MATH 2200 Homework #9 Solutions
Thomas Goller
April 16, 2013
General Comments:
• Thank you for your hard work!
• Please make your drawings of graphs BIG and neat!
• The singular form of “vertices” is “vertex”. Think of this as analogous to “matrices”
and “matrix”.
• The correct spelling of “bipartite” is “bipartite”.
• If you got a “2”, then please check in with me so I can give you tips on what to work
on for LC #5. I probably had trouble understanding your proofs.
• I only had time to grade some of the questions. Compare your other answers with the
solutions in this document. Many of the exercises have multiple “right” answers, so
your answer doesn’t have to match mine to be fully correct. In case of doubt, come
talk to me.
Exercise 7.3.10
Claim. Let n ∈ Z≥2 . The complete graph Kn has exactly
!n "
2
edges.
Proof #1. Kn has
every pair of distinct vertices.
!n" n vertices and exactly one edge between
!n "
Since there are 2 pairs of distinct vertices, Kn has 2 edges.
Proof #2. Since every vertex in Kn is adjacent to every other vertex, the degree of every
vertex is n − 1. By Exercise 7.1.9,
#
#
2 · |E| =
deg(v) =
(n − 1) = n(n − 1),
v∈V
so
v∈V
n(n − 1)
|E| =
=
2
1
$ %
n
.
2
Exercise 7.3.15
Claim. Let G be a graph. Then G is simple if and only if G has no cycles of length 1 or 2.
Proof. A loop together with its endpoint is a cycle of length 1. A pair of multiple edges
together with their endpoints are a cycle of length 2. Thus G has no loops and no multiple
edges if and only if it has no cycles of length 1 or 2.
Exercise 7.3.20
Claim. Let m, n ∈ Z≥0 . The complete bipartite graph Km,n has exactly mn edges.
Proof. Since Km,n is complete bipartite, we can write V = V1 # V2 , where |V1 | = m, |V2 | = n,
and Km,n has exactly one edge between every vertex in V1 and every vertex in V2 . Thus
there is exactly one edge for each choice of a vertex in V1 and a vertex in V2 , and by the
product rule there are |V1 | · |V2 | = mn ways to choose a vertex in V1 and a vertex in V2 .
Exercise 7.3.21
Claim (a). Let H be a subgraph of a graph G. If G is simple, then H is simple.
Proof. Since G has no loops or multiple edges, neither does H.
Claim (b). Let H be a subgraph of a graph G. If G is a cycle, then H is a cycle.
This claim is false! Counterexample: The subgraph • of the cycle • is not a cycle.
Claim. Let H be a subgraph of a graph G. If G is bipartite, then H is bipartite.
Proof. Since G = (V, E, φ) is bipartite, we can write V = V1 # V2 , where every edge in G has
one endpoint in V1 and the other endpoint in V2 . Then
VH = (V1 ∩ VH ) # (V2 ∩ VH ).
If e ∈ EH , then also e ∈ E, so e has one endpoint in V1 and the other endpoint in V2 . Since
both endpoints of e are in VH , we see that e has one endpoint in V1 ∩ VH and the other
endpoint in V2 ∩ VH . Thus H is bipartite.
Claim. If G is complete bipartite, then H is complete bipartite.
This claim is false! Counterexample: Let
•
G = K1,2 =
•
and
•
•
H=
•
2
•
Exercise 7.4.4
Claim. Let G be a bipartite graph. Then every walk in G that starts and ends at the same
vertex contains an even number of edges.
Proof. Since G = (V, E, φ) is bipartite, we can write V = V1 # V2 , such that every edge in E
has one endpoint in V1 and the other endpoint in V2 . Suppose
(v0 , e1 , v1 , e2 , . . . , vn−1 , en , v0 )
is a walk in G that starts and ends at the same vertex v0 . Since every edge has one endpoint
in V1 and the other in V2 , the vertices in the walk alternate between V1 and V2 . Since the
walk starts and ends at the same vertex, the number of vertices in the walk must be odd,
namely n + 1 is odd. Thus n, which is the number of edges in the walk, is even.
Exercise 7.4.10
Claim (a). Let G be a graph. If G is simple, then G is connected.
This claim is false! Counterexample: G = •
•.
Claim (b). Let G = (V, E, φ) be a graph. If G is complete, then G is connected.
Proof. Let v, w ∈ V . If v = w, then (v) is a walk in G from v to w. If v %= w, then since G
is complete, there is e ∈ E such that φ(e) = {v, w}. Then (v, e, w) is a walk in G from v to
w. So G is connected.
Claim (c). If G = (V, E, φ) is a cycle of length n, then G is connected.
Proof. Since G is a cycle of length n, we can write V = {v1 , . . . , vn } and E = {e1 , . . . , en },
such that φ(e1 ) = {vn , v1 } and φ(ei ) = {vi−1 , vi } for all 2 ≤ i ≤ n. Let vi , vj ∈ V . We may
assume i ≤ j. Then
(vi , ei+1 , vi+1 , . . . , ej , vj )
is a walk in G from vi to vj . So G is connected.
Claim (d). Let G be a graph. If G is bipartite, then G is connected.
This claim is false! Counterexample: Let G = K2,0 = •
•.
Claim (e). Let G be a graph. If G is complete bipartite, then G is connected.
This claim is false! Counterexample: Let G = K2,0 = •
3
•.
Claim (f). Let G = (V, E, φ) be a graph. If G is the complete bipartite graph Km,n for
m, n ≥ 1, then G is connected.
Proof. Since G is complete bipartite, we can write V = V1 # V2 , where |V1 | = m, |V2 | = n,
and G has exactly one edge between each vertex in V1 and each vertex in V2 . Let v, w ∈ V .
If v ∈ V1 and w ∈ V2 , then there is e ∈ E such that φ(e) = {v, w}, and (v, e, w) is a walk in
G from v to w. If instead v, w ∈ V1 , then choose any u ∈ V2 . There are e1 , e2 ∈ E such that
φ(e1 ) = {v, u} and φ(e2 ) = {w, u}, and (v, e1 , u, e2 , w) is a walk in G from v to w. Similar
arguments work when v ∈ V2 and w ∈ V1 or when v, w ∈ V2 . Thus G is connected.
Exercise 7.4.11
Claim (a). Let H1 and H2 be connected subgraphs of a graph G. Then H1 ∪ H2 is connected.
v
w
v
w
This claim is false! Counterexample: Let G = •
• , H1 = • , and H2 = • . Then
H1 and H2 are connected, but H1 ∪ H2 = G is not connected.
Claim (b). Let H1 and H2 be connected subgraphs of a graph G. Then H1 ∩ H2 is connected.
•
This claim is false! Counterexample: Let G = •
• . Then H1 ∩ H2 = •
• is not connected.
4
• , H1 = •
• , and H2 =