Applications of Euler’s Applications of Euler’s Formula for GraphsFormula for Graphs

Hannah StevensHannah Stevens

OutlineOutline

Important termsImportant terms Euler’s formula and proofEuler’s formula and proof Necessary parametersNecessary parameters Applications of parametersApplications of parameters Sylvester-Gallai TheoremSylvester-Gallai Theorem Pick’s TheoremPick’s Theorem

Important TermsImportant Terms Complete graphComplete graph: graph in which each edge is connected : graph in which each edge is connected

to every other edgeto every other edge Simple graphSimple graph: graph without loops or parallel edges: graph without loops or parallel edges

LoopLoop: edge connecting a vertex to itself: edge connecting a vertex to itself Parallel edgeParallel edge: two or more edges connecting the same two : two or more edges connecting the same two

verticesvertices Degree of a vertexDegree of a vertex: number of edges connected to a : number of edges connected to a

vertex (loops count as two)vertex (loops count as two) Connected graphConnected graph: graph in which we can get from one : graph in which we can get from one

vertex to any other vertex along a pathvertex to any other vertex along a path Cyclic graphCyclic graph: graph in which the first point on a path : graph in which the first point on a path

connects to the last pointconnects to the last point Planar graphPlanar graph: graph that can be drawn in the plane with : graph that can be drawn in the plane with

no edges crossingno edges crossing Bipartite graphBipartite graph: graph in which the vertices are split into : graph in which the vertices are split into

two disjoint sets such that no two vertices from the same two disjoint sets such that no two vertices from the same set are adjacentset are adjacent

Euler’s Formula for GraphsEuler’s Formula for Graphs

If G is a connected plane graph with v If G is a connected plane graph with v vertices, e edges, and f faces, thenvertices, e edges, and f faces, then

v – e + f = 2v – e + f = 2 Examples:Examples:

Proof of Euler’s FormulaProof of Euler’s Formula Basis step: Formula holds for e = 1Basis step: Formula holds for e = 1

Assume formula holds for e = nAssume formula holds for e = n Let G be a graph such that e = n + 1Let G be a graph such that e = n + 1

First consider if G has no cycles. Every edge goes to First consider if G has no cycles. Every edge goes to a new vertex, so there will be one vertex, a, with a new vertex, so there will be one vertex, a, with degree 1, connected to the graph via edge x. Delete degree 1, connected to the graph via edge x. Delete edge x and vertex a to create graph G` with n edges. edge x and vertex a to create graph G` with n edges. By assumption, the formula holds for G`.By assumption, the formula holds for G`.

ax

Proof cont.Proof cont. Second consider if G has a cycle. Let edge x Second consider if G has a cycle. Let edge x

be an edge with a cycle, then x is a boundary be an edge with a cycle, then x is a boundary for two faces. Delete x to create graph G` for two faces. Delete x to create graph G` with n edges. By assumption, the formula with n edges. By assumption, the formula holds for G`.holds for G`.

Therefore, the formula holds for graph G with Therefore, the formula holds for graph G with e = n + 1.e = n + 1.

x

Necessary ParametersNecessary Parameters We can count vertices by their degrees, We can count vertices by their degrees,

where vwhere vii is the number of vertices with is the number of vertices with

degree i.degree i. v = vv = v11 + v + v22 + v + v33 + … + …

Every edge has two ends, and contributes Every edge has two ends, and contributes two to the sum of all degrees.two to the sum of all degrees. 2e = v2e = v11 + 2v + 2v22 + 3v + 3v33 + … + …

The average degree of a graph isThe average degree of a graph is

v

ed

2

Necessary ParametersNecessary Parameters

We can count faces by the number of We can count faces by the number of edges bordering each face, where fedges bordering each face, where fkk is the is the

number of faces with k edges.number of faces with k edges. f = ff = f11 + f + f22 + f + f33 + … + …

Every edge borders two faces, soEvery edge borders two faces, so 2e = 1f2e = 1f11 + 2f + 2f22 + 3f + 3f33 + … + …

The average number of edges per face isThe average number of edges per face is

f

ef

2

Parameter ApplicationsParameter Applications

The complete graph of KThe complete graph of K55 is non-planar. is non-planar. v = 5v = 5 e = 10e = 10 f = e + 2 – v = 10 + 2 – 5 = 7f = e + 2 – v = 10 + 2 – 5 = 7

This implies the graph This implies the graph must have one vertex must have one vertex with degree at most 2, but with degree at most 2, but this is impossible (each vertex has degree 4).this is impossible (each vertex has degree 4).

