Checking Credit Cards (#122)

This quiz is super easy, of course. The reason I ran it though is that
I wanted
to see how people approached the Luhn algorithm implementation. It’s an
easy
enough process, but I found myself using an odd combination of Regexp
and eval()
when I was fiddling with it:

puts eval( ARGV.join.gsub(/(\d)?(\d)(?=(?:\d\d)*\d$)/) do
“#{$1 + ‘+’ if $1}#{($2.to_i * 2).to_s.split(’’).join(’+’)}+”
end ) % 10 == 0 ? “Valid” : “Invalid”

I knew that was ugly and wanted to see how you guys would pretty it up.

You have shown me the light and it tells me… Daniel M. is crazy.
I’ll
leave it to him to explain his own solution, as punishment for the time
it took
me to puzzle it out. I had to print that Array inside of the inject()
call
during each iteration to see how it built up the answer.

I do want to show you a slew of interesting tidbits though. First,
let’s get
the formality of a full solution out of the way. Here’s some code from
Drew
Olson:

class CreditCard
def initialize num
@number = num
end

# check specified conditions to determine the type of card
def type
  length = @number.size
  if length == 15 && @number =~ /^(34|37)/
    "AMEX"
  elsif length == 16 && @number =~ /^6011/
    "Discover"
  elsif length == 16 && @number =~ /^5[1-5]/
    "MasterCard"
  elsif (length == 13 || length == 16) && @number =~ /^4/
    "Visa"
  else
    "Unknown"
  end
end

# determine if card is valid based on Luhn algorithm
def valid?
  digits = ''
  # double every other number starting with the next to last
  # and working backwards
  @number.split('').reverse.each_with_index do |d,i|
    digits += d if i%2 == 0
    digits += (d.to_i*2).to_s if i%2 == 1
  end

  # sum the resulting digits, mod with ten, check against 0
  digits.split('').inject(0){|sum,d| sum+d.to_i}%10 == 0
end

end

if FILE == $0
card = CreditCard.new(ARGV.join.chomp)
puts “Card Type: #{card.type}”
if card.valid?
puts “Valid Card”
else
puts “Invalid Card”
end
end

As Drew shows, checking the type() is just a matter of verifying length
and
prefix against a known list. Nothing tricky here, as long as you don’t
get into
the more complicated cards discussed in the quiz thread.

The valid?() method is the Luhn algorithm I wanted to see. Drew
bypasses the
need for my crazy Regexp using a trick I’m always harping on: reverse
the data.
It won’t make any difference mathematically if the number is backwards
and then
you just need to double each second digit. Drew figures out when that
is by
combining each_with_index() with a little modulo test. From there, it’s
a
simple sum of the digits and the final modulo test to determine
validity.

The application code just runs those two methods and prints results.

Looking at Drew’s Luhn algorithm again, see how he declares the digits
variable
and then fills it up? That’s the classic inject() pattern, but inject()
wasn’t
an option there since the code needed each_with_index() over just plain
each().
This situation seems to call for an inject_with_index(), and look what
doug
meyer wrote:

class Array
def inject_with_index(injected)
each_with_index{|obj, index| injected = yield(injected, obj,
index) }
injected
end
end

I guess he felt the same way, though I would have added this method to
Enumerable instead of Array. Others used a hand rolled map_with_index()
in
similar ways.

Now, if you want to get away from all this index checking, you need to
take a
more functional approach and some solutions definitely did that. Here’s
the
same algorithm we’ve been examining from Ryan L.'s code:

require ‘enumerator’

def self.luhn_check(cc)
# I like functional-style code (though this may be a bit over the
top)
(cc.split(’’).reverse.enum_for(:each_slice, 2).inject(’’) do |s, (a,
b)|
s << a + (b.to_i * 2).to_s
end.split(’’).inject(0) {|sum, n| sum + n.to_i}) % 10 == 0
end

This process is interesting, I think, so let’s walk through it. The
card number
is split() into digits and reverse()d as we have been seeing.

Then an Enumerator for each_slice() is combined with inject() to build
up the
new digits. This passes the digits into inject() two at a time, so the
block
can just double the second one. This will give you an extra nil at the
end of
an odd card number, but to_i() turn that into a harmless zero.

The digits are again divided and this time summed to get a grand total.
Check
that for divisibility by ten and we have our answer.

Now Ryan chose to build up the digits and then sum, but you could do
both in the
same iterator. The first number is only ever one digit, so it’s only
the second
number that needs special handling:

require “enumerator”
puts ARGV.join.split("").reverse.enum_slice(2).inject(0) { |sum, (l,
r)|
sum + l.to_i +
(r.to_i * 2).to_s.split("").inject(0) { |s, d| s + d.to_i }
} % 10 == 0 ? “Valid” : “Invalid”

Ryan’s version is probably still a little cleaner though.

