Non-traditional Round Robin Tournaments Dalibor Froncek University of Minnesota Duluth Mariusz Meszka University of Science and Technology Kraków.

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 K 2n : 2n vertices, every two joined by an edge 1–factor: set of n independent edges 1–factorization: a partition of the edge set of K 2n into 2n–1 1–factors

1–factorization of complete graphs Most familiar 1–factorization of a complete graph K 2n : Kirkman, 1846 geometric construction labeling construction

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 K 2n 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

Bipartite fact K 8 R-B Another factorization: First decompose into two factors, K 4,4 a 2K 4.

Bipartite fact K 8 F 1 Another factorization: First decompose into two factors, K 4,4 a 2K 4. Then factorize K 4,4

Bipartite F 1 F 2 Another factorization: First decompose into two factors, K 4,4 a 2K 4. Then factorize K 4,4

Bipartite F 2 F 3 Another factorization: First decompose into two factors, K 4,4 a 2K 4. Then factorize K 4,4.

Bipartite F 3 F 4 Another factorization: First decompose into two factors, K 4,4 a 2K 4. Then factorize K 4,4

Bipartite 2K 4 Another factorization: First decompose into two factors, K 4,4 a 2K 4. Then factorize K 4,4 and finally factorize 2K 4.

Bipartite fact K 8 R-B Another factorization: First decompose into two factors, K 4,4 a 2K 4. Schedules of this type are useful for two-divisional leagues (like the (in)famous XFL scheduled by J. Dinitz and DF)

“Just run it through a computer!” Number of non-isomorphic 1-factorizations of the graph K n : n = 4, 6f = 1 n = 8f = 6 n = 10f = 396 n = 12f = 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: 10 58

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... AHAHAHAH...

Lemma: An RRT(2n, 2n–1) has at least 2n–2 breaks. Proof: Pigeonhole principle HAHAHAHA... AHAHAHAH...

Lemma: An RRT(2n, 2n–1) has at least 2n–2 breaks. We will now show that schedules with this number of breaks really exist.

Kirkman factorization of K 8 – Berger tables Round 1 – factor F 1 Round 2 – factor F 5 Round 3 – factor F 2 Round 4 – factor F 6 Round 5 – factor F 3 Round 6 – factor F 7 Round 7 – factor F 4

Berger tables with HAPs teamgames 1H 2H 3H 4H 5A 6A 7A 8A

Berger tables with HAPs teamgames 1HH 2HA 3HA 4HA 5AA 6AH 7AH 8AH

Berger tables with HAPs teamgames 1HHA 2HAH 3HAH 4HAH 5AAH 6AHA 7AHA 8AHA

Berger tables with HAPs teamgames 1HHAH 2HAHH 3HAHA 4HAHA 5AAHA 6AHAA 7AHAH 8AHAH

Berger tables with HAPs teamgames 1HHAHAHA 2HAHHAHA 3HAHAHHA 4HAHAHAH 5AAHAHAH 6AHAAHAH 7AHAHAAH 8AHAHAHA

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 1R 2R 3R 4R 5R 6R 7 1HHAHAHA 2HAHHAHA 3HAHAHHA 4HAHAHAH 5AAHAHAH 6AHAAHAH 7AHAHAAH 8AHAHAHA

HOME–AWAY PATTERNS WITH THE SCHEDULE R 1R 2R 3R 4 8–15–88–26–8 7–24–61–35–7 6–33–77–44–1 5–42–16–53–2 R 5R 6R 7 8–37–88–4 2–46–13–5 1–55–22–6 7–64–31–7 R 1R 2R 3R 4R 5R 6R 7 1HHAHAHA 2HAHHAHA 3HAHAHHA 4HAHAHAH 5AAHAHAH 6AHAAHAH 7AHAHAAH 8AHAHAHA

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

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

Problem: How to schedule an RRT(2n–1, 2n–1)? Equivalent problem: How to catch 2n–1 lions? Solution: Catch 2n of them and release one.

