Cent. Eur. J. Math. • 10(3) • 2012 • 1133-1140
DOI: 10.2478/s11533-012-0023-9
Central European Journal of Mathematics
Hypergraphs with large transversal number
and with edge sizes at least four
Research Article
Michael A. Henning1∗ , Christian Löwenstein1#
1 Department of Mathematics, University of Johannesburg, Auckland Park, 2006 South Africa
Received 3 August 2011; accepted 19 January 2012
Abstract: Let H be a hypergraph on n vertices and m edges with all edges of size at least four. The transversal number
τ(H) of H is the minimum number of vertices that intersect every edge. Lai and Chang [An upper bound for the
transversal numbers of 4-uniform hypergraphs, J. Combin. Theory Ser. B, 1990, 50(1), 129–133] proved that
τ(H) ≤ 2(n + m)/9, while Chvátal and McDiarmid [Small transversals in hypergraphs, Combinatorica, 1992, 12(1),
19–26] proved that τ(H) ≤ (n + 2m)/6. In this paper, we characterize the connected hypergraphs that achieve
equality in the Lai–Chang bound and in the Chvátal–McDiarmid bound.
MSC:
05C65, 05C69
Keywords: Transversal • 4-uniform hypergraph
© Versita Sp. z o.o.
1.
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 , E) is a finite set V = V (H) of elements, called vertices, together
with a finite multiset E = E(H) of subsets of V , called hyperedges or simply edges. We shall use the notation nH = |V |
and mH = |E|, and sometimes simply n and m without subscript if the actual H need not be emphasized, to denote the
order and size of H, respectively. In the problems studied here, one may assume that |e| ≥ 2 holds for all e ∈ E, and it
will also be possible to avoid multiple edges without loss of generality.
A k-edge in H is an edge of size k. The hypergraph H is said to be k-uniform if every edge of H is a k-edge. We remark
that the empty hypergraph with no edges is vacuously k-uniform. Every (simple) graph is a 2-uniform hypergraph. Thus
∗
#
E-mail: [email protected]
E-mail: [email protected]
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Hypergraphs with large transversal number and with edge sizes at least four
graphs are special hypergraphs. The hypergraph H is simple if no two edges contain exactly the same vertex set. The
degree of a vertex v in H, denoted by dH (v) or simply by d(v) if H is clear from the context, is the number of edges
of H which contain v. A vertex of degree k we call a degree-k vertex. The minimum degree among the vertices of H
is denoted by δ(H). Two edges in H are said to be overlapping if they intersect in at least two vertices. We define a
hypergraph H to be edge-minimal if every edge of H contains at least one vertex of degree 1 in H. If H 0 is a hypergraph
such that V (H 0 ) ⊆ V (H) and E(H 0 ) ⊆ E(H), then H 0 is called a subhypergraph of H and we write H 0 ⊆ H. Further if
V (H 0 ) = V (H), then H 0 is called a spanning subhypergraph of H.
Two vertices x and y of H are adjacent if there is an edge e of H such that {x, y} ⊆ e. The neighborhood of a vertex
v in H, denoted NH (v) or simply N(v) if H is clear from the context, is the set of all vertices different from v that are
adjacent to v. Two vertices x and y of H are connected if there is a sequence x = v0 , v1 , v2 . . . , vk = y of vertices of H in
which vi−1 is adjacent to vi for i = 1, 2, . . . , k. A connected hypergraph is a hypergraph in which every pair of vertices
are connected. A maximal connected subhypergraph of H is a connected component or simply a component of H. Thus,
no edge in H contains vertices from different components. The complement of H is the hypergraph H with V (H) = V (H)
and where e is an edge in H if and only if V (H) \ e is an edge in H.
If H denotes a hypergraph and X denotes a subset of vertices in H, then H − X will denote that hypergraph obtained
from H by removing the vertices X from H, removing all hyperedges that intersect X and removing all resulting isolated
vertices, if any. We remark that in the literature this is sometimes called strongly deleting the vertices in X .
If H = (V , E) denotes a hypergraph and E 0 ⊆ E, then by the hypergraph induced (or formed) by the edges in E 0 we
mean the hypergraph with edge set E 0 and with vertex set consisting of all vertices that belong to some edge in E 0 . The
hypergraph H − E 0 denotes the hypergraph with vertex set V and edge set E \ E 0 that is obtained from H by deleting
all edges in E 0 .
