Ian Wakeling

With mixed doubles there is a problem. Everyone has 10 possible partners and 10 possible opposite sex opponents, but only 9 possible same sex opponents. A common solution is to go for 9 games each, forbidding one specific opposite sex pairing per player. Usually this is presented as a schedule for 10 husbands (H) and 10 wives(W), with a spouse avoidance property, however it can of course be used for any group of 10 men and 10 women. There is an additional problem when there are 20 mixed doubles players. While it's possible to list all the games required: (H1 W3 v H2 W6) (H1 W9 v H3 W8) (H1 W7 v H4 W5) (H1 W0 v H5 W3) (H1 W8 v H6 W0) (H1 W2 v H7 W4) (H1 W6 v H8 W7) (H1 W5 v H9 W2) (H1 W4 v H0 W9) (H2 W4 v H3 W7) (H2 W1 v H4 W9) (H2 W8 v H5 W6) (H2 W0 v H6 W4) (H2 W9 v H7 W0) (H2 W3 v H8 W5) (H2 W7 v H9 W8) (H2 W5 v H0 W1) (H3 W5 v H4 W8) (H3 W2 v H5 W1) (H3 W9 v H6 W7) (H3 W0 v H7 W5) (H3 W1 v H8 W0) (H3 W4 v H9 W6) (H3 W6 v H0 W2) (H4 W6 v H5 W9) (H4 W3 v H6 W2) (H4 W1 v H7 W8) (H4 W0 v H8 W6) (H4 W2 v H9 W0) (H4 W7 v H0 W3) (H5 W7 v H6 W1) (H5 W4 v H7 W3) (H5 W2 v H8 W9) (H5 W0 v H9 W7) (H5 W8 v H0 W4) (H6 W8 v H7 W2) (H6 W5 v H8 W4) (H6 W3 v H9 W1) (H6 W9 v H0 W5) (H7 W9 v H8 W3) (H7 W6 v H9 W5) (H7 W1 v H0 W6) (H8 W1 v H9 W4) (H8 W2 v H0 W7) (H9 W3 v H0 W8) there is no known way of arranging them into 9 rounds of 5 matches each, where each player plays exactly once per round. If this is what you were looking for then it may be possible for you to rearrange the games above so you have rounds of three or four games each.
