660 THEOREM IN GEOMETRY OF POSITION. 
rule, part of the region of the figure in which these two polygons are contained. 
It is easy to see the connection of this method with the idea of a galvanic 
circuit of unit strength circulating (say right-handedly) in each of the poly- 
gons :—and the treating of any new or unused line as a conductor which can, 
when necessary, be split into two traversed by equal and opposite currents. 
It is probable that the known laws of such currents in a network may lead to 
the proof of the existence of a single polygon when the figure is a projection 
of a polyhedron. 
7. Another method is suggested by Mr Kempr’s solution of the map-colour- 
ing problem (Nature, vol. xxi. p. 399). As the number of districts is, necessarily, 
n+2, and the aggregate number of their sides 6, there must always be at least 
one district with fewer than six sides. Now, one side may be erased from a 
district of two or of three sides, and restored again, without altering the nomen- 
clature of the remaining lines. Similarly, either pair of opposite sides of a four- 
sided district may be erased, and afterwards restored. But when we erase any 
two non-adjacent sides of a five-sided district, a condition is thereby imposed 
on the nomenclature of the remaining lines, with which I do not yet see how 
generally to deal. ) 
8. An immediate consequence of the theorem is that, in any network of 
triangles (however many lines meet at a point) the sides of each triangle belong 
one to each of three groups into which the whole set of lines can be divided. 
The theorem itself follows, conversely, if this proposition be independently 
proved. 
9. In No. 494 of the Astronomische Nachrichten, CLAUSEN has a problem 
closely connected with the present subject. It refers to the minimum number 
of separate strokes of a pen by which a given figure consisting of lines can be 
drawn. LisTINnG, in his Vorstudien zur Topologie, has shown how to find this 
minimum number by counting the points at which an odd number of lines meet. 
In our present proposition, if one polygon can be found containing all the 
points, 7¢ and one of the unused lines together form one penstroke, and the 
remaining group of x—1 unused lines forms the rest. If there be two polygons, 
they and one of the unused lines together form one penstroke. And so on. 
10. To apply the result above to the problem of map-colouring, insert a new 
district surrounding each point of the map where more than three boundaries 
meet. Then divide the boundaries, which now meet in threes, into three groups 
as above. (The excepted case obviously cannot arise). Now let O separate 
the colours A and B, or C and D; |, A and C, or B and D; and II, A and D, 
or Band C; and the thing is done. For we may now suppose the inserted 
districts to become smaller, till they vanish. 
