Euler's formula in geometry: V-e+f = 2

Euler formula in geometry: v-e+f = 2,v, E, F represents the number of vertices, sides, and polygons of simple geometry.

Prove:

There are many proofs of it, and a recursive method is presented here.

For any simple geometry (the boundary of the geometry is not a curve), we look at each face of the geometry, set the edge into an N-shaped shape, we connect the other vertices from a fixed vertex, we are going to triangulate the N-shape from a vertex, we assume that each triangle is a polygon ( Because there are actually multiple triangles coplanar), you can see that the increments of E and F are the same in this process, so if the original geometry satisfies v-e+f = 2, then the geometry (depending on each triangle as a polygon) still satisfies the Euler formula.

When we randomly remove a polygon, the geometry after the split should meet V-e+f = 1.

Now we examine the geometry that was split by the triangle after removing a polygon, for a triangle, to examine its three edges (each edge is shared by two triangles), there are three cases:

(1) The face of the other triangle where one side is located is empty.

(2) The side of the other triangle where the two edges are located is empty.

(3) The side of the other triangle where the three edges are located is empty.

So below we start a "hollowed out process", for the convenience of analysis, we do not in a "hollowed out" of the interior of the area to "empty" a triangle, until the end of the remaining triangle, that is, we avoided the third case.

Faced with the situation (1), we hollowed out this triangle, found that the number of sides, the number of polygons minus 1,v-e+f value will not change.

Facing situation (2), we hollowed out this triangle, we found that there will be two cases, divided into the number of vertices minus 1 and the same situation (imagine), we very much like the previous situation, because this makes the number of edges minus 2, vertices minus 1, the number of polygons minus 1, which will make v-e+f unchanged, This is very helpful for us to continue the equivalent conversion of recursion.

So how to deal with this situation? Remember when we first randomly hollowed out the triangle, it is easy to see it must be surrounded by three triangles, that is, to share this hollowed out triangle of the three sides of the triangle, we label 1, 2, 3, and this three triangle is bound to be sandwiched between three triangles, we remember 4, 5, 6, we take the strategy is to first empty 1, 2, 3, and then hollowed out 4, 5, 6, this will ensure that v-e+f unchanged, and we will 1, 2, 3, 4, 5, 6 as the first layer of "fortress", for the second layer, think about whether it is the same situation (hollowed 4, 5, 6 will be three cases (1) Triangle in the box)? This guarantees the correctness of our recursion. (There is no need to consider the result of the number of triangles remaining 6, as this top-down "hollowing out" will eliminate the second case of the situation (2))

So let's take a look at the final state of recursion, for a triangle, V = 3, F = 1, E = 3, obviously there is v-e+f = 1, and then goes back to all previous states, and for any simple geometry, there is a Euler formula:

V-e+f = 2.

The certificate is completed.

Euler's formula in geometry: V-e+f = 2