Brain Bender Regatta Rotation

Discussion:

Gus Wirth

2011-09-01 14:29:11 UTC

Here's a little puzzle that I was recently sent that so far hasn't had a
response from the sailing crowd. So I thought maybe this could be viewed
as either a little math problem, or something that could be solved by an
appropriate program, hence its appearance here.

It kind of reminds me of the problems Martin Gardner would have in old
issues of "Scientific American".

Gus

-------- Original Message --------
Subject: Brain Bender Regatta Rotation
Date: Thu, 25 Aug 2011 09:49:36 -0700
From: John Fretwell

Get your geek hats on. Here's a little fun designing the format for the
MBYC Club Championship which is turning out just a bit trickier than I
envisioned. What we're facing is what I'm now calling "nested
round-robins":

* Assume 12 skippers. Let's call them ALPHA through LIMA. They
must all meet each other the same number of times in a fleet racing
format. I do not care whether this is done via a division into flights
(ie 4 groups of 3) or a s a pure competitor vs. competitor round robin.
But we are not desiring to have any eliminations during the series.

* Assume 6 boats are supplied by the regatta. All skippers must
sail all the boats, and the same number of times.

* This is a one day regatta, so assuming relatively short races
and time for rotations, I believe 12 total races sailed is about all we
can expect to fit in. Of course not all skippers need to sail in all 12
races, but they all have to sail the same number of races.

The heart of the problem seems to be that since two teams will sail
their first race in the same boat, and all teams must meet each other
the same number of times, all the schemes I've tried so far eventually
put teams on a collision course to be scheduled for the same boat in the
same race. The problem seems to exist whether or not you have the
competitors face off in flights or in a competitor vs. competitor round
robin.

I think that covers it. If the solution scenario has any possibility of
a tie on points (I think it does), we could defer to the head/head score
between tied skippers. If that is still tied, we could go with the
winner of the last head/head, or more satisfactorily go to a quick best
of 3 match race showdown with boats selected by the skippers in order of
the results of a coin toss.

Excel submissions preferred. Awe and ale offered in compensation.

And GO.......

John Fretwell

Junior Sailing Director

San Diego Yacht Club

1011 Anchorage Lane

San Diego, CA 92106

619.758.6320

www.sdyc.org/juniors <http://www.sdyc.org/juniors>

Carl Lowenstein

2011-09-07 19:34:44 UTC

Permalink

Post by Gus Wirth
Here's a little puzzle that I was recently sent that so far hasn't had a
response from the sailing crowd. So I thought maybe this could be viewed
as either a little math problem, or something that could be solved by an
appropriate program, hence its appearance here.
It kind of reminds me of the problems Martin Gardner would have in old
issues of "Scientific American".
Gus
-------- Original Message --------
Subject: Brain Bender Regatta Rotation
Date: Thu, 25 Aug 2011 09:49:36 -0700
From: John Fretwell
Get your geek hats on. Here's a little fun designing the format for the
MBYC Club Championship which is turning out just a bit trickier than I
envisioned. What we're facing is what I'm now calling "nested
* ? ? ? ? Assume 12 skippers. Let's call them ALPHA through LIMA. They
must all meet each other the same number of times in a fleet racing
format. I do not care whether this is done via a division into flights
(ie 4 groups of 3) or a s a pure competitor vs. competitor round robin.
But we are not desiring to have any eliminations during the series.
* ? ? ? ? Assume 6 boats are supplied by the regatta. All skippers must
sail all the boats, and the same number of times.
* ? ? ? ? This is a one day regatta, so assuming relatively short races
and time for rotations, I believe 12 total races sailed is about all we
can expect to fit in. Of course not all skippers need to sail in all 12
races, but they all have to sail the same number of races.
The heart of the problem seems to be that since two teams will sail
their first race in the same boat, and all teams must meet each other
the same number of times, all the schemes I've tried so far eventually
put teams on a collision course to be scheduled for the same boat in the
same race. The problem seems to exist whether or not you have the
competitors face off in flights or in a competitor vs. competitor round
robin.

I have been thinking about this intermittently, with no particular
good results. Even making a slot for one more imaginary team, thus
breaking the exact multiples. Anyone scheduled to compete against the
imaginary team would get a bye for that round.

I did find a good place to ask this question. Reading through the
forum Q&A it doesn't seem that anyone has come up with a situation
that could be mapped to this one.

<http://www.devenezia.com/round-robin/forum/YaBB.pl>

carl

--
? ? carl lowenstein? ? ? ?? marine physical lab? ?? u.c. san diego
? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?? ***@ucsd.edu

Andrew Lentvorski

2011-09-07 20:38:07 UTC

Permalink

Top-posting intentionally so as not to lose the full message.

I actually noodled on this a bit.

The problem is that in order to have the 12 skippers, 6 boats, same
number of races, same number of uses of boats, each skipper must sail
every boat, and 12 rounds max, you wind up with all 6 boats *must* be on
the water every round and each skipper sails each boat once. That means
that your rounds are either a group of 6, a round with two groups of 4
and 2, or a round with two groups of 3 and 3 and each skipper is on the
water 6 rounds of the 12.

I think that's overconstrained already by the fact that each skipper
must sail each boat once and only once. But add in the skippers must
face each other the same number of times, and it's death.