Problem: How to schedule an RRT(2n–1, 2n–1)? 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 1R 2R 3R 4 8–15–88–26–8 7–24–61–35–7 6–33–77–44–1 5–42–16–53–2 R 5R 6R 7 8–37–88–4 2–46–13–5 1–55–22–6 7–64–31–7 R 1R 2R 3R 4R 5R 6R 7 1HHAHAHA 2HAHHAHA 3HAHAHHA 4HAHAHAH 5AAHAHAH 6AHAAHAH 7AHAHAAH 8AHAHAHA

Dummy Team = 5 R 1R 2R 3R 4 8–15–88–26–8 7–24–61–35–7 6–33–77–44–1 5–42–16–53–2 R 5R 6R 7 8–37–88–4 2–46–13–5 1–55–22–6 7–64–31–7 R 1R 2R 3R 4R 5R 6R 7 1HHAHAHA 2HAHHAHA 3HAHAHHA 4HAHAHAH 5AAHAHAH 6AHAAHAH 7AHAHAAH 8AHAHAHA

Dummy Team = 5 R 1R 2R 3R 4 8–15–88–26–8 7–24–61–35–7 6–33–77–44–1 5–42–16–53–2 R 5R 6R 7 8–37–88–4 2–46–13–5 1–55–22–6 7–64–31–7 R 1R 2R 3R 4R 5R 6R 7 1HHAHHA 2HAHHAA 3HAHAHH 4AHAHAH 5 6AHAHAH 7AHAAAH 8AAHAHA

Dummy Team = 2 R 1R 2R 3R 4 8–15–88–26–8 7–24–61–35–7 6–33–77–44–1 5–42–16–53–2 R 5R 6R 7 8–37–88–4 2–46–13–5 1–55–22–6 7–64–31–7 R 1R 2R 3R 4R 5R 6R 7 1HAHAHA 2 3HAHHHA 4HAHAAH 5AAHAHH 6AHAAHA 7HAHAAH 8AHHAHA

Dummy Team = 8 R 1R 2R 3R 4 8–15–88–26–8 7–24–61–35–7 6–33–77–44–1 5–42–16–53–2 R 5R 6R 7 8–37–88–4 2–46–13–5 1–55–22–6 7–64–31–7 R 1R 2R 3R 4R 5R 6R 7 1HHAHAHA 2HAHHAHA 3HAHAHHA 4HAHAHAH 5AAHAHAH 6AHAAHAH 7AHAHAAH 8AHAHAHA

Dummy Team = 8 R 1R 2R 3R 4 7–24–61–35–7 6–33–77–44–1 5–42–16–53–2 R 5R 6R 7 2–46–13–5 1–55–22–6 7–64–31–7 A schedule with no breaks! R 1R 2R 3R 4R 5R 6R 7 1HAHAHA 2HAHAHA 3HAHAHA 4HAHAHA 5AHAHAH 6AHAHAH 7AHAHAH

Given the HAP, find a schedule R 1R 2R 3R 4 8–15–88–26–8 7–24–61–35–7 6–33–77–44–1 5–42–16–53–2 R 5R 6R 7 8–37–88–4 2–46–13–5 1–55–22–6 7–64–31–7 R 1R 2R 3R 4R 5R 6R 7 1HAHAHA 2HAHAHA 3HAHAHA 4HAHAHA 5AHAHAH 6AHAHAH 7AHAHAH

Look at the game between 1 and 2 R 1R 2R 3R 4 8–15–88–26–8 7–24–61–35–7 6–33–77–44–1 5–42–16–53–2 R 5R 6R 7 8–37–88–4 2–46–13–5 1–55–22–6 7–64–31–7 R 1R 2R 3R 4R 5R 6R 7 1HAHAHA 2HAHAHA 3HAHAHA 4HAHAHA 5AHAHAH 6AHAHAH 7AHAHAH

And so on… R 1R 2R 3R 4 8–15–88–26–8 7–24–61–35–7 6–33–77–44–1 5–42–16–53–2 R 5R 6R 7 8–37–88–4 2–46–13–5 1–55–22–6 7–64–31–7 R 1R 2R 3R 4R 5R 6R 7 1HAHAHA 2HAHAHA 3HAHAHA 4HAHAHA 5AHAHAH 6AHAHAH 7AHAHAH

One more step – teams 1 and 3 R 1R 2R 3R 4 8–15–88–26–8 7–24–61–35–7 6–33–77–44–1 5–42–16–53–2 R 5R 6R 7 8–37–88–4 2–46–13–5 1–55–22–6 7–64–31–7 R 1R 2R 3R 4R 5R 6R 7 1HAHAHA 2HAHAHA 3HAHAHA 4HAHAHA 5AHAHAH 6AHAHAH 7AHAHAH

One more step – teams 2 and 4 R 1R 2R 3R 4 8–15–88–26–8 7–24–61–35–7 6–33–77–44–1 5–42–16–53–2 R 5R 6R 7 8–37–88–4 2–46–13–5 1–55–22–6 7–64–31–7 R 1R 2R 3R 4R 5R 6R 7 1HAHAHA 2HAHAHA 3HAHAHA 4HAHAHA 5AHAHAH 6AHAHAH 7AHAHAH

