Hi Nick,
You can't satisfy the "everyone with everyone else" requirement with only 4 rounds, even a player such as A & E who are always in a foursome will only have 12 people with whom they play with, meaning there must be at least one other, from the 13 possible opponents, with whom they never meet. Your schedule also has pairs who meet twice, e.g. JM & KN.
There is a way to make a schedule without repeated pairs - for example take
the first 5 rounds corresponding to part 0 from here. Delete the first of the 5 rounds, and delete 14 & 15 from the remaining 4 rounds, which after some tidying up will leave you with something like this:
{ 8 9 2 4} { 0 1 10 12} {11 5 7} { 6 3 13}
{13 2 12 7} { 6 8 1 11} { 0 9 3} { 5 10 4}
{12 5 6 9} { 3 8 10 7} {11 0 2} { 4 13 1}
{ 3 4 11 12} { 5 0 13 8} { 2 6 10} { 1 7 9}
However, this is similar to your schedule as 8 and 12 are always in a foursome. My gut feeling is that you have to choose between having, no repeated pairs, or having everyone at least once in a threesome - you can't have both at the same time.
Ian
