Comments on: Kirkman’s schoolgirls and their friends

By: Futamarka

Futamarka — Wed, 27 Feb 2013 00:20:50 +0000

Он сбежал из дворца, и от женщин, и от богатств, и от роскоши, от всего… Поэтому я не против Зорбы-Грека, потому что Зорба-Грек — это сама основа Зорбы-Будды. Будда вырастает из этого опыта. Я целиком за этот мир, потому что я знаю, что другой мир может быть почувствован только через этот. Поэтому я не говорю, что его следует избегать, я не скажу вам: станьте монахом. Монах — это тот, кто пошел против Зорбы, он беглец, трус; он сделал что-то в спешке, неразумно. Он незрелый человек. Монах — неспелый, жадный — жаждущий иного мира, и хочет его слишком рано, а время еще не пришло, монах еще не созрел.

By: Ian Wakeling

Ian Wakeling — Mon, 23 Apr 2012 07:32:35 +0000

I have a solution to the (16,4) problem here:

http://www.devenezia.com/round-robin/forum/YaBB.pl?num=1312720662

I found it by testing all possible partitions of the 35 resolution classes of the 3-design. There are other partitions of the same design possible, but only the one shown can be described so simply.

By: Michael Steyer

Michael Steyer — Wed, 30 Nov 2011 08:04:34 +0000

Thanks a lot for the elegant solution and understandable explanation of factor 2-cover construction! This settles parts of a question that I posed myself 30 years ago when I designed soccer tournaments at high school. We were 8 boys and wanted to play a tournament with 4 teams each day so that each pair of players plays vs each other pair of players exactly once. I spent some classes with trying to find a design – of course in vain because there is no solution.

Many years later I rediscovered this problem, renamed it doubles tennis tournament design and found a solution for 10 players, first by something like brute force on a PC, later by design. I think that solution is not isomorphic to the one described here: I started with 8 players and designed 7 days by taking pairs of players from adjacent columns from the addition table of GF(8), adding (9,10) to each day. Then I left 10 at its place and permuted players 1-9 according to the rows of the addition table of GF(9). I wonder whether this is of any interest – in particular because this method seems to fail for n=18 – but I can provide details on request.

By: Randy Hudson

Randy Hudson — Fri, 25 Nov 2011 02:59:21 +0000

I found this blog while searching for a construction of a solution to an analog for the (16,4) case of Sylvester’s extension to Kirkman’s schoolgirls. That is, with 16 girls walking in groups of 4, they cam form different groups on each of the 5 weekdays. Can they do that in each of 7 weeks, such that no three of them are contained within such a group of 4 more than once?

I can find an arrangement of 140 S(3,4,16) into 35 collections of 4, so each set of 4 is a S(1,4,16). And of course the “Kirkman” problem, arranging the 20 elements of an S(2,4,16) into 5 partitions, each a S(1,4,16), is easily solved.

Has this problem been solved, or proven insoluble? If not, do you have any insight that might help me find a solution?

By: Gordon Royle

Gordon Royle — Fri, 03 Jun 2011 01:53:01 +0000

I don’t know the answer to your question, but is it possible that one factor 2-cover could give rise to different projective planes, by choosing a different factorisation to in order to “complete the plane”.

It would be nice to be able to make some new projective planes – the number of known planes of order 16 has been stuck for quite a while, but I am not sure whether anyone believes it is complete.