non-traditional round robin tournaments

Non-traditional Round Robin Tournaments

Dalibor FroncekUniversity of Minnesota Duluth

Mariusz MeszkaUniversity of Science and Technology Kraków

1–factorization of complete graphs

•the complete graph K2n:2n vertices, every two joined by an edge

•1–factor: set of n independent edges•1–factorization: a partition of the edge set of K2n into 2n–1 1–factors

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1–factorization of complete graphs

Most familiar 1–factorization of a complete graph K2n:

Kirkman, 1846

•geometric construction•labeling construction

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Round robin tournaments

Round robin tournament• 2n teams• every two teams play

exactly one game• tournament consists of

2n–1 rounds • each plays exactly one

game in each round

Complete graph• 2n vertices• every two vertices

joined by an edge• K2n is factorized into

2n–1 factors• factors are regular of

degree 1

STEINERAnother starter for labeling

•Kirkman:18, 27, 36, 45•Steiner:18, 26, 34, 57

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Bipartite fact K8 R-B

Another factorization:First decompose into two factors, K4,4 a 2K4.

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Bipartite fact K8 F1


Then factorize K4,4

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Bipartite F1 F2


Then factorize K4,4

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Bipartite F2 F3


Then factorize K4,4

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Bipartite F3 F4


Then factorize K4,4

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Bipartite 2K4


Then factorize K4,4

and finally factorize 2K4.1

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Bipartite fact K8 R-B


Schedules of this type are useful for two-divisional leagues(like the (in)famous XFLscheduled by J. Dinitz and DF)

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“Just run it through a computer!”

Number of non-isomorphic 1-factorizations of the graph Kn:

n = 4, 6 f = 1n = 8 f = 6n = 10 f = 396n = 12 f = 526 915 620 (Dinitz, Garnick, McKay, 1994)

Number of different schedules for 12 teams: 1 346 098 266 906 624 000

Estimated number of schedules for 16 teams: 1058

What is important:

• opponent – determined by factorization• in seasonal tournaments (leagues) – home

and away games (also determined by factorization)

Ideal home-away pattern (HAP):

Ideally either•HAHAHAHA... or•AHAHAHAH...

Unfortunately, there can be at most two teams with one of these ideal HAPs.

A subsequence AA or HH is called a break in the HAP.

Lemma: An RRT(2n, 2n–1) has at least 2n–2 breaks.


Proof:Pigeonhole principle•HAHAHAHA...•HAHAHAHA...•AHAHAHAH...


Proof:Pigeonhole principle

•HAHAHAHA...•AHAHAHAH...•AHAHAHAH...


We will now show that schedules with this number of breaks really exist.

Kirkman factorization of K8 – Berger tables

• Round 1 – factor F1







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Berger tables with HAPs

team games1 H2 H3 H4 H5 A6 A7 A8 A

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team games1 HH2 HA3 HA4 HA5 AA6 AH7 AH8 AH

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team games1 HHA2 HAH3 HAH4 HAH5 AAH6 AHA7 AHA8 AHA

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team games1 HHAH2 HAHH3 HAHA4 HAHA5 AAHA6 AHAA7 AHAH8 AHAH

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team games1 HHAHAHA2 HAHHAHA3 HAHAHHA4 HAHAHAH5 AAHAHAH6 AHAAHAH7 AHAHAAH8 AHAHAHA

Theorem 1: There exists an RRT(2n, 2n–1) with exactly 2n–2 breaks.

Proof: Generalize the example for 2n teams.

HOME–AWAY PATTERNS

R 1 R 2 R 3 R 4 R 5 R 6 R 71 H H A H A H A2 H A H H A H A3 H A H A H H A4 H A H A H A H5 A A H A H A H6 A H A A H A H7 A H A H A A H8 A H A H A H A

HOME–AWAY PATTERNS WITH THE SCHEDULE

R 1 R 2 R 3 R 48–1 5–8 8–2 6–87–2 4–6 1–3 5–76–3 3–7 7–4 4–15–4 2–1 6–5 3–2

R 5 R 6 R 78–3 7–8 8–42–4 6–1 3–51–5 5–2 2–67–6 4–3 1–7

R 1 R 2 R 3 R 4 R 5 R 6 R 7

1 H H A H A H A

2 H A H H A H A

3 H A H A H H A

4 H A H A H A H

5 A A H A H A H

6 A H A A H A H

7 A H A H A A H

8 A H A H A H A

Problem: How to schedule an RRT(2n–1, 2n–1)?


