ISSN 2590-9770 The Art of Discrete and Applied Mathematics 5 (2022) #P1.01 https://doi.org/10.26493/2590-9770.1380.1c9 (Also available at http://adam-journal.eu) Whitney’s connectivity inequalities for directed hypergraphs Anahy Santiago Arguello* Bioinformatics Group, Department of Computer Science; and Interdisciplinary Center of Bioinformatics, Leipzig University, Härtelstraße 16-18, D-04107 Leipzig, Germany Peter F. Stadler† Bioinformatics Group, Department of Computer Science and Interdisciplinary Center of Bioinformatics, Leipzig University, Härtelstraße 16-18, D-04107 Leipzig, Germany, and Max-Planck-Institute for Mathematics in the Sciences, Inselstraße 22, D-04103 Leipzig, Germany, and Department of Theoretical Chemistry, University of Vienna, Währingerstraß 17, A-1090 Wien, Austria, and Facultad de Ciencias, Universidad Nacional de Colombia, Sede Bogotá, Colombia Santa Fe Institute, 1399 Hyde Park Rd., Santa Fe NM 87501, USA Received 27 August 2020, accepted 26 Janury 2021, published online 7 March 2022 Abstract Whitney’s inequality established an important connection between vertex and edge con- nectivity, and the degree of a graph, which was later generalized to digraphs and undirected hypergraphs. Here we show, using the most common definitions of connectedness for di- rected hypergraphs, that an analogous result holds for directed hypergraphs. It relates the vertex connectivity under strong vertex elimination, edge connectivity under weak edge elimination, and a suitable degree-like parameter and it is a proper generalization of the situation in both digraphs and undirected hypergraphs. We furthermore relate the connec- tivity parameters of directed hypergraphs with those of its directed bipartite König repre- sentation. Keywords: Strong and weak vertex elimination, strong connectedness, unilateral connectedness, di- rected hypergraph, total degree, connectivity indices. Math. Subj. Class.: 05C65, 05C40, 05C20 *Consejo Nacional de Ciencia y Tecnologı́a, Mexico, CONACYT PostDoc grant 2019-000012-01EXTV †Corresponding author. E-mail addresses: jpscw@hotmail.com (Anahy Santiago Arguello), studla@bioinf.uni-leizpig.de (Peter F. Stadler) cb This work is licensed under https://creativecommons.org/licenses/by/4.0/ 2 Art Discrete Appl. Math. 5 (2022) #P1.01 1 Introduction Directed hypergraphs naturally arose as a model of dependencies e.g. in propositional logic, database theory, and model checking, see e.g. [3, 7] for reviews. Recently they also received increasing attention as models of biological [6, 9], chemical [2, 11], and transportation networks [10]. Connectivity parameters are one of the most fundamental characteristics of a network, and hence are also of directed practical relevance for applications of directed hypergraphs [6]. It is important to note that directed hypergraphs give rise to many different notions of connectedness. Here, we only consider the simplest, least restrictive, construction of hy- perpath, requiring only a single vertex in the overlap of the head of one directed hyperedge and the tail of the following one. In particular in chemical reactions networks, much more restrictive notions path and reachability are also of interest, see e.g. [1, 2, 6]. The concepts of connectivity explored here remain closely related to those of bipartite graphs representa- tion of directed hypergraphs [12] and, as we shall see, admit generalizations of well-known results for graphs and digraphs. The connectivity in an undirected graph G is described by two parameters, the vertex connectivity index κ and the edge connectivity index κ′. They are defined as the minimum number of vertices or edges, respectively, whose removal disconnects G or gives a triv- ial graph. Hassler Whitney [13] showed that all undirected graphs satisfy the inequality κ ≤ κ′ ≤ δ, where δ denotes the minimal vertex degree in G. Later Geller and Harary found a generalization to digraphs [8]. In hypergraphs, the situation becomes more compli- cated because there are different, natural ways to delete vertices and hyperedges and thus to derive sub-hypergraphs [4]. Nevertheless, Whitney’s inequalities for the connectivity parameters generalize to (undirected) hypergraphs [5]. In the present contribution we show that analogous results also hold for directed hy- pergraphs with respect to both strong and unilateral connectedness. In Section 2.1 we introduce the notation and give some simple preliminary results for later use. Section 3 introduces the various connectivity indices and established some universal inequalities be- tween them. For many pairs of indices, however, we show that they are not comparable in general. The main theorem, a generalization of Whitney’s inequalities, is the topic of Sec- tion 4. In the final Section 5 we explore relations between connectivities of Section 5 we explore relations between connectivities of directed hypergraphs and their bipartite digraph representation. 2 Notation and preliminaries 2.1 Directed hypergraphs A directed hypergraph H = (V,E) consists on a vertex set V and a set of directed hyper- edges or hyperarcs E = {(T (e), H(e))} | T (e) ⊆ V and H(e) ⊆ V }, where H(e) 6= ∅ and T (e) 6= ∅. The sets T (e) and H(e) are called the tail and the head of e, respectively. The support of a hyperedge e ∈ E is the pair supp(e) = T (e) ∪H(e). A directed hyper- graph is called k-uniform if |T (e)| = |H(e)| = k for all e ∈ E. Two edges e, e′ ∈ E are said to be parallel if T (e) = T (e′) and H(e) = H(e′). A directed hypergraph H = (V,E) is called simple if it has neither parallel hyperarcs and nor loops, that is, edges e with T (e)∩H(e) = ∅. A (directed) hypergraph is trivial if |V | = 1 and E = ∅, i.e., H consists of a single vertex. A. S. Arguello and P. F. Stadler: Whitney’s connectivity inequalities 3 We say that u, v ∈ V are adjacent if there exists a hyperarc e ∈ E such that u ∈ T (e) and v ∈ H(e). The neighborhood of a vertex v in a hypergraph (or graph) is the set of all the vertices adjacent to v not including v. The indegree of a vertex v, denoted as d−(v) in H , is defined as the number of hyperarcs that contain v in their head. The outdegree of a vertex v, denoted as d+(v) in H , is defined as the number of hyperarcs that contain v in their tail. The minimum indegree and minimum outdegree of H will be denoted by δ−(H) = min{d−(v)}v∈V and δ +(H) = min{d+(v)}v∈V , respectively. The number of arcs parallel to e (including e) is the multiplicity of e and it is denoted as mH(e). Every directed hypergraph H = (V,E) can be represented as a bipartite digraph G(H) with vertex set V ∪ E and directed arcs x → e iff x ∈ T (e) and e → x iff x ∈ H(e). The arcs of G(H) are called the bits of the directed hypergraph. The graph G(H) is called the incidence digraph, Levi digraph, or König digraph of H . There is a one-to-one correspondence between directed hypergraphs and bipartite graphs for which one partition (the one corresponding to the hyperarcs E) has neither sources nor sinks (since we do not allow hyperarcs with empty heads or tails.) For details we refer to [12]. 2.2 Subhypergraphs Substructures of directed hypergraphs can be constructed in two ways: In strong substruc- tures the hyperedges are either retained or removed as an entity. In weak substructures, hyperedges can be restricted to a subset of vertices as long as their heads and tails remain non-empty. More precisely, following [5] we define: A directed hypergraph H ′ = (V ′, E′) is a weak subhypergraph of the directed hyper- graph H = (V,E) if V ′ ⊂ V and E′ consists of edges e′ with T (e′) = {v | v ∈ T (e)∩V ′} and H(e′) = {v | v ∈ H(e) ∩ V ′} for some e ∈ E. A directed hypergraph H ′ = (V ′, E′) is a weak induced subhypergraph of the directed hypergraph H = (V,E) if V ′ ⊂ V and edge set E′ = {(T (e) ∩ V ′, H(e) ∩ V ′)|e ∈ E ∧ T (e) ∩ V ′ 6= ∅ ∧ H(e) ∩ V ′ 6= ∅}. A directed hypergraph H ′ = (V ′, E′) is called a strong subhypergraph of the directed hyper- graph H = (V,E) if V ′ ⊂ V and E′ ⊂ E. A strong subhypergraph H ′ = (V ′, E′) of H = (V,E), is induced by V ′ if supp(e) ⊆ V ′ and it is induced by E′ if V ′ = ⋃ e′∈E′ supp(e ′). H ′ = (V ′, E′) is a spanning subhypergraph of H = (V,E) if V ′ = V . The deletion of vertices and edges from a directed hypergraph will play a key role in this contribution. Just as the formation of subhypergraphs this can be done in two ways: Strong vertex deletion of v ∈ V removes v and all hyperarcs that are incident to v. Thus it creates the strong subhypergraph H ′ = (V ′, E′) of H = (V,E) with vertex set V ′ = V \ {v} and edge set E′ = {e ∈ E | v /∈ T (e) ∪H(e)}. For a subset X ⊂ V we write H\SX to denote the directed hypergraph formed by strongly deleting all the vertices of X from H . Weak vertex deletion of v ∈ V removes v from the vertex set, and all occurrences of v from the hyperarcs of the directed hypergraph H . This creates the hypergraph H ′ = (V ′, E′) where V ′ = V \{v} and E′ = {(T (e) ∩ V ′, H(e) ∩ V ′)|e ∈ E ∧ T (e) ∩ V ′ 6= ∅∧H(e)∩V ′ 6= ∅}. We use the notation H\W v to denote the directed hypergraph formed by weakly deleting the vertex v from H . For any subset X of V we write H\WX to denote the directed hypergraph formed by weakly deleting all the vertices of X from H . 4 Art Discrete Appl. Math. 5 (2022) #P1.01 Strong deletion of the hyperarc e ∈ E removes e from the hypergraph and weakly deletes all the vertices incident with e. Thus it produces the weak subhypergraph H ′ = (V ′, E′) with V ′ = V \supp(e) and E′ = {(T (e)∩ V ′, H(e)∩ V ′)|e ∈ E ∧ T (e)∩ V ′ 6= ∅∧H(e)∩V ′ 6= ∅}. We write H\Se to denote the hypergraph formed by strongly deleting the edge e from H . For any subset F of E, we use H\SF to denote the directed hypergraph formed by strongly deleting all the hyper arcs of F from H . Weak deletion of the hyperarc e ∈ E simply removes the hyperarc e without affecting the rest of the hypergraph. Thus it leads to the strong subhypergraph H ′ = (V,E′) with E′ = E \ {e}. We write H\W e to denote the directed hypergraph formed by weakly deleting the hyperarc e from H . For any subset F of E, we write H\WF to denote the directed hypergraph formed by weakly deleting all the hyperarcs of F from H . It follows directly from the definition that the order in which vertices or edges are deleted has no impact on the final result. Thus the hypergraphs H\SX , H\WX , H\SF , and H\WF are well-defined. 2.3 Connectedness A directed walk in a hypergraph H = (V,E) is a sequence Pv0,vk = (v0, e1, v1, e2, ..., ek, vk) where e1, ..., ek ∈ E and v0, ..., vk ∈ V , such that vi−1 6= vi, vi−1 ∈ T (ei) and vi ∈ H(ei). A directed p-path is a walk where the vertices v0, ..., vk are all distinct. A directed cycle is a directed walk with k distinct hyperarcs and k distinct vertices such that v0 = vk. The length of a directed walk, directed path, or cycle is the number of hyperarcs in the sequence; i.e., it is k in the foregoing definitions. Let e ∈ E where T (e) = {u1, ..., uk} and H(e) = {v1, ..., vl} then the reverse hyperarc of e is ē ∈ E such that H(ē) = {u1, ..., uk} and T (ē) = {v1, ..., vl}. Definition 2.1. We say that y is reachable from x in H if there is a directed p-path from x to y in H . For two hyperarcs e and e′ we say that e′ is reachable from e in H if there is x in H(e) and y ∈ T (e′) such that y is reachable from x. Furthermore, we say v is reachable from u in G(H) if there is a directed path from u to v. There are three natural notions of connectedness in digraphs: A digraph is said to be strongly connected if, for every pair of vertices x, y ∈ V , x is reachable from y and y is reachable from x. It is said to be unilaterally connected if, for every pair of vertices x, y ∈ V , x is reachable from y or from y is reachable from x. A bipartite graph with vertex set V1 ∪V2 is unilaterally connected on V1 if for every pair u, v ∈ V1, v is reachable from u or u is reachable from v. Finally, a digraph is weakly connected if its underlying graph, i.e without direction, is connected. These definitions can be generalized immediately to hypergraphs. Definition 2.2. A directed hypergraph H is strongly connected if for every pair of vertices u, v ∈ V , u is reachable from v and v is reachable from u. It is unilaterally connected if for every two pair of vertices u, v ∈ V , v is reachable from u or u is reachable from v. It is (weakly) connected if the underlying hypergraph, is connected. Corollary 2.3. For every directed hypergraph, “strongly connected” implies “unilaterally connected”, which in turn implies “weakly connected”. A. S. Arguello and P. F. Stadler: Whitney’s connectivity inequalities 5 Lemma 2.4. A directed hypergraph H is strongly, unilaterally, or weakly connected if and only if its incidence (di)graph G(H) is strongly connected, unilaterally connected on V , and weakly connected, respectively. Proof. An undirected hypergraph is connected if and only if its (undirected) incidence graph is connected, see e.g. [4, 5], hence the statement is true for weak connectedness. To show the statement for unilateral and strong connectedness we first show that for all x, y ∈ V there is hyperpath from x to y in H if and only if there is a path from x to y in G(H). First assume that a directed hyperpath x = v0, e1, v1, e2, . . . , vk−1, ek, vk = y in H exists. Then the bits (v0, e1), (e1, v2), . . . , (ek−1, vk) form a directed path for x to y in G(H). Conversely, suppose such a directed path exists in G(H). We note that the arcs (vi−1, ei) and (ei, vi) in G(H) by construction are bits induced by a hyperedge ei with vi−1 ∈ T (ei) and vi ∈ H(ei). Thus the sequence x = v0, e1, v1, e2, . . . , vk−1, ek, vk = y is a directed hyperpath in H . If G(H) is strongly connected then in particular there is a directed hyperpath in H between any pair of vertices, and thus H is strongly connected. Conversely, if H is strongly connected, we know that there is a directed path between any pair x, y in V . To see that every e ∈ E is reachable from every x ∈ V in G(H) we recall that every T (e) 6= ∅, i.e., there is u ∈ T (e). We already know that there is a directed path from x to u in G(H), which can be extended by the bit (u, e) to a directed path from x to e. Using that H(e) 6= ∅ we see that every x ∈ V is reachable from every e ∈ E. Concatenating a directed from e to x and from x to e′ we finally see that every e′ ∈ E is reachable from every e ∈ E, and thus G(H) is strongly connected. It follows immediately that H is unilaterally connected if for every x, y ∈ V there is a directed path from x to y or from y to x in G(H), i.e., if G(H) is unilaterally connected on V . Note that unilateral connectedness of H does not imply unilateral connectedness of G(H). As a counterexample consider the directed (hyper)graph H with V = {u, v, w, x} and hyperarcs e1 = (u, v), e2 = (v, w), e3 = (w, x), and e4 = (u, x). H is unilaterally but not strongly connected but there is no directed path from e2 to e4 or vice versa in G(H). In the following we say that H = (V,E) is C -connected with C ∈ {S ,U ,W } is strongly, unilaterally, or (weakly) connected. Correspondingly, we shall say that H is C - disconnected if it is not C -connected. 3 Connectivity in directed hypergraphs The degree of connectedness in an undirected hypergraph H is described by invariants describing the minimal number of vertices or edges that must be removed by either weak or strong elimination to disconnect the hypergraph or leave on a trivial hypergraph behind [5]. The situation becomes even more involved because each of these invariants or indices can be defined with respect to each of the three concepts of connectedness. We write κxC and κ′xC , where the prime refers to edge deletion, x ∈ {s, w} indicates strong or weak vertex/edge deletion and refers to strong, unilateral, or weak connectedness. The numbers κxC and κ ′ xC , are the minimum numbers of vertices and hyperedges, respectively, such that their x-elimination leaves a hypergraph C -disconnected or trivial. Let H = (V,E) be a directed hypergraph. A vertex v ∈ V is called a strong (weak) C - cut vertex of H if H\sv (H\wv) is C -disconnected or trivial. X is a strong (weak) vertex 6 Art Discrete Appl. Math. 5 (2022) #P1.01 Tk Tk v1 v2 v3 vk u1 u2 u3 uk v Tk Tk v u Figure 1: Left: Tk is the tournament of k vertices, where k is even. In this hypergraph κsS = 1, κwS = k 2 , κsU = 1, κwU = k, the minimum strong vertex, for sS and sU cut is {v} and a minimum weak vertex cut for wS is {u1, u3, ..., uk−1} and a weak vertex cut for wU is {u1, v2, u3, v4, ..., vk} . As k increases, an infinite family of hypergraphs for which this difference grows linearly is obtained. Right: Tk is the tournament of k vertices, where k is even. In this hypergraph κ′sS = κ ′ sU = 1 ((u, v) is a disconnecting arc), κ′wS = k 2 (all the arcs that have tail v), κ′wU = k + 1(all the arcs that have tail or head u). C -cut of H if H\sX (H\wX) is C -disconnected or trivial. We adopt the convention that κxC = 1 for trivial hypergraphs and κxC = 0 for null hypergraphs. A subset F ⊆ E is called a strong (weak) C -disconnecting set of H if H\sF (H\wF ) is C -disconnected or trivial. We set κ′xC = 1 for trivial hypergraphs and κ ′ xC = 0 for null hypergraphs. The following inequalities hold for all directed hypergraphs as an immediate conse- quence of the definition and the implications between the connectedness classes for both x = s and x = w. κxS ≤ κxU ≤ κxW κ ′ xS ≤ κ ′ xU ≤ κ ′ xW (3.1) Since W -connectedness coincides with the connectedness of undirected hypergraphs we focus on C ∈ {S ,U } in the following. The case of undirected hypergraphs is studied in detail in [5]. We first consider the relationships between strong and weak elimination: Lemma 3.1. Let H = (V,E) be a directed hypergraph. Then κsC ≤ κwC for C ∈ {S ,U }. Proof. If H is trivial or null, there is nothing to show. If H is C − disconnected, then κxC = κ ′ xC = 0, and the inequalities hold trivially. Now suppose that H is nontrivial and C -connected. We note that H\sX is a spanning strong subhypergraph of H \w X for all X ⊆ V . This implies immediately that κsC ≤ κwC for C ∈ {S ,U }. It is worth noting that κwC is a poor upper bound for κsC . Indeed, the difference between κwC −κsC can become arbitrarily large as shown in Figure 1(left). It is important to notice that not every strong vertex cut is contained in a weak vertex cut. The situation on the left hand side of Figure 1 is an example. Whitney’s inequalities [5] and their generalization to directed graphs [8] and undirected hypergraphs [13] relate the connectivity indices with each other. In the case of directed A. S. Arguello and P. F. Stadler: Whitney’s connectivity inequalities 7 v1 v2 v3 v v4 v5 v6 v1 v2 v3 ... vk u1 u2 u3 ... uk w1 w2 w3 ... wk Figure 2: Left: In this hypergraph κsS = κsU = 1 since v is a cut vertex; κwS = 2 since {v2, v4} is a minimum weak vertex cut; κ ′ sU = κ ′ sS = 2 since {(v5, v4), (v2, v3)} is a strong disconnecting set. Right: The hyperarcs in this hypergraph are e1 = ({v1, ..., vk}, {u1, ..., uk}), ē1, e2 = ({u1, ..., uk}, {w1, ..., wk}), ē2. We have κ ′ wS = 1 by removal of any hyperarc and κwS = k since {u1, ..., uk} is the minimal weak vertex. hypergraphs, however, some of these quantities do not fulfill universal inequalities. We give some simple counterexamples: κsU ⋚ κwS . In a directed cycle Cn of length n > 3 we have κsU = 2 and κwS = 1. Therefore κsU > κwS . On the other hand, the left hand side of figure 2 shows an example where κsU = 1 and κwS = 2, i.e., κsU < κwS . κ′sU ⋚ κ ′ wS . In a directed cycle Cn of length n > 4 we have κ ′ wS = 1 and κ ′ sU = 2. Therefore κ′sU > κ ′ wS . The hypergraph on the right hand side of Figure 1 has κ′sU = 1 and κ ′ wS = k 2 . For k > 3 we therefore have κ′sU < κ ′ wS . κsS ⋚ κ′sU . The hypergraph in Figure 2(left) satisfies κsS < κ ′ sU . Now consider the hypergraph in Figure 1(right) with all reverse hyperarcs added. Here, κ′sU = 1 and κsS = 2 since {u, v} is a vertex cut. Thus κ ′ sU < κsS . κsU ⋚ κ′sU . The hypergraph in Figure 2(left) satisfied κsU < κ ′ sU , while the hypergraph in Figure 1(right) satisfies κsU = 2 ({u, v} is a minimal strong the vertex cut) and κ′sU = 1. Therefore κ ′ sU < κsU . κ′sU ⋚ κwS . In a directed cycle Cn of length n > 4 we have κwS = 1 and κ ′ sU = 2, i.e., κwS < κ ′ sU . Again, we consider the hypergraph in Figure 1(right) with the reverse hyperarcs added. It satisfies κ′sU = 1 and κwS = 2, i.e., κ ′ sU < κwS . κwS ⋚ κ′wS . The hypergraph in Figure 2(right) satisfies κ ′ wS = 1 and κwS = k, i.e., κ′wS < κwS . For the hypergraph in Figure 1(right) we have κwS = 2 due to the weak way the vertex cut {u, v}. Furthermore, for k ≥ 4 we have κ′wS = k 2 and thus κwS < κ ′ wS . κ′wU ⋚ κwS . The hypergraph in Figure 1(right) satisfies κwS = 2, the set {u, v} being a minimum vertex cut. On the other hand, we have κ′wU = k+1 in the same example, hence, for k > 2, we have κwS < κ ′ wU . The hypergraph in Figure 2(right) satisfies κwS = k and κ ′ wU = 2, since {e1, ē1} is a minimal disconnecting set. Thus, for k > 2, we have κ′wU < κwS . κwS ⋚ κ′sU . The hypergraph in Figure 2(right) satisfied κ ′ sU = 1 since every hyperarc contains a strong cut vertex. On the other hand we have κwS = k and this, for 8 Art Discrete Appl. Math. 5 (2022) #P1.01 Tk Tk v1 v2 vk−1 vk u1 u2 uk−1 uk v Figure 3: In this hypergraph with even k we have κwS = κwU = 1 since the vertex v is a cut vertex; κ′sU = κ ′ sS = 1+ k 2 since {(v, {u1, ..., uk}), (v1, v2), (v3, v4), ..., (vk−1, vk)}) is a strong disconnecting set. k > 1, κ′sU < κwS . In a directed cycle Cn with n > 5 we have κwS = 1 and κsU = 2 and therefore κwS < κsU . κwU ⋚ κ′wS . In a directed cycle of length n > 3 we have κ ′ wS = 1 and κwU = 2 and this κ′wS < κwU . The hypergraph in Figure 1(right) satisfied κ ′ wS = k 2 and κwU = 2 since the set {u, v} is a weak vertex cut. Thus, for k > 4, we have κwU < κ ′ wS . κwU ⋚ κ′wU . The hypergraph in Figure 2(right) satisfies κ ′ wU = 2 and κwU = k. Thus, for k > 2 we have κ′wU < κwU . The hypergraph in Figure 1(right) satisfied κwU = 2 and κ′wU = k + 1. This, for k > 2 we have κwU < κ ′ wU . κsU ⋚ κ′wS . In a directed cycle Cn of length n > 3 we have κ ′ wS = 1 and κsU = 2, i.e., κ′wS < κsU . The hypergraph in Figure 1(right) satisfies κsU = 2 and κ ′ wS = k 2 . Thus, for k > 4 we have κsU < κ ′ wS . κ′sS ⋚ κsU . The hypergraph in Figure 1(right) satisfied κ ′ sS = 1 and κsU = k + 1, i.e., κ′sS < κsU . For the hypergraph in Figure 2(left) we have κsU = 1 and κ ′ sS = 2 and thus κsU < κ ′ sS . κ′sS ⋚ κsS . The hypergraph in Figure 1(right) satisfies κ ′ sS = 1 and κsS = 2, and thus κ′sS < κsS . For the hypergraph in Figure 2(left) we have κsS = 1 and κ ′ sS = 2, and therefore κsS < κ ′ sS . κ′sC ⋚ κwC . The hypergraph in Figure 3 satisfies that κ ′ sC = 1+ k 2 and κwC = 1, hence for k > 2 we have κ′sC > κwC . The hypergraph in Figure 2(right) satisfies κ ′ sC = 1 and κwC = k, thus for k > 1 we have κ ′ sC < κwC . κ′sS ⋚ κwU . The hypergraph in Figure 3 satisfies κ ′ sS = 1 + k 2 and κwU = 1 thus for k > 2 we have κ′sS > κwU . The hypergraph in Figure 2(right) satisfies κ ′ sS = 1 and κwU = k, hence for k > 1 we have κ ′ sS < κwU . κ′sC ⋚ κ ′ wC . The hypergraph in Figure 3 satisfies κ ′ sC = 1 + k 2 and κ′wC = 1, hence for k > 2 we have κ′sC > κwC . The hypergraph in Figure 1(left) satisfies κ ′ sC = 1 and κ′wC = k 2 , hence for k > 1 we have κ′sC < κ ′ wC . κ′sS ⋚ κ ′ wU . The hypergraph in Figure 3 satisfies κ ′ sS = 1 + k 2 and κ′wU = 1, hence for k > 2 we have κ′sS > κwU . The hypergraph in Figure 1(left) satisfies κ ′ sS = 1 and κ′wU = k 2 , hence for k > 1 we have κ′sS < κ ′ wU . A. S. Arguello and P. F. Stadler: Whitney’s connectivity inequalities 9 4 Whitney’s theorem for directed hypergraphs Let H = (V,E) a directed hypergraph (or digraph) and v ∈ V . The total degree of the vertex v is dt(v) = d+(v) + d−(v). Denote by δid, δod, and δW the minimum of d −, d+ and dt over all v ∈ V , respectively. Furthermore we introduce δUid = min v∈V {d−(v) + δid(H \w v)} and δ U od = min v∈V {d+(v) + δod(H \w v)}. With this notation we define δS = min{δid, δod} and δU = min{δ U id , δ U id }. These param- eters are direct generalizations of the corresponding quantities for directed hypergraphs, see e.g. [8]. The next theorem is a generalization of Whitney’s inequalities for directed hypergraphs. The proof follows ideas from [8] for the analogous result for digraphs. Theorem 4.1. Let H = (V,E) a directed hypergraph. Then κsU (H) ≤ κ ′ wU (H) ≤ δU (H) and κsS (H) ≤ κ ′ wS (H) ≤ δS (H). Proof. If H is trivial or null, the statements of the theorem are obviously valid. Let H be a U -connected hypergraph and let u, v ∈ V such that d−(u) = δid(H) and d−(v) = δid(H \w u). Weakly eliminate the hyperarcs such that their heads contain u and v; in this way, there is no (u, v)-directed path and there is no (v, u)-directed path on H . So κ′wU (H) ≤ δ U in . Applying the same dual argument we conclude that κ ′ wU (H) ≤ δ U od and so κ′wU (H) ≤ δU . On the other hand, if κ′wU (H) = 1, there is e ∈ E, such that H\we is not U - connected. If we eliminate in a strong way the vertex v ∈ T (e) ∪ H(e) then H\sv is not U -connected, so κsU (H) = 1 and the result is valid. Let κ ′ wU (H) > 1, for prov- ing that κsU (H) ≤ κ ′ wU (H) let weakly eliminate set of hyperarcs F , the cardinality of F is κ′wU (H) − 1 such that the directed hypergraph H ′ = (V ′, E′) = H\wF has κ′wU (H ′) = 1. Let e ∈ (E′) a hyperarc such that H ′\we is not unilaterally connected. Now we strongly eliminate the set of vertices X ∈ V such that each vertex on X is in ex- actly one hyperarc of F (there are enough vertices due to | e |> 1 for all e ∈ E), we denote this directed hypergraph H ′′ = (V ′′, E′′). If e /∈ E′′ then H ′′ is not unilaterally connected, so κsU (H) < κ ′ wU (H). If e ∈ E ′′ then κ′wU (H ′′) = 1 and κsU (H ′′) = κ′wU (H ′′) = 1 so κsU (H) = κ ′ wU (H). Let H be S -connected hypergraph and let v ∈ V such that d−(v) = δid(H). Weakly eliminate the hyperarcs such that their heads contain v; in this way, there is no (u, v)- directed path on H for all u ∈ V . So κ′wS (H) ≤ δin. Applying the same dual argument we conclude that κ′wS (H) ≤ δod and so κ ′ wS (H) ≤ δS . The proof that κsS (H) ≤ κ ′ wS (H) parallels the proof of the inequality κsU (H) ≤ κ′wU (H). Note that Theorem 4.1 reduces the corresponding statement for digraphs whenever all e ∈ E hyperarcs satisfy |T (e)| = |H(e)| = 1. Corollary 4.2. Let H = (V,E) a directed hypergraph. Then κsS ≤ κwU and κsS ≤ κ′wU . Proof. The first inequality follows from the note at the beginning of the previous section and Lemma 3.1. The second inequality follows from Theorem 4.1 and Equation 3.1. 10 Art Discrete Appl. Math. 5 (2022) #P1.01 T ′p T ′′ p Tq w1 w2 t arcs e arcs r arcs T ′p T ′′ p Tq Tet arcs r arcs Figure 4: Left: Construction (A); Right: Construction (B) We next show that the parameters κsC (H), κ ′ sC (H), κwC (H) and κ ′ wC (H) are inde- pendent for strongly or unilaterally connected directed hypergraphs satisfying κsC (H) ≤ min{κwC (H), κ ′ wC (H)}. Theorem 4.3. For every choice of natural numbers a, b, c, d, and e with a ≤ min{c, d}, b ≤ c and max{b, c, d} ≤ e and connectedness classes C ∈ {S ,U ,W } there exists a directed hypergraph such that κsC (H) = a, κ ′ sC (H) = b, κwC (H) = c, κ ′ wC (H) = d and δC (H) = e. Proof. We explicitly construct hypergraphs with the desired properties. Consider unilateral connectedness, i.e., C = U . We construct a hypergraph A form by five disjoint components: Two tournaments T ′p and T ′′ p on p vertices, a tournament Tq , and two single vertices w1 and w2; each arc in the tournaments T ′ p and T ′′ p has multiplicity e−p+1 and each arc in Tq has multiplicity e−q+1. We then insert r hyperarcs consisting of one vertex from T ′p and Tq in its tail and a vertex from T ′′ p in its head involved, with all reverse hyperarcs added and with one of these hyperarcs with multiplicity t. Finally, we insert t arcs from w1 to T ′ p and e arcs from w2 to T ′′ p with all reverse arcs added. The arcs are inserted in such a way that the vertices of the tournaments are covered as uniformly as possible. The construction is illustrated in Figure 4(left). Let a = q, b = r, c = p, d = t. The minimum strong vertex cut in A is V (Tq), the minimum strong disconnecting set is the r different hyperarcs, a minimum weak vertex cut is the set V (T ′p) and the minimum weak disconnecting set is the t arcs (recall that one of the r hyperarcs has multiplicity t); then q = κsU , r = κ ′ sU , p = κwU and t = κ ′ wU . Finally δU = e because d −(w1) = 0 and d −(w2) = e, since all the vertices in the tournaments have in-degree and out-degree at least e, because of the multiplicities of the arcs inside the tournaments. Next we consider strongly connectedness, C = S . We construct a hypergraph B form by five disjoint components: Two tournaments T ′p and T ′′ p on p vertices, a tournament Tq , and a tournament Te; each arc in the tournaments T ′ p and T ′′ p has multiplicity e − p + 1 and each arc in Tq has multiplicity e− q + 1. We then insert r hyperarcs consisting of one vertex from T ′p and Tq in its tail and a vertex from T ′′ p in its head involved, with all reverse hyperarcs added and with all of these hyperarcs with multiplicity t. The arcs are inserted in such a way that the vertices of the tournaments are covered as uniformly as possible. Finally, we insert t hyperarcs consisting of at least one vertex from Te in its tail and a A. S. Arguello and P. F. Stadler: Whitney’s connectivity inequalities 11 vertex from T ′′p in its head involved, all vertices in Te are in a head of these hyperarcs, all the reverse hyperarcs are added. The construction is illustrated in Figure 4(right). Let a = q, b = r, c = p, and d = t. The minimum strong vertex cut in B is V (Tq), the minimum strong disconnecting set is the r different hyperarcs, the minimum weak vertex cut is the set V (T ′′p ) and a minimum weak disconnecting set is the t arcs from Te to T ′′ p (don’t forget that each of the r hyperarcs has multiplicity t); then q = κsS , r = κ ′ sS , p = κwS and t = κ ′ wU . Finally δS = e because all the vertices in Te have indegree and outdegree e (all the vertices in the other tournaments have indegree and outdegree at least e, because of the multiplicities of the arcs inside the tournaments). Next, we consider weak connectedness, C = W . We construct a hypergraph C form by five disjoint components: Two complete graphs K ′p and K ′′ p on p vertices, a complete graph Kq , and a complete graph Ke; each edge in the complete graphs K ′ p and K ′′ p has multiplicity e−p+1 and each edge in Kq has multiplicity e−q+1. We then insert r edges consisting of one vertex from K ′p, Kq and K ′′ p , with all of these edges with multiplicity t. The edges are inserted in such a way that the vertices of the complete graphs are covered as uniformly as possible. Finally, we insert t edges consisting of at least one vertex from Te and one vertex from T ′′p , all vertices in Te are incident with these edges. The explanation of why this hypergraph has the desired parameters is analogous to the strong case. 5 König digraph of a directed hypergraph The connectivity invariants in digraphs are defined in the same way as in directed hy- pergraphs, the difference is that the weak or strong elimination of vertices or arcs is not relevant so we only have to consider a single connectivity index for each connectedness class, which we denote by κC and κ ′ C with C ∈ {S,U}. Lemma 5.1. Let H = (V,E) be a directed hypergraph and let G(H) = (V ∪E,A) be its König digraph. Then κsC (H) ≤ κC(G(H)) ≤ min{κwC (H), κ ′ wC (H)} (5.1) holds for C ∈ {U, S}. Proof. Let S ⊆ V ∪ E be a vertex cut in G(H) with |S| = κC(G(H)). Case 1: S ⊆ V , then S = {v1, ..., vk} is a weak vertex cut in H and so it is a strong vertex cut in H , so κsC (H) ≤ κC(G(H)). Case 2: Suppose S ⊆ E. We step-wisely construct a vertex cut S′ as follows by interacting over the hyperedges in S. In each step we add to S′ a single vertex vi ∈ ei that is not contained in ⋃ j 1, second multiplicative Zagreb index ∏ 2 (G) and sum lordeg index SL(G). The extremal graph is unique for k = 1, n = 4 and for k ≥ 2 and it consists from a path x1, x2, . . . , xn−1 and a new vertex xn adjacent with xk, xk+1 and xk+2. Keywords: Vertex-degree function index, general first Zagreb index, second multiplicative Zagreb index, sum lordeg index, k-generalized quasi-unicyclic graph, Jensen inequality. Math. Subj. Class.: 05C35, 05C75, 05C09 1 Introduction Let G be a simple graph. By V (G) and E(G) we denote the vertex set and the edge set of G, respectively. Let e(G) be the number of edges of G. For any x ∈ V (G), we denote by dG(x) the degree of x, i.e., the number of neighbors of x in G. If the graph G is clear under the context, then we use d(x) instead of dG(x). A vertex with degree one will also be referred as a pendant vertex. Suppose that V (G) = {v1, v2, . . . , vn} and the degree of vertex vi equals di for i = 1, 2, . . . , n, then π = (d1, d2, . . . , dn) is called the degree sequence of G. We always will enumerate the degrees in non-increasing order, i.e., d1 ≥ d2 ≥ . . . ≥ dn. Pn and Cn will denote the path and cycle on n vertices. *The author thanks the referees for valuable comments. E-mail address: ioan@fmi.unibuc.ro (Ioan Tomescu) cb This work is licensed under https://creativecommons.org/licenses/by/4.0/ 2 Art Discrete Appl. Math. 5 (2022) #P1.02 For S ⊂ V (G), the subgraph induced by S is denoted G[S]. For two vertex disjoint subsets S, T ⊂ V (G), the induced bipartite graph between S and T is denoted by G[S, T ]. For graph G and a subset X of V (G), G −X is the graph obtained from G by removing the vertices of X and all edges incident to any of them. In particular, when X consists of only one vertex v, G − {v} is always abbreviated to G − v. Similar notation is G − uv, where uv ∈ E(G). A unicyclic graph G of order n is connected and has n edges. It consists of a cycle Cr, where 3 ≤ r ≤ n and some vertex-disjoint trees having each a vertex common with Cr. A bicyclic graph G of order n is a connected graph of size e(G) = n+ 1. It has two linearly independent cycles which have a common vertex or a common path Pa with a ≥ 2 or they are connected by a path Pb with b ≥ 2. The quasi-tree is a graph G in which there exists a vertex v ∈ V (G) such that G−v is a tree. Similarly, if graph G has the property that G−v induces a unicyclic graph for a suitable vertex v, then G is called a quasi-unicyclic graph. In [14], Xu et. al. generalized the concept of quasi-tree to k-generalized quasi-tree as: Definition 1.1 ([14]). For any integer k ≥ 1, the connected graph G is called a k-generalized quasi-tree, if there exists a subset Vk ⊆ V (G) with cardinality k such that G− Vk is a tree but for every subset Vk−1 of cardinality k − 1 of V (G), the graph G− Vk−1 is not a tree. In [14], the authors pointed out that any tree is a quasi-tree since the deletion of any pendant vertex will produce another tree. Thus, they called any tree a trivial quasi-tree, and other quasi-tree graphs as non-trivial quasi-trees. With the similar reason, any unicyclic graph with at least one pendant vertex is also a quasi-unicyclic graph. Motivated from Definition 1.1, Javaid et. al. [7] generalized the concept of quasi-unicyclic graph to k- generalized quasi-unicyclic graph as: Definition 1.2 ([7]). For any integer k ≥ 1, the connected graph G is called a k-generalized quasi-unicyclic graph, if there exists a subset Vk ⊆ V (G) of cardinality k such that G−Vk is a unicyclic graph but for every subset Vk−1 of cardinality k − 1 of V (G), the graph G− Vk−1 is not unicyclic. It is easily checked that a quasi-unicyclic graph which is not unicyclic is just a 1- generalized quasi-unicyclic graph. In what follows, we call the vertex set Vk of Defini- tions 1.1 and 1.2 as a k-quasi-vertex set, and we use the symbol Ukn to denote the class of k-generalized quasi-unicyclic graphs with n vertices. Notice that a graph may by a k-generalized quasi-unicyclic graph for several non- consecutive values of k. For other notations in graph theory, we refer [13]. Among all (vertex) degree-based graph invariants, the first Zagreb index M1(G) [4] is a famous topological index. It is defined as M1(G) = ∑ v∈V (G) d2(v). The general first Zagreb index (sometimes referred as ”zeroth-order general Randić index” [8]), denoted by 0Rα(G) was defined [9] as 0Rα(G) = ∑ v∈V (G) d(v)α, I. Tomescu: Minimizing vertex-degree function index 3 where α is a real number, α 6∈ {0, 1}. For α = 2 it is the first Zagreb index M1(G). Extremal results concerning the first Zagreb index for trees, unicyclic and bicyclic graphs were obtained in [1, 3, 16]. The second multiplicative Zagreb index or modified Narumi-Katayama index [2, 6] is defined as ∏ 2 (G) = ∏ v∈V (G) d(v)d(v) = ∏ uv∈E(G) d(u)d(v). This index is minimum if and only if ln ∏ 2 (G) = ∑ v∈V (G) d(v) ln d(v) is minimum. The sum lordeg index is one of the Adriatic indices introduced in [12] and it is defined by SL(G) = ∑ v∈V (G) d(v) √ ln d(v) = ∑ v∈V (G):d(v)≥2 d(v) √ ln d(v). The vertex-degree function index Hf (G) was defined in [15] as Hf (G) = ∑ v∈V (G) f(d(v)) for a function f(x) defined on positive real numbers. In this paper we will impose to func- tion f(x) to be (i) strictly increasing and (ii) strictly convex. All indices mentioned above are vertex-degree function indices Hf (G): 0Rα(G) corresponds to f(x) = x α which satisfies (i)–(ii) for α > 1; the natural logarithm of the second multiplicative Zagreb in- dex ∏ 2 (G) to f(x) = x lnx which satisfies (i)–(ii) for x ≥ 1 and sum lordeg index to f(x) = x √ lnx satisfying (i)–(ii) for x ≥ 2 (see [5]). The rest of the paper is organized as follows. In Section 2, we introduce a function defined on the partitions of n and give some properties concerning the minimum of this function. In Section 3, we solve the problem of minimizing the vertex-degree function index Hf (G), where f(x) verifies (i)–(ii) for k-generalized quasi-unicyclic graphs of order n for k ≥ 1 and n ≥ 2k + 2. Section 4 includes an extremal result of this type for k-generalized quasi-trees when k ≥ 2 and n ≥ 3k. 2 Preliminary results In what follows we shall suppose that function f(x) satisfies requirements (i)–(ii). The following lemmas will be used in our proofs. Lemma 2.1. Let y > 0 and x ≥ y + 2. Then f(x) + f(y) > f(x− 1) + f(y + 1). Proof. The function f(x) being strictly convex, ϕ(x) = f(x + 1) − f(x) is a strictly increasing function. Since x − 1 ≥ y + 1 > y it follows that ϕ(x − 1) > ϕ(y), or f(x)− f(x− 1) > f(y + 1)− f(y). For integers n, p such that n ≥ 1 and p ≥ n denote by Dn,p the set of n-tuples x = (x1, x2, . . . , xn) such that x1 ≥ x2 ≥ . . . ≥ xn ≥ 1 and ∑n i=1 xi = p. Let the function F (x) = ∑n i=1 f(xi). By Lemma 2.1 the minimum of F (x) is reached if and only if 4 Art Discrete Appl. Math. 5 (2022) #P1.02 |xi − xj | ≤ 1 for every 1 ≤ i < j ≤ n, or equivalently, if and only if x1 + x2 + . . .+ xn is an equipartition of p, having almost equal parts. It follows that the point of minimum of F (x) on Dn,p is unique. Lemma 2.2. If q > p ≥ n then min x∈Dn,q F (x) > min x∈Dn,p F (x). Proof. If p = kn+ r, where 0 ≤ r ≤ n−1, then the point x∗ = (x∗1, . . . , x ∗ n) where F (x) reaches its minimum is x∗ = (k, k, . . . , k) for r = 0. Otherwise, x∗1 = . . . = x ∗ r = k + 1 and x∗r+1 = . . . = x ∗ n = k. If x ∗ = (x∗1, . . . , x ∗ n) and y ∗ = (y∗1 , . . . , y ∗ n) denote the points of minimum of F (x) in Dn,p and in Dn,p+1, respectively, it follows that there is an index t, 1 ≤ t ≤ n such that y∗t = x ∗ t + 1 and y ∗ i = x ∗ i for i 6= t. We get F (y ∗) > F (x∗), therefore min x∈Dn,q F (x) > min x∈Dn,q−1 F (x) > . . . > min x∈Dn,p F (x). For a natural number s, 1 ≤ s ≤ n − 1, denote by Dsn,p ⊂ Dn,p the set of n- tuples (x1, x2, . . . , xn) ∈ Dn,p such that the last s components are equal to 1: xn−s+1 = xn−s+2 = . . . = xn = 1. The following property also holds by Lemma 2.1: Lemma 2.3. If s < t ≤ n− 1 and p ≥ 2n− t+ 1 then min x∈Dsn,p F (x) < min x∈Dtn,p F (x). Proof. Let x∗ = (x∗1, . . . , x ∗ n) be the point of minimum of F (x) in D t n,p. It follows that x∗n−t+1 = . . . = x ∗ n = 1. Since p ≥ 2n − t + 1, we deduce that x ∗ 1 ≥ 3. By rearranging the numbers x∗1−1, x ∗ 2, . . . , x ∗ n−t, 2, 1, . . . , 1 in decreasing order we get the vector denoted by y∗, which belongs to Dt−1n,p . We get that F (x ∗) > F (y∗) ≥ minDsn,p F (x) and lemma holds. Since ∑n i=1 di = 2e(G) we get: Lemma 2.4. We have Hf (G) ≥ min x∈Dn,2e(G) F (x). Equality may hold only if the point of minimum (x∗1, x ∗ 2, . . . , x ∗ n) of F (x1, x2, . . . , xn) in Dn,2e(G) is graphical, i.e., if there exists a graph G with degrees di = x ∗ i for i = 1, . . . , n. 3 Main results Let k ≥ 1 and n ≥ 2k+2. By Fn,k we denote the graph consisting of a path x1, x2, . . . , xn−1 and a new vertex xn which is adjacent to xk, xk+1 and xk+2. For k ≥ 2 and n ≥ 2k + 2, this bicyclic graph belongs to the class Ukn of k-generalized quasi-unicyclic graphs with n vertices. For k = 1 and n = 4 this graph, denoted F4,1, consists of two cycles C3 having a common edge and belongs to the class U14 . For k ≥ 2 we have Hf (Fn,k) = 4f(3) + (n− 6)f(2) + 2f(1) and this expression does not depend on k. I. Tomescu: Minimizing vertex-degree function index 5 Theorem 3.1. If G ∈ U1n and n ≥ 4, then we have Hf (G) ≥ 2f(3) + (n− 2)f(2). The equality is reached if and only if G has two vertices of degree three and n− 2 vertices of degree two, i.e., it consists of two vertex disjoint cycles joined by a path Ps or two cycles having a common path Pt,where s, t ≥ 2. The extremal graph is unique only for n = 4, when it coincides to F4,1. Proof. Suppose that G ∈ U1n has minimum Hf (G). Since G ∈ U 1 n, there exists a vertex v0 such that G − v0 is a connected unicyclic graph with n − 1 vertices and n − 1 edges. We first will prove that dG(v0) = 2. Since k = 1, by Definition 1.2, G is not a unicyclic graph and so dG(v0) ≥ 2. We assume that dG(v0) ≥ 3. If v0u is an edge of G, denote G1 = G − v0u. Let H = G − v0 = G1 − v0, which is a unicyclic graph. G1 is not unicyclic because its number of edges equals |E(H)|+dG(v0)−1 ≥ n+1. It follows that G1 ∈ U 1 n. Since G1 = G− v0u we get Hf (G) > Hf (G1), a contradiction, since function f(x) is strictly increasing. Consequently, G has n − 1 + 2 = n + 1 edges. In Dn,2n+2 the minimum of the function F (x) is reached for the n-tuple (3, 3, 2, 2, . . . , 2), since 3+3+2+. . .+2 is an equipartition of 2n+ 2 with n almost equal parts. The degree sequence (32, 2n−2) is graphical and any graphical realization consists of two vertex disjoint cycles joined by a path Ps or two cycles having a common path Pt, where s, t ≥ 2 since G is connected. All these graphs belong to U1n and we are done. For n = 4 the extremal graph is F4,1, composed from two C3 with a common edge, for n = 5 there are two extremal graphs: G1 and G2, consisting of C4 and a new vertex adjacent to two adjacent and nonadjacent vertices, respectively, of C4. For n = 6 there exist five extremal graphs and so on. The following result was proved in [10]: Theorem 3.2 ([10]). If k ≥ 2, n ≥ 2k + 2 and G ∈ Ukn , then M1(G) ≥ 4n + 14, with equality if and only if G = Fn,k. An extension of this result is: Theorem 3.3. For k ≥ 2 and n ≥ 2k + 2, if G ∈ Ukn then we have Hf (G) ≥ 4f(3) + (n− 6)f(2) + 2f(1). The equality is reached if and only if G = Fn,k. Proof. Let G ∈ Ukn such that Hf (G) is minimum. By Definition 1.2, there exists a k- quasi-vertex set, which is a subset Vk ⊆ V (G) with cardinality k such that G − Vk is a unicyclic graph but for every subset Vk−1 of cardinality k−1 of V (G), the graph G−Vk−1 is not unicyclic. Let Wn−k = G − Vk. If there exists a vertex v ∈ Vk which is adjacent with a single vertex from Wn−k, then Vk−1 = Vk − v has the property that G − Vk−1 is unicyclic, which contradicts the Definition 1.2. It follows that every vertex of Vk is not adjacent to any vertex of Wn−k or is adjacent to at least two vertices from Wn−k. Suppose that the subgraph G[Vk] has r ≥ 1 connected components A1, A2, . . . , Ar. Since G is connected, we deduce that in each component there is at least one vertex which is adjacent with at least two vertices from Wn−k. Indeed, since G is connected, it follows that for every i, 1 ≤ i ≤ r, in component Ai of G[Vk] there is a vertex vi which is adjacent to at least one vertex in Wn−k. If vi would be adjacent to a single vertex in Wn−k, this would contradict the property that every vertex of Vk is not adjacent to any vertex or is adjacent to 6 Art Discrete Appl. Math. 5 (2022) #P1.02 at least two vertices of Wn−k. Each component i has at least |Ai| − 1 edges and equality holds if and only if this component is a tree. It follows that the number of edges of G having at least one end in Vk is at least 2r+ ∑r i=1(|Ai|−1) = 2r+k−r = k+r ≥ k+1. Since Wn−k is a unicyclic graph it has n − k edges. We get that the number of edges of G is at least k + 1 + n − k = n + 1. Equality holds only if r = 1 and G[Vk] has exactly |Vk| − 1 edges, i.e., G[Vk] is a tree. In other words, this happens when G[Vk] is a tree with k vertices and exactly one vertex, say w of this tree is adjacent with exactly two vertices of Wn−k. In this case G is bicyclic, having e(G) = n+ 1 edges. Now the proof splits into the following two cases: Case 1. e(G) = n + 1 and Case 2. e(G) ≥ n+ 2. Case 1. In this case G has at least one pendant vertex, since G[Vk] is a tree with k ≥ 2 vertices and exactly one vertex of this tree, denoted by w is adjacent to two vertices from Wn−k. Other four subcases may hold: Subcase 1.1. G has one pendant vertex. Subcase 1.2. G has two pendant vertices. Subcase 1.3. G has three pendant vertices. Subcase 1.4. G has at least four pendant vertices. Subcase 1.1. Since G is a connected bicyclic graph of size n + 1 with one pendant vertex one obtains that G[Vk] is a path w, y1, y2, . . . , yk−1 and dG(w) = 3, dG(y1) = 2, . . . , dG(yk−2) = 2 and dG(yk−1) = 1. The degree sequence may be π1 = (3 3, 2n−4, 1), π2 = (4, 3, 2 n−3, 1) or π3 = (5, 2 n−2, 1). We shall prove that in all cases G 6∈ Ukn . If the degree sequence is π1 then G contains two cycles Cr and Cs having a common path P = u, . . . , v where u 6= v or Cr and Cs are vertex disjoint and they are joined by P and w 6= u, v. In this case we can find a vertex z 6= w, y1, y2, . . . , yk−1, u, v such that if Vk−1 = {y2, . . . , yk−1, z} then G − Vk−1 is unicyclic, a contradiction. If the degree sequence of G is π2, then Cr and Cs have a common vertex and a similar conclusion as for π1 holds. In case of π3, w coincides with the common vertex of Cr and Cs, but in this situation G − Vk is not a unicyclic graph being disconnected, which contradicts the hypothesis. Subcase 1.2. In this case G has two pendant vertices. We further prove that the unique graph in this situation belonging to Ukn is Fn,k. We consider other two subcases: Subcase 1.2.1. G[Vk] is a path, as in the subcase 1.1. Subcase 1.2.2. The tree G[Vk] has two pendant vertices and the vertex w is adjacent to two vertices of Wn−k. Subcase 1.2.1. In this case, as in the subcase 1.1, G consists of two cycles, Cr and Cs, where r, s ≥ 3, a path P = u, . . . , v connecting cycles Cr and Cs or being a common path of Cr and Cs and a path w, y1, . . . , yk−1. Since G has two pendant vertices there exists another path having an end denoted by q 6= w, where w, q ∈ V (Cr) ∪ V (Cs) ∪ V (P ). If cycles Cr and Cs are disjoint or max{r, s} ≥ 4 and cycles have a common path or cycles have only a common vertex we can always find a vertex z such that G − Vk−1 is unicyclic, where Vk−1 = {y2, . . . , yk−1, z}, a contradiction. The remaining case is when Cr = Cs = C3 and the cycles have a common edge. In this last case G = Fn,k ∈ U k n . Subcase 1.2.2. In this situation G[Vk] consists of two paths having a common vertex w or it contains one vertex of degree three different from w. In both cases we can find a subset X ⊂ Vk of cardinality k−2 such that G[Vk−X] is an edge with an end w. As in the previous cases, G contains two cycles Cr and Cs having a common path P = u, . . . , v or Cr and Cs are vertex disjoint and they are joined by P , or the cycles have only a common vertex (when u = v), w ∈ V (Cr) ∪ V (Cs) ∪ V (P ) and w 6= u, v. In this case we can find a vertex z 6∈ Vk ∪ {u, v} such that G − Vk−1 is unicyclic, where Vk−1 = X ∪ {z}, a contradiction. I. Tomescu: Minimizing vertex-degree function index 7 Subcase 1.3. Since k ≥ 2 we have n ≥ 2k+2 ≥ 6. We shall consider the cases n = 6; n = 7 and n ≥ 8. If n = 6 then min x∈D3 6,14 F (x) is reached for x∗ = (42, 3, 13) by Lemma 2.1 and Hf (G) ≥ F (x ∗) = 2f(4)+ f(3)+ 3f(1) > Hf (F6,2) = 4f(3)+ 2f(1), a contradiction, since this inequality is equivalent to 2f(4) + f(1) > 3f(3). (3.1) Since f is strictly convex by Jensen inequality we get f(4) + f(2) > 2f(3) and by Lemma 2.1 we deduce f(4) + f(1) > f(3) + f(2). By summing up these inequalities we find (1). Similarly, for n = 7 we deduce Hf (G) ≥ minx∈D3 7,16 F (x). This minimum is reached for x∗ = (4, 33, 13) by Lemma 2.1 and F (x∗) = f(4) + 3f(3) + 3f(1) > Hf (F7,2) = 4f(3) + f(2) + 2f(1), a contradiction since this inequality is equivalent to f(4) + f(1) > f(3) + f(2). If n ≥ 8 then min x∈D3 n,2n+2 F (x) is reached for x∗ = (35, 2n−8, 13) by Lemma 2.1 and Hf (G) ≥ F (x ∗) = 5f(3) + (n − 8)f(2) + 3f(1) > Hf (Fn,k) = 4f(3) + (n − 6)f(2)+2f(1), a contradiction, since this inequality is equivalent to f(3)+f(1) > 2f(2). The last inequality follows by Jensen inequality since f is strictly convex. Subcase 1.4. Suppose that G has s ≥ 4 pendant vertices. By Lemma 2.3 we have minx∈Ds n,2n+2 F (x) > min x∈D3 n,2n+2 F (x), which implies Hf (G) > Hf (Fn,k), which is again a contradiction. Case 2. If e(G) = n + 2 then minx∈Dn,2n+4 F (x) is reached for x ∗ = (34, 2n−4) by Lemma 2.1. We get Hf (G) ≥ F (x ∗) = 4f(3) + (n − 4)f(2) > Hf (Fn,k) = 4f(3) + (n− 6)f(2) + 2f(1), or 2f(2) > 2f(1), which is true because f is strictly increasing and this fact contradicts the hypothesis about the minimality of G. If e(G) ≥ n + 3 Lemmas 2.4 and 2.2 yield that Hf (G) ≥ minx∈Dn,2n+6 F (x) > minx∈Dn,2n+4 F (x) > Hf (Fn,k), which contradicts the hypothesis. Thus, the theorem holds. 