Professor Kindred
Math 104, Graph Theory
Homework 9 Solutions
April 18, 2013
Introduction to Graph Theory, West
Section 6.1 33 (modified)
Grid graph problem
Section 6.3 6, 12, 32
Genus of complete graph problem
Problems you should be able to do: 6.2.1, 6.2.7, 6.3.3, 6.3.5, 6.3.34
DO NOT RE-DISTRIBUTE THIS SOLUTION FILE
6.1.33 Let G be a triangulation, and let ni be the number of vertices of degree i in G.
Prove that
∑(6 − i)ni = 12.
i
Use this fact to to show that any planar graph G must have a vertex of degree at most 5,
i.e., δ( G ) ≤ 5.
A triangulation of graph G with n vertices has 3n − 6 edges and hence degree-sum
∑v d(v) = 6n − 12. The degree-sum also equals ∑v d(v) = ∑i ini , where ni is the number
of vertices of degree i in G, since the vertices of degree i contribute ini to the total degree
sum. Therefore,
6n − 12 =
∑ ini
=⇒
i
12 = 6n − ∑ ini = 6 ∑ ni − ∑ ini =
i
i
i
∑ (6 − i ) n i
i
as desired.
Consider a planar graph H. We know there exists a maximal planar graph G such that H
is a spanning subgraph of G. Consider a planar embedding of G; by our previous
theorem, we know this is a triangulation. Thus, we have
∑(6 − i)ni = 12 ≥ 0
i
and
ni ≥ 0 ∀ i.
There must exist at least one value of i for which 6 − i > 0 and ni > 0; otherwise, all
terms in the sum are 0 or negative, which would contradict the fact that the total sum is
positive. This implies that there exists a value 0 ≤ i < 6 such that ni > 0, i.e., there exists
at least one vertex v with dG (v) < 6. Thus, δ( G ) ≤ 5. Because H is a subgraph of G, it
follows that δ( H ) ≤ δ( G ) ≤ 5.
1
Grid graph Consider the 3 × 3 grid graph G (as described at
http://mathworld.wolfram.com/GridGraph.html). Use the characterization for outerplanar graphs given in problem 6.2.7 to prove that G is not outerplanar in two different
ways: (a) using a subdivision of K4 , and (b) using a subdivision of K2,3 .
the 3 × 3 grid graph
subdivision of K4
subdivided edge
subdivision of K2,3
subdivided edge
6.3.6 Without using the Four Color Theorem, prove that every planar graph with at
most 12 vertices is 4-colorable. Use this to prove that every planar graph with at most
32 edges is 4-colorable.
First consider the case where G = (V, E) is a planar graph with 12 vertices and δ( G ) ≥ 5.
Then
1
1
5 · 12
≤ ∑ δ( G ) ≤ ∑ d(v) = | E| ≤ 3|V | − 6 = 30
30 =
2
2 v ∈V
2 v ∈V
so it must be that equality holds throughout the above series of inequalities. Thus, G is a
maximal planar graph that is 5-regular with 12 vertices and 30 edges. Since it is maximal
planar, a planar embedding of it is a triangulation and so all faces have degree (length) 3.
The only such graph is the icosahedral graph (see discussion of 6.1.28 on pages 242-243
of textbook), and an explicit 4-coloring of this graph is shown in Figure 1.
Figure 1: A 4-coloring of the icosahedral graph.
2
Hereafter, we assume G has less than 12 vertices or has δ( G ) ≤ 4. We prove a bound on
the minimum degree of a planar graph with less than 12 vertices.
Claim: A planar graph G = (V, E) with 3 ≤ |V | < 12 has minimum degree δ( G ) ≤ 4.
Since G is planar and |V | ≥ 3, we know
| E | ≤ 3|V | − 6
=⇒
2| E| ≤ 6|V | − 12.
Thus,
δ( G ) ≤ avg degree of vertex in G =
6|V | − 12
12
2| E |
≤
= 6−
< 6−1 = 5
|V |
|V |
|V |
Since the minimum degree of a graph is an integer value, it must be that δ( G ) ≤ 4.
