Question

I have the following method for doing a check digit on a tracking number, but it just feels lengthy/sloppy. Can it be refactored and just generally cleaned up?

I’m running Ruby 1.8.7.

def is_fedex(number)
  n = number.reverse[0..14]

  check_digit = n.first.to_i

  even_numbers = n[1..1].to_i + n[3..3].to_i + n[5..5].to_i + n[7..7].to_i + n[9..9].to_i + n[11..11].to_i + n[13..13].to_i

  even_numbers = even_numbers * 3

  odd_numbers = n[2..2].to_i + n[4..4].to_i + n[6..6].to_i + n[8..8].to_i + n[10..10].to_i + n[12..12].to_i + n[14..14].to_i

  total = even_numbers + odd_numbers

  multiple_of_ten = total + 10 - (total % 10)

  remainder = multiple_of_ten - total

  if remainder == check_digit
    true
  else
    false
  end
end

EDIT: Here are valid and invalid numbers.

Valid: 9612019950078574025848

Invalid: 9612019950078574025847

Answer 1

def is_fedex(number)
  total = (7..20).inject(0) {|sum, i| sum + number[i..i].to_i * ( i.odd? ? 1 : 3 ) }
  number[-1].to_i == (total / 10.0).ceil * 10 - total
end

I believe you should keep your code. While it’s not idiomatic or clever, it’s the one you will have the least trouble to understand a few months from now.