Four Color Map Theorem
Calvin Kim
Senior Thesis in Mathematics
Middlebury College
May 2016
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c Copyright by Calvin Kim, 2016.
All Rights Reserved
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Abstract
The history of the Four Color Map Theorem (or the Four Color Theorem) is interesting due to the fact that, despite how simple it is to state and understand the problem,
the theorem has yet to be proven by hand. It is the first theorem having a computerassisted proof, and thus has generated a lot of controversy. This paper outlines the
main ideas of the proof, beginning with a flawed demonstration by Kempe. After
outlining the proof by Appel and Haken, the paper then moves onto extensions of
the theorem, i.e. a generalized map coloring theorem for different surfaces. Finally,
it concludes with a discussion on the impact of the theorem and the philosophical
implications that came associated with such an unorthodox proof.
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Acknowledgements
I would like to thank the following:
My advisor, Professor Schumer, for helping me with my research and for meticulously editing my drafts.
The mathematics department, for giving me the opportunity to research a topic
of my interest and for supporting me through the process.
Jordan DuBeau and Eliot VanValkenburg, for making the thesis enjoyable and for
stopping me from writing it all in one night.
My suite-mates, for bearing with me and for pretending to be interested during
my often unintelligible ramblings.
And finally to all of my friends and family, for supporting me throughout this
process.
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In memory of Dr. Kenneth Appel:
For introducing me to the world of mathematics.
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Contents
Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
v
Acknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
vii
1 Introduction
3
1.1
History . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
1.2
Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
1.3
Graph Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
1.3.1
1.4
Euler’s Formula . . . . . . . . . . . . . . . . . . . . . . . . . .
10
Problem of the Five Princes . . . . . . . . . . . . . . . . . . . . . . .
13
2 Kempe Chains
2.1
2.2
15
Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15
2.1.1
20
Counterexample . . . . . . . . . . . . . . . . . . . . . . . . . .
Five Color Theorem
. . . . . . . . . . . . . . . . . . . . . . . . . . .
3 Proof of the Four Color Theorem
21
23
3.1
Unavoidable Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
24
3.2
Reducible Configurations . . . . . . . . . . . . . . . . . . . . . . . . .
27
4 Extensions
4.1
31
Heawood Conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . .
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31
xii
CONTENTS
4.1.1
5 Conclusion
Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
34
43
List of Figures
1.1
An example of a disconnected region in a map. . . . . . . . . . . . . .
5
1.2
A stereographic projection of a sphere onto a map [15]. . . . . . . . .
6
1.3
Converting a non-cubic map into a cubic map. . . . . . . . . . . . . .
7
1.4
K5 which is a non-planar graph. . . . . . . . . . . . . . . . . . . . . .
8
1.5
Two isomorphic, planar graphs. . . . . . . . . . . . . . . . . . . . . .
9
1.6
An example of a graph embedding. [16]. . . . . . . . . . . . . . . . .
10
1.7
Four mutually adjacent regions. . . . . . . . . . . . . . . . . . . . . .
14
2.1
A basic set of four unavoidable subgraphs. . . . . . . . . . . . . . . .
16
2.2
A case where the 4 neighbors of u are colored with four distinct colors.
16
2.3
A red-green Kempe chain where the neighbors of u are not linked by
the chain. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17
2.4
The result after flipping the colors of the Kempe chain from Figure 2.3. 18
2.5
A red-green Kempe chain where the neighbors of u are linked by the
chain. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18
2.6
A case where the 5 neighbors of u are colored with four distinct colors.
19
2.7
The case where deg(u) = 5 and two Kempe chains have been considered. 19
2.8
A counterexample to Kempe’s proof of the Four Color Theorem. . . .
2.9
The case where deg(u) = 5 and all of its neighbors are colored differently. 22
1
20
2
LIST OF FIGURES
3.1
A planar graph where every vertex has charge 6 − i. . . . . . . . . . .
25
3.2
The Birkhoff diamond. . . . . . . . . . . . . . . . . . . . . . . . . . .
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3.3
The Birkhoff diamond after it has been shrunk down to a smaller configuration. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
28
3.4
A case where the coloring for Birkhoff diamond cannot be extended. .
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3.5
The final coloring for the problematic case of the Birkhoff diamond. .
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4.1
A maximal coloring of a torus with 7 colors [14]. . . . . . . . . . . . .
34
4.2
A maximal coloring of a Möbius strip with 6 colors [17]. . . . . . . . .
34
Chapter 1
Introduction
The Four Color Map Theorem (or the Four Color Theorem) is an especially interesting
result in mathematics due to the fact that even though a child can understand the
problem, it took over 100 years, many incorrect proofs, and about 1,000 hours of
computer time to finally solve the theorem. Informally, the theorem states that any
map on a plane or a sphere can be colored using at most four colors such that no two
countries that are adjacent to each other share a color.
1.1
History
This problem is sometimes called Guthrie’s problem, named after the mathematician Francis Guthrie who first postulated this theorem in 1852 [11]. He posed the
question to his younger brother Frederick Guthrie who then submitted the theorem
to his professor, Augustus De Morgan. From here, it was introduced to the general
mathematics community. Alfred Bray Kempe published his “proof” of the Four Color
Theorem in the American Journal of Mathematics in 1879. In 1890, about 10 years
after Kempe had published his proof, Percy Heawood found an error in the proof
3
4
CHAPTER 1. INTRODUCTION
that could not be fixed readily. However, Heawood was able to prove the Five Color
Theorem using methods developed by Kempe.
From here on out, progress on this problem was gradual and fragmented. For example, in 1913, George Birkhoff introduced the idea of reducible configurations and
showed that the Birkhoff diamond is reducible. In 1948, Heinrich Heesch proposed
searching for an unavoidable set of reducible configurations. And finally in 1976, Kenneth Appel and Wolfgang Haken announced their proof of the Four Color Theorem
which involved generating and reducing 1482 configurations. Since then, this number has been reduced to 633 configurations through different generating techniques.
While this number can be reduced to 591, it comes at cost of more computing time.
[18] Finally in 2005, Georges Gonthier and Benjamin Werner formalized the proof of
the theorem within a logical computer system known as Coq [4].
