University of AberdeenDepartment of Mathematical Sciences

MX4020 Project Report

2009–2010

Random Topics in Graph Theory

Author : Juta Kawalerowicz

Supervisor : Dr Assaf Libman

Declaration

I hereby declare that this report has been composed by me. I alsodeclare that all sources of information have been specifically ac-knowledged and all quotations distinguished by quotation marks.

(signed) ...............................

Summary

As remarked by Cameron graph theory is a cuckoo in the combina-torial nest. Since the publication of Euler’s famous ’Seven Bridgesof Konigsberg’, graph theory emerged as a discipline on its ownrights which is often attributed to its extremely diverse applicationsranging from the obvious such as physics, computer sciences andbiochemistry to more suprisingly areas, for instance that of socialsciences.

The route taken for this project was initially aimed at learn-ing the basics of graph theory and reading the proof of Kempe-Heawood’s Five Colour Theorem, however with time the projectevolved to included more interesting aspects of graph theory such asrandom graphs and percolations discussed in the two last chapters.

Also, a decision was made to include a chapter on algebraic graphtheory in which one can find solutions to exercises on algabriac char-acterization of bipartite graph and a proof of the Friendship Theo-rem.

Contents

Chapter 1. Basic set-up 11. Graphs and subgraphs 12. Connectivity 33. Planar graphs 6

Chapter 2. Graph Colouring 101. Map colouring 132. Kempe-Heawood’s Five Colour Theorem 14

Chapter 3. Algebraic Graph Theory 161. Eigenvalues and Bipartite Graphs 172. Friendship Theorem 24

Chapter 4. Graph Theory meets Probability 271. Random graph 272. Erdos’ Proof 27

Chapter 5. What is percolation? 321. Set-up 322. Finding critical probabilities for k-branching infinite tree Tk 343. Bounds for Critical Probabilities on Square Lattice 36

Bibliography 39

0

CHAPTER 1

Basic set-up

In order to study graph theory we first need to formally define graphs together with

their parameters and structures appearing within. In the opening chapter we intro-

duce the notion of a graph and look at some well-known classes of graphs which will

be used later. Next, we will define what is meant by a planar graph and offer a proof

of the useful Euler’s Formula.

1. Graphs and subgraphs

A graph is a pair G = (V,E) where V is a set of vertices and E represents 2-subsets

of elements of V called edges.1 In order to avoid ambiguities we shall always assume

that V and E are disjoint.

Figure 1. A graph G with V = v1, v2, v3, v4, v5, v6 with edge setE = v1, v2, v1, v4, v1, v6, v2, v3, v3, v4, v3, v6.

If e = vx, vy ∈ E we say that vx, vy ∈ V are adjecent while e and vx are incident,

as are e and vy. Similarily, one calls edges e1 and e2 adjacent if they share a common

vertex. The cardinality of V is called the order of G, we will also sometimes say that

G is on V which means that it is a graph with a vertex set V . If the graph has order

1Note that our definition of a graph excludes certain kinds of edges: multiplie edges between twovertices and edges looped around one vertex are not allowed. Most of time we will be happy withsuch treatment, however occasionally we will refer to a pseudograph which is a finite set V togetherwith a multiset E which consists of one- or two-element subsets of V .

1

Chapter 1 Page 2

0 or 1 we call it a trivial graph. Furthermore, by degree of a vertex v we mean the

number of edges incident to v and if all vertices of G have the same degree k we say

that G is k-regular. In a graph G we denote the minimal vertex degree by δ(G) and

the maximal vertex degree by ∆. By convention, we will write q for |E| and p for |V |.

Below we present a result on the sum of degrees in a graph which is due to Euler

Lemma 1.1. (Handshaking Lemma) Let G be a finite graph. Then

∑v∈V

deg(v) = 2q (1.1)

Proof. Let i(v, e) = 1 if v ∈ e and else i(v, e) = 0.∑v∈V, e∈E

i(v, e) =∑v∈V

∑e∈E

i(v, e) =∑v∈V

deg(v)

∑v∈V, e∈E

i(v, e) =∑e∈E

∑v∈V

i(v, e) = 2q

The name of Lemma (1.1) comes from the fact that a handshake by default is being

exchanged between two people. Similarily, each edge of G is shared by 2 vertices.

The technique employed here is an example of double-counting, that is counting the

size of a set in 2 different ways and showing that the obtained expressions are equal.

Some important examples of graphs which we will use later on include:

• a complete graph Kn on n vertices in which every pair of vertices is connected

by an edge

• a bipartite graph G with vertex set divisible into two partitions U and V

such that every edge connects a vertex in U to one in V .

• a complete bipartite graph Ku,v which is a special kind of a bipartite graph

where every vertex in the partition U is connected to every vertex in the

partition V .

• a cycle graph Cn on n vertices which contain a single cycle through all vertices

Sometimes we want to investigate a graph H which is contained within a larger graph

G. If V (H) ⊆ V (G) and E(H) ⊆ E(G) we call H a subgraph of G. An important

class of subgraphs are the induced subgraphs. We define a vertex-induced subgraph

G′ = (V ′, E ′) to be a subgraph of G such that its edge set E ′ consists of all the edges

of G which have both ends in V ′.

Chapter 1 Page 3

Figure 2. (1) - complete graph on 5 vertices K5, (2) - bipartite graph,(3) complete bipartite graph K2,3, (4) - cycle graph C7

2. Connectivity

Intuitively, a graph is connected if it is possible to travel to all the vertices by the

means of the exisiting edge set. However, traveling can take different forms and

different routes within a graph can be distinguiushed.

Let v0 and vn be two vertices of a graph G. A walk is an alternating sequence

(v0, e1, v1, e2, . . . , en, vn) where ei is an edge between vi−1 and vi for i ∈ [1, . . . , n]. By

length l of v0 − vn walk we mean the number of edges in the alternating sequence.

Obviously, a walk can be either open which happens if v0 6= vn or closed which

happens if v0 = vn. A trial is a walk in which no edges are repeated and a path is a

walk with no vertex repetition.

The notion of a path is useful for defining a connected graph. We say that a graph G

is connected if there exists a path connecting every pair of vertices. A graph which is

not connected can be divided into disjoint connected subgraphs called components.

Both a trial and a path can be either closed or open; we define a circuit to be a closed

trail and a cycle as a closed path. Related with the notion of path is girth γ(G) of

a graph G, which is the length of the shortest closed path in G. If a graph G has

no cycles we say that γ(G) = ∞. Also, every v0 − vn walk contains a v0 − vn path,

for if we let k to be a positive integer such that there exists m < k with vm = vk

then deleting the sequence vm, vm+1, . . . , vk−1 gives a smaller walk. Repeating this

procedure, until there are no vertex repetitions, gives a v0 − vn path.

Definition 1.1. A tree is a connected graph with no cycles.

Chapter 1 Page 4

Figure 3. Examples of trees of order 5

Trees are a special class of graphs which due to their simple structure obey the

following principle

Theorem 1.1. A graph G on p vertices and q edges is a tree if and only if it is acyclic

and

p = q + 1 (1.2)

Proof. First, as a tree G is acyclic by definition. To show the equality we use

induction. For p = 1 the assertion holds. Assume that p = q + 1 holds for all trees

on p vertices and consider a tree Gp+1 on p + 1 vertices. Since every non-trivial tree

has at least two end vertices2 we can choose an end vertex vx ∈ V (Gp+1) and observe

that deletion of vx gives a tree Gp on p vertices. By the inductive hypothesis we have

p = |E(Gp)|+ 1 and

|V (Gp+1)| = p+ 1 = (|E(Gp)|+ 1) + 1 = |E(Gp+1)|+ 1

Conversely, assume that a graph G is acyclic and p = q+ 1. To show that G is a tree

we need to show that it is connected. Let G1, G2, . . . , Gk be the components of G.

Since each Gi is a tree in its own right, we have

p− 1 = q =k∑i=1

qi =k∑i=1

(pi − 1) = p− k

So k = 1 which means that G is connected, hence a tree.

Lemma 1.2. A circuit of an odd length contains an odd length cycle.

Proof. The shortest odd length circuit has length 3, since there is no circuit of

length 1. But then it is an odd cycle of length 3. Assume that all odd circuits of size

at most r contain an odd length cycle, for some odd integer r ≥ 3. We must show

that an odd circuit of lenght r + 2 contains an odd cycle. Consider a circuit of the

form v0, . . . , vr+2, v0 of odd lenght r + 2. If it is a cycle then we are done. Assume it

2this can be shown by considering u− v, the longest path on non-trivial tree graph T . Since itis the longest path, neither u nor v can be adjecent to some other vertex which is not on the path.Also, u is adjecent to a vertex proceeding it on the path, same for v. Finally, neither u nor v areadjecent to another vertex on the path u− v because a tree is acyclic.

