Transcript Prezentace aplikace PowerPoint

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
1
7
2
8
6
3
4
5
1–factorization of complete graphs
Most familiar
1–factorization of a
complete graph K2n:
Kirkman, 1846
•geometric construction
•labeling construction
1
7
2
8
6
3
4
5
1–factorization of complete graphs
Most familiar
1–factorization of a
complete graph K2n:
Kirkman, 1846
•geometric construction
•labeling construction
1
7
2
8
6
3
4
5
1–factorization of complete graphs
Most familiar
1–factorization of a
complete graph K2n:
Kirkman, 1846
•geometric construction
•labeling construction
1
7
2
8
6
3
4
5
1–factorization of complete graphs
Most familiar
1–factorization of a
complete graph K2n:
Kirkman, 1846
•geometric construction
•labeling construction
1
7
2
8
6
3
4
5
1–factorization of complete graphs
Most familiar
1–factorization of a
complete graph K2n:
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
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
Bipartite fact K8 R-B
Another factorization:
First decompose into two
factors, K4,4 a 2K4.
4
8
3
7
2
6
1
5
Bipartite fact K8 F1
Another factorization:
First decompose into two
factors, K4,4 a 2K4.
Then factorize K4,4
4
8
3
7
2
6
1
5
Bipartite F1 F2
Another factorization:
First decompose into two
factors, K4,4 a 2K4.
Then factorize K4,4
4
8
3
7
2
6
1
5
Bipartite F2 F3
Another factorization:
First decompose into two
factors, K4,4 a 2K4.
Then factorize K4,4
.
4
8
3
7
2
6
1
5
Bipartite F3 F4
Another factorization:
First decompose into two
factors, K4,4 a 2K4.
Then factorize K4,4
4
8
3
7
2
6
1
5
Bipartite 2K4
Another factorization:
First decompose into two
factors, K4,4 a 2K4.
Then factorize K4,4
and finally factorize 2K4.
4
8
4
8
4
8
3
7
3
7
3
7
2
6
2
6
2
6
1
5
1
5
1
5
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)
4
8
3
7
2
6
1
5
“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.
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
1
7
2
8
6
3
4
5
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
7
2
8
6
3
4
5
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
7
2
8
6
3
4
5
Berger tables with HAPs
team
1
2
3
4
5
6
7
8
games
H
H
H
H
A
A
A
A
1
7
2
8
6
3
4
5
Berger tables with HAPs
team
1
2
3
4
5
6
7
8
games
HH
HA
HA
HA
AA
AH
AH
AH
1
7
2
8
6
3
4
5
Berger tables with HAPs
team
1
2
3
4
5
6
7
8
games
HHA
HAH
HAH
HAH
AAH
AHA
AHA
AHA
1
7
2
8
6
3
4
5
Berger tables with HAPs
team
1
2
3
4
5
6
7
8
games
HHAH
HAHH
HAHA
HAHA
AAHA
AHAA
AHAH
AHAH
1
7
2
8
6
3
4
5
Berger tables with HAPs
team
1
2
3
4
5
6
7
8
games
HHAHAHA
HAHHAHA
HAHAHHA
HAHAHAH
AAHAHAH
AHAAHAH
AHAHAAH
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
1
2
3
4
5
6
7
8
R1 R2 R3 R4 R5 R6 R7
H H A H A H A
H A H H A H A
H A H A H H A
H A H A H A H
A A H A H A H
A H A A H A H
A H A H A A H
A H A H A H A
HOME–AWAY PATTERNS
1
2
3
4
5
6
7
8
R1 R2 R3 R4 R5 R6 R7
H H A H A H A
H A H H A H A
H A H A H H A
H A H A H A H
A A H A H A H
A H A A H A H
A H A H A A H
A H A H A H A
HOME–AWAY PATTERNS
1
2
3
4
5
6
7
8
R1 R2 R3 R4 R5 R6 R7
H H A H A H A
H A H H A H A
H A H A H H A
H A H A H A H
A A H A H A H
A H A A H A H
A H A H A A H
A H A H A H A
HOME–AWAY PATTERNS WITH THE SCHEDULE
R1 R2 R3 R4 R5 R6 R7
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
R1
8–1
7–2
6–3
5–4
R2
5–8
4–6
3–7
2–1
R3
8–2
1–3
7–4
6–5
R5
8–3
2–4
1–5
7–6
R6
7–8
6–1
5–2
4–3
R7
8–4
3–5
2–6
1–7
R4
6–8
5–7
4–1
3–2
Problem: How to schedule an RRT(2n–1, 2n–1)?