Equivalent problem: How to catch 2n–1 lions?


Equivalent problem: How to catch 2n–1 lions?

Solution: Catch 2n of them and release one.


Solution: Schedule a RRT(2n, 2n–1). Then select one team to be the dummy team. That means, whoever is scheduled to play the dummy team in a round i has a bye in that round.

We only need to be careful to select the right dummy team.

Select the Dummy Team

R 1 R 2 R 3 R 48–1 5–8 8–2 6–87–2 4–6 1–3 5–76–3 3–7 7–4 4–15–4 2–1 6–5 3–2

R 5 R 6 R 78–3 7–8 8–42–4 6–1 3–51–5 5–2 2–67–6 4–3 1–7

R 1 R 2 R 3 R 4 R 5 R 6 R 7

1 H H A H A H A

2 H A H H A H A

3 H A H A H H A

4 H A H A H A H

5 A A H A H A H

6 A H A A H A H

7 A H A H A A H

8 A H A H A H A

Dummy Team = 5

R 1 R 2 R 3 R 48–1 5–8 8–2 6–87–2 4–6 1–3 5–76–3 3–7 7–4 4–15–4 2–1 6–5 3–2

R 5 R 6 R 78–3 7–8 8–42–4 6–1 3–51–5 5–2 2–67–6 4–3 1–7

R 1 R 2 R 3 R 4 R 5 R 6 R 7

1 H H A H A H A

2 H A H H A H A

3 H A H A H H A

4 H A H A H A H

5 A A H A H A H

6 A H A A H A H

7 A H A H A A H

8 A H A H A H A

Dummy Team = 5

R 1 R 2 R 3 R 48–1 5–8 8–2 6–87–2 4–6 1–3 5–76–3 3–7 7–4 4–15–4 2–1 6–5 3–2

R 5 R 6 R 78–3 7–8 8–42–4 6–1 3–51–5 5–2 2–67–6 4–3 1–7

R 1 R 2 R 3 R 4 R 5 R 6 R 7

1 H H A H H A

2 H A H H A A

3 H A H A H H

4 A H A H A H

5

6 A H A H A H

7 A H A A A H

8 A A H A H A

Dummy Team = 2

R 1 R 2 R 3 R 48–1 5–8 8–2 6–87–2 4–6 1–3 5–76–3 3–7 7–4 4–15–4 2–1 6–5 3–2

R 5 R 6 R 78–3 7–8 8–42–4 6–1 3–51–5 5–2 2–67–6 4–3 1–7

R 1 R 2 R 3 R 4 R 5 R 6 R 7

1 H A H A H A

2

3 H A H H H A

4 H A H A A H

5 A A H A H H

6 A H A A H A

7 H A H A A H

8 A H H A H A

Dummy Team = 8

R 1 R 2 R 3 R 48–1 5–8 8–2 6–87–2 4–6 1–3 5–76–3 3–7 7–4 4–15–4 2–1 6–5 3–2

R 5 R 6 R 78–3 7–8 8–42–4 6–1 3–51–5 5–2 2–67–6 4–3 1–7

R 1 R 2 R 3 R 4 R 5 R 6 R 7

1 H H A H A H A

2 H A H H A H A

3 H A H A H H A

4 H A H A H A H

5 A A H A H A H

6 A H A A H A H

7 A H A H A A H

8 A H A H A H A

Dummy Team = 8

R 1 R 2 R 3 R 47–2 4–6 1–3 5–76–3 3–7 7–4 4–15–4 2–1 6–5 3–2

R 5 R 6 R 72–4 6–1 3–51–5 5–2 2–67–6 4–3 1–7

A schedule with no breaks!

