Round Robin Tournament Scheduling

Recent Posts

21
Requests / Re: 40 PLAYER, 6 ROUND EUCHRE TOURNAMENT
« Last post by Ian Wakeling on October 26, 2022, 09:34:41 AM »
There is a 40 player 13 round schedule in the top half of my reply #10 here. Use the first six rounds for your daughter's fundraiser.
23
Requests / 40 PLAYER, 6 ROUND EUCHRE TOURNAMENT
« Last post by Elizabeth-Anne Ward on October 24, 2022, 06:06:19 PM »
Would anyone be able to help me with a schedule for the above? I can't get my head wrapped around it and my daughter is running one as a fundraiser and I am trying to help her and can't find one online anywhere. Any help would be greatly appreciated. Thank you!
24
Requests / Re: Am I asking for the impossible?
« Last post by Ian Wakeling on October 23, 2022, 04:47:02 AM »
Ray,

I have been having another look at 26 players and using some statistical experimental design software, I can get an improvement.  Your excel file has a schedule with parameters (52,10,5) for (never meet, partner twice, partner/oppose once). The schedule below has parameters (46,0,9) where players 11 to 26 are the ones who get a bye.

(12  3 18  1) ( 7 23 25 14) (20 22 17 15) ( 8 21  2  4) (11  5 10 13) ( 6 19  9 24)  (16 26)
(23 12  4 10) ( 2 18 14 20) ( 1  9 25  5) ( 3 13  6 17) (19 26 15  8) (21 16  7 24)  (11 22)
( 2 22 10 16) (23 26  6  5) ( 4 17 19  1) (12 11 24 15) ( 9 14 21 13) (20  7  8  3)  (18 25)
( 6 10  8 20) (26 21 11  1) (24  3  2  5) (15 13 18 23) (25 19 12 16) (22  4  7  9)  (14 17)
(18 11  6  2) ( 3 16 15  9) ( 8 23 22  1) (26 13  4  7) (14  5 19 20) (21 17 25 10)  (12 24)
(25 11 22  3) ( 1 16 13 20) (24 17 14  8) ( 7 18 19 10) ( 6  4 15  5) (12  2 26  9)  (21 23)
( 4 14 16 11) (13 25  8  2) (10  1  9  6) ( 3 23 19 21) (12  5 17  7) (24 18 22 26)  (15 20)
( 2  1  7 15) (23 17  9 11) ( 3 14 26 10) (16  8 18  5) (21 12 22  6) ( 4 24 20 25)  (13 19)

as with the other schedules the match (A B C D) should be read as (A B) vs (C D).
25
Requests / Re: Am I asking for the impossible?
« Last post by raydog on October 18, 2022, 08:58:02 PM »
Certainly can't argue with your logic, Ian, and get another ideal solution! I'm updating my spreadsheet with each update, will repost once you get tired of optimizing my results.

Thanks again!

Ray
26
Requests / Re: "Whist" with two player pools, limited rounds, and 4 courts
« Last post by BigDonkey on October 18, 2022, 01:58:39 PM »
I do something similar, and if you're willing to make a compromise, you can make it work for any number of players as long as you play exactly 4 rounds. An even number of players always works out, while with an odd number of players you will have to have one match be all players from the B-pool play together. It's good enough for my purposes of keeping all the games somewhat evenly matched and getting everyone 4 games.

17 players:

Rd 1: A1B7 vs A3B4     A4B2 vs A7B3     A5B8 vs A8B6     A6B5 vs A2B9     (B1)

Rd 2: A4B3 vs A5B9     A1B2 vs A8B7     A2B8 vs A7B1     A3B5 vs A6B4     (B6)

Rd 3: A3B9 vs A7B2     A4B1 vs A6B7     A1B5 vs A5B6     A2B3 vs A8B8     (B4)

Rd 4: A2B5 vs A3B7     A5B4 vs A7B8     A6B1 vs A8B2     A1B6 vs A4B9     (B3)

