Space can be filled with a combination of tetrahedra and octahedra. This tetrahedral-octahedral honeycomb consists of alternating isohedral tetrahedrons and octahedrons in a ratio of 2:1.

In crystal structure terminology the vertices form an fcc structure. The vertices are at the corners and the face centers of the cubic unit cell. At each vertex corners of 8 tetrahedra and 6 octahedra meet. Edges ar common two 2 tetrahedra and 2 octahedra.

Elements having the fcc structure are aluminium, copper, silver and gold. Some alloys of these metals have an average fcc structure , for instance CuAu. The two types of metal atoms are more or less randomly ditributed over the equivalent sites. Locally the occupation of the sites may be ordered to some extent. Special cases of interest are the compositions or , because one may require that the composition of the tetrahedra is equal to the bulk composition. Similarly, the compositions , or may be satisfied by a corresponding occupation of the octahedra. In the following we will explore the consequences of the requirement that the cluster (t or o) composition is equal to the bulk composition for the long range order.

Because both polyhedra have an even number of vertices, both the tetrahedra and the octahedra may satisfy the bulk composition requirement. For the tetrahedron only a single configuration is possible, but for the octahedron there are two possibillities, all three atoms of the same kind occupy neighboring sites, or two atoms occupy neighboring sites and the third a site opposite to one of them.

Because the T and O have a face in common only the prototile denoted by prototiles shown in the figure are allowed. Selecting only the T(2,2) and O(3,3) tiles forces the ordered stucture shown below.

Similarly, if we select only the T(2,2) and the O(4,2) and O(2,4) another forced ordered structure, the socalled CuAu structure is obtained.

If both the O(2,4)/O(4,2) and the O(3,3) prototiles are admitted a disordered structure is obtained.

The question we now want to answer is, whether it is possible to find a subset of these O and T tiles and a substitution rule to build a *non-periodic* 3d tiling.

A big tetrahedron (BT) and a big octahedron (BO) twice as large in all directions may be constructed using four T and one O prototile for the BT and 8 T and 6 O prototiles for the BO.

If the composition of T is equal to the bulk composition, four different BT are possible, two BT using an O(3,3) and two using either an O(4,2) or an O(2,4). The BO have either a central A or B atom. In either case there are two possibble formations of the eight T around the central atoms. In the most symmetric case two O(4,2) or O(2,4) must be added at opposite positions, in the less symmetric case four O(3,3). The remaining four, resp. two O prototiles may be chosen to be either an O(3,3) or an O(4,2)/O(2,4) tile. The possible big tiles in terms of T prototiles are shown in the drawing below.

(still under construction)