MA010 Graph Theory: Lecture 3
We start with a characterization of planar graphs. A subdivision of a graph G is a
graph obtained from G by repeatedly subdividing some of its edges, i.e., replacing some
of its edges with paths, which stands for removing an edge and identifying its end vertices
with the first and the last vertices of a path. Observe that if G is a graph and G is a
subdivision of G, then G is planar if and only if G is planar. In particular, if G contains
a subdivision of K5 as a subgraph, then G is not planar.
Theorem (Kuratowski’s Theorem). A graph G is planar if and only if G does not contain
a subgraph that is a subdivision of K5 or K3,3.
Proposition (Euler’s Formula). Every connected plane graph G satisfies that
n + f = m + 2 (1)
where n, m and f are the numbers of vertices, edges and faces of G, respectively.
Proof. We proceed by induction on the number of edges of G. The base of the induction
is m = 0, when G consists of a single vertex. Hence, n = 1 and f = 1, and the identity (1)
holds.
We now present the induction step. Let G be a plane graph with m ≥ 1 edges, and let
n and f be the numbers of vertices and faces of G, respectively. We distinguish two cases
based on whether G has an edge e whose removal keeps G connected or G has no such
edge.
We start with the former case. Let ww be an edge such that removal of ww keeps the
graph G connected, and let G be the connected plane graph obtained from G by deleting
the edge ww . The graph G is connected and so there is a path P from w to w ; let C be
the cycle consisting of the edge ww and the path P. Note that the edge ww is incident
with two different faces of G: one is contained inside the area bounded by the cycle C
and the other outside. Hence, the number of faces of the graph G is f − 1 as the two
faces incident with ww gets merged to a single face by removal of the edge ww . Since the
number of vertices of G is n and the number of edges is m − 1, the induction assumption
implies that n + (f − 1) = (m − 1) + 2, which yields that (1) holds for G.
In the latter case, G is a tree (as we have proven in Lecture 3). Since G has at least
two vertices, it has a leaf and let v be any of its leaves. Let G be the graph obtained from
G by removing the leaf v (together with the incident edge). Observe that the number of
vertices of G is n − 1, the number of edges is m − 1 and the number of faces is f. The
induction assumption implies that (n − 1) + f = (m − 1) + 2, which yields that (1) holds
for G.
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Proposition. Every planar graph with n ≥ 3 has at most 3n − 6 edges.
Proof. Let G be a planar graph and G be a triangulation that contains G as a spanning
subgraph. Let n, m and f be the number of vertices, edges and faces of G , respectively.
Since each face is bounded by three edges and each edge is incident with two faces, the
number of incidences of edges and faces is 3f = 2m. It follows that f = 2m/3. Euler’s
Formula now yields that n + 2m/3 = m + 2, which implies that m = 3n − 6. Hence, the
number of edges of G is at most m = 3n − 6.
Note that the just proven proposition implies that K5 is not planar as K5 has 5 vertices
and 10 edges, which is larger than 3 · 5 − 6 = 9.
Proposition. Every planar graph has a vertex of degree at most five.
Proof. Let G be a planar graph and let n be its number of vertices. If n ≤ 6, then the
graph G trivially contains a vertex of degree at most five. Suppose that n ≥ 7 and the
degree of each vertex of G is at least six. Since the sum of the degrees of G, which is at
least 6n, is equal to twice the number of edges (see the proof of the Hand-shaking Lemma),
the number of edges of G is at least 3n. However, this is impossible as G can have at most
3n − 6 edges.
A proper k-vertex-coloring of a graph G is a function c : V (G) → {1, . . . , k} such that
c(u) = c(v) for every edge uv of G. We will often simply say a coloring or k-coloring
instead of a proper k-vertex-coloring unless we want to emphasize that the end vertices
of each edge have different colors, that we color the vertices or highlight the number of
colors. The chromatic number of a graph G is the smallest k such that G has a proper
k-vertex-coloring; the chromatic number of G is denoted by χ(G).
One of the most well-known results in graph theory is the Four Color Theorem, which
can be traced back to Francis Guthrie to 1852. The computer assisted proof was given
by Appel and Haken in 1976, and a simplified and refined (computer assisted) proof by
Robertson, Seymour and Thomas in 1996; Gonthier and Werner formalized the argument
in the Coq prover in 2005.
Theorem (Four Color Theorem). The chromatic number of every planar graph is at most
four.
We will prove the following weaker result.
Theorem (Five Color Theorem). The chromatic number of every planar graph is at most
five.
Proof. The proof proceeds by induction on the number of vertices of a planar graph G.
The base case is when G has a single vertex when χ(G) = 1. We now present the induction
step. Let G be an n-vertex planar graph with n ≥ 2.
Suppose that G has a vertex of degree at most four and let w be such a vertex. Consider
the (n − 1)-vertex graph G obtained from G by removing the vertex w. By induction, the
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vertices of G can be colored using five colors. Since this coloring can be extended to the
whole graph G by assigning w a color different from the colors of the neighbors of w, the
graph G can also be colored using five colors.
We assume in the rest of the proof that the minimum degree of G is at least five. As
we have shown in previous lectures, G is a spanning subgraph of a triangulation, say GT ,
and GT has a vertex of degree at most five, say w. Since the minimum degree of G is at
least five, the degree of w is actually equal to five. Let v1, . . . , v5 be the neighbors of w
listed in the cyclic order around w. Observe that both v1v3 and v2v4 cannot be edges of
GT (because of the planarity); by symmetry, we will assume that v1v3 is not an edge.
Let G be the (n − 2)-vertex graph obtained from GT as follows: Remove the vertices
w, v1 and v3, introduce a new vertex w drawn where w was, join each neighbor u of v1 to
w by drawing an edge along the edges wv1 and v1u, and similarly join each neighbor u of
v3 to w by drawing an edge along the edges wv3 and v3u (if a vertex u is a neighbor of
both v1 and v3, include only one of the edges). By induction, the vertices of the graph G
can be colored using five colors. We now define a coloring of the vertices of G that uses
five colors: all vertices different from w, v1 and v3 get the same color that they have in
G , and the vertices v1 and v3 get the color of the vertex w . Finally, the vertex w gets a
color that is not used on any of its neighbors; note that there are at most four colors used
on the neighbors of w as the vertices v1 and v3 have the same color. Hence, it is possible
to assign w one of the five colors and so we have obtained a coloring of the vertices of G
with five colors. This completes the proof of the induction step and so the proof of the
theorem.
We conclude with observing that
ω(G) ≤ χ(G) ≤ ∆(G) + 1
where ω(G) is the clique number of G, which is the order of the largest complete subgraph
of G, and ∆(G) is the maximum degree of a vertex of G. The first inequality is trivial as
vertices of any complete subgraph must get mutually distinct colors in any proper coloring.
To establish the inequality χ(G) ≤ ∆(G)+1, fix a graph G and consider vertices v1, . . . , vn
of G listed in any order. We now color the vertices in the order v1, . . . , vn: when the vertex
vi is to be colored, there are at most degG(vi) ≤ ∆(G) colors that are already assigned to
its neighbors, and so one of the ∆(G) + 1 colors can be assigned to vi without creating a
monochromatic edge.
Note that the inequality χ(G) ≤ ∆(G) + 1 is tight for odd cycles and complete graphs,
which we will discuss in more detail in the next lecture, and the difference can be arbitrarily
large as witnessed by complete bipartite graphs.
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