Question

What is a simple proof for Euler's polyhedra characteristic of V-E+F=2

Answer 1

A: Arrange the polyhedron in space so that no edge is horizontal -- in particular, so there is exactly one uppermost vertex U and lowermost vertex L.
Put a unit + charge at each vertex, a unit - charge at the center of each edge, and a unit + charge in the middle of each face. We will show that the charges all cancel except for those at L and at U. To do this, we displace all the vertex and edge charges into a neighboring face, and then group together all the charges in each face. The direction of movement is determined by the rule that each charge moves horizontally, counterclockwise as viewed from above.In this way, each face receives the net charge from an open interval along its boundary. This open interval is decomposed into edges and vertices, which alternate. Since the first and last are edges, there is a surplus of one -; therefore, the total charge in each face is zero. All that is left is +2, for L and for U.

Answer 2

Or,:This proof uses the fact that the planar graph formed by the polyhedron can be embedded so all edges form straight line segments.
Sum up the angles in each face of a straight line drawing of the graph (including the outer face); the sum of angles in a k-gon is (k-2)pi, and each edge contributes to two faces, so the total sum is (2E-2F)pi.
Now let's count the same angles the other way. Each interior vertex is surrounded by triangles and contributes a total angle of 2 pi to the sum. The vertices on the outside face contribute 2(pi - theta(v)). where theta denotes the exterior angle of the polygon. The total exterior angle of any polygon is 2 pi, so the total angle is 2 pi V - 4 pi.

Combining these two formulas and dividing through by 2 pi, we see that V - 2 = E - F, or equivalently V-E+F=2.

Answer 3

Or<:Define a height function on the surface of the polyhedron as follows: Choose arbitrary heights for each vertex. In each edge, choose a height for one interior point greater than that of the two endpoints, and interpolate the remainder of the edge linearly between the chosen point and the endpoints. In each face, choose a height for one interior point, greater than all heights on its boundary; interpolate the heights in the rest of the face linearly on line segments from the chosen point to the boundary. The result is a continuous function with V+F+E critical points: V local minima at the vertices, E saddles at the chosen points of edges, and F local maxima at chosen points of faces.
Now view the surface as an initially bone dry earth on which there is about to fall a deluge which ultimately covers the highest peak. We count the number of lakes and connected land masses formed and destroyed in this rainstorm to obtain the result.

For each local minimum there will be one lake formed. For each saddle there will either be two lakes joined or a single lake doubling back on itself and disconnecting one land mass from another (let J and D denote the number of times these events happen respectively). For each peak a land mass will be eliminated. Initially there is one land mass, and in the final sitation there is one global lake. Thus we have 1+D-F=0 and 0+V-J=1. Combining these two equations with the fact that D+J=E yields the result.

One can either view the rainfall as (unnaturally) causing the global water levels to always be at the same height, so that two lakes reach a saddle at the same time; or one can take a more realistic viewpoint and say that the rainfall may vary arbitrarily over the globe, but when one lake reaches a saddle the water will spill over it (and the lake will not rise) until the lake on the other side of the saddle reaches the same height.
This proof is close to self-dual, the biggest asymmetry being in the definition of the height function. As usual, the Jordan curve theorem is involved, in the fact that a lake doubling back on itself creates an island.

Answer 4

http://www.ics.uci.edu/~eppstein/junkyard/euler/

Answer 5

Yeah, i've seen them, don't get them though

Could you further explain the second post

Answer 6

A possibly simpler (persuasive) method is to show that the formula holds for a spanning tree and that adding an edge does not affect this; keep adding till you have a planar drawing.
Then with the idea that any convex polyhedron can be represented as a planar graph by stereographic projection on a plane.