Question

$5 \times 5$ 方格中的每个小方格要么被填充，要么为空；每个小方格最多有八个相邻小方格，相邻表示共边或共顶点。按以下规则变换方格：

• 任意一个已填充的小方格，如果有两个或三个已填充的相邻小方格，则保持填充。

• 任意一个空小方格，如果恰有三个已填充的相邻小方格，则变为填充。

• 所有其他小方格保持为空或变为空。

下图显示一个变换示例。

假设这个 $5 \times 5$ 方格有一圈空边框，围住一个 $3 \times 3$ 子方格。经过一次变换后，最终方格只在中心有一个已填充小方格。有多少种初始配置会产生这种结果？（旋转或翻折后相同的配置仍视为不同。）

Each square in a $5 \times 5$ grid is either filled or empty, and has up to eight adjacent neighboring squares, where neighboring squares share either a side or a corner. The grid is transformed by the following rules:

• Any filled square with two or three filled neighbors remains filled.

• Any empty square with exactly three filled neighbors becomes a filled square.

• All other squares remain empty or become empty.

A sample transformation is shown in the figure below.

Suppose the $5 \times 5$ grid has a border of empty squares surrounding a $3 \times 3$ subgrid. How many initial configurations will lead to a transformed grid consisting of a single filled square in the center after a single transformation? (Rotations and reflections of the same configuration are considered different.)

$\ 14$

$\ 18$

$\ 22$

$\ 26$

$\ 30$

答案：C

知识点：过程模拟分类讨论对称性

难度评级：2390

解答：

若中心格初始已填充，则它最终保持填充需要另外有 $2$ 或 $3$ 个已填充邻格，也就是相邻填充数为 $2$ 或 $3$ 。

这些邻格自己不能有过多已填充邻格。检查个角和 $3$ 类相邻关系时，要避免每个邻格拥有或个已填充邻格；因此可行时每个外侧填充格至多只有个相关邻格。

因此除中心外的填充格彼此不能相邻。可行情况只有选取一对相对角格，共 $4$ 种。假设中心格初始为空。那么中心格有 $4$ 个已填充邻格，而这些邻格每个最多只有 $4+4+4+8=20$ 个邻格。这说明没有小方格有两个已填充邻格。因此只可能有以下几种方式：与此同时，每个被选中的邻格最终必须变空，所以它们彼此之间不能使任何一个拥有两个或三个已填充邻格。图中前三类各有 $4$ 个旋转，最后一类有 $8$ 个旋转或翻折，共种；加上前面的两种，总数为 $20+2=22$ 。

所以正确答案是 C。

First suppose the center is initially filled. It must have exactly $2$ or $3$ filled neighbors to survive. Every such neighbor already touches the center, so to disappear it cannot touch any other filled neighbor. Checking these pairwise nonadjacent positions, the only choices that do not also give some empty square exactly $3$ filled neighbors are two opposite corners. There are $2$ such configurations.

Now suppose the center is initially empty. Exactly $3$ of its eight neighbors must be filled. Each of those three must disappear, so none may be adjacent to both of the others. Also, no empty square besides the center may be adjacent to all three. Applying these two tests gives the following four representative patterns:

Each of the first three patterns has $4$ distinct rotations. The last has $4$ rotations and their $4$ reflected images, for $8$ configurations. Thus the center-empty case contributes $4+4+4+8=20,$ and the total is $20+2=22.$

Thus, the answer is C .

2022 AMC 10B 第 19 题

答案：C

难度评级：2390

解答：

其他年份的第 19 题