Applications of Euler’s
Formula for Graphs
Hannah Stevens
Outline
 Important
terms
 Euler’s formula and proof
 Necessary parameters
 Applications of parameters
 Sylvester-Gallai Theorem
 Pick’s Theorem
Important Terms


Complete graph: graph in which each edge is connected
to every other edge
Simple graph: graph without loops or parallel edges



Loop: edge connecting a vertex to itself
Parallel edge: two or more edges connecting the same two
vertices
Degree of a vertex: number of edges connected to a
vertex (loops count as two)
 Connected graph: graph in which we can get from one
vertex to any other vertex along a path
 Cyclic graph: graph in which the first point on a path
connects to the last point
 Planar graph: graph that can be drawn in the plane with
no edges crossing
 Bipartite graph: graph in which the vertices are split into
two disjoint sets such that no two vertices from the same
set are adjacent
Euler’s Formula for Graphs
 If
G is a connected plane graph with v
vertices, e edges, and f faces, then
v–e+f=2
 Examples:
Proof of Euler’s Formula

Basis step: Formula holds for e = 1

Assume formula holds for e = n
 Let G be a graph such that e = n + 1

First consider if G has no cycles. Every edge goes to
a new vertex, so there will be one vertex, a, with
degree 1, connected to the graph via edge x. Delete
edge x and vertex a to create graph G` with n edges.
By assumption, the formula holds for G`.
x
a
Proof cont.

Second consider if G has a cycle. Let edge x
be an edge with a cycle, then x is a boundary
for two faces. Delete x to create graph G`
with n edges. By assumption, the formula
holds for G`.
x

Therefore, the formula holds for graph G with
e = n + 1.
Necessary Parameters
 We
can count vertices by their degrees,
where vi is the number of vertices with
degree i.

v = v1 + v 2 + v 3 + …
 Every
edge has two ends, and contributes
two to the sum of all degrees.

2e = v1 + 2v2 + 3v3 + …
 The
average degree of a graph is
2e
d
v
Necessary Parameters
 We
can count faces by the number of
edges bordering each face, where fk is the
number of faces with k edges.

f = f1 + f2 + f3 + …
 Every

edge borders two faces, so
2e = 1f1 + 2f2 + 3f3 + …
 The
average number of edges per face is
2e
f 
f
Parameter Applications
 The



complete graph of K5 is non-planar.
v = 5 e = 10
f = e + 2 – v = 10 + 2 – 5 = 7
20
f 
3
7
This implies the graph
must have one vertex
with degree at most 2, but
this is impossible (each vertex has degree 4).
Parameter Applications
 The
complete bipartite graph K3,3 is nonplanar.



v=6 e=9
f=e+2–v=9+2–6=5
18
f 
4
5
This implies the graph
has one face with at most 3,
but this is impossible (each face has 4).
Parameter Applications
 Proposition:
Let G by an simple graph with
v > 2 vertices. Then G has a vertex of
degree at most 5.
 Proof by Contradiction: Since G is simple,
every face will have at least three edges.



f = f3 + f4 + f5 + f6 + …
2e = 3f3 + 4f4 + 5f5 + 6f6 + …
Then 2e – 3f = f4 + 2f5 + 3f6 + … ≥ 0
Parameter Applications
 Assume
each vertex has degree at least 6,
then



v = v6 + v 7 + v 8 + v 9 + …
2e = 6v6 + 7v7 + 8v8 + 9v9 + …
Then 2e – 6v = v7 + 2v8 + 3v9 + … ≥ 0
 Combining


the previous inequalities
2e – 6v + 2(2e – 3f) = 6e – 6v – 6f ≥ 0
This contradicts Euler’s formula.
 Therefore
we must have a vertex with
degree at most 5.
The Sylvester-Gallai Theorem
Given a set of any v ≥ 3 nonlinear points, there
is always a line containing 2 of the points.
 Proof using Euler’s Formula:

If we embed the plane in
three-dimensional space,
then we can map the points
onto a sphere where each
point corresponds to a pair
of antipodal points on the
sphere, and the lines
correspond to great circles
on the sphere.
The Sylvester-Gallai Theorem
we have: Given any set of v ≥ 3 pairs
of antipodal points on the sphere, not all
on one great circle, there is always a great
circle containing exactly two of the pairs of
antipodal points.
 Now

If we dualize, we can
replace the pairs of
antipodal points with
the corresponding
great circle.
The Sylvester-Gallai Theorem

Now we have: Given any collection of v ≥ 3 great
circles on a sphere, not all of them passing through
one point, there is always a point that is on exactly
two of the great circles.

The arrangement of great
circles creates a simple plane
graph on the sphere whose
vertices are the intersection
points of two of the great circles.
This divides the great circles into
edges. The degree of each vertex
is even and at least four by construction.
Therefore, we have a vertex with degree less than five.
Pick’s Theorem

For this theorem, we
have polygons that lie
on a grid in which
each point in the grid
is integral and
equidistant from the
points above, below,
and beside that point.
The vertices of a
polygon fall only on
the points in the grid.
Pick’s Theorem
 Lemma:
Every triangle ABC has an area of
½.

Proof: Construct parallelogram ABCD by
rotating triangle ABC 180 degrees around the
midpoint of segment BC. By construction and
the properties of the
grid the parallelogram
C
is on, the area of the
parallelogram is 1.
B
The area of the triangle
A
is then ½.
D
Pick’s Theorem
 The
area of any polygon Q is given by
A(Q) = vint + ½vbd – 1 where vint is the
number of grid points inside the polygon,
and vbd is the number of grid points on the
border of the polygon.

Example:
vint = 8
vbd = 8
A = 8 + 4 – 1 = 11
Pick’s Theorem
 Proof:
We can use the interior grid points
to triangulate the interior. This can be
interpreted as a graph with f faces, where
f – 1 faces lie inside the polygon.
Each of these faces is
a triangle with an area
of ½. Then we have
A(Q) = ½(f – 1).
Pick’s Theorem
 Each
edge of the polygon Q appears in
one triangle, and each interior edge
appears in two triangles.

Let eint represent the
triangle edges inside
the polygon, and ebd
represent the triangle
edges that are edges
of the polygon.
Pick’s Theorem
3(f – 1) = 2eint + ebd which gives us
f = 2(e – f) – ebd + 3.
 Also ebd = vbd, so then we have
f = 2(e – f) – vbd + 3
 From Euler’s formula we know
e – f = v – 2.
 Then we have
f = 2(v – 2) – vbd + 3 = 2vint + vbd – 1.
 Therefore
A(Q) = ½(f – 1) = vint + ½vbd – 1.
 Then
Sources
11 Three applications of Euler’s
formula, Proofs from THE BOOK, by
Martin Aigner, Gunther M. Ziegler, and
K.H. Hofmann, pg. 65-70
 Discrete Mathematics, by Richard
Johnsonbaugh
 Ch.