3 Player Round Robin Tournament

[kubkarracer]

Hello,

We are having a pine car rally on a 3 car track.  I need to be able to generate a round robin tournament once the registration is complete on the day of the event.  We expect to have around 20 cars, but it could be a few more or less.  The tournament should have 3 cars in each race as often as possible and have every car participate in the same number of races and against every opponent once.

I haven't been able to find a generator that handles this case. I would appreciate any help you can give.

Thanks!

[Ian Wakeling]

I don't know of software that will solve the general problem, so I think you will have to build up a set of schedules for each number of players.

Look here for a 21 car solution, it the best number to have, since all your races can be with 3 cars and blocks of 7 races will have each car exactly once.  You could adapt this schedule for 24 cars, the missing pairs at the end would correspond to 12 races with 2 cars that I suggest you hold at the beginning of the tournament.  There is a similar solution for 20 cars and 70 races that I have given below.

R1  (1 - 14)
R2  (2 - 6)
R3  (3 - 9)
R4  (4 - 10)
R5  (5 - 13)
R6  (7 - 20)
R7  (8 - 17)
R8  (11 - 15)
R9  (12 - 19)
R10  (16 - 18)
R11  (4 15 18)
R12  (3 5 17)
R13  (2 13 19)
R14  (16 8 14)
R15  (10 18 3)
R16  (4 2 14)
R17  (11 2 1)
R18  (7 16 15)
R19  (8 1 10)
R20  (17 16 1)
R21  (7 3 13)
R22  (4 8 19)
R23  (15 13 9)
R24  (9 6 11)
R25  (14 20 5)
R26  (12 11 18)
R27  (9 20 12)
R28  (19 17 20)
R29  (5 10 6)
R30  (7 6 12)
R31  (19 1 7)
R32  (18 9 17)
R33  (6 19 15)
R34  (15 5 2)
R35  (1 3 6)
R36  (2 12 10)
R37  (13 4 20)
R38  (1 9 4)
R39  (8 18 7)
R40  (11 4 17)
R41  (3 19 16)
R42  (3 14 15)
R43  (10 14 13)
R44  (16 11 5)
R45  (9 5 8)
R46  (14 17 12)
R47  (8 20 6)
R48  (20 18 2)
R49  (10 7 11)
R50  (13 12 16)
R51  (14 7 9)
R52  (12 15 8)
R53  (20 11 3)
R54  (15 1 20)
R55  (12 4 3)
R56  (18 19 5)
R57  (16 9 2)
R58  (6 17 13)
R59  (17 2 7)
R60  (18 6 14)
R61  (17 15 10)
R62  (6 16 4)
R63  (2 3 8)
R64  (13 8 11)
R65  (19 10 9)
R66  (20 10 16)
R67  (5 7 4)
R68  (1 13 18)
R69  (5 12 1)
R70  (11 14 19)

Races 11 to 70 have all cars three times in each of the three columns (tracks).

If these look good, I will try to do a few more for you.

[kubkarracer]

Thank you Ian for the link and solution for 20 cars, I think with those we can make it work for us by throwing byes in there as needed.

I had another thought while reading your reply. It might be better to divide the field into pools of six cars each (or as close as we can get to it).  All cars will have the same number of races, and race each other car in the pool at least once.  We then use the results to seed the field into a traditional single elimination head to head draw to determine the winner.

If we do this then I would only need solutions for 5, 6 and 7 (or 6, 7 and 8 if that is better) cars to handle the variations that might occur with the number of participants.  Would you be able to find the solutions for this case?

Thank you again for your help, it is much appreciated.

[Ian Wakeling]

I think 6 or 7 cars per pool are the best choices.

With 6 cars you will need to race against every other car twice in a total of 10 races:

3 5 1
4 2 3
5 6 4
6 3 1
1 2 4
6 5 2
4 1 5
3 4 6
2 3 5
2 1 6

7 cars either or both of the following (either 7 or 14 races per pool):
1 2 4
2 3 5
3 4 6
4 5 7
5 6 1
6 7 2
7 1 3

1 2 6
2 3 7
3 4 1
4 5 2
5 6 3
6 7 4
7 1 5

Hope that helps.

[kubkarracer]

It helps a lot, thank you!