Orthogonal schedules in single round robin tournaments

A measure for the ﬂexibility of a Home-Away Pattern set (HAP-set) is the width. The width of a HAP-set equals the size of the largest set of schedules compatible with the HAP-set, for which no match is scheduled in the same round in any two schedules. We prove lower and upper bounds on the width, and identify HAP-sets with largest possible width when the number of teams is a power of 2. © 2023 The Author(s). Published by Elsevier B


Introduction
A basic and popular format for a competition is the well-known Single Round Robin (SRR) format.Given a set of teams T , with |T | even, an SRR format prescribes that each team plays against each other team once in a set of |T | − 1 rounds, such that each team plays once in each round.Moreover, our setting prescribes that in each match there is one team that plays home, and one team that plays away -sometimes abbreviated with H and A respectively.
When faced with the task of deciding upon the fixtures, i.e., to come up with a schedule that specifies which match is played in which round, and which team plays home and away in each match, various strategies have been described in literature that come up with a schedule; here, we simply refer to the surveys [6] and [1].We also mention [2] as an important source of references.
In this note we focus on a question that is relevant for a set of strategies that are known as first-break-then-schedule.These are strategies that follow a 2-step approach: first decide upon the home/away designation of each team in each round (thereby specifying a home-away pattern (HAP) for each team), then allocate all the matches in a way that is compatible with these HAPs, see [8] and [7] for early references.A key question is to what extent specifying the HAP-set in Step 1 impacts the set of possible schedules in Step 2. Or in other words, what is the diversity of schedules compatible with a given HAP-set?This issue has been investigated in [3] where various measures for the flexibility of a HAP-set are proposed and analyzed.One such measure is called the width.Informally, the width of a HAPset equals the size of the largest set of compatible schedules, such that no match is scheduled in the same round in any two schedules of the set (see Section 2 for precise definitions).Clearly, the larger the width, the more flexible the HAP-set.In [3], it is established that the width of a popular HAP-set known as the canonical HAP-set equals 1. Moreover there exists a match that, in every schedule compatible with the canonical HAP-set, is always scheduled in the same round.
Here we focus on the following questions: do there exist HAPsets that have large width?And how large can the width actually be?
Section 2 gives the preliminaries, and in Section 3 we give upper and lower bounds for the width.In Section 4, we prove that the upper bound on the width from Section 3 can be achieved for a particular HAP-set when the number of teams is a power of 2. Section 5 details a construction that allows one to combine HAPsets, and their corresponding schedules, in a way that preserves the width.We close with describing an extension in Section 6.

Preliminaries and notation
We consider a set of teams T , with 2n := |T |.To avoid trivialities, we assume n ≥ 2. A single round robin format uses 2n − 1 rounds that we denote by the set R. Throughout the paper, we take R = {1, 2, . . ., 2n − 1}.For each team t ∈ T , its Home-Away Pattern (HAP) is given by H(t) = (H r (t)) r∈R , with H r (t) ∈ {0, 1}, where H r (t) = 0 indicates team t playing Home and H r (t) = 1 in-dicates team t playing Away in round r ∈ R. We define a HAP-set H = {(H r (t)) r∈R : t ∈ T } to consist of a HAP for every team t ∈ T .Given two teams t, t , we define to be the number of rounds where t, t differ in Home/Awayallocation.
A partition M of the set of all matches into 2n − 1 rounds is given by M = ∪ r∈R S r where each S r is a matching of the 2n teams, in such a way that for every pair of distinct teams t, t ∈ T , there exists a round r ∈ R such that (t, t ) ∈ S r , ensuring that every team meets every other team once.A schedule S can be generated from such a partition by assigning a Home-team and Away-team in every match in the partition.As notation, we write S(t, t ) = r to indicate that in schedule S, the match between teams t, t is scheduled in round r ∈ R.
Thus, to any schedule S we can associate a corresponding HAPset H(S).Two distinct schedules can have equal HAP-sets.We say a HAP-set H is feasible if there exists a schedule S such that H = H(S).
We define H n to be the set of all feasible HAP-sets on 2n teams.As stated earlier, we define H(S) to be the HAP-set corresponding to schedule S. We also define S(H) = {S : H(S) = H} to be the set of all schedules S that have HAP-set H.We say that schedule S is compatible with HAP-set H if S ∈ S(H).In words, schedules S, S being orthogonal means that no match is scheduled in the same round for S, S .Definition 2.2.Two schedules S, S are rotational orthogonal -S⊥ rot S -if there is a permutation of the rounds σ : R → R out fixed elements, that is, σ is such that there is no round r ∈ R with σ (r) = r, and S r = S σ (r) ∀r ∈ R.
Clearly, any two schedules S, S that are rotational orthogonal are also orthogonal.Using these definitions of orthogonality, we define the following measures of a given HAP-set H.In Definition 2.3, we have defined three measures of a given HAP-set H. Our main goal here is to find HAP-sets with extremal values for these measures.Therefore, for each of these measures, we define o n , w n , x n as follows: Definition 2.4.For each 2n ≥ 4, we define: Simply put, o n , w n , x n equals the value of a HAP-set that scores best on the respective measure for a given n.
With these definitions, we get to the following fundamental question:

