'How to solve Luhn algoritm
there is a lot of information about how to write Luhn algortim. I'm trying it too and I think that I'am very close to succes but I have some mistake in my code and dont know where. The test card is VALID card but my algorithm says otherwise. Don't you know why? Thx for help
test = "5573497266530355"
kazde_druhe = []
ostatni = []
for i in test:
if int(i) % 2 == 0:
double_digit = int(i) * 2
if double_digit > 9:
p = double_digit - 9
kazde_druhe.append(p)
else:
kazde_druhe.append(double_digit)
else:
ostatni.append(int(i))
o = sum(ostatni)
k = sum(kazde_druhe)
total = o+k
if total % 10 == 0:
print(f"Your card is valid ")
else:
print(f"Your card is invalid ")
Finally! Thank you all for your help. Now it is working :-)
test = "5573497266530355" kazde_druhe = [] ostatni = []
for index, digit in enumerate(test):
if index % 2 == 0:
double_digit = int(digit) * 2
print(double_digit)
if double_digit > 9:
double_digit = double_digit - 9
kazde_druhe.append(double_digit)
else:
kazde_druhe.append(double_digit)
else:
ostatni.append(int(digit))
o = sum(ostatni)
k = sum(kazde_druhe)
total = o+k if total % 10 == 0:
print(f"Your card is valid ")
else:
print(f"Your card is invalid ")
Solution 1:[1]
This code works. :)
I fixed you code as much as i could.
test = "5573497266530355"
#test = "3379513561108795"
nums = []
for i in range(len(test)):
if (i % 2) == 0:
num = int(test[i]) * 2
if num > 9:
num -= 9
nums.append(num)
else:
nums.append(int(test[i]))
print(nums)
print((sum(nums) % 10) == 0)
I found where your code went wrong.
On the line:
for i in test:
if int(i) % 2 == 0:
It should be:
for i in range(len(test)):
if i % 2 == 0:
You should not be using the element of the string you should be using the index of the element.
Solution 2:[2]
2. With the payload, start from the rightmost digit. Moving left, double the value of every second digit (including the rightmost digit).
You have to check the digit position, not the number itself.
Change to this:
for i in range(len(test)):
if i % 2 == 0:
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | |
| Solution 2 | Teerapad Pipadboonyarat |