And so on… R 1R 2R 3R 4 8–15–88–26–8 7–24–61–35–7 6–33–77–44–1 5–42–16–53–2 R 5R 6R 7 8–37–88–4 2–46–13–5 1–55–22–6 7–64–31–7 R 1R 2R 3R 4R 5R 6R 7 1HAHAHA 2HAHAHA 3HAHAHA 4HAHAHA 5AHAHAH 6AHAHAH 7AHAHAH

One more time – teams 1 and 4 R 1R 2R 3R 4 8–15–88–26–8 7–24–61–35–7 6–33–77–44–1 5–42–16–53–2 R 5R 6R 7 8–37–88–4 2–46–13–5 1–55–22–6 7–64–31–7 R 1R 2R 3R 4R 5R 6R 7 1HAHAHA 2HAHAHA 3HAHAHA 4HAHAHA 5AHAHAH 6AHAHAH 7AHAHAH

One more time – teams 2 and 5 R 1R 2R 3R 4 8–15–88–26–8 7–24–61–35–7 6–33–77–44–1 5–42–16–53–2 R 5R 6R 7 8–37–88–4 2–46–13–5 1–55–22–6 7–64–31–7 R 1R 2R 3R 4R 5R 6R 7 1HAHAHA 2HAHAHA 3HAHAHA 4HAHAHA 5AHAHAH 6AHAHAH 7AHAHAH

…and we are done! R 1R 2R 3R 4 8–15–88–26–8 7–24–61–35–7 6–33–77–44–1 5–42–16–53–2 R 5R 6R 7 8–37–88–4 2–46–13–5 1–55–22–6 7–64–31–7 R 1R 2R 3R 4R 5R 6R 7 1HAHAHA 2HAHAHA 3HAHAHA 4HAHAHA 5AHAHAH 6AHAHAH 7AHAHAH

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

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 1R 2R 3R 4R 5R 6R 7 1HHAHAHA 2HAHHAHA 3HAHAHHA 4HAHAHAA 5AAHAHAH 6AHAAHAH 7AHAHAAH

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

Look at the teams starting HOME. 123456… 1HAHAHA… 2HAHAHA… 3HAHAHA…

Question: Can we find an RRT*(n,n) with the perfect HAP? Look at the teams starting HOME. They will never meet, no matter when they play their interdivisional games. 123456… 1HAHAHA… 2HAHAHA… 3HAHAHA…

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

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. Bye teams come in pairs, so suppose there are 4 of them. …AHABHAH… …AHABHAH… …HAHBAHA… …HAHBAHA…

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. Bye teams come in pairs, so suppose there are 4 of them. …HAHBAHA… …HAHBAHA…

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 2. There are at most two teams with a bye in any two consecutive rounds.. …AHABHAH… …HAHBAHA… …HAHABHA… AHAHBAH

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

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 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. BAHA…HAHA BHAH…AHAH AHAH…AHAB HAHA…HAHB

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 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. BAHA…HAHA BHAH…AHAH AHAH…AHAB HAHA…HAHB

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

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

But how?

Let us find a schedule for RRT(2n,2n) But how? The schedule for RRT(2n–1,2n–1) was uniquely determined by its HAP.

Let us find a schedule for RRT(2n,2n) 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 1R 2R 3R 4R 5R 6R 7R 8R 9 R10R11R12 1 HAHAHAHAHAH 2 HAHAHAHAHAH 3 HAHAHAHAHAH 4 HAHAHAHAHAH 5 HAHAHAHAHAH 6 HAHAHAHAHAH 7 AHAHAHAHAHA 8 AHAHAHAHAHA 9 AHAHAHAHAHA 10 AHAHAHAHAHA 11 AHAHAHAHAHA 12 AHAHAHAHAHA

Look at 1 and 2 R 1R 2R 3R 4R 5R 6R 7R 8R 9 R10R11R12 1 HAHAHAHAHAH 2 HAHAHAHAHAH 3 HAHAHAHAHAH 4 HAHAHAHAHAH 5 HAHAHAHAHAH 6 HAHAHAHAHAH 7 AHAHAHAHAHA 8 AHAHAHAHAHA 9 AHAHAHAHAHA 10 AHAHAHAHAHA 11 AHAHAHAHAHA 12 AHAHAHAHAHA

Look at 1 and 2 Both home: cannot play R 1R 2R 3R 4R 5R 6R 7R 8R 9 R10R11R12 1 HAHAHAHAHAH 2 HAHAHAHAHAH 3 HAHAHAHAHAH 4 HAHAHAHAHAH 5 HAHAHAHAHAH 6 HAHAHAHAHAH 7 AHAHAHAHAHA 8 AHAHAHAHAHA 9 AHAHAHAHAHA 10 AHAHAHAHAHA 11 AHAHAHAHAHA 12 AHAHAHAHAHA