R 1 R 2 R 3 R 4 R 5 R 6 R 7

1 H A H A H A

2 H A H A H A

3 H A H A H A

4 H A H A H A

5 A H A H A H

6 A H A H A H

7 A H A H A H

Given the HAP, find a schedule

R 1 R 2 R 3 R 48–1 5–8 8–2 6–87–2 4–6 1–3 5–76–3 3–7 7–4 4–15–4 2–1 6–5 3–2

R 5 R 6 R 78–3 7–8 8–42–4 6–1 3–51–5 5–2 2–67–6 4–3 1–7

R 1 R 2 R 3 R 4 R 5 R 6 R 7

1 H A H A H A

2 H A H A H A

3 H A H A H A

4 H A H A H A

5 A H A H A H

6 A H A H A H

7 A H A H A H

Look at the game between 1 and 2

R 1 R 2 R 3 R 48–1 5–8 8–2 6–87–2 4–6 1–3 5–76–3 3–7 7–4 4–15–4 2–1 6–5 3–2

R 5 R 6 R 78–3 7–8 8–42–4 6–1 3–51–5 5–2 2–67–6 4–3 1–7

R 1 R 2 R 3 R 4 R 5 R 6 R 7

1 H A H A H A

2 H A H A H A

3 H A H A H A

4 H A H A H A

5 A H A H A H

6 A H A H A H

7 A H A H A H


R 1 R 2 R 3 R 48–1 5–8 8–2 6–87–2 4–6 1–3 5–76–3 3–7 7–4 4–15–4 2–1 6–5 3–2

R 5 R 6 R 78–3 7–8 8–42–4 6–1 3–51–5 5–2 2–67–6 4–3 1–7

R 1 R 2 R 3 R 4 R 5 R 6 R 7

1 H A H A H A

2 H A H A H A

3 H A H A H A

4 H A H A H A

5 A H A H A H

6 A H A H A H

7 A H A H A H

And so on…

R 1 R 2 R 3 R 48–1 5–8 8–2 6–87–2 4–6 1–3 5–76–3 3–7 7–4 4–15–4 2–1 6–5 3–2

R 5 R 6 R 78–3 7–8 8–42–4 6–1 3–51–5 5–2 2–67–6 4–3 1–7

R 1 R 2 R 3 R 4 R 5 R 6 R 7

1 H A H A H A

2 H A H A H A

3 H A H A H A

4 H A H A H A

5 A H A H A H

6 A H A H A H

7 A H A H A H

One more step – teams 1 and 3

R 1 R 2 R 3 R 48–1 5–8 8–2 6–87–2 4–6 1–3 5–76–3 3–7 7–4 4–15–4 2–1 6–5 3–2

R 5 R 6 R 78–3 7–8 8–42–4 6–1 3–51–5 5–2 2–67–6 4–3 1–7

R 1 R 2 R 3 R 4 R 5 R 6 R 7

1 H A H A H A

2 H A H A H A

3 H A H A H A

4 H A H A H A

5 A H A H A H

6 A H A H A H

7 A H A H A H

One more step – teams 2 and 4

R 1 R 2 R 3 R 48–1 5–8 8–2 6–87–2 4–6 1–3 5–76–3 3–7 7–4 4–15–4 2–1 6–5 3–2

R 5 R 6 R 78–3 7–8 8–42–4 6–1 3–51–5 5–2 2–67–6 4–3 1–7

R 1 R 2 R 3 R 4 R 5 R 6 R 7

1 H A H A H A

2 H A H A H A

3 H A H A H A

4 H A H A H A

5 A H A H A H

6 A H A H A H

7 A H A H A H

And so on…

R 1 R 2 R 3 R 48–1 5–8 8–2 6–87–2 4–6 1–3 5–76–3 3–7 7–4 4–15–4 2–1 6–5 3–2

R 5 R 6 R 78–3 7–8 8–42–4 6–1 3–51–5 5–2 2–67–6 4–3 1–7

R 1 R 2 R 3 R 4 R 5 R 6 R 7

1 H A H A H A

2 H A H A H A

3 H A H A H A

4 H A H A H A

5 A H A H A H

6 A H A H A H

7 A H A H A H

One more time – teams 1 and 4

R 1 R 2 R 3 R 48–1 5–8 8–2 6–87–2 4–6 1–3 5–76–3 3–7 7–4 4–15–4 2–1 6–5 3–2

R 5 R 6 R 78–3 7–8 8–42–4 6–1 3–51–5 5–2 2–67–6 4–3 1–7

R 1 R 2 R 3 R 4 R 5 R 6 R 7

1 H A H A H A

2 H A H A H A

