For a cubic map:
4C2 + 3C3 + 2C4 + C5 - C7 - 2C8 - 3C9 ... = 12
represents the number of countries, or faces, in a cubic map that have n-sides. So,
C4 is the number
C5 the number of pentagons, etc...
This is an important consequence of Euler's Polyhedra formula.
To derive this result, begin by noting that the total number of countries is:
F = 2 + C3 + C4 + C5 + C6 + C7 + ... + Cn
The next step in deriving the formula is to count the number of vertices in a map. For each triangle, there
are 3, each square has 4, and in general
n vertices. So:
V = 2C2 + 3C3 + 4C4 + 5C5 + 6C6 + 7C7 + ... + nCn
But, as we are considering cubic maps, each vertex is shared by three countries. Therefore, the proper expression is:
3V = 2C2 + 3C3 + 4C4 + 5C5 + 6C6 + 7C7 + ... + nCn
V = 2/3C2 + C3 + 4/3C4 + 5/3C5 + 2C6 + 7/3C7 + ... + n/3Cn
The same can be done with the number of edges. However, this time, each edge is only shared between two countries.
2E = 2C2 + 3C3 + 4C4 + 5C5 + 6C6 + 7C7 + ... + nCn
E = C2 + 3/2C3 + 2C4 + 5/2C5 + 3C6 + 7/2C7 + ... + n/2Cn
With these expressions for
V, we can make use of Euler's Polyhedra formula:
2 = F - E + V
= (C2 + C3 + C4 + C5 + C6 + C7 + ... + Cn)
- (C2 + 3/2C3 + 2C4 + 5/2C5 + 3C6 + 7/2C7 + ... + n/2Cn)
+ (2/3C2 + C3 + 4/3C4 + 5/3C5 + 2C6 + 7/3C7 + ... + n/3Cn)
= 2/3C2 + 1/2C3 + 1/3C4 + 1/6C5 + 0C6 - 1/6C7 - ...
Although not immediately clear, this is the desired result. Multiplying through by 6 makes things clearer:
12 = 4C2 + 3C3 + 2C4 + C5 - C7 - 2C8 - 3C9 ...
This result does provide some interesting insight to the realm of cubic maps, and polyhedra.
Notice that only
C5 have positive coefficients. This means that
in any cubic map one of these shapes must appear. This is the Five neighbour theorem.