Implementation of Luhn Formula

Question

I was trying to implement the Luhn Formula in Python, here is my code:

import sys


def luhn_check(number):
    if number.isdigit():
        last_digit = int(str(number)[-1])
        reverse_sequence = list(int(d) for d in str(int(number[-2::-1])))

        for i in range(0, len(reverse_sequence), 2):
            reverse_sequence[i] *= 2

        for i in range(len(reverse_sequence)):
            if reverse_sequence[i] > 9:
                reverse_sequence[i] -= 9

        sum_of_digits = 0
        for i in range(len(reverse_sequence)):
            sum_of_digits += reverse_sequence[i]

        result = divmod(sum_of_digits, 10)

        if result == last_digit:
            print("[VALID] %s" % number)
        else:
            print("[INVALID] %s" % number)
        quit()

    print("[ERROR] \" %s \" is not a valid sequence." % number)
    quit()


def main():
    if len(sys.argv) < 2:
        print("Usage: python TLF.py <sequence>")
        quit()

    luhn_check(sys.argv[1])

if __name__ == '__main__':
    main()

But it isn't working properly:

[INVALID] 4532015112830366
[INVALID] 6011514433546201
[INVALID] 6771549495586802

and so on...

But the logic of the code seems OK to me. I followed this workflow:

The Luhn Formula:

1. Drop the last digit from the number. The last digit is what we want to check against Reverse the numbers

2. Multiply the digits in odd positions (1, 3, 5, etc.) by 2 and subtract 9 to all any result higher than 9

3. Add all the numbers together

4. The check digit (the last number of the card) is the amount that you would need to add to get a multiple of 10 (Modulo 10)

Answer 1

I think the algorithm is not correct.

The second step you need to sum the digits of the products instead of substract 9. Reference: Wikipedia.

In the Wikipedia you have this example:

def luhn_checksum(card_number):
    def digits_of(n):
        return [int(d) for d in str(n)]
    digits = digits_of(card_number)
    odd_digits = digits[-1::-2]
    even_digits = digits[-2::-2]
    checksum = 0
    checksum += sum(odd_digits)
    for d in even_digits:
        checksum += sum(digits_of(d*2))
    return checksum % 10

def is_luhn_valid(card_number):
    return luhn_checksum(card_number) == 0


result = is_luhn_valid(4532015112830366)
print 'Correct:' + str(result)
result = is_luhn_valid(6011514433546201)
print 'Correct:' + str(result)
result = is_luhn_valid(6771549495586802)
print 'Correct:' + str(result)

Result:

>>>Correct:True
>>>Correct:True
>>>Correct:True

Answer 2

The following might help some people to start with the Luhn algorithm in python.

num = list(input("Please enter the number to test (no space, no symbols, only \
numbers): "))

num = list(map(int, num))[::-1] #let's transform string into int and reverse it

for index in range(1,len(num),2):
    if num[index]<5:
        num[index] = num[index] *2
    else: #doubling number>=5 will give a 2 digit number
        num[index] = ((num[index]*2)//10) + ((num[index]*2)%10)

checksum=sum(num)

print("checksum= {}".format(checksum))

if checksum%10 !=0:
    print('the number is not valid')
else:
    print('the number is valid!')

Answer 3

check_numbers = ['49927398716', '4847352989263095', '79927398713', '5543352315777720']

def Luhn_Check(number):
    """
        this function checks a number by using the Luhn algorithm.

        Notes (aka - How to Luhn) :
            Luhn algorithm works in a 1 2 1 2 ... order.
            Therefore, in computer speak, 1 0 1 0 ... order

            step 1:
                -> reverse the # so we are not working from right to left (unless you want too)
                -> double every second number

            step 2:
                -> if the doubled number is greater then 9, add the individual digits of the number to get a single digit number
                    (eg. 12, 1 + 2 = 3)

            step 3:
               -> sum all the digits, if the total modulo is equal to 0 (or ends in zero) then the number is valid...
    """

    reverse_numbers = [int(x) for x in number[::-1]] # convert args to int, reverse the numbers, put into a list
    dbl_digits = list() # create empty list
    digits = list(enumerate(reverse_numbers, start=1)) # enumerate numbers starting with an index of 1

    for index, digit in digits:
        if index % 2 == 0:  # double every second (other) digit.

            doub_digit = digit * 2
            dbl_digits.append(doub_digit - 9) if doub_digit > 9 else dbl_digits.append(doub_digit)

        else:
            # if not '0' append to list (this would be the 1 in Luhn algo sequence (1 2 1 2 ...)
            dbl_digits.append(digit)

    return sum(dbl_digits) % 10


if (__name__ == "__main__"):
    print("Valid Numbers: %s " % [x for x in check_numbers if Luhn_Check(x) == 0])

Answer 4

This is the most concise python formula for the Luhn test I have found:

def luhn(n):
    r = [int(ch) for ch in str(n)][::-1]
    return (sum(r[0::2]) + sum(sum(divmod(d*2,10)) for d in r[1::2])) % 10 == 0

The above function and other Luhn implementations (in different programming languages) are available in https://www.rosettacode.org/wiki/Luhn_test_of_credit_card_numbers .

Answer 5

see this Python recipe

def cardLuhnChecksumIsValid(card_number):
    """ checks to make sure that the card passes a luhn mod-10 checksum """

    sum = 0
    num_digits = len(card_number)
    oddeven = num_digits & 1

    for count in range(0, num_digits):
        digit = int(card_number[count])

        if not (( count & 1 ) ^ oddeven ):
            digit = digit * 2
        if digit > 9:
            digit = digit - 9

        sum = sum + digit

    return ( (sum % 10) == 0 )

Answer 6

There are some errors in your code:

result = divmod(sum_of_digits, 10)

returns a tuple, you need only modulo, that is use

result = sum_of_digits % 10

Second, to check for validity, you don't omit last digit (that is checksum), but include it in computations. Use

reverse_sequence = list(int(d) for d in str(int(number[::-1]))) 

And check for result being zero:

if not result:
    print("[VALID] %s" % number)

Or if you insist on keeping this not needed complexity, check for last digit to be inverse of checksum modulo 10: keep

reverse_sequence = list(int(d) for d in str(int(number[-2::-1])))

but use

if (result + last_digit) % 10 == 0:
    print("[VALID] %s" % number)

For a simplier and shorter code, I can give you a reference to my older answer.