Tournament bracket for a 4-players game
$begingroup$
for Christmas a friend is trying to organize a tournament for 13 players. Each game will be played by 4 persons, each player will play 4 games and must play against everybody else.
I can find a solution, but I'd like to minimize the number of players playing two games in a row, and my solution is terrible. If someone can help us ... :)
The problem is similar to the social golfer and the schoolgirls, but the games will be played 1 by 1.
Thank you in advance.
combinatorics graph-theory algebraic-graph-theory combinatorial-designs
edited Dec 18 '18 at 21:35
Mike Earnest
23.5k12051
asked Dec 18 '18 at 20:57
Alcyon EldaraAlcyon Eldara
61
$endgroup$
add a comment |
$begingroup$
for Christmas a friend is trying to organize a tournament for 13 players. Each game will be played by 4 persons, each player will play 4 games and must play against everybody else.
I can find a solution, but I'd like to minimize the number of players playing two games in a row, and my solution is terrible. If someone can help us ... :)
The problem is similar to the social golfer and the schoolgirls, but the games will be played 1 by 1.
Thank you in advance.
combinatorics graph-theory algebraic-graph-theory combinatorial-designs
edited Dec 18 '18 at 21:35
Mike Earnest
23.5k12051
asked Dec 18 '18 at 20:57
Alcyon EldaraAlcyon Eldara
61
$endgroup$
$begingroup$
13 rounds of 1 single game. But, as much as possible, I'd like to avoid people playing twice in a row. My bracket seems to maximize it ...
$endgroup$
– Alcyon Eldara
Dec 18 '18 at 21:15
$begingroup$
Note that player 2 plays player 11 three times. Same for 3 and 12 and 4 and 13, so this cannot be what you want ... indeed, 2 does not play 13, and ...
$endgroup$
– Bram28
Dec 18 '18 at 21:20
$begingroup$
Yes, sorry, this was a copy/paste mistake. 1 5 6 7 - 1 8 9 10 - 1 11 12 13 - 2 5 8 11 - 2 6 9 12 - 2 7 10 13 - 3 5 9 13 - 3 6 10 11 - 3 7 8 12 - 4 5 10 12 - 4 6 8 13 - 4 7 9 11 - 1 2 3 4
$endgroup$
– Alcyon Eldara
Dec 18 '18 at 21:23
$begingroup$
This one has 1 player repeating each rpound ... and 12 different players repeat just once ... player 5 never repeats ..1 2 3 4 - 2 5 8 11 - 3 7 8 12 - 3 5 9 13 - 4 6 8 13 - 3 6 10 11 - 4 7 9 11 - 4 5 10 12 - 2 7 10 13 - 1 5 6 7 - 1 8 9 10 - 2 6 9 12 - 1 11 12 13 - I got this just by playing with your foursomes .. nothing systematic ... surely some kind of systematic way exists to 'spread' the play-times as much as possible (and not just in repeating, but also in terms of games in which you don't play between games you do play .. but I don;t know of any such method
$endgroup$
– Bram28
Dec 18 '18 at 21:52
$begingroup$
Thanks, I will keep it :)
$endgroup$
– Alcyon Eldara
Dec 20 '18 at 16:11
add a comment |
$begingroup$
for Christmas a friend is trying to organize a tournament for 13 players. Each game will be played by 4 persons, each player will play 4 games and must play against everybody else.
I can find a solution, but I'd like to minimize the number of players playing two games in a row, and my solution is terrible. If someone can help us ... :)
The problem is similar to the social golfer and the schoolgirls, but the games will be played 1 by 1.
Thank you in advance.
combinatorics graph-theory algebraic-graph-theory combinatorial-designs
edited Dec 18 '18 at 21:35
Mike Earnest
23.5k12051
asked Dec 18 '18 at 20:57
Alcyon EldaraAlcyon Eldara
61
$endgroup$
for Christmas a friend is trying to organize a tournament for 13 players. Each game will be played by 4 persons, each player will play 4 games and must play against everybody else.
I can find a solution, but I'd like to minimize the number of players playing two games in a row, and my solution is terrible. If someone can help us ... :)
The problem is similar to the social golfer and the schoolgirls, but the games will be played 1 by 1.
