Number of 4-player games in a tournament

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I have seen a lot of posts online about the number of games in a tournament for 2 player games. But what if it's a 4-player game (say Ludo).

Consider a scenario where there are 12 people participating in a Ludo tournament. Each game in the tournament is a 4-player game. Is it possible to schedule the games such that each player plays every other player just once? If yes, how many games will there be in total in the tournament?

Can this be generalized for m players in an n-player game tournament?

combinatorics

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asked Jul 18 at 17:01

user74207

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  Note that in a 4-player game, each player has three opponents in every game. So for each player to play every other player precisely once, certainly the number of other players must be a multiple of three. That is, the total number of players must be 1 more than a multiple of 3. So in general it is not possible to schedule the games in such a way.
  – Servaes
  Jul 18 at 17:10
  
  

  
  
  
  


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  @Servaes So the smallest tournament that we cannot immediately reject as impossible would be $16$, with 4 parallel games each round and 5 rounds. Is it possible? I don't know. Also, related post.
  – Arthur
  Jul 18 at 17:19
  
  

  
  
  
  


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  @Arthur this is analogous to Kirkman's schoolgirl problem (en.wikipedia.org/wiki/Kirkman%27s_schoolgirl_problem). In particular you seem to be looking for some sort of Steiner system (en.wikipedia.org/wiki/Steiner_system).
  – Michael Lugo
  Jul 18 at 17:33
  

  
  
  
  


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  @Servaes so it seems that in general for a n-player game, we need to have a (n-1)*x + 1 players? That's a necessary condition, is it also sufficient?
  – user74207
  Jul 18 at 17:51
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Arthur, it 'could' also be possible for 4 (obv), 7, 10, 13 as well, cant it? Why 16?
  – user74207
  Jul 18 at 17:51
  

  
  
  
  

 | 
show 5 more comments

up vote
1
down vote

favorite

3

I have seen a lot of posts online about the number of games in a tournament for 2 player games. But what if it's a 4-player game (say Ludo).

Consider a scenario where there are 12 people participating in a Ludo tournament. Each game in the tournament is a 4-player game. Is it possible to schedule the games such that each player plays every other player just once? If yes, how many games will there be in total in the tournament?

Can this be generalized for m players in an n-player game tournament?

combinatorics

share|cite|improve this question

asked Jul 18 at 17:01

user74207

1524

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Note that in a 4-player game, each player has three opponents in every game. So for each player to play every other player precisely once, certainly the number of other players must be a multiple of three. That is, the total number of players must be 1 more than a multiple of 3. So in general it is not possible to schedule the games in such a way.
  – Servaes
  Jul 18 at 17:10
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Servaes So the smallest tournament that we cannot immediately reject as impossible would be $16$, with 4 parallel games each round and 5 rounds. Is it possible? I don't know. Also, related post.
  – Arthur
  Jul 18 at 17:19
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Arthur this is analogous to Kirkman's schoolgirl problem (en.wikipedia.org/wiki/Kirkman%27s_schoolgirl_problem). In particular you seem to be looking for some sort of Steiner system (en.wikipedia.org/wiki/Steiner_system).
  – Michael Lugo
  Jul 18 at 17:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Servaes so it seems that in general for a n-player game, we need to have a (n-1)*x + 1 players? That's a necessary condition, is it also sufficient?
  – user74207
  Jul 18 at 17:51
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Arthur, it 'could' also be possible for 4 (obv), 7, 10, 13 as well, cant it? Why 16?
  – user74207
  Jul 18 at 17:51
  

  
  
  
  

 | 
show 5 more comments

up vote
1
down vote

favorite

3

up vote
1
down vote

favorite

3

3

I have seen a lot of posts online about the number of games in a tournament for 2 player games. But what if it's a 4-player game (say Ludo).

Consider a scenario where there are 12 people participating in a Ludo tournament. Each game in the tournament is a 4-player game. Is it possible to schedule the games such that each player plays every other player just once? If yes, how many games will there be in total in the tournament?

Can this be generalized for m players in an n-player game tournament?

combinatorics

share|cite|improve this question

asked Jul 18 at 17:01

user74207

1524

I have seen a lot of posts online about the number of games in a tournament for 2 player games. But what if it's a 4-player game (say Ludo).

Consider a scenario where there are 12 people participating in a Ludo tournament. Each game in the tournament is a 4-player game. Is it possible to schedule the games such that each player plays every other player just once? If yes, how many games will there be in total in the tournament?

