Long Drive Golf Tournament

[Orion]

I am hosting a long drive golf tournament where there are 20 competitors.

We will have four competitors at a time on the tee box hitting during each set.

I am trying to do a round robin tournament where every competitor competes against all of the other competitors only one time.  Each of the four competitors in a set will be awarded points depending on where they finish in the group of four, 200 points for 1st, 100 for 2nd, 50 for 3rd and 25 for fourth place.  At the end of the round robin we will advance the top 8 players to finish in a match play format to finish the competition.

I have tried to figure out how to arrange the pairings but I can only get about two rounds figured out and then it gets too complicated.

Can anyone help me with this problem?  I figure there will need to be 6 rounds to compete round robin.

Any help would be greatly appreciated.

Thank you,

Orion

[Ian Wakeling]

It is known that no six round schedule, which meets your 'only one time' criterion can exist, and even if it did, it would only cover 18 out of the 19 possible opponents...   unless you have room on the box for 5 players, in which case you can do it with a sixth round of fivesomes.

   (4 16  8 15) (1  7 11 19) (14  2  9 18) ( 6 12  3 17) (13  5 10 20)
   (5 17  9 11) (2  8 12 20) (15  3 10 19) ( 7 13  4 18) (14  1  6 16)
   (1 18 10 12) (3  9 13 16) (11  4  6 20) ( 8 14  5 19) (15  2  7 17)
   (2 19  6 13) (4 10 14 17) (12  5  7 16) ( 9 15  1 20) (11  3  8 18)
   (3 20  7 14) (5  6 15 18) (13  1  8 17) (10 11  2 16) (12  4  9 19)
       (1 2 3 4 5)  (6 7 8 9 10) (11 12 13 14 15) (16 17 18 19 20)

An alternative would be to eliminate the lowest scoring player from each fivesome before the sixth round, as I imagine that, given your proposed scoring system, it is most likely that the lowest scoring player in a fivesome will no longer be able to add enough points to reach the top 8.

[Orion]

Ian,

Thank you for your help with this, I really do appreciate it.  

One of the problems that I face is the # of competitors is not set at this point but will be in a few days.  I believe it will be either 19 or 20 but could be as high as 22.  We must use exactly 4 tee slots at a time, no more or less,  so the pace of play keeps up.  I would just use a ghost in the data you sent me for player #20 if only 19 players actually competed.

If I get more entries could you do the same thing for me with 21 or 22 players? I believe 22 players would be more ideal than 21 but I would just use a ghost for any non-players.

Also, did you use a round robin generator or calculator for that and if so how did you enter that fields? I tried to do it and couldn't figure it out at all.

Ryan

[Ian Wakeling]

Ryan,

Are you wanting to play in rounds?  Where everyone plays exactly once per round as in the schedule above.  If you do, then it makes life difficult, I can't see anything at all for 22 players, but 21 can be done with mixed threesomes and foursomes (4 of each per player).

  (14 19 17 13) (12 1 6  8) (16 20 3 4) (2 10 5) (7 21 15) ( 9 11 18)
  ( 8 20 18 14) (13 2 7  9) (17 21 4 5) (3 11 6) (1 15 16) (10 12 19)
  ( 9 21 19  8) (14 3 1 10) (18 15 5 6) (4 12 7) (2 16 17) (11 13 20)
  (10 15 20  9) ( 8 4 2 11) (19 16 6 7) (5 13 1) (3 17 18) (12 14 21)
  (11 16 21 10) ( 9 5 3 12) (20 17 7 1) (6 14 2) (4 18 19) (13  8 15)
  (12 17 15 11) (10 6 4 13) (21 18 1 2) (7  8 3) (5 19 20) (14  9 16)
  (13 18 16 12) (11 7 5 14) (15 19 2 3) (1  9 4) (6 20 21) ( 8 10 17)
  (1 11 19) (2 12 20) (3 13 21) (4 14 15) (5 8 16) (6 9 17) (7 10 18)

There is no software I can recommend to you that will make the schedules above, I am using a computer but it is part mathematical knowledge too.

[Orion]

Is it possible to do a similar format as above but with 25 players? We are hosting another long drive event next week and we were hoping for 32 entries so we could have two groups of 16 but it looks like we are only going to have 25. We will still have 4 players on the tee at one time and the goal is to have everyone hit against all other players just once.  If it is possible would it take 8 rounds?

[Ian Wakeling]

The only way I can see it working would be to have rounds where there are 4 groups of 4 and 3 groups of 3  (16 + 9 = 25).  In this format 9 rounds is not quite enough to have everyone hit against each other once,  so 10 rounds are required and there will be a 30 pairs who hit against each other twice.  There is an example schedule bellow.

