CS 177 Homework 1 Julian Panetta October, 009 1 Euler Characteristic 1.1 Polyhedral Formula We want to show for any polygonal disk consisting of vertex set V, edge set E, and face set F: V E + F = 1 First, let s show that this relationship holds for any individual polygon. It is clearly true for all triangles (n = ) since they have vertices, edges, and 1 face. Notice that any polygon with n > vertices can be turned into a polygon with n 1 vertices by removing a vertex and the two edges incident on that vertex then inserting a single new edge to close the polygon. This operation does not alter the Euler characteristic, since V reduces by 1, but E also reduces by 1 to counter its effect. So all polygons with n vertices have the same Euler characteristic as a polygon with n 1 vertices. Hence, if all polygons with n 1 vertices have an Euler characteristic of 1, all polygons with n vertices will also have an Euler characteristic of 1. Considering our base case that showed the Euler characteristic is 1 for all polygons with n = vertices, we see by induction that any polygon with n vertices has an Euler characteristic of 1 n. Thus the relationship is satisfied for all polygons. Now, we want to extend this to show that the relationship is satisfied for all polygonal disks. Because all polygons satisfy the relationship, clearly we have a base case that all polygonal disks made up of one polygon satisfy the relationship. Consider a polygonal disk made up of n > polygons. This disk can be turned into a polygonal disk (with the same silhouette!) by finding two adjacent polygons within the disk (these are guaranteed to exist since n > 1), and merging them by removing all edges they share in common. Say they share s edges. Then E is reduced by s, lowering the Euler characteristic. However, we obviously must remove a vertex when we remove both edges incident on it, so if s > 1, we remove s 1 vertices as we remove the s shared edges. Also, in merging the two faces into one, we reduce the face count, F by exactly one. Thus, the net effect on the Euler characteristic is V E + F = (s 1) ( s) 1 = s+1+s 1 = 0. Thus all polygonal disks of n polygons have the same Euler characteristic as some polygonal disk of n 1 polygons. That means if all polygonal disks made up of n 1 polygons have satisfy the relationship, all polygonal disks of n polygons must also satisfy the relationship. Because we already saw the base case where a polygonal disk of 1 polygon satisfies the relationship, by induction we see that all polygonal disks satisfy V E + F = 1. Furthermore, we can use this fact to prove something about the Euler characteristic of a polyhedron. Notice that if we remove a single face and no attached vertices or edges from any polyhedron, we get a polygonal disk (i.e., the resultant mesh with a boundary can be deformed into a disk without cutting or gluing). In performing this operation, we reduced the Euler characteristic by 1. Therefore the Euler characteristic of any polyhedron is exactly one more than the Euler characteristic of any polygonal disk, and we must have: 1
V E + F = 1. Euler-Poincaré Formula There is nothing to prove, but we are presented the Euler characteristic, V E + F = g. Tessellation.1 Regular Valence Consider a (connected, orientable) simplicial surface, K, with all vertices of regular valence. K is, by definition, made up only of triangular faces. We want to show that K s Euler characteristic is 0, which implies that g = 1. Let s begin by counting the edges, E, and faces, F in terms of V, the number of vertices in K. Count the edges by looping over each vertex and adding up the 6 edges incident on that vertex. After this procedure terminates, we have touched each edge twice (once for each endpoint), and we have therefore double-counted. It is easy to see that E = 6 V = V. Count the faces by again looping over each vertex and adding up the 6 triangles incident on that vertex. After this procedure terminates, we have touched each face three times (once for each of its three vertices), and we have therefore triple-counted. It is evident, then, that F = 6 V = V. Now we can compute the Euler characteristic: X = V E + F = V V + V = 0 = g = g = 1 Therefore, the only (connected, orientable) simplicial surface for which every vertex has regular valence is a torus (g = 1).. Minimally Irregular Valence We ve just seen a case with a torus that has all regular vertices, so clearly m(k) = 0 when g = 1. Now, let s consider the case where g = 0. By Euler-Poincaré: V E + F = g = To get an expression only in terms of vertex and face counts, let s count the edges in terms of faces. If we loop over all the faces and count three edges per face, we will end up counting each edge twice because each edge is shared by two faces. Therefore the number of edges is E = F, and V F + F = V F = = V = F + 4 Let I be the number of irregular vertices. Then V I is the number of regular vertices. Let s count the number of faces in terms of vertices. Each regular vertex touches 6 triangles, and each irregular vertex must touch at least triangles by the assumption that all vertices have at least valence. If we count faces by adding up all the triangles incident on all vertices, we will count each triangle times (once for each vertex belonging to the triangle), so:
