Doubles Pickle ball

[spdavey]

I need a double round robin for pickle ball for 9 and 10 players on two courts 

[Ian Wakeling]

I have put some schedules below that should work. 9 players works well as all pairs of players can partner once and oppose twice.  In the schedule for 10 players below all pairs of players, except A & F, partner once and oppose twice - this does mean that these two players have one less game than the others.

9 Players A to I
[(H A):(I F)] [(G C):(B D)]
[(A C):(F G)] [(I H):(E B)]
[(G D):(E I)] [(B A):(H C)]
[(G H):(C I)] [(D E):(A F)]
[(I B):(D A)] [(E G):(F H)]
[(F E):(B C)] [(H D):(G A)]
[(H B):(D F)] [(C E):(A I)]
[(B G):(A E)] [(D I):(C F)]
[(C D):(E H)] [(I G):(F B)]

10 players A to J
[(C F):(G J)] [(A D):(H I)]
[(D J):(B I)] [(E H):(C G)]
[(G I):(A C)] [(E F):(B J)]
[(I J):(E G)] [(C H):(B D)]
[(A I):(B E)] [(F J):(D H)]
[(B H):(A G)] [(D F):(C E)]
[(C J):(A B)] [(F H):(E I)]
[(B G):(D E)] [(C I):(H J)]
[(F I):(B C)] [(A J):(D G)]
[(G H):(B F)] [(A E):(C D)]
[(D I):(F G)] [(E J):(A H)]

[Foosball]

Hello, I've been looking for help with scheduling a smaller number of people, however, ensuring that everyone plays with one another to complete a season.
9 Players A to I
[(H A):(I F)] [(G C):(B D)]
[(A C):(F G)] [(I H):(E B)]
[(G D):(E I)] [(B A):(H C)]
[(G H):(C I)] [(D E):(A F)]
[(I B):(D A)] [(E G):(F H)]
[(F E):(B C)] [(H D):(G A)]
[(H B):(D F)] [(C E):(A I)]
[(B G):(A E)] [(D I):(C F)]
[(C D):(E H)] [(I G):(F B)]
I saw this response to a 9 person, however, "Team H A" only plays "Team I F", I'm looking for a schedule where "Team H A" plays all possible combinations of the remaining players (i.e., Team BC, Team BD, Team BE, etc) throughout the season . I'm only looking at 5 and 6 player leagues. I can generate the combinations but not necessarily "a schedule". I've been searching for help for a long time and I'm hoping I can find answers here. Thank you for you consideration.

[Ian Wakeling]

Hello,

Are you over-complicating things here, or is there some other property that you want the final schedule to have?  As I see it, if you can enumerate all the combinations, then that becomes the schedule.  For example with 5 players there are 15 possible doubles games that could be played as follows:

1  2 v 3  5
2  3 v 1  4
3  4 v 2  5
1  3 v 4  5
1  5 v 2  4
1  2 v 4  5
1  3 v 2  5
2  4 v 3  5
1  5 v 3  4
1  3 v 2  4
1  4 v 2  5
2  3 v 4  5
1  2 v 3  4
1  4 v 3  5
2  3 v 1  5

Would that work for you?

[Foosball]

I'm an idiot. I never added the last piece to my question. I can create the "schedule" for 5 and 6 as you shared - thank you. My real issue is creating the schedule where everyone sits out the same amount of time in a given session and no one sits out more than one game in a row. Over the entire season, it will equal out and everyone will play the same number of games and sit out the same number of times. I think I can do this manually for 5, but when it comes to 6 players, it gets more complicated to keep people from sitting out more than one game in a row.

My more difficult question is when I play foosball. For that, we have 5 or 6 players, however, the position of the player actually matters (offense or defense) as well as what team an individual player is on. For that, I need a schedule for 6 players, where everyone gets a chance to play offense and defense against everyone else in a round robin format. So, for 1 2 v 3 5, I have 2 1 v 3 5, then 2 1 v 5 3, and 1 2 v 5 3 and then I need to make sure each individual plays with all other players.

I think, however, you may just have the best point - maybe I'm over complicating both leagues and schedules and I should just simplify all this. I really appreciate you taking a look at this. I'm sure I'm not doing a good job at explaining

[Ian Wakeling]

I have looked at the easier problem of 6 players and the byes, and perfect solutions seem to be possible.  For n players there are n(n-1)(n-2)(n-3)/8 possible doubles games, so 45 games for n=6.  I believe I have arranged them below so that there are no consecutive byes.

