The $r$-matching sequencibility of complete graphs

Alspach [ Bull. Inst. Combin. Appl., 52 (2008), pp. 7-20] defined the maximal matching sequencibility of a graph $G$, denoted $ms(G)$, to be the largest integer $s$ for which there is an ordering of the edges of $G$ such that every $s$ consecutive edges form a matching. Alspach also proved that $ms(K_n) = \bigl\lfloor\frac{n-1}{2}\bigr\rfloor$. Brualdi et al. [ Australas. J. Combin., 53 (2012), pp. 245-256] extended the definition to cyclic matching sequencibility of a graph $G$, denoted $cms(G)$, which allows cyclical orderings and proved that $cms(K_n) = \bigl\lfloor\frac{n-2}{2}\bigr\rfloor$. In this paper, we generalise these definitions to require that every $s$ consecutive edges form a subgraph where every vertex has degree at most $r\geq 1$, and we denote the maximum such number for a graph $G$ by $ms_r(G)$ and $cms_r(G)$ for the non-cyclic and cyclic cases, respectively. We conjecture that $ms_r(K_n) = \bigl\lfloor\frac{rn-1}{2}\bigr\rfloor$ and ${\bigl\lfloor\frac{rn-1}{2}\bigr\rfloor-1}~ \leq cms_r(K_n) \leq \bigl\lfloor\frac{rn-1}{2}\bigr\rfloor$ and that both bounds are attained for some $r$ and $n$. We prove these conjectured identities for the majority of cases, by defining and characterising selected decompositions of $K_n$. We also provide bounds on $ms_r(G)$ and $cms_r(G)$ as well as results on hypergraph analogues of $ms_r(G)$ and $cms_r(G)$.


Introduction
The (maximal) matching sequencibility of a simple graph G, denoted ms(G), is the largest integer s for which there exists an ordering of the edges of G so that every s consecutive edges form a matching. Alspach [1] determined ms(K n ), as follows.
Brualdi, Kiernan, Meyer and Schroeder [3] considered the cyclic matching sequencibility cms(G) of a graph G, which is the natural analogue of the matching sequencibility for G when cyclic orders are allowed. They proved the cyclic analogue of Theorem 1, below.
Theorem 2 (Brualdi et al. [3]). For each integer n ≥ 4, The aim of this paper is to extend Theorem 1 and Theorem 2 by generalising the notion of matching sequencibility. In particular, for a graph G, ms r (G) denotes the analogue of ms(G) where consecutive edges form a subgraph whose vertices each has degree at most r. Similarly, cms r (G) is defined analogously to ms r (G) where we allow cyclic orderings of the edges of G.
Conjecture 3. Let n ≥ 3 and 1 ≤ r ≤ n − 2 be integers. Then The main results include the three to follow which verify the conjecture in many cases. In each result we assume n ≥ 3 and 1 ≤ r ≤ n − 2.
Theorem 4. If n or r is even, or n is odd and either r ≥ n−1 2 or gcd(r, n − 1) = 1, then ms r (K n ) = rn − 1 2 .
Theorem 5. If n is even, or n is odd and r = n−1 2 , then cms r (K n ) = rn − 1 2 .
Theorem 6. If n is odd and r is even, then One might ask which of the above bounds holds for which values of r and n. We discuss this question at the end of the paper and prove the following theorem which is the fourth and final of our main results.
The paper is organised as follows. In Section 2 we generalise the methods of [1] and [3], expressed as Propositions 9-11. The propositions allow us to reduce the problem of determining ms r (G) and cms r (G) to ordering subgraphs of G which are partially r-sequenceable for a smaller value of r. However, the parities of n and r play a crucial role in the effectiveness of Propositions 9-11: the case when n is odd is trickier and more so when r is also odd. Section 3 defines the Walecki decomposition [1] and other decompositions of K n . These are central to the proofs of Theorems 4-6; those are presented in Sections 4-7. Section 8 presents the proof of Theorem 7 and, as part of that proof, we consider sequencibility when certain general conditions are placed on consecutive edges of orderings of graphs. Section 9 concludes the paper with a discussion on Conjecture 3 and related open problems, and we provide some recursive bounds on ms r (G) and cms r (G) for general graphs G as well as for the complete k-graph K k n ; see Proposition 33 and Theorem 35, respectively.

