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4.4
CHAPTER 4. TREES
Problems and solutions
3.1.3 Show that a tree T of order n ≥ 2 is bipartite.
Obviously a tree contains no odd cycles, so by Theorem 1.3 a tree is
bipartite.
Proof by induction, with the first case n = 2 obvious. For n > 2, discard
a leaf v and the incident edge vw. What remains is a tree of lower order,
which is thus bipartite. Color the leaf v the opposite color of the adjacent
vertex w.
3.2.2 Suppose a tree T has an even number of edges. Show that at least
one vertex has even degree.
A tree of order N has N − 1 edges, so if T has an even number of edges
then it has an odd number of vertices.
Recall the formula
X
deg(v) = 2NE .
If all degrees were odd then
X
deg(v) mod 2 = 1,
but the sum is 0.
3.2.3 Let T be a tree with max degree ∆. Prove that T has at least ∆
leaves.
Use induction on the number of vertices, with the max degree equal to
∆. The initial case will be a tree T0 with ∆ + 1 vertices. It has a vertex v
with adjacent vertices w1 , . . . , w∆ . Since T0 is a tree, the only edges are vwn ,
and each wn is a leaf.
Assume the result is true for trees with fewer than N vertices. Let T
be a tree with max degree equal to ∆ and N > ∆ + 1 vertices. Suppose
deg(v) = ∆. Let w be a vertex chosen to maximize d(v, w). Then w is a leaf
which is not adjacent to v. By the induction hypothesis, T − w has at least
∆ leaves, and adding back w does not decrease the number of leaves.
3.2.5 Prove that a graph G is a tree if and only if for every vertex pair
u, v, there is exactly one path from u to v.
Suppose G is a tree. Pick two distinct vertices u1 and u2 . Since T is
connected there is path from u1 to u2. Suppose there are two different paths,
u1 = v1 , v2 , . . . , vM = u2 ,
u 1 = w1 , w2 , . . . , wN = u 2 .
4.4. PROBLEMS AND SOLUTIONS
51
Pick j as large as possible so that vi = wi for i ≤ j. If j < M, then
vj , vj+1, wj+1 are distinct vertices. The walk vj , . . . , vM , wN −1 , . . . , wj+1 contains a path from vj to wj+1, and so we have a cycle by adding the edge
wj+1 vj . But T has no cycles, so the path is unique.
Suppose G is not a tree. If G is not connected there are two vertices not
joined by any path. Suppose G is connected and has a cycle with vertices
v1 , . . . , vN , N ≥ 3. Then v1 and vN are joined by distinct paths v1 , . . . , vN
and the path consisting of the single edge v1 vN . Thus if a graph G is not a
tree, then there is a vertex pair u, v joined by a number of paths not equal
to one.
3.2.10 Let T be a tree of order N > 1. Show that the number of leaves is
X
2+
(deg(v) − 2).
deg(v)≥3
Use induction on the number of vertices N, with the first case N = 2
being trivial. Assume the result holds for trees with fewer than N vertices.
A tree T with N > 2 vertices has a leaf v with adjacent vertex w. By the
induction hypothesis the result holds for the tree S = T − v. Add back v.
The number of leaves does not increase if and only if w is a leaf of S, which
occurs if and only if deg(w) = 1 in S, or deg(w) = 2 in T . If deg(w) ≥ 2
in S, then the number of leaves in T is one more than in S, and deg(w) − 2
increases by one for the single vertex w which has degree at least 3 in T .
3.2.12 Let T be a tree such that every vertex adjacent to a leaf has degree
at least 3. Prove that some pair of leaves has a common neighbor.
The result is true if there is only a single vertex which is not a leaf. If
there are at least two vertices which are not leaves, pick one such vertex u,
and let v be the most distant vertex which is adjacent to a leaf w1 . The
unique path from u to v contains exactly one vertex w2 adjacent to v. By
the degree requirement, v is adjacent to another vertex w3 , which must be a
leaf by the ’most distant vertex’ requirement.
3.3.2 Prove that a graph G is a tree if and only if it is connected and has
exactly one spanning tree.
First suppose G is a tree with N vertices. Then G is connected and has
itself as a spanning tree. If G had another spanning tree, then G would have
at least N edges, which is impossible since G is a tree. Thus a tree has a
unique spanning tree.
Suppose G is not a tree, but is connected. Then G contains a cycle
with vertices v1 , . . . , vN and edges v1 v2 , . . . , vN v1 . The edges v1 v2 and v2 v3
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CHAPTER 4. TREES
are distinct. By removing edges different from v1 v2 from cycles we may
construct a spanning tree containing v1 v2 , and similarly for v2 v3 . Thus if G
is connected and contains a cycle, it has more than one spanning tree.
3.3.4 Let G be connected and let e be an edge of G. Prove that e is a
bridge if and only if it is in every spanning tree of G.
Recall that e = v1 v2 is a bridge in a connected graph G if and only if
G − e is disconnected. In particular v1 and v2 cannot be joined by a path in
G − e.
Suppose e is not a bridge, so that G − e is connected. Then G − e, as
well as G, contains a spanning tree not containing e. Thus e is not in every
spanning tree of G.
Suppose e is not in every spanning tree. Then G − e has a spanning tree,
so G − e is connected, and e is not a bridge.