Forum: Ruby Pinewood Derby Chart (#56)

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James G. (Guest)
on 2005-12-01 15:51
(Received via mailing list)
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
doing that very, not-so-fun thing, we all owe Bob a HUGE thank you!

This week's quiz failed to attract any submissions other than the rather
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
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
lanes and a number of cars to participate. The event would consist of a
of "heats" in which a car was assigned to each lane. The total number of
would be determined from the number of times each car would run in each

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 = []

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
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
	    h += 1

This method just loops through the heats in the @chart member and prints
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"
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
	    car = 0
	    (cars * rounds).times do |heat|
	      h = []
	      lanes.times do
	        h << car
	        car = (car + 1) % cars
	      chart << h


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,
cars, and so on. When we run out of cars, we start reassigning with the
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):

	   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
than a single heat gap between any two runs of a given car.

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

	   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
other a total of four times, while most of the other cars only meet each

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
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
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
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
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
last lane, we would want to favor those with fewer assignments against
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
	  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

	      # assigned heats by car, last heat by car
	      ah = { 0 }
	      lh =

	      # assignments by car/lane
	      al = { { 0 } }

	      # matchups by car pair
	      op =

	      # schedule by heat by lane

	      # 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]

	            # 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])

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


	          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

	          # slot car to current heat
	          h << car


	        # add current heat to chart
	        chart << h


	    rescue NoCarsException




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
current lane or that have already been scheduled to the current heat.

For the remaining cars, a "weight" factor is computed that considers the
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
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.
someone will take these starting points and come up with a technique for
creating optimal charts for any combinations of inputs.
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