CONSTRUCTING AN ANNUAL ROUND-ROBIN
TOURNAMENT PLAYED ON NEUTRAL GROUNDS
D.F. Robinson
(received 22 February, 1980)
Introduction
Seven golf clubs in North Canterbury, New Zealand, run an annual
round-robin tournament.
turn:
All seven teams meet at each of the courses in
while the home team sees to the hospitality the remaining six
teams play three matches of the tournament.
The assignment of matches
to courses is not difficult and there are many solutions, as discussed
below.
The seven clubs now wish to make sure that each pair of teams meets,
over the years on each of the five neutral grounds.
do this is to discover a five year cycle.
The neatest way to
The quest for and construc­
tion of such a cycle is the main theme of this paper.
The problem may be stated for any number of teams.
It is evident
that the scheme as a whole requires that the number of teams is odd.
There is no contest if there is only one team and there is only one way
of organising matches if there are three teams.
In this paper we shall
concentrate on the seven-team case but will incidentally dispose of the
five-team case.
Tournaments with more than seven teams have not been
investigated individually.
(2n - 1)
With
(2n + 1)
years.
Math. Chronicle 10(1981) 73-82.
73
teams the cycle would take
Single Year
We must begin by considering what happens in a single year.
A
table must be made showing which pairs of teams meet on which course.
Any assignment of all matches to courses so that three matches, no two
involving the same team, meet on the home course of the remaining team,
will be called a draw.
A table may be shown in several ways:
two
methods will be used in this paper, a polygon and a latin square.
A simple way of constructing a single draw is to mark seven points
round a circle representing the teams.
as in Figure 1.
Each pair of points is joined,
The lines fall into sets of three parallel lines.
Each set corresponds to the matches played at the seventh point.
example,
Bfl , CF
and DE
For
the seventh point is A
are parallel;
so these are the three matches played on course A.
t
This solution will
be called the standard draw.
A standard draw can clearly be constructed for any odd number of
t
teams. If the points of the standard draw are relabelled, while retain­
ing the sets of lines, the result will be a draw.
There are
ways of labelling the diagram, but not that many solutions.
labelling the diagram at any point with
tion labelling
7!(= 5040)
If we start
A , go round in either direc­
B , C , etc. on the consecutive points, or on alter­
nate points or on every third point we will obtain a draw which is in
fact the standard draw again.
done.
Hence there are
120
There are
42
ways in which this can be
different solutions which can be obtained
by applying the labels in different orders.
Are these all the draws
possible?
Any draw whatever may be described by a diagram such as Figure 1,
with the points labelled by the teams in a given order.
The lines are
assigned to sets of three in such a way that no point is incident with
two lines in the same set.
A more convenient form for most purposes
however is a latin square.
Table 1 shows the latin square representa­
tion of the standard draw.
74
Figure 1 :
The standard draw .
The letter by each edge indicates the
course on which the match between the
teams given by the end points is played.
75
A
E
B
P
C
G
D
E
B
F
C
G
D
A
B
P
C
G
D
A
E
F
C
0
D
A
E
B
C
G
D
A
E
B
F
G
D
A
E
B
F
C
D
A
E
B
F
C
G
Table 1:
Latin square for the standard draw
A
E
G
P
B
C
D
E
B
P
G
A
D
C
G
P
C
E
D
B
A
F
G
E
D
C
A
B
B
A
D
C
E
G
P
C
D
B
A
G
P
E
D
C
A
B
P
E
G
Table 2:
Latin square for another draw
76
To make the latin square complete we create the fictitious match of
each team against itself, played on its home course.
The latin square
thus has the names of the teams in fixed order down the main diagonal,
and the name of the course on which each genuine match is played in the
corresponding square.
Thus the top right hand corner square in Table 1
contains the name of the course on which
A
plays
G . Such a square
is automatically symmetric.
If in Figure 1 the edges of any two sets are taken, together they
form a single chain.
This is therefore a feature of every solution
formed from the standard draw by relabelling.
two sets, for example
G
However, if the edges in
F , are taken for the draw in Table 2,
and
four of the edges (AC,BC,BD,AD)
form a closed figure.
The feature can
be seen in the latin square representation by the presence of sets of
four entries, forming a rectangle whose opposite corners carry the same
symbol.
Thus Table 2 is not obtainable from Table 1 by relabelling.
In the case when there are five teams it is easy to show that the
six latin squares given in Table 3 are the only draws possible.
These
are all obtainable from the standard draw by relabelling.
Orthogonal draws
We will call two draws orthogonal if they never allot the same
match to the same ground.
The latin squares will thus differ at every
place except on the diagonal.
The search for a five-year cycle in which
every match is played on every neutral ground once is in fact a search
for five draws, each pair of which are orthogonal.
Given any draw it is not difficult to construct a draw orthogonal
to it, but given two draws already orthogonal it is very difficult and
often impossible to construct a draw orthogonal to them both.
The
prospect of success by constructing the draws one at a time is there­
fore small.
77
A
C
5
B
D
A
C
D
E
B
C
B
D
E
A
C
B
E
A
D
E
D
C
A
B
D
E
C
B
A
B
E
A
D
C
E
A
B
D
C
D
A
B
C
E
B
D
A
C
E
2.
1.
A
D
E
C
B
A
D
B
E
C
D
B
A
E
C
D
B
E
C
A
E
A
C
B
D
B
E
C
A
D
C
E
B
D
A
E
C
A
D
B
B
C
D
A
E
C
A
D
B
E
U.