Rd 5: B1B3 vs B4B6  
27
Requests / Re: Am I asking for the impossible?
« Last post by Ian Wakeling on October 18, 2022, 01:39:42 PM »
Consider the 18 player tournament, and count the pairs in the schedule in different ways.  With 8 rounds and 4 tables, there are 32 games.  Each game has 2 partner pairs and 4 opposition pairs, so over the tournament 64 partner pairs and 128 opposition pairs, or a total of 192 pairs.  With 18 players there are a possible 153 distinct player pairs, so the ideal arrangement, with no player pairs who never meet, or player pairs who meet 3 or more times, has:

114 pairs who meet once(x)
39 pairs who meet twice(y)

this follows because these numbers are the only solutions to:

x + y = 153
x*1 + y*2 = 192

As we have agreed, the best schedule will have all 39 pairs who meet twice of the type where 2 players partner once and oppose once. Therefore, by subtraction we can discover how the players pairs who meet once are distributed:

64 - 39 = 25 pairs who partner once
128 - 39 = 89 pairs who oppose once.

So putting this together the ideal parameters are as follows:
Never meet = 0
Partner once = 25
Oppose once = 89
Oppose twice = 0
Partner once + Oppose once = 39
Meet 3+ times = 0

This is not that far away from your Excel sheet.  So I went looking, and I can find such a schedule by computer search:

(13  6  5  1) (14  3  7 16) ( 2 10 15 11) (17 12 18  9)  ( 4  8)
( 6 11 12 15) ( 2  4 13 16) (17  8 10  3) ( 7 18  1  9)  (14  5)
(14  8  1 12) (11 17  7 13) (15  6 16  9) ( 5  2  4 18)  (10  3)
(13  1 10 16) ( 5 11  9 14) ( 3  4 18 12) ( 7  6  2  8)  (15 17)
(15 17  1  4) (11 13  8  3) (18  2 14 10) (16  5 12  7)  ( 9  6)
(18 14 15 13) ( 6  4 10  7) ( 5 17 16  8) ( 1  2  9  3)  (12 11)
( 3  5 10 15) (11 18 16  6) ( 8  9 14  4) ( 2 12  1 17)  ( 7 13)
( 3 17  6 14) ( 9 12 10 13) (15 18  7  8) ( 4  5 11  1)  ( 2 16)
28
Requests / Re: Am I asking for the impossible?
« Last post by raydog on October 17, 2022, 10:42:12 AM »
Ah, you caught me out! Those are more or less made up. I think I just applied some ratios I thought seemed reasonable. But there is in fact no reason to think those numbers are achievable (sorry if that was misleading - part of my hesitancy in posting this spreadsheet was the fact that there are numbers like that included!)

The cardinal rules are:
1) a pair of player never meets 3 times (which is now the case in all scenarios, thanks to your help with the 15-player scenario);
2) no player gets 2 byes until all other players have had 1 bye.

So the important numbers in that "ideal" scenario are the "never meet" number (which I strive to make 0 if there are more "possible meetings" than "possible pairings"), and the "meet 3 times" number (which I strive to make 0: solved). It's that first stipulation which may not actually be possible.

When I coded my program, I needed to specify what made a given solution better than any previous solution. So if "never meet" was the same for two scenarios [and "meet 3 times" was 0], I applied a 2nd order comparison, and prioritized "partner once + oppose once" over "oppose twice".  This just seemed reasonable, on a human level. So yes, you have interpreted correctly, but the 1st order priority is minimizing "never meet".
29
Requests / Re: Round Robin - 24 Players - 3 per board
« Last post by Ian Wakeling on October 17, 2022, 03:14:33 AM »
There is no perfect solution to this scheduling problem.  My reccomendation would be to use a schedule from Ed Pegg's page here.  Look for the one with the title "Can 24 golfers play in threesomes for 11 days?", this works but you must be prepared to have one round of 4 players per board.
30
Requests / Re: Round Robin - 24 Players - 3 per board
« Last post by Bandit on October 15, 2022, 01:24:00 PM »
Hi,

Yes, the tree players will play each other in a round robin. I require 8 groups of 3.