4 Concluding remarks In this paper we have solved a minimization problem concerning the vertex-degree function index Hf (G) in the class of k-generalized quasi-unicyclic graphs of order n for k ≥ 1 and n ≥ 2k + 2 if the function f is strictly increasing and strictly convex. This includes the case of general first Zagreb index 0Rα(G) if α > 1, second multiplicative Zagreb index∏ 2 (G) and sum lordeg index SL(G). For general first Zagreb index 0Rα(G) this problem was solved in [10] by other means for α = 2. By similar methods one can prove that for k ≥ 2 and n ≥ 3k the k-generalized quasi- trees of order n which reach the minimum of Hf (G) consist of three vertex-disjoint paths x1, . . . , xk; y1, . . . , yp and z1, . . . , zq , where p, q ≥ k and p+q = n−k and three additional edges x1y1, y1z1 and z1x1. This minimum equals 3f(3)+(n−6)f(2)+3f(1). The same result has been already obtained in [11] by using different arguments. ORCID iDs Ioan Tomescu https://orcid.org/0000-0002-4747-9843 8 Art Discrete Appl. Math. 5 (2022) #P1.02 References [1] H. Deng, A unified approach to the extremal Zagreb indices for trees, unicyclic graphs and bicyclic graphs, MATCH Commun. Math. Comput. Chem. 57 (2007), 597–616, https:// match.pmf.kg.ac.rs/content57n3.htm. [2] M. Ghorbani, M. Songhori and I. Gutman, Modified narumi-katayama index, Kragujevac J. Sci. 34 (2012), 57–64, http://www.pmf.kg.ac.rs/kjs/index.php?option= com_content&view=article&id=41&Itemid=4. [3] I. Gutman, Graphs with smallest sum of squares of vertex degrees, Kragujevac J. Math. 25 (2003), 51–54, https://imi.pmf.kg.ac.rs/kjm/en/index.php?page=vol_ 25. [4] I. Gutman and N. Trinajstić, Graph theory and molecular orbitals. total ϕ-electron energy of alternant hydrocarbons, Chem. Phys. Lett. 17 (1972), 535–538, doi:10.1016/0009-2614(72) 85099-1. [5] W. R. Inc., Wolfram alpha, champaign, IL, 2020, https://www.wolframalpha.com. [6] A. Iranmanesh, M. 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Chem. 55 (2006), 427–438, https: //match.pmf.kg.ac.rs/content55n2.htm. ISSN 2590-9770 The Art of Discrete and Applied Mathematics 5 (2022) #P1.03 https://doi.org/10.26493/2590-9770.1374.bf0 (Also available at http://adam-journal.eu) Distributions of restricted rotation distances Sean Cleary* Department of Mathematics, The City College of New York and the CUNY Graduate Center, City University of New York, New York, NY 10031, USA Haris Nadeem Department of Computer Science, The City College of New York, City University of New York, New York, NY 10031, USA Received 17 July 2020, accepted 26 October 2020, published online 21 March 2022 Abstract Rotation distances measure the differences in structure between rooted ordered binary trees. The one-dimensional skeleta of associahedra are rotation graphs, where two vertices representing trees are connected by an edge if they differ by a single rotation. There are no known efficient algorithms to compute rotation distance between trees and thus distances in rotation graphs. Limiting the allowed locations of where rotations are permitted gives rise to a number of notions related to rotation distance. Allowing rotations at a minimal such set of locations gives restricted rotation distance. There are linear-time algorithms to compute restricted rotation distance, where there are only two permitted locations for rota- tions to occur. The associated restricted rotation graph has an efficient distance algorithm. There are linear upper and lower bounds on restricted rotation distance with respect to the sizes of the reduced tree pairs. Here, we experimentally investigate the expected restricted rotation distance between two trees selected at random of increasing size and find that it lies typically in a narrow band well within the earlier proven linear upper and lower bounds. Keywords: Random binary trees, rotation distances. Math. Subj. Class.: 05C05, 68P05, 05C12 *Corresponding author. This material is based upon work supported by the National Science Foundation under Grant No. #1417820 http://cleary.ccnysites.cuny.edu E-mail addresses: cleary@sci.ccny.cuny.edu (Sean Cleary), haris.nadeem.bsc@gmail.com (Haris Nadeem) cb This work is licensed under https://creativecommons.org/licenses/by/4.0/ 2 Art Discrete Appl. Math. 5 (2022) #P1.03 1 Introduction Binary trees capture hierarchical relationships in a wide range of settings. For example, when there is an order on leaves, binary search trees have broad use, see Knuth [16]. Sim- ple local changes, called rotations, at nodes give rise to rotation distance and the rotation graph, where two trees are connected by an edge in the rotation graph if they differ by a single rotation. There are no known algorithms for computing rotation distance exactly in polynomial time, though there are some estimation algorithms which run in polynomial time of Baril and Pallo [1] and Cleary and St. John [11] and the problem is known to be fixed-parameter tractable, see Cleary and St. John [10]. But there is thus no known algo- rithm for calculating distances efficiently in rotation graphs. Given the apparent difficulty of computing rotation distance exactly, there are a number of related notions that have been considered, such as restricted rotation distance of Cleary [4], right-arm rotation distances of Cleary and Taback [12] and level-restricted rotation distances of Luccio, Pagli, and Mesa Enriquez [18]. In each of these, the locations where rotations are permitted is restricted in some way. If we only allow rotations either all along the right arm of the tree or only at the root and right child of the root, then there are linear-time algorithms for computing the resulting right-arm rotation and restricted rotation distances, see Cleary [4] and Cleary and Taback [12]. Thus we can explore the properties of distance in the related graph as- sociated to restricted rotation distance. Here, we experimentally study the distributions of restricted rotation distance between randomly selected trees of increasing size and find that the distances appear to grow on average quite linearly with size with a linear coefficient of between three and four, with the distances distributed centrally arranged near the average in relatively narrow spreads. This gives insight into the distribution of distances between pairs of trees in the re- stricted rotation graph which is not presently feasible at this scale for the rotation graph, and equivalently into the distribution of distances between vertices of the restricted rotation graph. 2 Background and definitions In the following, by tree we mean a rooted binary tree where each node has either zero or two children, a left child and a right child. Such trees are sometimes called 0-2 trees or proper binary trees, see Knuth [16]. A node with no children is a leaf, and a node with two children is an internal node. The size of a tree T is the number of internal nodes in T . We number the n+ 1 leaves in a tree with n internal nodes from left to right from 0 to n. We encode binary trees via the standard encoding of a preorder traversal where an internal node is denoted by 1 and a leaf node by 0. So the left hand tree in Figure 1 has encoding 1101100101000 and the right hand tree has encoding 1101110001000. A rotation at a node P is the operation depicted in Figure 1 where one grandchild of P is promoted to become a child of P , one child is demoted to become a grandchild, and where one grandchild’s parent node is switched in an order-preserving way. In terms of encodings, a left rotation at a node can be regarded as a string substitution of the form . . . 1x1yz . . . becoming . . . 11xyz . . . where x, y, and z are encodings of subtrees, with a right rotation the inverse string substitution operation. Given two trees S and T of size n, Culik and Wood [13] showed that there is always at least one sequence of rotations transforming S to T and thus defined rotation distance. Rotation distance between S and T , denoted d(S, T ), is the minimum number of rotations S. Cleary and H. Nadeem: Distributions of restricted rotation distances 3 P C C ′ P ′ Figure 1: An example of a left rotation at node P , with a rotation promoting child node C to C ′ and demoting parent node P to P ′. The left hand tree has encoding 1101100101000 and the right hand tree is 1101110001000. All other nodes are unaffected by the rotation at P . Right rotation at C ′ is the inverse operation, taking the tree on the right to the tree on the left. needed to transform S to T where the rotations are permitted at any nodes present. We need not have rotations permitted at every node to transform any tree to any other- a minimal set of permitted rotations has size 2, as described by Cleary [4]. We take those two locations to be the root and the right child of the root, giving restricted rotation distance between S and T , denoted dR(S, T ), as the minimum number of rotations needed to transform S to T where the rotations are permitted only at the root node (always present) and the right child of the root node, if present. The rotation graph RG(n) of size n is the graph whose vertices are rooted binary trees of size n and where two vertices are connected by an undirected edge if there is a single rotation transforming the one tree to the other. The rotation graph is the one-dimensional skeleton of the associahedron of the appropriate size. The notions foundational to the geometric realization of associahedra go back to Tamari [22] and Stasheff [21] and were first published concretely by Lee [17]. Here, we consider distances in the related restricted rotation graph RRG(n) where the vertices are again trees and an edge is present between trees S and T if they differ by a single rotation at either the root or the right child of the root. We note that the restricted rotation graph does not enjoy the same set of symmetries as the ordinary rotation graph- in fact, not even the valence is the same for every vertex. Most vertices have valence 4, corresponding to left and right rotations at the root and right child of the root, but some have smaller valence if the right child of the root is not present or if rotation in a direction is not possible at one of those two nodes. There are vertices of valence 1 in this graph, whereas the rotation graph has high symmetry, arising from the dihedral symmetries inherited from the full associahedron. A tree pair (S,T) is a pair of trees of the same size. A tree pair (S, T ) is unreduced if there are nodes in both S and T such that leaf node children numbered as i and i + 1, via preorder traversal of the tree, are the same in both trees. A reduction in a tree pair is the removal of such a pair of identically numbered siblings in each tree, replacing them with a single leaf i, and then renumbering to get a new tree pair (S′, T ′) of one smaller size, see Cannon, Floyd and Parry [3] for background as well as for connections with Thompson’s group F . A tree pair (S, T ) is said to be reduced if there are no possible reductions. Note that for both rotation distance and restricted rotation distances, the distances between S and T are the same as between the representatives of their reduced tree pair S′ and T ′ as the 4 Art Discrete Appl. Math. 5 (2022) #P1.03 5 0 0 3 1 1 2 2 3 4 4 5 6 Figure 2: A tree of size 6 with leaves numbered in red from 0 to 6 and with internal nodes numbered from 0 to 5. Nodes 0 and 5 are left nodes and all other internal nodes are interior nodes. same sequence of rotations will perform the required transformations, see [4]. The binary address of a node in a tree is a sequences of 0’s and 1’s representing the path from the root to the node with a 0 for each left child and 1 for each right child. For example, the address of node C in Figure 1 in the left hand tree is 011 as the path from the root to C is a left edge followed by two right edges. A right node of a tree is one whose binary address consists only of 1’s and has at least one 1. A left node is one whose binary address consists only of 0’s. The root node is thus a left node but not a right node. All non-right and non-left nodes of a tree are interior nodes. We number nodes with an in-order traversal of the tree, and a node pair from a tree pair (S, T ) is a pair of nodes numbered the same in such traversals. Figure 2 shows leaves and nodes numbered in the resulting left-to-right in-order traversals of leaves and interior nodes respectively. To calculate restricted rotation distance, we use the methods of Fordham [15]. His methods were designed to calculate word length exactly in Thompson’s group F with re- spect to the generating set {x0, x −1 0 , x1, x −1 1 }, and give minimal length representatives of a word with respect to that generating set. The generator x0 corresponds to right rotation at the root, with x−10 correspondingly the inverse which is a left rotation at the root. Similarly, x1 and its inverse correspond to rotations at the right child of the root. So word length in F translates into restricted rotation distance between trees, as described in [4, 12]. Fordham’s method takes as input two trees forming a reduced tree pair, and classifies each interior node as one of seven types as follows: • L0: The first node on the left side of the tree. • LL: Any left node other than the leftmost node. • I0: An interior node with no right child. • IR: An interior node with a right child. • RI : Any right node numbered k whose immediate successor node k+1 is an interior node. S. Cleary and H. Nadeem: Distributions of restricted rotation distances 5 R0 RNI RI Ll I0 IR R0 0 2 2 1 1 3 RNI 2 2 2 1 1 3 RI 2 2 2 1 3 3 Ll 1 1 1 2 2 2 I0 1 1 3 2 2 4 IR 3 3 3 2 4 4 Table 1: Weights for caret pairs by caret pair types. . • RNI : A right node which is not of type RI but for which there is some successor interior node. • R0: A right node with no successor interior node. A primary result of Fordham [15] is that the word length |w| in Thompson’s group F with respect to the standard finite generating set can be calculated by classifying node pairs into those seven types and summing the totals from the Table 1. Note that the first node pair is always of type (L0, L0) and adds weight 0, and the single L0 in each tree must necessarily be paired, so the caret type L0 is not listed in Table 1. As described [12], since all non-L0 carets contribute at least one to word length (and thus at least one to restricted rotation distance), and since a caret can contribute at most 4 to word length, analysis of caret types and configurations give that for two trees forming a reduced pair of size n, the restricted rotation distance between lies between n−1 and 4n−8 and is sharp for n ≥ 3. Fordham’s method goes further and can be in fact used to not only find restricted rotation distances, but also to find and enumerate all possible minimal length paths between the relevant trees. We note that there have been computations to calculate the number of words of Thompson’s group F of increasing word length with respect to the standard generating set (and thus restricted rotation distances) of increasing sizes by Burillo, Cleary, and Weist [2] and Elder, Fusy, and Rechnitzer [14], with the latter giving the first 1500 terms of the OEIS sequence A156945 [20] which are the number of elements of increasing word length size. The relationship between word length size and tree size is linear but knowing word length gives only linear bounds on the tree size. 3 Distributions of restricted rotation distance We study computationally the distribution of restricted rotation distance between rooted binary trees. This is equivalent to analyzing distances in the restricted rotation graph RRG(n) between vertices. Work of Cleary and Maio [6] analyzes distributions of ordinary rotation distances. Here, we address similar questions for restricted rotation distances. The general question is: given two trees of the same size n, what is the expected restricted ro- tation distance between them? We anticipate that on average, larger tree pairs have larger distances between them, but we would like to estimate the rates of growth as well as the dispersal. Work of Cleary and Taback [12] gave sharp lower and asymptotically sharp up- per bounds for restricted rotation distances, and we find that the vast majority of instances are clustered quite centrally and not near the bounds. 6 Art Discrete Appl. Math. 5 (2022) #P1.03 Tree size range # sampled Avg. red. frac. Avg. RRD ratio 10–19 138999 0.907533 2.24473 20–29 161500 0.917172 2.64333 30–39 150500 0.920593 2.83326 40–49 133000 0.922663 2.9421 50–59 144000 0.923896 3.00793 60–69 134500 0.924459 3.05513 70–79 129000 0.924884 3.08993 80–89 119000 0.925221 3.1151 90–99 118500 0.925659 3.13679 100–199 685191 0.92646 3.19676 200–299 509390 0.927268 3.24813 300–399 310962 0.927496 3.26887 400–499 111460 0.927678 3.27999 500–599 89580 0.92783 3.28727 600–699 100600 0.927795 3.29198 700–799 102600 0.9279 3.29606 800–899 43600 0.927921 3.29866 900–999 45450 0.928027 3.30121 1000–1249 89200 0.928008 3.30416 1250–1499 86000 0.928002 3.3069 1500–1749 99000 0.928071 3.30908 1750–1999 35600 0.928121 3.31039 2000–2249 20000 0.928089 3.31145 2250–2499 19800 0.928089 3.31235 2500–2749 18764 0.928117 3.31311 2750–2999 13900 0.928124 3.31386 3000–3249 12124 0.928094 3.31407 3250–3499 8044 0.928185 3.31517 3500–3999 3072 0.928024 3.31562 4000–4500 800 0.928023 3.31568 Table 2: Tree pair restricted rotation distances for unreduced tree pairs. Given are the average fractions of the reduced tree pairs size of the originally generated tree pair size and the average ratio of restricted rotation distance to the generated tree pair size. S. Cleary and H. Nadeem: Distributions of restricted rotation distances 7 Figure 3: Restricted rotation distance vs. raw size for randomly selected tree pairs of increasing sizes. Figure 4: Restricted rotation distance vs. reduced size for randomly selected tree pairs of increasing sizes, by the size of the resulting reduced tree pair after reduction. 8 Art Discrete Appl. Math. 5 (2022) #P1.03 Tree size range Number of tree pairs sampled Average RRD size 10–19 168846 2.609 20–29 166650 2.96244 30–39 145364 3.12548 40–49 152971 3.22228 50–59 144317 3.28264 60–69 139509 3.32627 70–79 132652 3.35818 80–89 126454 3.38269 90–99 94370 3.40162 100–199 700470 3.45925 200–299 504029 3.50732 300–399 272408 3.52717 400–499 116513 3.53867 500–599 97243 3.54577 600–699 107923 3.55041 700–799 74740 3.55356 800–899 48099 3.55662 900–999 40737 3.55859 1000–1249 94865 3.56172 1250–1499 100074 3.56451 1500–1749 77950 3.56616 1750–1999 22109 3.56769 2000–2249 21630 3.56872 2250–2499 20622 3.5695 2500–2749 15851 3.57045 2750–2999 13158 3.57073 3000–3249 8342 3.57162 3250–3499 2607 3.57268 3500–3999 821 3.57276 4000–4500 717 3.57285 Table 3: Tree pair restricted rotation distances divided by tree pair size, for reduced tree pairs of increasing size ranges. There are examples with ratios as small as 1 and approach- ing 4 for all n. S. Cleary and H. Nadeem: Distributions of restricted rotation distances 9 We sample rooted binary tree pairs at random using Remy’s algorithm [19] for each tree, which guarantees a uniform randomly generated tree of size n. Work on the asymp- totic density of isomorphism classes of subgroups of Thompson’s group F of Cleary, El- der, Rechnitzer and Taback [5] addresses the question of the expected fraction of tree pairs which are reduced, and later work of Cleary, Rechnitzer and Wong [9] describes the asymp- totics of the expected sizes of reduced components of tree pairs. Here, we study two main questions: • Given two trees selected at random of size n, what is the expected restricted rotation distance between them? • Given a reduced tree pair of size n, what is the expected restricted rotation distance between the pair? We also seek to understand the deviations from the means of these distances. We gen- erated tree pairs (S, T ) at random, then calculated the reduced representatives (S′, T ′) of each tree pair, then the corresponding restricted rotation distance, dR(S, T ) = dR(S ′, T ′), which are the same as the reductions reflect commonality which does not change the dis- tance. We note that generating reduced tree pairs of a specified size is not as feasible as gen- erating tree pairs generally. As described in [9] and [5], a tree pair selected at random is likely to have a number of reductions, and the resulting reduced representative is on average about 10% smaller. But of course there is a (increasingly small) chance that the generated tree is already reduced, and also a (vanishingly small) chance that it reduces all the way down to the empty tree pair. Cleary, Rechnitzer and Wong [9] analyze some properties of the distribution of the resulting sizes of reduced tree pairs. Cleary and Maio [8] have an algorithm which guarantees to produce not only a reduced tree pair of a specified size, but is difficult in an additional sense as well– not having any obvious initial first moves along minimal length paths. Unfortunately, though that algorithm is efficient, it does not choose uniformly from among the possible ones. The particular number of such difficult instances is not even known precisely, though Cleary and Maio [7] calculate the number of such cases exhaustively for small sizes and approximately for larger ones. By generating large families of trees across a range of sizes and then performing reduc- tions, we get a range of reduced tree pairs to consider and analyze. The resulting reduced tree pairs are necessarily smaller than the generated, possibly reducible, tree pairs, but since the number of reductions vary, there is a dispersal in the resulting sizes of the reduced tree pairs. That is, if we generate 1400 tree pairs of size 1000, the smallest resulting reduced pair may be 896 and the largest 955, with a mean and median of about 928 with the most commonly occurring being 929 with 73 occurrences. The tree pairs were generated of fixed sizes, often 500 apart. Thus, after reductions, these sizes would reduce to different extents which may lead to gaps in the resulting reduced sizes. So we generate many examples across a range of increasing sizes in an effort to get representative samples across a broad range. 4 Experiments and discussion For the computational experiments we described, we generated about 3.6 million tree pairs of sizes ranging from 10 to 4400. We reduced each tree pair to a reduced representative, and then calculated the restricted rotation distances using Fordham’s method. 10 Art Discrete Appl. Math. 5 (2022) #P1.03 To compare average restricted rotation distances across a range of sizes, we consider the RRD ratio, which for a tree pair (S, T ) of size n is dR(S, T )/n. This gives a somewhat normalized measure of the typical contribution of tree carets to the restricted rotation dis- tance and a sense of how quickly the restricted rotation distance grows with increased tree size. We note that trees realizing the lower bound of restricted rotation distance from [12] would have an RRD ratio limiting to 1, and those realizing the upper bound would have an RRD ratio limiting to 4. Table 2 tabulates the results across a range of unreduced sizes, with Figure 3 plotting the results for these unreduced sampled tree pairs. We see tight linear behavior of distance with respect to raw size, despite the fact that the amount of reductions varies considerably and the resulting sizes have a large influence on the corresponding distances. Owing to the time of computation, larger size tree pairs were not sampled as exten- sively as the smaller ones. In Figure 3 the sampling increments of size 500 are visible, and in Figure 4 the fact that those sizes have dispersed somewhat as the reductions in size vary is visible. The fraction of common edges in a more general sense was computed asymptot- ically by Cleary, Rechnitzer and Wong [9] to be 6− 16 π ∼ 0.907, so the observed fractions of reduced size from generated size of about 0.928 is consistent with that. That asymptotic analysis allowed reductions of internal common edges in addition to the peripheral ones relevant to the tree reductions considered here. In the remaining analyses, we restrict our attention to the resulting generated reduced tree pairs as the distances are more tightly related to the sizes after reduction. Table 3 tabulates the distances observed across a range of reduced tree pair sizes, and Figure 4 plots these results. We can again see tight linear behavior, where the reduced trees have on average larger rotation distances and a smaller spread in the observed reduced instances relative to the unreduced sizes. The examples from Cleary and Taback [12] giving the bounds of n− 1 ≤ dR(S, T ) ≤ 4n− 8 are clearly quite constrained, as the vast majority of the sampled lengths lie close to about 3.57n, well away from the upper and lower bounds. We note that in both cases, the maximum possible distances (about 4 times the size) and minimal possible distances (one less than the size) lie far away from the randomly-generated instances. This is not surprising as those examples to show the sharpness of the bounds were carefully constructed in a very specific manner to realize those bounds. We note that the only entries in Table 1 that contribute 4 to restricted rotation distance are (IR, I0) and (IR, IR) which involve interior carets being paired with interior carets. Given that the average distances are well above 3, such caret pairings are necessarily quite common and cannot occur in the examples realizing the lower bounds of n− 1. Not surprisingly, given the strong linear behavior observed, a fitted linear model agrees with the sampled data exceptionally well, giving dR(S, T ) ∼ 3.31941n − 17.0321 for re- stricted rotation distance in terms of unreduced tree pair sizes n, and dR(S, T ) ∼ 3.57612n− 16.1551 correspondingly for reduced tree pairs of size n. We see that the standard deviations of the observed RRD ratios of restricted rotation distance are relatively small and stable, dropping steadily from about 0.33 for the smallest size trees sampled, to about 0.025 for tree sizes in the hundreds, then dropping to about 0.01 for tree sizes in the hundreds, with an observed average standard deviation of ratios of 0.009 for the largest tree sizes sampled. These are for the normalized ratios- the standard deviations do increase with size, albeit somewhat more slowly. The distributions of restricted rotation for reduced tree pairs of a fixed size show an S. Cleary and H. Nadeem: Distributions of restricted rotation distances 11 Figure 5: Distribution of restricted rotation distances for 24,067 randomly-produced re- duced tree pairs of size 19. The sample mean is about 53.5 and the sample standard devia- tion is about 4.58. Figure 6: Distribution of restricted rotation distances for 19,307 randomly-produced re- duced tree pairs of size 29. The sample mean is about 88.5 and the sample standard devia- tion is about 5.45. 12 Art Discrete Appl. Math. 5 (2022) #P1.03 Figure 7: Distribution of restricted rotation distances for 17,196 randomly-produced re- duced tree pairs of size 47. The sample mean is about 152.3 and the sample standard deviation is about 6.36. Figure 8: Distribution of restricted rotation distances for 14,155 randomly-produced re- duced tree pairs of size 68. The sample mean is about 227.1 and the sample standard deviation is about 7.20. S. Cleary and H. Nadeem: Distributions of restricted rotation distances 13 Figure 9: Distribution of restricted rotation distances for 11,258 randomly-produced re- duced tree pairs of size 120. The sample mean is about 412.6 and the sample standard deviation is about 8.79. Figure 10: Distribution of restricted rotation distances for 8266 randomly-produced re- duced tree pairs of size 238. The sample mean is about 834.3 and the sample standard deviation is about 11.4. 14 Art Discrete Appl. Math. 5 (2022) #P1.03 Figure 11: Distribution of restricted rotation distances for 1200 randomly-produced re- duced tree pairs of size 714. The sample mean is about 2536.4. and the sample standard deviation is about 18.4. A normal distribution with the same mean and standard deviation is superimposed for comparison. approximately normal shape, slightly skewed to the left for smaller sizes but less so for larger sizes. Here, we chose a few sizes for which there were a reasonable number of observed instances, shown in Figures 5 to Figure 11. These distributions have characteristic normal shapes, and further suggest that the extremely short and extremely long cases shown earlier to be possible are exceptionally rare. The vast majority of randomly-selected cases lie in relatively narrow bands concentrated on a line well away from the lowest and highest possible bounds. For the largest million tree pairs sampled, less than 175,000 were more than 1% away from the distance predicted by the linear model, and all but 1054 were within 3% of the linear prediction, with the largest observed deviation from the linearly fitted model being less than 5% away from the predicted distance. For the size 714 case illustrated in Figure 11, the lower bound of 713 is nearly 100 standard deviations below the sample mean and the upper bound of 2848 is about 17 standard deviations above the sample mean of 2536. Thus we have developed some understanding of typical behavior of distances in re- stricted rotation graphs RRG(n) for a decent range of distances. Note that analyzing the corresponding questions for the rotation graphs RG(n) are not presently feasible beyond size about 20, even experimentally, due to the difficulty of computing ordinary rotation dis- tance exactly, where the best known algorithms have exponential running time in the size of the trees. ORCID iDs Sean Cleary https://orcid.org/0000-0002-3123-8658 Haris Nadeem https://orcid.org/0000-0001-9283-2805 S. Cleary and H. Nadeem: Distributions of restricted rotation distances 15 References [1] J.-L. Baril and J.-M. 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ISSN 2590-9770 The Art of Discrete and Applied Mathematics 5 (2022) #P1.04 https://doi.org/10.26493/2590-9770.1307.cb4 (Also available at http://adam-journal.eu) Product irregularity strength of graphs with small clique cover number* Daniil Baldouski University of Primorska, FAMNIT, Glagoljaška 8, 6000 Koper, Slovenia Received 19 July 2019, accepted 04 February 2021, published online 21 March 2022 Abstract For a graph X without isolated vertices and without isolated edges, a product-irregular labelling ω : E(X) → {1, 2, . . . , s}, first defined by Anholcer in 2009, is a labelling of the edges of X such that for any two distinct vertices u and v of X the product of labels of the edges incident with u is different from the product of labels of the edges incident with v. The minimal s for which there exists a product irregular labeling is called the product irregularity strength of X and is denoted by ps(X). Clique cover number of a graph is the minimum number of cliques that partition its vertex-set. In this paper we prove that connected graphs with clique cover number 2 or 3 have the product-irregularity strength equal to 3, with some small exceptions. Keywords: Product irregularity strength, clique-cover number. Math. Subj. Class.: 05C15, 05C70, 05C78 1 Introduction Throughout this paper let X be a simple graph, that is, a graph without loops or multiple edges, without isolated vertices and without isolated edges. Let V (X) and E(X) denote the vertex set and the edge set of X , respectively. Let ω : E(X) → {1, 2, . . . , s} be an integer labelling of the edges of X . Then the product degree pdX(v) of a vertex v ∈ V (X) in the graph X with respect to the labelling ω is defined by pdX(v) = ∏ v∈e ω(e). *The author would like to express his gratitude to an anonymous reviewer for carefully reading the manuscript, and for several helpful suggestions that improved the quality of this paper. E-mail address: d.baldovskiy@mail.ru (Daniil Baldouski) cb This work is licensed under https://creativecommons.org/licenses/by/4.0/ 2 Art Discrete Appl. Math. 5 (2022) #P1.04 If the graph X is clear from the context, then we will simply use pd(v). A labelling ω is said to be product-irregular, if any two distinct vertices u and v of X have different corresponding product degrees, that is, pdX(u) 6= pdX(v) for any u and v in V (X) (u 6= v). The product irregularity strength ps(X) of X is the smallest positive integer s for which there exists a product-irregular labelling ω : E(X) → {1, 2, . . . , s}. This concept was first introduced by Anholcer in [1] as a multiplicative version of the well-studied concept of irregularity strength of graphs introduced by Chartrand et al. in [4] and studied later quite extensively (see for example [3, 7, 8, 11]). A concept similar to product-irregular labelling is the product anti-magic labeling of a graph, where it is required that the labeling ω is bijective (see [9, 12]). It is clear that every product anti- magic labeling is product-irregular. Another related concept is the so-called multiplicative vertex-colouring (see [13, 14]), where it is required that pd(u) 6= pd(v) for every pair of adjacent vertices u and v, while non-adjacent vertices can have the same product degrees. It is easy to see that every product-irregular labelling is a multiplicative vertex-colouring. In [1] Anholcer gave upper and lower bounds on product irregularity strength of graphs. The main results in [1] are estimates for product irregularity strength of cycles, in particular it was proved that for every n > 2 ps(Cn) ≥ ⌈ √ 2n− 1 2 ⌉, and that for every ε > 0 there exists n0 such that for every n ≥ n0 ps(Cn) ≤ ⌈(1 + ε) √ 2n ln n⌉. Anholcer in [2] considered product irregularity strength of complete bipartite graphs and forests. Anholcer proved that for two integers m and n such that 2 ≥ m ≥ n it holds ps(Km,n) = 3 if and only if n ≥ ( m+2 2 ) . The main result in [2] is about product irregularity strength of almost all forests F such that △(F ) = D for arbitrary integer D ≥ 3, n2 = 0, n0 ≤ 0 and n2 = 0 of the forest F with all pendant edges removed, where nd denotes the number of vertices of degree d. Anholcer proved that in this case ps(F ) = n1. In [5], Darda and Hujdurović proved that for any graph X of order at least 4 with at most one isolated vertex and without isolated edges we have ps(X) ≤ |V (X)|−1. Connections between product irregularity strength of graphs and multidimensional multiplication table problem was established, see [6, 10] for some results on multidimensional multiplication problem. It is easy to see that the lower bound for the product irregularity strength of any graph is 3. In this paper we will give some sufficient conditions for a graph to have product irregularity strength equal to 3. In particular we will prove that graphs of order at least 3 with clique-cover number 2 have product irregularity strength 3 (see Corollary 3.5), where clique cover number of a graph is the minimum number of cliques that partition the vertex set of the graph. Moreover, we will prove that for a connected graph such that its vertex set can be partitioned into 3 cliques of sizes at least 4 then its product irregularity strength is 3 (see Corollary 4.14). The paper is organized as follows. In section 2 we rephrase the definition of product- irregular labellings in terms of the corresponding weighted adjacency matrices and give some constructions that will be used for proving our main results. In section 3 we will determine the product irregularity strength of graphs with clique cover number 2, while in section 4 we study product irregularity strength of graphs with clique cover number 3. D. Baldouski: Product irregularity strength of graphs with small clique cover number 3 2 Product-irregular matrices In this section we will rephrase the definition of product irregular labelling of graphs using weighted adjacency matrices. We start with the definition of a weighted adjacency matrix. Definition 2.1. Let w be an integer labelling of the edges of a graph X of order n with V (X) = {v1, v2, . . . , vn}. Weighted adjacency matrix of X is n × n matrix M where Mij = w({vi, vj}) if vi and vj are adjacent and Mij = 0 otherwise. Definition 2.2 (Product-irregular matrices and product degree for matrices). Assume that we have weighted adjacency n × n matrix M (n ≥ 2). Then for a k-th row of a matrix M , denoted Mk, define pd(Mk) := ∏ Mk,i 6=0 Mk,i to be the product of all non-zero elements of the k-th row. We say that M is product-irregular if ∀i, j ∈ {1, 2, . . . , n} for i 6= j pd(Mi) 6= pd(Mj). We will work with matrices with entries aij ∈ {0, 1, 2, 3} therefore to simplify reading for a row v from matrix M if pd(v) = 2a · 3b then we will use notation pd(v) := (a, b). Also define pd(v)[1] := a and pd(v)[2] := b. Observation 2.3. A graph labelling is product-irregular if and only if the corresponding weighted adjacency matrix is product-irregular. Let n ≥ 4 and let Mn(x, y, z) be n× n matrix such that Mn(x, y, z) = (mij) where mij =    0, if i = j x, if j ≤ n− i+ 1 and i 6= j z, if (i, j) = (k, n) or (i, j) = (n, k) for k = ⌈n 2 ⌉+ 1 y, otherwise For example: M7(x, y, z) =           0 x x x x x x x 0 x x x x y x x 0 x x y y x x x 0 y y y x x x y 0 y z x x y y y 0 y x y y y z y 0           . (2.1) We will denote with A⊕B the direct sum of matrices A and B, that is A⊕B = ( A 0 0 B ) , where 0 denotes the zero matrix of appropriate size. 2.1 Properties of Mn Let xi, yi and zi be the number of x, y and z respectively appearing in the i-th row of matrix Mn(x, y, z). For fixed n with k we denote k := ⌈ n 2 ⌉ + 1. Then the rows of the matrix Mn(x, y, z) can be separated into 3 types: 1st type: (xk, yk, zk) = (⌈ n−1 2 ⌉, ⌈n 2 ⌉ − 2, 1), 4 Art Discrete Appl. Math. 5 (2022) #P1.04 2nd type: (xi, yi, zi) = (n − i, i − 1, 0) for i < k and (xi, yi, zi) = (n − i + 1, i − 2, 0) for n > i > k, 3rd type: (xn, yn, zn) = (1, n− 3, 1). We denote by m(i)(M) a row of type i for i ∈ {1, 2, 3} of matrix M , where M is matrix Mn(x, y, z) (if the matrix Mn(x, y, z) is clear from the context, then we will simply use m(i)). We start by proving the following nice property of matrix Mn. Proposition 2.4. If {x, y, z} is a set of distinct pairwise relatively prime integers, then Mn(x, y, z) is product irregular matrix for any n ≥ 4. Proof. Suppose contrary, that is there exist mi and mj (that are rows of matrix Mn(x, y, z)) for some i 6= j such that pd(mi) = pd(mj). There are 3 types of rows therefore it is enough to check the equality above not for all rows, but for all types of rows. Observe that for every i ∈ {1, 2, . . . , n} the sum xi + yi + zi = n − 1 and pd(mi) = pd(mj) for some i 6= j if and only if xi = xj , yi = yj and zi = zj . It follows that: 1. If pd(m(1)) = pd(m(3)) then (⌈ n−1 2 ⌉, ⌈n 2 ⌉ − 2, 1) = (1, n − 3, 1), so n = 3 which is a contradiction. 2. Since rows of second type have value 0 at 3rd coordinate and rows of first and third types have value 1 at 3rd coordinate, then pd(m(2)) 6= pd(m(i)) for i ∈ {1, 3}. 3. It is clear that (xi, yi, zi) 6= (xj , yj , zj) for i < k and k < j < n i.e. product degrees of different rows of type 2 are different. We were considering different rows, that means we did not have to consider pd(m(i)) = pd(m(i)) for every i ∈ {1, 3}. We will define 3 matrices of class Mn(x, y, z) for specific x, y and z. Assign matrix An := Mn(1, 2, 3), Bn := Mn(2, 3, 1) and Cn := Mn(3, 1, 2). 2.2 Properties of An ⊕ Bm Lemma 2.5. For every m ≥ n ≥ 4, An⊕Bm is product irregular if (n,m) 6∈ {(4, 4), (5, 5), (6, 6)}. Proof. Suppose contrary, that is there exist ai and bj (that are rows of matrices An and Bm respectively) for some i and j such that pd(ai) = pd(bj). There are 3 types of rows therefore it is enough to check all of the 9 possibilities for different types of rows: 1. If pd(a(1)) = pd(b(1)) then (⌈ n 2 ⌉ − 2, 1) = (⌈m−1 2 ⌉, ⌈m 2 ⌉ − 2) which contradicts with m ≥ n. 2. If pd(a(1)) = pd(b(2)) then (⌈ n 2 ⌉ − 2, 1) = (m − j, j − 1) or (⌈n 2 ⌉ − 2, 1) = (m− j + 1, j − 2) which contradicts with m ≥ n ≥ 4. 3. If pd(a(1)) = pd(b(3)) then (⌈ n 2 ⌉ − 2, 1) = (1,m − 3), so (n,m) = (5, 4) or (n,m) = (6, 4) which contradicts with m ≥ n. 4. If pd(a(2)) = pd(b(1)) then (i − 1, 0) = (⌈ m−1 2 ⌉, ⌈m 2 ⌉ − 2) or (i − 2, 0) = (⌈m−1 2 ⌉, ⌈m 2 ⌉ − 2), so m = 3 or m = 4, thus m = n = 4 which is a contradiction. D. Baldouski: Product irregularity strength of graphs with small clique cover number 5 5. If pd(a(2)) = pd(b(2)) then we have that in both possible cases ((i − 1, 0 = (m − j, j − 1)) and (i − 2, 0) = (m − j + 1, j − 2)) we get i = m ≥ n which is a contradiction. 6. If pd(a(2)) = pd(b(3)) then pd(a(2))[2] = 0 and pd(b(3))[2] > 0 which is a contra- diction. 7. If pd(a(3)) = pd(b(1)) then (n − 3, 1) = (⌈ m−1 2 ⌉, ⌈m 2 ⌉ − 2), so (n,m) = (5, 5) or (n,m) = (6, 6) which is a contradiction. 8. If pd(a(3)) = pd(b(2)) then (n − 3, 1) = (m − j, j − 1) or (n − 3, 1) = (m − j + 1, j − 2), so n > m in both cases which is a contradiction. 9. If pd(a(3)) = pd(b(3)) then (n − 3, 1) = (1,m − 3), so (n,m) = (4, 4) which is a contradiction. This finishes the proof. For the next lemma we need to consider weighted adjacency matrix T :=   0 1 2 1 0 3 2 3 0   (2.2) Observe that pd(T1) = (1, 0), pd(T2) = (0, 1), pd(T3) = (1, 1) ⇒ ps(K3) = 3. Lemma 2.6. Let T be the matrix defined in (2.2). For every n ≥ 5 T ⊕ Bn is product irregular. Proof. Observe that {pd(Ti) : i ∈ {1, 2, 3}} ⊂ {pd((A4)i) : i ∈ {1, 2, 3, 4}} and we know from Lemma 2.5 that ∀n ≥ 5 A4 ⊕Bn is product irregular. 3 Graphs with clique-cover number 2 In this section we consider product irregularity strength of connected graphs with clique cover number two. Suppose that G is a graph with clique-cover number 2, that is the vertex set of G can be partitioned into two cliques C1 and C2, of sizes n and m respectively. Then it follows that G has a spanning subgraph isomorphic to Kn +Km, where for two graphs H1 and H2, H1 + H2 denotes the disjoint union of H1 and H2. Then by [5, Lemma 1] it follows that 3 ≤ ps(G) ≤ ps(Kn + Km). Hence we will start by considering product irregularity strength of Kn +Km. It can be proved that any 4 × 4 weighted adjacency matrix M (with weights 1, 2 and 3) is product irregular if and only if there exist row m ∈ M such that pd(m) = (1, 1). Therefore ps(K4 + K4) > 3. There are a lot of graphs of the form Kn + Km for some integers n and m with product irregularity strength greater than 3. But since such graphs are disconnected, we will define operation of adding an edge between components of these graphs, i.e. we will consider minimal connected graphs with clique cover number 2. Definition 3.1 (+edge). Let G1 and G2 be two graphs with disjoint vertex sets. With G1 + G2 + edge we denote a graph obtained by taking disjoint union of G1 and G2 and adding an edge between two vertices of G1 and G2. Lemma 3.2. ∀n ≥ 4, ps(K2 +Kn + edge) = 3. 6 Art Discrete Appl. Math. 5 (2022) #P1.04 Proof. Consider weighted adjacency (n+ 2)× (n+ 2) matrix L =             0 1 3 0 · · · 0 1 0 0 0 · · · 0 3 0 ... ... 0 0 Bn ... ... 0 0             (3.1) where L1,3 = L3,1 = 3. Clearly, L is weighted adjacency matrix of the graph K2 +Kn. We will show that L is product-irregular. Since we have that pd((Bn)i) = pd(Li+2) for every i ∈ {2, 3, . . . n} it is enough to show that product degrees of first 3 rows of matrix L are different and do not belong to the set {pd((Bn)i), i ∈ {2, 3, . . . , n}}. 1. It is clear that those rows are different and that first two rows of L are not in the set {pd((Bn)i), i ∈ {2, 3, . . . , n}}. 2. For the row L3 we have that pd(L3) = pd((Bn)1) + (0, 1) = (n− 1, 1). Therefore pd(L3)[1] + pd(L3)[2] = n − 1 + 1 > n − 1 ≥ pd((Bn)j)[1] + pd((Bn)j)[2] for any j ∈ {2, 3, . . . , n}. This finishes the proof. Corollary 3.3. For every n ≥ 4, ps(K1 +Kn + edge) = 3. Proof. Consider matrix L′ obtained from matrix L from (3.1) by deleting second row and column. Clearly, L′ is product-irregular. Theorem 3.4. For every positive integers n and m such that n+m > 2 we have ps(Kn+ Km + edge) = 3. Proof. Consider some cases that were not covered by previous Lemmas: (i) ps(K5 + K5) = 3. For proving this fact we can take direct sum of the following weighted adjacency matrices: T5 :=       0 3 1 1 1 3 0 1 3 2 1 1 0 1 1 1 3 1 0 2 1 2 1 2 0       and T̃5 :=       0 2 2 2 1 2 0 3 3 3 2 3 0 2 3 2 3 2 0 1 1 3 3 1 0       (3.2) (ii) ps(K6 + K6) = 3. For proving this fact we can take direct sum of the following weighted adjacency matrices: T6 :=         0 1 2 3 1 3 1 0 1 3 1 1 2 1 0 1 2 2 3 3 1 0 1 1 1 1 2 1 0 1 3 1 2 1 1 0         and T̃6 :=         0 1 2 3 3 3 1 0 2 3 3 2 2 2 0 2 1 2 3 3 2 0 3 1 3 3 1 3 0 3 3 2 2 1 3 0         (3.3) D. Baldouski: Product irregularity strength of graphs with small clique cover number 7 Also consider some cases that could not be proved without adding edges between cliques. (iii) ps(K4 + K4 + edge) = 3. For proving this fact we will consider the following product-irregular matrix:             0 1 1 1 0 0 0 0 1 0 1 2 0 0 0 0 1 1 0 3 0 0 0 0 1 2 3 0 3 0 0 0 0 0 0 3 0 2 2 2 0 0 0 0 2 0 2 3 0 0 0 0 2 2 0 1 0 0 0 0 2 3 1 0             (3.4) (iv) ps(K3 + K4 + edge) = 3. For proving this fact we will consider the following product-irregular matrix:           0 1 2 0 0 0 0 1 0 3 0 0 0 0 2 3 0 3 0 0 0 0 0 3 0 2 2 2 0 0 0 2 0 2 3 0 0 0 2 2 0 1 0 0 0 2 3 1 0           (3.5) Observe that this matrix is obtained from matrix (3.4) by deleting first row and col- umn. (v) ps(K3 + K3 + edge) = 3. For proving this fact we will consider the following product-irregular matrix:         0 1 1 0 0 0 1 0 3 0 0 0 1 3 0 0 0 3 0 0 0 0 2 2 0 0 0 2 0 3 0 0 3 2 3 0         (3.6) (vi) ps(K2 + K3 + edge) = 3. For proving this fact we will consider the following product-irregular matrix:       0 3 0 0 0 3 0 0 0 3 0 0 0 2 2 0 0 2 0 3 0 3 2 3 0       (3.7) Observe that this matrix is obtained from matrix (3.6) by deleting first row and col- umn. 8 Art Discrete Appl. Math. 5 (2022) #P1.04 (vii) ps(K1 + K3 + edge) = 3. For proving this fact we will consider the following product-irregular matrix:     0 0 0 3 0 0 2 2 0 2 0 3 3 2 3 0     (3.8) Observe that this matrix is obtained from matrix (3.7) by deleting first row and col- umn. We are left with some trivial cases and it is straihtforward to check that ps(K2 + K2 + edge) = ps(P4) = 3 and ps(K1 +K2 + edge) = ps(P3) = 3. The proof now follows by Lemmas 2.5, 2.6 and 3.2 and Corollary 3.3. Corollary 3.5. If G is a connected graph of order at least 3 with clique-cover number 2 then ps(G) = 3. Observe that K1 + K1 + edge = P2 is an isolated edge, for which product-irregular labelling is not defined, i.e. 2 is the lower bound of the sum n+m in Theorem 3.4. 4 Graphs with clique-cover number 3 In this section we consider the product irregularity strength of graphs with clique-cover number 3. Observe that a graph G has clique cover number 3, if and only if its complement has chromatic number equal to 3. If G is a graph with clique cover number 3, then its vertex set can be partitioned into three cliques, of sizes n, m and l. Then it follows that G has a spanning subgraph isomorphic to Kn +Km +Kl, hence we will first investigate the product irregularity strength of such graphs. 4.1 Properties of An ⊕ Bm ⊕ Cl Lemma 4.1. For every n ≥ 7 and m ≥ 4, An ⊕ Cm is product irregular. Proof. Suppose contrary, that is ∃ai and cj (that are rows of matrices An and Cm respec- tively) for some i and j such that pd(ai) = pd(cj). We will use the same type of proof as in the Lemma 2.5. 1. If pd(a(1)) = pd(c(1)) then (⌈ n 2 ⌉ − 2, 1) = (1, ⌈m−1 2 ⌉), so n = 5 or n = 6 and m = 2 or m = 3 which is a contradiction. 2. If pd(a(1)) = pd(c(2)) then (⌈ n 2 ⌉−2, 1) = (0,m−j) or (⌈n 2 ⌉−2, 1) = (0,m−j+1). In both cases n = 3 or n = 4 which is a contradiction. 3. If pd(a(1)) = pd(c(3)) then (⌈ n 2 ⌉ − 2, 1) = (1, 1), so n = 5 or n = 6 which is a contradiction. 4. If pd(a(2)) = pd(c(1)) then (i − 1, 0) = (1, ⌈ m−1 2 ⌉) or (i − 2, 0) = (1, ⌈m−1 2 ⌉). In both cases m = 1 which is a contradiction. 5. For pd(a(2)) = pd(c(2)) we have that pd(a(2))[2] = 0 and pd(c(2))[2] > 0 which is a contradiction. 6. If pd(a(2)) = pd(c(3)) then (i − 1, 0) = (1, 1) or (i − 2, 0) = (1, 1)) which is, clearly, a contradiction. D. Baldouski: Product irregularity strength of graphs with small clique cover number 9 7. If pd(a(3)) = pd(c(1)) then (n− 3, 1) = (1, ⌈ m−1 2 ⌉), so n = 4 which is a contradic- tion. 8. If pd(a(3)) = pd(c(2)) then (n− 3, 1) = (0,m− j) or (n− 3, 1) = (0,m− j + 1), so n = 3 which is a contradiction. 9. If pd(a(3)) = pd(c(3)) then (n− 3, 1) = (1, 1), so n = 4 which is a contradiction. This finishes the proof. Lemma 4.2. For every n ≥ m ≥ 5, Bn⊕Cm is product irregular if (n,m) 6∈ {(5, 5), (6, 6)}. Proof. Suppose contrary, that is there exist bi and cj (that are rows of matrices Bn and Cm respectively) for some i and j such that pd(bi) = pd(cj). We will use the same type of proof as in the Lemma 2.5. 1. If pd(b(1)) = pd(c(1)) then (⌈ n−1 2 ⌉, ⌈n 2 ⌉ − 2) = (1, ⌈m−1 2 ⌉), so n = 2 or n = 3 which is a contradiction. 2. If pd(b(1)) = pd(c(2)) then (⌈ n−1 2 ⌉, ⌈n 2 ⌉ − 2) = (0,m− j) or (⌈n−1 2 ⌉, ⌈n 2 ⌉ − 2) = (0,m− j + 1), so n = 1, a contradiction. 3. If pd(b(1)) = pd(c(3)) then (⌈ n−1 2 ⌉, ⌈n 2 ⌉ − 2) = (1, 1), so n = 2 or n = 3, a contradiction. 4. If pd(b(2)) = pd(c(1)) then (n − i, i − 1) = (1, ⌈ m−1 2 ⌉) or (n − i + 1, i − 2) = (1, ⌈m−1 2 ⌉). In the first case we have that ⌈m−1 2 ⌉ = i − 1 = n − 2, which implies that 2n − 4 ≤ m ≤ 2n − 3, so, in particular, 2n − 4 ≤ m ≤ n, therefore n ≤ 4, a contradiction. In the second case we have that n = i which is a contradiction. 5. For pd(b(2)) = pd(c(2)) we have that pd(b(2))[1] > 0 and pd(c(2))[1] = 0 which is a contradiction. 6. If pd(b(2)) = pd(c(3)) then (n− i, i− 1) = (1, 1) or (n− i+ 1, i− 2) = (1, 1), so n = 3 which is a contradiction. 7. If pd(b(3)) = pd(c(1)) then (1, n − 3) = (1, ⌈ m−1 2 ⌉), so m = 2(n − 3) or m = 2(n− 3)+1 which is a contradiction because for n ≥ 7 we have that m > n and for 5 ≤ n < 7 we have that (n,m) ∈ {(5, 5), (6, 6)}. 8. If pd(b(3)) = pd(c(2)) then (1, n− 3) = (0,m− j) or (1, n− 3) = (0,m− j + 1) which is a contradiction. 9. If pd(b(3)) = pd(c(3)) then (1, n− 3) = (1, 1), so n = 4 which is a contradiction. This finishes the proof. Theorem 4.3. For every n, m and l such that m ≥ l ≥ n ≥ 7 An ⊕ Bm ⊕ Cl is product irregular. Proof. Proof follows by Lemmas 2.5, 4.1 and 4.2. Corollary 4.4. For all positive integers n,m and l greater than or equal to 7 it holds that ps(Kn +Km +Kl) = 3. Lemma 4.5. For all positive integers n and m greater than 6 and k ∈ {4, 5, 6}, ps(Kn + Km +Kk) = 3. 10 Art Discrete Appl. Math. 5 (2022) #P1.04 Proof. Let m ≥ n and consider matrix An⊕Bm⊕Ck. From Lemmas 2.5, 4.1 and 4.2 we can conclude that this matrix is product-irregular. Lemma 4.6. For all positive integer n ≥ 7 ps(K6 +K6 +Kn) = 3. Proof. Consider T6⊕ T̃6⊕Bn which is product-irregular because for every row b of matrix Bn pd(b)[1] + pd(b)[2] ≥ 5, while for every row t of matrices T6 and T̃6 we have that pd(t)[1] + pd(t)[2] ≤ 4. Lemma 4.7. For all positive integer n ≥ 7 ps(K5 +K6 +Kn) = 3. Proof. Consider the following matrix: M :=         0 2 2 2 1 1 2 0 3 3 3 1 2 3 0 2 3 1 2 3 2 0 1 2 1 3 3 1 0 1 1 1 1 2 1 0         ⊕       0 3 1 1 1 3 0 1 3 2 1 1 0 1 1 1 3 1 0 2 1 2 1 2 0       ⊕Bn (4.1) M is product-irregular because for every row b of matrix Bn pd(b)[1]+pd(b)[2] ≥ 5, while for every row v of first two blocks of our matrix M we have that pd(v)[1] + pd(v)[2] ≤ 4. Lemma 4.8. For all positive integers n ≥ 6, ps(K5 +K5 +Kn) = 3. Proof. Consider weighted adjacency matrices T5 and T̃5 from (3.2) in the first item of the proof of Theorem 3.4: 1. ∀n ≥ 7 we have T5⊕ T̃5⊕Bn is product irregular because for every row b of matrix Bn pd(b)[1] + pd(b)[2] ≥ 5. 2. For n = 6 we have that T5 ⊕ T̃5 ⊕ P6 is product-irregular, where P6 :=         0 2 2 2 2 1 2 0 2 2 2 3 2 2 0 2 3 3 2 2 2 0 3 1 2 2 3 3 0 3 1 3 3 1 3 0         . (4.2) This finishes the proof. Consider the graph K4 +K4 +K4. Suppose that ps(K4 +K4 +K4) = 3. Then there exist a product-irregular adjacency matrix K of the form K = P1 ⊕ P2 ⊕ P3 of our graph K4 +K4 +K4, where Pi is a product-irregular adjacency matrix of a graph K4 for every i ∈ {1, 2, 3}. Therefore, we have that for every row v of matrix K pd(v)[1]+pd(v)[2] < 4. Also, it is clear that for every row v of matrix K we have pd(v)[1] < 4 and pd(v)[2] < 4. But there exist only 10 different pairs of the form (x, y) such that 0 ≤ x, y < 4 and x+y < 4, which implies that there exist two rows v and u of matrix K such that pd(v) = pd(u). Therefore, ps(K4 +K4 +K4) > 3. D. Baldouski: Product irregularity strength of graphs with small clique cover number 11 There are a lot of graphs of the form Kn + Km + Kk for some integers n, m and k with product irregularity strength greater than 3. But since such graphs are disconnected, we will define operation of adding 2 edges between components of these graphs such that the resulting graph will be connected, i.e. we will consider minimal connected graphs with clique cover number 3. Definition 4.9 (+2edges). Let +2edges for graphs G1 + G2 + G3 be the operation of adding edges, i.e. applying two times +edge between any 2 different pairs of different sets V (G1), V (G2) and V (G3). We will use the following notation for that operation: G1 +G2 +G3 + 2edges. Now we will describe this operation using matrix language. Consider weighted adja- cency matrices A,B,C of sizes n×n, m×m and l×l respectively. Let T12(A,B,C, i, j, w) be (n+m+l)×(n+m+l) matrix with all zeros except elements with coordinates (i, n+j) and (n+ j, i) of value w, where 1 ≤ i ≤ n and 1 ≤ j ≤ m. In a similar way we can define matrices T13(A,B,C, i, j, w) and T23(A,B,C, i, j, w) for which coordinates of non-zero elements are (i, n + m + j) and (n + m + j, i), where 1 ≤ i ≤ n and 1 ≤ j ≤ l and (n+ i, n+m+ j) and (n+m+ j, n+ i), where 1 ≤ i ≤ m and 1 ≤ j ≤ l respectively. For example one of the weighted adjacency matrices for graph Kn+Km+Kl+2edges where the edges between cliques are between vertices ai and bj of weight w1 and between vertices bj and ck of weight w2 where ai ∈ V (Kn), bj ∈ V (Km) and ck ∈ V (Kl) is An ⊕Bm ⊕ Cl + T12(An, Bm, Cl, i, j, w1) + T23(An, Bm, Cl, j, k, w2). Definition 4.10 (In-degree and in-edges). Consider graph G := G1 +G2 +G3 + 2edges. Let G′ := G1 + G2 + G3 be a subgraph of the graph G. Let g ∈ V (G) and let dG′(g) be the degree of the vertex g ∈ V (G′). Then define in-degree of vertex g ∈ V (G) to be d+(g) := d(g)− dG′(g). We say that for some i ∈ {1, 2, 3} Gi has t in-edges if and only if ∑ g∈V (Gi) d+(g) = t. For the next theorem we will define the following matrix. Let M̃n(x, y) := Mn(x, y, y) and matrices Ãn, B̃n and C̃n to be M̃n(1, 2), M̃n(2, 3) and M̃n(3, 1) respectively. Theorem 4.11. For all positive integers n, m and l that are greater than or equal to 5 we have that ps(Kn +Km +Kl + 2edges) = 3. Proof. Consider some cases that were not covered by previous Lemmas: 1. For (n,m, l) = (6, 6, 6) consider the following product-irregular matrix:         0 1 1 1 1 1 1 0 3 1 1 2 1 3 0 1 2 2 1 1 1 0 2 2 1 1 2 2 0 2 1 2 2 2 2 0         ⊕         0 2 2 2 2 2 2 0 1 2 2 3 2 1 0 2 3 3 2 2 2 0 3 3 2 2 3 3 0 3 2 3 3 3 3 0         ⊕         0 3 3 3 3 3 3 0 2 3 3 1 3 2 0 3 1 1 3 3 3 0 1 1 3 3 1 1 0 1 3 1 1 1 1 0         (4.3) 2. For (n,m, l) = (5, 6, 6) we can consider the same matrix as in (4.3) without first row (and column), i.e. without row (and column) v such that pd(v) = (0, 0). 12 Art Discrete Appl. Math. 5 (2022) #P1.04 3. For (n,m, l) = (5, 5, 5) we will consider Ã5 ⊕ B̃5 ⊕ C̃5 + 2edges. Let B̃5 to have 2 in-edges, then we have: (1) If B̃5 has 2 in-edges from one vertex, then we can take weighted adjacency matrix Ã5⊕B̃5⊕C̃5+T12(Ã5, B̃5, C̃5, 3, 3, 3)+T23(Ã5, B̃5, C̃5, 3, 3, 2) which is product-irregular. (2) If B̃5 has 2 in-edges from different vertices then we can take weighted adja- cency matrix Ã5⊕B̃5⊕C̃5+T12(Ã5, B̃5, C̃5, 3, 3, 3)+T23(Ã5, B̃5, C̃5, 1, 3, 2) which is product-irregular. The proof now follows by the above argumentation, together with Theorem 4.3 and Lem- mas 4.5, 4.6, 4.7 and 4.8. Lemma 4.12. For all positive integers n ≥ 7 and m ∈ {5, 6} we have that ps(K4+Kn+ Km) = 3. Proof. Consider three different cases for different m: 1. For m = 6 and n ≥ 8 consider matrix A4 ⊕ B6 ⊕ Bn which is product-irregular using Theorem 3.4. 2. For m = 6 and n = 7 consider matrix A4 ⊕ B7 ⊕ T̃6 which is product-irregular (where T̃6 is defined in (3.3)). 3. For m = 5 consider matrix A4 ⊕ Bn ⊕ T̃5 which is product-irregular (where T̃5 is defined in (3.2)). Theorem 4.13. For all positive integers n, m and l that are greater than or equal to 4 we have that ps(Kn +Km +Kl + 2edges) = 3. Proof. Consider some cases that were not covered by previous Lemmas and Theorems: 1. For (n,m, l) = (4, 5, 6) consider the following product-irregular matrix: A4 ⊕       0 2 2 2 1 2 0 3 1 3 2 3 0 2 3 2 1 2 0 1 1 3 3 1 0       ⊕B6 (4.4) Notice that the second block of this matrix is obtained from T̃5 from (3.2) by chang- ing the values t24 and t42 from 3 to 1. 2. For (n,m, l) = (4, 6, 6) consider the matrix C4+A6+ T̃6 which is product-irregular (where T̃6 is defined in (3.3)). Consider some cases for which we will add some edges between cliques: 3. For (n,m, l) = (4, 5, 5) we will consider Ã4 ⊕ B̃5 ⊕ C̃5 + 2edges. (B̃5) For the case when d +(B̃5) = 2 we have two options: (1) If B̃5 has 2 in-edges from one vertex, then we can take weighted adjacency matrix Ã4⊕ B̃5⊕ C̃5+T12(Ã4, B̃5, C̃5, 2, 3, 3)+T23(Ã4, B̃5, C̃5, 3, 3, 2) which is product-irregular. D. Baldouski: Product irregularity strength of graphs with small clique cover number 13 (2) If B̃5 has 2 in-edges from different vertices then we can take weighted ad- jacency matrix Ã5⊕B̃5⊕C̃5+T12(Ã4, B̃5, C̃5, 3, 3, 3)+T23(Ã4, B̃5, C̃5, 1, 3, 2) which is product-irregular. (Ã4) For the case when d +(Ã4) = 2 we have two options: (1) If Ã4 has 2 in-edges from one vertex then we can take weighted adjacency matrix Ã4⊕ B̃5⊕ C̃5+T12(Ã4, B̃5, C̃5, 2, 3, 2)+T13(Ã4, B̃5, C̃5, 2, 3, 2) which is product-irregular. (2) If Ã4 has 2 in-edges from different vertices then we can take weighted ad- jacency matrix Ã4⊕B̃5⊕C̃5+T12(Ã4, B̃5, C̃5, 2, 3, 2)+T13(Ã4, B̃5, C̃5, 4, 3, 2) which is product-irregular. 4. For (n,m) = (4, 4) and l ≥ 5 we will consider Ã4 ⊕ B̃l ⊕ C̃4 + 2edges. (B̃l) For the case when d +(B̃l) = 2 we have two options: (1) If B̃l has 2 in-edges from one vertex then we can take weighted adjacency matrix Ã4 ⊕ B̃l ⊕ C̃4 + T12(Ã4, B̃l, C̃4, 2, 3, 3)+ T23(Ã4, B̃l, C̃4, 3, 2, 2) which is product-irregular. (2) If B̃l has 2 in-edges from different vertices then we can take weighted adja- cency matrix Ã4⊕B̃l⊕C̃4+T12(Ã4, B̃l, C̃4, 2, 3, 3)+T23(Ã4, B̃l, C̃4, 1, 2, 2) which is product-irregular. (C̃4) For the case when d +(C̃4) = 2 we have two options: (1) If C̃4 has 2 in-edges from one vertex then we can take weighted adjacency matrix Ã4 ⊕ B̃l ⊕ C̃4 + T13(Ã4, B̃l, C̃4, 2, 2, 3)+ T23(Ã4, B̃l, C̃4, 3, 2, 3) which is product-irregular. (2) If C̃4 has 2 in-edges from different vertices then we can take weighted ad- jacency matrix Ã4⊕B̃l⊕C̃4+T13(Ã4, B̃l, C̃4, 2, 2, 3)+T23(Ã4, B̃l, C̃4, 3, 1, 3) which is product-irregular. 5. For (n,m, l) = (4, 4, 4) we will consider Ã4 ⊕ B̃4 ⊕ C̃4 + 2edges. Let C̃4 to have 2 in-edges. (1) If C̃4 has 2 in-edges from one vertex then we can take weighted adjacency matrix Ã4⊕B̃4⊕C̃4+T13(Ã4, B̃4, C̃4, 2, 2, 3)+T23(Ã4, B̃4, C̃4, 3, 2, 3) which is product-irregular. (2) If C̃4 has 2 in-edges from different vertices then we can take weighted adja- cency matrix Ã4⊕B̃4⊕C̃4+T13(Ã4, B̃4, C̃4, 2, 2, 3)+T23(Ã4, B̃4, C̃4, 3, 1, 3) which is product-irregular. The proof now follows by the above argumentation, together with Theorem 4.11 and Lemma 4.12. Corollary 4.14. If G is a connected graph such that its vertex set can be partitioned into 3 cliques of sizes at least 4 then ps(G) = 3. We would like to conclude the paper with proposing the following problem for possible further research. Problem 4.15. Are there only finitely many connected graphs with clique cover number 4 and product irregularity strength more than 3? 14 Art Discrete Appl. 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Ford, The distribution of integers with a divisor in a given interval, Ann. Math. (2) 168 (2008), 367–433, doi:10.4007/annals.2008.168.367. [7] A. Frieze, R. J. Gould, M. Karoński and F. Pfender, On graph irregularity strength, J. Graph Theory 41 (2002), 120–137, doi:10.1002/jgt.10056. [8] M. Kalkowski, M. Karoński and F. Pfender, A new upper bound for the irregularity strength of graphs, SIAM J. Discrete Math. 25 (2011), 1319–1321, doi:10.1137/090774112. [9] G. Kaplan, A. Lev and Y. Roditty, Bertrand’s postulate, the prime number theorem and product anti-magic graphs, Discrete Math. 308 (2008), 787–794, doi:10.1016/j.disc.2007.07.049. [10] D. Koukoulopoulos, Localized factorizations of integers, Proc. Lond. Math. Soc. (3) 101 (2010), 392–426, doi:10.1112/plms/pdp056. [11] P. Majerski and J. Przybyło, On the irregularity strength of dense graphs, SIAM J. Discrete Math. 28 (2014), 197–205, doi:10.1137/120886650. [12] O. Pikhurko, Characterization of product anti-magic graphs of large order, Graphs Comb. 23 (2007), 681–689, doi:10.1007/s00373-007-0748-6. [13] J. Skowronek-Kaziów, Multiplicative vertex-colouring weightings of graphs, Inf. Process. Lett. 112 (2012), 191–194, doi:10.1016/j.ipl.2011.11.009. [14] J. Skowronek-Kaziów, Graphs with multiplicative vertex-coloring 2-edge-weightings, J. Comb. Optim. 33 (2017), 333–338, doi:10.1007/s10878-015-9966-7. ISSN 2590-9770 The Art of Discrete and Applied Mathematics 5 (2022) #P1.05 https://doi.org/10.26493/2590-9770.1396.3c7 (Also available at http://adam-journal.eu) A tight relation between series–parallel graphs and bipartite distance hereditary graphs* Nicola Apollonio† Istituto per le Applicazioni del Calcolo, M. Picone, v. dei Taurini 19, 00185 Roma, Italy Massimiliano Caramia Dipartimento di Ingegneria dell’Impresa, Università di Roma “Tor Vergata”, v. del Politecnico 1, 00133 Roma, Italy Paolo Giulio Franciosa Dipartimento di Scienze Statistiche, Sapienza Università di Roma, p.le Aldo Moro 5, 00185 Roma, Italy Jean-François Mascari‡ Istituto per le Applicazioni del Calcolo, M. Picone, v. dei Taurini 19, 00185 Roma, Italy Received 25 September 2020, accepted 06 December 2020, published online 21 March 2021 Abstract Bandelt and Mulder’s structural characterization of bipartite distance hereditary graphs asserts that such graphs can be built inductively starting from a single vertex and by re- peatedly adding either pendant vertices or twins (i.e., vertices with the same neighborhood as an existing one). Dirac and Duffin’s structural characterization of 2–connected series– parallel graphs asserts that such graphs can be built inductively starting from a single edge by adding either edges in series or in parallel. In this paper we give an elementary proof that the two constructions are the same construction when bipartite graphs are viewed as the fundamental graphs of a graphic matroid. We then apply the result to re-prove known results concerning bipartite distance hereditary graphs and series–parallel graphs and to provide a new class of polynomially-solvable instances for the integer multi-commodity flow of maximum value. Keywords: Series-parallel graphs, bipartite distance hereditary graphs, binary matroids. Math. Subj. Class.: 05C *We are sincerely grateful to the referee for the careful reading of the paper and for his comments and detailed suggestions which helped us to improve considerably the manuscript. †Corresponding Author. Supported by the Italian National Research Council (C.N.R.) under national research project “MATHTECH”. ‡Supported by the Italian National Research Council (C.N.R.) under national research project “MATHTECH”. cb This work is licensed under https://creativecommons.org/licenses/by/4.0/ 2 Art Discrete Appl. Math. 5 (2022) #P1.05 1 Introduction Distance hereditary graphs are graphs with the isometric property, i.e., the distance func- tion of a distance hereditary graph is inherited by its connected induced subgraphs. This im- portant class of graphs was introduced and thoroughly investigated by Howorka in [24, 25]. A bipartite distance hereditary (BDH for short) graph is a distance hereditary graph which is bipartite. Such graphs can be constructed starting from a single vertex by means of the following two operations [6]: (BDH1) adding a pendant vertex, namely a vertex adjacent exactly to an existing vertex; (BDH2) adding a twin of an existing vertex, namely adding a vertex and making it adja- cent to all the neighbors of an existing vertex. Taken together the two operations above will be referred to as Bandelt and Mulder’s con- struction. A graph is series–parallel [7], if it does not contain the complete graph K4 as a mi- nor; equivalently, if it does not contain a subdivision of K4. This is Dirac’s [14] and Duffin’s [15] characterization by forbidden minors. Since both K5 and K3,3 contain a sub- division of K4, by Kuratowski’s Theorem any series–parallel graph is planar. Like BDH graphs, series–parallel graphs admit a constructive characterization which justifies their name: a connected graph is series–parallel if it can be constructed starting from a single edge by means of the following two operations: (SP1) adding an edge with the same end-vertices as an existing one (parallel extension); (SP2) subdividing an existing edge by the insertion of a new vertex (series extension). Taken together the two operations above will be referred to as Duffin’s construction. Here and throughout the rest of the paper we consider only 2–connected series–parallel graphs which can be therefore obtained by starting from a pair of a parallel edges rather than by starting from a single edge. The close resemblance between operations (BDH1) and (BDH2) and operations (SP1) and (SP2) is apparent. It becomes even more apparent after our Theorem 3.1, which estab- lishes that the constructions defining BDH and series–parallel graphs, namely, Bandelt and Mulder’s construction and Duffin’s construction, are the same construction when bipartite graphs are viewed as fundamental graphs of a graphic matroid (Theorem 3.1). Although this fact is fairly well known and short proofs can be given using the deep and refined notions of branch width and tree width of graphs and matroids1 (combined with classical results on graph minors), neither an elementary proof nor an explicit statement seem to be at hand. The intimate relationship between BDH graphs and series–parallel graphs was also already observed by Ellis-Monhagan and Sarmiento in [16]. The authors, motivated by the aim of finding polynomially computable classes of instances for the vertex–nullity in- terlace polynomial introduced by Arratia, Bollobás and Sorkin in [5], under the name of interlace polynomial, related the two classes of graphs via a topological construction in- volving the so called medial graph of a planar graph. By further relying on the relationships E-mail addresses: nicola.apollonio@cnr.it (Nicola Apollonio), caramia@disp.uniroma2.it (Massimiliano Caramia), paolo.franciosa@uniroma1.it (Paolo Giulio Franciosa), g.mascari@iac.cnr.it (Jean-François Mascari) 1In section 5, we give one of such a proof kindly supplied by an anonymous referee of an earlier version of the paper. N. Apollonio et al.: A tight relation between series–parallel graphs and bipartite distance . . . 3 between the Martin polynomial and the symmetric Tutte polynomial of a planar graph, they proved a relation between the the symmetric Tutte polynomial of a planar graph H , namely t(H;x, x)—recall that the Tutte polynomial is a two variable polynomial–and the interlace polynomial q(G;x) of a graph G derived from the medial graph of G (Theorem 4.1). Such a relation led to the following three interesting consequences: – the #P–completeness of the interlace polynomial of Arratia, Bollobás and Sorkin [5] in the general case; – a characterization of BDH graphs via the so-called γ invariant, (i.e., the coefficient of the linear term of the interlace polynomial); – an effective proof that the interlace polynomial is polynomial-time computable within BDH graphs. In view of a result due to Aigner and van der Holst (Theorem 4.6), the latter two con- sequences in the list above are straightforward consequences of Theorem 3.1 (see Sec- tion 4.1). Besides the new direct proofs of these results, Theorem 3.1 has some more applications. – Syslo’s characterization’s of series–parallel graphs in terms of Depth First Search (DFS) trees: the characterization asserts that a connected graph H is series–parallel if and only if every spanning tree of H is a DFS-tree of one of its 2–isomorphic copies. In other words, up to 2–isomorphism, series–parallel graphs have the characteristic property that their spanning trees can be oriented to become arborescences so that the corresponding fundamental cycles become directed circuits (cycles whose arcs are oriented in the same way). Recall that an arborescence is a directed tree with a single special node distinguished as the root such that, for each other vertex, there is a directed path from the root to that vertex. – New polynomially solvable instances for the problem of finding integer multi-commo- dity flow of maximum value: if the demand graph of a series–parallel graph is a co– tree, then the maximum value of a multi-commodity flow equals the minimum value of a multi-terminal cut; furthermore both a maximizing flow and a minimizing cut can be found in strongly polynomial time. Organization of the paper. The rest of the paper is organized as follows: in Section 2 we give the basic notions used throughout the rest of the paper. In Section 3 we prove our main result (Theorem 3.1) (two more proofs are given in Section 5) and discuss how it fits within circle graphs and how it relates with edge-pivoting. Theorem 3.1 is then applied in Section 4: in Section 4.1, we re-prove the previously mentioned couple of results in [16]; in Section 4.2 we re-prove Syslo’s characterization of series–parallel graphs and give a sort of hierarchy of characterizations of 2–connected planar graphs by the properties of their spanning trees; finally in Section 4.3, we give an application to multi-commodity flow in series–parallel graphs. 2 Preliminaries For a graph G the edge e with endvertices x and y will be denoted by xy. The graph induced by U ⊆ V (G) is denoted by G[U ]. If F ⊆ E(G), the graph G − F is the graph (V (G), E(G)− F ). 4 Art Discrete Appl. Math. 5 (2022) #P1.05 A digon is a pair of parallel edges, namely a cycle with two edges. A hole in a bipartite graph is an induced subgraph isomorphic to Cn for some n ≥ 6. A domino is a subgraph isomorphic to the graph obtained from C6 by joining two antipodal vertices by a chord. The domino is denoted by ⊟. A bipartite graph G is a chordal bipartite graph if G has no hole. Let F be a family of graphs. We say that G is F–free if G does not contain any induced copy of a member of F . If G is F–free and F = {G0}, then we say that G is G0–free. Graphs dealt with in this paper are, in general, not assumed to be vertex-labeled. How- ever, when needed, vertices are labeled by the first n naturals where n is the order of G. We denote labeled and unlabeled graphs with the same symbol. If u and v are two vertices of G, then a label swapping at u and v (or simply uv-swapping) is the labeled graph obtained by interchanging the labels of u and v. For a bipartite graph G with color classes A and B, let A ∈ {0, 1}A×B be the reduced adjacency matrix of G, namely, A is the matrix whose rows are indexed by the vertices of A, whose columns are indexed by the vertices of B and where Au,v = 1 if and only if u and v are adjacent vertices of G. The incidence graph of a matrix A ∈ {0, 1}A×B is the bipartite graph with color classes A and B and where u ∈ A and v ∈ B are adjacent if and only au,v = 1. We review very briefly some basic notions in matroid theory [28, 36, 37]. For a {0, 1}- matrix A the binary matroid generated by A, denoted by M(A), is the matroid whose elements are the indices of the columns of A and whose independent sets are those subsets of elements whose corresponding columns are linearly independent over GF (2). A binary matroid is a matroid isomorphic to the binary matroid generated by some {0, 1}-matrix A. If T is a basis of a binary matroid M and f 6∈ T , then T ∪ {f} contains a unique minimal non independent set C(f, T ). Thus, if F is a proper subset of C(f, T ), then F is an independent set of M . Such a set C(f, T ) is the so called fundamental circuit through f with respect to T and C(f, T ) − {f} is the corresponding fundamental path. A partial representation of a binary matroid M is a {0, 1}-matrix à whose columns are the incidence vectors over the elements of a basis of the fundamental circuits with respect to that basis. A fundamental graph of a binary matroid M is simply the incidence bipartite graph of any of its partial representations. Therefore a bipartite graph G is the fundamental graph of a binary matroid M if G is isomorphic to the graph BM (T ) with color classes T and T for some basis T and co-basis T (i.e., the complement of a basis) of M and where there is an edge of G between e ∈ T and f ∈ T if e ∈ C(f, T ), where C(f, T ) is the fundamental circuit through f with respect to T . If à is a partial representation of a binary matroid M , then M ∼= M([ I | à ]), that is M is isomorphic to the matroid generated by [ I | à ]. Clearly, à is a partial representation of M with rows and columns indexed by the elements of the basis T and of the co-basis T , respectively, if and only if à is the reduced adjacency matrix of BT (M), where the color class T indexes the rows of Ã. The cycle matroid (also known as graphic matroid) of a graph H , denoted by M(H), is the matroid whose elements are the edges of H and whose independent sets are the forests of H . If H is connected, then the bases of M(H) are precisely the spanning trees of H and its co-bases are precisely the co-trees, namely the subgraphs spanned by the complement of the edge–set of a spanning tree. A matroid M is a cycle matroid if it is isomorphic to the cycle matroid of some graph H . Cycle matroids are binary: if M is a cycle matroid, then there are a graph H and a spanning forest of H such that M ∼= M([ I | à ]) where à is the {0, 1}-matrix whose columns are the incidence vectors over the edges of a spanning forest of the fundamental cycles with respect to that spanning forest. N. Apollonio et al.: A tight relation between series–parallel graphs and bipartite distance . . . 5 A fundamental graph of a graph H is simply the fundamental graph of its cycle ma- troid M(H). For a graph H and one of its spanning forests T , we abridge the notation BM(H)(T ) into BH(T ) to denote the fundamental graph of H with respect to T (see Fig- ure 1, where H ∼= K4). If H is 2–connected, then BH(T ) is connected. Moreover, BH(T ) does not determine H in the sense that non-isomorphic graphs may have isomorphic funda- mental graphs. This because, while it is certainly true that isomorphic graphs have isomor- phic cycle matroids, the converse is not generally true (see Figure 2). Two graphs having isomorphic cycle matroids are called 2–isomorphic. α β γ a 1 1 0 b 1 1 1 c 0 1 1 a β γ α 1 1 0 b 1 0 1 c 0 1 1 a α b c γ β a α b c γ β T T ′ = T∆{a, α} α a b β γ c BK4(T ) a α b β γ c BK4(T ′) Figure 1: Two fundamental graphs of K4 with respect to two spanning trees T and T ′ along with the corresponding matrices and the respective fundamental graphs. The fundamental graph with respect to T ′ arises from the one with respect to T by pivoting along edge αa. u v u v a b c d e a � b � c � d � e � α β α� β� γ δ γ� δ� Figure 2: Two 2–isomorphic graphs that are not isomorphic: x 7→ x′ maps bijectively fundamental cycles of the graph on the left to fundamental cycles of the graph on the right. 6 Art Discrete Appl. Math. 5 (2022) #P1.05 3 BDH graphs are fundamental graphs of series parallel graphs In this section we prove our main result. Theorem 3.1. A connected bipartite graph G is a bipartite distance hereditary graph if and only if G is a fundamental graph of a 2–connected series–parallel graph. Proof. For a bipartite graph G let MG denote the binary matroid generated by the reduced adjacency matrix of G. Let us examine preliminarily the effect induced on a fundamental graph BH(T ) of a 2–connected graph H by series and parallel extensions and, conversely (and in a sense “dually”), the effect induced on MG by extending a connected bipartite graph G through the addition of violated vertices and twins. If MG is a graphic matroid and H is one of the 2–isomorphic graphs whose cycle matroid is isomorphic to MG, then Table 1 summarizes these effects. Operation on H Operation on BH(T ) Parallel extension on edge a of T ↔ adding a pendant vertex in color class T adjacent to a Series extension on edge a of T ↔ adding a twin of a in color class T Parallel extension on edge β of T ↔ adding a twin of β in color class T Series extension on edge β of T ↔ adding a pendant vertex in color class T adjacent to β. Table 1: The effects of series and parallel extension on H on its fundamental graph BH(T ). We can now proceed with the proof. The only if direction is proved by induction on the order of G. The assertion is true when G has two vertices because K2 is a BDH graph and at the same time is also the fundamental graph of a digon. Let now G have n ≥ 3 vertices and assume that the assertion is true for BDH graphs with n − 1 vertices. By Bandelt and Mulder’s construction G is obtained from a BDH graph G′ either by adding a pendant vertex or a twin. Let H ′ be a series–parallel graph having G′ as fundamental graph with respect to some spanning tree. Since, by Table 1, the last two operations correspond to series or parallel extension of H ′, the result follows by Duffin’s construction of series– parallel graphs. For the if direction, let G be the fundamental graph of a series–parallel graph H with respect to some tree T . By Duffin’s construction of series–parallel graphs and Table 1, G can be constructed starting from a single edge by either adding twins or pendant vertices. Therefore, G is a BDH graph by Bandelt and Mulder’s construction. Before going through applications, let us discuss how Theorem 3.1 relates to circle graphs, a thoroughly investigated class of graphs which we now briefly describe. A double occurrence word w over a finite alphabet Σ is a word in which each letter appears exactly twice, where w is cyclic word, namely, it is the equivalence class of a linear word modulo cyclic shifting and reversal of the orientation. Two distinct symbols of Σ in w are interlaced if one appears precisely once between the two occurrences of the other. By wrapping w along a circle and by joining the two occurrences of the same symbol of w by a chord labeled by the same symbols whose occurrences it joins, one obtains a pair (S, C) consisting of a circle S and a set C of chords of S. In knot theory terminology, such a pair is usually called a chord diagram and its intersection graph, namely the graph whose vertex set is C and where two vertices are adjacent if and only if the corresponding chords N. Apollonio et al.: A tight relation between series–parallel graphs and bipartite distance . . . 7 intersects, is called the interlacement graph of the chord diagram or the interlacement graph of the double occurrence word. A graph is an interlacement graph if it is the interlacement graph of some chord dia- gram or of some double occurrence words. Interlacement graphs are probably better known as circle graphs. The name interlacement graph comes historically from the Gauss Real- ization Problem of double occurrence words [13, 31, 34]. Distance hereditary graphs are circle graphs [8]. Thus BDH graphs form a proper subclass of bipartite circle graphs. De Fraysseix [11, 12] proved the following. Theorem 3.2 ([11, 12]). A bipartite graph is a bipartite circle graph if and only if it is the fundamental graph of a planar graph. Therefore Theorem 3.1 specializes de Fraysseix’s Theorem to the subclass of series– parallel graphs. 3.1 BDH graphs and edge–pivoting It follows from Theorem 3.1 that with every 2–isomorphism class of 2–connected series– parallel graphs one can associate all the BDH graphs that are fundamental graphs of each member in the class. Therefore BDH graphs that correspond to the same 2–isomorphism class are graphs in the same “orbit”. In this section we make precise the latter sentence and draw the graph-theoretical consequences of this fact. Given a {0, 1}-matrix A, pivoting A over GF (2) on a nonzero entry (the pivot element) means replacing à = ( 1 a b D ) by à = ( 1 a b D+ ba ) where a is a row vector, b is a column vector, D is a submatrix of A and the rows and columns of A have been permuted so that the pivot element is a1,1 ([10, p. 69], [32, p. 280]). If A is the partial representation of the cycle matroid of a graph H (or more generally a bi- nary matroid), then pivoting on a nonzero entry, C(e, f), say, yields a new tree (basis) with f in the tree (basis) and e in the co-tree (co-basis) and the matrix obtained after pivoting is a new partial representation matrix of the same matroid. Clearly the fundamental graphs associated with the two bases change accordingly so that pivoting on {0, 1}-matrices in- duces an operation on bipartite graphs whose concrete interpretation is a change of basis in the associated binary matroid. The latter operation on bipartite graph will be still referred to as edge–pivoting or simply to as pivoting in analogy to what happens for matrices (see also Figure 1). In the context of circle graphs, the operation of pivoting is a specialization to bipartite graph of the so called edge–local complementation. Since any bipartite graph is a fundamental graph of some binary matroid, the operation of pivoting can be described more abstractly as follows. Given a bipartite graph with color classes A and B, pivoting on edge uv ∈ E(G) is the operation that takes G into the graph Guv on the same vertex set of G obtained by complementing the edges between NG(u) \ {u} and NG(v) \ {v} and then by swapping the labels of u and v (if G is labeled). More formally, if ℓG : V (G) → N is a labeling of the vertices of G, then Guv = (V (G), E(G)∆((NG(u) \ {u})× (NG(v) \ {v}))) and ℓGuv is defined by ℓGuv (u) = ℓG(v), ℓGuv (v) = ℓG(u) and ℓGuv (w) = ℓG(w) if w 6∈ {u, v}. If e ∈ E(G) has endpoints uv, then we use Ge in place of Guv . 8 Art Discrete Appl. Math. 5 (2022) #P1.05 We say that a graph G̃ is pivot-equivalent to a graph G, written G̃ ∼ G, if for some k ∈ N, there is a sequence G1, . . . , Gk of graphs such that G1 ∼= G, Gk ∼= G̃ and, for i = 1, . . . , k− 1, Gi+1 ∼= G ei i , ei ∈ E(Gi). The orbit of G, denoted by [G], consists of all graphs that are pivot-equivalent to G. For later reference, we state as a lemma the easy though important facts discussed above. Figure 1 illustrates the contents of the lemma. Lemma 3.3. Let M be a connected graphic matroid. Then M determines both a class [G] of pivot-equivalent graphs and a class [H] of 2–isomorphic graphs. In particular, any graph in [G] is the fundamental graph of some 2–isomorphic copy of H and the fundamen- tal graph of any graph in [H] is pivot-equivalent to G. The operations of pivoting and of taking induced subgraphs commute in (bipartite) graphs. Lemma 3.4 (see [5]). Let G a bipartite graph, U ⊆ V (G) and e be an edge whose end- vertices are in U . Then Ge[U ] ∼= (G[U ])e. The next lemma relates in the natural way minors of a cycle matroid to the induced subgraphs of the fundamental graphs associated with the matroid. Lemma 3.5. Let M and N be cycle matroids. Let G be any of the fundamental graphs of M and let K be any of the fundamental graphs of N . Then N is a minor of M if and only if K is an induced subgraph in some bipartite graph in the orbit of G. Equivalently, N is a minor of M if and only if G contains some induced copy of a graph in the orbit of K. To get acquainted with pivoting, the reader may check Lemma 3.6 with the help of Figure 3. Refer to Section 2 for the definition of domino and hole. e f g e ẽ e G0 G e 0 G f 0 G g 0 G1 G e 1 G e,ẽ 1 G2 G e 2 Figure 3: The effect of pivoting a graph G along some of its edges when G ∼= ⊟, C8, C6. N. Apollonio et al.: A tight relation between series–parallel graphs and bipartite distance . . . 9 Lemma 3.6. Let k ≥ 6 be an even integer. – If either H ∼= ⊟ or H ∼= Ck, then for each uv ∈ E(H) there exists an induced subgraph H ′ of Huv such that either H ′ ∼= ⊟ or H ′ ∼= Ck. – If G ∼= Ck, then there is a graph G̃ in the orbit of H such that G̃ contains an induced copy of either ⊟ or C6. Proof. By inspecting Figure 3 one checks that if G ∼= ⊟, then either Ge ∼= ⊟ or Ge ∼= C6. If G ∼= C6, then G e ∼= ⊟ for every e ∈ E(G). If G ∼= Ck, k > 6, then by pivoting on uv ∈ E(G) and deleting u and v results in a graph G′ ∼= Ck−2. In particular, by repeatedly pivoting on new formed edges (like edge ẽ of graph Ge1 in Figure 3), one obtains a graph in the orbit of G which contains an induced copy of either ⊟ or C6. The second part of the proof is left to the reader. We are ready to extract the graph-theoretical consequence of Theorem 3.1. To this end let us recall that besides their constructive characterization, Bandelt and Mulder character- ized the class of BDH graphs also by forbidden induced subgraphs as follows. Theorem 3.7 ([6, Corollaries 3 and 4]). Let G be a connected bipartite graph. Then G is BDH if and only if G contains neither holes nor induced dominoes. The following two corollaries follow straightforwardly from Theorem 3.1 after Theo- rem 3.7 and assert that the class of BDH graphs–that is, of {hole, domino}–free graphs–is closed under pivoting, namely, that the orbit of a bipartite {hole, domino}–free graph con- sists of {hole, domino}–free graphs. Corollary 3.8. The following statements about a chordal bipartite graph G are equivalent: (i) G does not contain any induced domino; (ii) any graph in the orbit of G is a chordal bipartite graph. Corollary 3.9. Let G be a bipartite domino-free graph. If G is chordal, then so is any other graph in its orbit. 4 Applications 4.1 BDH graphs and the interlace polynomial As already mentioned, Ellis-Monaghan and Sarmiento related series–parallel graphs and BDH graphs topologically, via the medial graph. Let H be a plane graph (or even a 2-cell embedded graph in an oriented surface). For our purposes, we can assume that H is 2– connected. The medial graph m(H) of H is the graph obtained as follows: first place a vertex ve into the interior of each edge e of H . Then, for each face F of H , join ve to vf by an edge lying in F if and only if the edges e and f are consecutive on the boundary of F . Notice that if F is bounded by a digon {e, e′} or if e and e′ share a degree-2 endpoint in H , then vertices ve and ve′ are joined by two parallel edges. Let m(H) be the plane (2-cell embedded) graph obtained in this way. The graph underlying m(H) is the medial graph of H . The medial graph is clearly 4-regular, as each face creates two adjacencies for each edge on its boundary. Moreover, it can be oriented so that each vertex is entered by 2 arcs and left by 2 arcs. Given a 4-regular labeled graph N and one of its Eulerian circuits 10 Art Discrete Appl. Math. 5 (2022) #P1.05 C, we can associate with N a double occurrence word w which is the word consisting of the labels of the vertices of C cyclically met during the tour on C. The circle graph formed from C and chords between repeated pairs of letters of w is called the the circle graph of N . Ellis-Monaghan and Sarmiento, building also on the relations between the Martin polynomial and the symmetric Tutte polynomial, proved the following relation between the symmetric Tutte polynomial t(H;x, x) of a planar graph H and the vertex nullity interlace polynomial q(G;x) of a graph G derived, as described in the theorem below, from the medial graph of any of its plane embedding. Theorem 4.1 ([16]). If H is a plane embedding of a planar graph and G is the circle graph of some Eulerian circuit of the medial graph of H , then q(G;x) = t(H;x, x). The results were then specialized so as to give the following characterization of BDH graphs. Theorem 4.2 ([16]). G is a BDH graph with at least two vertices if and only if it is the circle graph of an Euler circuit in the medial graph of a plane embedding of a series– parallel graph H . Using Theorem 4.1 and Theorem 4.2, the authors deduced the following consequences stated below as Corollary 4.3, Corollary 4.4 and Corollary 4.5. Corollary 4.3. Computing the vertex-nullity interlace polynomial is #P-hard in general. Corollary 4.4. If G is a BDH graph, then q(G;x) is polynomial-time computable. Corollary 4.4 follows because the Tutte polynomial is polynomial-time computable for series–parallel graphs [29]. Corollary 4.5. A connected graph G is a BDH graph if and only if the coefficient of the linear term of q(G;x) equals 2. The latter coefficient referred to in Corollary 4.5, denoted by γ(G), is called the γ- invariant of G in analogy with the Crapo invariant β(G) which is the common value of the coefficients of the linear terms of t(G;x, y) where G has at least two edges. By a result due to Brylawski [9] (in the more general context of matroids) series–parallel graphs can be characterized by the value of the Crapo invariant as follows: a graph G is a series–parallel graph if and only if β(G) = 1. Both the corollaries above can be deduced directly by Theorem 3.1 after the following result due to Aigner and van der Holst [1]. Theorem 4.6 ([1]). If G is a bipartite graph, then q(G;x) = t(MG;x, x) where MG is the binary matroid generated by the reduced adjacency matrix of G and t(MG;x, x) is the Tutte polynomial of MG. Theorem 3.1 and Theorem 4.6 have the following straightforward consequence which re-proves directly Corollary 4.4 and Corollary 4.5. Corollary 4.7. If G is BDH graph, then q(G;x) = t(H;x, x) for some series–parallel graph H having G as fundamental graph and where t(H;x, x) is the Tutte polynomial of H , namely the Tutte polynomial of the cycle matroid of H . N. Apollonio et al.: A tight relation between series–parallel graphs and bipartite distance . . . 11 4.2 Characterizing series–parallel graphs by DFS-trees As credited by Syslo [35], Shinoda, Chen, Yasuda, Kajitani, and Mayeda, proved that series–parallel graphs can be completely characterized as in Theorem 4.8 by a property of their spanning trees, and Syslo himself gave a constructive algorithmic proof of the result [35]. Theorem 4.8 (S. Shinoda et al., 1981; Syslo, 1984). Every spanning tree of a connected graph H is a DFS-tree of one of its 2–isomorphic copies if and only if H is a series–parallel graph. When H is assumed to be 2–connected (an assumption that guarantees the connected- ness of its fundamental graphs), Theorem 4.8 will be equivalently stated as statement (1) below. Let T be a family of trees (or a family of oriented trees) and let G be a bipartite graph with color classes A and B. We say that G is a path/ T bipartite graph on A if there exist a member T of T and a bijection ξ : A → E(T ) such that, for each v ∈ B, {ξw | w ∈ NG(v)} is the edge–set (arc–set if T is oriented) of a simple cycle (directed circuit if T is oriented) in the (oriented) graph (V (T ), A∪B). Path/ T bipartite graphs on B are defined similarly. G is a path/ T bipartite graph if it is a path/ T bipartite graph on A or on B. G is a self–dual path/ T bipartite graph if it is a path/ T bipartite graph on both A and B. In any case T will be referred to as a supporting tree for G. For instance, if G ∼= K1,3 and G has color classes A = {a} and B = {α, β, γ} and if T is any family of paths containing paths of order 2 and order 4, then G is a path/ T bipartite graph: G is supported on A by a path of order 2 whose unique edge is labeled a and G is supported on B by a path of order 4 with three edges labeled α, β and γ. Recall that an arborescence is a directed tree with a single special node distinguished as the root such that, for each other vertex, there is a directed path from the root to that vertex. A DFS tree for a connected graph H (in the sense of [35]), is a pair (T, φ) consisting of a spanning tree T and an orientation φ of H , such that φT is a spanning arborescence of φH and for each f ∈ E(H) \ E(T ), φC(f, T ) is a directed circuit in φH (i.e, all arcs of φC(f, T ) are oriented in the same way). By choosing for T the class arborescence of arborescences, one can reformulate Theorem 4.8 in the following way (1) H is series–parallel if and only if for each spanning tree T of H the fundamental graph BT (H) is a self–dual path/arborescence bipartite graph. Indeed, if (T, φ) is a DFS-tree in a 2–isomorphic copy H ′ of H , then T is a spanning tree of graph H ′ whose cycle matroid is M(H); hence BH(T ) ∼= BH′(T ) and φT is a supporting arborescence for BH(T ). Conversely, suppose that G is a fundamental graph of H and that G is a path/arborescence bipartite graph. Let G have color classes A and B. Since G is a path/arborescence bipartite graph, then there is a supporting arborescence −→ T for G that induces an orientation φ of the graph H ′ = (V (T ), A ∪ B), T being the underlying undirected graph of −→ T . Clearly (T, φ) is a DFS tree in H ′ which in turn is 2–isomorphic to H because G is one of its fundamental graphs (i.e., H and H ′ have the same cycle matroid). Statement (1) is now a rather straightforward consequence of Corollary 3.8 and the fact that BDH graphs are self–dual path/arborescence bipartite graphs as shown by the following result proved in [4]. 12 Art Discrete Appl. Math. 5 (2022) #P1.05 Theorem 4.9 ([4]). Every connected BDH graph is a self–dual path/arborescence bipartite graph. Proof of (1). Let H be a 2–connected series–parallel graph. Then, by Theorem 3.1 BH(T ) is BDH for each spanning tree T of H . Hence, for every spanning tree T of H , BH(T ) is a self–dual path/arborescence bipartite graph by Theorem 4.9. Conversely, suppose that for every spanning tree T of a 2–connected graph H , the fundamental graph BH(T ) is a path/arborescence bipartite graph. Thus BH(T ) is chordal (see, e.g., [8]). Moreover, since if T ′ is any other spanning tree of H , then BH(T ′) is in the orbit of BH(T ), we conclude that each bipartite graph in the orbit of BH(T ) is a chordal bipartite graph. Therefore BH(T ) is a BDH graph by Corollary 3.8 and, consequently, H is a series–parallel graph. It is worth observing that, in the same way as Theorem 3.1 specializes de Fraysseix’s Theorem 3.2, Statement (1) specializes the following statement (see also [12]): (2) a bipartite graph is a bipartite circle graph if and only if it is a self–dual path/tree bipartite graph, tree being the class of trees. Proof. By Whitney’s planarity criterion [38] a graph is planar if and only if its cycle ma- troid is also co-graphic, namely, it is the dual matroid of another cycle matroid. Let now G be the fundamental graph of a 2–connected graph H with respect to some spanning tree T of H . Let A be the reduced adjacency matrix of G with rows indexed by the edges of T and columns indexed by the edges of T . Then, while [ I |A ] generates M(H), [ I |At ] generates M∗(H), the dual of M(H). Hence, when H is planar, by Whitney’s planarity criterion, M∗(H) is the cycle matroid of a 2–isomorphic copy of a plane dual H∗ of H . Therefore the neighbors of each vertex in the color class T spans a path in the co-tree T which is in turn a spanning tree of a 2–isomorphic copy of plane dual H∗ of H . In view of such a discussion it is reasonable to wonder whether there is a class of self dual path/ T0 bipartite graphs closed under edge–pivoting, where T0 is a family of trees sandwiched between trees and arborescences. The next result gives a negative answer in a sense. In what follows di-tree is the class of oriented trees. Theorem 4.10. If G is a connected bipartite graph whose orbit consists of self–dual path/di-tree bipartite graphs, then the orbit of G consists of path/arborescence bipartite graphs. Proof. Path/di-tree bipartite graphs are balanced (see [2]). Recall that a bipartite graph Γ is balanced if its reduced adjacency matrix does not contain the vertex-edge adjacency matrix of a chordless cycle of odd order. Equivalently, Γ is balanced if each hole of Γ has order congruent to zero modulo 4. Hence, since G and any other graph in its orbit is a self–dual path/di-tree bipartite graph, then G, and any other graph in its orbit must be balanced as well. Let G̃ be any member of [G] and suppose that G̃ contains a hole C. Let e ∈ E(C). The order t of C is at least eight, because G̃ is balanced. Nevertheless G̃e contains a hole of order t − 2 by Lemma 3.6. Since t − 2 ≡ 2 (mod 4) we conclude that any graph in the orbit of G must be hole-free. Therefore G is BDH by Corollary 3.9, and, by Theorem 3.1, it is the fundamental graph of a series–parallel graph. The thesis now follows by Statement (1). N. Apollonio et al.: A tight relation between series–parallel graphs and bipartite distance . . . 13 Remark 4.11. It is worth observing that by the proof above, if A is a class of balanced matrices closed under pivoting over GF (2), then A consists of totally balanced matrices, namely those matrices whose bipartite incidence graph is hole-free. Actually, and more sharply, in view of Corollary 3.9, every member of A is the incidence matrix of a γ-acyclic hypergraph [3]. 4.3 Packing paths and multi-commodity flows in series–parallel graphs In this section we give an application of Theorem 3.1 in Combinatorial Optimization. We show that a notoriously hard problem contains polynomially solvable instances when re- stricted to series–parallel graphs. Let H = (V,E) be a graph and let F ⊆ E be a set of prescribed edges of H called the nets of H . Following [19] a path P of H will be called F -admissible if it connects two vertices s, t of V with st ∈ F and E(P ) ⊆ E − F . Let U be the set of end-vertices of the nets. In the context of network-flow, vertices of U are thought of as terminals to be connected by a flow of some commodity (the nets are in fact also known as commodities). Let PF denote the family of all F -admissible paths of G and let PF,f ⊆ PF be the family of those F -admissible paths connecting the endpoints s,t of net f . An F -multiflow (see e.g. [33]), is a function λ : PF → R+, P 7→ λP . The multiflow is integer if λ is integer valued. The value of the F -multiflow on the net f is φf = ∑ P∈PF,f λP . The total value of the multiflow is the number φ = ∑ f∈F φf . Let w : E − F → Z+ be a function to be though of as a capacity function. An F -multiflow subject to w in H is an F -multiflow such that, ∑ P∈PF :E(P )∋e λP ≤ w(e), ∀e ∈ E − F (4.1) When w(e) = 1 for all e ∈ E − F , an integer multiflow is simply a collection of edge– disjoint F -admissible paths of H . The F -Max- Multiflow Problem is the problem of find- ing, for a given capacity function w, an F -multiflow subject to w of maximum total value. An F -multicut of H is a subset of B edges of E − F that intersects the edge–set of each F -admissible path. The name F -multicut is due to the fact that the removal of the edges of B from H leaves a graph with no F -admissible path: in the graph H − B it is not possi- ble to connect the terminals of any net. The capacity of the F -multicut B is the number∑ e∈B w(e). Multiflow Problems are very difficult problems (see [18], [19] and Vol. C, Chapter 70 in [33]). In [20] it has been shown that the Max-Multiflow Problem is NP-hard even for trees and even for {1, 2}-valued capacity functions. The problem though is shown to be polyno- mial time solvable for constant capacity functions by a dynamic programming approach. However, even for constant functions, the linear programming problem of maximizing the value of the multiflow over the system of linear inequalities (4.1) has not even, in general, 1 2 Z-valued optimal solutions. In [26], the NP-completeness of the Edge–Disjoint–Multi commodity Path Problem for series–parallel graphs (and partial 2–trees) has been estab- lished while, previously in [39], the polynomial time solvability of the same problem for partial 2–trees was proved under some restriction either on the number of the commodities (required to be a logarithmic function of the order of the graph) or on the location of the nets. Theorem 4.12. Let H = (V,E) be a 2–connected series–parallel graph and let F be the edge–set of any of its spanning co-trees. Then the maximum total value of an F -multiflow 14 Art Discrete Appl. Math. 5 (2022) #P1.05 equals the minimum capacity of an F -multicut. Furthermore, both a maximizing multiflow and a minimizing multicut can be found in strongly polynomial time. Proof. Let A be a {0, 1}m×n−valued matrix and b ∈ Zm+ be a vector. Let 1d be the all ones vector in Rd. Consider the linear programming problem max x∈Rn + { 1Tnx | Ax ≤ b } (4.2) and its dual min y∈Rm + { bT y | AT y ≥ 1n } . (4.3) By the results of Hoffman, Kolen and Sakarovitch [23] and Farber [17], if A is a totally balanced matrix (i.e., A is the reduced adjacency matrix of a bipartite chordal graph), then both the linear programming problems above have integral optimal solutions and, by linear programming duality, the two problems have the same optimum value. Furthermore, an integral optimal solution x∗ to the maximization problem in (4.2) satisfying the additional constraint x∗ ≤ 1n (4.4) and an integral optimal solution y∗ to the minimization problem in (4.3) satisfying the additional constraint y∗ ≤ 1n (4.5) can be found in strongly polynomial time. Let now H be a 2–connected graph and let F be the edge–set of a co–tree T of some spanning T tree of H . By giving a total order on the edge–set of T , one can define a vector b whose entries are the values of the capacity function w : E(H) − F → Z+. If A is the incidence matrix of PF , namely the matrix whose columns are the incidence vectors of the F -admissible paths of H , then A is a partial representation of M(H). Moreover, if H is series–parallel, then A is totally balanced: by Theorem 3.1, A is the reduced adjacency ma- trix of a BDH graph which is chordal being hole-free (by Theorem 3.7). On the other hand, integral solutions to the problem in (4.2) satisfying constraint (4.4) and to the problem in (4.3) satisfying constraint (4.5) are incidence vectors of F -multiflows and F -multicuts, respectively. Hence, both an F -multiflow of maximum value and an F -multicut of mini- mum capacity can be found in strongly polynomial-time by solving the linear programming problems above. Moreover, linear programming duality implies that the maximum value of an F -multiflow and the minimum capacity of an F -multicut coincide. 5 Two more proofs of Theorem 3.1 In this section, we give two more proofs of Theorem 3.1: one is due to an anonymous referee of an earlier version of the paper and it relies on the deep and refined notion of branch- and rank-width of a matroid (for the undefined terms given in the proof we address the reader to the references therein); the other fits the theory of double occurrences words and relies on a result in [5]. Second proof of Theorem 3.1. Suppose that a connected bipartite graph G is a fundamental graph of a 2-connected series parallel graph H . Since 2-connected graphs of branch-width at most 2 are exactly 2-connected series parallel graphs ([30]), the branch-width of H is N. Apollonio et al.: A tight relation between series–parallel graphs and bipartite distance . . . 15 at most 2. As proved in [22], the branch-width of a graph equals that of its cycle matroid. Hence, the branch-width of H equals the branch-width of M(H). By a result in [27], the branch-width of a binary matroid (in particular of a cycle matroid) equals the rank-width of any of its fundamental graphs plus 1. By definition, G is a fundamental graph of M(H) and thus rw(G)+1 = bw(M(H)) = bw(H) ≤ 2, where rw(·) and bw(·) denote the rank- width and branch-width parameters, respectively. Hence the rank-width of G is at most 1 and we conclude that G is bipartite distance hereditary because, still by a result in [27], distance hereditary graphs are precisely the graphs of rank-width at most 1. For the other direction, suppose that a connected bipartite graph G is distance-hereditary. Let MG be the binary matroid generated by the reduced adjacency matrix of G. By the same reasons (and the same notation) given above, it holds that bw(MG) = rw(G)+1 ≤ 2. By a result in [21], MG is a series parallel matroid (see [36] for the definition) and any such a matroid is the cycle matroid of a series parallel graph (see Lemma 4.2.12 in [36]). Hence MG = M(H) for some series parallel graph H . Furthermore, H is 2-connected, otherwise, G is disconnected. The third proof will require a result in [5]. Let C be an Eulerian cycle in a 4-regular labeled graph H and let w be the double occurrence word it induces (Section 3, following the first proof of Theorem 3.1). Recall that two vertices, say labeled a and b, are interlaced in w if w = uaxbyaz for some (possibly empty) intervals u, x, y and z of w. For two vertices u and v, labeled a and b, respectively, the uv-transposition of w is the word w uv = uaybxaz [5]. Thus a uv-transposition of w amounts to replace one of the subpaths of C connecting u and v with the other one. The relation between uv-transposition and uv pivoting is given in the next lemma which specializes a more general result in [5] (see also [13]). Lemma 5.1. Let H be a 4-regular graph and let w be any of the double occurrence words it induces. Further, let G(H,w) denote the interlacement graph of w. Suppose that G(H,w) is a bipartite graph. Then, for any edge uv of G(H,w) of H , one has G(H,w)uv = G(H,wuv). Third proof of Theorem 3.1. If G is a fundamental graph of a series–parallel graph, then MG is a binary matroid with no M(K4) minor by Dirac and Duffin’s characterization. Dominoes are fundamental graphs of K4 and holes can be pivoted to either dominoes or C6 (recall Lemma 3.6)—notice that C6 is a fundamental graph of K4 as well (Figure 1)–it follows that G is BDH-free by Lemma 3.3. Conversely, if G is BDH, then by Theorem 4.2 (in the language of Lemma 5.1), G ∼= G(m(H),w) for some series–parallel graph H (observe that m(H) is a 4-regular graph) and some code w. By Lemma 5.1, pivoting on edges G affects neither H nor m(H). Consequently, every graph in [G] is a BDH. Therefore MG has no M(K4) minor by Lemma 3.3 and Lemma 3.5 and G is a fundamental graph of such a matroid and therefore the fundamental graph of a series–parallel graph. ORCID iDs Nicola Apollonio https://orcid.org/0000-0001-6089-1333 Massimiliano Caramia https://orcid.org/0000-0002-9925-1306 Paolo Giulio Franciosa https://orcid.org/0000-0002-5464-4069 Jean-François Mascari https://orcid.org/0000-0002-0210-3375 16 Art Discrete Appl. Math. 5 (2022) #P1.05 References [1] M. 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ISSN 2590-9770 The Art of Discrete and Applied Mathematics 5 (2022) #P1.06 https://doi.org/10.26493/2590-9770.1402.d50 (Also available at http://adam-journal.eu) Sierpiński products of r-uniform hypergraphs Mark Budden* Department of Mathematics and Computer Science, Western Carolina University Cullowhee, NC, 28723, USA Josh Hiller Department of Mathematics and Computer Science, Adelphi University Garden City, NY 11530-0701, USA Received 8 August 2020, accepted 9 February 2021, published online 4 April 2022 Abstract If H1 and H2 are r-uniform hypergraphs and f is a function from the set of all (r− 1)- element subsets of V (H1) into V (H2), then the Sierpiński product H1 ⊗f H2 is defined to have vertex set V (H1)× V (H2) and hyperedges falling into two classes: (g, h1)(g, h2) · · · (g, hr), such that g ∈ V (H1) and h1h2 · · ·hr ∈ E(H2), and (g1, f({g2, g3, . . . , gr}))(g2, f({g1, g3, . . . , gr})) · · · (gr, f({g1, g2, . . . , gr−1})), such that g1g2 · · · gr ∈ E(H1). We develop the basic structure possessed by this product and offer proofs of numerous extremal properties involving connectivity, clique numbers, and strong chromatic numbers. Keywords: Hypergraph products, cliques, chromatic numbers. Math. Subj. Class.: 05C65, 05C15, 05C40 1 Introduction Sierpiński graphs were first introduced in 1997 by Klavžar and Milutinović [8] stemming from their work on the Tower of Hanoi problem. Since then, numerous properties and gen- eralizations of Sierpiński graphs have been extensively studied (e.g., see [6, 7, 9, 10, 12, 13], and [14]). Recently, Kovič, Pisanski, Zemljič, and Žitnik [11] have used Sierpiński graphs *Corresponding author. E-mail addresses: mrbudden@email.wcu.edu (Mark Budden), johiller@adelphi.edu (Josh Hiller) cb This work is licensed under https://creativecommons.org/licenses/by/4.0/ 2 Art Discrete Appl. Math. 5 (2022) #P1.06 as a motivation for a graph product structure, which they referred to as a Sierpiński product. Their introductory work on this product included proofs of the product’s basic properties involving connectivity, planarity, automorphism groups, and a consideration of the prod- uct with multiple factors. The present paper seeks to generalize the Sierpiński product to the setting of r-uniform hypergraphs and to describe its structure, with an emphasis on extremal properties. We begin with the construction of a Sierpiński product in the setting of graphs. Given graphs G1 and G2, and a function f : V (G1) −→ V (G2), the Sierpiński product G1⊗fG2 is defined to have vertex set V (G1)× V (G2) and edge set consisting of edges that fall into two classes: (g, h)(g, h′), such that g ∈ V (G1) and hh ′ ∈ E(G2), (g, f(g′))(g′, f(g)), such that gg′ ∈ E(G1). Edges in these classes are referred to as inner and connecting edges, respectively. Ob- serve that regardless of the choice of function f , the graph G1 ⊗f G2 is a subgraph of the lexicographic product G1[G2], defined to have vertex set V (G1)× V (H2) and edge set E(G1[G2]) = {(g, h)(g ′, h′) | (g = g′ and hh′ ∈ E(G2)) or gg ′ ∈ E(G1)}. Like the lexicographic product, the Sierpiński product is not commutative in general. For each vertex g ∈ V (G1), the subgraph induced by the set gG2 = {(g, h) | h ∈ V (G2)} is isomorphic to G2. It follows that when |V (G1)| = 1, the Sierpiński product G1 ⊗f G2 is isomorphic to G2, regardless of the choice of f . It is also easily confirmed that when |V (G2)| = 1, the function f must be constant and the Sierpiński product G1 ⊗f G2 is isomorphic to G1. Among these properties, it was proven in [11] that G1⊗fG2 is connected if and only if G1 and G2 are both connected. In Section 2, we consider a generalization of the Sierpiński product to r-uniform hy- pergraphs and prove several properties regarding connectivity. In Section 3, we turn our attention to clique numbers and the strong chromatic number. We note that in the case of the strong chromatic number, Theorems 3.2 and 3.4 and Corollary 3.5 are stated to include the case r = 2, offering new results involving the chromatic number of Sierpiński products of graphs. In Section 4, we conclude by offering some directions for future research and an alternate generalization of the Sierpiński product of r-uniform hypergraphs. 2 The Sierpiński product of r-uniform hypergraphs Recall that an r-uniform hypergraph H consists of a nonempty vertex set V (H) and a hyperedge set E(H), consisting of unordered r-tuples of distinct elements from V (H). For our purposes, we assume that all r-uniform hypergraphs are simple (i.e., we do not allow multi-hyperedges). When r = 2, this definition coincides with that of simple graphs. If H1 and H2 are r-uniform hypergraphs, then denote by ( V (H1) r−1 ) the set of all un- ordered (r − 1)-tuples of elements in V (H1). For a function f : ( V (H1) r − 1 ) −→ V (H2), M. Budden and J. Hiller: Sierpiński products of r-uniform hypergraphs 3 the Sierpiński product H1 ⊗f H2 has vertex set V (H1) × V (H2). The hyperedges in E(H1 ⊗f H2) have the following forms: (g, h1)(g, h2) · · · (g, hr), such that g ∈ V (H1) and h1h2 · · ·hr ∈ E(H2), and (g1, f({g2, g3, . . . , gr}))(g2, f({g1, g3, . . . , gr})) · · · (gr, f({g1, g2, . . . , gr−1})), such that g1g2 · · · gr ∈ E(H1). Hyperedges in the first class are called inner hyperedges, while those in the second class are called connecting hyperedges. This product agrees with the definition in Section 1 in the case where r = 2. For each g ∈ V (H1), the subhypergraph of H1 ⊗f H2 induced by gH2 = {(g, h) | h ∈ V (H2)} is isomorphic to H2. Among any r distinct g1H2, g2H2, . . . , grH2 chosen, there exists at most a single connecting hyperedge. In total, we find that H1 ⊗f H2 contains |V (H1)| · |E(H2)| inner hyperedges and |E(H1)| connecting hyperedges. Before considering examples and properties involving connectivity, we must recall some definitions. Recall that a Berge path of length ℓ is a sequence of ℓ + 1 distinct vertices v1, v2, . . . , vℓ+1 and distinct hyperedges e1, e2, . . . , eℓ such that vi, vi+1 ∈ ei for all i ∈ {1, 2, . . . , ℓ}. We write such a path as P = v1e1v2e2 · · · eℓvℓ+1 and observe that although the hyperedges are distinct, each pair of hyperedges may have numerous vertices in common. A Berge path P = v1e1v2e2 · · · eℓvℓ+1 forms a loose path if all vertices in P other than v2, v3, . . . , vℓ have degree 1. In this case, all vertices are necessarily distinct and P has order r+(r−1)(ℓ−1). While we have defined Berge paths and loose paths as “stand alone” hypergraphs, we also refer to subhypergraphs isomorphic to these hypergraph constructions by the same names. An r-uniform hypergraph H is called connected if for any distinct pair of vertices, there exists a Berge path that contains them both. An r-uniform hypergraph that is not connected is called disconnected. When an r-uniform hypergraph is connected, but the removal of any hyperedge (while retaining all vertices) disconnects it, then it is called minimally connected (e.g., see [2]). Given a Berge path P = v1e1v2e2 · · · eℓvℓ+1, if there exists a hyperedge eℓ+1 distinct from e1, e2, . . . , eℓ such that v1, vℓ+1 ∈ eℓ+1, then we say that C = v1e1v2e2 · · · eℓvℓ+1eℓ+1v1 is a Berge cycle. An r-uniform hypergraph is an r-uniform tree if it is connected and does not contain any Berge cycles. Other equivalent definitions for an r-uniform tree are given in [2] and [3]. In particular, note that every r-uniform tree is minimally connected, but not every minimally connected r-uniform hypergraph is an r-uniform tree. Example 2.1. For example, consider K (3) 4 , the complete 3-uniform hypergraph of order 4, and denote by P the 3-uniform loose path of size 2. Then if f : ( V (K (3) 4 ) 2 ) −→ V (P ) 4 Art Discrete Appl. Math. 5 (2022) #P1.06 is any constant function that maps to a vertex of degree 1 in P , the Sierpiński product K (3) 4 ⊗f P is given in Figure 1. Observe that each copy of gP is isomorphic to P and the hypergraph spanned by the connecting hyperedges (dashed in Figure 1) is isomorphic to K (3) 4 . Figure 1: The Sierpiński Product K (3) 4 ⊗f P , where P is a 3-uniform loose path of length 2 and f is a constant function whose range consists of a single vertex of degree 1 in P . The inner hyperedges are solid while the connecting hyperedges are dashed. Example 2.2. Let C denote the 3-uniform Berge cycle of size 2 and order 4 containing exactly two vertices of degree 1. Suppose that V (C) = {x1, x2, x3, x4}, where x1 and x4 have degree 1. Also, let P be the loose path described in Example 2.1, with vertex set V (P ) = {y1, y2, y3, y4, y5} such that y3 is the unique vertex of degree 2. Define the function f : ( V (C) 2 ) −→ V (P ) by f({x1, x2}) = y1, f({x1, x3}) = y2, f({x1, y4}) = y3, f({x2, x3}) = y3, f({x2, x4}) = y4, f({x3, x4}) = y5. Then the connecting hyperedges in C ⊗f P are given by e1 = (x1, y3)(x2, y2)(x3, y1) and e2 = (x2, y5)(x3, y4)(x4, y3). Since f is nonconstant, such a hypergraph becomes more difficult to illustrate. So in Fig- ure 2, we represent the connecting hyperedges by drawing segments from each hyperedge to the vertices they include. Also, note that the vertex (xi, yj) is labeled ij in this figure. Examples 2.1 and 2.2 provide illustrations of some 3-uniform Sierpiński products when the underlying hypergraphs are connected. We note that when f is a constant function (as M. Budden and J. Hiller: Sierpiński products of r-uniform hypergraphs 5 11 12 21 2322 31 33 32 4143 42 e e 1 2 34 35 4445 15 14 13 2524 Figure 2: The Sierpiński Product C⊗f P , where C is a 3-uniform Berge cycle of size 2 and order 4 containing exactly two vertices of degree 1, P is a 3-uniform loose path of length 2, and f is the surjective function described in Example 2.2. in Example 2.1), the resulting Sierpiński product may be considered a hypergraph general- ization of a rooted product graph (for example, see [5]). In Proposition 2.10 of [11], it was shown that when G1 and G2 are graphs, then G1 ⊗f G2 is connected if and only if G1 and G2 are connected. The following theorem considers connectivity for higher uniformity. Theorem 2.3. Assume that r ≥ 3, H1 and H2 are r-uniform hypergraphs, and f :( V (H1) r−1 ) −→ V (H2) is a function. If H1 and H2 are connected, then H1 ⊗f H2 is con- nected. If H1 ⊗f H2 is connected, then H1 is connected. If H1 ⊗f H2 is connected and f is a constant function, then H2 is connected. Proof. First, suppose that H1 and H2 are connected and let (g, h) and (g ′, h′) be vertices in H1 ⊗f H2. If g = g ′, then there exists a Berge path that contains both (g, h) and (g, h′) since gH2 is isomorphic to H2. Otherwise, suppose that g 6= g ′. Since H1 is connected, there exists a Berge path P = ge0g1e1g2e2 · · · gℓ−1eℓ−1g ′ in H1 (and we may write g = g0 and g ′ = gℓ). Each hyperedge ei in P corresponds with a unique hyperedge Ei in H1 ⊗f H2. Suppose that (gi, hi) ∈ Ei−1 while (gi, ki) ∈ Ei. If hi = ki, then Ei−1 and Ei are adjacent. If hi 6= ki, then there must exist a Berge path Qi connecting (gi, hi) to (gi, ki) in giH2. Thus, we are able to form a Berge path from (g, h) to (g′, h′) in H1 ⊗f H2 by following along the hyperedges E0, E1, . . . , Eℓ−1 and taking a detour along the Berge path Qi in giH2 whenever Ei−1 ∩ giH2 6= Ei ∩ giH2. 6 Art Discrete Appl. Math. 5 (2022) #P1.06 Finally, if (g′, k) ∈ Eℓ−1 and k 6= h ′, then we again follow the Berge path connecting (g′, k) to (g′, h′) in g′H2. Thus, H1 ⊗f H2 is connected. Now assume that H1 ⊗f H2 is connected. For any pair g, g′ ∈ V (H1), there exists a Berge path from gH2 to g ′H2 in H1 ⊗f H2 that corresponds with a Berge path from g to g ′ in H1. Thus, H1 is connected. Now assume that H1 ⊗f H2 is connected, f is a constant function, and k, k ′ ∈ E(H2) are distinct. Then there exists a Berge path from (g, k) to (g, k′) that does not contains any of the connecting hyperedges in H1⊗f H2 since all such hyperedges intersect gH2 at a single vertex. Such a Berge path necessarily corresponds with a Berge path in gH2, from which it follows that H2 must be connected. Theorem 2.3 is not as strong as in the case of graphs. This is demonstrated in Exam- ple 2.4, where a case is given in which H1 ⊗f H2 is connected, but H2 is disconnected. Example 2.4. Consider the Sierpiński product K (3) 4 ⊗f 2K (3) 3 , where 2K (3) 3 is the disjoint union of two 3-uniform hyperedges and f : ( V (K (3) 4 ) 2 ) −→ V (2K (3) 3 ) by f({x1, x2}) = y6, f({x1, x3}) = y1, f({x1, y4}) = y1, f({x2, x3}) = y1, f({x2, x4}) = y6, f({x3, x4}) = y6. Here, V (K (3) 4 ) = {x1, x2, x3, x4} and 2K (3) 3 consists of the hyperedges y1y2y3 and y4y5y6. The connecting hyperedges are given by e1 = (x1, y1)(x2, y1)(x3, y6) e2 = (x1, y6)(x2, y1)(x4, y6) e3 = (x1, y6)(x3, y1)(x4, y1) e4 = (x2, y6)(x3, y6)(x4, y1). From Figure 3, it is clear that K (3) 4 ⊗f 2K (3) 3 is connected even though 2K (3) 3 is discon- nected. Consider the case where H1 ⊗f H2 is minimally connected and H2 is assumed to be connected. Then by Theorem 2.3, H1 is also connected. When an inner hyperedge of H1 ⊗f H2 is removed, the removal of the corresponding hyperedge in H2 disconnects H2. When a connecting hyperedge is removed, the removal of the corresponding hyperedge in H1 disconnects H1. We obtain the following corollary. Corollary 2.5. Assume that r ≥ 3, H1 and H2 are r-uniform hypergraphs, and f :( V (H1) r−1 ) −→ V (H2) is a function. If H1 ⊗f H2 is minimally connected and H2 is con- nected, then H1 and H2 are minimally connected. In the more restrictive class of r-uniform trees, we obtain the following theorem. Theorem 2.6. Assume that r ≥ 3, H1 and H2 are r-uniform hypergraphs, and f :( V (H1) r−1 ) −→ V (H2) is a function. If H1 ⊗f H2 is an r-uniform tree and H2 is connected, then, H2 is an r-uniform tree. Proof. Assume that H2 is connected. Since H1 ⊗f H2 contains a subhypergraph isomor- phic to H2, H1 ⊗f H2 will contain a Berge cycle whenever H2 contains a Berge cycle. It follows that H2 is an r-uniform tree whenever H1 ⊗f H2 is an r-uniform tree. M. Budden and J. Hiller: Sierpiński products of r-uniform hypergraphs 7 11 12 21 2322 31 33 32 4143 42 e e e e 1 4 2 34 35 4445 15 14 13 2524 16 26 36 46 3 Figure 3: The Sierpiński Product K (3) 4 ⊗f 2K (3) 3 , where f is the function described in Example 2.4. Observe that K (3) 4 ⊗f 2K (3) 3 is connected even though 2K (3) 3 is disconnected. 3 Cliques and strong chromatic numbers In this section, we focus on the clique numbers and chromatic numbers of Sierpiński prod- ucts. These parameters serve as measures of connectivity and they play important roles in various extremal aspects of the study of hypergraphs. If H is any r-uniform hypergraph, then the clique number ω(H) is the maximum order of a complete subhypergraph of H . When r = 2, it is well-known that ω(G1[G2]) = ω(G1)ω(G2) (e.g., see [4]), and since G1 ⊗f G2 is a subgraph of G1[G2] for all f , it follows that ω(G1 ⊗f G2) ≤ ω(G1)ω(G2). When r ≥ 3, we obtain the following theorem. Theorem 3.1. Let r ≥ 3. If H1 and H2 are r-uniform hypergraphs and f : ( V (H1) r−1 ) −→ V (H2) is a function, then ω(H1 ⊗f H2) ≤ max{ω(H1), ω(H2)}. If ω(H2) ≥ ω(H1), then ω(H1 ⊗f H2) = ω(H2). If f is a constant function, then ω(H1 ⊗f H2) = max{ω(H1), ω(H2)}. Proof. The statement ω(H1 ⊗f H2) ≤ max{ω(H1), ω(H2)} 8 Art Discrete Appl. Math. 5 (2022) #P1.06 follows from Theorem 3 of [1], where it was proved that the lexicographic product satisfies ω(H1[H2]) = max{ω(H1), ω(H2)}, and the observation that H1 ⊗f H2 is a subhypergraph of H1[H2]. Since each gH2 con- tained in H1 ⊗f H2 is isomorphic to H2, we find that H1 ⊗f H2 contains a complete subhypergraph at least as large as a clique in H2. It follows that ω(H1 ⊗f H2) = ω(H2) whenever ω(H2) ≥ ω(H1). Finally, if f is a constant function, then the subhypergraph induced by H1h = {(g, h) | g ∈ V (H1)} is isomorphic to H1 for the unique vertex h in the image of f . So, H1 ⊗f H2 contains complete subgraphs of orders equal to both ω(H1) and ω(H2), giving us ω(H1 ⊗f H2) = max{ω(H1), ω(H2)} in this case. In the setting of r-uniform hypergraphs, there are many ways to generalize chromatic numbers. In this paper, we will focus on the strong chromatic number of an r-uniform hypergraph H . First, define a strong proper vertex coloring of an r-uniform hypergraph H to be a map c : V (H) −→ {1, 2, . . . , n} such that no two adjacent vertices in H receive the same color. Then the least n for which a strong proper vertex coloring exists is called the strong chromatic number of H , and is denoted χs(H). Our reasoning for focusing on this generalization is due to the re- lationship between the strong chromatic number and the existence of certain hypergraph homomorphisms. Recall that if H1 and H2 are two r-uniform hypergraphs, then a homo- morphism is a function φ : V (H1) −→ V (H2) such that if x1x2 · · ·xr ∈ E(H1), then φ(x1)φ(x2) · · ·φ(xr) ∈ E(H2). For any strong proper vertex coloring c : V (H) −→ {1, 2, . . . , n}, there is a natural homomorphism φ : V (H) −→ V (K (r) n ) given by mapping each vertex h ∈ V (H) to a vertex φ(h) ∈ V (K (r) n ) identified with the color class of h under c. This identification of strong proper vertex colorings of r-uniform hypergraphs with homomorphisms leads to a useful property. Specifically, if H1 and H2 are any r-uniform hypergraphs and if φ : V (H1) −→ V (H2) is a homomorphism, then χs(H1) ≤ χs(H2), (3.1) since any strong proper vertex coloring of H2 can be applied to H1 under φ. Theorem 3.2. For r ≥ 2, let H1 and H2 be r-uniform hypergraphs such that χs(H2) = n. Let φ : V (H2) −→ V (K (r) n ) be a homomorphism. For any function f : ( V (H1) r−1 ) −→ V (H2), χs(H1 ⊗f H2) ≤ χs(H1 ⊗f∗ K (r) n ), where f∗ := φ ◦ f . M. Budden and J. Hiller: Sierpiński products of r-uniform hypergraphs 9 Proof. Let c : V (H2) −→ {1, 2, . . . , n} be a strong proper vertex coloring of H2 such that χs(H2) = n. Note that c is necessarily surjective. Such a coloring naturally extends to the surjective homomorphism φ : V (H2) −→ K (r) n given by sending each vertex in h ∈ V (H2) to a vertex in K (r) n identified with the color class of h under c. Consider the map φ∗ : V (H1 ⊗f H2) −→ V (H1 ⊗f∗ K (r) n ) given by φ∗(g, h) = (g, φ(h)). We claim that φ∗ is a homomorphism. To prove this claim, let (g1, h1)(g2, h2) · · · (gr, hr) ∈ E(H1 ⊗f H2) and consider (g1, φ(h1))(g2, φ(h2)) · · · (gr, φ(hr)) ∈ E(H1 ⊗f∗ K (r) n ). Then either g1 = g2 = · · · = gr (in which case, φ(h1)φ(h2) · · ·φ(hr) ∈ E(K (r) n ) since h1h2 · · ·hr ∈ E(H2) and φ is a homomorphism) or g1g2 · · · gr ∈ E(H1) and φ(h1) = φ(f({g2, g3, . . . , gr})) = f ∗({g2, g3, . . . , gr}) φ(h2) = φ(f({g1, g3, . . . , gr})) = f ∗({g1, g3, . . . , gr}) ... φ(hr) = φ(f({g1, g2, . . . , gr−1})) = f ∗({g1, g2, . . . , gr−1}). It follows that φ∗ is a homomorphism, from which we conclude that χs(H1 ⊗f H2) ≤ χs(H1 ⊗f∗ K (r) n ) by (3.1). Note that in the previous theorem, the case r = 2 is included. In this case, χs is the usual chromatic number for graphs. We find that in general, Theorem 3.2 is the strongest statement that can be made, as the following example demonstrates a case where a strict inequality is satisfied. Example 3.3. Consider the complete 3-uniform hypergraph K (3) 4 with vertex set V (K (3) 4 ) = {x1, x2, x3, x4} and the 3-uniform loose path P of length 2 with vertex set V (P ) = {y1, y2, y3, y4, y5}, where y3 is the unique vertex in P with degree 2. Let f : ( V (K (3) 4 ) 2 ) −→ V (P ) be the function f({x1, x2}) = y1, f({x1, x3}) = y2, f({x1, y4}) = y3, f({x2, x3}) = y4, f({x2, x4}) = y5, f({x3, x4}) = y5. Then the Sierpiński product K (3) 4 ⊗f P is given in Figure 4, with vertex (xi, yi) labelled ij. The connecting hyperedges are given by e1 = (x1, y4)(x2, y2)(x3, y1) e2 = (x1, y5)(x2, y3)(x4, y1) e3 = (x1, y5)(x3, y3)(x4, y2) e4 = (x2, y5)(x3, y5)(x4, y4). 10 Art Discrete Appl. Math. 5 (2022) #P1.06 Since every strong proper vertex coloring of a hypergraph containing at least one hyperedge requires at least 3 colors, the coloring given in Figure 4 implies that χs(K (3) 4 ⊗f P ) = 3. Note that χs(P ) = 3, and we can identify a strong proper vertex coloring of P with the 11 12 21 2322 31 33 32 4143 42 e e e e 1 3 2 4 34 35 4445 15 14 13 2524 Figure 4: The Sierpiński product K (3) 4 ⊗f P , where P is a 3-uniform loose path of length 2 and f is the function given in Example 3.3. The strong proper vertex coloring given shows that this hypergraph has a strong chromatic number of 3. homomorphism φ : V (P ) −→ K (3) 3 that maps φ(y1) = φ(y5) = z1, φ(y2) = φ(y4) = z2, and φ(y3) = z3, where V (K (3) 3 ) = {z1, z2, z3}. Then f ∗ := φ ◦ f and the connecting hyperedges in K (3) 4 ⊗f∗ K (3) 3 are given by e′1 = (x1, z2)(x2, z2)(x3, z1) e′2 = (x1, z1)(x2, z3)(x4, z1) e′3 = (x1, z1)(x3, z3)(x4, z2) e′4 = (x2, z1)(x3, z1)(x4, z2). The resulting hypergraph K (3) 4 ⊗f∗ K (3) 3 is given in Figure 5. To obtain a strong proper vertex coloring, we begin by focusing on the connecting hyperedges e′2 and e ′ 3. Without loss of generality, suppose that (x1, z1) is red and (x4, z2) is blue. This forces (x4, z1) and (x3, z3) to be green and (x2, z3) to be blue. Then (x3, z1) must be red and (x2, z1) must be green. At this point, no color is available for (x2, z2) as (x2, z1) and (x2, z3) require it to be different from blue and green, but e ′ 1 already contains a red vertex. So, χs(K (3) 4 ⊗f∗ K (3) 3 ) ≥ 4, and one can continue with this process to obtain M. Budden and J. Hiller: Sierpiński products of r-uniform hypergraphs 11 11 13 12 21 2322 31 33 32 4143 42 e' e' e' e' 1 2 3 4 Figure 5: The Sierpiński product K (3) 4 ⊗f∗ K (3) 3 given in Example 3.3. a strong proper vertex 4-coloring of K (3) 4 ⊗f∗ K (3) 3 , showing that χs(K (3) 4 ⊗f∗ K (3) 3 ) = 4. Thus, our example demonstrates that there are cases where equality is not obtained in Theorem 3.2. While we can not be precise with the evaluation of the strong chromatic number in general, an exact evaluation can be found when f is assumed to be constant. Theorem 3.4. Let r ≥ 2 and suppose that H1 and H2 are r-uniform hypergraphs. If f : ( V (H1) r−1 ) −→ V (H2) is a constant function, then χs(H1 ⊗f H2) = max{χs(H1), χs(H2)}. Proof. When f : ( V (H1) r−1 ) −→ V (H2) is a constant function, the subhypergraph spanned by the connecting hyperedges is isomorphic to H1 and each gH2 is isomorphic to H2. It follows that χs(H1 ⊗f H2) ≥ max{χs(H1), χs(H2)}. To prove the opposite inequality, observe that all connecting hyperedges include at most one vertex from each gH2. Begin with a strong proper vertex coloring of the vertices spanned by the connecting hyperedges using at most χs(H1). The specific color assigned for at most one vertex in each gH2 does not affect the number of colors needed to form a strong proper vertex coloring of each gH2. Hence, it is possible to color H1 ⊗f H2 using max{χs(H1), χs(H2)} colors, completing the proof. An immediate consequence of this theorem is the following corollary. 12 Art Discrete Appl. Math. 5 (2022) #P1.06 Corollary 3.5. Let r ≥ 2 and suppose that H1 and H2 are r-uniform hypergraphs with χs(H2) = n. If f : ( V (H1) r−1 ) −→ V (H2) is a constant function, then χs(H1 ⊗f H2) = χs(H1 ⊗f∗ K (r) n ). 4 Conclusion We conclude our investigation of Sierpiński products of r-uniform hypergraphs by iden- tifying numerous directions for future study. Our primary focus has been on measures of connectivity, but there are many additional parameters (e.g., independence numbers, di- ameters, etc...) and applications worthy of inquiry. As subhypergraphs of lexicographic products, Sierpiński products may offer new results in Ramsey theory or bounds for certain Turán numbers (e.g., see [1]). Several of the topics studied in Kovič, Pisanski, Zemljič, and Žitnik’s paper [11], such as automorphism groups and products involving more than two factors, have not been considered here and should be considered for hypergraphs. Finally, the generalization of Sierpiński products to r-uniform hypergraphs that we have used seemed like the natural choice, but there are other ways in which one can make such a generalization. For example, let H1 and H2 be r-uniform hypergraphs. For a function f : V (H1) −→ V (H2), define the product H1 ⊗ f H2 to have vertex set V (H1)× V (H2). The hyperedges in E(H1 ⊗ f H2) have the following forms: (g, h1)(g, h2) · · · (g, hr), such that g ∈ V (H1) and h1h2 · · ·hr ∈ E(H2), and (g1, f(π(g1)))(g2, f(π(g2))) · · · (gr, f(π(gr))), such that g1g2 · · · gr ∈ E(H1) and π is any nontrivial permutation on {g1, g2, . . . , gr}. Observe that we have denoted this generalization of the Sierpiński product by writing f as a superscript rather than a subscript. Perhaps many of the results proved in this paper hold for this product as well. We leave its investigation for future work. ORCID iDs Mark Budden https://orcid.org/0000-0002-4065-6317 Josh Hiller https://orcid.org/0000-0001-5747-4061 References [1] M. Bruce, M. Budden and J. Hiller, Lexicographic products of r-uniform hypergraphs and some applications to hypergraph Ramsey theory, Australas. J. Comb. 70 (2018), 390–401, https: //ajc.maths.uq.edu.au/?page=get_volumes&volume=70. [2] M. Budden, J. Hiller and A. Penland, Minimally connected r-uniform hypergraphs, Aus- tralas. J. Comb. 82 (2022), 1–20, https://ajc.maths.uq.edu.au/?page=get_ volumes&volume=82. [3] M. Budden and A. Penland, Trees and n-good hypergraphs, Australas. J. Comb. 72 (2018), 329–349, https://ajc.maths.uq.edu.au/?page=get_volumes&volume=72. [4] D. Geller and S. Stahl, The chromatic number and other functions of the lexicographic product, J. Comb. Theory Ser. B 19 (1975), 87–95, doi:10.1016/0095-8956(75)90076-3. [5] C. D. Godsil and B. D. McKay, A new graph product and its spectrum, Bull. Austral. Math. Soc. 18 (1978), 21–28, doi:10.1017/s0004972700007760. M. Budden and J. Hiller: Sierpiński products of r-uniform hypergraphs 13 [6] A. M. Hinz, S. Klavžar and S. S. Zemljič, Sierpiński graphs as spanning subgraphs of Hanoi graphs, Cent. Eur. J. Math. 11 (2013), 1153–1157, doi:10.2478/s11533-013-0227-7. [7] M. Jakovac and S. Klavžar, Vertex-, edge-, and total-colorings of Sierpiński-like graphs, Dis- crete Math. 309 (2009), 1548–1556, doi:10.1016/j.disc.2008.02.026. [8] S. Klavžar and U. Milutinović, Graphs S(n, k) and a variant of the Tower of Hanoi problem, Czechoslov. Math. J. 47(122) (1997), 95–104, doi:10.1023/a:1022444205860. [9] S. Klavžar, U. Milutinović and C. Petr, 1-perfect codes in Sierpiński graphs, Bull. Austral. Math. Soc. 66 (2002), 369–384, doi:10.1017/s0004972700040235. [10] S. Klavžar and S. S. Zemljič, On distances in Sierpiński graphs: almost-extreme vertices and metric dimension, Appl. Anal. Discrete Math. 7 (2013), 72–82, doi:10.2298/aadm130109001k. [11] J. Kovič, T. Pisanski, S. Zemljič and A. Žitnik, The Sierpiński product of graphs, Ars Math. Contemp. (2022), doi:10.26493/1855-3974.1970.29e. [12] D. Parisse, On some metric properties of the Sierpiński graphs S(n, k), Ars Comb. 90 (2009), 145–160. [13] B. Xue, L. Zuo and G. Li, The Hamiltonicity and path t-coloring of Sierpiński-like graphs, Discrete Appl. Math. 160 (2012), 1822–1836, doi:10.1016/j.dam.2012.03.022. [14] B. Xue, L. Zuo, G. Wang and G. Li, Shortest paths in Sierpiński graphs, Discrete Appl. Math. 162 (2014), 314–321, doi:10.1016/j.dam.2013.08.029. ISSN 2590-9770 The Art of Discrete and Applied Mathematics 5 (2022) #P1.07 https://doi.org/10.26493/2590-9770.1373.60a (Also available at http://adam-journal.eu) Groups for which it is easy to detect graphical regular representations* Dave Witte Morris†, Joy Morris Department of Mathematics and Computer Science, University of Lethbridge, Lethbridge, Alberta. T1K 3M4, Canada Gabriel Verret‡ Department of Mathematics, The University of Auckland, Private Bag 92019, Auckland 1142, New Zealand Received 08 July 2020, accepted 09 December 2020, published online 4 April 2022 Abstract We say that a finite group G is DRR-detecting if, for every subset S of G, either the Cayley digraph Cay(G,S) is a digraphical regular representation (that is, its automorphism group acts regularly on its vertex set) or there is a nontrivial group automorphism ϕ of G such that ϕ(S) = S. We show that every nilpotent DRR-detecting group is a p-group, but that the wreath product Zp ≀Zp is not DRR-detecting, for every odd prime p. We also show that if G1 and G2 are nontrivial groups that admit a digraphical regular representation and either gcd ( |G1|, |G2| ) = 1, or G2 is not DRR-detecting, then the direct product G1 × G2 is not DRR-detecting. Some of these results also have analogues for graphical regular representations. Keywords: Cayley graph, GRR, DRR, automorphism group, normalizer. Math. Subj. Class.: 05C25,20B05 *This work was supported in part by the Natural Science and Engineering Research Council of Canada (grant RGPIN-2017-04905). The authors thank two anonymous referees for helpful comments. †Corresponding author. ‡The author is grateful to the N.Z. Marsden Fund for its support (via grant UOA1824). E-mail address: dave.morris@uleth.ca (Dave Witte Morris), joy.morris@uleth.ca (Joy Morris), g.verret@auckland.ac.nz (Gabriel Verret) cb This work is licensed under https://creativecommons.org/licenses/by/4.0/ 2 Art Discrete Appl. Math. 5 (2022) #P1.07 1 Introduction All groups and graphs in this paper are finite. Recall [1] that a digraph Γ is said to be a digraphical regular representation (DRR) of a group G if the automorphism group of Γ is isomorphic to G and acts regularly on the vertex set of Γ. If a DRR of G happens to be a graph, then it is also called a graphical regular representation (GRR) of G. Other terminology and notation can be found in Section 2. It is well known that if Γ is a GRR (or DRR) of G, then Γ must be a Cayley graph (or Cayley digraph, respectively), so there is a subset S of G such that Γ ∼= Cay(G,S) (and S is inverse-closed if Γ is a graph). It is traditional [5, p. 243] to let Aut(G,S) = {ϕ ∈ Aut(G) | ϕ(S) = S }. Since Aut(G,S) ⊆ Aut ( Cay(G,S) ) , it is obvious (and well known) that if Aut(G,S) is nontrivial, then Cay(G,S) is not a GRR (or DRR). In this paper, we discuss groups for which the converse holds: Definition 1.1. We say that a group G is GRR-detecting if, for every inverse-closed subset S of G, Aut(G,S) = {1} implies that Cay(G,S) is a GRR. Similarly, a group G is DRR-detecting if for every subset S of G, Aut(G,S) = {1} implies that Cay(G,S) is a DRR. Remark 1.2. Every Cayley graph is a Cayley digraph, so every DRR-detecting group is GRR-detecting. Definition 1.3. We say that a Cayley (di)graph Γ = Cay(G,S) on a group G witnesses that G is not GRR-detecting (respectively, not DRR-detecting) if Aut(G,S) = {1} but Γ is not a GRR (respectively, not a DRR) for G. An important class of DRR-detecting groups was found by Godsil. His result actually deals with vertex-transitive digraphs, rather than only the more restrictive class of Cayley graphs, but here is a special case of his result in our terminology: Theorem 1.4 (Godsil, cf. [5, Corollary 3.9]). Let G be a p-group and let Zp be the cyclic group of order p. If G admits no homomorphism onto the wreath product Zp ≀Zp then G is DRR-detecting (and therefore also GRR-detecting). Since Zp ≀ Zp is nonabelian, the following statement is an immediate consequence: Corollary 1.5. Every abelian p-group is DRR-detecting (and therefore also GRR-detecting). Remark 1.6. It is obvious (without reference to Theorem 1.4) that most abelian p-groups are GRR-detecting. Indeed, it is well known that every abelian group is GRR-detecting (unless it is an elementary abelian 2-group), because the nontrivial group automorphism x 7→ x−1 is an automorphism of Cay(G,S). The following result shows that the bound in Godsil’s theorem is sharp, in the sense that Zp ≀ Zp cannot be replaced with a larger p-group (when p is odd): Theorem 1.7. If p is an odd prime, then the wreath product Zp ≀ Zp is not GRR-detecting (and is therefore also not DRR-detecting). D. W. Morris, J. Morris and G. Verret: Groups for which it is easy to detect GRRs 3 Remark 1.8. The conclusion of Theorem 1.7 is not true for p = 2, because Z2 ≀ Z2 is GRR-detecting. This is a special case of the fact that if a group has no GRR, then it is GRR-detecting [4, Theorem 1.4]. The following two results provide additional examples, by showing that direct products often yield groups that are not DRR-detecting: Theorem 1.9. If G1 and G2 are nontrivial groups that admit a DRR (a GRR, respectively) and gcd ( |G1|, |G2| ) = 1, then G1×G2 is not DRR-detecting (not GRR-detecting, respec- tively). Theorem 1.10. If G1 admits a DRR (a GRR, respectively) and G2 is not DRR-detecting (not GRR-detecting, respectively), then G1×G2 is not DRR-detecting (not GRR-detecting, respectively). These two results are the main ingredients in the proof of the following theorem: Theorem 1.11. Every nilpotent DRR-detecting group is a p-group. Remark 1.12. The phrase “DRR-detecting” in Theorem 1.11 cannot be replaced with “GRR-detecting.” For example, every abelian group is GRR-detecting (unless it is an ele- mentary abelian 2-group), as was pointed out in Remark 1.6. Here is an outline of the paper. A few definitions and basic results are recalled in Sec- tion 2. Theorem 1.7 is proved in Section 3. A generalization of Theorem 1.9 is proved in Section 4, by using wreath products of digraphs. In Section 5, we recall some funda- mental facts about cartesian products of digraphs and use them to prove Theorem 1.10. Theorem 1.11 is proved in Section 6. 