Chapter 1 Page 5

is not a cycle and there exist positive integers m, k such that m < k and vm = vk. We

can consider two separate circuits v0, . . . , vm, vk+1, . . . , vr+2, v0 and vm+1, . . . , vk. The

sum of their lengths is odd, hence one of the circuits must be odd and of length less

than r. By the hypothesis this circuit contains an odd cycle, hence v0, . . . , vr+2, v0

contains an odd cycle.

Coming back to the connected graph G we can define an eqivalence relation on the

vertex set V by the rule: v0 ≡ vn ⇔ there is a walk from v0 to vn. Each subgraph

induced by the vertices in an equivalence class is called a component of G. Note that

the relation satisfies reflexivity since there exists a trivial vx − vx walk, symmetry

since the existance of vx − vy walk implies existance of vy − vx walk and transitivity

since the existance of walks vx − vz and vz − vy implies that there is vx − vy walk.

Armed with the means of this section we can now give an important characterization

of previously introduced bipartite graphs.

Theorem 1.2. A graph G is bipartite if and only if it contains no odd cycles.

Proof. Since G is bipartite if and only if every component is bipartite it is enough

to give a proof for a single component H. First, we will show that if H is bipartite

then it contains no odd length cycles. Note that trees are acyclic bipartite graphs,

hence certainly they have no odd length cycles. Assume that H is not acyclic and

within H consider an arbitrary vertex v0 ∈ H and a cycle v0, v1, . . . , vr, v0. Note that

vi and vi+1 are adjacent so by definition of bipartite graph they must lie in different

partitions. Since v0 and vk are adjacent they must belong to different partitions,

hence k is even and the cycle has even length.

Conversely, we will show that if H contains no odd length cycles then it is bipartite.

Let H contain no odd length cycles and define an equvalence relation R on H as

follows: for v, u in the vertex set of H u ∼ v ⇔ every u− v walk is of even length.As

shown before it is easy to check that the relation satisfies 3 properties of an equivalence

relation, furthermore note that the relation is well-defined - if there is one u− v walk

of even length then all such walks have to be even because there are no odd cycles in

H.3 Select arbitrary vertex va and let va ∈ A, where A is a partition of V induced by

the equivalence relation R. Next, observe that vb, the vertex adjacent to va does not

belong to A becasue va− vb constitues a walk of an odd length, hence let it belong to

B, where B is the partition of V induced by the equivalence relation R. This way we

3For suppose that was not the case and we had two u− v walks, w1 being a walk of even lengthand w2 a walk of odd length. Then combining walk w1 and reverted walk w2 gives a circuit of an oddlength and it follows from Lemma (1.2) this implies that there exists an odd cycle, a contradiction.

Chapter 1 Page 6

accomplished dividing the vertex set of H into two partitions of vertices according to

the parity of the path joining va with vertices of H, hence arriving at a bipartition of

H.

Now that we defined what is meant by connected graph G, one can wonder what

is the measure of how well connected G is. For instance we would think that Cn

belongs to weakly connected graphs, since removing any two edges would disconnect

it, while Kn would pass as a strongly connected graph with a minimum of n − 1

edges resulting in disconnection. We can introduce edge-connectivity κ(G) to be the

minimal number of edges which need to be removed from G in order to disconnect

it. Although knowing which edges to remove does not always prove to be easy, one

notices that in the worst case we can always achieve disconnection by isolating the

vertex of the smallest degree.

Lemma 1.3. Let G be a graph, then

κ(G) ≤ δ(G) (1.3)

3. Planar graphs

Graphs come in a wide range of shapes and sizes and there is a special class of the so

called planar graphs. A planar graph is a graph that can be embedded in the plane,

that is it can be drawn on the plane in such a way that no edges cross each other.

A short survey of the graphs we met so far reveals that all the cycle graphs Cn are

trivially planar, while complete graphs Kn are not for n ≥ 5. As for the bipartite

graphs the story is more interesting, while some bipartite gaphs like K2,3 are planar,

anybody who has ever tried to solve the 3 Utilities Puzzle knows that K3,3 is not

planar4.

Since every planar graph divides the plane into a set of disjoint regions, we can

distinguish a new feature emerging within planar graphs. Any cycle which surrounds

a region forms a face fi. The portion of the graph lying outside of the graph G is

called the infinte face f0. If f0, f1, . . . , fn is the set of all faces of a planar graph

G then (f0 ∪ . . . ∪ fn) is the entire plane. We say that two faces are adjacent if they

share an edge e with the shortest cycle which bounds a face having length 3. If we

denote the number of faces by r and the number of edges by q, double counting yields

3r ≤ 2q (1.4)

4A classic result due to Kuratowski states a graph G is planar if and only if it contains neitherK5 nor K3,3 as a topological minor

Chapter 1 Page 7

Figure 4. Both K4 and K2,3 are planar while neither K3,3 nor K5 is.

which becomes strict if at least one of the regions is not triangular. Our next result

proves to be even more convinient because it elegantly relates the number of faces,

edges and vertices of a planar graph G.

Theorem 1.3. (Euler’s Formula) Let G be a connected planar graph, and let p, q and

r denote, respectively, the numbers of vertices, edges, and faces in a plane drawing of

G. Then p− q + r = 2.

Proof. We use induction on q and Theorem (1.1). For q = 0 the result holds

since we have p = 1 and r = 1, where r is an infinite region. Assume the result holds

for all connected plane graphs with less than q edges and examine G with q edges.

This is where trees come handy - first observe that if G is a tree then p = q + 1

and the formula holds. If G is not a tree then it contains a cycle with some e being

an edge of this cycle. Let us remove e and consider G − e, a connected graph on p

vertices with q − 1 edges and r − 1 regions. The equality p − (q − 1) + (r − 1) = 2

holds by the inductive hypothesis, hence p− q + r = 2.

The two results presented above provide a playground for problems dealing with

restriction on the degree of vertices in a planar graph. In a planar graph G what is

the restriction on δ, the minimal vertex degree?

Corollary 1.1. Every simple planar graph G contains a vertex of degree at most 5.

Proof. Combining equation (1.4) with Euler’s Formula yiels that in a planar

graph G we have q ≤ 3p− 6 which implies that

p∑i=1

deg vi = 2q ≤ 6p− 12

Chapter 1 Page 8

Recall, that the first equality is given by Lemma (1.1). Suppose that every vertex in

G has degree 6 or more. In this case∑p

i=1 deg(vi) ≥ 6p, a contradiction. Hence, G

contains a vertex of degree at most 5.

Can we come up with a planar graph which contains no vertex of degree less than 5?

An example of such is found in icosahedron, one of five platonic solids. Note that the

exterior of icosahedron should be through of as one of 20 triangular faces and we can

convince ourselves that A and B in Figure 5 are isomorphic 5.

Figure 5. Icosahedron

Bringing our interest in the degree problems one step further, let us prepare grounds

for the final example. In order to solve the problem we need the following side lemma:

Lemma 1.4. Let G be a planar graph in which there is a vertex v0 ∈ V such that

deg(v0)=2 and every edge e ∈ G is contained in two regions. Then either G = K3 or

there exists a face fi such that fi is not a triangle.

Proof. Let us consider a vertex such that deg(v0)=2 and its adjecent vertices v1

and v2. Since every edge is contained in two regions v0v1 and v0v2 ⊂ R1, R2. If either

R1 or R2 are not triangle we are done. Otherwise both R1 and R2 are bounded by

v0v1, v0v2 and v1v2 hence G = K3

Example 1.1. (Exercise 7.3 in [2]) Every connected planar graph of order p ≥ 4 has

at least 4 vertices with degree less than or equal to 5.

Proof. The strategy is to obtain a contradiction and show that there does not

exist a graph G of order p ≥ 4 with at most 3 vertices of order less than or equal

to 5. Let p be the number of vertices, p5 number of vertices of degree at most 5, q

number of edges, q1 number of edges which bound only one region, V≥6 vertex subset

5In fact every simple polyhedron can be turned into a planar graph by using the polyhedron’svertices as vertices of the graph and the polyhedron’s edges as edges of the graph. The faces of theresulting planar graph then correspond to the faces of the polyhedron

Chapter 1 Page 9

containing vertices of degree at least 6, V≤5 vertex subset containing vertices of degree

at most 5.

Assume that |V≤5| ≤ 3. Recall Lemma (1.1) to find a lower boundary for∑

v∈V deg(v):

2q =∑v∈V

deg(v) =∑v∈V≥6

deg(v) +∑v∈V≤5

deg(v) ≥ 6(p− p5) + 2(p5) (1.5)

At this point a careful reader should wonder why vertices of degree 1 were not included

in the above boundary. This is because we are only interested in connected planar

graphs where vertices of degree 1 do not occur. Second, observe that the boundary of

any face f of G is the union of the edges e of G such that e ⊂ f and e ∈ E2. Let us

define Ω, the set of pairs consisting of a face R and an edge e lying on the boundary

of R as Ω = (R, e) | e ⊂ R, e ∈ E2. Then

| Ω | =∑R

e ∈ E2 : e ⊂ R ≥ 3r

| Ω | =∑e∈E2

e = 2(q − q1)

Therefore

3r ≤ 2(q − q1) (1.6)

Where the inequality (1.6) becomes strict if at least one region of G is not a triangle.

