As a couple of solvers noted, this is an easy problem that’s really just
about
counting. In fact, you can solve the entire problem without ever
rolling a die.
The trick is really in how you think about the problem. Basically we
want to
enumerate all of the possible rolls from:
[1, 1, 1, 1, 1, 1]
to:
[6, 6, 6, 6, 6, 6]
Let me show you that one more time though, minus some Ruby syntax
characters:
111111 to 666666
Now it starts to look like counting, right? It almost is. If we
counted
straight between those two numbers though, we would pick up some digits
that
aren’t on our dice. But if we drop the digits on both sides by one:
000000 to 555555
Now that is counting, in base six, and it’s just a quick transliteration
to
adjust the digits up. We’re going to look at a solution that does just
that.
While I put the code together that I’m going to show you, it’s heavily
inspired
from many of the submissions I liked. Let’s just say that this week’s
showcased
solution belongs to everyone.
Here’s the argument processing code:
#!/usr/bin/env ruby -wKU
MODE = if ARGV.first =~ /-([sv])/
ARGV.shift
$1
end
unless ARGV.size == 2
abort “Usage: #{File.basename($PROGRAM_NAME)} [-s|-v] NUM_DICE
MIN_FIVES”
end
NUM_DICE, MIN_FIVES = ARGV.map { |n| Integer(n) }
…
This code should be pretty easy to break down. We check to see if the
first
argument looks like a switch and store it in the MODE constant when it
does.
Note that the code doesn’t provide an else branch for this check, but
Ruby will
default the constant to nil for us when the if condition fails to match.
The next bit of code just ensures that we have the two required
arguments,
converts them or throws an error if they are not what we expected, and
stores
them in constants for future use.
We’re now ready for the counting code:
…
desirable_outcomes = 0
POSSIBLE_OUTCOMES = 6 ** NUM_DICE
POSSIBLE_OUTCOMES.times do |i|
outcome = i.to_s(6).rjust(NUM_DICE, “0”)
if desirable = outcome.count(“4”) >= MIN_FIVES
desirable_outcomes += 1
end
if MODE == “v” or (MODE == “s” and (i % 50_000).zero?)
puts “%#{NUM_DICE}d [%s]#{’ <==’ if desirable}” %
[i + 1, outcome.tr(“0-5”, “1-6”).gsub(/(\d)(\d)/, ‘\1,\2’)]
end
end
…
This is the heart of the code and, as you can see, still quite simple.
We begin
by defining the range we will iterate over, from zero to the maximum
count on
the requested number of dice.
The times() iterator does the actual count. Remember that we really
want these
numbers in base six. The first line of the iterator does the
conversion,
padding with zeros to ensure we have the right number of “dice.”
The first if branch is our test for the minimum number of fives. When
found, we
bump a global counter that can later be used to generate results.
The final if statement handles printing for either of the inspection
modes. If
we’re in verbose mode, or sample mode and on a 50,000th iteration, the
code
prints the result index, the dice values, and an indicator arrow for
rolls that
were flagged as desirable. This is where you see the code to bump all
of the
digits so they will match pips on a die.
The final bit of code, just prints the results:
…
puts if MODE
puts “Number of desirable outcomes is #{desirable_outcomes}”
puts “Number of possible outcomes is #{POSSIBLE_OUTCOMES}”
puts
puts “Probability is #{desirable_outcomes / POSSIBLE_OUTCOMES.to_f}”
This is as easy as it gets. We show the desirable, the possible, and
calculate
the odds from the two of them. The first line just sneaks in a line of
padding
between any sampling and the end results.
Again, this code combined things I liked from many of the solutions, so
they are
definitely worth a look. Especially don’t miss Alex LeDonne’s second
solution
which uses some math to skip counting at all, when it can get away with
it.
As always, my thanks to all who took the time to solve this quiz. I
feel like
we always get creative results for the simple problems.
Tomorrow we will tackle one of the all time famous tough problems, and
see if we
just can’t solve it reasonably well with minimal knowledge of the
domain…