Eulers Formula

The Euler characteristic for a surface built of polygons is much more conveniently calculated according to the Euler formula [33]... 

The Adams Open Formulas are a class of muliistep formulas such that the first-order formula reproduces the Euler formula. The second-order Adams open formula is given by... 

This relation can be easily verified by employing the Euler formula c = cos 6 i sin 0 for the pure imaginary part of the exponential function and by observing the definite integrals ... 

The Euler characteristic is usually calculated in two ways by using the Gauss-Bonnet theorem [7,85,207,210,222] [Eq. (8)] or by combining the Cartesian and Gauss theorems [Eqs. (122) and (123)], which is also called the Euler formula [23,76,224]. 

The computation of the Euler characteristic based on Eq. (8) is not practical, particularly when the system is represented by a set of points on the lattice. The practical way of computing % is related to the coverage of the surface with polygons. Then, the calculation of the Euler characteristic is straightforward when it is based on the Euler formula ... 

The disagreement of those two methods can be used to characterize quantitatively the double-separated morphology. One should simply calculate the Euler characteristic once by using the Euler formula (130), and then calculate it a second time by using the digital pattern method. The difference between obtained quantities will give the number of double-separated structures. 

A k-valent sphere, whose faces have gonality a orb, is called a ( a, b), k)-sphere (see Chapter 2). We call the parameters ( a, b), k) elliptic, parabolic, hyperbolic, according to the sign cfa(2, b, k). This sign has a consequence for the finiteness and growth of the number of graphs in the class of ( a, b], k)-spheres. Here, the link is provided by the Euler formula (1.1). 

We can interpret the quantity 2k — b(k — 2) as the curvature of the faces of gonality b Euler formula is the condition that the total curvature is a constant, equal to 4k, for fc-valent plane graphs. This curvature has an interpretation and applications in Computational Group Theory, see [Par06] and [LySc77, Chapter 9]. 

Proof The proof is based on curvature estimates. In fact, it is the counting of vertex-face incidences plus using the Euler formula. We choose to use the curvature viewpoint (see Section 4.4) since it express nicely ellipticity. 

This formula is called Gauss-Bonnet formula (see [Ale50]). Expressed differently, Euler formula v — e + / = 2 is for plane graphs with no boundaries, but it can be extended to plane graphs with boundaries. 

Consider a particular case of a finite polycycle P in which each vertex has degree q. In this case, 2e = qn, where n is the total number of vertices and e is the total number of edges of P in view of the equality 2e = rpr + k, we can rewrite the Euler formula n — e + pr = 1 as follows ... 

We classify only the strictly face-regular spheres and strictly face-regular normal balanced planes. The plane case contains the toms case as a subcase. For the plane, the Euler formula does not hold, but the condition of normality, discussed thereafter,... 

The 6-cycle Cq has two domains outer and inner. There are two cases either two 3-cycles lie in distinct domains, or both lie in the same, say, outer domain. In the first case, by symmetry, we can consider only the outer domain. The Euler formula shows that the boundary circuit of the cycle of 5-gons should have three tails. It is easy to verify that it is not possible to form a net of 6-gons using three tails of the 6-cycle and six tails of the 3-cycle. 

Thkea( 5, b, 3)-sphere or torus that is 5i j. Euler formula reads 12/ = p — (b — 6)ph- There are np = pairs of 5-gons. Every such pair defines AnP pattens b5b and 2nP patterns b55b in the boundary sequences of fe-gons. A packing argument yields ... 

Proof Euler formula in Theorem 13.1. l(iii) implies the relation ... 

Proof We will first treat the simpler toroidal case (i). First, we have the relation ps = 3 i + 4n2. By the Euler formula, we also have ps =(b — 6)pb, which implies the relation ... 

Assume that a vertex, say Cf, has degree 1. The connected component C, is incident to the face Fj along a cycle of length h. Consider now the plane graph formed by C/ only. So, its feces are only 4-, 9- and i-gons with i > 30 and the h-gon. The Euler formula then reads ... 

Every connected component Cj is incident to two faces Fi and jFJ+i (mod 0 along cycles of length k and which are the numbers of (5,3)-polycycles Fi in those cycles. The Euler formula for the plane graph C,- reads ... 

Assume b < 11. Given a ( 5, b], 3)-plane, the gonality of a face in the major skeleton is equal to the number of (5,3)-polycycles E and C3 to which it is incident. Clearly, there are at most five such incidences for each face. Since the major skeleton is 3-valent, we reach a contradiction by Euler formula (1.1) and (i) holds. 

Euler formula implies also, that the number ea a of edges of adjacency of two a-gons in any ( a, b], 3)-sphere bRz is ... 

Also, by Euler formula (1.1), we obtain p = 2p%. The total number e of edges is equal to ... 

Every vertex, which is incident to three fi-gonal faces, corresponds to a 3-gonal face of b(G). Every (4,3)-polycycle 4, 3 — v also corresponds to a 3-gonal face. Every (4,3)-polycycle 4,3 — e corresponds to a 2-gonal face. On the other hand, all P2 x Pk correspond to 4-gonal faces. A 3-valent map, whose faces have gonality at most four, does not exist on the torus and, clearly, has at most 8 vertices on the sphere by Euler Formula 1.1. 

Proof, (i) By standard arguments (double counting and Euler formula), we obtain ... 

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