Round Robin Tournament Scheduling

Round Robin - 8 Player Doubles

tkam · 19 · 3906

tkam

  • Newbie
  • *
    • Posts: 0
on: September 10, 2023, 05:03:57 PM
Hi Ian,

Looking for help with a balanced schedule for my weekly badminton session of 8 player doubles on 2 courts.

We play 2 consecutive games before we swap players as our courts are booked in different halls in the same venue so single game swapping isn't practical. We play a total of 20 games across the 2 courts (10 per player).

I've tried to construct a round robin that ensures players play with each other min 1 to max 2 times (3 default playe WITH repetitions games due to 10 games), and playing vs others players the same amount of times (ideally 3 max).

I'm starting to think because of the 2 game rotation per swap, its not possible to get the outcome I'm looking for.

From what I've read, I'm looking for a Whist design, but with a caveat that a 4 player court plays 2 games before swapping with the players from the other court.

I.e. the games flows as follows:
Court 1 game 1: A/B vs C/D
Court 1 game 2: A/D vs B/C
Court 2 game 1: E/F vs G/H
Court 2 game 2: E/H vs F/G

Then a rotation will occur between courts and so on.

The most balanced distribution I've been able to achieve is:

PLAY WITH: 2:2:2:1:1:1:1
PLAY AGAINST: 5:4:3:3:2:2:1

Desired:

PLAY WITH: 2:2:2:1:1:1:1
PLAY AGAINST: 3:3:3:3:3:3:2

Is this logically possible given the constraints of my round robin setup?

Here's my solution for the 2222111 / 5433221 ratio:




Thanks in advance...

« Last Edit: September 12, 2023, 04:49:51 PM by tkam »


tkam

  • Newbie
  • *
    • Posts: 0
Reply #1 on: September 20, 2023, 09:03:23 AM
Hi Ian, any chance you had a chance to look at this please mate? 


Ian Wakeling

  • Forum Moderator
  • God Member
  • *****
    • Posts: 1141
Reply #2 on: September 21, 2023, 02:09:03 AM
I think you have the best possible arrangement.  I can see that it's unavoidable that each player will have one other player whom they oppose only once, so this rules out 3:3;3:3:3:3:2.   I can get 4:4;3:3:3:2:1 but only by allowing repeated games at a court.


tkam

  • Newbie
  • *
    • Posts: 0
Reply #3 on: September 21, 2023, 04:37:20 AM
Cheers Ian, I can put this to rest finally and just enjoy the session know its the best it can be.

Tej



tkam

  • Newbie
  • *
    • Posts: 0
Reply #4 on: September 21, 2023, 10:13:22 AM
I think you have the best possible arrangement.  I can see that it's unavoidable that each player will have one other player whom they oppose only once, so this rules out 3:3;3:3:3:3:2.  I can get 4:4;3:3:3:2:1 but only by allowing repeated games at a court.

On second thoughts Ian, could you provide the 4433321 solution with the repeat games please?


Ian Wakeling

  • Forum Moderator
  • God Member
  • *****
    • Posts: 1141
Reply #5 on: September 21, 2023, 12:13:40 PM
Tej,

I had another go and I think the schedule below has 4:4:3:3:3:2:1 opponents with no repeated games.  It's has a slightly different order to your example, the first two rows are played on the same court, same for the next two rows, etc.. Hope that helps,

Ian

6  5 v 4  1
6  1 v 5  4
8  3 v 7  2
8  7 v 2  3
1  8 v 7  6
1  7 v 8  6
5  2 v 3  4
5  3 v 2  4
2  1 v 5  8
2  8 v 5  1
3  6 v 4  7
3  7 v 4  6
4  2 v 6  8
4  6 v 8  2
7  3 v 1  5
7  5 v 3  1
3  1 v 6  2
3  2 v 1  6
4  8 v 5  7
4  5 v 8  7


tkam

  • Newbie
  • *
    • Posts: 0
Reply #6 on: September 21, 2023, 12:39:16 PM
Amazing, that's an upgrade for sure.

Thanks you Ian!