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
R1 R2 R3 R4 R5 R6 R7
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
R1
8–1
7–2
6–3
5–4
R2
5–8
4–6
3–7
2–1
R3
8–2
1–3
7–4
6–5
R5
8–3
2–4
1–5
7–6
R6
7–8
6–1
5–2
4–3
R7
8–4
3–5
2–6
1–7
R4
6–8
5–7
4–1
3–2
Select the Dummy Team
R1 R2 R3 R4 R5 R6 R7
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
R1
8–1
7–2
6–3
5–4
R2
5–8
4–6
3–7
2–1
R3
8–2
1–3
7–4
6–5
R5
8–3
2–4
1–5
7–6
R6
7–8
6–1
5–2
4–3
R7
8–4
3–5
2–6
1–7
R4
6–8
5–7
4–1
3–2
Dummy Team = 5
R1 R2 R3 R4 R5 R6 R7
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
R1
8–1
7–2
6–3
5–4
R2
5–8
4–6
3–7
2–1
R3
8–2
1–3
7–4
6–5
R5
8–3
2–4
1–5
7–6
R6
7–8
6–1
5–2
4–3
R7
8–4
3–5
2–6
1–7
R4
6–8
5–7
4–1
3–2
Dummy Team = 5
R1 R2 R3 R4 R5 R6 R7
1
H
H
A
H
2
H
A
H
H
A
3
H
A
H
A
H
4
A
H
H
A
A
A
H
7
A
H
A
H
A
H
A
H
A
H
A
A
H
8
A
A
H
A
A
A
H
R2
5–8
4–6
3–7
2–1
R3
8–2
1–3
7–4
6–5
R5
8–3
2–4
1–5
7–6
R6
7–8
6–1
5–2
4–3
R7
8–4
3–5
2–6
1–7
H
5
6
R1
8–1
7–2
6–3
5–4
R4
6–8
5–7
4–1
3–2
Dummy Team = 2
R1 R2 R3 R4 R5 R6 R7
1
H
A
H
A
H
A
H
H
A
2
3
H
A
H
4
H
A
H
A
5
A
A
H
A
H
6
A
H
A
A
H
A
H
A
H
A
A
H
H
A
H
A
7
8
A
H
A
H
H
R1
8–1
7–2
6–3
5–4
R2
5–8
4–6
3–7
2–1
R3
8–2
1–3
7–4
6–5
R5
8–3
2–4
1–5
7–6
R6
7–8
6–1
5–2
4–3
R7
8–4
3–5
2–6
1–7
R4
6–8
5–7
4–1
3–2
Dummy Team = 2
R1 R2 R3 R4 R5 R6 R7
1
H
A
H
A
H
A
H
H
A
2
3
H
A
H
4
H
A
H
A
5
A
A
H
A
H
6
A
H
A
A
H
A
H
A
H
A
A
H
H
A
H
A
7
8
A
H
A
H
H
R1
8–1
7–2
6–3
5–4
R2
5–8
4–6
3–7
2–1
R3
8–2
1–3
7–4
6–5
R5
8–3
2–4
1–5
7–6
R6
7–8
6–1
5–2
4–3
R7
8–4
3–5
2–6
1–7
R4
6–8
5–7
4–1
3–2
Dummy Team = 8
R1 R2 R3 R4 R5 R6 R7
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
R1
8–1
7–2
6–3
5–4
R2
5–8
4–6
3–7
2–1
R3
8–2
1–3
7–4
6–5
R5
8–3
2–4
1–5
7–6
R6
7–8
6–1
5–2
4–3
R7
8–4
3–5
2–6
1–7
R4
6–8
5–7
4–1
3–2
Dummy Team = 8
R1 R2 R3 R4 R5 R6 R7
1
H
2
H
A
3
H
A
A
A
H
4
H
5
A
6
A
H
A
7
A
H
A
H
A
H
A
H
A
H
A
H
A
A
H
A
H
A
H
A
H
A
H
H
A
H
H
A
H
R1
7–2
6–3
5–4
R2
4–6
3–7
2–1
R3
1–3
7–4
6–5
R5
2–4
1–5
7–6
R6
6–1
5–2
4–3
R7
3–5
2–6
1–7
A schedule with
no breaks!
