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the path. Then by definition, no edge of
S
is in the path, hence all of them
are transitive, and
S
is not a guarding set. Conversely, suppose that
S
is not
guarding. Then there exists an acyclic orientation of
G
in which all edges of
S
are transitive. The corresponding ordering of the vertices in
V
yields a Hamilton
path in
H
.
4.2 Special Cases and an Upper Bound
Lemma 1.
The minimum size of a guarding set in a complete graph on
n
ver-
tices is
n
−
1
.
Proof.
First note that there always exists a guarding set of size
n
−
1 that consists
of all edges incident to one vertex.
Now we need to show that any other edge subset of smaller size is not a
guarding set. This amounts to stating that every graph with at least
2
−
2)
edges has a Hamilton path. To see this, proceed by induction on
n
. Suppose this
holds for graphs with less than
n
vertices. Consider a set
S
of at most
n
(
n
−
−
2
edges of the complete graph. Let
u, v
be two vertices with
uv
∈
S
. One of the
two vertices, say
v
, is incident to at most
(
n
−
2)
/
2
edges of
S
. Consider a
Hamilton path on the
n
=
v
, which exists by induction hypothesis.
Then vertex
v
must have one or two incident edges that do not belong to
S
and
connect
v
to the first, or the last, or two consecutive vertices of this path. Hence
we can integrate
v
in the Hamilton path.
−
1 vertices
Since acyclic orientations of the complete graph
K
n
correspond to the permu-
tations of
S
n
, this is in fact the solution to the puzzle “hitting a consecutive
pair” in the introduction. The dual graph of the arrangements of hyperplanes
corresponding to the complete graph (graphic hyperplane arrangement of
K
n
)
is the skeleton graph of the permutohedron. Hence the above result can also be
stated in the following form.
Corollary 1.
The minimum size of a set of zones covering the vertices of the
skeleton graph of the
n
-dimensional permutohedron is
n
−
1
.
We now give a simple, polynomial-time computable upper bound on the size of
a guarding set. A set
C
ↆ
E
is an
edge cut
whenever the graph (
V,E
\
C
)isnot
connected.
Theorem 4.
Every edge cut of
G
is a guarding set of
G
.
Proof.
Consider an edge cut
C
E
and an acyclic orientation
A
G
of
G
. This
acyclic orientation can be used to define a partial order on
V
. Let us consider a
total ordering of
V
that extends this partial order, and pick an edge
e
=
uv
ↆ
C
that minimizes the rank difference between
u
and
v
. We claim that
e
is flippable.
Suppose for the sake of contradiction that
e
is not flippable. Then
e
must be
transitive and there exists a directed path
P
in
A
G
between
u
and
v
that does
not use
e
. Since
C
is a cut,
u
and
v
belong to distinct connected components of
∈
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