Combining Euler’s Formula with inequality (1.6) and equation (1.5) gives:

2 = p− q + r ≤ p− q +2

3q − 2

3q1 = p− 1

3q − 2

3q1

≤ p− ((p− p5) +1

3(p5))− 2

3q1 =

2

3p5 −

2

3q1 (1.7)

Our strategy is to show the impossibility of the above inequality to hold. Since we

assume that p5 ≤ 3 it follows that the inequality can hold only if p5 = 3 and q1 = 0.

Importantly, deg(v) ≥ 2 for all v ∈ V . Now we have two cases to consider

Case 1: There exists v0 ∈ V≤5 such that deg(v0) > 2. In this case inequality (1.5)

becomes strict hence (1.7) becomes a strict inequality and we arrive at a contradiction.

Case 2: For every v0 ∈ V≤5 deg(v0) = 2. Since q1 = 0 we can apply Lemma

(1.4) and conclude that since p ≥ 4 and so G 6= K3, there exists a face fi which is

not a triangle. In this case (1.6) becomes a strict inequality and the contradiction

follows.

CHAPTER 2

Graph Colouring

Now that we defined a graph G let us assume that we want to colour its vertices using

colour set C = c1, . . . , cn where each element ci is referred to as a colour. By vertex

colouring we mean a map θ : V → C such that θ(vx) 6= θ(vy) whenever vx and vy

are adjecent. A natural question arises of what is the minimal number of colours we

need to use. Mathematics does not like waste - a valid colouring of Cn can surely be

obtained by the means of n colours but in fact the maximum of 3 colours will always

do.

Definition 2.1. The smallest number of colours needed to colour the vertices of G

is called the chromatic number of G and is denoted χ(G). Furthermore, we say that

G is critically n-chromatic if χ(G) = n and χ(G− v) = n− 1 for all v ∈ V (G).

In some cases the chromatic number can be easily established, for instance χ(Kn) = n,

χ(Cn) = 2 if n is even and χ(Cn) = 3 if n is odd and Ku,v = 2. Connected to problems

of colouring is the notion of an independent set, which is a set of vertices of which no

pair is adjacent. The independence number α(G) of a graph G is the size of the largest

independent set of G. Since in a colouring of G all colour classes are independent we

immediately get the following result

Lemma 2.1. Let G be a graph on n vertices. Then

χ(G)α(G) ≥ n (2.1)

Chromatic number proves to be a a difficult parameter with no general formula for

an arbitrary graph known. What we know, however, are the bounds, first of which

we are going to present below and second in the chapter on algebraic graph theory.

Let us prepare the grounds by presenting the connection between chromatic number

χ(G) and edge-connectivity κ(G) as presented in [2].

Definition 2.2. The edge-connectivity κ(G) of a connected graph G is the smallest

number of edges whose removal results in disconnection of G or a trivial graph. Hence,

a graph is n-edge-connected if it remains connected whenever fewer than n edges are

removed.

10

Chapter 2 Page 11

Theorem 2.1. For every critically n-chromatic graph G the following holds

κ(G) ≥ n− 1 (2.2)

Remark. We will provide an outline of the proof by contradiction as given in [2].

Start with an assumption that G is critically n-chromatic and not (n− 1)-edge con-

nected. The key lies in the fact that a graph G is n-edge connected if and only if

there is no partition into sets W,U ∈ V (G) such that the number of edges between W

and U is less than n. If G is not (n− 1)-edge connected we can find the above W,U

partition. Both induced subgraphs on W and U are (n− 1)-colourable. However, an

assertion that the edges joining W and U are incident to vertices of different colours

leads to G being (n-1)-colorable, a contradiction. What is left is to show that we can

always permute colours so that the edges connecting W and U have end assigned to

different colours, which produced a contradiction.

Moreover, Lemma (1.3) and Theorem (2.1) give that if G is critically n-chromatic then

δ(G) ≥ n − 1. We are now ready to present a useful bound for chromatic number

χ(G) due to Szekeres and Wilf as stated in [2].

Theorem 2.2. For any graph G we have χ(G) ≤ 1 +max δ(G′) where the maximum

is taken over all induced subgraphs G′ of G.

Proof. Assume that G is n-chromatic graph, where n ≥ 2. Let H be a critically

n-chromatic subgraph of G and use the fact that δ(H) ≥ n − 1. Observe that H is

an induced subgraph of itself hence

δ(H) ≤ max δ(H ′)

where H ′ is an induced subgraph of H. But H is a subgraph of G hence every induced

subgraph of H is also an induced subgraph of G which gives

max δ(H ′) ≤ max δ(G′)

Thus

χ(G)− 1 = χ(H)− 1 = κ(H) ≤ δ(H) ≤ max δ(G′)

Which gives that

χ(G) ≤ 1 +max δ(G′)

Finally, let us offer solutions to two exercises, first found in assigment problems of

Jacob Fox’s notes on combinatorics [5] and second coming from Problem Set 15.1 in

[2].

Chapter 2 Page 12

Example 2.1. Let G1 and G2 be two graphs on the same vertex set V . Show that

the chromatic number of their union G1 ∪ G2 (we take the union of the edge sets of

both graphs) satisfies

χ(G1 ∪G2) ≤ χ(G1)χ(G2)

Proof. Let χ(G1) = n and χ(G2) = m. Clearly if (E(G1)∩E(G2)) = ∅ then we

can colour G1 ∪ G2 with k = max(n,m). How to show that mn is always enough?

Since G1 is coloured using colours a1, a2, . . . , an and G2 is coloured with b1, b2, . . . , bm.

We can define mn new colours as ordered pais (ai, bj) and label a verex v ∈ G1 ∪G2

with (ai, bj) colour if v was coloured with ai in G1 and bj in G2. What remains is to

give an example of graphs G1 and G2 where χ(G1)∪χ(G2) = χ(G1)χ(G2) is attained.

Consider the complete graph Knm and divide its vertex set V into desjoint sets

V1, . . . , Vn each with m vertices each. Let G1 be a graph on vertex set V such that vx

is adjecent to vy ⇔ x−y ≡ 0 mod m. Then χ(G1) = n since it is the disjoint union of

m complete graphs on n vertices. Furthermore, let G2 be the the graph on vertex set

V where two vertices are adjecent⇔ they are not adjecent in G1. Observe that G2 is

a complete m-bipartite graph on V with partition V1, . . . , Vn and χ(G2) = m. Clearly,

G1 ∪G2 = Knm, hence χ(G1 ∪G2) = nm. Hence, χ(G1)χ(G2) = χ(G1 ∪G2) = nm.

Example 2.2. Show that for any graph G on p vertices χ(G) +χ(G) ≤ p+ 1, where

G is a graph on the same vertex set as G where two vertices are adjecent in G if and

only if they are not adjecent in G. Thus, V (G) = V (G) and G ∪ G = Kp.

Proof. Choose H, a subgraph of G with max δ(H) and denote maximal δ(H)

as m. It follows from Theorem (2.2)that χ(G) ≤ 1 +m.

Let us find a similar bound for χ(G). Since δ(H) denotes the smallest vertex degree

within H it follows that |V (H)| ≥ m + 1. Also, for all v ∈ H we have degH(v) ≥ m

where by degH(v) we mean the degree of v in H. Since H is a subgraph of G,

degG(v) ≥ degH(v) ≥ m and from the definition of G the degree of v as a vertice

of G is equal to degG(v) = p − degG(v) − 1 ≤ p −m − 1. Consider K, any induced

subgraph of G. We distinguish between the case that K and H are disjoint and the

case that K and H have a vertex in common.

Case 1: K ∩H = ∅.Since no vertex of H is shared with K we have |V (K)| ≤ p − |V (H)| ≤ p −m − 1

which implies degK(u) ≤ p−m− 2 for any u ∈ K and hence δ(K) ≤ p−m− 2. So

χ(G) ≤ p−m− 1 by Theorem (2.2)

Case 2: There exists a vertex v ∈ K ∩H

Chapter 2 Page 13

We established previously that since v ∈ H, p−m− 1 ≥ degG(v) ≥ degK(v) ≥ δ(K).

Then max δ(K) ≤ p−m− 1 and hence δ(G) ≤ p−m− 1 for each K ⊆ G which by

Theorem (2.2) implies that χ(G) ≤ p − m With both bounds we can complete the

proof: χ(G) + χ(G) ≤ (1 +m) + (p−m) = 1 + p

1. Map colouring

In this section we will investigate the least number of colours needed for colouring

of any map in a plane. The map colouring problem has notable history on its own

and over the decades it grown to be one of the well known problems in mathematics.

The statement that in order to colour any map one needs at most 4 colours, is

usually attributed to Guthrie, who was to be the first mathematican to conjecture

the so called 4 Colour Theorem in 1852. What initially seemed like an easy problem,

turned out to unsolvale for a long time, despite numerous attempts by distingushed

mathematicians. The proof of the 4-colour conjecture was eventually announced by

Appel and Haken in 1977 and to some disliking of mathematical community it was

the first major proof aided by a computer.

To relate map colouring with graph theory we first need to define a dual G∗ of graph

G and then show that region colouring of G is equavalent to vertex coloruing of the

dual graph of G.

Definition 2.3. Let G be a connected, planar graph. Its dual G∗ is a pseudograph

which has a vertex for each region of G. Two vertices in G∗ are connected by an

edge if and only if the corresponding ragions in G have an edge in common. Dual

Figure 1. For each face of G, the dual graph G∗ (drawn with dashedlines and empty nodes) has a vertex. Two vertices of G∗ are adjecentif and only if the corresponsing faces of G are adjecent. Graph H is theunderlying graph of G

pseudographs have a number of interesting properties:

Chapter 2 Page 14

• The property of being a dual is symmetric. If G∗ is a dual of G then G is a

dual of G∗.

• If G has p vertices and r regions then G∗ has r vertices and p regions. The

number of edges remains unchanged.

An underlying graph H of G is obtained from a dual G∗ in which each set of multiple

edges is repleaced by single edge and all loops are deleted.

From this point it seems like problems of region colouring are equivalent to vertice

colouring. Let us prove it

Lemma 2.2. A graph G is n-region colourable if and only if it is n-colourable.

Proof. Let G be an arbitrary n-region colourable plane graph. Consider G∗, a

dual of G and note that each region of G∗ contains one vertex of G. Two vertices of G

are adjecent if and only if the corresponding regions of G∗ are adjecent. Now because

G∗ is n-region colourable, G is n-colourable. Conversely, let G be n-colourable planar

graph and consider the underlying graph H. Two regions of G are adjecent if and

only if the corresponding vertices of H are adjecent. Since H is planar it follows that

it is n-colourable, hence G must be n-region colourable.

2. Kempe-Heawood’s Five Colour Theorem

As remarked by Soifer in [3] the numerous attempts to produce a compelling proof

of 4 Colour Theorem establish something of a modern version of Shakespeares Com-

edy of Errors. Most notably, in 1879 it was announced in the American Journal of

Mathematics that the 4 Colour Theorem had been proven by an amateour mathe-

matician, Alfred Kempe. Although largely acclaimed, Kempe’s proof contained an

oversight which was pointed out by Percy Heawood only 11 years later. There was a

constructive side of finding the error - in 1980 Heawood published a paper in which

he stated that Kempe’s argument actually proves that five colours are always enough

to colour any map.

Theorem 2.3. Any planar graph is 6-colourable.

Proof. Every planer graph has a minimum vertex degree equal to at most 5 as

we shown in Corollary (1.1) Every subgraph of a planar graph is planar and hence

also has a minimum vertex degree of at most 5. Applying Theorem (2.2) we obtain

that χ(G) ≤ 6.

Below is a proof of a 5 Colour Theorem as presented by Heawood.

Chapter 2 Page 15

Theorem 2.4. (Kempe-Heawood’s Five Colour Theorem) Any planar graph is 5-

colourable

Proof. Proof by induction on the number of vertices. For p = 1 the result

holds. Assume that every graph of order p − 1 is 5-colourable. Let G be a graph

of order p. The trick is to delete a vertex v of degree at most 5, which we know

to exist by Corollary 1.1. By doing so we arrive at 5-colourable graph G′ = G − v

Figure 2. Delete vertex v and by the inductive hypothesis arrive atvalid 5-colouring of G− v

of order p − 1. If deg(v) < 5 then some colour ci has not been used in colouring

of the vertices adjecent to v and by assigning it to v we arrive at 5-colouring of G.

Assume deg(v) = 5 and v1, . . . , v5 are the vertices adjectent to v assigned distinct

colours c1, . . . , c5 respectively. Consider H1,3, a subgraph of G′ induced by all vertices

coloured c1 or c3. If the vertices v1 and v3 do not belong to the same component of

H1,3 we can interchange 1 the colours 1 and 3 in the component containing v1 and in

such a way obtain a spare colour which we then assign to v. Assume that v1 and v3

belong to the same component H1,3. Then there exists a path v1 − v3 in G′. Now

restore the deleted vertex v and note that v − v1 − v3 produces a cycle which closes

v2. Therefore v2 and v4 belong to different components of a subgraph H2,4 induced by

all the vertices coloured c2 and c4 so by interchanging the colours in the component

containing v2 we gain a spare colour c2 with which we then colour v. Hence we arrived

at valid 5-clouring of G.

1Why can we do it? Since we work on induced subgraph which contains all the vertices colouredc1 and c3 an interchange does not alter the valid colouring of G′. All that could go wrong withchanging vertex colouring is possible ending up with at least two adjecent vertices assigned thesame colour. This cannot happen with the above interchange because we do not alter neighbours ofvertices coloured c1 and c3 and by swaping the colour classes we make sure they remain coloured ina valid way.

CHAPTER 3

Algebraic Graph Theory

We begin by introducing the n × n adjecency matrix A of a graph G on n vertices

whose aij entry is given by

aij =

1 if vi, vj ∈ E(G)

0 otherwise

Theorem 3.1. Let G be a graph with adjecency matrix A = [aij]. The (i, j) entry of

Ak matrix is equal to the number of different walks of length k from vi to vj.

Proof. For k = 1 the result holds from the definition of adjecency matrix because

a walk of length 1 is just an edge hence if vi and vj are adjecent then aij = 1. Notice

that by the properties of matrix multiplication in the matrix A2 the a2ij entry is given

by

a2ij = ai1a1j + ai2a2j + . . .+ aikakj =

∑k

aikakj

Let An−1 = [an−1ij ] and assume an−1

ij is the number of different vi − vj walks of length

n− 1 in G. Let An = [anij]. By the properties of matrix multiplication we have

anij =∑k

an−1ik akj

Every vi − vj walk of length n consists of vi − vk walk of length n − 1 where vk is

adjecent to vj, followed by edge vk, vj and the vertex vj. Thus by induction the

assertion holds.

So far we came across some pretty straightforward parameters of graphs such as degree

of vartex, chromatic number, connectivity or girth. As pointed by [5] eigenvalues of

adjecency matrix A(G), although less obvious, are very useful parameter of graphs.

In this chapter we aim to present context in which eigenvalues are used.

Definition 3.1. For n× n matrix A a scalar λ is an eigenvalue if for some vector x,

not equal to zero

Ax = λx (3.1)

16

Chapter 3 Page 17

The vector x is called eigenvector and the set of eigenvalues is called the spectrum of

A.

Before venturing further into algebraic graph theory let us recall some highlights of

linear algebra which will be used later.

Theorem 3.2. The sum of eigenvalues is equal to∑n

i=1 λi = Tr(A) =∑n

i=1 aii.

Proof. This is because A is similar to some upper triangular matrix U . We have

A = P−1UP for some invertible matrix P and using the fact that Tr(AB) = Tr(BA)

we obtain

Tr(A) = Tr(P−1UP ) = Tr(P−1PU) = Tr(U)

which shows that the trace of similar matrices is equal. But U has λ1, . . . , λn on the

main diagonal.

Using the same technique we find two further properties of eigenvalues:

Theorem 3.3. The product of eigenvalues is equal to∏n

i=1 λi = det(A)

Theorem 3.4. The number of non-zero eigenvalues is equal to the rank of A.

The fact that the adjecency matrix of a graph G has all diagonal entries equal to zero

gives the following

Lemma 3.1. The sum of all eigenvalues of a graph is equal to zero.

1. Eigenvalues and Bipartite Graphs

To gain a better understading of algebraic properties of graphs and see the eigenvalues

in action, we set out to investigate the discuss the eigenvalues of a bipartite graph.

Interestingly, the adjecency matrix of a bipartite graph has always the following form:

AKn,m =

(0 BT

B 0

)How does this property translate in terms of eigenvalues? In order to find properties

of bipartite graphs we will proceed with a number of smaller steps. It is convenient

to start by considering a complete bipartite graphs Kn,m.

Lemma 3.2. Let Kn,m be a complete bipartite graph. The only non-zero eigenvalues

λ1, λn are ±√mn.

