On upper transversals in 3-uniform hypergraphs

A set S of vertices in a hypergraph H is a transversal if it has a nonempty intersection with every edge of H. The upper transversal number Υ(H) of H is the maximum cardinality of a minimal transversal in H. We show that if H is a connected 3-uniform hypergraph of order n, then Υ(H) > 1.4855 3 √ n − 2. For n sufficiently large, we construct infinitely many connected 3-uniform hypergraphs, H, of order n satisfying Υ(H) < 2.5199 3 √ n. We conjecture that sup n→∞ ( inf Υ(H) 3 √ n ) = 3 √ 16, where the infimum is taken over all connected 3-uniform hypergraphs H of order n. Mathematics Subject Classifications: 05C88, 05C89


Introduction
In this paper, we continue the study of transversals in hypergraphs.Hypergraphs are systems of sets which are conceived as natural extensions of graphs.A hypergraph H = (V (H), E(H)) is a finite set V (H) of elements, called vertices, together with a finite multiset E(H) of nonempty subsets of V (H), called hyperedges or simply edges.A k-edge in H is an edge of size k.The hypergraph H is k-uniform if every edge of H is a k-edge.Every 2-uniform hypergraph is a graph.Thus graphs are special hypergraphs.The degree of a vertex v in H, denoted by d H (v), is the number of edges of H which contain v.The minimum and maximum degrees among the vertices of H is denoted by δ(H) and ∆(H), respectively.
A subset T of vertices in a hypergraph H is a transversal (also called hitting set or vertex cover or blocking set in many papers) if T has a nonempty intersection with every edge of H.A vertex hits or covers an edge if it belongs to that edge.The transversal number τ (H) of H is the minimum size of a transversal in H, while the upper transversal number Υ(H) of H is the maximum cardinality of a minimal transversal in H.In hypergraph theory the concept of transversal is fundamental and well studied.The major monograph [1] of hypergraph theory gives a detailed introduction to this topic.Transversals in hypergraphs are well studied in the literature (see, for example, [3,4,11,12,13] for recent results and further references).
A set S of vertices in a graph G is a dominating set of G if each vertex in V (G) \ S has a neighbor in S. A set is independent if no two vertices in it are adjacent.An independent dominating set of G is a set that is both dominating and independent in G.
The independent domination number of G, denoted by i(G), is the minimum cardinality of an independent dominating set.Domination is well studied in graph theory and we refer the reader to the monographs [9,10] which detail and survey many results on the topic.A survey of known results on independent domination in graphs can be found in [8].

Main Results
We have two immediate aims in this paper.First to provide a sharp lower bound on the upper transversal number of graphs.Secondly to present a lower bound on the upper transversal number of 3-uniform hypergraphs, and to show that this bound is a sense asymptotically best possible.More precisely, we prove the following results, where we use the notation n H = |V (H)| to denote the order of H. Proofs of Theorem 1 and Theorem 2 are given in Section 3 and Section 4, respectively.Theorem 1.If H is a connected graph with δ(H) δ, then and this bound is sharp.
Further, there exist infinitely many connected 3-uniform hypergraphs H of sufficiently large order n H satisfying

Proof of Theorem 1
Recall that a transversal in a graph is a set of vertices covering all the edges of the graph, where a vertex covers an edge if it is incident with it.Theorem 1 can be restated as follows.
Theorem 1.If G is a connected graph of order n with δ(G) δ, then Υ(G) 2 √ δn−2δ, and this bound is sharp.
In order to prove Theorem 1, we first establish a relationship between the upper transversal number and independent domination number of a graph.
Proof.Let G be an isolate-free graph.Let S be an independent dominating set in G of minimum cardinality, and so |S| = i(G).Let T = V (G) \ S and note that T is a transversal in G as S is an independent set.Since every vertex in T has a neighbor in S, we furthermore note that T is a minimal transversal, which implies that Υ(G) Conversely, let T be a minimal transversal in G of maximum cardinality, and so |T | = Υ(G).Let S = V (G)\T and note that S is an independent set as T is a transversal.Since T is a minimal transversal, every vertex in T has a neighbor in S, implying that S is an independent dominating set in G.
Favaron [5] was the first to prove the following upper bound on the independent domination of a graph with no isolated vertex: If G is an isolate-free graph of order n, then i(G) n + 2 − 2 √ n.We remark that this result also follows from a result due to Bollobás and Cockayne [2] (and was also proved in [6]).Sun and Wang [14] proved the following more general result, which was originally posed as a conjecture by Favaron [5] and was proved for δ = 2 by Glebov and Kostochka [7].
Theorem 1 is an immediate consequence of Theorem 3 and Theorem 4. That this bound is sharp, follows from a result of Favaron [5] who showed that for every positive integer δ, the bound in Theorem 4 is attained for infinitely many graphs.The same graphs achieve equality in the bound for Theorem 1.For example, for c 2, let G c be the connected graph constructed as follows.Let F c be the complete graph of order c, and so √ n − 2, noting that c 2. Thus, the bound of Theorem 1 when δ = 1 is sharp.For every δ 2 one can similarly show that Theorem 1 is tight.
As a special case of Theorem 1, if H is a connected 2-uniform hypergraph of order n 2, then Υ(H) 2 √ n − 2. When n 3, we observe that 2 √ n − 2 1 2 n.Further, we observe that when n = 2, Υ(H) = 1 = 1 2 n.Thus, as an immediate consequence of Theorem 1 we observe that if H is a connected graph, then Υ(H)