Upper and lower bounds for the width
We establish the following lower and upper bounds on the width: Proof.The inequalities x n ≤ w n ≤ o n are immediate, as for each H ∈ H n , we have opp(H) ≥ width(H) ≥ rotw(H).
We now argue that o n ≤ n.Consider any HAP-set H ∈ H n .As H is feasible, it follows that in every round, there are n teams that play at home and n teams that play away, so there are n 2 pairs of teams with a different Home/Away-allocation, leading to a total sum of different Home/Away-allocations equal to As there are 2n 2 = n(2n − 1) pairs of teams, there must be a pair of teams t, t ∈ T with (t, t ) ≤ n.Thus, for any HAP-set H ∈ H n , we have opp(H) ≤ n, which implies o n ≤ n.
We now show that 2 ≤ x n .We prove this inequality by first considering a partition of the set of all matches into 2n − 1 rounds; next, we construct a HAP-set such that there exist two schedules compatible with it, such that all matches in round r in one schedule, are scheduled in round r + 1 in the other schedule, where we work modulo 2n − 1.
There are many ways to partition the set of 2n 2 matches into 2n − 1 rounds S 1 , . . ., S 2n−1 such that each round consists of n matches featuring each team exactly once; let's call such a partition a feasible partition of the matches.One possibility to find a feasible partition is the well-known Circle Method, see e.g.[5] and [9]; it is given by any proper edge coloring of K 2n with 2n − 1 colors.Thus, we can assume we are given a set of rounds S 1 , . . ., S 2n−1 ; notice that the Home/Away assignment for the matches in these rounds has not been specified.We will give a procedure that constructs a HAP-set H in a round-by-round fashion, in such a way that there exist two schedules compatible with H that are rotational orthogonal.
Fix a round r ∈ R, and construct a simple undirected graph G r,r+1 = (V = T , E r,r+1 ), where (t, t ) ∈ E r,r+1 iff match (t, t ) is played in round r or r + 1 (indices are read modulo 2n − 1, thus 2n = 1).Clearly, G r,r+1 is a regular graph of degree 2, where every node (team) is incident to an edge corresponding to its match-up in round r and to an edge corresponding to its match-up in round r + 1, r ∈ R. As G r,r+1 is the edge-union of two perfect matchings, it can be seen as a collection of even cycles.For every cycle, define an orientation (clockwise or counter-clockwise), traverse every cycle, and for every edge (match) that is scheduled in round r + 1, assign the first node (with respect to the orientation) to be the Home playing team and the second node the team that plays Away in round r + 1.
Notice that this Home/Away assignment accommodates the matches in S r+1 as the nodes corresponding to each pair of teams that are matched in S r+1 are connected in G r,r+1 , and the construction ensures that these nodes receive a different Home/Away assignment.It is also true that this Home/Away assignment simultaneously accommodates the matches in S r .Indeed, again the nodes corresponding to each pair of teams that are matched in S r are connected in G r,r+1 , and hence receive a different Home/Away assignment.
When we perform this procedure sequentially for r = 1, 2, . . ., 2n − 1, we thereby specify a HAP-set H.By construction, this HAPset is compatible with the schedule S, and it is also compatible with the schedule S , where S is obtained from S by moving every match one round earlier (modulo 2n − 1).
Concluding, we have constructed a HAP-set H, and we have shown that there two schedules S and S that are compatible with it while S⊥ rot S .This concludes the proof.
Remark 3.1.Notice that, if n is odd, the upper bound from Theorem 3.1 can be improved.Indeed, any feasible HAP-set H on 2n teams has exactly n(2n − 1) Home assignments, which is an odd number when n is odd.When two teams t, t both have an even number of Home's assigned, they have a different Home/Awayallocation in an even number of rounds, so (t, t ) is even.Also, when two teams t, t both have an odd number of Home's assigned, (t, t ) is even.When there are more than 2 teams, there must be a pair of teams t, t that have the same parity number of Home games, so they will have (t, t ) even.As n is presumed odd, and the only way to make a HAP-set with opp(H) = n is if every pair of teams t, t has (t, t ) = n, we see that opp(H) < n.
Notice also that the procedure sketched in the second part of the proof of Theorem 3.1 starts from any feasible partition of the matches into rounds.In case one would start from the partition that is generated by the well-known circle method, we can, using this procedure identify an explicit HAP-set.In fact, we claim that the HAP-set from Table 1 arises, and it follows that this HAPset has rotational width at least 2. One might comment that this HAP-set is unbalanced in the sense that Team 0 only has Away matches, whereas each other team plays only n − 1 away matches.However, as the operation of inverting all Home/Away assignments in a single round does not impact the (rotational) width, one can improve this balance by inverting the Home/Away assignments in n 2 rounds, leading to a HAP-set such that no team plays more than 3 2 n − 1 matches away.

