Transversals and bipancyclicity in bipartite graph families
aa r X i v : . [ m a t h . C O ] J a n TRANSVERSALS AND BIPANCYCLICITY IN BIPARTITE GRAPH FAMILIES
PETER BRADSHAW
Abstract.
A bipartite graph is called bipancyclic if it contains cycles of every even length from four up tothe number of vertices in the graph. A theorem of Schmeichel and Mitchem states that for n ≥
4, everybalanced bipartite graph on 2 n vertices in which each vertex in one color class has degree greater than n andeach vertex in the other color class has degree at least n is bipancyclic. We prove a generalization of thistheorem in the setting of graph transversals. Namely, we show that given a family G of 2 n bipartite graphson a common set X of 2 n vertices with a common balanced bipartition, if each graph of G has minimumdegree greater than n in one color class and minimum degree at least n in the other color class, then thereexists a cycle on X of each even length 4 ≤ ℓ ≤ n that uses at most one edge from each graph of G . Wealso show that given a family G of n bipartite graphs on a common set X of 2 n vertices meeting the samedegree conditions, there exists a perfect matching on X that uses exactly one edge from each graph of G . Introduction
One of the oldest problems in graph theory is that of determining conditions that guarantee a cycle of agiven length in a graph. The classic Dirac’s theorem [7] gives such a condition, stating that if every vertexof a graph G on n vertices has a degree of at least n , then G contains a Hamiltonian cycle. A later result ofBondy [4] shows that the degree condition of Dirac’s theorem guarantees more than just a Hamiltonian cycle;Bondy proves that under these same degree conditions, G contains a cycle of every length ℓ , for 3 ≤ ℓ ≤ n ,except when G ∼ = K n/ ,n/ . A graph on n vertices that contains a cycle of every length ℓ , for 3 ≤ ℓ ≤ n , iscalled pancyclic . Therefore, Bondy’s result can be concisely summarized by the statement that every graphon n vertices with minimum degree at least n is pancyclic, except for the complete balanced bipartite graph.The problem of finding conditions that guarantee cycles of given lengths in graphs has been considerednot only for general graphs, but also specifically for bipartite graphs. Interestingly, the results of Dirac andBondy mentioned above have close analogues in the bipartite setting. In particular, Moon and Moser [12]prove the following theorem for bipartite graphs, which closely resembles Dirac’s theorem. Here, we use d ( v )to denote the degree of a vertex v . Theorem 1.1.
Let n ≥ . Let G be a balanced bipartite graph on n vertices with a red and a blue colorclass. Suppose that for each red vertex v ∈ V ( G ) , d ( v ) > n . Suppose further that for each blue vertex w ∈ V ( G ) , d ( w ) ≥ n . Then G contains a Hamiltonian cycle. Additionally, not only Dirac’s theorem has a bipartite analogue, but Bondy’s strengthened version ofDirac’s theorem also has a close analogue for bipartite graphs. In particular, the following result of Schmeicheland Mitchem [15], which generalizes Theorem 1.1 for n ≥
4, closely resembles the result of Bondy mentionedabove.
Theorem 1.2.
Let n ≥ . Let G be a balanced bipartite graph on n vertices with a red and a blue colorclass. Suppose that for each red vertex v ∈ V ( G ) , d ( v ) > n . Suppose further that for each blue vertex w ∈ V ( G ) , d ( w ) ≥ n . Then G contains a cycle of every even length ℓ , for ≤ ℓ ≤ n . A balanced bipartite graph on 2 n vertices that contains a cycle of every even length ℓ , for 4 ≤ ℓ ≤ n , iscalled bipancyclic . Therefore, Theorem 1.2 may be summarized by stating that a balanced bipartite graphmeeting certain minimum degree conditions is bipancyclic. In Theorem 1.2, the condition that n ≥ n = 3, the cycle on six vertices meets the minimum degree conditions of the theorembut contains no cycle of length four. We note that the authors of [12] and [15] also give stronger versions The author of this work has been partially supported by a supervisor’s grant from the Natural Sciences and EngineeringResearch Council of Canada (NSERC). f Theorems 1.1 and 1.2, but we only consider the weaker versions stated here in order to highlight the factthat these results for bipartite graphs closely resemble the results of Dirac and Bondy for general graphs.In this paper, we will also seek graph conditions that guarantee cycles of given lengths, but we will considerthis question in the context of graph transversals , a special type of transversals , which are defined as follows.Suppose we have a universal set E and a family F = { F i : i ∈ I } of subsets of E , where I is some index. Thena transversal on F is a set { f i : i ∈ I } such that f i ∈ F i for each i ∈ I , and f i = f j for i = j . In other words,a transversal is a system of distinct representatives for a family of sets. Transversals frequently appear ininfinitary combinatorics under several similar definitions (see, for example, [8] and [3]), and transversals arealso extensively studied in the context of Latin squares (see [16] for a survey).The notion of a transversal may be applied to a family of graphs as follows. Suppose we have a set X of n vertices, and suppose that G = { G , . . . , G s } is a family of s graphs, each with X as its set of vertices.Then, we define a transversal on G to be a a set of s edges E ⊆ (cid:0) X (cid:1) for which there exists a bijectivefunction φ : E → [ s ] such that for all e ∈ E , it holds that e ∈ E ( G φ ( e ) ). If | E | < s and there exists aninjective function φ : E → [ s ] satisfying the same property, then we say that E is a partial transversal on G .Informally, if we imagine that each graph G i ∈ G has its edges colored with a single distinct color, then atransversal on G is a set of edges in which each color appears exactly once, and a partial transversal on G isa set of edges in which each color appears at most once.Recently, certain classical results about cycles of specified lengths in graphs have been extended to thesetting of graph transversals. For example, Aharoni et al. [1] give a transversal analogue of Mantel’s theorem,showing that given a graph set G = { G , G , G } on a common set of n vertices in which each graph G i hasroughly at least 0 . n edges, there must exist a partial transversal on G isomorphic to K . Additionally,Joos and Kim [11] show that given a graph family G = { G , . . . , G n } on a common vertex set of n vertices,if the minimum degree of each graph G i is at least n , then G has a transversal isomorphic to a Hamiltoniancycle, which gives a generalization of Dirac’s theorem. Cheng, Wang and Zhao [5] furthermore show thatgiven a graph family G = { G , . . . , G n } on a common vertex set X of n vertices, if the minimum degree ofeach graph G i is at least n +12 , then G contains a partial transversal of length ℓ for every length 3 ≤ ℓ ≤ n − Theorem 1.3.
