The objective is to find a Cellular Automata rule that can form a 2D point pattern with a maximum number of points (1-cells). Points are not allowed to touch each other, they have to be separated by 0-cells, and every 0-cell can find at least one point in its Moore-neighborhood. Probabilistic rules are designed that can solve this task with asynchronous updating and cyclic boundary condition. The task is considered as a tiling problem, where point tiles are used to cover the space with overlaps. A point tile consists of a center pixel (the kernel with value 1) and 8 surrounding pixels forming the hull with value 0. The term pixel is used to distinguish the cells of a tile from the cells of a cellular automaton. For each of the 9 tile pixels a so-called template is defined by a shift of the point tile. In the rule application, the 9 templates are tested at the actual cell position. If all template pixels (except the central reference pixel) of a template match with the corresponding neighbors of the actual cell under consideration, the cell's state is adjusted to the reference pixel's value. Otherwise the cell is set to the random value 0 or 1 with a certain probability. The hull pixels are allowed to overlap. In order to evolve a maximum of points, the overlap between tiles has to be maximized. To do that, the number of template hits is counted. Depending on the hit-number, additional noise is injected with certain probabilities. Thereby optimal patterns with the maximum number of points can be evolved. The behavior and performance of the designed rules is evaluated for different parameter settings.
翻译:目标是找到一个 Cellulal Automata 规则, 该规则可以形成 2D 点模式, 以最大数目的点数构成 2D 点模式( 1- 单元格) 。 点不允许相互触碰, 点必须用 0 个单元格分隔, 每个 0 个单元格可以在 Moore 附近找到至少一个点 。 每个 0 单元格可以在 Moore 的单元格中找到至少一个点 。 概率规则的设计可以用无同步更新和循环边界条件来解决这个问题 。 任务被视为一个小问题, 即点砖砖用重叠来覆盖空间 。 一个点是像素中心像素( 内核带有值 1 ) 和 8 围绕像素以 值 0 组成船体的像素模式。 名像素规则用于将一个小点像素单元格的单元格与细胞的每只量更新值区分为零 。 在规则应用程序中, 最高级的比值中, 最高级的比值可以被测试 。 在实际的单元格的比值中, 的比值中, 值比值比值中, 的比值比值比值比值比值比值中, 的比值比值比值比值更值要高。 。 的比值比值比值比值比值比值比值比值比值比值的比值比值比值要高。