R4
5–7
4–1
3–2
Given the HAP, find a schedule
R1 R2 R3 R4 R5 R6 R7
1
H
2
H
A
3
H
A
A
A
H
4
H
5
A
6
A
H
A
7
A
H
A
H
A
H
A
H
A
H
A
H
A
A
H
A
H
A
H
A
H
A
H
H
A
H
H
A
H
R1
8–1
7–2
6–3
5–4
R2
5–8
4–6
3–7
2–1
R3
8–2
1–3
7–4
6–5
R5
8–3
2–4
1–5
7–6
R6
7–8
6–1
5–2
4–3
R7
8–4
3–5
2–6
1–7
R4
6–8
5–7
4–1
3–2
Look at the game between 1 and 2
R1 R2 R3 R4 R5 R6 R7
1
H
2
H
A
3
H
A
A
A
H
4
H
5
A
6
A
H
A
7
A
H
A
H
A
H
A
H
A
H
A
H
A
A
H
A
H
A
H
A
H
A
H
H
A
H
H
A
H
R1
8–1
7–2
6–3
5–4
R2
5–8
4–6
3–7
2–1
R3
8–2
1–3
7–4
6–5
R5
8–3
2–4
1–5
7–6
R6
7–8
6–1
5–2
4–3
R7
8–4
3–5
2–6
1–7
R4
6–8
5–7
4–1
3–2
Look at the game between 1 and 2
R1 R2 R3 R4 R5 R6 R7
1
H
2
H
A
3
H
A
A
A
H
4
H
5
A
6
A
H
A
7
A
H
A
H
A
H
A
H
A
H
A
H
A
A
H
A
H
A
H
A
H
A
H
H
A
H
H
A
H
R1
8–1
7–2
6–3
5–4
R2
5–8
4–6
3–7
2–1
R3
8–2
1–3
7–4
6–5
R5
8–3
2–4
1–5
7–6
R6
7–8
6–1
5–2
4–3
R7
8–4
3–5
2–6
1–7
R4
6–8
5–7
4–1
3–2
Look at the game between 1 and 2
R1 R2 R3 R4 R5 R6 R7
1
H
2
H
A
3
H
A
A
A
H
4
H
5
A
6
A
H
A
7
A
H
A
H
A
H
A
H
A
H
A
H
A
A
H
A
H
A
H
A
H
A
H
H
A
H
H
A
H
R1
8–1
7–2
6–3
5–4
R2
5–8
4–6
3–7
2–1
R3
8–2
1–3
7–4
6–5
R5
8–3
2–4
1–5
7–6
R6
7–8
6–1
5–2
4–3
R7
8–4
3–5
2–6
1–7
R4
6–8
5–7
4–1
3–2
Look at the game between 1 and 2
R1 R2 R3 R4 R5 R6 R7
1
H
2
H
A
3
H
A
A
A
H
4
H
5
A
6
A
H
A
7
A
H
A
H
A
H
A
H
A
H
A
H
A
A
H
A
H
A
H
A
H
A
H
H
A
H
H
A
H
R1
8–1
7–2
6–3
5–4
R2
5–8
4–6
3–7
2–1
R3
8–2
1–3
7–4
6–5
R5
8–3
2–4
1–5
7–6
R6
7–8
6–1
5–2
4–3
R7
8–4
3–5
2–6
1–7
R4
6–8
5–7
4–1
3–2
Look at the game between 1 and 2
R1 R2 R3 R4 R5 R6 R7
1
H
2
H
A
3
H
A
A
A
H
4
H
5
A
6
A
H
A
7
A
H
A
H
A
H
A
H
A
H
A
H
A
A
H
A
H
A
H
A
H
A
H
H
A
H
H
A
H
R1
8–1
7–2
6–3
5–4
R2
5–8
4–6
3–7
2–1
R3
8–2
1–3
7–4
6–5
R5
8–3
2–4
1–5
7–6
R6
7–8
6–1
5–2
4–3
R7
8–4
3–5
2–6
1–7
R4
6–8
5–7
4–1
3–2
Look at the game between 2 and 3
R1 R2 R3 R4 R5 R6 R7
1
H
2
H
A
3
H
A
A
A
H
4
H
5
A
6
A
H
A
7
A
H
A
H
A
H
A
H
A
H
A
H
A
A
H
A
H
A
H
A
H
A
H
H
A
H
H
A
H
R1
8–1
7–2
6–3
5–4
R2
5–8
4–6
3–7
2–1
R3
8–2
1–3
7–4
6–5
R5
8–3
2–4
1–5
7–6
R6
7–8
6–1
5–2
4–3
R7
8–4
3–5
2–6
1–7
R4
6–8
5–7
4–1
3–2
Look at the game between 2 and 3
R1 R2 R3 R4 R5 R6 R7
1
H
2
H
A
3
H
A
A
A
H
4
H
5
A
6
A
H
A
7
A
H
A
H
A
H
A
H
A
H
A
H
A
A
H
A
H
A
H
A
H
A
H
H
A
H
H
A
H
R1
8–1
7–2
6–3
5–4
R2
5–8
4–6
3–7
2–1
R3
8–2
1–3
7–4
6–5
R5
8–3
2–4
1–5
7–6
R6
7–8
6–1
5–2
4–3
R7
8–4
3–5
2–6
1–7
R4
6–8
5–7
4–1
3–2
Look at the game between 2 and 3
R1 R2 R3 R4 R5 R6 R7
1
H
2
H
A
3
H
A
A
A
H
4
H
5
A
6
A
H
A
7
A
H
A
H
A
H
A
H
A
H
A
H
A
A
H
A
H
A
H
A
H
A
H
H
A
H
H
A
H
R1
8–1
7–2
6–3
5–4
R2
5–8
4–6
3–7
2–1
R3
8–2
1–3
7–4
6–5
R5
8–3
2–4
1–5
7–6
R6
7–8
6–1
5–2
4–3
R7
8–4
3–5
2–6
1–7
R4
6–8
5–7
4–1
3–2
Look at the game between 2 and 3
R1 R2 R3 R4 R5 R6 R7
1
H
2
H
A
3
H
A
A
A
H
4
H
5
A
6
A
H
A
7
A
H
A
H
A
H
A
H
A
H
A
H
A
A
H
A
H
A
H
A
H
A
H
H
A
H
H
A
H
R1
8–1
7–2
6–3
5–4
R2
5–8
4–6
3–7
2–1
R3
8–2
1–3
7–4
6–5
R5
8–3
2–4
1–5
7–6
R6
7–8
6–1
5–2
4–3
R7
8–4
3–5
2–6
1–7
R4
6–8
5–7
4–1
3–2
Look at the game between 2 and 3
R1 R2 R3 R4 R5 R6 R7
1
H
2
H
A
3
H
A
A
A
H
4
H
5
A
6
A
H
A
7
A
H
A
H
A
H
A
H
A
H
A
H
A
A
H
A
H
A
H
A
H
A
H
H
A
H
H
A
H
R1
8–1
7–2
6–3
5–4
R2
5–8
4–6
3–7
2–1
R3
8–2
1–3
7–4
6–5
R5
8–3
2–4
1–5
7–6
R6
7–8
6–1
5–2
4–3
R7
8–4
3–5
2–6
1–7
R4
6–8
5–7
4–1
3–2
Look at the game between 3 and 4
R1 R2 R3 R4 R5 R6 R7
1
H
2
H
A
3
H
A
A
A
H
4
H
5
A
6
A
H
A
7
A
H
A
H
A
H
A
H
A
H
A
H
A
A
H
A
H
A
H
A
H
A
H
H
A
H
H
A
H
R1
8–1
7–2
6–3
5–4
R2
5–8
4–6
3–7
2–1
R3
8–2
1–3
7–4
6–5
R5
8–3
2–4
1–5
7–6
R6
7–8
6–1
5–2
4–3
R7
8–4
3–5
2–6
1–7
R4
6–8
5–7
4–1
3–2
And so on…
R1 R2 R3 R4 R5 R6 R7
1
H
2
H
A
3
H
A
A
A
H
4
H
5
A
6
A
H
A
7
A
H
A
H
A
H
A
H
A
H
A
H
A
A
H
A
H
A
H
A
H
A
H
H
A
H
H
A
H
R1
8–1
7–2
6–3
5–4
R2
5–8
4–6
3–7
2–1
R3
8–2
1–3
7–4
6–5
R5
8–3
2–4
1–5
7–6
R6
7–8
6–1
5–2
4–3
R7
8–4
3–5
2–6
1–7
R4
6–8
5–7
4–1
3–2
One more step – teams 1 and 3
R1 R2 R3 R4 R5 R6 R7
1
H
2
H
A
3
H
A
A
A
H
4
H
5
A
6
A
H
A
7
A
H
A
H
A
H
A
H
A
H
A
H
A
A
H
A
H
A
H
A
H
A
H
H
A
H
H
A
H
R1
8–1
7–2
6–3
5–4
R2
5–8
4–6
3–7
2–1
R3
8–2
1–3
7–4
6–5
R5
8–3
2–4
1–5
7–6
R6
7–8
6–1
5–2
4–3
R7
8–4
3–5
2–6
1–7
R4
6–8