Chapter 3 Page 18

0 0 1 0 00 0 1 1 1 1 1 0 0 0 0 1 0 0 0 0 1 0 0 0

v

v

v

v v

1 2

3 4 5

Figure 1. Bipartite graph G together with its adjecency matrix A(G)

Proof. By the remark on the general form of adjecency matrix the rank of Kn,m

is 2. Thus, by Theorem (3.4) we know that Kn,m has 2 non-zero eigenvalues and 0-

eigenvalues of multiplicity n−2. Furthermore, Lemma (3.1) implies that the non-zero

eigenvalues are equal to ±λ, because the sum of eigenvalues must be equal to 0. To

find λ we need to solve Ax = λx:

AKn,m

x1

...

xn

xn+1

...

xn+m

=

y1 = xn+1 + . . .+ xn+m

...

yn = xn+1 + . . .+ xn+m

yn+1 = x1 + . . .+ xn...

yn+m = x1 + . . .+ xn

= λ

x1

...

xn

xn+1

...

xn+m

(3.2)

which leads to a crucial observation regarding the eigenvector x, that is

xT = (α, . . . , α︸ ︷︷ ︸n

β, . . . , β︸ ︷︷ ︸m

)

and (Ax)T = (mβ, . . . ,mβ, nα, . . . , nα). What is left is solving two equations

mβ = λα

nα = λβ

which leads to

mnβ = λ2β

λ =√mn

Chapter 3 Page 19

The drawback of the above result (given by Fox in [5] is in that it holds only for

complete bipartite graphs Kn,m. Ideally, we would like to learn more about a general

bipartitie graph. The next 5 Lemmas and Theorem (3.5) were a part of Jacob Fox’s

Assigment Set [5].

Lemma 3.3. Let λ1 ≥ . . . ≥ λn be the eigenvalues of adjecency matrix A and W (m)

denote the number of closed walks of length m in G. For each positive m, W (m) =

Tr(Am) =∑n

i=1 λmi .

Proof. By Theorem (3.1) for each positive integer m

W (m) = Tr(Am) =n∑i=1

λmi

The first equality follows straight from the definition of a closed walk. However, there

is some work to be done to show the second equality. Suppose that λ is an eigenvalue

of A, which means that λ satisfies Ax = λx. Assume λk−1 is an eigenvalue of Ak−1,

then Ak−1x = λk−1x. Multiplying both sides by A yields

Akx = A(λk−1x)

Akx = λk−1(Ax)

Akx = λk−1(λx)

Ak = λkx

Hence, if λ is an eigenvalue of A then λk is an eigenvalue of Ak.

Lemma 3.4. If A is the adjecency matrix with eigenvalues λ1 ≥ . . . ≥ λn then

λ1 ≥ −λn (3.3)

Proof. Proof by contradiction. Assume that |λn| > λ1 and choose an odd m.

By the Lemma (3.3)

W (m) = Tr(Am) =n∑i=1

λmi = |λn|mn∑i=1

λi|λn|m

In order to safeguard ourselves from the possible equalities let λn = λn−1 = . . . =

λn−k+1. Evaluating limit of W (m) as m tends to infinity gives

limm→∞

W (m) = limm→∞

|λn|mn−k∑i=1

(λi|λn|

)m + (−1)mk = −∞

Chapter 3 Page 20

for limm→∞( λi

|λn|)m = 0. Hence, following the |λn| > λ1 assumption we proved that

fom m large enough W (m) < 0. However, the number of closed walks cannot be a

negative integer, hence a contradiction.

The spectrum of adjecency matrix is scattered around zero and there exists a leading

eigenvalue λ1. Moreover, we know that in terms of absolute values λ1 can be either

leading or equal to the least eigenvalue, λn. Can we say exactly in what cases (3.3)

becomes an equality?

Lemma 3.5. G is bipartite if and only if λ1 = λn.

Proof. First assume G is bipartite. By Theorem (1.2) G is bipartite if and only

if it contains no walks of odd length. We will show that this condition is breached if

λ1 > −λn. Building on an argument similar to that in Lemma (3.4):

W (m) = Tr(Am) = |λ1|mn∑i=1

(λi|λ1|

)m = |λ1|mn∑i=k

(λi|λ1|

)m + k (3.4)

Taking a limit out of (3.4) gives limm→∞W (m) =∞, which implies that there exists

some G and some odd m for which W (m) 6= 0, hence it cannot be a bipartite graph.

Conversely, we claim that if G is not bipartite then for all large enough odd numbers m

there is a walk of length m from any vertex vx to any vertex vy. By Perron-Frobenious

Theorem 1 this claim suffices to show that λ1 > λn.

Proposition 3.1. Let G be a connected, non-bipartite graph. Then for all large

enough odd numbers m there is a walk of length m between any pair of vertices.

Proof. Since G is not bipartite by Theorem (1.2) there exists an odd cycle of

length k around some vertex x. For any vertex y ∈ G and y 6= x fix a walk w′yifrom

yi to x and let l1, . . . , ln−1 be the lengths of w′yi(see Figure 2). We can choose such

a walk wyithat k divides wyi

in the following way:

wyi= w′yi

· w′yi· . . . · w′yi︸ ︷︷ ︸

k times

where w′yiis the inverted yi − x walk and wyi

= lik. Take L = max li and set

m = (2L + 1)k. Now what is left is to show that there exists a walk of length m

1Wikipedia [9] gives the following formulation of the Perron Frobenius Theorem: Let A be a realn× n matrix with strictly positive entries aij > 0, then there is a positive real number r, called thePerronFrobenius eigenvalue, such that r is an eigenvalue of A and any other eigenvalue λ is strictlysmaller than r in absolute value, that is |λ| < r

Chapter 3 Page 21

x

y1

y2

w'y1

w'y2

odd cycle of length k

Figure 2. If G is not bipartite there exists a cycle of odd length k

between any two vertices of G. If ya and yb are two distinct vertices different from x,

define a walk yi − yj as follows:

yjlik−→ x

loop 2L+1−li−lj−−−−−−−−−−→times

xljk−→ yj

The length of this walk is lik+ (2L+ 1)k− lik = ljk+ ljk = (2L+ 1)k as we wanted.

If yj = x we proceed in a similar fashion:

yilik−→ x

loop 2L+1−li−−−−−−−−→times

x

Which also gives a walk of length (2L+ 1)k.

By Perron Frobenius applied with Proposition (3.1) and Lemma (3.4) we see that if

G is not bipartite then λ1 > |λn| which completes the proof.

In the next results we will use previous finding to explore the notion of eigenvalue

as a limit of W (m)1/m. Also, we will show that the condition for the graph to be

bipartite is that this limit does not exist.

Lemma 3.6. limm→∞W (2m)1

2m = λ1, where m ranges over positive integers

Proof. As previously we need to safeguard the case of equalities λ1 = λi. Let us

say that for k of the eigenvalues we have λ1 = . . . = λk. Then expand W (2m)1

2m as

below

Chapter 3 Page 22

W (2m)1

2m = Tr(2m)1

2m =n∑i=1

(λ2mi )

12m

=

(λ2m

1

n∑i=1

(λiλ1

)2m) 1

2m

= λ1

(k +

n∑i=k+1

(λiλ1

)2m) 1

2m

= λ1k1

2m

Evaluating the limit of W (2m)1

2m as m tends to infinity yields

limm→∞

W (2m) = limm→∞

λ1k1

2m = λ1 (3.5)

Lemma 3.7. If G is bipartite then limm→∞W (m)1m does not exists.

Proof. Intuitively, W (m)1m will oscillate between positive and negative limit,

depending on the parity of m. Assume that G is bipartite, then as previously shown

W (m)1m = Tr(m)

1m =

n∑i=1

(λmi )1m =

(λm1

n∑i=1

(λiλ1

)m) 1m

Now use that G is bipartite if and only if λ1 = −λ1 to show where the problem is

limm→∞

((−λn)m

n∑i=1

(λiλ1

)m) 1m

=

λn if m is even

−λn if m is odd

Thus if G is bipartite, limm→∞W (m)1m does not exist.

Next, let us turn to the bounding problem. Although eigenvalues are pretty not-

straight forward features of graphs we can bound them with more feasible parameters

such as minimum and maximum degree.

Theorem 3.5. If G is a connected graph with the minimum degree δ and maximum

degree ∆ then

δ ≤ λ1 ≤ ∆ (3.6)

Chapter 3 Page 23

Proof. For the first inequality use the fact that λ1 = maxvTAv|v|2 . If we denote the

vector with all n entries equal to 1 by x1 we can write

xT1Ax1

n≤ λ1

xT1Ax1 ≤ λ1n (3.7)

But now we note that xT1Ax1 equals to the sym of degrees of n vertices of G and

n∑i=1

deg(vi) ≥ δn (3.8)

Combining (3.7) and (3.8) yiels

δ ≤ λ1 (3.9)

For the second inequality observe that the total number of walks of length 2m is at

most n ·∆2m hence

λ1 = limm→∞

W (2m)1

2m ≤ limm→∞

(n ·∆1

2m )1

2m = ∆ (3.10)

Finally, we combine the above results to obtain a useful upper bound on chromatic

number χ(G) of G.

Theorem 3.6. Let G be a simple graph, then χ(G) ≤ 1 + λ1.

