6 Jan
2013
6 Jan
'13
10:23 a.m.
"Henry Baker" <hbaker1@pipeline.com> Q. Is it possible to "quadrangulate" a 3D surface with _only_ (planar) quadrilaterals "hinged" together ? (A quadrilateral has to have 4 non-trivial sides, so no triangles as limiting cases.)
--First answer: A triangle can be divided into 3 quads using 4 extra vertices at edge midpoints and some central point. And any convex polyhedron can be regarded as having triangle faces only (happens automatically by perturbing all vertices by epsilon randomly). --Additional remark: Let F=#quads, E=#edges, V=#vertices. F-E+V=2 (Euler) and E=2*F for a quadrangle-faced polyhedron with same topology as a sphere. So V = 2+F.