Going one step further is to think about that second digit some more.
It’s the
source of the need for tricky code, because it might be a two digit
number.
However, it doesn’t have to be. Nine doubled is 18, but the sum of the
digits
of 18 is 9. In fact, if you subtract nine from any double over ten you
will get
the same sum. One way to put this to use is with a trivial lookup
table, as
Brad E. did:

Returns true if Luhn check passes for this number

def luhn_valid?
# a trick: double_and_sum[8] == sum_digits(8*2) == sum_digits(16) ==
# 1 + 6 == 7
double_and_sum = [0, 2, 4, 6, 8, 1, 3, 5, 7, 9]
split(//).reverse.
mapn{|a,b| a.to_i + double_and_sum[b.to_i]}.sum % 10 == 0
end

Skipping over the sum() and mysterious mapn() methods for now, just take
in the
usage of double_and_sum. Those are the sums of all possible doubles of
a single
digit. Given that, Brad’s solution has to do a lot less busy work than
many of
the others we saw. I thought this was a clever reduction of the
problem.

I’m sure you can guess what that sum() method does, but let’s do take a
quick
peek at mapn(), which is another bit of clever code:

require ‘enumerator’
module Enumerable
# Maps n-at-a-time (n = arity of given block) and collects the
results
def mapn(&b)
r = []
each_slice(b.arity) {|*args| r << b.call(*args) }
r
end

def sum; inject(0){|s, i| s + i} end

end

You can see that mapn() is essentially, enum_slice(N).map(). The
interesting
part is that N is chosen from the arity of your provided block. If we
ask for
two at a time, we get two; ask for three and we get three:

(1…10).mapn { |a, b| [a, b] }
=> [[1, 2], [3, 4], [5, 6], [7, 8], [9, 10]]

(1…10).mapn { |a, b, c| [a, b, c] }
=> [[1, 2, 3], [4, 5, 6], [7, 8, 9], [10, nil, nil]]

That’s a pretty smart iterator, I say.

This completes our tour of some of the wild and wacky ideas for applying
the
Luhn algorithm to card numbers. My thanks to all who shared them.

Ruby Q. will now take a one week break. Work has been rough this week
and I
need some down time. I’ll be back next week, rested, and with new
quizzes…

I came to this quiz late, but came up with the following, before
reading the solution, which uses String.tr to do the double and sum
instead of an array.

require ‘enumerator’
def luhn(string)
string.scan(/./).reverse.enum_for(:each_with_index).inject(0) do
|sum, (digit, index)|
digit = digit.tr(‘0123456789’, ‘0246813579’) if index % 2 == 1
sum += digit.to_i
end # % 10 == 0
end

After seeing the neat use of each_slice instead of each_with_index

def luhn(string)
string.scan(/./).reverse.enum_for(:each_slice, 2).inject(0) do |sum,
(d1, d2)|
sum += d1.to_i + (d2||‘0’).tr(‘0123456789’, ‘0246813579’).to_i
end % 10 == 0
end

No need to define any extra Enumerator methods.


Rick DeNatale

My blog on Ruby
http://talklikeaduck.denhaven2.com/

Ruby Q. will now take a one week break. Work has been rough this week and I
need some down time. I’ll be back next week, rested, and with new quizzes…

Means that: no quiz tomorrow ?!?


greets

                        (
                        )
                     (
              /\  .-"""-.  /\
             //\\/  ,,,  \//\\
             |/\| ,;;;;;, |/\|
             //\\\;-"""-;///\\
            //  \/   .   \/  \\
           (| ,-_| \ | / |_-, |)
             //`__\.-.-./__`\\
            // /.-(() ())-.\ \\
           (\ |)   '---'   (| /)
            ` (|           |) `
      jgs     \)           (/

one must still have chaos in oneself to be able to give birth to a
dancing star

On 5/3/07, Rick DeNatale [email protected] wrote:

After seeing the neat use of each_slice instead of each_with_index

def luhn(string)
string.scan(/./).reverse.enum_for(:each_slice, 2).inject(0) do |sum, (d1, d2)|
sum += d1.to_i + (d2||‘0’).tr(‘0123456789’, ‘0246813579’).to_i
end % 10 == 0
end

No need to define any extra Enumerator methods.

Nice, I like it. When I first thought about using
enum_for(:each_slice, 2) I was surprised that it even worked.
Enumerator is a powerful library.

It seems like I learn something new every time I do one of these
quizzes, either from making my own solution or reading someone else’s.

