MATH20902: Discrete Maths, Problem Set 7
These problems are also available from http://bit.ly/DiscreteMancMaths. Some
refer to Dieter Jungnickel’s book, Graphs, Networks and Algorithms, an online copy
of which is available from within the university at http://bit.ly/Jungnickel2.
(1) (Euler’s theorem and graphs from regular polyhedra).
We have often thought of the cube graph I3 as being produced by squashing a wireframe model of a cube flat. Draw a planar diagram of I3 and find n, the number of
vertices; f the number of faces and m, the number of edges and use them to verify
Euler’s theorem
n − m + f = 2.
Now do the same thing for the tetrahedron, the octahedron, the dodecahedron
(the necessary planar diagram appears in Problem Set 5) and the icosahedron.
(2) (Edges and bridges).
Recall that an edge e in a connected graph G is a bridge if the graph G\e formed
deleting e form G is not connected.
(a) Sketch two planar graphs, one that includes a bridge and one that does not.
(b) Prove that if an edge is not a bridge, then it is part of at least one cycle.
Subdivisions and Graph Homeomorphism
Suppose that G is a graph with vertex set V and edge set E. Recall that a subdivision
of G is a graph H that can be derived from G by, perhaps repeatedly, performing the
following operation: replace an edge (a, b) ∈ E with a path (a, x1 , . . . , xk , b) where
the x1 , . . . , xk are an arbitrary number of new vertices (that is, vertices not added
in a previous round of subdivision). By convention, a graph G is also considered to
be a subdivision of itself.
Two graphs H and H 0 are said to be homeomorphic if they are isomorphic to
subdivisions of the same graph G.
(3). Draw three non-isomorphic graphs that are all homeomorphic to K3,3 and, for
each, compute the difference |E| − |V |. That is, find the difference between the
number of edges and the number of vertices.
(4) (After Jungnickel’s exercise 1.5.6). Let H be a graph with vertex set V and
edge set E. Similarly, let H 0 be a second graph with vertex set V 0 and edge set E 0 .
Prove that if H and H 0 are homeomorphic, then
|E| − |V | = |E 0 | − |V 0 |.
(5) (After Jungnickel’s exercise 1.5.9). The Petersen graph is shown below:
Prove that it is not a planar graph in the following two ways:
(a) using the bound on m proved in lecture, which is restated below;
(b) using Kuratowski’s theorem, which is also restated below.
(6) (After Jungnickel’s exercise 1.5.12). What is the minimum number of edges that
one has to remove from Kn to get a planar graph? For each n, construct a planar
graph having as many edges as possible. Hint: use the Corollary below.
(7) (After Jungnickel’s exercise 1.5.13). Let G be a planar graph on n ≥ 3 vertices
and denote the number of vertices of degree at most d by nd . Prove
nd ≥
n(d − 5) + 12
d+1
Theorem. If G(V, E) is a connected planar graph with n = |V | vertices and m = |E|
edges then either:
(a) G is acyclic and m = n − 1;
(b) G has at least one cycle and so has a well-defined girth g. In this case
m ≤
g(n − 2)
.
g−2
Corollary. If G is a connected, planar graph with n vertices and m edges then
m ≤ 3n − 6.
Theorem (Kuratowski). A graph G is planar if and only if it does not contain a
subgraph homeomorphic to either K3,3 or K5 .
Two more problems about planar graphs
The last two problems both come from Chapter 10 of Bondy and Murty’s Graph
Theory.
(8) (Direct proofs that K5 and K3,3 aren’t planar). We proved that K5 isn’t planar
using a hard-won inequality that bounds the number edges in a planar graph in
terms of the graph’s girth and number of vertices, but it is also possible to prove
that K5 is non-planar more directly, by contradiction.
Take the vertex set to be {v1 , v2 , v3 , v4 , v5 } and assume that we have a planar
diagram. Consider the cycle C ≡ (v1 , v2 , v3 , v1 ): as the diagram is planar, the points
and arcs corresponding to this cycle must form a Jordan curve. Without loss of
generality, we can assume that v4 is in the interior of this curve. And from this
assumption it follows that, except for their endpoints, the arcs corresponding to the
edges (v1 , v4 ), (v2 , v4 ) and (v3 , v4 ) must also lie in the interior of C: see the figure
below.
v1
v2
v4
v3
Now define C1 ≡ (v2 , v3 , v4 , v2 ), C2 ≡ (v1 , v3 , v4 , v1 ) and C3 ≡ (v1 , v2 , v4 , v1 ) and
note that vj is in the exterior of Cj for j = 1, 2 and 3. Then proceed as follows:
(a) Prove that the planarity of the diagram implies that v5 lies in the intersection
of the exteriors of the Cj .
(b) Show that this implies that the arc (v4 , v5 ) must cross C, which contradicts
planarity.
(c) Invent a proof in the same style to show that K3,3 can’t have a planar diagram.
(9) (A problem related to a conjecture by John Conway). The diagrams studied in
this problem are a kind of opposite to planar diagrams. In a planar diagram the
arcs representing edges aren’t allowed to cross at all, but in the diagrams discussed
here they must do so.
A diagram for a graph is said to be a thrackle 1 or a thrackle embedding if:
• the arcs representing edges do not cross themselves, but
• every pair of distinct arcs cross exactly once, either at their endpoints (that
is, both members of the pair are incident on a common vertex) or somewhere
in their interiors.
The diagrams below illustrate two thrackles (left and center) and one diagram
that isn’t a thrackle (at right).
Conway conjectured that if a graph has a thrackle embedding then m ≤ n. We’ll
consider some cases where either m = n or m = n − 1.
Prove that:
(a) the cycle graph C3 has a thrackle embedding;
(b) all cycle graphs Ck for which k ≥ 3 is an odd number have a thrackle embedding;
(c) the cycle graph C4 does not have a thrackle embedding.
1
Apparently “thrackle” is a somewhat obscure dialect word for a tangle, such as a tangled-up
fishing line. Dan Archdeacon at the University of Vermont reports (http://bit.ly/thrackle)
that Conway has said:
When I was a teenager, on holiday with my parents in Scotland, we once stopped to
ask directions of a man who was fishing by the side of a lake. He happened to mention
that his line was thrackled. I’d previously called this kind of drawing a tangle, but
since I’d just found a knot-theoretical use for that term, I changed this to thrackle.
Several people have told me that they’ve searched in vain for this word in dialect
dictionaries, but since I quizzed the fisherman about it, I’m sure I didn’t mishear it;
he really did use it.