The Math Forum



Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.


Math Forum » Discussions » sci.math.* » sci.math

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: Kirkman schoolgirl problem redux
Replies: 2   Last Post: Sep 21, 2007 4:38 PM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
Bill Daly

Posts: 69
Registered: 12/8/04
Kirkman schoolgirl problem redux
Posted: Sep 20, 2007 3:00 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

I can't account for this, so maybe somebody on this forum can provide
an explanation.

The Kirkman schoolgirl problem has been widely reported to have
exactly 7 independent solutions. I'm interested in "movements" as they
relate to tournaments, so I constructed a solution of my own based on
a movement. For me, it's easier to describe the solution in a
tournament context, so I'll restate the problem in these terms. There
are to be 15 players playing at 5 tables in groups of three (could be
skat, or maybe single-dummy bridge); the event will consist of 7
rounds, in such a way that no player faces another more than once,
which in turn implies that each player faces each other player exactly
once. The movement consists of three sets of players, one stationary,
and the remainder divided into two sets of 7 who move cyclically. This
description assumes that player 15 is stationary, and that the cyclic
sets are the odd numbers {1,3,5,7,9,11,13} and the even numbers
{2,4,6,8,10,12,14}, with the convention that within each cycle, a
player "follows" the player with the next lower number within the same
set (1 follows 13 and 2 follows 14). The rounds are:

15,1,2 3,5,9 7,6,12 11,4,14
13,8,10
15,3,4 5,7,11 9,8,14 13,6,2
1,10,12
15,5,6 7,9,13 11,10,2 1,8,4
3,12,14
15,7,8 9,11,1 13,12,4 3,10,6
5,14,2
15,9,10 11,13,3 1,14,6 5,12,8 7,2,4
15,11,12 13,1,5 3,2,8 7,14,10 9,4,6
15,13,14 1,3,7 5,4,10 9,2,12
11,6,8

(I apologize if this is unreadable on your computer; it works best
with a fixed width font.)

I'm pretty sure this works (I've stared at it a lot). So the natural
question is, which of the 7 standard solutions is it equivalent to?
Well, I can't find any isomorphism from this solution to one of the
standard solutions. Either it is an 8th solution independent of the
others, or I have horribly misunderstood something. Any opinions?

I can describe my procedure for finding equivalent solutions, but
this message is already long enough, so I'll leave that aside for the
moment.




Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum at NCTM 1994-2018. All Rights Reserved.