1 Solution of Test - UNC Charlotte

Math 3116 Name:
Dr. Franz Rothe
June 24, 2013

08SUM\3116_2013t1.tex

1 Solution of Test

Lemma 1 (Return lemma). In a graph all vertices of which have degree at least two, a
cycle exists.

10 Problem 1.1. Prove the Lemma.

Answer. One can continue a path until it is maximal. Let u be its end vertex. Because
the degree of u is at least two, it has another neighbor w besides its predecessor in the
path. Since the path is maximal, this neighbor w must be an earlier vertex of the path.
Thus we complete a cycle.

Definition 1. A vertex of degree one of a tree is called end-vertex or leaf.

10 Problem 1.2. Show that every tree has at least two leaves.

Answer. For d = 1, 2, 3, . . . let ni count the vertices of a tree of degree d. We use the
handshaking lemma in the form

2m = dnd

d≥1

the fact that a tree has m = n − 1 edges, and the obvious count

n = nd

d≥1

Hence

2 = 2n − 2m = (2 − d)nd = n1 − (d − 2)nd

d≥1 d≥3

which implies n1 ≥ 2 for the number of leaves.

Second proof. Let P = v1, v2 . . . vk be a maximal path in the given tree T . The vertex
v1 cannot have any other neighbor except v2 among the vertices of the path, because
T has no cycles. Too, the vertex v1 cannot have any neighbor among the remaining
vertices of T \ P , since the path is maximal. Hence v2 is the only neighbor of v1 and
deg(v1) = 1. Similar we see that deg(vk) = 1 for the vertex vk at the other end of the
maximal path.

Hence both end-vertices of any maximal path have degree one. Especially, in a tree

there exist at least two vertices of degree one.

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10 Problem 1.3. For which polyhedral graphs is the number n of vertices and
number of faces f given by the pairs (n, f ) = (4, 4), (5, 5), (6, 6). Draw the corresponding
planar graphs.

Figure 1: The three pyramides with 4, 5, 6 vertices and faces.

Figure 2: A less obvious polyhedron with 6 vertices and 6 faces.
Answer. The obvious solutions are shown above: the tetrahedron for n = f = 4, and
the four and five sided pyramides for n = f = 5 and n = f = 6, respectively. There
indeed exist a not so obvious solution for n = f = 6 shown the next figure on page 2.

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The automorphism group Aut(G) of a graph consists of all isomorphisms of the graph
to itself. The group orbit of a vertex v of the graph is the set of all images gv of that
vertex, where the automorphism g takes as values all group elements. The number of
items in a group orbit is always a divisor of the order of the group. It turns out to
depend on the vertex v.

Figure 3: The pseudo-rhomb-cuboctahedron.
10 Problem 1.4. How many elements has the automorphism group for the
pseudo-rhomb-cuboctahedron from the figure on page 3. Find the three group orbits
of the vertices and mark them with different colors.
Answer. The automorphism group is the semidirect product D4 ×s S2, which has the
order 16. There are 24 vertices, and the number of vertices in a group orbit is a divisor
of the order of the group. We see from this information that several group orbits have
to occur. Actually, there are three orbits of 8 vertices, marked red white and black in
the figure on page 4.

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Figure 4: There are three group orbits for the pseudo-rhomb-cuboctahedron.
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Proposition 1. For a simple planar graph which is not acyclic, and all cycles of which
have length at least l, the number of edges and faces are at most

(1.1) m ≤ l l (n − 2)
(1.2) − 2

f ≤ l 2 (n − 2)
− 2

10 Problem 1.5. Use the proposition 1 to show directly that the complete bipartite
K3,3 is not planar.

Answer. The complete bipartite graph K3,3, has n = 6 vertices and l = 4 as minimal

length of any cycle, since and cycles have even length.

Hence l (n − 2) = 8. This is less than the number of edges m = 9. Thus the
l−2

estimate in proposition 1 is not true and hence the graph is not planar.

10 Problem 1.6. Use the proposition 1 to show directly that the Petersen graph
is not planar.

Answer. The Petersen graph has n = 10 vertices and l = 5 as minimal length of any

cycle.

Hence l (n − 2) = 40 . This is less than the number of edges m = 15. Thus the
l−2 3

estimate in proposition 1 is not true and hence the graph is not planar.

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10 Problem 1.7. The rhombus-dodecahedron is the dual Archimedean polyhedron
for which all faces have length 4 and are surrounded by vertices of degrees (3, 4, 3, 4).
Draw the planar graph for the rhombus-dodecahedron and four-color its edges in a proper
way.

Figure 5: The rhombus-dodecahedron with colored edges

Answer.

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10 Problem 1.8. A polyhedra has 9 vertices and 7 faces, of which 6 are square
faces. Determine the length of the seventh face and draw its planar graph,—or explain
why it is not possible.

Answer. By Euler’s formula, there are m = n + f − 2 = 14 edges. The handshaking
lemma and counting yield

28 = 2m = {deg(v) of all vertices v} = 3n3 + 4n4 + 5n5
9 = n3 + n4 + n5

Subtracting three times the second equation from the first yields 1 = n4 + 2n5. We see
that there exists exactly one vertex v of degree four and hence n4 = 1 and n3 = 8.

It is assumed that six of the seven faces are squares. Let x be the length of the

remaining face. By the dual handshaking lemma,

28 = 2m = {deg(f ) of all faces f } = 6 · 4 + x

and hence x = 4. Hence all faces of the polyhedra have to be squares.
We begin the drawing with the unique vertex v of degree four. The resulting planar

Figure 6: A polyhedra with 9 vertices and 7 faces all squares does not exist.

graph is already uniquely constructible, as shown in the figure on page 7. But one gets
a polyhedra with the following faces: 2 triangles, 4 squares and one six-face! Check:
2 · 3 + 4 · 4 + 6 = 28.

A polyhedra with 9 vertices and 7 faces which are all squares does not exist,— in spite
of this choice being compatible with Euler’s formula and both handshaking lemmas.

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10 Problem 1.9. A planar, bipartite graph for which all vertices have degree
three or six is 3 face-colorable. Give two examples for such a graph, convince yourself
of this proposition and illustrate it for your examples.

Figure 7: The faces of these graphs can be properly colored with three colors.
Answer. A proper 3-coloring of the faces is done shown in the figure on page 8. These
examples are illustrations of the proposition mentioned above,—which gives the method
how to color the faces of a planar, three-regular and bipartite graph with three colors.

One colors the vertices black and white, and orders the three face colors. They
are used in that order to color the faces around any black vertex clockwise. We go
on coloring the faces around an adjacent white vertex in counterclockwise order, and
around any adjacent black vertex in clockwise order. One can go on until all faces are
colored.

10 Problem 1.10. Give a proof that a planar, bipartite graph for which the degree
of all vertices is divisible by three is 3 face-colorable.
Answer. One can indeed use the same procedure as for a 3-regular graph. We order
the three given colors and use them in that order to color the faces around any black
vertex clockwise. While one surrounds a vertex of degree 6, 9, . . . , all three colors ap-
pear 2, 3, . . . times. We go on coloring the faces around an adjacent white vertex in
counterclockwise order, and around any adjacent black vertex in clockwise order. One
can go on until all faces are colored.

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