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Published 2002 Revised 2019
There are 2 well-known types of tournament formats that sport schedulers use.
In this format, one defeat is enough to eliminate a team from the tournament.
Scheduling a single elimination tournament is relatively easy. The first step is to get the number of teams to a power of two: 2, 4, 8, 16, 32...
The schedule for 8 teams is shown below.
With 20 teams, you might select eight of those teams to have a preliminary round. The four winners would then join the other twelve to fill out the sixteen-team field.
If you have 24 teams, you could have everyone compete in a preliminary round; this would leave 12 teams. Then, 4 of those teams (randomly selected) could get a bye and the other 8 teams would play to find the other 4 contestants for the next round.
In a Round Robin tournament every team plays every other team .
There is a systematic approach to scheduling a Round Robin tournament. This method assumes that there are enough fields / pitches / courts so that all the games in a round can be played simultaneously. The technique is called the polygon method .
Let N = number of teams in the tournament. There will be N -1 rounds (each team will play N-1 games). Since each team will play every other team once, no team will be idle during any of the rounds.
Let us schedule a round-robin tournament for 8 teams numbered from 1 to 8.
Draw a regular (N -1) sided polygon (i.e., a heptagon for 8 teams). Each vertex and the centre point represents one team.
Draw horizontal stripes as shown below. Then, join the vertex that has been left out to the centre. Each segment represents teams playing each other in the first round.
So (7, 6), (1, 5), (2, 4) and (3, 8) play in the first round.
Rotate the polygon 1/(N-1)th of a circle (i.e. one vertex point). The new segments represent the pairings for round two.
So (6, 5), (7, 4), (1, 3) and (2, 8) play the second round.
Continue rotating the polygon until it returns to its original position.
One more rotation will bring the polygon back to its original position.
If A, B, C and D are the fields / pitches / courts, the schedule could look like this:
Round | A | B | C | D |
---|---|---|---|---|
I | 7, 6 | 1, 5 | 2, 4 | 3, 8 |
II | 6, 5 | 7, 4 | 1, 3 | 2, 8 |
III | 5, 4 | 6, 3 | 7, 2 | 1, 8 |
IV | 4, 3 | 5, 2 | 6, 1 | 7, 8 |
V | 3, 2 | 4, 1 | 5, 7 | 6, 8 |
VI | 2, 1 | 3, 7 | 4, 6 | 5, 8 |
VII | 1, 7 | 2, 6 | 3, 5 | 4, 8 |
We can also rotate the teams around so that each team plays in every field / pitch / court at least once (at present team 8 always plays in D).
Round | A | B | C | D |
---|---|---|---|---|
I | 7, 6 | 1, 5 | 2, 4 | 3, 8 |
II | 6, 5 | 7, 4 | 1, 3 | 2, 8 |
III | 1, 8 | 6, 3 | 7, 2 | 5, 4 |
IV | 4, 3 | 5, 2 | 7, 8 | 6, 1 |
V | 3, 2 | 4, 1 | 5, 7 | 6, 8 |
VI | 2, 1 | 5, 8 | 4, 6 | 3, 7 |
VII | 1, 7 | 2, 6 | 3, 5 | 4, 8 |
Let N = number of teams in the tournament. There will be N rounds (since each team will play every other team once, and will be idle for exactly one round ).
Let us work out the schedule for 7 teams, numbering the teams from 1 to 7. Draw a regular N-gon (heptagon for 7 teams). Each vertex represents one team.
Draw horizontal stripes as shown below. The vertex that has been left out gives the idle team. Each segment represents teams playing each other in the first round.
So (7, 6), (1, 5) and (2, 4) play in the first round.
Rotate the polygon 1/Nth of a circle (i.e. one vertex point.) The new segments represent the pairings for round two.
Continue rotating the polygon until it returns to its original position.
One more rotation will bring the polygon back to its original position. Therefore, the schedule could look like this:
Round | A | B | C |
---|---|---|---|
I | 7, 6 | 1, 5 | 2, 4 |
II | 6, 5 | 7, 4 | 1, 3 |
III | 5, 4 | 6, 3 | 7, 2 |
IV | 4, 3 | 5, 2 | 6, 1 |
V | 3, 2 | 4, 1 | 5, 7 |
VI | 2, 1 | 3, 7 | 4, 6 |
VII | 1, 7 | 2, 6 | 3, 5 |
The restriction that no vertex has more than one segment drawn to/from it ensures that no team is scheduled for more than one game in each round.
Restricting ourselves to horizontal stripes ensures that no segment is a rotation or reflection of another segment. This means that no pairing will be repeated in a future round.
Notice that in the case where N (no. of teams) was odd, by having only one idle team in each round, the tournament can be completed in the minimum number of rounds.
Arunachalam Y. is a member of the HeyMath! team