5–7
4–1
3–2
One more step – teams 1 and 3
R1 R2 R3 R4 R5 R6 R7
1
H
2
H
A
3
H
A
A
A
H
4
H
5
A
6
A
H
A
7
A
H
A
H
A
H
A
H
A
H
A
H
A
A
H
A
H
A
H
A
H
A
H
H
A
H
H
A
H
R1
8–1
7–2
6–3
5–4
R2
5–8
4–6
3–7
2–1
R3
8–2
1–3
7–4
6–5
R5
8–3
2–4
1–5
7–6
R6
7–8
6–1
5–2
4–3
R7
8–4
3–5
2–6
1–7
R4
6–8
5–7
4–1
3–2
One more step – teams 1 and 3
R1 R2 R3 R4 R5 R6 R7
1
H
2
H
A
3
H
A
A
A
H
4
H
5
A
6
A
H
A
7
A
H
A
H
A
H
A
H
A
H
A
H
A
A
H
A
H
A
H
A
H
A
H
H
A
H
H
A
H
R1
8–1
7–2
6–3
5–4
R2
5–8
4–6
3–7
2–1
R3
8–2
1–3
7–4
6–5
R5
8–3
2–4
1–5
7–6
R6
7–8
6–1
5–2
4–3
R7
8–4
3–5
2–6
1–7
R4
6–8
5–7
4–1
3–2
One more step – teams 1 and 3
R1 R2 R3 R4 R5 R6 R7
1
H
2
H
A
3
H
A
A
A
H
4
H
5
A
6
A
H
A
7
A
H
A
H
A
H
A
H
A
H
A
H
A
A
H
A
H
A
H
A
H
A
H
H
A
H
H
A
H
R1
8–1
7–2
6–3
5–4
R2
5–8
4–6
3–7
2–1
R3
8–2
1–3
7–4
6–5
R5
8–3
2–4
1–5
7–6
R6
7–8
6–1
5–2
4–3
R7
8–4
3–5
2–6
1–7
R4
6–8
5–7
4–1
3–2
One more step – teams 2 and 4
R1 R2 R3 R4 R5 R6 R7
1
H
2
H
A
3
H
A
A
A
H
4
H
5
A
6
A
H
A
7
A
H
A
H
A
H
A
H
A
H
A
H
A
A
H
A
H
A
H
A
H
A
H
H
A
H
H
A
H
R1
8–1
7–2
6–3
5–4
R2
5–8
4–6
3–7
2–1
R3
8–2
1–3
7–4
6–5
R5
8–3
2–4
1–5
7–6
R6
7–8
6–1
5–2
4–3
R7
8–4
3–5
2–6
1–7
R4
6–8
5–7
4–1
3–2
And so on…
R1 R2 R3 R4 R5 R6 R7
1
H
2
H
A
3
H
A
A
A
H
4
H
5
A
6
A
H
A
7
A
H
A
H
A
H
A
H
A
H
A
H
A
A
H
A
H
A
H
A
H
A
H
H
A
H
H
A
H
R1
8–1
7–2
6–3
5–4
R2
5–8
4–6
3–7
2–1
R3
8–2
1–3
7–4
6–5
R5
8–3
2–4
1–5
7–6
R6
7–8
6–1
5–2
4–3
R7
8–4
3–5
2–6
1–7
R4
6–8
5–7
4–1
3–2
One more time – teams 1 and 4
R1 R2 R3 R4 R5 R6 R7
1
H
2
H
A
3
H
A
A
A
H
4
H
5
A
6
A
H
A
7
A
H
A
H
A
H
A
H
A
H
A
H
A
A
H
A
H
A
H
A
H
A
H
H
A
H
H
A
H
R1
8–1
7–2
6–3
5–4
R2
5–8
4–6
3–7
2–1
R3
8–2
1–3
7–4
6–5
R5
8–3
2–4
1–5
7–6
R6
7–8
6–1
5–2
4–3
R7
8–4
3–5
2–6
1–7
R4
6–8
5–7
4–1
3–2
One more time – teams 2 and 5
R1 R2 R3 R4 R5 R6 R7
1
H
2
H
A
3
H
A
A
A
H
4
H
5
A
6
A
H
A
7
A
H
A
H
A
H
A
H
A
H
A
H
A
A
H
A
H
A
H
A
H
A
H
H
A
H
H
A
H
R1
8–1
7–2
6–3
5–4
R2
5–8
4–6
3–7
2–1
R3
8–2
1–3
7–4
6–5
R5
8–3
2–4
1–5
7–6
R6
7–8
6–1
5–2
4–3
R7
8–4
3–5
2–6
1–7
R4
6–8
5–7
4–1
3–2
…and we are done!