Tej


tkam

  • Newbie
  • *
    • Posts: 0
Reply #7 on: January 14, 2024, 06:20:33 PM
Hi Ian,

Following on from the 2 court solution you helped me with above, please could you help me with an optimal round robin structure for 3 courts, 12 players, with the exact same requirements as before?


Ian Wakeling

  • Forum Moderator
  • God Member
  • *****
    • Posts: 1141
Reply #8 on: January 15, 2024, 12:54:50 PM
Hi Tej,

12 is not a nice number!  I have thought about it, with this format of play, I think each player has to play 16 games in order to have all partners at least once, and all opponents at least twice.  Is that too many games?  If so how many were your thinking of?


tkam

  • Newbie
  • *
    • Posts: 0
Reply #9 on: January 15, 2024, 03:02:53 PM
Sorry, I should have clarified.

- 10 games per person

- playing with: partnering with everyone one once only, with the exception of 1 person

- playing against: as balanced as is possible given the above constraints

Everything else as before i.e. 2 games before a court change.


Ian Wakeling

  • Forum Moderator
  • God Member
  • *****
    • Posts: 1141
Reply #10 on: January 16, 2024, 05:49:38 PM
I believe the partnering-with-everyone-once-with-the-exception-of-one-person part is not possible - at least I can't find a schedule like this - as I end up with 10 or 11 pairs who don't partner.   Then the opponent balance is terrible, it varies from zero times per pair to 4 times per pair.  You would be far better advised to pick a different format.


tkam

  • Newbie
  • *
    • Posts: 0
Reply #11 on: January 17, 2024, 04:52:16 AM
Thanks for having a look Ian.

Actually your 10-11 pair gap and 0 to 4 against is much better than what i've discovered already. Would you mind sharing that?

Otherwise, what is the most evenly distributed possible schedule if the partner with everyone requirement is relaxed?

Currently, the best i've been able to get is the below, where partnering with is max 2, min 0 and playing against is more extreme with max 10, min 0 (it seems you have already got one a lot better than this).

Ultimately i'm happy with any schedule which ends up smoothing out the extremes and results in the same ratios for everyone for both WITH and AGAINST, albeit with instances of zero partnering occurrences...



« Last Edit: January 17, 2024, 07:31:36 AM by tkam »


Ian Wakeling

  • Forum Moderator
  • God Member
  • *****
    • Posts: 1141
Reply #12 on: January 17, 2024, 12:34:57 PM
Hi Tej,

Here is my schedule which has 11 pairs who never partner, and opposition pairs that vary from zero to 4.


 2  3  8 10
 2  8  3 10
 5  7 12  6
 5 12  7  6
11  4  1  9
11  9  4  1
 3  4  9  6
 3  9  4  6
 1  8 12  7
 1 12  8  7
11 10  5  2
11  5  2 10
 4  5 10  7
 4 10  5  7
 2  9  8 12
 2 12  9  8
11  1  3  6
11  6  1  3
 5  1  6  8
 5  6  1  8
 3 10 12  9
 3 12 10  9
11  2  4  7
11  7  2  4
 1  2  7  9
 1  7  2  9
 4  6 12 10
 4 12  6 10
11  8  3  5
11  3  5  8


My version of your two symmetric matrices are immediately below (partners and opponents counts), followed by the sum of the two matrices.  I have put zeroes down the main diagonals.

 0  1  1  1  1  0  1  2  1  0  1  1
 1  0  1  1  1  0  0  1  2  1  1  1
 1  1  0  1  1  1  0  0  1  2  1  1
 1  1  1  0  1  2  1  0  0  1  1  1
 1  1  1  1  0  1  2  1  0  0  1  1
 0  0  1  2  1  0  1  1  1  1  1  1
 1  0  0  1  2  1  0  1  1  1  1  1
 2  1  0  0  1  1  1  0  1  1  1  1
 1  2  1  0  0  1  1  1  0  1  1  1
 0  1  2  1  0  1  1  1  1  0  1  1
 1  1  1  1  1  1  1  1  1  1  0  0
 1  1  1  1  1  1  1  1  1  1  0  0

