Maths Item of the Month – May 2013
The two colour theorem?
Given n circles in the plane, prove that no matter how these circles are arranged, the map
which they form may be properly coloured with two colours.
This problem is taken from a 2012 MEI Conference session. For details of the 2013 MEI
conference see: www.conference.mei.org.uk.
Solution
An explanation of this can be shown by viewing the map as a graph. We take each
intersection of lines to be a node, and each sub-line between intersections to be an edge.
Call this graph G .
We can assume G to be a connected graph, as any seperate components can be regarded
as isolated cases, independent of surroundings. For this reason we can also ignore any nonintersecting circles.
Notice that every node has an even degree. This proof applies to any graph with nodes of an
exclusively even degree, so a map made entirely from closed shapes, or infinite lines/curves
will also be two-colourable.
Consider G ' , the dual of graph G . As every node of G has an even degree, every face of
G ' will have an even number of edges. Also, every node in G ' will have a degree of at least
2, as we are dealing with at least two connected circles. The result of these two points is that
every node can be described as part of a cycle, and that every possible cycle contains an
even number of nodes.
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We can therefore pick any node, and alternately colour all of the nodes of G ' , without
reaching contradictions (as even cycles will always result in the starting colour). This then
corresponds to a proper two-colour of G .
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