Appendix: Towards Goldberg-Coxeter operation on mixed polyhedra¶
The methods discussed up to now only handle polyhedra with a single type of face, either 3- or 4-sided. This section discusses the extension of the Goldberg-Coxeter operation to polyhedra with mixed triangle and quad faces. It’s possible to divide a polyhedron by applying a certain \(\Delta\) operator to the triangle faces, a certain \(\Box\) operator to the quadrilateral faces, and dealing with what happens over the edges of the original polyhedron.
Two operators are ‘’compatible’’ if:
- The same number of vertices lie along the base edge.
- The same number of edges cross the base edge, excluding ones that meet a vertex at the edge.
- The same number of edges lie on the base edge.
- Edges that cross the base edge do not create digons (faces with 2 edges).
For the operators \(\Delta(a,b)\) and \(\Box(c,d)\), the first 3 requirements can be distilled to these equations:
- \(\gcd(a, b) = \gcd(c, d) = g\)
- \(c + d + g = 2(a + b)\). This equation has no solution if c and d are both odd numbers.
The 4th requirement is more complicated to put in equational form.
Consequences of these rules include:
- \(\Box(1,0)\) is compatible with \(\Delta(1,0)\)
- Iff \(\Box(a,b)\) is compatible with \(\Delta(c,d)\), then \(\Box(na,nb)\) is compatible with \(\Delta(nc,nd)\) for all positive integers n.
- \(\Box(n,0)\) is compatible with \(\Delta(n,0)\) (Class I)
- Iff \(\Box(a,b)\) is compatible with \(\Delta(c,d)\), then \(\Box(b,a)\) is compatible with \(\Delta(d,c)\).
- \(\Delta(n,n)\) is compatible with \(\Box(n,2n)\) and \(\Box(2n,n)\) (Class II triangles)
- \(\Box(n,n)\) is not compatible with any \(\Delta\). (Class II squares)
- If \(\Box(a_1,b_1)\) is compatible with \(\Delta(c_1,d_1)\), and \(\Box(a_2,b_2)\) is compatible with \(\Delta(c_2,d_2)\), it does not follow that the composed subdivision \(\Box(a_1,b_1)\Box(a_2,b_2)\) is compatible with \(\Delta(c_1,d_1)\Delta(c_2,d_2)\). \(\Delta(1,1)\Delta(1,1) = \Delta(3,0)\); \(\Box(1,2)\Box(1,2) = \Box(4,3)\); \(\Box(1,2)\Box(2,1) = \Box(5,0)\).