R1 R2 R3 R4 R5 R6 R7
1
H
2
H
A
3
H
A
A
A
H
4
H
5
A
6
A
H
A
7
A
H
A
H
A
H
A
H
A
H
A
H
A
A
H
A
H
A
H
A
H
A
H
H
A
H
H
A
H
R1
8–1
7–2
6–3
5–4
R2
5–8
4–6
3–7
2–1
R3
8–2
1–3
7–4
6–5
R5
8–3
2–4
1–5
7–6
R6
7–8
6–1
5–2
4–3
R7
8–4
3–5
2–6
1–7
R4
6–8
5–7
4–1
3–2
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.
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.
R1 R2 R3 R4 R5 R6 R7
1
H
A
H
A
H
A
H
A
H
A
H
A
2
H
A
3
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.
R1 R2 R3 R4 R5 R6 R7
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?
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.
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.
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)
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.
R1
1
2
3
4
5
6
H
H
H
H
H
7
8
9
10
11
12
A
A
A
A
A
R2
H
A
A
A
A
A
A
H
H
H
H
H
R3
R4
A H
H
H A
H A
H A
H A
H A
A
A H
A H
A H
A H
R5
R6
R7
A H A
A H A
H A
H A
H A H
H A H
H A H
H A H
A H
A H
A H A
A H A
R8
H
H
H
H
A
A
A
A
A
A
H
H
R9
A
A
A
A
H
H
H
H
H
A
R10
H
H
H
H
H
A
A
A
A
A
A
H
R11
A
A
A
A
A
H
H
H
H
H
R12
H
H
H
H
H
H
A
A
A
A
A
A
By our Theorem 3,
a schedule for 12
teams looks like
this.
R1
R2
R3
R4
R5
R6
R7
R8
R9
R10
R11
R12
Look at 1 and 2
1
2
3
4
5
6
H
H
H
H
H
7
8
9
10
11
12
A
A
A
A
A
H
A
A
A
A
A
A
H
H
H
H
H
A H
H
H A
H A
H A
H A
H A
A
A H
A H
A H
A H
A H A
A H A
H A
H A
H A H
H A H
H A H
H A H
A H
A H
A H A
A H A
H
H
H
H
A
A
A
A
A
A
H
H
A
A
A
A
H
H
H
H
H
A
H
H
H
H
H
A
A
A
A
A
A
H
A
A
A
A
A
H
H
H
H
H
H
H
H
H
H
H
A
A
A
A
A
A
R1
R2
R3
R4
R5
R6
R7
R8
R9
R10
R11
R12
Look at 1 and 2
1
2
3
4
5
6
H
H
H
H
H
7
8
9
10
11
12
A
A
A
A
A
H
A
A
A
A
A
A
H
H
H
H
H
A H
H
H A
H A
H A
H A
H A
A
A H
A H
A H
A H
A H A
A H A
H A
H A
H A H
H A H
H A H
H A H
A H
A H
A H A
A H A
H
H
H
H
A
A
A
A
A
A
H
H
A
A
A
A
H
H
H
H
H
A
H
H
H
H
H
A
A
A
A
A
A
H
A
A
A
A
A
H
H
H
H
H
H
H
H
H
H
H
A
A
A
A
A
A
Both home:
cannot play
R1
R2
R3
R4
R5
R6
R7
R8
R9
R10
R11
R12
Look at 1 and 2
1
2
3
4
5
6
H
H
H
H
H
7
8
9
10
11
12
A
A
A
A
A
H
A
A
A
A
A
A
H
H
H
H
H
A H
H
H A
H A
H A
H A
H A
A
A H
A H
A H
A H
A H A
A H A
H A
H A
H A H
H A H
H A H
H A H
A H
A H
A H A
A H A
H
H
H
H
A
A
A
A
A
A
H
H
A
A
A
A
H
H
H
H
H
A
H
H
H
H
H
A
A
A
A
A
A
H
A
A
A
A
A
H
H
H
H
H
H
H
H
H
H
H
A
A
A
A
A
A
Both home:
cannot play
Both away:
cannot play
R1
R2
R3
R4
R5
R6
R7
R8
R9
R10
R11
R12
Look at 1 and 2
1
2
3
4
5
6
H
H
H
H
H
7
8
9
10
11
12
A
A
A
A
A
H
A
A
A
A
A
A
H
H
H
H
H
A H
H
H A
H A
H A
H A
H A
A
A H
A H
A H
A H
A H A
A H A
H A
H A
H A H
H A H
H A H
H A H
A H
A H
A H A
A H A
H
H
H
H
A
A
A
A
A
A
H
H
A
A
A
A
H
H
H
H
H
A
H
H
H
H
H
A
A
A
A
A
A
H
A
A
A
A
A
H
H
H
H
H
H
H
H
H
H
H
A
A
A
A
A
A
Both home:
cannot play
Both away:
cannot play
So there is just one
round left.