This means that: 6( V I) + I F = V I + I = V I V = F + 4 V I + 4 = I 4 In other words, a simplicial surface of genus g = 0 must have at least 4 irregular vertices. To show that this bound is tight, consider a tetrahedron; this is a simplicial surface with 4 irregular vertices. Therefore m(k) = 4 when g = 0. Finally, let s consider the case where g. We already know from.1 there must be at least one irregular vertex because we showed that the only case where all vertices are regular is when g = 1. However, we can show that having just a single vertex of irregular valence is sufficient to bring down the Euler characteristic to any negative value resulting from g. Let X be the valence of our single irregular vertex. There are V 1 regular vertices if only one is irregular, so we can count the faces to be: F = 6( V 1) + X = V + X So, since we already found E = F, the Euler characteristic of the simplicial surface is: V E + F = V F = V V + X = V V + 1 + x 6 = 1 X 6 Our only restriction is that X, so we see that regardless of how low the RHS of the Euler-Poincaré formula goes as g is increased, we can always increase X, the valence of the single irregular vertex, to lower the LHS so equality is maintained. Thus, when g, m(k) = 1. Collecting everything we ve shown, we see: 4, g = 0 m(k) = 0, g = 1 1, g Discrete Gaussian Curvature.1 Area of a Spherical Triangle Consider the spherical triangle and its diangles shown in figure 1. Because each diangle covers ( α i ) π = α i π of the unit sphere s surface, the surface area of each diangle is: ( A i = (4πr αi ) ( αi ) ) = (4 π) = 4α i π π We can find the spherical triangle s area by combining A 1, A, and A. Note that all three diangles overlap exactly at the shaded spherical triangle and at the identical spherical triangle on the opposite side of the sphere to cover them each three times. Furthermore, the diangles collectively cover all the remaining sphere surface exactly once. If we take A s to be the area of the sphere, and A = A to be the area of the spherical triangle on the opposite side, we find:
.1 Area of a Spherical Triangle Show that the area of a spherical triangle on the unit sphere with interior angles α 1, α, α is (Hint: consider the areas A = α 1 + α + α π. A 1, A, A of the three shaded regions (called diangles ) pictured below.) 1 Figure 1: The diangles of spherical triangle with angles α 1 α α. A 1 corresponds to α 1, A to α, and A to α. Note that there is an identical version of the spherical triangle in the back of the sphere (it has the same angles as the original triangle.) A 1 + A + A = A + A + (A s A A ) = A + A + 4π r = 4A + 4π A = 1 4 (A 1 + A + A ) π = 1 4 (4α 1 + 4α + 4α ) π = α 1 + α + α π. Area of a Spherical Polygon Using the standard ear trimming algorithm (or another more efficient algorithm), one can triangulate any n-gon, P, into n polygons. This algorithm, which forms and removes a triangle by taking two outside edges and creating one internal edge, will clearly work on the surface of a sphere. Because a triangulation of a polygon will have the same area as the original polygon, we can measure the area of any spherical polygon by adding up the area of the individual triangular components of the triangularization, T which we know how to measure from the last problem: A(P ) = A(t) = t,1 + α t, + α t, π) t T t T(α ( ) = α t,1 + α t, + α t, (n )π = ( n)π + α α T t T 4
Where α T means all angles in the triangulation. The important thing to note here is that the all the angles in the triangularization of the n-gon can be partitioned into n nonempty partitions whose members add up to a polygon angle β i i {1,..., n}. This partition is constructed by grouping all the interior angles at polygon vertex i of all the triangles incident on i. As a result of this fact: And so we have found: α = α T i=1 β i A(P ) = ( n)π + i=1 β i. Angle Defect Figure : Here we see a polygonal surface and two orthogonal planes that are spanned by face f s normal and one of each of the two neighboring face normals. The face normals are shown in green lines, the two angles we want to relate are shown in blue, and the perpendicular relationship between the edges of face f and the planes is also noted. If we can find the angles of the polygon defined by the intersection of the face normals with the unit sphere, we can use the result of. to find the area of this polygon, the Gaussian curvature. For consistency with., let s call these angles β f, where f is the face number of the face whose normal defines the polygon vertex at this angle. Then, if we number the faces so that f F v = {1,..., n}, we can use our formula from. verbatim to express the Gaussian curvature: K = ( n)π + β i (1) i=1 5
We want to express this curvature in terms of f (v), the interior angle of face f at vertex v. In figure, this is shown as f. Also, the β f shown in this figure is in fact the angle at vertex f of the spherical polygon. This is clear since the edges of the spherical polygon are the great circle arcs defined by the intersection with the unit sphere of the planes spanned by two adjacent face normals, and β f is the angle between two such adjacent planes. To find β f in terms of f (v), we note that the edges of face f are actually normal to the planes. This is because the normal to the edges of face f can be thought range from face f s normal to the adjacent face s normal. In other words, the normal of the edge lies in a sector on the plane spanned by the two adjacent face normals, so the edge itself must be perpendicular to that plane. This is illustrated by the blue lines on figure. From that figure, we see that β f and f (v) therefore belong to a quadrilateral whose other two angles are both π. So that means: β f + f (v) + π + π = π = β f = π f (v) Plugging this angle into (1) we see: K = ( n)π + (π i=1 K = π i(v)) = ( n)π + nπ f (v) = d(v) f (v) And we have shown that the Gaussian curvature is equal to the angle defect, d(v)..4 Discrete Gauss-Bonnet Theorem First, we count the number of edges in terms of the number of faces. If we loop over all the faces and count three edges per face, we will end up counting each edge twice because each edge is shared by two faces. Therefore the number of edges is E = F. That means we can write the Euler characteristic as: And so: X = V E + F = V F + F = V F πx = π V π F Now looking at the sum over all vertices of the angle defect, we see: d(v ) = π f (v) = π V f (v) This double summation adds up all the angles incident on every vertex, so it will add up each angle in the surface exactly once. Since each face s angles adds up to π, the sum of all angles in the simplicial surface is π F. Thus: 6
d(v ) = π V π F = πx 7