2 Preliminaries Definition 2.1. Recall that if S is a subset of a group G, then the Cayley digraph of G (with respect to the connection set S) is the digraph Cay(G,S) whose vertex set is G, such that there is a directed edge from g1 to g2 if and only if g2 = sg1 for some s ∈ S. If S is closed under inverses, then Cay(G,S) is a graph, and is called a Cayley graph. See Remark 4.4 for a general definition of the wreath product of two groups. The following special case is less complicated: Definition 2.2. Let Zp ≀Zp = Zp⋉ (Zp) p, where Zp acts on (Zp) p by cyclically permuting the coordinates: for (v1, v2, . . . , vp) ∈ (Zp) p and g ∈ Zp, we have (v1, v2, . . . , vp) g = (vg+1, vg+2, . . . , vp, v1, v2, . . . , vg). We will use the following well-known results. Theorem 2.3 (Babai [1, Theorem 2.1]). If a finite group does not admit a DRR, then it is isomorphic to Q8, (Z2) 2, (Z2) 3, (Z2) 4, or (Z3) 2, where Q8 is the quaternion group of order 8, which means Q8 = 〈 i, j, k | i 2 = j2 = k2 = −1, ij = k 〉. 4 Art Discrete Appl. Math. 5 (2022) #P1.07 Lemma 2.4. Let Ĝ be the right regular representation of G. Then: 1. Ĝ is contained in Aut ( Cay(G,S) ) for every subset S of G. 2. The normalizer of Ĝ in Aut ( Cay(G,S) ) is Aut(G,S)⋉ Ĝ. The latter has the following simple consequence: Lemma 2.5. If Γ is a Cayley digraph on G (a Cayley graph on G, respectively), then Γ witnesses that G is not DRR-detecting (not GRR-detecting, respectively) if and only if the regular representation of G is a proper self-normalizing subgroup of Aut(Γ). 3 Proof of Theorem 1.7 Let p be an odd prime. In this section, we show that Zp ≀ Zp is not GRR-detecting. (This proves Theorem 1.7.) To do this, we will construct a Cayley graph Γ on Zp ≀ Zp such that Γ is not a GRR, but the regular representation of Zp ≀ Zp is self-normalizing in Aut(Γ). In order to construct this graph, we first construct a certain group G that properly contains Zp ≀ Zp. We will then define Γ in such a way that G is contained in Aut(Γ). Let A ∼= Zp be a cyclic group of order p, and choose an irreducible representation of A on a vector space Q ∼= (Z2) n over the finite field with 2 elements, such that n ≥ 2. Now construct the corresponding semidirect product A ⋉ Q, which is a nonabelian group of order 2np. Choose a nontrivial 1-dimensional representation χ : Q → {±1} ⊆ Z×p (where Z × p de- notes the multiplicative group of nonzero elements of Zp), and induce it to a representation of A⋉Q on a vector space V over Zp [10, §3.3, pp. 28–30]. Since Q has index p in A⋉Q, the vector space V has dimension p, so V ∼= (Zp) p. Let G = (A⋉Q)⋉ V. Since the representation of A ⋉ Q on V is induced from a one-dimensional repre- sentation of the normal subgroup Q, the restriction to Q decomposes as a direct sum of one-dimensional representations: V = V1 ⊕ · · · ⊕ Vp, where each Vi is a subgroup of order p that is normalized by Q (cf. [10, Proposition 22, p. 58]). (More precisely, for each i ∈ {1, . . . , p}, there is some a ∈ A, such that the representation of Q on Vi is given by χa, where χa(g) = χ(ga −1 ) for g ∈ Q, and gh = h−1gh for g, h ∈ G.) Note that, since A normalizes Q, it must (cyclically) permute the Q-irreducible summands V1, . . . , Vp, so the Sylow p-subgroup A⋉ V of G is isomorphic to Zp ≀ Zp. Fix a nonidentity element a of A. Since A normalizes Q, we know that the coset Qa is fixed by the action of Q on the space Q\G of right cosets of Q. Also fix some nonzero v1 ∈ V1. Then, for each i ∈ {1, . . . , p}, let vi = v ai−1 1 , so vi is a nonzero element of Vi, and define z = v1 + v2 + · · ·+ vp, so z is a generator of the center Z(A⋉ V ). Now let S = ( 〈v1, v2〉r 〈v1〉 ) ∪ (a zQ)±1 ⊆ A⋉ V ⊆ G, and let Γ = Cay(A⋉ V, S). Since Q normalizes 〈v1〉 and 〈v2〉, and fixes the coset Qa in Q\G, it is clear that SQ = QS. Therefore, after identifying the vertex set A⋉ V of Γ with Q\QAV = Q\G in the natural way, we have G ⊆ Aut(Γ), via the natural action of G on Q\G. (Note that the action D. W. Morris, J. Morris and G. Verret: Groups for which it is easy to detect GRRs 5 of G on Q\G is faithful, because Q does not contain any nontrivial, normal subgroup of G. Otherwise, since the action of A on Q is irreducible, the entire subgroup Q would have to be normal, which would mean that Q acts trivially on Q\G. But this is false, because the representation of Q on V is nontrivial.) So Γ is not a GRR. Therefore, in order to show that Zp ≀ Zp ∼= A⋉ V is not GRR-detecting, it will suffice to show that Aut(A⋉ V, S) is trivial. To this end, let ϕ be an automorphism of A⋉ V that fixes S. We will show that ϕ is trivial. Since V is characteristic in A ⋉ V (for example, it is the only abelian subgroup of order pp), we know that ϕ(V ∩ S) = V ∩ S = 〈v1, v2〉r 〈v1〉 ⊆ 〈v1, v2〉. So ϕ ( 〈v1, v2〉 ) = ϕ ( 〈v1v2, v2〉 ) = 〈ϕ(v1v2), ϕ(v2)〉 ⊆ 〈ϕ(V ∩ S)〉 ⊆ 〈v1, v2〉. Since ϕ is injective, we conclude that ϕ fixes 〈v1, v2〉 (setwise). Then ϕ also fixes 〈v1, v2〉r S = 〈v1〉. We have ϕ(a) /∈ V (because a /∈ V and ϕ fixes V ), which means ϕ(a) = akv′ for some k ∈ Z×p and v ′ ∈ V . Then (since v′ centralizes V , because V is abelian) we have 〈v1, v2〉 = ϕ ( 〈v1, v2〉 ) ∋ ϕ(v2) = ϕ(v a 1 ) = ϕ(v1) ϕ(a) ∈ 〈v1〉 ak = 〈vk+1〉, so k ∈ {0, 1} ∩ Z×p = {1}, which means ϕ(a) = av′. Note that (since ϕ(V ) = V ) this implies ϕ(aV ) = aV. Since ϕ fixes 〈v1〉, we have ϕ(v1) = ℓv1 for some ℓ ∈ Z × p . For every i ∈ {1, . . . , p}, this implies ϕ(vi) = ϕ(v ai−1 1 ) = ϕ(v1) ϕ(ai−1) = (ℓv1) ai−1 = ℓvi. Since {v1, . . . , vp} generates V , we conclude that ϕ(v) = ℓv for all v ∈ V . To complete the proof, we will show that v′ is trivial and ℓ = 1. (This means that ϕ fixes a, and also fixes every element of V . So ϕ is the trivial automorphism, as desired.) For all z0 ∈ z Q, we have a · (v′ + ℓz0) = a v ′ · (ℓz0) = ϕ(a)ϕ(z0) = ϕ(a z0) ∈ ϕ(S ∩ aV ) = ϕ(S) ∩ ϕ(aV ) = S ∩ aV = a zQ. Therefore, if we write v′ = ∑p i=1 sivi (with si ∈ Zp) and z0 = ∑p i=1 tivi (with ti ∈ {±1}), then we have si + ℓti ∈ {±1} (mod p) for every i. 6 Art Discrete Appl. Math. 5 (2022) #P1.07 For any given i, the representation of Q on Vi is nontrivial, so we may choose z0 so that ti = −1. Therefore, we have si − ℓ ≡ ±1 (mod p). On the other hand, by letting z0 = z (and noting that si− ℓ 6≡ si+ ℓ (mod p)) we see that we also have si+ ℓ ≡ ∓1 (mod p). Adding these two equations and dividing by 2 yields si = 0 (for all i). So v ′ is trivial (which means ϕ(a) = a). All that remains is to show that ℓ = 1 (which means that ϕ acts trivially on V ). Suppose this is not true. (That is, suppose ℓ 6= 1.) We have ±ℓ = 0 + ℓ(±1) = si + ℓti ∈ {±1} (mod p), so this implies ℓ = −1. For convenience, let Z = 〈z〉 = Z(A⋉ V ). Note that, since ϕ(a) = a, we have a · (−z) = a · (ℓz) = ϕ(az) ∈ ϕ(S ∩ aV ) = S ∩ aV = a zQ, so there is some g ∈ Q, such that zg = −z. Since Z = 〈z〉, this implies that g is an element of the normaliser NQ(Z) of Z in Q. Also note that g is nontrivial, because z g = −z 6= z. Then, since NQ(Z) is normalized by A (because A normalizes Q and Z), the irreducibility of the representation of A on Q implies that NQ(Z) = Q. Hence, Q acts on Z by conjugation, so Q/CQ(Z) embeds in the cyclic group Aut(Z) ∼= Z × p . Since Q is an elementary abelian 2-group, this implies that |Q/CQ(Z)| ≤ 2. It is clear that |Q| ≥ 4 (because Q ∼= (Z2) n and n ≥ 2), so we conclude that CQ(Z) is nontrivial. Using once again the fact that the representation of A on Q is irreducible, we conclude that CQ(Z) = Q, which means that Q centralizes Z. However, since Z = 〈z〉 = 〈v1 + v2 + · · ·+ vp〉, and each 〈vi〉 = Vi is a Q-invariant subspace, this implies that Q centralizes each vi, and is therefore trivial on V . On the other hand, we have zg = −z 6= z, and g ∈ Q. This is a contradiction. 4 Using wreath products to construct witnesses In this section, we prove Corollary 4.11, which is a generalization of Theorem 1.9. Notation 4.1. In this section, N always denotes a normal subgroup of a group G. We let G = G/N , and use : G → G to denote the natural homomorphism. Notation 4.2. For each c ∈ G and each function f : G → N , we let ϕc,f be the permutation on G that is defined by ϕc,f (x) = xc f(x) for x ∈ G. Let W (G,N) be the set of all such permutations of G. Remark 4.3. Informally speaking, an element of W (G,N) is defined by choosing an ele- ment of G (or, more accurately, by choosing a coset representative) to permute the cosets of N , and then choosing an element of N to act on each coset. (The elements of N can be chosen independently on each coset.) We have ϕc,f = ϕc′,f ′ if and only if there is some n ∈ N , such that c ′ = cn and f ′(x) = n−1f(x) for all x. From this, it follows that |W (G,N)| = |G| · |N ||G|. D. W. Morris, J. Morris and G. Verret: Groups for which it is easy to detect GRRs 7 Remark 4.4. The usual definition of the wreath product of two groups K and H is essen- tially: K ≀H = W (K ×H, {1} ×H). Definition 4.5. Recall that the wreath product X ≀ Y of two (di)graphs X and Y is the (di)graph whose vertex set is the cartesian product V (X) × V (Y ), and with a (directed) edge from (x1, y1) to (x2, y2) if and only if either there is a (directed) edge from x1 to x2 or x1 = x2 and there is a (directed) edge from y1 to y2. This is also known as the lexico- graphic product of X and Y . The following two observations are well known (and fairly immediate from the defini- tions). The first is a concrete version of the Universal Embedding Theorem, which states that G is isomorphic to a subgroup of (G/N) ≀N . Lemma 4.6. W (G,N) is a subgroup of the symmetric group on G. It is isomorphic to the wreath product G ≀N , and contains the regular representation of G. Lemma 4.7. Suppose Cay(G,S1) is a loopless Cayley digraph on G, and Cay(N,S2) is a Cayley digraph on N . Let S1 = { g ∈ G | g ∈ S1 }. Then Cay(G,S1 ∪ S2) ∼= Cay(G,S1) ≀ Cay(N,S2), and W (G,N) is contained in the automorphism group of Cay(G,S1 ∪ S2). The following result is a special case of the general principle that the automorphism group of a wreath product of digraphs is usually the wreath product of the automorphism groups. We have stated it only for DRRs, making use of some straightforward observations about the automorphism group of a DRR on more than 2 vertices, but the much more general statement in [3] applies to all vertex-transitive digraphs. Lemma 4.8 (cf. Dobson-Morris [3, Theorem 5.7]). Assume that the graphs Cay(G,S1) and Cay(N,S2) are loopless DRRs, and let S1 be as in Lemma 4.7. If either |G| 6= 2 or |N | 6= 2, then Aut ( Cay(G,S1 ∪ S2) ) = W (G,N). In light of Lemmas 2.5 and 4.8, it is of obvious interest to us to determine when the regular representation of G is self-normalizing in W (G,N). Our next result is the answer to this question. Recall that the abelianization of a group H is the largest abelian quotient of H , or, in other words, the quotient group H/[H,H], where [H,H] is the commutator subgroup of H . Theorem 4.9. Let N be a normal subgroup of G. Then the regular representation of G is self-normalizing in W (G,N) if and only if 1. Z(N) ≤ Z(G), and 2. the order of the abelianization of G/N is relatively prime to |Z(N)|. Proof. (⇒) We prove the contrapositive. (1) If Z(N) 6≤ Z(G), then there exists n ∈ Z(N) such that n /∈ Z(G). Conjugation by n is an element of W (G,N) that normalizes the right regular representation of G, but is not in the right regular representation of G. (2) If the order of the abelianization of G/N is not relatively prime to |Z(N)|, then there is a 8 Art Discrete Appl. Math. 5 (2022) #P1.07 nontrivial homomorphism f : G → Z(N). We may assume that hypothesis (1) is satisfied, and then it is straightforward to verify that the corresponding element ϕf,1 of W (G,N) normalizes the right regular representation of G: ϕf,1(xg) = xg f(xg) (definition of ϕf,1) = x f(xg) g (f(xg) ∈ f(G) ⊆ Z(N) ⊆ Z(G)) = x f(x) f(g) g (f is a homomorphism) = ϕf,1(x) · f(g) g (definition of ϕf,1). (⇐) By Lemma 2.4, it suffices to show that Aut(G) ∩W (G,N) is trivial. To this end, let ϕ ∈ Aut(G) ∩ W (G,N). Since ϕ ∈ W (G,N), there exist c ∈ G and f : G → N , such that ϕ(x) = xc f(x) for all x ∈ G. Since ϕ is a group automorphism we know ϕ(1) = 1 ∈ N , so we may assume c = 1, after multiplying c on the right by an element of N . Then we must have f(1) = 1. Now, for each n ∈ N , we have n = 1, so ϕ(n) = n · f(n) = n · f(1) = n · 1 = n. Therefore, for all g ∈ G and n ∈ N , we have gn · f(g) = gn · f(gn) = ϕ(gn) = ϕ(g)ϕ(n) = g f(g) · n, so n · f(g) = f(g) · n. Since this is true for all n ∈ N , we conclude that f(g) ∈ Z(N). Since Z(N) ⊆ Z(G), this implies f(g) ∈ Z(G) for all g. Therefore, for all g, h ∈ G, we have gh · f(gh) = ϕ(gh) = ϕ(g)ϕ(h) = g f(g) · h f(h) = gh · f(g) f(h). So f is a group homomorphism. Since f(G) is contained in Z(N), which is abelian, we see from (2) that f must be trivial. Since c is also trivial, we conclude that ϕ(x) = x for all x. Since ϕ is an arbitrary element of Aut(G)∩W (G,N), this completes the proof. Remark 4.10. A slight modification of the proof of Theorem 4.9 shows that if Ĝ is the right regular representation of G, then the normalizer of Ĝ in W (G,N) is { ϕc,f | c ∈ G, f ∈ Z 1 ( G,Z(N) ) } , where Z1 ( G,Z(N) ) = { f : G → Z(N) | f(gh) = f(g)h f(h) for all g, h ∈ G } is the set of all “1-cocycles” or “crossed homomorphisms” from G to Z(N) (in the termi- nology of group cohomology [12]). This fact is presumably known. It may also be of interest to note that hypotheses (1) and (2) in Theorem 4.9 are obvi- ously satisfied when Z(N) is trivial. Combining the results of this section, we obtain the following. D. W. Morris, J. Morris and G. Verret: Groups for which it is easy to detect GRRs 9 Corollary 4.11. Let N be a nontrivial, proper, normal subgroup of G, such that N and G/N each admit a DRR (or, respectively, a GRR). If 1. Z(N) ≤ Z(G), and 2. the order of the abelianization of G/N is relatively prime to |Z(N)|, then G is not DRR-detecting (respectively, not GRR-detecting). More precisely, if we let Γ1 be a DRR (respectively, GRR) on G/N and Γ2 be a DRR (respectively, GRR) on N , then Γ1 ≀Γ2 witnesses that G is not DRR-detecting (respectively, not GRR-detecting). Proof. Clearly, either |G| 6= 2 or |N | 6= 2. It then follows by Lemma 4.7 and Lemma 4.8 that Aut ( Γ1 ≀ Γ2) = W (G,N). By Theorem 4.9, the regular representation of G is self- normalizing in W (G,N), therefore Γ1 ≀Γ2 witnesses that G is not DRR-detecting (respec- tively, not GRR-detecting). Note that Theorem 1.9 can be obtained from Corollary 4.11 by letting G = G1 × G2 and N = G2. 5 Using cartesian products to construct witnesses Definition 5.1. Recall that the cartesian product X  Y of two (di)graphs X and Y is the (di)graph whose vertex set is the cartesian product X × Y , such that there is a (directed) edge from (x1, y1) to (x2, y2) if and only if either x1 = x2 and there is a (directed) edge from y1 to y2, or y1 = y2, and there is a (directed) edge from x1 to x2. We say that a (di)graph is prime (with respect to cartesian product) if it has more than one vertex, and is not isomorphic to the cartesian product of two (di)graphs, each with more than one vertex. It is well known that every (di)graph can be written uniquely as a cartesian product of prime factors (up to a permutation of the factors), but we do not need this fact. To avoid the need to consider permutations of the factors, the following result includes the hypothesis that the factors are pairwise non-isomorphic. (This is not assumed in [11], which also considers isomorphisms between two different cartesian products, instead of only automorphisms of a single digraph.) The upshot is that, in this situation, the automor- phism group of the cartesian product is the direct product of the automorphism groups. Theorem 5.2 (Walker, cf. [11, Theorem 10]). Let Γ1, . . . ,Γk be weakly connected prime digraphs that are pairwise non-isomorphic. If ϕ is an automorphism of Γ1  · · ·  Γk, then for each i, there is an automorphism ϕi of Γi such that, for every vertex (v1, . . . , vk) of Γ1  · · ·  Γk, we have ϕ(v1, . . . , vk) = ( ϕ1(v1), . . . , ϕk(vk) ) . Prime graphs are quite abundant: Theorem 5.3 (Imrich [7, Theorem 1]). If Γ is a graph (with more than one vertex), such that neither Γ nor its complement Γ is prime, then Γ is one of the following: 1. the cycle of length 4 or its complement (two disjoint copies of K2); 2. the cube or its complement (the graph K2 ×K4); 10 Art Discrete Appl. Math. 5 (2022) #P1.07 3. K3  K3 (which is self-complementary); or 4. K2  ∆, where ∆ is the graph obtained by deleting an edge from K4 (which is self-complementary). The following is an analogous result for digraphs. (Recall that a digraph is proper if it is not a graph.) Theorem 5.4 (Grech-Imrich-Krystek-Wojakowski [6, Theorem 1.2] and Morgan-Morris- Verret [8, Theorem 2.2]). If Γ is a proper digraph, then at least one of Γ or Γ is prime. Corollary 5.5. If a nontrivial group G admits a DRR (respectively, GRR), then it admits a DRR (respectively, GRR) that is prime (and weakly connected). Furthermore, if G is not DRR-detecting (respectively, not GRR-detecting), then there is a witness that is prime (and weakly connected). Proof. First, note that Γ and Γ have the same automorphism group, so Γ is a DRR (GRR, respectively) for G if and only if Γ is. Similarly, Γ is a witness that G is not DRR-detecting (GRR-detecting, respectively) if and only if Γ is. Also note that if a prime digraph Γ is not weakly connected, and is either a DRR or a witness that some group is not DRR-detecting, then Γ = K2 (so Γ is prime and weakly connected). This is because any vertex-transitive digraph is isomorphic to Γ0 Kn, where Γ0 is a weakly connected component of the digraph, and n is the number of components. Suppose that Γ is a GRR for G. By Theorem 5.3, at least one of Γ or Γ is prime with respect to cartesian product, unless Γ is one of the graphs listed in that theorem, but none of those graphs is a GRR, because the automorphism group does not act regularly on the set of vertices: 1. the automorphism group of a cycle of length 4 (or its complement) is the dihedral group of order 8; 2. the automorphism group of the cube (or its complement) is Z2 × Sym(4), of order 48; 3. the automorphism group of K3  K3 is Z2 ≀ Sym(3), of order 72; and 4. the graph K2 ∆ is not vertex-transitive (it is not even true that all vertices have the same valency). Now, suppose that Γ is a DRR for G. We may assume that Γ is a proper digraph. (Otherwise, Γ is a GRR, so the preceding paragraph applies.) Then, by Theorem 5.4, either Γ or Γ is prime with respect to cartesian product. Finally, suppose Γ is a witness that G is not DRR-detecting (or not GRR-detecting, respectively), such that neither Γ nor Γ is prime. This implies that Γ is one of the graphs listed in Theorem 5.3. (So G is not GRR-detecting.) However, it is easy to see that none of the graphs listed in Theorem 5.3 is a witness. First, recall that a p-subgroup of a group cannot be self-normalizing unless it is a Sylow subgroup. Therefore (by Lemma 2.5), if a graph Γ of prime-power order pk is a witness that some group is not GRR-detecting, then pk must be the largest power of p that divides Aut(Γ). This shows that the graphs in (1) and (2) are not witnesses. If Γ is as described in (3), then the only regular subgroup of Aut(Γ) is the unique (Sylow) subgroup of order 9, which is normal, and is therefore obviously not self-normalizing. Finally, as noted above, the graphs in (4) are not vertex-transitive. D. W. Morris, J. Morris and G. Verret: Groups for which it is easy to detect GRRs 11 Proof of Theorem 1.10. For simplicity, we consider only DRRs (because the proof is the same for GRRs). Let Γ1 = Cay(G1, S1) be a DRR for G1, and let Γ2 = Cay(G2, S2) be a witness that G2 is not DRR-detecting. By Corollary 5.5, we may assume that Γ1 and Γ2 are prime with respect to cartesian product (and are weakly connected). Since Γ1 is a DRR, but Γ2 is not, we know that Γ1 6∼= Γ2. Therefore, we see from Theorem 5.2 that Aut(Γ1  Γ2) = Aut(Γ1)×Aut(Γ2). Since Γ2 is not a DRR, Γ1Γ2 is not a DRR. Similarly, since the regular representation of G1 is all of Aut(Γ1) and the regular representation of G2 is self-normalizing in Aut(Γ2), the regular representation of G1 ×G2 is self-normalizing in Aut(Γ1  Γ2). So Γ1  Γ2 is a witness that G1 ×G2 is not DRR-detecting. 6 Nilpotent DRR-detecting groups are p-groups In this section, we prove Theorem 1.11, which states that if a nilpotent group is not a p- group, then it is not DRR-detecting. In most cases, this follows easily from Theorems 1.9 and 1.10, but there is one special case that requires a different proof: Lemma 6.1. If H is a nontrivial group of odd order and H 6∼= Z3 × Z3, then Q8 × H is not DRR-detecting. Proof. From Theorem 2.3, we see that H admits a DRR (because it has odd order, but is not Z3 × Z3), so we may let Cay(H,S1) be a DRR. Let S = S1 ∪ {i} ∪ jH ⊆ G. It suffices to show that Aut(G,S) = {1} and that Cay(G,S) is not a DRR. Let ϕ ∈ Aut(G,S). We can characterise S1 as the set of all elements of S that have odd order. Thus, we must have ϕ(S1) = S1, so H = 〈S1〉 is fixed setwise by ϕ. Since the identity vertex is also fixed and the induced subgraph on H is a DRR, every element of H must be fixed by ϕ. We can use this fact to distinguish i from the elements of jH (all of which differ from each other by elements of H), so i is fixed by ϕ. Finally, j is the unique element of order 4 in jH , so it too is fixed by ϕ. We now know that ϕ is an automorphism of G that fixes every element of a generating set for G. So ϕ must be trivial. All that remains is to show that Cay(G,S) is not a DRR. Fix a nontrivial element h ∈ H , and define a permutation τ of G by τ(x) = { x if x ∈ 〈H, i〉; xh if x ∈ j〈H, i〉. Note that τ is a permutation of G, because right multiplication by h is a permutation of G that fixes 〈H, i〉 setwise. We claim that τ is an automorphism of Cay(G,S). First, note that a directed edge of the form g → s1g or g → ig either has both of its endpoints in 〈H, i〉, or has both of its endpoints in j〈H, i〉. Since right multiplication by h is an automorphism of Cay(G,S), it is clear that τ preserves such directed edges. The remaining directed edges are of the form g → gjh′ for some h′ ∈ H . Multiplying either g or gjh′ on the right by h results in another such directed edge. This completes the proof that τ is an automorphism of Cay(G,S). Proof of Theorem 1.11. Let G be a nilpotent group, and assume that G is not a p-group. (Note that |G| is divisible by at least two distinct primes.) We will show that G is not DRR-detecting. 12 Art Discrete Appl. Math. 5 (2022) #P1.07 Case 1. |G| is divisible by at least three distinct primes. Let p be the largest prime divisor of |G| and let P be a Sylow p-subgroup of G. Since G is nilpotent, we may write G = P × H for some subgroup H with gcd ( |P |, |H| ) = 1. Since p is the largest of at least three primes dividing |G|, neither P nor H is a 2-group or a 3-group, so we see from Theorem 2.3 that P and H each admit a DRR. Therefore, Theorem 1.9 implies that G = P ×H is not DRR-detecting. Case 2. |G| is divisible by precisely two distinct primes p and q. Since G is nilpotent, we have G = P ×Q, where P is a Sylow p-subgroup and Q is a Sylow q-subgroup of G. If P and Q each admit a DRR, then Theorem 1.9 implies that G = P ×Q is not DRR-detecting. We may thus assume, without loss of generality, that P does not admit a DRR. Using Theorem 2.3 and Lemma 6.1 and interchanging P and Q if necessary, we may assume that P is isomorphic to one of (Z2) 2, (Z2) 3, (Z2) 4, or (Z3) 2. Thus, we may write P = (Zp) r, with r ≥ 2. Since (Zp) r−1 × Q is not a p-group, we may assume, by induction on |G|, that it is not DRR-detecting. Also note that Zp admits a DRR. (Take the directed p-cycle −→ Cp if p ≥ 3; or take K2 if p = 2.) Therefore, by applying Theorem 1.10 with G1 = Zp and G2 = (Zp) r−1 × Q, we see that the group G = Zp × ( (Zp) r−1 × Q ) is not DRR- detecting. ORCID iDs Joy Morris https://orcid.org/0000-0003-2416-669X Gabriel Verret https://orcid.org/0000-0003-1766-4834 References [1] L. Babai, Finite digraphs with given regular automorphism groups, Period. Math. Hungar. 11 (1980), 257–270, doi:10.1007/bf02107568. [2] J. Dixon and B. Mortimer, Permutation Groups, volume 163 of Graduate Texts in Mathematics, Springer-Verlag, New York, 1996, doi:10.1007/978-1-4612-0731-3. [3] E. Dobson and J. Morris, Automorphism groups of wreath product digraphs, Electron. J. Com- bin. 16 (2009), Research Paper 17, 30, doi:10.37236/106. [4] C. D. Godsil, GRRs for nonsolvable groups, in: Algebraic methods in graph theory, Vol. I, II (Szeged, 1978), North-Holland, Amsterdam-New York, volume 25 of Colloq. Math. Soc. János Bolyai, pp. 221–239, 1981, https://www.elsevier.com/books/ algebraic-methods-in-graph-theory/sos/978-0-444-85442-1. [5] C. D. Godsil, On the full automorphism group of a graph, Combinatorica 1 (1981), 243–256, doi:10.1007/bf02579330. [6] M. Grech, W. Imrich, A. D. Krystek and L. u. J. Wojakowski, Direct product of automorphism groups of digraphs, Ars Math. Contemp. 17 (2019), 89–101, doi:10.26493/1855-3974.1498. 77b. [7] W. Imrich, On products of graphs and regular groups, Israel J. Math. 11 (1972), 258–264, doi:10.1007/bf02789317. [8] L. Morgan, J. Morris and G. Verret, Digraphs with small automorphism groups that are Cayley on two nonisomorphic groups, Art Discrete Appl. Math. 3 (2020), Paper No. 1.01, 11, doi: 10.26493/2590-9770.1254.266. D. W. Morris, J. Morris and G. Verret: Groups for which it is easy to detect GRRs 13 [9] J. Morris and P. Spiga, Classification of finite groups that admit an oriented regular representa- tion, Bull. Lond. Math. Soc. 50 (2018), 811–831, doi:10.1112/blms.12177. [10] J. Serre, Linear Representations of Finite Groups, volume 42 of Graduate Texts in Mathematics, Springer-Verlag, New York, 1977, doi:10.1007/978-1-4684-9458-7. [11] J. W. Walker, Strict refinement for graphs and digraphs, J. Combin. Theory Ser. B 43 (1987), 140–150, doi:10.1016/0095-8956(87)90018-9. [12] Wikipedia, Group cohomology — Wikipedia, the free encyclopedia, https://en. wikipedia.org/wiki/Group_cohomology, 2021. ISSN 2590-9770 The Art of Discrete and Applied Mathematics 5 (2022) #P1.08 https://doi.org/10.26493/2590-9770.1414.f67 (Also available at http://adam-journal.eu) Circulant almost cross intersecting families Michal Parnas* The Academic College of Tel-Aviv-Yaffo, Tel-Aviv, Israel Received 16 July 2020, accepted 29 April 2021, published online 4 April 2022 Abstract Let F and G be two t-uniform families of subsets over [k] = {1, 2, ..., k}, where |F| = |G|, and let C be the adjacency matrix of the bipartite graph whose vertices are the subsets in F and G, where there is an edge between A ∈ F and B ∈ G if and only if A ∩ B 6= ∅. The pair (F ,G) is q-almost cross intersecting if every row and column of C has exactly q zeros. We further restrict our attention to q-almost cross intersecting pairs that have a circulant intersection matrix Cp,q , determined by a column vector with p > 0 ones followed by q > 0 zeros. This family of matrices includes the identity matrix in one extreme, and the adjacency matrix of the bipartite crown graph in the other extreme. We give constructions of pairs (F ,G) whose intersection matrix is Cp,q , for a wide range of values of the parameters p and q, and in some cases also prove matching upper bounds. Specifically, we prove results for the following values of the parameters: (1) 1 ≤ p ≤ 2t− 1 and 1 ≤ q ≤ k− 2t+1. (2) 2t ≤ p ≤ t2 and any q > 0, where k ≥ p+ q. (3) p that is exponential in t, for large enough k. Using the first result we show that if k ≥ 4t − 3 then C2t−1,k−2t+1 is a maximal isolation submatrix of size k × k in the 0, 1-matrix Ak,t, whose rows and columns are labeled by all subsets of size t of [k], and there is a one in the entry on row x and column y if and only if subsets x, y intersect. Keywords: Circulant matrix, intersecting sets, Boolean rank, isolation set. Math. Subj. Class.: 05D05, 15B34 1 Introduction One of the fundamental results of extremal set theory is the theorem of Erdős, Ko and Rado [8], which shows that the size of an intersecting t-uniform family of subsets over *The author would like to thank the referee for the useful comments that helped improve the presentation of the paper. E-mail address: michalp@mta.ac.il (Michal Parnas) cb This work is licensed under https://creativecommons.org/licenses/by/4.0/ 2 Art Discrete Appl. Math. 5 (2022) #P1.08 [k] = {1, 2, ..., k} is bounded above by ( k−1 t−1 ) . Numerous variations of the original problem have been suggested and studied over the years. Among them is the problem of cross intersecting families of subsets (e.g. [4, 10, 14, 17, 21]). Specifically, if F and G are two t-uniform families of subsets over [k], then the pair (F ,G) is cross intersecting if every subset in F intersects with every subset in G and vice versa. Pyber [21] proved that in this case |F| · |G| ≤ ( k−1 t−1 )2 . Many of the extremal combinatorial problems considered so far can be inferred as re- sults about maximal submatrices of the 0, 1-matrix Ak,t of size ( k t ) × ( k t ) , whose rows and columns are labeled by all subsets of size t of [k], and there is a one in the entry on row x and column y if and only if subsets x, y intersect. Hence, in this setting, the re- sult of Erdős, Ko and Rado can be inferred as stating the size of the largest all-one square principal submatrix of Ak,t, and the result of Pyber states the size of the largest all-one submatrix of Ak,t. We note that considering the classic results of extremal combinatorics as maximal submatrices of Ak,t, allows us to employ tools from algebra in addition to the combinatorial techniques. Another variation of the problem of cross intersecting families was introduced by Gerb- ner et al. [11], which defined the notion of a q-almost cross intersecting pair (F ,G). Here every subset in F does not intersect with exactly q subsets in G and vice versa. If F = G then F is called a q-almost intersecting family of subsets. Hence, if C is the submatrix of Ak,t whose rows are indexed by the subsets of F and columns by the subsets of G, then every row and column of C has exactly q zeros. Using a classic theorem of Bollobás [3], it is possible to prove that the largest square submatrix C of Ak,t, representing a 1-almost cross intersecting pair, is of size ( 2t t ) × ( 2t t ) . A theorem proved in [11] shows that if C is a submatrix of size n × n of Ak,t, with exactly q zeros in each row and column, then n ≤ (2q − 1) ( 2t t ) . In this paper we consider the problem of finding maximal circulant submatrices of Ak,t, representing an almost cross intersecting pair, for a range of parameters. A circulant matrix is a matrix in which each row is shifted one position to the right compared to the preceding row (or alternatively, each column is shifted one position compared to the preceding col- umn). Therefore, such a matrix C is determined completely by its first row or first column. Circulant matrices were studied extensively in the context of the multiplicative commuta- tive semi-group of circulant Boolean matrices and also when discussing Cayley graphs of cyclic groups (see e.g. [1, 5, 6, 7, 22]). However, they were not studied in the context of extremal combinatorics, besides some special cases that will be discussed shortly. C4,4 =             1 0 0 0 0 1 1 1 1 1 0 0 0 0 1 1 1 1 1 0 0 0 0 1 1 1 1 1 0 0 0 0 0 1 1 1 1 0 0 0 0 0 1 1 1 1 0 0 0 0 0 1 1 1 1 0 0 0 0 0 1 1 1 1             Figure 1: The circulant matrix Cp,q , where p = q = 4. Our focus will be on circulant matrices that are determined by a column vector with M. Parnas: Circulant almost cross intersecting families 3 p ones followed by q zeros. Such a matrix will be denoted by Cp,q . See Figure 1 for an example. Thus, in one extreme, if p = 1 and q > 0, then Cp,q is the identity matrix. The other extreme is q = 1 and p > 0, and then Cp,q is the adjacency matrix of a crown graph (where a crown graph is a complete bipartite graph from which the edges of a perfect matching have been removed). Hence, the structure of the circulant matrix Cp,q forms a bridge which connects these two extreme cases, and it is interesting to find a unifying theorem which determines the maximal size of the matrix Cp,q as a function of p, q, k and t. We note that two trivial examples of circulant submatrices of Ak,t include the case of q = 0, where we get an all-one submatrix of Ak,t of maximal size ( k−1 t−1 ) × ( k−1 t−1 ) , and the case of p = 0, where we get an all-zero submatrix of Ak,t of maximal size ( k/2 t ) × ( k/2 t ) . Hence, the problem of studying the size of circulant submatrices Cp,q of Ak,t is interesting only if both p, q > 0. Furthermore, we must require that k ≥ 2t, as otherwise, Ak,t is the all-one matrix itself. As we discuss shortly, one of our results also provides a simple construction of maximal isolation submatrices of Ak,t, thus providing simple small witnesses to the Boolean rank of Ak,t. The Boolean rank of a 0, 1-matrix A of size n×m is equal to the smallest integer r, such that A can be factorized as a product of two 0, 1-matrices, X · Y = A, where X is a matrix of size n× r and Y is a matrix of size r×m, and all additions and multiplications are Boolean (that is, 1+1 = 1, 1+0 = 0+1 = 1, 1 ·1 = 1, 1 ·0 = 0 ·1 = 0). A 0, 1-matrix B of size s × s is called an isolation matrix, if we can select s ones in B, so that no two of the selected ones are in the same row or column of B, and no two of the selected ones are contained in a 2 × 2 all-one submatrix of B. It is well known that if B is an isolation submatrix of size s × s in a given 0, 1-matrix A, then s bounds below the Boolean rank of A (see [2, 16]). However, computing the Boolean rank or finding a maximal isolation submatrix in general is an NP-hard problem (see [13, 18, 24]). Hence, it is interesting to find and characterize families of maximal isolation sets for specific given matrices. 1.1 Our results Our main goal is to determine the range of parameters, p and q, for which Cp,q is a sub- matrix of Ak,t. The constructions and upper bounds we present differ in their structure and proof methods according to the size of p, q compared to t, k. We first consider the range of values of relatively small p, that is 1 ≤ p ≤ 2t − 1, and prove in Section 2 the following positive result. Theorem 1.1. Let k ≥ 2t, 1 ≤ p ≤ 2t − 1 and 1 ≤ q ≤ k − 2t + 1. Then Cp,q is a submatrix of Ak,t. In the extreme case of p = 1 and q = k − 2t + 1, this construction gives the identity submatrix of size (k−2t+2)× (k−2t+2). Recently, [20] proved that this is the maximal size of an identity submatrix in Ak,t. The other extreme is p = 2t− 1 and q = k − 2t+ 1, in which case we get a circulant submatrix of size k × k. As we show in Section 2, if k ≥ 4t − 3 then C2t−1,k−2t+1 is a maximal isolation submatrix of size k × k in Ak,t. Since the Boolean rank of Ak,t is k for k ≥ 2t (see [19]), then the size of a maximal isolation submatrix of Ak,t is upper bounded by k × k, and thus, our result is optimal in this case. Furthermore, for k = 2t+ p− 2 and p ≥ 2, the construction described in Theorem 1.1 provides an isolation submatrix of size (2p − 1) × (2p − 1). We note that [20] gave 4 Art Discrete Appl. Math. 5 (2022) #P1.08 constructions of isolation submatrices in Ak,t, of the same size as achieved here. However, the constructions described in [20] are quite complex, and thus, the result described in Theorem 1.1 provides an alternative simpler construction of a maximal isolation submatrix in Ak,t, for large enough k. We then prove the following upper bound that matches the size of the construction given in Theorem 1.1, for the range of values of 1 ≤ p ≤ 2t − 1 and q ≥ p − 1. The proof of this result characterizes the structure of the Boolean decompositions of Cp,q for this range of parameters. Theorem 1.2. Let Cp,q be a submatrix of Ak,t, where k ≥ 2t, 1 ≤ p ≤ 2t− 1 and q > 0. If q ≥ p− 1 then q ≤ k − 2t+ 1. In Section 3 we address the range of slightly larger values of p, that is, 2t ≤ p ≤ t2, and provide a different construction of circulant submatrices of Ak,t of the form Cp,q . As we show, for this range of values of p, there is no upper bound on the size of q, as we had in Theorem 1.1 and Theorem 1.2, as long as k ≥ p+ q. Furthermore, the proof for this range of parameters provides a decomposition of Cp,q into a product of two Boolean circulant matrices X,Y , where X has t ones in each row and Y has t ones in each column. If we view the rows of X and the columns of Y as the characteristic vectors of subsets of size t, then X and Y each represents a circulant t-uniform family. Thus, the construction used in the proof of the next theorem, uses a pair F ,G of circulant families to construct Cp,q . Theorem 1.3. Let 2t ≤ p ≤ t2 and q > 0. Then Cp,q is a submatrix of Ak,t for k ≥ q+ p. Finally, in Section 4, we consider the range of large p. Using the result of [11] stated above, we know that if Cp,q is a submatrix of Ak,t of size n × n, then n ≤ (2q − 1) ( 2t t ) , and [23] proved a conjecture of [12] and showed that for large enough q and t, the size of a q-almost intersecting family F is bounded by (q+1) ( 2t−2 t−1 ) . Note that this last result refers to q-almost cross intersecting pairs (F ,G) in which F = G. Furthermore, the constructions presented in [23], which achieve this bound, do not have a circulant intersection matrix. Indeed, we can get a better upper bound for circulant submatrices of the form Cp,q . Using a theorem of Frankl [9] and Kalai [15] about skew matrices, it is possible to show that p ≤ ( 2t t ) −1. Hence, if Cp,q is a submatrix of size n×n of Ak,t then n ≤ ( 2t t ) +q−1. In the extreme case of p = ( 2t t ) − 1 and q = 1, the simple construction that takes all subsets of size t of [2t] as row and column indices, results in a submatrix Cp,q of size( 2t t ) × ( 2t t ) . This is optimal, as it matches the upper bound of ( 2t t ) + q − 1. For larger q, we give a simple construction of Cp,q for p = q · ( ( 2t/q t/q ) − 1), when t mod q = 0 and k is large enough. Note that there is a relatively large gap between the size of Cp,q stated here, and the upper bound of ( 2t t ) + q− 1. As we prove, this gap can be slightly narrowed for q = 2: Theorem 1.4. Let q = 2 and p = 2t +2t−2 − 2, where t > 2. Then Cp,q is a submatrix of Ak,t for large enough k. We conclude by considering the case of t = 2 and p = ( 2t t ) − 1 = 5 and fully characterize it. As we show, in this case, Cp,q is a submatrix of Ak,t, for q = 1 and k ≥ 5, or for q = 3 and k ≥ 6. Thus, for t = 2, p = 5 and q = 1, 3, we get a result which matches the upper bound of ( 2t t ) + q − 1. However, as we prove, for t = 2, p = 5 and M. Parnas: Circulant almost cross intersecting families 5 q > 0, q 6= 1, 3, there is no k for which Cp,q is a submatrix of Ak,t. This implies that the upper bound of ( 2t t ) +q−1 is not tight in general. It remains an open problem to determine for what values of q > 1 is Cp,q a submatrix of Ak,t, given that p = ( 2t t ) − 1 and t > 2. 2 The range of 1 ≤ p ≤ 2t − 1 In this section we prove Theorems 1.1 and 1.2, which address the range of small p, that is, 1 ≤ p ≤ 2t− 1. As stated above, this range of values includes the identity matrix, as well as allows us to provide a simple construction of maximal isolation sets for large enough k. It will be useful to identify subsets of [k] with their characteristic vectors. Thus, a subset of size t of [k] will be represented by a 0, 1-vector of length k with exactly t ones. Furthermore, in order to show that some matrix C of size n×m is a submatrix of Ak,t, it will be enough to show that there exists a Boolean decomposition C = X · Y , where X is a Boolean matrix of size n× k with exactly t ones in each row, and Y is a Boolean matrix of size k ×m with exactly t ones in each column, and all operations are Boolean. 2.1 A construction of Cp,q for 1 ≤ p ≤ 2t − 1 The following lemma will be useful in proving Theorem 1.1. It shows that it is possible to decompose a matrix of the form Cp,q into a product of two circulant matrices of the same type, for a wide range of parameters. Lemma 2.1. Let i, j, z be three integers, such that i, j ≥ 1 and i+ j − 1 ≤ z. Then Ci,z−i · Cj,z−j = Ci+j−1,z−i−j+1. Proof. It is well known that the product of two circulant Boolean matrices is a circulant Boolean matrix (where all operations are Boolean). Thus, it is enough to determine the first column c = (c1, c2, ..., cz) of the product matrix Ci,z−i · Cj,z−j , and to show that it has i + j − 1 ones, followed by z − i − j + 1 zeros. The proof follows directly from the definition of matrix multiplication using the Boolean operations. Specifically, it is clear that cs = 1 for 1 ≤ s ≤ i, since the first element in each of the first i rows of Ci,z−i is a 1, and the first element of the first column of Cj,z−j is also a 1 (since i, j ≥ 1). Next consider element ci+s for 1 ≤ s ≤ j − 1. Note that row i + s of Ci,z−i begins with s zeros and then has i ones, and the first j elements of the first column of Cj,z−j are ones. Since s ≤ j− 1, then the result of multiplying row i+ s of Ci,z−i with the first column of Cj,z−j , is a one. It remains to show that the remaining elements of c are all zeros. But the last z−i−j+1 rows of Ci,z−i begin with at least j zeros. Therefore, multiplying any of these rows with the first column of Cj,z−j , results in a zero. Using Lemma 2.1, we can now prove Theorem 1.1. Proof of Theorem 1.1. Let 1 ≤ i, j ≤ t such that i + j − 1 = p. Let Jn,m be the all-one matrix of size n×m, and On,m the all-zero matrix of size n×m. Define, two matrices X and Y as follows: X = [Ci,p+q−iOp+q,t−jJp+q,t−i], Y =   Cj,p+q−j Jt−j,p+q Ot−i,p+q   . 6 Art Discrete Appl. Math. 5 (2022) #P1.08 Using Lemma 2.1, where z = p+ q, we have that X · Y = Ci,p+q−i · Cj,p+q−j = Ci+j−1,p+q−i−j+1 = Cp,q. Furthermore, each row of X and each column of Y is a vector with exactly t ones, whose length is: (p+ q) + (t− j) + (t− i) = p+ q + 2t− i− j = p+ q + 2t− (p+ 1) = q + 2t− 1. Therefore, if k ≥ q + 2t − 1, then we can view the rows of X and columns of Y as the characteristic vectors of subsets in ( [k] t ) . Thus, X · Y = Cp,q is a submatrix of Ak,t as claimed. As we show next, if k ≥ 4t − 3 then the construction described in the proof of Theo- rem 1.1, provides a maximal isolation submatrix of size k×k in Ak,t. This result is optimal since the Boolean rank of Ak,t is k for k ≥ 2t (see [19]). Corollary 2.2. Let 2 ≤ p ≤ 2t − 1 and let k = 2t + p − 2. Then Cp,p−1 is an isolation submatrix of size (2p−1)×(2p−1) in Ak,t. Furthermore, if k ≥ 4t−3 then C2t−1,k−2t+1 is an isolation submatrix of size k × k in Ak,t. Proof. Let k = 2t+p−2. If we set q = k−2t+1 = (2t+p−2)−2t+1 = p−1, then by Theorem 1.1, Cp,q is a submatrix of Ak,t of size (2p− 1)× (2p− 1) since p+ q = 2p− 1. It is easy to verify that in this case, since q = p− 1, the ones on the main diagonal of Cp,q form an isolation set of size p+ q. In the extreme case of p = 2t−1, and if k ≥ 4t−3, then q = k−2t+1 ≥ 2t−2 ≥ p−1, and we get an isolation matrix Cp,q of size k × k, since p+ k − 2t+ 1 = k. 2.2 Upper bounds on the size of Cp,q for 1 ≤ p ≤ 2t − 1 We now turn to prove Theorem 1.2, which provides a matching upper bound to the size of the construction given in Theorem 1.1, for 1 ≤ p ≤ 2t − 1 and q ≥ p − 1. We note that if q ≥ p − 1 then p + q ≤ k (for any value of p), since in this case Cp,q is an isolation submatrix of Ak,t. Thus, its Boolean rank, which is p + q, is bounded above by k, which is the Boolean rank of Ak,t. However, the proof of Theorem 1.2, which provides a tight upper bound on p+ q, will require a more elaborate proof. The following simple claim is easy to verify, and will be needed for the proof of Theo- rem 1.2. Claim 2.3. Let B be an all-one submatrix of size i × j of Cp,q , where p, q > 0. Then, 1 ≤ i, j ≤ p and i+ j ≤ p+ 1. The next lemma is a generalization of a claim proved in [19], which characterizes the Boolean decompositions of the identity matrix. Here we characterize the Boolean decom- positions of circulant isolation matrices of the form Cp,q . Denote by |x| the number of ones in a vector x, and let x⊗ y denote the outer product of a column vector x and a row vector y, where both x, y are of length n. That is, x⊗ y is a matrix of size n× n. Lemma 2.4. Let p, q > 0 and n = p + q. Let X · Y = Cp,q be a Boolean decomposition of Cp,q , where X is an n × r Boolean matrix and Y is an r × n Boolean matrix. Denote by x1, . . . , xr the columns of X , and by y1, . . . , yr the rows of Y . Then: M. Parnas: Circulant almost cross intersecting families 7 1. For each i ∈ [r], if xi has more than p ones then yi is the all-zero vector, and if yi has more than p ones then xi is the all-zero vector. 2. For each i ∈ [r], if |xi|, |yi| > 0, then |xi|+ |yi| ≤ p+ 1. 3. If q ≥ p − 1, then there exist n indices i1, ..., in, such that |xij |, |yij | > 0 for every j ∈ [n]. Proof. For simplicity, denote C = Cp,q . If we write the decomposition X · Y = C with outer products, then C = ∑r i=1 xi ⊗ yi. First note that if we have an i such that xi has more than p ones, and yi is not the all-zero vector, then xi ⊗ yi has a column with more than p ones. Since the addition is the Boolean addition, then ∑r i=1 xi ⊗ yi 6= C. A similar argument shows that if yi has more than p ones then xi is the all-zero vector. Thus, item (1) follows. Assume now, by contradiction, that item (2) does not hold. Thus, there exists an i, such that |xi|, |yi| > 0 and |xi|+|yi| > p+1. Let |xi| = s and |yi| = ℓ, where by our assumption ℓ ≥ p− s+2. Thus, the matrix xi ⊗ yi has an all-one submatrix B of size s× ℓ. Since the addition is Boolean, Cp,q , also has an all-one submatrix of size s× ℓ ≥ s× (p− s+2), in contradiction to Claim 2.3. It remains to prove item (3). Since q ≥ p− 1, then C is an isolation matrix. Therefore, its Boolean rank is n = p + q. Assume by contradiction that there are strictly less than n pairs xi, yi such that |xi|, |yi| > 0. Note that if xi or yi is the all-zero vector then xi ⊗ yi is the all-zero matrix. Thus, we can remove from X any column xi which is the all-zero vector, and remove the corresponding row yi from Y , and similarly, remove from Y any row yj which is the all-zero vector, and remove the corresponding column xj from X . We get two new matrices X ′, Y ′, such that X ′ ·Y ′ = C, where the size of X ′ is n× ℓ, the size of Y ′ is ℓ × n, and by our assumption ℓ < n. Therefore, the Boolean rank of C is strictly less than n, and we get a contradiction. Lemma 2.5. Let p, q > 0 and q ≥ p−1, and let n = p+q. Let X ·Y = Cp,q be a Boolean decomposition of Cp,q , where X is an n × r Boolean matrix and Y is an r × n Boolean matrix. Then the sum of the number of ones in X and the number of ones in Y is at most (p+ 1)n+ (r − n)n. Proof. Let x1, . . . , xr be the columns of X , and y1, . . . , yr the rows of Y . By Lemma 2.4, there exist n indices i1, ..., in, such that |xij |, |yij | > 0 for every j ∈ [n]. Furthermore, for these indices it holds that |xij | + |yij | ≤ p + 1. Assume, without loss of generality, that these are indices 1, ..., n. As for the remaining pairs, xi, yi, for n < i ≤ r: by Lemma 2.4, if |xi|, |yi| > 0 then |xi| + |yi| ≤ p + 1, and if |xi| ≥ p + 1 then yi is the all-zero vector, and similarly if |yi| ≥ p+ 1 then xi is the all-zero vector. Thus, |xi|+ |yi| is maximized when xi or yi is the all-zero vector and the other is the all-one vector, since in this case |xi| + |yi| = n = p+ q ≥ p+ 1. Hence, the sum of the number of ones in X and the number of ones in Y is at most (p+ 1)n+ (r − n)n. Proof of Theorem 1.2. Consider the Boolean decomposition X · Y = Ak,t, where X is a matrix of size ( k t ) × k and Y is a matrix of size k × ( k t ) , and X and Y each contain all characteristic vectors of subsets in ( [k] t ) . 8 Art Discrete Appl. Math. 5 (2022) #P1.08 Since Cp,q is a submatrix of Ak,t then there exist two matrices X ′ ⊆ X,Y ′ ⊆ Y , such that X ′ · Y ′ = Cp,q . Notice that X ′ is an n × k matrix and Y ′ is an k × n matrix, where n = p + q, and the sum of the number of ones in X and the number of ones in Y is exactly 2nt. But, by Lemma 2.5, the total number of 1’s in both X ′ and Y ′ is at most (p+1)n+(k−n)n. Thus, 2nt ≤ (p+1)n+(k−n)n. Hence, p+q = n ≤ k−2t+p+1, as claimed. 