Preliminaries
In this paper, graphs will always be simple. A matching of a graph G is a subgraph M in which each vertex has degree 1. A graph G is (≤ r)-regular if each of its vertices has degree at most r. If every vertex has degree equal to r, then G is r-regular. In particular, a matching of a graph is a 1-regular subgraph. We define ms r (G) to be the maximum value of ms r (ℓ) over all orderings ℓ of G. In particular, the special case ms 1 (G), which we also denote as ms(G), is the same number as presented in the Introduction. The edges e 0 , . . . , e s−1 of a graph G = (V, E) are cyclically consecutive in ℓ if the labels of e 0 , . . . , e s−1 are consecutive integers modulo |E|. We define cms r (ℓ) and cms r (G) analogously to ms r (ℓ) and ms r (G), respectively, where we allow cyclically consecutive edges. If G is a (≤ r)-regular graph, then, by definition, cms r (G) = ms r (G) = |E(G)|. So for the remainder of the paper, we only consider the more interesting case in which r is strictly less than the maximum degree of a vertex of G, denoted by ∆(G).
For disjoint graphs G 0 , . . . , G a−1 on the same vertex set V , with labellings ℓ 0 , . . . , ℓ a−1 respectively, let ℓ 0 ∨ · · · ∨ ℓ a−1 denote the ordering ℓ of G = (V, a−1 i=0 E(G i )) defined by ℓ(e ij ) = ℓ j (e ij ) + j−1 l=0 |E(G l )| where e ij ∈ E(G j ) for all i and j. Let s be an integer and G and G ′ be disjoint graphs on the same vertex set V and each having at least s − 1 edges. Also, let G and G ′ have labellings ℓ and ℓ ′ , respectively, and let G s be the subgraph of (V, E(G) ∪ E(G ′ )) that consists of the last s − 1 edges of ℓ and the first s − 1 edges of ℓ ′ . Then we will let ℓ ∨ s ℓ ′ denote the ordering of G s for which the edges of G s appear in the same order as they do in ℓ ∨ ℓ ′ . Now we define ms r (ℓ, ℓ ′ ) to be the largest integer s such that ℓ ∨ s ℓ ′ has r-matching sequencibility s.
An r-regular decomposition of a graph G is a set of edge-disjoint r-regular subgraphs of G that partition the edge set of G. A (≤ r)-regular graph decomposition and a matching decomposition are defined analogously.
The main method used to prove Theorems 4-6 is to decompose K n into regular parts (regular in the sense that every vertex has the same degree), then order the edges in each part, and concatenate the parts to obtain an ordering for K n . The following propositions will facilitate this, under certain conditions. The propositions are given in more generality than we will require them, as they may be useful for other matching sequencibility problems. In each proposition, the subscripts of the orderings ℓ i are taken modulo t: Proposition 9. Let G be a graph that decomposes into matchings M 0 , . . . , M t−1 , each with n edges and orderings ℓ 0 , . . . , ℓ t−1 , respectively. Suppose, for some ǫ ∈ [n] and r < ∆(G), that ms(ℓ i , ℓ i+r ) ≥ n − ǫ for all i ∈ [t − r]. Then ms r (G) ≥ rn − ǫ, and if ms(ℓ i , ℓ i+r ) ≥ n − ǫ for all i ∈ [t], then cms r (G) ≥ rn − ǫ.
Proof of Proposition 11. The cyclic and non-cyclic cases are similar so we only show the cyclic case. Let ℓ be the ordering corresponding to L ℓ 0 (R 0 ) ∨ · · · ∨ L ℓ t−1 (R t−1 ). Consider a set E of ⌊ rn+1 2 ⌋ − ǫ consecutive edges of ℓ. The edges of E, in order, will always be of the form e 1 , . . . , e j , and a. Without loss of generality, we can assume that E∩E(R i ) and E ∩ E(R i+u+2−a ) are non-empty and so 0 < j < an − ⌊ n 2 ⌋ − ǫ. There are 0 < an − ⌊ n 2 ⌋ − ǫ ≤ 2n edges in E ∩ (E(R i ) ∪ E(R i+u+2−a )). Therefore, a = 1 or a = 2.
If a = 1, then the first j edges and last ⌈ n 2 ⌉ − j − ǫ edges of E are the last j edges of ℓ i and the first ⌈ n 2 ⌉ − j − ǫ edges of ℓ i+u+1 , respectively. Thus, the j + ⌈ n and, by assumption, form a matching. The remaining edges of E are from the u = r−1 2 (≤ 2)-regular graphs R i+1 , . . . , R i+u and, hence, the edges of E form a (≤ r)-regular graph.
If a = 2, then the first j edges and last ⌈ 3n 2 ⌉ − j − ǫ edges of E are the last j edges of ℓ i and the first ⌈ n 2 ⌉ − j − ǫ edges of ℓ i+u , respectively. Therefore, the j + ⌈ 3n 2 ⌉−ǫ ℓ i+u and, by assumption, form a (≤ 3)-regular graph. The remaining edges of E are from the u − 1 = r−3 2 (≤ 2)-regular graphs R i+1 , . . . , R i+u−1 , and, thus, the edges of E form a (≤ r)-regular graph.
