10 team league with film trade

[rattler77]

I am trying to put together a ten team schedule in which each team sees a double scout of a team x number of times throughout the season on film.  We have nine consecutive weeks in the season, and each team trades the two previous game films prior to their match-up.  Obviously, if A is playing B, A will see B on the 2 trades (single scout), but I am concerned about the other team on the films (team C aka double scout).  I believe that a schedule can be created in which every team sees the same number of films (not including single scouts) on each team.  I just can't quite put the pencil to it.

[rattler77]

I am going to rephrase this question.  I realized it is somewhat loaded.  I am in need of a ten team round robin schedule.  Let's say I am in the middle of a season.  Team A is going to play team B.  Team A watches team B's last two games vs. teams C and D.  Using this model, can a schedule be created where A only sees C one time on film?  Keep in mind that we only see the previous two films on an opponent.  Maybe this is more clear.

[wbport]

I'm sure you have seen the shades of pink model on Richard's Main RR site, or perhaps Porter-Berger.  As it happens the Porter-Berger example was already made for 10 players, er, teams and you can move forwards or backwards in the schedule as required.

(Ignore the Crenshaw side in this paragraph) Round 3, or week 3 is the first time you can have two games behind you.  On "Board 2" teams with consecutive pairing numbers will be matched up and this is the only place I found where that is a problem you described.  In the 3rd round, teams 6 and 7 will play.  6 has played 5 and 10 and 7 has played 4 & 5.  *All teams will be on "Board 2" twice during the season.  Note the two teams will not see the same game 5 played in.  If you are limited to only the two previous weeks, I don't think you can get out of showing one game involving 5 to both teams.

BTW, the black background indicates an away game--white or yellow is the home team.

HTH

* Correction, Team 10 stays on the end.

[Ian Wakeling]

If I understand the problem correctly, then by the time of the last round of the tournament, each team will be in possession of 15 films, this is assuming that one film is handed over before the second round is played, and two films for each subsequent round.  As each film contains one double scout, and 15 is not divisble by 9 (the number of possible opponents), then the schedule as a whole can not have the balance that you are looking for.  The only point at which it might be possible, is just before the 6th round is played when each team should have 9 films available - one double scout per opponent.  Is this what you are looking for?  I should warn you that at the moment, I can't see any way of constructing such a schedule.

PS: Now that I re-read the above, I think I must be wrong, since there is no interest in seeing teams on film that you have already played.  Accounting for that will make the problem harder still!

[rattler77]

The Porter-Berger schedule and the Crenshaw schedule are both complete opposites of what I have in mind, but it may be the most fair way to construct the schedule.  I have been working on this for some time now, and I had never thought about the fact that 15 is not divisible by 9.  Anyway, thanks for the responses.  I now have other avenues to explore.