(1 2 v 3 5)  [4 6]
(1 4 v 3 6)  [2 5]
(2 5 v 3 6)  [1 4]
(1 2 v 4 5)  [3 6]
(1 3 v 2 6)  [4 5]
(3 5 v 4 6)  [1 2]
(1 4 v 2 6)  [3 5]
(2 3 v 5 6)  [1 4]
(1 5 v 4 6)  [2 3]
(1 3 v 2 5)  [4 6]
(1 3 v 4 6)  [2 5]
(2 3 v 4 5)  [1 6]
(1 6 v 2 5)  [3 4]
(2 3 v 4 6)  [1 5]
(1 5 v 2 3)  [4 6]
(2 6 v 4 5)  [1 3]
(1 2 v 3 4)  [5 6]
(2 5 v 4 6)  [1 3]
(1 5 v 3 4)  [2 6]
(1 2 v 4 6)  [3 5]
(2 5 v 3 4)  [1 6]
(1 5 v 3 6)  [2 4]
(1 6 v 2 4)  [3 5]
(2 6 v 3 5)  [1 4]
(1 4 v 3 5)  [2 6]
(1 2 v 3 6)  [4 5]
(3 6 v 4 5)  [1 2]
(1 2 v 5 6)  [3 4]
(2 6 v 3 4)  [1 5]
(1 3 v 4 5)  [2 6]
(1 5 v 2 6)  [3 4]
(1 6 v 3 4)  [2 5]
(1 4 v 2 5)  [3 6]
(1 6 v 3 5)  [2 4]
(2 4 v 3 5)  [1 6]
(1 6 v 4 5)  [2 3]
(2 4 v 3 6)  [1 5]
(1 3 v 5 6)  [2 4]
(1 5 v 2 4)  [3 6]
(3 4 v 5 6)  [1 2]
(1 6 v 2 3)  [4 5]
(2 4 v 5 6)  [1 3]
(1 3 v 2 4)  [5 6]
(1 4 v 5 6)  [2 3]
(1 4 v 2 3)  [5 6]

Surely there are going to be too many games if you look at offense and defense, I think you are asking for 4 times as many rounds as the schedule above.

[Ian Wakeling]

I have been thinking some more, and for 6 players it is possible to balance offense and defense within a 45 round schedule.  So what follows is like the schedule above but with additional properties.

In both schedules, any given pair of players (A, B) , whether they play together as partners or opponents, appear in exactly 18 games. However, in the schedule below these 18 games are balanced for offense and defense because all such pairs (A,B) play:

3 times as partners with A in offense and B in defense.
3 times as partners with B in offense and A in defense.
3 times as opponents with A in offense and B in defense.
3 times as opponents with B in offense and A in defense.
3 times as opponents with both A & B in offense.
3 times as opponents with both A & B in defense.

 O D - O D    bye
(3 5 v 1 2)  [4 6]
(4 1 v 3 6)  [2 5]
(6 2 v 5 1)  [3 4]
(4 2 v 5 3)  [1 6]
(5 6 v 1 4)  [2 3]
(3 2 v 5 4)  [1 6]
(2 6 v 3 1)  [4 5]
(2 1 v 4 5)  [3 6]
(2 6 v 3 5)  [1 4]
(3 4 v 6 1)  [2 5]
(4 6 v 5 2)  [1 3]
(1 4 v 3 2)  [5 6]
(6 4 v 1 5)  [2 3]
(6 5 v 3 2)  [1 4]
(4 5 v 1 6)  [2 3]
(1 2 v 6 3)  [4 5]
(5 3 v 6 4)  [1 2]
(6 1 v 5 2)  [3 4]
(3 1 v 5 4)  [2 6]
(6 2 v 3 4)  [1 5]
(1 5 v 4 2)  [3 6]
(3 1 v 5 6)  [2 4]
(4 3 v 2 1)  [5 6]
(5 6 v 2 4)  [1 3]
(1 3 v 6 4)  [2 5]
(2 5 v 3 4)  [1 6]
(3 6 v 1 5)  [2 4]
(2 5 v 1 4)  [3 6]
(5 3 v 1 6)  [2 4]
(4 6 v 1 2)  [3 5]
(4 5 v 6 3)  [1 2]
(2 4 v 1 3)  [5 6]
(5 4 v 2 6)  [1 3]
(2 3 v 1 6)  [4 5]
(4 3 v 6 5)  [1 2]
(6 1 v 2 4)  [3 5]
(4 1 v 3 5)  [2 6]
(6 3 v 2 5)  [1 4]
(4 1 v 6 2)  [3 5]
(5 1 v 2 3)  [4 6]
(3 6 v 4 2)  [1 5]
(5 2 v 1 3)  [4 6]
(2 3 v 4 6)  [1 5]
(2 1 v 6 5)  [3 4]
(5 1 v 4 3)  [2 6]

The schedule also has balance for the end of the table being used, so each player appears 15 times on the left of the 'v' and 15 times on the right of the 'v'.  Does that work for you?

[Foosball]

You are amazing! I need to digest this a little (the math), AND you answered my very poorly worded question and solved my riddle of the schedule!!!
Thank you so much!