Remark 12. Proposition 11 holds for r = 1 by replacing the assumption ms 3 Note that Proposition 11 requires two conditions on the orderings ℓ 0 , . . . , ℓ t−1 , namely ms (ℓ i , ℓ i+u+1 ) ≥ ⌈ n 2 ⌉−ǫ and ms 3 (ℓ i , ℓ i+u ) ≥ ⌈ 3n 2 ⌉−ǫ (for the relevant i), whereas Proposition 9 and Proposition 10 require only one condition. This makes Proposition 11 harder to use and largely explains why Theorems 4-6 cover more cases when r and n are not both odd. Also, the requirement of two conditions in Propositions 11 may suggest that the case when r and n are odd is inherently more difficult for K n (or even any graph of odd order).
Here and throughout the paper, we will allow orderings to be defined on sets of integers; that is, a bijection α : A → [|A|] on a set E of integers will also be considered an ordering. However, we will only use such orderings for re-indexing. The following auxiliary lemma guarantees the existence of an ordering of integers with particular useful properties. It will find repeated use in later sections, so it is given here for easy reference.
Lemma 13. Let t and u be integers with t > u and set d := gcd(u, t). Define a i,j := i (mod t d ) + j t d (mod t) for all integers i and j. Then there exists an ordering α of [t] with the property that α(a i+1,j ) = α(a i,j ) + u (mod t) for all i ∈ t d and j ∈ [d]. Proof. We check that the function α : [t] → [t] defined by α(a i,j ) = iu + j (mod t) for i ∈ t d and j ∈ [d] will suffice. First, we will show that α is a bijection. Suppose that iu As d divides t and u, any multiple of u modulo t is also a multiple of d. Thus, j − j ′ is a multiple of d, while 0 ≤ |j − j ′ | ≤ d − 1. This is only possible if j = j ′ and so (i − i ′ )u ≡ 0 (mod t). As 0 ≤ |i ′ − i| ≤ t d − 1 and lcm(t, u) = tu d , we must also have that i = i ′ . Thus, α is injective and so bijective; α is thus an ordering of [t]. For any i ∈ t d and j ∈ [d], α(a i+1,j ) = (i + 1)u + j (mod t) = iu + j + u (mod t) = α(a i,j ) + u (mod t) .
Hence, α has the required properties.
The function α in the proof can be used to show a non-cyclic version of the lemma: Lemma 14. Let t > u. Then there exists an ordering α of [t] with the property that, if α(a) ≤ t − u − 1, then α(a + 1) = α(a) + u.
x 0 1 2 3 4 5 6 7 8 9 α(x) 0 4 8 2 6 1 5 9 3 7 3 Decompositions of the complete graph K n To prove Theorems 4-6, we will require matching decompositions of K n when n is even and 2-regular decompositions of K n when n is odd, so that we can apply the applicable proposition from Section 2. Here we present the required decompositions.

Decompositions of K n for even n
The following sets (with the singleton excluded) are given in [6]. For an integer x and odd integer y, let P x,y = {x + l, x − l} : l ∈ [ y+1 2 ] , where the elements of the members of P x,y are taken modulo y. We also note the following useful fact.
Remark 16. Each family P x,y forms a partition of [y] into pairs and a singleton, and the set {P x,y : x ∈ [y]} partitions the pairs and singletons of [y].
These are indeed matchings, as the edge incident to v ∞ in each M i,j is unique and a vertex v d ; also by the above remark, the choice for such an a 2 and b 2 is unique.
Remark 16, such i and j, and therefore M i,j exist and are uniquely given. Thus, the edge {v Note that this decomposition is the same for different values of c, just indexed differently. Indeed, the bijection τ c,d : is an isomorphism, showing that the decomposition for a particular value of c is isomorphic to the decomposition for c = 2m − 1. Note that the Walecki decomposition (see [1]) decomposes K 2m into Hamiltonian cycles and a complete matching, from which the matching decomposition for c = 2m − 1, given above, can be easily obtained.

Decompositions of K n when n is odd
Let n = 2m + 1. We will present two different decompositions for K 2m+1 . The first is the Walecki decomposition [1] where σ acts on the vertices of V . Alspach [1] proved the following lemma, and we give a similar proof for completeness.   The second decomposition overcomes this problem but does not exist for all n. Recall that n = 2m + 1 and let m be odd.
For an integer x and an odd integer y, let P x,y be as defined in Subsection 3.1.
