Quasi-Eulerian Hypergraphs

We generalize the notion of an Euler tour in a graph in the following way. An Euler family in a hypergraph is a family of closed walks that jointly traverse each edge of the hypergraph exactly once. An Euler tour thus corresponds to an Euler family with a single component. We provide necessary and sufficient conditions for the existence of an Euler family in an arbitrary hypergraph, and in particular, we show that every 3-uniform hypergraph without cut edges admits an Euler family. Finally, we show that the problem of existence of an Euler family is polynomial on the class of all hypergraphs. This work complements existing results on rank-1 universal cycles and 1-overlap cycles in triple systems, as well as recent results by Lonc and Naroski, who showed that the problem of existence of an Euler tour in a hypergraph is NP-complete.


Introduction
As first claimed by Euler in 1741 [6], and proved by Hierholzer and Wiener in 1873 [9], it is now well known that a connected graph admits an Euler tour -that is, a closed walk traversing each edge exactly once -if and only if it has no vertices of odd degree.In this paper, we are concerned with the analogous problem for hypergraphs.As we shall see, this extended problem is much more complex; in fact, there is more than one natural way to generalize the notion of an Euler tour to hypergraphs.We shall consider three such natural extensions, one being the main focus.
An Euler tour of a hypergraph H is a closed walk traversing every edge of H exactly once, and a hypergraph is called eulerian if it admits and Euler tour.Not much has been previously known about eulerian properties of hypergraphs.The most in-depth treatment to date can be found in [12], where Euler tours in k-uniform hypergraphs are considered.In particular, the authors of [12] determine some necessary conditions for the existence of an Euler tour in a hypergraph, and show that these are also sufficient for certain classes of k-uniform hypergraphs.They also show that the problem of determining whether or not a given hypergraph is eulerian is NP-complete on the class of k-uniform hypergraphs (for each k 3), and even just on the class of 3-uniform hypergraphs with a connected "skeleton" [12,Theorem 7].Thus, there is little hope to characterize eulerian (uniform) hypergraphs in general.
The results of this paper will also be of interest to those who study universal cycles for set systems.This concept was introduced by Chung, Diaconis, and Graham [4], and generalizes de Bruijn cycles to various combinatorial structures.In this context, an Euler tour in a 3-uniform hypergraph has also been called a rank-1 universal cycle [5] and 1-overlap cycle [11].Thus, the results of [5,11] show that certain classes of triple systems are eulerian.In a recent work, Wagner and the second author applied our results (Theorem 5.2) to show that in fact all triple systems with at least two blocks are eulerian.
We generalize the notion of an Euler tour as follows.An Euler family of a hypergraph H is a family of closed walks of H such that each edge of H lies in exactly one component (member) of the family.A hypergraph is called quasi-eulerian if it admits an Euler family.Clearly, every eulerian hypergraph is quasi-eulerian, but the converse does not hold.
The main contributions of this paper are as follows.In Sections 3 and 4, respectively, we present some necessary and some sufficient conditions for a hypergraph to be eulerian or quasi-eulerian.In Section 5, we show that every 3-uniform hypergraph without cut edges is quasi-eulerian.In Section 6, we exhibit a close relationship between the rich theory of factors in graphs and quasi-eulerian hypergraphs, and give necessary and sufficient conditions for the existence of an Euler family in an arbitrary hypergraph.Finally, in Section 7 we show that the problem of existence of an Euler family is polynomial on the class of all hypergraphs.We conclude the paper with a brief discussion of closely related concepts such as flag-traversing tours and cycle decompositions of hypergraphs.

Preliminaries
For basic hypergraph terminology, we refer the reader to [2] or [3]; the latter being a longer version of the present article.
