A polyhedron mesh consists convex polyhedrons. Is this formula right: F+1=E+K, F=number of the interior faces, E=number of the interior edges, K=number of the polyhedrons in the mesh? If yes answer, references?
Many thanks.
Depends on the topology of the mesh. Euler's formula states that V-E+F = χ(S), where V=number of vertices, E=number of edges and F=number of faces and where χ(S) is the Euler characteristic, equal to 2(1-g), where g=genus of the surface, assuming it's orientable, cf. http://people.fas.harvard.edu/~ivogt/Exposition/EulerChar_PrimesTalk.pdf
and, for a more general discussion, cf. http://www.math.toronto.edu/~mat1300/oldnotes/classifsurfaces.pdf
Let the total number of vertices, edges, faces and polyhedrons be N0, N1, N2, N3 respectively. If you count the exterior space as a polyhedron (whose vertices , edges and faces are the exterior vertices, edges and faces) then
N0 − N1 + N2 − N3 = 0.
That formula is due to Schläfli but I’m sorry I have no reference.
So to get a formula for the interior vertices, edges, faces and polyhedrons we simply have to remove the exterior polyhedron. It satisfies Euler’s formule V − E + F = 2. Subtracting 2 from N0 − N1 + N2 and 1 from N3 we get, for the interior elements
n0 − n1 + n2 − n3 = −1.
Stam ~
Yes that's right.
I see now that I should have used the Euler characteristic of the external surface instead of presuming it to be of genus 0. Thus
n0 − n1 + n2 − n3 = 1 − χ
(Assuming Xiu Yi is thinking of his polyhedron cluster ("mesh") as free of self-intersection, orientability is not an issue...)
Not if the mesh has toroidal topology (e.g. from periodic boundary conditions) or has twists, however. So the question, as stated, can't be answered unambiguously.
Stam ~ I took it for granted that Xiu Yi was asking about a cluster of face-sharing polyhedra with surfaces of genus 0 in Euclidean space E3 (like a foam...). I gave a simple answer to a simple question. To go into all the possible generalizations would amount to a substantial research project!
Orientability is an issue-and a very interesting one, that can be studied, precisely, through such simulations.
Eric is right. I am a finite element person and don't know much about geometry which is tooooo hard for me. What we are doing is to divide a 3D convex polyhedron domain to many small convex polyhedrons shared faces. Then define piecewise polynomials associated with these partitions (or meshes) to approximate solutions of PDE.
Let me make it straight that for my situation, the formula is
n_0 − n_1 + n_2 − n_3 = −1.
with
n_0=total number of interior vertices,
n_1= total number interior edges,
n_2=total number interior faces
n_3=total number polyhedrons (not just interior).
Is the above statement right?
Many thanks to Stam and Eric.
I think the formula you want is different from that. Suppose you have a convex body divided into convex pieces, which we assume are polyhedra. Let V be the number of vertices, E the number of edges, F the number of faces, and K the number of polyhedra. Then the Euler-Poincare formula says V-E+F-K=1. Now V=V_in + V_ex, the number of internal vertices plus the number of external vertices. Likewise, E=E_in + E_ex and F=F_in + F_ex. Since the overall object is topologically a cell, so that its boundary is a sphere, then the Euler formula says V_ex-E_ex+F_ex=2. Subtracting, we get
V_in-E_in+F_in-K=-1, or V_in+F_in+1=E_in+K.
To illustrate, take a cube subdivided into eight smaller cubes. There are 27 vertices,, 54 edges, 36 (square) faces. 27-54+36-8=1. There are 6 internal edges, 12 internal faces, and one internal vertex.
1+12 +1=V+F+1=E+K=6+8
Dear Eric,
So it is! I did not see the full set of answers, or I could have saved myself the trouble of replying.
@Eric Lord, @Xiu Ye.
The formula for the TOTAL number of Vertices, Edges, Faces and Polyhedra reads N_0 - N_1 + N_2 - N_3 = Topological Euler Characteristic of the whole body
= 1 (and not 0 as Eric claims ) for a contractible (e.g. starshaped) body
This is a combination of three well known result in homology theory
(*) The topological Betti numbers b_0, b_1, .., b_n are the same as the cellular homology Betti numbers b_0^(cell), b_1^(cell),... in particular topological Euler characteristic = b_0 - b_1 + b_2 - b_3 + b_4, ... = Cellular Euler Characteristic = b_0^(cell), - b_1^(cell) + b_2^(cell - ..
(**) The Cellular Euler characterisic = #0cells - #1-cells + ...
(***) If two spaces are homotopy equivalent they have the same Betti-numbers. In particular if a space is contractible to a point, then the space has the same betti numbers as a point i.e. b_0 = 1, b_1 = b_2 = b_3, .. = 0.
Luckily for Xiu that makes his formula
n_0 - n_1 + n_2 - N_3 = -1
come out right for a convex polyhedron of dimension 3.
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