Ryan

On May 3, 2007, at 9:15 AM, anansi wrote:

Ruby Q. will now take a one week break. Work has been rough
this week and I
need some down time. I’ll be back next week, rested, and with new
quizzes…

Means that: no quiz tomorrow ?!?

Yes. I need a short break. We will have one the following Friday
though.

James Edward G. II

James Edward G. II wrote:

Yes. I need a short break. We will have one the following Friday though.

James Edward G. II

Maybe someone is interested or has an idea for an unoffical Ruby Q.
for this week. I’m bored , so If someone has any idea, feel free to drop
it :wink:


greets

                        (
                        )
                     (
              /\  .-"""-.  /\
             //\\/  ,,,  \//\\
             |/\| ,;;;;;, |/\|
             //\\\;-"""-;///\\
            //  \/   .   \/  \\
           (| ,-_| \ | / |_-, |)
             //`__\.-.-./__`\\
            // /.-(() ())-.\ \\
           (\ |)   '---'   (| /)
            ` (|           |) `
      jgs     \)           (/

one must still have chaos in oneself to be able to give birth to a
dancing star

anansi wrote:

Maybe someone is interested or has an idea for an unoffical Ruby Q.
for this week. I’m bored , so If someone has any idea, feel free to drop
conway’s game of life?

On 5/3/07, Ryan L. [email protected] wrote:

Enumerator is a powerful library.

Yep. And personally, I like the fact that Ruby 1.9 is on the path to:

  1. Move enumerator to the core classes.
  2. Have enumerator methods return an enumerator if they are called
    without a block.

The second one seems to be slightly controversial, but I love it.


Rick DeNatale

My blog on Ruby
http://talklikeaduck.denhaven2.com/

Puria Nafisi Azizi wrote:

anansi wrote:

Maybe someone is interested or has an idea for an unoffical Ruby Q.
for this week. I’m bored , so If someone has any idea, feel free to drop
conway’s game of life?

Never heard about it but read now wiki and sounds interesting, so I
would try it. Would you suggest a fixed count of columns and rows for
that, if yes which?


greets

                        (
                        )
                     (
              /\  .-"""-.  /\
             //\\/  ,,,  \//\\
             |/\| ,;;;;;, |/\|
             //\\\;-"""-;///\\
            //  \/   .   \/  \\
           (| ,-_| \ | / |_-, |)
             //`__\.-.-./__`\\
            // /.-(() ())-.\ \\
           (\ |)   '---'   (| /)
            ` (|           |) `
      jgs     \)           (/

one must still have chaos in oneself to be able to give birth to a
dancing star

anansi wrote:

Never heard about it but read now wiki and sounds interesting, so I
would try it. Would you suggest a fixed count of columns and rows for
pass it as an argument

Puria Nafisi Azizi wrote:

anansi wrote:

Never heard about it but read now wiki and sounds interesting, so I
would try it. Would you suggest a fixed count of columns and rows for
pass it as an argument

k so far I would suggest:

(I)
passing row and col count as arguments
(II)
rules for cell stati:

  1. life : a cell has 2 or 3 alive neighbours
  2. death: a cell has <2 or >3 alive neighbours
  3. birth: an empty field has 3 alive neighbours

but what’s about the initial cells to begin with ?
any suggestions? randomly with a percentage Input how many cells,
dependent on fields at all shall be filled ? by the hand of the user ?
fixed patterns to fill them?


greets

                        (
                        )
                     (
              /\  .-"""-.  /\
             //\\/  ,,,  \//\\
             |/\| ,;;;;;, |/\|
             //\\\;-"""-;///\\
            //  \/   .   \/  \\
           (| ,-_| \ | / |_-, |)
             //`__\.-.-./__`\\
            // /.-(() ())-.\ \\
           (\ |)   '---'   (| /)
            ` (|           |) `
      jgs     \)           (/

one must still have chaos in oneself to be able to give birth to a
dancing star

anansi wrote:

passing row and col count as arguments
(II)
rules for cell stati:

  1. life : a cell has 2 or 3 alive neighbours
  2. death: a cell has <2 or >3 alive neighbours
  3. birth: an empty field has 3 alive neighbours

but what’s about the initial cells to begin with ?
any suggestions? randomly with a percentage Input how many cells,
dependent on fields at all shall be filled ? by the hand of the user ?
pass as an argument to star with:

  • a glider
  • a glider gun
  • a 10x10 input file
  • a 10x10 random cells
  • an exploder

pah, is not yet friday

On May 3, 2007, at 9:35 AM, anansi wrote:

though.
James Edward G. II

Maybe someone is interested or has an idea for an unoffical Ruby
Quiz for this week.