Example 20. The following graph depicts R 0 , R 1 and R 2 for K 7 .
Clearly R 0 , R 1 and R 2 form a 2-regular decomposition of K 7 . Now we will show that the same is true in general.
Example 22. When c = 5 and d = 3, the matching M 0,0 is labelled as in Figure 2.
Note that the matchings M i,0 are obtained by rotating the above graph but the orderings ℓ i,0 are not. We will use the following notation.
The crucial component of the proof of Theorems 4 and 5 for when n is even is the following lemma which allows us to apply Proposition 9.
. First, suppose that y = 0. If e 1 in M i,j and e 2 in M i+1,j are both incident to ∞, then ℓ i,j (e 1 ) − ℓ i+1,j (e 2 ) = 0 − 0 = 0 < 2. We therefore only need to check the remaining vertices Therefore, x − x ′ is 1 or −1 and in particular less than 2. The case in which v = v i−x,j for some x ∈ [ c+1 2 ] can be treated in a similar fashion and is left to the reader. Now suppose that y = 0. Let v = v i+2x,j+y for some x ∈ [c]. Let e 1 and e 2 be the edges incident to v in M i,j and M i+1,j , respectively. Let v = v i+1+2x ′ ,j+y for some x ′ ∈ [c]. Then i + 2x ≡ i + 1 + 2x ′ (mod c). As gcd(2, c) = 1, this reduces to We will now check that ℓ i,j (e 1 ) − ℓ i+1,j (e 2 ) < 2. The labels of e 1 and e 2 are, respectively, By (1), the difference between these two labels is The right term is c − 1 more than the left term before taking modulo c. So the difference is either 1 or −(c − 1) and in particular less than 2. Hence, ℓ i,j (e 1 ) − ℓ i+1,j (e 2 ) < 2. The case in which v = v i−2x,j−y is similar and therefore omitted. Thus, E forms a matching.
Proof of Theorems 4 and 5 when n is even. Theorem 4 when n is even follows from Theorem 5 when n is even, so we only prove the latter. Let α and a i,j be as defined in Lemma 13 for u = r and t = 2m . If x = α(a i,j ), then, by Lemmas 13 and 23, ms(ℓ x , ℓ x+r ) = ms(ℓ i,j , ℓ i+1,j ) ≥ m − 1 (where x + r in ℓ x+r is taken modulo 2m − 1). Thus, Proposition 9 yields cms r (K 2m ) ≥ rm − 1, using the matchings M ′ 0 , . . . , M ′ 2m−1 ordered by ℓ ′ 0 , . . . , ℓ ′ 2m−1 , respectively. The reverse inequality, cms r (K 2m ) ≤ rm − 1, follows from Lemma 8. This completes the proof.  The first m edges of ℓ i+u form a matching in which i + u + m is the only isolated vertex. Therefore, the vertices W 2 are those incident to edges with labels between m and m + l, excluding i + u + m. In particular, , by re-indexing. Similarly, the vertices in W 1 are those incident to an edge in E(H i ) − E; these edges are labelled between 0 and l by ℓ i . Thus, Finally, we will show that if i ∈ [m − u − 1], then ms(ℓ i , ℓ i+u+1 ) ≥ m. Let ℓ = ℓ i ∨ m ℓ i+u+1 and consider a set E of m consecutive edges in ℓ. Suppose that there are m − l edges in E ∩ E(H i ), and so l edges in E ∩ E(H i+u+1 ) for some non-zero l ∈ [m]. The first m edges of ℓ a form a matching as do the last m edges of ℓ a for a ∈ [m]. Thus, E does not form a matching only if a vertex is incident to an edge in E ∩ E (H i ) and an edge in E ∩ E(H i+u+1 ). Let W 0 be the vertices incident to no edges in E ∩ E (H i ) and W 1 be the vertices incident to an edge in E ∩ E(H i+u+1 ). To prove that E forms a matching, it suffices to show that W 1 ⊆ W 0 .
The last m edges of ℓ i form a matching in which ∞ is the only vertex not incident to an edge. Thus, the members of W 0 are the vertices incident to an edge with a label in between m + 1 and m + l, along with ∞. Hence, [l]}. The members of W 1 are the vertices incident to one of the first l edges of ℓ i+u+1 , so This completes the proof.
