SAMDRRlogician series

This is another puzzle from the field of design theory. The problem here is reportedly taken from real life. Brayton, Coppersmith and Hoffman (`Self-orthogonal Latin squares of all orders n not equal to 2, 3 or 6’, Bulletin of the American Mathematical Society vol. 80, 1974) report that they were asked by the Briarcliff Raquet Club in Briarcliff, New York to schedule a tennis tournament known as a “spouse-avoiding mixed doubles round robin” or SAMDRR.

In tennis, a mixed doubles match consists of two teams each of which comprises one man and one woman. In a spouse-avoiding mixed doubles tournament, players enter in man-woman couples (referred to as husband and wife). The tournament is a round robin (everyone plays against everyone else exactly once, and everyone partners everyone of the opposite sex exactly once) except that husband and wife are kept apart. Thus:

Husband and wife never appear in the same match, either as partners or opponents.
Each pair of players of the same sex oppose each other in exactly one match.
Each pair of players of the opposite sex, if not married to each other, play in exactly one match as partners and exactly one match as opponents.

Find a SAMDRR tournament schedule for 4, 5 and 7 couples. Note that such a schedule cannot exist for 2, 3 or 6 couples.

HARDER VERSION: A SAMDRR is “resolvable” if it can be partitioned into rounds such that, if the number of couples is even, everyone plays in each round (the matches in a round taking place in parallel). If the number of couples is odd, each couple sits out for one round while everyone else plays.

Find resolvable SAMDRR schedules for the same numbers of couples as before.

It is known that a SAMDRR exists for all possible numbers of couples except for 2, 3 and 6. It is known that a resolvable SAMDRR exists for all numbers except for 2, 3, 6 and 10 and with the possible exception of 14. This result is quite recent (Burger, Kidd and van Vuuren, ‘There is no unipotent SOLSSOM of order 10’, 2009).

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