Round Robin Tournament Scheduling

Euchre Tables, 28 and 32 players, 12 Rounds

k48038 · 9 · 1734

k48038

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on: September 10, 2017, 09:43:53 AM
I am trying to put together a schedule for a 12 round euchre tournament for 28 or 32 players (I do not have a final count, yet).

I am looking for a schedule that will have each player partner with a new person each round while attempting to minimize the number of times any individual player is at the same table with each other player, while trying to get most players to be seated at the same table with most other players at least once.  (Try to get as much "mingling" of players as possible over the course of the 12 rounds).

Similar to another posting from this site that was solved very well for that case: here.

What I am in the process of doing is using the output from your tournament scheduler (https://www.devenezia.com/downloads/round-robin/rounds.php) for 28 players, then randomly selecting 12 of the 27 rounds provided by the scheduler, and then going through a tedious process (using lots of excel functions and hit or miss manual manipulation of the pairing and partners) of getting a mix of players, partners, and opponents.  

I wonder if there is a simpler, more structured approach than what I've done so far.  I've spent hours on this for the 28 player tournament and will have to do the same for the 32 player.  Still, after hours, I have multiple players sitting at the same table with other players as many five times, while not sitting at the same table with some other players even once (so, for example, player 12 plays at the same table as player 18 five times, but never sits at the same table with players 1, 4, 17, 18, 24, 25, or 32) .

Thank you in advance.


Ian Wakeling

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Reply #1 on: September 11, 2017, 10:37:49 AM
Here is a schedule for 28 players.  The mingling is not perfect, there are 15 pairs who never meet, but otherwise I think the schedule is good enough to use.  I believe all partner pairs are unique and all opposition pairs are unique.  I am still thinking about 32 players - this is a slightly harder problem.

  (22  8 v 15  2) (23 10 v  7 17) (11  9 v 27 24) (12  5 v 28 26) ( 6 16 v 18 21) ( 3 14 v  4 25) (20  1 v 19 13)
  (23  9 v 13  3) (24 11 v  8 18) (12  7 v 25 22) (10  6 v 28 27) ( 4 17 v 16 19) ( 1 15 v  5 26) (21  2 v 20 14)
  (24  7 v 14  1) (22 12 v  9 16) (10  8 v 26 23) (11  4 v 28 25) ( 5 18 v 17 20) ( 2 13 v  6 27) (19  3 v 21 15)
  (22 17 v 26 27) (16 20 v  9  8) ( 6 21 v 25 24) ( 7 14 v  3  5) (18 12 v 13  1) ( 4 28 v 23 15) ( 2 10 v 19 11)
  (23 18 v 27 25) (17 21 v  7  9) ( 4 19 v 26 22) ( 8 15 v  1  6) (16 10 v 14  2) ( 5 28 v 24 13) ( 3 11 v 20 12)
  (24 16 v 25 26) (18 19 v  8  7) ( 5 20 v 27 23) ( 9 13 v  2  4) (17 11 v 15  3) ( 6 28 v 22 14) ( 1 12 v 21 10)
  (21 24 v 15 13) ( 9  1 v  6 23) (11 27 v 14 19) ( 8  5 v 12  4) ( 2 25 v 28 17) (16 26 v  3  7) (20 18 v 10 22)
  (19 22 v 13 14) ( 7  2 v  4 24) (12 25 v 15 20) ( 9  6 v 10  5) ( 3 26 v 28 18) (17 27 v  1  8) (21 16 v 11 23)
  (20 23 v 14 15) ( 8  3 v  5 22) (10 26 v 13 21) ( 7  4 v 11  6) ( 1 27 v 28 16) (18 25 v  2  9) (19 17 v 12 24)
  (28  8 v 21  9) (17  3 v  6 13) ( 7 26 v 27  2) (10 19 v  5 25) (22 11 v  1 23) (12 15 v 18 14) ( 4 16 v 24 20)
  (28  9 v 19  7) (18  1 v  4 14) ( 8 27 v 25  3) (11 20 v  6 26) (23 12 v  2 24) (10 13 v 16 15) ( 5 17 v 22 21)
  (28  7 v 20  8) (16  2 v  5 15) ( 9 25 v 26  1) (12 21 v  4 27) (24 10 v  3 22) (11 14 v 17 13) ( 6 18 v 23 19)


Ian Wakeling

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Reply #2 on: September 11, 2017, 11:13:37 AM
And here is a similar solution for 32 players.

