Edge-connectivity of undirected and directed hypergraphs

Section 6.3. Orientation with supermodular requirements 101
to finding a (k, l)-edge-connected orientation of H (the notion of (k, l)-edge-connectivity of
directed hypergraphs was defined in Chapter 4). Any local edge-connectivity requirement
can also be formulated in this way.
One difference from the graph case is that in a directed hypergraph the role of the head
node and that of the tail nodes are not symmetric, therefore an orientation cannot be
reversed. It should also be mentioned that if we consider the bipartite graph associated
to a hypergraph as in Chapter 5, then hypergraph orientation can be seen as a kind of
constrained orientation problem for bipartite graphs, where every node in one class should
have out-degree exactly 1. So these problems could be described in the graph framework;
however, the hypergraph formulation is much more comfortable.
6.3.2 Orientation Lemma
The easiest orientation problem is when there is no connectivity requirement, but the
in-degree of every node is specified. The following hypergraph orientation lemma, which
characterizes the existence of such an orientation, is a straightforward extension of the
corresponding graph result (see e.g. [44]). Note that this is not a real generalization,
since it follows from the graph case by considering the bipartite graph associated to the
hypergraph. Here we give a direct proof.
Lemma 6.10. Given a hypergraph H and an in-degree specification mi : V → Z+, there is
an orientation H of H such that ϱ H (v) = mi(v) for every v ∈ V if and only if mi(V ) = |E|
and mi(Y ) ≥ iH(Y ) for every Y ⊆ V .
Proof. The necessity is straightforward. We prove the sufficiency by induction on the
number of hyperedges. Call a set Y tight if mi(Y ) = iH(Y ). Let e ∈ E be an arbitrary
hyperedge, and let Z be the set of nodes in e. Then mi(Z − X) ≥ 1 for any tight set
X for which Z ⊆ X (including X = ∅), otherwise Z ∪ X would violate the condition. If
there is a node v ∈ Z with mi(v) > 0 such that Z ⊆ X for every tight set X containing
v, then we can remove the hyperedge e, decrease mi(v) by one, find a feasible orientation
of the resulting hypergraph by induction, and add the oriented hyperedge (e, h(e)) with
h(e) = v. Otherwise, since a single tight set X ⊇ Z cannot contain every node v ∈ Z with
mi(v) > 0, we can choose tight sets X1 and X2 that are both maximal among the tight
sets that separate Z. Then X1 ∪ X2 is tight and dH(X1, X2) = 0 because of (1.3), therefore
Z − (X1 ∪ X2) = ∅, which contradicts the maximality of X1 and X2.
To see the importance of this Lemma, observe that the in-degrees of an orientation
determine whether the orientation covers a given set function p or not. More precisely, an