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81

Requests / Create a "balanced" schedule for 10 players and 2 courts over 25 weeks

« Last post by karyn on October 21, 2024, 04:07:08 PM »

I am trying to create a schedule for 10 tennis players over 25 weeks.

Every week, there will be 2 courts of 4 people.  Two people will have a bye each week.  

Once on the court, the 4 people will decide amongst themselves who wants to play with whom.

I would like to balance the schedule, as much as is possible with 10 people, so that no two people are on the same court together a disproportionate amount of time.  To be clear, I don't care if anyone is always playing on court 1 or mostly on court 2; I just don't want, for example, Person 1 and Person 2 being on the same court together 15 times and Person 1 and Person 8 only ever being on the same court together 3 times.

Any help would be appreciated.  I've gotten pretty close with modifying some of what I've seen here, but find myself thinking that for those with skills, it's probably a short and elegant piece of code...

82

Requests / Re: Am I asking for the impossible?

« Last post by markyb610 on October 19, 2024, 06:29:01 AM »

Ian,

That's amazing - Perfect !!

Many thanks   - and so quick too !
Mark

83

Requests / Home teams clash

« Last post by angus1964 on October 19, 2024, 01:50:19 AM »

Years ago I had a formula for our darts league which ensured when two teams shared a board, they were both not drawn at home.
I used to write out the numbers like this.
1 2 3 4 5
6 7 8 9 10
Then rotate round to form the schedule. 
But now I cant remember how I did it, and all attempts are no where near.

Any pointers would be greatly appreciated.

84

Requests / Re: Am I asking for the impossible?

« Last post by Ian Wakeling on October 18, 2024, 11:44:50 AM »

Mark,

Here is a schedule that might be useful.

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

All the partners are different, and all the opposition pairs are different, but there are about 17 pairs who meet twice.  I am sure it's possible to do better, but the problem is that my software is not optimized for this scenario.

85

Requests / Re: Am I asking for the impossible?

« Last post by markyb610 on October 18, 2024, 05:35:04 AM »

Ian,

Quite right !  I can see that 10 rounds would require a mix of 7 or 8 to be played by all players. My mistake.

So I think my (corrected) requirement is  . . . 

6 courts          12 rounds         32 players - playing in random doubles each round

Total matches played = 72 ( 6 courts x 12 rounds)

Therefore 72 x 4 = 288 = individual scores/results

3 byes per player - so 9 matches per player (12 rounds - 3 byes) will count towards the result

In each round there will be 8 byes ( Total players less doubles on all courts = 32 - (4 x 6) )


Many thanks
Mark

86

Requests / Re: Am I asking for the impossible?

« Last post by Ian Wakeling on October 18, 2024, 04:07:46 AM »

"(So presumably with 10 matches per court - 8 matches per player and two byes per person) ?"

I am confused about the numbers here.  There are 8 byes per round. So for everyone to have the same number of games, there needs to be a multiple of 4 matches per court.  For example 8 matches per court would be two byes per person.  Is that what you want?

87

Requests / Re: Am I asking for the impossible?

« Last post by markyb610 on October 17, 2024, 05:11:37 AM »

Ian,

I should have said this is for doubles (or perhaps it was obvious !).

Other solutions on the web don't seem to provide the goal of as few "duplicates" (partners & opponents) as you have managed so well !

Many thanks
Mark 

88

Requests / Re: Am I asking for the impossible?

« Last post by markyb610 on October 17, 2024, 04:47:25 AM »

Ian,

I can see you've done some fantastic work here. I was looking at your 32 player round robin and it's so nearly what I'm looking for. I can see you rely on there being eight courts.

Would it be possible to generate a six court solution ? (So presumably with 10 matches per court - 8 matches per player and two byes per person) ??

Many thanks
Mark

89

Requests / Re: Doubles schedule for 12 players,

« Last post by Ian Wakeling on October 17, 2024, 03:12:24 AM »

Here is a six round schedule - there are three pairs of players who meet three times (4,7) (5,6) & (9,11), other wise all partnerships are different, and players play against all others once or twice.

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

90

Requests / Re: Doubles schedule for 12 players,

« Last post by MCR426 on October 16, 2024, 08:42:11 AM »

What you are asking is not possible - there needs to be 16 players or more if you want no pair of players to be on court together more than once.

I think the schedule below is the best that you can do.  There are 18 pairs of players who meet twice, and 12 pairs who never play together.

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

Just discovered this schedule and it will work GREAT for me....but can you please add two more rounds?  THANK YOU IN ADVANCE!!