However, you can probably brute force an attempt with a program to
check. Since you *know* that all 6 boats *must* be on the water each
round and the skippers can only be on the water 6 times, the solution
space is probably feasible to brute force.

Sadly, I don't have a Prolog system readily to hand. This seems like a
problem custom made for one. And it should be able to tell you the
moment the space became overconstrained.

-a

Post by Carl Lowenstein

Post by Gus Wirth
Here's a little puzzle that I was recently sent that so far hasn't had a
response from the sailing crowd. So I thought maybe this could be viewed
as either a little math problem, or something that could be solved by an
appropriate program, hence its appearance here.
It kind of reminds me of the problems Martin Gardner would have in old
issues of "Scientific American".
Gus
-------- Original Message --------
Subject: Brain Bender Regatta Rotation
Date: Thu, 25 Aug 2011 09:49:36 -0700
From: John Fretwell
Get your geek hats on. Here's a little fun designing the format for the
MBYC Club Championship which is turning out just a bit trickier than I
envisioned. What we're facing is what I'm now calling "nested
* Assume 12 skippers. Let's call them ALPHA through LIMA. They
must all meet each other the same number of times in a fleet racing
format. I do not care whether this is done via a division into flights
(ie 4 groups of 3) or a s a pure competitor vs. competitor round robin.
But we are not desiring to have any eliminations during the series.
* Assume 6 boats are supplied by the regatta. All skippers must
sail all the boats, and the same number of times.
* This is a one day regatta, so assuming relatively short races
and time for rotations, I believe 12 total races sailed is about all we
can expect to fit in. Of course not all skippers need to sail in all 12
races, but they all have to sail the same number of races.
The heart of the problem seems to be that since two teams will sail
their first race in the same boat, and all teams must meet each other
the same number of times, all the schemes I've tried so far eventually
put teams on a collision course to be scheduled for the same boat in the
same race. The problem seems to exist whether or not you have the
competitors face off in flights or in a competitor vs. competitor round
robin.

I have been thinking about this intermittently, with no particular
good results. Even making a slot for one more imaginary team, thus
breaking the exact multiples. Anyone scheduled to compete against the
imaginary team would get a bye for that round.
I did find a good place to ask this question. Reading through the
forum Q&A it doesn't seem that anyone has come up with a situation
that could be mapped to this one.
<http://www.devenezia.com/round-robin/forum/YaBB.pl>
carl

Stewart C. Strait

2011-09-08 03:35:35 UTC

Permalink

Post by Carl Lowenstein

I have been thinking about this intermittently, with no particular
good results. Even making a slot for one more imaginary team, thus
breaking the exact multiples. Anyone scheduled to compete against the
imaginary team would get a bye for that round.
I did find a good place to ask this question. Reading through the
forum Q&A it doesn't seem that anyone has come up with a situation
that could be mapped to this one.
<http://www.devenezia.com/round-robin/forum/YaBB.pl>

...

Post by Carl Lowenstein
? ? carl lowenstein? ? ? ?? marine physical lab? ?? u.c. san diego

If I'm understanding this correctly:

Each team has 11 opponents, and must face each opponent the same number of
times. Unless the number, say x, of opponents faced by a particular team
per race varies, the number of times each opponent is faced is

N=x*(number of races for the particular team)/11.

With only six boats, x<=5 and thus x!=11 (x not equal to 11).
So N can only be an integer if each team races 11 times (or a multiple of 11).
Each team must use each boat the same number of times, so for 6 boats each
team must race a number of times divisible by 6 also. Combining these, each
team must race 66 times, or a multiple of 66.

Since 66 races is too many, some constraint should be relaxed, I think.

I'm not sure I understand what variable numbers of opponents per race
or "flights" mean, however. Any explanations, criticism, etc would be
welcome.

Thanks
Stewart Strait

Gus Wirth

2011-09-08 04:25:48 UTC

Permalink

On 09/07/2011 10:36 PM, Stewart C. Strait wrote:
[snip]

Post by Stewart C. Strait
Each team has 11 opponents, and must face each opponent the same number of
times. Unless the number, say x, of opponents faced by a particular team
per race varies, the number of times each opponent is faced is
N=x*(number of races for the particular team)/11.
With only six boats, x<=5 and thus x!=11 (x not equal to 11).
So N can only be an integer if each team races 11 times (or a multiple of 11).
Each team must use each boat the same number of times, so for 6 boats each
team must race a number of times divisible by 6 also. Combining these, each
team must race 66 times, or a multiple of 66.
Since 66 races is too many, some constraint should be relaxed, I think.
I'm not sure I understand what variable numbers of opponents per race
or "flights" mean, however. Any explanations, criticism, etc would be
welcome.

A flight is a fixed group of competitors treated as inseparable. The
example given by Mr. Fretwell would be a flight consisting of three
boats. Since there are 12 skippers total, there would be four flights
(12/3). It is customary to assign colors to each flight and to then
combine the flights to get all possible combination's. So for example:

Red - Green
Red - Blue
Red - Yellow
Green - Blue
Green - Yellow
Blue - Yellow

This kind of scheme is used to manage large fleets of boats by causing
essentially two races at a time to be run. You then use the results to
do a split (seed) who races in the Championship fleet versus the "Thanks
for coming" fleet. For the purposes of seeding you don't really care
that you have to race some competitors twice as often as the others.

For some reason this reminds me of a pseudo-random number generator.

Gus