Massachusetts Institute of Technology
6.042J/18.062J, Spring ’10
: Mathematics for Computer Science
March 19
Prof. Albert R. Meyer
revised March 22, 2010, 587 minutes
Solutions to InClass Problems Week 7, Fri.
Problem 1.
Figures 1–4 show different pictures of planar graphs.
1
a
b
c
d
Figure 1
a
b
c
d
Figure 2
a
b
c
d
e
Figure 3
a
b
c
d
e
Figure 4
Creative Commons
2010,
Prof. Albert R. Meyer
.
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2
Solutions
to
InClass
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.
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(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.
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(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:
There is one vertex and no edges, so this case holds vacuously.
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 Spring '11
 Prof.AlbertR.Meyer
 Computer Science, Graph Theory, line graph, Planar graph, planar embedding

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