A subset T of vertices in a hypergraph H is a transversal (also called vertex cover or hitting set in many papers) if T has
a nonempty intersection with every edge of H. The transversal number τ(H) of H is the minimum size of a transversal
in H. A transversal of size τ(H) is called a τ(H)-set.
1.1.
Known results on transversals in hypergraphs
Transversals in hypergraphs are well studied in the literature. Let H = (V , E) be a hypergraph on n vertices and m
edges with all edges of size at least two. If H only consists of isolated vertices, then τ(H) = m = 0. If ∆(H) ≥ 1, let
v be a vertex of maximum degree ∆(H) in H and let Sv be the set consisting of the vertex v and one vertex from every
edge that does not contain v. Then, Sv is a transversal in H, and so τ(H) ≤ |Sv | ≤ 1 + (m − ∆(H)) ≤ m, with strict
inequality if ∆(H) ≥ 2. Hence we have the following trivial bound on the transversal number.
Observation 1.1.
If H is a hypergraph with m edges, then τ(H) ≤ m with equality if and only if ∆(H) ≤ 1.
If all edges of H have size at least three, then Chvátal and McDiarmid [1] and Tuza [5] independently established that
4τ(H) ≤ n + m. A short proof of this result can be found in [4]. The extremal connected hypergraphs that achieve
equality in this bound were characterized in [2]. We remark that there are four infinite families of such hypergraphs that
achieve equality in this bound, as well as two exceptional hypergraphs, one of order 7 and one of order 8.
If all edges of H have size at least four, then we have the following results due to Chvátal and McDiarmid [1], Lai and
Chang [3] and Thomassé and Yeo [4].
Theorem 1.2.
Let H be a hypergraph on n vertices and m edges with all edges of size at least four. Then the following holds.
(a) τ(H) ≤ m (Observation 1.1),
(b) τ(H) ≤ n/6 + m/3 [1],
1134
M.A. Henning, Chr. Löwenstein
(c) τ(H) ≤ 2n/9 + 2m/9 [3],
(d) τ(H) ≤ 5n/21 + 4m/21 [4].
For k ≥ 2, let Hk denote the class of all k-uniform hypergraphs. For H ∈ Hk , we define the transversal-order ratio
of H, denoted τord (H), to be the ratio of the transversal number and the order of H. Further we denote the average
degree of H by dav (H). Hence,
τord (H) =
τ(H)
|V (H)|
and
As a consequence of Theorem 1.2 we have the following result.
dav (H) = k ·
|E(H)|
.
|V (H)|
Corollary 1.3.
Let H ∈ H4 . Then the following holds.
(a) τord (H) ≤ dav (H)/4 (Observation 1.1),
(b) τord (H) ≤ dav (H)/12 + 1/6 [1],
(c) τord (H) ≤ dav (H)/18 + 2/9 [3],
(d) τord (H) ≤ dav (H)/21 + 5/21 [4].
The four straight lines corresponding to equality in the four equations in Corollary 1.3 are illustrated in Figure 1 (i).
We observe that the best upper bound on the transversal-ratio of H for 0 ≤ dav (H) ≤ 1 is the trivial bound (see
Corollary 1.3 (a)). For 1 ≤ dav (H) ≤ 2, the Chvátal–McDiarmid bound (see Corollary 1.3 (b)) is the best, while for
dav (H) ≥ 2 the Thomassé–Yeo bound (see Corollary 1.3 (d)) is the best bound. For dav (H) = 2, the Chvátal–McDiarmid
bound, the Lai–Chang bound and the Thomassé–Yeo bound are all equal. For dav (H) > 4, a tight upper bound has yet
to be established.
τord (H)
(a)
3
7
(b)
1
3
1
4
1
Figure 1.
2
(i)
4
(c)
(d)
dav (H)
τord (H)
3
7
1
3
1
4
R4
1
2
(ii)
4
dav (H)
Comparing the transversal-order ratio and the average degree
Let H4 ∈ H4 and H6 ∈ H4 denote the two hypergraphs shown in Figure 2. Further, let H7 ∈ H4 denote the complement
of the Fano plane, where the Fano plane is shown in Figure 2. We now consider the set of points (x, y), where x denotes
the average degree and y the transversal-order ratio. We observe that the empty hypergraph with no edges is sharp
for (0, 0), while the hypergraphs H4 and H6 shown in Figure 2 are sharp for (1, 1/4) and (2, 1/3), respectively. Moreover
1135
Hypergraphs with large transversal number and with edge sizes at least four
+
the complement, H7 , of the Fano plane is sharp for (4, 3/7). Let Q+
0 and R0 be the set of all rational numbers and real
numbers, respectively, greater or equal to 0. Let
U4 =
(a, b) : a, b ∈ Q+
0 and there exists H ∈ H4 with dav (H) = a satisfying τord (H) ≥ b .
Then, U4 ⊂ R4 , where R4 is the convex set illustrated in Figure 1 (ii) consisting of all real numbers a, b ∈ R+
0 satisfying
b ≤ min {a/4, (a + 2)/12, (a + 5)/21}.
H4
Figure 2.
2.
H6
Fano plane
The hypergraphs H4 , H6 and the Fano plane
Main results
Our aim in this paper is to characterize the extremal hypergraphs that achieve equality in the bounds in Theorem 1.2.
Equivalently, we characterize the hypergraphs that are sharp for each of the points (0, 1), (1, 1/4), (2, 1/3) and (4, 3/7)
in Figure 1 (ii). It suffices for us to restrict attention to connected hypergraphs H on n vertices and m edges with all
edges of size at least 4. By Observation 1.1, the trivial bound is only achieved for hypergraphs with maximum degree
at most 1. To avoid this trivial case, we shall assume that m ≥ 2 from now on. We shall prove the following main result.
Theorem 2.1.
Let H be a connected hypergraph on n vertices and m edges with all edges of size at least 4. Then the following holds.
(a) If τ(H) = 2(n + m)/9, then H = H6 .
(b) If τ(H) = (n + 2m)/6, then H ∈ {H4 , H6 }.
2.1.
Proof of Theorem 2.1 (a)
In order to prove Theorem 2.1 (a), we shall need the following result due to Lai and Chang [3]. Following their notation,
we call an edge an end-edge if it contains a vertex of degree 1 and we denote the number of end-edges in a hypergraph
H by t(H), or simply by t if H is clear from the context.
Theorem 2.2 (Lai, Chang [3]).
If H ∈ H4 has n vertices, m edges and t end-edges, then τ(H) ≤ (2n + 2m − t)/9.
Recall that if e ∈ E(H) and v ∈ V (e), then shrinking e by deleting v produces the hypergraph obtained from H by
deleting the edge e and adding the edge V (e) \ {v}. We are now in a position to present a proof of Theorem 2.1 (a).
1136
M.A. Henning, Chr. Löwenstein
Proof of Theorem 2.1 (a). Assume that theorem is false. Then there exist connected hypergraphs H on n vertices
and m edges with all edges of size at least 4 such that τ(H) = 2(n + m)/9 but H 6= H6 . Among all such counterexamples,
P
let H = (V , E) be chosen so that v∈V dH (v) is a minimum. We proceed further with the following two claims.
Claim A.
H is 4-uniform.
Suppose that H is not 4-uniform (and still H is connected). Then, H has at least one edge
e of size at least 5. Among all vertices in e, let v ∈ V (e) be chosen to have minimum degree in H. We now
shrink e by deleting v to produce the hypergraph H 0 of order n0 = n and size m0 = m with all edges of size at
least 4. By Theorem 1.2 (c), τ(H 0 ) ≤ 2(n0 + m0 )/9 = 2(n + m)/9. Every transversal in H 0 is a transversal in H, and so
2(n + m)/9 = τ(H) ≤ τ(H 0 ) ≤ 2(n + m)/9. Therefore, we must have equality throughout this inequality chain. Hence,
τ(H 0 ) = 2(n0 + m0 )/9, implying that every component H ∗ of H 0 satisfies τ(H ∗ ) = 2(|V (H ∗ )| + |E(H ∗ )|)/9. By the minimality
P
of v∈V dH (v), we note that every component in H 0 is an H6 . Every vertex in the edge V (e) \ {v} therefore has degree 2
in both H 0 and H, while the vertex v has degree 2 in H 0 and degree 3 in H. But this contradicts the choice of the vertex
v to be a vertex of smallest degree in the edge e. Hence, H is 4-uniform.
Proof of Claim A.
By Claim A, we have that H ∈ H4 .
Claim B.
H is 2-regular.
If t ≥ 1, then by Theorem 2.2, τ(H) < (2n + 2m)/9, a contradiction. Therefore, t = 0, implying
P
that δ(H) ≥ 2. Since H ∈ H4 , we have that m = 1/4 v∈V dH (v) ≥ n/2. If m > n/2, then, by Theorem 1.2 (d),
τ(H) ≤ (5n + 4m)/21 < 2(n + m)/9, a contradiction. Hence, m ≤ n/2. Consequently, n = 2m and H is 2-regular.
Proof of Claim B.
Claim C.
H contains overlapping edges.
Assume, to the contrary, that H has no overlapping edges. Let v ∈ V and let e = {v, v1 , v2 , v3 }
and f = {v, u1 , u2 , u3 } be the two edges containing v. Let H 0 = H − v have n0 vertices, m0 edges and t 0 end-edges. Then,
n0 = n − 1 and m0 = m − 2. By the 2-regularity of H, each vertex adjacent to v has degree 1 in H 0 and therefore belongs
to an end-edge of H 0 . For i = 1, 2, 3, let ei be the edge in H 0 that contains vi and note that ei is an end-edge in H 0 .
Since H has no overlapping edges, ei =
6 ej for 1 ≤ i < j ≤ 3. Thus, t 0 ≥ 3. By Theorem 2.2, τ(H 0 ) ≤ (2n0 + 2m0 − t 0 )/9.
0
Hence, 2(n + m)/9 = τ(H) ≤ τ(H ) + 1 ≤ (2n0 + 2m0 − t 0 )/9 + 1 ≤ 2(n + m)/9 − (2 + 4 + 3)/9 + 1 = 2(n + m)/9. Therefore
we must have equality throughout this inequality chain. In particular, t 0 = 3. This implies that each vertex in V (f) \ {v}
belongs to one of the edges ei , 1 ≤ i ≤ 3. Since H has no overlapping edges, we may assume, renaming the vertices
u1 , u2 , u3 if necessary, that ui ∈ V (ei ) for i = 1, 2, 3.
Proof of Claim C.
Suppose that V (ei ) ∩ V (ej ) =
6 ∅ for some i and j, where 1 ≤ i < j ≤ 3. For notational convenience, we may assume that
the edges e1 and e2 intersect and that x is the vertex common to these two edges. Let H ∗ = H − {v, x} have n∗ vertices,
m∗ edges and t ∗ end-edges. Then, n∗ = n−|{u1 , u2 , v, v1 , v2 , x}| = n−6, m0 = m−4 and t ∗ ≥ 1 (in fact, t ∗ ≥ 2), we have
that τ(H) ≤ τ(H ∗ ) + 2 ≤ (2n∗ + 2m∗ − t ∗ )/9 + 2 ≤ 2(n + m)/9 − (12 + 8 + 1)/9 + 2 < 2(n + m)/9, a contradiction. Hence the
edges e1 , e2 , e3 are pairwise non-intersecting edges. Let e1 = {v1 , u1 , w, x} and let ew be the edge containing w that is
different from e1 . Let H 00 = H −{v, x} have n00 vertices, m00 edges and t 00 end-edges. Then, n00 = n−|{v, v1 , u1 , x}| = n−4
and m00 = m − 4. Further, H 00 contains at least three end-edges, namely the edges e2 , e3 and ew , and so t 00 ≥ 3. Hence,
τ(H) ≤ τ(H 00 ) + 2 ≤ (2n00 + 2m00 − t 00 )/9 + 2 ≤ 2(n + m)/9 − (8 + 8 + 3)/9 + 2 < 2(n + m)/9, a contradiction.
We now return to the proof of Theorem 2.1 (a). By Claims A, B and C, we have that H is a 4-uniform 2-regular
hypergraph with overlapping edges. Let e and f be two overlapping edges in H, and let {u, v} ⊆ V (e) ∩ V (f). If e = f,
then n = 4, m = 2 and τ(H) = 1 < 2(n + m)/9, a contradiction. Hence, e =
6 f. Let H 0 = H − v have n0 vertices, m0
0
0
edges and t end-edges. If e and f intersect in three vertices, then n = n − 3, m0 = m − 2 and t 0 ≥ 1, implying that
1137
Hypergraphs with large transversal number and with edge sizes at least four
τ(H) ≤ τ(H 0 ) + 1 ≤ (2n0 + 2m0 − t 0 )/9 + 1 ≤ 2(n + m)/9 − (6 + 4 + 1)/9 + 1 < 2(n + m)/9, a contradiction. Therefore, every
two edges of H intersect in at most two vertices. Let e = {u, v, x, y} and f = {u, v, w, z}. Note that each of x, y, w, z has
degree 1 in H 0 , and so t 0 ≥ 1 with equality if and only if {x, y, w, z} is an edge of H. Since n0 = n − 2 and m0 = m − 2,
we have that 2(n + m)/9 = τ(H) ≤ τ(H 0 ) + 1 ≤ (2n0 + 2m0 − t 0 )/9 + 1 ≤ 2(n + m)/9 − (4 + 4 + 1)/9 + 1 = 2(n + m)/9.
Therefore we must have equality throughout this inequality chain. In particular, t 0 = 1, implying that {x, y, w, z} is an
edge of H. But then H = H6 .
2.2.
Proof of Theorem 2.1 (b)
Assume that the theorem is false. Then there exist connected hypergraphs H on n
vertices and m edges with all edges of size at least 4 such that τ(H) = (n + 2m)/6 but H ∈
/ {H4 , H6 }. Among all such
P
counterexamples, let H = (V , E) be chosen so that v∈V dH (v) is minimal. Let V1 and V2 denote the set of vertices in
H of degree 1 and degree 2, respectively. For a vertex v ∈ V , we let Hv = H − v and we let Hv have nv vertices and
mv edges. We proceed further with the following three claims.
Proof of Theorem 2.1 (b).
Claim I.
The following hold in the hypergraph H.
(a) δ(H) ≥ 1.
(b) ∆(H) ≤ 2.
(c) No edge contains two degree-1 vertices and a degree-2 vertex.
(a) If H has isolated vertices, then let H 0 be obtained from H by deleting all isolated vertices.
Let H 0 have n0 vertices and m0 edges. Then, n0 ≤ n − 1 and m0 = m. Every transversal in H 0 is a transversal in H,
and so τ(H) ≤ τ(H 0 ). Applying the Chvátal–McDiarmid bound in Theorem 1.2 (b) to the hypergraph H 0 , we have that
τ(H) ≤ τ(H 0 ) ≤ (n0 + 2m0 )/6 < (n + 2m)/6, a contradiction. Hence, δ(H) ≥ 1.
Proof of Claim I.
(b) Suppose that ∆(H) ≥ 3. Let v be a vertex of maximum degree ∆(H) in H. Then, nv ≤ n − 1 and mv ≤ m − 3.
Applying the Chvátal–McDiarmid bound in Theorem 1.2 (b) to the hypergraph Hv , we have that τ(Hv ) ≤ (nv + 2mv )/6.
Every transversal in Hv can be extended to a transversal in H by adding to it the vertex v, and so τ(H) ≤ τ(Hv ) + 1 ≤
(n + 2m)/6 − (1 + 6)/6 + 1 < (n + 2m)/6, a contradiction. Hence, ∆(H) ≤ 2.
(c) It suffices to show that if ∆(H) = 2, then every edge of H contains at most one vertex of degree 1. Suppose ∆(H) = 2.
Let e ∈ E and assume that e contains two or more vertices of degree 1. Since H is connected and ∆(H) = 2, every
edge of H contains at least one vertex of degree 2. Let v be a vertex in V (e) of degree 2 in H. Then, nv ≤ n − 3 and
mv ≤ m − 2. Hence, τ(H) ≤ τ(Hv ) + 1 ≤ (nv + 2mv )/6 + 1 ≤ (n + 2m)/6 − (3 + 4)/6 + 1 < (n + 2m)/6, a contradiction.
Hence the edge e contains at most one vertex of degree 1.
Claim II.
H is 1-regular or 2-regular.
