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Total Domination of Graphs and Small Transversals of
Hypergraphs
Stéphan Thomassé, Anders Yeo
To cite this version:
Stéphan Thomassé, Anders Yeo. Total Domination of Graphs and Small Transversals of Hypergraphs.
Combinatorica, Springer Verlag, 2007, 27, pp.473-487. �lirmm-00250084�
Total domination of graphs and small transversals of
hypergraphs.
St´ephan Thomass´e
∗
and
Anders Yeo
†
Abstract
The main result of this paper is that every 4-uniform hypergraph on n vertices and m edges has
a transversal with no more than (5n + 4m)/21 vertices. In the particular case n = m, the transversal
has at most 3n/7 vertices, and this bound is sharp in the complement of the Fano plane. Chv´atal and
McDiarmid [5] proved that every 3-uniform hypergraph with n vertices and edges has a transversal
of size n/2. Two direct corollaries of these results are that every graph with minimal degree at least
3 has total domination number at most n/2 and every graph with minimal degree at least 4 has total
domination number at most 3n/7. These two bounds are sharp.
1 Introduction.
Given a graph G = (V, E), a total dominating set is a subset S of the vertices of G such that every vertex
of G has a neighbour in S. The minimum size of a total dominating set is the total domination number
of G. It was proved by Favaron et al. [6] that a graph with n vertices and minimum degree at least 3
has total domination number at most 7n/13. This result has b ee n recently extended to n/2 [4], [2], and
a fractional approach can be found in [7]. A transversal in a hypergraph is a subset of vertices which
intersects every edge. We are mainly concerned here with transversals of k-uniform hypergraphs with the
same number of e dges and vertices. Precisely, we raise the problem to find the minimum c
k
for which
every k-uniform hypergraph with n vertices and n edges has a transversal of size c
k
n. It directly follows
that every graph G with minimum degree k and n vertices has a total dominating set with c
k
n vertices,
since in the hypergraph whose edges are the neighb ourhoods of the vertices of G, a transversal is a total
dominating set. The main advantage of considering hypergraphs instead of graphs is that the structure
is easier to handle - for instance we can limit ourselves to k-uniform structures. When H is a hypergraph
on the vertex set V and X ⊆ V , we denote by H \ X the induced subhypergraph on V \ X - that is, we
delete all the vertices of X, and all the edges having a vertex in X.
To fix the ideas, and a bound, let us calculate c
2
:
Lemma 1 Every 2-uniform hypergraph H has a transversal T such that 3|T | ≤ n + m
Proof. By induction on the number of vertices. If some vertex x has degree at least 2, we put it in the
transversal, and apply the induction hypothesis to H \ x. Otherwise no vertex has degree 2 and then the
edges form a matching, and the formula holds.
∗
LIRMM, 16 1 rue Ada, 34392 Montpellier Cedex 5, France, thomasse@lirmm.fr
†
Department of Computer Science, Royal Holloway, University of London, Egham; Surrey TW20 0EX, UK (email:
yeo@dcs.rhbnc.ac.uk)
1
Corollary 1 c
2
= 2/3
When n = 3k, the extremal graph for c
2
is certainly a disjoint union of triangles, and this is the
sole example by the unicity of the extremal Tur´an graphs. This kind of example generalizes for all k: If
one considers a disjoint union of complete k-uniform hypergraphs on k + 1 vertices, this clearly forms a
k-uniform hypergraph with n edges and n vertices and the minimum transversal has exactly 2n/(k + 1)
vertices. By this, we directly have that c
k
≥ 2/(k + 1). The interesting fact is that equality holds also
for k = 3, that is c
3
= 1/2, but many extremal examples arise: consider for instance the hypergraph
on vertex set {1, . . . , n, 1
0
, . . . , n
0
} and edges {i, i
0
, i + 1} and {i, i
0
, (i + 1)
0
} where i + 1 is understood
modulo n. However, it turns out that c
4
is not equal to 2/5, as we will observe. But let us now prove that
c
3
= 1/2. The following result can be found in [5], but we will give an alternative pro of as an introduction
to the proof of our main result.
Theorem 1 Let H = (V, E) be a 3-uniform hypergraph. There exists a transversal T of H such that
4|T | ≤ |V | + |E|.
