Graph Theory
Spring 2012
Prof. Gábor Elek
Assist. Filip Morić
Exercise sheet 1: Solutions
Caveat emptor: These are merely extended hints, rather than complete solutions.
1. Let T be a tree. Show that the number of leaves is at least ∆(T ).
Solution. Let v be a vertex with neighbors v1 , . . . , v∆(T ) . For every
i = 1, . . . , ∆(T ) choose a maximal path of the form vvi . . . ui . The
endpoints u1 , . . . , u∆(T ) of these paths are clearly pairwise distinct and
of degree 1.
2. Prove that any connected graph G has a closed walk that traverses
each edge exactly twice.
Solution. Consider a multigraph G0 obtained by ’doubling’ the edges
of G. Since G0 is an even multigraph, it has an Eulerian tour, which
corresponds to a walk in G that traverses every edge twice.
Alternative solution: It is easy to see that the statement holds for trees.
Now, given any connected graph G, we can consider its spanning tree
T . There is a closed walk that double-traverses T and we can easily
insert additional steps in that walk so as to cover the remaining edges.
3. Prove that an Eulerian graph does not have a cut edge.
Solution. Let G be Eulerian and suppose e ∈ E(G) is a bridge. Then
G − e has two components and in each component there is exactly one
vertex of odd degree. Contradiction.
Alternative solution: Note that removing an edge from a (closed) Eulerian walk leaves a non-closed walk that visits all vertices.
4. (a) Is there a graph with degree sequence: 3, 3, 3, 3, 5, 6, 6, 6, 6, 6, 6?
(b) Is there a graph with degree sequence: 1, 1, 3, 3, 3, 3, 5, 6, 8, 9?
Graph Theory
Spring 2012
Prof. Gábor Elek
Assist. Filip Morić
Solution. (a) No, since the number of vertices of odd degree should
be even.
(b) No, since otherwise deleting the vertex of degree 9 would result in
a graph with degree sequence 2, 2, 2, 2, 4, 5, 7, which has 7 vertices and
a vertex of degree 7.
5. Integer weights are written on the edges of a complete graph. Prove
that the total weight of every cycle is even if and only if the total
weight of every triangle is even.
Solution. Suppose the weight of every triangle is even and let v1 . . . vk
be a cycle. Then
w(v1 . . . vk ) ≡
k−1
X
w(v1 vi vi+1 ) ≡ 0
(mod 2) .
i=2
6. Let T be a tree and f : V (T ) → V (T ) be a map that maps two
adjacent vertices into two adjacent vertices OR into one single point.
Prove that f fixes either a vertex or an edge.
Solution. Suppose the contrary. Let v0 ∈ V (T ) be such that d(v0 , f (v0 ))
is minimum. Let v0 . . . vk be the path between v0 and f (v0 ) = vk . Set
vk+1 = f (v1 ), vk+2 = f (v2 ) and so on. Then due to the choice of v0
we can show by induction that d(v0 , vk+i ) = k + i for all i ≥ 1. Since
the tree is finite, this is a contradiction.
Alternative solution: Let T 0 be a tree obtained by removing a leaf x of
T . Then we can define a function f 0 on T 0 that does the same as f
except that the vertices that were sent to x by f are now sent to the
neighbor of x. Now we can apply the induction hypothesis to T 0 and f 0 .
Yet another solution: The sequence V (T ), f (V (T )), f (f (V (T ))), . . .
eventually stabilizes at U ⊆ V (T ). Note that T 0 = T [U ] is a tree and
f |V (T 0 ) is a bijection. In particular, f preserves the degrees. Thus if
we remove the leaves of T 0 to get a tree T 00 , we have that f fixes an
edge or a vertex of T 00 by induction.
If you spot any mistakes on this sheet, please drop an email to [email protected].