Euler’s formula and colou rings of gra

Euler's formula states v − e + f = 2. This together with the fact that each edge is shared by two regions, 2e = 3f, can be used to show 6v − 2e = 12. Now, the degree of a vertex is the number of edges abutting it. If vn is the number of vertices of degree n and D is the maximum degree of any vertex,

But since 12 > 0 and 6 − i ≤ 0 for all i ≥ 6, this demonstrates that there is at least one vertex of degree 5 or less.

The intuitive idea underlying discharging is to consider the planar graph as an electrical network. Initially positive and negative "electrical charge" is distributed amongst the vertices so that the total is positive.

Recall the formula above:

One can also consider the coloring problem on surfaces other than the plane (Weisstein). The problem on the sphere or cylinder is equivalent to that on the plane. For closed (orientable or non-orientable) surfaces with positive genus, the maximum number p of colors needed depends on the surface's Euler characteristic χ according to the formula

where the outermost brackets denote the floor function.

(Weisstein).

This formula, the Heawood conjecture, was conjectured by P.J. Heawood in 1890 and proven by Gerhard Ringel and J. T. W. Youngs in 1968. The only exception to the formula is the Klein bottle, which has Euler characteristic 0 (hence the formula gives p = 7) and requires 6 colors, as shown by P. Franklin in 1934 (Weisstein).

For example, the torus has Euler characteristic χ = 0 (and genus g = 1) and thus p = 7, so no more than 7 colors are required to color any map on a torus

V-E+F=15-20+7=2

Euler's formula states that for a map on the sphere, , where is the

number of vertices, is the number of faces, and is the number of edges.

This Demonstration shows a map in the plane (so the exterior face counts as

a face). The formula is proved by deleting edges lying in a cycle (which

causes and to each decrease by one) until there are no cycles left. Then

one has a tree, and one can delete vertices of degree one and the edges

connected to them until only a point is left. Each such move decreases and

by one. So all the moves leave unchanged, but at the end and are

each 1 and is 0, so must have been 2 at the start.

Euler’s formula and planar graphs