Download HW Solutions for Week IV

AMS 550.472/672: Graph Theory
Homework Problems - Week IV
Problems to be handed in on Wednesday, Feb 24: 4, 5, 8.
1. Recall that a tree is always bipartite. Show that a tree always has a leaf in its larger partite
set.
Solution: If there is no leaf in the larger partite set, then every vertex in that set has degree
at least 2. But then the number of edges is at least 2 ∗ |V1 | where V1 is the larger partite set.
Since it is the larger partite set, V1 has at least |V (G)|/2 vertices. Thus, we have at least
|V (G)| = 2 ∗ (|V (G)|/2) edges in the graph, which is a contradiction to the fact that every
tree has exactly |V (G)| − 1 edges.
2. Let the maximum degree of a vertex in a tree be ∆. Show that there are at least ∆ leaves in
the tree.
Solution: Consider the vertex v with degree ∆ in the tree T , and consider T \ {v}. Note that
T \ {v} has ∆ components. Now each component is a tree (it is a connected, acyclic subgraph
because removing vertices cannot create cycles) and so each component has at least 2 leaves.
Adding back v to get T can remove at most one leaf from each component, leaving at least
∆ leaves.
3. Let T be a tree such that every leaf is adjacent to a vertex of degree at least 3. Show that
there are two leaves with a common neighbor.
Solution: Suppose no two leaves have a common neighbor, then the graph obtained by removing all the leaves has degree at least 2. Now removing a leaf creates a new tree, and so
the new graph obtained by removing all the leaves (one by one) should be a tree. However,
if every vertex has degree at least 2, we saw in class that the graph contains a cycle using a
maximal path argument. This is a contradiction.
4. Let d1 , d2 , . . . , dn be n strictly positive integers with n ≥ 2. Show that there exists a tree
with vertex degrees d1 , d2 , . . . , dn if and only if d1 + d2 + . . . + dn = 2n − 2.
Solution: Necessity of the condition follows from the degree-sum formula. We show that if
d1 , d2 , . . . , dn be n strictly positive integers (n ≥ 2) such that d1 + d2 + . . . + dn = 2n − 2,
there exist a tree on n vertices with vertex degrees d1 , d2 , . . . , dn . We proceed by induction
on n.
If n = 2, then d1 + d2 = 2 and since they are strictly positive integers, d1 = d2 = 1. The tree
on 2 vertices with one edge between them satisfies the conditions.
So we assume the statement holds for for all integers 2, . . . , n and consider the statement
for n + 1. So we have strictly positive integers d1 , . . . dn+1 such that d1 + d2 + . . . + dn+1 =
2(n + 1) − 2 = 2n. We also assume, without loss of generality, that the numbers are ordered
in decreasing order: d1 ≥ d2 ≥ . . . ≥ dn+1 .
Claim: dn+1 = 1 and d1 ≥ 2. If dn+1 > 1, then since dn+1 is the smallest number in the
list, the sum of the numbers is at least 2(n + 1) which contradicts the given sum of 2n. Also,
if d1 < 2, that is, d1 = 1, since d1 is the largest number in the list, the sum is at most n + 1
such is strictly less than the required 2n when n ≥ 2. This concludes the proof of the claim.
Consider the following set of n strictly positive integers d01 , . . . , d0n defined as follows: d01 =
d1 − 1 and d0i = di for i = 2, . . . , n. They are strictly positive because d01 = d1 − 1 ≥
2 − 1 = 1 from the claim, and these rest of the numbers are not changed. Also, the sum is
1
d1 − 1 + d2 + d3 + . . . + dn = (d1 + d2 + . . . dn+1 ) − 2 (since dn+1 = 1 by the claim) which
is equal to 2n − 2. by the induction hypothesis, we have a tree on n vertices with vertex
degrees d01 , . . . , d0n . Now add a new vertex to the vertex with degree d01 . This new vertex has
degree 1 = dn ; the degree of its neighbor now becomes d01 + 1 = d1 , and the other degrees are
d2 , . . . , dn . This is exactly what we wanted.
5. Show that a tree with no vertex of degree 2, has more leaves than non-leaf vertices.
Solution: Consider any tree T on n vertices with no vertex of degree two. Let there be k
leaves and n − k non-leaves. Since every non-leaf vertex has at least degree three, we have
P
P
2|E(G)| =
x is a leaf deg(x) +
x is a non-leaf deg(x)
≥ k + 3(n − k)
= 3n − 2k
Since |E(G)| = n − 1, we thus have 2n − 2 ≥ 3n − 2k, which after rearrangement yields
k ≥ n − k + 2 and therefore, k ≥ n − k. This is exactly what we need.
6. Oriented Trees (Arborescences). Let T be a undirected tree and let T̂ be an orientation
of T such that every vertex is the head of at most one edge. Such an oriented tree is called
an arborescence.
(i) Show that there exists a vertex that is not the head for any edge. This vertex is called
a root of T̂ .
Solution: Clearly, T̂ is a directed acyclic graph. Thus, there is a source by part (i) of
the previous problem.
(ii) Show that for every vertex there is a unique directed path to it from a root. Thus
conclude that T̂ has a unique root.
Solution: For any vertex v, there is an undirected path from the root to v. If all the
edges are directed away from the root on this path, then we are done. Otherwise, since
the first edge in the path is directed away from the root (the root is not the head of
any edge), there must be a vertex where the edge directions change. But this would be
a vertex which has two incoming edges, contradicting the assumption that every vertex
has at most 1 incoming edge. Thus, we are done.
7. Let G be a tree with 2k vertices of odd degree. Prove that we can find k paths P1 , P2 , . . . , Pk
such that E(G) = E(P1 ) ] E(P2 ) ] . . . ] E(Pk ).