Thank you in advance.
combinatorics graph-theory algebraic-graph-theory combinatorial-designs
combinatorics graph-theory algebraic-graph-theory combinatorial-designs
edited Dec 18 '18 at 21:35
Mike Earnest
23.5k12051
asked Dec 18 '18 at 20:57
Alcyon EldaraAlcyon Eldara
61
edited Dec 18 '18 at 21:35
Mike Earnest
23.5k12051
asked Dec 18 '18 at 20:57
Alcyon EldaraAlcyon Eldara
61
edited Dec 18 '18 at 21:35
Mike Earnest
23.5k12051
edited Dec 18 '18 at 21:35
Mike Earnest
23.5k12051
edited Dec 18 '18 at 21:35
Mike Earnest
23.5k12051
23.5k12051
asked Dec 18 '18 at 20:57
Alcyon EldaraAlcyon Eldara
61
asked Dec 18 '18 at 20:57
Alcyon EldaraAlcyon Eldara
61
asked Dec 18 '18 at 20:57
Alcyon EldaraAlcyon Eldara
61
61
$begingroup$
13 rounds of 1 single game. But, as much as possible, I'd like to avoid people playing twice in a row. My bracket seems to maximize it ...
$endgroup$
– Alcyon Eldara
Dec 18 '18 at 21:15
$begingroup$
Note that player 2 plays player 11 three times. Same for 3 and 12 and 4 and 13, so this cannot be what you want ... indeed, 2 does not play 13, and ...
$endgroup$
– Bram28
Dec 18 '18 at 21:20
$begingroup$
Yes, sorry, this was a copy/paste mistake. 1 5 6 7 - 1 8 9 10 - 1 11 12 13 - 2 5 8 11 - 2 6 9 12 - 2 7 10 13 - 3 5 9 13 - 3 6 10 11 - 3 7 8 12 - 4 5 10 12 - 4 6 8 13 - 4 7 9 11 - 1 2 3 4
$endgroup$
– Alcyon Eldara
Dec 18 '18 at 21:23
$begingroup$
This one has 1 player repeating each rpound ... and 12 different players repeat just once ... player 5 never repeats ..1 2 3 4 - 2 5 8 11 - 3 7 8 12 - 3 5 9 13 - 4 6 8 13 - 3 6 10 11 - 4 7 9 11 - 4 5 10 12 - 2 7 10 13 - 1 5 6 7 - 1 8 9 10 - 2 6 9 12 - 1 11 12 13 - I got this just by playing with your foursomes .. nothing systematic ... surely some kind of systematic way exists to 'spread' the play-times as much as possible (and not just in repeating, but also in terms of games in which you don't play between games you do play .. but I don;t know of any such method
$endgroup$
– Bram28
Dec 18 '18 at 21:52
$begingroup$
Thanks, I will keep it :)
$endgroup$
– Alcyon Eldara
Dec 20 '18 at 16:11
add a comment |
$begingroup$
13 rounds of 1 single game. But, as much as possible, I'd like to avoid people playing twice in a row. My bracket seems to maximize it ...
$endgroup$
– Alcyon Eldara
Dec 18 '18 at 21:15
$begingroup$
Note that player 2 plays player 11 three times. Same for 3 and 12 and 4 and 13, so this cannot be what you want ... indeed, 2 does not play 13, and ...
$endgroup$
– Bram28
Dec 18 '18 at 21:20
$begingroup$
Yes, sorry, this was a copy/paste mistake. 1 5 6 7 - 1 8 9 10 - 1 11 12 13 - 2 5 8 11 - 2 6 9 12 - 2 7 10 13 - 3 5 9 13 - 3 6 10 11 - 3 7 8 12 - 4 5 10 12 - 4 6 8 13 - 4 7 9 11 - 1 2 3 4
$endgroup$
– Alcyon Eldara
Dec 18 '18 at 21:23
$begingroup$
This one has 1 player repeating each rpound ... and 12 different players repeat just once ... player 5 never repeats ..1 2 3 4 - 2 5 8 11 - 3 7 8 12 - 3 5 9 13 - 4 6 8 13 - 3 6 10 11 - 4 7 9 11 - 4 5 10 12 - 2 7 10 13 - 1 5 6 7 - 1 8 9 10 - 2 6 9 12 - 1 11 12 13 - I got this just by playing with your foursomes .. nothing systematic ... surely some kind of systematic way exists to 'spread' the play-times as much as possible (and not just in repeating, but also in terms of games in which you don't play between games you do play .. but I don;t know of any such method
$endgroup$
– Bram28
Dec 18 '18 at 21:52
$begingroup$
Thanks, I will keep it :)
$endgroup$
– Alcyon Eldara
Dec 20 '18 at 16:11
$begingroup$
13 rounds of 1 single game. But, as much as possible, I'd like to avoid people playing twice in a row. My bracket seems to maximize it ...
