If you want more symmetry, put them at the vertices of a regular 2n-gon and pair i with 2n-1-i, giving n parallel chords, which rotate much as before. For an odd number, 0 doesn't get to play -- his opponent having the bye. In fact, the `box' description is really for this method. R. On Mon, 15 Sep 2003, Michael Kleber wrote:
On Monday, September 15, 2003, at 02:51 PM, Richard Guy wrote:
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.
Yes, I didn't think I was the first to come up with that :-). But let me stress that I had been hoping for more symmetry: this scheme has a distinguished player
Oh, and to preseve NJAS's honor, let me note that the EIS contains the counts if you leave off the symmetry part of my question:
ID Number: A000474 Sequence: 1,1,1,6,396,526915620 Name: Non-isomorphic 1-factorizations of K_{2n}.
and
ID Number: A000438 Sequence: 1,1,6,6240,1225566720,252282619805368320 Name: 1-factorizations of K_{2n}. Extension: For K_14 the answer is approx 9.8 x 10^28, for K_16 1.48 x 10^44 and for K_18 1.52 x 10^63. - Dinitz et al.
both with references:
CRC Handbook of Combinatorial Designs (see pages 655, 720-723). Dinitz, Jeffrey H.; Garnick, David K.; McKay, Brendan D.; There are 526,915,620 nonisomorphic one-factorizations of K_{12}. J. Combin. Des. 2 (1994), no. 4, 273-285. W. D. Wallis, 1-Factorizations of complete graphs, pp. 593-631 in J. H. Dinitz and D R. Stinson, Contemporary Design Theory, Wiley, 1992.
--Michael Kleber kleber@brandeis.edu
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