Math problem regarding rotating teams playing a game - need help please!

[jackie112]

Hi all!
I have a question I need some help with. I'm organizing four rounds of games for 18 teams to play. Two teams play each other for each round. There are 9 numbered game tables in the room. I need to have the teams rotate around the room so that no two teams will ever play each other twice for any of the four rounds of games.

For the first game, I just assign a random table to each team. (So two teams play each other at each of the 9 tables).

For the second game, the winner from the first round of games stays at the table they played at for the first round. The loser moves up a table (ie: if the loser was on table 1 for the first round, they will move up to table 2 for the second round).

For the third game, the winner from the second round of games stays at the table they played at for the second round. The loser moves up *TWO* tables (ie: if the loser was on table 2 for the second round, they will move up to table 4 for the third round). I have checked this manually and it seems to work that no team will ever play a team they have played already for the third round.

Here is the tricky part - for the fourth round, the winner from the third round of games stays at the table they played at for the third round. The loser moves up *FOUR* tables (ie: if the loser was on table 2 for the third round, they will move up to table 6 for the fourth round). DOES THIS WORK??? Will no team ever play each other twice? It hurts by brain to figure it out!! hehe. Please help! I know it doesn't work out if the loser moves up three tables for the last round, but does it work with moving four tables? Any help is greatly appreciated!

 

[JeffM]

You need to clarify what you are trying to do. If you simply want to ensure that no team plays another twice, then this will work. Assign 9 teams to a table. Those teams will not move. Call them the stayers. Assign the remaining teams, the movers, to a table for the first round. On the next round, each mover moves to the next higher numbered table except for the movers at table 9, who move to table 1. Follow the same procedure for rounds three and four. It appears, however, that you are trying to achieve something more complex than just ensuring that no team plays another team twice. What is it? (Once I know what it is, I do not guarantee that I shall be able to answer it, but someone will.)

 

[jackie112]

Thanks so much for the reply Jeff! I am trying to have it so that the winner always stays on the table they are playing at and the loser moves tables (and not just have a random team always stay at a table). I guess that is the part that makes it a bit more complicated. Sorry, I didn't clarify that part.
So in this case, for example, it didn't work when I tried to move the teams over three tables for the fourth round. Because if I had the losing teams move three tables over for round four, then they could end up playing the first team they had already played in round one. (Of course this would depend on the win/loss record of the team they played for round 1 - ie: if Team A's record was Win,Win,Loss and Team B's record was Loss, Loss, Win, then these two teams would play each other again for the last round with losing teams shifting three tables over for the last round). So I changed it to having the losing teams move four tables over for the last round and I think it might work but am not certain. I think it works for the case where a team will not play another team from the first round, but am racking my brain trying to make sure they won't ever play one of the other teams they've come across for round 2 (depending on their win/loss records). I hate proofs but isn't there some mathematical proof or something that I can do to figure this out? Thanks so much for the help, I greatly appreciate it!!!

 

[JeffM]

Thanks so much for the reply Jeff! I am trying to have it so that the winner always stays on the table they are playing at and the loser moves tables (and not just have a random team always stay at a table). I guess that is the part that makes it a bit more complicated. Sorry, I didn't clarify that part.
So in this case, for example, it didn't work when I tried to move the teams over three tables for the fourth round. Because if I had the losing teams move three tables over for round four, then they could end up playing the first team they had already played in round one. (Of course this would depend on the win/loss record of the team they played for round 1 - ie: if Team A's record was Win,Win,Loss and Team B's record was Loss, Loss, Win, then these two teams would play each other again for the last round with losing teams shifting three tables over for the last round). So I changed it to having the losing teams move four tables over for the last round and I think it might work but am not certain. I think it works for the case where a team will not play another team from the first round, but am racking my brain trying to make sure they won't ever play one of the other teams they've come across for round 2 (depending on their win/loss records). I hate proofs but isn't there some mathematical proof or something that I can do to figure this out? Thanks so much for the help, I greatly appreciate it!!!

I still do not think I have all the desired constraints in my mind. I can give to you (and prove) a simple algorithm that has losers move, winners stay put, and none of the eighteen teams play each other twice in k + 1 rounds where k < 9. But it looks to me as though you are in essence also trying to seed the players; that is if two teams have each won three games in the first three rounds, you want them to play against each other in the fourth and final round. If you have some sort of scheme as that in mind, there are situations in which it is impossible to have all winning teams always stay put. To put it a different way, if you have some sort of laddering in mind, it probably makes sense to start there.

 

[jackie112]

Yes, that is what I'm looking for - the losers move, winners stay put, and none of the eighteen teams play each other twice in 4 rounds of games. So does it work for the last round by rotating the losers 4 tables up after round 3? Will none of the teams play each other twice? And is k which you referred to equal to the number of rounds of games? If so, k would be < 4, right? Or are you using k as the number of tables (which is 9).

Ideally, yes, I would like to have it so that if two teams have each won three games in the first three rounds - then I do want them to play against each other in the fourth/final round. However, I thought it would be impossible to figure out what to do with the remaining teams after pulling out the undefeated teams to play each other, as I want the other teams to still rotate around the tables on their own and continue playing a team they haven't played before in the fourth round. So my plan was just to not account for the undefeated teams in figuring out the rotation - and then to manually swap the undefeated teams to play each other (ie: have one of the undefeated teams move tables so that they play other undefeated team - and the 2 losing teams from those two tables will just play each other). Unless is there a better solution than just doing that last part manually? But then there is also the issue where there could be more than 2 teams who are undefeated. So I think I'll have to end up with some manual maneuvering regardless I am fine settling for the initial solution of just making sure none of the teams play each other by having them rotate on their own as I listed in the first paragraph and then do the 'undefeated team swap'. (As long as you think it works shifting 4 tables for the losing team after round 3). Thanks again for all your help!

 

[JeffM]

Yes, that is what I'm looking for - the losers move, winners stay put, and none of the eighteen teams play each other twice in 4 rounds of games. So does it work for the last round by rotating the losers 4 tables up after round 3? Will none of the teams play each other twice? And is k which you referred to equal to the number of rounds of games? If so, k would be < 4, right? Or are you using k as the number of tables (which is 9).

Ideally, yes, I would like to have it so that if two teams have each won three games in the first three rounds - then I do want them to play against each other in the fourth/final round. However, I thought it would be impossible to figure out what to do with the remaining teams after pulling out the undefeated teams to play each other, as I want the other teams to still rotate around the tables on their own and continue playing a team they haven't played before in the fourth round. So my plan was just to not account for the undefeated teams in figuring out the rotation - and then to manually swap the undefeated teams to play each other (ie: have one of the undefeated teams move tables so that they play other undefeated team - and the 2 losing teams from those two tables will just play each other). Unless is there a better solution than just doing that last part manually? But then there is also the issue where there could be more than 2 teams who are undefeated. So I think I'll have to end up with some manual maneuvering regardless I am fine settling for the initial solution of just making sure none of the teams play each other by having them rotate on their own as I listed in the first paragraph and then do the 'undefeated team swap'. (As long as you think it works shifting 4 tables for the losing team after round 3). Thanks again for all your help!

I think what you are proposing is complicated. Here is what I would do. Each winning team stays put. Each losing team goes to the first table with a higher number that has only one team sitting that the moving team has not yet played. If the moving team has not found such a table by table 9, they start over with table 1.