  (29 24 v 20 14) (11  6 v  4 23) ( 9  8 v  3 30) ( 7 17 v 27 22) (19 31 v 25 21) (13  1 v  5 32) ( 2 26 v 28 10) (16 15 v 12 18)
  (30 22 v 21 15) (12  4 v  5 24) ( 7  9 v  1 28) ( 8 18 v 25 23) (20 31 v 26 19) (14  2 v  6 32) ( 3 27 v 29 11) (17 13 v 10 16)
  (28 23 v 19 13) (10  5 v  6 22) ( 8  7 v  2 29) ( 9 16 v 26 24) (21 31 v 27 20) (15  3 v  4 32) ( 1 25 v 30 12) (18 14 v 11 17)
  ( 8  3 v  5 26) (32  9 v 18 10) (22 16 v 19 29) (14  6 v  7 31) ( 1 17 v 21 23) ( 4 24 v 25 30) (13 12 v  2 27) (11 28 v 15 20)
  ( 9  1 v  6 27) (32  7 v 16 11) (23 17 v 20 30) (15  4 v  8 31) ( 2 18 v 19 24) ( 5 22 v 26 28) (14 10 v  3 25) (12 29 v 13 21)
  ( 7  2 v  4 25) (32  8 v 17 12) (24 18 v 21 28) (13  5 v  9 31) ( 3 16 v 20 22) ( 6 23 v 27 29) (15 11 v  1 26) (10 30 v 14 19)
  (29 32 v 30  5) (27  4 v 18 16) (28 20 v  6 25) ( 1 23 v 10 31) (22  8 v 11 24) ( 7 21 v  3  9) (13  2 v 15 17) (14 26 v 19 12)
  (30 32 v 28  6) (25  5 v 16 17) (29 21 v  4 26) ( 2 24 v 11 31) (23  9 v 12 22) ( 8 19 v  1  7) (14  3 v 13 18) (15 27 v 20 10)
  (28 32 v 29  4) (26  6 v 17 18) (30 19 v  5 27) ( 3 22 v 12 31) (24  7 v 10 23) ( 9 20 v  2  8) (15  1 v 14 16) (13 25 v 21 11)
  (15 19 v  6 24) ( 2 23 v  3  5) (20 29 v 18  1) (16 12 v 28 31) (10 21 v  8 11) ( 7 13 v 26 30) (17 14 v  9  4) (27 32 v 25 22)
  (13 20 v  4 22) ( 3 24 v  1  6) (21 30 v 16  2) (17 10 v 29 31) (11 19 v  9 12) ( 8 14 v 27 28) (18 15 v  7  5) (25 32 v 26 23)
  (14 21 v  5 23) ( 1 22 v  2  4) (19 28 v 17  3) (18 11 v 30 31) (12 20 v  7 10) ( 9 15 v 25 29) (16 13 v  8  6) (26 32 v 27 24)


I feel there might be room for improvement.  For an alternative schedule you could start with the plan here for 32 golfers playing in foursomes for 10 days and then add on two extra rounds.
« Last Edit: September 11, 2017, 11:14:29 AM by Ian »


k48038

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Reply #3 on: September 12, 2017, 09:46:15 PM
The 28 player schedule you sent is a HUGE improvement over what I was able to come up with using my trial and error manual method.

For example, my result was that on average, each individual player had 6.2 other players that he did not share a table with over the course of the 12 rounds.  So, after 12 rounds there were 6 other players that I never sat at the same table as.  AND, everyone had at least 4 players that they did not play with and some had as many as 8 players that they never played with (this range of 4 to 8 resulted in the 6.2 average).

The pairings that you sent greatly improved both of those numbers.  The average went down from 6.2 to 2.1.  Meaning that there are only, on average, 2 other players that I do not play at the same table with all night.  And, in your scenario some people actually played at the same table as every other player at least once, and the worst-case was that there were 4 people they did not play with.  My BEST case was 4 and your WORST case was 4. Your range was 1 to 4 giving me the 2.1 average.

Thanks so much.

I have not had the chance to run the same analysis on the 32 player pairings that you've provided, but I hope and expect improvement there, too.

