Balanced Equitable Mixed Doubles Round-Robin
Tournaments
David R. Berman
Department of Computer Science
University of North Carolina Wilmington
Wilmington, NC 28403
[email protected]
Douglas D. Smith
Department of Mathematics and Statistics
University of North Carolina Wilmington
Wilmington, NC 28403
[email protected]
Abstract
For mixed doubles tournaments with varying partnerships for ranked men and
women, we consider fairness in two senses: a better player should not have a
more difficult schedule than any lesser player, and no player of any rank should
have a more difficult schedule than the player of the opposite sex who shares
that rank. Tournament schedules with a variety of round-robin characteristics are
considered.
AMS subject classification: 05B30, 94C30
1. Introduction
A mixed doubles tournament is a set of games or matches between two teams
of two players, where each team consists of one male and one female player, as
in mixed doubles tennis. More generally, for disjoint sets A and B, both of size
n, a mixed doubles tournament consists of a collection of games between two
teams such that each team contains exactly one member of each set, as for
example in a pro-am golf tournament. We are concerned here with the situation
in which the teams are not fixed, but vary throughout the tournament, unlike,
say, the usual arrangement in a bridge tournament, where the same two players
form a team in every match they play. Also, we are not concerned with
elimination tournaments in which (usually fixed) teams are removed from
contention after one or sometimes two losses. Rather, we deal with any of the
types of tournaments for mixed doubles that have round-robin properties.
A round-robin tournament (for individual players or for teams considered as a
unit) is a set of games in which each player or team competes exactly once
against every other. For versions of mixed doubles tournaments, we consider
similar conditions such as: every male and female are partners exactly once or
some other specified number of times; every player opposes every other player
of the same sex exactly once or some specified number of times; or every player
opposes every player of the opposite sex exactly once or some other specified
number of times. A variation that has been widely studied, the spouse-avoiding
mixed doubles round-robin tournament for married couples (defined below),
requires the further condition that no husband and wife ever partner or oppose
each other in a match.
To begin, we assume that all the n male players are linearly ranked (seeded)
from 0 to n – 1, so that the highest seeding, 0, is assigned to the best male
player, the lowest seeding, n – 1, goes to the worst male player, and that the
female players are similarly ranked. These seedings may have been derived
either from previous mixed doubles competition or some form of singles
competitions, or by any other means.
In the following, we compute the combined strength of teams to project a win,
a loss, or a tie for each game a team plays. We deem the following to be a
desirable property for a tournament schedule: higher seeded players should not
have more difficult schedules than lower seeded players. This is standard
practice in single elimination tournaments, where the highest seeded team
begins play against the lowest seeded team. This condition, that a tournament
schedule be “fair” or “equitable,” for both men and women players has been
previously studied for spouse-avoiding mixed-doubles [3, 4] and also for whist
tournaments, which are not mixed doubles but do involve constantly shifting
pairings of two-player teams [5].
We introduce here a new desirable condition for a mixed-doubles tournament:
the concept of gender equity. We say that a mixed doubles tournament is
“balanced” when for every rank, neither the male nor the female who share that
ranking has a schedule that is more difficult than the other. That is, their
projected scores are identical.
Our first example in the next section illustrates how the fairness and balance
properties may be violated. This schedule for a spouse-avoiding tournament
with 4 couples fails the gender equality criterion because the male seeded 0 has
an easier schedule (with 3 projected wins) than the woman seeded 0 (who has
only two projected wins and a projected tie), and the schedule is unfair to men
because the male seeded 2 has a projected win and a tie, while the male seeded
higher at 1 has only a projected win.
Given a number n of men and women, our goal is to find a mixed doubles
tournament schedule with the desired round-robin conditions that is both
balanced and equitable. One could consider a full round-robin schedule in which
every male-female pair plays once against every other male-female pair, but this
requires a very large number of games and thus is impractical even for small n.
For example, such a tournament with 8 men and 8 women would require 1,568
games. A spouse-avoiding mixed doubles round-robin tournament for 8 married
couples would require only 28 games, but if we restrict our attention to this type
of tournament we would be forced to identify a “spouse” for each participant,
even when the players are not married couples or there is no readily available
justification for determining which pairs of players should never be contestants
in the same game.
