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/lastSuccessfulBuild/artifact/book-result/Regex/Regex.html
>
> Saludos, Pablo.
> El 30 de abr. de 2020 10:11 -0300, St��phane Ducasse <stephane.ducasse@inria.fr>, escribi��:
>
> 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
>
>
>
>
>
> --------------------------------------------
> St��phane 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
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>