$endgroup$
– Alcyon Eldara
Dec 18 '18 at 21:15
$begingroup$
13 rounds of 1 single game. But, as much as possible, I'd like to avoid people playing twice in a row. My bracket seems to maximize it ...
$endgroup$
– Alcyon Eldara
Dec 18 '18 at 21:15
$begingroup$
Note that player 2 plays player 11 three times. Same for 3 and 12 and 4 and 13, so this cannot be what you want ... indeed, 2 does not play 13, and ...
$endgroup$
– Bram28
Dec 18 '18 at 21:20
$begingroup$
Note that player 2 plays player 11 three times. Same for 3 and 12 and 4 and 13, so this cannot be what you want ... indeed, 2 does not play 13, and ...
$endgroup$
– Bram28
Dec 18 '18 at 21:20
$begingroup$
Yes, sorry, this was a copy/paste mistake. 1 5 6 7 - 1 8 9 10 - 1 11 12 13 - 2 5 8 11 - 2 6 9 12 - 2 7 10 13 - 3 5 9 13 - 3 6 10 11 - 3 7 8 12 - 4 5 10 12 - 4 6 8 13 - 4 7 9 11 - 1 2 3 4
$endgroup$
– Alcyon Eldara
Dec 18 '18 at 21:23
$begingroup$
Yes, sorry, this was a copy/paste mistake. 1 5 6 7 - 1 8 9 10 - 1 11 12 13 - 2 5 8 11 - 2 6 9 12 - 2 7 10 13 - 3 5 9 13 - 3 6 10 11 - 3 7 8 12 - 4 5 10 12 - 4 6 8 13 - 4 7 9 11 - 1 2 3 4
$endgroup$
– Alcyon Eldara
Dec 18 '18 at 21:23
$begingroup$
This one has 1 player repeating each rpound ... and 12 different players repeat just once ... player 5 never repeats ..1 2 3 4 - 2 5 8 11 - 3 7 8 12 - 3 5 9 13 - 4 6 8 13 - 3 6 10 11 - 4 7 9 11 - 4 5 10 12 - 2 7 10 13 - 1 5 6 7 - 1 8 9 10 - 2 6 9 12 - 1 11 12 13 - I got this just by playing with your foursomes .. nothing systematic ... surely some kind of systematic way exists to 'spread' the play-times as much as possible (and not just in repeating, but also in terms of games in which you don't play between games you do play .. but I don;t know of any such method
$endgroup$
– Bram28
Dec 18 '18 at 21:52
$begingroup$
This one has 1 player repeating each rpound ... and 12 different players repeat just once ... player 5 never repeats ..1 2 3 4 - 2 5 8 11 - 3 7 8 12 - 3 5 9 13 - 4 6 8 13 - 3 6 10 11 - 4 7 9 11 - 4 5 10 12 - 2 7 10 13 - 1 5 6 7 - 1 8 9 10 - 2 6 9 12 - 1 11 12 13 - I got this just by playing with your foursomes .. nothing systematic ... surely some kind of systematic way exists to 'spread' the play-times as much as possible (and not just in repeating, but also in terms of games in which you don't play between games you do play .. but I don;t know of any such method
$endgroup$
– Bram28
Dec 18 '18 at 21:52
$begingroup$
Thanks, I will keep it :)
$endgroup$
– Alcyon Eldara
Dec 20 '18 at 16:11
$begingroup$
Thanks, I will keep it :)
$endgroup$
– Alcyon Eldara
Dec 20 '18 at 16:11
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
An easy way to set up such a tournament makes use of the fact that the projective plane $PG(2,3)$ consists of $13$ points on $13$ lines, each line containing $4$ points (for a graphic of that plane, see e.g. https://tex.stackexchange.com/questions/336897/).
