It's well known to those who well know it that you can put the players, 0 at the centre, and 1, ... ,2n-1 at the vertices of a regular (2n-1)-gon. Pair player i with player 2n-i and 0 with n [i.e. a radius and n-1 parallel chords]. Now rotate these n lines as a rigid conguration about the centre, giving n rounds. For an odd number of players, leave out 0 and let n have a bye. In practice this is implemented with 2n-1 unit cubes in a 2 x n box. R. On Mon, 15 Sep 2003, Michael Kleber wrote:
On Monday, September 15, 2003, at 12:05 PM, Thane Plambeck wrote:
I think they're called one factorizations
Yes K_2n always has a one-factorization, with 2n-1 "factors", that's a folk theorem. I'm not sure about cyclic ones
Ah, thank you Thane. Of course -- I knew the definition of k-factorizations, but it didn't occur to me to look things up that way. I now find:
--------- Hartman, Alan; Rosa, Alexander Cyclic one-factorization of the complete graph. European J. Combin. 6 (1985), no. 1, 45--48.
It is shown that the complete graph $K_n$ has a cyclic 1-factorization if and only if $n$ is even and $n\neq2^t, t\geq3$. Further, the number of cyclic 1-factorizations of $K_n$ for $n\leq16$ is determined. ---------
I assume that a "cyclic 1-factorization" is the symmetry I was asking about.
Since my first email, I noticed that the method I used for n=6 generalizes to any n=2*odd, but I haven't had the time to think about other cases.
That still leaves the counting problem, which I haven't thought about at all. Maybe tomorrow I'll grab a copy of the above paper and see what they did up to n=16. (NJAS fodder, I'm sure...)
--Michael Kleber kleber@brandeis.edu
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun