Homework Number 8 Solutions
HW pp. 449 −450:
#3 No because the graph has an odd number of odd-degree vertices (namely one).
#4 No. Since 3 of the 5 vertices have maximum degree (4), this means these three vertices are
adjacent to each vertex. Therefore, the degree of the other two vertices is at least 3.
#5 For each vertex v, we have 0 ≤deg(v)≤n−1.This means we have npossibilities for the
degree of each vertex and nvertices. If no two vertices have the same degree, then for each ksuch
that 0 ≤k≤n−1,there exists a vertex vwith deg(v) = k. This means there is a vertex of degree
0 and a vertex of degree n−1.This can’t be since the vertex of degree 0 is adjacent to no other
vertex and the vertex of degree n−1 is adjacent to all the vertices. Therefore, there are only n−1
possibilities for the degree of each vertex(there are nvertices). By the pigeonhole principle, there
exists two vertices with the same degree.
#9
#12 Note that the 1st graph is not isomorphic to the 3rd graph because the 3rd has a 4-cycle, but
the 1st graph does not. Similarly for the 2nd and 3rd graphs. Below, find an isomorphism for the
1st and 2nd graphs.
#30 Knhas an Eulerian Circuit (closed Eulerian trails) if the degree of each vertex is even. This
means nhas to be odd, since the degree of each vertex in Knis n−1.
Knhas an Eulerian trail (or an open Eulerian trail) if there exists exactly two vertices of odd
degree. Since each of the nvertices has degree n−1,we need n= 2.
