Not balanced round robin ?

[rftt]

We have a weekly round robin pickleball mixer ( similar to tennis).  Typically we have about 12-16 people show up,  enough to fill 3-4 courts.  ( If I can get help how to solve the 3 court issue then I'll move on to higher number combinations !).  The challenge I'm having is meeting 4 desired criteria:
1.  we only play 6 games,
2.you don't play with the same partner more than once
3. we want to assure everyone has the opportunity to play with everyone else ( not necessarily on the same team) and
4. that we minimize the number of time you play with any other person.   Ideally we'd like to play with everyone once but no more than twice.

Playing with 3 other people every game,  you have 18 People you play with during 6 games.  With only 11 other people there, it sounded easy to play with 4 people once and 7 people twice.............. but I can't seem to find the balance key.

I tried just selecting 6 lines of pairings out of the 11 typical round robin games  but always seem to end up with some people playing other people 3 or 4 times while others aren't played.

Suggestions regarding how to proceed ?

[Ian Wakeling]

The best strategy would be to persuade 16 people to turn up.  Then you can use a social square, see the schedule here.  This arrangement for 5 rounds on 4 courts has everybody playing with each other exactly once, note that this will always be the case no matter how you assign the foursomes in the schedule to two teams of two.  Then for round 6 just play any random assigment of the 16 players to the courts, the pairings here will be the only ones that occur twice.

For 12 players on 3 courts there is an unavoidable problem which is the same as I describe here.

Hope that helps.

[rftt]

Thank you very very much for the quick understandable response.  The web site is easy to navigate and a wealth of information. I've read through the social schedule and whist pages a couple times now and have learned these problems are much more complicated than I originally thought.

I fully understand the benefit of having 16 people at my mixer but the circumstances of the mixer stack the odds against me.   When we have 17,18, or 19 people ,   we have 1,2,or 3 people  sitting out each game.  Can I just substitute these numbers for other numbers in the matrix.  Example,   have 17 play for #1 in game 1,  #2 in game two etc.  and not be too far off optimum ?  

I understand the problem with numbers less than 16.  I assume there is way to use a computer to look at the combinations and minimize the F value for 6 games------- any chance this has been done and is sitting in a table somewhere ( ie:CRC Handbook of Combinatorial Designs?)

When I get up to 20 players,  Is there a way to logically expand the 16 player matrix ,  maybe using your cyclic algorithm  or is it back to a computer optimizing solution ?

[Ian Wakeling]

The problem with the substitution strategy is that there is no guarantee that the players meeting #1 in round 1, #2 in round 2, etc... are all distinct, in other words player 17 will end up meeting some players twice. You may well be able to live with this, but substituting a second time to create an 18 player schedule will most likely make a mess of things.

In general combinatorial texts are only likely to give plans for more symmetric problems like the 16 player social square.  For the 17, 18 and 19 player schedules mentioned, the number of matches per player varies. Since there are so many more of these asymmetric problems, it usually means that they have not been considered as a combinatorial problem and, as I think you suspect, comupter optimization is probably the best way forward.

I have a way of optimizing pairs within foursomes, which will work provided that people are never required to meet the same person twice, either as partner or opponent.  This requirement simplifies the problem since partners and opponents do not have to be considered separately.  For example I can find the solution below for 19 players (A to S), 4 courts and 6 rounds.

  Court 1    Court 2    Court 3    Court 4     Byes
(I M B R)  (G J Q H)  (E D C O)  (P A K S)   [F L N]
(N I O J)  (H R D S)  (A C M G)  (F Q L E)   [B K P]
(O G S B)  (H L A N)  (R P Q C)  (M F J K)   [D E I]
(D A I Q)  (B H P F)  (S J C L)  (R N E K)   [G M O]
(N G F D)  (C K H I)  (J B E A)  (L O M P)   [Q R S]
(P E G I)  (Q S N M)  (K L D B)  (A O R F)   [C H J]

A is the only person to play in every round and meets all 18 other players, B to S have one bye each meeting 15 out of the other 18 players.  As before, you can assign the foursomes to partners and opponents in any way.