Is it possible to create a schedule where we have 6 teams play 5 different games with each team playing each game and each team only once? Or do we have to increase the amount of games to make this possible? The amount of teams cannot change, however. I have attempted to use the room square method with arrays but haven't figured out the best way to solve this issue and am stumped. Any help is appreciated.
$\endgroup$2 Answers
$\begingroup$Week $0$:
$0-5$
$1-4$
$2-3$
Week 1:
$0-1$
$2-4$
$3-5$
Week 2:
$0-2$
$3-4$
$1-5$
Week 3:
$0-3$
$1-2$
$4-5$
Week 4:
$0-4$
$1-3$
$2-5$
The way this works is the following: First, suppose there were not $6$ teams, but $5$ teams (labeled $0$ through $4$), and still $5$ weeks (labeled $0$ through $4$). In week $n$, we match team $k$ with team $n-k$, modulo $5$.
So in week $0$, team $0$ is matched with itself, and then team $1$ plays team $4$ while team $2$ plays team $3$ (since $0+0 = 1+4 = 2+3 = 0$, mod $5$). Then in Week $1$, we'll have $0+1 = 2+4 = 3+3 = 1$, so $3$ is matched with itself. If a team is matched with itself, then it doesn't play.
There is one "unmatched" team each week.
Now expand to $6$ teams. Leave the schedule as-is, but each week pair the "unmatched" team with the new team.
Note: this solution generalizes to an arbitrary number of teams. Let $M>0$ be odd. Then in week $n$ we'll match team $k$ with team $n-k$, modulo $M$, yielding efficient round-robin schedules for leagues of either $M$ (odd) or $M+1$ (even) teams.
$\endgroup$ 4 $\begingroup$There is quite a nice pattern that works for any number of teams. I'll show it for 6 teams but it generalises in the obvious way. First imagine arranging them around a table:
1..2..3
| |
6..5..4so you start with 1v6, 2v5, 3v4. Now keep 1 fixed, and rotate all the others by one position:
1..3..4
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2..6..5which gives you the second pairings, then
1..4..5
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3..2..6...and keep going until you get back to the initial position. If you have an odd number of teams, treat the team who plays no. 1 as sitting out on that iteration.
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