Balanced Equitable Mixed Doubles Round-Robin Tournaments David R. Berman Department of Computer Science University of North Carolina Wilmington Wilmington, NC 28403 bermand@uncw.edu Douglas D. Smith Department of Mathematics and Statistics University of North Carolina Wilmington Wilmington, NC 28403 smithd@uncw.edu 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 s i 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 s 0, s 1, s n-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 n 2 (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 v 13 06 v 25 01 v 32 03 v 46 04 v 51 02 v 64 35 v 40 10 v 24 14 v 36 12 v 43 16 v 50 15 v 62 53 v 61 30 v 52 26 v 31 21 v 45 23 v 54 20 v 63 42 v 56 41 v 60 34 v 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 n 2 /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 Z n. 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, k v 0, 0 : : : : k 1, 2k 1 v 2k 1, k 1 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 0,2 v k 2,0 0, k 1 v 1, 0 : : : : : : k + 1, k + 2 v 0, k + 1 k + 2, k + 4 v 0, k + 2 k + 1, 0 v k + 2, k + 1 : : : : : : 2k 2,0 v k 4, 2k 2 2k 1, 0 v k 2, 2k 1 2k 1,1 v k 3,2k 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 Z n, and for males x, z and females y, w in Z n, 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 Z n, 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 Z n. Then male i and female r are partners in games opposing male j and female s for all j and s in Z n 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 Z n 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 Z n ; 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 Z n 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 n 2 (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 k 2 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.