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 <pablo...@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, Stéphane Ducasse > <stephane.duca...@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 >