3 The range of 2t ≤ p ≤ t2 The circulant decomposition given in Lemma 2.1 is not suitable for p ≥ 2t, since if we take the decomposition Ci+j−1,z−i−j+1 = Ci,z−i · Cj,z−j , and let p = i + j − 1 and p ≥ 2t, then i+ j ≥ 2t+ 1. Thus, either i or j are strictly larger than t, and therefore, the rows of Ci,z−i or the columns of Cj,z−j cannot represent subsets of size t of [k]. However, Theorem 1.3 stated in Section 1.1 and proved next, shows that when 2t ≤ p ≤ t2, there exists a different circulant decomposition Cp,q = X · Y , in which each row of X and each column of Y has exactly t ones as required. See Figure 2 for an illustration, and note also that since 2t ≤ p ≤ t2 then t ≥ 2. Figure 2: The construction described in Theorem 1.3 for t = 3, p = 6, q = 2. The matrices presented are X · Y = W , where W is achieved from Cp,q by moving the last row of Cp,q as defined in the introduction to be first. The first row x1 of X begins with q + t − 1 zeros, and then every t positions there is a one outlined with a rectangle. The remaining positions of x1 contain zeros and ones for a total of t ones in x1. The matrix Y = Ct,p+q−t. The ones in the columns of Y that intersect the outlined ones in x1 are also outlined with rectangles. Proof of Theorem 1.3. Let W be the matrix achieved from Cp,q by moving the last row of Cp,q as defined in the introduction to be first. Hence, the first row of W has q zeros followed by p ones. It is enough to show that there exist two square matrices X,Y , of size p+ q, such that each row of X and each column of Y has exactly t ones, and X · Y = W . The matrix Y is just the matrix Ct,p+q−t as defined in the introduction, and thus each column has t ones as required. The matrix X is also circulant and is defined as follows. The first row x1 of X begins with q+ t−1 zeros. The remaining p− t+1 ≥ 2t− t+1 = t+1 coordinates of x1 start with a one and then there is a one every t positions, and a one in the last position of x1. The remaining positions have ones and zeros in an arbitrary order, for a total of t ones in x1. Note that there are at most t− 1 zeros between every two consecutive ones in x1, and in the extreme case of p = t 2, there are exactly t − 1 zeros between every two consecutive ones. M. Parnas: Circulant almost cross intersecting families 9 Finally, we must show that X ·Y = W . Since both X and Y are circulant, the resulting product matrix X · Y is circulant. Thus, it is sufficient to prove that the first row of X · Y is equal to the first row (0, 0, ..., 0, 1, 1, ..., 1) of W . Let y1, ..., yp+q be the columns of Y . • Columns y1, ..., yq have ones only in positions at most q + t − 1, whereas x1 has zeros in the first q + t− 1 positions. Hence, x1 · yi = 0, for 1 ≤ i ≤ q, as required. • Now consider the last p columns of Y . Note that for each such column yj , at least one of the t consecutive ones of yj avoids the q + t − 1 consecutive zeros of x1. Furthermore, every t consecutive columns of Y have a one in a common row. It is easy to verify that columns yj , for q + 1 ≤ j ≤ q + t, intersect with the first one in x1, as they all have a one in position q + t. The next t columns of Y all have a one in position q + 2t, and since x1 has a one every t coordinates, these columns also intersect with x1, and so on. The last t columns of Y intersect with the last one of x1. Hence, X · Y = W as claimed, and the theorem follows. 4 The range of large p Bollobás [3] proved that for any m pairs of subsets (Ai, Bi), such that |Ai| = a, |Bi| = b for 1 ≤ i ≤ m, and Ai ∩ Bj = ∅ if and only if i = j, it holds that m ≤ ( a+b a ) . An immediate corollary of this theorem is that the largest circulant submatrix Cp,q of Ak,t, for q = 1, is of size ( 2t t ) × ( 2t t ) . This result is tight. This theorem has several generalizations. Among them is a result of Frankl [9] and Kalai [15] that considered the skew version of the problem, and showed that the same bound holds even under the following relaxed assumptions: Let (Ai, Bi) be pairs of sets, such that |Ai| = a, |Bi| = b for 1 ≤ i ≤ m, Ai ∩ Bi = ∅ for every 1 ≤ i ≤ m, and Ai ∩Bj 6= ∅ if i > j. Then m ≤ ( a+b a ) . We immediately get the following corollary, where here and throughout this section we assume that the first row of Cp,q has q zeros followed by p ones. Corollary 4.1. Let Cp,q be a submatrix of Ak,t of size n × n, for a given fixed q. Then, n ≤ ( 2t t ) + q − 1, that is, p ≤ ( 2t t ) − 1. Proof. Consider the submatrix B of Cp,q that is defined by the first p+1 rows and columns of Cp,q . The matrix B maintains the conditions of the Theorem of Frankl and Kalai, and, thus, its size is at most ( 2t t ) × ( 2t t ) . Hence, n ≤ ( 2t t ) +q−1 as claimed, and p ≤ ( 2t t ) −1. The following lemma presents a simple construction of a large circulant submatrix Cp,q of Ak,t for a given fixed q. See also Figure 3 for an illustration. Lemma 4.2. Let q > 0, t ≥ q, where t mod q = 0. Then Cp,q is a submatrix of Ak,t, for p = q · ( ( 2t/q t/q ) − 1) and k ≥ 3t− t/q. Proof. Let n = ( 2t/q t/q ) . The matrix Cp,q , where p + q = q · n, can be partitioned into q disjoint submatrices of size n× (p+ q), as follows. The ith submatrix, 1 ≤ i ≤ q, contains rows i+j ·q, 0 ≤ j ≤ n−1, of Cp,q . Each such submatrix is a blowup of Cn−1,1, since we can partition each row of these q submatrices into blocks of q consecutive entries, where the blocks of the ith submatrix are shifted by one position compared to the blocks of the 10 Art Discrete Appl. Math. 5 (2022) #P1.08 Figure 3: The construction described in Lemma 4.2, for t = 4, p = 10, q = 2. The matrix C10,2 is composed of two submatrices, one containing the odd rows and one the even rows. Each row of these submatrices is partitioned into n = 6 blocks of size q = 2, as outlined by rectangles. The first submatrix is the intersection matrix of all subsets of size 2 of [4], and the second submatrix is the intersection matrix of all subsets of size 2 of {a, b, c, d}, where in both cases each of the subsets assigned to the columns appears twice (in columns belonging to the same block). Since the subsets assigned to the first submatrix are disjoint from the subsets assigned to the second submatrix, and the blocks in the two submatrices are shifted, this defines an assignment of different subsets of {1, 2, 3, 4, a, b, c, d}, each of size at most 4, to the columns and rows of C10,2. previous submatrix (in a circulant way). Thus, the entries in each block are identical (either all ones or all zeros). For example, the first row of the first submatrix starts with a block of q zeros, followed by n− 1 blocks of q ones. Hence, we can view each of these q submatrices as the intersection matrix of the two families of all subsets of size t/q of [2t/q] (since each subset intersects with all subsets but one), where columns belonging to the same block in a submatrix are assigned the same subset. Now, if we take disjoint copies of the subsets assigned to each submatrix, and label each column in Cp,q with the subset that is the union of all subsets of size t/q assigned to this column, then we get q · n different subsets, each of size t, assigned to the columns (the subsets are different because the blocks in each submatrix are shifted compared to the other submatrices). As to the rows, each row is assigned a different subset of size t/q, M. Parnas: Circulant almost cross intersecting families 11 and therefore, we can define t − t/q additional new elements that do not belong to any of the subsets, and add them to each row, so that the rows are also assigned subsets of size t. Finally, by taking the union of all subsets, we get that k ≥ 2t+ t− t/q = 3t− t/q. The size of the construction given in Lemma 4.2 is quite far from the upper bound given in Corollary 4.1. As we show in the next subsection, there exists a slightly larger construction for q = 2. Finally, in Subsection 4.2, we show that the upper bound of Corollary 4.1 is tight for t = 2, p = ( 2t t ) − 1 = 5 and q = 1, 3, but there is no k for which C5,2 is a submatrix of Ak,2 when q 6= 1, 3. 4.1 The values q = 2 and p = 2t + 2t−2 − 2 We now prove Theorem 1.4 and show a construction of Cp,q for q = 2 and p that is exponential in t. The construction we present is recursive in nature, and exploits the fact that Cp,2 has two blocks on the main diagonal, such that each one of these blocks is half the size of Cp,2, and the structure of each block is almost identical to that of Cp,q , where the only difference is that there is a 1 in the first position of the last row instead of a zero in Cp,q . This small difference complicates the recursive argument. The details of the proof follow. See Figure 4 for an illustration of the proof of Theorem 1.4. Figure 4: The construction of Cp,q described in Theorem 1.4, for q = 2 and p = 2 t + 2t−2 − 2, where t = 3. The column indices are written above the matrix C8,2 and the row indices to the right of the matrix. Proof of Theorem 1.4. Let h = (p + q)/2. We prove by induction on t that Cp,q is the 12 Art Discrete Appl. Math. 5 (2022) #P1.08 intersection matrix of two families of t-subsets Fa,b = {F1, ..., Fp+q}, Ga,b = {G1, ..., Gp+q}, where the subsets in Fa,b are the row indices and the subsets in Ga,b are the column indices, and a, b are two integers with the following properties: • a ∈ G1, ..., Gh, and a ∈ Fh+1, ..., Fp+q−1. • b ∈ Gh+1, ..., Gp+q , and b ∈ F1, ..., Fh−1. • a, b appear only in the subsets specified above. In particular, a, b 6∈ Fp+q . Let C̃p,q be the matrix that is achieved from Cp,q , by modifying to 1 the first position of the last row of Cp,q , and let F̃a,b be a family that is identical to Fa,b with one difference: the subset Fp+q also contains the element a. It is not hard to verify that if Cp,q is the intersection matrix of Fa,b and Ga,b, then C̃p,q is the intersection matrix of F̃a,b and Ga,b. The base of the induction is t = 3, and the construction of C8,2 is given in Figure 4, where in this case a = 8, b = 9. Note that if we modify the last row index {2, 6, 3} to be {2, 6, 3, 8}, then we get a construction of C̃8,2 as claimed. Assume now that t > 3, let pt = 2 t + 2t−2 − 2 and pt−1 = 2 t−1 + 2t−3 − 2, and consider Cpt,q . Then it has the following structure: there are two matrices of the form C̃pt−1,q on the main diagonal, and two blocks of size (pt + q)/2 that are all one, but the leftmost entry on the bottom row of each of these blocks that is a 0. By the induction hypothesis there exist, as specified above, two families of (t − 1)- subsets Fa,b = {F1, ..., Fpt−1+q}, Ga,b = {G1, ..., Gpt−1+q}, whose intersection matrix is Cpt−1,q . Let F ′b,a = {F ′ 1, ..., F ′ pt−1+q} be a family of subsets that is identical to Fa,b, but a, b are interchanged in all subsets. That is, for 1 ≤ i ≤ pt−1 + q: F ′i =    Fi \ {a} ∪ {b}, if a ∈ Fi, Fi \ {b} ∪ {a}, if b ∈ Fi, Fi, if a, b 6∈ Fi. Similarly define G′b,a = {G ′ 1, ..., G ′ pt−1+q}, which is identical to Ga,b, but a, b are inter- changed in all subsets. Note that since a, b appear only in subsets as specified above, then it also holds that Cpt−1,q is the intersection matrix of the two families F ′ b,a and G ′ b,a. Now let c, d be two new elements that do not appear in any of the above families, and define the following families: Fd = {F1 ∪ {d}, F2 ∪ {d}, ..., Fpt−1+q−1 ∪ {d}, Fpt−1+q ∪ {a}}, Fc = {F ′ 1 ∪ {c}, F ′ 2 ∪ {c}, ..., F ′ pt−1+q−1 ∪ {c}, F ′ pt−1+q ∪ {b}}, Gc = {G1 ∪ {c}, G2 ∪ {c}, ..., Gpt−1+q ∪ {c}}, Gd = {G ′ 1 ∪ {d}, G ′ 2 ∪ {d}, ..., G ′ pt−1+q ∪ {d}}. Finally, define the families Fc,d,Gc,d as follows: Fc,d = Fd ∪ Fc, Gc,d = Gc ∪ Gd. M. Parnas: Circulant almost cross intersecting families 13 It is clear that Fc,d,Gc,d are two families of t-sets, each of size pt + q, and their structure is as claimed above, where c and d are in the role of a and b, respectively. It remains to prove that Cpt,q is the intersection matrix of Fc,d,Gc,d. First note that by the induction hypothesis, and using the structure of the subsets we defined, C̃pt−1,q is the intersection matrix of Fd,Gc, as well as the intersection of Fc,Gd. Consider now the matrix C which is the intersection matrix of Fd,Gd. It is clear that the first pt−1 + q− 1 rows of C are all ones, since the first pt−1 + q− 1 families of Fd,Gd all contain d. We next show that the last row of C is of the form (0, 1, 1, ..., 1). By the induction hypothesis, the intersection of Fpt−1+q with all subsets of Ga,b gives a vector of the form (0, 1, 1, ...1, 0). Thus, since a, b 6∈ Fpt−1+q and G ′ b,a is identical to Ga,b, but a, b are interchanged in all subsets, then the intersection of of Fpt−1+q with all subsets of G′b,a results also with the vector (0, 1, 1, ...1, 0). Since the last subset of Fd is defined as Fpt−1+q ∪ {a} and the last subset of Gd is G ′ pt−1+q ∪ {d}, and a ∈ G ′ pt−1+q , then we get that Fpt−1+q ∪ {a} and G ′ pt−1+q ∪ {d} also intersect as required. A similar argument shows that the intersection matrix of Fd,Gd is also a matrix that is all one, but the first element on the last row of this matrix, which is a zero. This completes the proof of the theorem. 4.2 The values t = 2, p = ( 2t t ) − 1, q > 0 Finally, we address the range of values of t = 2 and p = ( 2t t ) − 1 = 5. We first show that Cp,q is a submatrix of Ak,t for these values of p and t, and for q = 1, 3. Lemma 4.3. Let t = 2 and p = ( 2t t ) − 1 = 5. Then Cp,q is a submatrix of Ak,t for q = 1 and k ≥ 5, or for q = 3 and k ≥ 6. Proof. If t = 2, p = 5, q = 1, then C5,1 is a submatrix of A5,2. Simply take as row/column indices all subsets of size 2 of [4]. As to the case of t = 2, p = 5, q = 3, Figure 5 shows that C5,3 is a submatrix of Ak,2, for any k ≥ 6. Figure 5: A construction of Cp,q for t = 2, p = 5, q = 3. We conclude by proving that Cp,q is not a submatrix of Ak,t for t = 2, p = 5 and q 6= 1, 3. Unfortunately, this proof cannot be generalized to the case of p = ( 2t t ) − 1 and t > 2. Thus, it remains an open problem to determine for what values of q > 1 is Cp,q a submatrix of Ak,t, when p = ( 2t t ) − 1 and t > 2. Lemma 4.4. Let t = 2, p = ( 2t t ) − 1 = 5, q 6= 1, 3, q > 0. Then Cp,q is not a submatrix of Ak,t for any k. 14 Art Discrete Appl. Math. 5 (2022) #P1.08 Proof. Assume by contradiction that C5,q is a submatrix of Ak,2 for some k, where the first row of C5,q starts with q 6= 1, 3 zeros, followed by p ones. Let n = p+ q = q + 5 ≥ 7 be the size of C5,q , and let Ai, Bi, 0 ≤ i ≤ n − 1, be the 2-uniform subsets defining the row and column indices, respectively, of C5,q . Assume first that there exists some i such that Bi ∩ B(i+1) mod n = ∅, that is, two consecutive column indices are disjoint. Since C5,q is circulant, then we can assume that i = 0, that is, B0 ∩B1 = ∅. Since B0 and B1 both intersect with A3, A4, A5, then each of these three subsets contains one element from each of B0, B1. Furthermore, as all subsets are different and of size 2, then each element of B0, B1 is contained in at most two of these three subsets. Next consider B2. It also intersects with A3, A4, A5, and since there is no common element of B0, B1 in these three subsets, then B2 also includes two elements from B0∪B1 (although here B2 can contain two elements from the same subset B0 or B1). Now, consider A7 mod n, where if q = 2 then A7 mod n = A0 and otherwise, A7 mod n = A7. In both cases, since p = 5, the row labeled by A7 mod n starts with two zeros followed by a one. Thus, since A7 mod n ∩ B2 6= ∅, then A7 mod n contains an element from B0∪B1, in contradiction to the fact that A7 mod n∩B0 = A7 mod n∩B1 = ∅. Hence, we can assume from now on that Bi ∩ B(i+1) mod n 6= ∅, and similarly that Ai ∩A(i+1) mod n 6= ∅, for 0 ≤ i ≤ n− 1. There are two cases: • There exists an i such that Bi ∩ B(i+1) mod n ∩ B(i+2) mod n 6= ∅. Since Cp,q is circulant, then assume that i = 0, and let b ∈ B0, B1, B2. Thus, B0 = {b0, b}, B1 = {b1, b}, B2 = {b2, b}. From this and the structure of C5,q , we can deduce the fol- lowing: 1. b0 ∈ A1, A2 = {b0, b1}, and b ∈ A3, A4, A5. 2. The row labeled by A6 starts with a zero followed by 5 ones, and so b 6∈ A6. But A6 ∩B1 6= ∅, A6 ∩B2 6= ∅. Thus, A6 = {b1, b2}. 3. b 6∈ B3 as B3 ∩A3 = ∅. But B3 ∩B2 6= ∅, and therefore, b2 ∈ B3. 4. b, b0, b1 6∈ B3 as also B3 ∩ A2 = ∅. But B3 ∩ A5 6= ∅ and A5 ∩ A6 6= ∅. Therefore, b2 ∈ A5. 5. Since b, b0, b1 6∈ B3 and B3 ∩ A4 6= ∅, then there exists a new element b3 ∈ B3 ∩A4. 6. A4 ∩ B4 = ∅ and so b 6∈ B4. Hence, b2 ∈ B4 since B4 ∩ A5 6= ∅. In a similar way, b1 ∈ B5. Hence, the subsets defining the first seven rows and columns of C5,q have the fol- lowing structure so far, where they are written to the left and above the submatrix: b0 b1 b2 b2 b2 b1 b b b b3 0 0 b0 1 0 0 b0, b1 1 1 0 0 b 1 1 1 0 0 b, b3 1 1 1 1 0 0 b, b2 1 1 1 1 1 0 0 b1, b2 0 1 1 1 1 1 0 M. Parnas: Circulant almost cross intersecting families 15 Now, if q ≥ 4, we already get a contradiction, since in C5,q it holds that A2∩B5 = ∅, whereas here b1 ∈ A2 ∩B5. Therefore, assume that q = 2, and so all remaining entries in the submatrix above are ones. From the structure of the submatrix and the information we have so far, we can deduce that A1 = {b0, b3} and hence B5 = {b1, b0} (since B5 ∩ A1 6= ∅ and B5 ∩ A4 = ∅ and so b3 6∈ B5). But then since b0, b1 6∈ A0, we get a contradiction since A0 ∩B5 6= ∅. • Bi∩B(i+1) mod n∩B(i+2) mod n = ∅, but Bi∩B(i+1) mod n 6= ∅, for 0 ≤ i ≤ n−1. Thus, Bi = {bi, b(i+1) mod n}, where some of the bi’s may be identical. If all bi’s in the subsets Bq, Bq+1, Bq+2, Bq+3, Bq+4 are different, then A0 cannot intersect with these subsets, since |A0| = 2. Hence, there exist 0 ≤ i 6= j ≤ 4 such that bq+i = bq+j . Assume, without loss of generality, that i = 0 (as the matrix is circulant). Since the intersection of every three consecutive subsets is empty, and each subset contains two different elements, then j 6= 1, 2. If j = 3 then bq = bq+3, and since A2 does not intersect with Bq, Bq+1 then bq, bq+1, bq+2 6∈ A2. But A2 intersects with Bq+2 = {bq+2, bq+3 = bq}, and we get a contradiction. A similar contradiction is achieved if j = 4 when considering A5. Thus, in all cases we get a contradiction and the lemma follows. ORCID iDs Michal Parnas https://orcid.org/0000-0003-0189-6999 References [1] B. Alspach, Isomorphism and cayley graphs on abelian groups, in: Graph Symmetry, Springer, volume 497 of NATO Adv. Sci. Inst. Ser. C Math. Phys. Sci., pp. 1–22, 1997, doi:10.1007/ 978-94-015-8937-6 1. [2] L. B. Beasley, Isolation number versus Boolean rank, Linear Algebra Appl. 436 (2012), 3469– 3474, doi:10.1016/j.laa.2011.12.013. [3] B. Bollobás, On generalized graphs, Acta Math. Acad. Sci. Hungar. 16 (1965), 447–452, doi: 10.1007/bf01904851. [4] P. 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Schwarz, Circulant Boolean relation matrices, Czechoslovak Math. J. 24 (1974), 252–253. [23] A. Scott and E. Wilmer, Hypergraphs of bounded disjointness, SIAM J. Discrete Math. 28 (2014), 372–384, doi:10.1137/130925670. [24] Y. Shitov, On the complexity of Boolean matrix ranks, Linear Algebra Appl. 439 (2013), 2500– 2502, doi:10.1016/j.laa.2013.06.033. ISSN 2590-9770 The Art of Discrete and Applied Mathematics 5 (2022) #P1.09 https://doi.org/10.26493/2590-9770.1377.8e9 (Also available at http://adam-journal.eu) On avoiding 1233 Toufik Mansour* Department of Mathematics, University of Haifa, 3498838 Haifa, Israel Mark Shattuck Department of Mathematics, University of Tennessee, 37996 Knoxville, TN Received 12 August 2020, accepted 16 June 2021, published online 11 April 2022 Abstract In this paper, we establish a recurrence relation for finding the generating function for the number of k-ary words of length n that avoid 1233 for arbitrary k. Comparable generating function formulas may also be found counting words where a single permutation pattern of length three is avoided in addition to 1233. Keywords: k-ary words, Kernel method, Avoiding 1233. Math. Subj. Class.: 05A15, 05A05 1 Introduction We denote the set of all words of length n over the alphabet [k] = {1, . . . , k} by [k]n and refer to members of [k]n as k-ary words. Let π = π1 · · ·πn ∈ [k] n and τ = τ1 · · · τm ∈ [ℓ]m such that each letter from [ℓ] appears at least once in τ (possibly with repetitions). We say that π contains τ if there exist indices 1 ≤ i1 < · · · < im ≤ n such that πiaΦπib if and only if τaΦτb, for any relation Φ ∈ {<,=, >} and a, b ∈ [m]. In this context, the word τ is called a pattern, and it is said that π avoids τ if π fails to contain τ per the preceding definition. The area of permutation pattern avoidance has received considerable attention in recent decades; see, e.g., [13] and references therein. Alon and Friedgut [2] extended this study to avoidance on k-ary words in obtaining an upper bound on the number of permutations of length n that avoid a given pattern. The question of pattern avoidance on permutations was initiated by Knuth [6], who found that the number of permutations of length n that avoid the pattern τ for any τ ∈ S3 is given by the n-th Catalan number 1 n+1 ( 2n n ) . Later, Simion and Schmidt [12] extended this result by determining the number of permutations *Corresponding author. E-mail addresses: tmansour@univ.haifa.ac.il (Toufik Mansour), mshattuc@utk.edu (Mark Shattuck) cb This work is licensed under https://creativecommons.org/licenses/by/4.0/ 2 Art Discrete Appl. Math. 5 (2022) #P1.09 of length n that avoid all patterns in any subset T of S3. Comparable results involving k-ary words were found by Burstein [4] and Albert et al. [1], and later by Burstein and Mansour [5], who allowed patterns to contain repeated letters. See also [11, 14] concerning the avoidance of 123 by words as well as [9] for general enumeration schemes for words avoiding a permutation pattern. Concerning avoidance of patterns of length four by k-ary words, only the following more general results are known: • Regev [10] showed that the number of k-ary words of length n that avoid 12 · · · (ℓ+ 1) is asymptotic to nℓ(k−ℓ)ℓn ℓℓ(k−ℓ) ∏ℓ i=1 ∏k−ℓ j=1(i+ j − 1) . This result was re-derived by Brändén and Mansour [3]. • The patterns 11 · · · 1 and 11 · · · 121 · · · 11 have been considered in [5]. • Brändén and Mansour [3] (see also [8]) suggested an automaton for the enumeration of k-ary words of length n that avoid a fixed pattern for a given k. We remark that it is a challenging problem in general to enumerate the k-ary words of length n that avoid a given pattern where k is arbitrary. Even in the case of a pattern of length four, the task at hand is still not a simple one. Here, we consider the problem of enumerating the members of [k]n that avoid 1233 for arbitrary k. The main purpose of this paper is to provide a recurrence relation on k for finding the number k-ary words of length n that avoid 1233, see Theorem 2.2 below. This recurrence represents an improvement in the case of 1233 over the general procedure described in [3, 8], which was derived using automata theory. Some further results are found involving avoidance of 1233 and a single pattern of length three. 2 k-ary words that avoid 1233 Let an,k denote the number of k-ary words π of length n that avoid 1233. In order to write recurrences, we must refine an,k according to the prefix of a word π. Given s ≥ 1 and i1, . . . , is ∈ [k], let an,k(i1, . . . , is) denote the number of 1233-avoiding k-ary words π of length n having the form π = i1 · · · isπ ′, where π′ is possibly empty. Clearly, we have an,1 = 1 and an,2 = 2 n. Henceforth, we may assume k ≥ 3. By the definitions, for all 1 ≤ i ≤ k − 2, an,k(i) = an,k(i, k) + i∑ j=1 an,k(i, j) + k−1∑ j=i+1 an,k(i, j). Note that an,k(i, k) = an−1,k(i) and an,k(i, j) = an−1,k(j) for all 1 ≤ j ≤ i ≤ k − 2. Thus, an,k(i) = an−1,k(i) + i∑ j=1 an−1,k(j) + k−1∑ j=i+1 an,k(i, j). T. Mansour and M. Shattuck: On avoiding 1233 3 Next observe that a k-ary word of the form π = ijπ′ with 1 ≤ i < j ≤ k − 1 must have any letters from [j + 1, k] = {j + 1, j + 2, . . . , k} distinct in order to avoid 1233. If we assume that exactly ℓ letters of π belong to [j + 1, k], then there are ( k−j ℓ ) choices for these letters, ( n−2 ℓ ) choices for their positions within π and ℓ! ways in which to order these letters within their positions. Thus, an,k(i, j) = k−j∑ ℓ=0 ℓ! ( n− 2 ℓ )( k − j ℓ ) an−1−ℓ,j(i). Hence, for all 1 ≤ i ≤ k − 2, an,k(i) = an−1,k(i) + i∑ j=1 an−1,k(j) + k−1∑ j=i+1 k−j∑ ℓ=0 ℓ! ( n− 2 ℓ )( k − j ℓ ) an−1−ℓ,j(i), (2.1) with an,k(k) = an,k(k − 1) = an−1,k. In order to study the sequence determined by recurrence (2.1), we define the distribution polynomial An,k(v) = k∑ i=1 an,k(i)v i−1, n, k ≥ 1, with A0,k(v) = 1 and the generating function Ak(x, v) = ∑ n≥0 An,k(v)x n, k ≥ 1. Multiplying both sides of (2.1) by vi−1, and summing over i = 1, 2, . . . , k − 2, yields for n, k ≥ 3 the recurrence An,k(v)− an−1,k(v k−1 + vk−2) = An−1,k(v)− an−2,k(v k−1 + vk−2) + k−1∑ i=1 an−1,k(i) vi−1 − vk−2 1− v + k−1∑ j=2 k−j∑ ℓ=0 ℓ! ( n− 2 ℓ )( k − j ℓ ) (An−1−ℓ,j(v)− an−2−ℓ,jv j−1), which, by an,k = An,k(1), leads to An,k(v) = 1 1− v (An−1,k(v)− v kAn−1,k(1)) + k∑ j=2 k−j∑ ℓ=0 ℓ! ( n− 2 ℓ )( k − j ℓ ) (An−1−ℓ,j(v)− v j−1An−2−ℓ,j(1)), with A1,k(v) = αk(v) = ∑k i=1 v i−1 and A2,k(v) = kαk(v). 4 Art Discrete Appl. Math. 5 (2022) #P1.09 Multiplying both sides of the last equation by xn, and summing over n ≥ 3, we obtain Ak(x, v) = 1 + αk(v)x(1 + kx) + x 1− v (Ak(x, v)− αk(v)x− 1− v kAk(x, 1) + v k + kvkx) + x(Ak(x, v)− 1− αk(v)x)− v k−1x2(Ak(x, 1)− 1) + k−1∑ j=2 ( x(Aj(x, v)− 1− αj(v)x)− v j−1x2(Aj(x, 1)− 1) ) + k−1∑ j=2 k−j∑ ℓ=1 xℓ+2 ( k − j ℓ ) ∂ℓ ∂xℓ ( xℓ−1Aj(x, v)− v j−1xℓAj(x, 1) ) . (2.2) Example 2.1 (Case k = 3). Note that A1(x, v) = 1+ x 1−x and A2(x, v) = 1+ x(1+v) 1−2x , by the definitions. For k = 3, we have A3(x, v) = 1 + (1 + v + v 2)x(1 + 3x) + x 1− v (A3(x, v)− (1 + v + v 2)x− 1− v3A3(x, 1) + v 3 + 3v3x) + x(A3(x, v)− 1− (1 + v + v 2)x)− v2x2(A3(x, 1)− 1) + x(A2(x, v)− 1− (1 + v)x)− vx 2(A2(x, 1)− 1) + x 3 ∂ ∂x (A2(x, v)− vxA2(x, 1)) . To solve this functional equation, we make use of the kernel method and take v = 1−2x1−x to obtain 1 + (1 + v + v2)x(1 + 3x) + x 1− v (−(1 + v + v2)x− 1− v3A3(x, 1) + v 3 + 3v3x) + x(−1− (1 + v + v2)x)− v2x2(A3(x, 1)− 1) + x(A2(x, v)− 1− (1 + v)x)− vx 2(A2(x, 1)− 1) + x3 ∂ ∂x (A2(x, v)− vxA2(x, 1)) = 0. Hence, A3(x, 1) = (1−x)(1−4x+5x2) (1−2x)4 . Substituting this expression into the one above for A3(x, v) then yields A3(x, v) = (1− x)(1− 4x+ 5x2)(1 + (v − 1)(v + 2)x) (1− 2x)4 . Following Example 2.1, to solve the functional equation (2.2), we use the kernel method. Taking v = v0 = 1−2x 1−x in (2.2) yields Ak(x, 1) = (1− x)k−3 (1− 2x)k−1 · ( 1− (k − 1)x+ k−1 ∑ j=2 k−j ∑ ℓ=0 x ℓ+2 ( k − j ℓ ) ∂ℓ ∂xℓ (xℓ−1Aj(x, v)− v j−1 x ℓ Aj(x, 1)) |v=v0 ) . T. Mansour and M. Shattuck: On avoiding 1233 5 Substituting this expression for Ak(x, 1) into (2.2), and observing the identity x k∑ j=2 αj−1(v) = (1 + kx)αk(x)− 1 + xαk(v)− v k(1 + kx) 1− v , k ≥ 2, we obtain our main result. Theorem 2.2. The generating function Ak(x, v) for k ≥ 3 is given by Ak(x, v) = (1− v)(1− (k − 1)x) 1− 2x− v(1− x) − (x+ v(1− x))xvk−1 1− 2x− v(1− x) Ak(x, 1) + (1− v) k−1∑ j=2 k−j∑ ℓ=0 xℓ+2 ( k − j ℓ ) ∂ℓ ∂xℓ (xℓ−1Aj(x, v)− v j−1xℓAj(x, 1)) 1− 2x− v(1− x) , where A1(x, v) = 1 1−x , A2(x, v) = 1 + x(1+v) 1−2x and Ak(x, 1) = (1− x)k−3 (1− 2x)k−1 · ( 1− (k − 1)x+ k−1 ∑ j=2 k−j ∑ ℓ=0 x ℓ+2 ( k − j ℓ ) ∂ℓ ∂xℓ (xℓ−1Aj(x, v)− v j−1 x ℓ Aj(x, 1)) |v=v0 ) , where v0 = 1−2x 1−x . Note that Theorem 2.2 provides a recurrence formula for finding the generating function Ak(x, 1) (even, more generally, for finding Ak(x, v)). For instance, upon making use of software such as Maple or Mathematica, one can obtain from Theorem 2.2 the following explicit formulas for k = 3, 4, 5, 6: A3(x, 1) = (1− x)(1− 4x+ 5x2) (1− 2x)4 , A4(x, 1) = 1− 10x+ 44x2 − 104x3 + 140x4 − 100x5 + 31x6 (1− 2x)7 , A5(x, 1) = 1− 15x+ 105x2 − 435x3 + 1175x4 − 2129x5 + 2595x6 − 2041x7 +946x8 − 190x9 (1− 2x)10 , A6(x, 1) = 1− 20x+ 192x2 − 1136x3 + 4604x4 − 13380x5 + 28599x6 − 45154x7 +52338x8 − 43320x9 + 24401x10 − 8386x11 + 1391x12 (1− 2x)13 . Remarks: By Theorem 2.2 and induction on k, one can show that the generating function Ak(x, v) for k ≥ 2 may be expressed in the form Pk(x, v)/(1− 2x) αk , where αk ≥ 1 and Pk(x, v) is a polynomial in x and v (and not divisible by 1 − 2x). Upon taking v = 1, it is seen that there exists a constant ck such that the number of k-ary words of length n that avoid 1233 is asymptotic to ckn βk2n for some 1 ≤ βk < αk, which was also shown in [8]. We conjecture that αk = 3k − 5 = βk + 1 for all k, the fact of which is demonstrated by programming for 3 ≤ k ≤ 15. Note that Theorem 2.2 provides a recurrence relation for 6 Art Discrete Appl. Math. 5 (2022) #P1.09 finding an explicit formula for the generating function Ak(x, 1) and is an improvement in the case of 1233 over the more general procedure described in [3, 8]. We close this section with some remarks concerning avoidance of the general pattern 123m, where m ≥ 2. Let a (m) n,k denote the number of k-ary words of length n that avoid 123m, with a (m) n,k (i1, . . . , is) defined analogously as before. If 1 ≤ i ≤ k − 2, then a (m) n,k (i) = a (m) n−1,k(i) + i∑ j=1 a (m) n−1,k(j) + k−1∑ j=i+1 a (m) n,k (i, j). To determine a formula for a (m) n,k (i, j), we consider enumerating a restricted class of finite functions as follows. Given a, b, c ≥ 0, let d (c) a,b denote the number of functions f : [b] → [a] such that |{x ∈ [b] : f(x) = i}| ≤ c for all i ∈ [a]. Such functions could be described as being at most c-to-1. Upon considering the number ℓ of letters in a word belonging to [j + 1, k], we have a (m) n,k (i, j) = k−j∑ ℓ=0 d (m−1) k−j,ℓ ( n− 2 ℓ ) a (m) n−1−ℓ,j(i), if i < j < k. Note that the d (m−1) k−j,ℓ factor accounts for the number of ways in which to select and arrange the elements of [j + 1, k] within ℓ preselected positions such that none of these elements occur m or more times. Hence, for all 1 ≤ i ≤ k − 2, a (m) n,k (i) = a (m) n−1,k(i) + i∑ j=1 a (m) n−1,k(j) + k−1∑ j=i+1 k−j∑ ℓ=0 d (m−1) k−j,ℓ ( n− 2 ℓ ) a (m) n−1−ℓ,j(i), with a (m) n,k (k) = a (m) n,k (k − 1) = a (m) n−1,k. To write a recurrence for the array d (c) a,b, consider the number j of elements in [a] whose pre-image cardinality is exactly c. This implies for a, b, c ≥ 1, d (c) a,b = t∑ j=0 ( a j )( b c, . . . , c, b− jc ) d (c−1) a−j,b−jc, where t = min{a, ⌊b/c⌋} and the c index appears exactly j times in the multinomial coefficient of order j + 1. One may verify the initial conditions d (c) a,0 = 1 for all a, c ≥ 0 and d (c) a,b = 0 if ac = 0 and b ≥ 1. Note that from the recurrence when c = 1, we have d (1) a,b = 0 if b > a, which is in agreement with the pigeonhole principle, whereas if b ≤ a, then d (1) a,b = b! ( a b ) = a(a−1) · · · (a−b+1), as it should. Finding a simple explicit formula for d (c) a,b in general appears not to be an easy task. Note that by induction on c using the recurrence, one has the following multi-sum expression for d (c) a,b: d (c) a,b = min{a,⌊ b c ⌋}∑ jc=0 min{a−jc,⌊ b−cjc c−1 ⌋} ∑ jc−1=0 · · · min{a− ∑c p=3 jp,⌊ b− ∑c p=3 pjp 2 ⌋} ∑ j2=0 Ra,b(j2, . . . , jc), T. Mansour and M. Shattuck: On avoiding 1233 7 where Ra,b(j2, . . . , jc) = b! 2!j2 · · · c!jc ( a− ∑c p=2 jp b− ∑c p=2 pjp ) c∏ i=2 ( a− ∑c p=i+1 jp ji ) . 3 Further results As the previous section illustrates, it is challenging in general to ascertain formulas, either explicitly or by a recurrence, for the number of k-ary words for all k that avoid a single fixed pattern of length four (or of arbitrary length). Another possible direction to pursue is that of enumerating words which avoid 1233 and a second pattern τ . Here, we present two cases when τ is of length three demonstrating that even this problem is highly non-trivial. In particular, we consider the cases when τ = 132 or τ = 213 and leave the remaining cases when τ is a permutation pattern of length three as exercises for the interested reader (the patterns 231 and 321 apparently requiring a lengthier analysis than the others). 3.1 Case 132 Let Ak(x) denote the generating function (g.f.) for the number of k-ary words of length n that avoid {132, 1233} for each k ≥ 1 and define A(x, y) = ∑ k≥0 Ak(x)y k, where A0(x) = 1. In order to find a formula for A(x, y), we let A ′(x, y) = (1−y)A(x,y)−1 y and A′′(x, y) = (1−y)A(x,y)−1 y(1−y) , in accordance with [7, Notation 2.2]. Note that yA ′′(x, y) represents the restriction of the g.f. A(x, y) to nonempty k-ary words, whereas yA′(x, y) is the further restriction to such words that contain 1. We wish to write an equation for A(x, y). Let π be a nonempty k-ary word that avoids {132, 1233}. We represent π by π = π(1)k · · ·π(s)kπ(s+1), where each π(j) is (k− 1)-ary and s ≥ 0. Proceeding according to [7, Proposition 2.1], we consider the cases s = 0, s = 1 and s ≥ 2. This yields the following: • If s = 0, then one has a contribution of yA(x, y). • If s = 1, then xy1−y + xy2A′′(x,y) A′(x,y) ((A ′(x, y) + 1)2 − 1). • If s ≥ 2, then ∑ s≥2 xsy 1− y + ∑ s≥2 s−1∑ d=1 xsy2 ( s− 1 d ) B′d−1B′′ + 2 ∑ s≥2 s−1∑ d=0 xsy2 ( s− 1 d ) B′dA′′(x, y) + ∑ s≥2 s−1∑ d=0 xsy2 ( s− 1 d ) B′dA′(x, y)A′′(x, y), where B′ = (1−y)B(x,y)−1 y , B′′ = (1−y)B(x,y)−1 y(1−y) , and B(x, y) = 1−x 1−x−y is the g.f. for the number of k-ary words of length n that avoid 12 for all n, k ≥ 0. To realize the last two cases above, first note that yB′′ is seen to enumerate nonempty, weakly decreasing k-ary words of length n, whereas yB′ counts such words that contain 8 Art Discrete Appl. Math. 5 (2022) #P1.09 1. Observe further that various sections π(i) of π are accounted for by B′ in the s ≥ 2 case above, instead of by yB′, since one must divide by y to compensate for the fact the minimum letter of one section can coincide with the maximum letter of the subsequent nonempty section. The same also applies when considering the π(i) blocks accounted for by A′(x, y). Combining all of the above contributions, and simplifying, we have that A(x, y) satis- fies (1− y)A(x, y) = 1 + xy (1− x)(1− y) − x2y2 (1− x)(1− y) + x2y2 1− 2x− y + xy + x((y − 1)A(x, y) + 1)((y − 1)A(x, y)− 2y + 1)(1− x− y) (1− y)(1− 2x− y + xy) . Solving for A(x, y) in the last equation, and simplifying, yields the following result. Theorem 3.1. The generating function for the number of k-ary words of length n that avoid both 132 and 1233 for all n, k ≥ 0 is given by 1− 2x2 − y − xy − √ (1−2x−y+xy)((1−x−y−xy)2−x(1−x)(1−y)) (1−x)(1−y) 2x(1− x− y) . For example, extracting the coefficient of yk in the formula for A(x, y) in Theorem 3.1 yields the following formulas for Ak(x) where 1 ≤ k ≤ 5: A1(x) = 1 1− x , A2(x) = 1 1− 2x , A3(x) = 1− 3x+ 4x2 − x3 (1− x)2(1− 2x)2 , A4(x) = 1− 4x+ 9x2 − 6x3 + 2x4 (1− x)2(1− 2x)3 , A5(x) = 1− 6x+ 21x2 − 34x3 + 32x4 − 16x5 + 4x6 (1− x)3(1− 2x)4 . 3.2 Case 213 By the reverse complement operation, the number of k-ary words of length n that avoid {213, 1233} is the same as the number that avoid {132, 1123}. Here, it is more convenient to enumerate the latter. Let Bk(x) denote the g.f. for the number of k-ary words π of length n that avoid {132, 1123} for each k ≥ 1, with B0(x) = 1. Consider cases based on whether π can be expressed as π = kℓπ′, where ℓ ≥ 0 and π′ is (k − 1)-ary, or as π = kℓπ′′kπ′′′, where π′′ is a word on the alphabet [i, k− 1] for some i ∈ [k− 1] such that i occurs at least once and π′′′ is (i + 1)-ary on [i] ∪ {k}. Note that π′ and π′′′ both avoid {132, 1123}, whereas π′′ avoids {132, 112}. This implies Bk(x) = 1 1− x Bk−1(x) + 1 1− x k∑ j=2 (Mj(x)−Mj−1(x))Bk+2−j(x), k ≥ 1, T. Mansour and M. Shattuck: On avoiding 1233 9 where Mk(x) is the g.f. for the number of k-ary words of the form γk such that γ is (k−1)- ary and avoids {132, 112}. Note that the Mj(x) − Mj−1(x) factor accounts for the fact that the letter i must occur at least once in the section π′′ of π above. We now must determine Mk(x). Note that M1(x) = x, so assume k ≥ 2. Then ρ enumerated by Mk(x) is either of the form ρ = (k − 1) ℓρ′k, where ℓ ≥ 0 and ρ′ contains no k − 1, or of the form ρ = (k − 1)ℓρ′′(k − 1)ρ′′′k, where ρ′′ is a word on [i, k − 2] for some i ∈ [k − 2] containing at least one i and ρ′′′ is (i + 1)-ary on [i] ∪ {k − 1}. Note that ρ′ and ρ′′′ both avoid {132, 112}, whereas ρ′′ avoids {132, 11}. Furthermore, observe that within ρ′′′, any (k − 1)’s must occur prior to any i’s, for otherwise there would be a 1123 in ρ of the form ii(k − 1)k, where the first i occurs in ρ′′. Concerning ρ′′′, we therefore consider additional cases based on whether ρ′′′ contains (i) neither i nor k − 1, (ii) exactly one of i, k − 1 or (iii) both i and k − 1. Note that all (k − 1)’s in ρ′′′ occur as an initial run in case (iii), for otherwise a 132 would occur. Hence, we get contributions of Mi(x), 2(Mi+1(x)−Mi(x)) and x 1−x (Mi+1(x)−Mi(x)) for (i), (ii) and (iii), respectively. Considering all possible i, and replacing i with k−i, then gives for all k ≥ 2 the recurrence Mk(x) = 1 1− x Mk−1(x) + 1 1− x k−1∑ i=2 (Li(x)− Li−1(x)) ( 2− x 1− x Mk+1−i(x)− 1 1− x Mk−i(x) ) , where Lk(x) is the g.f. for the number of k-ary words of the form γk that avoid {132, 11}. Since such words correspond to 132-avoiding permutations whose largest letter is last, we have Lk(x) = ∑k−1 j=0 Cj ( k−1 j ) xj+1 for k ≥ 1. Define the bivariate g.f.’s by B(x, y) = ∑ k≥0 Bk(x)y k, M(x, y) = ∑ k≥1 Mk(x)y k and L(x, y) = ∑ k≥1 Lk(x)y k. Then the recurrences above for Bk(x) and Mk(x) imply ( 1− y 1− x ) B(x, y) = 1+ 1 (1− x)y2 ((1− y)M(x, y)− xy) ( B(x, y)− 1− y 1− x ) and ( 1− y 1− x ) M(x, y) = xy+ ((1− y)L(x, y)− xy)((2− x− y)M(x, y)− (2− x)xy) (1− x)2y , where L(x, y) = xy1−yC ( xy 1−y ) and C(z) = 1− √ 1−4z 2z = ∑ n≥0 Cnz n denotes the g.f. for the Catalan number sequence. Solving the preceding equations for B(x, y) yields after several algebraic steps the following result. Theorem 3.2. The generating function for the number of k-ary words of length n that avoid both 213 and 1233 (132 and 1123) for all n, k ≥ 0 is given by 4(1− 2x)(1− x)2 − 2(1− x)(4− 7x+ 4x2)y + (4− 7x+ 4x2)y2 + xy2 √ 1− 4xy1−y 2(2(1− 2x)(1− x)2 − (1− x)(2− x)(3− 4x)y + 2(1− x)(3− 2x)y2 − (2− x)y3) . By Theorem 3.2, we have for example the following formulas for Bk(x) where 1 ≤ k ≤ 5: 10 Art Discrete Appl. Math. 5 (2022) #P1.09 B1(x) = 1 1− x , B2(x) = 1 1− 2x , B3(x) = 1− 3x+ 4x2 − x3 (1− x)2(1− 2x)2 , B4(x) = 1− 3x+ 6x2 + 2x4 (1− x)(1− 2x)3 , B5(x) = 1− 6x+ 21x2 − 34x3 + 32x4 − 26x5 + 13x6 + 8x7 − 8x8 (1− x)3(1− 2x)4 . ORCID iDs Toufik Mansour https://orcid.org/0000-0001-8028-2391 Mark Shattuck https://orcid.org/0000-0001-8473-2505 References [1] M. H. Albert, R. E. L. Aldred, M. D. Atkinson, C. Handley and D. Holton, Permutations of a multiset avoiding permutations of length 3, European J. Combin. 22 (2001), 1021–1031, doi:10.1006/eujc.2001.0538. [2] N. Alon and E. Friedgut, On the number of permutations avoiding a given pattern, J. Combin. Theory Ser. A 89 (2000), 133–140, doi:10.1006/jcta.1999.3002. [3] P. Brändén and T. Mansour, Finite automata and pattern avoidance in words, J. Combin. Theory Ser. A 110 (2005), 127–145, doi:10.1016/j.jcta.2004.10.007. [4] A. Burstein, Enumeration of words with forbidden patterns, ProQuest LLC, Ann Arbor, MI, 1998, thesis (Ph.D.)–University of Pennsylvania, http://gateway.proquest.com/ openurl?url_ver=Z39.88-2004&rft_val_fmt=info:ofi/fmt:kev:mtx: dissertation&res_dat=xri:pqdiss&rft_dat=xri:pqdiss:9829868. [5] A. Burstein and T. Mansour, Words restricted by patterns with at most 2 distinct letters, Elec- tron. J. Combin. 9 (2002/03), Research paper 3, 14, doi:10.37236/1675, permutation patterns (Otago, 2003). [6] D. E. Knuth, The art of computer programming. Volume 3, Addison-Wesley Series in Com- puter Science and Information Processing, Addison-Wesley Publishing Co., Reading, Mass.- London-Don Mills, Ont., 1973, sorting and searching. [7] T. Mansour, Restricted 132-avoiding k-ary words, Chebyshev polynomials, and continued frac- tions, Adv. in Appl. Math. 36 (2006), 175–193, doi:10.1016/j.aam.2005.04.003. [8] T. Mansour, R. Rastegar and A. Roitershtein, Finite automata, probabilistic method, and occur- rence enumeration of a pattern in words and permutations, SIAM J. Discrete Math. 34 (2020), 1011–1038, doi:10.1137/19M1262206. [9] L. Pudwell, Enumeration schemes for words avoiding permutations, in: Permutation patterns, Cambridge Univ. Press, Cambridge, volume 376 of London Math. Soc. Lecture Note Ser., pp. 193–211, 2010, doi:10.1017/CBO9780511902499.010. [10] A. Regev, Asymptotics of the number of k-words with an l-descent, Electron. J. Combin. 5 (1998), Research Paper 15, 4, doi:10.37236/1353. T. Mansour and M. Shattuck: On avoiding 1233 11 [11] N. Shar and D. Zeilberger, The (ordinary) generating functions enumerating 123-avoiding words with r occurrences of each of 1, 2, . . . , n are always algebraic, Ann. Comb. 20 (2016), 387–396, doi:10.1007/s00026-016-0308-y. [12] R. Simion and F. W. Schmidt, Restricted permutations, European J. Combin. 6 (1985), 383– 406, doi:10.1016/S0195-6698(85)80052-4. [13] H. S. Wilf, The patterns of permutations, Discrete Math. 257 (2002), 575–583, doi:10.1016/ S0012-365X(02)00515-0, kleitman and combinatorics: a celebration (Cambridge, MA, 1999). [14] D. Zeilberger, A snappy proof that 123-avoiding words are equinumerous with 132-avoiding words, 2005. ISSN 2590-9770 The Art of Discrete and Applied Mathematics 5 (2022) #P1.10 https://doi.org/10.26493/2590-9770.1389.fa2 (Also available at http://adam-journal.eu) Cayley graphs of order 6pq and 7pq are Hamiltonian Farzad Maghsoudi* Carleton University, Ottawa, Canada Received 25 September 2020, accepted 8 May 2021, published online 11 May 2022 Abstract Assume G is a finite group, such that |G| “ 6pq or 7pq, where p and q are distinct prime numbers, and let S be a generating set of G. We prove there is a Hamiltonian cycle in the corresponding connected Cayley graph CaypG;Sq. Keywords: Cayley graphs, Hamiltonian cycles. Math. Subj. Class.: 05C25, 05C45 1 Introduction Arthur Cayley [1] introduced the definition of Cayley graph in 1878. All graphs in this paper are undirected (graphs without loops and direction on the edges). Definition 1.1 ([16, Definition 1.1], cf. [11, p. 34]). Let S be a subset of a finite group G. The Cayley graph CaypG;Sq is the graph whose vertices are elements of G, with an edge joining g and gs, for every g P G and s P S. Since then, the theory of Cayley graphs has developed into an important branch of alge- braic graph theory. It is an interesting topic to work on because not only is it related to pure mathematics problems, but it is connected to fascinating problems studied by computer scientists, molecular biologists, and coding theorists (see [15] for more information). *Theorem 1.3 and Proposition 1.4 are the main results of this paper. I would like to express my sincere gratitude to my supervisor, professor Joy Morris who always supported me throughout my graduate journey. I am especially grateful to my co-supervisor, professor Dave Morris, for the patient guidance and advice he has provided during my graduate study. I have been extremely lucky to have a co-supervisor who cared so much about my research, and who responded to my questions so promptly. I am also thankful to professor Hadi Kharaghani and professor Amir Akbary and cannot forget their valuable help and motivation during my graduate years. I am truly grateful to my family for their immeasurable love and care. E-mail address: FARZADMAGHSOUDI@cmail.carleton.ca (Farzad Maghsoudi) cb This work is licensed under https://creativecommons.org/licenses/by/4.0/ 2 Art Discrete Appl. Math. 5 (2022) #P1.10 Recall that a Hamiltonian cycle is a cycle that visits every vertex of a graph. Finding Hamiltonian cycles is a fundamental question in graph theory, but in general, it is extremely difficult. To be precise, it is an NP-complete problem, which means most mathematicians do not believe there exists an efficient algorithm to determine whether an arbitrary graph contains such a cycle. Because the general case is so hard, it is natural to look at special cases. Cayley graphs are one of these cases that mathematicians are interested in working on. There have been many papers on the topic of Hamiltonian cycles in Cayley graphs, but it is still an open question whether every connected Cayley graph has a Hamiltonian cycle. (See survey papers [5, 24, 21] for more information. We ignore the trivial counterexamples on 1 or 2 vertices.) The following result combines the main result of this paper with the previous work of several authors (C. C. Chen and N. Quimpo [2], S. J. Curran, J. Morris and D. W. Morris [6], E. Ghaderpour and D. W. Morris [9, 10], D. Jungreis and E. Friedman [13], Kutnar et al. [16], K. Keating and D. Witte [14], D. Li [17], D. W. Morris and K. Wilk [20], and D. Witte [23]). Theorem 1.2 ([16, 20, 23]). Let G be a finite group. If |G| has any of the forms below (where p, q, and r are distinct primes), then every connected Cayley graph on G has a Hamiltonian cycle. 1. kp, where 1 ď k ď 47, 2. kpq, where 1 ď k ď 7, 3. pqr, 4. kp2, where 1 ď k ď 4, 5. kp3, where 1 ď k ď 2, 6. pk, where 1 ď k ă 8. Previously, part (2) of Theorem 1.2 was only known for 1 ď k ď 5, but we improve this condition: we show that 5 can be replaced with 7. This is the new part of the above theorem which is our result. The hard part is when k “ 6: Theorem 1.3. Assume G is a finite group of order 6pq, where p and q are distinct prime numbers. Then every connected Cayley graph on G contains a Hamiltonian cycle. This generalizes [10], which considered only the case where q “ 5. The proof takes up all of Section 3, after some preliminaries in Section 2. Unlike Theorem 1.3, the following observation follows easily from known results, and may be known to experts. The proof is on page 8. Proposition 1.4. Assume G is a finite group of order 7pq, where p and q are distinct prime numbers. Then every connected Cayley graph on G contains a Hamiltonian cycle. The Introduction of the author’s masters thesis [18] provides additional background and a description of the methods that are used in the proof of the main theorem. F. Maghsoudi: Cayley graphs of order 6pq and 7pq are Hamiltonian 3 2 Preliminaries This section establishes basic terminology and notation, and proves a number of technical results that will be used in the proof of Theorem 1.3. In particular, it is shown we may assume that |G| is square-free (note |G| “ 6pq in Theorem 1.3), so the Sylow subgroups of G are C2, C3, Cp, and Cq , and that |G 1| has precisely 2 prime factors, so G1 is either Cp ˆ Cq or C3 ˆ Cp. 2.1 Basic notation and definitions Throughout the paper, we have used standard terminology of graph theory and group theory that can be found in textbooks, such as [11, 12]. The following notation is used throughout the paper: • The commutator ghg´1h´1 of g and h is denoted by rg, hs. • We will always let G1 “ rG,Gs be the commutator subgroup of G. • We define G “ G{G1, g “ gG1 for any g P G, and S “ tg; g P Su for any S Ď G. • CG1 pSq denotes the centralizer of S in G 1. • G ˙ H denotes a semidirect product of groups G and H , where H is normal. • D2n denotes the dihedral group of order 2n. • e denotes the identity element of G. • For S Ď G, a sequence ps1, s2, . . . , snq of elements of S Y S ´1 specifies the walk in the Cayley graph CaypG;Sq that visits the vertices: e, s1, s1s2, . . . , s1s2 ¨ ¨ ¨ sn. Also, ps1, s2, . . . , snq ´1 “ ps´1n , s ´1 n´1, . . . , s ´1 1 q. • We use ps1, s2, . . . , snq to denote the image of the walk ps1, s2, . . . , snq in the CaypG{G1;Sq “ CaypG;Sq which is a Cayley graph on the quotient group G{G1. • For k P Z`, we use ps1, s2, . . . , smq k to denote the concatenation of k copies of the sequence ps1, s2, . . . , smq. • p and q are distinct prime numbers. • Cn denotes the cyclic group of order n. • pG “ G{Cp, when Cp is a normal subgroup, we also let qG “ G{Cq when Cq is a normal subgroup, and let ÐÑ G “ G{C3 when C3 is a normal subgroup. Also, pg “ gCp, qg “ gCq , for any g P G, and pS “ tpg; g P Su, qS “ tqg; g P Su for any S Ď G. • We let a2, a3, γp, and aq be elements of G that generate C2, C3, Cp, and Cq , respec- tively. Remark 2.1. When |G| “ 6pq and it is square free (as is usually the case in Section 3), the Sylow subgroups are C2, C3, Cp, and Cq . Also, the commutator subgroup G 1 will usually be either Cp ˆ Cq or C3 ˆ Cp, so Cp is a normal subgroup and either Cq or C3 is also a normal subgroup. 