A hypergraph H = (V, E) is called trivial if it has only one vertex.The degree of a vertex v ∈ V (denoted by deg H (v) or simply deg(v) if no ambiguity can arise) is the number of edges e ∈ E such that v ∈ e.A vertex of degree 0 is called isolated, and a vertex of degree 1 is called pendant.A hypergraph H = (V, E) is called even if all of its vertices are of even degree, and r-regular on a set E) is an alternating sequence of vertices and edges with {v i−1 , v i } ⊆ e i and v i−1 = v i for all i = 1, . . ., k.Such a walk W is called a strict trail if e 1 , . . ., e k are pairwise distinct, and closed if k 2 and v 0 = v k .Thus, each component in an Euler family is a closed strict trail.A closed strict trail Let c(H) denote the number of connected components of a hypergraph H = (V, E).A cut edge of H is an edge e ∈ E such that c(H − e) > c(H).A graph with a cut edge obviously does not admit an Euler tour.The issue is more complex for hypergraphs.First, we need to distinguish between two types of cut edges.If e is a cut edge in a hypergraph H = (V, E), then c(H − e) c(H) + |e| − 1 [2, Lemma 3.15].A cut edge that achieves the upper bound in this inequality is called strong; all other cut edges are called weak.Observe that a cut edge has cardinality at least two, and that any cut edge of cardinality two (and hence any cut edge in a graph) is necessarily strong.
The incidence graph I(H) of a hypergraph H = (V, E) is a bipartite simple graph with bipartition {V, E} and edge set {ve : v ∈ V, e ∈ E, v ∈ e}.The elements of V and E are called the v-vertices and e-vertices, respectively.
The following lemma will be one of our main tools in the study of quasi-eulerian hypergraphs.We leave the proof to the reader.Lemma 1.Let H = (V, E) be a hypergraph and G its incidence graph.Then H is quasi-eulerian if and only if G has an even subgraph that is 2-regular on E.Moreover, H is eulerian if and only if G has an even subgraph with a single non-trivial connected component that is 2-regular on E.

Necessary Conditions
It is easy to see that a quasi-eulerian hypergraph has at least two edges.Here is a more interesting result.
Lemma 2. Let H = (V, E) be a quasi-eulerian hypergraph, and let V odd be the set of odd-degree vertices in H. Then the following hold.To prove (d), suppose to the contrary that e is a strong cut edge of H. Since H admits an Euler family, the edge e lies in a closed strict trail, and consequently in a cycle of H.However, by [2, Theorem 3.18], no strong cut edge lies in a cycle -a contradiction.
It is easy to see that for graphs, the necessary conditions in Lemma 2 are also sufficient (Note that some of these conditions alone are not sufficient.In fact, conditions (b) and (c) alone are sufficient, but conditions (a) and (d) alone are not).This is not the case for general hypergraphs.Figures (A) and (B) show the incidence graphs of hypergraphs that satisfy conditions (a)-(d) of Lemma 2, but are not quasi-eulerian.(Note that in all figures, black dots represent the v-vertices.)Observe that the hypergraph on the left (A) has only weak cut edges, while the hypergraph on the right (B) has no cut edges.
Very recently, Stamplecoskie [15] showed that for every c 2 and m 5, there exists a connected hypergraph with m edges, each of size at least c, that has no cut edges, satisfies the necessary conditions from Lemma 2, but is not quasi-eulerian.In Theorem 5.2 we shall see that every 3-uniform hypergraph without cut edges is quasi-eulerian.However, Figure (C) shows that not all 3-uniform hypergraphs without cut edges are eulerian.Proof.Suppose, to the contrary, that H − e has at least two non-trivial connected components.Since e is a cut edge of H, it is a cut e-vertex in its incidence graph G = I(H) [2, Theorem 3.23], and the connected components of the vertex-deleted graph G\e are the incidence graphs of the connected components of H − e [2, Lemma 2.8 and Corollary 3.12].Hence, by assumption, G\e has at least two connected components with e-vertices.Since H has an Euler tour, G has a closed trail T traversing each e-vertex (including e) exactly once.Hence T \e is a trail that traverses every e-vertex in G\e, contradicting the above.
We conclude this section with the following open question.Question 3.2.Does there exist a connected k-uniform hypergraph with k 4 that has no cut edges and that satisfies the necessary conditions from Lemma 2 but is not quasi-eulerian?Using the techniques of [12,Theorem 3] in the proof of Case k > 3 of Theorem 4.1, we extend this result to quasi-eulerian hypergraphs, as well as a to a larger family of eulerian (not necessarily uniform) hypergraphs; see Theorems 4.2 and 4.3 below, respectively.