So, you can identify the players with the points and the games with the lines, and run them in any order: Since you have a projective plane, any two lines will intersect in a single point, i.e. any two games will have one player in common. That will add up to $12$ players having to play two games in a row — I don't know how far this is from the minimum you are trying to achieve.
answered Dec 20 '18 at 14:23
jpveejpvee
2,55521427
$endgroup$
add a comment |
$begingroup$
This one has 1 player repeating each round ... and 12 different players repeating just once (player 5 never repeats) ... which also implies that no one three-peats:
1 2 3 4
2 5 8 11
3 7 8 12
3 5 9 13
4 6 8 13
3 6 10 11
4 7 9 11
4 5 10 12
2 7 10 13
1 5 6 7
1 8 9 10
2 6 9 12
1 11 12 13
I got this just by playing with your foursomes ... I didn't use any systematic method, though I would think that such a method exists.
There are schedules where no one repeats during some round ... but now you will have multiple people repeating some other round ... so I think having exactly 1 person repeat each round, and having those 12 repeats be of 12 different people is about as fair as possible as far as repeating goes.
Still, in addition to worrying about repeats, you could also worry about the spread of games more general. For example, in the schedule above, player 1 plays in games 1, 10, 12, and 13 ... that seems like it is a lot less 'evenly distributed' than, say, player 9, who plays in games 4,7,11, and 12, although at the same time it is more 'spread out' than a player like 3, who plays in games 1,3,4, and 6.
So, you could try and come up with some 'measure' of how 'evenly distributed' or 'spread out' the games are, and a measure of 'fairness' on top of that. Again, I assume that some work has already been done on that, though I am not an expert. Also, for your purposes of having some fun at a Christmas party, this would seem to going far beyond what you were looking for :)
answered Dec 20 '18 at 16:31
Bram28Bram28
63.1k44793
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
An easy way to set up such a tournament makes use of the fact that the projective plane $PG(2,3)$ consists of $13$ points on $13$ lines, each line containing $4$ points (for a graphic of that plane, see e.g. https://tex.stackexchange.com/questions/336897/).
So, you can identify the players with the points and the games with the lines, and run them in any order: Since you have a projective plane, any two lines will intersect in a single point, i.e. any two games will have one player in common. That will add up to $12$ players having to play two games in a row — I don't know how far this is from the minimum you are trying to achieve.
answered Dec 20 '18 at 14:23
jpveejpvee
2,55521427
$endgroup$
add a comment |
$begingroup$
An easy way to set up such a tournament makes use of the fact that the projective plane $PG(2,3)$ consists of $13$ points on $13$ lines, each line containing $4$ points (for a graphic of that plane, see e.g. https://tex.stackexchange.com/questions/336897/).
So, you can identify the players with the points and the games with the lines, and run them in any order: Since you have a projective plane, any two lines will intersect in a single point, i.e. any two games will have one player in common. That will add up to $12$ players having to play two games in a row — I don't know how far this is from the minimum you are trying to achieve.
answered Dec 20 '18 at 14:23
jpveejpvee
2,55521427
$endgroup$
add a comment |
$begingroup$
An easy way to set up such a tournament makes use of the fact that the projective plane $PG(2,3)$ consists of $13$ points on $13$ lines, each line containing $4$ points (for a graphic of that plane, see e.g. https://tex.stackexchange.com/questions/336897/).
So, you can identify the players with the points and the games with the lines, and run them in any order: Since you have a projective plane, any two lines will intersect in a single point, i.e. any two games will have one player in common. That will add up to $12$ players having to play two games in a row — I don't know how far this is from the minimum you are trying to achieve.
answered Dec 20 '18 at 14:23
jpveejpvee
2,55521427
$endgroup$
An easy way to set up such a tournament makes use of the fact that the projective plane $PG(2,3)$ consists of $13$ points on $13$ lines, each line containing $4$ points (for a graphic of that plane, see e.g. https://tex.stackexchange.com/questions/336897/).
