20 Golfers 2 Groups of 10

[darcimer]

Hey guys,

I am still trying to figure this out for a golf outing.  I have 10 players in Group A.  They will play each other member of Group A 1x for 9 weeks.  Easy. 
Then I have Group B, who will also play each other simultaneously (along with Group A) 1x each for 9 weeks. Still easy using 5 foursomes each week.


How do you schedule it so that group A players match up with group B players 2x during the same 9 week span?  I'm lost.
The goal is to not have any golfer match up with a player from the opposite group more than twice.  I realize there will be 2 players that will only get 1 match up (or 1 player not matched up up at all since 9 weeksx2 players=18 matchups, not 20 needed to be perfect.)

Thanks in advance,

Todd A.

[Ian Wakeling]

Hi Todd,

I have moved your post here as I thought it deserved it's own thread.

There is good news and bad news.  The good: there is a way to do what you want if you go for the option where a player is not matched up at all with one player from the opposite group.  The schedules here will do this if they are rearranged slightly.  For each doubles game such as [(H1 W2):(H5 W4)] consider H to be group A, and W to be group B, so this says that A1 plays A5, and B2 plays B4 in the same foursome.  If you build the schedule like this then it will have the properties you want, and the pairs that never oppose will be (A1,B1), (A2,B2), etc.

The bad news is that a schedule for 20 players does not exist.  There is a link that shows the doubles games necessary, however it is not possible to arrange them in rounds of 5 foursomes.

Hope that is of some help.

Ian

[darcimer]

Hi Todd,

I have moved your post here as I thought it deserved it's own thread.

There is good news and bad news.  The good: there is a way to do what you want if you go for the option where a player is not matched up at all with one player from the opposite group.  The schedules here will do this if they are rearranged slightly.  For each doubles game such as [(H1 W2):(H5 W4)] consider H to be group A, and W to be group B, so this says that A1 plays A5, and B2 plays B4 in the same foursome.  If you build the schedule like this then it will have the properties you want, and the pairs that never oppose will be (A1,B1), (A2,B2), etc.

The bad news is that a schedule for 20 players does not exist.  There is a link that shows the doubles games necessary, however it is not possible to arrange them in rounds of 5 foursomes.

Hope that is of some help.

Ian

Hi Ian,
Thanks, but I am not sure I understand 100%.  your link goes to:

(H1 W3 v H2 W6) (H1 W9 v H3 W8) (H1 W7 v H4 W5) (H1 W0 v H5 W3) (H1 W8 v H6 W0)
(H1 W2 v H7 W4) (H1 W6 v H8 W7) (H1 W5 v H9 W2) (H1 W4 v H0 W9) (H2 W4 v H3 W7)
(H2 W1 v H4 W9) (H2 W8 v H5 W6) (H2 W0 v H6 W4) (H2 W9 v H7 W0) (H2 W3 v H8 W5)
(H2 W7 v H9 W8) (H2 W5 v H0 W1) (H3 W5 v H4 W8) (H3 W2 v H5 W1) (H3 W9 v H6 W7)
(H3 W0 v H7 W5) (H3 W1 v H8 W0) (H3 W4 v H9 W6) (H3 W6 v H0 W2) (H4 W6 v H5 W9)
(H4 W3 v H6 W2) (H4 W1 v H7 W8) (H4 W0 v H8 W6) (H4 W2 v H9 W0) (H4 W7 v H0 W3)
(H5 W7 v H6 W1) (H5 W4 v H7 W3) (H5 W2 v H8 W9) (H5 W0 v H9 W7) (H5 W8 v H0 W4)
(H6 W8 v H7 W2) (H6 W5 v H8 W4) (H6 W3 v H9 W1) (H6 W9 v H0 W5) (H7 W9 v H8 W3)
(H7 W6 v H9 W5) (H7 W1 v H0 W6) (H8 W1 v H9 W4) (H8 W2 v H0 W7) (H9 W3 v H0 W8)

So what you are saying is that these are the matchups that have to happen, but there is no way to schedule them?  I see that H1 plays every match in the top row, which is Week 1 in my example.

Group A matching up with every other A opponent works perfect, as does Group B.  So in my scenario, what is the optimal way to pair up the B's with A's, knowing that there is no possible balance?  What if Group A skips a group B player, could it be done then?  I'm not concerned if it's unbalanced, as long as it's close.  My attempts so far always end up with 4+ pairings with same opponent when I start getting deeper into the schedule.

Thanks again, I had no idea there were others out there taking this seriously!!

Todd

[Ian Wakeling]

Todd,

Yes, you are correct, those are the 45 matchups that need to happen, but there is no way of arranging them so they can be played in 9 rounds of 5 matchups each (this has been proven mathematically).  I have a version which is 9 rounds of 4, and 3 rounds of 3, but clearly this is 3 weeks extra and a lot of byes - I can post the schedule if you are interested.

If you can wait 3 or 4 days, I may be able to find a schedule which can better your 4+ pairings.

Ian

[darcimer]

Todd,

Yes, you are correct, those are the 45 matchups that need to happen, but there is no way of arranging them so they can be played in 9 rounds of 5 matchups each (this has been proven mathematically).  I have a version which is 9 rounds of 4, and 3 rounds of 3, but clearly this is 3 weeks extra and a lot of byes - I can post the schedule if you are interested.

If you can wait 3 or 4 days, I may be able to find a schedule which can better your 4+ pairings.

Ian

Thanks Ian, I'd love to see what you come up with.  No urgency on my end.  Thankfully, I have given up trying to force excel to do something that's mathematically impossible.  I will keep tinkering with it in the meantime.  I have read quite a bit on this site and have learned a lot, thank you..  I noticed many requests for similar schedules, so hopefully a better solution benefits the group..

Todd A

[Ian Wakeling]

Todd,

I wrote some code to optimize your 20 golfers.  At the moment the best schedule I am getting looks like this:

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

Above 1-10 represent Group A, and 11-20 represent Group B. There are no 4+ pairings for any pair of players from different groups.  There are precisely 6 pairs from different groups that occur 3 times, the remaining pairs all occur once or twice.  If I get anything better, then I will let you know.  Hope that helps,

Ian

[darcimer]

Todd,

I wrote some code to optimize your 20 golfers.  At the moment the best schedule I am getting looks like this:

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

Above 1-10 represent Group A, and 11-20 represent Group B. There are no 4+ pairings for any pair of players from different groups.  There are precisely 6 pairs from different groups that occur 3 times, the remaining pairs all occur once or twice.  If I get anything better, then I will let you know.  Hope that helps,

Ian

 You are too kind Ian, thanks for your help.  It is definitely useful..

Todd