Solutions to Homework 2
1.3.3
Theorem 1,3,4. Every tree with at least one edge has a least two vertices.
Proof by contradiction. Suppose that at most one edge in the tree has degree one. We saw
that by Theorem 1.3.1, that there was at least one vertex of degree1. Using Theorem 1.1.1
p
and Theorem 1.3.2, we have 2( p  1)  2q   di  2( p  1)  1 . This is a contradiction.
i 1
1.3.7.
We will use Theorem 1.1.1.
5+4+3+2+k = 2q, where k is the number of vertices of degree 1. That is, 14 + k = 2q. By
Theorem 1.3.2, q  p  1 and p  4  k  q  3  k  14  k  2(3  k )  6  2k  k  8.
1.3.8
We will use Theorem 1.1.1.
p
d
i 1
i
 2q where all the di are odd. Since the sum of an odd number of di would result in
an odd number, we must have p be even. By Theorem 1.3.2, q = p – 1 and q must be odd.
2.1.2
3 if n is even
 4 if n is odd
 (Wn )  
2.1.7
10; every vertex is at a distance 1 or 2 from every other vertex other than itself.
2.1.14
Every vertex in the graph below is on 4 different 3-cycles. We show that it does not
contain a subgraph K3,3. With the deletion of one edge from an incident vertex, that
vertex remains on 2 3-cycles. Therefore, two edges must be deleted from each of the 6
vertices. But, we would then have a graph in which every vertex has degree 2. Every
vertex in K3,3, has degree 3.
2.1.18
n+1
Every cycle of length n either includes the hub or does not include the hub. There is
exactly one n-cycle that does not include the hub. Any n-cycle that does include the hub
does not include an edge on the outer rim. There are n such edges.