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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.

Deusovi
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MattShuey
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2 Answers2

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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.

mathmandan
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  • Why should "modulo" and the like be needed ? If it is 5 weeks, simply schedule such that all 6 teams play each week, 3 matches/week. – true blue anil Sep 15 '15 at 11:54
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    Well, it's not enough to have every team play every week--you have to decide WHOM each team plays each week (so that by the end, every team has played every other team exactly once). Indeed, this is the point of the OP's question! Also, I include the "modulo" stuff because (1) people like to know how a solution was generated, and (2) you can see that this solution generalizes to $M$ teams: you can get an efficient (minimum-time) round-robin tournament for any number of teams, even or odd, by doing some simple arithmetic. – mathmandan Sep 15 '15 at 15:05
  • Ok, I understand. What I meant was that for a small number of 6 teams, schedules can be worked out without using a system. – true blue anil Sep 15 '15 at 15:52
  • @Morgan: I agree. Deleting my answer. – true blue anil Sep 15 '15 at 18:00
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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..4

so you start with 1v6, 2v5, 3v4. Now keep 1 fixed, and rotate all the others by one position:

1..3..4
|     |
2..6..5

which gives you the second pairings, then

1..4..5
|     |
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.

Julia Hayward
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