To avoid these deficiencies, we consider alternate forms of mixed doubles
tournaments, each of which has interesting round-robin characteristics. One of
these (see Section 4) allows for each male–female pair to play, not against every
other pair, but in n – 1 games, thus reducing the total number of games from that
of the full round-robin. We give a general construction for balanced equitable
tournaments of this form that, in the case n = 8, yields a total of 224 games.
While this is a significant improvement, a tournament constructed in this way
for 30 men and 30 women would call for an improbable 13,050 games. In
Section 3 we discuss one other form of tournament, in which each male and
female play exactly once as partners and every player opposes every other
player of the same sex at least once. Such a tournament has the imperfection,
from a strictly round-robin point of view, that each player must oppose some
player of the same sex exactly twice. However, we can find balanced equitable
schedules of this type of tournament for all even n, and we need only a relatively
modest number of games (32 when n = 8 and 450 when n = 30) to do so. For
some n, there is a tournament of this form such that every player opposes every
player of the opposite sex once, but balanced equitable schedules are more
difficult to find.
2. Definitions; Spouse-Avoiding Mixed Doubles Tournaments
Assume that n men have been ranked from 0 (the best player) to n – 1 (the
worst), and that n women are similarly ranked.
A mixed doubles tournament for n male and n female players is usually
described as a “schedule” of games between two teams such that each team
consists of one man and one woman. For example
Ab v Cd
Ba v Dc
Ac v Db
Bd v Ca
Ad v Bc
Cb v Da
represents a tournament with 4 men A, B, C, D and 4 women a, b, c, d; in the
first game man A and woman b compete against man C and woman d. In this
paper we think of the array above as specifying the structure of the tournament
but we use the word schedule to refer to the listing of games that result when
ranked players are assigned positions in the structure. For example if the men
ranked 0, 1, 2, 3 are designated as players A, B, C, D respectively and women
ranked 0, 1, 2, 3 are designated as players a, b, d, c, then the schedule consists of
the games below.
01 v 22
10 v 33
03 v 31
12 v 20
02 v 13
21 v 30
Let x, y, z, w be in {0, 1, . . . , n – 1} and x y v z w be a game in a mixed
doubles schedule. We say that the game is a projected win for players x and y
and a loss for z and w when x + y < w + z and a projected tie when x + y = w + z;
otherwise it is a projected loss for x and y and a win for z and w. For example in
the schedule above male 0 has projected wins in all 3 games and male 1 has one
projected win and two projected losses. The projected score si for the player
ranked i is the number of projected wins plus half the number of projected ties
for that player. The score vector for men (women) is the corresponding sequence
s0, s1, … sn-1 of projected scores. The score vectors for the example above are:
3, 1, 1.5, .5 for men and 2.5, 1.5, 1, 1 for women.
A schedule is fair for men (women) if the score vector for men (women) is
decreasing, and we say a schedule is fair or equitable if it is fair for both sexes.
Thus the example above is not fair, because it is not fair for men. We prefer to
find a strictly decreasing sequence whenever possible, but we shall see that this
cannot always be achieved. The results of a fair schedule allow for separate
unbiased comparisons of players with other players of the same gender.
A schedule is balanced when for every rank i, the two players of rank i have
the same score vector. The example above is an extreme case of a schedule that
is not balanced – for every rank, one of the two players has a more difficult
schedule than the other. The results of a balanced schedule allow for an unbiased
comparison of equally ranked players of opposite sex.
A full round-robin schedule, in which every male-female pair plays once
against every other male-female pair, is easily seen to be balanced and equitable.
However, a full round-robin consists of n2(n – 1)2/2 games, and this number
quickly becomes unmanageable. Our goal is to find balanced equitable (BE)
schedules with appealing round-robin characteristics that do not require an
excessive number of games.
The best-known form of a mixed doubles tournament in which partners are
not fixed is the Spouse-Avoiding Mixed Doubles Round-Robin tournament for n
married couples (a SAMDRR(n)), which is a set of matches in which each
player opposes every other player of the same sex exactly once, opposes each
player of the opposite sex except their spouse exactly once, and partners every
player of the opposite sex except their spouse exactly once. A SAMDRR(n)
exists for all n ≠ 2, 3, 6, has n(n – 1)/2 games, and is specified by an idempotent
self-orthogonal Latin square of order n [1]. The structure presented above is a
SAMDRR(4) in which the spouses of A, B, C, and D are, respectively, a, b, c,
and d.
