Question

一个扫描码由 $7 \times 7$ 的方格组成，其中一些小方格涂成黑色，其余涂成白色。在这个 $49$ 个小方格的网格中，必须至少有一个小方格为每种颜色。如果把整个正方形绕中心逆时针旋转 $90^\circ$ 的倍数，或沿连接相对顶点的直线、连接相对边中点的直线反射后，外观都不改变，则称这个扫描码是对称的。可能的对称扫描码总数是多少？

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 $90^\circ$ 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?

$510$

$1022$

$8190$

$8192$

$65{,}534$

答案：B

知识点：对称性基本计数

难度评级：1800

解答：

在正方形的对称群作用下， $49$ 个方格分成若干轨道，同一轨道中的每个方格必须颜色相同。以中心为原点，给方格坐标 $(i,j)$ ，其中 $-3\le i,j\le3.$ 旋转和翻折可以改变坐标符号并交换两个坐标，所以每个轨道都有唯一代表满足 $0\le i\le j\le3.$ 这样的数对共有 $1+2+3+4=10$ 个。每个轨道可选黑色或白色，得到 $2^{10}$ 种着色，但要排除全黑和全白的网格。因此共有 $2^{10} - 2 = 1022$ 个对称扫描码。

因此，正确答案是 B。

Under the symmetry group of the square, the $49$ cells break into orbits, and every cell in an orbit must have the same color. Give a cell coordinates $(i,j)$ relative to the center, where $-3\le i,j\le3.$ Rotations and reflections can change signs and interchange the coordinates, so each orbit has one representative with $0\le i\le j\le3.$ There are $1+2+3+4=10$ such pairs. Each orbit is black or white, giving $2^{10}$ colorings, but the all-black and all-white grids are excluded. So there are $2^{10} - 2 = 1022$ symmetric scanning codes.

Thus, the correct answer is B.

2018 AMC 12A 第 15 题

答案：B

难度评级：1800

解答：

其他年份的第 15 题