Planar graphs
Typically a drawing of a graph is simply a notational shorthand or a more
visual way to capture the structure of the graph. Now we focus on the
drawings themselves.
Definition A drawing of a graph G = (V, E) in the plane is a pair of
functions f and g such that
1. f : V → R2 is one-to-one;
(no two vertices are drawn on top of one another)
2. for an edge uv, g(uv) is a simple curve whose endpoints are f (u) and
f (v), but for any other vertex w, f (w) is not on g(uv).
(no edge crosses itself and the only vtcs on it are its endpts)
Furthermore, a drawing is crossing-free if g(uv) ∩ g(xy) = ∅ and
g(uv) ∩ g(uw) = {f (u)} for distinct vtcs u, v, w, x, y.
(distinct edges do not intersect except at a shared endpoint)
A crossing-free drawing of G in the plane is also referred to as a planar
embedding of G.
The study of planar graphs necessarily involves topology of the plane, but
we do not attempt to be strictly rigorous in topological matters, so that we
do not obscure the combinatorial aspects of the topic.
Definition A curve is the image of a
continuous map from [0, 1] to plane R2.
• A curve is closed if its first and last
points (endpts) are the same.
• It is simple if it has no repeated
points (mapping is one-to-one) except possibly the first and last pts.
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Theorem (Jordan curve theorem). Let C be a simple closed curve in
the plane, R2. Then R2 − C contains exactly two regions, each having
C is a boundary.
The regions are classified as “inside” (bounded) and “outside” (unbounded).
The statement of the Jordan curve thm seems obvious but is hard to prove.
Easy to establish the result for simple curves such as polygonal lines —
the problem came in generalizing it to all kind of curves, which included
nowhere differentiable curves such as the Koch snowflake.
To give you some idea of the difficulty, consider the simple, closed curve
below – is the house in the interior or exterior of the curve?
Definition A graph is planar if it has a planar embedding.
Primary examples of nonplanar graphs K5 and K3,3
We can consider embedding, or drawing, graphs on other surfaces besides
the plane.
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Question What kinds of graphs can be drawn without crossings on the
surface of a sphere?
Theorem. A graph can be embedded on the surface of a sphere without crossings if and only if it can be embedded in the plane without
crossings.
n
Proof. Stereographic projection of a sphere on a plane. The projection
maps each point x 6= n of the sphere to a point x0 in the plane. (For point
n (the North pole), projection is undefined.)
The closer x is to n, the more distant its image is from (0, 0) in the plane.
For this reason, it is common to speak of n as mapping to “infinity” in
the plane, and of the sphere as completing the plane by adding a “point at
infinity”.
There is a one-to-one correspondence between points of the sphere and the
finite points on the plane; points at infinity in the plane correspond to the
point n on the sphere.
Given a spherical embedding of G (no crossings), where the point n lies on
no curve/edge of the drawing, the projection yields a planar embedding of
G. Conversely, a planar embedding yields a crossing-free drawing on the
sphere by the inverse projection.
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Definition A crossing-free drawing separates the plane R2 into faces.
(Cut out the graph, and pick up the pieces.)
Both planar embeddings above have four faces.
Question Given a planar graph, does the number of faces vary based on
the choice of planar embedding?
Theorem (Euler’s formula). If a connected planar graph G with n vtcs
and e edges has a planar embedding with f faces, it follows that
n − e + f = 2.
We give a proof by induction on # of edges, but there are several other
possible proofs, including a proof by induction on # of faces.
Proof. Proof by induction on e = |E(G)|. Fix the number of vtcs in G as n.
Base case: Since G is connected, it must have at least n − 1 edges, so
base case is e = n − 1. Then G is a tree. Since it is acyclic, it has exactly
one face in any planar embedding. So
n − e + f = n − (n − 1) + 1 = 2
as desired.
Induction hypothesis: Suppose the result holds for any connected,
planar graph with n vtcs and e ≥ n − 1 edges.