3 H A H A H A

4 H A H A H A

5 A H A H A H

6 A H A H A H

7 A H A H A H

One more time – teams 2 and 5

R 1 R 2 R 3 R 48–1 5–8 8–2 6–87–2 4–6 1–3 5–76–3 3–7 7–4 4–15–4 2–1 6–5 3–2

R 5 R 6 R 78–3 7–8 8–42–4 6–1 3–51–5 5–2 2–67–6 4–3 1–7

R 1 R 2 R 3 R 4 R 5 R 6 R 7

1 H A H A H A

2 H A H A H A

3 H A H A H A

4 H A H A H A

5 A H A H A H

6 A H A H A H

7 A H A H A H

…and we are done!

R 1 R 2 R 3 R 48–1 5–8 8–2 6–87–2 4–6 1–3 5–76–3 3–7 7–4 4–15–4 2–1 6–5 3–2

R 5 R 6 R 78–3 7–8 8–42–4 6–1 3–51–5 5–2 2–67–6 4–3 1–7

R 1 R 2 R 3 R 4 R 5 R 6 R 7

1 H A H A H A

2 H A H A H A

3 H A H A H A

4 H A H A H A

5 A H A H A H

6 A H A H A H

7 A H A H A H

Theorem 2: There exists an RRT(2n –1, 2n–1) with no breaks. Moreover, this schedule is unique.

Proof: Use the ideas from the example.

RRT(2n,2n) – a schedule for 2n teams in 2n weeks.Every team has exactly one bye.


Who needs such a schedule anyway??

University of Vermont — Men’s Basketball 2002–2003

JAN Thu 2 at Maine* Sun 5 STONY BROOK* Wed 8 NORTHEASTERN* Sat 11 at Boston University* Mon 13 at CornellWed 15 at Albany*Wed 22 MAINE* Sat 25 HARTFORD* Wed 29 at New Hampshire*

FEB Sun 2 at Binghamton* Tue 4 MIDDLEBURY Sat 8 at Stony Brook* Wed 12 BINGHAMTON* Sat 15 at Northeastern*Wed 19 NEW HAMPSHIRE* Sat 22 BOSTON UNIVERSITY* Wed 26 at Hartford* MAR Sun 2 ALBANY*

Definition: An extended round robin tournament RRT*(n,k) is a tournament RRT(n,k) (with n k) where every bye is replaced by an interdivisional game.


R 1 R 2 R 3 R 4 R 5 R 6 R 7

1 H A H A H A

2 H A H A H A

3 H A H A H A

4 H A H A H A

5 A H A H A H

6 A H A H A H

7 A H A H A H


R 1 R 2 R 3 R 4 R 5 R 6 R 7

1 H H A H A H A

2 H A H H A H A

3 H A H A H H A

4 H A H A H A A

5 A A H A H A H

6 A H A A H A H

7 A H A H A A H

Question: Can we find an RRT*(n,n) with the perfect HAP?


Look at the teams starting HOME.

1 2 3 4 5 6 …

1 H A H A H A …

2 H A H A H A …

3 H A H A H A …


Look at the teams starting HOME. They will never meet, no matter when they play their interdivisional games.

1 2 3 4 5 6 …

1 H A H A H A …

2 H A H A H A …

3 H A H A H A …


We want to prove the following

Theorem 3: In RRT(2n,2n) without breaks there are exactly two teams with a bye in weeks 1, 3,…, 2n–1 or in weeks 2, 4,…, 2n.

Proof:Claim 1. There are at most two teams with a bye in a round. Moreover, these two teams have complementary HAPs.