Proof of Theorem 4 when n is odd and gcd(r, n − 1) = 1. By Lemma 25 and Proposition 11, using H 0 , . . . , H m−1 with labellings ℓ 0 , . . . , ℓ m−1 , respectively, we have that ms r (K n ) ≥ rn−1 2 . The reverse inequality follows from Lemma 8. 6 Proof of Theorem 6 and the remaining cases of Theorem 4 Let n = 2m + 1, r ∈ [2m] − {0} and let V 2m and H i be as defined in Section 3.2. Also, let ℓ i be the labelling of H i defined as follows: Example 26. When n = 11, the Hamiltonian cycle H 0 is labelled as follows: The other H i 's and their labellings are obtained by rotating the above graph. Therefore, ℓ i is just the labelling that takes alternating edges of the Hamilton cycle H i starting from {∞, i}. Thus, it is clear that ms(ℓ i ) = m for i ∈ [m].

Proof of Theorem 4 for when n is odd and r is even
We will require the following lemma to apply Proposition 10.
Proof. Let s = 2m + 1 − |j − i| and consider a set E of s consecutive edges of ℓ = ℓ i ∨ s ℓ j . As ms(ℓ a ) ≥ m for all a ∈ [m], the last m edges of ℓ i form a matching as do the first m edges of ℓ j . Thus, if both E ∩ E(H i ) and E ∩ E(H j ) contain fewer than m + 1 edges, then E ∩ E(H i ) and E ∩E(H j ) both form matchings. Hence, no vertex would have degree greater than 2 in E. So, consider a set of s consecutive edges E in ℓ where either E ∩ E(H i ) or E ∩ E(H j ) contains at least m + 1 edges. Let a ∈ {i, j} be the integer for which |E ∩ E(H a )| ≥ m + 1. Let W 2 be the vertices of degree 2 in E ∩ E(H a ) and W 1 be the vertices of degree 1 in E ∩ E(H a ′ ), where a ′ ∈ {i, j} and a ′ = a. As ms(ℓ a ′ ) ≥ m, E ∩ E(H a ′ ) forms a matching. Thus, there exists a vertex incident to more than two edges in E only if W 1 ∩ W 2 = ∅. Hence, to show that E forms a (≤ 2)-regular graph, it suffices to prove that W 1 ∩ W 2 = ∅.
If a = i, then E ∩ E(H i ) has s − l edges and E ∩ E(H j ) has l edges for some non-zero l ∈ [s − m]. The last m edges of ℓ i form a matching in which ∞ is the only vertex not incident to an edge. Therefore, the vertices in W 2 are the vertices incident to an edge labelled x by ℓ i such that l + |j − i| ≤ x ≤ m, excluding ∞. In particular, W 2 contains i + m and i + x and i − x for l + |j − i| ≤ x ≤ m − 1. The vertices in W 1 are the vertices incident to any of the first l edges of ℓ j ; hence, W 1 contains ∞, j, and j + x ′ and j − x ′ for x ′ ∈ [l] − {0}. Then W 2 and W 1 can be expressed (modulo 2m) as As i − |j − i| < j + 1 and i + |j − i| > j − 1, W 1 ∩ W 2 = ∅. The case in which a = j is similar and we omit the details. Thus, the s consecutive edges of E form a (≤ 2)-regular graph. We will require the following two lemmas.
Proof. We show that the ordering α u : [t] → [t], defined delow, will suffice: It is easy to check that α u is injective and thus bijective; α u is thus an ordering of [t]. For each Let W a be the vertices incident to two edges in E ∩ E(H a ) for a = i, j. To show that E forms a (≤ 3)-regular graph, it suffices to prove that W i ∩ W j = ∅, as H i and H j are 2-regular. The last m edges of ℓ i form a matching that covers every vertex except ∞. Therefore, W i contains the vertices that are incident to one of the edges with labels between |j − i| + l + ǫ − 1 and m, apart from ∞. Thus, the vertices in W i are i+m and i+x and i−x for |j−i|+l+ǫ−1 ≤ x ≤ m − 1. Similarly, the first m edges of ℓ j form a matching that covers all vertices except j + m. Hence, W j contains the vertices that are incident to one of the edges with labels between m and m + l − 1, except j + m. Therefore, the vertices in W j are ∞ and j + x ′ and j − x ′ + 1 for x ∈ [l] − {0}. So W i ∩ W j is clearly empty when l = 1. In the remaining cases, we can then express W i and W j as follows modulo 2m: We see that W i ∩ W j = ∅. Thus, the s consecutive edges of E form a (≤ 3)-regular graph. Now set ℓ := ℓ i ∨ s ℓ j with s = m + 1 − |j − i| − ǫ where ǫ = 0 if i < j and ǫ = 1 otherwise. We want to show that ms(ℓ) ≥ s. Consider a set of s consecutive edges E of ℓ. Suppose that there are s − l edges in E ∩ E(H i ) and so l edges in E ∩ E(H j ) for some non-zero l ∈ [s]. Let W a be the vertices incident to an edge in E ∩ E(H a ) for a = i, j. As ms(ℓ b ) ≥ m for all b ∈ [m], the last m edges of ℓ i form a matching as do the first m edges of ℓ j . In particular, E ∩ E(H i ) and E ∩ E(H j ) are matchings. Thus, to show that E forms a matching, it suffices to prove that W i ∩ W j = ∅. The vertices in W i and W j are the vertices incident to one of the last s − l edges of ℓ i and one of the first l edges of ℓ j , respectively. Hence, the vertices in W i are i + x and i − x + 1 for l + |j − i| + ǫ ≤ x ≤ m, while the vertices in W j are ∞, j, and j + x ′ and j − x ′ for x ′ ∈ [l] − {0}. We can then express W i and W j as follows modulo 2m: We see that W i ∩ W j = ∅. Thus, the s consecutive edges in E form a matching.

