Math CentralQuandaries & Queries


Question from Tyler:

I'm scheduling a social curling league where all skips will play with all thirds, all seconds, all leads. And all thirds will do the same and so on. We have 6 teams of 4 and will be doing 6 rotations. Is it possible that all skips will play with all other players from other positions without anyone doubling up?
My initial thoughts were (1,1,1,1) (2,2,2,2) (3,3,3,3) (4,4,4,4)(5,5,5,5)(6,6,6,6)... then rotate the thirds up and seconds down but don't know what to do with the leads and even with just 3 positions there's doubling (1,2,6,x)(2,3,5,x)(3,4,4,x)...4s have doubled (4,5,3,x)(5,6,2,x)(6,1,6,x) as you can see I'm having problems
If you could let me know if this is even possible it would be greatly appreciated


It is not possible to have every pair together in the way you describe. What you are looking for corresponds to a mathematical object known as a resolvable transversal design with 4 groups of size 6. This is known not to exist. (Because it is equivalent to the existence of 3 mutually orthogonal Latin squares of side 6, which were shown not to exist by the Gaston Tarry in around 1900.)

The problem arises mainly because of the number 6. If, for example, there were 7 teams, then you could do it. That opens the door to an approximate solution that is not perfect, but might not be too bad.

Imagine you have 7 teams in round 1. (Team 7 is a fake.) In round 2, shift the leads by 1 (the lead from 7 wraps around to team 1), the seconds by 2 (wrapping around similarly, 6-> 1, 7 -> 2) and the thirds by 3 (again, wrapping around similarly. Now take the lead from team 7 and use it to replace lead #7 wherever that player occurs. Do the same thing with the second and third from team 7. (Hope this makes sense.) Now delete the seventh team in each round, and all of round 2.

Good luck.

About Math Central


Math Central is supported by the University of Regina and the Imperial Oil Foundation.
Quandaries & Queries page Home page University of Regina