MATH20150 Graphs and Networks
Tutorial 9 Solutions
Graph isomorphism and symmetry. Planar graphs.
1. a) The trees are nonisomorphic, though they have the same
degree sequences (1, 1, 1, 1, 1, 1, 2, 2, 3, 3, 4). To see they are
nonisomorphic, observe that the vertices of degree 3 are adjacent to vertices of degrees (1, 1, 4) and (1, 2, 2) respectively
in the left-hand tree and of degrees (1, 2, 4) and (1, 1, 2) in
the right-hand tree.
b) The picture below shows all nonisomorphic trees of order
6. One can easily see that each tree there has at least one
nontrivial symmetry.
Can you construct a tree without nontrivial symmetries?
2
2. a) Yes, the following two graphs are isomorphic.
An isomorphism can be given as follows:
0 7→ a
5 7→ i
1 7→ b
7 7→ d
2 7→ c
9 7→ e
3 7→ g
6 7→ h
4 7→ f
8 7→ j
b) Here is a planar embedding of the pentagonal prism.
It has 7 faces, so the Euler formula holds:
|V | − |E| + |F | = 10 − 15 + 7 = 2.
The degrees of the infinite face and of the internal face are
5, five other faces have degree 4. Therefore
X
d(f ) = 2 · 5 + 5 · 4 = 30 = 2 · |E|
f ∈F
in accordance with the Face Shaking Lemma.
This graph is not isomorphic to either one in part a). One
way to see this is that it has cycles of length 4 and graphs
in part a) don’t. Try to invent other arguments to see this!
3. a) Since in any simple graph there are no cycles of length less
than 3, in a planar the degree of any face is at least 3 and
the FSL gives
X
2 |E| =
d(f ) ≥ 3 |F | .
f ∈F
.
b) Substituting |F | = |E| + 2 − |V | (Euler’s formula) in the
above inequality, we get |E| ≤ 3|V | − 6.
c) Any graph with “too many” edges, e.g. K5 .
3
4. If Petersen’s graph was planar it would have 2 + |E| − |V | =
2 + 15 − 10 = 7 faces. But it has no cycles of length less than 5,
so every face would have degree at least 5. This contradicts the
FSL:
X
30 = 2|E| =
d(f ) ≥ 5|F | = 35 .
f ∈F