HAP-sets with maximum width
In this section, we identify a family of HAP-sets whose width equals the upper bound established in Section 3 in case the number of teams is a power of 2. Theorem 4.1.When n = 2 ( ∈ N), there is a HAP-set H * ∈ H n with width(H * ) = n.
Proof.We prove this by constructing HAP-set H * and providing n orthogonal schedules that are compatible with H * .
Constructing H * Let P be the set of all 2 +1 subsets of {1, 2, . . ., + 1}, and let P 0 = P \{∅}.The 2n teams are represented by T := {T A : A ∈ P}, and the rounds by R := {R B : B ∈ P 0 }.For each A ∈ P and for each B ∈ P 0 we choose: When H R B (T A ) = 0, this implies a Home match for team T A ∈ T in round R B ∈ R, otherwise when H R B (T A ) = 1 this implies an away match.We claim that the HAP-set H * = {(H R B (T A )) B∈P 0 : A ∈ P} is a HAP-set with (rotational) width n.

Generating the schedules Note that two teams T A 1 and T A
Therefore, the following is true: Next, for each K ∈ P 0 , we define a set of (unordered) pairs of teams, each being a perfect matching on the teams: Crucially, all the matches in M K can be scheduled at round R B if and only if |K ∩ B| is odd by (1) -and there are exactly n rounds R B for which this holds.
To find the n rotational orthogonal schedules, construct the bi- We find the four pairwise orthogonal schedules for the HAP-set in Table 3.
Proof of Theorem 4.1 gives a HAP-set on 2n teams with complete rotational width in the case n = 2 .Next, we will show that this complete rotational width is reserved only for powers of two.

Table 3
Four pairwise orthogonal schedules for 8 teams.
S c h e d u l e R o u n d1 2 3 4 5 6 7 A i be the set of rounds in which team i plays its game at Home.On the other hand, as rot w(H) = n, the set of all pairs of teams can be partitioned into 2n − 1 perfect matchings M 1 , M 2 , . . ., M 2n−1 such that each M i can be played in at least n rounds.Then, by (2), we can conclude that each M i can be played at exactly n rounds.On the other hand, if (x, y), (z, w) ∈ M i for some i, then A x A y and A z A w contain the rounds in which all the matches M i can be played.In other words, we obtain

By
On the other hand, we know (y, z) ∈ M k for some k / ∈ {i, j}, then we obtain S i S j = S k for some k ∈ R. As a result, if x n = n, we can find S 1 , S 2 , . . ., S 2n−1 ⊆ R such that For any i, j ∈ R with i = j, there exists k ∈ R with S i S j = S k . (3) We will show that such sets can be found only if n is a power of two.For each i ∈ R, consider the indicator vector v i of S i that is defined as Let v 0 be the zero vector of dimension 2n − 1, and write C = {v i : 0 ≤ i ≤ 2n − 1}.We can think of C as a subset of the vector space , under usual addition and scalar multiplication.On the other hand, it can be easily seen that ( 3) is equivalent to the following: For any i, j ∈ R, we have This implies that C is closed under addition.Moreover, it is closed under scalar multiplication, as this is trivial for F 2n−1 2 .So we can conclude that C is a subspace of F 2n−1 2 .Therefore, its order 2n should be a power of 2, which implies that n is a power of 2, which completes the proof.

