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Subject: Setting up a tournament (Puerto Rico) rss

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Guillaume G.
Switzerland
Genève
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I'd like to build a 5-players Puerto Rico tournament so that :
* each player meets each other players twice
* each player plays 5 times (one in every game position : 1st player, 2nd player, 3rd player, 4th player, 5th player)

Knowing those constraints, I guess 11 players could meet the requirements but I still have difficulties setting up a correct 5x11 grid (identifying who would play against who and in which game) which meets those constraints. I tried to set up an Excel spreadsheet with the solver but couldn't make it work.

Would anybody know how to build such a grid or know where I could find one meeting those constraints ?
I'm sure many people have faced this kind of question but despite of multiple research over the internet, I couldn't get a satisfying solution.

Many thanks in advance,

-- Guillaume
 
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C.K. Au
Malaysia
Kuala Lumpur
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Hmm.. I'm watching this thread as I intend to set up a Puerto Rico tournament in Malaysia too.
 
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Guillaume G.
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Let's offer 3 GG to the first to provide me with a 5x11 table meeting filled with integer values from 1 to 11 meeting the constraints named above :
* 1 line of the table = 5 different values
* each value appears 5 times in the table
* each combination of 2 values exists on exactly 2 lines of the table
* each column of the table is made of each integer from 1 to 11

I'm even providing a partial solution on which swapping needs to be done to reach a solution :

A B C D E
1 2 3 4 5
1 2 9 6 7
1 4 11 8 9
1 3 6 10 11
1 8 7 10 5
2 5 9 8 11
2 3 4 8 10
2 6 7 10 11
3 5 7 9 x
3 6 8 x x
4 5 6 x x

values 4, 7, 9 and 11 need to replace the "x" but it implies moving figures around in the table.

Good luck to the braves who will try to solve this problem.
 
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Werner Bär
Germany
Karlsruhe
Baden
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Actorios wrote:
1 2 3 4 5
1 4 9 11 9
1 2 6 8 7
1 3 6 10 11
1 5 7 8 10


Well, player 1 plays twice against player 9. But if would be better if this would happen in two different games.
 
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Guillaume G.
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Quick Edit... It should work better this way. Sorry. blush
 
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Richard Irving
United States
Salinas
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There's an even more obvious problem: You have a system with the most annoying & inconvenient schedule possible--you can't even run 2 games at the same time.

Here is your partial list:

Quote:
A B C D E
1 2 3 4 5
1 2 9 6 7
1 4 11 8 9
1 3 6 10 11
1 8 7 10 5
2 5 9 8 11
2 3 4 8 10
2 6 7 10 11
3 5 7 9 x
3 6 8 x x
4 5 6 x x


Game 1 features players 1 2 3 4 5--but every other game features one or more players from this game. The same is true with game 2, etc. I don't think you can't play ANY two games at the same time--at least one player would have to play in 2 games at once.

So basically you are going to have to schedule every game in advance--what if real life intervenes? What if one of your players can't make it at the scheduled date & time due to a family emergency, or had to work late, or simply forgot, etc.

If running this as a party or at a game convention or one day or weekend event, you have most of the players sitting around wasting time the next game can start. And what if a player has to sit out, two or three games in a row... Hard to maintain the interest of your players with that sort of scheduling.

If you doing this as simultaneous PBeM games--be prepared for a lot of confusion--everyone trying to keep track of multiple games at the same time.
 
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Guillaume G.
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I entirely agree with you. The fact is this tournament is an annual tournament at a club with regular sessions (1 every two weeks). It is interesting because it offers a continuous tournament where everyone meets everyone and where the total number of games / player is reasonable (5 here).
It will not be scheduled as such but the available players of one evening will start there game (potential others would play something else outside of the tournament, which they won't mind). Only the last games (with the most unfrequent players) will have to be specifically scheduled.
Of course this system is of no use for a "one-day tournament".
 
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Keldon Jones
United States
Fort Collins
Colorado
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Here you go:

1 2 3 4 5
2 1 6 7 8
3 4 7 6 9
5 3 8 10 6
7 5 1 9 10
4 9 10 8 2
8 7 11 5 4
9 6 5 2 11
11 8 9 1 3
6 10 4 11 1
10 11 2 3 7

I hope this meets all your criteria. Had to write some really ugly code to find this. There are many more arrangements, of course.
 
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Guillaume G.
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Well done, Mr Keldon !

Just for the sake of it, I wanted to show those who followed this post, the answer I got from a french forum where I asked the same question.
No ugly code, just "pattern recognition" :

1 B A 8 5
2 1 B 9 6
3 2 1 A 7
4 3 2 B 8
5 4 3 1 9
6 5 4 2 A
7 6 5 3 B
8 7 6 4 1
9 8 7 5 2
A 9 8 6 3
B A 9 7 4

Impressively simple !
 
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Marshall P.
United States
Wichita
Kansas
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"Nothing in Biology Makes Sense Except in the Light of Evolution" - Theodosius Dobzhansky
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Looks like Keldon beat me to it. If you are interested I believe the below schedule is the minimum number of players required such that each player plays every other exactly once in 5 player games. If you have this many players you could just randomize which player is assigned to which number and run through the schedule again so that they each play against everyone twice.


A B C D E
1 2 3 4 5
1 6 7 8 9
1 10 11 12 13
1 14 15 16 17
1 18 19 20 21
2 6 10 14 18
2 7 11 15 19
2 8 12 16 20
2 9 13 17 21
3 6 11 16 21
3 7 10 17 20
3 8 13 14 19
3 9 12 15 18
4 6 12 17 19
4 7 13 16 18
4 8 10 15 21
4 9 11 14 20
5 6 13 15 20
5 7 12 14 21
5 8 11 17 18
5 9 10 16 19
 
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Guillaume G.
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mdp4828 wrote:
Looks like Keldon beat me to it. If you are interested I believe the below schedule is the minimum number of players required such that each player plays every other exactly once in 5 player games. If you have this many players you could just randomize which player is assigned to which number and run through the schedule again so that they each play against everyone twice.


A B C D E
1 2 3 4 5
1 6 7 8 9
1 10 11 12 13
1 14 15 16 17
1 18 19 20 21
2 6 10 14 18
2 7 11 15 19
2 8 12 16 20
2 9 13 17 21
3 6 11 16 21
3 7 10 17 20
3 8 13 14 19
3 9 12 15 18
4 6 12 17 19
4 7 13 16 18
4 8 10 15 21
4 9 11 14 20
5 6 13 15 20
5 7 12 14 21
5 8 11 17 18
5 9 10 16 19


Thank you Marshall... I knew you would have the answer
The number of players available for Puerto Rico is more between 10 and 15 so it was to be eleven. But this 21x5 table is very interesting.
 
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