Forbidden patches

When we are covering the plane with patterns emerged from $lca(2,1)_{110}$ , we can not put triangles in arbitrary positions, so we are accepting the local evolution rules. In this way it is easiest to work with legal patterns and to forget those that never occur. So, when we are talking about patches, we are referring only to those patches that are legal graphical patterns in $lca(2,1)_{110}$ .

A forbidden patch[6] is a 2-space graphical region of cells in a cellular automata evolution that doesn't obey $lca(2,1)_{110}$ local evolution rules, there are observations that we can say to build legal patches constructions.

Observation 1   Two triangles can not be aligned in their top sides

The reason for the above observation is the sequence of 1-cells formed with two triangles aligned in its top sides, between two triangles is formed a sequence $111\to 1$ , that is not in the local evolution rule of $lca(2,1)_{110}$ , so it is a forbidden pattern. This condition is shown in figure 3.

Figure 3: Ilegal positions in a patch

Observation 2   Both overlaps and rotations are not allowed.

When more than one equal patches are full overlapped, we can see just one of them, we consider that there are not any copies; in the same way, when one large patch covers completely more than one minors patches, we can see only the bigger at top of them, in this cases, we consider only the triangle that is in the top. Furthermore, due to the emergent shapes of $lca(2,1)_{110}$ evolutions, there is only one orientation of the triangles. A remark is the interrelation of different subspaces full-covered (legally) by triangles. Precisely when two or more different subspaces occurs in the same evolution space, a glider occurs.

Let be now $\mathcal{F}\subset\mathcal{X}^{(\ast)}$ the subset of all covered subspaces that are forbidden in $lca(2,1)_{110}$ evolutions, for short, we can say forbidden patches. For every set of forbidden patches $\mathcal{F}$ , we call $\mathcal{X}_{\setminus\mathcal{F}}$ to the set of full-covers that have no element of $\mathcal{F}$ .

For short, we can omit both super-index and subindex on $\mathcal{X}^{(\ast)}_{\setminus\mathcal{F}}$ , because from now on, we work only with legal full-covers in $lca(2,1)_{110}$ ; therefore:

$$\mathcal{X} \leftrightharpoons \mathcal{X}^{(\ast)}_{\setminus\mathcal{F}}=\bigcup_{i\in\mathbb{Z}^0}\mathcal{X}^{(i)}_{\setminus\mathcal{F}}$$ (4)