MATH 682
Problem Set #3 Solutions
1.
(15 points)
Demonstrate the following facts about a directed graph
D
.
(a)
(5 points)
Prove that
∑
v
∈
V
(
D
)
d

D
(
v
) =
∑
v
∈
V
(
D
)
d
+
D
(
v
)
. Recall that
d

and
d
+
represent the indegree and outdegree respectively.
Note that
d

(
v
) =
{
(
v, u
)
∈
E
(
G
)
}
=
∑
u
←
v
1, and similarly with opposite
directionality for
d
+
(
v
). Thus:
X
v
∈
V
(
D
)
d

D
(
v
) =
X
v
∈
V
(
D
)
X
u
←
v
1 =
X
u,v
∈
V
(
G
)
(
v,u
)
∈
E
(
G
)
1 =

E
(
G
)

Likewise,
∑
v
∈
V
(
D
)
d
+
D
(
v
) =

E
(
G
)

.
(b)
(10 points)
Prove that a directed Eulerian tour (i.e.
a directed closed trail
traversing every edge) on a strictly connected graph
D
exists if and only if
d

(
v
) =
d
+
(
v
)
for all
v
∈
V
(
D
)
.
Let us start by supposing
D
has an eulerian tour: we shall show that all vertices
of
D
have equal indegree and outdegree.
Let us denote our eulerian tour by
v
1
→
v
2
→
v
3
→
v
4
→ · · · →
v
m
→
v
1
, and let
e
i
= (
v
i
, v
i
+1
), with the special
case
e
m
= (
v
m
, v
1
).
Then, by the definition of an eulerian tour, all the
e
i
are
distinct and
E
(
G
) =
{
e
1
, e
2
, . . . , e
m
}
.
Now, consider a vertex
u
in
G
.
Some
number
k
of the
v
i
are equal to
u
: let us denote
u
=
v
i
1
=
v
i
2
=
· · ·
=
v
i
k
. Then,
for any
j
, note that
v
i
j
±
1
(using the wraparound convention that
v
0
=
v
k
and
v
k
+1
=
v
1
where applicable) must
not
be identical to
u
, since
v
i
j
±
1
is adjacent
to
u
and
G
is a simple digraph. Thus, since none of the
i
j
are consecutive, the
indices
{
i
1

1
, i
i
, i
2

1
, i
2
, . . . , i
k

1
, i
k
}
are all distinct. Since
u
is the head of
e
i
if and only if either
i
=
i
j
and
u
is the tail if and only if
i
=
i
j

1, it follows
that the above set of indices is exactly the indices of the edges incident on
u
,
and exactly half of them are incoming, and half of them outgoing. Since every
element of
E
(
G
) is associated with exactly one index, the set of edges incident on
u
is equinumerous with this index set, so
d
+
(
u
) =
d

(
u
) =
k
.
To prove that equal indegree and outdegree at every vertex is sufficient, we dis
pense with the trivial case where
D
consists only of an isolated vertex and prove
the nontrivial digraphs are eulerian by contradiction. Consider a nontrivial con
nected digraph
D
in which every vertex has equal indegree and outdegree. Since
D
is strictly connected and nontrivial,
δ
+
(
D
)
≥
1 and
δ

(
D
)
≥
1, since there
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 Spring '09
 WILDSTROM
 Math, Graph Theory, Vertex, NJ, Bipartite graph, eulerian tour

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