MATH 663
S PRING 2012
H OMEWORK #3 S OLUTIONS
Section 1.3
22. (a) Let C be an odd cycle of minimum length l. Let x be a vertex not on C. By way of contradiction,
assume x has at least three neighbors on C. Since the graph is triangle-free, these three neighbors cannot
be adjacent on the cycle. Thus, they split the cycle into three parts and between each adjacency there is
a path of length at least 2 on the cycle. Note that since all three paths together equal l, an odd number,
we know at least one of these paths must be odd. This odd path plus the edges to x from an odd cycle of
smaller length, which is a contradiction. Thus no vertex can have more than 2 neighbors on C.
(b) Let m be the number of edges between C and G − C. Counting edges “out” of C, implies that
m ≥ (k − 2)l, since all vertices have degree at least k. Counting edges “in” from G − C, implies that
m ≥ 2(n − l) using part (a) above.
(c) Construct G by making k/2 copies of Cl . and labelling the vertices of each copy as vi for i = 1 to l.
For each vi add edges to the corresponding vertices vi and vi+2 for all the other cycles, where indices are
taken modulo l.
31. a. The left side counts e(Kn ). The right side counts the same thing by partitioning the vertex set into
two groups: one with k vertices and the other with n − k vertices. Then counting the edges (i) among the
k vertices, (ii) between the two sets, and (iii) among the n − k vertices.
(b) This follows from the observation that if the vertex set of Kn is partitioned into (perhaps many) sets
P ni where the ith set has ni vertices, then
counts the number of edges in each partition. Since each
2
edge of Kn gets counted at most once in this method, the sum is bounded above by e(Kn ).
40. a. e(G) ≤ n2 − a2
P
P
b. Let ki=1 ni = n. So e(G) ≤ ki=1 n2i . Assume n1 ≤ n2 ≤ · · · ≤ nk and component Ci has ni
vertices. Let v ∈ V (Ci ) for i 6= k. If v is moved to Ck , we lose ni − 1 edges but gain nk edges. Thus, if
nk − ni + 1 > 0 we will gain edges by redistributing vertices from smaller components to Ck . Specifically,
that last inequality implies that we gain edges by moving vertices from smaller components into a single
large component. In order to maintain exactly k components, we must leave one vertex in each compo
edges.
nent. Thus, we get n−k+1
2
c. Part b shows we want only two components such that one component is trivial. The number of edges in
this graph is n−1
.
2
41. If G is not connected, then G must have at least two components. Call the first two components C1
1
and C2 . Since ∆(G) = dn/2e, without loss of generality, we can assume there exists a vertex v in C1 such
that d(v) = ∆. Thus, |V (C1 )| ≥ ∆ + 1 = dn/2e + 1. But δ(G) = bn/2c − 1 implies |V (C2 )| ≥ bn/2c.
Thus, |V (C1 )| + |V (C2 )| ≥ dn/2e + 1 + bn/2c = n + 1, a contradiction. So G must be connected.
Section 1.4
4. Let W be a closed odd walk in digraph D. We will proceed by induction on m, the number of edges is
W.
Basis Step: If l = 1, then the walk is a single loop. This is a cycle of length 1 and the proposition holds.
Inductive Step: Assume W has length l > 1 and all closed walks of length less than l contain an odd
cycle. If W has no repeated vertices, then W is itself a cycle of odd the length and the proposition holds.
If W does have a repeated vertex, say x, then we can split W into two smaller closed walks by starting
at x. Each of these has length smaller than W. Since W has odd length, one of the smaller walks has odd
length and so by the inductive hypothesis contains an odd cycle. Thus, W contains an odd cycle. [This is
precisely the argument from Lemma 1.2.15]
10. (⇒:) Assume the digraph G is strongly connected. Let S and T be a nonempty partition of V (G).
Let a ∈ S and b ∈ T. Since G is strongly connected, there exists a path from a to b. Thus, there exists an
edge from S to T.
(⇐:) Assume that for every nonempty partition S and T of the vertex set of the digraph G there exists an
edge from S to T . Let a and b be vertices in G. We need to show there exists an a, b-path. Define S as the
set of vertices, x, such that there exists an ax-path. If b is in this set, we’re done. If not, then b ∈ S = T.
So S and T form a nonempty parition of V (G). Thus there exists an edge from S to T . But this implies
there is at least one vertex in T that is in S, a contradiction. Thus, T must be empty and G is strongly
connected.
11. Let G be a diagraph. Let D be a digraph constructed from G by letting the vertices of D be the
strong components of G. For C1 , C2 ∈ V (D), we say C1 C2 ∈ E(D) if and only if there exists an edge
from some vertex in C1 to some vertex in C2 . From Lemma 1.4.23, we know that if every vertex of D has
outdegree at least 1 (or indegree at least 1), then D contains a cycle. This means that the set of components
on the cycle in D are in the same component in G, a contradiction. Thus, there must be at least one vertex
of D with indegree 0 and a vertex in D with outdegree 0. The result follows.
29. Let G be a graph with an odd cycle C = (v1 , v2 , · · · , vk ) and let D be an orientation of G that is
strongly connected. If every edge on C is oriented vi vi+1 then D contains an (oriented) odd cycle. If not,
then there exists an edge oriented vi+1 vi . Since G is strongly connected, there must exist a vi , vi+1 -path in
D. If this path has an even number of edges, then this path along with the edge of C forms an odd cycle.
But now using edges of C or paths, we can form a (directed) closed odd trail starting at v1 and going
2
through vertices of C. Every odd closed trail contains an odd cycle.
Section 2.1
Pk
4. Let Gi for i = 1 to k, be the distinct components of G. Then e(G) =
i=1 e(Gi ) and n(G) =
Pk
i=1 n(Gi ). Since, e(G) < n(G), we know there exists at least one i such that e(Gi ) < n(Gi ). So
e(Gi ) ≤ n(Gi ) − 1. But Gi is a component and therefor connected. Hence e(Gi ) ≥ n(Gi ) − 1. So Gi is a
connected graph with one less edge than vertex. It must be a tree.
18. Let v be a vertex such that d(v) = ∆(T ) for tree T. Let vwi i = 1, 2, · · · , ∆ be the edges incident
to v. Since T is a tree, for each i, the longest path starting at v in the direction of wi ends in a vertex of
degree 1 and is distinct for distinct i. Thus, T has at least ∆ vertices of degree 1.
The example can be constructed from a star with center of degree ∆. Then insert m = n − ∆ − 1 vertices
of degree 2 into the edges of the star.
26. Assume G is a graph on n vertices with the property that for every v ∈ V (G), G − v is a tree. By
P
counting the edges in every such tree we get: ni=1 (n − 2) = n(n − 2). This sum must count every edge
in G n − 2 times since an edge will appear in the count unless one of its end points are deleted which can
only happen twice. So, e(G) = n. Then n − 2 = e(G − v) = e(G) − d(v) = n − d(v). But this implies
that G is regular of degree 2. Since G − v is connected, G must be a cycle.
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