Now suppose that the spouses of the men ranked 0, 1, 2, 3 are ranked,
respectively, 2, 0, 3, and 1. Consider the schedule
00 v 21
12 v 33
03 v 30
11 v 22
01 v 13
20 v 32
Then the score vectors are 2.5, 2, 1, .5 for both males and females so this is a
balanced equitable schedule with the structure of a SAMDRR(4). If we
interchange the men and women in each game (so that game Ab v Cd in the
structure becomes Ba v Dc), we get another BE schedule with the same score
vectors such that the spouses of the men ranked 0, 1, 2, 3 are ranked,
respectively, 1, 3, 0, 2.
Unfortunately, these are the only possible BE schedules based on a
SAMDRR(4) that have strictly decreasing score vectors. There are four other BE
schedules, but for 18 of the 24 possible ways the rankings of spouses might be
paired, there is no BE schedule with the structure of a SAMDRR(4). It is worth
noting that no BE schedule with the structure of a SAMDRR(4) is possible when
spouses are ranked identically.
Given the rankings of n married couples, we can define the permutation ρ on
{0, 1, . . . n – 1} by ρ(i) = the rank of the spouse of the male ranked i. For n = 5,
with ρ = (1243), the following is a BE schedule with a SAMDRR(5) structure
and score vectors 3, 2.5, 2, 1.5, 1.
04 v 13
22 v 40
03 v 21
30 v 44
02 v 34
11 v 20
01 v 42
10 v 33
14 v 41
23 v 32
There is a BE schedule for the permutations (0132), (0214), (0341), and (0423)
and for the inverses of each of these five, all with the same score vector. For the
permutations (0341) and (0143) there is a different BE schedule with
SAMDRR(5) structure and score vectors 3.5, 2.5, 2, 1.5, .5.
There are 26 other permutations that admit BE schedules with score vectors
that are not strictly decreasing. The important observation here is that for n = 5
there are only 36 of the 120 possible pairings of ranked spouses for which there
exists a BE schedule. In particular, there is no BE schedule for the case in which
all spousal pairs share the same ranking.
These results suggest that most schedules for SAMDRR tournaments are
likely to be unfair either for the men or women players or both, or to lack gender
equality. It would be extremely useful, however, to identify balanced equitable
SAMDRR schedules in which all spouses are ranked identically. Although this
is an unlikely situation for randomly selected married couples, such a schedule
could be used for mixed doubles tournaments in general, simply by specifying
that players with the same rank never play in the same game. This condition
may be viewed as somewhat of a mixed blessing: the best male and female
players would never meet as opponents, but on the other hand no other team
would have to face their partnership. The advantage of this arrangement is that it
would satisfy the fairness conditions with fewer games than other schedules we
will consider.
For n = 7, there are 3840 idempotent SOLS of order 7 [7]. Of the 3840
corresponding schedules in which all spouses have the same rank, there are six
that are balanced and equitable, although none have strictly decreasing score
vectors. One of these, with score vector 5, 5, 3, 3, 3, 1, 1, is
05
03
35
12
53
21
42
v
v
v
v
v
v
v
13
46
40
43
61
45
56
06
04
10
16
30
23
41
v
v
v
v
v
v
v
25
51
24
50
52
54
60
01
02
14
15
26
20
34
v
v
v
v
v
v
v
32
64
36
62
31
63
65
Open Questions: Is there a balanced equitable schedule with the structure of a
SAMDRR(n) for every n ≥ 8? If there is a BE schedule, is there one for which
the score vector is strictly decreasing? Most importantly, for which n is there a
BE schedule with the structure of a SAMDRR(n) such that all spouses have
identical ranks?