37

20f

Parameter ApplicationsParameter Applications

The complete bipartite graph KThe complete bipartite graph K3,33,3 is non- is non-planar.planar. v = 6v = 6 e = 9e = 9 f = e + 2 – v = 9 + 2 – 6 = 5f = e + 2 – v = 9 + 2 – 6 = 5

This implies the graphThis implies the graphhas one face with at most 3,has one face with at most 3,but this is impossible (each face has 4).but this is impossible (each face has 4).

45

18f

Parameter ApplicationsParameter Applications

PropositionProposition: Let G by an simple graph with : Let G by an simple graph with v > 2 vertices. Then G has a vertex of v > 2 vertices. Then G has a vertex of degree at most 5.degree at most 5.

Proof by ContradictionProof by Contradiction: Since G is simple, : Since G is simple, every face will have at least three edges.every face will have at least three edges. f = ff = f33 + f + f44 + f + f55 + f + f66 + … + …

2e = 3f2e = 3f33 + 4f + 4f44 + 5f + 5f55 + 6f + 6f66 + … + …

Then 2e – 3f = fThen 2e – 3f = f44 + 2f + 2f55 + 3f + 3f66 + … + … ≥ 0≥ 0

Parameter ApplicationsParameter Applications

Assume each vertex has degree at least 6, Assume each vertex has degree at least 6, thenthen v = vv = v66 + v + v77 + v + v88 + v + v99 + … + … 2e = 6v2e = 6v66 + 7v + 7v77 + 8v + 8v88 + 9v + 9v99 + … + … Then 2e – 6v = vThen 2e – 6v = v77 + 2v + 2v88 + 3v + 3v99 + … + … ≥ 0≥ 0

Combining the previous inequalitiesCombining the previous inequalities 2e – 6v + 2(2e – 3f) = 6e – 6v – 6f ≥ 02e – 6v + 2(2e – 3f) = 6e – 6v – 6f ≥ 0 This contradicts Euler’s formula.This contradicts Euler’s formula.

Therefore we must have a vertex with Therefore we must have a vertex with degree at most 5.degree at most 5.

The Sylvester-Gallai TheoremThe Sylvester-Gallai TheoremGiven a set of any v Given a set of any v ≥ 3 nonlinear points, there ≥ 3 nonlinear points, there is always a line containing 2 of the points.is always a line containing 2 of the points.

Proof using Euler’s Formula:Proof using Euler’s Formula: If we embed the plane inIf we embed the plane in

three-dimensional space, three-dimensional space,

then we can map the points then we can map the points

onto a sphere where each onto a sphere where each

point corresponds to a pair point corresponds to a pair

of antipodal points on the of antipodal points on the

sphere, and the lines sphere, and the lines

correspond to great circles correspond to great circles

on the sphere.on the sphere.

The Sylvester-Gallai TheoremThe Sylvester-Gallai Theorem Now we have: Now we have: Given any set of v ≥ 3 pairs Given any set of v ≥ 3 pairs

of antipodal points on the sphere, not all of antipodal points on the sphere, not all on one great circle, there is always a great on one great circle, there is always a great circle containing exactly two of the pairs of circle containing exactly two of the pairs of antipodal points.antipodal points. If we dualize, we can If we dualize, we can

replace the pairs of replace the pairs of

antipodal points with antipodal points with

the corresponding the corresponding

great circle.great circle.

The Sylvester-Gallai TheoremThe Sylvester-Gallai Theorem Now we have: Now we have: Given any collection of v ≥ 3 great Given any collection of v ≥ 3 great

circles on a sphere, not all of them passing through circles on a sphere, not all of them passing through one point, there is always a point that is on exactly one point, there is always a point that is on exactly two of the great circles.two of the great circles. The arrangement of great The arrangement of great

circles creates a simple plane circles creates a simple plane graph on the sphere whose graph on the sphere whose vertices are the intersection vertices are the intersection points of two of the great circles. points of two of the great circles. This divides the great circles into This divides the great circles into edges. The degree of each vertex edges. The degree of each vertex is even and at least four by construction. is even and at least four by construction. Therefore, we have a vertex with degree less than five.Therefore, we have a vertex with degree less than five.

