Pinewood Derby Chart (#56)


#1

by Bob S.

[Ediitor’s Note: I hope everyone noticed that Bob just ran, solved, and
summarized a Ruby Q. solo. Take it from me, a guy with a little
experience
doing that very, not-so-fun thing, we all owe Bob a HUGE thank you!
–JEG2]

This week’s quiz failed to attract any submissions other than the rather
weak
ChaoticChart generator that I posted. (This code was actually adapted
from a
Perl version I created a couple of years ago.)

The problem with scheduling a derby is balancing fun with fairness. The
factors
to consider are:

* All the boys should be participating throughout the event (fun)
* Each boy should race the same number of times (fair)
* All lanes of the track should be used evenly (fair)
* Cars should race against the other cars as evenly as possible (fair)

Our challenge was to construct a chart given a track with a fixed number
of
lanes and a number of cars to participate. The event would consist of a
number
of “heats” in which a car was assigned to each lane. The total number of
heats
would be determined from the number of times each car would run in each
lane.

My solution starts with a simple Chart class:

# Pinewood Derby chart
class Chart

  attr_reader :lanes, :cars, :rounds, :chart

  # create a new empty chart with given lanes, cars, and rounds.
  def initialize(lanes, cars, rounds)
    raise "Need at least #{lanes} cars" unless cars >= lanes
    raise "Need at least 1 round" unless rounds >= 1
    @lanes = lanes
    @cars = cars
    @rounds = rounds
    @chart = []
  end

Here we just take the three factors that control the chart’s size
(lanes, cars,
and rounds) and put them into member variables (after doing some
rudimentary
sanity checking). A read-only accessor is created for each member.

The @chart member variable is initialized to an empty array. This will
eventually receive the heats as they are assigned.

We need a way to print out a chart:

# prints the chart
def print_chart(io = $stdout)
  io.puts "Chart:"
  h = 0
  chart.each do |heat|
    io.printf "%4d: ", h
    heat.each do |car|
      io.printf "%4d", car
    end
    io.puts
    h += 1
  end
end

This method just loops through the heats in the @chart member and prints
them
onto the supplied object (which defaults to standard output, but could
be a File
object or a StringIO object, or any object that acts like an IO object).

One simple way to assign cars to heats is through a “round robin”
approach.
Let’s create a class that can generate such a chart. We’ll do that by
subclassing the Chart class and adding a generate method:

class RoundRobinChart < Chart

  # generate chart via simple round-robin assignment
  def generate
    chart.clear
    car = 0
    (cars * rounds).times do |heat|
      h = []
      lanes.times do
        h << car
        car = (car + 1) % cars
      end
      chart << h
    end
  end

end

The algorithm here is extremely simple. The first four cars are assigned
to the
first heat. The second heat is assigned starting with the fifth, sixth,
etc.
cars, and so on. When we run out of cars, we start reassigning with the
first
car again.

Here’s the output from generating a round robin chart for 1 round of 5
cars on 4
lanes (heats and cars are numbered from zero):

Chart:
   0:    0   1   2   3
   1:    4   0   1   2
   2:    3   4   0   1
   3:    2   3   4   0
   4:    1   2   3   4

This is actually a perfect chart. Each car runs in four heats, and runs
in each
lane exactly once. Each car faces all the other cars exactly three
times. Also,
the assignments are evenly distributed through the event; there is never
more
than a single heat gap between any two runs of a given car.

However, the round robin algorithm breaks down when you begin to
consider
different combinations of inputs. For example, here’s the output for 6
cars on 4
lanes for 1 round:

Chart:
   0:    0   1   2   3
   1:    4   5   0   1
   2:    2   3   4   5
   3:    0   1   2   3
   4:    4   5   0   1
   5:    2   3   4   5

If you’ll notice, car 0 runs twice in the first and third lanes, but not
at all
in the second and fourth lanes. Further more, cars 0 and 1 run against
each
other a total of four times, while most of the other cars only meet each
other
twice.

If you checked any of the resources under “Pope’s Pinewood Pages
Portal”, you
may have come across the “Perfect-N” method. This is a variation on the
naive
round robin approach above that can achieve a “perfect” chart in terms
of lane
assignments and opponent matchups. Unfortunately, the Perfect-N method
does not
work for all combinations of inputs.

Let’s consider another way of generating a chart. I’ll call this a
“chaotic”
method, because we’re going to assign cars to heats at random, while
trying to
maximize our fairness and fun criteria as we go.