 0  1  1  1  1  4  3  2  3  0  3  1
 1  0  1  1  1  0  4  3  2  3  3  1
 1  1  0  1  1  3  0  4  3  2  3  1
 1  1  1  0  1  2  3  0  4  3  3  1
 1  1  1  1  0  3  2  3  0  4  3  1
 4  0  3  2  3  0  1  1  1  1  1  3
 3  4  0  3  2  1  0  1  1  1  1  3
 2  3  4  0  3  1  1  0  1  1  1  3
 3  2  3  4  0  1  1  1  0  1  1  3
 0  3  2  3  4  1  1  1  1  0  1  3
 3  3  3  3  3  1  1  1  1  1  0  0
 1  1  1  1  1  3  3  3  3  3  0  0

 0  2  2  2  2  4  4  4  4  0  4  2
 2  0  2  2  2  0  4  4  4  4  4  2
 2  2  0  2  2  4  0  4  4  4  4  2
 2  2  2  0  2  4  4  0  4  4  4  2
 2  2  2  2  0  4  4  4  0  4  4  2
 4  0  4  4  4  0  2  2  2  2  2  4
 4  4  0  4  4  2  0  2  2  2  2  4
 4  4  4  0  4  2  2  0  2  2  2  4
 4  4  4  4  0  2  2  2  0  2  2  4
 0  4  4  4  4  2  2  2  2  0  2  4
 4  4  4  4  4  2  2  2  2  2  0  0
 2  2  2  2  2  4  4  4  4  4  0  0

Note that the 12 players fall into two distinct groups with different patterns of partners and opponents (Group A = 1-5 & 12, Group B = 6-11).  Finally note that there are 6 pairs of players who never meet either as partners or as opponents.  Hope that makes some sense.


tkam

  • Newbie
  • *
    • Posts: 0
Reply #13 on: January 17, 2024, 06:57:58 PM

Ian, thanks a lot, thatís fantastic!

Could we tweak the schedule to ensure the games are closely matched in terms of skill? Ideally, we want each team to have a similar aggregate skill level, so that player 12, being the strongest through to player 1, being the least skilled, are distributed in a way that makes each game competitive and balanced.

I've approached this by assigning each player a numerical ranking, with the top player receiving 12 points and the lowest-ranked player receiving 1 point. I then added up the points for each team in a game and determined the absolute difference in points between the teams. This value indicates the level of skill balance for each game; a lower total signifies a more evenly matched setup. My best attempt is the below which maintains the structure you provided but with player numbers swapped around. This leads to an overall total of 132. Is it possible to arrange the players to achieve an even more balanced total?

6  1  2  5
6  2  1  5
7  4  8  3
7  8  4  3
9  11  10  12
9  12  11  10
1  11  12  3
1  12  11  3
10  2  8  4
10  8  2  4
9  5  7  6
9  7  6  5
11  7  5  4
11  5  7  4
6  12  2  8
6  8  12  2
9  10  1  3
9  3  10  1
7  10  3  2
7  3  10  2
1  5  8  12
1  8  5  12
9  6  11  4
9  4  6  11
10  6  4  12
10  4  6  12
11  3  8  5
11  8  3  5
9  2  1  7
9  1  7  2

« Last Edit: January 18, 2024, 11:19:17 AM by tkam »


Ian Wakeling

  • Forum Moderator
  • God Member
  • *****
    • Posts: 1141
Reply #14 on: January 18, 2024, 12:30:25 PM
I believe the schedule below has the same partner/opponent properties as before but has a skill difference score of 104.

 6  8 12 10
 6 12  8 10
 9  3 11  1
 9 11  3  1
 5  2  4  7
 5  7  2  4
 8  2  7  1
 8  1  7  2
 4 12 11  3
 4 11 12  3
 5 10  9  6
 5  9  6 10
 2  9 10  3
 2 10  9  3
 6 12 11  7
 6 11  7 12
 5  1  4  8
 5  4  8  1
 9  1  4 12
 9  4  1 12
 8  7 11 10
 8 11 10  7
 5  3  6  2
 5  6  2  3
 4  3  6  7
 4  6  3  7
 2 11  1 10
 2  1 11 10
 5  8  9 12
 5 12  8  9

I am doing the same thing that you described, just swapping around the player numbers, but the starting point was a similar schedule to the one in my previous post (generated by the same algorithm).