I received the following very interesting E-Mail on 22 September 1997. Dr. Young describes his very valuable extensions to the Perfect-N Chart structure.
A flurry of E-Mail followed, and in the course of the next two months we produced the theorems which established the general basis and, then, a web page for producing a wide range of charts. All in all, it was a very exciting and very satisfying experience.
Subj: Perfect N-theory
Date: 97-09-22 11:06:34 EDT
From: cyoung@dcn.att.com (now cory_young@yahoo.com)
To: Stan Pope
Stan,
[Personal references omitted]
Until last year, our pack used double-elimination to run our Pinewood Derby. The results were predictable. Only two or three races for most of the scouts, crowd control problems, and a general feeling that luck of the draw had a lot to do with the outcome.
Last winter, I stuck my neck out (my wife is the Scout leader in the family, not me) and converted our pack to a "Final Standing" method, with much better results. Each scout got at least six races and I think all involved were satisfied with the fairness of the methods. We ran about 300 heats in less than 4 hours. Some of the Scout leaders who used to struggle to run 100 heats in 3 hours were more than a bit surprised.
At first, I had planned on using Pat Peronto's BASIC program to generate my racing charts. But as you are likely aware, the charts generated by this software do not stand up to the highest of standards of perfection, such as the Perfect-N criteria.
So I created my own charts from 3-3 to 9-3, which was enough to run all of our intra-den heats and intra-grade championships. Some of my charts, such as the 7-3 chart, were Perfect-N. For some of the others, I chose non-Perfect-N approaches which I thought were better for specific situations.
For example, for the 6-3 chart, I noted that if I ran 20 races, I could get every possible combination of cars in exactly 1 race. In other words, "6 choose 3" = C(6,3) = 6!/(3!3!) = 20. This gives every car 10 races, which meant every car races 4 times in one lane, and 3 times each in the other two lanes. This violates Perfect-N, but our lanes are equal enough that I felt that this was okay.
My main worry in creating my charts was that some astute parent would find some small inequity somewhere, and it might have made a small difference in choosing a winner. Thankfully, all of the Dens and grades (we only send one car per grade to the District races) had a dominant car which could win in any lane. But I still worry about the day when we have two nearly equal cars in the same set of heats.
That having been said, I have a comment about the Perfect 7-3(2) on your web page. Here it is:
Heat Ln 1 Ln 2 Ln 3 ---- ----- ----- ----- 1 1 2 4 2 2 3 5 3 3 4 6 4 4 5 7 5 5 6 1 6 6 7 2 7 7 1 3 8 1 3 4 9 2 4 5 10 3 5 6 11 4 6 7 12 5 7 1 13 6 1 2 14 7 2 3
Here is an alternative Perfect 7-3(2) chart:
Heat Ln 1 Ln 2 Ln 3 ---- ----- ----- ----- 1 1 2 4 2 2 3 5 3 3 4 6 4 4 5 7 5 5 6 1 6 6 7 2 7 7 1 3 8 1 7 5 9 2 1 6 10 3 2 7 11 4 3 1 12 5 4 2 13 6 5 3 14 7 6 4
In your chart, you used generators {1,2} and {2,1}. My chart uses generators {1,2} and {6,5}.
Note that in my chart, whenever two cars race for the second time, they have SWAPPED LANES!
In your chart, this is not the case. For example, if Car 1 is a tiny bit faster than Car 4, but Lane 3 is quite a bit faster than Lane 1, then Car 4 will beat Car 1 both times they race. If these two cars happen to be by far the fastest two cars, then you've got the wrong winner.
In my chart, they will split their two races. Assuming they then finish tied for first (because they're the fastest two cars), you now have the opportunity to break the tie more equitably (e.g. with elapsed time, if you have a timer) instead of by an arbitary numbering of the cars.
Note that if you add my generators together, element by element, you get {7,7}, and 7 is the number of cars. I don't fully understand this yet, but intuitively, it seems significant.
Your Perfect 4-3 (2) can also be improved. Here is your chart:
Heat Ln 1 Ln 2 Ln 3 ---- ----- ----- ----- 1 1 2 3 2 2 3 4 3 3 4 1 4 4 1 2
This chart uses generator {1,1}. Note that, for example, if Lane 1 is fastest, and Lane 2 is second fastest, then it's possible for Car 1 to beat Car 2 in both their races, even if Car 2 is the faster car.
Here is an alternate Perfect 4-3 (2) chart:
Heat Ln 1 Ln 2 Ln 3 ---- ----- ----- ----- 1 1 2 4 2 3 4 2 3 2 1 3 4 4 3 1
This uses generator {1,2} starting at 1 and skipping 2, and generator {3,2} starting at 2 and skipping 2. Note again that the sum of my generators is {4,4}, and 4 is the number of cars.
Again, all head-to-head rematches occur by swapping lanes.
The 4-3 chart I ended up using was a Perfect 4-3 (6) chart. Each pair of cars raced each other in all 6 of the possible lane pairings. Here it is:
Heat Ln 1 Ln 2 Ln 3
---- ----- ----- -----
1 1 2 3
2 3 4 1
3 1 3 4
4 3 1 2
5 1 4 2
6 3 2 4
7 2 3 1
8 4 1 3
9 2 1 4
10 4 3 2
11 2 4 3
12 4 2 1
The generators are: {1,1} starting at 1, skipping 2
{2,1} starting at 1, skipping 2
{3,2} starting at 1, skipping 2
{1,2} starting at 2, skipping 2
{3,3} starting at 2, skipping 2
{2,3} starting at 2, skipping 2
Note that the generators from the "starting at 1" group can be be paired off with generators from the "starting at 2" group in such a way that each pair of generators adds up to {4,4}.
I present all of this to you as food for thought. These two notions, that of a complimentary generator and that of alternating starting points between a given generator and its complimentary generator, seem to be useful in creating even better Perfect-N charts. I haven't spent any time rigorously studying the ideas. These are simply facts that I have noted in applying your generator ideas to the charts I created "manually".
I will study them as time permits, but I thought you might be interested in them right away.
[Computer Software description omitted]
[Personal references omitted]
Cory Young
Pinewood Derby Organizer
Pack 146
St. Timothy School
Chantilly, VA
B.A. in Mathematics, Davidson College, 1977, Ph.D. in Mathematics, University of Virginia, 1983. Studied algebraic topology under Bob Stong, who is quite well-known in that field. Presently a computer programmer for AT&T in northern Virginia.