So, you can identify the players with the points and the games with the lines, and run them in any order: Since you have a projective plane, any two lines will intersect in a single point, i.e. any two games will have one player in common. That will add up to $12$ players having to play two games in a row — I don't know how far this is from the minimum you are trying to achieve.
answered Dec 20 '18 at 14:23
jpveejpvee
2,55521427
answered Dec 20 '18 at 14:23
jpveejpvee
2,55521427
answered Dec 20 '18 at 14:23
jpveejpvee
2,55521427
answered Dec 20 '18 at 14:23
jpveejpvee
2,55521427
2,55521427
add a comment |
add a comment |
$begingroup$
This one has 1 player repeating each round ... and 12 different players repeating just once (player 5 never repeats) ... which also implies that no one three-peats:
1 2 3 4
2 5 8 11
3 7 8 12
3 5 9 13
4 6 8 13
3 6 10 11
4 7 9 11
4 5 10 12
2 7 10 13
1 5 6 7
1 8 9 10
2 6 9 12
1 11 12 13
I got this just by playing with your foursomes ... I didn't use any systematic method, though I would think that such a method exists.
There are schedules where no one repeats during some round ... but now you will have multiple people repeating some other round ... so I think having exactly 1 person repeat each round, and having those 12 repeats be of 12 different people is about as fair as possible as far as repeating goes.
Still, in addition to worrying about repeats, you could also worry about the spread of games more general. For example, in the schedule above, player 1 plays in games 1, 10, 12, and 13 ... that seems like it is a lot less 'evenly distributed' than, say, player 9, who plays in games 4,7,11, and 12, although at the same time it is more 'spread out' than a player like 3, who plays in games 1,3,4, and 6.
So, you could try and come up with some 'measure' of how 'evenly distributed' or 'spread out' the games are, and a measure of 'fairness' on top of that. Again, I assume that some work has already been done on that, though I am not an expert. Also, for your purposes of having some fun at a Christmas party, this would seem to going far beyond what you were looking for :)
answered Dec 20 '18 at 16:31
Bram28Bram28
63.1k44793
$endgroup$
add a comment |
$begingroup$
This one has 1 player repeating each round ... and 12 different players repeating just once (player 5 never repeats) ... which also implies that no one three-peats:
1 2 3 4
2 5 8 11
3 7 8 12
3 5 9 13
4 6 8 13
3 6 10 11
4 7 9 11
4 5 10 12
2 7 10 13
1 5 6 7
1 8 9 10
2 6 9 12
1 11 12 13
I got this just by playing with your foursomes ... I didn't use any systematic method, though I would think that such a method exists.
There are schedules where no one repeats during some round ... but now you will have multiple people repeating some other round ... so I think having exactly 1 person repeat each round, and having those 12 repeats be of 12 different people is about as fair as possible as far as repeating goes.
Still, in addition to worrying about repeats, you could also worry about the spread of games more general. For example, in the schedule above, player 1 plays in games 1, 10, 12, and 13 ... that seems like it is a lot less 'evenly distributed' than, say, player 9, who plays in games 4,7,11, and 12, although at the same time it is more 'spread out' than a player like 3, who plays in games 1,3,4, and 6.
So, you could try and come up with some 'measure' of how 'evenly distributed' or 'spread out' the games are, and a measure of 'fairness' on top of that. Again, I assume that some work has already been done on that, though I am not an expert. Also, for your purposes of having some fun at a Christmas party, this would seem to going far beyond what you were looking for :)
answered Dec 20 '18 at 16:31
Bram28Bram28
63.1k44793
$endgroup$
add a comment |
$begingroup$
This one has 1 player repeating each round ... and 12 different players repeating just once (player 5 never repeats) ... which also implies that no one three-peats:
1 2 3 4
2 5 8 11
3 7 8 12
3 5 9 13
4 6 8 13
3 6 10 11
4 7 9 11
4 5 10 12
2 7 10 13
1 5 6 7
1 8 9 10
2 6 9 12
1 11 12 13
I got this just by playing with your foursomes ... I didn't use any systematic method, though I would think that such a method exists.
There are schedules where no one repeats during some round ... but now you will have multiple people repeating some other round ... so I think having exactly 1 person repeat each round, and having those 12 repeats be of 12 different people is about as fair as possible as far as repeating goes.