3. Mitchell Mixed Doubles Round-Robin Tournaments
When Ian Anderson [2] discovered that J.T Mitchell’s 1897 study [9] of a
form of mixed doubles tournament designs made use of examples of what are
now called spouse-avoiding mixed doubles tournaments, he asked about the
existence of other designs of this type. A Mitchell Mixed Doubles Round-Robin
(MMDRR) tournament for n men and n women is a set of matches such that
every man and woman partner exactly once and every pair of players of the
same sex oppose at least once. Berman and Smith [6] constructed examples of
Mitchell designs, showing first that a MMDRR(n) exists for all even n. We now
observe that the schedules constructed there are balanced and equitable.
Note first that since every player appears in n games, every player must
oppose one person of the same sex exactly twice. Also, the number of games in
a MMDRR(n) is n2/2. It follows that this tournament structure can be considered
only when n is even.
Theorem 1. For every even n there exists a balanced and equitable schedule
with a Mitchell mixed doubles round-robin structure.
Proof: Let n = 2k, for k ≥ 1. We designate both male and female players by their
rank i, for i Zn. For each i, the player of the same sex that i will oppose twice
is i + k (mod n).
The first 2k games are given by two cycles, each of length k, as follows:
0, k v k, 0
:
:
k – 1, 2k – 1 v 2k – 1, k – 1
k, k v 0, 0
:
:
2k –1, 2k – 1 v k –1, k – 1
For k > 1, the remaining games are given by k – 1 cycles of length 2k:
0,1 v k – 1,0
:
:
k + 1, k + 2 v 0, k + 1
0,2 v k – 2,0
:
:
…
0, k – 1 v 1, 0
:
:
k + 2, k + 4 v 0, k + 2
:
:
2k – 1, 0 v k – 2, 2k – 1
:
:
2k – 2,0 v k – 4, 2k – 2
2k – 1,1 v k – 3,2k – 1
k + 1, 0 v k + 2, k + 1
:
:
…
2k – 1, k – 2 v 0, 2k – 1
This schedule has the property that if xy opposes zw then yx opposes wz. Thus
the schedule has the strongest possible form of gender equality – not only do the
male and female players with the same rank have the same projected score
vectors, they have identical margins of victory or loss in corresponding games.
To show that the schedule is fair, it suffices to show that the score vector is
decreasing for males. To begin, we note that in the games generated from 0, k v
k, 0 all players have exactly one tie. If k is even, every player has two additional
projected ties in the games generated from 0, k/2 v k/2, 0. In all other games
there is a winning and a losing team. Notice that a lower ranked male cannot
gain an advantage in terms of games won over a higher ranked male among the
games generated by 0, 0 v k, k because all players 0, 1, . . . k – 1 win and all
lower ranked players lose one of these games.
Let 1 ≤ r < k/2, i < j, and suppose male j wins more games than i among those
games generated from 0, r v k – r, 0. In these games male players 0, 1, . . . , k –
2r – 1 win two games and males k – 2r, . . . , k + r – 1 all win one game. Males
2k – 1, 2k – 2, . . . , 2k – r also win once in these games; the only male players
who do not win a game are k + r, k + r + 1, . . . , 2k – r – 1. Thus i is in the set
{k + r, k + r + 1,. . ., 2k – r – 1}, j is in the set {2k – 1, 2k – 2, . . . , 2k – r}, the
relevant games are i, i + r v i – k – r, i and j, j + r – 2k v j – k – r, j, and we
have 2i + r > 2i – k – r and 2j + r – 2k < 2j – k – r. Then among the games
generated by 0, k – r v r, 0 we have the games i, i – k – r v i + r, i and
j, j – k – r v j + r – 2k, j. Since 2i – k – r < 2i + r and 2j – k – r > 2j + r – 2k,
male i wins and j loses in these games. We conclude that a lower ranked male
cannot gain an advantage over a higher ranked male in the games considered
thus far.
Now suppose 1 ≤ r < k/2, i < j, and male j wins more games than i among
those games generated from 0, k – r v r, 0. Since male players 0, 1, …,k + r – 1
all win exactly one of these games and male j can win at most twice, we have j ≥
k + r and j wins two of these games. Specifically, male k + r wins game
k, 2k – r v k + r, k and game k + r, 0 v k + 2r, k + r. However in the games
generated by 0, r v k – r, 0 all male players ranked above k + r win at least one
game and male k + r loses the game k + r, k + 2r v 0, k + r. The same
reasoning applies for all j such that k + r < j < k + 2r: each such male j wins two
games among those generated by 0, k – r v r, 0 but among the games generated
by 0, r v k – r, 0 male j loses twice and every higher ranked male wins at least
once. Thus the lower ranked player never gains an advantage in projected wins
in comparison with a higher ranked player.
