non-traditional round robin tournaments dalibor froncek university of minnesota duluth mariusz...

Non-traditional Round Robin Tournaments

Dalibor Froncek

University of Minnesota Duluth

Mariusz Meszka

University 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

STEINER

Another starter for labeling

•Kirkman:

18, 27, 36, 45•Steiner:

18, 26, 34, 57

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8

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

.1

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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 XFL

scheduled 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 = 1

n = 8 f = 6

n = 10 f = 396

n = 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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4 5

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team games1 HHAH2 HAHH3 HAHA4 HAHA5 AAHA6 AHAA7 AHAH8 AHAH

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4 5

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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 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

HOME–AWAY PATTERNS WITH THE SCHEDULE

R 1 R 2 R 3 R 4

8–1 5–8 8–2 6–8

7–2 4–6 1–3 5–7

6–3 3–7 7–4 4–1

5–4 2–1 6–5 3–2

R 5 R 6 R 7

8–3 7–8 8–4

2–4 6–1 3–5

1–5 5–2 2–6

7–6 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 4

8–1 5–8 8–2 6–8

7–2 4–6 1–3 5–7

6–3 3–7 7–4 4–1

5–4 2–1 6–5 3–2

R 5 R 6 R 7

8–3 7–8 8–4

2–4 6–1 3–5

1–5 5–2 2–6

7–6 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 4

8–1 5–8 8–2 6–8

7–2 4–6 1–3 5–7

6–3 3–7 7–4 4–1

5–4 2–1 6–5 3–2

R 5 R 6 R 7

8–3 7–8 8–4

2–4 6–1 3–5

1–5 5–2 2–6

7–6 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 4

8–1 5–8 8–2 6–8

7–2 4–6 1–3 5–7

6–3 3–7 7–4 4–1

5–4 2–1 6–5 3–2

R 5 R 6 R 7

8–3 7–8 8–4

2–4 6–1 3–5

1–5 5–2 2–6

7–6 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 4

8–1 5–8 8–2 6–8

7–2 4–6 1–3 5–7

6–3 3–7 7–4 4–1

5–4 2–1 6–5 3–2

R 5 R 6 R 7

8–3 7–8 8–4

2–4 6–1 3–5

1–5 5–2 2–6

7–6 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 4

8–1 5–8 8–2 6–8

7–2 4–6 1–3 5–7

6–3 3–7 7–4 4–1

5–4 2–1 6–5 3–2

R 5 R 6 R 7

8–3 7–8 8–4

2–4 6–1 3–5

1–5 5–2 2–6

7–6 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 4

7–2 4–6 1–3 5–7

6–3 3–7 7–4 4–1

5–4 2–1 6–5 3–2

R 5 R 6 R 7

2–4 6–1 3–5

1–5 5–2 2–6

7–6 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 4

8–1 5–8 8–2 6–8

7–2 4–6 1–3 5–7

6–3 3–7 7–4 4–1

5–4 2–1 6–5 3–2

R 5 R 6 R 7

8–3 7–8 8–4

2–4 6–1 3–5

1–5 5–2 2–6

7–6 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 4

8–1 5–8 8–2 6–8

7–2 4–6 1–3 5–7

6–3 3–7 7–4 4–1

5–4 2–1 6–5 3–2

R 5 R 6 R 7

8–3 7–8 8–4

2–4 6–1 3–5

1–5 5–2 2–6

7–6 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 4

8–1 5–8 8–2 6–8

7–2 4–6 1–3 5–7

6–3 3–7 7–4 4–1

5–4 2–1 6–5 3–2

R 5 R 6 R 7

8–3 7–8 8–4

2–4 6–1 3–5

1–5 5–2 2–6

7–6 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 4

8–1 5–8 8–2 6–8

7–2 4–6 1–3 5–7

6–3 3–7 7–4 4–1

5–4 2–1 6–5 3–2

R 5 R 6 R 7

8–3 7–8 8–4

2–4 6–1 3–5

1–5 5–2 2–6

7–6 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 4

8–1 5–8 8–2 6–8

7–2 4–6 1–3 5–7

6–3 3–7 7–4 4–1

5–4 2–1 6–5 3–2

R 5 R 6 R 7

8–3 7–8 8–4

2–4 6–1 3–5

1–5 5–2 2–6

7–6 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 4

8–1 5–8 8–2 6–8

7–2 4–6 1–3 5–7

6–3 3–7 7–4 4–1

5–4 2–1 6–5 3–2

R 5 R 6 R 7

8–3 7–8 8–4

2–4 6–1 3–5

1–5 5–2 2–6

7–6 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 4

8–1 5–8 8–2 6–8

7–2 4–6 1–3 5–7

6–3 3–7 7–4 4–1

5–4 2–1 6–5 3–2

R 5 R 6 R 7

8–3 7–8 8–4

2–4 6–1 3–5

1–5 5–2 2–6

7–6 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 4

8–1 5–8 8–2 6–8

7–2 4–6 1–3 5–7

6–3 3–7 7–4 4–1

5–4 2–1 6–5 3–2

R 5 R 6 R 7

8–3 7–8 8–4

2–4 6–1 3–5

1–5 5–2 2–6

7–6 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 4

8–1 5–8 8–2 6–8

7–2 4–6 1–3 5–7

6–3 3–7 7–4 4–1

5–4 2–1 6–5 3–2

R 5 R 6 R 7

8–3 7–8 8–4

2–4 6–1 3–5

1–5 5–2 2–6

7–6 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 4

8–1 5–8 8–2 6–8

7–2 4–6 1–3 5–7

6–3 3–7 7–4 4–1

5–4 2–1 6–5 3–2

R 5 R 6 R 7

8–3 7–8 8–4

2–4 6–1 3–5

1–5 5–2 2–6

7–6 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 Cornell

Wed 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


Proof:

Claim 1. There are at most two teams with a bye in a round.

Moreover, these two teams have complementary HAPs.



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

or in weeks 2, 4,…, 2n.



Proof:







WRONG!!!



Proof:







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 teams

with 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:


With 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:






with 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 A

10 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 home:

cannot play

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 home:

cannot play

Both away:

cannot play

So there is just one

round 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 pairs

k, 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 another

game 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 each

other when another

game has already

been scheduled

Cannot play when

both have a home

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

Cannot play each

other when another

game has already

been scheduled

Cannot play when

both have a home

game

or both have an

away 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

Cannot play each

other when another

game has already

been scheduled

Cannot play when

both have a home

game

or both have an

away 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 4

7–2 4–6 1–3 5–7

6–3 3–7 7–4 4–1

5–4 2–1 6–5 3–2

R 5 R 6 R 7

2–4 6–1 3–5

1–5 5–2 2–6

7–6 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 4

8–2 5–6 1–3 6–7

7–3 4–7 8–4 5–8

6–4 3–8 7–5 4–1

2–1 3–2

R 5 R 6 R 7 R 8

2–4 7–8 3–5 8–1

1–5 6–1 2–6 7–2

8–6 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 get

the 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 dalibor froncek university of minnesota duluth mariusz...

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