Proof. Let G be a connected n-chromatic graph on p vertices and let G′ be a

critically n-chromatic subgraph of G on p′ vertices. Note that δ(G′) ≥ n − 1. Let

A(G) = [aij] and A(G′) = [a′ij] be adjacency matrices of G and G′. Also, let A∗ be a

p × p matrix obtained from A(G) by putting zero rows and columns in the place of

vertices deleted from G in a process of obtaining G′. Observe that the eigenvalues of

A∗ are those of A(G′) and an additional p− p′ zero-eigenvalues. Since the maximum

eigenvalue of G is greater or equal to a maximum eigenvalue of subgraph G′ we have

λ1(G′) = λ1(A∗) ≤ λ1(G) (3.11)

Furthermore we have that for G’:

λ1(G′) = maxvTA(G′)v

|v|2

hence for xT = (1, . . . , 1)

λ1(G′) = maxvTA(G′)v

|v|2≥ xTA(G′)x

|x|2=

∑p′

j=1(∑p′

i=1 a′ij)

p′

Chapter 3 Page 24

where the RHS is equal to the average of the row sums of the matrix A(G’). Since

δ(G′) ≥ n− 1, the minimum row of A(G′) must be at least n− 1 and hence

λ1(G′) ≥ n− 1 (3.12)

Putting (3.11) and (3.12) together we arrive at

χ(G) ≤ 1 + λ1(G) (3.13)

The above upper bound on the chromatic number χ(G) was obtained by Wilf. Impor-

tantly,it is an improvement of Brook’s upper bound χ(G) ≤ ∆ + 1 obtained through

the application of the greedy algorithm. This follows directly from Theorem (3.5) in

which we have shown that λ1 ≤ ∆.

Last but not least let us offer a proof of a well-known Friendship Theorem as im-

mortalized in [1]. In its simplest version the statement goes as follows: In a group

of people where any pair of people have precisely one common friend there is always

a person (also referred to as a politician) who is a friend of everybody. Since the

statement was initially introduced as combinatorial problem, much efforts were put

into conducting a proof relying on combinatorial argument only. Interestingly, the

statement can be shown to be true using an argument based partly on combinatorial

arguement and partly on an application of algebra. The proof presented below is due

to Erdos, Renyi and Sos.

2. Friendship Theorem

If we translate friendship theorem into the language of graph theory we arrive at the

following statement

Theorem 3.7. (Friendship Theorem)Suppose that G is a finite graph in which any

two vertices have precisely one common neighbour. Then there exists a vertex which

is adjecent to all other vertices.

The proof is by contradiction. Assume that G satisfies the conditions of the theorem

but there does not exist any vertex v which is connected with all other vertices.

First we are going to show that the condition of sharing precisely one neighbour by

each pair of vertices forces G to be a k − regular graph. Note that the condition

of the theorem requires that there are no cycles of length 4 in G. We will call this

requirement a C4 condition.

Chapter 3 Page 25

Figure 3. The C4 condition demands that the above situation doesnot occur.

Lemma 3.8. If G is finite graph in which any two vertices have precisely one common

neighbour then G has to be regular of degree k ≥ 4.

Proof. We choose two arbitrary non-adjecent vertices u and v and let µ(u) and

µ(v) denote the sets of neighbours of u and v. Let us construct a function ϕ : µ(u)→µ(v) as follows: for any w ∈ µ(u) let ϕ(w) be the unique friend of w and v. Note that

the function is well-defined since v /∈ µ(u) so v 6= w. Suppose ϕ is not injective which

means that there exist w1, w2 ∈ µ(u) such that ϕ(w1) = ϕ(w2) = z where z ∈ µ(v).

We have z 6= u becasue u /∈ µ(v) and z ∈ mu(v). Next, z 6= w1 becasue if z = w1 then

w1 = ϕ(w1) ∈ µ(w)1) which is impossible, similar argument applies to w2. Hence,

z, w1, w2 and u are distinct and we have the following situation occuring:

uv z

w

1

2

w

By the C4 condition we arrived at a contradiction. Therefore

deg(u) = |µ(u)| ≤ |µ(v)| = deg(v)

and by symmetry deg(u) = deg(v) = k.

Now we would like to show that G is k-regular. Let w0 be a vertex common for

u and v. If w is a vertex other then w0 then, by the C4 condition w cannot be

a friend of both u and v. So either w is not adjecent to u or w is not adjecent

to v and hence deg(w) = deg(u). By the hypothesis w0 is not connected to all

vetices in G hence there exists some z which is not adjacent to w0 which implies that

deg(w0) = deg(z) = deg(u) = k and G is k-regular. Furthermore, w0 has u, v ∈ µ(w0)

but also z1 ∈ µ(w0) where z1 is a common friend of w0 and u and z2 ∈ µ(w0), where

z2 is a common friend of w0 and v. Clearly, z1 6= u, v but also z1 6= z2 by the C4

condition. Hence w0 has at least 4 neighbours and k ≥ 4.

Chapter 3 Page 26

With this result we progress to algebraic part of the proof which will use Therorem

(3.1) and results on eigenvalues. Fix v and look at walks of length 2 from v. We can

distinguish two kinds of such walks:

Case 1: The walk is closed. In such a walk we basically go forth to one of the

neighbours of v and then back to v. There are deg(v) = k such walks

Case 2: The walk is not closed. Note that the walk is completely determined

by its endpoint u (because v and u have a unique common neighbour) which implies

that for all u 6= v there exists unique walk of length 2 which terminates at u hence

there exists n− 1 such walks.

Summing over the degrees of the k neighbours of v we get

W (2) = k2 = k + n− 1

Consider the adjecency matrix A of graph G. By Theorem (3.1) A2 is as follows

A2 =

k 1 · · · 1

1 k · · · 1...

.... . .

...

1 1 · · · k

= (k − 1)I + J

where I is the identity matrix and J is the matrix with all entries equal to one. Next,

we would like to find eigenvalues of matrix A2 and A. The rank of J is 1, so there

is one nonzero eigenvalue equal to n, similarily the rank of I is n and 1 is the only

eigenvalue with multiplicity n. It follows that A2 has eigenvalues k − 1 + n = k2 of

multiplicity 1 and k−1 of multiplicity n−1. This implies that A has eigenvalues k or

multiplicity 1 and ±√k − 1. Suppose that r of the eigenvalues are equal to

√k − 1

and s of eigenvalues are equal to −√k − 1 where r + s = n − 1. Since the sum of

eigenvalues is equal to zero by Theorem (3.1) we have

k + r√k − 1− s

√k − 1 = 0

since r 6= s√k − 1 =

k

s− rAt this point we need to use Dedakin’s result which says that if a square root

√m of

natural number m is rational, then it is an integer. Let h =√k − 1 ∈ N then

h(s− r) = k = h2 + 1 (3.14)

Now since h divides h2 +1 and obviously h divides h2, we conclude that h can only be

equal to 1 and therefore k = 2 which we ruled out, hence a contradiction is obtained.

CHAPTER 4

Graph Theory meets Probability

In the following two chapters we will explore the areas where graph theory intersects

probability theory. Our aim is to introduce the notion of a random graph. Next,

in the context of probabilistic method we will to show that there are graphs with

arbitrarily large girth and chromatic number.

1. Random graph

A random graph is a graph generated by some random process. To be more precise let

G be a random graph on n vertices and let us flip a coin which comes up heads with

probability p independently for each of the(n2

)potential edges. If the coin comes

up head then the edge e ∈ E(G), otherwise the edge e /∈ E(G). The probability

space G(n, p) consists of all graphs on n vertices where the individual edge e appears

with probability p, independently from each other. For instance, the probability of a

complete graph Kn would be P (Kn) = p(nk), since all the possible edges would need

to belong to G. Generally, we have P (G) = pn(1 − p)(n2)−m if there are exactly m

edges in E(G).

2. Erdos’ Proof

Recall that girth γ(G) is the length of the shortest cycle contained in the graph G and

chromatic number χ(G) is the smallest number of colours needed to colour a graph G

so that no two adjecent vertices are assigned the same colour. Observe that choosing

a graph with high chormatic number or high girth alone does not pose a probelm

and is the question of existance of arbitrarily high chromatic number and high girth

that is interesting. For example in Kn, a complete graph on n vertices, we have

χ(Kn) = n, but γ(Gn) = 3. On the other hand in Cn, a cycle graph on n vertices, we

have γ(Cn) = n, but χ(Cn) ≤ 3. Intuitively, it may seem that high chromatic number

goes at the price of low girth and vice versa. The assumption that we can either

achieve high chromatic number or high girth seems plausible. Surprisingly, such an

intuition in this case is wrong and a proof due to Erdos presented below shows that

there exists a graph G with arbitrarily high chromatic number and girth.

27

Chapter 4 Page 28

The proof is based on probabilistic method, a nonconstructive approach pioneered by

Paul Erdos which says that

if in a given set of objects, the probability that an object does not

have a certian property is less than 1, then there must exist an object

with this property [3]

The elegance of this method lies in the fact that it allows to prove existence of difficult

objects (such as a graph in question), without creating or providing a method of

constructing the object directly. Indeed, anyone who tried to give an example of

graph with high girth and high chromatic number will find that this is not an easy

task. In the light of this discussion the following theorem comes as a surprise.