Let n ≥ . Let X be a set of n red vertices and n blue vertices, and let G = { G , . . . , G n } be a set of n bipartite graphs on the vertex set X . Suppose • For each G i ∈ G , the red vertices of X and the blue vertices of X both form independent sets in G i ; • For each G i ∈ G and for each red vertex v ∈ V ( G i ) , d ( v ) > n ; • For each G i ∈ G and for each blue vertex w ∈ V ( G i ) , d ( w ) ≥ n .Then G contains a transversal isomorphic to a Hamiltonian cycle. If we let G = G = · · · = G n in Theorem 1.3, then any Hamiltonian cycle in G is a Hamiltoniantransversal on G , so Theorem 1.3 is a generalization of Theorem 1.1. After proving Theorem 1.3, we willalso be able to prove the following stronger theorem, which generalizes Theorem 1.2 in the setting of graphtransversals. Theorem 1.4.
Let n ≥ . Let X be a set of n red vertices and n blue vertices, and let G = { G , . . . , G n } be a set of n bipartite graphs on the vertex set X . Suppose • For each G i ∈ G , the red vertices of X and the blue vertices of X both form independent sets in G i ; • For each G i ∈ G and for each red vertex v ∈ V ( G i ) , d ( v ) > n ; • For each G i ∈ G and for each blue vertex w ∈ V ( G i ) , d ( w ) ≥ n .Then G contains a (partial) transversal isomorphic to a cycle of length ℓ for each even length ℓ , ≤ ℓ ≤ n . Again, if we let G = G = · · · = G n in Theorem 1.4, then any cycle of length ℓ in G is a partialtransversal on G , so Theorem 1.4 is a generalization of Theorem 1.2. Theorem 1.4 is also a generalization ofTheorem 1.3 except for small values of n , and hence these results may be summarized by saying that if agraph family G satisfies the conditions of Theorems 1.3 and 1.4, then G is bipancyclic in the sense of graphtransversals. The main tool used in our proofs is an auxiliary digraph technique introduced by Joos andKim in [11]. e note that Theorems 1.3 and 1.4 are best possible in that we may not relax the minimum degreecondition. Indeed, for any even n , by letting each graph G i ∈ G be an identical copy of a disjoint unionof two complete graphs K n/ ,n/ , we find a graph family with minimum degree n and no Hamiltoniantransversal. A similar example shows that the minimum degree condition for odd n may not be relaxedeither.Lastly, we will consider graph transversals isomorphic to perfect matchings, which we denote as perfectmatching transversals . Joos and Kim consider perfect matching transversals in [11] and establish minimumdegree conditions that guarantee perfect matching transversals in general graphs. Our last theorem showsthat under the degree conditions of Theorems 1.3 and 1.4, a family of n balanced bipartite graphs on acommon set of 2 n vertices contains a perfect matching transversal. Theorem 1.5.
Let n ≥ . Let X be a set of n red vertices and n blue vertices, and let G = { G , . . . , G n } bea set of n bipartite graphs on the vertex set X . Suppose • For each G i ∈ G , the red vertices of X and the blue vertices of X both form independent sets in G i ; • For each G i ∈ G and for each red vertex v ∈ V ( G i ) , d ( v ) > n ; • For each G i ∈ G and for each blue vertex w ∈ V ( G i ) , d ( w ) ≥ n .Then G contains a perfect matching transversal. Theorem 1.5 is equivalent to a special case of a matching theorem for r -partite r -uniform hypergraphsby Aharoni, Georgakopoulos, and Spr¨ussel in [2], for the case r = 3. We nevertheless include a proof ofTheorem 1.5 in order to demonstrate the power of Joos and Kim’s auxiliary digraph from [11] techniquefor investigating graph transversals. We will also give a construction showing that the degree condition ofTheorem 1.5 is, in a sense, best possible.2. A Hamiltonian transversal: Proof of Theorem 1.3
This section will be dedicated to proving Theorem 1.3. When n = 2, the theorem is trivial; thus we willassume throughout the proof that n ≥
3. We will let X and G be defined as in Theorem 1.3. We define a Hamiltonian transversal on G as a transversal on G isomorphic to a Hamiltonian cycle.Throughout this section, we will let X have a blue vertex set { p , . . . , p n } and a red vertex set { q , . . . , q n } .We will assume throughout this section that G does not contain a Hamiltonian transversal, and we will arriveat a contradiction. This strategy will allow us to write certain steps of the proof more concisely. The firstgoal in our proof will be to establish the following claim. Claim 2.1. G contains a partial transversal isomorphic to the disjoint union of a cycle of length n − anda K . In order to prove Claim 2.1, we will first prove two auxiliary claims. The first of these auxiliary claimsshows that Claim 2.1 holds whenever G contains a partial transversal isomorphic to a Hamiltonian path. Claim 2.2. If G contains a partial transversal isomorphic to a Hamiltonian path, then G contains a partialtransversal isomorphic to the disjoint union of a cycle of length n − and a K .Proof. Let P be a partial transversal in G isomorphic to a Hamiltonian path. We assume without loss ofgenerality that P has a vertex sequence ( q , p , q , p , . . . , p n , q n , p ). We let P have an associated injectivefunction φ : E ( P ) → [2 n ]. We write φ ( p q ) = m and [2 n ] \ im( φ ) = { m } . We show the key parts of G in Figure 1. If p q ∈ E ( G m ), then G contains a Hamiltonian transversal, and if p q ∈ E ( G m ), then theclaim is proven; hence, we may assume that no such edge exists in G m . (Note that this immediately provesthe claim when n = 3, and hence we may assume that n ≥ p q E ( G m ).By our minimum degree conditions on red vertices and the assumption that p q E ( G m ), there mustexist at least n − edges of the form p j q ∈ E ( G m ) with j ∈ [4 , n ]. Similarly, by our minimum degreeconditions on blue vertices and our assumption that p q , p q E ( G m ), there must exist at least n − j ∈ [4 , n ] for which p q j − ∈ E ( G m ). As there exist a total of n − j ∈ [4 , n ], it follows bythe pigeonhole principle that there exists a value j ∈ [4 , n ] for which p j q ∈ E ( G m ) and p q j − ∈ E ( G m ).Thus, G contains a partial transversal C isomorphic to a cycle of length 2 n − q , p j , q j , p j +1 , q j +1 , . . . , p n , q n , p , q j − , p j − , . . . , q , p , q ) , n q p n p m q j − p j q p m m Figure 1.
The figure shows the key parts of the graph family G considered in Claim 2.2. TheHamiltonian path partial transversal P is represented by the broken circle with endpoints q and p . The edge p q belongs to G m , and P “misses” the graph G m . If there existtwo edges belonging to G m and G m in the form of the dotted edges, then G must containa partial transversal containing the single edge q p as well as a cycle of length 2 n − X .as shown in Figure 1, and with an injective function ψ : C → [2 n ] satisfying [2 n ] \ im( ψ ) = { φ ( q p ) , φ ( q j − p j ) } .Therefore, C ∪ { q p } gives a partial transversal isomorphic to the disjoint union of a cycle of length 2 n − K , and the claim is proven. (cid:3) In the next claim, we show that G must contain a partial transversal isomorphic to a cycle of length 2 n − Claim 2.3. G contains a partial transversal isomorphic to a cycle of length n − .Proof. Suppose the claim does not hold. Let G be an edge-maximal counterexample to the claim; that is, let G be a family of graphs such that after adding any edge to any graph G i ∈ G , the resulting family containseither a Hamiltonian transversal or a partial transversal isomorphic to a cycle of length 2 n −
2. If each graph G i ∈ G is a complete bipartite graph, then G certainly contains a Hamiltonian transversal. Therefore, we mayassume without loss of generality that G is not a complete bipartite graph and that we may add some edge e to G with a red endpoint and a blue endpoint. We may further assume that G + e = { G + e, G , . . . , G n } contains a partial transversal C isomorphic to a cycle of length 2 n −
2, since if G + e contains a Hamiltoniantransversal, then G must contain a partial transversal isomorphic to a Hamiltonian path, and we are doneby Claim 2.1.We say without loss of generality that e = p q . We also say without loss of generality that C has a vertexsequence ( p , q , p , q , p , . . . , p n − , q n − , p ). Let φ : E ( C ) → [2 n ] be the injective function associated withthe partial transversal C , and let [2 n ] \ im( φ ) = { m , m } . By our minimum degree conditions on the redvertices and the assumption that p q E ( G ), there must exist at least (cid:6) n − (cid:7) values j ∈ [2 , n −
1] forwhich q p j ∈ E ( G ). Similarly, by our minimum degree conditions on the blue vertices, there must exist atleast (cid:6) n − (cid:7) values j ∈ [2 , n −
1] for which p q j − ∈ E ( G m ). Then, one of two cases must occur, and wewill see that in both cases, our minimal counterexample G is not actually a counterexample to the claim,which will complete the proof.(1) There exists a value j ∈ [2 , n −
1] for which q p j ∈ E ( G ) and p q j − ∈ E ( G m ). Then, G has apartial transversal isomorphic to a cycle of length 2 n − q , p , q , . . . , q j − , p , q n − , p n − , . . . , p j , q ) , and G is not a counterexample to the claim.
2) Otherwise, as (cid:6) n − (cid:7) + (cid:6) n − (cid:7) = n −
2, it must follow from the pigeonhole principle that there existexactly (cid:6) n − (cid:7) values j ∈ [2 , n −
1] for which p q j ∈ E ( G ) and exactly (cid:6) n − (cid:7) values j ∈ [2 , n − q p j − ∈ E ( G m ). Then, it must follow that p q n ∈ E ( G ) and q p n ∈ E ( G m ). Then, G contains a partial transversal isomorphic to a Hamiltonian path, namely one beginning at p n andending at q n , and by Claim 2.2, G is not a counterexample to the claim. (cid:3) We are now ready to prove Claim 2.1, after which we will be ready for the main method of our proof.
Proof of Claim 2.1:
By Claim 2.3, we may assume that G contains a partial transversal C isomorphicto a cycle of length 2 n −
2. Let φ : E ( C ) → [2 n ] be the injective function associated with C , and let[2 n ] \ im( φ ) = { m , m } . Let C have a vertex sequence ( p , q , p , q , . . . , p n − , q n − , p ). If p n q n ∈ E ( G m )or p n q n ∈ E ( G m ), then the claim is proven; otherwise, for each edge e ∈ E ( G m ) ∪ E ( G m ), e has anendpoint in V ( C ).We aim to show that G contains a partial transversal P isomorphic to a Hamiltonian path. Let A ⊆ V ( C )be the set of vertices of V ( C ) adjacent to q n via G m . Recall that A is a set of blue vertices. As | A | ≥ n + ,it follows that at most n − red vertices of C are not adjacent in C to a vertex of A . Therefore, as p n has at least n red neighbors in V ( C ) via G m , there must exist a blue vertex a ∈ A , a red vertex b ∈ V ( C )adjacent to a in C , and an edge bp n ∈ E ( G m ). Then G contains a partial transversal isomorphic to aHamiltonian path beginning at p n and ending at q n . Then, by Claim 2.2, we may find a partial transversalin G isomorphic to the disjoint union of a cycle of length 2 n − K . (cid:3) Now, with Claim 2.1 in place, we are ready for the main idea of Theorem 1.3. We will follow a methodof Joos and Kim [11] used for proving a transversal version of Dirac’s theorem. The method of Joos andKim fits our proof very closely, and therefore the remainder of our proof uses the ideas of [11] with very fewchanges.We let G have a partial transversal isomorphic to the disjoint union of a cycle C of length 2 n − K ∼ = K . We will rename our vertices; we say that V ( K ) = { x, y } , and we let C have a vertexsequence ( v , v , . . . , v n − , v ). We let x, v , v , . . . , v n − be red vertices, and we let y, v , v , . . . , v n − beblue vertices. For each vertex v i ∈ V ( C ), we say that φ ( v i v i +1 ) = i (where v n − is identified with v ). Welet φ ( xy ) = 2 n −
1. With these assignments, φ “misses” the value 2 n .We define an auxiliary digraph H on X . For every red vertex v i ∈ V ( C ) and for every edge v i v j ∈ E ( G i )with j = i + 1, we let H include the arc v i v j . Furthermore, if v i y ∈ E ( G i ), we let H contain the arc v i y .We write d + ( v ) and d − ( v ) respectively for the out-degree and in-degree of a vertex v ∈ V ( H ). Note that foreach red vertex v i ∈ V ( H ), v i has at least n + incident edges in the graph G i , and hence d + ( v i ) ≥ n − . Claim 2.4. d − ( y ) ≤ n − .Proof. Suppose d − ( y ) ≥ n − . Let v j ∈ V ( C ) be a neighbor of x via G n . For every in-neighbor v i of y , wemust have v i +1 v j +1 E ( G j ), as otherwise, G contains a Hamiltonian transversal with the vertex sequence( v j , x, y, v i , v i − , . . . , v j +2 , v j +1 , v i +1 , v i +2 , . . . , v j ) , as shown in Figure 2. Therefore, | N G j ( v j +1 ) ∩ V ( C ) | ≤ ( n − − (cid:18) n − (cid:19) = n − , and hence we must have v j +1 y ∈ E ( G j ). Then, G contains a Hamiltonian transversal with a vertex sequence( v , . . . , v j , x, y, v j +1 , v j +2 , . . . , v n − , v ), a contradiction. Therefore, we conclude that d − ( y ) ≤ n − (cid:3) Next, we show that some blue vertex w ∈ V ( C ) satisfies d − ( w ) ≥ n − . Recall that each red vertexof V ( C ) has out-degree at least n − . Let E − ( C ) denote the set of arcs in H that end in V ( C ). As d − ( y ) ≤ n −
1, we have that( ⋆ ) | E − ( C ) | ≥ (cid:18) n − (cid:19) ( n − − (cid:16) n − (cid:17) > ( n − (cid:18) n − (cid:19) . i +1 v j v i v j +1 G j G n G i xy Figure 2.
In the figure, the broken circle represents the edges of the partial transversal C considered in the proof of Theorem 1.3. The figure shows one form of Hamiltoniantransversal that may be found in G using the argument in Claim 2.4 if d − ( y ) > n − w ∈ V ( C ) must satisfy d − ( w ) ≥ n −
1. Furthermore,if all blue vertices w ′ ∈ V ( C ) satisfy d − ( w ′ ) ≤ n −
1, then d − ( w ∗ ) = n − w ∗ , andhence n is even. However, in this case, each red vertex of V ( C ) has an out-degree of at least n , and we canobtain a better lower bound for | E − ( C ) | : | E − ( C ) | ≥ n n − − (cid:16) n − (cid:17) > ( n − (cid:18) n − (cid:19) . This second inequality implies that some blue vertex of V ( C ) has in-degree at least n . Therefore, we mayproceed with the assumption that there exists a blue vertex w ∈ V ( C ) for which d − ( w ) ≥ n − .Now, we consider two cases. In both cases, we will find a Hamiltonian transversal on G , which will com-plete the proof. Case 1:
There exists a vertex w ∈ V ( C ) for which d − ( w ) ≥ n .We will attempt to find a partial transversal on G isomorphic to a Hamiltonian path and satisfying certainconditions. Without loss of generality, we assume d − ( v ) ≥ n . We claim that the set N ∗ := ( N G ( x ) ∪ N G n ( y )) ∩ V ( C )is not an independent set in the cycle C . Indeed, suppose that N ∗ is an independent set in C . The set N G ( x ) ∩ V ( C ) must contain at least n − vertices, and thus there must exist at least n + verticesadjacent in C to N G ( x ) ∩ V ( C ). Then, if N ∗ is an independent set, N G n ( y ) ∩ V ( C ) contains at most( n − − (cid:0) n + (cid:1) < n − v j , v j +1 suchthat either v j ∈ N G ( x ) and v j +1 ∈ N G n ( y ), or v j +1 ∈ N G ( x ) and v j ∈ N G n ( y ) (where v n − is identifiedwith v ). If j = 1, then G contains a Hamiltonian transversal, which may be observed in Figure 3 byidentifying ( v j , v j +1 ) with ( v , v ). Otherwise, we have a partial transversal P isomorphic to a Hamiltonianpath with an injective function ψ and a vertex sequence of the form( v , v , . . . , v j , σ, τ, v j +1 , . . . , v n − , v ) , where either ( σ, τ ) = ( x, y ) or ( σ, τ ) = ( y, x ). We rename this vertex sequence of P as ( v , v , . . . , v n ). Weobserve that [2 n ] \ im( ψ ) = { j } .Finally, we consider the following sets, again identifying v n − with v : I j := { i ∈ [2 n −
2] : v i +1 v ∈ E ( G j ) } ,I − := { i ∈ [2 n −
2] : v i ∈ N − ( v ) } . As d − ( v ) ≥ n and d + ( x ) = 0, we know that | I − | ≥ n , and as d G j ( v ) ≥ n + , we know that | I j | ≥ n − .Furthermore, if v v ∈ E ( G j ), then G contains a Hamiltonian transversal; hence we may assume that I j ⊆ [2 n − ∩ Z . Similarly, as v does not have an in-neighbor v n − , we may also assume that I − ⊆ [2 n − ∩ Z .Therefore, I j , I − are both subsets of [2 n − ∩ Z , which contains n − | I j | + | I − | = n , k v j v k +1 v j +1 (i) v v G j G k G G n xy v k v j v k +1 v j +1 (ii) G k G j G n yx Figure 3.
In both figures, the broken circle represents the edges of the partial transversal C considered in the proof of Theorem 1.3. Figure (i) shows one form of a Hamiltoniantransversal found in Case 1 of the proof, and Figure (ii) shows one form of a Hamiltoniantransversal found in Case 2 of the proof.we have at least two elements in the intersection I j ∩ I − , and in particular we have an element k ∈ I j ∩ I − for which k = j . We may write v l = v k , and as k = j , v k +1 = v l +1 . Then, we see that( v , v , . . . , v l , v n − , v n − , . . . , v l +1 , v )gives a vertex sequence for a Hamiltonian transversal on G , which we illustrate in Figure 3 (i). Thus, in thiscase, the proof is complete. Case 2:
Every vertex w ∈ V ( C ) satisfies d − ( w ) ≤ n − .In this case, as there must exist a vertex in V ( C ) with an in-degree of exactly n − , we know that n isodd. Let a be the number of vertices w ∈ V ( C ) satisfying d − ( w ) = n − . Then, by ( ⋆ ), we have a (cid:18) n − (cid:19) + ( n − − a ) (cid:18) n − (cid:19) ≥ (cid:18) n − (cid:19) ( n − − (cid:16) n − (cid:17) , or equivalently, a ≥ n . Therefore, since | N G n ( y ) ∩ V ( C ) | ≥ n − , we may choose a neighbor v j +1 of y via G n such that d − ( v j ) = n − . Note that since v j is a blue vertex, j is odd.Finally, we consider the following sets, again identifying v n − with v : I j := { i ∈ [2 n −
2] : v i +1 x ∈ E ( G j ) } ,I − := { i ∈ [2 n −
2] : v i ∈ N − ( v j ) } . If j − ∈ I j , then G must contain a Hamiltonian transversal, so we may assume that j − I j . Furthermore,by construction of H , j − I − . Therefore, both I j and I − are subsets of [2 n − ∩ (2 Z ) \ { j − } , a set of n − d G j ( x ) ≥ n + , we know that | I j | ≥ n − . Furthermore, as d − ( v j ) = n − and d + ( x ) = 0,we know that | I − | = n − . Therefore, | I j | + | I − | = n − , and hence there must exist a value k ∈ I j ∩ I − .Therefore, there exists a Hamiltonian transversal with a vertex sequence( v k +1 , v k +2 , . . . , v j − , v j , v k , v k − , . . . , v j +1 , y, x, v k +1 ) , as shown in Figure 3 (ii). Thus, in this case as well, the proof is complete.3. Bipancyclicity: Proof of Theorem 1.4
This section will be dedicated to proving Theorem 1.4. We will continue to use our definitions of X and G from the previous section. One technique that we will frequently use is that of “augmenting” a cycle, asin Figure 4. We will establish two lemmas that will be useful when using this technique. C Figure 4.