Can this be generalized for m players in an n-player game tournament?

combinatorics

share|cite|improve this question

asked Jul 18 at 17:01

user74207

1524

share|cite|improve this question

share|cite|improve this question

asked Jul 18 at 17:01

user74207

1524

asked Jul 18 at 17:01

user74207

1524

asked Jul 18 at 17:01

user74207

1524

1524

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Note that in a 4-player game, each player has three opponents in every game. So for each player to play every other player precisely once, certainly the number of other players must be a multiple of three. That is, the total number of players must be 1 more than a multiple of 3. So in general it is not possible to schedule the games in such a way.
  – Servaes
  Jul 18 at 17:10
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Servaes So the smallest tournament that we cannot immediately reject as impossible would be $16$, with 4 parallel games each round and 5 rounds. Is it possible? I don't know. Also, related post.
  – Arthur
  Jul 18 at 17:19
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Arthur this is analogous to Kirkman's schoolgirl problem (en.wikipedia.org/wiki/Kirkman%27s_schoolgirl_problem). In particular you seem to be looking for some sort of Steiner system (en.wikipedia.org/wiki/Steiner_system).
  – Michael Lugo
  Jul 18 at 17:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Servaes so it seems that in general for a n-player game, we need to have a (n-1)*x + 1 players? That's a necessary condition, is it also sufficient?
  – user74207
  Jul 18 at 17:51
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Arthur, it 'could' also be possible for 4 (obv), 7, 10, 13 as well, cant it? Why 16?
  – user74207
  Jul 18 at 17:51
  

  
  
  
  

 | 
show 5 more comments

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Note that in a 4-player game, each player has three opponents in every game. So for each player to play every other player precisely once, certainly the number of other players must be a multiple of three. That is, the total number of players must be 1 more than a multiple of 3. So in general it is not possible to schedule the games in such a way.
  – Servaes
  Jul 18 at 17:10
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Servaes So the smallest tournament that we cannot immediately reject as impossible would be $16$, with 4 parallel games each round and 5 rounds. Is it possible? I don't know. Also, related post.
  – Arthur
  Jul 18 at 17:19
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Arthur this is analogous to Kirkman's schoolgirl problem (en.wikipedia.org/wiki/Kirkman%27s_schoolgirl_problem). In particular you seem to be looking for some sort of Steiner system (en.wikipedia.org/wiki/Steiner_system).
  – Michael Lugo
  Jul 18 at 17:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Servaes so it seems that in general for a n-player game, we need to have a (n-1)*x + 1 players? That's a necessary condition, is it also sufficient?
  – user74207
  Jul 18 at 17:51
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Arthur, it 'could' also be possible for 4 (obv), 7, 10, 13 as well, cant it? Why 16?
  – user74207
  Jul 18 at 17:51
  

  
  
  
  

2

2

Note that in a 4-player game, each player has three opponents in every game. So for each player to play every other player precisely once, certainly the number of other players must be a multiple of three. That is, the total number of players must be 1 more than a multiple of 3. So in general it is not possible to schedule the games in such a way.
– Servaes
Jul 18 at 17:10

Note that in a 4-player game, each player has three opponents in every game. So for each player to play every other player precisely once, certainly the number of other players must be a multiple of three. That is, the total number of players must be 1 more than a multiple of 3. So in general it is not possible to schedule the games in such a way.
– Servaes
Jul 18 at 17:10

1

1

@Servaes So the smallest tournament that we cannot immediately reject as impossible would be $16$, with 4 parallel games each round and 5 rounds. Is it possible? I don't know. Also, related post.
– Arthur
Jul 18 at 17:19

@Servaes So the smallest tournament that we cannot immediately reject as impossible would be $16$, with 4 parallel games each round and 5 rounds. Is it possible? I don't know. Also, related post.
– Arthur
Jul 18 at 17:19

@Arthur this is analogous to Kirkman's schoolgirl problem (en.wikipedia.org/wiki/Kirkman%27s_schoolgirl_problem). In particular you seem to be looking for some sort of Steiner system (en.wikipedia.org/wiki/Steiner_system).
– Michael Lugo
Jul 18 at 17:33

@Arthur this is analogous to Kirkman's schoolgirl problem (en.wikipedia.org/wiki/Kirkman%27s_schoolgirl_problem). In particular you seem to be looking for some sort of Steiner system (en.wikipedia.org/wiki/Steiner_system).
– Michael Lugo
Jul 18 at 17:33

@Servaes so it seems that in general for a n-player game, we need to have a (n-1)*x + 1 players? That's a necessary condition, is it also sufficient?
– user74207
Jul 18 at 17:51

@Servaes so it seems that in general for a n-player game, we need to have a (n-1)*x + 1 players? That's a necessary condition, is it also sufficient?
– user74207
Jul 18 at 17:51

@Arthur, it 'could' also be possible for 4 (obv), 7, 10, 13 as well, cant it? Why 16?
– user74207
Jul 18 at 17:51

@Arthur, it 'could' also be possible for 4 (obv), 7, 10, 13 as well, cant it? Why 16?
– user74207
Jul 18 at 17:51

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