CHANGING TRIPLES ROUND ROBIN

[LJG1937]

WE HAVE 12 PLAYERS -numbered 1 to 12- and they will always play IN FOUR GROUPS OF THREE [TRIPLETS].  Each series of MATCHES will involve all 4 triplets: playing in 2 simultaneous matches, i.e. 1,2,3 versus 4,5,6 AND 7,8,9 versus 10,11,12. After MATCH 1, the 4 triplets all change and MATCH 2 then takes place. Every Series of Matches - say up to 6 or if possible up to 12 - must involve all 12 players [the 4 triplets] playing at the same time, as above. What I am seeking is for a series of combinations for a total (unspecified, but at least 5) Series of Matches such that, at the end everyone will have played WITH and AGAINST every other player. With the minimum - or optimum- duplication.  In theory, after 5 matches, every player should/could have played WITH 10 of the 11 opponents, and in the 6th game, they will be partnered by the 11th player and only doubles up with one player for a second game.  At the same time - i.e. over 6 MATCHES - every player will play AGAINST the other players - AT BEST - with 4 opponents once and 7 opponents twice. And no one THREE TIMES! The first question is to determine if this is actually a possibility? Quite likely, the two conditions governing the choice of the triplets, means that the above 2 theories are impossible in practice.  Therefore, what I'm really seeking is a Round Robin Scheduling for those 12 players in their triples, over an unspecified number of matches with the minimum of duplication. Thus far, using Trial-and-Error and complex 12x12 Grids, I have failed to complete 6 (or even 5) "perfect" combinations. Can anyone out there HELP ME please ?

[Ian Wakeling]

I think it likely that you will need 11 series to get a balanced schedule and that the optimal schedule you are looking for with 6 series may well be impossible (but I have not proof of that).  With 11 series you can have all 11 partners exactly twice, and all 11 opponents exactly three times.

(11  0  1 v  2  4  7)   ( 3  8  10 v  5  6  9)
(11  1  2 v  3  5  8)   ( 4  9   0 v  6  7 10)
(11  2  3 v  4  6  9)   ( 5 10   1 v  7  8  0)
(11  3  4 v  5  7 10)   ( 6  0   2 v  8  9  1)
(11  4  5 v  6  8  0)   ( 7  1   3 v  9 10  2)
(11  5  6 v  7  9  1)   ( 8  2   4 v 10  0  3)
(11  6  7 v  8 10  2)   ( 9  3   5 v  0  1  4)
(11  7  8 v  9  0  3)   (10  4   6 v  1  2  5)
(11  8  9 v 10  1  4)   ( 0  5   7 v  2  3  6)
(11  9 10 v  0  2  5)   ( 1  6   8 v  3  4  7)
(11 10  0 v  1  3  6)   ( 2  7   9 v  4  5  8)

The schedule above is Example 18.2 from page 9 of this paper by Abel et al on Generalized Whist Designs which is freely downloadable.  I have used 0 to 11 for the players so you can see more easily how the first series given in the paper is translated into the full schedule.

Hope that helps.

Ian.

[LJG1937]

HELLO IAN:

Thank you for your rapid response to my request regarding the combinations for the 12 players over 11 series of matches.  The game we play is actually the French game of Boules or Petanque - the most popular -  as our version is called.  Have had no chance yet to examine the arrangements, but will come back to you as quickly as I can having checked out everything.  
Kindest regards

Len Griffiths

[LJG1937]

Hello Ian:

