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.