Proof of Theorem 6
To prove Theorem 6, we will need the following ordering of the integers in {l, l+1 . . . , l+t−1}.
Proof of Theorem 5 when r = n−1 2 for even r. Let H i be the Hamiltonian cycle defined in Subsection 3.2. Also, let ℓ i be the ordering of H i defined in Section 6. Let α and a i,j be as defined in Lemma 13 for u = r 2 and t = m. Let H ′ α(a i,j ) := H a i,j and ℓ ′ α(a i,j ) := ℓ a i,j for i ∈ [2] and j ∈ [ r 2 ] = [ m 2 ]. As gcd(r, m) = m 2 , it follows that t u = 2. Thus, |a i+1,j − a i,j | = 1 for all i, j. Lemmas 13 and 27 therefore imply that, for x = α(a i,j ), for all i, j. By applying Proposition 10 to H ′ 0 , . . . , H ′ m−1 ordered by ℓ ′ 0 , . . . , ℓ ′ m−1 , respectively, we see that ms r (K n ) ≥ rn−1 2 . The reverse inequality follows from Lemma 8, completing the proof. Now let r = m be odd and let R i be the 2-regular graph defined in Section 3.2 for i ∈ [m]. Let ℓ i be the ordering of R i defined as follows: Example 30. When n = 7, R 0 is ordered as It is easy to check that ℓ i is indeed a valid ordering of R i and that the first m edges of ℓ i form a matching as do the last m edges of ℓ i .
Proof. First, we will show that ms 3 (ℓ i , ℓ i+1 ) ≥ 3m + 1 for all i. Set ℓ := ℓ i ∨ 3m+1 ℓ i+1 and consider a set of 3m + 1 consecutive edges E of ℓ. A vertex v has degree more than 3 in E only if v has degree two in both E ∩ E(R i ) and E ∩ E(R i+1 ). The first m edges of ℓ j form a matching as do the last m edges of ℓ j for all j. Therefore, if there are m or fewer edges in either E ∩ E(R i ) or E ∩ E(R i+1 ), then E forms a (≤ 3)-regular graph. So, suppose that there are 2m + 1 − l edges in E ∩ E(R i ) and so m + l edges in E ∩ E(R i+1 ) for some non-zero l ∈ [m + 1]. Let W 2 be the vertices of degree 2 in E ∩ E(R i+1 ) and W 1 be the vertices of degree at most 1 in E ∩ E(R i ). To show that E forms a (≤ 3)-regular graph, it suffices to show that W 2 ⊆ W 1 .
As R i is 2-regular and the first m edges of ℓ i form a matching, W 1 contains the vertices incident to one of the first l edges of ℓ i . Thus, The first m edges of ℓ i+1 form a matching which covers every vertex except v i,1 . Therefore, W 2 the vertices incident to an edge with label between m and m + l − 1, excluding v i,1 : . By simplifying and re-indexing, we see that Therefore, E forms a (≤ 3)-regular graph and ms 3 (ℓ i , ℓ i+1 ) ≥ 3m + 1.
Finally, we show that ms(ℓ i , ℓ i−1 ) ≥ m for all i. Set ℓ := ℓ i ∨ m ℓ i−1 and consider a set of m consecutive edges E of ℓ. Suppose that there are m − l edges in E ∩ E(R i ) and thus l edges in E ∩ E(R i−1 ) for some non-zero l ∈ [m]. Let W 1 be the vertices incident to an edge in E ∩ E(R i−1 ) and W 0 be the vertices not incident to any edge in E ∩ E(R i ). The last m edges of ℓ i form a matching as do the first m edges of ℓ i−1 . Thus, a vertex v is incident to 2 or more edges of E only if v ∈ W 1 and v is incident to an edge in E ∩ E(R i ). Therefore, it suffices to show that W 1 ⊆ W 0 .
