2.6 Minimum Diameter of Orientations of Multigraphs
63
2.6 Minimum Diameter of Orientations of Multigraphs
The same complexity result holds for the following problem: find a minimum
diameter orientation of a graph. Indeed, the following assertion holds.
Theorem 2.6.1 (Chv´
atal and Thomassen)
[164] It is
NP
complete to
decide whether an undirected graph admits an orientation of diameter 2.
For a bridgeless multigraph
G
, let diam
min
(
G
) denote the minimum di
ameter of an orientation of
G
. We will present a minor modification of the
original proof of Theorem 2.6.1 by Chv´
atal and Thomassen [164]. The main
difference is in the use of Lemma 2.6.2 (which is applied to two different
results in this section). Define a bipartite tournament
BT
s
, with partite
sets
U, W
, each of cardinality
s
, as follows. Let
U
=
{
u
1
, u
2
, . . . , u
s
}
and
W
=
{
w
1
, w
2
, . . . , w
s
}
. The vertex
u
i
dominates only vertices
w
i
, w
i
+1
, . . . ,
w
i
+
b
s/
2
c
1
(the subscripts are taken modulo
s
) for every
i
= 1
,
2
, . . . , s
.
Lemma 2.6.2
Let
s
≥
2
. The diameter
diam(
BT
s
)
equals 3. In particular,
dist(
U, U
) = dist(
W, W
) = 2
.
Proof:
Clearly, it suffices to show that dist(
U, U
) = dist(
W, W
) = 2
.
This
follows from the fact that, for every
i
6
=
j
, we have
N
+
(
u
i
)

N
+
(
u
j
)
6
=
∅
and, hence, there is a vertex
w
∈
W
such that
u
i
→
w
→
u
j
.
ut
Lov´
asz [520] proved that it is
NP
hard to decide whether a hypergraph of
rank
3
3 is 2colourable. By the result of Lov´
asz, Theorem 2.6.1 follows from
the next theorem.
Theorem 2.6.3
Given a hypergraph
H
of rank 3 and order
n
, one can con
struct in polynomial time (in
n
) a graph
G
such that
diam
min
(
G
) = 2
if and
only if
H
is 2colourable.
Proof:
Let
k
be the integer satisfying 8
≤
k
≤
11 and
n
+
k
is divisible by 4.
Let
H
0
be a hypergraph obtained from
H
by adding
k
new vertices
v
1
, . . . , v
k
.
Moreover, append three new edges
{{
v
i
, v
i
+1
}
:
i
= 1
,
2
,
3
}
to
H
0
if
H
has an
odd number of edges, and add four new edges
{{
v
i
, v
i
+1
}
:
i
= 1
,
2
,
3
,
4
}
to
H
0
otherwise. Observe that
H
0
has an even number of edges, which is at least
four. To construct
G
, take disjoint sets
R
and
Q
such that the elements of
R
(
Q
) are in a onetoone correspondence with the vertices (the edges) of
H
0
.
Let
G
h
R
i
and
G
h
Q
i
be complete graphs, and
p
∈
R
and
q
∈
Q
be adjacent if
and only if the vertex corresponding to
p
belongs to the edge corresponding
to
q
(in
H
0
).
Append four new vertices
w
1
, w
2
, w
3
, w
4
and join each of them to all the
vertices in
R
∪
Q
. Finally, add a new vertex
x
and join it to all the vertices
3
Recall that the rank of a hypergraph is the cardinality of its largest edge.