Math 575
Problem Set #1
1. Suppose that G is a graph and the degrees of its vertices are
1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3 5, 5, 5, 7.
How many edges does G have?
Solution: 25
2. A graph is cubic if every vertex has degree 3. Suppose that G is a cubic graph on
n vertices. How many edges does G have?
3n
Solution:
.
2
3. (a). Explain why every cubic graph must contain an even number of vertices.
(b). Give an example of a cubic graph on each of 4, 6 and 8 vertices.
(c). Show that for every even integer n, there is a cubic graph that contains n
vertices.
4. Suppose that G is a graph that contains 14 vertices and 25 edges. Given that every
vertex of G has degree 3 or 5. How many vertices of G have degree 3?
Solution: There are four of degree 5 and ten of degree 3.
A vertex of a graph G is said to be isolated if its degree is 0.
5. Suppose that G is a graph that had no isolated vertex. Show that there must be two
vertices of G that have the same degree.
Solution: Suppose that G has n vertices. There are only n – 1 possible degrees for
the vertices. Since there are n vertices, the Pigeon Hole Principle guarantees that
some two must have the same dregree.
6. (a). Suppose that for every two non-adjacent vertices v and u of the graph G,
deg(v) + deg(u) ! n " 1 . Show that G is connected.
Solution: Suppose that G is disconnected and A and B are separate components
of G. Let a be a vertex in A and b be a vertex in B. Then
deg(a)!!!| A| "1 and deg(b) !!| B | !"1. Hence,
deg(a) + deg(b) !!| A| "1+ | B| "1 ! n " 2. But this contradicts the assumption
made in the problem.
(b). Suppose that G is a graph on n vertices and for every vertex v of G,
n "1
. Show that G is connected.
deg(v) !
2
Hint: use part (a).
7. Let G be a cubic graph on n vertices.
Suppose that the edge e = xy is a bridge in G. Show that the two components
of G – e each contain an odd number of vertices.
Solution: Each of the two components has exactly one vertex of degree 2 and the
others of degree 3. Since there are an even number of the degree 3 vertices, there
are an odd number of vertices in the component.
8. Prove that if G is a connected graph with radius r and diameter d, then r ≤ d ≤ 2r.
Hint: You may assume that d(v, w) ! d(v,u) + d(u, w) for all vertices v, u and w.
For the right-hand inequality, begin by choosing two vertices v and u for which
d(v, u) = diam(G) = d, and a vertex w that belongs to the center of G.
Solution: The left-hand inequality is immediate from the definitions of the terms,
i.e., r = min{e(v) : v !V (G)} " max{e(v) : v !V (G)} = d .
For the right-hand inequality, choose two vertices v and u for which
d(v, u) = diam(G) = d, and a vertex w that belongs to the center of G.
Then we have, d = diam(G) = d(v,u) ! d(v, w) + d(w,u) ! e(w) + e(w) ! 2r.
Here we used:
d(v, w) ! e(w) and d(u, w) ! e(w) and e(w) = r since w is in the center.