Solution: We know by the result about Eulerian trails, that the graph decomposes into k
Eulerian trails. Since in a tree, we have no cycles, a trail must be a path. Thus, the graph
decomposes into k paths.
8. Common Intersection property. Let G be a simple graph. We say that a finite collection
of subgraphs G1 , G2 , . . . , Gk has the common intersection property if the following logical
implication is true:
(∀i 6= j V (Gi ) ∩ V (Gj ) 6= ∅) =⇒ (V (G1 ) ∩ V (G2 ) ∩ . . . ∩ V (Gk ) 6= ∅).
[In other words, if we have nonempty pairwise intersections, there is a common vertex.]
(a) Suppose T is a tree, and T1 , T2 , . . . , Tk are subgraphs of T that are also trees (but not
necessarily spanning). Show that T1 , . . . , Tk have the common intersection property.
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Solution: First, if any of T1 , . . . Tk consists of a single vertex, then the assertion is trivial.
So we may assume that each Ti is a tree with at least two vertices. We now prove the
result by induction on the number of vertices n of T .
First, if T is just one vertex, i.e., n = 1, then the result is obvious: each of T1 , T2 , . . . , Tk
are in fact the single vertex T . So the base case of the induction is done.
So assume the statement holds for n = 1, . . . , m and we prove it for n = m + 1. Consider
any leaf v ∈ T . Define T 0 := T \ {v} which is a tree on m vertices by a result from class,
and define Ti0 := Ti \ {v} which are also subtrees of T 0 . We now check that for each i 6= j,
Ti0 ∩ Tj0 6= ∅. We know by assumption that Ti ∩ Tj 6= ∅; if this common vertex is not the
leaf v, then this is a common vertex for Ti0 and Tj0 . If the common vertex for Ti and Tj
is v, then the neighbor u of v is also a common vertex of Ti and Tj since we assume each
of these have at least 2 vertices. Thus, in this case, u is a common vertex for Ti0 and Tj0 .
So we have pairwise nonempty intersection in T10 , . . . , Tk0 . By the induction hypothesis,
there is a common vertex in V (T10 ) ∩ V (T20 ) ∩ . . . ∩ V (Tk0 ). Since V (Ti0 ) ⊆ V (Ti ), we have
a common vertex in V (T1 ) ∩ V (T2 ) ∩ . . . ∩ V (Tk ).
(b) Show that a simple graph G is a forest if and only if every collection of subpaths (subgraphs that are paths) G1 , . . . , Gk has the common intersection property.
Solution: (⇐) If G is not a forest, then there is a cycle and since G is simple, the cycle
is of length at least 3. Now consider any 3 distinct vertices x, y, z on the cycle. Let
P1 be the path on the cycle from x to y, P2 be the path from y to z, and P3 be the
path from y to z. These paths intersect pairwise, but have no common vertex, which
is a contradiction to the common intersection property. Thus, G has no cycles and is a
forest.
(⇒) This direction is similar to the proof of (a).
[Hint: Induction is your best friend !]
9. Let G be a simple n-vertex graph having n − 2 edges. Show that either G has an isolated
vertex, or has two components each of which is a tree with at least 2 vertices.
Solution: Since we have n − 2 edges, we have at least two components. Let the components
be C1 , C2 , . . . , Ck and let ni be the number of vertices in component Ci . We assume that G
has no isolated vertices and show there are at least two trees with at least 2 vertices. This
will prove the statement.
If G has no isolated vertices, each component has at least two vertices. Suppose at most one
of them is a tree, and without loss of generality let that component be Ck (if at all there
is a tree). Then for i = 1, . . . , k − 1, Ci has at least ni edges (otherwise it is a tree), and
of course, Ck has at least nk − 1 edges since it is connected. But then we have at least
n1 + n2 + . . . + nk − 1 = n − 1 edges, which is a contradiction. So we must have at least two
trees amongst C1 , . . . , Ck .
10. Simultaneous exchange property. Let T, T 0 be two distinct spanning trees of a simple
graph G. For any e ∈ E(T ) \ E(T 0 ), show that there exists e0 ∈ E(T 0 ) \ E(T ) such that both
(T \ {e}) ∪ {e0 } and (T 0 \ {e0 }) ∪ {e} are spanning trees of G.
Solution: We fix any edge e ∈ E(T ) \ E(T 0 ). We know removing e from T creates exactly
two components T1 and T2 . Define E = {e0 ∈ E(T 0 ) : endpoints of e0 are in T1 and T2 }. Sice
T 0 is connected, E is nonempty because otherwise, the vertices of T1 are disconnect from the
vertices of T2 on T 0 . Also, e is the unique edge between T1 and T2 in T . Therefore, E has no
edges from E(T ).
We claim that for every e0 ∈ E, T \ {e} ∪ {e0 } is also a tree. This is because adding e0 to T1
and T2 combines them into a single component and by our discussion in class, this means e0
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does not create a cycle when added to T \ {e}. Thus, T \ {e} ∪ {e0 } has no cycles and has
n − 1 edges and is therefore a tree.
Let the endpoints of e be x and y, and without loss of generality assume x ∈ T1 and y ∈ T2 .
Now, there exists a path from x to y in T 0 and therefore one of these edges must have an
endpoint in T1 and the other endpoint in T2 . Such an edge e0 is therefore in E. By the
discussion above, T \ {e} ∪ {e0 } is a spanning tree. We need to check that T 0 \ {e0 } ∪ {e} is a
spanning tree also. Note that T 0 ∪ {e} creates a (unique) cycle which also contains e0 since e0
was on the unique path from x to y in T 0 . Thus, removing e0 from T 0 ∪ {e} leaves a connected
graph with n − 1 edges, which is therefore a tree.
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