how to create conditional loop for Luhn object to check if index[0] is not 0

Question

I am using a Luhn lib from Github. Luhn class has more methods but I am intersted in only "generate" and "valid?" methods.

  m = Luhn.generate(9) # generates a random number like [123456789] or [098765432]
  Luhn.valid?(m)   # returns true or false based on Luhn checksum passing

But I don't want the generated number starts with "0".

so how do I put in my ruby script:

1. Discard the generated number that starts with "0".
2. Generate another number and check again until m[0] is non-zero.
3. run valid?() on generated number: if false, generate another non-zero number until true.
4. return the valid 9 digits number to the caller code

Answer 1

Will this work for you? (Very basic recursion)

require 'luhn'
def generate_non_zero_luhn(size=9)
  m = Luhn.generate(size)
  m[0] != "0" && Luhn.valid?(m) ? m : generate_non_zero_luhn(size)
end

Example Outputs

10.times.map { generate_non_zero_luhn }
#=> ["582428355", "353646136", "821379013", "818135220", "696652767", 
     "934766437", "885229401", "226319044", "682121645", "688197078"]

Although I think the validity check is not needed because Luhn is generating the number so it will always be valid. e.g.

1_000_000.times.map { Luhn.valid?(Luhn.generate(9))}.any?{|val| !val}
#=> false

So then the method can become

def generate_non_zero_luhn(size=9)
  m = Luhn.generate(size)
  m[0] != "0" ? m : generate_non_zero_luhn(size)
end