Proof. Induction on the number of vertices. If some vertex v has degree at least 3, we find, by the
induction hypothesis, a transversal T
0
of H \ v. In turn, T
0
∪ {v} satisfies our condition. We can now
suppose that every vertex has degree at most 2. If some edge abc satisfies that the degree of a is 2 and
the degree of b is 1, we apply the induction hypothesis to H \ {a, b} in order to find a transversal T
0
:
the transversal T := T
0
∪ {a} still satisfies the hypothesis. If one edge abc has no vertex of degree 2, we
simply apply the induction hypothesis to H \ {a, b, c}. Now the hypergraph is 2-regular. If the same edge
abc appears twice in H, we simply apply the induction hypothes is to H \ {a, b, c}. If not, we can find the
following subset of edges
x
1
y
1
x
2
, x
2
y
2
x
3
, x
3
y
3
x
4
, . . . , x
k
y
k
x
1
where all the vertices x
1
, . . . , x
k
, y
1
, . . . , y
k
are distinct, because of the following: If two edges intersect
on two vertices, then these two edges are of the form x
1
y
1
x
2
, x
2
y
2
x
1
. If no two edges intersect on
two points, the above sequence is simply obtained by considering a minimum cyclic sequence of edges
where only adjacent edges intersect. Now, two cases can happen: If k = 2l, choose a transversal T
0
of
H
0
:= H \{x
i
: i = 1..k} which satisfies the induction hypothesis: the transversal T := T
0
∪{x
2i
: i = 1..l}
satisfies the hyp othesis for H since we deleted 2l vertices and 2l edges to get H
0
from H. If k = 2l + 1,
choose a transversal T
0
of H
0
:= H \ ({y
k
} ∪ {x
i
: i = 1..k}) which satisfies the induction hypothesis: the
transversal T := T
0
∪ {x
2i
: i = 1..l} ∪ {y
k
} satisfies the hypothesis for H since we deleted 2l + 2 vertices
and 2l + 2 edges to get H
0
from H.
Corollary 2 Every 3-uniform hypergraph H = (V, E) with |V | = |E| = n has a transversal with at most
n/2 vertices. In other words, c
3
= 1/2.
Finally we would like to mention a result of Seymour [9]: every minimal non bipartite hypergraph
has at least as many edges as vertices. From this result follows a short proof of the existence of a n/2
transversal in a connected 3-uniform 3-regular hyp e rgraph H with 2l vertices: remove an edge e of H
for which H − e is connected (e exists since there exists a connected spanning subhypergraph with at
most n − 2 edges). Now every proper induced subhypergraph of H − e has less edges than vertices, in
particular, by Seymour’s result, H − e is two colourable. If e belongs to both colours, we are done. If the
color class of e has at most n/2 vertices, it is our transversal. Finally, if it has more than l + 1 vertices,
take the other color class and a vertex of e as a transversal.
2
2 The 4-uniform hypergraphs.
The case of 4-uniform hypergraphs is much more complicated since what seems to be the extremal case
is not the complete 4-uniform hypergraph on 5 vertices but the complement of the Fano plane. Precisely,
we consider the hypergraph F on the vertex set {1, 2, 3, 4, 5, 6, 7} and edge s et {Q + i : i := 1..7}, where
+ is understood modulo 7 and Q := {0, 1, 2, 4} is the set of quadratic residues. In this case we have
7 vertices and we need 3 vertices to hit all the edges. This example provides a graph with 14 vertices,
minimum degree 4 and total domination number 6, simply by considering the incidence bipartite graph
of F . The main result of this section is to prove that these two examples are extremal, in other words
we have the following result:
Theorem 2 c
4
= 3/7
We prove a more general formula for 4-uniform hypergraphs which directly implies the value of c
4
.
An analog formula was proved in [3], where they established that every 4-uniform hypergraph has a
transversal with no more than (2n + 2m − d)/9 vertices, where d is the number of edges which contain a
vertex of degree one. This gives the upper bound c
4
≤ 4/9.
An edge of a hypergraph is overlapping if it intersects another edge on at least 2 vertices. An edge is
special if it is overlapping and has exactly one vertex of degree 1. For k > 0, an edge is k-degenerated if it
is not special and has exactly k vertices of degree 1. An edge which is not sp e cial and not k-degenerated
is plain.
Theorem 3 Let H be a 4-uniform hypergraph with n vertices, p plain edges, s special edges and d
degenerated edges. There exists a transversal T such that 21|T | ≤ 5n + 4p + 3s/2 + d.
Proof. Let us consider a counterexample H, minimum with respect to its degree sequence, ordered
lexicographically (that is, for each hypergraph we order its degree sequence in the decreasing order and
we compare two hyp e rgraphs using the lexicographical order on their respective sequences.)
Claim 1 The maximum degree in H is at most three.
Proof. Assume not and remove a vertex v of degree at least 4. If all its incident edges are plain, the
removal of these edges gives at least -16 and the removal of v gives -5. In all, we get at least -21, and
adding v in the transversal T
0
of H \ v, the total count is +21-21, so this contradicts the minimality of
H. Now, if some edges incident with v contain degree one vertices, we delete these vertices as well, in
this case, the count for removing one edge is at least -6 (that is at least -1 for the removed edge and -5
for the isolated ve rtex). Again removing all the edges incident with v (plus possibly the now isolated
vertices) gives again at least -16, so we conclude as previously.
In the following of the proof, we will indicate the counting argument we did in the proof of Claim 1 as
a sum +x − y − z − t where x is the value of the vertices in the transversal, y is the value of the removed
edges, z is the value of the removed vertices, and t is the value change of the edges (since some of the
remaining edges can become special or degenerated.)
Claim 2 Every degree two vertex x of H is in a non plain edge.
Proof. If x is in two plain edges e
1
, e
2
, we split the vertex x into two vertices x
1
, x
2
, letting x
1
∈ e
1
and
x
2
∈ e
2
. There is a +5 for the new vertex, but at least 2.(-2,5) since e
1
, e
2
become at least special. Observe
that every transversal of this new hype rgraph is also a transversal of H, contradicting the minimality of
the degree sequence of H.
Claim 3 Every vertex x of degree three is incident to at least two plain edges.
3
Proof. If not, removing x gives at least +21-6-15.
Claim 4 There is no 4-degenerated edge. Sum +21-1-20.
Claim 5 There is no 3-degenerated edge. Sum +21-5-20, in all -4.
Claim 6 Two edges of H do not intersect on 3 vertices.
Proof. Let e and f be two edges intersecting on x, y, z. If e is not plain (its degree one vertex being
t), we simply remove the edge f and find a transversal T
0
. If T
0
does not contain t, it is a transversal of
H. If it contains t, we replace t by any of x, y, z. So both are plain. We add then a new vertex ω to H,
remove the edges e, f and add the edge xyzω. Observe that this new graph is lexicographically smaller
than H, and that the count is -6,5+5 in the worst case. Let T
0
be a transversal of this new graph. If
T
0
does not contain ω, it is a transversal of H. If it contains ω, we replace ω by any of x, y, z to get a
transversal of H.
Claim 7 There is no 2-degenerated edge. Sum -2.
Proof. Let e be a 2-degenerated edge. If some vertex of e has degree three, removing it gives at least
+21-9-15, that is -3. So there are exactly two vertices x, y of degree two, incident with two edges e
x
, e
y
,
and two vertices of degree one. If e
x
= e
y
, removing x gives at least +21-5-20, so -4. By Claim 6, e
x
and
e
y
do not intersect on three vertices. If e
x
and e
y
intersect on two vertices, we remove all the vertices of
e, we delete e
x
and e
y
, and add the new edge (e
x
∪ e
y
) \ {x, y}. This new graph has a lexicographically
smaller degree sequence than H, and any transversal T
0
can be extended by either x or y to form a
transversal of H. This gives us 21-5-20 if e
x
and e
y
are plain. If one is special and the other plain, the
sum is 21-5-20. If both e
x
and e
y
are special, 21-3-20. Observe that this exhausts all possibilities. Now
e
x
and e
y
intersect on at mos t one vertex. If e
x
and e
y
are both plain, we remove all the vertices of e, we
delete e
x
and e
y
, and add a new edge containing any four vertices of (e
x
∪ e
y
) \ {x, y}. As previously, this
gives 21-5-20. If they are not both plain, one vertex z of, say, e
x
has degree 1. In this case, we remove
all the vertices of e, we delete z, we delete e
x
and e
y
, and add a new edge containing any four vertices of
(e
x
∪ e
y
) \ {x, y, z}. This gives 21+1-25.
We will widely use Claim 7 in our proofs. As soon as removing a vertex yields a 2-degenerated edge,
we will substract 2 to our sum.
We define the intersection graph of H as the multigraph G on vertex set V and edge set E in the
following way: for all pairs e, f of edges of H such that e ∩ f = {x, y}, we add the edge xy to G.
Claim 8 The graph G does not have 2-cycles.
Proof. If there are three edges e, f, g pairwise intersecting on x, y, removing x gives at least +21-12-10.
This calculation holds when all the edges are plain. If they are not, we get a better bound.
Claim 9 The graph G has maximum degree 2.
Proof. Suppose we have three edges e, f, g such that e ∩ f = {y, x}, f ∩ g = {y, z} and e ∩ g = {y, t}.
If one of x, z, t has degree two, removing y gives at least 21-12-10. Thus, x, z, t have degree 3, and we
let e
x
, e
z
, e
t
their incident other edges. If e is special, by Claim 3, e
x
is plain, and removing y gives
21-9,5-10-2,5. So e, f, g are plain. If e
z
is non plain, deleting z gives 21-9-10-2,5. Thus all are plain, and
we delete y: If e
x
= e
y
= e
z
, we have 21-12-5 and an additional -4, since e
x
becomes 3-degenerated. If
e
x
6= e
y
, we have 21-12-5 and -5, since e
x
and e
y
become at least special.
4