From: "semmons99@..." <semmons99@...>
Date: 2006-03-28T22:38:56+09:00
Subject: Re: B & E (#72)

You are right that there are 10,000 unique codes, but each of those
codes can have any of the three unique modifier keys attached to it so
10,000 * 3 = 30,000 unique combinations you must press. And, since
there are 5 keys in the 4 key code + 1 modifier key, it is 30,000 * 5 =
150,000. At first I didn't realize this until I wrote the loop to
generate all of the unique code/modifier key combinations. The one key
to this I believe, is that you have to treat the modifier key (1,2,3)
as part of the code not a seperate entity. Here is some code to prove
the solution.

<code>

output, num_codes = '', 0

10.times do |x1| 10.times do |x2| 10.times do |x3|
10.times do |x4|   3.times do |x5|
    output += x1.to_s + x2.to_s + x3.to_s + x4.to_s + x5.to_s
    num_codes += 1
end end
end end end

print "number codes:  ", num_codes.to_s, "\n"
print "output length: ", output.length, "\n"

</code>