4 Art Discrete Appl. Math. 5 (2022) #P1.10 2.2 Basic methods In this subsection, we explain some of the key ideas in the proof of our main result (Theo- rem 1.3). It is easy to see that CaypG;Sq is connected if and only if S generates G ([11, Lemma 3.7.4]). Also, if S is a subset of S0, then CaypG;Sq is a subgraph of CaypG;S0q that con- tains all of the vertices. Therefore, in order to show that every connected Cayley graph on G contains a Hamiltonian cycle, it suffices to consider CaypG;Sq, where S is a generating set that is minimal, which means that no proper subset of S generates G. The following well known (and easy) result handles the case of Theorem 1.3 where G is abelian. Lemma 2.2 ([3, Corollary on page 257]). Assume G is an abelian group. Then every connected Cayley graph on G has a Hamiltonian cycle. Note CaypC2; tauq is a Cayley graph with two vertices, where C2 “ xay. We consider pa, aq as its Hamiltonian cycle which is: e a Ñ a a Ñ a2 “ e. Although graph theorists would not typically consider this a cycle, it satisfies the basic property of visiting each vertex exactly once. In some of our inductive proofs, we require a Hamiltonian cycle in a Cayley graph on a quotient group. When this quotient group is C2, this Hamiltonian cycle provides the structure we need for our inductive arguments to work. Theorem 2.3 (Marušič [19], Durnberger [7, 8], and Keating-Witte [14]). If the commutator subgroup G1 of G is a cyclic p-group, then every connected Cayley graph on G has a Hamiltonian cycle. Theorem 2.4 (Chen-Quimpo [4]). Let v and w be two distinct vertices of a connected Cayley graph CaypG;Sq. Assume G is abelian, |G| is odd, and the valency of CaypG;Sq is at least 3. Then CaypG;Sq has a Hamiltonian path that starts at v and ends at w. The following lemma (and its corollary) often provide a way to lift a Hamiltonian cy- cle in CaypG{N ;Sq to a Hamiltonian cycle in CaypG;Sq. Before stating the results, we introduce a useful piece of notation. Notation 2.5. Suppose N is a normal subgroup of G, and C “ ps1, s2, . . . , snq is a walk in CaypG;Sq. If the walk ps1N, s2N, . . . , snNq in CaypG{N ;SN{Nq is closed, then its voltage is the product VpCq “ s1s2 ¨ ¨ ¨ sn. This is an element of N . In particular, if C “ ps1, s2, . . . , snq is a Hamiltonian cycle in CaypG,Sq, then VpCq “ s1s2 ¨ ¨ ¨ sn. Factor Group Lemma 2.6 ([24, Section 2.2]). Suppose: • S is a generating set of G, • N is a cyclic normal subgroup of G, • G “ G{N , • C “ ps1, s2, . . . , snq is a Hamiltonian cycle in CaypG{N ;Sq, and • the voltage VpCq generates N . F. Maghsoudi: Cayley graphs of order 6pq and 7pq are Hamiltonian 5 Then there is a Hamiltonian cycle in CaypG;Sq. Corollary 2.7 ([10, Corollary 2.3]). Suppose: • S is a generating set of G, • N is a normal subgroup of G, such that |N | is prime, • sN “ tN for some s, t P S with s ‰ t, and • there is a Hamiltonian cycle in CaypG{N ;Sq that uses at least one edge labeled s. Then there is a Hamiltonian cycle in CaypG;Sq. Lemma 2.8. Assume G “ H ˙ pCp ˆ Cqq, where G 1 “ Cp ˆ Cq , and let S be a generating set of G. As usual, let G “ G{G1 – H . Assume there is a unique element c of S that is not in H ˙ Cq , and C is a Hamiltonian cycle in CaypG;Sq such that c occurs precisely once in C. Then the subgroup generated by VpCq contains Cp. Proof. Write C “ ps1, s2, ¨ ¨ ¨ , snq, and let H ` “ H ˙ Cq . By assumption, there is a unique k, such that sk “ c, and all other elements of S are in H `. Therefore, VpCq “ s1s2...sn P H ` ¨ H` ¨ ¨ ¨H` ¨ c ¨ H` ¨ H` ¨ ¨ ¨H` “ H`cH`. Since c R H`, we conclude that VpCq R H`. On the other hand, since VpCq is an element of G1 “ Cp ˆCq , we have VpCq “ a i qγ j p P H`γjp. Since VpCq R H `, this implies j ı 0 pmod pq, so xaiqγ j py contains Cp. Definition 2.9. The Cartesian product X1 ˝ X2 of graphs X1 and X2 is a graph such that the vertex set of X1 ˝ X2 is V pX1q ˆ V pX2q “ tpv, v 1q; v P V pX1q, v 1 P V pX2qu, and two vertices pv1, v2q and pv 1 1, v 1 2q are adjacent in X1 ˝ X2 if and only if either • v1 “ v 1 1 and v2 is adjacent to v 1 2 in X2 or • v2 “ v 1 2 and v1 is adjacent to v 1 1 in X1. Lemma 2.10 ([4, Lemma 5 on page 28]). The Cartesian product of a path and a cycle is Hamiltonian. Corollary 2.11 (cf. [4, Corollary on page 29]). The Cartesian product of two Hamiltonian graphs is Hamiltonian. Lemma 2.12 ([16, Lemma 2.27]). Let S generate the finite group G, and let s P S, such that xsy Ÿ G. If CaypG{xsy;Sq has a Hamiltonian cycle, and either 1. s P ZpGq , or 2. ZpGq X xsy “ teu, then CaypG;Sq has a Hamiltonian cycle. 6 Art Discrete Appl. Math. 5 (2022) #P1.10 2.3 Some facts from group theory In this subsection, we state some facts in group theory, which are used to prove our main result. The following lemma often makes it possible to use Factor Group Lemma 2.6 for finding Hamiltonian cycles in connected Cayley graphs of G. Lemma 2.13 ([6, Corollary 4.4]). Assume G “ xa, by and G1 is cyclic. Then G1 “ xra, bsy. Corollary 2.14. Assume G “ xa, by and gcdpk, |a|q “ 1, where k P Z, and G1 is cyclic. Then G1 “ xrak, bsy. Proposition 2.15 ([12, Theorem 9.4.3 on page 146], cf. [10, Lemma 2.11]). Assume |G| is square-free. Then: 1. G1 and G{G1 are cyclic, 2. ZpGq X G1 “ teu, 3. G – Cn ˙ G 1, for some n P Z`, 4. If b and γ are elements of G such that xbG1y “ G{G1 and xγy “ G1, then xb, γy “ G, and there are integers m, n, and τ , such that |γ| “ m, |b| “ n, bγb´1 “ γτ , mn “ |G|, gcdpτ ´ 1,mq “ 1, and τn ” 1 pmod mq. Lemma 2.16. Assume • G “ pCp ˆ Cqq ˙ pCr ˆ Ctq, • G1 “ pCr ˆ Ctq, • a P G, • p, q, r, and t are distinct primes. If |a| “ pq, then |a| “ pq. Proof. Suppose |a| ‰ pq. Without loss of generality, assume |a| is divisible by r. Then (after replacing a by a conjugate) the abelian group xay contains Cp ˆ Cq and Cr, so Cr centralizes Cp ˆ Cq . Since Cr also centralizes Ct, this implies that Cr Ď ZpGq. This contradicts the fact that G1 X ZpGq “ teu (see Proposition 2.15(2)). Lemma 2.17 ([22, Exercise 19 on page 43]). Assume |G| “ 2k, where k is odd. Then G has a subgroup of index 2. Corollary 2.18. Assume |G| “ 2k, where k is odd. Then |G1| is odd. Proof. By Lemma 2.17, there is a normal subgroup H of G such that rG : Hs “ 2. Now since G{H has order 2, then G{H is abelian, so G1 Ď H . Therefore, |G1| is odd. Notation 2.19. For τ as defined in Proposition 2.15(4), we use τ´1 to denote the inverse of τ modulo m (so τ´1 ” τn´1 pmod mq). F. Maghsoudi: Cayley graphs of order 6pq and 7pq are Hamiltonian 7 2.4 Cayley graphs that contain a Hamiltonian cycle We show, throughout this subsection, that there exists a Hamiltonian cycle in some con- nected Cayley graphs with additional assumptions. The following proposition shows that in our proof of Theorem 1.3 we can assume |G| is square-free, since the cases where |G| is not square-free have already been dealt with. At the end of this subsection we prove Proposition 1.4. Proposition 2.20. Assume: • |G| “ 6pq, where p and q are distinct prime numbers, and • |G| is not square-free (i.e. tp, qu X t2, 3u ‰ H). Then every connected Cayley graph on G has a Hamiltonian cycle. Proof. Without loss of generality we may assume q P t2, 3u. Then |G| P t12p, 18pu. Therefore, Theorem 1.2(1) applies. Proposition 2.21 ([25, Proposition 5.5]). If n is divisible by at most 3 distinct primes, then every Cayley diagram (directed Cayley graph) in D2n has a Hamiltonian cycle. The following proposition demonstrates that we can assume |G1| in Theorem 1.3 is a product of two distinct prime numbers. Proposition 2.22. Assume |G| “ 2pqr, where p, q and r are distinct odd prime num- bers. If |G1| P t1, pqru or |G1| is prime, then every connected Cayley graph on G has a Hamiltonian cycle. Proof. If |G1| “ 1, then G1 “ teu. So G is an abelian group. Therefore, Lemma 2.2 applies. If |G1| is prime, then Theorem 2.3 applies. Finally, if |G1| “ pqr, then G “ C2 ˙ pCp ˆ Cq ˆ Crq – D2pqr. So Proposition 2.21 applies. The following lemmas show that some special Cayley graphs have a Hamiltonian cycle, and we use these facts in Section 3 in order to prove our main result. Lemma 2.23. Assume G “ pC2 ˆ Crq ˙ G 1, and G1 “ Cp ˆ Cq , where p, q and r are distinct odd prime numbers and let S “ ta, bu be a generating set of G. Additionally, assume |a| P t2, 2ru, |b| “ r and gcdp|b|, r ´ 1q “ 1. Then CaypG;Sq contains a Hamiltonian cycle. Proof. We have C “ pb r´1 , a, b ´pr´1q , a´1q as a Hamiltonian cycle in CaypG;Sq. Now we calculate its voltage VpCq “ br´1ab´pr´1qa´1 “ rbr´1, as. Since gcdp|b|, r ´ 1q “ 1, then by Lemma 2.14 we have rbr´1, as “ G1. Therefore, Factor Group Lemma 2.6 applies. Lemma 2.24 (cf. [10, Case 2 of proof of Theorem 1.1, pages 3619-3620]). Assume 8 Art Discrete Appl. Math. 5 (2022) #P1.10 • G “ pC2 ˆ Crq ˙ pCp ˆ Cqq, • |S| “ 3, • pS is a minimal generating set of pG “ G{Cp, • Cr centralizes Cq , • C2 inverts Cq . Then, CaypG;Sq contains a Hamiltonian cycle. Lemma 2.25 ([10, Lemma 2.6]). Assume: • G “ xay ˙ xS0y, where xS0y is an abelian subgroup of odd order, • |pS0 Y S ´1 0 q| ě 3, and • xS0y has a nontrivial subgroup H , such that H Ÿ G and H X ZpGq “ teu. Then CaypG;S0 Y tauq has a Hamiltonian cycle. Lemma 2.26 ([10, Lemma 2.9]). If G “ D2pq ˆ Cr, where p, q and r are distinct odd primes, then every connected Cayley graph on G has a Hamiltonian cycle. Now we prove Proposition 1.4 which is on page 2. Proof of Proposition 1.4. If p ‰ 7 and q ‰ 7, then Theorem 1.2(3) applies. So we may assume q “ 7, which means |G| “ 49p (and p ‰ 7). We may also assume that G is not abelian, for otherwise Lemma 2.2 applies. If a Sylow p-subgroup P of G is normal, then |G{P | “ 49, so the quotient G{P is abelian. (Because if q is prime, then every group of order q2 is abelian). Therefore, since P is normal and G{P is abelian, then G1 is contained in P . So |G1| “ p. Therefore, Theorem 2.3 applies. Now we may assume P is not normal in G. Then by Sylow’s Theorem, np|49 and np ” 1 pmod pq, where np is the number of Sylow p-subgroups in G. Thus, p P t2, 3u, so |G| P t14q, 21qu. Therefore, Theorem 1.2(1) applies. 2.5 Some specific sets that generate G This Subsection presents a few results that provide conditions under which certain 2- element subsets generate G. Obviously, no 3-element minimal generating set can contain any of these subsets. Lemma 2.27. Assume G “ pC2ˆC3q˙G 1, and G1 “ CpˆCq . Also, assume CG1 pC3q “ Cq and Cq Ę CG1 pC2q. If pa, bq is one of the following ordered pairs 1. pa3aq, a2a j 3a k qγpq, 2. pa2a3, a j 3a k qγpq, where k ı 0 pmod qq, 3. pa2a3aq, a j 3a k qγpq, where k ı 0 pmod qq, 4. pa2a3aq, a2a j 3a k qγpq, where k ı 1 pmod qq, F. Maghsoudi: Cayley graphs of order 6pq and 7pq are Hamiltonian 9 then xa, by “ G. Proof. It is easy to see that pa, bq “ G, so it suffices to show that xa, by contains Cp and Cq . Thus, it suffices to show that Ğ and qG are nonabelian, where Ğ “ G{pC3 ˙Cpq – D2q and qG “ G{Cq . Since a3 does not centralize Cp, it is clear in each of p1q ´ p4q that qa does not central- ize γp (and γp is one of the factors in qb), so qG is not abelian. The pair pă, b̆q is either paq, a2a k q q, pa2, a k q q where k ı 0 pmod qq, pa2aq, a k q q where k ı 0 pmod qq, or pa2aq, a2a k q q where k ı 1 pmod qq. Each of these is either a reflec- tion and a nontrivial rotation or two different reflections, and therefore generates the (non- abelian) dihedral group D2q “ Ğ. Lemma 2.28. Assume G “ pC2ˆC3q˙G 1, and G1 “ CpˆCq . Also, assume CG1 pC3q “ teu. If pa, bq is one of the following ordered pairs 1. pa2a3, a i 2a j 3a k qγpq, where k ı 0 pmod qq, 2. pa3aq, a2a j 3γpq, where j ı 0 pmod 3q, 3. pa3, a2a j 3a k qγpq, where k ı 0 pmod qq, 4. pa2a3aq, a i 2a j 3γpq, where j ı 0 pmod 3q, then xa, by “ G. Proof. It is easy to see that pa, bq “ G, so it suffices to show that xa, by contains Cp and Cq . we need to show that pG and qG are nonabelian, where pG “ G{Cp and qG “ G{Cq , as usual. As in the proof of Lemma 2.27, since a3 does not centralize Cp, it is clear in each of p1q´p4q that qa does not centralize γp (and γp is one of the factors in qb), so qG is not abelian. In p1q ´ p4q, aq appears in one of the generators in ppa,pbq, but not the other, and the other generator does have an occurrence of a3. Since a3 does not centralize aq , this implies that pG is not abelian. Lemma 2.29. Assume G “ pC2ˆCqq˙G 1, and G1 “ C3ˆCp. Also, assume CG1 pCqq “ C3 and C3 Ę CG1 pC2q. If pa, bq is one of the following ordered pairs 1. pa2aq, a i 2a j qa k 3γpq, where k ı 0 pmod qq, 2. paqa3, a2a j qa k 3γpq, 3. pai2a m q a3, a2a j qγpq, where m ı 0 pmod qq, then G “ xa, by. Proof. It is easy to see that pa, bq “ G, so it suffices to show that xa, by contains Cp and C3. We need to show that Ğ and ÐÑ G are nonabelian, where Ğ “ G{pCq ˙ Cpq – D6 and ÐÑ G “ G{C3. In each of p1q ´ p4q, aq appears in ÐÑa , and γp appears in ÐÑ b (but not in ÐÑa ). Since aq does not centralize γp, this implies that ÐÑ G is not abelian. In each of p1q ´ p4q, pÐÑa , ÐÑ b q consists of either a reflection and a nontrivial rotation or two different reflections, so it generates the (nonabelian) dihedral group D6 “ ÐÑ G . 10 Art Discrete Appl. Math. 5 (2022) #P1.10 3 Proof of the main result In this section, we prove Theorem 1.3, which is the main result. We are given a generating set S of a finite group G of order 6pq, where p and q are distinct prime numbers, and we wish to show CaypG;Sq contains a Hamiltonian cycle. The proof is a long case-by-case analysis (see Figures 1, 2 and 3 for outlines of the many cases that are considered). Here are our main assumptions throughout the whole section. Assumption 3.1. We assume: 1. p, q ą 7, otherwise Theorem 1.2(1) applies. 2. |G| is square-free, otherwise Proposition 2.20 applies. 3. G1 X ZpGq “ teu, by Proposition 2.15(2). 4. G – Cn ˙ G 1, by Proposition 2.15(3). 5. |G1| P tpq, 3pu, by Corollary 2.18. 6. For every element s P S, |s| ‰ 1. Otherwise, if |s| “ 1, then s P G1, so G1 “ xsy or |s| is prime. In each case CaypG{xsy;Sq has a Hamiltonian cycle by part 2 or 3 of Theorem 1.2. By Assumption 3.1(3), xsy X ZpGq “ teu, therefore, Lemma 2.12(2) applies. 7. S is a minimal generating set of G. Note that S must generate G, for otherwise CaypG;Sq is not connected. Also, in order to show that every connected Cayley graph on G contains a Hamiltonian cycle, it suffices to consider CaypG;Sq, where S is a generating set that is minimal. 3.1 Assume |S| “ 2 and G1 “ Cp ˆ Cq In this subsection, we prove the part of Theorem 1.3 where, |S| “ 2 and G1 “ Cp ˆ Cq . Recall G “ G{G1 and pG “ G{Cp. Proposition 3.2. Assume • G “ pC2 ˆ C3q ˙ pCp ˆ Cqq, • |S| “ 2. Then CaypG;Sq contains a Hamiltonian cycle. Proof. Let S “ ta, bu. For every s P S, |s| ‰ 1, by Assumption 3.1(6). Case 1. Assume S is minimal. Then |a|, |b| P t2, 3u. When |a| “ |b| “ 2 or |a| “ |b| “ 3, then G ‰ xa, by. Therefore, G ‰ xa, by which contradicts the fact that G “ xa, by. So we may assume |a| “ 2 and |b| “ 3. Since |b| P t3, 3p, 3q, 3pqu, then gcdp|b|, 2q “ 1. Thus, Lemma 2.23 applies. Case 2. Assume S is not minimal. Then t|a|, |b|u is either t6, 2u, t6, 3u, or t6u. We may assume |a| “ 6. F. Maghsoudi: Cayley graphs of order 6pq and 7pq are Hamiltonian 11 I. ⑤S⑤ ✏ 2 A. G✶ ✏ Cp ✂ Cq (Section 3.1). 1. S is a minimal generating set. 2. S is not a minimal generating set. B. G✶ ✏ C3 ✂ Cp (Section 3.2). 1. ⑤a⑤ ✏ ⑤b⑤ ✏ 2q. 2. ⑤a⑤ ✏ q. 3. ⑤a⑤ ✏ 2q and ⑤b⑤ ✏ 2. 4. None of the previous cases apply. Figure 1: Outline of the cases in the proof of Theorem 1.3 where |S| “ 2 II. ⑤S⑤ ✏ 3. A. G✶ ✏ Cp ✂ Cq . a. CG✶♣C3q ✘ te✉ or ♣S is minimal. i. CG✶♣C3q ✘ te✉ (Section 3.3). 1. a ✏ a2 and b ✏ aqa3. 2. a ✏ a2 and b ✏ a2aqa3. 3. a ✏ a2a3 and b ✏ a2aq . 4. a ✏ a2a3 and b ✏ aqa3. 5. a ✏ a2a3 and b ✏ a2a3aq . ii. ♣S is minimal (Section 3.4). 1. CG✶♣C2q ✏ Cp ✂ Cq . 2. CG✶♣C2q ✏ Cq . 3. CG✶♣C2q ✏ Cp. 4. CG✶♣C2q ✏ te✉. b. CG✶♣C3q ✏ te✉ and ♣S is not mini- mal. i. CG✶♣C2q ✏ Cp ✂ Cq (Section 3.5). 1. a ✏ a3 and b ✏ a2aq . 2. a ✏ a3 and b ✏ a2a3aq . 3. a ✏ a2a3 and b ✏ a3aq . 4. a ✏ a2a3 and b ✏ a2aq . 5. a ✏ a2a3 and b ✏ a2a3aq . ii. CG✶♣C2q ✘ te✉ (Section 3.6). 1. a ✏ a2a3 and b ✏ a2a3aq . 2. a ✏ a2a3 and b ✏ a2aq . 3. a ✏ a2a3 and b ✏ a3aq . 4. a ✏ a3 and b ✏ a2aq . iii. CG✶♣C2q ✏ te✉ (Section 3.7). 1. a ✏ a2a3 and b ✏ a2a3aq . 2. a ✏ a2a3 and b ✏ a2aq . 3. a ✏ a2a3 and b ✏ a3aq . 4. a ✏ a3 and b ✏ a2aq . B. G✶ ✏ C3 ✂ Cp. (Section 3.8). 1. a ✏ a2aq and b ✏ a2a m q a3. 2. a ✏ a2aq and b ✏ a2a3. 3. a ✏ a2aq and b ✏ a m q a3. 4. a ✏ a2 and b ✏ aqa3. Figure 2: Outline of the cases in the proof of Theorem 1.3 where |S| “ 3 12 Art Discrete Appl. Math. 5 (2022) #P1.10 III. ⑤S⑤ ➙ 4 (Section 3.9). This part of the proof applies whenever ⑤G⑤ ✏ pqrt with p, q, r, and t distinct primes. 1. ⑤G✶⑤ has only two prime factors. 2. ⑤G✶⑤ has three prime factors. Figure 3: Outline of the cases in the proof of Theorem 1.3 where |S| ě 4 Subcase 2.1. Assume |b| “ 2. So we have b “ a3, then b “ a3γ, where G1 “ xγy for otherwise xa, by “ xa, a3γy “ xa, γy ‰ G which contradicts the fact that G “ xa, by. Now by Proposition 2.15(4), we have τ P Z` such that aγa´1 “ γτ and τ6 ” 1 pmod pqq, also gcdpτ ´ 1, pqq “ 1. This implies that τ ı 1 pmod pq and τ ı 1 pmod qq. We have C1 “ pa 2, b, a´2, b ´1 q as a Hamiltonian cycle in CaypG;Sq. Now we calculate its voltage. VpC1q “ a 2ba´2b´1 “ a2a3γa´2γ´1a´3 “ γτ 5´τ3 “ γτ 3pτ2´1q. We may assume gcdpτ2 ´ 1, pqq ‰ 1 for otherwise Factor Group Lemma 2.6 applies. Without loss of generality let τ2 ” 1 pmod qq, then τ ” ´1 pmod qq. We may assume τ ı ´1 pmod pq, for otherwise G – D2pq ˆ C3, so Lemma 2.26 applies. Consider pG “ G{Cp “ C6˙Cq . Since |a| “ 6, then by Lemma 2.16 |a| “ 6, so |pa| “ 6. We may assume |pb| “ 2, for otherwise Corollary 2.7 applies with s “ b and t “ b´1 since xpay ‰ pG, so any Hamiltonian cycle must use an edge labeled pb. Thus, pb “ pa3aq , where xaqy “ Cq . Since τ ” ´1 pmod qq, then C3 centralizes Cq and C2 inverts Cq . Therefore, pG – D2q ˆ C3. Now we have C2 “ pppa5,pb,pa´5,pbqpq´3q{2, ppa5,pbq3q as a Hamiltonian cycle in Cayp pG; pSq. The picture in Figure 4 on page 13 shows the Hamiltonian cycle when q “ 7. If in C2 we change one occurrence of ppa5,pb,pa´5,pbq to ppa´5,pb,pa5,pbq we have another Hamiltonian cycle. Note that, a5ba´5b “ a5 ¨ a3γ ¨ a´5 ¨ a3γ “ a2γa´2γ “ γτ 2`1, and a´5ba5b “ a´5 ¨ a3γ ¨ a5 ¨ a3γ “ a´2γa2γ “ γτ ´2`1. Since τ4 ı 0 pmod pq we see that τ2 ` 1 ı τ´2 ` 1 pmod pq. Therefore, the voltages of these two Hamiltonian cycles are different, so one of these Hamiltonian cycles has a nontrivial voltage. Thus, Factor Group Lemma 2.6 applies. Subcase 2.2. Assume |b| “ 3. Since |b| “ 3, then |b| P t3, 3p, 3q, 3pqu. Since |a| “ 6, then by 2.16 |a| “ 6. Since gcdp|b|, 2q “ 1, then Lemma 2.23 applies. Subcase 2.3. Assume |b| “ 6. Then we have a “ b or a “ b ´1 . Additionally, by Lemma 2.16 we have |a| “ |b| “ 6. We may assume a “ b by replacing b with its inverse F. Maghsoudi: Cayley graphs of order 6pq and 7pq are Hamiltonian 13 if necessary. Then b “ aγ, where G1 “ xγy, because G “ xa, by. We have C “ pa5, bq as a Hamiltonian cycle in CaypG,Sq. Now we calculate its voltage VpCq “ a5b “ a5aγ “ a6γ “ γ which generates G1. Therefore, Factor Group Lemma 2.6 applies. 3.2 Assume |S| “ 2 and G1 “ C3 ˆ Cp In this subsection, we prove the part of Theorem 1.3 where, |S| “ 2 and G1 “ C3 ˆ Cp. Recall G “ G{G1 and pG “ G{Cp. Proposition 3.3. Assume • G “ pC2 ˆ Cqq ˙ pC3 ˆ Cpq, • |S| “ 2. Then CaypG;Sq contains a Hamiltonian cycle. 1 1 2 2 3 3 44 5 5 pe Figure 4: The Hamiltonian cycle C1: pa edges are solid and pb edges are dashed. 14 Art Discrete Appl. Math. 5 (2022) #P1.10 Proof. Let S “ ta, bu. Since the only non-trivial automorphism of C3 is inversion, Cq centralizes C3. Since G 1 X ZpGq “ teu (see Proposition 2.15(4)), C2 does not centralize C3. Case 1. Assume |a| “ |b| “ 2q. Then b “ am, where 1 ď m ď q ´ 1 by replacing b with its inverse if needed. Therefore, b “ amγ, where G1 “ xγy. Also, gcdpm, 2qq “ 1. So, by Proposition 2.15(4) we have aγa´1 “ γτ where τ2q ” 1 pmod 3pq and gcdpτ ´ 1, 3pq “ 1. Consider G “ C2q . Subcase 1.1. Assume m ą 3. Then we have C “ pb ´2 , a´2, b, a, b, a´pm´2q, b ´1 , am´4, b ´1 , a´p2q´2m´3qq as a Hamiltonian cycle in CaypG;Sq. Now we calculate its voltage. VpCq “ b´2a´2baba´pm´2qb´1am´4b´1a´p2q´2m´3q “ γ´1a´mγ´1a´ma´2amγaamγa´m`2γ´1a´mam´4γ´1a´ma´2q`2m`3 “ γ´1a´mγ´1a´2γam`1γa´m`2γ´1a´4γ´1am`3 “ γ´1´τ ´m`τ´m´2`τ´1´τ´m`1´τ´m´3 “ γ´1`τ ´1´τ´m`1´τ´m`τ´m´2´τ´m´3 . We may assume VpCq does not generate G1 “ C3 ˆ Cp. Therefore, the subgroup generated by VpCq either does not contain C3, or does not contain Cp. We already know τ ” ´1 pmod 3q, then we have ´1 ` τ´1 ´ τ´m`1 ´ τ´m ` τ´m´2 ´ τ´m´3 ” ´4 “ ´1 pmod 3q. This implies that the subgroup generated by VpCq contains C3. So we may assume the subgroup generated by VpCq does not contain Cp, then 0 ” ´1 ` τ´1 ´ τ´m`1 ´ τ´m ` τ´m´2 ´ τ´m´3 pmod pq. (1.1A) Multiplying by ´τm`3 we have 0 ” τm`3 ´ τm`2 ` τ4 ` τ3 ´ τ ` 1 pmod pq. (1.1B) Replacing ta, bu with ta´1, b ´1 u replaces τ with τ´1. Therefore, applying the above ar- gument to ta´1, b ´1 u establishes that 1.1A holds with τ´1 in the place of τ , which means we have 0 ” ´τm`3 ` τm`2 ´ τm ´ τm´1 ` τ ´ 1 pmod pq. (1.1C) By adding 1.1B and 1.1C we have 0 ” ´τm ´ τm´1 ` τ4 ` τ3 “ τ3pτ ` 1qp1 ´ τm´4q pmod pq. If τ ” ´1 pmod pq, then C2q inverts C3p, so Cq centralizes Cp. This implies that G – D6p ˆ Cq , so Lemma 2.26 applies. The only other possibility is τ m´4 ” 1 pmod pq. Multiplying by τ4, we have τm ” τ4 pmod pq. We also know that τ2q ” 1 pmod pq. So τd ” 1 pmod pq, where d “ gcdpm ´ 4, 2qq. Since m is odd and m ă q, then d “ 1. This contradicts the fact that gcdpτ ´ 1, 3pq “ 1. F. Maghsoudi: Cayley graphs of order 6pq and 7pq are Hamiltonian 15 Subcase 1.2. Assume m ď 3. Therefore, either m “ 1 or m “ 3. If m “ 1, then a “ b and b “ aγ. So we have C1 “ pa 2q´1, bq as a Hamiltonian cycle in CaypG;Sq. Now we calculate its voltage. VpC1q “ a 2q´1b “ a2q´1aγ “ γ which generates G1. Therefore, Factor Group Lemma 2.6 applies. Now if m “ 3, then b “ a3γ and we have C2 “ pb 2 , a´1, b ´1 , a´1, b 3 , a´2, b, a2q´11q as a Hamiltonian cycle in CaypG;Sq. We calculate its voltage. VpC2q “ b 2a´1b´1a´1b3a´2ba2q´11 “ a3γa3γa´1γ´1a´3a´1a3γa3γa3γa´2a3γa´11 “ a3γa3γa´1γ´1a´1γa3γa3γaγa´11 “ γτ 3`τ6´τ5`τ4`τ7`τ10`τ11 “ γτ 11`τ10`τ7`τ6´τ5`τ4`τ3 We may assume VpC2q does not generate G 1 “ C3ˆCp. Therefore, the subgroup generated by VpCq does not contain either C3, or Cp. We already know τ ” ´1 pmod 3q, then τ11 ` τ10 ` τ7 ` τ6 ´ τ5 ` τ4 ` τ3 ” ´1 ` 1 ´ 1 ` 1 ` 1 ` 1 ´ 1 “ 1 pmod 3q. This implies that the subgroup generated by VpC2q contains C3. So we may assume the subgroup generated by VpC2q does not contain Cp, for otherwise Factor Group Lemma 2.6 applies. Then we have 0 ” τ11 ` τ10 ` τ7 ` τ6 ´ τ5 ` τ4 ` τ3 pmod pq “ τ3pτ8 ` τ7 ` τ4 ` τ3 ´ τ2 ` τ ` 1q. This implies that 0 ” τ8 ` τ7 ` τ4 ` τ3 ´ τ2 ` τ ` 1 pmod pq. (1.2A) We can replace τ with τ´1 in the above equation, by replacing ta, bu with ta´1, b ´1 u if necessary. Then we have 0 ” τ´8 ` τ´7 ` τ´4 ` τ´3 ´ τ´2 ` τ´1 ` 1 pmod pq. Multiplying τ8, then we have 0 ” 1 ` τ ` τ4 ` τ5 ´ τ6 ` τ7 ` τ8 pmod pq “ τ8 ` τ7 ´ τ6 ` τ5 ` τ4 ` τ ` 1. Now by subtracting the above equation from 1.2A we have 0 ” τ6 ´ τ5 ` τ3 ´ τ2 pmod pq 16 Art Discrete Appl. Math. 5 (2022) #P1.10 “ τ2pτ ´ 1qpτ3 ` 1q. This implies that τ ” 1 pmod pq or τ3 ” ´1 pmod pq. If τ ” 1 pmod pq, then it contradicts the fact that gcdpτ ´ 1, 3pq “ 1. Now if τ3 ” ´1 pmod pq, then τ6 ” 1 pmod pq. We already know τ2q ” 1 pmod pq. Then τd ” 1 pmod pq, where d “ gcdp2q, 6q. Since gcdp2, 6q “ 2 and gcdpq, 6q “ 1, then d “ 2. This implies that τ2 ” 1 pmod pq, which means Cq centralizes Cp. Then we have G “ Cq ˆ pC2 ˙ C3pq – Cq ˆ D6p. So Lemma 2.26 applies. Case 2. Assume |a| “ q. Then |b| P t2, 2qu. Thus |b| P t2, 2q, 2p, 2pqu. If |b| “ 2pq, then Cq centralizes Cp. This implies that G “ Cq ˆ pC2 ˙ C3pq – Cq ˆ D6p so, Lemma 2.26 applies. Therefore, we may assume Cq does not centralize Cp, so |a| is not divisible by p. If |b| “ 2p, then Corollary 2.7 applies with s “ b and t “ b´1, because we have a Hamiltonian cycle in Cayp pG; pSq by Theorem 1.2(3). Since b is the only generator whose order is even, then any Hamiltonian cycle in Cayp pG; pSq must use some edge labeled pb. We may now assume |b| P t2, 2qu. We have C “ paq´1, b, a´pq´1q, b ´1 q as a Hamilto- nian cycle in CaypG;Sq. Now if |a| “ q, then by Lemma 2.14 we have G1 “ xraq´1, bsy. Therefore, Factor Group Lemma 2.6 applies. So, we may assume |a| “ 3q. Since Cq does not centralize Cp, then after conjugation we can assume a “ a3aq and b “ a2a j qγp, where 0 ď j ď q ´ 1. We already know that C is a Hamiltonian cycle in CaypG;Sq. So we can assume gcdp3q, q ´ 1q ‰ 1 for otherwise Lemma 2.14 applies, which implies that Factor Group Lemma 2.6 applies. This implies that gcdp3, q ´ 1q ‰ 1 which means q ” 1 pmod 3q. Consider pG “ G{Cp. Then pa “ a3aq and pb “ a2ajq . Therefore, there exists 0 ď k ď 3q ´ 1 such that pb´1papb “ pak. Since pb inverts a3 and centralizes aq , then we must have pa “ pbpakpb´1 “ a´k3 akq , so k ” ´1 pmod 3q and k ” 1 pmod qq. Since q ” 1 pmod 3q, then k “ q ` 1. Additionally, we have aγpa ´1 “ γpτp , where pτ q ” 1 pmod pq. We also have pτ ı 1 pmod pq, because Cq does not centralize Cp. Now we have b´1ab “ γ´1p a ´j q a2aa2a j qγp “ γ ´1 p a q`1γp. This implies that b´1aib “ pb´1abqi “ pγ´1p a q`1γpq i “ γ´1p a ipq`1qγp. Therefore, b´1aib “ γ´1p a ipq`1qγp ” γ ´1 p a iγp pmod C3q. We have C1 “ ppaq´3,pb´1,pa´pq´2q,pb,pa´1,pb´1,pa,pb,paq´2,pb´1, F. Maghsoudi: Cayley graphs of order 6pq and 7pq are Hamiltonian 17 pe paq´3 pb´1pa2q´3 paq´1 pb´1pa3q´2 pa2q´1 pb´1pa3q´3 paq pb´1pa2q´2 pa3q´1 pb´1paq´2pb´1 Figure 5: The Hamiltonian cycle C1: pa edges are solid and pb edges are dashed. pe paq´1 pb´1paq´1 pa2q`2 pb´1pa2q`1 pa2q´1 pb´1paq paq`2 pb´1pa2q´2 pa2q`3 pb´1pa2q´1 paq pb´1papb´1 Figure 6: The Hamiltonian cycle C2: pa edges are solid and pb edges are dashed. pa´pq´3q,pb,paq´2,pb´1,pa,pb,pa´1,pb´1,pa´pq´2q,pbq as our first Hamiltonian cycle in Cayp pG; pSq. The picture in Figure 5 on page 17 shows the Hamiltonian cycle. In addition, C2 “ ppaq´1,pb´1,pa´pq´3q,pb,pa´1,pb´1,paq´2,pb,pa,pb´1,pa2,pb, paq´4,pb´1,pa´pq´5q,pb,paq´4,pb´1,pa,pb,pa,pb´1,pa´1,pbq is the second Hamiltonian cycle in Cayp pG; pSq. The picture in Figure 6 on page 17 shows the Hamiltonian cycle. We calculate the voltage of C1 in ÐÑ G “ G{C3. Since a q ” e pmod C3q, we have VpC1q ” a ´3pb´1a2bqa´1pb´1abqa´2pb´1a3bqa´2pb´1abqa´1pb´1a2bq pmod C3q “ a´3pγ´1p a 2γpqa ´1pγ´1p aγpqa ´2pγ´1p a 3γpqa ´2pγ´1p aγpqa ´1pγ´1p a 2γpq “ a´3pγpτ 2´1 p a 2qa´1pγpτ´1p aqa ´2pγpτ 3´1 p a 3qa´2pγpτ´1p aqa ´1pγpτ 2´1 p a 2q “ a´3γpτ 2´1 p aγ pτ´1 p a ´1γpτ 3´1 p aγ pτ2`pτ´2 p a 2 “ γpτ ´3ppτ2´1q`pτ´2ppτ´1q`pτ´3ppτ3´1q`pτ´2ppτ2`pτ´2q p “ γ´2pτ ´3´3pτ´2`3pτ´1`2 p . We may assume this does not generate Cp, so 0 ” ´2pτ´3 ´ 3pτ´2 ` 3pτ´1 ` 2 pmod pq. Multiplying by pτ3, we have 0 ” 2pτ3 ` 3pτ2 ´ 3pτ ´ 2 “ ppτ ´ 1qppτ ` 2qp2pτ ` 1q pmod pq. 18 Art Discrete Appl. Math. 5 (2022) #P1.10 Since pτ ı 1 pmod pq, then we may assume pτ ” ´2 pmod pq, by replacing pa with pa´1 if needed. Now we calculate the voltage of C2 in ÐÑ G “ G{C3. VpC2q ” a ´1pb´1a3bqa´1pb´1a´2bqapb´1a2bq ¨ a´4pb´1a5bqa´4pb´1abqapb´1a´1bq pmod C3q “ a´1pγ´1p a 3γpqa ´1pγ´1p a ´2γpqapγ ´1 p a 2γpq ¨ a´4pγ´1p a 5γpqa ´4pγ´1p aγpqapγ ´1 p a ´1γpq “ a´1pγpτ 3´1 p a 3qa´1pγpτ ´2´1 p a ´2qapγpτ 2´1 p a 2q ¨ a´4pγpτ 5´1 p a 5qa´4pγpτ´1p aqapγ pτ´1´1 p a ´1q “ a´1γpτ 3´1 p a 2γpτ ´2´1 p a ´1γpτ 2´1 p a ´2γpτ 5´1 p aγ pτ´1 p a 2γpτ ´1´1 p a ´1 “ γpτ ´1ppτ3´1q`pτppτ´2´1q`pτ2´1`pτ´2ppτ5´1q`pτ´1ppτ´1q`pτppτ´1´1q p “ γpτ 3`2pτ2´2pτ`1´pτ´1´pτ´2 p . We may assume this does not generate Cp, so 0 ” pτ3 ` 2pτ2 ´ 2pτ ` 1 ´ pτ´1 ´ pτ´2 pmod pq. Multiplying by pτ2, we have 0 ” pτ5 ` 2pτ4 ´ 2pτ3 ` pτ2 ´ pτ ´ 1 pmod pq. We already know pτ ” ´2 pmod pq. By substituting this in the equation above, we have 0 ” p´2q5 ` 2p´2q4 ´ 2p´2q3 ` p´2q2 ´ p´2q ´ 1 “ 21 “ 3 ¨ 7 pmod pq. Since p ą 7, then 21 ı 0 pmod pq. This is a contradiction. Case 3. Assume |a| “ 2q and |b| “ 2. Since |a| “ 2q, then by Lemma 2.16 |a| “ 2q. We have b “ aqγ where G1 “ xγy. By Proposition 2.15(4) we have aγa´1 “ γτ , where τ2q ” 1 pmod 3pq and gcdpτ ´ 1, 3pq “ 1. This implies that τ ı 0, 1 pmod pq and τ ” ´1 pmod 3q. Suppose, for the moment, that τ ” ´1 pmod pq. Then G – D6p ˆ Cq , so CaypG;Sq has a Hamiltonian cycle by Lemma 2.26. We may now assume that τ ı ´1 pmod pq. Recall that pG “ G{Cp “ C2q ˙ C3. We may assume pa “ a2aq and pb “ a2a3. We have C1 “ pppa,pb,pa,pb,pa´1,pb,pa,pb,pa´1,pb,pa,pbqpq´5q{2,pa,pb,pa4, pb,pa´3,pb,pa´1,pb,pa2,pb,pa2,pb,pa´1,pb,pa´3,pb,pa4,pbq as the first Hamiltonian cycle in Cayp pG; pSq. The picture in Figure 7 on page 19 shows the Hamiltonian cycle. We also have C2 “ pppa,pb,pa´1,pb,pa,pbqq´5,pa3,pb,pa2,pb,pa´1,pb,pa´3,pb,pa3,pb,pa´3,pb,pa´1,pb,pa2,pb,pa3,pbq F. Maghsoudi: Cayley graphs of order 6pq and 7pq are Hamiltonian 19 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 11 11 12 12 13 13 14 14 pe aq a 2 q a q´5 q a2 a3 a2a ´1 3 a ´1 3 a2a3 a2a q´1 q a3a q´1 q a2a ´1 3 a q´1 q a ´1 3 a q´1 q a2a3a q´1 q a q´1 q Figure 7: The Hamiltonian cycle C1: pa edges are solid and pb edges are dashed. as the second Hamiltonian cycle in Cayp pG; pSq. The picture in Figure 8 on page 21 shows the Hamiltonian cycle. Now we calculate the voltage of C1. VpC1q “ ppababa ´1bqpaba´1babqqpq´5q{2paba4ba´3ba´1ba2ba2ba´1ba´3ba4bq “ ppaaqγaaqγa´1aqγqpaaqγa´1aqγaaqγqqpq´5q{2 ¨ paaqγa4aqγa´3aqγa´1aqγa2aqγa2aqγa´1aqγa´3aqγa4aqγq “ ppaq`1γaq`1γaq´1γqpaq`1γaq´1γaq`1γqqpq´5q{2 ¨ paq`1γaq`4γaq´3γaq´1γaq`2γaq`2γaq´1γaq´3γaq`4γq “ ppγτ q`1`τ2`τq`1aq`1qpγτ q`1`1`τq`1aq`1qqpq´5q{2 ¨ pγτ q`1`τ5`τq`2`τ`τq`3`τ5`τq`4`τ`τq`5aq`5q “ ppγ2τ q`1`τ2aq`1qpγ2τ q`1`1aq`1qqpq´5q{2 ¨ pγτ q`5`τq`4`τq`3`τq`2`τq`1`2τ5`2τaq`5q “ ppγ2τ q`1`τ2`τq`1p2τq`1`1qa2qqpq´5q{2 ¨ pγτ q`5`τq`4`τq`3`τq`2`τq`1`2τ5`2τaq`5q “ pγ3τ q`1`3τ2a2qpq´5q{2pγτ q`5`τq`4`τq`3`τq`2`τq`1`2τ5`2τaq`5q “ pγp3τ q`1`3τ2qpτq´5´1q{pτ2´1qaq´5qpγτ q`5`τq`4`τq`3`τq`2`τq`1`2τ5`2τaq`5q “ γp3τ q`1`3τ2qpτq´5´1q{pτ2´1q`τq´5pτq`5`τq`4`τq`3`τq`2`τq`1`2τ5`2τq. Since τ2q ” 1 pmod pq, we have τ q ” ˘1 pmod pq. Let us now consider the case where τ q ” 1 pmod pq, then by substituting this in the formula for the voltage of C1 we have VpC1q “ γ p3τ`3τ2qpτ´5´1q{pτ2´1q`τ´5pτ5`τ4`τ3`τ2`τ`2τ5`2τq “ γ3τp1`τqpτ ´5´1q{pτ`1qpτ´1q`p1`τ´1`τ´2`τ´3`τ´4`2`2τ´4q “ γ3τpτ ´5´1q{pτ´1q`p3`τ´1`τ´2`τ´3`3τ´4q 20 Art Discrete Appl. Math. 5 (2022) #P1.10 “ γp´2`2τ ´3q{pτ´1q. We may assume this does not generate Cp, then 0 ” ´2 ` 2τ´3 pmod pq. Multiplying by τ3, we have 0 ” ´2τ3 ` 2 pmod pq. This implies that τ3 ” 1 pmod pq, which contradicts the fact that τ q ” 1 pmod pq but τ ı 1 pmod pq. Now we may assume τ q ” ´1 pmod pq, then substituting this in the formula for the voltage of C1 we have VpC1q “ γ p´3τ`3τ2qp´τ´5´1q{pτ2´1q´τ´5p´τ5´τ4´τ3´τ2´τ`2τ5`2τq “ γ3τpτ´1qp´τ ´5´1q{pτ`1qpτ´1q`p1`τ´1`τ´2`τ´3`τ´4´2´2τ´4q “ γ3τp´τ ´5´1q{pτ`1q`p´1`τ´1`τ´2`τ´3´τ´4q “ γp´4τ`2τ ´1`2τ´2´4τ´4q{pτ`1q. We may assume this does not generate Cp, then 0 ” ´4τ ` 2τ´1 ` 2τ´2 ´ 4τ´4 pmod pq. Multiplying by p´τ4q{2, we have 0 ” 2τ5 ´ τ3 ´ τ2 ` 2 “ pτ ` 1qp2τ4 ´ 2τ3 ` τ2 ´ 2τ ` 2q pmod pq. Since we assumed τ ı ´1 pmod pq, then the above equation implies that 0 ” 2τ4 ´ 2τ3 ` τ2 ´ 2τ ` 2 pmod pq. (3A) Now we calculate the voltage of C2. VpC2q “ paba ´1babqpq´5qpa3ba2ba´1ba´3ba3ba´3ba´1ba2ba3bq “ paaqγa´1aqγaaqγqpq´5qpa3aqγa2aqγ ¨ a´1aqγa´3aqγa3aqγa´3aqγa´1aqγa2aqγa3aqγq “ paq`1γaq´1γaq`1γqpq´5qpaq`3γaq`2γaq´1 ¨ γaq´3γaq`3γaq´3γaq´1γaq`2γaq`3γq “ pγτ q`1`1`τq`1aq`1qpq´5qpγτ q`3`τ5`τq`4`τ`τq`4`τ`τq`τ2`τq`5aq`5q “ pγ2τ q`1`1aq`1qpq´5qpγτ q`5`2τq`4`τq`3`τq`τ5`τ2`2τaq`5q “ pγp2τ q`1`1qppτq`1qpq´5q´1q{pτq`1´1qapq`1qpq´5qq ¨ pγτ q`5`2τq`4`τq`3`τq`τ5`τ2`2τaq`5q F. Maghsoudi: Cayley graphs of order 6pq and 7pq are Hamiltonian 21 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10 1111 12 12 pe aq a 2 q a q´5 q a2 a3 a2a ´1 3 a ´1 3 a2a3 a2a q´1 q a3a q´1 q a2a ´1 3 a q´1 q a ´1 3 a q´1 q a2a3a q´1 q a q´1 q Figure 8: The Hamiltonian cycle C2: pa edges are solid and pb edges are dashed. “ γp2τ q`1`1qppτq`1qpq´5q´1q{pτq`1´1q`τ pq`1qpq´5qpτq`5`2τq`4`τq`3`τq`τ5`τ2`2τq. Since we are assuming τ q ” ´1 pmod pq, then by substituting this in the above formula we have VpC2q “ γ p´2τ`1qpp´τq´5´1q{p´τ´1q´τ´5p´τ5´2τ4´τ3´1`τ5`τ2`2τq “ γp2τ ´4`2τ´τ´5´1q{p´τ´1q`1`2τ´1`τ´2`τ´5´1´τ´3´2τ´4 “ γp2τ´3´3τ ´1`3τ´3`3τ´4´2τ´5q{p´τ´1q. We may assume this does not generate Cp, then 2τ ´ 3 ´ 3τ´1 ` 3τ´3 ` 3τ´4 ´ 2τ´5 ” 0 pmod pq. Multiplying by τ5, we have 0 ” 2τ6 ´ 3τ5 ´ 3τ4 ` 3τ2 ` 3τ ´ 2 “ pτ2 ´ 1qp2τ4 ´ 3τ3 ´ τ2 ´ 3τ ` 2q pmod pq. Since τ2 ı 1 pmod pq, then the above equation implies that 0 ” 2τ4 ´ 3τ3 ´ τ2 ´ 3τ ` 2 pmod pq. Therefore, by subtracting the above equation from 3A, we have 0 ” pτ3 ` 2τ2 ` τq “ τpτ ` 1q2 pmod pq. This is a contradiction. Case 4. Assume none of the previous cases apply. Since xa, by “ G, we may assume |a| is divisible by q, which means |a| is either q or 2q. Since Case 2 applies when |a| “ q, we must have |a| “ 2q. Then |b| “ q, since Cases 1 and 3 do not apply. So Case 2 applies after interchanging a and b. 22 Art Discrete Appl. Math. 5 (2022) #P1.10 3.3 Assume |S| “ 3, G1 “ Cp ˆ Cq and CG1 pC3q ‰ teu In this subsection, we prove the part of Theorem 1.3 where, |S| “ 3, G1 “ Cp ˆ Cq and CG1 pC3q ‰ teu. Recall G “ G{G 1, qG “ G{Cq and pG “ G{Cp. Proposition 3.4. Assume • G “ pC2 ˆ C3q ˙ pCp ˆ Cqq, • |S| “ 3, • CG1 pC3q ‰ teu. Then CaypG;Sq contains a Hamiltonian cycle. Proof. Let S “ ta, b, cu. If CG1 pC3q “ Cp ˆ Cq , then since G 1 X ZpGq “ teu (see Proposition 2.15(2)), we conclude that CG1 pC2q “ teu. So we have G “ C3 ˆ pC2 ˙ Cpqq – C3 ˆ D2pq. Therefore, Lemma 2.26 applies. Since CG1 pC3q ‰ teu, then we may assume CG1 pC3q “ Cq by interchanging q and p if necessary. Since G1 X ZpGq “ teu, then C2 inverts Cq . Since C3 centralizes Cq and ZpGq X G1 “ teu (by Proposition 2.15(2)), then C2 inverts Cq . Thus, pG “ pC2 ˆ C3q ˙ Cq – pC2 ˙ Cqq ˆ C3 “ D2q ˆ C3. Now if pS is minimal, then Lemma 2.24 applies. Therefore, we may assume pS is not minimal. Choose a 2-element subset ta, bu of S that generates pG. From the minimality of S, we see that xa, by “ D2q ˆ C3 after replacing a and b by conjugates. The projection of pa, bq to D2q must be of the form pa2, aqq or pa2, a2aqq, where a2 is reflection and aq is a rotation. Also note that pb ‰ aq because S X G1 “ H by Assumption 3.1(6). Therefore, pa, bq must have one of the following forms: 1. pa2, a3aqq, 2. pa2, a2a3aqq, 3. pa2a3, a2aqq, 4. pa2a3, a3aqq, 5. pa2a3, a2a3aqq. Let c be the third element of S. We may write c “ ai2a j 3a k qγp with 0 ď i ď 1, 0 ď j ď 2 and 0 ď k ď q ´ 1. Note since S X G1 “ H, we know that i and j cannot both be equal to 0. Additionally, we have a3γpa ´1 3 “ γ pτ p where pτ3 ” 1 pmod Cpq. Also, pτ ı 1 pmod pq since CG1 pC3q “ Cq . Therefore, we conclude that pτ2 ` pτ ` 1 ” 0 pmod pq. Note that this implies pτ ı ´1 pmod pq. Case 1. Assume a “ a2 and b “ a3aq . F. Maghsoudi: Cayley graphs of order 6pq and 7pq are Hamiltonian 23 Subcase 1.1. Assume i ‰ 0. Then, c “ a2a j 3a k qγp. Thus, by Lemma 2.27(1) xb, cy “ G which contradicts the minimality of S. Subcase 1.2. Assume i “ 0. Then j ‰ 0. We may assume j “ 1, by replacing c with c´1 if necessary. Thus c “ a3a k qγp. Consider G “ C2 ˆ C3. We have a “ a2, b “ a3 and c “ a3. Therefore, b “ c “ a3. We have pa, b 2 , a, b ´2 q as a Hamiltonian cycle in CaypG;Sq. Since we can replace each b by c, then we consider C1 “ pa, b 2 , a, b ´1 , c´1q and C2 “ pa, b 2 , a, c´2q as Hamiltonian cycles in CaypG;Sq. Now since there is one occurrence of c in C1, then by Lemma 2.8 the subgroup generated by VpC1q contains Cp. Also, VpC1q “ ab 2ab´1c´1 ” a2 ¨ a 2 3a 2 q ¨ a2 ¨ a ´1 q a ´1 3 ¨ a ´k q a ´1 3 pmod Cpq “ a´2q a3a ´1´k q a ´1 3 “ a´3´kq . We can assume this does not generate Cq , for otherwise Factor Group Lemma 2.6 applies. Therefore, ´3 ´ k ” 0 pmod qq. Thus, k ” ´3 pmod qq. Now we calculate the voltage of C2. VpC2q “ ab 2ac´2 ” a2 ¨ a 2 3 ¨ a2 ¨ γ ´1 p a ´1 3 γ ´1 p a ´1 3 pmod Cqq “ a23γ ´1 p a ´1 3 γ ´1 p a ´1 3 “ γ´pτ 2´pτ p . Since pτ2 ` pτ ` 1 ” 0 pmod pq, then ´pτ2 ´ pτ ” 1 pmod pq. Thus, γ´pτ2´pτp “ γp generates Cp. VpC2q “ ab 2ac´2 ” a2 ¨ a 2 3a 2 q ¨ a2 ¨ a ´k q a ´1 3 a ´k q a ´1 3 pmod Cpq “ a´2q a 2 3a ´k q a ´1 3 a ´k q a ´1 3 “ a´2pk`1qq . We know k ” ´3 pmod qq, therefore, ´2pk`1q ” 4 pmod qq, so Factor Group Lemma 2.6 applies. Case 2. Assume a “ a2 and b “ a2a3aq . Subcase 2.1. Assume i “ 0, then j ‰ 0. If k ‰ 0, then c “ aj3a k qγp. Thus, by Lemma 2.27(3) xb, cy “ G which contradicts the minimality of S. Therefore, we may assume k “ 0. We may also assume j “ 1, by replacing c with c´1 if necessary. Then c “ a3γp. 24 Art Discrete Appl. Math. 5 (2022) #P1.10 Consider G “ C2 ˆ C3, thus a “ a2, b “ a2a3 and c “ a3. Therefore, |a| “ 2, |b| “ 6 and |c| “ 3. Consider C “ pb 2 , c, b, c´1, aq as a Hamiltonian cycle in CaypG;Sq. Now we calculate its voltage. VpCq “ b2cbc´1a ” a2a3aqa2a3aq ¨ a3 ¨ a2a3aq ¨ a ´1 3 ¨ a2 pmod Cpq “ a´1q which generates Cq . By considering the fact that C2 might centralize Cp or not, we have VpCq “ b2cbc´1a ” a2a3a2a3 ¨ a3γp ¨ a2a3 ¨ γ ´1 p a ´1 3 ¨ a2 pmod Cqq “ γpa3γ ¯1 p a ´1 3 “ γ1¯pτp . which generates Cp. Therefore, the subgroup generated by VpCq is G 1. So, Factor Group Lemma 2.6 applies. Subcase 2.2. Assume j “ 0. Then i ‰ 0. If k ‰ 1, then c “ a2a k qγp. Thus, by Lemma 2.27(4) xb, cy “ G which contradicts the minimality of S. We may therefore assume k “ 1. Then c “ a2aqγp. Consider G “ C2 ˆ C3, then a = c “ a2 and b “ a2a3. Thus, |a| “ |c| “ 2 and |b| “ 6. We have C “ pb 2 , c, b ´2 , aq as a Hamiltonian cycle in CaypG;Sq. Since there is one occurrence of c in C, and it is the only generator of G that contains γp, then by Lemma 2.8 we conclude that the subgroup generated by VpCq contains Cp. Also, VpCq “ b2cb´2a ” a2a3aqa2a3aq ¨ a2aq ¨ a ´1 q a ´1 3 a2a ´1 q a ´1 3 a2 ¨ a2 pmod Cpq “ a´1q a3aqa3a ´1 q a ´1 3 aqa ´1 3 a ´1 q “ a´1q . which generates Cq . Therefore, the subgroup generated by VpCq is G 1. So, Factor Group Lemma 2.6 applies. Subcase 2.3. Assume i ‰ 0 and j ‰ 0. We may assume j “ 1, by replacing c with c´1 if necessary. So c “ a2a3a k qγp. If k ‰ 1, then by Lemma 2.27(4) xb, cy “ G which contradicts the minimality of S. We may now assume k “ 1. Then c “ a2a3aqγp. Consider G “ C2 ˆ C3. Then a “ a2 and b “ c “ a2a3. Therefore, |b| “ |c| “ 6 and |a| “ 2. We have C “ pc, a, pb, aq2q as a Hamiltonian cycle in CaypG;Sq. Since there is one occurrence of c in C, and it is the only generator of G that contains γp, then by Lemma 2.8 we conclude that the subgroup generated by VpCq is Cp. Also, VpCq “ capbaq2 ” a2a3aq ¨ a2 ¨ a2a3aq ¨ a2 ¨ a2a3aq ¨ a2 pmod Cpq “ a3a ´2 q a3a ´1 q a3 “ a´3q F. Maghsoudi: Cayley graphs of order 6pq and 7pq are Hamiltonian 25 which generates Cq . Therefore, the subgroup generated by VpCq is G 1. So, Factor Group Lemma 2.6 applies. Case 3. Assume a “ a2a3 and b “ a2aq . Since b “ a2aq is conjugate to a2 via an element of Cq (which centralizes C3), then ta, bu is conjugate to ta2a3a m q , a2u for some nonzero m. So Case 2 applies (after replacing aq with a m q ). Case 4. Assume a “ a2a3 and b “ a3aq . Subcase 4.1. Assume i ‰ 0. Then c “ a2a j 3a k qγp. Thus, by Lemma 2.27(1) xb, cy “ G which contradicts the minimality of S. Subcase 4.2. Assume i “ 0. Then j ‰ 0 and c “ aj3a k qγp. If k ‰ 0, then by Lemma 2.27(2) xa, cy “ G which contradicts the minimality of S. So we may assume k “ 0. We may also assume j “ 1, by replacing c with c´1 if necessary. Then c “ a3γp. Consider G “ C2 ˆ C3. Therefore, a “ a2a3 and b “ c “ a3. In addition, |a| “ 6 and |b| “ |c| “ 3. We have C “ pc, b, a, b ´2 , a´1q as a Hamiltonian cycle in CaypG;Sq. Since there is one occurrence of c in C, and it is the only generator of G that contains γp, then by Lemma 2.8 we conclude that the subgroup generated by VpCq contains Cp. Also, VpCq “ cbab´2a´1 ” a3 ¨ a3aq ¨ a2a3 ¨ a ´2 q a ´2 3 ¨ a ´1 3 a2 pmod Cpq “ a3aqa 2 3a 2 q “ a3q which generates Cq . Therefore, the subgroup generated by VpCq is G 1. Thus, Factor Group Lemma 2.6 applies. Case 5. Assume a “ a2a3, b “ a2a3aq . Subcase 5.1. Assume i “ 0. Then j ‰ 0 and c “ aj3a k qγp. If k ‰ 0, then by Lemma 2.27(3) xb, cy “ G which contradicts the minimality of S. So we may assume k “ 0. We may also assume j “ 1, by replacing c with c´1 if necessary. Then c “ a3γp. Consider G “ C2 ˆ C3. Therefore, a “ b “ a2a3 and c “ a3. Thus, |a| “ |b| “ 6 and |c| “ 3. We have C “ pa, c2, b ´1 , c´2q as a Hamiltonian cycle in CaypG;Sq. Now we calculate its voltage. VpCq “ ac2b´1c´2 ” a2a3 ¨ a 2 3 ¨ a ´1 q a ´1 3 a2 ¨ a ´2 3 pmod Cpq “ a´13 aqa ´2 3 “ aq which generates Cq . Also VpCq “ ac2b´1c´2 ” ac2a´1c´2 pmod Cqq pbecause a ” b pmod Cqqq “ ac´1a´1c pbecause |c| “ 3q “ ra, c´1s. 26 Art Discrete Appl. Math. 5 (2022) #P1.10 This generates Cp, because ta, cu generates G{Cq . Therefore, the subgroup generated by VpCq is G1. So, Factor Group Lemma 2.6 applies. Subcase 5.2. Assume i ‰ 0. Then c “ a2a j 3a k qγp. If k ‰ 1, then by Lemma 2.27(4) xb, cy “ G which contradicts the minimality of S. So we may assume k “ 1. Then c “ a2a j 3aqγp. We show that xa, cy “ G. Now, we have xa, cy “ xa2, a3, cy pbecause xay “ xa2a3y “ xa2, a3yq “ xa2, a3, a2a j 3aqγpy “ xa2, a3, aqγpy “ xa2, a3, aq, γpy “ G, which contradicts the minimality of S. 3.4 Assume |S| “ 3, G1 “ Cp ˆ Cq and pS is minimal In this subsection, we prove the part of Theorem 1.3 where, |S| “ 3, G1 “ Cp ˆ Cq and CG1 pC3q “ teu. Recall G “ G{G 1 and pG “ G{Cp. Proposition 3.5. Assume • G “ pC2 ˆ C3q ˙ pCp ˆ Cqq, • |S| “ 3, • pS is minimal. Then CaypG;Sq contains a Hamiltonian cycle. Proof. Let S “ ta, b, cu. If CG1 pC3q ‰ teu, then Proposition 3.4 applies. Hence we may assume CG1 pC3q “ teu. Then we have four different cases. Case 1. Assume CG1 pC2q “ Cp ˆ Cq , thus G “ C2 ˆ pC3 ˙ Cpqq. Since pS is minimal, then all three elements belonging to pS must have prime order. There is an element pa P pS, such that |pa| “ 2, otherwise all elements of S belong to a subgroup of index 2 of G, so xa, b, cy ‰ G which is a contradiction. If |a| “ 2p, then Corollary 2.7 applies with s “ a and t “ a´1, because there is a Hamiltonian cycle in Cayp pG; pSq (see Theorem 1.2(3)) which uses at least one labeled edge pa because pS is minimal. Now we may assume |a| “ 2. Replacing a by a conjugate we may assume xay “ C2. Thus, xb, cy “ C3 ˙ Cpq . By Theorem 1.2(3), there is a Hamiltonian path L in CaypC3 ˙ Cpq, tb, cuq. Therefore, LaL ´1a´1 is a Hamiltonian cycle in CaypG;Sq. Case 2. Assume CG1 pC2q “ Cq . Therefore, pG “ G{Cp “ C6 ˙ Cq – C2 ˆ pC3 ˙ Cqq. There is some a P S such that |pa| “ 2. Thus, we can assume |a| “ 2, for otherwise Corollary 2.7 applies with s “ a and t “ a´1. (Note since pS is minimal, then pa must F. Maghsoudi: Cayley graphs of order 6pq and 7pq are Hamiltonian 27 be used in any Hamiltonian cycle in Cayp pG; pSq.) We may assume a “ a2. Since pS is minimal, S XG1 “ H (see Assumption 3.1(6)) and each element belonging to pS has prime order, then |pb| “ |pc| “ 3. We may assume pa “ a2, pb “ a3 and pc “ a3aq . We have the following two Hamiltonian paths in CaypC3 ˙ Cq; tpb,pcuq: L1 “ pppc,pb2qq´1,pc,pbq and L2 “ pppb,pc,pbqq´1,pb,pcq. These lead to the following two Hamiltonian cycles in Cayp pG; pSq: C1 “ pL1,pa, L´11 ,paq and C2 “ pL2,pa, L´12 ,paq. Then if we let ź L1 “ pcb 2qq´1cb “ pcb2qqb´1 P a´13 Cp and ź L2 “ pbcbq q´1bc “ pbcbqqb´1 “ bpcb2qqb´2 “ bp ź L1qb ´1 then it is clear that V pCiq “ r ś Li, as for i “ 1, 2. Therefore, we may assume a cen- tralizes ś L1 and ś L2, for otherwise Factor Group Lemma 2.6 applies. Now, since a centralizes ś L1, and ś L1 P a ´1 3 Cp, we must have ś L1 “ a ´1 3 . So ś L2 “ ba ´1 3 b ´1. If b does not centralize a3, then VpC1q ‰ VpC2q, so the voltage of C1 or C2 cannot both be equal to identity. Therefore, Factor Group Lemma 2.6 applies. Now if b centralizes a3, then we can assume b “ a3. Therefore, c “ a3aqγp. We calculate the voltage of C1. We have VpC1q “ pcb 2qqb´1appcb2qqb´1q´1a “ pa3aqγp ¨ a 2 3q q ¨ a´13 ¨ a2 ¨ ppa3aqγp ¨ a 2 3q q ¨ a´13 q ´1 ¨ a2 “ pa3aqγpa ´1 3 q qa´13 a2ppa3aqγpa ´1 3 qa ´1 3 q ´1a2 “ a3a q qγ q pa ´1 3 a ´1 3 a2pa3a q qγ q pa ´1 3 a ´1 3 q ´1a2 “ a3γ q pa ´2 3 a2pa3γ q pa ´2 3 q ´1a2 “ a3γ q pa ´2 3 a2a 2 3γ ´q p a ´1 3 a2 “ a3γ 2q p a ´1 3 which generates Cp. Thus, Factor Group Lemma 2.6 applies. Case 3. Assume CG1 pC2q “ Cp. Therefore, qG “ G{Cq “ C6 ˙ Cp – C2 ˆ pC3 ˙ Cpq. 28 Art Discrete Appl. Math. 5 (2022) #P1.10 Now since S XG1 “ H (see Assumption 3.1(6)) and C3 does not centralize Cp, then for all a P S, we have |qa| P t2, 3, 6, 2pu. If |qa| “ 6, then |pa| is divisible by 6 which contradicts the minimality of pS. Note that every element belong to pS has prime order. If |qa| “ 2p, then |pa| “ 2 (because pS is minimal). Therefore, Corollary 2.7 applies with s “ a and t “ a´1. Note that since pS is minimal, then there is a Hamiltonian cycle in Cayp pG; pSq uses at least one labeled edge pa. Thus, |qa| P t2, 3u for all a P S. This implies that qS is minimal, because we need an a2 and an a3 to generate C2 ˆ C3 and two elements whose order divisible by 2 or 3 to generate Cp. So by interchanging p and q the proof in Case 2 applies. Case 4. Assume CG1 pC2q “ teu. Consider pG “ G{Cp “ pC2 ˆ C3q ˙ Cq. Now since pS is minimal, every element of pS has prime order. Since S X G1 “ H (see Assumption 3.1(6)), then for every ps P pS, we have |ps| P t2, 3u. Since CG1 pC2q “ teu and CG1 pC3q “ teu, this implies that for every s P S, we have |s| P t2, 3u. From our assumption we know that S “ ta, b, cu. Now we may assume |a| “ 2 and |b| “ 3. Also, we know that |c| P t2, 3u. If |c| “ 2, then c “ aγ, where γ P G1. Suppose, for the moment, xγy ‰ G1. Since xγy Ÿ G, then we have G “ xa, b, cy “ xa, b, γy “ xa, byxγy. Now since pS is minimal, xa, by does not contain Cq . So this implies that xγy contains Cq . Since xγy does not contain G1, then xγy “ Cq . Thus, we may assume that a “ a2 (by conjugation if necessary), b “ a3γp and c “ a2aq . So xb, cy “ xa3γp, a2aqy “ G (since a3γp and a2aq clearly generate G and do not commute modulo p or modulo q, they must generate G). This contradicts the minimality of S. Therefore, xγy “ G1. Consider G “ C2 ˆ C3. Then a “ c. We have |a| “ |c| “ 2 and |b| “ 3. We also have C1 “ pc ´1, b ´2 , a, b 2 q as a Hamiltonian cycle in CaypG;Sq. Now we calculate its voltage. VpC1q “ c ´1b´2ab2 “ γ´1a´1b´2ab2. Now, a´1b´2ab2 P G1. Since xa, by ‰ G, we have a´1b´2ab2 P te, γpu. If a ´1b´2ab2 “ e, then a and b2 commute, so a and b commute. Hence b “ a3, so xb, cy “ G, a contra- diction. So a´1b´2ab2 “ γp, and VpC1q “ γ ´1γp which generates G 1. Therefore, Factor Group Lemma 2.6 applies. Now we can assume |c| “ 3. Then c “ bγ, where γ P G1 (after replacing c with its inverse if necessary). Suppose, for the moment, xγy ‰ G1. Since xγy Ÿ G, then we have G “ xa, b, cy “ xa, b, γy “ xa, byxγy. Now since pS is minimal, then xa, by does not contain Cq . So this implies that xγy contains Cq . Since xγy does not contain G 1, then xγy “ Cq . Therefore, we may assume that a “ a2γp (by conjugation if necessary), b “ a3 and c “ a3aq . So xa, cy “ xa2γp, a3aqy “ G (since a2γp and a3aq clearly generate G and do not commute modulo p or modulo q, they must generate G). This contradicts the minimality of S. So xγy “ G1. F. Maghsoudi: Cayley graphs of order 6pq and 7pq are Hamiltonian 29 Consider G “ C2 ˆ C3. Then b “ c. We have |a| “ 2 and |b| “ |c| “ 3. We also have C2 “ pc ´1, b ´1 , a´1, b 2 , aq as a Hamiltonian cycle in CaypG;Sq. Now we calculate its voltage. VpC2q “ c ´1b´1a´1b2a “ γ´1b´1b´1a´1b2a. Now, b´2a´1b2a P G1. Since xa, by ‰ G, we have b´2a´1b2a P te, γpu. If b ´2a´1b2a “ e, then a and b2 commute, so a and b commute. Hence a “ a2, so xa, cy “ G, a contra- diction. So b´2a´1b2a “ γp, and VpC2q “ γ ´1γp which generates G 1. Therefore, Factor Group Lemma 2.6 applies. 