WHAT'S THE "SIMPLEST" POLYHEDRAL g-HOLED TORUS? If you wish to make a g-holed torus with V vertices, E edges, and F triangular faces, then comparing boundaries gives 3F=2E while Euler's formula is 2-2g = V-E+F so that E=3(V-2+2g). On the other hand, with V vertices, the largest number of edges you could draw would be to connect every possible pair, giving E <= ( V choose 2 ), so there is an inequality relating V and g: V^2-7V +12(1-g) >= 0, i.e. V is larger than the positive root V >= (7/2) + sqrt(48g-1)/2. It can be shown that the minimal number of vertices is integral, that is, using the minimum number of vertices requires joining every pair of them, iff g is of one of these forms: [proof: write 48g+1=x^2; find x=1,7,17,23 mod 24] g=12k^2+k, V=12k+4 (V,g) = (4,0), (16, 13), ... g=12k^2+7k+1, V=12k+7 (V,g) = (7,1), (19,20), ... g=12k^2-7k+1, V=12k (V,g) = [(0,1),] (12,6), (24, 35),... g=12k^2-k, V= 12k+3 (V,g) = [(3,0),] (15,11), (27,46),... So nicest genera are g=0, 1, 6, 11, 13, 20, ... The case g=0 is topologically a sphere; the standard tetrahedon shows it is possible to build one with the minimum number of vertices (4). The case g=1 is discussed in the other files in this directory. As far as I know it is unknown whether it is possible to glue 44 triangles onto the 66 edges joining 12 points in R^3 in such a way that the triangles do not intersect (except at the edges); this would realize the minimal configuration for g=6. TABLES FOR SMALL GENUS: For small g we compute the minimal V, the corresponding number of total edges (V choose 2) and edges to be used E=3(V-2+2g), with the difference (=possible edges not used): g: 0 1 2 3 4 5 6 Vmin 4 7 9 10 11 12 12 Eavail 6 21 36 45 55 66 66 Eused 6 21 33 42 51 60 66 Eunused 0 0 3 3 4 6 0