Four-colour problem/ Four-colour conjecture → Four-colour theorem
NOTES
Can be expressed in a variety of different ways, lets begin with an informal version:
A map consists of countries bounded by simple closed curves, two countries are adjacent if they
have a common border which contains at least a segment of a curve, not just a finite collection of
isolated points.
We wish to colour the countries so that any two adjacent countries have different colours.
The problem is how many colours we need in order to be able to colour all maps this way.
In order that the problem should have an answer, need to make a few restrictions on what
constitutes a map, for example:
- Each country must be connected, else there is no bound on the number of colours which might be
required.
- Outside region of the map, also considered a country.
- Countries have to be contiguous. If not contiguous, it is simple to draw a map that needs 5 colours.
Officially this question was first raised by Francis Guthrie in 1852.
Before 1852, never been raised by map-makers
F. Guthrie was trying to colour a map of the counties of England, he found it helpful to have a map
where every bordering county had a different colour, and wondered how few colours he could use
and still do this.
Three colours wouldn’t work. Easy to draw a map where 4 colours are certainly necessary; draw 4
countries that are adjacent to each other of the other three.
He found he could make do with four. So he wondered, would 4 colours be enough for any map??
He asked his brother Frederick, who explained the problem to his math teacher De Morgan,
University College London, who showed it to Cayley. –NO PROOF!!!!
Nothing happened in 26 years.
Problem raised again in 1878. Cayley posed a question to the London mathematical Society if it had
been solved. Answer-NO!
In 1879 he made a paper on the difficulties that lie in attempting to prove the conjecture.
In 1879 Kempe provided a clever argument which purported that 4 colours were sufficient, it was
published in American Journal of Mathematics that same year.
Another proof was given by Peter Tait in 1880.
Both proofs were thought to be true and were unchallenged for some years.
1890: Heawood discovered an error in Kempe’s proof.
1891: Proof by P. Tait was incomplete.
Both false proofs did have some value. Kempe discovered what became known as
“Kempe chains”: Largest set of countries you can get to from a given place by keeping to countries
of a particular two colours. Tait found an equivalent formulation of the four CT in terms of threeedge colouring.
Heawood showed that Kempe’s proof only showed that 5 colours were sufficient.
The problem remained open. It was generally thought that the problem was not seriously hard, and
would soon be solved.
To give an idea of the attitude: A renowned mathematicians, H. Minkowski once told his students
the problem remained unsolved because third-rate mathematicians has worked on it, only to admit,
later, that: “heaven is angered at my arrogance; my proof is also defective”.
All the attempted proofs of the four colour problem were reduced to checking many, many specially
constructed maps. Determining which special maps need to be checked and then checking them
leaves enormous room for mistakes.
1976: K. Appel, W. Haken, Koch developed the first proof of the problem that withstood the test of
time. They were able to reduce the number of special maps to something manageable. Required a
computer + 1200 hours of computing time to complete it.
Two main ingredients in the proof: discharging: producing a long list of local
obstructions/unavoidable configurations and reducibility; configuration can be reduced to a smaller
case, which we by induction can assume to be 4-colourable.
1996: researchers were set out to verify the Appel-Haken proof. Soon gave up. They developed
their own greatly simplified proof. Still required a computer.
In 1976: thought of using a computer in a proof was considered outrageous. Evoked heated debate.
“The death of proof”. As mathematicians have grown more comfortable with computers, the
concern about their use in proofs has subsided.

To understand the principles of 4CT, we must know some basic graph theory.
- A graph consists of a set of vertices and a set of edges.
- Associated to each edge is two distinguished vertices, endpoints + they are said to be adjacent.
- Two ends are allowed to coincide, the edge is called a loop.
- Each vertex is represented by a point in the plane
- Each edge is represented by a continuous curve between its two ends.
- An edge connects two ends.
- Two edges which join the same pair of vertices are called parallel.
- 2 graphs are considered to be the same if there is a one-to-one correspondence between their
vertices and ends, preserving the ends.
- Degree of a vertex: number of edges which are incident to it.
- A complete graph on n vertices, denoted Kn, has one and only one edge between any pair of
distinct vertices.
- Planar graphs; graphs that can be represented in the plane without any edges intersecting away
from the vertices.
- A graphical representation of a planar graph without intersections is called a plane drawing.
- Faces of a plane drawing of a planar graph G are the regions of the plane that are separated from
each other by the edges of G.
- Unbounded face
- A graph with no edges has a single face.
Theorem, Eulers formula: G is a plane drawing of a connected planar graph. v, e & f denote the
number of vertices, edges and faces. Then, v – e + f = 2.
- Corollary: any planar graph contains a vertex of degree at most 5.
- The notion of duality is very important because it allows us to relate the colouring of faces to
colouring of vertices.
- The dual graph is constructed by placing exactly one vertex in every face of the graph.
The four colour conjecture is equivalent to prove that no planar graph (graph must be planar!)
requires more than four colours to colour its vertices. Can also be expressed in terms of the dual
graph: any map can have its faces coloured with at most four different colours, such that no two
faces that share an edge are the same colour. (A map is a planar graph with no parallel edges)
THEOREM: All maps/ all planar graphs are 6-colourable:
- proceed by induction on the number of vertices n.
- G is a planar graph with n vertices.
- If n ≤ 5, the G is 6-colourable since there are less than six vertices to colour….theorem holds for
the base case.
- Assume all graphs with n-1 vertices are 6-colourable.
- According to corollary, G contains a vertex v of degree most five.
- remove v and all edges connected to it.
- The graph left has n-1 vertices, and therefore is 6-colurable.
- Return the vertex and all the edges that were previously present (at most 5) to the graph and colour
v a different colour from vertices connected to it.
- Thus there exists a colouring of G with at most 6 colours.
Heawood’s five-colour theorem:
Lemma: Let G be 5-colourable. If a country has five neighbours (five edges which are connected to
five vertices) then the map can 5-coloured in such a way that only 4 colours are used for these 5
neighbours.
Proof: (Using Kempe chains!). Five colours we need: red, orange, blue, black and green in cyclic
order. If the red and blue countries are not in the same red-blue chain, then we can swap the colours
in one of these chains, and achieve the desired colouring. But if they are in the same red-blue chain ,
then this separates the orange and black countries from each other, which are, therefore, not in the
same orange-black chain, so one of them can be re-coloured.
THEOREM: All maps are five-colourable.
Proof: We can choose a country with five or fewer (using corollary). We can remove this vertex.
Then we obtain a map with fewer countries, which can be five-coloured, by induction. Then we put
back the vertex and choose a colour for it. If it has four or fewer neighbours, then these vertices use
up at most four colours, so there is always one left. Similarly, if there are five neighbours, then the
lemma says you only need four colours for them, so there is still one left.
Generalizations:
- consider colouring problem on surfaces other than the plane.
- Problem on sphere or cylinder is equivalent to that on the plane.
- For closed (orientable or non-orientable) surfaces with positive genus, the max. number k of
colours needed depends on the surface’s Euler characteristic χ according to the formula:
k = [(7 + √ (49 - 24χ)) / 2],
where the outermost brackets denote the floor function. The only exception to the formula is the
Klein bottle. Has Euler characteristic 0 and requires 6 colours.
Alternatively, for an orientable surface the formula can be given in terms of the genus of a surface,
g:
k = [(7 + √ (1 + 48g)) / 2].
For example, the torus has Euler characteristic χ = 0 (and genus, g = 1) and thus k = 7, so no more
than 7 colours are required to colour any map on a torus.