Petersen Graph Double-star snark Hoffman-Singleton Graph

Figure 1. Examples of graphs with γ(G) ≥ 5 and χ(G) ≥ 3: Petersengraph is the smallest 3-regular graph with γ = 5 and χ = 3, Hoffman-Singleton graph has γ = 5 and χ(G) = 4. Peterson graph belongs toa special class of graphs called snarks which all have girth ≥ 5 andχ = 3. Credits to [9] and [8]

Theorem 4.1. For every k ≥ 2 there exists a graph G with χ(G) > k and γ(G) > k.

By the probabilistic method, in a certain probability space G(n, p) on all graphs with

n vertices the probability that a graph has either χ(G) ≤ k or γ(G) ≤ k is strictly less

than 1, therefore there exists a graph G ∈ G(n, p) with χ(G) > k and γ(G) > k. The

strategy is to show that on a certain probability space, which we will choose wisely,

the probability for χ(G) ≤ k is smaller than 12

and the probability for γ(G) ≤ k is

smaller than 12. Hence there exist a graph G with desired properties.

2.1. Chromatic number. We start by noting that vertices which belong to the

same colour class must by definition be pairwise non-adjecent and hence they form

an indepenent set.

Chapter 4 Page 29

By (2.1) α(G) is inversely proportional to χ(G), hence if α(G) is small as compared

to n, then χ(G) must be large. For 2 ≤ r ≤ n the probability that a graph G contains

a fixed set of size r, which we will call r-set R as an independent set is (1 − p)(r2),

because there are(r2

)edges which we deselect. If we denote G such that G contains

R as an independent set by CR the following holds

P (G : α(G) ≥ r) = P (⋃R⊆V

CR) ≤∑R⊆V

P (CR)

=

(n

r

)(1− p)(

r2) ≤ nr(1− p)(

r2)

= (n(1− p)r−12 )r ≤ (ne

−p(r−1)2 )r (4.1)

Where the last inequality holds because 1− p ≤ e−p for all p.

At this point comes the first part of the magic, a clever choice of p which makes the

proof work. For every n let us chose p = n−k

k+1 and proceed to show that for all n

large enough

P (G : α(G) ≥ n

2k) < 1

2(4.2)

What comes next may seem surprising but in fact it is nothing more than using our

special set-up to validate (4.2). Firstly, we have

n1

k+1 ≥ 6k lnn (4.3)

for all n large enough. 1 We can divide both sides by n > 0 to obtain

n1

k+1

n= n

1k+1− k+1

k+1 = n−kk+1 = p ≥ 6k lnn

n

For r := n2k

this gives

pr ≥ 3 lnn (4.4)

Having (4.3) in mind we continue evaluating (4.1) as follows

ne−p(r−1)

2 = ne−pr2

+ p2 ≤ ne−

32

lnne12 = n−

12 e

12 =

√e√n→ 0 as n →∞ (4.5)

1To see this use l’Hopital’s rule, which gives

limn→∞

6k lnn

n1

k+1= limn→∞

(6k lnn)′

(n1

k+1 )′= limn→∞

6kn−1

1k+1n

− kk+1

= 0

Chapter 4 Page 30

Note that ep2 ≤ e

12 since p ≤ 1. Hence, (4.2) holds for all n ≥ n1 for some large n1.

2.2. Girth. Let us turn to the second parameter, girth γ(G). For a given k our

aim is not to have too many short cycles of length less or equal to k. Let 3 ≤ i ≤ k

and consider a fixed i-set I ⊆ V . The number of possible i-cycles on I is equal to(i−1)!

2, where the division by 2 accounts for the fact that each cycle can be travelled

in two directions. The total number of possible i-cycles in V is therefore(ni

) (i−1)!2

and

every such cycle on I appears with probability pi.

Let X be the random variable on G(n, p) where p = n−k

k+1 , counting the number of

cycles of length less or equal k in a graph G ∈ G(n, p). For every i-cycle I where i ≤ k

with vertices in V , the vertex set of the graphs in G(n, p), let XI : G(n, p) → 0, 1be the indicator random variable of I:

XI : G 7→

1 if I ⊆ G

0 if I * G

For the expectation of XI we have

E(XI) = 0 · P (XI = 0) + 1 · P (XI = 1) = P (G : G ⊇ I) = pi

Because of the linearily of the expectation we obtain

E(X) =∑I

E(XI) =∑I

pi =k∑i=3

(n

i

)(i− 1)!

2pi ≤ 1

2

k∑i=3

nipi ≤ 1

2(k − 2)nkpk

Note that (np)i ≤ (np)k because np = n1

k+1 ≥ 1.

In particular we infer that there exists n2 such that P (X ≥ n2) < 1

2for all n ≥ n2.

2.3. Putting it together. Let an event A = G : α(G) ≥ n2k and event

B = G : γ(G) ≤ k. Up to this point it has been shown that for all n >> 0

P (A ∪B) ≤ P (A) + P (B) <1

2+

1

2= 1 (4.6)

Therefore the probability of complimentary event P ((A ∪ B)C) > 0. Hence there

exists a graph H ∈ AC ∩ BC such that α(H) < n2k

and H contains less than n2

short

cycles of length less or equal to k. If we delete one vertex from each of the short cycles

then for the resulting graph G we would have γ(G) > k. The question is whether

we can delete vertices without ruining the high chromatic number of H in the same

time. Suprisingly, in this particular set-up the answer is yes. Deleting vertices does

affect the chromatic number but not at the pace that would spoil the final result.

Chapter 4 Page 31

We certainly deleted less than n2

vertices from the graph H so the resulting graph G

contains more than n2

vertices and satisfies α(G) ≤ α(H) < n2k

. Finally, we arrive at

χ(G) ≥ n/2

α(G)≥ n

2α(H)>

n

n/k= k

which finishes the proof.

CHAPTER 5

What is percolation?

Percolation is a probabilistic model exhibiting a phase transition, originally intro-

duced by Broadbent and Hammersley to study the flow of fluids through a porous

medium. Percolation theory builds up of the notion of random graphs where all edges

of a graph G are chosen to be open with probability p and closed with probability

1 − p. A basic question in this model is what is the probability that there is an

infinite open path, that is a path whose all edges are open? This probability is called

percolation probability which we will denote θ(p).

1. Set-up

Although percolation theory can be largely treated as an extension of random graphs

there are some terminological differences. If Λ is a graph, then V (Λ) denotes the set

of vertices which we refer to as sites and E(Λ) denotes the set of edges which we refer

to as bonds. Within the random graph Λ we can distingush site percolation which

is obtained by random selection of vertices or bond percolation which is obtaned by

random selection of edges. The sites or bonds which are selected are called open and

those which are not selected are called closed. In material to follow in this section we

will concentrate on the bond problems.

For sites x and y we write x → y for the event that there exists an open path from

x to y with P (x → y) being the probability of this event. Following this convention

we write x → ∞ for the event that there exists an infinite open path starting at x.

We denote Cx = y ∈ Λsuch thatx→ y, the set of sites y for which there is an open

x − y path. An open cluster is a component of an open subgraph, and θ(p) is the

probability Cx is infinite, that is for every site in the cluster the event x→∞ holds.

Clearly, if all bonds are closed θ(0) = 0, on the other hand if all bonds are opened

θ(1) = 1. It is intuitively clear that θ(p) is non-decreasing, a fact which we will not

prove here 1. Thus, there is a critical probability 0 ≤ pH ≤ 1 such that if p < pH

then θx(p) = 0 for every site x and if p > pH then θx(p) > 0 for all x.

1the proof is attained using a coupling technique

32

Chapter 5 Page 33

Figure 1. Graph of θ.

By application of Kolmogorov’s 1-0 law2, which we will not present here, we have

an interesting situation: if p < pH then the probability of the even E that there is

an infinite open cluster is 0, while if p > pH this probability is 1. What this means

is that the component structure of Λ is quite different for p < pH and p > pH .

Terminology-wise, in a given model we say that percolation occurs if θx > 0 and

P (E) = 1.

Although for p < pH the open cluster Cx is finite, the expected size of Cx does not

to be finite. The second critical probability, pT provides information about when the

expected size of Cx is no longer finite. Let

χx(p) = Ep(|Cx|)

χ(p) is increasing with p and is finite for some site x if and only if it is finite for all

sites. We define pT as follows

pT = supp : χx(p) <∞ = infp : χx(p) =∞

By definition pT ≤ pH , however for many graphs (including Zd) pT and pH are equal.

Although initial attempts were made to establish critical probabilities for various

graphs, this porgramme proved to be not too successful as there is only a handful of

graphs, such as k-branching tree and square lattice, for which precise values of critical

probabilities are known. Meanwhile it was observed thtat a number of interesting

phenomena occurs as p → pc, or as other parameters tend to infinity while the

2Let X = (X1, X2, . . .) be a sequence of independent random variables and let A be an event inthe σ-field generated by X. Suppose that for evey n, the event A, called tail event, is independentof X1, . . . , Xn. Then P (A) is either 0 or 1. An example of tail event is a sequence of 100 consecutiveheads occurring infinitely many times.