Proof:Claim 1. There are at most two teams with a bye in a round. Moreover, these two teams have complementary HAPs.Bye teams come in pairs, so suppose there are 4 of them.

… A H A B H A H …


… H A H B A H A …



Proof:Claim 1. There are at most two teams with a bye in a round. Moreover, these two teams have complementary HAPs.Bye teams come in pairs, so suppose there are 4 of them.




Proof:Claim 2. There are at most two teams with a bye in any two consecutive rounds..



… H A H A B H A …

A H A H B A H



Claim 2. There are at most two teams with a bye in any two consecutive rounds..

So we are done, and have byes in weeks 1, 3,…, 2n–1 or in weeks 2, 4,…, 2n.






WRONG!!!






WRONG!!! What about rounds 1, 3, … , 2n–2, 2n?


Proof:Claim 3. If there are teams with a bye in Round 1 then there are no teamswith a bye in Round 2n and vice versa.

B A H A … H A H A

B H A H … A H A H

A H A H … A H A B

H A H A … H A H B


Proof:Claim 3. If there are teams with a bye in Round 1 then there are no teamsWith a bye in Round 2n and vice versa.

B A H A … H A H A

B H A H … A H A H

A H A H … A H A B

H A H A … H A H B




Claim 3. If there are teams with a bye in Round 1 then there are no teamswith a bye in Round 2n and vice versa.

Now we are really done.

Let us find a schedule for RRT(2n,2n)


• But how?


• But how?• The schedule for RRT(2n–1,2n–1) was uniquely

determined by its HAP.


• But how?• The schedule for RRT(2n–1,2n–1) was uniquely

determined by its HAP.

• So let us try what the HAP yields.

By our Theorem 3, a schedule for 12 teams looks like this.

R 1 R 2 R 3 R 4 R 5 R 6 R 7 R 8 R 9 R10 R11 R12

1 H A H A H A H A H A H2 H A H A H A H A H A H3 H A H A H A H A H A H4 H A H A H A H A H A H5 H A H A H A H A H A H6 H A H A H A H A H A H7 A H A H A H A H A H A8 A H A H A H A H A H A9 A H A H A H A H A H A10 A H A H A H A H A H A11 A H A H A H A H A H A12 A H A H A H A H A H A

Look at 1 and 2R 1 R 2 R 3 R 4 R 5 R 6 R 7 R 8 R 9 R10 R11 R12


Look at 1 and 2

Both home: cannot play

R 1 R 2 R 3 R 4 R 5 R 6 R 7 R 8 R 9 R10 R11 R12


Look at 1 and 2


Both away:cannot play

R 1 R 2 R 3 R 4 R 5 R 6 R 7 R 8 R 9 R10 R11 R12


Look at 1 and 2


Both away:cannot play

So there is just oneround left.

R 1 R 2 R 3 R 4 R 5 R 6 R 7 R 8 R 9 R10 R11 R12


This way it works for all pairsk, k+1

R 1 R 2 R 3 R 4 R 5 R 6 R 7 R 8 R 9 R10 R11 R12


Now teams 1 and 3R 1 R 2 R 3 R 4 R 5 R 6 R 7 R 8 R 9 R10 R11 R12


Now teams 1 and 3

Cannot play each other when anothergame has already been scheduled

R 1 R 2 R 3 R 4 R 5 R 6 R 7 R 8 R 9 R10 R11 R12


Now teams 1 and 3


Cannot play whenboth have a homegame

R 1 R 2 R 3 R 4 R 5 R 6 R 7 R 8 R 9 R10 R11 R12


Now teams 1 and 3



or both have anaway game

R 1 R 2 R 3 R 4 R 5 R 6 R 7 R 8 R 9 R10 R11 R12


Now teams 1 and 3



or both have anaway game.

So again just one choice.

R 1 R 2 R 3 R 4 R 5 R 6 R 7 R 8 R 9 R10 R11 R12


Similarly…R 1 R 2 R 3 R 4 R 5 R 6 R 7 R 8 R 9 R10 R11 R12


And again…R 1 R 2 R 3 R 4 R 5 R 6 R 7 R 8 R 9 R10 R11 R12


Theorem 3: There exists a RRT(2n, 2n) with no breaks. Moreover, this schedule is unique.