Thank you for the very well balanced schedule you sent me last month [ 08/14/14 at 10:32:31] which solved my problem practically, earlier this week, for a round robin tournament for 12 players organised into triples. As you stated, we needed to play 11 series of triples matches [ n-1 where n = no.of players] to get this balanced schedule.  You added that "with 11 series you can have have all 11 partners exactly twice"  and that is so, I've checked with a neat spreadsheet and that works. Congrats. You further state in same sentence that  "all opponents are played exactly three times". My same spreadsheet found that although the average was three times, every player actually played against their respective opponents a little differently.
Each player's pattern of opponents followed the same format. They each were scheduled against other players [the ideal of] three times for five of their eleven opponents -  [that's 5 out of the 11]. In addition, they actually played three opponents just twice and - therefore - against each of the remaining three opponents, four times. So that's the average of three times and that's 5 + 3 + 3 = 11 games, obviously. - but not exactly as suggested - within a total of just the 11 series of games.
My follow-up question therefore, is as follows .   Obviously [ me thinks ] it has to be a multiple of "n -1" series , for my chosen 12 players [ 11 in this instance ] to obtain absolute equality,  But do you know how many multiples of 11 it has to be before one arrives at (Elysian Fields / Heaven of) exactly equal numbers of games versus all 11 opponents whilst preserving the equal numbers of games with those same 11 partners?

regards

LEN GRIFFITHS  / LJG1937 USER NAME

[Ian Wakeling]

Hello Len,

I have checked and I stand by my original statement that every pair of players opposes exactly three times in the schedule above.

Ian.

[LJG1937]

HELLO IAN:

Thank you very much for the help you have provided thus far.  We have been using the round robin tournament scheduled you Posted on: 08/14/14 at 10:32:31 and it has been very successful.  We now wish to get access to - or be directed towards - the algorithm; code and/or edge graphs that will enable us to write something to accommodate other numbers of players. If possible, to write something that can be operated from a smart-phone outdoors as people arrive for a game.  Always for triples v triples or including one triple v a double if an odd number of players arrive.  For 12 players you indicated we will need 11 series to get a balanced schedule. With those 11 series we can have all 11 partners twice and all 11 opponents exactly three times. So, it looks like the series length is (n-1) where n= number of players? Plus - a link to the same for games of doubles v doubles, changing partners and opponents each new game for a varied no.of players from say 6 through 48. If an odd number arrives, they should form a triple - for one match only, with everyone getting the one match as a triple v a double or pair.   Sadly, I'm not savvy with coding - but merely the organiser for the weekly games and monthly competitions.  But I have friends who seem to be coders etc..one works here in London for Firefox

Kindest regards

Len Griffiths   LJG1937

[Ian Wakeling]

Here are some more balanced schedules for tournaments with games of 3 vs 3. They all have an odd number of players, with the same number of rounds as players. In each round there is one player with a bye (player 0 is the bye in the first round, player 1 is the bye in the 2nd round, etc..)

In general 3 vs 3 schedules will be possible for 6n players (like the 12 player schedule above), or 6n+1 players. Unfortunately no such schedule exists for 6 players and the existence of the 18 player schedule is an open problem.

I have sent you an e-mail regarding software/algorithms.

7 players
( 1  2  4 v 3  5  6)
( 2  3  5 v 4  6  0)
( 3  4  6 v 5  0  1)
( 4  5  0 v 6  1  2)
( 5  6  1 v 0  2  3)
( 6  0  2 v 1  3  4)
( 0  1  3 v 2  4  5)

13 players
(12  4  1 v  3 10  9) ( 2  7 11 v  6  8  5)
( 0  5  2 v  4 11 10) ( 3  8 12 v  7  9  6)
( 1  6  3 v  5 12 11) ( 4  9  0 v  8 10  7)
( 2  7  4 v  6  0 12) ( 5 10  1 v  9 11  8)
( 3  8  5 v  7  1  0) ( 6 11  2 v 10 12  9)
( 4  9  6 v  8  2  1) ( 7 12  3 v 11  0 10)
( 5 10  7 v  9  3  2) ( 8  0  4 v 12  1 11)
( 6 11  8 v 10  4  3) ( 9  1  5 v  0  2 12)
( 7 12  9 v 11  5  4) (10  2  6 v  1  3  0)
( 8  0 10 v 12  6  5) (11  3  7 v  2  4  1)
( 9  1 11 v  0  7  6) (12  4  8 v  3  5  2)
(10  2 12 v  1  8  7) ( 0  5  9 v  4  6  3)
(11  3  0 v  2  9  8) ( 1  6 10 v  5  7  4)