Chapter 5 Page 34

probability p is kept at pc. According to Kesten [7], many functions obey power

laws3.

2. Finding critical probabilities for k-branching infinite tree Tk

In general the critical probabilities introduced in the previous section are not easy

to find. However, in the example to follow we can obtain both pT and pH at an

affordable price.

Theorem 5.1. (k-branching tree) Let Tk be an infinite rooted tree in which each

vertex has k children.

pT (Tk) = pH(Tk) =1

k(5.1)

v0

Figure 2. A finite 2-branching tree of height 3

Proof. Let πn be a probability that there exists an open path from v0 to n

levels below. By default, π0 = 1. Assume there is some path going through v1. The

probability that an open path from v0 to n levels below goes through v1 is pπn−1,

which is the probability p that the v0 − v1 bond is open times the probability πn−1

that a path from v1 to a vertex n− 1 levels below is open. Similarily, the probability

that there is no open path from v0 to n levels below going through v1 is 1 − pπn−1.

Now observe that any path which goes n levels below has to go through one of the

children of v0, hence the probability that there is no open path from v0 to n levels

below is (1− (pπn−1))k, which implies that

πn = 1− (1− pπn−1)k (5.2)

Let us define a function fk,p = 1−(1−px)k, then (πn) = fk,p(πn−1). We are interested

in what values fk,p takes on the interval [0, 1]. First observe that we can differentiate

3For example it is believed for the expected number of vertices in Cxχ(p), the following holds:− lnχ(p)ln (p−pc)

→ γ for a suitable constant γ.

Chapter 5 Page 35

using Chain Rule:

f ′k,p(x) = pk(1− px)k−1

f ′′k,p(x) = −(p2k)(k − 1)(1− px)k−2

hence fk,p is increasing and concave. Furthermore, fk,p(0) = 0 and fk,p(1) < 1, hence

there exists some x0 ∈ (0, 1) such that fk,p(x0) = x0 if and only if f ′k,p = kp > 1.

y=f (x)

k,p

x0

f' (0) = kp > 1k,p

Bearing in mind that (πn) = fk,p(πn−1) we see from the graph that if p > 1k

and

x > x0 then x0 < fk,p(x) < x. Since π0 = 1 we have that

x0 < π1 = fk,p(π0) < π0 = 1

x0 < π2 = fk,p(π1) < π1

. . .

x0 < πn = fk,p(n− 1) < πn−1

and hence πn > x0 for all n. It follows that

θv0(Tk;p) = limn→∞

πn = x0

implying that

pH(Tk) ≤1

kOn the other hand if p < 1

kthen πn converges to 0 and θv0(Tk; p) = 0. The critical

probability pT must be 1k.

For the pT , observe that the probability that a site y which is at distance l from vo

belongs to Cv0 is equal pl and therefore

Chapter 5 Page 36

χv0(Tk; p) = E(|Cv0|) =∞∑l=0

klpl (5.3)

where (5.3) if finite for p < 1k

and infinite for p ≥ 1k. Therfore, the critical probability

pT (Tk) = 1k

3. Bounds for Critical Probabilities on Square Lattice

Next, we are going to discuss the bounds on the critical probabilities pH , pT on square

lattice Z2. Recall that the dual Λ∗ of a graph Λ drawn in the plane has a vertex for

each face of Λ and an edge e∗ for each e. An edge e∗ ∈ E(Λ∗) joins two vertices of Λ∗

which correspond to two faces of Λ in whose boundary edge e ∈ E(Λ) lies. In studying

Z2 the convention is to take v∗ = (x+ 12, y+ 1

2) as the dual vertex which correspons to

face bounded by following vertices:(x, y), (x+ 1, y), (x+ 1, y+ 1), (x, y+ 1). The sites

and bonds of Λ∗ are called dual sites and dual bonds, with the dual bond e∗ open if

e is closed and vice versa.

Before turning out attention to the critical probabilities on square lattice, one more

result which will be used in the boundary proof is needed.

Definition 5.1. A self-avoiding walk µn(Λ;x) on a square lattice is a path which

never intersects itself.

Let Λ = Z2. We can provide the following bound on the number of self-avoiding

walks on Z2:

µn(Λ;x) ≤ 4 · 3n−1 (5.4)

The above inequality comes from the fact that the number of self-avoiding walks

must be smaller than a number of the so called nonreversing walks, which are ran-

dom walks whose alternating sequence does not have any subsequence of the form

vn, en, vn+1, en, vn. Hence, for the nonrevesing walk there are 4 choices for the firts

step and three choices for each step therafter.

If H is a finite, connected subgraph of Λ with vertex set C, it follows that there

exists an infinite component C∞ of Λ−C, which is the subgraph of Λ induced by the

vertices outside H. Let σ∞C be the set of bonds of Λ∗ dual to bonds of Λ joining C

and C∞.

Proposition 5.1. If C is the vertex set of a finite connected subgraph H of Λ = Z2,

then σ∞C is a cycle with C in its interior, hence it is a boundary.

Chapter 5 Page 37

Figure 3. Subgraph H together with its external boundary σ∞C

We will not give a detailed proof but outline the technique of the prove given in [3].

Denote−→F to be the set of the C−C∞ bonds, oriented from C to C∞. For

−→f =

−→ab ∈

−→F

orient to dual bond−→f ∗ = −→uw so that a is on its left. Let

−−−→σ∞C =

−→f ∗ :−→f ∈

−→F . Then

through a straight-forward consideration it can be shown that if−→f ∗ = −→uv ∈

−−−→σ∞C then

there is unique bond of−−−→σ∞C leaving v, which proves that the underlying unoriented

graph σ∞C consist of a single cycle S, with C∞ being outside of S and C inside S.

Lemma 5.1. For bond percolation in Z2 we have

1

3≤ pT ≤ pH ≤

2

3(5.5)

Proof. As remarked earlier pT ≤ pH , hence we complete the proof by showing

that 13≤ pT and pH ≤ 2

3. Assume that p < 1

3. Let C0 be the open cluster of the origin

on square lattice where each bond is open with probability p. From the definition of

C0, it follows that for every site x ∈ C0 there exists at least one 0−x path. So |C0| isat most the number X of open paths starting at 0. Since a path of length n is open

with probability pn

χ(p) = E(|C0|) ≤ E(X) =∑n≥0

µnpn ≤ 4

3

∑n≥0

(3p)n <∞

And we see that pT ≥ p, hence pT ≥ 13.

Suppose that p > 23

and consider Λ together with its dual Λ∗. Let Lk be a line segment

which joins the orgin with the point (k, 0) and let S be a dual cycle surrounding Lk

of length 2l. The idea for the upper inequality comes from an observation that C0

is finite precisely when it contains some Lk and there is an open cycle in Λ∗ which

surrounds Lk.

Chapter 5 Page 38

Since any dual cycle S must cross the x-axis at some coordinate x∗0 such that k+ 12≤

x∗0 ≤ 2l−32

, there is fewer then l choices for e∗. If we split S into e∗ and the remaining

path of length 2l − 1 it becomes clear that there are at most lµ2l−1 choices for S.

Denote the number of open dual cycles surrounding Lk by Yk. For any closed cycles

in Λ∗ surrounding Lk let zS be the random variable such that

zs(G) =

1 if Λ∗ contains S

0 otherwise

And Yk(G) =∑

S zS(G) =∑

S⊆Λ∗ 1. Keeping in mind that a dual edge e∗ ∈ E(Λ∗) is

open precisely when an edge e ∈ E(Λ) is closed we evaluate E(Yk) to obtain

Ep(Yk) =∑S

E(zS) ≤∑l≥k+2

lµ2k−1(1−p)2l ≤∑l≥k+2

l·4·32l−3(1−p)2l =∑l≥k+2

4l

9(3(1−p))2l

(5.6)

By the initial choice of p we get (3(1−p)) < 1, hence (5.6) is convergent and E(Yk)→ 0

as k →∞ which implies that there is some k for which Ep(Yk) < 1.

Let Ak be the event that there are no open dual cycles surrounding Lk, since Ep(Yk) <

1 we have that Pp(Ak) > 0. Let Bk be the event that all k bonds of Lk are open.

Note that Ak and Bk are independent. Importantly, (Ak ∩Bk) is the event of all the

graphs which contain Lk and there is no cycle S in the dual surrounding 0. Hence, if

both Ak and Bk hold, by Proposition (5.1) the open cluster C0 is infinite.

θ0(p) ≥ Pp(Ak ∩Bk) = Pp(Ak)Pp(Bk) = Pp(Ak)pk > 0

Hence pH ≤ p. By our arbitrary choice of p, pH ≤ 23.

Bibliography

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University Press[5] Fox, J., MAT 307 Combinatorics (Spring 2009) [Course notes, available from

http://www.princeton.edu/ jacobfox/MAT307.html Accessed on February 02, 2010][6] Grunbaum, B., Soifer, A., Rousseau, C. 2008. The Mathematical Coloring Book: Mathematics

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