1.2
Basics
First, we need some basic definitions for maps.
Definition 1. A map is a plane that is divided into different regions by boundary
lines.
Definition 2. A meeting point is the point where two or more boundary lines meet.
Definition 3. Two regions are said to be neighbors or adjacent to each other if they
share a boundary line rather than simply a meeting point.
If two regions were considered to be adjacent at a meeting point, it is easy to
construct a map where an arbitrarily large number of regions share a meeting point
and would therefore require an arbitrarily large number of colors (think of wedges of a
pie) [1]. In addition, every region must be a connected region because if a region were
1.2. BASICS
5
Figure 1.1: An example of a disconnected region in a map.
allowed to be disconnected (such as Michigan is in the United States), it is fairly easy
to construct a map where five regions are all adjacent to each other. For example,
in Figure 1.1, if both regions labeled A are one region and thus must be colored the
same color, then there is no way to color this map in just four colors.
It is also important to note that proving the Four Color Theorem on the surface
of spheres is equivalent to proving it on a 2-D map. Given a map on a sphere, we
can convert it into a planar map by a method known as stereographic projection (see
Figure 1.2) and vice versa [18]. The idea is to place the plane below the sphere such
that sphere is tangent to the plane. Then from the North Pole N draw a line through
a point on the sphere p and project this to the point q where the line intersects on
the plane. This is one way we project our 3-D Earth onto a 2-D map. Notice that
this does not affect the coloring of the map or the adjacency between regions.
Although it may seem inconsequential, it is also important to note that exactly
two regions share a boundary line. In other words, there are no “bridges” that span
between countries that do not share a boundary line.
6
CHAPTER 1. INTRODUCTION
Figure 1.2: A stereographic projection of a sphere onto a map [15].
In addition, for reasons that will become apparent later on, we will be working
with cubic maps.
Definition 4. A cubic map is a map in which there are exactly three boundary lines
at each meeting point.
Every map can be converted to a cubic map and we can show that if the cubic
map can be four colored, then the original map can be four colored. If a meeting
point has more than three boundary lines, as shown in Figure 1.3 then we can replace
the meeting point with a new region in order to make the map cubic. Note that if
we can color the cubic map with four colors, then we can remove the new region and
it will not affect the rest of the map’s coloring. If a meeting point has exactly two
boundary lines, then we can simply remove the meeting point without changing the
map. Therefore, since proving that the cubic map can be colored with four colors is
a harder problem, proving the theorem on cubic maps also proves that the all maps
can also be colored with four colors.
1.3. GRAPH THEORY
7
Figure 1.3: Converting a non-cubic map into a cubic map.
1.3
Graph Theory
Now that we we’ve defined what maps are, we will be defining graphs. Graphs are
better defined than maps and are therefore easier to work with. In addition, there is
an easy and intuitive conversion from maps to graphs which we will be using.
Definition 5. A graph G contains a finite set of vertices V (G), and a finite set edges
E(G).
Vertices are connected to one another by edges.
Definition 6. A vertex u is said to be a neighbor of another vertex v if there exists
a uv-edge between the two vertices.
Definition 7. The degree of a vertex u, denoted deg(u) is the number of edges that
are incident with that vertex.
A vertex with degree i will be denoted as an i-vertex.
Definition 8. A connected graph is a graph in which there exists a path between
any two vertices in the graph.
8
CHAPTER 1. INTRODUCTION
Figure 1.4: K5 which is a non-planar graph.
Definition 9. Two graphs G and G0 are said to be isomorphic iff there exists a
mapping f of vertices such that if uv ∈ E(G) then f (u)f (v) ∈ E(G0 ).
Definition 10. A complete graph Kn is a graph on n vertices where for all pairs of
vertices u and v, uv ∈ E(Kn ).
Figure 1.4 is an example of a complete K5 graph.
Definition 11. A planar graph is a graph that can be embedded onto a plane such
that the edges only intersect at the vertices.
In other words, there exists an embedding of the graph on a plane such that no
edges cross each other. Figure 1.5 is an example of a planar graph. Notice that
the two graphs are isomorphic, and although the first graph does not appear to be
planar, there exists an isomorphic graph that is planar; so the graph is considered to
be planar. On the other hand, Figure 1.4 is not planar since there is no way to draw
this graph such that no edges cross each other.
We can convert any map into a planar graph by inserting a vertex into every
region, and adding an edge between two vertices if the two regions that contain the
vertices share a boundary. This is known as a graph embedding and results in a dual
1.3. GRAPH THEORY
9
Figure 1.5: Two isomorphic, planar graphs.
graph. An example of this is given in Figure 1.6 where the blue represents the graph
and the red represents the map. A blue vertex is added to every region in the map,
and these vertices are connected by an edge if the regions are adjacent to each other.
Notice that the outside area is considered a region.
Definition 12. Faces of a planar graph G are the smallest regions that are enclosed
by edges.
Faces can be thought of as the “regions” of a graph. Note that the area outside
of the graph is also a face, called the unbounded face.
Definition 13. A maximal planar graph is a planar graph in which no more edges
can be added to the graph while still maintaining the planarity condition.
It is fairly easy to see that a maximal planar graph will consist of faces which are
all triangles. If any of the faces are not triangles, then we can add edges in until the
face is divided into triangles. For example, if a face is a square we can add in an edge
as the diagonal in order to convert the square into two triangles. It is also easy to
see that if a map is cubic, then the embedded planar graph will be a maximal planar
graph because at each meeting point in the map, three regions intersect so those three
regions will become three vertices that are mutually connected, i.e. a triangle. One
property of maximal planar graphs G is that if u is a vertex in a G with deg(u) = i,
then the neighbors of u create a ring of i vertices around u.
10
CHAPTER 1. INTRODUCTION
Figure 1.6: An example of a graph embedding. [16].
We need one last definition for graphs.
Definition 14. An n-coloring of G is an assignment of n colors to vertices subject
to certain constraints.
In our case, the constraint will be that if u is a neighbor of v then u and v cannot
share a color. In this way, the Four Color Theorem can be rephrased as the following:
Theorem 1 (Four Color Theorem). There exists a 4-coloring of every planar graph
G such that no two adjacent vertices share a color.
1.3.1
Euler’s Formula
Euler’s Formula is an important result in graph theory, and we will need it to discuss
the proof of the Four Color Theorem.
Theorem 2 (Euler’s Formula). Given a planar graph G,
V −E+F =2
1.3. GRAPH THEORY
11
where V is the number of vertices, E is the number of edges, and F is the number of
faces.
Proof. The proof follows by induction on the number of edges.
The base case when E = 1 is fairly trivial. Consider a graph with one edge. Then
V = 2, E = 1, and F = 1. The formula V − E + F = 2 holds. Now assume that the
theorem holds for any graph G0 with E − 1 edges. Let G be a graph with E edges.
Now we arbitrarily pick an edge e ∈ E(G) and remove it.
Case 1: If this disconnects the graph then we have to consider both of the connected components. Say the first component has V1 vertices, E1 edges and F1 faces
while the second component has V2 vertices, E2 edges and F2 faces. Since we haven’t
removed any vertices, V = V1 + V2 . We removed one edge, so E = E1 + E2 + 1. And
finally, notice that removing the edge didn’t affect the number of faces. However, the
outside face is double counted by the two components, so F = F1 + F2 − 1. We know
from the induction hypothesis that V1 − E1 + F1 = 2 and V2 − E2 + F2 = 2. Add
the two equations to get V1 + V2 − E1 − E2 + F1 + F2 = 2 + 2 which simplifies to
V − (E − 1) + (F + 1) = 4 which further simplifies to V − E + F = 2.
Case 2: If this doesn’t disconnect the graph, then we now have one less face and
edge than before, but the same number of vertices. By the induction hypothesis,
V − (E − 1) + (F − 1) = 2 which simplifies to V − E + F = 2.
From this result we get another important result for planar graphs:
Corollary 2.1. Given a planar map G, E ≤ 3V − 6. If G is maximal, then E =
3V − 6.
Proof. Notice that every face is bounded by at least 3 edges and every edge is part
of 2 faces. Therefore,
3F ≤ 2E
12
CHAPTER 1. INTRODUCTION
which we can rewrite as
2
F ≤ E
3
Plug this into Euler’s Formula (Theorem 2) to get
2
V −E+ E ≥2
3
which simplifies to
E ≤ 3V − 6
Notice that if the graph is maximal, 3F = 2E which results in F = 32 E. We plug this
into Euler’s Formula to get V − E + 32 E = 2 which simplifies to E = 3V − 6.
The following theorem also follows from the above.
Theorem 3. Given a planar graph G, there exists at least one vertex u such that
deg(u) ≤ 5.
Proof. Given a planar graph G, we have that E ≤ 3V − 6 by Corollary 2.1. Now
assume, for contradiction, that every vertex in G has degree 6 or greater. Then
2E ≥ 6V since 2E will give you the total sum of the vertex degrees. This simplifies
to E ≥ 3V which is a contradiction. Thus there has to be at least one vertex of
degree less than or equal to 5.
This in turn leads to another important theorem.
Theorem 4. Given a maximal planar graph G,
∞
X
(6 − i)vi = 12
i=1
where vi is the number of vertices with degree i.
1.4. PROBLEM OF THE FIVE PRINCES
13
Proof. Given a maximal planar graph G, let vi be the number of i-vertices. Then
v1 + 2v2 + 3v3 + . . . = 2E. By Corollary 2.1, 2E = 6V − 12 so we get
v1 + 2v2 + 3v3 + . . . = 6V − 12
. Since V = v1 + v2 + v3 + . . ., the equation above simplifies to
12 = 5v1 + 4v2 + 3v3 + 2v4 + v5 − v7 − 2v8 − 3v9 . . .
which can be rewritten as
∞
X
(6 − i)vi = 12
i=1
1.4
Problem of the Five Princes
Before we begin discussing the proof of the Four Color Theorem, it is important to
note the difference between the Four Color Theorem and the problem of the five
princes. The problem of the five princes is the following:
There was once a king in India who had a large kingdom and five sons.
In his last will, the king said that after his death the sons should divide
the kingdom among themselves in such a way that the region belonging
to each son should have a borderline (not just a point) in common with
the remaining four regions. How should the kingdom be divided [18]?
Notice that if the king had four sons, this problem is possible and it is fairly trivial
as we can see in Figure 1.7. However, since the king has five sons, the problem is
impossible by the inequality we proved above.
14
CHAPTER 1. INTRODUCTION
Figure 1.7: Four mutually adjacent regions.
Proof. In terms of graph theory, this problem asks whether it is possible for a graph
with 5 vertices to be mutually connected and planar. Notice that the resulting graph
is a complete graph K5 which has 5 vertices and 10 edges. By the Corollary 2.1 we
get 10 ≤ (3 ∗ 5) − 6 = 9 which is false. Thus by contradiction, the problem above is
impossible.
Another way to prove this is to notice that the problem is equivalent to a planar
K5 graph. Since we know that this is impossible, the problem is also impossible.
Unfortunately, this proof isn’t sufficient to prove the Four Color Theorem. If the
Four Color Theorem is true, then the solution for this problem naturally follows, but
the converse doesn’t hold. Just because it is impossible to construct a map with
five mutually adjacent regions, it doesn’t necessarily follow that only four colors are
needed to color any map.
Chapter 2
Kempe Chains
In 1879, Alfred Bray Kempe produced the most famous incorrect proof of the Four
Color Theorem. His error was subtle enough that it took around ten years before a
mistake was discovered by Percy John Heawood. Although the error was substantial
enough that it could not be fixed easily, Kempe’s proof introduced many of the ideas
and foundations for the actual proof to come in about a hundred years.
2.1
Proof
The “proof” will follow from induction on the number of vertices. Recall from Theorem 3 that every graph must have at least one vertex, u, of degree 5 or less. This
means that every graph must contain a subgraph containing one of the elements in
Figure [?]. Now consider a graph G with n vertices. By the induction hypothesis,
we assume that every graph with less than n vertices can be colored with four colors.
The base cases are trivial since a graph with 4 or less vertices can be colored with 4
colors. Now consider the vertex, u, with degree 5 or less.
If deg(u) = 2, then we remove u from G. By the induction hypothesis, G − {u}
15
16
CHAPTER 2. KEMPE CHAINS
Figure 2.1: A basic set of four unavoidable subgraphs.
Figure 2.2: A case where the 4 neighbors of u are colored with four distinct colors.
can be colored with four colors since G − {u} contains n − 1 vertices. Now reinsert u
into G. Since deg(u) = 2, there is at least one color that is unused by the neighbors
of u so u can be colored with that color.
If deg(u) = 3, then we remove u from G. By the induction hypothesis, G − {u}
can be colored with four colors. Now reinsert u into G. Since deg(u) = 3, there is
at least one color that is unused by the neighbors of u so u can be colored with that
color.
If deg(u) = 4, then we remove u from G. By the induction hypothesis, G − {u}
can be colored with four colors. Now reinsert u into G. Here, it is possible that the
neighbors of u are already colored using four colors as seen in Figure 2.2.
This is where Kempe introduced the method known as the method of Kempe chains
[18]. Consider two of the neighbors of u that are not adjacent to each other, say the
red and green ones. Starting from the green vertex, consider all of its red neighbors,
and all of those green neighbors, and so on. We will end up with a subgraph of all the
2.1. PROOF
17
Figure 2.3: A red-green Kempe chain where the neighbors of u are not linked by the
chain.
green and red vertices in which there is a green-red path to the original green vertex.
The set of all of these vertices is the Kempe chain (which is usually not a “chain,”
but may contain “branches” and even “loops”) [18]. Formally:
Definition 15. Given a graph G, a vertex u, and two colors a and b, an a-b Kempe
chain is the maximal connected subset of V (G) containing u whose vertices are all
colored with either a or b.
If this green-red Kempe chain does not link up with the red neighbor of u, then
we can flip the coloring of this chain such that the green vertices become red and the
red vertices become green, thus freeing up the color green for u as shown in Figures
2.3 and 2.4.
Otherwise, if the green-red Kempe chain from green vertex connects to the red
vertex as seen in Figure 2.5, we consider the blue-yellow Kempe chain starting from
the blue neighbor of u.
There is no way for this chain to link up to the yellow neighbor of u since the
18
CHAPTER 2. KEMPE CHAINS
Figure 2.4: The result after flipping the colors of the Kempe chain from Figure 2.3.
Figure 2.5: A red-green Kempe chain where the neighbors of u are linked by the
chain.
2.1. PROOF
19
Figure 2.6: A case where the 5 neighbors of u are colored with four distinct colors.
Figure 2.7: The case where deg(u) = 5 and two Kempe chains have been considered.
green-red Kempe chain encircles one of these vertices. Therefore, the blue-yellow
Kempe chain can flip its colors and allow u to be colored blue.
If deg(u) = 5, then we remove u from G. By the induction hypothesis, G − {u}
can be colored with four colors. Now reinsert u into G. Here, it is possible that the
neighbors of u are already colored using four colors as seen in Figure 2.6.
We take a similar approach to the problem as we did with the previous case.
Consider the red-yellow Kempe chain and the red-green Kempe chain. If either or
both of these chains do not link up with the corresponding vertex, then we can
alternate the coloring of Kempe chain beginning with the red vertex such that u can
be colored red. Otherwise, we end up with the case illustrated in Figure 2.7.
20
CHAPTER 2. KEMPE CHAINS
Figure 2.8: A counterexample to Kempe’s proof of the Four Color Theorem.
From here, we consider the blue-yellow Kempe chain and the blue-green Kempe
chain starting at the blue vertices. By Figure 2.7 we can see that it is impossible
for both of these chains to link up with its respective vertiex so we can recolor the
Kempe chains starting with the blue vertices and thus u can be colored blue.
This then concludes the “proof”; we’ve shown that one of these four configurations
must exist in every maximal planar graph and that they can all be reduced to a 4coloring [18].
2.1.1
Counterexample
About ten years later, this purported proof was disproven by Percy John Heawood
with a simple example. Figure 2.8 isn’t the example that Heawood himself gave, but
it’s an easier example to see why Kempe’s proof fails.
It we apply Kempe’s method to this example, we would first begin by checking the
red-yellow and green-yellow Kempe chains. Since they both link up, we would then
consider the blue-red and blue-green Kempe chains. By Kempe’s proof we should then
alternate the blue-green Kempe chain coloring on the left and the blue-red Kempe
chain coloring on the right. However, once we do this, we end up with two blue
vertices that are adjacent to each other (the previously green and red vertices on the
2.2. FIVE COLOR THEOREM
21
far left and far right of the figure). This is a contradiction to Kempe’s proof.
Unfortunately, Kempe was unable to fix this error and the Four Color Theorem
remained unproven. However, his work with the theorem helped introduce many
important ideas that were eventually used in the actual proof about a hundred years
later.
2.2
Five Color Theorem
One consequence of Kempe’s proof is the proof of the Five Color Theorem, proven
by Heawood.
Theorem 5 (Five Color Theorem). Given a planar graph G, every vertex can be
colored using at most 5 colors such that no two adjacent vertices share a coloring.
Proof. The proof follows by induction on the number of vertices and is nearly equivalent to Kempe’s proof of the Four Color Theorem. We know from Theorem 3 that
there has to be at least one vertex of degree 5 or less. Let u be that vertex. If
deg(u) < 5, then since it has less than five neighbors, there will always be at least
one color available after the induction step.
If deg(u) = 5 then after removing it, recoloring the graph, and reinserting it,
we may end up with Figure 2.9 where each of the neighbors of u are colored with
five colors such there are no colors remaining for u. In this case, consider the bluegreen Kempe chain. If these chains do not connect, then the blue-green Kempe chain
starting from the blue neighbor of u can be flipped in color such that u can now be
colored blue. Otherwise, do the same thing with the yellow-red Kempe chain. This
Kempe chain cannot cannot be connected it the blue-green Kempe chain is connected
so the yellow-red Kempe chain starting from the yellow neighbor of u can be flipped
in color such that u can now be colored yellow.
22
CHAPTER 2. KEMPE CHAINS
Figure 2.9: The case where deg(u) = 5 and all of its neighbors are colored differently.
Chapter 3
Proof of the Four Color Theorem
In this chapter we will finally go through the main ideas of the proof of the Four
Color Theorem which was proven by Wolfgang Haken and Kenneth Appel on July
22, 1976 with extensive computer assistance [18]. Due to the unorthodox nature of the
proof, there was a lot of skepticism about its validity, but eventually mathematicians
began to accept the proof. Besides a few minor errors, which were quickly corrected,
and a few optimizations, the Appel-Haken proof is now accepted by the mathematics
community.
The proof attempts to find a set of subgraphs that must exist in every maximal
planar graph and then show that each of these can be four colored. In other words,
we will be searching for an unavoidable set of reducible configurations. Before we
begin, note that since the actual proof involved about a 1,000 hours of computer time
that cannot be fully replicated here, we will be going through an outline of the proof
in which all of the main ideas are given.
Definition 16. An unavoidable set is a set of configurations (or subgraphs) such that
at least one of these configurations must exist in every maximal planar graph.
Definition 17. A reducible configuration is a configuration of vertices and edges such
23
24
CHAPTER 3. PROOF OF THE FOUR COLOR THEOREM
that if a graph contains a reducible configuration, then any coloring of the rest of the
graph with four colors can be extended into a coloring of the whole graph.
The first step in Kempe’s alleged proof was to show that every maximal planar
graph must contain a vertex of degree 5 or less. This is an unavoidable set. The next
step was to show that each of these configurations are reducible. Kempe’s proof failed
because although he managed to find an unavoidable set, the configuration with a
5-vertex was not reducible.
3.1
Unavoidable Sets
The method used to find an unavoidable set is called discharging. Assign a charge of
6 − i to every vertex u where the deg(u) = i. For example, a 5-vertex will have charge
1, an 8-vertex will have charge -2, etc. Figure 3.1 gives an example of a planar graph
with its charges. Recall from Theorem 4 that the equation
∞
X
(6 − i)vi = 12
i=1
holds for maximal planar graphs. Notice that due to this equation, the total charge
on the graph must be equal to 12. Also recall that if there exists a vertex of degree
less than 5, then the graph is reducible. Therefore, we only consider graphs where
the minimum vertex degree is 5.
The next step is to “discharge” all of the 5-vertices by sending
1
5
of its charge
to each of its neighbors. Notice that since we’re only moving charges around, no
charge is gained or lost so the total charge on the graph is still 12. Now consider
what happens to the charges of individual vertices in the graph. For example, if a
7-vertex is adjacent to two 5-vertices, then after the discharging, the 7-vertex will
3.1. UNAVOIDABLE SETS
25
Figure 3.1: A planar graph where every vertex has charge 6 − i.
have a charge of −1 + 2( 51 ) = − 35 .
We now want to discover all vertices that will have a positive charge after this
process. A 5-vertex which has 5-vertex neighbors will definitely have a positive charge
since it starts with a charge of 0 and recieves
1
5
charge from each of its 5-vertex
neighbors. Therefore, we can now disregard any cases where a 5-vertex has any
5-vertex neighbors since we know these will end up with a positive charge.
After eliminating these cases, how does the charge change for an arbitrary vertex?
A 5-vertex will give up all of its charge and won’t receive any since it’s not adjacent
to any other 5-vertices, so it will end up with a charge of 0.
If i is even, then an i-vertex can have at most
i
2
5-vertex neighbors. This is
because the neighbors of the vertex form a ring, and we assumed that no 5-vertex
was adjacent to another 5-vertex, therefore every other vertex in the ring cannot be
a 5-vertex. Thus the charge on an even i-vertex after discharging is at most
i 1
(6 − i) + ( )( )
2 5
26
CHAPTER 3. PROOF OF THE FOUR COLOR THEOREM
which simplifies to
6−(
9
)i
10
If i is odd, then an i-vertex can have at most
(3.1)
i−1
2
5-vertex neighbors for similar
reasons to the even i. Thus the charge on an odd i-vertex after discharging is at most
(6 − i) + (
i−1 1
)( )
2
5
which simplifies to
59
9
− ( )i
10
10
(3.2)
For what values of i will the charge on an i-vertex be positive after discharging?
By Equation 3.1
6−(
9
)i > 0
10
9i < 60
i<
20
3
and since i must be an integer, i ≤ 6.
By Equation 3.2
59
9
− ( )i > 0
10
10
9i < 59
i<
59
9
and since i must be an integer, once again i ≤ 6.
Assuming that there are no vertices of degree less than 5, this means that every
3.2. REDUCIBLE CONFIGURATIONS
27
Figure 3.2: The Birkhoff diamond.
graph has to have a 5-vertex adjacent to a 5-vertex or a 5-vertex adjacent to a 6-vertex
since these are the only configurations that will maintain a positive charge after
discharging. Therefore, using this technique, we’ve obtained another unavoidable
set: 1-vertex, 2-vertex, 3-vertex, 4-vertex, 5-vertex adjacent to a 5-vertex, 5-vertex
adjacent to a 6-vertex.
By using other, more complicated methods of discharging, the final proof by Appel and Haken found an unavoidable set with 1482 configurations. All in all, 487
discharging rules were used to generate this set. Through optimizations, this number
was eventually reduced to 633 configurations and only used 32 discharging rules.
3.2
Reducible Configurations
The next step is to prove that all of these configurations are reducible. The best way
to show how more complicated reductions are done is through an example. We will
be working with the Birkhoff diamond illustrated in Figure 3.2.
Let G be the graph with n vertices in which the Birkhoff diamond is a subgraph.
28
CHAPTER 3. PROOF OF THE FOUR COLOR THEOREM
Figure 3.3: The Birkhoff diamond after it has been shrunk down to a smaller configuration.
Case v1
1
R
2
R
3
R
4
R
5
R
6
R
v2
G
G
G
G
G
G
v3
B
B
B
R
R
R
P
Y
Y
Y
Y
Y
Y
v5
R
G
B
R
G
B
Table 3.1: A table of all 6 possible colorings of the shrunk Birkhoff diamond.
To show that the Birkhoff diamond is reducible, we shrink it down to a smaller
configuration as shown in Figure 3.3. By induction, this new graph has less than n
vertices so it is four colorable. Essentially, the shrinking is done by combining the
vertices v4 , v6 , a, b, c, d into one vertex P . Whatever we end up coloring P will be what
we end up coloring v4 and v6 in G since they will be adjacent to the same neighbors.
Now there are 6 different ways to color this configuration (up to isomorphism) as
demonstrated in Table 3.1
Now after bringing back the original Birkhoff diamond, five of these colorings can
be extended fairly easily without a need for recoloring as shown in Table 3.2. In the
fourth case (as illustrated in Figure 3.4), it is impossible to color a, b, c, d such that
3.2. REDUCIBLE CONFIGURATIONS
Case v1
1
R
2
R
3
R
4
R
5
R
6
R
v2
G
G
G
G
G
G
v3
B
B
B
R
R
R
v4
Y
Y
Y
Y
Y
Y
29
v5
R
G
B
R
G
B
v6
Y
Y
Y
Y
Y
Y
a b
B Y
G Y
B Y
c
R
R
R
d
G
B
G
B
B
G
G
R
R
Y
Y
Table 3.2: A table of all 6 possible colorings of the original Birkhoff diamond.
Figure 3.4: A case where the coloring for Birkhoff diamond cannot be extended.
no two adjacent vertices share a color.
This is easy to see by looking at the a-c-d-triangle. All three of these vertices are
adjacent to red and yellow vertices, so they have to be colored green or blue. However,
these vertices form a triangle and are therefore mutually adjacent and require 3 colors
which is impossible. Therefore, some modification has to be made.
The modification can be made by a Kempe chain argument. First consider the
red-blue Kempe chain from v5 . If this chain does not connect to either v1 or v3 then
we can alternate the coloring of this chain such that v5 is now blue. Then we are in
case 6 which has already been solved. Otherwise, consider the yellow-green Kempe
chain from v4 . If the previous chain exists, then this chain cannot be connected to
30
CHAPTER 3. PROOF OF THE FOUR COLOR THEOREM
Figure 3.5: The final coloring for the problematic case of the Birkhoff diamond.
v6 , so we can alternate the coloring of this chain such that v4 is now green.
This new coloring can now be extended by coloring a green, b blue, c yellow, and
d blue as shown in Figure 3.5. Therefore, since all possible colorings of the shrunk
Birkhoff diamond can be extended into a good coloring, we have shown that the
Birkhoff diamond is reducible. [6]
By the use of this technique and other techniques similar to this, the 1482 (or
633) unavoidable configurations were proven to be reducible. The job of checking
whether configurations were reducible or not was mainly left for the computer to
solve [18]. This concluded the proof of the Four Color Theorem since an unavoidable
set of reducible configurations was found.
Chapter 4
Extensions
This section will focus on extensions of the Four Color Theorem. While an obvious
extension to this problem seems to be working in 3-D space with 3-D regions, such
as spheres and cubes, this problem turns out to be trivial. It is possible, to construct
a 3-D map where an arbitrarily large number of regions are all mutually adjacent to
one another. One way to visualize this is to think of the regions as pipe cleaners.
With enough coiling and twisting, an arbitrary number of pipe cleaners can be made
to be mutually adjacent to each other.
However, it turns out that coloring 2-D maps on different surfaces is a nontrivial
question. For example, how many colors do you need to color a map on a torus? The
following extensions generalize the Four Color Theorem into a map coloring theorem
which applies to maps (and therefore graphs) on all surfaces.
4.1
Heawood Conjecture
Before we begin, we need to define a few terms.
Definition 18. Given a surface S, the genus n is the largest number of noninter31
32
CHAPTER 4. EXTENSIONS
secting simple closed curves that can be drawn on the surface without separating it,
typically denoted Sn . [12]
In other words, the genus can be thought of as the number of handles or holes in
a surface. For example, a sphere has genus 0 while a torus has genus 1.
Definition 19. Given a planar graph G embedded onto a surface Sn , the Euler
characteristic χ of G is defined as χ = V − E + F where V is the number of vertices,
E is the number of edges, and F is the number of faces. [10]
Definition 20. A surface is considered orientable if when a direction is carried around
a closed path on the surface, the direction is always consistent.
Definition 21. A surface is considered nonorientable if there exists a closed path on
the surface such that if a direction is carried around the loop, the direction is reversed
when moved around this path.
A sphere or a torus are orientable because if you draw a loop around the surface,
the direction is always consistent. However, a Möbius strip is nonorientable because
if you draw a loop down the middle of the band, the direction changes after one
loop around. Another way to think about this is that an orientable surface has an
“outside” and an “inside” while a nonorientable surface does not.
Given an orientable surface with genus n
χ = 2 − 2n
Given a nonorientable surface with genus n
χ=2−n
4.1. HEAWOOD CONJECTURE
33
Since a plane (or sphere) is an orientable surface of genus 0, a graph G on a plane
has χ = 2 − 2(0) = 2, which means that V − E + F = 2 and is consistent with Euler’s
Formula from before.
Two more quick definitions before the theorem.
Definition 22. The floor of n, denoted bnc is the largest integer less than or equal
to n.
Definition 23. The ceiling of n, denoted dne is the smaller integer greater than or
equal to n.
And finally, the big theorem.
Theorem 6 (Heawood Conjecture). Given a planar graph G on an orientable surface
Sn with genus n, G can be colored with
γ(n) =
7+
√
1 + 48n
2
colors where γ(n) is called the chromatic number of n.
This can be generalized to orientable and nonorientable surfaces by replacing the
genus with the Euler characteristic for orientable surfaces to get
γ(χ) =
7+
√
49 − 24χ
2
The proof of this follows almost directly from the first equation so we will just be
proving the orientable case in this paper.
For example, consider the torus. The torus is an orientable surface with genus
1. Therefore, by the conjecture, the torus can be colored with 7 colors and such a
coloring is shown in Figure 4.1. The Möbius strip is a nonorientable surface of genus
34
CHAPTER 4. EXTENSIONS
Figure 4.1: A maximal coloring of a torus with 7 colors [14].
Figure 4.2: A maximal coloring of a Möbius strip with 6 colors [17].
1. Therefore it has an Euler characteristic of 1 and by the conjecture, it can be colored
with 6 colors as shown in Figure 4.2.
4.1.1
Proof
The proof is twofold. We will first prove sufficiency and then prove necessity.
Sufficiency. Let G be a maximal graph on an orientable surface with genus n > 0
(the n = 0 is a special case to be explained later). Let D be the average degree of
the vertices in G. That is
PV
D=
i=1
deg(vi )
V
4.1. HEAWOOD CONJECTURE
35
where vi ∈ V (G). Now since G is maximal
DV = 2E = 3F
And since G is on an orientable surface
χ = V − E + F = 2 − 2n
We the combine these equations to get
V −
DV
DV
+
= 2 − 2n
2
3
Which simplifies to
D =6+
12(n − 1)
V
Now notice that D ≤ V − 1 since every vertex can be at most connected to every
other vertex. Thus
6+
12(n − 1)
≤V −1
V
Which simplifies to
V 2 − 7V + 12(1 − n) ≥ 0
(4.1)
Now by the quadratic equation, the roots for V are
7±
p
√
49 − 4(12)(1 − n)
7 ± 1 + 48n
=
2
2
Recall that the genus n > 0. Since
√
7− 1+48n
2
< 0 when n > 0 and having a negative
number of vertices doesn’t make any sense, we only need to consider the case where
36
V ≥
CHAPTER 4. EXTENSIONS
√
7+ 1+48n
.
2
And since we only care about integers, we end up with
V ≥
For simplicity, let M =
√
7+ 1+48n
.
2
7+
√
1 + 48n
2
If V = bM c, then we have the same number of
colors as vertices and the coloring is trivial. If V > bM c then it follows that V > M
(since V is an integer). Recall that D = 6 +
D =6+
12(n−1)
.
V
By substitution, we get
12(n − 1)
12(n − 1)
<6+
V
M
(4.2)
By substituting into the Equation 4.1, we also get
M 2 − 7M + 12(1 − n) ≥ 0
Which simplifies to
M −7≥
12(n − 1)
M
Which implies that
6+
12(n − 1)
≤6+M −7=M −1
M
Once we combine this with Equation [?] we end up with
D <M −1
Recall that D was the average degree of a graph. This means that there exists a
4.1. HEAWOOD CONJECTURE
37
vertex u in G such that deg(u) < M − 1 or
deg(u) ≤ bM c − 1
Now let G0 = G − {u}. Notice that G0 has V − 1 vertices. If V − 1 = bM c then
we can color G0 and when we reinsert u, u was adjacent to at most bM c − 1 colors
so there remains at least one color for u. If V − 1 > bM c, then we repeat the same
steps in order to get a vertex v in G0 such that deg(v) ≤ bM c − 1 We can then let
G00 = G0 − {v} and repeat the process. Eventually we’ll end up with a graph H with
bM c vertices and can build up the graph by replacing one vertex at a time. Since
each reinserted vertex has degree less than or equal to bM c − 1, there will be a color
remaining for every vertex. This proves sufficiency.
The proof of necessity requires a theorem that we won’t prove in this paper. It
involves proving a number of cases and the proof of this theorem is probably worthy
of being its own thesis (for more on this proof, see Ringel and Young). In order to
demonstrate how complicated this proof is, consider the work it took to finally prove
it.
In 1891, L. Heffter proved the formula for all n ≤ 12: he did this by
exhibiting maps that require the number of colors given by the formula in
these cases. In 1952, Ringel proved it for n = 13, and in 1954 for all n of
the form 12k + 5. In 1961, Ringel proved it for all n of the form 12k + j
for j = 7, 10 and 3. In 1963, G. Gustin proved it for 12k + 4. In 1963,
C. M. Terry, L. R. Welch and J. W. T. Youngs solved it for the case 12k.
Between 1963 and 1965, Gustin and Youngs did the cases 12k + 1 and
12k + 9. In 1966 Youngs did the case 12k + 6. In 1967, Ringel and Youngs
38
CHAPTER 4. EXTENSIONS
solved the remaining cases: 12k + j for j = 2, 8 and 11.
But wait! that’s not the end of the story. There were some exceptional
cases that were not covered in the above solutions. There were still the
cases n = 18, 20, 23, 30, 35, 46 and 59. Cases 18, 20 and 23 were disposed
of in 1967 by Jean Mayer, a professor of French literature. In 1968, Ringel
and Youngs gave a lecture on the subject at a graph theorist’s meeting in
Michigan. One of those attending, Richard Guy, solved the case n = 59
the night after the lecture. The cases n = 35 and 47 were done later by
Ringel and Youngs. The final case, n = 30, was solved independently by
Mater and Youngs in 1968. [2]
Note that these proofs are only for the orientable surfaces. A similar amount of
work had to be done to prove the nonorientable cases. For reasons of time and space
the following theorem will be left unproven in this paper.
Theorem 7. The genus of a complete graph KV is
l
(V −3)(V −4)
12
m
.
Now we’re ready to prove necessity.
Necessity. In order to prove necessity, we want to show that given a surface Sn , there
j √
k
exists a planar graph G on that surface such that at least γ(n) = 7+ 1+48n
colors
2
are needed to color the graph. We know that graphs are hard to color when there are
lots of edges. Therefore, we can try to find the largest complete graph KV such that
KV can be embedded onto a surface of genus n. Notice that KV requires V colors to
color since every vertex is mutually connected to every other vertex. By Theorem 7
l
m
l
m
−4)
(V −3)(V −4)
we know that the genus of KV = (V −3)(V
.
Therefore,
n
≥
. Assume
12
12
4.1. HEAWOOD CONJECTURE
39
that V is the largest V for which this inequality holds. Since
(V − 3)(V − 4)
(V − 3)(V − 4)
≤
≤n
12
12
It follows that
n≥
(V − 3)(V − 4)
12
In addition, since V is “maximal,” this means that KV +1 cannot be embedded as a
planar graph onto Sn . Therefore
((V + 1) − 3)((V + 1) − 4)
(V − 2)(V − 3)
n<
=
12
12
Which implies that
n<
(V − 2)(V − 3)
12
Combining these two inequalities, we end up with
(V − 2)(V − 3)
(V − 3)(V − 4)
≤n<
12
12
Which we can simplify to
V 2 − 7V + 12 ≤ 12n < V 2 − 5V + 6
By applying the quadratic formula to the left-hand side of the inequality, we end up
with
V ≤
7+
p
√
49 − 4(12 − 12n)
7 + 1 + 48n
=
2
2
40
CHAPTER 4. EXTENSIONS
Doing the same on the right-hand side of the inequality, we end up with
V >
5+
p
√
√
25 − 4(6 − 12n)
5 + 1 + 48n
7 + 1 + 48n
=
=
−1
2
2
2
We can combine these two inequalities to get
7+
√
√
1 + 48n
7 + 1 + 48n
−1<V ≤
2
2
Or in other words,
V =
7+
√
1 + 48n
2
Since we’re working with complete graphs, KV requires V colors so this proves necessity.
This concludes the proof of the Heawood Conjecture since both sufficiency and
necessity have been proven. [6]
However, there’s a slight problem with this proof; it only holds for surfaces Sn
where n > 0. It fails for n = 0 in the middle part of Equation 4.2
6+
12(n − 1)
12(n − 1)
<6+
V
M
Where if n = 0,
6−
12
12
<6−
V
M
This simplifies to
V <M
This is a contradiction since we showed that V > M . However, the Appel-Haken
proof shown in Chapter 3 of the Four Color Theorem proves the case when n = 0.
4.1. HEAWOOD CONJECTURE
41
Therefore the Heawood Conjecture is now true for all surfaces with genus n ≥ 0.
Unfortunately, there is one sole exception to the Heawood conjecture which is the
Klein bottle. The conjecture predicts the the Klein bottle, a nonorientable surface of
genus 2 and χ = 0, to require 7 colors, but actually, it only needs 6. The reasons for
this has to do with the way Theorem 7 was proven. For more on this, see Ringel and
Young.
42
CHAPTER 4. EXTENSIONS
Chapter 5
Conclusion
In this chapter, we will explore the ramifications, both positive and negative, that
the theorem and proof has had on the mathematical community.
While there are very few, if any, practical uses for the theorem, the numerous
attempts to solve it has advanced mathematics, especially the field of graph theory.
It has led to discussions on the nature of a proof and has changed the way mathematicians view proofs.
The main criticisms come from mathematicians who believe that the proof of the
Four Color Theorem is not valid since it has not been fully verified by humans. While
there are only a handful of skeptics who deny that the theorem is true, there are some
who claim that the proof is not as good as the proof of, say, Fermat’s Last Theorem.
There is concern that a proof using computers is more akin to a scientific experiment:
valid and almost certainly true, but without the same degree of certainty that a pencil
and paper proof has. As John Horgan put it, a ”“proof which offers a probability of
truth - and not certainty - is an oxymoron [5].”
Since a lot of the debate seems to be focused on whether or not the proof of the
Four Color Theorem is a valid proof, it is useful to first define what a proof means
43
44
CHAPTER 5. CONCLUSION
for mathematicians. The Concise Oxford Dictionary of Mathematics defines a proof
as a “chain of reasoning, starting from axioms, usually also with assumptions on
which the conclusion then depends, that leads to a conclusion and which satisfies the
logical rules of inference [3].” By this definition, the Four Color Theorem seems to
be proven. There is some question as to whether using a computer for computations
still qualifies as “a chain of reasoning,” but otherwise the rest of the criteria seem
to be met. Note that for practical reasons, this definition is not necessarily followed
strictly by mathematicians. It is impractical, and sometimes impossible, to express a
proof based on axioms simply due to the fact that the proof may be too long.
Another big criticism is that even if the theorem has been proven and the proof
is accepted, nothing was gained from the proof. Since the proof was completed in
a computer, there is very little insight on what was happening within the computer
and questions remain such as: Why four colors? Why not seven? There are no
insights to be gained since there is no way to understand what the computer was
“thinking” when it proved the theorem. In other words, we still do not know why the
Four Color Theorem is true, we simply know that it is. A large motivation for many
mathematicians is the moment of clarity when everything clicks and all the pieces fall
into place. The fact that this proof denies this clarity to mathematicians is probably
a large reason for the pushback against it.
It is interesting to note that the proof of the classification of simple finite groups
is approximately 15,000 pages in length, yet there is barely any criticism against the
proof. Even though it is next to impossible for any one individual to fully verify the
proof, since the individual sections were completed by humans and not by computers,
it is considered to be more “valid” than the proof of the Four Color Theorem.
Another way to look at this debate is to ask: What is it that mathematicians
do? Paul Erdös once said that a “mathematician is a device for turning coffee into
45
theorems.” While this whimsical saying is almost certainly true, it remains a bit
vague. Wilson Thurston explains that mathematicians usually aren’t looking for the
“answers,” as one might expect, but rather they are looking for understanding [9].
If you tell mathematicians that “Theorem X” is true, without a doubt, they would
still want to know why the theorem is true. This is related to the clarity argument
above. In a similar vein, having a computer output that the Four Color Theorem is
true doesn’t explain why it’s true.
There is another argument that computers can “make mistakes,” but this argument is not as persuasive since the same argument could be made for human proofs.
A perfect example of this is Kempe’s “proof’ which contained a mistake that was
undiscovered for about ten years.
There are still mathematicians who are attempting to find a more standard proof
of the Four Color Theorem. Due to the simplicity in which the theorem can be
stated and proved, there is an intuition that a simpler proof exists. However, many
doubt that such a proof exists, and even if it did, the controversy surrounding a
computer-assisted proof would still exist since many other theorems have been proven
using computers since then. Whether or not one accepts the proof of the Four Color
Theorem, it appears to have become a turning point that may shape the structure of
future mathematics.
46
CHAPTER 5. CONCLUSION
Bibliography
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[2] David Barnette. Map Coloring, Polyhedra, and the Four-Color Problem. Mathematical Association of America, Washington D.C., 1983.
[3] Christopher Clapham and James Nicholson. ”proof”. The Concise Oxford Dictionary of Mathematics, 2009.
[4] Georges Gonthier. Formal Proof – The Four-Color Theorem. Notices of the
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