R1
1
2
3
4
5
6
H
H
H
H
H
7
8
9
10
11
12
A
A
A
A
A
R2
H
A
A
A
A
A
A
H
H
H
H
H
R3
R4
A H
H
H A
H A
H A
H A
H A
A
A H
A H
A H
A H
R5
R6
R7
A H A
A H A
H A
H A
H A H
H A H
H A H
H A H
A H
A H
A H A
A H A
R8
H
H
H
H
A
A
A
A
A
A
H
H
R9
A
A
A
A
H
H
H
H
H
A
R10
H
H
H
H
H
A
A
A
A
A
A
H
R11
A
A
A
A
A
H
H
H
H
H
R12
H
H
H
H
H
H
A
A
A
A
A
A
This way it works
for all pairs
k, k+1
R1
R2
R3
R4
R5
R6
R7
R8
R9
R10
R11
R12
Now teams 1 and 3
1
2
3
4
5
6
H
H
H
H
H
7
8
9
10
11
12
A
A
A
A
A
H
A
A
A
A
A
A
H
H
H
H
H
A H
H
H A
H A
H A
H A
H A
A
A H
A H
A H
A H
A H A
A H A
H A
H A
H A H
H A H
H A H
H A H
A H
A H
A H A
A H A
H
H
H
H
A
A
A
A
A
A
H
H
A
A
A
A
H
H
H
H
H
A
H
H
H
H
H
A
A
A
A
A
A
H
A
A
A
A
A
H
H
H
H
H
H
H
H
H
H
H
A
A
A
A
A
A
R1
R2
R3
R4
R5
R6
R7
R8
R9
R10
R11
R12
Now teams 1 and 3
1
2
3
4
5
6
H
H
H
H
H
7
8
9
10
11
12
A
A
A
A
A
H
A
A
A
A
A
A
H
H
H
H
H
A H
H
H A
H A
H A
H A
H A
A
A H
A H
A H
A H
A H A
A H A
H A
H A
H A H
H A H
H A H
H A H
A H
A H
A H A
A H A
H
H
H
H
A
A
A
A
A
A
H
H
A
A
A
A
H
H
H
H
H
A
H
H
H
H
H
A
A
A
A
A
A
H
A
A
A
A
A
H
H
H
H
H
H
H
H
H
H
H
A
A
A
A
A
A
Cannot play each
other when another
game has already
been scheduled
R1
R2
R3
R4
R5
R6
R7
R8
R9
R10
R11
R12
Now teams 1 and 3
1
2
3
4
5
6
H
H
H
H
H
7
8
9
10
11
12
A
A
A
A
A
H
A
A
A
A
A
A
H
H
H
H
H
A H
H
H A
H A
H A
H A
H A
A
A H
A H
A H
A H
A H A
A H A
H A
H A
H A H
H A H
H A H
H A H
A H
A H
A H A
A H A
H
H
H
H
A
A
A
A
A
A
H
H
A
A
A
A
H
H
H
H
H
A
H
H
H
H
H
A
A
A
A
A
A
H
A
A
A
A
A
H
H
H
H
H
H
H
H
H
H
H
A
A
A
A
A
A
Cannot play each
other when another
game has already
been scheduled
Cannot play when
both have a home
game
R1
R2
R3
R4
R5
R6
R7
R8
R9
R10
R11
R12
Now teams 1 and 3
1
2
3
4
5
6
H
H
H
H
H
7
8
9
10
11
12
A
A
A
A
A
H
A
A
A
A
A
A
H
H
H
H
H
A H
H
H A
H A
H A
H A
H A
A
A H
A H
A H
A H
A H A
A H A
H A
H A
H A H
H A H
H A H
H A H
A H
A H
A H A
A H A
H
H
H
H
A
A
A
A
A
A
H
H
A
A
A
A
H
H
H
H
H
A
H
H
H
H
H
A
A
A
A
A
A
H
A
A
A
A
A
H
H
H
H
H
H
H
H
H
H
H
A
A
A
A
A
A
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
R1
R2
R3
R4
R5
R6
R7
R8
R9
R10
R11
R12
Now teams 1 and 3
1
2
3
4
5
6
H
H
H
H
H
7
8
9
10
11
12
A
A
A
A
A
H
A
A
A
A
A
A
H
H
H
H
H
A H
H
H A
H A
H A
H A
H A
A
A H
A H
A H
A H
A H A
A H A
H A
H A
H A H
H A H
H A H
H A H
A H
A H
A H A
A H A
H
H
H
H
A
A
A
A
A
A
H
H
A
A
A
A
H
H
H
H
H
A
H
H
H
H
H
A
A
A
A
A
A
H
A
A
A
A
A
H
H
H
H
H
H
H
H
H
H
H
A
A
A
A
A
A
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.
R1
1
2
3
4
5
6
H
H
H
H
H
7
8
9
10
11
12
A
A
A
A
A
R2
R3
R4
H
A
A
A
A
A
A
H
H
H
H
H
A H
H
H A
H A
H A
H A
H A
A
A H
A H
A H
A H
R7
R8
R9
R10
R11
R12
A H A
A H A
H A
H A
H A H
H A H
H A H
H A H
A H
A H
A H A
A H A
H
H
H
H
A
A
A
A
A
A
H
H
A
A
A
A
H
H
H
H
H
A
A
A
A
A
A
H
A
A
A
A
A
H
H
H
H
H
H
A
A
A
A
A
A
R5
R6
H
H
H
H
H
A
H
H
H
H
H
Similarly…
R1
1
2
3
4
5
6
H
H
H
H
H
7
8
9
10
11
12
A
A
A
A
A
R2
R3
R4
H
A
A
A
A
A
A
H
H
H
H
H
A H
H
H A
H A
H A
H A
H A
A
A H
A H
A H
A H
R7
R8
R9
R10
R11
R12
A H A
A H A
H A
H A
H A H
H A H
H A H
H A H
A H
A H
A H A
A H A
H
H
H
H
A
A
A
A
A
A
H
H
A
A
A
A
H
H
H
H
H
A
A
A
A
A
A
H
A
A
A
A
A
H
H
H
H
H
H
A
A
A
A
A
A
R5
R6
H
H
H
H
H
A
H
H
H
H
H
And again…
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)
1
1
2
3
4
5
6
7
R1
7–2
6–3
5–4
R2
4–6
3–7
2–1
R3
1–3
7–4
6–5
R5
2–4
1–5
7–6
R6
6–1
5–2
4–3
R7
3–5
2–6
1–7
R4
5–7
4–1
3–2
2
3
4
5
6
aij = k
7
the game between i and j is
played in Round k
Another look at our RRT(7, 7)
1
1
2
2
3
4
5
6
7
2
3
4
5
6
7
4
5
6
7
1
2
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
7
7
1
2
3
4
5
5
R1
7–2
6–3
5–4
R2
4–6
3–7
2–1
R3
1–3
7–4
6–5
R5
2–4
1–5
7–6
R6
6–1
5–2
4–3
R7
3–5
2–6
1–7
R4
5–7
4–1
3–2
aij = k
the game between i and j is
played in Round k
Another look at our RRT(7, 7)
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
R1
7–2
6–3
5–4
R2
4–6
3–7
2–1
R3
1–3
7–4
6–5
R5
2–4
1–5
7–6
R6
6–1
5–2
4–3
R7
3–5
2–6
1–7
R4
5–7
4–1
3–2
aii = k
team i has a bye in Round k
Another look at RRT(8, 8)
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
R1
8–2
7–3
6–4
R2
5–6
4–7
3–8
2–1
R3
1–3
8–4
7–5
R4
6–7
5–8
4–1
3–2
R5
2–4
1–5
8–6
R6
7–8
6–1
5–2
4–3
R7
3–5
2–6
1–7
R8
8–1
7–2
6–3
5–4
A general RRT(n, n)
…
n–1
n
3
n–1
n
3
4
n
1
4
5
1
2
1
2
3
1
1
2
2
2
3
3
…
…
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)
…
n–1
n
3
n–1
n
3
4
n
1
4
5
1
2
1
2
3
1
1
2
2
2
3
3
…
…
n–1
n–1
n
1
…
n–4
n–3
n–2
n
n
1
2
…
n–3
n–2
n–1
Substitute i  i–1
A general RRT(n, n)
…
…
n–3
n–2
n–1
2
n–3
n–2
n–1
2
3
n–2
n–1
0
3
4
n–1
0
1
0
1
2
0
0
1
1
1
2
2
…
…
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
Substitute i  i–1
to get the group Zn !!
A general RRT(n, n)
…
…
n–3
n–2
n–1
2
n–3
n–2
n–1
2
3
n–2
n–1
0
3
4
n–1
0
1
0
1
2
0
0
1
1
1
2
2
…
…
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
Substitute i  i–1
to get the group Zn !!
In other words, we get
the table of addition mod n
!!!!!!!!!!!!!! THE END !!!!!!!!!!!!