▄
For k > 1, the schedules constructed above lack a round-robin characteristic
that may be desired in a tournament: they do not have the property that each
player opposes every player of the opposite sex once.
Theorem 2. In the schedules described above for k > 1, (i) if k is even, then
male players with even rank never oppose females with odd rank and males with
odd rank never oppose females with even rank, and (ii) if k is odd, then male
players with even rank j never oppose females with even rank other than j and
males with odd rank j never oppose females with odd rank other than j.
Proof: First, note that in the games generated from 0, k v k, 0 and from
0, 0 v k, k, the only female opponents of male 0 are 0 and k, respectively.
(i) Let k be even and 1 ≤ r < k. In each of the games 0, r v k – r, 0, the female
opponent of male 0 is 0. In the games generated from these games we find the
games k + r, k + 2r v 0, k + r. As r ranges from 1 to k – 1 the female opponent
(that is, k + 2r mod 2k) of male 0 is always even and takes on values from 0 to
2k – 2 except k. Thus we have male 0 opposing female 0 a total of k + 1 times
and every other female with even rank once.
(ii) Let k be odd and 1 ≤ r < k. Then female 0 is the opponent of male 0 in each
of the games 0, r v k – r, 0, and in the games generated from these games we
have the games k + r, k + 2r v 0, k + r. As r ranges from 1 to k – 1 the female
opponent (that is, k + 2r mod 2k) of male 0 is always odd and takes on values
from 1 to 2k – 1 except k. Thus we have male 0 opposing female 0 a total of k
times and every female with odd rank once.
The cyclic nature of the construction ensures the conclusions of parts (i) and
(ii).
▄
We call a MMDRR tournament structure strict if every player opposes every
player of the opposite sex exactly once. It is shown in [6] that there exists a strict
MMDRR(n) for all n such that (i) n = 2, 4, 6, 10, or 14 (ii) n is a multiple of 16,
and (iii) n ≥ 52 and n ≡ 4 (mod 16). The complete spectrum for strict MMDRRs
is not known.
The following 8 games exhibit a strict MMDRR structure for n = 4 with men
A, B, C, D and women a, b, c, d.
Aa v Bd
Ba v Cb
Bb v Ac
Ca v Dc
Cc v Ad
Ab v Da
Dd v Bc
Db v Cd
There are 4!∙4! ways ranked players could be assigned positions in this structure
to create a schedule, but because (AB)(ab)(CD)(cd), (ADBC)(acbd), and
(ACBD)(1423) are automorphisms of this structure we may assume that player
A is ranked 0. Of the 144 schedules with this structure, eight are balanced and
equitable, but none has strictly decreasing score vectors. For example the
schedule in which A, B, C, D, a, b, c, d have ranks 0, 1, 2, 3, 0, 3, 2, 1
respectively has these games:
00 v 11
10 v 23
13 v 02
20 v 32
22 v 01
03 v 30
31 v 12
33 v 21
All eight BE schedules have 3.5, 2, 2, 0.5 for both score vectors.
An example of a strict MMDRR(6) is given in [6]. Of the 64,800 distinct
schedules based on this structure, there are 120 that are balanced and equitable,
including 12 with strictly decreasing score vectors. There are exactly three
distinct strictly decreasing score vectors: 5.5, 4, 3, 2.5, 2, 1; 5, 4, 3.5, 2.5, 2, 1;
and 5, 4, 3.5, 3, 2, .5. A schedule with score vector 5.5, 4, 3, 2.5, 2, 1 is:
00 v 33 01 v 22 02 v 44 03 v 30 04 v 15 05 v 51
10 v 21 11 v 43 12 v 35 13 v 52 14 v 20 23 v 45
24 v 53 25 v 32 31 v 54 34 v 41 40 v 55 42 v 50
Open Questions: For which n > 6 does there exist a balanced equitable schedule
with the structure of a strict MMDRR(n), and for which of these is there a
schedule with a strictly decreasing score vector?