In each figure, the cycle C is “augmented” by the bolded edges; that is, whenthe bolded edges are added to the cycle C , a cycle longer than C may be found. Lemma 3.1.
Let n be an integer, and let ( Z n , +) be the cyclic group of n elements. Let d ∈ [1 , n ] , and let A ⊆ Z n . Let B = ( A + d ) ∪ ( A − d ) . If | A | = | B | , then A = A + 2 d .Proof. If d = n , then 2 d = 0, and A = A + 2 d . Otherwise, we define an auxiliary bipartite graph H withpartite sets A and B . For an element a ∈ A , we let a share an edge with a + d and with a − d . By construction,each element a ∈ A is of degree 2 in H . By a handshaking argument, if | A | = | B | , then each element b ∈ B must also have degree 2 in H .We consider an element a ∈ A . The element a has a neighbor a + d ∈ B . As a + d has degree 2 in H ,it follows that a + 2 d ∈ A ; otherwise, no other element of A could be a second neighbor of a + d in H .Therefore, A = A + 2 d . (cid:3) Lemma 3.2.
Let n ≥ be an integer, and let ( Z n , +) be the cyclic group of n elements. Let k ≤ n , andlet A ⊆ Z n be a set of k odd integers. Let B = ( A + 1) ∪ ( A − . Then • | B | ≥ | A | + 1 . • If | B | = | A | + 1 , then A is a set of the form { a, a + 2 , a + 4 , . . . , a + 2( k − } for some a ∈ Z n .Proof. For each a ∈ A , clearly a + 1 ∈ A + 1, so | B | ≥ | A | . If | B | = | A | , then by Lemma 3.1, A = A + 2,from which it follows that | A | = n , a contradiction.Suppose | B | = | A | + 1. Suppose there exists a set S ( A of size t of the form { a, a + 2 , . . . , a + 2( t − } ,and suppose that for each element x ∈ ( S + 1) ∪ ( S − { x + 1 , x − } ∩ A ⊆ S . Then | ( S + 1) ∪ ( S − | = t + 1,and letting A ′ = A \ S , we have that | ( A ′ + 1) ∪ ( A ′ − | = | A ′ | . However, then it follows again from Lemma3.1 that | A ′ | = n , a contradiction. (cid:3) We now consider the graph family G . Our first aim will be to show that G contains a partial transversalisomorphic to a cycle of length 2 n −
2, which will be established by the next two claims.
Claim 3.3. G contains a partial transversal isomorphic to either a cycle of length n − or the disjointunion of a cycle of length n − and a K .Proof. By Theorem 1.3, G contains a partial transversal C isomorphic to a Hamiltonian cycle. Let C havea vertex sequence ( v , v , . . . , v n , v ) and an associated bijective function φ : E ( C ) → [2 n ]. Again, we let v , v , . . . , v n be red vertices, and we let v , v , . . . , v n − be blue vertices.Without loss of generality, we may assume that for each v i ∈ V ( C ), φ ( v i v i +1 ) = i , where v n +1 is identifiedwith v . We will again define an auxiliary digraph H on V ( C ). For each red vertex v i ∈ V ( C ), and for eachedge v i v j ∈ E ( G i ) with j = i + 1, we add the arc v i v j to H . As each red vertex of H has out-degree at least n − , it follows that some blue vertex of H has in-degree at least n − . We may assume without loss ofgenerality that d − ( v ) ≥ n − . We note that if v n − v is an arc of H , then G contains a partial transversalisomorphic to a cycle of length 2 n − v , v , v , . . . , v n − , v n − , v ). Otherwise, weassume that v n − v is not an arc of H . Similarly, we may assume that v n − v is not an arc of H .Let k be the largest integer for which v k v is an arc of H . We may assume that k is even and k ≤ n − v v k +3 ∈ E ( G ) or v v k +5 ∈ E ( G ), then G contains a partial transversal isomorphic toone of our desired graphs. Furthermore, for each other arc v j v in H , if v v j +3 ∈ E ( G ), then G containsa partial transversal isomorphic to a cycle of length 2 n −
2. As d − ( v ) ≥ n − , there exist at least n + vertices v j ∈ V ( C ) for which the edge v v j ∈ E ( G ) implies a partial transversal isomorphic to one of our esired graphs. As v has at least n + neighbors via E ( G ), it then follows that there must exist someedge v v j ∈ E ( G ) that belongs to a partial transversal in G isomorphic to a cycle of length 2 n − n − K . (cid:3) Claim 3.4. G contains a partial transversal isomorphic to a cycle of length n − .Proof. We may assume from Claim 3.3 that G contains a partial transversal isomorphic to the disjoint unionof a cycle of length 2 n − K .Let G contain a partial transversal isomorphic to the disjoint union of a cycle C of length 2 n − K ∼ = K . Let C have a vertex sequence ( v , . . . , v n − , v ). (Throughout the entire argument, we will identify v n − and v .) Let V ( K ) = { x, y } , and let X \ ( V ( C ) ∪ V ( K )) = { w, z } . Let [2 n ] \ im( φ ) = { m , m , m } .We assume without loss of generality that x and z are red vertices and that y and w are blue vertices. Case 1: n = 4In this first case, we seek a partial transversal isomorphic to a cycle of length 6. We assume without lossof generality that our red vertices are v , v , x, z ; hence, our blue vertices are v , v , y, w . For our injectivefunction φ : E ( C ) ∪ E ( K ) → [8], we assume without loss of generality that φ ( v v ) = 1, φ ( v v ) = 2, φ ( v v ) = 3, φ ( v v ) = 4, and φ ( xy ) = 5. Hence, [8] \ im( φ ) = { , , } . We seek a partial transversal over G isomorphic to a cycle of length 6.For this case, we define a wildcard edge as an edge e that belongs to E ( G i ) for at least 2 n − i ∈ [8]. The idea behind this definition is that a wildcard edge may extend any partial transversal with fewerthan six edges. We aim to show that if no partial transversal isomorphic to a 6-cycle exists in G , then eachedge in Figure 5 is a wildcard edge. We observe that x has a neighbor in { v , v } via each of G , G , G .If v y ∈ E ( G ) ∪ E ( G ) ∪ E ( G ), then G contains a partial transversal isomorphic to a 6-cycle; hence, weassume that v y E ( G ) ∪ E ( G ) ∪ E ( G ). We may similarly assume that v y E ( G ) ∪ E ( G ) ∪ E ( G ).It follows that for any edge e ∈ E ( C ), we may arbitrarily redefine φ ( e ) = m for each m ∈ { , , } whileletting φ still satisfy the conditions of a partial transversal’s injective function. Then, by redefining φ andrepeating the same argument at some edge e ∈ E ( C ) for each value m ∈ { , , } , we may conclude that eachedge in V ( C ) is a wildcard edge, along with the two edges incident to w in Figure 5. Additionally, if thereexist two distinct i, j ∈ [8] for which v z ∈ E ( G i ) and v z ∈ E ( G j ), then G contains a partial transversalisomorphic to a 6-cycle, so we may assume that yz is also a wildcard edge. We may similarly conclude that xy is a wildcard edge. Hence, every edge in Figure 5 is a wildcard edge.Now, we recall that for each i ∈ [8], x has a neighbor in { v , v } via G i , and we hence note that due toour wildcard edges, if wz ∈ E ( G j ) for any j ∈ [8], then G contains a partial transversal isomorphic to a cycleof length 6. Therefore, for each value j ∈ [8], we assume that zv , zv ∈ E ( G j ). However, then G contains apartial transversal isomorphic to a 6 cycle using the vertices x , y , z , and three vertices of V ( C ). Thus, theclaim holds in this case. Case 2: n ≥ n −
2. We let A := { v ∈ V ( C ) : vx ∈ E ( G m ) } , and we let N C ( A ) ⊆ V ( C ) denote the set of vertices adjacent to A in C . We observe that | A | ≥ n − .If | N C ( A ) | ≥ n + , then at most n − red vertices belong to V ( C ) \ N C ( A ). As y has at least n − V ( C ) via the graph G m , there must exist vertices v j , v j +1 ∈ V ( C ) such that x has a neighborin { v j , v j +1 } via the graph G m and y has a neighbor in { v j , v j +1 } via the graph G m . Then we may find apartial transversal over G isomorphic to a cycle of length 2 n − V ( C ) ∪ { x, y } .Otherwise, suppose | N C ( A ) | ≤ n . If y has at least n − neighbors in V ( C ) via G m , then by a similarargument, we may find a partial transversal over G isomorphic to a cycle of length 2 n − V ( C ) ∪ { x, y } . Hence, we assume instead that y has exactly n − V ( C ) via G m andhence that yz ∈ E ( G m ). Then we apply Lemma 3.2, which tells us that A is of the form A = { v i , v i +2 , v i +4 , . . . , v i − ⌈ n − ⌉ } , v v v wxyz Figure 5.
The figure shows the vertices of X as considered in Lemma 3.4 in the case n = 4.The aim in the proof of this case of Lemma 3.4 is to show that each edge shown in the figureis a wildcard edge —that is, an edge that belongs to at least 2 n − G . As thesewildcard edges may extend any partial transversal of length less than 6, they are useful forfinding partial transversals over G .for some v i ∈ V ( C ). Then, as n ≥
5, it follows that there exist at least | A | + 2 ≥ n + vertices in C at adistance of exactly 2 from a vertex of A . Hence, as z has at least n − neighbors in V ( C ) via the graph G m , z must have some neighbor in C via G m that is at distance exactly 2 from a vertex of A . Therefore,there exist two vertices v j , v j +2 ∈ V ( C ) such that x has a neighbor in { v j , v j +2 } via the graph G m and z has a distinct neighbor in { v j , v j +2 } via G m . Therefore, G contains a partial transversal isomorphic to acycle of length 2 n − V ( C ) \ { v j +1 } ∪ { x, y, z } . (cid:3) We have finally established that G contains a partial transversal isomorphic to a cycle of length 2 n − G does not contain a partial transversal isomorphic toa cycle of length ℓ , for some even integer 4 ≤ ℓ ≤ n −
4. Let C be a partial transversal over G isomorphicto a cycle of length 2 n − φ : E ( C ) → [2 n ]. We let C have a vertexsequence ( v , . . . , v n − , v ). We let X \ V ( C ) = { x, y } . For each i ∈ [2 n − φ ( v i v i +1 ) = i , with v n − and v identified. Then it follows that [2 n ] \ im( φ ) = { n − , n } .We again let our red vertices be x, v , v , . . . , v n − , and we let our blue vertices be y, v , v , . . . , v n − .For the sake of convenience, we will endow V ( C ) with some group operations. For a vertex v i ∈ V ( C )and an integer k , we define v i + k = v i + k , again with v n − “overflowing” to v . By this definition, V ( C ) isisomorphic as a group to ( Z n − , +). For a set A ⊆ V ( C ) and an integer k , we define A + k = { v + k : v ∈ A } .We make several observations. Claim 3.5.