Assume this is false. Then there exists an edge e = {x1 , x2 , . . . , x` } ∈ E(H), where ` ≥ 4, such
that d(x1 ) = 1 and d(x2 ) = 2. By Claim I, d(xi ) = 2 for i = 2, 3, . . . , `. Since nx2 ≤ n − 2 and mx2 ≤ m − 2, we note
that if τ(Hx2 ) < (nx2 + 2mx2 )/6 or nx2 < n − 2, then τ(H) ≤ τ(Hx2 ) + |{x2 }| < (n + 2m)/6, a contradiction. Hence,
P
τ(Hx2 ) = (nx2 + 2mx2 )/6 and nx2 = n − 2. By the minimality of v∈V (H) dH (v), we note that every component in Hx2 is
either an H4 or an H6 . This is true for every vertex xi , 2 ≤ i ≤ `. Furthermore, if there exists an edge f ∈ E(H) with
|V (e) ∩ V (f)| ≥ 2, then considering a vertex v ∈ V (e) ∩ V (f) instead of x2 would lead to a contradiction with nv = n − 2.
For i ∈ {3, 4}, let fi be the edge of H, distinct from e, that contains xi . Then, f3 =
6 f4 and both f3 and f4 are edges
in H4 -components in Hx2 . In particular, we note that f3 and f4 are vertex disjoint. By Claim I, the edge f4 contains a
degree-2 vertex, say x40 , in H distinct from x4 . Since f3 and f4 are vertex disjoint, we note that neither the edge e nor
the edge f3 contain the vertex x40 , implying that x40 is also a degree-2 vertex in Hx3 . This contradicts the fact that every
component in Hx3 is either an H4 or an H6 .
Proof of Claim II.
1138
M.A. Henning, Chr. Löwenstein
Recall that if e ∈ E(H) and v ∈ V (e), then shrinking e by deleting v produces the hypergraph obtained from H by
deleting the edge e and adding the edge V (e) \ {v}.
Claim III.
H is 4-uniform.
Proof of Claim III. Assume, to the contrary, that H contains an edge e of size at least 5. If some vertex v ∈ V (e)
has d(v) = 1, then shrink e by deleting v. Otherwise if there is an edge f distinct from e such that |V (e) ∩ V (f)| ≥ 2,
then let v ∈ V (e) ∩ V (f) and shrink e by deleting v. Otherwise let v ∈ V (e) be an arbitrary vertex in e and shrink e by
deleting v. We note that the resulting hypergraph H 0 has n vertices and m edges with all edges of size at least 4. Further,
τ(H) ≤ τ(H 0 ). By Theorem 1.2 (b), τ(H 0 ) ≤ (n + 2m)/6. If τ(H 0 ) < (n + 2m)/6, then τ(H) < (n + 2m)/6, a contradiction.
Hence, τ(H 0 ) = (n + 2m)/6, implying that every component H ∗ of H 0 satisfies τ(H ∗ ) = (|V (H ∗ )| + 2|E(H ∗ )|)/6. By the
P
minimality of v∈V dH (v), we note that every component in H 0 is either an H4 or an H6 . But this is a contradiction
since we either obtain a vertex of degree zero in H 0 or an edge with both degree-1 and degree-2 vertices or a degree-2
vertex that does not belong to overlapping edges. Hence, H is 4-uniform.
We now return to the proof of Theorem 2.1 (b). By Claim II, either H is 1-regular or 2-regular. By Claim III, H is
4-uniform. If H is 1-regular, then H = H4 . If H is 2-regular, then n = 2m. But then τ(H) = (n + 2m)/6 = 2(n + m)/9.
Therefore, by Theorem 2.1 (a), H = H6 .
3.
Concluding remarks
In this paper, we characterize the hypergraphs that achieve equality in the Chvátal–McDiarmid bound of Theorem 1.2 (b)
and in the Lai–Chang bound of Theorem 1.2 (c). A natural problem is to characterize the hypergraphs that achieve
equality in the Thomassé–Yeo bound of Theorem 1.2 (d). We remark that a characterization of the 4-uniform hypergraphs
that achieve equality in the Thomassé–Yeo bound has recently been given by Anders Yeo [6] who proved the following
result.
Theorem 3.1 (Yeo [6]).
