Where angels fear to tread. [Tho now I've finished, I guess that this must be well known to those who well know it] It's clear that each of the 2m men can oppose at most 2m - 1 others, so there can't be more than 2m - 1 rounds. The men, considered separately, play a round-robin tournament, as do the women. Ways of constructing such tournaments are well known. What is needed is a pair of `orthogonal' tournaments. Here are solutions for m = 1 and m = 2, with men odd-numbered and women even-numbered, and the extra condition that 01, 23, 45, ... are never partners. 03.21 03.65 21.47 05.27 41.63 07.43 61.25 At this point I began to suspect that a solution for m = 3 might translate (by playing two matches in each round with ends swapped, and adding a round in which couples played against themselves) into a solution of Euler's officers problem. However, we can get by with pairs of orthogonal latin squares of odd order, which always exist. I illustrate with m = 4. The general pattern should be clear. Take two orthogonal latin squares of order 2m - 1 and append a column [2m,2m,...,2m] to the right of one and to the left of the other: 1 3 5 7 2 4 6 8 8 1 2 3 4 5 6 7 2 4 6 1 3 5 7 8 8 3 4 5 6 7 1 2 3 5 7 2 4 6 1 8 8 5 6 7 1 2 3 4 4 6 1 3 5 7 2 8 8 7 1 2 3 4 5 6 5 7 2 4 6 1 3 8 8 2 3 4 5 6 7 1 6 1 3 5 7 2 4 8 8 4 5 6 7 1 2 3 7 2 4 6 1 3 5 8 8 6 7 1 2 3 4 5 The Left one comprises Ladies, say, and the Right one Men. Since each Lady opposes 2m - 1 others, she partners all but one of the Men. We can pair these in any we like, so if Alice is averse to alf, Beth bothers bill, Carol crosses colin, Daisy dislikes dick, Evie execrates ed, Fay fights fred, Grace grudges george and Hattie hates harry, then we can number them A1 B2 C3 D4 E5 F6 G7 H8 a7 b2 c4 d6 e1 f3 g5 h8 and the partners for Sun are Ah Bc Ce Dg Eb Fd Gf Ha Mon are Ag Bh Cd Df Ea Fc Ge Hb Tue are Af Ba Ch De Eg Fb Gd Hc Wed are Ae Bg Cb Dh Ef Fa Gc Hd Thu are Ad Bf Ca Dc Eh Fg Gb He Fri are Ac Be Cg Db Ed Fh Ga Hf Sat are Ab Bd Cf Da Ec Fe Gh Hg Now arrange an ordinary round-robin for the ladies (or the men); for example A v B C v H D v G E v F A v C D v B E v H F v G A v D E v C F v B G v H A v E F v D G v C H v B A v F G v E H v D B v C A v G H v F B v E C v D A v H B v G C v F D v E and let the men (or the ladies) tag along Sun Ah.Bc Ce.Ha Dg.Gf Eb.Fd Mon Ag.Cd Df.Bh Ea.Hb Fc.Ge Tue Af.De Eg.Ch Fb.Ba Gd.Hc Wed Ae.Ef Fa.Dh Gc.Cb Hd.Bg Thu Ad.Fg Gb.Eh He.Dc Bf.Ca Fri Ac.Ga Hf.Fh Be.Ed Cg.Db Sat Ab.Hg Bd.Gh Cf.Fe Da.Ec R. On Mon, 5 Jan 2004, R. Hess wrote:
A happy new year problem for math funsters.
Tennis -- Mixed Doubles Round Robin 5 courts in play 10 men(M), 10 women(W) (say the players are numbered 1-20, the men being odd and the women even) All matches are M/W vs M/W No person partners any person more than once. No person opposes any person (M or W) more than once.
I have a solution for 5 rounds (25 matches) obeying the above requirements. Is it possible to do more rounds? What is the general limit for 2m men and 2m women playing mixed doubles?
Happy puzzling, Dick Hess