4. A Construction with Repeated Partnerships and
Oppositions
Finally, we consider a mixed doubles structure that provides a balanced
equitable schedule for n men and n women, for every n > 1. The practical
disadvantage of this format is that it requires more games to be played than the
SAMDRR(n) or MMDRR(n).
Represent each player by their rank in Zn, and for males x, z and females y, w
in Zn, define the games by x y v z w if and only if x + w = y + z (mod n) and
x ≠ z, y ≠ w.
Theorem 3. In the structure defined above, (i) every male-female partnership
occurs in n – 1 games, and (ii) every player opposes every player of the same
sex n times and every player of the opposite sex n – 1 times. Furthermore the
schedule is balanced and equitable with strictly decreasing score vectors.
Proof: Note first that x y v z w is a game if and only if y x v w z is a game, so for
every x in Zn, male x has exactly the same score vector as female x, and in fact
every property of male players holds for females as well. Let i, r be in Zn. Then
male i and female r are partners in games opposing male j and female s for all j
and s in Zn such that j ≠ i and s = j + r – i (mod n). For fixed i and r, male
opponent j takes on n – 1 values, so i and r are partners n – 1 times. For fixed i
and j, j ≠ i, r takes on n values, so players of the same sex are opponents n times.
For fixed i and s, female s is an opponent of i exactly when j + r = i + s for some
j ≠ i. Male j takes on n – 1 values, so players of the opposite sex are opponents
n – 1 times.
Since the score vectors for men and women are identical, it remains to show
that score vectors for men are strictly decreasing. Suppose i, j Zn and i < j. If j
wins in any game j r v k s, then i wins in the game i r v h s, where h = k + i – j
(mod n) and h Zn; if j ties in game j r v k s, then i ties or wins game i r v h s.
Thus the projected score for i is at least as large as the projected score for j.
Finally, we show that if i < j, then male i wins at least one more game than j.
For i ≤ n – 2, let r = n – i – 2 and s Zn be such that s = n – 2i – 3 (mod n).
Consider the games i r v (n – 1) s and (i + 1) r v 0 s. Note that these games exist
because i + s = i + (n – 2i – 3) = (n – i – 2) + (n – 1) = r + (n – 1) (mod n) and
(i + 1) + s = (i + 1) + (n – 2i – 3) = (n – i – 2) + 0 = r + 0 (mod n). Male i wins
the first of these games because i + r = i + (n – i – 2) = n – 2 and (n – 1) + s is at
least as large as n – 1. Male i + 1 loses or ties the second of these games because
(i + 1) + (n – i – 2) = n – 1 and 0 + s ≤ n – 1. Since male i + 1 does not win this
game, we conclude that for every j > i, male j does not win the corresponding
game in which j partners with female r against female s.
▄
Because there are so many repeated partnerships and oppositions, a
tournament with this structure requires n2(n – 1)/2 games to complete.
Even though the projected scores are the same, it might be argued that a
schedule with say, 3 projected wins and 10 projected ties is more (or less)
desirable than a schedule with 8 projected wins and no ties. The structure
presented above makes such debate unnecessary when the number n of men and
women is odd. This observation follows from the fact that the conditions x + w =
y + z (mod n), x ≠ z and y ≠ w, and x + y = z + w for a game to be a tie are
inconsistent for odd n. (For n = 2k there are k2 tie games in the schedule, all of
the form i, k + j v k + i. j where 0 ≤ i, j < k.)
5. Conclusions
The properties of the mixed doubles tournament structures we have considered
are summarized in the following table. The entry “all” indicates that the
structure contains all possible partnerships or oppositions and “one-half”
indicates that each player in the (non-strict) MMDRR(n) described in Section 3
opposes half of all the opposite-sex players. On the existence side of the chart,
the number of games in a schedule is given when a balanced equitable structure
exists, and “X” indicates either that the structure does not exist for the given size
or that there is no BE schedule. The notation “(?)” indicates that, while the
structure exists with the indicated number of games, it is not known whether
there is a BE schedule.
