An Efficient Exhaustive Search Algorithm for the Escherization Problem

Abstract

In the Escherization problem, given a closed figure in the plane, the objective is to find a closed figure that is as close as possible to the input figure that tiles the plane. Koizumi and Sugihara’s formulation reduces this problem to an eigenvalue problem. In their formulation, only one of the potentially numerous templates is used to parameterize the possible tile shapes for each isohedral type. In this research, we try to search for the best tile shape using all possible templates for the nine most general isohedral types. This extension provides a considerable flexibility in possible tile shapes and improves the quality of the obtained tile shapes. However, the exhaustive search of all possible templates is computationally unrealistic using conventional calculation methods, and we develop an efficient algorithm to perform this in a reasonable computation time.

Appendix: Parameterization of Tile Shapes for IH7

For any $i \in I$ and $k \in K_i$, the matrix $B_{ik}$ can be constructed in O(n) time, as described in Sect. 3.3. However, this may be nontrivial if a template has two adjacent J edges that must form a specified angle. We present another example of how the matrix $B_{ik}$ is constructed in O(n) time for the template of IH7, which includes such J edges.

Figure 12 shows the template of IH7 represented in the same way as in Fig. 3. The constraint conditions imposed on the tiling vertices are expressed by

$$\begin{aligned} \left\{ \begin{array}{l} x^v_{3} - x^v_{2} = \cos \theta (x^v_{1} - x^v_{2}) - \sin \theta (y^v_{1} - y^v_{2}) \\ y^v_{3} - y^v_{2} = \sin \theta (x^v_{1} - x^v_{2}) + \cos \theta (y^v_{1} - y^v_{2}) \\ x^v_{5} - x^v_{4} = \cos \theta (x^v_{3} - x^v_{4}) - \sin \theta (y^v_{3} - y^v_{4}) \\ y^v_{5} - y^v_{4} = \sin \theta (x^v_{3} - x^v_{4}) + \cos \theta (y^v_{3} - y^v_{4}) \\ x^v_{1} - x^v_{6} = \cos \theta (x^v_{5} - x^v_{6}) - \sin \theta (y^v_{5} - y^v_{6}) \\ y^v_{1} - y^v_{6} = \sin \theta (x^v_{5} - x^v_{6}) + \cos \theta (y^v_{5} - y^v_{6}) \end{array} \right. , \end{aligned}$$

(43)

where $\theta = 120^{\circ }$. Then, we can obtain the matrix $B_v$ in the same way as in Eq. (16).

The xy-coordinates of the n points are constrained by the following equations (only the constraints for the first and second tiling edges are shown):

$$\begin{aligned} \left\{ \begin{aligned} x_{h(2)+i}-x_{h(2)}&= \cos \theta (x_{h(2)-i}-x_{h(2)}) - \sin \theta (y_{h(2)-i}-y_{h(2)}) \ (i = 1, \dots , k_1)\\ y_{h(2)+i}-y_{h(2)}&= \sin \theta (x_{h(2)-i}-x_{h(2)}) + \cos \theta (y_{h(2)-i}-y_{h(2)}) \ (i = 1, \dots , k_1)\\&\vdots \end{aligned} \right. . \end{aligned}$$

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Fig. 12

Template of IH7 for a specific assignment of the points to the tiling edges

By parameterizing $x_{h(2)-i}-x_{h(2)}$ and $y_{h(2)-i}-y_{h(2)}$ as $\xi _{2i}^s$ and $\xi _{2i+1}^s$, respectively, the tile shape U is then parameterized as follows (only the xy-coordinates of the first and second tiling edges are shown):

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where the blank elements in the column vectors are zero. The matrix $B'_d$ (see Eq. 22) is then given by the following formula:

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where the row vectors of $B'_d$ are given by

$$\begin{aligned} \begin{bmatrix} {\varvec{d^b_2}}^{\top } \\ {\varvec{d^f_2}}^{\top } \\ {\varvec{d^b_{n_v+2}}}^{\top } \\ {\varvec{d^f_{n_v+2}}}^{\top } \end{bmatrix}= & {} \begin{bmatrix} {\varvec{d_2}}^{\top } \\ {\varvec{d_2}}^{\top } \\ {\varvec{d_{n_v+2}}}^{\top } \\ {\varvec{d_{n_v+2}}}^{\top } \end{bmatrix} - \dfrac{1}{2} { \begin{bmatrix} 1 \\ \cos \theta \\ 0 \\ \sin \theta \end{bmatrix} }^{\top } \begin{bmatrix} {\varvec{d_2}}^{\top } \\ {\varvec{d_2}}^{\top } \\ {\varvec{d_{n_v+2}}}^{\top } \\ {\varvec{d_{n_v+2}}}^{\top } \end{bmatrix} \begin{bmatrix} 1 \\ \cos \theta \\ 0 \\ \sin \theta \end{bmatrix}\nonumber \\&- \dfrac{1}{2} { \begin{bmatrix} 0 \\ -\sin \theta \\ 1 \\ \cos \theta \end{bmatrix} }^{\top } \begin{bmatrix} {\varvec{d_2}}^{\top } \\ {\varvec{d_2}}^{\top } \\ {\varvec{d_{n_v+2}}}^{\top } \\ {\varvec{d_{n_v+2}}}^{\top } \end{bmatrix} \begin{bmatrix} 0 \\ -\sin \theta \\ 1 \\ \cos \theta \end{bmatrix} . \end{aligned}$$

(47)

Obviously, we can construct the matrix $B_{ik}$ in O(n) time because the matrix $B_s$ is sparse, all column vectors of $B_s$ are mutually orthogonal, and the matrix $B'_d$ is obtained in O(n) time.

In fact, we must consider $\theta =-120^{\circ }$ as well in the template of IH7. This can be easily implemented by considering the reverse numbering scheme of the goal polygon. The same process is necessary for IH21 and IH28.