Timeline for Optimal schedule for a soccer tournament
Current License: CC BY-SA 4.0
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 26, 2023 at 11:15 | comment | added | Peter Taylor | No. The definition of $\text{ovmax}$ doesn't count the matches before a team's first match as a break. | |
| Apr 26, 2023 at 11:04 | comment | added | HenrikRüping | Isn't it pretty much the same argument for BESTMAX ? If $ovmax(\sigma)=k$ then no team can play twice in the first $k$ games (since then there would be one team which does not play and it would have an initial break of $k+1$ games) and then in the $k+1$-st games the two teams from the first match cannot play against each other, so that one of them has a break of length $k+1$. | |
| Apr 26, 2023 at 7:25 | comment | added | Peter Taylor | Correction: I didn't claim that it should be easy to prove the bound on BESTMAX. For BESTMIN it's a simple contradiction: if $n=2k$ and $\text{ovmin}(\sigma) \ge k$ then no team can play twice in the first $k$ matches, and the only teams which can play in the $(k+1)$st match (and that only if $\text{ovmin}(\sigma) = k$) are the teams which already met in the first match, violating the tournament structure. For BESTMAX we can show that the teams which play in the first match can't play in the last $n-2$ matches, but that doesn't look restrictive enough. | |
| Apr 26, 2023 at 6:09 | history | answered | HenrikRüping | CC BY-SA 4.0 |