AMC12 2018 A

AMC12 2018 A · Q15

AMC12 2018 A · Q15. It mainly tests Counting with symmetry / Burnside (rare), Symmetry.

A scanning code consists of a $7 \times 7$ grid of squares, with some of its squares colored black and the rest colored white. There must be at least one square of each color in this grid of 49 squares. A scanning code is called symmetric if its look does not change when the entire square is rotated by a multiple of 90° counterclockwise around its center, nor when it is reflected across a line joining opposite corners or a line joining midpoints of opposite sides. What is the total number of possible symmetric scanning codes?

一个扫描码由$7 \times 7$的方格网格组成，其中一些方格涂黑，其余涂白。该49个方格中必须至少有一个方格是每种颜色。扫描码被称为对称的，如果整个方格绕中心逆时针旋转90°的倍数时外观不变，也不改变当反射穿过连接对角线的线或连接对边中点的线时。可能的对称扫描码总数是多少？

(A) 510 510

(B) 1022 1022

(D) 8192 8192

(E) 65,534 65,534

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): None of the squares that are marked with dots in the sample scanning code shown below can be mapped to any other marked square by reflections or non-identity rotations. Therefore these 10 squares can be arbitrarily colored black or white in a symmetric scanning code, with the exception of “all black” and “all white”. On the other hand, reflections or rotations will map these squares to all the other squares in the scanning code, so once these 10 colors are specified, the symmetric scanning code is completely determined. Thus there are $2^{10}-2=1022$ symmetric scanning codes.

答案（B）：在下面给出的示例扫描码中，用点标记的那些方格，在反射或非恒等旋转下都不能映射到任何其他被标记的方格。因此，在一个对称的扫描码中，这 10 个方格可以任意涂成黑色或白色，但要排除“全黑”和“全白”这两种情况。另一方面，反射或旋转会把这些方格映射到扫描码中的所有其他方格，所以一旦这 10 个颜色被确定，这个对称扫描码就完全确定了。因此，对称扫描码的数量为 $2^{10}-2=1022$。

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