3.5 Assume |S| “ 3, G1 “ Cp ˆ Cq and CG1 pC2q “ Cp ˆ Cq In this subsection, we prove the part of Theorem 1.3 where, |S| “ 3, G1 “ Cp ˆ Cq , CG1 pC2q “ Cp ˆCq , and neither CG1 pC3q ‰ teu nor pS is minimal holds. Recall G “ G{G1, qG “ G{Cq and pG “ G{Cp. Proposition 3.6. Assume • G “ pC2 ˆ C3q ˙ pCp ˆ Cqq, • |S| “ 3, • CG1 pC2q “ Cp ˆ Cq . Then CaypG;Sq contains a Hamiltonian cycle. Proof. Let S “ ta, b, cu. If CG1 pC3q ‰ teu, then Proposition 3.4 applies. So we may assume CG1 pC3q “ teu. Now if pS is minimal, then Proposition 3.5 applies. So we may assume pS is not minimal. Consider pG “ G{Cp “ pC2 ˆ C3q ˙ Cq – pC3 ˙ Cqq ˆ C2. Choose a 2-element ta, bu subset of S that generates pG. From the minimality of S, we see that xa, by “ pC3 ˙ Cqq ˆ C2, after replacing a and b by conjugates. The projection of pa, bq to C3 ˙ Cq must be of the form pa3, aqq or pa3, a3aqq (perhaps after replacing a and/or b with its inverse; also note that pb ‰ aq because S XG1 “ H). Therefore, pa, bq must have one of the following forms: 1. pa3, a2aqq, 2. pa3, a2a3aqq, 3. pa2a3, a3aqq, 4. pa2a3, a2aqq, 5. pa2a3, a2a3aqq. 30 Art Discrete Appl. Math. 5 (2022) #P1.10 Let c be the third element of S. We may write c “ ai2a j 3a k qγp with 0 ď i ď 1, 0 ď j ď 2 and 0 ď k ď q ´ 1. Note since S X G1 “ H, we know that i and j cannot both be equal to 0. Additionally, we have a3γpa ´1 3 “ γ pτ p where pτ3 ” 1 pmod pq and pτ ı 1 pmod pq. Thus pτ2 ` pτ ` 1 ” 0 pmod pq. Note that this implies pτ ı ´1 pmod pq. Also we have a3aqa ´1 3 “ a qτ q . By using the same argument we can conclude that qτ ı 1 pmod qq and qτ2 ` qτ ` 1 ” 0 pmod qq. Note that this implies qτ ı ´1 pmod qq. Combining these facts with pτ3 ” 1 pmod pq and qτ3 ” 1 pmod qq, we conclude that pτ2 ı ˘1 pmod pq, and qτ2 ı ˘1 pmod qq. Case 1. Assume a “ a3 and b “ a2aq . Subcase 1.1. Assume i “ 0. Then j ‰ 0 and c “ aj3a k qγp. For future reference in Subcase 4.1 of Proposition 3.7, we note that the argument here does not require our current assumption that C2 centralizes Cp. We may assume j “ 1, by replacing c with c ´1 if necessary. Then c “ a3a k qγp. Consider G “ C2 ˆ C3. Then we have a “ c “ a3, b “ a2. We have C1 “ pc, a, b, a ´2, bq and C2 “ pc 2, b, a´2, bq as Hamiltonian cycles in CaypG;Sq. Since there is one occurrence of c in C1, then by Lemma 2.8 we conclude that the subgroup generated by VpC1q contains Cp. Also, VpC1q “ caba ´2b ” a3a k q ¨ a3 ¨ a2aq ¨ a ´2 3 ¨ a2aq pmod Cpq “ akqτ`qτ 2`1 q “ aqτ 2`kqτ`1 q . We may assume this does not generate Cq , for otherwise Factor Group Lemma 2.6 applies. Therefore, 0 ” qτ2 ` kqτ ` 1 pmod qq. (1.1A) We also have 0 ” qτ2 ` qτ ` 1 pmod qq. (1.1B) By subtracting the above equation from 1.1A, we have 0 ” pk´1qqτ pmod qq. This implies that k “ 1. Now we calculate the voltage of C2. VpC2q “ c 2ba´2b ” a3γpa3γp ¨ a2 ¨ a ´2 3 ¨ a2 pmod Cqq “ γpτ`pτ 2 p which generates Cp. Also VpC2q “ c 2ba´2b ” a3aq ¨ a3aq ¨ a2aq ¨ a ´2 3 ¨ a2aq pmod Cpq “ aqτ`qτ 2`qτ2`1 q “ a2qτ 2`qτ`1 q . F. Maghsoudi: Cayley graphs of order 6pq and 7pq are Hamiltonian 31 We may assume this does not generate Cq , for otherwise Factor Group Lemma 2.6 applies. Then 0 ” 2qτ2 ` qτ ` 1 pmod qq. By subtracting 1.1B from the above equation we have 0 ” qτ2 pmod qq which is a contradiction. Subcase 1.2. Assume j “ 0. Then i ‰ 0 and c “ a2a k qγp. For future reference in Subcase 4.2 of Proposition 3.7, we note that the argument here does not require our current assumption that C2 centralizes Cp. If k ‰ 0, then by Lemma 2.28(3) xa, cy “ G which contradicts the minimality of S. So we can assume k “ 0. Then c “ a2γp. Consider G “ C2 ˆ C3. Then we have a “ a3 and b “ c “ a2. This implies that |a| “ 3 and |b| “ |c| “ 2. We have C “ pc´1, a2, b, a´2q as a Hamiltonian cycle in CaypG;Sq. Since there is one occurrence of c in C, and it is the only generator of G that contains γp, then by Lemma 2.8 we conclude that the subgroup generated by VpCq contains Cp. Similarly, since there is one occurrence of b in C, and it is the only generator of G that contains aq , then by Lemma 2.8 we conclude that the subgroup generated by VpCq contains Cq . Therefore, the subgroup generated by VpCq is G1. So, Factor Group Lemma 2.6 applies. Subcase 1.3. Assume i ‰ 0 and j ‰ 0. Then c “ a2a j 3a k qγp. If k ‰ 0, then by Lemma 2.28(3) xa, cy “ G which contradicts the minimality of S. So we can assume k “ 0. We may also assume j “ 1, by replacing c with c´1 if necessary. Then c “ a2a3γp. Consider G “ C2 ˆ C3. Then we have a “ a3, b “ a2 and c “ a2a3. This implies that |a| “ 3, |b| “ 2 and |c| “ 6. We have C “ pc, b, a, c, a ´1, cq as a Hamiltonian cycle in CaypG;Sq. Now we calculate its voltage. VpCq “ cbaca´1c ” a2a3 ¨ a2aq ¨ a3 ¨ a2a3 ¨ a ´1 3 ¨ a2a3 pmod Cpq “ a3aqa 2 3 “ aqτq which generates Cq . Also VpCq “ cbaca´1c ” a2a3γp ¨ a2 ¨ a3 ¨ a2a3γp ¨ a ´1 3 ¨ a2a3γp pmod Cqq “ a3γpa 2 3γ 2 p “ γpτ`2p . We may assume this does not generate Cp, for otherwise Factor Group Lemma 2.6 applies. Then pτ ” ´2 pmod pq. By substituting this in 0 ” pτ2 ` pτ ` 1 pmod pq, 32 Art Discrete Appl. Math. 5 (2022) #P1.10 we have 0 ” 4 ´ 2 ` 1 pmod pq “ 3. This contradicts the fact that p ą 3. Case 2. Assume a “ a3 and b “ a2a3aq . Subcase 2.1. Assume i ‰ 0 and j ‰ 0. Then c “ a2a j 3a k qγp. If k ‰ 0, then by Lemma 2.28(3) xa, cy “ G which contradicts the minimality of S. So we can assume k “ 0. Then c “ a2a j 3γp. Thus, by Lemma 2.28(4) xb, cy “ G which contradicts the minimality of S. Subcase 2.2. Assume i “ 0. Then j ‰ 0. We may assume j “ 1, by replacing c with c´1 if necessary. Then c “ a3a k qγp. Suppose, for the moment, that k ‰ 1. Then c “ a3a k qγp. We have xb, cy “ xa2a3, a3y “ G. Consider tpb,pcu “ ta2a3aq, a3akqu. Since C2 centralizes Cq , then ra2a3aq, a3a k q s “ ra3aq, a3a k q s “ a3aqa3a k qa ´1 q a ´1 3 a ´k q a ´1 3 “ a qτ`kqτ2´qτ2´kqτ q “ aqτpk´1qpqτ´1qq which generates Cq . Now consider tqb,qcu “ ta2a3, a3γpu. Since C2 centralizes Cp, then ra2a3, a3γps “ ra3, a3γps “ a3a3γpa ´1 3 γ ´1 p a ´1 3 “ γ pτ2´pτ p “ γ pτppτ´1q p which generates Cp. Therefore, xb, cy “ G which contradicts the minimality of S. Now we can assume k “ 1. Then c “ a3aqγp. Consider G “ C2 ˆ C3. We have a “ c “ a3 and b “ a2a3. This implies that |a| “ |c| “ 3 and |b| “ 6. We have C “ pc, b, a2, b, aq as a Hamiltonian cycle in CaypG;Sq. Since there is one occurrence of c in C, and it is the only generator of G that contains γp, then by Lemma 2.8 we conclude that the subgroup generated by VpCq is Cp. Also, VpCq “ cba2ba ” a3aq ¨ a2a3aq ¨ a 2 3 ¨ a2a3aq ¨ a3 pmod Cpq “ a3aqa3a 2 qa3 “ aqτ`2qτ 2 q “ aqτp1`2qτqq . We may assume this does not generate Cq , for otherwise Factor Group Lemma 2.6 applies. Therefore, 1 ` 2qτ ” 0 pmod qq. This implies that qτ ” ´1{2 pmod qq. By substituting qτ ” ´1{2 pmod qq in qτ2 ` qτ ` 1 ” 0 pmod qq, then we have 3{4 ” 0 pmod qq, which contradicts Assumption 3.1(1). F. Maghsoudi: Cayley graphs of order 6pq and 7pq are Hamiltonian 33 Subcase 2.3. Assume j “ 0. Then i ‰ 0 and c “ a2a k qγp. If k ‰ 0, then by Lemma 2.28(3) xa, cy “ G which contradicts the minimality of S. So we can assume k “ 0. Then c “ a2γp. Consider G “ C2 ˆ C3. Then we have a “ a3, b “ a2a3 and c “ a2. This implies that |a| “ 3, |b| “ 6 and |c| “ 2. We have C “ pc, a, b, a´1, b 2 q as a Hamiltonian cycle in CaypG;Sq. Since there is one occurrence of c in C, and it is the only generator of G that contains γp, then by Lemma 2.8 we conclude that the subgroup generated by VpCq contains Cp. Also, VpCq “ caba´1b2 ” a2 ¨ a3 ¨ a2a3aq ¨ a ´1 3 ¨ a2a3aqa2a3aq pmod Cpq “ a23a 2 qa3aq “ a2qτ 2`1 q . We may assume this does not generate Cq , for otherwise Factor Group Lemma 2.6 applies. Thus, qτ2 ” ´1{2 pmod qq. By substituting this in qτ2 ` qτ ` 1 ” 0 pmod qq, we have qτ ” ´1{2 pmod qq which contradicts qτ2 ” ´1{2 pmod qq. Case 3. Assume a “ a2a3 and b “ a3aq . Since b “ a3aq is conjugate to a3 via an element of Cq , then ta, bu is conjugate to ta2a3a m q , a3u for some nonzero m. So Case 2 applies (after replacing aq with a m q ). Case 4. Assume a “ a2a3 and b “ a2aq . Subcase 4.1. Assume i “ 0. Then j ‰ 0 and c “ aj3a k qγp. If k ‰ 0, then by Lemma 2.28(1) xa, cy “ G which contradicts the minimality of S. So we can assume k “ 0. We may also assume j “ 1, by replacing c with c´1 if necessary. Then c “ a3γp. Consider G “ C2 ˆ C3. Thus, a “ a2a3, b “ a2 and c “ a3. This implies that |a| “ 6, |b| “ 2 and |c| “ 3. We have C “ pa2, b, c, a, c´1q as a Hamiltonian cycle in CaypG;Sq. Since there is one occurrence of b in C, and it is the only generator of G that contains aq , then by Lemma 2.8 we conclude that the subgroup generated by VpCq contains Cq . Also, VpCq “ a2bcac´1 ” a23 ¨ a2 ¨ a3γp ¨ a2a3 ¨ γ ´1 p a ´1 3 pmod Cqq “ γpa3γ ´1 p a ´1 3 “ γ1´pτp which generates Cp. Therefore, the subgroup generated by VpCq is G 1. So, Factor Group Lemma 2.6 applies. Subcase 4.2. Assume j “ 0. Then i ‰ 0 and c “ a2a k qγp. If k ‰ 0, then by Lemma 2.28(1) xa, cy “ G which contradicts the minimality of S. So we can assume k “ 0. Then c “ a2γp. Consider G “ C2 ˆ C3, then a “ a2a3 and b “ c “ a2. This implies that |a| “ 6 and |b| “ |c| “ 2. We have C “ ppa, bq 2, a, cq as a Hamiltonian cycle in CaypG;Sq. Since there is one occurrence of c in C, and it is the 34 Art Discrete Appl. Math. 5 (2022) #P1.10 only generator of G that contains γp, then by Lemma 2.8 we conclude that the subgroup generated by VpCq contains Cp. Also, VpCq “ pabq2ac ” pa2a3 ¨ a2aqq 2 ¨ a2a3 ¨ a2 pmod Cpq “ a3aqa3aqa3 “ aqτ`qτ 2 q . which generates Cq . Therefore, the subgroup generated by VpCq is G 1. Thus, Factor Group Lemma 2.6 applies. Subcase 4.3. Assume i ‰ 0 and j ‰ 0. Then c “ a2a j 3a k qγp. If k ‰ 0, then by Lemma 2.28(1) xa, cy “ G which contradicts the minimality of S. So we can assume k “ 0. We may also assume j “ 1, by replacing c with c´1 if necessary. Then c “ a2a3γp. Consider G “ C2 ˆ C3. Thus, a “ c “ a2a3 and b “ a2. This implies that |a| “ |c| “ 6 and |b| “ 2. We have C “ pa, c, b, a´2, bq as a Hamiltonian cycle in CaypG;Sq. Since there is one occurrence of c in C, and it is the only generator of G that contains γp, then by Lemma 2.8 we conclude that the subgroup generated by VpCq contains Cp. Also, VpCq “ acba´2b ” a2a3 ¨ a2a3 ¨ a2aq ¨ a ´2 3 ¨ a2aq pmod Cpq “ a23aqa ´2 3 aq “ aqτ 2`1 q which generates Cq , because qτ2 ı ´1 pmod qq. Therefore, the subgroup generated by VpCq is G1. So, Factor Group Lemma 2.6 applies. Case 5. Assume a “ a2a3 and b “ a2a3aq . If k ‰ 0, then by Lemma 2.28(1) xa, cy “ G which contradicts the minimality of S. So we can assume k “ 0. Also, if j ‰ 0, then by Lemma 2.28(4) xb, cy “ G which contradicts the minimality of S. So we may also assume j “ 0. Then i ‰ 0. Therefore, c “ a2γp. So Case 4 applies, after interchanging b and c, and interchanging p and q. 3.6 Assume |S| “ 3, G1 “ Cp ˆ Cq and CG1 pC2q ‰ teu In this subsection, we prove the part of Theorem 1.3 where, |S| “ 3, G1 “ Cp ˆ Cq , CG1 pC2q ‰ teu, and neither CG1 pC2q “ CpˆCq nor CG1 pC3q ‰ teu nor pS is minimal holds. Recall G “ G{G1, qG “ G{Cq and pG “ G{Cp. Proposition 3.7. Assume • G “ pC2 ˆ C3q ˙ pCp ˆ Cqq, • |S| “ 3, • CG1 pC2q ‰ teu. F. Maghsoudi: Cayley graphs of order 6pq and 7pq are Hamiltonian 35 Then CaypG;Sq contains a Hamiltonian cycle. Proof. Let S “ ta, b, cu. If CG1 pC3q ‰ teu, then Proposition 3.4 applies. Therefore, we may assume CG1 pC3q “ teu. Now if CG1 pC2q “ Cp ˆ Cq , then Proposition 3.6 applies. Since CG1 pC2q ‰ teu, then we may assume CG1 pC2q “ Cq , by interchanging q and p if necessary. This implies that C2 inverts Cp. Now if pS is minimal, then Proposition 3.5 applies. So we may assume pS is not minimal. Consider pG “ G{Cp “ pC2 ˆ C3q ˙ Cq. Choose a 2-element subset ta, bu in S that generates pG. From the minimality of S, we see that xa, by “ pC2 ˆ C3q ˙ Cq after replacing a and b by conjugates. We may assume |a| ě |b| and (by conjugating if necessary) a is an element of C2 ˆ C3. Then the projection of pa, bq to C2 ˆ C3 has one of the following forms after replacing a and b with their inverses if necessary. • pa2a3, a2a3q, • pa2a3, a2q, • pa2a3, a3q, • pa3, a2q. So there are four possibilities for pa, bq: 1. pa2a3, a2a3aqq, 2. pa2a3, a2aqq, 3. pa2a3, a3aqq, 4. pa3, a2aqq. Let c be the third element of S. We may write c “ ai2a j 3a k qγp with 0 ď i ď 1, 0 ď j ď 2 and 0 ď k ď q ´ 1. Note since S X G1 “ H, we know that i and j cannot both be equal to 0. Additionally, we have a3γpa ´1 3 “ γ pτ p where pτ3 ” 1 pmod pq and pτ ı 1 pmod pq. Thus pτ2 ` pτ ` 1 ” 0 pmod pq. Note that this implies pτ ı ´1 pmod pq. Also we have a3aqa ´1 3 “ a qτ q . By using the same argument we can conclude that qτ ı 1 pmod qq and qτ2 ` qτ `1 ” 0 pmod qq. Note that this implies qτ ı ´1 pmod qq. Therefore, we conclude that pτ2 ı ˘1 pmod pq, and qτ2 ı ˘1 pmod qq. Case 1. Assume a “ a2a3 and b “ a2a3aq . If k ‰ 0, then by Lemma 2.28(1), xa, cy “ G which contradicts the minimality of S. So we can assume k “ 0. Now if j ‰ 0, then by Lemma 2.28(4), xb, cy “ G which contradicts the minimality of S. Therefore, we may assume j “ 0. Then i ‰ 0 and c “ a2γp. Consider G “ C2 ˆ C3. Thus a “ b “ a2a3 and c “ a2. Therefore, |a| “ |b| “ 6 and |c| “ 2. We have C “ pa, b, c, a ´2, cq as a Hamiltonian cycle in CaypG;Sq. Since there is one occurrence of b in C, and it is the 36 Art Discrete Appl. Math. 5 (2022) #P1.10 only generator of G that contains aq , then by Lemma 2.8 we conclude that the subgroup generated by VpCq contains Cq . Also, VpCq “ abca´2c ” a2a3 ¨ a2a3 ¨ a2γp ¨ a ´2 3 ¨ a2γp pmod Cqq “ a23γ ´1 p a ´2 3 γp “ γ´pτ 2`1 p which generates Cp. Therefore, the subgroup generated by VpCq is G 1. So, Factor Group Lemma 2.6 applies. Case 2. Assume a “ a2a3 and b “ a2aq . Subcase 2.1. Assume i “ 0. Then j ‰ 0 and c “ aj3a k qγp. If k ‰ 0, then by Lemma 2.28(1), xa, cy “ G which contradicts the minimality of S. So we can assume k “ 0. We may also assume j “ 1, by replacing c with c´1 if necessary. Then c “ a3γp. Consider G “ C2 ˆ C3. Thus, a “ a2a3, b “ a2 and c “ a3. Therefore, |a| “ 6, |b| “ 2 and |c| “ 3. We have C “ pa2, b, c, a, c´1q as a Hamiltonian cycle in CaypG;Sq. Since there is one occurrence of b in C, and it is the only generator of G that contains aq , then by Lemma 2.8 we conclude that the subgroup generated by VpCq contains Cq . Also, VpCq “ a2bc´1ac ” a23 ¨ a2 ¨ a3γp ¨ a2a3 ¨ γ ´1 p a ´1 3 pmod Cqq “ γ´1p a3γ ´1 p a ´1 3 “ γ´1´pτp which generates Cp. Therefore, the subgroup generated by VpCq is G 1. So, Factor Group Lemma 2.6 applies. Subcase 2.2. Assume j “ 0. Then i ‰ 0 and c “ a2a k qγp. If k ‰ 0, then by Lemma 2.28(1), xa, cy “ G which contradicts the minimality of S. So we can assume k “ 0. Then c “ a2γp. Consider G “ C2 ˆ C3, then a “ a2a3 and b “ c “ a2. We have C “ ppa, bq 2, a, cq as a Hamiltonian cycle in CaypG;Sq. Since there is one occurrence of c in C, and it is the only generator of G that contains γp, then by Lemma 2.8 we conclude that the subgroup generated by VpCq contains Cp. Now we calculate its voltage. Also, VpCq “ pabq2ac ” pa2a3 ¨ a2aqq 2 ¨ a2a3 ¨ a2 pmod Cpq “ a3aqa3aqa3 “ aqτ`qτ 2 q . which generates Cq . Therefore, the subgroup generated by VpCq generates G 1. So, Factor Group Lemma 2.6 applies. F. Maghsoudi: Cayley graphs of order 6pq and 7pq are Hamiltonian 37 Subcase 2.3. Assume i ‰ 0 and j ‰ 0. If k ‰ 0, then c “ a2a j 3a k qγp. Thus, by Lemma 2.28(1), xa, cy “ G which contradicts the minimality of S. So we can assume k “ 0. We may also assume j “ 1, by replacing c with c´1 if necessary. Then c “ a2a3γp. Consider G “ C2 ˆ C3. Thus, a “ c “ a2a3 and b “ a2. Therefore, |a| “ |c| “ 6 and |b| “ 2. We have C “ pa, c, b, a´2, bq as a Hamiltonian cycle in CaypG;Sq. Since there is one occurrence of c in C, and it is the only generator of G that contains γp, then by Lemma 2.8 we conclude that the subgroup generated by VpCq contains Cp. Also, VpCq “ acba´2b ” a2a3 ¨ a2a3 ¨ a2aq ¨ a ´2 3 ¨ a2aq pmod Cpq “ a23aqa ´2 3 aq “ aqτ 2`1 q . Since qτ2 ı ´1 pmod qq, Factor Group Lemma 2.6 applies. Case 3. Assume a “ a2a3 and b “ a3aq . Subcase 3.1. Assume i ‰ 0 and j ‰ 0. If k “ 0, then c “ a2a j 3γp. Thus, by Lemma 2.28(2), xb, cy “ G which contradicts the minimality of S. So we can assume k ‰ 0. Then c “ a2a j 3a k qγp. Thus, by Lemma 2.28(1), xa, cy “ G which contradicts the minimality of S. Subcase 3.2. Assume i “ 0. Then j ‰ 0 and c “ aj3a k qγp. If k ‰ 0, then by Lemma 2.28(1), xa, cy “ G which contradicts the minimality of S. So we can assume k “ 0. We may also assume j “ 1, by replacing c with c´1 if necessary. Then c “ a3γp. Consider G “ C2 ˆ C3, then a “ a2a3, b “ c “ a3. Therefore, |a| “ 6 and |b| “ |c| “ 3. We have C “ pc, b, a, b ´2 , a´1q as a Hamiltonian cycle in CaypG;Sq. Since there is one occurrence of c in C, and it is the only generator of G that contains γp, then by Lemma 2.8 we conclude that the subgroup generated by VpCq contains Cp. Also, VpCq “ cbab´2a´1 ” a3 ¨ a3aq ¨ a2a3 ¨ a ´1 q a ´1 3 a ´1 q a ´1 3 ¨ a ´1 3 a2 pmod Cpq “ a23aqa3a ´1 q a ´1 3 a ´1 q a ´2 3 “ aqτ 2´1´qτ´1 q “ aqτ 2´1´qτ2 q “ a´1q which generates Cq . Therefore, the subgroup generated by VpCq is G 1. So, Factor Group Lemma 2.6 applies. Subcase 3.3. Assume j “ 0. Then i ‰ 0 and c “ a2a k qγp. If k ‰ 0, then by Lemma 2.28(1), xa, cy “ G which contradicts the minimality of S. So we can assume k “ 0. Then c “ a2γp. Consider G “ C2 ˆ C3, then a “ a2a3, b “ a3 and c “ a2. Therefore, |a| “ 6, |b| “ 3 and |c| “ 2. We have C “ pa, c, b, a, b ´1 , aq 38 Art Discrete Appl. Math. 5 (2022) #P1.10 as a Hamiltonian cycle in CaypG;Sq. Since there is one occurrence of c in C, and it is the only generator of G that contains γp, then by Lemma 2.8 we conclude that the subgroup generated by VpCq contains Cp. Also, VpCq “ acbab´1a ” a2a3 ¨ a2 ¨ a3aq ¨ a2a3 ¨ a ´1 q a ´1 3 ¨ a2a3 pmod Cpq “ a23aqa3a ´1 q “ aqτ 2´1 q . Since qτ2 ı 1 pmod qq, Factor Group Lemma 2.6 applies. Case 4. Assume a “ a3 and b “ a2aq . Subcase 4.1. Assume i “ 0. Then j ‰ 0 and c “ aj3a k qγp. Thus, the argument in Subcase 1.1 of Proposition 3.6 applies. Subcase 4.2. Assume j “ 0. Then i ‰ 0 and c “ a2a k qγp. Thus, the argument in Subcase 1.2 of Proposition 3.6 applies. Subcase 4.3. Assume i ‰ 0 and j ‰ 0. Then c “ a2a j 3a k qγp. If k ‰ 0, then by Lemma 2.28(3) xa, cy “ G which contradicts the minimality of S. So we can assume k “ 0. We may also assume j “ 1, by replacing c with c´1 if necessary. Then c “ a2a3γp. Consider G “ C2 ˆ C3. Then we have a “ a3, b “ a2 and c “ a2a3. This implies that |a| “ 3, |b| “ 2 and |c| “ 6. We have C “ pc, b, a, c, a ´1, cq as a Hamiltonian cycle in CaypG;Sq. Since there is one occurrence of b in C, and it is the only generator of G that contains aq , then by Lemma 2.8 we conclude that the subgroup generated by VpCq contains Cq . Also, since a2 inverts Cp VpCq “ cbaca´1c ” a2a3γp ¨ a2 ¨ a3 ¨ a2a3γp ¨ a ´1 3 ¨ a2a3γp pmod Cqq “ a3γ ´1 p a 2 3 “ γ´pτp which generates Cp. Therefore, the subgroup generated by VpCq is G 1. So, Factor Group Lemma 2.6 applies. 3.7 Assume |S| “ 3, G1 “ Cp ˆ Cq and CG1 pC2q “ teu In this subsection, we prove the part of Theorem 1.3 where, |S| “ 3, G1 “ Cp ˆ Cq , CG1 pC2q “ teu, and neither CG1 pC3q ‰ teu nor pS is minimal holds. Recall G “ G{G1, qG “ G{Cq and pG “ G{Cp. Proposition 3.8. Assume • G “ pC2 ˆ C3q ˙ pCp ˆ Cqq, • |S| “ 3, • CG1 pC2q “ teu. F. Maghsoudi: Cayley graphs of order 6pq and 7pq are Hamiltonian 39 Then CaypG;Sq contains a Hamiltonian cycle. Proof. Let S “ ta, b, cu. If CG1 pC3q ‰ teu, then Proposition 3.4 applies. So we may assume CG1 pC3q “ teu. Now if pS is minimal, then Proposition 3.5 applies. So we may assume pS is not minimal. Consider pG “ G{Cp “ pC2 ˆ C3q ˙ Cq. Choose a 2-element subset ta, bu in S that generates pG. From the minimality of S, we see xa, by “ pC2 ˆ C3q ˙ Cq. after replacing a and b by conjugates. We may assume |a| ě |b| and (by conjugating if necessary) a is in C2 ˆ C3. Then the projection of pa, bq to C2 ˆ C3 is one of the following forms after replacing a and b with their inverses if necessary. • pa2a3, a2a3q, • pa2a3, a2q, • pa2a3, a3q, • pa3, a2q. There are four possibilities for pa, bq: 1. pa2a3, a2a3aqq, 2. pa2a3, a2aqq, 3. pa2a3, a3aqq, 4. pa3, a2aqq. Let c be the third element of S. We may write c “ ai2a j 3a k qγp with 0 ď i ď 1, 0 ď j ď 2 and 0 ď k ď q ´ 1. Note since S X G1 “ H, we know that i and j cannot both be equal to 0. Additionally, we have a3γpa ´1 3 “ γ pτ p where pτ3 ” 1 pmod pq and pτ ı 1 pmod pq. Thus pτ2 ` pτ ` 1 ” 0 pmod pq. Note that this implies pτ ı ´1 pmod pq. We have a3aqa ´1 3 “ a qτ q . By using the same argument we can conclude that qτ ı 1 pmod qq and qτ2 ` qτ ` 1 ” 0 pmod qq. Note that this implies qτ ı ´1 pmod qq. Therefore, we conclude that pτ2 ı ˘1 pmod pq, and qτ2 ı ˘1 pmod qq. Case 1. Assume a “ a2a3 and b “ a2a3aq . If k ‰ 0, then by Lemma 2.28(1) xa, cy “ G which contradicts the minimality of S. So we can assume k “ 0. Now if j ‰ 0, then by Lemma 2.28(4), xb, cy “ G which contradicts the minimality of S. Therefore, we may assume j “ 0. Then i ‰ 0 and c “ a2γp. We have xb, cy “ xa2a3, a2y “ G. Consider tqb,qcu “ ta2a3, a2γpu. Therefore, ra2a3, a2γps “ a2a3a2γpa ´1 3 a2γ ´1 p a2 “ a3γpa ´1 3 γp “ γ pτ`1 p . which generates Cp. Now consider tpb,pcu “ ta2a3aq, a2u, then ra2a3aq, a2s “ a2a3aqa2a ´1 q a ´1 3 a2a2 “ a3a ´2 q a ´1 3 “ a ´2qτ q which generates Cq . Therefore, xb, cy “ G which contradicts the minimality of S. 40 Art Discrete Appl. Math. 5 (2022) #P1.10 Case 2. Assume a “ a2a3 and b “ a2aq . If k ‰ 0, then by Lemma 2.28(1), xa, cy “ G which contradicts the minimality of S. So we can assume k “ 0. Subcase 2.1. Assume j ‰ 0. We may also assume j “ 1, by replacing c with c´1 if necessary. Then c “ ai2a3γp. We have xb, cy “ xa2, a i 2a3y “ G. Consider t pb,pcu “ ta2aq, a i 2a3u. We have ra2aq, a i 2a3s “ a2aqa i 2a3a ´1 q a2a ´1 3 a i 2 “ a ´1 q a i`1 2 a3a ´1 q a ´1 3 a i`1 2 “ a´1q a3a ¯1 q a ´1 3 “ a ´1¯qτ q which generates Cq . Now consider tqb,qcu “ ta2, ai2a3γpu. We have ra2, a i 2a3γps “ a2a i 2a3γpa2γ ´1 p a ´1 3 a i 2 “ a i`1 2 a3γ 2 pa ´1 3 a i`1 2 “ γ ˘2pτ p which generates Cp. Therefore, xb, cy “ G which contradicts the minimality of S. Subcase 2.2. Assume j “ 0. Then i ‰ 0 and c “ a2γp. Consider G “ C2 ˆ C3, then a “ a2a3 and b “ c “ a2. Thus, |a| “ 6 and |b| “ |c| “ 2. We have C “ ppa, bq 2, a, cq as a Hamiltonian cycle in CaypG;Sq. Since there is one occurrence of c in C, and it is the only generator of G that contains γp, then by Lemma 2.8 we conclude that the subgroup generated by VpCq contains Cp. Also, VpCq “ pabq2pacq ” a2a3 ¨ a2aq ¨ a2a3 ¨ a2aq ¨ a2a3 ¨ a2 pmod Cpq “ a3aqa3aqa3 “ aqτ`qτ 2 q which generates Cq . Therefore, the subgroup generated by VpCq is G 1. So, Factor Group Lemma 2.6 applies. Case 3. Assume a “ a2a3 and b “ a3aq . If k ‰ 0, then by Lemma 2.28(1), xa, cy “ G which contradicts the minimality of S. So we can assume k “ 0. Subcase 3.1. Assume i ‰ 0 and j ‰ 0. Then c “ a2a j 3γp. Thus, by Lemma 2.28(2), xb, cy “ G which contradicts the minimality of S. Subcase 3.2. Assume j “ 0. Then i ‰ 0 and c “ a2γp. We have xb, cy “ xa3, a2y “ G. Consider tqb,qcu “ ta3, a2γpu. Then we have ra3, a2γps “ a3a2γpa ´1 3 γ ´1 p a2 “ a3γ ´1 p a ´1 3 γp “ γ ´pτ`1 p which generates Cp. Now consider tpb,pcu “ ta3aq, a2u. Thus, ra3aq, a2s “ a3aqa2a ´1 q a ´1 3 a2 “ a3a 2 qa ´1 3 “ a 2qτ q which generates Cq . Therefore, xb, cy “ G which contradicts the minimality of S. F. Maghsoudi: Cayley graphs of order 6pq and 7pq are Hamiltonian 41 Subcase 3.3. Assume i “ 0. Then j ‰ 0. We may also assume j “ 1, by replacing c with c´1 if necessary. Then c “ a3γp. Consider G “ C2 ˆ C3, then we have a “ a2a3, b “ c “ a3. Thus, |a| “ 6 and |b| “ |c| “ 3. We have C “ pc, b, a, b ´2 , a´1q as a Hamiltonian cycle in CaypG;Sq. Since there is one occurrence of c in C, and it is the only generator of G that contains γp, then by Lemma 2.8 we conclude that the subgroup generated by VpCq contains Cp. Also, VpCq “ cbab´2a´1 ” a3 ¨ a3aq ¨ a2a3 ¨ a ´1 q a ´1 3 a ´1 q a ´1 3 ¨ a ´1 3 a2 pmod Cpq “ a23aqa3aqa ´1 3 aqa ´2 3 “ aqτ 2`1`qτ´1 q “ aqτ 2`1´qτ2 q “ aq which generates Cq . Therefore, the subgroup generated by VpCq is G 1. So, Factor Group Lemma 2.6 applies. Case 4. Assume a “ a3 and b “ a2aq . Subcase 4.1. Assume i “ 0. Then j ‰ 0. We may also assume j “ 1, by replacing c with c´1 if necessary. Then c “ a3a k qγp. Consider G “ C2 ˆ C3. Then we have a “ c “ a3 and b “ a2. This implies that |a| “ |c| “ 3 and |b| “ 2. We have C “ pc ´2, b, a2, bq as a Hamiltonian cycle in CaypG;Sq. Now we calculate its voltage. VpCq “ c´2ba2b ” γ´1p a ´1 3 γ ´1 p a ´1 3 ¨ a2 ¨ a 2 3 ¨ a2 pmod Cqq “ γ´1p a ´1 3 γ ´1 p a3 “ γ´1´pτ ´1 p which generates Cp. Also VpCq “ c´2ba2b ” a´kq a ´1 3 a ´k q a ´1 3 ¨ a2aq ¨ a 2 3 ¨ a2aq pmod Cpq “ a´kq a ´1 3 a ´k q a ´1 3 a ´1 q a 2 3aq “ a´k´kqτ ´1´qτ´2`1 q . If k “ 2, then a´k´kqτ ´1´qτ´2`1 q “ a ´2´2qτ´1´qτ´2`1 q “ a ´pqτ´1`1q2 q which generates Cq . So we may assume k ‰ 2 and the subgroup generated by VpCq does not contain Cq , for otherwise Factor Group Lemma 2.6 applies. Therefore, 0 ” ´k ´ kqτ´1 ´ qτ´2 ` 1 pmod qq “ p1 ´ kq ´ kqτ´1 ´ qτ´2. 42 Art Discrete Appl. Math. 5 (2022) #P1.10 Multiplying by qτ2, we have 0 ” p1 ´ kqqτ2 ´ kqτ ´ 1 pmod qq. (4.1A) We can replace qτ with qτ´1 in the above equation, by replacing a3,a and c with their inverses. 0 ” p1 ´ kqqτ´2 ´ kqτ´1 ´ 1 pmod qq. Multiplying by qτ2, then 0 ” p1 ´ kq ´ kqτ ´ qτ2 pmod qq. By subtracting 4.1A from the above equation, we have 0 ” pk ´ 2qqτ2 ` p2 ´ kq pmod qq. This implies that qτ2 ” 1 pmod qq, a contradiction. Subcase 4.2. Assume j “ 0. Then i ‰ 0. If k ‰ 0, then c “ a2a k qγp. Thus, by Lemma 2.28(3), xa, cy “ G which contradicts the minimality of S. So we can assume k “ 0. Then c “ a2γp. Consider G “ C2 ˆ C3, then a “ a3 and b “ c “ a2. We have C “ pa2, b, a´2, cq as a Hamiltonian cycle in CaypG;Sq. Since there is one occurrence of c in C, and it is the only generator of G that contains γp, then by Lemma 2.8 we conclude that the subgroup generated by VpCq contains Cp. Similarly, since there is one occurrence of b in C, and it is the only generator of G that contains aq , then by Lemma 2.8 we conclude that the subgroup generated by VpCq contains Cq . Therefore, the subgroup generated by VpCq is G1. So, Factor Group Lemma 2.6 applies. Subcase 4.3. Assume i ‰ 0 and j ‰ 0. If k ‰ 0, then c “ a2a j 3a k qγp. Thus, by Lemma 2.28(3), xa, cy “ G which contradicts the minimality of S. So we can assume k “ 0. We may also assume j “ 1, by replacing c with c´1 if necessary. Then c “ a2a3γp. We have xb, cy “ xa2, a2a3y “ G. Consider tpb,pcu “ ta2aq, a2a3u. Then we have ra2aq, a2a3s “ a2aqa2a3a ´1 q a2a ´1 3 a2 “ a ´1 q a3a ´1 q a ´1 3 “ a ´1´qτ q which generates Cq . Now consider tqb,qcu “ ta2, a2a3γpu. Then ra2, a2a3γps “ a2a2a3γpa2γ ´1 p a ´1 3 a2 “ a3γ 2 pa ´1 3 “ γ 2pτ p which generates Cp. Therefore, xb, cy “ G which contradicts the minimality of S. 3.8 Assume |S| “ 3 and G1 “ C3 ˆ Cp In this subsection, we prove the part of Theorem 1.3 where, |S| “ 3 and G1 “ C3 ˆ Cp. Recall G “ G{G1, pG “ G{Cp and ÐÑG “ G{C3. Proposition 3.9. Assume • G “ pC2 ˆ Cqq ˙ pC3 ˆ Cpq, F. Maghsoudi: Cayley graphs of order 6pq and 7pq are Hamiltonian 43 • |S| “ 3. Then CaypG;Sq contains a Hamiltonian cycle. Proof. Let S “ ta, b, cu. Since Cq centralizes C3 and ZpGq X G 1 “ teu (by Proposi- tion 2.15(2)), then C2 inverts C3. Now if pS is minimal, then Lemma 2.24 applies. So we may assume pS is not minimal. Consider pG “ G{Cp “ pC2 ˆ Cqq ˙ C3. Choose a 2-element subset ta, bu in S that generates pG. From the minimality of S we see xa, by “ pC2 ˆ Cqq ˙ C3. after replacing a and b with conjugates. Then the projection of pa, bq to C2 ˆ Cq has one of the following forms: • pa2aq, a2a m q q, where 1 ď m ď q ´ 1, • pa2aq, a2q, • pa2aq, a m q q, where 1 ď m ď q ´ 1, • pa2, aqq. Thus, there are four different possibilities for pa, bq after assuming, without loss of gener- ality, that a P C2 ˆ Cq: 1. pa2aq, a2a m q a3q, 2. pa2aq, a2a3q, 3. pa2aq, a m q a3q, 4. pa2, aqa3q. Let c be the third element of S. We may write c “ ai2a j qa k 3γp with 0 ď i ď 1, 0 ď j ď q´1 and 0 ď k ď 2. Since Cq centralizes C3, we may assume Cq does not centralize Cp, for otherwise Lemma 2.26 applies. Now we have aqγpa ´1 q “ γ pτ p , where pτ q ” 1 pmod pq. We also have pτ ı 1 pmod pq. Since pτ q ” 1 pmod pq, this implies pτ q´1 ` pτ q´2 ` ¨ ¨ ¨ ` 1 ” 0 pmod pq. Note that this implies pτ ı ´1 pmod pq. Case 1. Assume a “ a2aq and b “ a2a m q a3. If k ‰ 0, then by Lemma 2.29(1) xa, cy “ G which contradicts the minimality of S. So we can assume k “ 0. Now if i ‰ 0, then by Lemma 2.29(3) xb, cy “ G which contradicts the minimality of S. Therefore, we may assume i “ 0. Then j ‰ 0 and c “ ajqγp. Consider G “ C2 ˆ Cq . Then we have a “ a2aq , b “ a2a m q and c “ a j q . We may assume m is odd by replacing b with b´1 (and m with q ´ m) if necessary. Note that this implies b “ am. Also, we have |a| “ |b| “ 2q and |c| “ q. 44 Art Discrete Appl. Math. 5 (2022) #P1.10 Subcase 1.1. Assume m “ 1. Then a “ b. We have C “ pcq´1, b, c´pq´1q, a´1q as a Hamiltonian cycle in CaypG;Sq. Since there is one occurrence of b in C, and it is the only generator of G that contains a3, then by Lemma 2.8 we conclude that the subgroup generated by VpCq contains C3. Now by considering the fact that C2 might centralize Cp or not we have VpCq “ cq´1bc´pq´1qa´1 ” pajqγpq q´1 ¨ a2aq ¨ pa j qγpq ´pq´1q ¨ a´1q a2 pmod C3q “ γpτ j`pτ2j`¨¨¨`pτ pq´1qj p a pq´1qj q a2aqa ´pq´1qj q γ ´ppτj`pτ2j`¨¨¨`pτ pq´1qjq p a ´1 q a2 “ γpτ jp1`pτj`¨¨¨`pτ pq´2qjq p aqγ ¯pτjp1`pτj`¨¨¨`pτ pq´2qjq p a ´1 q . Now if pτ j ı 1 pmod pq, then VpCq “ γpτ jp1`pτj`¨¨¨`pτ pq´2qjq p aqγ ¯pτjp1`pτj`¨¨¨`pτ pq´2qjq p a ´1 q “ γpτ jpppτjqq´1´1q{ppτj´1q¯pτj`1pppτjqq´1´1q{ppτj´1q p “ γpτ jpppτ´jq´1q{ppτj´1q¯pτj`1pppτ´jq´1q{ppτj´1q p “ γp1´pτ jqp1¯pτq{ppτj´1q p “ γ´p1¯pτqp . We may assume this does not generate Cp, for otherwise Factor Group Lemma 2.6 applies. Therefore, pτ j ” 1 pmod pq or pτ ” ˘1 pmod pq. The second case is impossible. So we must have pτ j ” 1 pmod pq. We also know that pτ q ” 1 pmod pq. So pτd ” 1 pmod pq, where d “ gcdpj, qq. Since 1 ď j ď q ´ 1, then d “ 1, which contradicts the fact that pτ ı 1 pmod pq. Subcase 1.2. Assume m ‰ 1 and j “ 2. Then c “ a2qγp. We have C “ pb, c´pm´1q{2, a, cpm´1q{2, a2q´m´1q as a Hamiltonian cycle in CaypG;Sq. Since there is one occurrence of b in C, and it is the only generator of G that contains a3, then by Lemma 2.8 we conclude that the subgroup generated by VpCq contains C3. Considering the fact that C2 might centralize Cp or not we have VpCq “ bc´pm´1q{2acpm´1q{2a2q´m´1 ” a2a m q ¨ pa 2 qγpq ´pm´1q{2 ¨ a2aq ¨ pa 2 qγpq pm´1q{2 ¨ a2q´m´1q pmod C3q “ a2a m q pγ pτ2`ppτ2q2`¨¨¨`ppτ2qpm´1q{2 p a pm´1q q q ´1a2aq ¨ pγpτ 2`ppτ2q2`¨¨¨`ppτ2qpm´1q{2 p a pm´1q q qa ´m´1 q “ a2a m q a ´m`1 q γ ´pτ2p1`pτ2`¨¨¨`ppτ2qpm´3q{2q p a2aqγ pτ2p1`pτ2`¨¨¨`ppτ2qpm´3q{2q p a ´2 q “ aqγ ˘pτ2p1`pτ2`¨¨¨`pτ2qpm´3q{2 p aqγ pτ2p1`pτ2`¨¨¨`pτ2qpm´3q{2 p a ´2 q F. Maghsoudi: Cayley graphs of order 6pq and 7pq are Hamiltonian 45 “ γ˘pτ 3ppτm´1´1q{ppτ2´1q`pτ4ppτm´1´1q{ppτ2´1q p “ γpτ 3ppτm´1´1qp˘1`pτq{ppτ2´1q p . We may assume this does not generate Cp, for otherwise Factor Group Lemma 2.6 applies. Therefore, pτm´1 ” 1 pmod pq. We also know that pτ q ” 1 pmod pq. So pτd ” 1 pmod pq, where d “ gcdpm ´ 1, qq. Since 2 ď m ď q ´ 1, then d “ 1, which contradicts the fact that pτ ı 1 pmod pq. Subcase 1.3. Assume m ‰ 1 and j ‰ 2. We may also assume j is an even number, by replacing c with its inverse and j with q´j if necessary. This implies that c “ aj . We have C “ pb, c, a, c´1, b ´1 , am´2, c, a´pj´3q, c, a2q´m´j´2q as a Hamiltonian cycle in CaypG;Sq. Now we calculate its voltage. VpCq “ bcac´1b´1am´2ca´pj´3qca2q´m´j´2 ” a2a3 ¨ a2 ¨ a ´1 3 a2 ¨ a m´2 2 ¨ a ´pj´3q 2 ¨ a 2q´m´j´2 2 pmod Cq ˙ Cpq “ a2a3a2a ´1 3 “ a´23 which generates C3. Also considering the fact that C2 might centralize Cp or not we have VpCq “ bcac´1b´1am´2ca´pj´3qca2q´m´j´2 ” a2a m q ¨ a j qγp ¨ a2aq ¨ γ ´1 p a ´j q ¨ a ´m q a2 ¨ a2a m´2 q ¨ a j qγp ¨ a ´j`3 q a2 ¨ a j qγp ¨ a2a 2q´m´j´2 q pmod C3q “ am`jq γ ˘1 p aqγ ´1 p a ´2 q γpa 3 qγ ˘1 p a ´m´j´2 q “ γ˘pτ m`j´pτm`j`1`pτm`j´1˘pτm`j`2 p “ γpτ m`j´1p˘pτ3´pτ2˘pτ`1q p . So we may assume this does not generate Cp, for otherwise Factor Group Lemma 2.6 applies. Then we have 0 ” ˘pτ3 ´ pτ2 ˘ pτ ` 1 pmod pq. Let t “ pτ if C2 centralizes Cp and t “ ´pτ if C2 inverts Cp. Then 0 ” t3 ´ t2 ` t ` 1 pmod pq. (1.3A) We can replace t with t´1 in the above equation after replacing ta, b, cu with their inverses, then 0 ” t´3 ´ t´2 ` t´1 ` 1 pmod pq. Multiplying by t3, we have 0 ” 1 ´ t ` t2 ` t3 pmod pq 46 Art Discrete Appl. Math. 5 (2022) #P1.10 “ t3 ` t2 ´ t ` 1. By subtracting 1.3A from the above equation, we have 0 ” 2t2 ´ 2t pmod pq “ 2tpt ´ 1q This implies that t ” 1 pmod pq which contradicts the fact that pτ ı ˘1 pmod pq. Case 2. Assume a “ a2aq and b “ a2a3. If k ‰ 0, then by Lemma 2.29(1) xa, cy “ G which contradicts the minimality of S. So we can assume k “ 0. Subcase 2.1. Assume i “ 0. Then j ‰ 0 and c “ ajqγp. We may assume j is an odd number, by replacing c with its inverse and j with q´j if necessary. Consider G “ C2ˆCq . Then we have a “ a2aq , b “ a2 and c “ a j q . Also, we have |a| “ 2q, |b| “ 2 and |c| “ q. Now if j ‰ 1, then we have C “ pc, a´1, b, a2, b, c´1, aj´3, b, a´pq´4q, b, aq´j´2q as a Hamiltonian cycle in CaypG;Sq. Now we calculate the voltage of C. VpCq “ ca´1ba2bc´1aj´3ba´pq´4qbaq´j´2 ” a2 ¨ a2a3 ¨ a 2 2 ¨ a2a3 ¨ a j´3 2 ¨ a2a3 ¨ a ´pq´4q 2 ¨ a2a3 ¨ a q´j´2 2 pmod Cq ˙ Cpq “ a3a2a3a2a3a2a2a3 “ a23 which generates C3. By considering the fact that C2 might centralize Cp or not, we have VpCq “ ca´1ba2bc´1aj´3ba´pq´4qbaq´j´2 ” ajqγp ¨ a ´1 q a2 ¨ a2 ¨ a 2 q ¨ a2 ¨ γ ´1 p a ´j q ¨ a j´3 q ¨ a2 ¨ a2a ´q`4 q ¨ a2 ¨ a q´j´2 q pmod C3q “ ajqγpaqγ ¯1 p a ´j´1 q “ γpτ j¯pτj`1 p “ γpτ jp1¯pτq p which generates Cp. Therefore, the subgroup generated by VpCq is G 1. Thus, Factor Group Lemma 2.6 applies. So we may assume j “ 1, then c “ aqγp and c “ aq . We have C1 “ ppb, cq q´1, b, aq as a Hamiltonian cycle in CaypG;Sq. Now we calculate its voltage. VpC1q “ pbcq q´1ba ” pa2a3q q´1 ¨ a2a3 ¨ a2 pmod Cq ˙ Cpq “ a´13 F. Maghsoudi: Cayley graphs of order 6pq and 7pq are Hamiltonian 47 which generates C3. If C2 centralizes Cp, then VpC1q “ pbcq q´1ba ” pa2 ¨ aqγpq q´1 ¨ a2 ¨ a2aq pmod C3q “ paqγpq q´1aq “ γpτ`pτ 2`¨¨¨`pτq´1 p “ γ´1p which generates Cp. So in this case, the subgroup generated by VpC1q is G 1. Thus, Factor Group Lemma 2.6 applies. Now if C2 inverts Cp, then VpC1q “ pbcq q´1ba ” pa2 ¨ aqγpq q´1 ¨ a2 ¨ a2aq pmod C3q “ γ´pτ`pτ 2´¨¨¨´pτq´2`pτq´1 p . Since pτ ı ´1 pmod pq, then VpC1q “ γ ´pτ`pτ2´¨¨¨´pτq´2`pτq´1 p “ γppτ q`1q{ppτ`1q´1 p . We may assume this does not generate Cp, for otherwise Factor Group Lemma 2.6 applies. Therefore, since pτ q ” 1 pmod pq, then 0 ” ppτ q ` 1q{ppτ ` 1q ´ 1 pmod pq “ 2{ppτ ` 1q ´ 1. This implies that pτ ” 1 pmod pq, which is impossible. Subcase 2.2. Assume j “ 0. Then i ‰ 0 and c “ a2γp. Consider G “ C2 ˆ Cq . Then we have a “ a2aq and b “ c “ a2. This implies that |a| “ 2q and |b| “ |c| “ 2. We have C “ pc, aq´1, b, a´pq´1qq as a Hamiltonian cycle in CaypG;Sq. Since there is one occurrence of b in C, and it is the only generator of G that contains a3, then by Lemma 2.8 we conclude that the subgroup generated by VpCq contains C3. Similarly, since there is one occurrence of c in C, and it is the only generator of G that contains γp, then by Lemma 2.8 we conclude that the subgroup generated by VpCq contains Cp. Therefore, the subgroup generated by VpCq is G1. So, Factor Group Lemma 2.6 applies. Subcase 2.3. Assume i ‰ 0 and j ‰ 0. Then c “ a2a j qγp. Consider G “ C2 ˆ Cq . Then we have a “ a2aq , b “ a2 and c “ a2a j q . This implies that |a| “ |c| “ 2q and |b| “ 2. We may assume j is even by replacing c with its inverse and j with q ´ j if necessary. Suppose, for the moment, that j “ q ´ 1, then c “ a2a ´1 q γp and c “ a ´1. We have C1 “ pc, b, pa ´1, bqq´1q 48 Art Discrete Appl. Math. 5 (2022) #P1.10 as a Hamiltonian cycle in CaypG;Sq. Since there is one occurrence of c in C, and it is the only generator of G that contains γp, then by Lemma 2.8 we conclude that the subgroup generated by VpC1q contains Cp. Also, VpC1q “ cbpa ´1bqq´1 ” a2 ¨ a2a3 ¨ pa2 ¨ a2a3q q´1 pmod Cq ˙ Cpq “ aq3 which generates C3. Therefore, the subgroup generated by VpC1q contains G 1. Thus, Factor Group Lemma 2.6 applies. So we may assume j ‰ q ´ 1. Then we have C2 “ pc, a q´j´1, b, a´q`j`1, pa´1, bqjq and C3 “ pc, a q´j´2, b, a´q`j`2, pa´1, bqj´1, a´2, b, aq as Hamiltonian cycles in CaypG;Sq. Since there is one occurrence of c in C2, and it is the only generator of G that contains γp, then by Lemma 2.8 we conclude that the subgroup generated by VpC2q contains Cp. Also, VpC2q “ ca q´j´1ba´q`j`1pa´1bqj ” a2 ¨ a q´j´1 2 ¨ a2a3 ¨ a ´q`j`1 2 ¨ a j 3 pmod Cq ˙ Cpq “ aj`13 . We may assume this does not generate C3, for otherwise Factor Group Lemma 2.6 applies. Then j ” ´1 pmod 3q. Since there is one occurrence of c in C3, and it is the only generator of G that contains γp, then by Lemma 2.8 we conclude that the subgroup generated by VpC3q contains Cp. Also, VpC3q “ ca q´j´2ba´q`j`2pa´1bqj´1a´2ba ” a2 ¨ a q´j´2 2 ¨ a2a3 ¨ a ´q`j`2 2 ¨ a j´1 3 ¨ a ´2 2 ¨ a2a3 ¨ a2 pmod Cq ˙ Cpq “ a2a3a2a j´1 3 a2a3a2 “ aj´33 “ aj3 Since j ” ´1 pmod 3q, this generates C3. So, Factor Group Lemma 2.6 applies. Case 3. Assume a “ a2aq and b “ a m q a3. If k ‰ 0, then by Lemma 2.29(1) xa, cy “ G which contradicts the minimality of S. So we can assume k “ 0. Now if i ‰ 0, then by Lemma 2.29(3) xb, cy “ G which contradicts the minimality of S. Therefore, we may assume i “ 0. Then j ‰ 0 and c “ ajqγp. Consider G “ C2 ˆ Cq . Then we have a “ a2aq , b “ amq and c “ a j q . Suppose, for the moment, that m “ j. Then b “ c. We have C1 “ pc ´1, b ´pq´2q , a´1, b q´1 , aq F. Maghsoudi: Cayley graphs of order 6pq and 7pq are Hamiltonian 49 as a Hamiltonian cycle in CaypG;Sq. Since there is one occurrence of c in C1, and it is the only generator of G that contains γp, then by Lemma 2.8 we conclude that the subgroup generated by VpC1q contains Cp. Also, VpC1q “ c ´1b´pq´2qa´1bq´1a ” a ´pq´2q 3 ¨ a2 ¨ a q´1 3 ¨ a2 pmod Cq ˙ Cpq “ a´2q`33 “ a´2q3 which generates C3, because gcdp2q, 3q “ 1. So, the subgroup generated by VpC1q is G 1. Therefore, Factor Group Lemma 2.6 applies. So we may assume m ‰ j. We may also assume m and j are even, by replacing tb, cu with their inverses, m with q ´ m, and j with q ´ j if necessary. Now suppose, for the moment, j “ 2. Then we have c “ a2qγp. We also have C2 “ pb, c ´pm´2q{2, a´1, cm{2, a2q´m´1q as a Hamiltonian cycle in CaypG;Sq. Since there is one occurrence of b in C2, and it is the only generator of G that contains a3, then by Lemma 2.8 we conclude that the subgroup generated by VpC2q contains C3. Now by considering the fact that C2 might centralize Cp or not, we have VpC2q “ bc ´pm´2q{2a´1cm{2a2q´m´1 ” amq ¨ pa 2 qγpq ´pm´2q{2 ¨ a´1q a2 ¨ pa 2 qγpq m{2 ¨ a2q´m´12 a 2q´m´1 q pmod C3q “ amq pγ pτ2`ppτ2q2`¨¨¨`ppτ2qpm´2q{2 p a pm´2q q q ´1a´1q a2 ¨ pγpτ 2`ppτ2q2`¨¨¨`ppτ2qm{2 p a m q qa2a ´m´1 q “ amq a ´pm´2q q γ ´pτ2p1`pτ2`¨¨¨`ppτ2qpm´4q{2q p ¨ a´1q γ ˘pτ2p1`pτ2`¨¨¨`ppτ2qpm´2q{2q p a m q a ´m´1 q . Since pτ2 ´ 1 ı 0 pmod pq, then VpC2q “ a 2 qγ ´pτ2ppτm´2´1q{ppτ2´1q p a ´1 q γ ˘pτ2ppτm´1q{ppτ2´1q p a ´1 q “ γ´pτ 4ppτm´2´1q{ppτ2´1q˘pτ3ppτm´1q{ppτ2´1q p “ γpτ 3p1¯pτqp´pτm´1¯1q{ppτ2´1q p . We may assume this does not generate Cp, for otherwise Factor Group Lemma 2.6 applies. Therefore, pτ ” ˘1 pmod pq or pτm´1 ” ˘1 pmod pq. The first case is impossible. So we may assume pτm´1 ” ˘1 pmod pq. Thus, pτ2pm´1q ” 1 pmod pq. We also know that pτ q ” 1 pmod pq. So we have pτd ” 1 pmod pq, where d “ gcdp2pm ´ 1q, qq. Since gcdp2, qq “ 1 and 2 ď m ď q ´ 1, then d “ 1, which contradicts the fact that pτ ı 1 pmod pq. So we may assume j ‰ 2. We have C3 “ pb, c, a, c ´1, b ´1 , am´2, c, a´pj´3q, c, a2q´m´j´2q 50 Art Discrete Appl. Math. 5 (2022) #P1.10 as a Hamiltonian cycle in CaypG;Sq. Now we calculate its voltage. VpC3q “ bcac ´1b´1am´2ca´pj´3qca2q´m´j´2 ” a3 ¨ a2 ¨ a ´1 3 ¨ a m´2 2 ¨ a ´j`3 2 ¨ a 2q´m´j´2 2 pmod Cq ˙ Cpq “ a23 which generates C3. Also, by considering the fact that C2 might centralize Cp or not, we have VpC3q “ bcac ´1b´1am´2ca´pj´3qca2q´m´j´2 ” amq ¨ a j qγp ¨ a2aq ¨ γ ´1 p a ´j q ¨ a ´m q ¨ a m´2 2 a m´2 q ¨ ajqγp ¨ a ´j`3 q a ´j`3 2 ¨ a j qγp ¨ a 2q´m´j´2 2 a 2q´m´j´2 q pmod C3q “ am`jq γpa2aqγ ´1 p a ´2 q γpa 3 qa2γpa ´m´j´2 q “ am`jq γpaqγ ¯1 p a ´2 q γ ˘1 p a 3 qγpa ´m´j´2 q “ γpτ m`j¯pτm`j`1˘pτm`j´1`pτm`j`2 p “ γpτ m`j´1ppτ3¯pτ2`pτ˘1q p . We may assume this does not generate Cp, for otherwise Factor Group Lemma 2.6 applies. Therefore, 0 ” pτ3 ¯ pτ2 ` pτ ˘ 1 pmod pq. If C2 centralizes Cp, then 0 ” pτ3 ´ pτ2 ` pτ ` 1 pmod pq. (3A) We can replace pτ with pτ´1 in the above equation after replacing ta, b, cu with their inverses in the Hamiltonian cycle, then 0 ” pτ´3 ´ pτ´2 ` pτ´1 ` 1 pmod pq. Multiplying by pτ3, we have 0 ” 1 ´ pτ ` pτ2 ` pτ3 pmod pq “ pτ3 ` pτ2 ´ pτ ` 1. Subtracting 3A from the above equation we have 0 ” 2pτ2 ´ 2pτ pmod pq “ 2pτppτ ´ 1q which is impossible, because pτ ı 1 pmod pq. Now if C2 inverts Cp, then 0 ” pτ3 ` pτ2 ` pτ ´ 1 pmod pq. (3B) F. Maghsoudi: Cayley graphs of order 6pq and 7pq are Hamiltonian 51 We can replace pτ with pτ´1 in the above equation after replacing ta, b, cu with their inverses. Then 0 ” pτ´3 ` pτ´2 ` pτ´1 ´ 1 pmod pq. Multiplying by pτ3, then 0 ” 1 ` pτ ` pτ2 ´ pτ3 pmod pq “ ´pτ3 ` pτ2 ` pτ ` 1. By adding 3B and the above equation, we have 0 ” 2ppτ2 ` pτq pmod pq “ 2pτppτ ` 1q which is also impossible, because pτ ı ´1 pmod pq. Case 4. Assume a “ a2 and b “ aqa3. Subcase 4.1. Assume i ‰ 0. Then c “ a2a j qa k 3γp. By Lemma 2.29(2) xb, cy “ G which contradicts the minimality of S. Subcase 4.2. Assume i “ 0. Then j ‰ 0 and c “ ajqa k 3γp. We may assume j is even by replacing c with its inverse and j with q ´ j if necessary. Consider G “ C2 ˆ Cq . Then we have a “ a2, b “ aq and c “ a j q . This implies that |a| “ 2 and |b| “ |c| “ q. We have C1 “ pc, b q´j´1 , c, b ´pj´2q , a, b q´1 , aq as a Hamiltonian cycle in CaypG;Sq. Now we calculate its voltage. VpC1q “ cb q´j´1cb´pj´2qabq´1a ” ajqγp ¨ a q´j´1 q ¨ a j qγp ¨ a ´j`2 q ¨ a2 ¨ a q´1 q ¨ a2 pmod C3q “ ajqγpa ´1 q γpa ´j`1 q “ γpτ j´1ppτ`1q p which generates Cp. Also VpC1q “ cb q´j´1cb´pj´2qabq´1a ” ak3 ¨ a q´j´1 3 ¨ a k 3 ¨ a ´j`2 3 ¨ a2 ¨ a q´1 3 ¨ a2 pmod Cq ˙ Cpq “ ak`q´j´1`k´j`2´q`13 “ a 2pk´j`1q 3 . We may assume this does not generate C3, for otherwise Factor Group Lemma 2.6 applies. Then 0 ” k ´ j ` 1 pmod 3q. (4.2A) 52 Art Discrete Appl. Math. 5 (2022) #P1.10 We also have C2 “ pc, a, pb, aq q´j´1, b j , a, b ´pj´1q q as a Hamiltonian cycle in CaypG;Sq. We calculate its voltage. Since there is one occur- rence of c in C2, and it is the only generator of G that contains γp, then by Lemma 2.8 we conclude that the subgroup generated by VpC2q contains Cp. Also, VpC2q “ capbaq q´j´1bjab´pj´1q ” ak3 ¨ a2 ¨ pa3a2q q´j´1 ¨ aj3 ¨ a2 ¨ a ´j`1 3 pmod Cq ˙ Cpq “ ak´2j`13 . We may assume this does not generate C3, for otherwise Factor Group Lemma 2.6 applies. Therefore, 0 ” k ´ 2j ` 1 pmod 3q. By subtracting the above equation from 4.2A we have j ” 0 pmod 3q. Now we have C3 “ pc, a, b q´j´1 , a, b ´pq´j´2q , c´1, b j´2 , a, b ´pj´1q , aq as a Hamiltonian cycle in CaypG;Sq. We calculate its voltage. VpC3q “ cab q´j´1ab´pq´j´2qc´1bj´2ab´pj´1qa ” ajqγp ¨ a2 ¨ a q´j´1 q ¨ a2 ¨ a ´q`j`2 q ¨ γ ´1 p a ´j q ¨ a j´2 q ¨ a2 ¨ a ´j`1 q ¨ a2 pmod C3q “ ajqγpaqγ ´1 p a ´j´1 q “ γpτ jp1´pτq p . which generates Cp. Also VpC3q “ cab q´j´1ab´pq´j´2qc´1bj´2ab´pj´1qa ” ak3 ¨ a2 ¨ a q´j´1 3 ¨ a2 ¨ a ´q`j`2 3 ¨ a ´k 3 ¨ a j´2 3 ¨ a2 ¨ a ´j`1 3 ¨ a2 pmod Cq ˙ Cpq “ ak´q`j`1´q`j`2´k`j´2`j´13 “ a´2q`4j3 . We may assume this does not generate C3, for otherwise Factor Group Lemma 2.6 applies. Then 0 ” ´2q ` 4j pmod 3q “ q ` j We already know j ” 0 pmod 3q. By substituting this in the above equation, we have q ” 0 pmod 3q which contradicts the fact that gcdpq, 3q “ 1. F. Maghsoudi: Cayley graphs of order 6pq and 7pq are Hamiltonian 53 3.9 Assume |S| ě 4 In this subsection, we prove the following general result that includes the part of The- orem 1.3, where |S| ě 4 (see Assumption 3.1). Unlike in the other subsections of this section, we do not assume |G| “ 6pq. Proposition 3.10. Assume |G| is a product of four distinct primes and S is a minimal generating set of G, where |S| ě 4. Then CaypG;Sq contains a Hamiltonian cycle. Proof. Suppose S “ ts1, s2, ..., sku and let Gi “ xs1, s2, ..., siy for i “ 1, 2, ..., k. Since S is minimal, we know teu Ă G1 Ă G2 Ă ...Gk “ G. Therefore, the number of prime factors of |Gi| is at least i. Since |G| “ p1p2p3q is the product of only 4 primes, and k “ |S| ě 4, we can conclude that |Gi| has exactly i prime factors, for all i. This implies that |S| “ 4. This also implies every element of S has prime order. Since |G| is square-free, we know that G1 is cyclic (see Proposition 2.15(1)), so G1 ‰ G. We may assume |G1| ‰ 1, for otherwise G is abelian, so Lemma 2.2 applies. Also, if |G1| is equal to a prime number, then Theorem 2.3 applies. So we may assume |G1| has at least two prime factors. Therefore, the number of prime factors of |G1| is either 2 or 3. Case 1. Assume |G1| has only two prime factors. This implies |G| “ p1p2, where p1 and p2 are two distinct primes. Suppose s P S, then s P S. We know that |s| ‰ 1 (see Assumption 3.1(6)). Now since every element of S has prime order, then |s| is either p1 or p2. Also, every element of order p1 must commute with every element of order p2, because the subgroup H generated by any element of S that has order p1, together with any element of S that has order p2 has exactly two prime factors, so |H| “ p1p2, H 1 Ď G1, and |G1| “ p3p4. Thus, |H 1| “ 1. Let Sp1 be the elements of order p1 in S, and let Sp2 be the elements of order p2. Also let Hp1 and Hp2 be the subgroups generated by Sp1 and Sp2 , respectively. This implies that CaypG;Sq – CaypGp1 ;Sp1q ˝ CaypGp2 ;Sp2q. Therefore, CaypG;Sq contains a Hamiltonian cycle (see Corollary 2.11). Case 2. Assume |G1| has three prime factors. We may write (see Proposition 2.15(3)) G “ Cq ˙ G 1 “ Cq ˙ pCp1 ˆ Cp2 ˆ Cp3q, where p1, p2, p3 and q are distinct primes. Note that G 1 X ZpGq “ teu (see Proposi- tion 2.15(2)). Now we may assume xs4y “ Cq . Since |xsi, s4y| has only two prime factors (for 1 ď i ď 3), we must have si “ s ki 4 api (after permuting p1, p2, p3), where api is a generator of Cpi . We may also assume SXG 1 “ H (see Lemma 2.12), so ki ı 0 pmod qq. Now consider G2 “ xs1, s2y “ xs k1 4 ap1 , s k2 4 ap2y. Since Cp1 is a normal subgroup in G, we can consider G2 “ G2{Cp1 , then ts1, s2u “ tsk14 , s k2 4 ap2u. We have s k ´1 2 4 “ ps k1 4 q k ´1 1 k ´1 2 “ s k ´1 1 k ´1 2 1 . Multiplying by s2, then ap2 “ s k ´1 2 4 ¨ s k2 4 ap2 “ s k ´1 1 k ´1 2 1 s2 P G2. 54 Art Discrete Appl. Math. 5 (2022) #P1.10 Since ap2 generates Cp2 , this implies |G2| is divisible by p2. Similarly, we can show that |G2| is divisible by p1. Also, |s1| “ q, so |G2| is divisible by q. Therefore, |G2| has three prime factors, which is a contradiction. ORCID iDs Farzad Maghsoudi https://orcid.org/0000-0002-0482-319X References [1] A. Cayley, The theory of groups: Graphical representation, Am. J. Math. 1 (1878), 174–176, doi:10.2307/2369306. [2] C. 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