HOMEWORK 2
SOLUTIONS
(1) Let G be a simple graph where the vertices correspond to each of the squares of an
8
×
8 chess board and where two squares are adjacent if, and only if, a knight can
go from one square to the other in one move. What is/are the possible degree(s)
of a vertex in
G
? How many vertices have each degree? How many edges does
G
have?
Solution:
A knight moves two squares horizontally and one square vertically, or
two squares vertically and one square horizontally, on a chessboard. By examining
all possibilities, we see that; the four “corner” vertices of the graph have degree
2; the eight “edge” vertices that are next to the corners have degree 3; twenty
vertices, the remaining 16 edge vertices plus four more that are next to the corners
but not on the edge, have degree 4; sixteen vertices have degree 6; the remaining
16 “interior” vertices have degree 8.
The preceding paragraph answers the first two questions. For the last question
we appeal to the HandShaking Theorem.

E
(
G
)

=
1
2
X
v
∈
V
(
G
)
d
G
(
v
) =
1
2
(4
×
2 + 8
×
3 + 20
×
4 + 16
×
6 + 16
×
8) = 168
(2) Let
G
be a graph with
n
vertices and exactly
n

1 edges. Prove that
G
has either
a vertex of degree 1 or an isolated vertex.
Solution:
Another way to state this problem would be: Prove that
G
has at
least one vertex of degree less than 2. We will prove this by contradiction, that is
we will start out by assuming that the degree of each of the vertices of
G
is greater
than or equal to 2.
Under this assumption, the HandShaking Theorem gives us
2
n

2 = 2(
n

1) = 2

E
(
G
)

=
X
v
∈
V
(
G
)
d
G
(
v
)
≥
2 + 2 +
· · ·
+ 2

{z
}
n
times
= 2
n
But this is clearly impossible. Therefore our assumption must be false and there
must be at least one vertex that has degree less than 2.
(3) Prove that if a graph
G
has exactly two vertices
u
and
v
of odd degree, then
G
has
a
u, v
path.
Solution:
We will prove the statement by contradiction. That is, we first assume
that
G
is a graph with exactly two vertices of odd degree,
u
and
v
, and that there
is
no
u, v
path in
G
.
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2
SOLUTIONS
By definition,
G
is a disconnected graph and the vertices
u
and
v
must lie in
separate connected components of
G
.
Thus the vertex
u
lies in
H
, a connected
subgraph of
G
, and all other vertices in
H
(which, of course, are also vertices in
G
)
have even degree. Therefore, the sum of the degrees of the vertices in
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 Spring '11
 ADEBOYE
 Graph Theory, Planar graph, vertices, complete graph, Complete bipartite graph

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