The last m edges of ℓ i form a matching in which v ∞ is the only isolated vertex. Therefore, W 0 contains the vertex v ∞ along with the vertices incident to an edge with label between m + 1 and m + l; that is, . Now, W 1 contains the vertices that are incident to one of the first l edges of ℓ i−1 . In other words, Hence, E forms a matching and ms(ℓ i , ℓ i−1 ) ≥ m, as required.

General conditions and the proof of Theorem 7
In the process of proving Theorem 7, we develop some notions of sequencibility where an arbitrary condition is placed on the subgraphs formed by consecutive edges. We express such a condition by letting C be an arbitrary family of graphs on a fixed set of vertices V with some fixed vertex labelling. A ordering ℓ of some graph is cyclically (s, C)-sequenceable if all s cyclically consecutive edges in ℓ form a graph in C. A graph G is cyclically (s, C)-sequenceable if there exists a cyclically (s, C)-sequenceable ordering ℓ of G. Note that s is not maximised here: for an arbitrary set of conditions C, maximising s may be trivial or otherwise not of interest. For a graph G = (V, E), let where E(C)∆E(G) is the symmetric difference of E(C) and E(G).
Lemma 32. Let C be a set of conditions on vertex-labelled graphs; let G be a graph, and let s be an integer. Then for an ordering ℓ of G, ℓ is cyclically (s, C)-sequenceable if and only if ℓ is cyclically (|E(G)| − s, C ∁ G )-sequenceable.
Proof. Let k = |E(G)| and let ℓ be an ordering of G which is cyclically (s, C)-sequenceable. Also, let e i be the edge of G labelled i by ℓ. Consider a set of (k − s)-cyclically consecutive edges E in ℓ, namely e j , e j+1 , . . . , e j+k−s−1 for some j ∈ [k], where the subscripts are taken modulo k. The edges of G not in E are e j+k−s , e j+k−s+1 , . . . , e j−1 , and they are in this order in ℓ. By assumption, the s edges of E(G) − E form a graph in C. Thus, the (k − s)-cyclically consecutive edges e j , e j+1 , . . . , e j+k−s−1 must form a member of C ∁ G . Hence, ℓ is cyclically (k − s, C ∁ G )-sequenceable. If ℓ is cyclically (|E(G)| − s, C ∁ G )-sequenceable, then it follows, from the above argument and the identities (C ∁ G ) ∁ G = C and k − (k − s) = s, that ℓ is cyclically (s, C)-sequenceable.
Proof of Theorem 7. Let s a = an−1 2 for each a ∈ [n − 1]. Also, let C r be the set of all vertexlabelled (≤ r)-regular graphs on n vertices. Suppose that cms r (ℓ) = s r and, in particular, suppose that ℓ is cyclically (s r , C r )-sequenceable for some ordering of K n . Then, by Lemma 32, ℓ is n(n−1) 2 − s r , C ∁ Kn r -sequenceable. Set s ′ := n(n−1) 2 − s r = s n−1−r + 1. Then C ∁ Kn r is the family of all vertex-labelled subgraphs of K n whose vertices each have degree at least n−1−r. The minimum number of edges in a member of C ∁ Kn r is s ′ . Also, any member of C ∁ Kn r with s ′ edges must be a graph in which each vertex has degree n − 1 − r except one vertex which has degree n − r.
Consider s ′ +1 cyclically consecutive edges e 0 , . . . , e s ′ in ℓ. Since ℓ is (s ′ , C . Also as s ′ < n(n−1) 2 , e 0 = e s ′ . I claim that this ensures that the edges E ′ := {e 1 , . . . , e s ′ −1 } form a (≤ n − 1 − r)-regular graph. Assume otherwise; then some vertex v is incident to at least n − r of the edges in E ′ . Let v 0 and v 1 be the endpoints of e 0 . The edges of E 0 must form a graph in C C Kn r , i.e., a graph whose vertices each has degree n − 1 − r except one which has degree n − r. As v is incident to at least n − r of the edges in E ′ ⊆ E 0 , v must be incident to exactly n − r of the edges in E 0 . In particular, v must be distinct from v 0 and v 1 . So, v 0 and v 1 are each incident to n − 1 − r of the edges in E 0 . However, this means that v 0 and v 1 are each incident to n − 2 − r of the edges in E ′ . Thus, either v 0 or v 1 is incident to only n − 2 − r of the edges in E 1 , since e s ′ = e 0 . Therefore, the graph formed by the edges of E 1 is not in C ∁ Kn r , a contradiction. Any set E ′ of s ′ − 1 = s n−1−r cyclically consecutive edges in ℓ is a consecutive subsequence of some s ′ + 1 cyclically consecutive edges in ℓ of the form e 0 ∨L ℓ (E ′ ) ∨e s ′ . Thus, by the above argument, every vertex must have degree at most n − 1 − r in E ′ . Hence, cms(ℓ) ≥ s n−1−r and, therefore, cms n−1−r (K n ) ≥ s n−1−r . By Lemma 8, cms n−1−r (K n ) = s n−1−r . The reverse direction, namely that cms n−1−r (K n ) = s n−1−r implies cms r (K n ) = s r , follows by applying the above argument with r replaced by n − 1 − r.
