Activity 2-7:

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www.carom-maths.co.uk
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 wont worry about the sea...)
Worksheet here...
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It turns out that FOUR will do.
This section shows us that we certainly need
four.
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Actually, including the sea is no problem.
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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.
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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.
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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.
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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.
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We will prove here a much easier result that
every map is sixcolourable.
We need a helping hand or two with this, starting
with a result called Eulers 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. Eulers Theorem tells us that V F
E 2 (including the outer region as a face).
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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 Eulers 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.
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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.
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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.
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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 n1 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.
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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?
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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.
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It is possible to run through this argument
againwith added precision and a number of extra
remarksto 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!
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With thanks toWikipedia, 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_at_jonny-griffiths.net