Round Robin Tournament Scheduling

Round Robin - 8 Player Doubles

tkam · 7 · 173

tkam

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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

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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

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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

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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

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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

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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

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

Thanks you Ian!

Tej