2
Solutions
to
In-Class
Problems
Week
7,
Fri.
(a)
For each picture, describe its discrete faces (simple cycles that deﬁne the region borders).
Solution.
Figs 1 & 2: abda, bcdb, abcda. Fig 3: abcdea, adea,abda,bcdb. Fig 4: abcda, abdea, bdcb,
adea.
(b)
Which of the pictured graphs are isomorphic? Which pictures represent the same
planar
em-
bedding
? – that is, they have the same discrete faces.
Solution.
Figs 1 & 2 have the same faces, so are different pictures of the
same
planar drawing. Figs
3 & 4 both have four faces, but they are different, for example, Fig 3 has a face with 5 edges, but
the longest face in Fig 4 has 4 edges.
(c)
Describe a way to construct the embedding in Figure 4 according to the recursive Deﬁni-
tion
12.3.1
of planar embedding. For each application of a constructor rule, be sure to indicate the
faces (cycles) to which the rule was applied and the cycles which result from the application.
Solution.
Here’s one way. (By Lemma
12.7.1
, the constructor steps could be done in any order.)
recursive step
faces
vertex
a
(base case)
a
vertex
b
(base)
b
a
—
b
(bridge)
aba
vertex
c
(base)
c
b
—
c
(bridge)
abcba
vertex
d
(base)
d
c
—
d
(bridge)
abcdcba
a
—
d
(split)
dabcd,
dabcd
b
—
d
(split)
dabd,
dbcd,
abcda
vertex
e
(base)
e
d
—
e
(bridge)
dedabd,
dbcd,
abcda
a
—
e
(split)
abdea,
adea,
dbcd,
abcda
Problem 2.
Prove the following assertions by structural induction on the deﬁnition of planar embedding.
(a)
In a planar embedding of a graph, each edge is traversed a total of two times by the faces of
the embedding.
Solution.
Proof.
The induction hypothesis is that if
E
is a planar embedding of a graph, then each
edge is traversed exactly twice by the faces of
E
.
Base case: