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MATH 239
Assignment 8
DUE: NOON Friday 3 April 2009
in the drop boxes opposite the Math Tutorial Centre
MC 4067.
1. Prove that a planar graph without a 3cycle has a vertex of degree 3 or less.
Solution:
Let
G
be a planar graph without a 3cycle. There are two cases to consider.
Case 1:
G
has no cycles. Thus the components of
G
are trees. If there are any com
ponents with no edges (i.e. single vertices), then we have a vertex of degree 0, and the
result follows. Otherwise, we know from Theorem 5.1.5 that every tree with at
p
≥
2
has at least two vertices of degree 1, and the result also follows.
Case 2:
G
has a cycle, so its girth is deﬁned. Let us suppose to the contrary that
G
has
all vertices of degree at least 4 (i.e. deg(
v
)
≥
4 for any vertex
v
). From Theorem 4.1.7,
we know that 2
q
=
∑
v
deg(
v
)
≥
4
p
, and thus
q
≥
2
p
.
Since
G
has no 3cycles, we know the girth is at least 4 (and
G
has at least four vertices).
Thus from Theorem 6.4.6, we know that
q
≤
2
p

4, which contradicts
q
≥
2
p
. Thus we
know that
G
cannot have all vertices of degree 4 or more (i.e. has at least one vertex of
degree at most 3). The result follows.
2. Prove that if G is a planar graph with no cycles of length less than 4, then G is 4
colourable. (Do not use the 4colour theorem).
Solution:
We can prove this result by induction (
note that this closely mimics the proof of the
6
colour theorem
). Any graph with
p
≤
4 is trivially 4colourable (one can assign a
diﬀerent colour to every vertex).
(Inductive Hypothesis:) Suppose that any planar graph
G
on
p
vertices with no cycles
of length less than 4 is 4colourable.
Consider a planar graph on
p
+ 1 vertices with no cycles of length less than 4. From
question 1, we know that such a graph has a vertex
v
of degree at most 3. Consider
the graph
G
0
=
G

v
, the graph formed by removing
v
and its incident edges from
G
. Note that
G

v
is still planar, and still has no cycles of length less than 4 (i.e. we
still satisfy the assumptions of the inductive hypothesis); thus
G

v
is 4colourable.
Consider a proper 4colouring of
G

v
, and use the same colour assignment to colour
all the vertices except
v
in
G
; in this case, all edges not containing
v
have each endpoint
coloured diﬀerently. We only need to colour
v
in such a way that
v
is coloured with
a colour diﬀerent from each of its neighbours. Since
v
has at most 3 neighbours, then
there will be at least one colour available to colour
v
so that the complete colouring is
proper (i.e. no adjacent vertices are coloured the same). Thus
G
is also 4colourable,
and the induction follows.
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View Full Document3. Prove that a graph with maximum degree Δ can be (Δ + 1)coloured.
Solution 1:
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 Spring '09
 M.PEI
 Math, Combinatorics

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