3.
A
s
D
B
C
A
E
B
C
D
E
B
A
C
D
E
B
D
A
C
D
A
C
E
B
B
D
C
E
A
B
C
E
D
A
C
A
E
D
B
C
D
B
A
E
D
C
A
B
E
5.
Table 3:
6.
The six draws for five teams
Examination of the six tables in Table 3 for the five-team problem
shows that any pair of them have exactly one letter appearing in the
same four off-diagonal squares.
Hence no two draws are othogonal.
There is however complete symmetry between the six.
In terms of the
original problem, to make all pairs meet on neutral grounds equally
often, the best solution is to use a six-year cycle, taking all the
draws in any order.
Then in each cycle each pair meets on each course
twice.
If we look into the details, we find that if the draws are taken
in the order
then twice at
AC
12 3 4 5 6
then the match
D , then twice at
AB
is played twice at
C ,
E . On the other hand the match
is never played on the same course two years running.
clearly not be possible by choosing the order of the draws
It will
1
to
6
to eliminate matches being played on the same course in consecutive
years.
Indeed, there are always exactly two matches being played on
the same course as in the previous year.
By reordering the draws we can however distribute the
rences over a cycle more evenly between the
in order
12
3 4 5 6,
10
matches.
12
recur­
For example,
two matches recur three times, one recurs
twice, four recur once and three not at all, while in the order
14 3 5 6,
none recur three times, three recur twice, six recur once,
and one not at all.
It is not possible to find an order for which
every match has either one or two recurrences.
Orthogonal Patterns
Consider the off-diagonal positions which are occupied by a letter
in the latin square form in a particular draw.
With seven teams there
are six occurrences, in a symmetrical pattern.
Such a set of six
squares, one in each row and column aprt from those belonging to the
corresponding letter, will form a pattern for the letter.
In a pair of
orthogonal draws, the patterns for the same letter must occupy diffe­
rent squares.
In a set of five mutually orthogonal solutions, the five
79
patterns for a single letter must exactly cover the non-diagonal posi­
tions.
G , we can list the fifteen possible patterns
Choosing the letter
for
G . These are:
AB AB
AB AC
CD
CE
CF
EF
DF
AC AC
BD
BE
DE EF
DF
BF
DE
AD AD
AD
AE
AE AE
AF AF
AF
BC
BF
BC
BD
BF
BC
BD
BE
CE
DF
CF
CD
DE
CE
CD
EF
BE
CF
We have to select from these five mutually-orthogonal patterns.
We
find that each pattern is orthogonal to eight others, since each pair
appears in three patterns, and two patterns cannot agree in exactly two
pairs.
AB ,
Any set of five orthogonal patterns will contain one beginning
one beginning
AC
and so on.
It is then easy to check the possible
combinations and show that six sets of mutually orthogonal patterns for
G
exist, of which one is:
AB
AC
AD
AE
AF
CD
BE
BF
BD
BC
EF
DF
CE
CF
DE
F.ach pattern occurs in exactly two such sets of five, complete
symmetry being preserved.
The eight patterns orthogonal to a given one
can be divided into two sets of four, each of which forms a set of five
orthogonal patterns with the pattern originally chosen.
On the other
hand some pairs of patterns in different sets of four may be orthogonal.
It proved possible to construct a set of five orthogonal draws
starting in this way.
was selected at random.
First a set of five orthogonal patterns for
The
15
patterns for
80
A
G
were then constructed,
A
G
B
E
D
C
F
A
D
G
B
F
E
C
G
B
A
P
C
E
D
D
B
F
A
G
C
E
B
A
C
G
P
D
E
G
F
C
E
p
A
D
E
P
G
D
A
B
C
B
A
E
D
C
G
F
D
C
P
A
E
G
B
F
0
B
C
E
D
A
C
E
D
B
G
F
A
E
C
A
G
D
P
R
P
D
E
C
B
A
G
C
E
D
F
A
B
G
A
E
F
G
C
D
B
A
F
D
C
G
B
E
E
B
D
C
A
G
F
F
B
E
G
P
D
C
B
G
E
A
D
E
C
F
G
C
B
D
F
A
E
C
G
F
C
A
G
F
E
B
D
G
D
D
G
E
A
B
F
C
B
B
P
A
E
D
C
G
E
,
«
!
i
Table 4:
A
C
A
G
B
D
B
E
A
A
B
E
C
F
A
G
E
C
F
D
C
B
A
F
D
G
A
C
E
F
B
G
D
C
B
G
E
F
D
A
E
G
C
A
D
B
F
F
E
A
D
G
C
B
B
F
D
G
E
A
C
G
D
B
C
A
F
E
D
A
F
B
C
E
G
A cycle of five orthogonal draws for
seven teams.
81
A
and it proved possible to select a set of five patterns for
which
G
were orthogonal and which when assigned to particular patterns for
were consistent with them.
The process was continued for
the number of choices being much reduced.
and
E
B
and
The final patterns for
C ,
D
were inserted into the remaining spaces in the five latin
squares and were found on testing to have the desired properties.
Thi
solution is shown in Table 4.
It seems likely that such an approach should work for any larger
odd number of teams.
With
2n + 1
teams there are
(In - 1) (2n - 3) ... 7.5.3.1
patterns for each letter, so the amount of labour will increase very
rapidly with
n .
University of Canterbury
82