Proof: Use the ideas from the example.

Another look at our RRT(7, 7)

R 1 R 2 R 3 R 47–2 4–6 1–3 5–76–3 3–7 7–4 4–15–4 2–1 6–5 3–2

R 5 R 6 R 72–4 6–1 3–51–5 5–2 2–67–6 4–3 1–7

aij = k the game between i and j is played in Round k

1 2 3 4 5 6 7

1

2

3

4

5

6

7


R 1 R 2 R 3 R 47–2 4–6 1–3 5–76–3 3–7 7–4 4–15–4 2–1 6–5 3–2

R 5 R 6 R 72–4 6–1 3–51–5 5–2 2–67–6 4–3 1–7

aij = k the game between i and j is played in Round k

1 2 3 4 5 6 7

1 2 3 4 5 6 7

2 2 4 5 6 7 1

3 3 4 6 7 1 2

4 4 5 6 1 2 3

5 5 6 7 1 3 4

6 6 7 1 2 3 5

7 7 1 2 3 4 5


R 1 R 2 R 3 R 47–2 4–6 1–3 5–76–3 3–7 7–4 4–15–4 2–1 6–5 3–2

R 5 R 6 R 72–4 6–1 3–51–5 5–2 2–67–6 4–3 1–7

aii = k

team i has a bye in Round k

1 2 3 4 5 6 7

1 1 2 3 4 5 6 7

2 2 3 4 5 6 7 1

3 3 4 5 6 7 1 2

4 4 5 6 7 1 2 3

5 5 6 7 1 2 3 4

6 6 7 1 2 3 4 5

7 7 1 2 3 4 5 6

Another look at RRT(8, 8)

R 1 R 2 R 3 R 48–2 5–6 1–3 6–77–3 4–7 8–4 5–86–4 3–8 7–5 4–1

2–1 3–2

R 5 R 6 R 7 R 82–4 7–8 3–5 8–11–5 6–1 2–6 7–28–6 5–2 1–7 6–3

4–3 5–4

1 2 3 4 5 6 7 8

1 1 2 3 4 5 6 7 8

2 2 3 4 5 6 7 8 1

3 3 4 5 6 7 8 1 2

4 4 5 6 7 8 1 2 3

5 5 6 7 8 1 2 3 4

6 6 7 8 1 2 3 4 5

7 7 8 1 2 3 4 5 6

8 8 1 2 3 4 5 6 7

A general RRT(n, n)

1 2 3 … n–1 n

1 1 2 3 n–1 n

2 2 3 4 n 1

3 3 4 5 1 2

… …

n–1 n–1 n 1 … n–4 n–3 n–2

n n 1 2 … n–3 n–2 n–1

A general RRT(n, n)

Substitute i i–11 2 3 … n–1 n

1 1 2 3 n–1 n

2 2 3 4 n 1

3 3 4 5 1 2

… …

n–1 n–1 n 1 … n–4 n–3 n–2

n n 1 2 … n–3 n–2 n–1

A general RRT(n, n)

Substitute i i–1

to get the group Zn !!

0 1 2 … … n–3 n–2 n–1

0 0 1 2 n–3 n–2 n–1

1 1 2 3 n–2 n–1 0

2 2 3 4 n–1 0 1

… …

n–2 n–2 n–1 0 … … n–5 n–4 n–3

n–1 n–1 0 1 … … n–4 n–3 n–2

A general RRT(n, n)

Substitute i i–1

to get the group Zn !!

In other words, we getthe table of addition mod n

0 1 2 … … n–3 n–2 n–1

0 0 1 2 n–3 n–2 n–1

1 1 2 3 n–2 n–1 0

2 2 3 4 n–1 0 1

… …

n–2 n–2 n–1 0 … … n–5 n–4 n–3

n–1 n–1 0 1 … … n–4 n–3 n–2

!!!!!!!!!!!!!! THE END !!!!!!!!!!!!

non-traditional round robin tournaments

Documents

2n1 factorsfactors

bipartite 2k4

regular of degree

edge set of k2n

steineranother starter