I proposed as alternative to Luhn check. In special to validation problems like credit card number validation. El jue., 30 abr. 2020 11:11, Richard O'Keefe <raoknz@gmail.com> escribió:
I'm sure you could fit regular expressions into the Luhn check, but I am rather mystified as to what one could possibly gain by doing so. How do you do the equivalent of #(0 2 4 6 8 1 3 5 7 9) at: char digitValue + 1 in a regular expression and why would you want to?
On Fri, 1 May 2020 at 01:31, Pablo Navarro <pablo1n7@gmail.com> wrote:
Hello!. For this problem is possible to use Regular Expressions too.
https://ci.inria.fr/pharo-contribution/job/UpdatedPharoByExample/lastSuccess...
Saludos, Pablo. El 30 de abr. de 2020 10:11 -0300, Stéphane Ducasse <
stephane.ducasse@inria.fr>, escribió:
It looks like a cool problem from where did you take it?
I hope I can discuss my approch to this problem :
Given a number determine whether or not it is valid per the Luhn formula.
The Luhn algorithm is a simple checksum formula used to validate a
variety of identification numbers, such as credit card numbers and Canadian Social Insurance Numbers.
The task is to check if a given string is valid.
I like people that are always thinking in string as if a collection of
number would not make it :)
Validating a Number
Strings of length 1 or less are not valid. Spaces are allowed in the
input, but they should be stripped before checking. All other non-digit characters are disallowed.
Example 1: valid credit card number
4539 1488 0343 6467
The first step of the Luhn algorithm is to double every second digit,
starting from the right. We will be doubling
4_3_ 1_8_ 0_4_ 6_6_
If doubling the number results in a number greater than 9 then subtract
9 from the product. The results of our doubling:
8569 2478 0383 3437
Then sum all of the digits:
8+5+6+9+2+4+7+8+0+3+8+3+3+4+3+7 = 80
If the sum is evenly divisible by 10, then the number is valid. This
number is valid!
my idea was to do these steps
1) reverse the input.
2) use this to double every second digit and calculate the sum of all
the numbers :
checkNumber := (collection reverse selectwith index: [:item :index |
(index % 2 == 0) . IfTrue: [item *2]] ) sumNumbers
you can also 1 to: xx by: 2 do:
3) check if its a valid number by doing this :
^ (checkNumber % 10 == 0)
is this a good game plan or has it flaws or can I do it better ?
Regards,
Roelof
-------------------------------------------- Stéphane Ducasse http://stephane.ducasse.free.fr / http://www.pharo.org 03 59 35 87 52 Assistant: Julie Jonas FAX 03 59 57 78 50 TEL 03 59 35 86 16 S. Ducasse - Inria 40, avenue Halley, Parc Scientifique de la Haute Borne, Bât.A, Park Plaza Villeneuve d'Ascq 59650 France