Let H be a connected 4-uniform hypergraph on n vertices and m edges. If H ∈
/ {H6 , H7 }, then
1
τ(H) ≤
21
40
5−
283
39
n+ 4+
283
m .
As a consequence of Theorem 3.1, we have a characterization of the extremal hypergraphs that achieve equality in the
bound of Theorem 1.2 (d). We provide the short proof for completeness.
Corollary 3.2 (Yeo [6]).
If H is a connected 4-uniform hypergraph on n vertices and m edges satisfying τ(H) = 5n/21+4m/21, then H ∈ {H6 , H7 }.
Let v be a vertex of maximum degree ∆(H). Let H 0 = H − v have n0 vertices and m0 edges. Then, n0 ≤ n − 1
and m = m − ∆(H). Every transversal in H 0 can be extended to a transversal in H by adding to it the vertex v, and so
(5n + 4m)/21 = τ(H) ≤ τ(H 0 ) + 1 ≤ (5n0 + 4m0 )/21 + 1 ≤ (5n + 4m)/21 − (5 + 4∆(H))/21 + 1, implying that ∆(H) ≤ 4.
P
Since H is 4-uniform, we therefore have that m = 1/4 v∈V dH (v) ≤ n. If H ∈
/ {H6 , H7 }, then by Theorem 3.1,
Proof.
0
τ(H) ≤
5
4
1
n+
m+
21
21
21
a contradiction. Hence, H ∈
/ {H6 , H7 }.
39
40
m−
n
283
283
<
5
4
n+
m,
21
21
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Hypergraphs with large transversal number and with edge sizes at least four
As a consequence of Corollary 3.2, we have the following slightly stronger result where we relax the edge size from
exactly 4 to at least 4.
Corollary 3.3.
If H is a connected hypergraph on n vertices and m edges with all edges of size at least 4 satisfying τ(H) = 5n/21+4m/21,
then H ∈ {H6 , H7 }.
Proof. Assume that the theorem is false. Then there exist connected hypergraphs H on n vertices and m edges with
all edges of size at least 4 such that τ(H) = 5n/21 + 4m/21 but with H ∈
/ {H6 , H7 }. Among all such counterexamples,
P
let H = (V , E) be chosen so that v∈V dH (v) is a minimum. By Yeo’s Corollary 3.2, H is not 4-uniform. Let e be an
edge of size at least 5 in H. Among all vertices in e, let v ∈ V (e) be chosen to have minimum degree in H. We now
shrink e by deleting v to produce the hypergraph H 0 order n0 = n and size m0 = m with all edges of size at least 4. By
Theorem 1.2 (d),
5 0
4 0
5
4
τ(H 0 ) ≤
n +
m =
n+
m.
21
21
21
21
Every transversal in H 0 is a transversal in H, and so
5
4
5
4
n+
m = τ(H) ≤ τ(H 0 ) ≤
n+
m.
21
21
21
21
Therefore, we must have equality throughout this inequality chain. Hence, τ(H 0 ) = 5n0 /21 + 4m0 /21, implying that every
P
component H ∗ of H 0 satisfies τ(H ∗ ) = 5|V (H ∗ )|/21 + 4|E(H ∗ )|/21. By the minimality of v∈V dH (v), we note that every
component in H 0 is an H6 or an H7 . Let e0 = V (e) \ {v}. On the one hand, if the edge e0 belongs to an H6 -component
in H 0 , then every vertex in e0 therefore has degree 2 in both H 0 and H, while the vertex v has degree 2 in H 0 and
degree 3 in H. On the other hand, if the edge e0 belongs to an H7 -component in H 0 , then every vertex in e0 therefore
has degree 4 in both H 0 and H, while the vertex v has degree 4 in H 0 and degree 5 in H. In both cases we contradict
the choice of the vertex v to be a vertex of smallest degree in the edge e.
Acknowledgements
The authors wish to thank the referees for their very helpful comments and insight which resulted in shorter and more
elegant proofs than the authors’s original proofs of Theorem 2.1 (a), Theorem 2.1 (b) and Corollary 3.3.
Research of the first author was supported in part by the South African National Research Foundation, and research of
the second author supported by the Deutsche Forschungsgemeinschaft (GZ: LO 1758/1-1).
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