum 10 mod 0 = ;
    
    : make-luhn ( n -- n )
        10 * dup luhn-sum 10 mod 10 swap - 10 mod + ;
    

    ( scratchpad ) 2771 luhn? .
    f
    ( scratchpad ) 2771 make-luhn dup . luhn? .
    27714
    t
    ( scratchpad ) 49927398716 luhn? .
    t
    ( scratchpad ) 49927398716 make-luhn dup . luhn? .
    499273987168
    t

  8. […] or Canadian Social¬†Insurance¬†Numbers are validated you can take a look at this Programming Praxis article . It’s all about a simple, tiny, patented (now public domain) algorithm invented by […]

  9. My python 3.x solution:

    def luhn_check(num):
        ''' Number - List of reversed digits '''
        digits = [int(x) for x in reversed(str(num))]
        check_sum = sum(digits[::2]) + sum((dig//10 + dig%10) for dig in [2*el for el in digits[1::2]])
        return check_sum%10 == 0
     
    if __name__ == "__main__":
        print(luhn_check(543298376
    
  10. rohit said

    nice solution

  11. Rainer said

    My try in REXX

    
    numeric digits 30
    debug = 0
    
    number = '49927398716'
    
    say number 'is' check_luhn(number)
    
    number = '4992739871'
    say number 'plus check_digit =' add_check_digit(number)
    
    exit
    
    check_luhn: procedure
        parse arg test
        test = reverse(test)
        summe = 0
        ind = 0
        do i = 1 to length(test)
            ind = ind + 1
            ziffer = substr(test,i,1)
            say 'Ziffer' ziffer
            if even(ind) then do
                ziffer = ziffer * 2
                if length(ziffer) > 1 then summe = summe + left(ziffer,1)
            end
            summe = summe + right(ziffer,1)
            say 'Summe' summe
        end
        if (summe//10) > 0 then return 'wrong'
                           else return 'correct'
    
    add_check_digit: procedure
        parse arg urtest
        test = reverse(urtest)	
        summe = 0
        ind = 0
        do i = 1 to length(test)
            ind = ind + 1
            ziffer = substr(test,i,1)
            if odd(ind) then do
                ziffer = ziffer * 2
                if length(ziffer) > 1 then summe = summe + left(ziffer,1)
            end
            summe = summe + right(ziffer,1)
            say 'Summe' summe
        end
        if (summe//10) > 0 then return urtest||(summe%10*10)+10-summe
                           else return urtest||0
    
    even: procedure
        parse arg number
        return \ odd(number)
    
    odd: procedure
        parse arg number
        return (number//2)
    
    
  12. Khanh Nguyen said
    open System
    
    let rec sum_digit inp =
        if inp < 10 then inp else sum_digit (inp / 10) + (inp % 10)
    
    let f (inp: string) =    
        inp.ToCharArray() 
        |> List.ofArray 
        |> List.map (fun x -> (int x) - (int '0')) 
        |> List.rev 
        |> List.zip [1.. String.length inp]
        |> List.map (fun (i, v) -> if i % 2 = 1 then v*2 else v)
        |> List.map sum_digit
        |> List.sum 
    
    let add_check_digit (inp : string) =
        inp + string(10 - (f inp) % 10)
    let isValid (inp : string) = 
        inp = add_check_digit (inp.Remove(String.length inp - 1))
    
    add_check_digit "4992739871"
    isValid "49927398715"
    isValid "49927398716"
    
    
  13. Ben Atwell said

    My solution in Python 3.2

    def addCCDigits(number):
    number = int(str(number)[::-1])
    numString = str(number)
    check = False
    mySum = 0
    for ch in numString:
    num = int(ch)
    if (check):
    num = num * 2
    strNum = str(num)
    for c in strNum:
    num2 = int(c)
    mySum += num2
    else:
    mySum += num
    check = not check
    return mySum

    def validate(number):
    mySum = addCCDigits(number)
    if (mySum % 10 == 0):
    return True
    else:
    return False

    def addCheckDigit(number):
    mySum = addCCDigits(number * 10 + 1) – 1
    digit1 = int(mySum / 10)
    goal = (digit1 + 1) * 10;
    diff = goal – mySum;
    mySum = int(mySum / 10)
    strNumber = str(number) + str(diff)
    number = int(strNumber)
    return number
    [\code]

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Follow

Get every new post delivered to your Inbox.

Join 634 other followers

%d bloggers like this: