6.4.1 Combination of Reflective Symmetries

Figure 6.14: Combination of reflective symmetries

The set $A$ has two reflective symmetries indicated by the hyperplanes with the (orthogonal) normal vectors $\bm{n}_1$ and $\bm{n}_2$. The reconstruction process uses the subset which is on the positive side of both hyperplanes $A^+$ and the reflection functions $\operatorname{refl}_{\bm{n}_1,d_1}$ and $\operatorname{refl}_{\bm{n}_2,d_2}$. The missing diagonal set is obtained by applying the combination of these reflection functions $\operatorname{refl}_{\bm{n}_1,d_1} \circ \operatorname{refl}_{\bm{n}_2,d_2}$ on $A^+$.

An object with multiple reflective symmetries, where the reflecting hyperplanes are pairwise orthogonal to each other, can benefit from each reflective symmetry by recursively applying the approaches proposed in Section 6.2 as visualized in Figure 6.14. This is ensured by Lemma 6.2.

Lemma 2 (Combination of reflective symmetries)   Let $A \subseteq {\mathbb{R}}^n$ be a set with two reflective symmetries, each represented by its reflecting hyperplane $\mathbb{H}_{\bm{n}_1,d_1}, \mathbb{H}_{\bm{n}_2,d_2}$. If the hyperplanes are orthogonal to each other, i.e., their normal vectors are orthogonal to each other, then $A^{+,\bm{n}_1,d_1}$ has a reflective symmetry with hyperplane $\mathbb{H}_{\bm{n}_2,d_2}$ with

$$A^{+,\bm{n}_1,d_1} := \{ \bm{x} \in A \vert \bm{n}_1 \cdot \bm{x} \geq d_1 \}$$ (6.13)

Proof. For $x \in A^{+,\bm{n}_1,d_1}$, the following holds:

$$\bm{n}_1 \cdot \operatorname{refl}_{\bm{n}_2,d_2}(\bm{x}) = \bm{n... ...2) \underbrace{\bm{n}_1 \cdot \bm{n}_2}_{ = 0} = \bm{n}_1 \cdot \bm{x} \geq d_1$$ (6.14)

Therefore, for every $\bm{x} \in \operatorname{refl}_{\bm{n}_2,d_2}(A^{+,\bm{n}_1,d_1})$, the inner product of $\bm{x}$ with $\bm{n}_1$ is larger or equal to $d_1$ and $\operatorname{refl}_{\bm{n}_2,d_2}(A^{+,\bm{n}_1,d_1}) = A^{+,\bm{n}_1,d_1}$. $\qedsymbol$

This lemma also works for sets which have multiple reflective symmetries. Note, that for $A \subseteq {\mathbb{R}}^n$, $A$ can have a maximum number of $n$ reflective symmetries which are pairwise orthogonal to each other. Let $A \subseteq {\mathbb{R}}^n$ be a set with two reflective orthogonal symmetries with their reflecting hyperplanes $\mathbb{H}_{\bm{n}_1,d_1}, \mathbb{H}_{\bm{n}_2,d_2}$. The resulting templated geometry is described by using one geometry template being

$$A^+ := \{ \bm{x} \in A \vert \bm{x} \cdot \bm{n}_1 \geq d_1 \land \bm{x} \cdot \bm{n}_2 \geq d_2 \}$$ (6.15)

and all possible compositions of the reflection functions ${\mathbb{I}}, \operatorname{refl}_{\bm{n}_1, d_1}, \operatorname{refl}_{\bm{n}... ...}, \operatorname{refl}_{\bm{n}_1, d_1} \circ \operatorname{refl}_{\bm{n}_2,d_2}$. The resulting templated geometry therefore is

$${\Lambda}= ((A^+, ({\mathbb{I}}, \operatorname{refl}_{\bm{n}_1, d... ...e{refl}_{\bm{n}_1, d_1} \circ \operatorname{refl}_{\bm{n}_2,d_2}), (1,1,1,1))).$$ (6.16)

The same approach can also be applied to objects which have more than two (pairwise orthogonal) reflective symmetries (cf. Figure 6.14).

The approaches and algorithms presented in Section 6.2 can be modified to work with multiple reflections as well. The property that for reflective symmetries the instance interfaces are always conforming, also holds for multiple reflective symmetries. Therefore, Algorithm 6.1 can easily be adapted to scenarios with multiple reflective symmetries. The modified algorithm is presented in Algorithm 6.4. Instead of using the positive side of just one hyperplane, the positive side of all hyperplanes is used as the geometry template (Line 3). Additionally, multiple transformation functions have to be created for every composition of reflections and identities (Lines 6-8). The algorithm will generate a total number of $2^k$ transformation functions. Therefore, a theoretical improvement in memory usage and algorithm runtime of a factor of $2^k$ can be achieved in memory and runtime optimization.