Proof of Theorem 2
We first prove the lower bound in Theorem 2.
Proof.Let H be a connected 3-uniform hypergraph of order n H and let T be a minimal transversal of maximum size.Let T = {t 1 , t 2 , . . ., t c }, and so Υ(H) = |T | = c.For all i and j where 1 i < j c and for all k ∈ [c], define Z i,j , E k and Y k as follows.
be a minimum set of vertices in T that covers all edges that are completely within T (i.e., all edges e with V (e) ⊆ V (T )).Possibly, Q = ∅.Let q = |Q|.Renaming vertices of T if necessary, we may assume that Q = {t 1 , . . ., t q }.Let We note that I = I q ∪ I >q .We proceed further with the following claims.Claim 6. |Z i,j | c − q for all (i, j) ∈ I.
Proof.Suppose, to the contrary, that |Z i,j | > c − q for some i and j, where 1 i < j c.Let R = V (H) \ {t i , t j }.Clearly, R is a transversal in H as it contains all vertices in H except two and H is 3-uniform.Let R be obtained from R by removing vertices until we get a minimal transversal in H.We note that Z i,j ⊆ R since each vertex z ∈ Z i,j is needed in order to cover the edge {t i , t j , z}.Further, R contains at least q vertices from T in order to cover the edges that are contained entirely within T .Hence, Υ(H) |R | |Z i,j | + q > c, contradicting the fact that Υ(H) = c.
Proof.Suppose, to the contrary, that Let R = V (H) \ {t q+1 , . . ., t c }.By definition of the set Q, every edge of H intersects {t q+1 , . . ., t c } in at most two vertices, implying that R is a transversal in H. Let R be obtained from R by removing vertices from R until we get a minimal transversal in H.We note that since each vertex z ∈ Z i,j where q + 1 i < j c is needed in order to cover the edge {t i , t j , z}.Further, R contains at least q vertices from Q in order to cover the edges that are contained entirely within T .Hence, Proof.As observed earlier, I = I q ∪ I >q .By Claim 6, |Z i,j | c − q for all (i, j) ∈ I q .Since there are c 2 − c−q 2 pairs (i, j) ∈ I q where 1 i q and i < j c, we note by Claim 6 and Claim 7 that Claim 9. |Y i | ( c−q 2 + 1) 2 for all i ∈ [c].Proof.Suppose, to the contrary, that |Y i | > ((c − q)/2 + 1) 2 for some i ∈ [c].Let H be the graph with vertex set V (H ) = Y i and with edge set and note that R is a transversal in H. Let R be obtained from R by removing vertices from R until we get a minimal transversal in H.In order to cover the edges E i , we must have T ⊆ R , noting that T is a minimal transversal in H . Further, R contains at least q vertices from T \ {t i } in order to cover the edges that are contained entirely within T .Therefore, the electronic journal of combinatorics 25(4) (2018), #P4.27 contradicting the fact that Υ(H) = c.
Since T is a transversal in H, we note that Let β be defined such that (c − q) = βc.We note that 0 β 1.By Equation (1) and by Claim 8 and 9, we therefore get the following.
The maximum value of f (β) when 0 β 1 is obtained when β = (1 + √ 13)/6, noting that 0 = f (β) = −6β 2 + 2β + 2 implies β = (1 ± √ 13)/6.Therefore, f (β) f 1+ We remark that3 1 0.305 > 1.48559, and so as a consequence of Theorem 5, if H is a connected 3-uniform hypergraph of order n 3, then Υ(H) > 1.4855 3 √ n − 2. When n 17, we observe that 1.4855 3 √ n − 2 > 3 1 3 n.Further, we observe that when n = 3, Υ(H) = 1 = 3 1 3 n, while for 4 n 16, Υ(H) 2 > 3 1 3 n.Thus, as an immediate consequence of Theorem 5, we observe that if H is a connected 3-uniform hypergraph, the electronic journal of combinatorics 25(4) (2018), #P4.27 > 0, we can choose n sufficiently large so that n < , implying by Proposition 10 that the connected 3-uniform hypergraph H = H n satisfies denote the resulting graph of order n = c 2 .We note that all the new vertices added to F c have degree 1 in G c .Every transversal in G c must contain all except possibly one vertex of F c in order to cover all the edges of F c .If a minimal transversal in G c contains exactly c − 1 vertices of F c , say all vertices of F c except for v, then the transversal contains exactly c − 1 vertices not in F c , namely v 1 , . . ., v c−1 , in order to cover the edges vv i for all i ∈ [c − 1].Such a minimal transversal therefore has size exactly 2(c − 1).If a minimal transversal in G c contains all c vertices of F c , then it contains no other vertex of G c and therefore has size exactly c.Therefore, the connected graph G of order n = c 2 satisfies Υ(G) = max{2(c − 1), c} = 2(c − 1) = 2