Applications of Euler’s Formula for Graphs

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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.

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### 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:

x

a

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

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`.
• 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 + v2 + v3 + …
• 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
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
Parameter Applications
• The complete graph of K5 is non-planar.
• v = 5 e = 10
• f = e + 2 – v = 10 + 2 – 5 = 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 non-planar.
• v = 6 e = 9
• f = e + 2 – v = 9 + 2 – 6 = 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 + v7 + v8 + v9 + …
• 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
• 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
• Now 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.
• 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.

D

C

B

A

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

is on, the area of the

parallelogram is 1.

The area of the triangle

is then ½.

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
• Then 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.

Sources
• Ch. 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