Graph Theory
Excercise Set 1
1.12 Determine κ(G) and λ(G) for the following graphs:
P m , m ≥ 3 : Path of m vertices.
A path with m vertices is connected therefore λ(P m ) > 0 and
κ(P m ) > 0 . The path connecting the two leaves in P m can be cut
by removing any edge therefore λ(P m ) ≤ 1 which together implies
λ(P m ) = 1 . Since κ(P m ) ≤ λ(P m ) it follows that κ(P m ) = 1 as
well.
C n , n ≥ 3 : Cycle of n vertices.
Given any pair of vertices we can describe at least two independent
paths connecting them, one moving clockwise one moving counterclockwise. Therefore 2 ≤ κ(C n ) ≤ λ(C n ) . Since it is 2 -regular we
know that the minimal degree is δ(C n ) = 2 and hence λ(C n ) ≤ 2 .
This implies that κ(C n ) = λ(C n ) = 2 .
K n , n ≥ 3 : Complete graph on n vertices.
We use induction over n . We see that for n = 3 removing any vertex
leaves us with a K 2 which is still connected. Removing a second
vertex results in a K 1 . By definition κ(K 1 ) = 0 and therefore
κ(K 3 ) = 2 . Now when removing any vertex from K n the remaining
graph has n − 1 vertices all of which are directly connected by edges,
a K n−1 . Therefore κ(K n ) = n − 1 . Since κ(K n ) ≤ λ(K n ) ≤
δ(K n ) = n − 1 we see that λ(K n ) = n − 1 .
Km,n , m, n ≥ 1 : Complete bipartite graph on m and n vertices.
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Graph Theory
Excercise Set 1
We see that Km,n is connected for m, n > 0 . Removing a vertex
leaves either a Km−1,n or Km,n−1 and these are connected as long
as m − 1 6= 0 or n − 1 6= 0 . Therefore at least min(m, n) vertices
need to be removed, i.e. κ(Km,n ) = min(m, n) . Since κ(Km,n ) ≤
λ(Km,n ) ≤ δ(Km,n ) and δ(Km,n ) = min(m, n) we see that also
λ(Km,n ) = min(m, n) .
Zd2 , d ≥ 3 : d -dimensional cube. To find a lower bound on κ(Zd2 ) we
d=1
d=2
d=3
enumerate independent paths between two vertices x and y . We
need to differentiate two cases.
i.) If x and y differ in all coordinates we construct d independent
paths by changing the coordinates of x in cyclic order till y is
reached. We can start with any of the d different coordinates of
x and each of these will lead to an independent path.
ii.) We see that for d = 2 there are exactly 2 paths between any
pair of points. If x and y agree on at least one coordinate,
they can be embedded into a Z2d−1 by removing the common
coordinate. By the induction assumption there are at least d − 1
independent paths in this embedding. One more path can be
constructed by changing the common coordinate in x and y , so
they become x0 and y 0 and then using any of the d − 1 paths
in the embedding of x0 and y 0 into Zd−1
.
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This shows that d ≤ κ(Z2d ) . But since δ(Z2d ) = d we have κ(Z2d ) =
λ(Zd2 ) = d .
1.18 Let G be a tree without degree 2 vertices. Let L ⊂ V be the subset of
leaves of G . Since d(v) ≥ 3 for all v ∈ G \ L we see that
X
d(v) = 2|E| = 2|V | − 2
v∈V
|L| + 3|G \ L| ≤ 2|L| + 2|G \ L| − 2
|G \ L| ≤ |L| − 2
Hence there are more leaves than other vertices.
3.7 The vertices of a block graph B of a graph G are the maximal 2 connected subgraphs, the bridges, and the cutvertices of G . If G is
connected then so is B since it can be constructed from G through contractions and subdivisions. We show that B is acyclic by contradiction.
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Graph Theory
Excercise Set 1
Assume there is a cycle in B , then there are at least two vertices that
have 2 or more independent paths between them. This is a contradiction
to the maximality of the 2 -connectedness of the blocks.
Therefore B is connected and acyclic, hence a tree.
3.15 The proof claims to use an induction, this requires GA and GB to be
’smaller’ in some sense. We construct an easy counterexample.
y
z
x
Now let A = {y, z} , B = {x, y} , and choose as a minimal cut set X = B .
It then is easy to see that GA = G and no induction is possible.
3.25 We need to show that any k -linked graph is 2k−1 -connected. We assume
G is not 2k − 1 -connected and then prove that G can not be k -linked.
If G has less than 2k vertices it can not be k -linked. We can therefore
assume it has 2k or more vertices and since it is not 2k − 1 connected
there exists a cut set X (not necessarily minimal) of size 2k − 2 . There
are at least two verticesn a, b ∈ G \ X that are separated by the cutset
X . Partition X = XA t XB into two sets of equal size k − 1 . We then
see that the sets (a, XA ) and (b, XB ) cannot be linked since there exists
no path disjoint from X between a and b . Therefore G is not k -linked.
An easy example of a graph that is k -linked but not 2k -connected is the
K 4 , which is 2 -linked but only 3 -connected.
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