For each v j ∈ V ( C ) , v j v j +1 ∈ E ( G n − ) or v j v j +1 ∈ E ( G n ) .Proof. To prove this claim, we will apply an idea of Bondy [4] which is commonly used in proving pancyclicityresults (c.f. [5], [15]). Suppose that v j v j +1 E ( G n − ) and v j v j +1 E ( G n ). For edges in C incident to v j and v j +1 , we will create the following pairings for each k ∈ [2 n − \ { j + 1 } of parity opposite j : • If j
6∈ { k − ℓ + 3 , k − ℓ + 4 , . . . , k − , k − } , then we pair v j v k with v j +1 v k − ℓ +3 . • If j ∈ { k − ℓ + 3 , k − ℓ + 4 , . . . , k − , k − } , then we pair v j v k with v j +1 v k − ℓ +1 .It is clear from Figure 6 that if G does not contain a partial transversal isomorphic to a cycle of length ℓ ,then for any pair ( a, b ) given above, if a ∈ E ( G n − ), then b E ( G n ). One may also check that each edgeincident to v j is paired with exactly one edge incident to v j +1 , and one may also check that no edge v j v k ispaired with v j +1 v j . Therefore, for the pair v j , v j +1 , | N G n − ( v j ) ∩ V ( C ) | + | N G n ( v j +1 ) ∩ V ( C ) | ≤ n − . Then, as each of v j , v j +1 has at most one neighbor outside of V ( C ) via any graph, it follows that d G n − ( v j )+ d G n ( v j +1 ) ≤ n , a contradiction. (cid:3) Using Claim 3.5, for any edge v i v i +1 ∈ E ( C ) for which φ ( v i v i +1 ) = i , we may redefine φ so that φ ( v i v i +1 ) ∈{ n − , n } . As a result, we may define our injective function φ so that φ “misses” any given value i ∈ [2 n ]. k − ℓ +3 ℓ − v k v j v j +1 k − j − j − k + ℓ − v k − ℓ +1 v k v j v j +1 Figure 6.
In both figures, the circle represents a partial transversal C isomorphic to acycle of length 2 n − φ satisfying [2 n ] \ im( φ ) = { n − , n } .In both figures, if one dashed edge belongs to E ( G n − ) and the other belongs to E ( G n ),then G contains a partial transversal isomorphic to a cycle of length ℓ containing the boldededges and the dashed edges. The number of edges represented by each bolded arc, whenmore than one, is indicated.This observation will help us prove our next claim, which shows that xy is a wildcard edge , where an edge e is said this time to be a wildcard edge if e ∈ E ( G i ) for all i ∈ [2 n ]. Claim 3.6.
For every graph G i ∈ G , xy ∈ E ( G i ) .Proof. Recall that x is a red vertex. First, we note that by Claim 3.5, given any graph G i ∈ G , we mayfind a partial transversal over G on E ( C ) with an associated injective function ψ such that i ∈ [2 n ] \ im( ψ ).Therefore, it suffices to consider our original partial transversal C and associated function φ and to showthat xy ∈ E ( G n − ) and xy ∈ E ( G n ).Suppose for the sake of contradiction that xy E ( G n ). Then all neighbors of x via G n belong to V ( C ).Let A denote the set of neighbors of x via the graph G n . In particular, the set B = A + ( ℓ −
2) has atleast n + elements. As x has at least n − neighbors in V ( C ) via the graph G n − , and as at most( n − − ( n + ) = n − blue vertices of V ( C ) do not belong to B , it follows that some neighbor of x via G n − belongs to B . Then it follows that G in fact contains a partial transversal isomorphic to a cycleof length ℓ . Therefore, we may assume that xy ∈ E ( G n ), and by a similar argument, we may assume that xy ∈ E ( G n − ). (cid:3) Claim 3.6 shows us that xy is a wildcard edge in G , which gives us the last piece that we need to finishour proof. Let A ⊆ V ( C ) be the set of vertices in C adjacent to x via G n − . Clearly | A | ≥ n − . Let B = ( A + ( ℓ − ∪ ( A − ( ℓ − | B | ≥ n + , then by the argument in Claim 3.6, x has a neighbor in B via the graph G n , and then G contains a partial transversal isomorphic to a cycle of length ℓ . Otherwise, | A | = | B | ≤ n , and then by Lemma 3.1, A = A + (2 ℓ − B ′ = ( A + ( ℓ − ∪ ( A − ( ℓ − xy is an edge of all graphs in G , if y has a neighbor in B ′ by G n , then G has a partial transversal isomorphic to a cycle of length ℓ . If | B ′ | ≥ n + , then y has aneighbor in B ′ via G n , and we find our partial transversal in G of length ℓ . Otherwise, | A | = | B ′ | ≤ n , andby Lemma 3.1, A = A + (2 l − A = A + (2 ℓ − A = A + (2 ℓ − A = A + 2. Then it follows that | A | = | B | = n −
1, and x has a neighbor in B via G n . Hence, in all cases, G contains a partial transversalisomorphic to a cycle of length ℓ . This completes the proof. (cid:3) Perfect matchings: Proof of Theorem 1.5
This section will be dedicated to the proof of Theorem 1.5. In this section, we define X as a set of n bluevertices { p , . . . , p n } and n red vertices { q , . . . , q n } . This time, we assume G = { G , . . . , G n } is a set of n ipartite graphs on the vertex set X , such that for each G i ∈ G , the red vertices of X make up one colorclass of G i , and the blue vertices of X make up the other color class of G i . We will show that under theconditions of Theorem 1.5, G contains a transversal isomorphic to a perfect matching.Supposing that Theorem 1.5 does not hold, we consider an edge-maximal counterexample G to the theorem.That is, we let G be a family of graphs such that after adding any edge to any graph G i ∈ G , the resultingfamily contains a perfect matching transversal. If each graph G i ∈ G is a complete bipartite graph, then G certainly contains a perfect matching transversal. Therefore, we may assume without loss of generality that G n is not a complete bipartite graph and that we may add some edge e to G n with a red endpoint and ablue endpoint. As G is an edge-maximal counterexample, G + e = { G , G , . . . , G n + e } contains a perfectmatching transversal M with an associated bijective function φ : E ( M ) → [ n ]. We may assume withoutloss of generality that e = p n q n , and that M consists of edges p q , p q , . . . , p n q n , and that for each edge p i q i , φ ( p i q i ) = i . It follows that G contains a partial transversal p q , . . . , p n − q n − for which each edge p i q i (1 ≤ i ≤ n −
1) belongs to E ( G i ).The remainder of the proof will rely heavily on ideas of Joos and Kim from [11]. We define an auxiliarydigraph H for which V ( H ) = X . For each value i , 1 ≤ i ≤ n −
1, and for each edge e ∈ E ( G i ) incident to q i , e = p i q i , we add an arc e to H directed away from q i . We claim that in H , d − ( p n ) ≤ n − . Indeed,suppose that d − ( p n ) ≥ n −
1. We have assumed that p n q n E ( G n ), and hence we may assume that q n has at least n + neighbors among p , . . . , p n − via G n . As p n has at least n − H among q , . . . , q n − , and as ( n + ) + ( n − > n −
