Originally Posted by Fou-Lu
I can't follow most of what you just said. I haven't a clue how these sports games are played.
This re-introduces the exact same problem as before. With 8 teams there is not enough days to cover 2x of every other team, so we have a remainder of 12 games to play when we need 28 and completely excludes these "bye" games.
So we're back to sqrt(1); after every team plays every other team, there remains 12 games left to play. How do we select whom plays for duplicates and factor in these "bye" games? There are not enough games available to play every team twice, so that's not an option unless you bring it up to 56 games in a 10 week period to allot playing every team twice.
You need to figure out the ruleset involved in dealing with remaining days. Standard combinations will not work as there is simply not enough containers to cover the entire set of options.
Fou look at it this way if you have 8 teams you'll have 40 games over 10 weeks,
8 /40 = 5 that's 5 games the same teams will wind up playing each other
during the season that happens all the time,
problem I'm having is coding things so they are spread out evenly based on start times
match ups and byes if needed.
if you have 12 teams you'll then have 60 games total, that's normal and so on