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Let G be a connected planar graph with n vtcs and e + 1 edges. Consider
a planar embedding of G with f faces. Since G is not a tree (it has at least
n edges), there exists an edge uv of G that is in a cycle and hence is not a
cut edge.
Let G0 = G − uv. Then G0 has n0 = n vtcs and e0 = e − 1 edges. As a
consequence of the Jordan curve thm, there are two distinct faces in the
planar embedding of G which have edge uv as a boundary; when uv is
removed, these two faces become one face, so this embedding of G0 has one
less face that that of G, i.e., f 0 = f − 1.
We have
n − e + f = n0 − (e0 + 1) + (f 0 + 1) = n0 − e0 + f 0 = 2
by applying the induction hypothesis for the planar embedding of G0.
Corollary. Let G be a planar graph with n vtcs, e edges, and c components. Suppose a planar embedding of G has f faces. Then
n − e + f − c = 1.
Proof. For a planar embedding G of a planar graph, let ψ(G) denote the
number of faces in the embedding. If H1, . . . , Hc are components of a
planar embedding of G, then
n = |V (H1)| + · · · + |V (Hc)|
e = |E(H1) + · · · + |E(Hc)|
f = ψ(H1) + (ψ(H2) − 1) + · · · + (ψ(Hc) − 1)
= ψ(H1) + · · · + ψ(Hc) − (c − 1).
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Thus,
n−e+f =
X
[|V (Hi) − |E(Hi) + φ(Hi)] − (c − 1)
i
= 2c − (c − 1)
=c+1
and so the result follows.
Corollary. Let G be a planar graph. Then the number of faces in any
planar embedding of G is the same.
Definition The degree of a face in a planar embedding is the number
of edges on the boundary of the face. If both “sides” of an edge are on the
face, we count it twice in the degree.
The degree of a face can also be thought of as the total length of the closed
walk in G bounding the face.
Examples
3
3
6
4
7
5
Proposition. Any planar embedding of a planar graph G has the sum
of the degrees of its faces equal to 2|E(G)|.
Remark For a simple planar graph, the degree of a face is always ≥ 3
unless the embedding has at most one edge.
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We can use Euler’s formula to derive upper bounds on the number of edges
in planar graphs.
Theorem. Let G be a simple planar graph with n ≥ 3 vertices and e
edges. Then e ≤ 3n − 6. If we know further that G is triangle-free,
then e ≤ 2n − 4.
Exercise for students Construct a proof of this result.
Remark Notice that this necessary condition is not a sufficient condition
for a simple graph to be planar. For example, the triangle-free graph G
resulting from adding one vertex and incident edge to K3,3 has 7 vtcs and
10 edges. Then
|E(G)| = 10 ≤ 2 · 7 − 4 = 2|V (G)| − 4
but this graph contains K3,3, which is nonplanar by a previous remark.
An immediate consequence of this result is that a planar graph cannot have
all vertices be of high degree.
Corollary. If G is a planar graph, then δ(G) ≤ 5.
Proof. If, for sake of contradiction, δ(G) ≥ 6, then
X
2|E(G)| =
d(v) ≥ 6|V (G)|.
v∈V (G)
Thus, |E(G)| ≥ 3|V (G)| but by previous theorem, it must be that |E(G)| ≤
3|V (G)| − 6. ⇒⇐
Question What do the upper bounds on numbers of edges in planar
graphs imply about K3,3 and K5?
Theorem. K5 and K3,3 are nonplanar.
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Proof. For K5, we have
|E(K5)| = 10 > 9 = 3 · 5 − 6 = 3|V (K5)| − 6
and for the triangle-free graph K3,3, we have
|E(K3,3)| = 9 > 8 = 2 · 6 − 4 = 2|V (K3,3)| − 4
so by the contrapositive statements of previous theorem, neither graph is
planar.
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