Non-traditional Round Robin Tournaments Dalibor Froncek University of Minnesota Duluth Mariusz...
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Transcript of 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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4 5
6
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8
1–factorization of complete graphs
Most familiar
1–factorization of a complete graph K2n:
Kirkman, 1846
•geometric construction•labeling construction
1
2
3
4 5
6
7
8
1–factorization of complete graphs
Most familiar
1–factorization of a complete graph K2n:
Kirkman, 1846
•geometric construction•labeling construction
1
2
3
4 5
6
7
8
1–factorization of complete graphs
Most familiar
1–factorization of a complete graph K2n:
Kirkman, 1846
•geometric construction•labeling construction
1
2
3
4 5
6
7
8
1–factorization of complete graphs
Most familiar
1–factorization of a complete graph K2n:
Kirkman, 1846
•geometric construction•labeling construction
1
2
3
4 5
6
7
8
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
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
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
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
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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Bipartite fact K8 R-B
Another factorization:
First decompose into two factors, K4,4 a 2K4.
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Bipartite fact K8 F1
Another factorization:
First decompose into two factors, K4,4 a 2K4.
Then factorize K4,4
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Bipartite F1 F2
Another factorization:
First decompose into two factors, K4,4 a 2K4.
Then factorize K4,4
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4
5
6
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Bipartite F2 F3
Another factorization:
First decompose into two factors, K4,4 a 2K4.
Then factorize K4,4
.1
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3
4
5
6
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Bipartite F3 F4
Another factorization:
First decompose into two factors, K4,4 a 2K4.
Then factorize K4,4
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Bipartite 2K4
Another factorization:
First decompose into two factors, K4,4 a 2K4.
Then factorize K4,4
and finally factorize 2K4.1
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3
4
5
6
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8
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8
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5
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8
Bipartite fact K8 R-B
Another factorization:
First decompose into two factors, K4,4 a 2K4.
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
“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
“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
“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
“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
“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
“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...
Lemma: An RRT(2n, 2n–1) has at least 2n–2 breaks.
Proof:
Pigeonhole principle
•HAHAHAHA...•AHAHAHAH...•AHAHAHAH...
Lemma: An RRT(2n, 2n–1) has at least 2n–2 breaks.
Proof:
Pigeonhole principle
•HAHAHAHA...•AHAHAHAH...•AHAHAHAH...
Lemma: An RRT(2n, 2n–1) has at least 2n–2 breaks.
Proof:
Pigeonhole principle
•HAHAHAHA...•AHAHAHAH...•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 K8 – Berger tables
• Round 1 – factor F1
• Round 2 – factor F5
• Round 3 – factor F2
• Round 4 – factor F6
• Round 5 – factor F3
• Round 6 – factor F7
• Round 7 – factor F4
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4 5
6
7
8
Kirkman factorization of K8 – Berger tables
• Round 1 – factor F1
• Round 2 – factor F5
• Round 3 – factor F2
• Round 4 – factor F6
• Round 5 – factor F3
• Round 6 – factor F7
• Round 7 – factor F4
1
2
3
4 5
6
7
8
Kirkman factorization of K8 – Berger tables
• Round 1 – factor F1
• Round 2 – factor F5
• Round 3 – factor F2
• Round 4 – factor F6
• Round 5 – factor F3
• Round 6 – factor F7
• Round 7 – factor F4
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Berger tables with HAPs
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
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
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?
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 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
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 = 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
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
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
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
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
Look at the game between 2 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
Look at the game between 2 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
Look at the game between 2 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
Look at the game between 2 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
Look at the game between 2 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
Look at the game between 3 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 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 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 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 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*
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.
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
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 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 …
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.
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 …
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.
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 …
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.
… A H A B H A H …
… A H A B H A H …
… H A H B A H A …
… H A H B A H A …
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.
… H A H B A H A …
… H A H B A H A …
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.
… H A H B A H A …
… H A H B A H A …
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.
… H A H B A H A …
… H A H B A H A …
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..
… A H A B H A H …
… H A H B A H A …
… H A H A B H A …
A H A H B A H
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..
… A H A B H A H …
… H A H B A H A …
… H A H A B H A …
A H A H B A H
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..
… A H A B H A H …
… H A H B A H A …
… H A H A B H A …
A H A H B A H
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.
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.
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.
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
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.
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
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.
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
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?• 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 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
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 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
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 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
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 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
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
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
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
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
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
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
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
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
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
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
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
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
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
Similarly…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
And again…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
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
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
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
Another look at our RRT(7, 7)
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