Homework 6 Solutions - People.stat.sfu.ca

Homework 6 Solutions
1. Prove that a set of edges in a connected plane graph G form a spanning tree if and only
if the duals of the remaining edges form a spanning tree in G ∗ .
Solution see problem 3 from homework 5.
2. Use Exercise 1 to obtain a new proof of Euler’s Formula.
Solution: Let G be a connected plane graph with v vertices, e edges, and f faces. Let G ∗ be
a dual of G. Let S ⊆ E(G) be the set of edges of a spanning tree in G, let T = E(G) \ S,
and let T ∗ be the edges in G ∗ dual to those in T . Using the above exercise, we now have
2 = (|V (G)| − |S|) + (|V (G ∗ )| − |T |) = v − e + f as desired. ✷
3. Let G be a connected 3-regular plane graph with n vertices in which every vertex lies on
one face of length 4, one face of length 6, and one face of length 10.
(a) In terms of n, determine the number of faces of each length.
(b) Use Euler’s Formaula to determine n.
Solution: Let f denote the total number of faces and let fi denote the number of faces of
length i for i ≥ 1. By counting the number of times a face of size i is incident with a vertex
for i = 4, 6, 10 we have n = 4f4 = 6f6 = 10f10. Thus f4 = n
4 , f6 = n
6 , and f10 = n . This
10
solves (a). For (b), observe that G is 3-regular, so setting e = |E(G)| we find 2e = 3n so
e = 3n.
Now, by Euler’s formula we get the following equation.
2
2 = n − e + f
= n − 3n
2
= n
60
n n n
+ + + 4 6 10
So we conclude that n = 120 (in fact, in this case G must be isomorphic to the great rhombic
icosadodecahedron).
Peripheral Cycles: A cycle C ⊆ G is called peripheral if C is an induced subgraph of G
and G − V (C) is connected.
4. Let G be a 3-connected graph, let C be a cycle of G, and let H1, H2, . . . , Hk be the
components of G − V (C) ordered so that |V (Hi)| ≥ |V (Hi+1)| for 1 ≤ i ≤ k − 1. Show that
if C is chosen so that

(i) |V (H1)| is maximum
(ii) |V (H2)| is maximum subject to (i)
(iii) |V (H3)| is maximum subject to (i) and (ii)
.
then the cycle C is peripheral. (Hint: consider Hk)
Solution: We shall prove a slightly stronger result with a very similar argument. Namely, we
prove that for every edge e and every vertex x not incident with e, there exists a peripheral
cycle which contains e and does not contain x. Note that this implies the existence of at
least two peripheral cycles through every edge. Let C be a cycle which contains the edge
e and does not contain x (one must exist by 3-connectivity) and order the components
H0, H1, H2, . . . , Hk so that x ∈ V (H0) and |V (Hi)| ≥ |V (Hi+1| for 1 ≤ i ≤ k − 1. Suppose
further that C is chosen so that
(i) |V (H0)| is maximum
(ii) |V (H1)| is maximum (subject to (i))
.
We claim that C is peripheral. If C is not an induced subgraph, then there is a cycle
with fewer vertices in G[V (C)] containing e and this contradicts our choice. Thus, C is an
induced subgraph. Suppose (for a contradiction) that k ≥ 1 and consider the component
Hk. Let y, y ′ ∈ V (C) be the two vertices closest to e on the cycle in the clockwise and
counterclockwise directions, and let P ⊆ C be the path of C from y to y ′ which does not
contain e, and let P ′ ⊆ C be the path from y to y ′ which does contain e. It follows from
our connectivity that y = y ′ . Further, there must exist a component Hj with j < k which
contains a vertex adjacent to a point in the interior of the path P , since otherwise {y, y ′ } is a
cutset. Now, we may choose a path Q from y to y ′ with all interior vertices in Hk. Replacing
C by the cycle P ′ ∪ Q now gives us a contradiction, since the size of the component Hj will
increase. Thus, k = 0 and C is peripheral.
5. A slight extension of the previous exercise shows that for every 3-connected graph G and
every edge e ∈ E(G), there exist at least two peripheral cycles containing e. Observe that if
G is planar, there must be exactly two peripheral cycles containing e. Use this fact to prove
that any two duals of a 3-connected planar graph are isomorphic.
Solution Since every cycle in a plane graph separates the plane, every peripheral cycle in a
2

plane graph must be the boundary of a face. It follows that every planar graph has ≤ 2
peripheral cycles through any given edge. Combining this with the previous problem, we find
that for a 3-connected planar graph G, every edge of G is contained in exactly two peripheral
cycles. Thus, for any drawing of G in the plane, the boundaries of the faces must be exactly
the peripheral cycles. We conclude that any two graphs which are dual to drawings of G
must be isomorphic.
3