1, it follows from the pigeonhole principle that there exists apair p i , q i such that p i q n is an arc of H and p i q n ∈ E ( G n ). This implies that a perfect matching transversalexists over G containing the edges p i q n and p n q i . Therefore, we assume that d − ( p n ) ≤ n − .We first consider the case that n is even. In this case, the total number of arcs in H not ending at p n isat least ( n − n ) − ( n − > ( n − n − p i , 1 ≤ i ≤ n −
1, for which d − ( p i ) ≥ n . As q n has at least n + 1 neighbors among p , . . . , p n − via G n , and as p n has at least n − q , . . . , q n − via G i , there exists a pair p j , q j for which q j p n ∈ E ( G i ) and p j q n ∈ E ( G n ).We may assume that i = j , as otherwise there exists a perfect matching transversal over G containing p j q n and q j p n . Finally, as we may assume that p i q i E ( G j ), we see that q i has at least n neighbors among p , . . . , p i − , p i +1 , . . . , p n − via G j , and p i has at least n in-neighbors in H among q , . . . , q i − , q i +1 , . . . , q n − .Therefore, we may choose a pair p k , q k , k = j for which p i q k is an arc of H and p k q i ∈ E ( G j ). Then itfollows that there exists a perfect matching transversal over G containing the edges p k q i and q k p i , and theproof is complete.Next, we consider the case that n is odd. In this case, the total number of arcs in H not ending at p n isat least ( n − n − ) − ( n − ) = n − n + 2. We further consider two cases. Case 1:
There exists a vertex q i , 1 ≤ i ≤ n −
1, with in-degree at least n + .As q n has at least n + neighbors among p , . . . , p n − via G n , and as p n has at least n − neighbors among q , . . . , q n − via G i , there exists a pair p j , q j for which p j q n ∈ E ( G n ) and q j p n ∈ E ( G i ). We may assume that i = j , as otherwise there exists a perfect matching transversal over G containing p j q n and q j p n . Finally, as wemay assume that p i q i E ( G j ), we see that q i has at least n − neighbors among p , . . . , p i − , p i +1 , . . . , p n − via G j , and p i has at least n + in-neighbors in H among q , . . . , q i − , q i +1 , . . . , q n − . Therefore, we maychoose a pair p k , q k , k = j for which p i q k is an arc of H and p k q i ∈ E ( G j ). Then it follows that there existsa perfect matching transversal over G containing the edges p k q i and q k p i , and the proof is complete. Case 2:
There exists no vertex p i , 1 ≤ i ≤ n −
1, with in-degree at least n + .In this case, we let a denote the number of vertices among p , . . . , p n − with in-degree n − . By countingthe arcs of H not ending at p n , we see that a ( n −
12 ) + ( n − − a )( n −
32 ) ≥ n − n + 2 , from which it follows that a ≥ n + . As p n has at least n + neighbors among q , . . . , q n − , there mustexist a pair p j , q j for which p n q j ∈ E ( G n ) and d − ( p j ) = n − . Finally, we may assume that p j q n E ( G j ),as otherwise, there would exist a perfect matching transversal over G containing p j q n and q j p n . Therefore,as p j has n − in-neighbors among q , . . . , q j − , q j +1 , . . . , q n − , and as q n has at least n − neighbors mong p , . . . , p j − , p j +1 , . . . , p n − via G j , there must exist a pair p k , q k for which q k p j is an arc of H and p k q n ∈ E ( G j ). Then it follows that there exists a perfect matching transversal over G containing the edges p j q k and p k q n , and the proof is complete. (cid:3) We give a construction to show that the degree condition of Theorem 1.5 is in some sense best possible.First, we consider the case that n is even, and we show that a minimum degree of n for all vertices does notimply the existence of a perfect matching transversal. We partition our red vertices of X into two equallysized parts A, B , and we partition our blue vertices of X into two equally sized parts C, D . For 1 ≤ i ≤ n − G i be the union of the complete graphs on the pairs ( A, C ) and (
B, D ). We let G n be the union ofthe complete graphs on the pairs ( A, D ) and (
B, C ). We let G = { G , . . . , G n } , and we see that for each G i ∈ G , δ ( G i ) = n .Suppose that G contains a perfect matching transversal M . One of the pairs ( A, C ) or (
B, D ) mustcontain n edges of M from the graphs G , . . . , G n − . This implies, however, that every edge of E ( G n ) mustbe incident to an edge e from a graph G i , 1 ≤ i ≤ n −
1, for which e belongs to M . It follows that M is nota perfect matching, giving a contradiction.In the case that n is odd, we may use a similar construction in which | A | = | C | = n + and | B | = | D | = n − to show that a minimum degree of n − is not sufficient to guarantee a perfect matching transversal.Thus the minimum degree condition of Theorem 1.5 is in a way best possible.5. Conclusion
The condition that we give in Theorem 1.3 for a Hamiltonian transversal is a bipartite analogue to a resultof Joos and Kim [11], which generalizes Dirac’s theorem into the language of graph transversals. Similarly,the condition given in Theorem 1.4 for bipancyclicity with respect to partial transversals is a bipartiteanalogue to a result of Cheng, Wang and Zhao [5], which asymptotically generalizes a pancyclic version ofDirac’s theorem into the language of transversals. It is well known, however, that Dirac’s condition for bothHamiltonicity and pancyclicity can be extended in many different ways to more general conditions on vertexdegree sums and degree sequences (see, for example, [13], [9], [4]). Similarly, for bipartite graphs, Theorems1.1 and 1.2 have extensions to more general conditions for vertex degree sums and degree sequences (see, forexample, [12], [6], [10]).A question that remains open is whether these more general conditions for Hamiltonicity and (bi)pancyclicitycan be translated into the language of graph transversals. For example, is there a theorem for graph transver-sals that directly generalizes Ore’s theorem [13] or Fan’s condition [9] for Hamiltonicity? While such theoremsfor graph transversals may exist, their proofs seem to be much more difficult than their traditional counter-parts. 6.
Acknowledgements
I am grateful to Ladislav Stacho for making a suggestion to consider problems of pancyclicity in thesetting of graph transversals and for carefully reading this paper. I am also grateful to Kevin Halasz forexplaining key concepts and tools used in the study of graph transversals and for many helpful discussions.Additionally, I am grateful to Jaehoon Kim for pointing out the connection between Theorem 1.5 and themain result from [2], which ultimately led to a slight strengthening of the main three theorems of this paper.Finally, I am grateful to the referees who offered many helpful suggestions that improved the overall qualityof this paper.
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Department of Mathematics, Simon Fraser University, Vancouver, Canada
Email address : [email protected]@sfu.ca