Question

一个扫描码由一个 $7 \times 7$ 方格构成，其中一些小方格涂黑，其余小方格涂白。在这个由 $49$ 个小方格组成的网格中，两种颜色都必须至少出现一次。

如果把整个正方形绕中心逆时针旋转 $90^{\circ}$ 的任意整数倍，或关于连接两个对角顶点的直线、连接两条对边中点的直线反射后，扫描码的外观都不改变，就称它是对称的。

一共有多少种可能的对称扫描码？

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：对称性乘法原理

难度评级：1970

小提示：

按照正方形的对称性把小方格分成若干轨道。

Classify grid squares by symmetry orbits

大提示：

每个轨道选择一种颜色，再排除两种纯色涂法。

Choose one color per orbit, excluding the two constant colorings

解答：

把行和列从 $-3$ 编号到 $3$ ，中心为 $(0,0)$ 。旋转和反射可以改变坐标的符号，也可以交换两个坐标，所以 $(x,y)$ 所在的轨道由 $(|x|,|y|)$ 从小到大排序后得到的数对决定。这样的数对是满足 $0\le u\le v\le3$ 的 $(u,v)$ ，共有 $4+3+2+1=10$ 个。

每个轨道选定一个小方格的颜色后，该轨道内其余方格的颜色都由对称性确定。

因此，在考虑两种颜色都必须出现的条件之前，共有 $2^{10}$ 种对称涂法。

全黑和全白两种涂法不合要求，所以有效的对称扫描码共有 $2^{10}-2=1022$ 种。因此正确答案是 B。

Number rows and columns from $-3$ through $3,$ with the center at $(0,0).$ Rotations and reflections can change signs and exchange coordinates, so the orbit of $(x,y)$ is determined by the ordered pair obtained by sorting $(|x|,|y|).$ These pairs are $(u,v)$ with $0\le u\le v\le3,$ of which there are $4+3+2+1=10.$

Once one square in each orbit is colored, symmetry forces the colors of all other squares in that orbit.

There are therefore $2^{10}$ symmetric colorings before the condition about using both colors. The all-black and all-white colorings are not allowed.

The total number of valid symmetric scanning codes is $2^{10}-2=1022$ . Thus, B is the correct answer.

2018 AMC 10A 第 20 题

答案：B

难度评级：1970

小提示：

大提示：

解答：

其他年份的第 20 题