Have you worked all 122 quizzes? :wink:

James Edward G. II

One more way to do it: mutually recursive functions that each strip
off the last character and alternate the doubling:

http://pastie.caboo.se/58669

or

def luhn_sum s
val = s.slice!(-1)
if val==nil
return 0
end
val.chr.to_i + luhn_helper( s)
end

def luhn_helper s
val = s.slice!(-1)
if val==nil
return 0
end
(2 * val.chr.to_i).to_s.split("").inject(0){|sum,x|sum + x.to_i} +
luhn_sum( s)
end

def luhn? s
luhn_sum(s) % 10 == 0
end

Regards,

Paul.

BTW we missed the score of Pascal’s half of a hexagon by the numbers
of edges of a pentagon (no politics involved here). I express myself
like this because I am a square head of course;)
If you get my point you will miss the line under the triangle but I
could not come up with this.
Maybe I should just have submitted ten solutions :wink:

Take advantage of your break James.

Cheers
Robert

Ruby Q. [email protected] writes:

You have shown me the light and it tells me… Daniel M. is crazy. I’ll
leave it to him to explain his own solution, as punishment for the time it took
me to puzzle it out. I had to print that Array inside of the inject() call
during each iteration to see how it built up the answer.

As if you needed further evidence after my third solution to pp
Pascal.

And the one in my solution is not so bad. The really nasty Luhn
implementation was what I posted as a follow-up
(http://blade.nagaokaut.ac.jp/cgi-bin/scat.rb/ruby/ruby-talk/249669)
That algorithm was:

def luhn(s)
s.scan(/\d/).map{|x|x.to_i}.inject([0,0]){
|(a,b),c|[b+c%9,a+2*c%9]}[0]%10 == s.scan(/9/).size%10
end

Now, here’s an explanation of the version in my posted solution:

First, the code in my solution was this:

def luhn(s)
s.scan(/\d/).inject([0,0]){|(a,b),c|[b+c.to_i, a+c.to_i*2%9+(c==‘9’ ? 9 : 0)]}[0]%10 == 0
end

The main issue with doing the Luhn algorithm as a straight-forward
inject command is that you don’t know on each digit whether this digit
is one that should be doubled or not.

Now, there are a few ways around this:

  1. have a “should_double” variable that you initialize to
    should_double = (s.length % 2 == 0)
    and then in your block do
    should_double = !should_double
  2. reverse the data, and use something like each_index to get
    you the index each time, and double when the index is even
    (most people did this)
  3. like choice (1), but reverse the data and initialize
    should_double to false.

I chose a different path: at each stage, compute both possibilities.
That is, do an inject loop with two running totals - one in which the
number we’re dealing with now should be doubled, and one in which it
shouldn’t. At each stage, swap the two totals. At the end, pick the
total that did not involve doubling the last digit.

If the Luhn algorithm just involved adding the doubled digits (and
didn’t involve the added complication of adding their digits),
generating both sums would be just:

s.scan(/\d/).map{|x|x.to_i}.inject([0,0]){
|(sum2, sum1),x| [sum1 + x, sum2 + 2*x]
}

Note how I swapped sum1 and sum2 in the process.

Then, to take the sum that did not involve doubling the last digit,
just do:

s.scan(/\d/).map{|x|x.to_i}.inject([0,0]){
|(sum2, sum1),x| [sum1 + x, sum2 + 2*x]
}[0]

Now, adding digits of numbers together is the well-known process of
“casting out nines”, and essentially what you’re doing with that sum
is taking the residue modulo 9. (in other language: “finding the
remainder when dividing by 9”) That is, adding the digits together
turns 2x into 2x%9, except that if x is 9 then you get 9, not 0.
This makes our sum algorithm into:

s.scan(/\d/).map{|x|x.to_i}.inject([0,0]){
|(sum2, sum1),x| [sum1 + x, sum2 + 2*x%9 + (x==9 ? 9 : 0)]
}[0]

This is almost my luhn method above, except that I used unhelpful
one-letter variable names and didn’t do the map call, preferring
instead to use .to_i in the body of my inject loop.

Now, go back and puzzle out the version I mentioned at the top of this
email. That version was born out of a desire to avoid the ? :
operator, since I tend to find it ugly.

On 5/4/07, Robert D. [email protected] wrote:

BTW we missed the score of Pascal’s half of a hexagon by the numbers
of edges of a pentagon (no politics involved here). I express myself

I don’t think half a hexagon is what you think it is :slight_smile:

martin

On 5/6/07, Martin DeMello [email protected] wrote:

On 5/4/07, Robert D. [email protected] wrote:

BTW we missed the score of Pascal’s half of a hexagon by the numbers
of edges of a pentagon (no politics involved here). I express myself

I don’t think half a hexagon is what you think it is :slight_smile:
Hmm you think to know what I think, I think you know more than me than
:wink:

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