Sufficient Conditions
For any hypergraph H = (V, E), define a digraph D 3 (H) as follows: its vertex set is E and its arc set is Recall that an arborescence is a directed graph whose underlying undirected graph is a tree, and whose arcs are all directed towards a root.Theorem 4.2.Let H = (V, E) be a hypergraph such that its digraph D 3 (H) has a spanning subdigraph that is a vertex-disjoint union of non-trivial arborescences.Then H is quasieulerian.
Proof.For convenience, we say that a digraph satisfies Property P if has a spanning subdigraph that is a vertex-disjoint union of non-trivial arborescences.We shall prove by induction on the number of edges that every hypergraph H whose digraph D 3 (H) satisfies Property P possesses an Euler family.Observe that such a hypergraph necessarily has at least two edges.
First, let H = (V, E) be a hypergraph with E = {e, f } such that its digraph D 3 (H) satisfies Property P. Then D 3 (H) must have a spanning arborescence.Moreover, we have that |e ∩ f | 3. Take any u, v ∈ e ∩ f such that u = v.Then T = uevf u is an Euler tour of H. Thus H possesses an Euler family as claimed.
Assume that for some m 2, every hypergraph H with at least m edges whose digraph D 3 (H) satisfies Property P possesses an Euler family.Let H = (V, E) be a hypergraph with |E| = m + 1 such that its digraph D 3 (H) has a spanning subdigraph D that is a vertex-disjoint union of non-trivial arborescences.If each arborescence in D is of order the electronic journal of combinatorics 24(3) (2017), #P3.30 2, then (just as in the base case above) each gives rise to a closed strict trail of length 2 in H, and the union of all these strict trails is an Euler family in H.
Hence assume that D has a weakly connected component A that is an arborescence of order at least 3. Let e ∈ E be a leaf (that is, vertex of indegree 0 and outdegree 1) of A and f its outneighbour in A. Then |f − e| = 1 and |e ∩ f | 3. Now D 3 (H − e) has a spanning digraph D \ e that is a vertex-disjoint union of non-trivial arborescences.Hence by the induction hypothesis, the hypergraph H − e possesses an Euler family F. Let T = uf vW -where u and v are distinct vertices in f , and W is an appropriate (v, u)-walk -be a closed strict trail in F. We now reroute T to include the edge e, resulting in a closed strict trail T of H, as follows.
Since |f − e| = 1, at least one of u and v -say v without loss of generality -is also in e, and since |e ∩ f | 3, there exists w ∈ e ∩ f such that w = u, v. Then T = uf wevW is a closed strict trail of H. Finally, replace T in F by T to obtain an Euler family of H.
The result follows by induction.
With very minor changes to the above proof we obtain the following.
Theorem 4.3.Let H = (V, E) be a hypergraph such that its digraph D 3 (H) has a nontrivial spanning arborescence.Then H is eulerian.
Using Lemma 1, certain families of hypergraphs can be easily seen to be eulerian or quasi-eulerian.The first of the following corollaries is immediate.Proof.The incidence graph G of H is an r-regular bipartite graph with r 2. Therefore, as a corollary of Hall's Theorem [8], G admits two edge-disjoint perfect matchings, and hence a 2-factor.Thus H is quasi-eulerian by Corollary 4.4.
Proof.Let G be the incidence graph of H.In G, every e-vertex has degree 2k, and every v-vertex has even degree.A result by Hilton [10,Theorem 8] then shows that G has an evenly equitable k-edge colouring; that is, a k-edge colouring such that (i) every vertex is incident with an even number of edges of each colour, and (ii) for each vertex, the numbers of edges of any two colours that are incident with this vertex differ by at most two.Hence the i-th colour class, for i = 1, 2, . . ., k, induces an even subgraph G i of G that is 2-regular on E. By Lemma 1, each G i corresponds to an Euler family F i of H, and H is quasi-eulerian.

3-uniform Hypergraphs
Recall from Lemma 2 that a quasi-eulerian hypergraph cannot have strong cut edges, while a hypergraph with weak cut edges may or may not be quasi-eulerian.The main result of this section (Theorem 5.2 below) completes the picture for 3-uniform hypergraphs.The key ingredient in the proof is the following result by Fleischner.
Theorem 5.1.[7] Every graph without cut edges and of minimum degree at least 3 has a spanning even subgraph without isolated vertices.