19 players
( 2  5  6 v  1 18 10) (11 15  8 v  3  9 17) (12  7 13 v 14  4 16)
( 3  6  7 v  2  0 11) (12 16  9 v  4 10 18) (13  8 14 v 15  5 17)
( 4  7  8 v  3  1 12) (13 17 10 v  5 11  0) (14  9 15 v 16  6 18)
( 5  8  9 v  4  2 13) (14 18 11 v  6 12  1) (15 10 16 v 17  7  0)
( 6  9 10 v  5  3 14) (15  0 12 v  7 13  2) (16 11 17 v 18  8  1)
( 7 10 11 v  6  4 15) (16  1 13 v  8 14  3) (17 12 18 v  0  9  2)
( 8 11 12 v  7  5 16) (17  2 14 v  9 15  4) (18 13  0 v  1 10  3)
( 9 12 13 v  8  6 17) (18  3 15 v 10 16  5) ( 0 14  1 v  2 11  4)
(10 13 14 v  9  7 18) ( 0  4 16 v 11 17  6) ( 1 15  2 v  3 12  5)
(11 14 15 v 10  8  0) ( 1  5 17 v 12 18  7) ( 2 16  3 v  4 13  6)
(12 15 16 v 11  9  1) ( 2  6 18 v 13  0  8) ( 3 17  4 v  5 14  7)
(13 16 17 v 12 10  2) ( 3  7  0 v 14  1  9) ( 4 18  5 v  6 15  8)
(14 17 18 v 13 11  3) ( 4  8  1 v 15  2 10) ( 5  0  6 v  7 16  9)
(15 18  0 v 14 12  4) ( 5  9  2 v 16  3 11) ( 6  1  7 v  8 17 10)
(16  0  1 v 15 13  5) ( 6 10  3 v 17  4 12) ( 7  2  8 v  9 18 11)
(17  1  2 v 16 14  6) ( 7 11  4 v 18  5 13) ( 8  3  9 v 10  0 12)
(18  2  3 v 17 15  7) ( 8 12  5 v  0  6 14) ( 9  4 10 v 11  1 13)
( 0  3  4 v 18 16  8) ( 9 13  6 v  1  7 15) (10  5 11 v 12  2 14)
( 1  4  5 v  0 17  9) (10 14  7 v  2  8 16) (11  6 12 v 13  3 15)

[LJG1937]

HELLO IAN:

We thank you for the latest series of triples schedules where tournaments with games of 3 v 3 have to accommodate an odd number of players. Along the lines of 6n+1 with 7, 13 and 19 competitors.  The detailed series of triples allowing for one more player to sit out for one game has been tried but it falls down in our situation as a game of French Boules / Petanque Triples can take up to an hour to reach the 13 points required for a winning score. And leaving out players for that length of time was not appreciated.  
Odd numbers arriving for Round Robin Tournaments involving Doubles are easier to handle as extra numbers of players merely convert doubles into triples for one game only. What is still a problem with games of doubles is to fairly allocate the extra players over a series of games.  So if we have 12 players, that's 6 doubles obviously. If 13 arrive, there are 5 doubles and one triple; 14 players need two triples; 15 begets three triples. When we get 16 players - of course - that's easy as its a perfect number for doubles.  Because there is a distinct advantage in boules [ as in green bowls ] of a double playing against a triple, we're seeking the fairest allocation of the extra players into triples. Fair means everyone ( over a series of say n-1 or 2n-1 perhaps)  playing in the same number of games in a double and/or a triple with every other player.  We are hosting a series of such competitions across June and I'm providing  the scheduling. Unfortunately, we never know how many players will register on the day, until the very last minute.
Similarly, for a triples (conventionally the chosen game in petanque) tournament, we seek a similar arrangement to accommodate extra players. Am therefore seeking equality - in terms of playing in doubles or triples - for all 5 numbers between ideal groups. So the ideal numbers for games of triples are 6,12,18, 24 and 30 players. We need schedules to cover the 5 numbers in between of n+1; n+2; n+3; n+4 and n+5. In every case, changing partners and opponents as per your earlier schedules and optimising the number of times that each player experiences during any series up to 6n -1 possibly.
Finally, finally, in your most recent email you state various codes, namely Б; а; н etc including numbers 1086; 1074; 1089. Please what are those references?  Are they the previous postings from you?