Pick’s TheoremPick’s Theorem

For this theorem, we For this theorem, we have polygons that lie have polygons that lie on a grid in which on a grid in which each point in the grid each point in the grid is integral and is integral and equidistant from the equidistant from the points above, below, points above, below, and beside that point. and beside that point. The vertices of a The vertices of a polygon fall only on polygon fall only on the points in the grid.the points in the grid.

Pick’s TheoremPick’s Theorem Lemma:Lemma: Every triangle ABC has an area Every triangle ABC has an area

of ½.of ½. ProofProof: Construct parallelogram ABCD by : Construct parallelogram ABCD by

rotating triangle ABC 180 degrees around the rotating triangle ABC 180 degrees around the midpoint of segment BC. By construction and midpoint of segment BC. By construction and the properties of the the properties of the

grid the parallelogram grid the parallelogram

is on, the area of the is on, the area of the

parallelogram is 1. parallelogram is 1.

The area of the triangle The area of the triangle

is then ½. is then ½. A

B

C

D

Pick’s TheoremPick’s Theorem The area of any polygon Q is given by The area of any polygon Q is given by

A(Q) = vA(Q) = vintint + ½v + ½vbdbd – 1 where v – 1 where vintint is the is the

number of grid points inside the polygon, number of grid points inside the polygon, and vand vbdbd is the number of grid points on the is the number of grid points on the

border of the polygon.border of the polygon.

ExampleExample::

vvintint = 8 = 8 vvbdbd = 8 = 8

A = 8 + 4 – 1 = 11 A = 8 + 4 – 1 = 11

Pick’s TheoremPick’s Theorem Proof:Proof: We can use the interior grid points We can use the interior grid points

to triangulate the interior. This can be to triangulate the interior. This can be interpreted as a graph with f faces, where interpreted as a graph with f faces, where

f – 1 faces lie inside the polygon. f – 1 faces lie inside the polygon.

Each of these faces is Each of these faces is

a triangle with an areaa triangle with an area

of ½. Then we have of ½. Then we have

A(Q) = ½(f – 1).A(Q) = ½(f – 1).

Pick’s TheoremPick’s Theorem Each edge of the polygon Q appears in Each edge of the polygon Q appears in

one triangle, and each interior edge one triangle, and each interior edge appears in two triangles.appears in two triangles. Let eLet eintint represent the represent the

triangle edges inside triangle edges inside

the polygon, and ethe polygon, and ebdbd

represent the triangle represent the triangle

edges that are edges edges that are edges

of the polygon.of the polygon.

Pick’s TheoremPick’s Theorem Then 3(f – 1) = 2eThen 3(f – 1) = 2eintint + e + ebdbd which gives us which gives us

f = 2(e – f) – ef = 2(e – f) – ebdbd + 3. + 3.

Also eAlso ebdbd = v = vbdbd, so then we have , so then we have

f = 2(e – f) – vf = 2(e – f) – vbdbd + 3 + 3

From Euler’s formula we know From Euler’s formula we know

e – f = v – 2.e – f = v – 2. Then we have Then we have

f = 2(v – 2) – vf = 2(v – 2) – vbdbd + 3 = 2v + 3 = 2vintint + v + vbdbd – 1. – 1.

Therefore Therefore

A(Q) = ½(f – 1) = vA(Q) = ½(f – 1) = vintint + ½v + ½vbdbd – 1. – 1.

SourcesSources

Ch. 11 Three applications of Euler’s Ch. 11 Three applications of Euler’s formula, formula, Proofs from THE BOOK,Proofs from THE BOOK, by by Martin Aigner, Gunther M. Ziegler, and Martin Aigner, Gunther M. Ziegler, and K.H. Hofmann, pg. 65-70K.H. Hofmann, pg. 65-70

Discrete MathematicsDiscrete Mathematics, by Richard , by Richard JohnsonbaughJohnsonbaugh