Proof.First, we show that since H has no cut edges, its incidence graph G = I(H) has no cut edges either.Suppose, to the contrary, that ve is a cut edge of G (where v ∈ V and e ∈ E), and let G v and G e be the connected components of G − ve containing vertex v and e, respectively.Since |e| > 1, the component G e must contain a v-vertex w, and v and w are disconnected in G − ve.Hence they are disconnected in H − e, showing that e is a cut edge of H, a contradiction.Therefore G has no cut edges as claimed.
Note that we may assume that H, and hence G, has no isolated vertices.Clearly, G has no vertices of degree 1, since the edge incident with such a vertex would necessarily be a cut edge.Suppose G has a vertex of degree 2. Then it must be a v-vertex, since |e| = 3 for all e ∈ E. Obtain a graph G * from G by replacing, for every vertex v of degree 2, the 2-path e 1 ve 2 in G with an edge e 1 e 2 .Observe that in G * , all e-vertices have degree 3, and all v-vertices have degree at least 3.Moreover, G * has no cut edges since G does not.Therefore, by Theorem 5.1, G * has a spanning even subgraph G * 1 without isolated vertices.We construct a subgraph G 1 of G as follows: for any vertex v of degree 2 in G, and its incident edges e 1 and e 2 , if e 1 e 2 is an edge of G * 1 , then replace it with the 2-path e 1 ve 2 .The resulting graph G 1 is an even subgraph of G without isolated e-vertices.Since every e-vertex of G has degree 3 in G, it has degree 2 in G 1 .Thus, by Lemma 1, G 1 gives rise to an Euler family of H.
The reader may have noticed that if H has no isolated vertices, then the Euler family of H constructed in the proof of Theorem 5.2 traverses every vertex of H except possibly some of the vertices of degree 2. Observe also that Theorem 5.2 does not generally hold for graphs -for example, a cycle with a chord -nor for hypergraphs in which every edge has size 2 or 3 -an example is given in Figure (D).
the electronic journal of combinatorics 24(3) (2017), #P3.30Corollary 5.3.Let H = (V, E) be a 3-uniform hypergraph with at least two edges such that each pair of vertices lie together in at least one edge.Then H is quasi-eulerian.Proof.By Theorem 5.2, it suffices to show that H has no cut edges.If |V | = 3, then clearly none of the edges are cut edges.Hence assume |V | 4, and suppose that e ∈ E is a cut edge of H. Let u 1 , u 2 ∈ V be vertices of e that lie in distinct connected components of H − e, and consider any vertex w ∈ e.Then there exist edges e 1 , e 2 such that w, u i ∈ e i for i = 1, 2. Since obviously e 1 , e 2 = e, vertex w must lie in the same connected component of H − e as both u 1 and u 2 , a contradiction.
Recall that a triple system TS(n,λ) is a 3-uniform hypegraph of order n such that every pair of vertices lie together in exactly λ edges.
Very recently, applying the results of this section, Wagner and the second author [14] proved that all triple systems TS(n,λ), except for TS (3,1), are in fact eulerian, thus extending various results on rank-1 universal cycles [5] and 1-overlap cycles [11].
The proof of Theorem 4.1 [12,Theorem 2] for the case k = 3 is very long and technical; as another corollary of our Theorem 5.2, we show that every 3-uniform hypergraph with a connected 2-intersection graph is eulerian provided that it has no pendant vertices.Corollary 5.5.Every 3-uniform hypergraph with a connected 2-intersection graph and without pendant vertices is eulerian.
Proof.Let H = (V, E) be a 3-uniform hypergraph with a connected 2-intersection graph L and without pendant vertices.Hence H has at least 2 edges.Suppose it has a cut edge e.Since L is connected, e shares exactly two of its vertices with another edge; consequently, these two vertices lie in the same connected component of H −e. Thus H −e has exactly two connected components; let H 1 be the connected component containing a single vertex, w, of e.By assumption, w is not a pendant vertex in H, so E(H 1 ) = ∅.Take any e 1 ∈ E(H 1 ) and any e 2 ∈ E − E(H 1 ).Then e 1 ∩ e 2 ⊆ {w}, whence e 1 e 2 ∈ E(L).It follows that L is disconnected, a contradiction.