Look at 1 and 2 Both home: cannot play Both away: cannot play R 1R 2R 3R 4R 5R 6R 7R 8R 9 R10R11R12 1 HAHAHAHAHAH 2 HAHAHAHAHAH 3 HAHAHAHAHAH 4 HAHAHAHAHAH 5 HAHAHAHAHAH 6 HAHAHAHAHAH 7 AHAHAHAHAHA 8 AHAHAHAHAHA 9 AHAHAHAHAHA 10 AHAHAHAHAHA 11 AHAHAHAHAHA 12 AHAHAHAHAHA

Look at 1 and 2 Both home: cannot play Both away: cannot play So there is just one round left. R 1R 2R 3R 4R 5R 6R 7R 8R 9 R10R11R12 1 HAHAHAHAHAH 2 HAHAHAHAHAH 3 HAHAHAHAHAH 4 HAHAHAHAHAH 5 HAHAHAHAHAH 6 HAHAHAHAHAH 7 AHAHAHAHAHA 8 AHAHAHAHAHA 9 AHAHAHAHAHA 10 AHAHAHAHAHA 11 AHAHAHAHAHA 12 AHAHAHAHAHA

This way it works for all pairs k, k+1 R 1R 2R 3R 4R 5R 6R 7R 8R 9 R10R11R12 1 HAHAHAHAHAH 2 HAHAHAHAHAH 3 HAHAHAHAHAH 4 HAHAHAHAHAH 5 HAHAHAHAHAH 6 HAHAHAHAHAH 7 AHAHAHAHAHA 8 AHAHAHAHAHA 9 AHAHAHAHAHA 10 AHAHAHAHAHA 11 AHAHAHAHAHA 12 AHAHAHAHAHA

Now teams 1 and 3 R 1R 2R 3R 4R 5R 6R 7R 8R 9 R10R11R12 1 HAHAHAHAHAH 2 HAHAHAHAHAH 3 HAHAHAHAHAH 4 HAHAHAHAHAH 5 HAHAHAHAHAH 6 HAHAHAHAHAH 7 AHAHAHAHAHA 8 AHAHAHAHAHA 9 AHAHAHAHAHA 10 AHAHAHAHAHA 11 AHAHAHAHAHA 12 AHAHAHAHAHA

Now teams 1 and 3 Cannot play each other when another game has already been scheduled R 1R 2R 3R 4R 5R 6R 7R 8R 9 R10R11R12 1 HAHAHAHAHAH 2 HAHAHAHAHAH 3 HAHAHAHAHAH 4 HAHAHAHAHAH 5 HAHAHAHAHAH 6 HAHAHAHAHAH 7 AHAHAHAHAHA 8 AHAHAHAHAHA 9 AHAHAHAHAHA 10 AHAHAHAHAHA 11 AHAHAHAHAHA 12 AHAHAHAHAHA

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 1R 2R 3R 4R 5R 6R 7R 8R 9 R10R11R12 1 HAHAHAHAHAH 2 HAHAHAHAHAH 3 HAHAHAHAHAH 4 HAHAHAHAHAH 5 HAHAHAHAHAH 6 HAHAHAHAHAH 7 AHAHAHAHAHA 8 AHAHAHAHAHA 9 AHAHAHAHAHA 10 AHAHAHAHAHA 11 AHAHAHAHAHA 12 AHAHAHAHAHA

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 1R 2R 3R 4R 5R 6R 7R 8R 9 R10R11R12 1 HAHAHAHAHAH 2 HAHAHAHAHAH 3 HAHAHAHAHAH 4 HAHAHAHAHAH 5 HAHAHAHAHAH 6 HAHAHAHAHAH 7 AHAHAHAHAHA 8 AHAHAHAHAHA 9 AHAHAHAHAHA 10 AHAHAHAHAHA 11 AHAHAHAHAHA 12 AHAHAHAHAHA

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 1R 2R 3R 4R 5R 6R 7R 8R 9 R10R11R12 1 HAHAHAHAHAH 2 HAHAHAHAHAH 3 HAHAHAHAHAH 4 HAHAHAHAHAH 5 HAHAHAHAHAH 6 HAHAHAHAHAH 7 AHAHAHAHAHA 8 AHAHAHAHAHA 9 AHAHAHAHAHA 10 AHAHAHAHAHA 11 AHAHAHAHAHA 12 AHAHAHAHAHA

Similarly… R 1R 2R 3R 4R 5R 6R 7R 8R 9 R10R11R12 1 HAHAHAHAHAH 2 HAHAHAHAHAH 3 HAHAHAHAHAH 4 HAHAHAHAHAH 5 HAHAHAHAHAH 6 HAHAHAHAHAH 7 AHAHAHAHAHA 8 AHAHAHAHAHA 9 AHAHAHAHAHA 10 AHAHAHAHAHA 11 AHAHAHAHAHA 12 AHAHAHAHAHA

And again… R 1R 2R 3R 4R 5R 6R 7R 8R 9 R10R11R12 1 HAHAHAHAHAH 2 HAHAHAHAHAH 3 HAHAHAHAHAH 4 HAHAHAHAHAH 5 HAHAHAHAHAH 6 HAHAHAHAHAH 7 AHAHAHAHAHA 8 AHAHAHAHAHA 9 AHAHAHAHAHA 10 AHAHAHAHAHA 11 AHAHAHAHAHA 12 AHAHAHAHAHA

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 1R 2R 3R 4 7–24–61–35–7 6–33–77–44–1 5–42–16–53–2 R 5R 6R 7 2–46–13–5 1–55–22–6 7–64–31–7 a ij = k the game between i and j is played in Round k 1234567 1 2 3 4 5 6 7

Another look at our RRT(7, 7) R 1R 2R 3R 4 7–24–61–35–7 6–33–77–44–1 5–42–16–53–2 R 5R 6R 7 2–46–13–5 1–55–22–6 7–64–31–7 a ij = k the game between i and j is played in Round k 1234567 1234567 2245671 3346712 4456123 5567134 6671235 7712345

Another look at our RRT(7, 7) R 1R 2R 3R 4 7–24–61–35–7 6–33–77–44–1 5–42–16–53–2 R 5R 6R 7 2–46–13–5 1–55–22–6 7–64–31–7 a ii = k team i has a bye in Round k 1234567 11234567 22345671 33456712 44567123 55671234 66712345 77123456

Another look at RRT(8, 8) R 1R 2R 3R 4 8–25–61–36–7 7–34–78–45–8 6–43–87–54–1 2–13–2 R 5R 6R 7R 8 2–47–83–58–1 1–56–12–67–2 8–65–21–76–3 4–35–4 12345678 112345678 223456781 334567812 445678123 556781234 667812345 778123456 881234567

A general RRT(n, n) 123… n–1n–1 n 1123 n–1n–1 n 2234n1 334512 …… n–1n–1n–1n–1n 1… n–4n–4n–3n–3n–2n–2 n n12… n–3n–3n–2n–2n–1n–1

Substitute i  i–1 123… n–1n–1 n 1123 n–1n–1 n 2234n1 334512 …… n–1n–1n–1n–1n 1… n–4n–4n–3n–3n–2n–2 n n12… n–3n–3n–2n–2n–1n–1

A general RRT(n, n) Substitute i  i–1 to get the group Z n !! 012…… n–3n–3n–2n–2n–1n–1 0012 n–3n–3n–2n–2n–1n–1 1123 n–2n–2n–1n–1 0 2234 n–1n–1 01 …… n–2n–2n–2n–2n–1n–1 0…… n–5n–5n–4n–4n–3n–3 n–1n–1n–1n–1 01…… n–4n–4n–3n–3n–2n–2

A general RRT(n, n) Substitute i  i–1 to get the group Z n !! In other words, we get the table of addition mod n 012…… n–3n–3n–2n–2n–1n–1 0012 n–3n–3n–2n–2n–1n–1 1123 n–2n–2n–1n–1 0 2234 n–1n–1 01 …… n–2n–2n–2n–2n–1n–1 0…… n–5n–5n–4n–4n–3n–3 n–1n–1n–1n–1 01…… n–4n–4n–3n–3n–2n–2

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

Non-traditional Round Robin Tournaments Dalibor Froncek University of Minnesota Duluth Mariusz Meszka University of Science and Technology Kraków.

Similar presentations

Presentation on theme: "Non-traditional Round Robin Tournaments Dalibor Froncek University of Minnesota Duluth Mariusz Meszka University of Science and Technology Kraków."— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Non-traditional Round Robin Tournaments Dalibor Froncek University of Minnesota Duluth Mariusz Meszka University of Science and Technology Kraków.

Similar presentations

Presentation on theme: "Non-traditional Round Robin Tournaments Dalibor Froncek University of Minnesota Duluth Mariusz Meszka University of Science and Technology Kraków."— Presentation transcript:

Similar presentations

About project

Feedback