Activity 2-7: The Four Colour Theorem

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You want to shade

the map of countries

on the right so that no two

countries sharing a

borderare shaded

with the same colour.

How many colours do you need?

(We won’t worry about the sea...)Worksheet here...

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 countriesso that five colours are essential? Six?

Note: the boundary that two countries sharehas to be more than a set of single points…

otherwise, this would require nine colours.

In fact, it requires three.

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 ituntil 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…

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!

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).

So every standard map must have a face with 5 or fewer edges.

For our next step, we can use Proof by Induction (again!)

We can now remove this face, while

keeping the map as standard.

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).

There can be at most five colours surrounding it,

so we can use the sixth to colour it.

We now have an n-faced map, which we know is six-colourable.

Carry out the colouring, and then replace the region.

But … what if our map is not standard?

So we have by Induction that every standard map

is six-colourable.

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 againwith added precision and a number of extra remarks

to prove that every map is five-colourable.

A simple proof that every map is four-colourable,however, remains tantalisingly out of reach!

If you feel like a challenge, visit

The five-colour theorem

This involves converting our map into a graph first.

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