We conclude that H has no cut edges, and hence is quasi-eulerian by Theorem 5.2.Let F = {T 1 , . . ., T k } be an Euler family of H with a minimum number of components, and suppose k 2. Let G be the incidence graph of H, let G be the subgraph of G corresponding to F, and G 1 , . . ., G k the connected components of G corresponding to the closed strict trails T 1 , . . ., T k of H. Since L is connected, without loss of generality, there exist e-vertices e 1 of G 1 and e 2 of G 2 that are adjacent in L, and hence in G have two common neighbours, say v 1 and v 2 .Since e 1 and e 2 are of degree 3 in G, and of degree 2 in G , each is adjacent to at least one of v 1 and v 2 in G , and since they lie in distinct connected components of G , we may assume without loss of generality that v 1 e 1 , v 2 e 2 ∈ E(G ).Obtain G by replacing these two edges of G with edges v 1 e 2 and v 2 e 2 .Then G is an even subgraph of G that is 2-regular on E, so it corresponds to an Euler family of H.But since G has one fewer connected component than G , it contradicts the minimality of F.
We conclude that k = 1, that is, H admits an Euler tour.
the electronic journal of combinatorics 24(3) (2017), #P3.30 6 Necessary and Sufficient Conditions via (g, f )-factors Recall that by Lemma 1, a hypergraph H = (V, E) admits an Euler family if and only if its incidence graph I(H) has an even subgraph that is 2-regular on E. We shall now combine this observation with Lovasz's (g, f )-factor Theorem [13] to give more easily verifiable necessary and sufficient conditions.For a graph G and functions f, g : ) denote the number of edges of G 1 with one end in V 1 and the other in V 2 .Theorem 6.1.[13] Let G be a graph and let f, g : V (G) → N be functions such that g(x) f (x) and g(x) ≡ f (x) (mod 2) for all x ∈ V (G).Then G has a (g, f )-factor F such that deg F (x) ≡ f (x) (mod 2) for all x ∈ V (G) if and only if all disjoint subsets S and T of V (G) satisfy where q(S, T ) is the number of connected components C of G\(S∪T ) such that ε G (V (C), T ) is odd.
Proof.By Lemma 1, H has an Euler family if and only if G has an even subgraph G that is 2-regular on E. Define functions f, g : V ∪ E → N as follows: where K is a sufficiently large even integer.Observe that f and g satisfy the assumptions of Theorem 6.1.Moreover, a subgraph G of G with the required properties is a (g, f )factor F of G with deg F (x) ≡ f (x) (mod 2) for all x ∈ V (G), and conversely.For any subsets S and T of V ∪ E, if S ∩ V = ∅, then x∈S f (x) is very large, and Condition (1) clearly holds for S and T .Thus Theorem 6.1 asserts that G has an (f, g)-factor if and only if Condition (1) holds for all disjoint sets S ⊆ E and T ⊆ V ∪ E of V (G).
Observing that x∈V (C) f (x)+ε G (V (C), T ) ≡ ε G (V (C), T ) (mod 2), it is then straightforward to show that Condition (1) in Theorem 6.1 is equivalent to Condition (2) in the statement of this corollary.The result follows as claimed.
(a) Each edge of H contains at least two non-pendant vertices.(b) |E| v∈V deg H (v)/2 .(c) |V odd | e∈E (|e| − 2).(d) H has no strong cut edges.Moreover, statements (b) and (c) are equivalent.Proof.Statement (a) follows immediately from the definition of an Euler family.Statements (b) and (c), and their equivalence, are straightforward extensions of a result by Lonc and Naroski [12, Proposition 1 ].
Of course, for a hypergraph to be eulerian, conditions (a)-(d) of Lemma 2 are also necessary.The following result elaborates on (d).Proposition 3.1.Let H = (V, E) be an eulerian hypergraph with a cut edge e.Then H − e has at most one non-trivial connected component.

For
any positive integer , we define the -intersection graph L (H) of the hypergraph H = (V, E) as the graph with vertex set E and edge set {ee : e, e ∈ E, e = e , |e∩e | = }.Lonc and Naroski [12, Theorem 2] showed the following.

Corollary 4 . 4 .
Let H be a hypergraph with the incidence graph G.If G has a 2-factor, then H is quasi-eulerian.If G is hamiltonian, then H is eulerian.Corollary 4.5.Let H be an r-regular r-uniform hypergraph for r 2. Then H is quasieulerian.