Golf Foursomes: Twelve Players Question: Is it possible to arrange a multi-round tournament involving twelve players in such a way that each player gets matched in a foursome with the other 11 players at least once? Answer: Yes, it is possible to make foursome assignments such that every players gets to golf with all others, but we must be willing to allow repeats and “three-peats” This problem is surprisingly harder to solve than the 16, 20, or 24 player situation. For those, I used an approach based upon Latin squares. In the 12 player case, however, the number of possible 3x3 Latin squares are too few; there are only two orthogonal 3x3 Latin squares, plus the degenerate square, leaving us one short of what we would need for 12 players. In a way, this suggests that we must rely upon “three-peats” to accomplish our goal anyway. (For an explanation of Latin squares, see the 16 player case.) The methodology for twelve players After hunting around, I encountered an approach that appeared promising: one based upon graphs. In this case, a symmetrical graph based upon 12 regions seemed to be a good starting point. The graph I chose contains regions that are bounded by four other regions. This graph seems promising in that there appear some options for exploring disconnected regions as forming the foursomes. Let's investigate this approach, next. The starting point using set of non-adjacent regions There are three unique arrangements where a given region can be included in a set of other non-bounding regions, as shown in the three following color-coded graphs: The mappings above are equivalent to the following foursome assignments, where each figure represents the foursome assignment for a single round: As is evident here, we have already incurred repetition of players (AB, CD, and FG). Repetition at such an early stage is an expected outcome since it is impossible to assign completely unique foursomes with 12 players after the first round. The above three graphs represent the only options of non-adjacent arrangements, yet we have not achieved our goal of pairing every player with every other player. So, at this stage, we have no choice but to consider sets containing neighboring regions. Neighboring regions If you look carefully at the graph, you will see that it is possible to select pairs of adjoining regions. One such selection is shown, next. This results in the following foursome assignment. It is tempting to continue following this approach, following the symmetry involving separated pairs of neighboring regions. However, I found that I started incurring repeated pairings rather quickly. For instance, simply rotating the coloring is an option, yet that choice results in a repeated pairing of the central region with the outermost region. Therefore, we have exhausted the unconnected pairs as an option. That leaves us with one remaining option: breaking symmetry. Breaking symmetry Here, I could not find a steadfast rule in selecting the regions. Through trial and error experimentation, I found the following three mappings that, in combination with the previous four mappings yielded my desire goal - that of pairing every region with every other yet no more often than three times throughout. These mappings yields three more unique foursome assignments: The end result is the following table of 7 rounds: Synopsis Every player get paired with every other player at least once. The number of repeats and "three-peats" is fairly evenly distributed across all players. No player is paired with another player four times. Next: Sixteen Players