Hi all!

New in the group. Always loved math but I am not a "calculus" expert.

I have a problem to solve to arrange a special tennis tournament.

I have 16 players of 4 different skill levels. Basically, there will be 4 players at each skill level.

Each player will be assigned a player number from 1 to 16.

Level 1 will be players 1-4

Level 2 will be players 5-8

Level 3 will be players 9-12

Level 4 will be players 13-16

The idea is that each player will play doubles with 8 different partners and different combinations of opponents.

Each match is one full tennis set and we will use 4 different tennis courts at the same time. We will play 4 rounds per day for 2 days, a total of 32 doubles matches. That means each player have played 8 matches with 8 different partners. The player with the highest win marginal (won games minus lost games) will be pronounced winner of the entire tournament. In case of a tie, we will first count won serve games and secondly, the number game-winning true serve aces.

The first day players in level 1 partner together with players from level 4, and players in level 2 partner with players in level 3

The second day players in level 1 partner with players from level 3 and players in level 2 partner with players in level 4

Each player must see every other player as an opponent at least once

Here is the "Sudoku"-trick that would be cool to accomplish if possible. Since number 1 through 16 in combinations of four can sum up to 34, (1+2+3+4+5...+16)/4=34) it would be cool if that would be possible to accomplish.

So if player 1 and 16 play against player 2 and 15 their combined player numbers sum up to 34.

I have been sitting with a Google spreadsheet trying to find a logical "algorithm" way to move numbers around in a matrix but I have found it to complex for me. But I am not the sharpest knife in this "math-drawer" either...

I do use Google sheets but that's not a requirement for presenting your suggestion for this game schedule.

## 1-16 in 32 different 4 number variations all sums up to 34

**Moderators:** jestingrabbit, Moderators General, Prelates

### Re: 1-16 in 32 different 4 number variations all sums up to 34

Each player can be considered to have a level (ranging from 0 to 3) and a positional ranking (also ranging from 0 to 3). Their number is just their positional ranking plus one plus four times their level. Once you decide the levels of each matchup much of the way to the thirty-four target number is already there. You already specified the levels of partners, but not of opponents. On the first day we have pairs of 0/3 and 1/2. Either way that gives the team a total of three therefore the match has a total of six, meaning the positional rankings need to add to six.

The second day you have teams of 1/3 and 0/2. This gives a team sum of two or four, this a match sum of four, six, or eight. This means the positional rankings need to add up to fourteen, six, or negative two. Uh, problem? So you can't do 1/3 vs 1/3. And you can't do 0/2 vs 0/2 either, since four positional rankings add up to at most twelve (actually ten, since pulling opponents from the same level limits duplicate numbers).

So suppose I am a player with level 0. On the second day, all four of my matches are against level 1/3 players. Since I don't have room for overlap (16 opponents and 15 unique opponents), that means I face all eight level 1/3 players on the second day, and must face all seven other level 0/2 players on the first day. On the first day I have four matches, and each team has exactly 1 player from levels 0/2. Thus, there will be at least three players that I do not face.

Tl;dr: the problem is overconstrained, and cannot be solved. The number game requires you to balance high level players with low level players, but the restriction on teams that you have specified don't leave room for that if you want to face every opponent in just eight matches.

The second day you have teams of 1/3 and 0/2. This gives a team sum of two or four, this a match sum of four, six, or eight. This means the positional rankings need to add up to fourteen, six, or negative two. Uh, problem? So you can't do 1/3 vs 1/3. And you can't do 0/2 vs 0/2 either, since four positional rankings add up to at most twelve (actually ten, since pulling opponents from the same level limits duplicate numbers).

So suppose I am a player with level 0. On the second day, all four of my matches are against level 1/3 players. Since I don't have room for overlap (16 opponents and 15 unique opponents), that means I face all eight level 1/3 players on the second day, and must face all seven other level 0/2 players on the first day. On the first day I have four matches, and each team has exactly 1 player from levels 0/2. Thus, there will be at least three players that I do not face.

Tl;dr: the problem is overconstrained, and cannot be solved. The number game requires you to balance high level players with low level players, but the restriction on teams that you have specified don't leave room for that if you want to face every opponent in just eight matches.

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