A note on the cyclic matching sequencibility of graphs

In this note we present answers to the open problems posed by Brualdi, Kiernan, Meyer and Schroeder in [Cyclic matching sequencibility of graphs, Australas. J. Combin. 53 (2012), 245–256]. 1 Discussion and response Let G ⊆ Kn be a graph of order n with m edges. The matching number of G is the maximum number of edges in a matching. The matching number of a linear ordering e1, e2, . . . , em of the edges of G is the largest number d such that every d consecutive edges in the ordering form a d-matching of G. The matching sequencibility of G, denoted ms(G), is the maximum matching number of a linear ordering of the edges of G. The cyclic matching sequencibility of G, denoted cms(G), is the largest integer d such that there exists a cyclic ordering of the edges so that every d consecutive edges in the ordering form a matching of G. In [1] Brualdi, Kiernan, Meyer, and Schroeder pose three questions concerning the relationship between ms(G) and cms(G). In this note we use the graph Yn in Figure 1 to provide answers to each of these questions. If G is any simple graph, kG denotes the multi-graph in which every edge of G is replicated k times. If we consider the linear ordering α as a function α : E(G) 7→ {1, ...,m} we can define the linear distance in α between two edges ei, ej as: dα(ei, ej) = |α(ei)− α(ej)| Similarly if we consider the cyclic ordering β as a function: β : E(G) 7→ {1, ...,m} D.L. KREHER ET AL. / AUSTRALAS. J. COMBIN. 61 (2) (2015), 142–146 143 v1 v2 v3 v4 vn . . . vn+1 vn+2 Figure 1: The graph Yn we can define the cyclic distance in β between two edges ei, ej as: dβ(ei, ej) = min{|β(ei)− β(ej)|,m− |β(ei)− β(ej)|} Question 1: Given a graph G with matching number p, is there a positive integer k such that ms(kG) = p (cms(kG) = p)? The graph Yn has diameter n and hence because no two of the edges {v1, v2}, {v1, vn+1}, {v1vn+2} are in a matching it is easy to see the matching number of Yn is n/2 if n is even; (n+ 1)/2 if n is odd. However when n is odd the largest matching containing v1v2 is (n−1)/2. Hence, ms(Yn) ≤ (n−1)/2. Now consider kG = kYn, n odd. Any of the k edges between the vertices {v1}, {v2} can be in a matching of size at most (n− 1)/2. Thus, ms(kYn) ≤ (n− 1)/2 for any k. The answer to Question 1 is no. Question 2: For a graph G, we have ms(G) ≥ cms(G). How large can ms(G) − cms(G) be? Is cms(G) ≥ ms(G)− 1? Consider the graph Yn, when n is even. Label the edges according to the following table: Edge Label {v2i, v2i+1} 7→ i+ 1, 1 ≤ i < n2 {v2i+1, v2i+2} 7→ n2 + 1 + i, 1 ≤ i < n 2 {vn+2, v1} 7→ n2 + 1 {v1, v2} 7→ n+ 1 {vn+1, v1} 7→ 1 This labelling gives us ms(Yn) ≥ n2 − 1. Let β be a cyclic ordering of Yn and let e0, e1, e2 be the three edges incident to the vertex v1 ordered such that 1 ≤ β(e0) < β(e1) < β(e2). For any i ∈ Z3 consider the set {e ∈ E(G) : β(ei) ≤ β(e) ≤ β(ei+1)}. This is a set of size dβ(ei, ei+1) + 1 which is not a matching. Therefore the β-distance between D.L. KREHER ET AL. / AUSTRALAS. J. COMBIN. 61 (2) (2015), 142–146 144 edges ei and ei+1 is an upper bound to the matching number of β. The sum of these distances is: dβ(e0, e1) + dβ(e1, e2) + dβ(e2, e0) = n+ 1. Taking the average we obtain: dβ(e0, e1) + dβ(e1, e2) + dβ(e2, e0) 3 = n+ 1 3 . Hence for some i we have dβ(ei, ei+1) ≤ n+1 3 and the matching number of β is at most n+1 3 . Therefore cms(Yn) ≤ n+1 3 . Thus, ms(Yn)− cms(Yn) ≥ n 2 − 1− n+ 1 3 = n− 4 6 . Thus, we see that lim n→∞ ms(Yn)− cms(Yn) n ≥ 1 6 . Consequently our answer to Question 2 is that the difference ms(G) − cms(G) can be made as large as desired. Question 3: Given a graph G, is cms(2G) = ms(G)? From our answer to Question 2 we know that ms(Y2k) ≥ k − 1, where n = 2k. Now consider 2Y2k. Let β be a cyclic ordering of 2Y2k and let e0, e1, e2 . . . , e5 be the six edges incident to the vertex v1 ordered such that 1 ≤ β(e0) < β(e1) < · · · < β(e5). For any i ∈ Z6 consider the set {e ∈ E(G) : β(ei) ≤ β(e) ≤ β(ei+1)}. This is a set of size dβ(ei, ei+1) + 1 which is not a matching. Therefore the β-distance between edges ei and ei+1 is an upper bound to the matching number of β. The sum of these distances is: dβ(e0, e1) + dβ(e1, e2) + · · ·+ dβ(e5, e0) = 4k + 2. Taking the average we obtain: dβ(e0, e1) + dβ(e1, e2) + · · ·+ dβ(e5, e0) 6 = 4k + 2 6 . Hence for some i we have dβ(ei, ei+1) ≤ 4k+2 6 and the matching number of β is at most 4k+2 6 . Therefore cms(2Y2k) ≤ 4k+2 6 ≤ k − 1 ≤ ms(2Y2k), and the answer to the Question 3 is no. D.L. KREHER ET AL. / AUSTRALAS. J. COMBIN. 61 (2) (2015), 142–146 145