A formula due to Euler (in fact, a special case of the Euler characteristic in R3). Take any 3 dimensional polytope (i.e. a solid with planar faces) with no holes in it. Count its 0-, 1-, and 2-dimensional features: Then V-E+F=2. Always.

For instance, consider a regular dodecahedron. It has 12 faces (F=12), each a pentagon, for a total of 30 edges (recall that each edge is shared by 2 faces, so E=30). 3 faces meet at each vertex, so there are 20 vertices (V=20). Amazingly, 12-30+20=2, as promised.

See also Euler formula.

The most general form of Euler's formula for polyhedra includes the genus:

V - E + F = 2 - 2g,

where 2 - 2g = χ is the Euler characteristic.

In the case of the plane, g = 0, so we get the original V - E + F = 2. For a torus or the Klein bottle, g = 1, so V - E + F = 0.

Euler's Formula for Polyhedra

For any polyhedron:
   (number of faces:F) - (number of edges:E) + (number of verticies:V) = 2

On 14 November 1750, Euler wrote a letter to Christian Goldbach (of the Goldbach's Conjecture) describing some work he had been doing on polyhedra. Included in this letter was the formula given above, but he was unable to provide a suitable proof.

Euler did produce a proof by dissection, but it has been shown to not be complete. His argument was to take a polyhedron, and remove successive tetrahedron-shaped pieces such that F-E+V remains unchanged. This is done until a single tetrahedron remains. This holds for F-E+V=2, so for all the previous steps, as F-E+V has remained unchanged, it must also equal 2 for the original shape.

This is not complete, as it assumes that it is always possible to remove the tetrahedral piece.

The first complete proof was provided in 1794 by Adrien-Marie Legendre in his book Eléments de géométrie, using concepts of angle and area.

The proof given here however, is the one provided by Augustin-Louis Cauchy in 1813. The first step is to obtain the graph equivalent of a polyhedron. This is done by projecting the points of the polyhedron onto a plane. For example, the cube:

      o--------o            o---------o
     /|       /|            | \     / |
    / |      / |            |  o---o  |
   o--------o  |            |  |   |  |
   |  o-----|--o            |  o---o  |
   | /      | /             | /     \ |
   |/       |/              o---------o
   o--------o  

It is clear to see that the number of faces, edges and vertices remains unchanged - if you count the area completely unbound by edges as a face.

Given this, we can then remove edges, one at a time, with the only constraint that we keep the diagram in one piece. In doing this, there are two possible cases; either we remove an edge that is seperating two faces (A) or we remove a 'hanging' edge (B).

            Case A             |            Case B
  o---------o    o---------o   |  o---------o    o---------o
  | \     / |    | \       |   |  | \     /      | \     /
  |  o---o  |    |  o---o  |   |  |  o---o       |  o---o
  |  |   |  | => |  |   |  |   |  |  |   |    => |  |   |
  |  o---o  |    |  o---o  |   |  |  o---o       |  o---o
  | /     \ |    | /     \ |   |  | /            | /
  o---------o    o---------o   |  o---------o    o
                               |

In Case A, we have reduced the number of edges (E) by 1. We have also collapsed two faces into one, reducing the number of faces (F) by 1. Assuming F-E+V=2 prior this step, once we carry it out the equation can be written as:
   (F - 1) - (E - 1) + V = 2
   F - 1 - E + 1 + V = 2
   F - E + V = 2
Hence, Case A maintains Euler's formula.

In Case B, we have again reduced the number of edges by 1, but this time we have also reduced the number of vertices (V) by 1. As before, Euler's formula can be written:
   F - (E - 1) + (V - 1) = 2
   F - E + 1 + V - 1 = 2
   F - E + V = 2
Case B also maintains Euler's Formula.

Once all the edges have been removed, we are left with a single vertex and a single face - remember that we are counting the area unbound by the diagram as a face. Putting these values in Euler's formula:
   F - E + V = 2
   1 - 0 + 1 = 2
   2 = 2
shows that it still holds.

Any polyhedron can be reduced to a single vertex, all the while maintaining Euler's formula. Therefore, the formula holds for the original polyhedron.

Euler's formula does fail on certain polyhedra with particular properties. This was shown by Simon-Antoine-Jean Lhuilier. One case in particular is where the polyhedron has a hole through it - for example a torus. A more general form of Euler's formula has been found, assuming a polyhedra with K holes through it:
   F - E + V = 2 - 2K

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