Thanks again!


k48038

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Reply #4 on: October 12, 2017, 07:25:04 PM
Since your two schedules for 28 and 32 players were such an improvement over what I could come up with (even using the table generators on your website, then modifying myself), I wonder if you could take a stab at a 24 player rotation, subject to the same "rules" as the others you did for me already (that is, no one plays with the same person as partner more than once over the twelve rounds, try to minimize players playing at the same table with one another as much as possible, and attempt to get everyone to play at the same table at least once with everyone else--which is VERY hard to do).

Thanks!


Ian Wakeling

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Reply #5 on: October 13, 2017, 08:07:01 AM
I think the schedule below will work.  All 12 partners should be different and a player should get all possible opponents at least once - in other words a player has one person whom they oppose twice.  The repeated opposition pairs are generated by the columns marked with an x.

                                    x       x                           x    x
  (20 19 v 23 12) (17  4 v 11  7) (18  3 v 24  5) ( 1 15 v 13 16) ( 6  8 v  2 22) ( 9 21 v 14 10)
  (21 20 v 24  1) (18  5 v 12  8) (19  4 v 13  6) ( 2 16 v 14 17) ( 7  9 v  3 23) (10 22 v 15 11)
  (22 21 v 13  2) (19  6 v  1  9) (20  5 v 14  7) ( 3 17 v 15 18) ( 8 10 v  4 24) (11 23 v 16 12)
  (23 22 v 14  3) (20  7 v  2 10) (21  6 v 15  8) ( 4 18 v 16 19) ( 9 11 v  5 13) (12 24 v 17  1)
  (24 23 v 15  4) (21  8 v  3 11) (22  7 v 16  9) ( 5 19 v 17 20) (10 12 v  6 14) ( 1 13 v 18  2)
  (13 24 v 16  5) (22  9 v  4 12) (23  8 v 17 10) ( 6 20 v 18 21) (11  1 v  7 15) ( 2 14 v 19  3)
  (14 13 v 17  6) (23 10 v  5  1) (24  9 v 18 11) ( 7 21 v 19 22) (12  2 v  8 16) ( 3 15 v 20  4)
  (15 14 v 18  7) (24 11 v  6  2) (13 10 v 19 12) ( 8 22 v 20 23) ( 1  3 v  9 17) ( 4 16 v 21  5)
  (16 15 v 19  8) (13 12 v  7  3) (14 11 v 20  1) ( 9 23 v 21 24) ( 2  4 v 10 18) ( 5 17 v 22  6)
  (17 16 v 20  9) (14  1 v  8  4) (15 12 v 21  2) (10 24 v 22 13) ( 3  5 v 11 19) ( 6 18 v 23  7)
  (18 17 v 21 10) (15  2 v  9  5) (16  1 v 22  3) (11 13 v 23 14) ( 4  6 v 12 20) ( 7 19 v 24  8)
  (19 18 v 22 11) (16  3 v 10  6) (17  2 v 23  4) (12 14 v 24 15) ( 5  7 v  1 21) ( 8 20 v 13  9)
« Last Edit: October 13, 2017, 08:07:14 AM by Ian »


k48038

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Reply #6 on: September 30, 2019, 07:07:28 PM
You helped me TREMENDOUSLY a couple years ago, hoping you can do it again.

You can read prior posts for detailed explanation, but I am looking to seat 20 different players over the course of a twelve round euchre tournament.

20 players (5 tables, 4 players per table)
At each table, one team of 2 against another team of 2
No player can be the partner of any other player more than one time over the 12 rounds
Players should be at the same table (not necessarily as partners) at least once with as many other players as possible
But, players should be at a table with the same players as few times as possible

Trying to get as many people to play with or against as many other players as possible, while reducing or equalizing the number of times any one player sits at the same table as another. 

You can see from previous posts that your algorithm was far superior to my "method" (which was a lot of manual trial and error).

Thank you.


Ian Wakeling

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Reply #7 on: October 01, 2019, 04:52:01 AM
My method will work for your tournament, with a cyclic structure within each block of 4 rounds.  Below, everyone should have 12 different partners and should oppose the other players either once or twice.

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

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

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


k48038

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Reply #8 on: October 04, 2019, 10:22:13 AM
Again, super improvement over what I was able to do myself manually.

I had 13 people that did not get a chance to play at the same table with every other player at least once...with your method, everyone played at the same table as every other player at least once.

My method had 16 instances where one player at a table with the same other player 4 times over the twelve rounds.   Your method had zero cases.

Mine had 78 instances where one player played at the same table with another player 3 times...your method only had 24.

So, much more mingling of the players, which is what I was after.

Thank you!!!