  ( 4  1 17  9) (11 13  6 18) (19 16 24  2) (21 14 20 12) (25 15  5) (22  8 23)  (3  7 10)
  ( 5  2 18 10) (12 14  7 19) (20 17 25  3) (22 15 16 13) (21 11  1) (23  9 24)  (4  8  6)
  ( 1  3 19  6) (13 15  8 20) (16 18 21  4) (23 11 17 14) (22 12  2) (24 10 25)  (5  9  7)
  ( 2  4 20  7) (14 11  9 16) (17 19 22  5) (24 12 18 15) (23 13  3) (25  6 21)  (1 10  8)
  ( 3  5 16  8) (15 12 10 17) (18 20 23  1) (25 13 19 11) (24 14  4) (21  7 22)  (2  6  9)
  (12 25  9  8) (11 20 22  5) (23  4 10 19) ( 1 14 15  2) (17 16  6) ( 7 18 13)  (3 24 21)
  (13 21 10  9) (12 16 23  1) (24  5  6 20) ( 2 15 11  3) (18 17  7) ( 8 19 14)  (4 25 22)
  (14 22  6 10) (13 17 24  2) (25  1  7 16) ( 3 11 12  4) (19 18  8) ( 9 20 15)  (5 21 23)
  (15 23  7  6) (14 18 25  3) (21  2  8 17) ( 4 12 13  5) (20 19  9) (10 16 11)  (1 22 24)
  (11 24  8  7) (15 19 21  4) (22  3  9 18) ( 5 13 14  1) (16 20 10) ( 6 17 12)  (2 23 25)

[Orion]

Would it make a difference if we limited the number of rounds to 5 or 6 and just did a random number generator to decide the order?  I think that what you have worked out would just be too many rounds to go through. I really appreciate your efforts in working the numbers for me, thank you.

[Ian Wakeling]

With random allocation, I would think that the players would notice that they get to play against others two or three times and others not al all.

Arranging the groups into rounds where everyone plays once is the reason that the problem is hard. Relax this requirement and then it is easy.  For example if you play the following 50 groups of 4, then everyone hits against each other exactly once.

( 1  2  6 25)  (11 13 21  4)
( 2  3  7 21)  (12 14 22  5)
( 3  4  8 22)  (13 15 23  1)
( 4  5  9 23)  (14 11 24  2)
( 5  1 10 24)  (15 12 25  3)
( 6  7 11  5)  (16 18  1  9)
( 7  8 12  1)  (17 19  2 10)
( 8  9 13  2)  (18 20  3  6)
( 9 10 14  3)  (19 16  4  7)
(10  6 15  4)  (20 17  5  8)
(11 12 16 10)  (21 23  6 14)
(12 13 17  6)  (22 24  7 15)
(13 14 18  7)  (23 25  8 11)
(14 15 19  8)  (24 21  9 12)
(15 11 20  9)  (25 22 10 13)
(16 17 21 15)  ( 1  3 11 19)
(17 18 22 11)  ( 2  4 12 20)
(18 19 23 12)  ( 3  5 13 16)
(19 20 24 13)  ( 4  1 14 17)
(20 16 25 14)  ( 5  2 15 18)
(21 22  1 20)  ( 6  8 16 24)
(22 23  2 16)  ( 7  9 17 25)
(23 24  3 17)  ( 8 10 18 21)
(24 25  4 18)  ( 9  6 19 22)
(25 21  5 19)  (10  7 20 23)

Can you use that?

[Orion]

It looks like our group is set at 24 players, any ideas of how we could do 4 on the tee at the same time?  We could possibly split into the groups of 12.

[Ian Wakeling]

If you really do want rounds (I am still not sure about this), then I can't find anything in 8 rounds, so I suggest you try the 9 round schedule below, which gives all pairs at least once and at most twice.

  (15  7 14 10) (12 21 19 16) (22  4  1 20) (24 11 17 23) ( 6  5  8 13) ( 9 18  2  3)
  (13  8 15 11) (10 19 20 17) (23  5  2 21) (22 12 18 24) ( 4  6  9 14) ( 7 16  3  1)
  (14  9 13 12) (11 20 21 18) (24  6  3 19) (23 10 16 22) ( 5  4  7 15) ( 8 17  1  2)
  (13 17  7 18) (12  9 10  1) ( 2  6 16 24) (23  8  4 19) (15 20 21  3) (14  5 11 22)
  (14 18  8 16) (10  7 11  2) ( 3  4 17 22) (24  9  5 20) (13 21 19  1) (15  6 12 23)
  (15 16  9 17) (11  8 12  3) ( 1  5 18 23) (22  7  6 21) (14 19 20  2) (13  4 10 24)
  (22  2 13 15) (24 17 14 21) (16 11 18  4) ( 9 19 23  7) ( 5 10  3  8) (20  1 12  6)
  (23  3 14 13) (22 18 15 19) (17 12 16  5) ( 7 20 24  8) ( 6 11  1  9) (21  2 10  4)
  (24  1 15 14) (23 16 13 20) (18 10 17  6) ( 8 21 22  9) ( 4 12  2  7) (19  3 11  5)

[Orion]

I think that will work for us. I really do appreciate your help with this. You are very kind to be willing to help me.

Thank you, again.