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Activity 2-7:
The Four Colour Theorem
You want
to shade
the map of
countries
on the right
so that no two
countries
sharing a
border
are shaded
with the
same colour.
How many colours do
you need?
Worksheet here...
(We won’t worry about the sea...)
It turns out that
FOUR will do.
This section shows us that
we certainly need four.
Actually,
including the
sea is no
problem.
But will four colours do for ANY map?
Might there be some configuration of countries
so that five colours are essential? Six?
Note: the boundary that two countries share
has to be more than a set of single points…
In fact, it requires three.
otherwise, this would
require nine colours.
This problem has a fascinating history...
In 1852, Frances Guthrie (left) discovered
that he seemed to be able to four-colour
any map of countries.
In 1879, a mathematician called Kempe
claimed to have proved this, but in 1890,
Heawood found a fatal error in his argument.
In 1922, Franklin proved that four colours would suffice for
every map with 25 or fewer countries. This number was
gradually increased over the years, to 95 in 1976.
The big breakthrough came in
1976, from Appel and Haken.
Their proof was remarkable,
being the first to rely heavily
on the use of a machine. They
reduced the problem to a
large number of special cases,
which were then checked by a
computer, in a calculation
that would be completely
unfeasible for a human being
to carry out.
Philosophically, this was a problem
for the mathematical community.
With every previous proof, it had been possible to check it
until it was validated by mathematicians in the field.
This was impossible here –
the computer gave us a simple ‘Yes’ or ‘No’,
leaving us with the option
to accept or reject this answer.
In the time since then, their proof has been replicated
by much more powerful computer methods,
so we can be sure now that the result is true.
We will prove here a much easier result –
that every map is six–colourable.
We need a helping hand or two with this, starting with a result
called Euler’s Theorem (he had a lot of these!).
Suppose we have a connected graph
(a collection of vertices and edges
so that you can travel from any vertex
to any other vertex along edges).
Let V = # of vertices, F = # faces,
and E = # edges.
Euler’s Theorem tells us that
V + F – E = 2 (including the
outer region as a face).
Certainly if we have no edges then this is true,
since V = 1, F = 1, E = 0  V + F – E = 2.
Suppose we know that Euler’s Theorem holds
for all connected graphs with n edges.
Take any connected graph with n + 1 edges.
If there is an edge that connects two vertices,
amalgamate the vertices.
V decreases by 1, E decreases by 1, F stays the same.
If the only edges available are loops,
then remove the loop.
E decreases by one, F decreases by 1, and V stays the same.
Either way, V + F – E stays the same, and so must be 2.
To help us further with our six-colourable proof,
We will define a standard map.
This is one where
every vertex is of
degree three.
A non-standard map…
Lemma: every standard map has a face with 5 or fewer edges.
Suppose every face has 6 or more edges.
So 6F  2E (since each edge is in two faces).
But if every vertex is of degree 3, then 3V = 2E,
since each edge gets counted twice in the 3V.
So V + F – E  2E/3 + E/3 – E = 0, a contradiction!
So every standard map must have
a face with 5 or fewer edges.
For our next step, we can use Proof by Induction (again!)
Clearly a map with one face is six-colourable.
Suppose all standard maps with n faces are six-colourable.
Pick any standard map with n+1 faces.
One of these faces must have 5 edges or fewer
(by our lemma).
We can now remove
this face, while
keeping the map as
standard.
We now have an n-faced map, which we know is six-colourable.
Carry out the colouring, and then replace the region.
There can be at most five colours
surrounding it,
so we can use the sixth to colour it.
So we have by Induction that
every standard map
is six-colourable.
But … what if our map is not standard?
This is easy.
Replace every vertex of degree greater than 3
with a small circle, as above.
We now have a standard map – six-colour it.
Now remove the circles – this will not damage the six-colouring.
So we are done – every map can be six-coloured.
It is possible to run through this argument again
with added precision and a number of extra remarks
to prove that every map is five-colourable.
If you feel like a challenge, visit
The five-colour theorem
This involves converting our map into a graph first.
A simple proof that every map is four-colourable,
however, remains tantalisingly out of reach!
With thanks to:
Wikipedia, for a very helpful article.
Worldatlas for their map of Africa.
William Tross for his Nature of Mathematics site.
The Proofwiki site.
David Eppstein for his Geometry Junkyard site.
Carom is written by Jonny Griffiths, mail@jonny-griffiths.net