regards Len Griffiths

[LJG1937]

HELLO IAN:

We all thank you very much for your schedules which we adopted for our recent series of triples competitions. This may be the 2nd time you see this request but in case the 1st one went astray, I will attempt to reconstruct our new requirement.  We are going to play an Autumn series of petanque matches between several teams of 6 players. Let's assume we have 4 teams {TEAMS A, B, C, D } taking part - but it could be six {adding Teams E, F} if we're lucky.  Once a week or every month, sets of two teams play each other over a series of 5 [or 6] games. So A v B; C v D; etc.. Already I can see how that Round Robin is organised from your excellent web-site . Now, within every game the 6 [ numbered 1,2,3,4,5 6 ] players on each team are split up into a triple [ 3 players] a double [ 2 players ] and a singleton [ 1 player ] . However after each game all 6 players change places: they do not remain in the same triple/double/singleton format but switch around, but remain as Team A. Same format inside other Teams B,C,D, E, F. What we are looking for is the format to cover the series of 5 or 6 matches between Team A versus Team B etc.  At the conclusion of the series of the 5 / 6 games and I suspect that 5 ( or n-1 ) is best, we seek an optimum solution such that everyone of the 6 players play once as a singleton, twice in a double, but with different partners and three times in a triple - again with different partners. To clarify our request: we only require one schedule to cover the series of matches between teams and can then leave it to individual Team managers to organise their teams accordingly.  Incidentally, the winners are rewarded with points: 4 for winning the triple; 3 for winning the double and 2 for the singles. The points are added together after the series of 5 /6 games to determine the winning Team - on the day. And overall at the completion of the Teams round-robin schedule. That would be 3 series if we have 4 teams participating and 5 should we manage to get 6 teams to take part.

Regards

LEN GRIFFITHS  LJG 1937  

[Ian Wakeling]

Hello Len,

The problem you are describing is highly asymmetric and I am afraid that I don't have any suggestions for how to schedule for this.  Having experimented a little, my feeling here is that there may not be any good schedules at all.

I can see the intuitive appeal.  2 teams of 6, 6 rounds, where each round is one singles match, one doubles match and one triples match.  I think 6 rounds is better than 5, as then everyone can get exactly 1 singles match, 2 doubles matches and 3 triples matches.

I tried a few ideas, but have found nothing you would want to use, as there may be some pairs who oppose only once, while other pairs oppose four times.  Partner balance is equally hard to obtain, in the scenario described you have 8 partners, but I still found it impossible to get all 5 possible partners either once or twice.

Ian.

[LJG1937]

Hello  Ian

Thank you for answer: like you, I've also tried intuitively to devise the perfect distribution of 6 players into one single, one double and one triple; each time different, over a series of 6 matches, without success. Plus, I realised that I need two schedules that are different: one for the six-man team A and another totally different for six-man team B. Otherwise each member of team A will go on meeting his equal [but opposite] opponent in Team B, unless the 2 schedules are completely different !  I hope I've explained that clearly?  
As I said, using a slightly modified whist-shuffle technique, I have a 6 round [ I agree 6 better than 5 ] schedule that does change both partners and opponents; but not exactly equal [ as you'd discovered ] which is better than nothing.  And I use that schedule for Team A  which has players in order 1,2.3,4,5, 6 and substitute 6 for 1; 5 for 2; 4 for 3 etc... 1 for 6 inside Team B, to provide variety, even if it is a slightly unbalanced shuffle.  Am going to see if our Committee will allow me to run a one-day tournament between 2 or even 4 teams.  And then ask the players if they'd like to run it as a longer tournament with more teams and the round robin approach etc
If it works - or I can improve my schedule a little - I will let you know.  Thus far my petanque group love the way we've changed our series of matches - with your help - switching both the partners and opponents, every match.  The main reason for this almost universal acclamation is that each series of games was now being won [plus the 2nd & 3rd spots] by different individuals. Using the changing partners & opponents approach, it became almost impossible to forecast the end result: happiness shared ......

kindest regards

LEN.