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Map Coloring and
Some of Its
Applications
MD. SHAH ALAM
DEPARTMENT OF MATHEMATICS UNIVERSITY OF DHAKA
ABSTRACT
Map coloring more precisely graph coloring is an important topic of graph theory. In this
project we have studied the basics of graph theory and some of its applications in map coloring.
First chapter will introduce the concept of graphs and graph theory. In chapter two, we describe
some basic terminologies of graphs with important theorems and corollaries related to this
topic. Finally, the chapter three introduces the idea of map coloring where the famous Four
Color Theorem is discussed and some real life problems have been solved by using the very
important concept.
Contents
Chapter One Introduction Page
Basic 1
1.1 Graph 2
1.2 Graph Theory 2
1.3 Importance of Graph Theory 2-3
Chapter Two Basic Concept of Graph Theory
2.1 Basic terminology of Graphs 4-9
2.2 Theorems and Corollaries on Graph Theory 9-12
2.3 Some Problems on Graph Theory 12-14
Chapter Three Graph Coloring
3.1 Graph Coloring 15
3.2 Importance Graph Coloring 15
3.3 Chromatic Number of a Graph 15
3.4 Problems on Chromatic Number 16
3.5 Subgraph 17
3.6 Theorems and Corollaries on Graph
Coloring
17-20
3.7 Four Color Theorem 21
3.8 History of Four Color Theorem 21-24
3.9 An Example of Four Color Theorem 25-27
3.10 Applications on Graph Coloring 27-32
Conclusion 33
Reference 34
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Chapter 1
Introduction
Basic
Graphs are discrete structures consisting of vertices and edges that connect these vertices.
There are different kinds of graphs, depending on whether edges have directions or not, whether
multiple edges can connect the same pair of vertices, and whether loops are allowed or not.
Problems in almost every conceivable discipline can be solved using graph models. Many
structures involving real world situation can be conveniently present on paper by means of a
diagram consists of a set of points together with lines joining some or all pairs of these points.
Graphs are used to represent the competition of different species in an ecological niche, to
represent who influences whom in an organization, to represent the outcomes of round-robin
tournaments. Graphs can be used to model acquaintanceships between people, collaboration
between researchers, telephone calls between telephone numbers, and links between websites.
They can be used to model roadmaps and the assignment of jobs to employees of an
organization. Using graph models, we can determine whether it is possible to walk down all
the streets in a city without going down a street twice, and we can find the number of colors
needed to color the regions of a map. Graphs help to determine whether a circuit can be
implemented on a planar circuit board. We can distinguish between two chemical compounds
with the same molecular formula but different structures using graphs. Without the basic idea
of graph theory it will be an unsuccessful attempt to understand map coloring. By learning the
various properties of graph theory we can know that map coloring is nothing but a very useful
application of it. In map coloring problem we mainly deal with the undirected graph. We can
determine whether two computers are connected by a communications link using graph models
of computer networks. Graphs with weights assigned to their edges are also used to solve
problems such as finding the shortest path between two cities in a transportation network. We
can also use graphs to schedule exams and assign channels to television stations.
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1.1 Graph: The set or collection of finite number of vertices (V) and edges (E) is called a graph
and is denoted by G= (V, E). More specifically in graph theory, a graph is a representation of
a set of objects where some pairs of objects are connected by links. Typically, a graph is
depicted in diagrammatic form as a set of dots for the vertices, joined by lines or curves for the
edges.
1.2 Graph Theory: Graph theory is the study of graphs, which are mathematical structures
used to model pairwise relations between objects. A graph in this context is made up of vertices,
nodes, or points which are connected by edges, arcs, or lines. Graph theory concerns the
relationship among lines and points. Graph theory is a growing area in mathematical research,
and has a large specialized vocabulary.
1.3 Importance of Graph Theory: Graph theory is very important in many branches of
science such as mathematics, physics, computer science, chemistry, life science etc.
1. Graphs are used in the field of chemistry to model chemical compounds. In computational
biochemistry some sequences of cell samples have to be excluded to resolve the conflicts
between two sequences. This is modeled in the form of graph where the vertices represent the
sequences in the sample. An edge will be drawn between two vertices if and only if there is a
conflict between the corresponding sequences. The aim is to remove possible vertices,
(sequences) to eliminate all conflicts. [2]
2. The major role of graph theory in computer applications is the development of graph
algorithms. Numerous algorithms are used to solve problems that are modeled in the form of
graphs. These algorithms are used to solve the theoretical concepts which in turn used to solve
the corresponding computer science application problems.
3. Graph theory is a very natural and powerful tool in combinatorial operations research. A
network called transport network where a graph is used to model the transportation of
commodity from one place to another. The objective is to maximize the flow or minimize the
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cost within the prescribed flow. The graph theoretic approach is found to be more efficient for
these types of problems though they have more constraints.
4. Graphs are extensively used to model social structures based on different kinds of
relationships between people or groups of people. These social structures, and the graphs that
represent them, are known as social networks. In these graph models, individuals or
organizations are represented by vertices; relationships between individuals or organizations
are represented by edges. The study of social networks is an extremely active multidisciplinary
area, and many different types of relationships between people have been studied using them.
[1]
5. The World Wide Web can be modeled as a directed graph where each Web page is
represented by a vertex and where an edge starts at the Web page a and ends at the Web page
b if there is a link on a pointing to b. Because new Web pages are created and others removed
somewhere on the Web almost every second, the Web graph changes on an almost continual
basis. [3]
6. Graphs can be used to model telephone calls made in a network, such as a long distance
telephone network. In particular, a directed multigraph can be used to model calls where each
telephone number is represented by a vertex and each telephone call is represented by a directed
edge. The edge representing a call starts at the telephone number from which the call was made
and ends at the telephone number to which the call was made. We need directed edges because
the direction in which the call is made matters. We need multiple directed edges because we
want to represent each call made from a particular telephone number to a second number. [1]
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Chapter 2
Basic Concepts of Graph theory
2.1 Basic terminology of Graphs
We will introduce some basic terms of graph and graph theory.
2.1.1 Vertex: The point or node in a graph G is known as vertex.
2.1.2 Edge: The connection or line in a graph G is called edge, i.e. the line joining two vertices
is an edge.
Figure: 2.1
In the figure 2.1 A, B, C are three vertices and E1, E2, E3 are edges.
2.1.3 Adjacent vertices: In an undirected graph G if two vertices u and v are end points of an
edge then they are called adjacent or neighbors.
2.1.4 Incident edge: If an edge or line associated with the vertices u and v of a graph then the
edge is called incident with the vertices u and v. It is also said to connect u and v.
Figure: 2.2
In the figure 2.2 A and B, A and C, C and D, D and B are adjacent vertices and E1, E2, E3, E4
are incident with the vertices A and B, A and C, C and D, D and B respectively.
A B
E1
C
A B
E2E3
D
E1
E2
E3
E4
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2.1.5 Directed Graph: A directed graph (or digraph) is a graph, or set of vertices connected
by edges, where the edges have a direction associated with them. So, if a graph is directed it
has distinct direction along all of its vertices.
Figure: 2.3
Figure 2.3 is a directed graph.
2.1.6 Undirected Graph: If there is no direction among the edges and the vertices of the graph
G= (V, E), then G is called an undirected graph.
Figure: 2.4
In 2.4, the graph is an undirected graph, because they have no direction through the edges E1,
E2, E3, E4
2.1.7 Degree of a vertex: In an undirected graph G the degree of a vertex is defined as the
number of edges connected or incident to that vertex. It is denoted by deg() or simply d.
Figure: 2.5
In figure 2.5 deg(A)= 2, deg(B)=3, deg(C)= 3, deg(D)=1, deg(E)= 2, deg(F)=3 and
deg(H)=0. [1]
A B
E1
A B
E1
DCB
C
E2E3
D
E2
E3
E4
A F E
H
’
’
’
’
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2.1.8 Isolated vertex: A vertex of degree zero is known as isolated. H is isolated vertex in the
figure 2.5
2.1.9 Pendant: A vertex is pendant if and only if it has degree one. In figure 2.5, D is a pendant.
2.1.10 Simple Graph: If there is no direction among the edges of the graph G= (V, E) then G
is called a simple graph. It is also known as undirected graph.
2.1.11 Complete Graph: If a graph G contains exactly one edge between each pair of distinct
vertices then it is called a complete graph and is denoted by Kn, where n=1, 2, 3, 4,….
For n=1, 2, 3, 4 the graphs are given below-
Figure: 2.6
2.1.12 Bipartite Graph: A simple graph G is called bipartite if its vertex set V can be
partitioned into two disjoint sets V1 and V2 such that every edge in the graph connects a vertex
in V1 and a vertex in V2 (so that no edge in G connects either two vertices in V1 or two vertices
in V2). When this condition holds, we call the pair (V1, V2) a bipartition of the vertex set V of
G.
Figure: 2.7
Figure 2.7 shows a bipartite graph where the vertices are partitioned into two sets say
V1= {a, b, c} and V2= {m, n, o, p} [6]
a b c
m on p
K1
K2 K3 K4
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2.1.13 Isomorphism of Graphs: The simple graphs G1 = (V1, E1) and G2 = (V2, E2) are
isomorphic if there exists a one to-one and onto function f from V1 to V2 with the property that
a and b are adjacent in G1 if and only if f (a) and f (b) are adjacent in G2, for all a and b in V1.
Such a function f is called an isomorphism. In other words, when two simple graphs are
isomorphic, there is a one-to-one correspondence between vertices of the two graphs that
preserves the adjacency relationship.
Figure: 2.8
Figure 2.8 shows that the graphs G1 = (V, E) and G2 = (W, F) are isomorphic. Because the
function f with f (u1) = v1, f (u2) = v4, f (u3) = v3, and f (u4) = v2 is a one to-one correspondence
between V and W. Adjacent vertices in G1 are u1 and u2, u1 and u3, u2 and u4, and u3 and u4,
and each of the pairs f (u1) = v1 and f (u2) = v4, f (u1) = v1 and f (u3) = v3, f (u2) = v4 and f (u4)
= v2, and f (u3) = v3 and f (u4) = v2 consists of two adjacent vertices in G2. [1]
2.1.14 Path: A path is a sequence of edges that begins at a vertex of a graph and travels from
vertex to vertex along edges of the graph.
Formally, Let n be a nonnegative integer and G be an undirected graph. A path of length n from
u to v in G is a sequence of n edges e1, e2,…, en of G for which there exists a sequence x0 = u,
x1, . . . , xn−1, xn = v of vertices such that ei has, for i= 1, . . . , n, the endpoints xi−1 and xi . When
the graph is simple, we denote this path by its vertex sequence x0, x1,…… ,xn
Figure: 2.9
In the simple graph shown in Figure 2.9 A, D, C, F, E, is a simple path of length 4, because
{A, D}, {D, C}, {C, F }, and {F, E} are all edges. However, D, E, C, A is not a path, because
{E, C} is not an edge.
u1
u2
u3 u4
v1 v2
v3 v4
G1 G2
A
B
C
D E F
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2.1.15 Circuit: A path is a circuit if it begins and ends at the same vertex, that is, if u = v, and
has length greater than zero. The path or circuit is said to pass through the vertices x1, x2, ….,
xn−1 or traverse the edges e1, e2, . . . , en.
In 2.9 B, C, F, E, B is a circuit of length 4 because {B, C}, {C, F}, {F, E}, and {E, B} are
edges, and this path begins and ends at B.
2.1.16 Simple Path: A path is simple if it does not contain the same edge more than once.
In the figure 2.9 the path A, B, E, D, A, B which is of length 5, is not simple because it contains
the edge {A, B} twice. A, E, F is a simple path in figure 2.9
2.1.17 Connected Graph or Connectivity of a Graph: If there is a path between every pair
of distinct vertices of an undirected graph then it is called the connectivity of that graph or more
simply we say the graph is connected.
Figure: 2.10(a) Figure: 2.10(b)
In the figure 2.10(a) the graph G1 is connected because every pair of its vertices are connected.
But the figure 2.10(b) of G2 is not, because there is no path between A and D, B and D, C and
D, A and E etc.
2.1.18 Planar Graph: A graph G= (V, E) drawn in a plane with no crossing among the edges
is said to be a planar graph.
Figure: 2.11
The graph in the figure 2.11 is a planar graph. [2]
B
A
B
C
D E
A B
C
D
E
G1 G2
A
D
C
E
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2.1.19 Region: In a planar graph a connected component of the subset of the plane that is
disjoint from the graph is termed as a region.
At 2.11 {A, B, C, D, A} is a region.
2.1.20 Tree: A tree is a connected undirected graph with no simple circuits. [6]
2.2 Theorems and corollaries on Graph Theory
Theorem 2.1: Handshaking theorem: The sum of the degrees of the vertices of an undirected
graph is an even number equal to twice the number of edges. In symbols, if G (V, E) is an
undirected graph, then
∑ deg(𝑣) = 2𝐸
𝑣∈𝑉
Proof: Adding the degrees of all the vertices involves counting one for each edge incident with
each vertex. If the edge is not a loop, it is incident with two different vertices and so gets
counted twice exactly, once at each vertex. On the other hand, a loop at a vertex is also counted
twice, by convention, in the degree of that vertex. Hence the proof is completed.
Theorem 2.2: An undirected graph has an even number of vertices of odd degree.
Proof: Let V1 and V2 be the set of vertices of even degree and the set of vertices of odd degree,
respectively, in an undirected graph G = (V,E) with e edges. Then
2𝐸 = ∑ deg(𝑣) =
𝑣∈𝑉
∑ deg⁡( 𝑣1)
𝑣∈𝑉1
+ ∑ 𝑑𝑒𝑔⁡(𝑣2)
𝑣∈𝑉2
Because deg(v) is even for v ∈ V1, the first term in the right-hand side of the last equality is
even. Furthermore, the sum of the two terms on the right-hand side of the last equality is even,
because this sum is 2E. Hence, the second term in the sum is also even. Because all the terms
in this sum are odd, there must be an even number of such terms. Thus, there are an even
number of vertices of odd degree. [1]
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Theorem 2.3: A simple graph is bipartite if and only if it is possible to assign one of two
different colors to each vertex of the graph so that no two adjacent vertices are assigned the
same color.
Proof: First, we suppose that G = (V, E) is a bipartite simple graph. Then V = V1 ∪ V2, where
V1 and V2 are disjoint sets and every edge in E connects a vertex in V1 and a vertex in V2. If
we assign one color to each vertex in V1 and a second color to each vertex in V2, then no two
adjacent vertices are assigned the same color. Now suppose that it is possible to assign colors
to the vertices of the graph using just two colors so that no two adjacent vertices are assigned
the same color. Let V1 be the set of vertices assigned one color and V2 be the set of vertices
assigned the other color. Then, V1 and V2 are disjoint and V = V1 ∪ V2. Furthermore, every
edge connects a vertex in V1 and a vertex in V2 because no two adjacent vertices are either both
in V1 or both in V2. Consequently, G is bipartite.
Theorem 2.4: There is a simple path between every pair of distinct vertices of a connected
undirected graph.
Proof: Let u and v be two distinct vertices of the connected undirected graph G = (V, E).
Because G is connected, there is at least one path between u and v. Let x0, x1, ……, xn, where
x0 = u and xn = v, be the vertex sequence of a path of least length. This path of least length is
simple. To see this, suppose it is not simple. Then xi = xj for some i and j with 0 ≤ i < j. This
means that there is a path from u to v of shorter length with vertex sequence x0, x1,…., xi−1, xj
, . . . , xn obtained by deleting the edges corresponding to the vertex sequence xi, . . . , xj−1.
Theorem 2.5: (Euler’s formula)
Let G be a connected planar graph. Let v, e, and r denote the number of vertices, edges, and
regions. Then
v –e + r = 2
Proof: The proposition will be proved by induction on e. If e = 0, then G has one vertex, no
edges, and one region that is v = 1, e = 0, r = 1. So the theorem holds for the base case. Suppose
the theorem is true for any graph with at most e−1 edges. If G is a tree (a tree is a graph with
no cycles), then e = v−1, and r =1; so the theorem holds. If G is not a tree, let e' be an edge of
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some cycle of G. Then G with the edge e' removed is a connected plane graph with v vertices,
e−1 edges, and r−1 regions.
So, v−(e−1) + (r −1) = 2 (by the induction hypothesis).
Hence v – e + r = 2.
Corollary 2.1: Let G be a connected planar simple graph. Let v and e denote the number of
vertices and edges with v ≤ 3. Then
e≤ 3v−6
Proof: Let r be the number of regions. Every region is surrounded by at least 3 edges (otherwise
the result is clearly true). Every edge has one and only one region on either of its two sides. If
we draw a planar graph with r regions and count the number of edges on the boundary of each
region, sum of these numbers denoting by n will be n < 2e. Also, since each boundary contains
at least three edges, n > 3r. So have 3r ≤2e. Plugging the inequality into Theorem 2.5 (Euler’s
formula) we get,
3v-3e+3r=6 (multiplying v–e+r=2 by 3)
=>3v-3e-6=-3r≥-2e
=>3v-3e-6≥-2e
=>3v-6≥e
so, e ≤ 3v-6
Corollary 2.2.: If G is a connected planar simple graph then G has a vertex of degree not
exceeding 5.
Proof: Suppose deg(vi) > 6 for every vertex vi. Since Σ deg(vi) = 2e (By Handshaking Theorem
2.1), we would have 2e > 6v and hence e ≥ 3v > 3v -6. This contradicts e < 3v -6. Which is a
contradiction for all v (By the Theorem 2.5). Therefore, there must be at least a single vertex
with degree at most 5.
So, deg(vi) ≤ 5.
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Corollary 2.3: If G is a connected planar simple graph with v vertices and e edges, with v≥3
and no circuits of length 3 then, e ≤ 2v - 4.
Proof: For graph G with r regions, it follows from Theorem 2.1 (Handshaking theorem) for
planar graphs that 2e ≥ 4r (because the degree of each region of a simple graph without triangles
or circuits of length 3 is at least 4), so that r ≤ 1/2 e.
Combining this with Euler's formula we get,
v - e + r = 2
=> e -v + 2 = r
=> e - v + 2 ≤ 1/2 e
Hence e ≤ 2v - 4
2.3 Some problems on Graph Theory
We will solve a few problems using the above theorems and corollaries.
Problem---2.1: Show that K3,3 is non-planar.
Proof: Suppose that K3,3 is a planar graph. Since K3,3 has 6 vertices and 9 edges and no triangles
that is no circuits of length 3 (this is easy to see because it is bipartite), it follows from Corollary
2.3 that
e≤ 2v-4
i.e. 9≤ (2×6) – 4
=> 9 ≤ 8,
which is a contradiction. Hence K3,3 is non-planar.
Figure: 2.12
The above figure shows non-planar graph K3,3
4 2 6
1 3
5
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Problem---2.2: Show that K5 is non-planar.
Proof: Suppose that K5 is a planar graph. Since K5 has 5 vertices and 10 edges it follows from
Corollary---2.1 that
e≤ 3v-6
i.e. 10≤ (3 × 5) - 6
=> 10 ≤ 9,
which is a contradiction and this shows that K5 is non-planar.
Figure: 2.13
Figure 2.13 shows that K5 is non-planar.
Problem---2.3: Suppose that a connected planar graph has 30 edges. If a planar representation
of this graph divides the plane into 20 regions, how many vertices does this graph have?
Solution: Given that,
Number of edges, e=30
Number of regions, r=20
Number of vertices, v=?
From the Theorem---2.1 (Euler’s formula) we know that,
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v –e + r =2
=> v-30 +20 =2
So, v=12.
Hence the given connected planar graph has 12 vertices with 30 edges in 20 regions.
Figure: 2.14
Figure 2.14 represents the graph with 12 vertices A, B, C, D, E, F, G, H, I, J, K, L respectively,
30 edges and 20 regions.
A B C D E
L
K
J I
F
G
H
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Chapter 3
Graph Coloring
3.1 Graph Coloring: The task to color the vertices (V) of a graph G= (V, E) such that no two
adjacent vertices assigned the same color is known as graph coloring. This is also known as
map coloring, because maps are colored using the graph coloring rules.
3.2 Importance of Graph Coloring: There are many applications of graph coloring in
Mathematics, Computer science, Information technology and Communication.
1. To finalize the schedule of exam at a university or at any educational institution.
2. Frequency assignments of television channels to operate.
3. To create a computer network with multiple links between the data centers.
4. To visualize the ecosystem of wildlife. [1]
5. To show the relationship between the people. [2]
6. It can be used to model the roadmaps where the vertices represent the intersections and
the edges represent roads.
3.3 Chromatic Number of a Graph: The least number of colors needed to color a graph G is
termed as chromatic number of that graph so that no two adjacent vertices have the same color.
It is represented by (G).
Figure: 3.1
Let us color the above graph. Let A be colored blue while B, C and D can’t be blue. So, if B is
colored green, C is red, then we can color D as green, because D and B are not connected. At
last E can be colored blue. So the chromatic number, (G) =3.
Blue
A
Green
B
Red
C
Green
D
Blue
E
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3.4 Problems on Chromatic Number
Problem 3.1: Find the chromatic number of the given graph.
Figure: 3.2
Solution: First we color A with red. Since A is directly connected with B, C, E, D so we can
not color these vertices with red. Suppose B assigns green. Since B is not directly connected
with C so C is also assigned green. E and D can be blue while F can be red. Because F is not
adjacent to A. Hence the chromatic number of the given graph is 3. (G)= 3.
Problem 3.2: Find the chromatic number of the given graph.
Figure: 3.3
Solution: First we color A with red. Then B and E should be different say green. Since C is
not connected to A so it can be red and D has no connection with B and C thus D can assign
green. So the given graph has chromatic number, (G) = 2.
Red
A
Green
B
Green
C
Blue
D
Blue
E
Red
F
Red
A
Green
B
Green
E
Red
C
Green
D
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3.5 Subgraph: If G and H are two graphs and each vertex and edge of H are contained in G
then H is called a subgraph of G. If H ≠ G then H is a proper subgraph of G.
Figure: 3.4
Figure 3.4 represents a graph and its subgraph. [6]
3.6 Theorems and Corollaries on Graph Coloring
Theorem 3.1: Let ∆ (G) be the maximum of the degrees of the vertices of a graph 5. Then
(G)≤ 1 + ∆ (G).
Proof: We will prove this theorem by the method of induction. Let V be the vertices of a graph
G and if V=1 then ∆ (G) =0 and (G) = 1. So the results clearly holds. Let K be an integer and
K≥1 and assume that the result holds for all graphs for V=K. Now let G be a graph with K+1.
Let G0=G-{V} be a subgraph. So ∆ (G0) ≤ ∆ (G). Now GO can be colored with (G0). Since
G0 has K vertices so by the induction method we have (G0) ≤ 1 + ∆ (G0). So G0 can be colored
with at most 1+∆ (G0) colors. Since there are at most ∆ (G) vertices adjacent to V, one of the
available 1 + ∆ (G) colors remains for V. Thus, G can be colored with at most 1 + ∆ (G) colors.
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Theorem 3.2 (Five Color Theorem): Every planar graph can be colored with 5 colors. That
is
(G)≤ 5
Proof: The proof is by induction on the number of vertices n; when n≤5 this is trivial. Now
suppose G is planar on more than 5 vertices; by Corollary 2.2 some vertex v has degree at most
5.
By the induction hypothesis, G − v can be colored with 5 colors. Color the vertices of G, other
than v, as they are colored in a 5-coloring of G − v. If d(v)≤4, then v can be colored with one
of the 5 colors to give a proper coloring of G with 5 colors. So we now suppose d(v)=5. If the
five neighbors of v are colored with four or fewer of the colors, then again v can be colored to
give a proper coloring of G with 5 colors.
Now we suppose that all five neighbors of v have a different color, as indicated in following
figure
Figure: 3.5
Figure 3.5 shows that five neighbors of v colored with 5 colors: v1 is red, v2 is purple, v3 is
green, v4 is blue, v5 is orange.
Suppose that in G there is a path from v1 to v3, and that the vertices along this path are
alternately colored red and green; call such a path a red-green alternating path. Then together
with v, this path makes a cycle with v2 on the inside and v4 on the outside, or vice versa. This
means there cannot be a purple-blue alternating path from v2 to v4. Supposing that v2 is inside
the cycle, we change the colors of all vertices inside the cycle colored purple to blue, and all
V1
red
V2
purple
V3
green
V5
orange
V4
blue
blue vertices are recolored purple. This is still a proper coloring of all vertices of G except v,
and now no neighbor of v is purple, so by coloring v purple we obtain a proper coloring of G.
If there is no red-green alternating path from v1 to v3, then we recolor vertices as follows:
Change the color of v1 to green. Change all green neighbors of v1 to red. Continue to change
the colors of vertices from red to green or green to red until there are no conflicts, that is, until
a new proper coloring is obtained. Because there is no red-green alternating path from v1 to v3,
the color of v3 will not change. Now no neighbor of v is colored red, so by coloring v red we
obtain a proper coloring of G.
Corollary 3.1: Show that (Cn) = 2 if n is even positive integer with n ≥ 4.
Proof: We want to show that two colors are needed to color Cn when n is even. To construct
such a coloring, we pick a vertex and color it red. Proceed around the graph in a clockwise
direction (using a planar representation of the graph) coloring the second vertex blue, the third
vertex red, and so on. The nth vertex can be colored blue, because the two vertices adjacent to
it, namely the (n − 1)st
and the first vertices, are both colored red.
For example if we take n = 4 then the graph C4 can be colored with 2 colors red and blue and
if n = 6 then the graph C6 can also be colored with same way.
Hence (Cn) = 2 if n is even positive integer with n ≥ 4.
Figure: 3.6(a) Figure: 3.6(b)
Figure 3.6(a) and 3.6(b) shows C4 and C6
A
Red
B
Blue
D
Blue
C
Red
A
Red
B
Blue
C
Red
E
Red
D
Blue
F
Blue
P a g e | 20
Corollary 3.2: Show that (Cn) = 3 if n is odd positive integer with n ≥ 3.
Proof: When n is odd and n > 1, the chromatic number of Cn is 3. To see this, we pick an initial
vertex. To use only two colors, it is necessary to alternate colors as the graph is traversed in a
clockwise direction. The nth
vertex reached is adjacent to two vertices of different colors,
namely, the first and (n − 1)st
. For example let n = 5 and consider C5. Let us pick a vertex and
color it red. Proceeding clockwise, it is necessary to assign a second color, say blue, to the next
vertex reached. Continuing in the clockwise direction, the third vertex can be colored red, and
the fourth vertex can be colored blue. The fifth vertex cannot be colored either red or blue,
because it is adjacent to the fourth vertex and the first vertex. Consequently, a third color is
required for this vertex. Thus, the chromatic number of C5 is 3. Hence, a third color must be
used and we can say that (Cn) = 3 if n is odd positive integer with n ≥ 3.
Figure: 3.7
Figure 3.7: C5 needs 3 colors.
A
Red
B
Blue
C
Red
D
Blue
E
Yellow
P a g e | 21
3.7 Four Color Theorem
There is an important theorem regarding map coloring which is known as Four Color theorem.
Statement of Four Color Theorem: Four colors are sufficient to color a planar map (with five
regions).
3.8 History of Four Color Theorem: The Four Color Theorem has fascinated people for
almost a century and a half. The Four Color Conjecture first seems to have been made
by Francis Guthrie. He was a student at University College London where he studied under De
Morgan. After graduating from London he studied law but by this time his brother Frederick
Guthrie had become a student of De Morgan. Francis Guthrie showed his brother some results
he had been trying to prove about the coloring of maps and asked Frederick to ask De
Morgan about them.
De Morgan was unable to give an answer but, on 23 October 1852, the same day he was asked
the question, he wrote to Hamilton in Dublin. De Morgan wrote:-
A student of mine asked me today to give him a reason for a fact which I did not know was a
fact - and do not yet. He says that if a figure be anyhow divided and the compartments
differently colored so that figures with any portion of common boundary line are differently
colored - four colors may be wanted, but not more - the following is the case in which four
colors are wanted. Query cannot a necessity for five or more be invented. ...... If you retort with
some very simple case which makes me out a stupid animal, I think I must do as the Sphinx
did....
Hamilton replied on 26 October 1852 (showing the efficiency of both himself and the postal
service):-
I am not likely to attempt your quaternion of color very soon.
However De Morgan kept asking if anyone could find a solution to Guthrie's problem and
several mathematicians worked on it. Charles Peirce in the USA attempted to prove the
Conjecture in the 1860's and he was to retain a lifelong interest in the problem. Cayley also
learnt of the problem from De Morgan and on 13 June 1878 he posed a question to the London
mathematical society asking if the Four Color Conjecture had been solved. Shortly
afterwards Cayley sent a paper on the coloring of maps to the Royal Geographical Society and
P a g e | 22
in was published in 1879. The paper explains where the difficulties lie in attempting to prove
the Conjecture.
On 17 July 1879 Alfred Brey Kempe announced in Nature that he had a proof of the Four Color
Conjecture. Kempe was a London barrister who had studied mathematics under Cayley at
Cambridge and devoted some of his time to mathematics throughout his life. At Cayley's
suggestion Kempe submitted the Theorem to the American Journal of Mathematics where it
was published in 1879. Story read the paper before publication and made some simplifications.
Story reported the proof to the Scientific Association of Johns Hopkins University in November
1879 and Charles Peirce, who was at the November meeting, spoke at the December
meeting of the Association of his own work on the Four Color Conjecture.
Kempe used an argument known as the method of Kempe chains. If we have a map in which
every region is colored red, green, blue or yellow except one, say X. If this final region X is not
surrounded by regions of all four colors there is a color left for X. Hence suppose that regions
of all four colors surround X. If X is surrounded by regions A, B,C, D in order, colored red,
yellow, green and blue then there are two cases to consider.
(i) There is no chain of adjacent regions from A to C alternately colored red and green.
(ii) There is a chain of adjacent regions from A to C alternately colored red and green.
If (i) holds there is no problem. Change A to green, and then interchange the color of the
red/green regions in the chain joining A. Since C is not in the chain it remains green and there
is now no red region adjacent to X. Color X red.
If (ii) holds then there can be no chain of yellow/blue adjacent regions from B to D. [It could
not cross the chain of red/green regions.] Hence property (i) holds for B and D and we change
colors as above. [6]
Kempe received great acclaim for his proof. He was elected a Fellow of the Royal Society and
served as its treasurer for many years. He was knighted in 1912. He published two improved
versions of his proof, the second in 1880 aroused the interest of P.G. Tait, the Professor of
Natural Philosophy at Edinburgh. Tait addressed the Royal Society of Edinburgh on
the subject and published two papers on the (what we should now call) Four Color Theorem.
They contain some clever ideas and a number of basic errors.
P a g e | 23
The Four Color Theorem returned to being the Four Color Conjecture in 1890. Percy John
Heawood, a lecturer at Durham England, published a paper called Map coloring theorem. In it
he states that his aim is
Rather destructive than constructive, for it will be shown that there is a defect in the now
apparently recognized proof.
Although Heawood showed that Kempe's proof was wrong he did prove that every map can be
5-coloured in this paper. Kempe reported the error to the London Mathematical
Society himself and said he could not correct the mistake in his proof. In 1896 De La Vallee
Pousin also pointed out the error in Kempe’s paper, apparently unaware of Heawood’s work.
Heawood was to work throughout his life on map coloring, work which spanned nearly 60
years. He successfully investigated the number of colors needed for maps on other surfaces and
gave what is known as the Heawood estimate for the necessary number in terms of the Euler
characteristic of the surface.
Heawood was to make further contributions to the Four Color Conjecture. In 1898 he proved
that if the number of edges around each region is divisible by 3 then the regions are 4-
colourable. He then wrote many papers generalizing this result.
To understand the later work we need to define some concepts.
Clearly a graph can be constructed from any map the regions being represented by the vertices
and two vertices being joined by an edge if the regions corresponding to the vertices are
adjacent. The resulting graph is planar, that is can be drawn in the plane without any edges
crossing. The Four Color Conjecture now asks if the vertices of the graph can be colored with
4 colors so that no two adjacent vertices are the same color.
From the graph a triangulation can be obtained by adding edges to divide any non-triangular
face into triangles. A configuration is part of a triangulation contained within a circuit. An
unavoidable set is a set of configurations with the property that any triangulation must contain
one of the configurations in the set. A configuration is reducible if it cannot be contained in a
triangulation of the smallest graph which cannot be 4-coloured. The search for avoidable sets
began in 1904 with work of Weinicke. Renewed interest in the USA was due to Veblen who
published a paper in 1912 on the Four Color Conjecture generalizing Heawood's work.
P a g e | 24
Franklin in 1922 published further examples of unavoidable sets and used Birkhoff's idea of
reducibility to prove, among other results, that any map with ≤ 25 regions can be 4-coloured.
The number of regions which resulted in a 4-colourable map was slowly increased. Reynolds
increased it to 27 in 1926, Winn to 35 in 1940, Ore and Stemple to 39 in 1970 and Mayer to 95
in 1976.
However the final ideas necessary for the solution of the Four Color Conjecture had been
introduced before these last two results. Heesch in 1969 introduced the method of discharging.
This consists of assigning to a vertex of degree i the charge 6 - i. Now from Euler's formula
we can deduce that the sum of the charges over all the vertices must be 12. A given set S of
configurations can be proved unavoidable if for a triangulation T which does not contain a
configuration in S we can redistribute the charges (without changing the total charge) so that
no vertex ends up with a positive charge.
Heesch thought that the Four Color Conjecture could be solved by considering a set of around
8900 configurations. There were difficulties with his approach since some of his configurations
had a boundary of up to 18 edges and could not be tested for reducibility. The tests for
reducibility used Kempe chain arguments but some configurations had obstacles to prevent
reduction.
The year 1976 saw a complete solution to the Four Color Conjecture when it was to become
the Four Color Theorem for the second, and last, time. The proof was achieved by Appel and
Haken, their methods based on reducibility using Kempe chains. They carried through the ideas
of Heesch and eventually they constructed an unavoidable set with around 1500 configurations.
They managed to keep the boundary ring size down to ≤ 14 making computations easier that
for the Heesch case. There was a long period where they essentially used trial and error together
with unbelievable intuition to modify their unavoidable set and their discharging procedure.
Appel and Haken used 1200 hours of computer time to work through the details of the final
proof. Koch assisted Appel and Haken with the computer calculations.
The Four Color Theorem was the first major theorem to be proved using a computer, having a
proof that could not be verified directly by other mathematicians. Despite some worries about
this initially, independent verification soon convinced everyone that the Four Color Theorem
had finally been proved. Details of the proof appeared in two articles in 1977. Recent work has
led to improvements in the algorithm. [1, 6, 8]
P a g e | 25
3.9 An Example of Four Color Theorem
First let us take a planar graph of 30 regions and try to color it with the help of Four Color
Theorem. Our first graph in figure 3.8 will be of no color which has been given below---
Figure: 3.8
Now we want to color the graph with the first color red in figure 3.9.
Figure: 3.9
P a g e | 26
In figure 3.10 we will use the green color with the red which was previously used in figure 3.9.
Figure: 3.10
Then the figure 3.11 can be formed by using three colors red, green and blue.
Figure: 3.11
P a g e | 27
At the last stage in figure 3.12 we will use the color yellow and get our desired graph which
assigns four color and thus the four color theorem is established by this particular example.
Figure: 3.12
3.10 Applications of Graph Coloring
3.10.1 Scheduling problem for final examination using four color theorem: The scheduling
problem can be solved using a graph model, where the vertices represent courses and an edge
between two vertices if there is a common student in the courses they represent. Each time slot
for a final exam is represented by a different color. A scheduling of the exams corresponds to
a coloring of the associated graph.
For instance, let us consider that there are 10 finals to be scheduled. Suppose the courses are
labeled as 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 .Suppose that the following pairs of courses have common
students: 1 and 2, 1 and 3, 1 and 4, 1 and 7, 1 and 10, 2 and 3, 2 and 4, 2 and 5, 2 and 7, 3 and
4, 3 and 6, 3 and 7, 4 and 5, 4 and 6, 5 and 6, 5 and 7, 6 and 7, 6 and 9, 7 and 8, 7 and 10, 8 and
P a g e | 28
10, 8 and 5, 8 and 9, 9 and 10 and 10 and 6. The graph associated with this set of classes is
shown in the table given below. A scheduling consists of a coloring of this graph. Because the
chromatic number of this graph is 4 (By the assumption of the Four Color Theorem), four time
slots are needed. A coloring of the graph using four colors and the associated schedule are
shown in Figure 3.13.
Figure: 3.13
Time Period Courses
I 1, 6, 8
II 2, 9, 10
III 3, 5
IV 4, 7
Now we want to solve this problem using computer programming by the help of Mathematica.
Our goal is to color the above graph using four color theorem that is the chromatic number
should not exceed the number 4. The required inputs and the corresponding outputs are given
below.
1
Red
10
Blue
2
Blue
3
Green
4
Yellow
5
Green6
Red
7
Yellow
8
Red
9
Green
P a g e | 29
Input 1: Needs["Combinatorica`"]
Input 2: s1 = {{1, 2}, {1, 3}, {1, 4}, {1, 7}, {2, 3}, {2, 4}, {2, 5}, {2,
7}, {3, 4}, {3, 6}, {3, 7}, {4, 5}, {4, 6}, {5, 6}, {5, 7}, {6,
7}, {6, 9}, {7, 10}, {8, 10}, {8, 5}, {10, 6}};
Input 3: ShowLabeledGraph[FromUnorderedPairs[s1]]
Output 1:
Input 4:
ColorableQ[G_, k_]:
= Module[{𝑐, 𝑑, 𝑖}, (Expand[Apply[Times, Apply[Subtract, Map[𝑐, 𝐺, {2}],1]]]/. 𝑐[i_]d_.
⧴ 𝑐[𝑖]𝑑mod𝑘) = ! = 0]
Input 5: ColorableQ[s1,4]
Output 2: True
Input 6: ColorableQ[s1,3]
Output 3: False
Since the output is true that is the given graph is colorable with at most four color so the four
color theorem holds and the chromatic number is 4. [12]
1
23
4
5
6
7 8
9
10
P a g e | 30
3.10.2 Frequency assignment of television channel using four color theorem: Suppose
there are 12 television channels to assign the frequency in such a way that there will be no two
channels have same frequency which are within 500 kilometers. [1]
This problem can be solved easily by map coloring or graph coloring. Let us construct a graph
by assigning a vertex to each station. Two vertices are connected by an edge if they are located
within 500 kilometers of each other.
We label the channels as 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 where the adjacent vertices within
500 kilometers are given as follows---
Channel name adjacent channels
1 9, 12
2 3, 8, 9
3 2, 8, 10
4 5, 6, 11
5 4, 6, 7, 8
6 4, 5, 11, 12
7 5, 8, 10
8 2, 3, 5, 7, 9, 10
9 1, 2, 5, 6, 8
10 3, 7, 8
11 6, 4, 12
12 1, 6, 11
Now if we color the station 1 with red then 9 and 12 must be different say green and blue
respectively.
2 can be red since it is 530 kilometers away from 1. Since it is connected with 9 which is green
so 3 may be blue while 8 can be yellow.
For channel 3, 2 and 8 have color red and yellow respectively so 10 will be red.
Channel 4 is not associated with anyone of the previous so it can be red and 6 can be green
while 5 is blue.
Channel 5, we color only 7 with green.
P a g e | 31
For Channel 6, we color only 11 with yellow. The distance between 6 and 9 is 505 kilometers.
Similarly if we color the rest of the channel stations we shall get the figure 3.14 which
represents the frequency allotment of the channels where R, G, B, Y stand for red, green, blue
and yellow respectively.
Figure: 3.14
This problem can also be checked whether it follows the four color theorem or not by the
Mathematica codes which are given below---
Input 1: Needs["Combinatorica`"]
Input 2: s2 = {{1, 9}, {1, 12}, {2, 3}, {2, 8}, {2, 9}, {3, 8}, {3, 10}, {4,
5}, {4, 6}, {4, 11}, {5, 6}, {5, 7}, {5, 8}, {6, 11}, {6, 12}, {7,
8}, {7, 10}, {8, 9}, {8, 10}, {5, 9}, {6, 9}, {11, 12}};
Input 3: ShowLabeledGraph[FromUnorderedPairs[s2]]
1-R 2-R530 370
3-B
4-R
5-B
9-G
12-B
10-R
7-G11-Y
360
180
360
180 350
P a g e | 32
Output 1:
Input 4:
ColorableQ[G_, k_]:
= Module[{𝑐, 𝑑, 𝑖}, (Expand[Apply[Times, Apply[Subtract, Map[𝑐, 𝐺, {2}],1]]]/. 𝑐[i_]d_.
⧴ 𝑐[𝑖] 𝑑mod𝑘
) = ! = 0]
Input 5: ColorableQ[s2,4]
Output 2: True
Input 6: ColorableQ[s2,3]
Output 3: False
Since the output is true that is the given graph is colorable with at most four color so the four
color theorem holds and the chromatic number is 4. [12]
1
2
3
4
5
6
7
8
9
10
11
12
P a g e | 33
Conclusion
In this project paper, at first the idea of graph, graph theory and its importance have been given.
Then in the next chapter there are basic terminologies of graph theory, useful theorems and
corollaries which are prerequisite for the graph theory. There are Euler’s theorem, some
problems solved by using the theorem and corollaries associated by the theorem with some
definitions. At the last chapter there are the history, statement, some applications of famous
Four Color Theorem. This paper will help us to understand how a critical problem can be solved
using map or graph coloring.
There are many applications of graph theory and map or graph coloring in the real world and
our attempt was to show some of them.
Due to limitation of time and volume of the project, this report could not cover extensive
applications of map coloring. But it is hoped that Graph Theory and its applications would be
studied further whenever the opportunity comes.
P a g e | 34
References
1. Discrete Mathematics and Applications with Combinatorics and Graph Theory,
Kenneth H. Rossen (Seventh Edition)
2. Discrete Mathematics with Graph Theory, Edgar G. Goodaire and Michael M.
Parmenter (Second Edition)
3. Graph Theory, V.K. Balakrishnan (Schaum’s outline series)
4. Discrete Mathematics, Seymour Lipschutz and Marc Lipson (Schaum’s outline series)
5. Graph Theory, Tero Harju, Department of Mathematics University of Turku FIN-20014
Turku, Finland 1994 – 2012
6. http://en.Wikipedia.org/wiki/Map-coloring
7. Graph Theory, R. Diestel, Springer–Verlag (2005)
8. Graph Theory and Its Applications, J. Gross & J. Yellen, CRC Press (2006)
9. Introduction to Graph Theory, D. B. West, Prentice–Hall (1996)
10. http://Mathworld.wolfram.com
11. Mathematica 9.0, Combinatorica
12. How to color a graph with computer algebra---Yuri Matiyasevich Steklov Institute of
Mathematics at Saint-Petersburg 27 Fontanka, Saint-Petersburg, 191011, Russia E-
mail: yumat@pdmi.ras.ru URL: http://logic.pdmi.ras.ru/~yumat

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Map Coloring and Some of Its Applications

  • 1. P a g e | 0 Map Coloring and Some of Its Applications MD. SHAH ALAM DEPARTMENT OF MATHEMATICS UNIVERSITY OF DHAKA
  • 2. ABSTRACT Map coloring more precisely graph coloring is an important topic of graph theory. In this project we have studied the basics of graph theory and some of its applications in map coloring. First chapter will introduce the concept of graphs and graph theory. In chapter two, we describe some basic terminologies of graphs with important theorems and corollaries related to this topic. Finally, the chapter three introduces the idea of map coloring where the famous Four Color Theorem is discussed and some real life problems have been solved by using the very important concept.
  • 3. Contents Chapter One Introduction Page Basic 1 1.1 Graph 2 1.2 Graph Theory 2 1.3 Importance of Graph Theory 2-3 Chapter Two Basic Concept of Graph Theory 2.1 Basic terminology of Graphs 4-9 2.2 Theorems and Corollaries on Graph Theory 9-12 2.3 Some Problems on Graph Theory 12-14 Chapter Three Graph Coloring 3.1 Graph Coloring 15 3.2 Importance Graph Coloring 15 3.3 Chromatic Number of a Graph 15 3.4 Problems on Chromatic Number 16 3.5 Subgraph 17 3.6 Theorems and Corollaries on Graph Coloring 17-20 3.7 Four Color Theorem 21 3.8 History of Four Color Theorem 21-24 3.9 An Example of Four Color Theorem 25-27 3.10 Applications on Graph Coloring 27-32 Conclusion 33 Reference 34
  • 4. P a g e | 1 Chapter 1 Introduction Basic Graphs are discrete structures consisting of vertices and edges that connect these vertices. There are different kinds of graphs, depending on whether edges have directions or not, whether multiple edges can connect the same pair of vertices, and whether loops are allowed or not. Problems in almost every conceivable discipline can be solved using graph models. Many structures involving real world situation can be conveniently present on paper by means of a diagram consists of a set of points together with lines joining some or all pairs of these points. Graphs are used to represent the competition of different species in an ecological niche, to represent who influences whom in an organization, to represent the outcomes of round-robin tournaments. Graphs can be used to model acquaintanceships between people, collaboration between researchers, telephone calls between telephone numbers, and links between websites. They can be used to model roadmaps and the assignment of jobs to employees of an organization. Using graph models, we can determine whether it is possible to walk down all the streets in a city without going down a street twice, and we can find the number of colors needed to color the regions of a map. Graphs help to determine whether a circuit can be implemented on a planar circuit board. We can distinguish between two chemical compounds with the same molecular formula but different structures using graphs. Without the basic idea of graph theory it will be an unsuccessful attempt to understand map coloring. By learning the various properties of graph theory we can know that map coloring is nothing but a very useful application of it. In map coloring problem we mainly deal with the undirected graph. We can determine whether two computers are connected by a communications link using graph models of computer networks. Graphs with weights assigned to their edges are also used to solve problems such as finding the shortest path between two cities in a transportation network. We can also use graphs to schedule exams and assign channels to television stations.
  • 5. P a g e | 2 1.1 Graph: The set or collection of finite number of vertices (V) and edges (E) is called a graph and is denoted by G= (V, E). More specifically in graph theory, a graph is a representation of a set of objects where some pairs of objects are connected by links. Typically, a graph is depicted in diagrammatic form as a set of dots for the vertices, joined by lines or curves for the edges. 1.2 Graph Theory: Graph theory is the study of graphs, which are mathematical structures used to model pairwise relations between objects. A graph in this context is made up of vertices, nodes, or points which are connected by edges, arcs, or lines. Graph theory concerns the relationship among lines and points. Graph theory is a growing area in mathematical research, and has a large specialized vocabulary. 1.3 Importance of Graph Theory: Graph theory is very important in many branches of science such as mathematics, physics, computer science, chemistry, life science etc. 1. Graphs are used in the field of chemistry to model chemical compounds. In computational biochemistry some sequences of cell samples have to be excluded to resolve the conflicts between two sequences. This is modeled in the form of graph where the vertices represent the sequences in the sample. An edge will be drawn between two vertices if and only if there is a conflict between the corresponding sequences. The aim is to remove possible vertices, (sequences) to eliminate all conflicts. [2] 2. The major role of graph theory in computer applications is the development of graph algorithms. Numerous algorithms are used to solve problems that are modeled in the form of graphs. These algorithms are used to solve the theoretical concepts which in turn used to solve the corresponding computer science application problems. 3. Graph theory is a very natural and powerful tool in combinatorial operations research. A network called transport network where a graph is used to model the transportation of commodity from one place to another. The objective is to maximize the flow or minimize the
  • 6. P a g e | 3 cost within the prescribed flow. The graph theoretic approach is found to be more efficient for these types of problems though they have more constraints. 4. Graphs are extensively used to model social structures based on different kinds of relationships between people or groups of people. These social structures, and the graphs that represent them, are known as social networks. In these graph models, individuals or organizations are represented by vertices; relationships between individuals or organizations are represented by edges. The study of social networks is an extremely active multidisciplinary area, and many different types of relationships between people have been studied using them. [1] 5. The World Wide Web can be modeled as a directed graph where each Web page is represented by a vertex and where an edge starts at the Web page a and ends at the Web page b if there is a link on a pointing to b. Because new Web pages are created and others removed somewhere on the Web almost every second, the Web graph changes on an almost continual basis. [3] 6. Graphs can be used to model telephone calls made in a network, such as a long distance telephone network. In particular, a directed multigraph can be used to model calls where each telephone number is represented by a vertex and each telephone call is represented by a directed edge. The edge representing a call starts at the telephone number from which the call was made and ends at the telephone number to which the call was made. We need directed edges because the direction in which the call is made matters. We need multiple directed edges because we want to represent each call made from a particular telephone number to a second number. [1]
  • 7. P a g e | 4 Chapter 2 Basic Concepts of Graph theory 2.1 Basic terminology of Graphs We will introduce some basic terms of graph and graph theory. 2.1.1 Vertex: The point or node in a graph G is known as vertex. 2.1.2 Edge: The connection or line in a graph G is called edge, i.e. the line joining two vertices is an edge. Figure: 2.1 In the figure 2.1 A, B, C are three vertices and E1, E2, E3 are edges. 2.1.3 Adjacent vertices: In an undirected graph G if two vertices u and v are end points of an edge then they are called adjacent or neighbors. 2.1.4 Incident edge: If an edge or line associated with the vertices u and v of a graph then the edge is called incident with the vertices u and v. It is also said to connect u and v. Figure: 2.2 In the figure 2.2 A and B, A and C, C and D, D and B are adjacent vertices and E1, E2, E3, E4 are incident with the vertices A and B, A and C, C and D, D and B respectively. A B E1 C A B E2E3 D E1 E2 E3 E4
  • 8. P a g e | 5 2.1.5 Directed Graph: A directed graph (or digraph) is a graph, or set of vertices connected by edges, where the edges have a direction associated with them. So, if a graph is directed it has distinct direction along all of its vertices. Figure: 2.3 Figure 2.3 is a directed graph. 2.1.6 Undirected Graph: If there is no direction among the edges and the vertices of the graph G= (V, E), then G is called an undirected graph. Figure: 2.4 In 2.4, the graph is an undirected graph, because they have no direction through the edges E1, E2, E3, E4 2.1.7 Degree of a vertex: In an undirected graph G the degree of a vertex is defined as the number of edges connected or incident to that vertex. It is denoted by deg() or simply d. Figure: 2.5 In figure 2.5 deg(A)= 2, deg(B)=3, deg(C)= 3, deg(D)=1, deg(E)= 2, deg(F)=3 and deg(H)=0. [1] A B E1 A B E1 DCB C E2E3 D E2 E3 E4 A F E H ’ ’ ’ ’
  • 9. P a g e | 6 2.1.8 Isolated vertex: A vertex of degree zero is known as isolated. H is isolated vertex in the figure 2.5 2.1.9 Pendant: A vertex is pendant if and only if it has degree one. In figure 2.5, D is a pendant. 2.1.10 Simple Graph: If there is no direction among the edges of the graph G= (V, E) then G is called a simple graph. It is also known as undirected graph. 2.1.11 Complete Graph: If a graph G contains exactly one edge between each pair of distinct vertices then it is called a complete graph and is denoted by Kn, where n=1, 2, 3, 4,…. For n=1, 2, 3, 4 the graphs are given below- Figure: 2.6 2.1.12 Bipartite Graph: A simple graph G is called bipartite if its vertex set V can be partitioned into two disjoint sets V1 and V2 such that every edge in the graph connects a vertex in V1 and a vertex in V2 (so that no edge in G connects either two vertices in V1 or two vertices in V2). When this condition holds, we call the pair (V1, V2) a bipartition of the vertex set V of G. Figure: 2.7 Figure 2.7 shows a bipartite graph where the vertices are partitioned into two sets say V1= {a, b, c} and V2= {m, n, o, p} [6] a b c m on p K1 K2 K3 K4
  • 10. P a g e | 7 2.1.13 Isomorphism of Graphs: The simple graphs G1 = (V1, E1) and G2 = (V2, E2) are isomorphic if there exists a one to-one and onto function f from V1 to V2 with the property that a and b are adjacent in G1 if and only if f (a) and f (b) are adjacent in G2, for all a and b in V1. Such a function f is called an isomorphism. In other words, when two simple graphs are isomorphic, there is a one-to-one correspondence between vertices of the two graphs that preserves the adjacency relationship. Figure: 2.8 Figure 2.8 shows that the graphs G1 = (V, E) and G2 = (W, F) are isomorphic. Because the function f with f (u1) = v1, f (u2) = v4, f (u3) = v3, and f (u4) = v2 is a one to-one correspondence between V and W. Adjacent vertices in G1 are u1 and u2, u1 and u3, u2 and u4, and u3 and u4, and each of the pairs f (u1) = v1 and f (u2) = v4, f (u1) = v1 and f (u3) = v3, f (u2) = v4 and f (u4) = v2, and f (u3) = v3 and f (u4) = v2 consists of two adjacent vertices in G2. [1] 2.1.14 Path: A path is a sequence of edges that begins at a vertex of a graph and travels from vertex to vertex along edges of the graph. Formally, Let n be a nonnegative integer and G be an undirected graph. A path of length n from u to v in G is a sequence of n edges e1, e2,…, en of G for which there exists a sequence x0 = u, x1, . . . , xn−1, xn = v of vertices such that ei has, for i= 1, . . . , n, the endpoints xi−1 and xi . When the graph is simple, we denote this path by its vertex sequence x0, x1,…… ,xn Figure: 2.9 In the simple graph shown in Figure 2.9 A, D, C, F, E, is a simple path of length 4, because {A, D}, {D, C}, {C, F }, and {F, E} are all edges. However, D, E, C, A is not a path, because {E, C} is not an edge. u1 u2 u3 u4 v1 v2 v3 v4 G1 G2 A B C D E F
  • 11. P a g e | 8 2.1.15 Circuit: A path is a circuit if it begins and ends at the same vertex, that is, if u = v, and has length greater than zero. The path or circuit is said to pass through the vertices x1, x2, …., xn−1 or traverse the edges e1, e2, . . . , en. In 2.9 B, C, F, E, B is a circuit of length 4 because {B, C}, {C, F}, {F, E}, and {E, B} are edges, and this path begins and ends at B. 2.1.16 Simple Path: A path is simple if it does not contain the same edge more than once. In the figure 2.9 the path A, B, E, D, A, B which is of length 5, is not simple because it contains the edge {A, B} twice. A, E, F is a simple path in figure 2.9 2.1.17 Connected Graph or Connectivity of a Graph: If there is a path between every pair of distinct vertices of an undirected graph then it is called the connectivity of that graph or more simply we say the graph is connected. Figure: 2.10(a) Figure: 2.10(b) In the figure 2.10(a) the graph G1 is connected because every pair of its vertices are connected. But the figure 2.10(b) of G2 is not, because there is no path between A and D, B and D, C and D, A and E etc. 2.1.18 Planar Graph: A graph G= (V, E) drawn in a plane with no crossing among the edges is said to be a planar graph. Figure: 2.11 The graph in the figure 2.11 is a planar graph. [2] B A B C D E A B C D E G1 G2 A D C E
  • 12. P a g e | 9 2.1.19 Region: In a planar graph a connected component of the subset of the plane that is disjoint from the graph is termed as a region. At 2.11 {A, B, C, D, A} is a region. 2.1.20 Tree: A tree is a connected undirected graph with no simple circuits. [6] 2.2 Theorems and corollaries on Graph Theory Theorem 2.1: Handshaking theorem: The sum of the degrees of the vertices of an undirected graph is an even number equal to twice the number of edges. In symbols, if G (V, E) is an undirected graph, then ∑ deg(𝑣) = 2𝐸 𝑣∈𝑉 Proof: Adding the degrees of all the vertices involves counting one for each edge incident with each vertex. If the edge is not a loop, it is incident with two different vertices and so gets counted twice exactly, once at each vertex. On the other hand, a loop at a vertex is also counted twice, by convention, in the degree of that vertex. Hence the proof is completed. Theorem 2.2: An undirected graph has an even number of vertices of odd degree. Proof: Let V1 and V2 be the set of vertices of even degree and the set of vertices of odd degree, respectively, in an undirected graph G = (V,E) with e edges. Then 2𝐸 = ∑ deg(𝑣) = 𝑣∈𝑉 ∑ deg⁡( 𝑣1) 𝑣∈𝑉1 + ∑ 𝑑𝑒𝑔⁡(𝑣2) 𝑣∈𝑉2 Because deg(v) is even for v ∈ V1, the first term in the right-hand side of the last equality is even. Furthermore, the sum of the two terms on the right-hand side of the last equality is even, because this sum is 2E. Hence, the second term in the sum is also even. Because all the terms in this sum are odd, there must be an even number of such terms. Thus, there are an even number of vertices of odd degree. [1]
  • 13. P a g e | 10 Theorem 2.3: A simple graph is bipartite if and only if it is possible to assign one of two different colors to each vertex of the graph so that no two adjacent vertices are assigned the same color. Proof: First, we suppose that G = (V, E) is a bipartite simple graph. Then V = V1 ∪ V2, where V1 and V2 are disjoint sets and every edge in E connects a vertex in V1 and a vertex in V2. If we assign one color to each vertex in V1 and a second color to each vertex in V2, then no two adjacent vertices are assigned the same color. Now suppose that it is possible to assign colors to the vertices of the graph using just two colors so that no two adjacent vertices are assigned the same color. Let V1 be the set of vertices assigned one color and V2 be the set of vertices assigned the other color. Then, V1 and V2 are disjoint and V = V1 ∪ V2. Furthermore, every edge connects a vertex in V1 and a vertex in V2 because no two adjacent vertices are either both in V1 or both in V2. Consequently, G is bipartite. Theorem 2.4: There is a simple path between every pair of distinct vertices of a connected undirected graph. Proof: Let u and v be two distinct vertices of the connected undirected graph G = (V, E). Because G is connected, there is at least one path between u and v. Let x0, x1, ……, xn, where x0 = u and xn = v, be the vertex sequence of a path of least length. This path of least length is simple. To see this, suppose it is not simple. Then xi = xj for some i and j with 0 ≤ i < j. This means that there is a path from u to v of shorter length with vertex sequence x0, x1,…., xi−1, xj , . . . , xn obtained by deleting the edges corresponding to the vertex sequence xi, . . . , xj−1. Theorem 2.5: (Euler’s formula) Let G be a connected planar graph. Let v, e, and r denote the number of vertices, edges, and regions. Then v –e + r = 2 Proof: The proposition will be proved by induction on e. If e = 0, then G has one vertex, no edges, and one region that is v = 1, e = 0, r = 1. So the theorem holds for the base case. Suppose the theorem is true for any graph with at most e−1 edges. If G is a tree (a tree is a graph with no cycles), then e = v−1, and r =1; so the theorem holds. If G is not a tree, let e' be an edge of
  • 14. P a g e | 11 some cycle of G. Then G with the edge e' removed is a connected plane graph with v vertices, e−1 edges, and r−1 regions. So, v−(e−1) + (r −1) = 2 (by the induction hypothesis). Hence v – e + r = 2. Corollary 2.1: Let G be a connected planar simple graph. Let v and e denote the number of vertices and edges with v ≤ 3. Then e≤ 3v−6 Proof: Let r be the number of regions. Every region is surrounded by at least 3 edges (otherwise the result is clearly true). Every edge has one and only one region on either of its two sides. If we draw a planar graph with r regions and count the number of edges on the boundary of each region, sum of these numbers denoting by n will be n < 2e. Also, since each boundary contains at least three edges, n > 3r. So have 3r ≤2e. Plugging the inequality into Theorem 2.5 (Euler’s formula) we get, 3v-3e+3r=6 (multiplying v–e+r=2 by 3) =>3v-3e-6=-3r≥-2e =>3v-3e-6≥-2e =>3v-6≥e so, e ≤ 3v-6 Corollary 2.2.: If G is a connected planar simple graph then G has a vertex of degree not exceeding 5. Proof: Suppose deg(vi) > 6 for every vertex vi. Since Σ deg(vi) = 2e (By Handshaking Theorem 2.1), we would have 2e > 6v and hence e ≥ 3v > 3v -6. This contradicts e < 3v -6. Which is a contradiction for all v (By the Theorem 2.5). Therefore, there must be at least a single vertex with degree at most 5. So, deg(vi) ≤ 5.
  • 15. P a g e | 12 Corollary 2.3: If G is a connected planar simple graph with v vertices and e edges, with v≥3 and no circuits of length 3 then, e ≤ 2v - 4. Proof: For graph G with r regions, it follows from Theorem 2.1 (Handshaking theorem) for planar graphs that 2e ≥ 4r (because the degree of each region of a simple graph without triangles or circuits of length 3 is at least 4), so that r ≤ 1/2 e. Combining this with Euler's formula we get, v - e + r = 2 => e -v + 2 = r => e - v + 2 ≤ 1/2 e Hence e ≤ 2v - 4 2.3 Some problems on Graph Theory We will solve a few problems using the above theorems and corollaries. Problem---2.1: Show that K3,3 is non-planar. Proof: Suppose that K3,3 is a planar graph. Since K3,3 has 6 vertices and 9 edges and no triangles that is no circuits of length 3 (this is easy to see because it is bipartite), it follows from Corollary 2.3 that e≤ 2v-4 i.e. 9≤ (2×6) – 4 => 9 ≤ 8, which is a contradiction. Hence K3,3 is non-planar. Figure: 2.12 The above figure shows non-planar graph K3,3 4 2 6 1 3 5
  • 16. P a g e | 13 Problem---2.2: Show that K5 is non-planar. Proof: Suppose that K5 is a planar graph. Since K5 has 5 vertices and 10 edges it follows from Corollary---2.1 that e≤ 3v-6 i.e. 10≤ (3 × 5) - 6 => 10 ≤ 9, which is a contradiction and this shows that K5 is non-planar. Figure: 2.13 Figure 2.13 shows that K5 is non-planar. Problem---2.3: Suppose that a connected planar graph has 30 edges. If a planar representation of this graph divides the plane into 20 regions, how many vertices does this graph have? Solution: Given that, Number of edges, e=30 Number of regions, r=20 Number of vertices, v=? From the Theorem---2.1 (Euler’s formula) we know that,
  • 17. P a g e | 14 v –e + r =2 => v-30 +20 =2 So, v=12. Hence the given connected planar graph has 12 vertices with 30 edges in 20 regions. Figure: 2.14 Figure 2.14 represents the graph with 12 vertices A, B, C, D, E, F, G, H, I, J, K, L respectively, 30 edges and 20 regions. A B C D E L K J I F G H
  • 18. P a g e | 15 Chapter 3 Graph Coloring 3.1 Graph Coloring: The task to color the vertices (V) of a graph G= (V, E) such that no two adjacent vertices assigned the same color is known as graph coloring. This is also known as map coloring, because maps are colored using the graph coloring rules. 3.2 Importance of Graph Coloring: There are many applications of graph coloring in Mathematics, Computer science, Information technology and Communication. 1. To finalize the schedule of exam at a university or at any educational institution. 2. Frequency assignments of television channels to operate. 3. To create a computer network with multiple links between the data centers. 4. To visualize the ecosystem of wildlife. [1] 5. To show the relationship between the people. [2] 6. It can be used to model the roadmaps where the vertices represent the intersections and the edges represent roads. 3.3 Chromatic Number of a Graph: The least number of colors needed to color a graph G is termed as chromatic number of that graph so that no two adjacent vertices have the same color. It is represented by (G). Figure: 3.1 Let us color the above graph. Let A be colored blue while B, C and D can’t be blue. So, if B is colored green, C is red, then we can color D as green, because D and B are not connected. At last E can be colored blue. So the chromatic number, (G) =3. Blue A Green B Red C Green D Blue E
  • 19. P a g e | 16 3.4 Problems on Chromatic Number Problem 3.1: Find the chromatic number of the given graph. Figure: 3.2 Solution: First we color A with red. Since A is directly connected with B, C, E, D so we can not color these vertices with red. Suppose B assigns green. Since B is not directly connected with C so C is also assigned green. E and D can be blue while F can be red. Because F is not adjacent to A. Hence the chromatic number of the given graph is 3. (G)= 3. Problem 3.2: Find the chromatic number of the given graph. Figure: 3.3 Solution: First we color A with red. Then B and E should be different say green. Since C is not connected to A so it can be red and D has no connection with B and C thus D can assign green. So the given graph has chromatic number, (G) = 2. Red A Green B Green C Blue D Blue E Red F Red A Green B Green E Red C Green D
  • 20. P a g e | 17 3.5 Subgraph: If G and H are two graphs and each vertex and edge of H are contained in G then H is called a subgraph of G. If H ≠ G then H is a proper subgraph of G. Figure: 3.4 Figure 3.4 represents a graph and its subgraph. [6] 3.6 Theorems and Corollaries on Graph Coloring Theorem 3.1: Let ∆ (G) be the maximum of the degrees of the vertices of a graph 5. Then (G)≤ 1 + ∆ (G). Proof: We will prove this theorem by the method of induction. Let V be the vertices of a graph G and if V=1 then ∆ (G) =0 and (G) = 1. So the results clearly holds. Let K be an integer and K≥1 and assume that the result holds for all graphs for V=K. Now let G be a graph with K+1. Let G0=G-{V} be a subgraph. So ∆ (G0) ≤ ∆ (G). Now GO can be colored with (G0). Since G0 has K vertices so by the induction method we have (G0) ≤ 1 + ∆ (G0). So G0 can be colored with at most 1+∆ (G0) colors. Since there are at most ∆ (G) vertices adjacent to V, one of the available 1 + ∆ (G) colors remains for V. Thus, G can be colored with at most 1 + ∆ (G) colors.
  • 21. P a g e | 18 Theorem 3.2 (Five Color Theorem): Every planar graph can be colored with 5 colors. That is (G)≤ 5 Proof: The proof is by induction on the number of vertices n; when n≤5 this is trivial. Now suppose G is planar on more than 5 vertices; by Corollary 2.2 some vertex v has degree at most 5. By the induction hypothesis, G − v can be colored with 5 colors. Color the vertices of G, other than v, as they are colored in a 5-coloring of G − v. If d(v)≤4, then v can be colored with one of the 5 colors to give a proper coloring of G with 5 colors. So we now suppose d(v)=5. If the five neighbors of v are colored with four or fewer of the colors, then again v can be colored to give a proper coloring of G with 5 colors. Now we suppose that all five neighbors of v have a different color, as indicated in following figure Figure: 3.5 Figure 3.5 shows that five neighbors of v colored with 5 colors: v1 is red, v2 is purple, v3 is green, v4 is blue, v5 is orange. Suppose that in G there is a path from v1 to v3, and that the vertices along this path are alternately colored red and green; call such a path a red-green alternating path. Then together with v, this path makes a cycle with v2 on the inside and v4 on the outside, or vice versa. This means there cannot be a purple-blue alternating path from v2 to v4. Supposing that v2 is inside the cycle, we change the colors of all vertices inside the cycle colored purple to blue, and all V1 red V2 purple V3 green V5 orange V4 blue
  • 22. blue vertices are recolored purple. This is still a proper coloring of all vertices of G except v, and now no neighbor of v is purple, so by coloring v purple we obtain a proper coloring of G. If there is no red-green alternating path from v1 to v3, then we recolor vertices as follows: Change the color of v1 to green. Change all green neighbors of v1 to red. Continue to change the colors of vertices from red to green or green to red until there are no conflicts, that is, until a new proper coloring is obtained. Because there is no red-green alternating path from v1 to v3, the color of v3 will not change. Now no neighbor of v is colored red, so by coloring v red we obtain a proper coloring of G. Corollary 3.1: Show that (Cn) = 2 if n is even positive integer with n ≥ 4. Proof: We want to show that two colors are needed to color Cn when n is even. To construct such a coloring, we pick a vertex and color it red. Proceed around the graph in a clockwise direction (using a planar representation of the graph) coloring the second vertex blue, the third vertex red, and so on. The nth vertex can be colored blue, because the two vertices adjacent to it, namely the (n − 1)st and the first vertices, are both colored red. For example if we take n = 4 then the graph C4 can be colored with 2 colors red and blue and if n = 6 then the graph C6 can also be colored with same way. Hence (Cn) = 2 if n is even positive integer with n ≥ 4. Figure: 3.6(a) Figure: 3.6(b) Figure 3.6(a) and 3.6(b) shows C4 and C6 A Red B Blue D Blue C Red A Red B Blue C Red E Red D Blue F Blue
  • 23. P a g e | 20 Corollary 3.2: Show that (Cn) = 3 if n is odd positive integer with n ≥ 3. Proof: When n is odd and n > 1, the chromatic number of Cn is 3. To see this, we pick an initial vertex. To use only two colors, it is necessary to alternate colors as the graph is traversed in a clockwise direction. The nth vertex reached is adjacent to two vertices of different colors, namely, the first and (n − 1)st . For example let n = 5 and consider C5. Let us pick a vertex and color it red. Proceeding clockwise, it is necessary to assign a second color, say blue, to the next vertex reached. Continuing in the clockwise direction, the third vertex can be colored red, and the fourth vertex can be colored blue. The fifth vertex cannot be colored either red or blue, because it is adjacent to the fourth vertex and the first vertex. Consequently, a third color is required for this vertex. Thus, the chromatic number of C5 is 3. Hence, a third color must be used and we can say that (Cn) = 3 if n is odd positive integer with n ≥ 3. Figure: 3.7 Figure 3.7: C5 needs 3 colors. A Red B Blue C Red D Blue E Yellow
  • 24. P a g e | 21 3.7 Four Color Theorem There is an important theorem regarding map coloring which is known as Four Color theorem. Statement of Four Color Theorem: Four colors are sufficient to color a planar map (with five regions). 3.8 History of Four Color Theorem: The Four Color Theorem has fascinated people for almost a century and a half. The Four Color Conjecture first seems to have been made by Francis Guthrie. He was a student at University College London where he studied under De Morgan. After graduating from London he studied law but by this time his brother Frederick Guthrie had become a student of De Morgan. Francis Guthrie showed his brother some results he had been trying to prove about the coloring of maps and asked Frederick to ask De Morgan about them. De Morgan was unable to give an answer but, on 23 October 1852, the same day he was asked the question, he wrote to Hamilton in Dublin. De Morgan wrote:- A student of mine asked me today to give him a reason for a fact which I did not know was a fact - and do not yet. He says that if a figure be anyhow divided and the compartments differently colored so that figures with any portion of common boundary line are differently colored - four colors may be wanted, but not more - the following is the case in which four colors are wanted. Query cannot a necessity for five or more be invented. ...... If you retort with some very simple case which makes me out a stupid animal, I think I must do as the Sphinx did.... Hamilton replied on 26 October 1852 (showing the efficiency of both himself and the postal service):- I am not likely to attempt your quaternion of color very soon. However De Morgan kept asking if anyone could find a solution to Guthrie's problem and several mathematicians worked on it. Charles Peirce in the USA attempted to prove the Conjecture in the 1860's and he was to retain a lifelong interest in the problem. Cayley also learnt of the problem from De Morgan and on 13 June 1878 he posed a question to the London mathematical society asking if the Four Color Conjecture had been solved. Shortly afterwards Cayley sent a paper on the coloring of maps to the Royal Geographical Society and
  • 25. P a g e | 22 in was published in 1879. The paper explains where the difficulties lie in attempting to prove the Conjecture. On 17 July 1879 Alfred Brey Kempe announced in Nature that he had a proof of the Four Color Conjecture. Kempe was a London barrister who had studied mathematics under Cayley at Cambridge and devoted some of his time to mathematics throughout his life. At Cayley's suggestion Kempe submitted the Theorem to the American Journal of Mathematics where it was published in 1879. Story read the paper before publication and made some simplifications. Story reported the proof to the Scientific Association of Johns Hopkins University in November 1879 and Charles Peirce, who was at the November meeting, spoke at the December meeting of the Association of his own work on the Four Color Conjecture. Kempe used an argument known as the method of Kempe chains. If we have a map in which every region is colored red, green, blue or yellow except one, say X. If this final region X is not surrounded by regions of all four colors there is a color left for X. Hence suppose that regions of all four colors surround X. If X is surrounded by regions A, B,C, D in order, colored red, yellow, green and blue then there are two cases to consider. (i) There is no chain of adjacent regions from A to C alternately colored red and green. (ii) There is a chain of adjacent regions from A to C alternately colored red and green. If (i) holds there is no problem. Change A to green, and then interchange the color of the red/green regions in the chain joining A. Since C is not in the chain it remains green and there is now no red region adjacent to X. Color X red. If (ii) holds then there can be no chain of yellow/blue adjacent regions from B to D. [It could not cross the chain of red/green regions.] Hence property (i) holds for B and D and we change colors as above. [6] Kempe received great acclaim for his proof. He was elected a Fellow of the Royal Society and served as its treasurer for many years. He was knighted in 1912. He published two improved versions of his proof, the second in 1880 aroused the interest of P.G. Tait, the Professor of Natural Philosophy at Edinburgh. Tait addressed the Royal Society of Edinburgh on the subject and published two papers on the (what we should now call) Four Color Theorem. They contain some clever ideas and a number of basic errors.
  • 26. P a g e | 23 The Four Color Theorem returned to being the Four Color Conjecture in 1890. Percy John Heawood, a lecturer at Durham England, published a paper called Map coloring theorem. In it he states that his aim is Rather destructive than constructive, for it will be shown that there is a defect in the now apparently recognized proof. Although Heawood showed that Kempe's proof was wrong he did prove that every map can be 5-coloured in this paper. Kempe reported the error to the London Mathematical Society himself and said he could not correct the mistake in his proof. In 1896 De La Vallee Pousin also pointed out the error in Kempe’s paper, apparently unaware of Heawood’s work. Heawood was to work throughout his life on map coloring, work which spanned nearly 60 years. He successfully investigated the number of colors needed for maps on other surfaces and gave what is known as the Heawood estimate for the necessary number in terms of the Euler characteristic of the surface. Heawood was to make further contributions to the Four Color Conjecture. In 1898 he proved that if the number of edges around each region is divisible by 3 then the regions are 4- colourable. He then wrote many papers generalizing this result. To understand the later work we need to define some concepts. Clearly a graph can be constructed from any map the regions being represented by the vertices and two vertices being joined by an edge if the regions corresponding to the vertices are adjacent. The resulting graph is planar, that is can be drawn in the plane without any edges crossing. The Four Color Conjecture now asks if the vertices of the graph can be colored with 4 colors so that no two adjacent vertices are the same color. From the graph a triangulation can be obtained by adding edges to divide any non-triangular face into triangles. A configuration is part of a triangulation contained within a circuit. An unavoidable set is a set of configurations with the property that any triangulation must contain one of the configurations in the set. A configuration is reducible if it cannot be contained in a triangulation of the smallest graph which cannot be 4-coloured. The search for avoidable sets began in 1904 with work of Weinicke. Renewed interest in the USA was due to Veblen who published a paper in 1912 on the Four Color Conjecture generalizing Heawood's work.
  • 27. P a g e | 24 Franklin in 1922 published further examples of unavoidable sets and used Birkhoff's idea of reducibility to prove, among other results, that any map with ≤ 25 regions can be 4-coloured. The number of regions which resulted in a 4-colourable map was slowly increased. Reynolds increased it to 27 in 1926, Winn to 35 in 1940, Ore and Stemple to 39 in 1970 and Mayer to 95 in 1976. However the final ideas necessary for the solution of the Four Color Conjecture had been introduced before these last two results. Heesch in 1969 introduced the method of discharging. This consists of assigning to a vertex of degree i the charge 6 - i. Now from Euler's formula we can deduce that the sum of the charges over all the vertices must be 12. A given set S of configurations can be proved unavoidable if for a triangulation T which does not contain a configuration in S we can redistribute the charges (without changing the total charge) so that no vertex ends up with a positive charge. Heesch thought that the Four Color Conjecture could be solved by considering a set of around 8900 configurations. There were difficulties with his approach since some of his configurations had a boundary of up to 18 edges and could not be tested for reducibility. The tests for reducibility used Kempe chain arguments but some configurations had obstacles to prevent reduction. The year 1976 saw a complete solution to the Four Color Conjecture when it was to become the Four Color Theorem for the second, and last, time. The proof was achieved by Appel and Haken, their methods based on reducibility using Kempe chains. They carried through the ideas of Heesch and eventually they constructed an unavoidable set with around 1500 configurations. They managed to keep the boundary ring size down to ≤ 14 making computations easier that for the Heesch case. There was a long period where they essentially used trial and error together with unbelievable intuition to modify their unavoidable set and their discharging procedure. Appel and Haken used 1200 hours of computer time to work through the details of the final proof. Koch assisted Appel and Haken with the computer calculations. The Four Color Theorem was the first major theorem to be proved using a computer, having a proof that could not be verified directly by other mathematicians. Despite some worries about this initially, independent verification soon convinced everyone that the Four Color Theorem had finally been proved. Details of the proof appeared in two articles in 1977. Recent work has led to improvements in the algorithm. [1, 6, 8]
  • 28. P a g e | 25 3.9 An Example of Four Color Theorem First let us take a planar graph of 30 regions and try to color it with the help of Four Color Theorem. Our first graph in figure 3.8 will be of no color which has been given below--- Figure: 3.8 Now we want to color the graph with the first color red in figure 3.9. Figure: 3.9
  • 29. P a g e | 26 In figure 3.10 we will use the green color with the red which was previously used in figure 3.9. Figure: 3.10 Then the figure 3.11 can be formed by using three colors red, green and blue. Figure: 3.11
  • 30. P a g e | 27 At the last stage in figure 3.12 we will use the color yellow and get our desired graph which assigns four color and thus the four color theorem is established by this particular example. Figure: 3.12 3.10 Applications of Graph Coloring 3.10.1 Scheduling problem for final examination using four color theorem: The scheduling problem can be solved using a graph model, where the vertices represent courses and an edge between two vertices if there is a common student in the courses they represent. Each time slot for a final exam is represented by a different color. A scheduling of the exams corresponds to a coloring of the associated graph. For instance, let us consider that there are 10 finals to be scheduled. Suppose the courses are labeled as 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 .Suppose that the following pairs of courses have common students: 1 and 2, 1 and 3, 1 and 4, 1 and 7, 1 and 10, 2 and 3, 2 and 4, 2 and 5, 2 and 7, 3 and 4, 3 and 6, 3 and 7, 4 and 5, 4 and 6, 5 and 6, 5 and 7, 6 and 7, 6 and 9, 7 and 8, 7 and 10, 8 and
  • 31. P a g e | 28 10, 8 and 5, 8 and 9, 9 and 10 and 10 and 6. The graph associated with this set of classes is shown in the table given below. A scheduling consists of a coloring of this graph. Because the chromatic number of this graph is 4 (By the assumption of the Four Color Theorem), four time slots are needed. A coloring of the graph using four colors and the associated schedule are shown in Figure 3.13. Figure: 3.13 Time Period Courses I 1, 6, 8 II 2, 9, 10 III 3, 5 IV 4, 7 Now we want to solve this problem using computer programming by the help of Mathematica. Our goal is to color the above graph using four color theorem that is the chromatic number should not exceed the number 4. The required inputs and the corresponding outputs are given below. 1 Red 10 Blue 2 Blue 3 Green 4 Yellow 5 Green6 Red 7 Yellow 8 Red 9 Green
  • 32. P a g e | 29 Input 1: Needs["Combinatorica`"] Input 2: s1 = {{1, 2}, {1, 3}, {1, 4}, {1, 7}, {2, 3}, {2, 4}, {2, 5}, {2, 7}, {3, 4}, {3, 6}, {3, 7}, {4, 5}, {4, 6}, {5, 6}, {5, 7}, {6, 7}, {6, 9}, {7, 10}, {8, 10}, {8, 5}, {10, 6}}; Input 3: ShowLabeledGraph[FromUnorderedPairs[s1]] Output 1: Input 4: ColorableQ[G_, k_]: = Module[{𝑐, 𝑑, 𝑖}, (Expand[Apply[Times, Apply[Subtract, Map[𝑐, 𝐺, {2}],1]]]/. 𝑐[i_]d_. ⧴ 𝑐[𝑖]𝑑mod𝑘) = ! = 0] Input 5: ColorableQ[s1,4] Output 2: True Input 6: ColorableQ[s1,3] Output 3: False Since the output is true that is the given graph is colorable with at most four color so the four color theorem holds and the chromatic number is 4. [12] 1 23 4 5 6 7 8 9 10
  • 33. P a g e | 30 3.10.2 Frequency assignment of television channel using four color theorem: Suppose there are 12 television channels to assign the frequency in such a way that there will be no two channels have same frequency which are within 500 kilometers. [1] This problem can be solved easily by map coloring or graph coloring. Let us construct a graph by assigning a vertex to each station. Two vertices are connected by an edge if they are located within 500 kilometers of each other. We label the channels as 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 where the adjacent vertices within 500 kilometers are given as follows--- Channel name adjacent channels 1 9, 12 2 3, 8, 9 3 2, 8, 10 4 5, 6, 11 5 4, 6, 7, 8 6 4, 5, 11, 12 7 5, 8, 10 8 2, 3, 5, 7, 9, 10 9 1, 2, 5, 6, 8 10 3, 7, 8 11 6, 4, 12 12 1, 6, 11 Now if we color the station 1 with red then 9 and 12 must be different say green and blue respectively. 2 can be red since it is 530 kilometers away from 1. Since it is connected with 9 which is green so 3 may be blue while 8 can be yellow. For channel 3, 2 and 8 have color red and yellow respectively so 10 will be red. Channel 4 is not associated with anyone of the previous so it can be red and 6 can be green while 5 is blue. Channel 5, we color only 7 with green.
  • 34. P a g e | 31 For Channel 6, we color only 11 with yellow. The distance between 6 and 9 is 505 kilometers. Similarly if we color the rest of the channel stations we shall get the figure 3.14 which represents the frequency allotment of the channels where R, G, B, Y stand for red, green, blue and yellow respectively. Figure: 3.14 This problem can also be checked whether it follows the four color theorem or not by the Mathematica codes which are given below--- Input 1: Needs["Combinatorica`"] Input 2: s2 = {{1, 9}, {1, 12}, {2, 3}, {2, 8}, {2, 9}, {3, 8}, {3, 10}, {4, 5}, {4, 6}, {4, 11}, {5, 6}, {5, 7}, {5, 8}, {6, 11}, {6, 12}, {7, 8}, {7, 10}, {8, 9}, {8, 10}, {5, 9}, {6, 9}, {11, 12}}; Input 3: ShowLabeledGraph[FromUnorderedPairs[s2]] 1-R 2-R530 370 3-B 4-R 5-B 9-G 12-B 10-R 7-G11-Y 360 180 360 180 350
  • 35. P a g e | 32 Output 1: Input 4: ColorableQ[G_, k_]: = Module[{𝑐, 𝑑, 𝑖}, (Expand[Apply[Times, Apply[Subtract, Map[𝑐, 𝐺, {2}],1]]]/. 𝑐[i_]d_. ⧴ 𝑐[𝑖] 𝑑mod𝑘 ) = ! = 0] Input 5: ColorableQ[s2,4] Output 2: True Input 6: ColorableQ[s2,3] Output 3: False Since the output is true that is the given graph is colorable with at most four color so the four color theorem holds and the chromatic number is 4. [12] 1 2 3 4 5 6 7 8 9 10 11 12
  • 36. P a g e | 33 Conclusion In this project paper, at first the idea of graph, graph theory and its importance have been given. Then in the next chapter there are basic terminologies of graph theory, useful theorems and corollaries which are prerequisite for the graph theory. There are Euler’s theorem, some problems solved by using the theorem and corollaries associated by the theorem with some definitions. At the last chapter there are the history, statement, some applications of famous Four Color Theorem. This paper will help us to understand how a critical problem can be solved using map or graph coloring. There are many applications of graph theory and map or graph coloring in the real world and our attempt was to show some of them. Due to limitation of time and volume of the project, this report could not cover extensive applications of map coloring. But it is hoped that Graph Theory and its applications would be studied further whenever the opportunity comes.
  • 37. P a g e | 34 References 1. Discrete Mathematics and Applications with Combinatorics and Graph Theory, Kenneth H. Rossen (Seventh Edition) 2. Discrete Mathematics with Graph Theory, Edgar G. Goodaire and Michael M. Parmenter (Second Edition) 3. Graph Theory, V.K. Balakrishnan (Schaum’s outline series) 4. Discrete Mathematics, Seymour Lipschutz and Marc Lipson (Schaum’s outline series) 5. Graph Theory, Tero Harju, Department of Mathematics University of Turku FIN-20014 Turku, Finland 1994 – 2012 6. http://en.Wikipedia.org/wiki/Map-coloring 7. Graph Theory, R. Diestel, Springer–Verlag (2005) 8. Graph Theory and Its Applications, J. Gross & J. Yellen, CRC Press (2006) 9. Introduction to Graph Theory, D. B. West, Prentice–Hall (1996) 10. http://Mathworld.wolfram.com 11. Mathematica 9.0, Combinatorica 12. How to color a graph with computer algebra---Yuri Matiyasevich Steklov Institute of Mathematics at Saint-Petersburg 27 Fontanka, Saint-Petersburg, 191011, Russia E- mail: yumat@pdmi.ras.ru URL: http://logic.pdmi.ras.ru/~yumat