Still, in addition to worrying about repeats, you could also worry about the spread of games more general. For example, in the schedule above, player 1 plays in games 1, 10, 12, and 13 ... that seems like it is a lot less 'evenly distributed' than, say, player 9, who plays in games 4,7,11, and 12, although at the same time it is more 'spread out' than a player like 3, who plays in games 1,3,4, and 6.
So, you could try and come up with some 'measure' of how 'evenly distributed' or 'spread out' the games are, and a measure of 'fairness' on top of that. Again, I assume that some work has already been done on that, though I am not an expert. Also, for your purposes of having some fun at a Christmas party, this would seem to going far beyond what you were looking for :)
answered Dec 20 '18 at 16:31
Bram28Bram28
63.1k44793
$endgroup$
This one has 1 player repeating each round ... and 12 different players repeating just once (player 5 never repeats) ... which also implies that no one three-peats:
1 2 3 4
2 5 8 11
3 7 8 12
3 5 9 13
4 6 8 13
3 6 10 11
4 7 9 11
4 5 10 12
2 7 10 13
1 5 6 7
1 8 9 10
2 6 9 12
1 11 12 13
I got this just by playing with your foursomes ... I didn't use any systematic method, though I would think that such a method exists.
There are schedules where no one repeats during some round ... but now you will have multiple people repeating some other round ... so I think having exactly 1 person repeat each round, and having those 12 repeats be of 12 different people is about as fair as possible as far as repeating goes.
Still, in addition to worrying about repeats, you could also worry about the spread of games more general. For example, in the schedule above, player 1 plays in games 1, 10, 12, and 13 ... that seems like it is a lot less 'evenly distributed' than, say, player 9, who plays in games 4,7,11, and 12, although at the same time it is more 'spread out' than a player like 3, who plays in games 1,3,4, and 6.
So, you could try and come up with some 'measure' of how 'evenly distributed' or 'spread out' the games are, and a measure of 'fairness' on top of that. Again, I assume that some work has already been done on that, though I am not an expert. Also, for your purposes of having some fun at a Christmas party, this would seem to going far beyond what you were looking for :)
answered Dec 20 '18 at 16:31
Bram28Bram28
63.1k44793
answered Dec 20 '18 at 16:31
Bram28Bram28
63.1k44793
answered Dec 20 '18 at 16:31
Bram28Bram28
63.1k44793
answered Dec 20 '18 at 16:31
Bram28Bram28
63.1k44793
63.1k44793
add a comment |
add a comment |
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13 rounds of 1 single game. But, as much as possible, I'd like to avoid people playing twice in a row. My bracket seems to maximize it ...
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– Alcyon Eldara
Dec 18 '18 at 21:15
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Note that player 2 plays player 11 three times. Same for 3 and 12 and 4 and 13, so this cannot be what you want ... indeed, 2 does not play 13, and ...
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– Bram28
Dec 18 '18 at 21:20
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Yes, sorry, this was a copy/paste mistake. 1 5 6 7 - 1 8 9 10 - 1 11 12 13 - 2 5 8 11 - 2 6 9 12 - 2 7 10 13 - 3 5 9 13 - 3 6 10 11 - 3 7 8 12 - 4 5 10 12 - 4 6 8 13 - 4 7 9 11 - 1 2 3 4
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– Alcyon Eldara
Dec 18 '18 at 21:23
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This one has 1 player repeating each rpound ... and 12 different players repeat just once ... player 5 never repeats ..1 2 3 4 - 2 5 8 11 - 3 7 8 12 - 3 5 9 13 - 4 6 8 13 - 3 6 10 11 - 4 7 9 11 - 4 5 10 12 - 2 7 10 13 - 1 5 6 7 - 1 8 9 10 - 2 6 9 12 - 1 11 12 13 - I got this just by playing with your foursomes .. nothing systematic ... surely some kind of systematic way exists to 'spread' the play-times as much as possible (and not just in repeating, but also in terms of games in which you don't play between games you do play .. but I don;t know of any such method
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– Bram28
Dec 18 '18 at 21:52
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Thanks, I will keep it :)
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– Alcyon Eldara
Dec 20 '18 at 16:11