We prove by induction on n = |V | that a planar graph G is 4-colorable when δ( G ) ≤ 4.
Base case(s): For n ≤ 4, it is clear that G is 4-colorable (by assigning a unique color to
each vertex).
Induction hypothesis: Suppose that a planar graph G with n vertices and δ( G ) ≤ 4 is
4-colorable.
Let G be a planar graph with n + 1 vertices and δ( G ) ≤ 4.
Because δ( G ) ≤ 4, we know there exists a vertex v in G such that d(v) ≤ 4. Consider the
graph G − v. By the induction hypothesis, we know that G − v is 4-colorable since G − v
has n vertices, is still planar, and δ( G − v) ≤ δ( G ) ≤ 4. Consider a 4-coloring of G − v,
where the set of colors is {1, 2, 3, 4}.
Now, we replace the vertex v in the graph to obtain G. The worst case is that v has four
neighbors, u1 , u2 , u3 , and u4 , in G and each neighbor has a distinct color, say colors 1, 2,
3, and 4, respectively. We use the notion of Kempe chains, as in the proof of the 5-Color
Theorem, to complete the proof.
WLOG, suppose there exists an alternating 1-3 path (Kempe chain) from u1 to u3 . Then,
by the planarity of G, there cannot exist simultaneously an alternating 2-4 path from u2
to u4 . Consider the graph induced by the vertices colored with colors 2 and 4. On the
component of G [color 2,4 vtcs] that contains vertex u2 , swap colors of all color 2 and
color 4 vertices. Note that the result is a proper 4-coloring of G − v, and observe also that
v is then not adjacent to a vertex of color 2. Extend the 4-coloring to include v which is
assigned the leftover color 2. The result follows by induction.
Finally, we consider any planar graph G = (V, E) with | E| ≤ 32. If |V | ≤ 12, then by
proof above, G is 4-colorable. Otherwise, |V | ≥ 13, and we have
13δ( G ) ≤
∑ d(v) = 2|E| ≤ 64
v ∈V
3
=⇒
δ( G ) ≤
64
< 5.
13
Hence, in this case, too, we have that δ( G ) ≤ 4, and so by the same inductive argument
as that above, we have that G is 4-colorable.
6.3.12 Without using the Four Color Theorem, prove that every outerplanar graph is
3-colorable. Apply this to prove the Art Gallery Theorem:
If an art gallery is laid out as a simple polygon with n sides, then it is possible
to place b n3 c guards such that every point of the interior is visible to some
guard.
For n ≥ 3, construct a polygon that requires b n3 c guards.
Let G be an outerplanar graph with n vertices. We prove that G is 3-colorable via
induction on n.
Base case: Any outerplanar graph with at most 3 vertices is clearly 3-colorable.
Induction hypothesis: Suppose the result is true for any outerplanar graph with n
vertices.
Let G be an outerplanar graph with n + 1 vertices. Every simple outerplanar graph has a
vertex of degree at most 2 (by Proposition 6.1.20). We can delete such a vertex v, 3-color
G − v by the induction hypothesis, and extend the coloring to v since there must be an
unused color amongst the neighbors of v in G. Thus, the result holds by induction.
Next
claim that any art gallery laid out as a polygon with n segments can be guarded
nwe
by 3 guards so that every point of the interior is visible to some guard. The proof is
comprised of the following steps:
1. Show that the polygon can be triangulated.
2. Show that for such a triangulation of the polygon, its vertices can be colored with
three colors such that all three colors are present in every triangle of the
triangulation.
3. Choose the vertices of the polygon assigned the least frequent color.
The art gallery is a drawing of an n-cycle in the plane. We add straight-line segments to
obtain a maximal outerplane graph with n vertices. To do this, observe that polygons
with three sides are already triangulated without adding segments. For n > 3, some
corner can see some other corner across the interior of the polygon, i.e., there exists a
diagonal of the polygon that lies entirely in the interior of the polygon.1 We add this
segment and proceed inductively on the two resulting polygons with fewer corners.
1 To
see that this is true, pick any corner vertex A and an adjacent vertex B. Consider a ray with the end
point at A originally in the direction of B. Let it rotate from its initial position towards the interior of the
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onaid
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C oloring O u terplanar G raphs
F irst note that it is sufficient to show that each 2-connected component of an outerplanar graph
is 3-colorable. We prove the 3-colorability of 2-connected outerplanar graphs by induction on the
number of vertices n. It is clear that every outerplanar graph with at most 3 vertices is 3-colorable.
Let G be a 2-connected outerplanar graph with n vertices and assume that we have an outerplanar
embedding of G. If G is a cycle of length n, it is obviously 3-colorable. O therwise, G consists of
a cycle of length n (the outer face) and of at least one additional edge that connects two vertices
of the n-cycle. Let e = {u, v} be such an edge. T he edge e separates G into two (2-connected)
outerplanar graphs G1 and G2 with at most n − 1 vertices. B y the induction hypothesis, G1
and G2 are 3-colorable. Assume that we know 3-colorings of G1 and G2 such that u and v have
the same colors in both colorings. T his can be achieved by renaming the colors of two arbitrary
1
2 of G
21 and G22 . Since
2
2
2 edges
3-colorings
there2are2no edges connecting
G1 and 2G2 except for the
16 to u and v, the 3-colorings of G1 and G2 together form a 3-coloring of G.
adjacent
4 the polygon, we
n with
To prove the A rt G allery T heorem we proceed as follows. Starting
add diagonals (edges connecting two vertices of the polygon) until we3obtain a triangulation of
the polygon, i.e., until all inner faces of the resulting outerplanar graph are triangles. We can
color
n this outerplanar graph with 3 colors. In such a coloring, every triangle contains every color
exactly
once. Because it is sufficient to place a guard on one of the vertices of every triangle of
3
the triangulation, it is sufficient to place a guard at all the vertices with the same color fornone of
3
the colors. For one of the three colors, this set of vertices has size at most "n/3#.
T he following art gallery needs exactly n/3 guards because every shaded triangle needs one
guard (the upper corners can only be guarded from within the shaded triangles).
Since all vertices are on the boundary of the unbounded face and all bounded faces are
triangles, it is a maximal outerplanar graph. Consider a proper 3-coloring of this graph,
which
previous
proof. rity
Since each bounded face is a triangle, its vertices 336
are
Sectionexists
6.3: by
Paour
rameters
of Plana
pairwise adjacent and receive distinct colors. Thus each color class contains a vertex of
each triangle. Any vertex in a triangle sees all points in the triangle. Thus, guards placed
this
(l m class
+ l can
n see
+ mthenentire
+ 2lgallery.
mn +Since
2m ln
+ 2n lm) , plus lower
at
thesums
verticestoof a color
the three color classes
order terms.
l = the
m=
n , thisclass
becomes
( 9 / 16 )n vertices.
.
partition
the set W
of hen
vertices,
smallest
has at most
For the other
const r uction, begin with an optimal drawing of K n . T u r n
The
bound
of
guards
is best possible.
alcoves inofthe
polygon
require
their
each ver tex into an independent
set The
consisting
one
ver texbelow
from
each par
t.
alcoves.
When
own
guards;
no
guard
can
see
into
more
than
one
of
them.
There
are
W hen there are th ree par ts, each edge of the original drawing has now
n is not divisible by 3, we can add the extra vertex (or two) anywhere.
become a bundle consisting of 6 edges. For each crossing in the drawing
of K n , we get 36 crossings between the two bundles. For each edge in the
drawing of K n , we get at most 15 crossings within the bundle. N ear a ver tex
of the original drawing, we get at most 36 crossings (actually slightly
!n " less)
between the bundles cor responding to incident edges. T here are 2 edges
! "
in K n , and n Cn−1
pairs of incident
edges, Gbut
always
(n 4 ) crossings, so the
haracterization
O u terplanar
raphs
2 Figure 2: Anof
art gallery that requires n3 guards.
other cont ributions
areG of
order.
herefore,
we have
only
For a given graph
let Gsmaller
denote the graph
that weTobtain
from G by adding
a new vertex
to 36 (K n ) +
it and connecting all other vertices to the new vertex. T he following statements are equivalent:
3
4
O(n ) crossings. With the best k nown bound of (K n ) ≤ n / 64 + O(n 3 ) , we
!r "2
(a) G is outerplanar.
get the same(b)constant
9
/
16.
T
his
generalizes
easily
to
/ 16n 4 for the
G is planar.
2
6.3.32 Construct
an embedding
of
a 3-regular
nonbipartite
complete
multipar
tite
graph
with
size n . simple graph on the torus
(c) G does
not contain
a subdivision
ofrKpar
or K tsasof
a subgraph.
+
+
+
so that every face
has even length.
(d) G does not contain a subdivision of K
5
3,3
6.3.32. A n embed d i ng of a 3-regu l a r nonbip a rtite si mple gr a ph on the tor us
such
th ato
t every
h as even
length.
I t two
suffices
useof K
It
suffices
use K4face
as shown
in Figure
3. It has
faces –toone
degree/length
and
4 as shown 4below.
one
of degree/length
8. be obtained from this.
L arger
examples can
4
or K2,3 as a subgraph.
•
•
•
•
6.3.33.
If n3: is
t least 9 a nofdKn4 on
is not
a pr iso
methat
or every
tw iceface
a pr
i me,
then
there is
Figure
Ana embedding
the torus
has
even
length.
a 6-regu l a r toroi d a l gr a ph w ith n vertices. G iven these conditions, express
n as rs with r and s both at least 3. Now form C r Cs ; this 4-regular graph
polygon. Eventually, the ray will pass through another of the polygon’s vertices. Let C be the first vertex
embeds
natu rally
on the
tor
facediagonal.
havingBy length
4. there
O n could
the
when
this happens.
Then either
AC or
BCus
(or with
both) iseach
the desired
construction,
not
possibly
be
any
vertices
inside
the
triangle
ABC.
Had
there
been
any,
they
would
have
been
met
earlier
combinatorial description of the tor us as a rectangle, the embedding looks
than C by the rotating ray.
li ke the interior of Pr+1 Ps+1 , but the top and bot tom rows of ver tices are
the same, and the left and right columns of ver tices are the same.
Now add a chord in each face from its lower left cor ner in this pictu re
to its upper right cor ner. T he resulting graph is 6-regular, toroidal, and
has n ver tices.
5
6.3.34. Regu l a r embed d i ngs of K 4,4 , K 3,6 , a n d K 3,3 on the tor us. T he number of faces times the face-length is twice the number of edges, and the
Genus of complete graph Use the Euler characteristic for closed, orientable surfaces to
prove that
(n − 3)(n − 4)
γ ( Kn ) ≥
12
for n ≥ 3.
Assume n ≥ 3. Let g = γ(Kn ). Using the Euler characteristic, we know that
n
2 − 2g = n − e + f = n −
+ f,
(?)
2
where e = | E(Kn )| = (n2 ). Furthermore, since n ≥ 3, each face in an embedding of Kn has
degree at least 2. Thus, we have
2 n
2
.
3 f ≤ ∑ d(face) = 2e =⇒ f ≤ e =
3
3 2
faces
Substituting this expression for f into (?) gives us
n
2 n
n
+ f ≤ n−
+
2 − 2g = n −
2
2
3 2
1 n
= n−
3 2
n ( n − 1)
= n−
6
7n − n2
=
.
6
Solving for g, we have
g≥
n2 − 7n + 12
(n − 4)(n − 3)
=
.
12
12
Thus,
γ ( Kn ) = g ≥
(n − 3)(n − 4)
.
12
6