Note that a similar result to Lemma 32 could be can proved for non-cyclic sequences. However, the notion of sequencibility would have to be generalised to allow partially cyclical sequences. A result analogous to Theorem 7 follows from a similar proof, but there is not an equivalence between ms r (K n ) = rn−1 2 and ms n−1−r (K n ) = (n−1−r)n−1 2 for odd r and n.

Concluding remarks
When r is even and n is odd, we expect that cms r (K n ) = rn−1 2 . Theorem 5 confirms this for even r = n−1 2 . By computer search, we were able to find the following two orderings for K 7 that show that cms 2 (K 7 ) = 6 = 2n−1 2 and cms 4 (K 7 ) = 13 = 4n−1 2 , respectively. The orderings are represented by the sequence of edges that has corresponding ordering value sequence 0, . . . , 20. We also found an ordering for K 9 , showing that cms 2 (K 9 ) = 8 = 2n−1 2 , as given below. When r and n are both odd, Theorem 5 implies that cms r (K n ) = rn−1 2 for r = n−1 2 . If there are other cases for which cms r (K n ) = rn−1 2 with r and n odd, then, by Theorem 7, the condition for which cms(K n ) = rn−1 2 holds must be invariant under replacing r with n − 1 − r.
Note that this proposition and Theorems 1 and 2 together imply that ms r (K n ) ≥ r n−1 2 and cms r (K n ) ≥ r n−3 2 , respectively, when n is odd. Proof. Let s = cms r 1 (G) and let ℓ be a labelling of G for which cms r 1 (ℓ) = cms r 1 (G). Any set of r 2 s cyclically consecutive edges E of ℓ are just r 2 sets of s cyclically consecutive edges of ℓ and in each set, every vertex has degree at most r 1 . Thus, every vertex has degree at most r 1 r 2 in E and, hence, cms r 1 r 2 (G) ≥ cms r 1 r 2 (ℓ) ≥ r 2 s. The non-cyclic case is similar.
A hypergraph is a pair (V, E) where V is a set and E is a family of subsets of V . A k-graph is a hypergraph (V, E) for which each member e ∈ E has cardinality |e| = k. For instance, each graph is a 2-graph. The notion of matching sequencibility naturally extends to hypergraphs, as do Proposition 33 and the propositions of Section 2, using analogous proofs. For example the natural hypergraph analogue of Proposition 9 is as follows.
constructed, that the first l edges of ℓ i+1,j are e 0 , . . . , e l−1 , and that any b consecutive edges of L ℓ i,j (M i,j ), e 0 , . . . , e l−1 form a matching. Let e ′ 0 , . . . , e ′ b−2 be the last b − 1 edges of ℓ i,j . If l ≤ b − 2, then there are at least n k − (b − l − 1)k − l > 0 edges in M i+1,j − {e 0 , . . . , e l−1 } which do not share a common vertex with any of the edges e ′ l , . . . , e ′ b−2 . Thus, we can choose an edge e l in M i+1,j − {e 0 , . . . , e l−1 } such that e ′ l , . . . , e ′ b−2 , e 0 , . . . , e l−1 , e l forms a matching and let ℓ i+1,j (e l ) = l. If l > b − 2, then for an arbitrary edge e l in M i+,j − {e 0 , . . . , e l−1 }, let ℓ i+1,j (e l ) = l. Now let e 0 , . . . , e b−2 be the last b − 1 edges of ℓ 0,j . We are free to permute the ordering ℓ 0,j and maintain the identity ms(ℓ 0,j , ℓ 1,j ) ≥ b so long as the labels of e 0 , . . . , e b−2 remain unchanged. Therefore, we can apply a similar argument to the above to permute ℓ 0,j in such a way that ensures that ms(ℓ c−1,j , ℓ 0,j ) ≥ b, Kühn and Osthus [5] offer an alternate decomposition of K k n than those given by Baranyai's Theorem, into Berge cycles which broadly generalise cycles in graphs. Their decomposition would however not be likely to be useful for proving a matching sequencibility result.