I think I’ve found a way for you to work out your five-player matrix without too much trouble. I tried to see if there was a reasonably systematic / programmatic way to work out the answer for the simpler case of a three-player matrix; there is, and you should be able to scale it up. It’s a manual process that involves checking things – but it uses a process of elimination, so some parts of the process get shorter as you progress, which makes it manageable.

It also breaks the problem down into stages, so that you don’t have to try to figure out all the variables at once by an excruciating process of trial and error. Crucially, it first determines only the issue of “who plays with whom in each game”, and it leaves until a second stage the issue of who plays which turn order position.

Here’s the approach I took, after studying the photo you posted (which answered a lot of my earlier questions).

For clarity, I designated the players with numbers (1, 2, 3, etc.), and I designated the turn order positions with capital letters (A, B and C). You might want to give Roman numerals to the games.

I started with Player 1 (hereafter just 1). The game has three turn order positions, and therefore also has three players per game. 1 has to play once in each turn order position, which means three games. There are three players per game, so at each game 1 plays against two opponents. The opponents are required to all be different at every game, so this means that 1 will play against six other people, numbered from 2 to 7. Thus:

1 plays against 2, 3, 4, 5, 6, and 7

Since it’s a three-person game, this translates into:

1 plays against (2 & 3) and (4 & 5) and (6 and 7)

All three of 1’s match-us are now spoken for (though note that we haven’t yet nailed down which turn order position he plays in each pairing; that will come later), so we can turn to the next player to consider. Here’s where the process of elimination comes in. The next player to consider is 2. If we were replicating the pattern of this line…

1 plays against 2, 3, 4, 5, 6, and 7

…you’d think this that corresponding “2” line would look like this…

2 plays against 1, 3, 4, 5, 6, and 7…

but in actuality it looks like this…

2 plays against 4, 5, 6, and 7

…with two numbers (1 and 3) eliminated because 2 has already played against both of them. (The systematic check to apply at every step of this procedure is the question, “has this been used before?”)

At first glance, if we follow what we did with 1, this…

2 plays against 4, 5, 6, and 7

…translates into…

2 plays against (4 & 5) and (6 and 7)

…but if we apply the “has this been used before?” test we see that (4 & 5) and (6 and 7) were both used with 1, so we can’t use them with 2. Switching two of the digits, however, gives this…

2 plays against (4 & 6) and (5 and 7)

…which works because 4 hasn’t previously played against 6, and 5 hasn’t previously played against 7.

At this point I put together a little table to see if anything was missing. The results I have so far are:

The “1” match-ups

1 2 3

1 4 5

1 6 7

The “2” match-ups:

2 4 6

2 5 7

This leaves the “3” match-ups to work out. 3 has already been used with 1 and 2, but has not been used with 4, so I wrote the partial line:

3 4

We need another number to complete the line. 5 comes after 4, but it’s been used with 4 previously. Next comes 6, but that’s also been used with 4 previously. Next is 7, which has no previous use with 4, so the line becomes:

3 4 7

I then went back over all the previous lines to check whether there are any remaining numbers with which 3 had not yet been used. Yes indeed: 5 and 6. This generated the following line:

3 5 6

I don’t know if the forum has a posting length limit, so I’m going to break off here and continue in a second post in a few moments.