Taking our 6 cars/4 lanes example from above, we could assign the first
heat by
just choosing four cars at random. For example:

   0:    3   1   5   0

Now, let’s start assigning cars to the second heat. Our objective is to
run each
car in each lane one time (one round), so car 3 is not a candidate for
the first
lane. Of the remaining cars, 1, 5, and 0 have run recently, while 2 and
4 have
not, so we should prefer 2 or 4 over the others. Between 2 and 4, does
it
matter? No, so let’s choose one at random:

   0:    3   1   5   0
   1:    2

Now, for assigning to the second lane, car 2 is obviously out (since a
car can’t
run in two lanes in the same heat). Of the five remaining cars, all have
run
recently except for 4, so let’s choose him:

   0:    3   1   5   0
   1:    2   4

The candidates for the third lane are 0, 1, and 3 (2 and 4 are already
in this
heat, and 5 has already run in this lane). Let’s choose one at random:

   0:    3   1   5   0
   1:    2   4   0

The last lane can take 1 or 3. Again, we can just choose at random:

   0:    3   1   5   0
   1:    2   4   0   3

As we increase the number of cars, the matchups between cars will be an
increasingly important factor. Given a number of cars to choose from for
the
last lane, we would want to favor those with fewer assignments against
the
opponents already slotted to the current heat.

So the algorithm is:

* rule out any cars already scheduled in this heat
* rule out any cars that have already run the maximum number of times 

in
this lane
* favor cars with fewer matchups against these opponents
* favor cars that have not been scheduled recently

Here’s a ChaoticChart class that implements this logic:

class ChaoticChart < Chart

  # coefficients for weighting formula for lane assignment.
  # these were derived by experimentation.
  FL = 3.0
  FP = 1.0
  FD = 3.0

  # generates the chart by assigning cars to heats
  def generate

    begin
      # assigned heats by car, last heat by car
      ah = Array.new(cars) { 0 }
      lh = Array.new(cars)

      # assignments by car/lane
      al = Array.new(cars) { Array.new(lanes) { 0 } }

      # matchups by car pair
      op = Matchups.new

      # schedule by heat by lane
      chart.clear

      # generate each heat
      (cars * rounds).times do |heat|

        # current car assignments by lane
        h = []

        # slot each lane
        lanes.times do |lane|

          # computed weights for each car
          w = {}

          # assign weights to each car for this slot
          cars.times do |car|

            # skip car if it's already been slotted to this heat
            next if h.include? car

            # skip car if it's already run max heats in this lane
            next if al[car][lane] >= @rounds

            # weight factor 1: no. of times slotted to this lane
            f1 = FL * al[car][lane]

            # weight factor 2: no. of times against these opponents
            f2 = FP * h.inject(0) do |f, opp|
              f + op[car, opp]
            end

            # weight factor 3: no. of heats since last scheduled
            # (distribute cars through the heats)
            f3 = 0
            if lh[car]
              f3 = FD * (cars / lanes) / (heat - lh[car])
            end

            # total weight for this car
            w[car] = f1 + f2 + f3

          end

          raise NoCarsException if w.empty?

          # sort by weight and get the lowest weight(s)
          w = w.sort_by { |k, v| v }
          w.pop while w[-1][1] > w[0][1]

          # randomly choose a car and slot it
          car = w[rand(w.size)][0]

          # accumulate statistics
          ah[car] += 1
          lh[car] = heat
          al[car][lane] += 1
          h.each do |opp|
            op[car, opp] += 1
          end

          # slot car to current heat
          h << car

        end

        # add current heat to chart
        chart << h

      end

    rescue NoCarsException
      retry

    end

  end

end

The generate method tracks several things as it works:

* The number of heats each car has been assigned to
* The last heat each car was assigned to
* The number of times each car has been assigned to each lane
* The number of times each car has been matched against every other car

The algorithm steps through each heat and “slots” a car to each lane.
For each
slot, it first eliminates any cars that have run their maximum times in
the
current lane or that have already been scheduled to the current heat.

For the remaining cars, a “weight” factor is computed that considers the
factors
mentioned above. The “weight” value for each car acts as a bias against
selecting that car; i.e. the car(s) with the lowest weights will be
considered
for slotting. If multiple cars have the lowest weight, a random choice
is made
from among them.

Daniel S. provided some code to analyze a generated chart to see
how fair
it is. My second submission contains some analysis routines as well.
Perhaps
someone will take these starting points and come up with a technique for
creating optimal charts for any combinations of inputs.