* For spouse-avoiding tournaments, BE schedules exist only for certain
patterns of rankings for spouses. Only in the case n = 7 is there known to be a
schedule based on a SAMDRR(n) for which spouses are all ranked identically.
For a given n, what structure should one use to create a balanced equitable
schedule? For n = 2 or 3, there is only one choice. For n = 4 or 6, the strict
Mitchell design is available, unless a spouse-avoiding tournament is preferred
for 4 couples and the rankings of spouses are fortuitously arranged. The same
issue applies regarding spouse-avoiding tournaments for n = 5 and all n ≥ 8, but
so far as we know (i) with one exception, when n is odd the only other option is
the design with repeated partnerships and oppositions and (ii) when n is even the
best option is the non-strict Mitchell design. The exceptional case is n = 7. If a
spouse-avoiding tournament is not required, one should take advantage of the
existence of schedules based on a SAMDRR(7) for which spouses are all ranked
the same to make a schedule in which players of equal rank never play in the
same game. Such SAMDRR schedules for n ≥ 8 are greatly to be desired.
Finally, we consider resolvability of the mixed doubles tournament schedules
presented here. A mixed doubles tournament for n men and n women, n even, is
resolvable when the schedule can be arranged into rounds so that every player
appears once in every round. For n odd, rounds must be arranged so that in each
round, exactly one male and one female do not appear in a game. Except for the
SAMDRR(7) example which cannot be resolved., all the examples in Section 2
are presented as resolved tournaments. The strict Mitchell designs for n = 4 and
6 in Section 3 are not resolvable. Among the non-strict MMDRR(n) schedules,
the cases n = 4, 8, and 16 are resolvable by splitting each long cycle into two
resolution classes. The case n = 14 can be resolved by using the games:
01 v 60
10 12 v 1 10
9 13 v 12 9
7 8 v 13 7
35 v 83
26 v 52
4 11 v 11 4
as one round. The cases n = 6, 10, and 12 are not resolvable. The examples in
Section 4 are all resolvable. For a given n, we first form rounds for a roundrobin tournament on n individual players [see 8]. Each of these rounds is then
used to form a round in the mixed doubles tournament by changing each game x
v y to the game xx v yy. From an existing round of the mixed doubles
tournament, we get another round by adding 1 (mod n) to the male player in
each partnership of each game in the round, leaving the female players
unchanged. By iterating this step, each initial round generates n rounds of
games. It is routine to check that this process generates the entire tournament,
and that each round includes all players when n is even and all but one male and
female when n is odd.
6. Acknowledgments
We are grateful to the referee for several helpful suggestions.
References
[1] I. Anderson, Combinatorial Designs and Tournaments, Oxford University
Press, Oxford, 1997.
[2] I. Anderson, Early examples of spouse avoidance, Bull. Inst. Combin. Appl.
54 (2008), 47-52.
[3] D.R. Berman, S.C. McLaurin, and D.D. Smith, Fairness in spouse-avoiding
mixed doubles round-robin tournaments, Congr. Numer. 119 (1996) 65-72.
[4] D.R. Berman, S.C. McLaurin, and D.D. Smith, Additional results on
fairness in spouse-avoiding mixed doubles round-robin tournaments for n
couples, Congr. Numer. 123 (1997) 181-191.
[5] D.R. Berman, S.C. McLaurin, and D.D. Smith, Fairness in whist
tournaments, Congr. Numer. 130 (1998) 189-198.
[6] D.R. Berman and D.D. Smith, Mitchell tournaments, Bull. Inst. Combin.
Appl. 65 (2012), 33-42.
[7] A.P. Burger, M.P. Kidd, and J.H. van Vuuren, 2010. Enumeration of
isomorphism classes of self-orthogonal Latin squares, Ars Combinatoria, 97
(2010), 143-152.
[8] J.H. Dinitz, D. Froncek, E.R. Lamken, W.D. Wallis, Scheduling a
tournament, in: C.J. Colbourn and J.H. Dinitz (eds.), Handbook of
Combinatorial Designs, second edition, CRC Press, Boca Raton, FL, 2007,
591-606.
[9] J.T. Mitchell, Duplicate Whist, 2nd edition, Ihling Bros., Kalamazoo, 1897.