Maximum opposing HAP-sets
We call a HAP-set H on 2n teams a maximum opposing HAP-set if opp(H) = n.In this section we introduce a procedure to create a HAP-set H that is maximum opposing, provided that we have two other HAP-sets, H 1 , H 2 on 2n 1 , 2n 2 teams with n = 2n 1 n 2 and H 1 , H 2 both maximum opposing.
For the matrices A m×n and B p×q , the Kronecker product of A and B, denoted by A ⊗ B, is defined as the pm × qn block matrix: Lemma 5.1.Suppose A m×m and B n×n are orthogonal matrices.Then, A ⊗ B is also orthogonal.
Observe that all the entries of M are ±1 The procedure in the proof of Theorem 5.1 preserves the property of having maximum opposing rounds of two HAP-sets.So far we have only shown for HAP-sets where the number of teams equals a power of two that they can have maximum opposing rounds.And in those cases, we can even explicitly construct the HAP-set, without using smaller sized HAP-sets, as shown in Section 4.
However, the property of having maximum opposing rounds, is not reserved for HAP-sets on 2 teams, as can be seen by the HAP-set H on 12 teams in Table 4.It has opp(H) = 6, which is maximum.It is also the smallest possible number of teams n for which n = 2 and the HAP-set is maximum opposing.
The largest known set of orthogonal schedules compatible with the HAP-set in Table 4, obtained using a mixed integer program, has order 4.

An extension
It is clear that the width is a measure indicating to what extent a particular HAP-set can accommodate distinct schedules where an individual match is played in different rounds in distinct schedules.This is clearly relevant in a first-break-then-schedule approach, see Section 1. From a scheduling point of view, not only the round in which a match is scheduled matters, but also the other matches in that round.It could very well be that there is a preference to have pairs of matches in separate rounds.To capture this idea, we introduce match-pair disjointness.Given a HAP-set, two schedules are match-pair disjoint if each pair of matches that are in the same round in one schedule, are not in the same round in the other schedule.

Definition 2.1. Two
schedules S, S are orthogonal-S⊥S -if for every pair of distinct teams t, t ∈ T , the round S(t, t ) is different from S (t, t ).
• rotw(H): The maximum number of schedules with HAP-set H that are pairwise rotational orthogonal.It is not difficult to see that opp(H) ≥ width(H) ≥ rotw(H) for each feasible H ∈ H. Remark 2.1.For notational purposes, we define the rotational width of a schedule S = (S r ) r compatible with H, denoted width rot (S) instead of width rot (H) as follows.width rot (S) equals the cardinality of the largest set of pairwise orthogonal schedules that can be obtained by permuting the rounds S r , with all schedules compatible to H. Obviously width rot (S) ≤ width rot (H).

Table 1 A
HAP-set with rotational width at least 2.
2 can play each other at round R B if and only if |A 1 ∩ B| and |A 2 ∩ B| have different parities.This is equivalent to the condition that and only if |K ∩ B| is odd.Since each nonempty set has equal number of odd sized and even sized subsets, G is n-regular.Hence, there exists a 1-factorization of G -each 1-factor is a perfect matching giving one of the n rotational orthogonal schedules.
2n} such that H r (t) = M r+1,t holds for all 1 ≤ r ≤ 2n − 1 and 1 ≤ t ≤ 2n.Let H 1 , H 2 be HAP-sets on 2n 1 , 2n 2 teams and opp(H i ) = n i for i = 1, 2. Then, there is a HAP-set on 4n 1 n 2 teams that is maxi-It is important to recall that if H is a HAP-set with opp(H) = n, then (t, t ) = n holds for all teams t = t , which shows M H is orthogonal.Conversely, if M 2n×2n is a matrix with entries {1, −1} such that its rows are orthogonal to each other and all the entries in its first column are the same, then opp(H(M)) = n holds.Suppose HAP-sets H 1 , H 2 are given by: otherwise, for 1 ≤ i, j ≤ 2n.For a given matrix M 2n×2n with entries {1, −1}, define the HAP-set H(M) = (H r (t)) 1≤r≤2n−1 : t ∈ {1, . . ., and any two rows of M are orthogonal to each other.Moreover, each entry in the first column of M is 1.Then, the HAP-set H(M) on 4n 1 n 2 teams satisfies opp(H(M)) = 2n 1 n 2 .When o n = n, o 2n = 2n.Proof.If o n = n, there must exist a HAP-set H on 2n teams, such that opp(H) = n.Notice that the trivial HAP-set on 2n = 2 teams, H * , has opp(H * ) = 1 = n 1 .By Theorem 5.1, the result follows.

Table 5 A
HAP-set that allows two feasible match-pair disjoint schedules.

Table 6
Two match-pair disjoint orthogonal schedules compatible with the HAP-set from Table6.