A square is drawn in the Cartesian coordinate plane with vertices at $(2, 2)$, $(-2, 2)$, $(-2, -2)$, and $(2, -2)$. A particle starts at $(0, 0)$. Every second it moves with equal probability to one of the eight lattice points (points with integer coordinates) closest to its current position, independently of its previous moves. In other words, the probability is $1/8$ that the particle will move from $(x, y)$ to each of $(x, y + 1)$, $(x + 1, y + 1)$, $(x + 1, y)$, $(x + 1, y - 1)$, $(x, y - 1)$, $(x - 1, y - 1)$, $(x - 1, y)$, or $(x - 1, y + 1)$. The particle will eventually hit the square for the first time, either at one of the 4 corners of the square or at one of the 12 lattice points in the interior of one of the sides of the square. The probability that it will hit at a corner rather than at an interior point of a side is $m/n$, where $m$ and $n$ are relatively prime positive integers. What is $m + n$?
在笛卡尔坐标平面中画一个正方形，顶点为 $(2, 2)$、$(-2, 2)$、$(-2, -2)$ 和 $(2, -2)$。一个粒子从 $(0, 0)$ 开始。每秒它以相等概率移动到当前位置最近的八个格点（具有整数坐标的点）之一，与之前的移动独立。换句话说，从 $(x, y)$ 移动到 $(x, y + 1)$、$(x + 1, y + 1)$、$(x + 1, y)$、$(x + 1, y - 1)$、$(x, y - 1)$、$(x - 1, y - 1)$、$(x - 1, y)$ 或 $(x - 1, y + 1)$ 的概率均为 $1/8$。粒子最终会第一次击中正方形，要么在 $4$ 个顶点之一，要么在 $4$ 条边的 $12$ 个边内格点之一。击中顶点的概率比击中边内点的概率为 $m/n$，其中 $m$ 和 $n$ 互质。求 $m + n$？
Answer (E): Let $A=\{(1,0),(0,1),(-1,0),(0,-1)\}$, let $C=\{(0,0)\}$, and let $I=\{(1,1),(-1,1),(-1,-1),(1,-1)\}$. A particle in $A$ will move to $A$ with probability $\frac{2}{8}$, to $C$ with probability $\frac{1}{8}$, to $I$ with probability $\frac{2}{8}$, and to an interior point of a side of the square with probability $\frac{3}{8}$. Similarly, a particle in $C$ will move to $A$ with probability $\frac{4}{8}$ and to $I$ with probability $\frac{4}{8}$; and a particle in $I$ will move to $A$ with probability $\frac{2}{8}$, to $C$ with probability $\frac{1}{8}$, to a corner of the square with probability $\frac{1}{8}$, and to an interior point of a side of the square with probability $\frac{4}{8}$. Let $a$, $c$, and $i$ be the probabilities that the particle will first hit the square at a corner, given that it is currently in $A$, $C$, and $I$, respectively. The transition probabilities noted above lead to the following system of equations.
$$
\begin{array}{l}
a=\frac{2}{8}a+\frac{1}{8}c+\frac{2}{8}i\\[4pt]
c=\frac{4}{8}a+\frac{4}{8}i\\[4pt]
i=\frac{2}{8}a+\frac{1}{8}c+\frac{1}{8}
\end{array}
$$
This system can be solved by elimination to yield $a=\frac{1}{14}$, $c=\frac{4}{35}$, and $i=\frac{11}{70}$. The required fraction is $c$, whose numerator and denominator sum to $39$.
答案（E）：令 $A=\{(1,0),(0,1),(-1,0),(0,-1)\}$，令 $C=\{(0,0)\}$，并令 $I=\{(1,1),(-1,1),(-1,-1),(1,-1)\}$。位于 $A$ 的粒子以概率 $\frac{2}{8}$ 移动到 $A$，以概率 $\frac{1}{8}$ 移动到 $C$，以概率 $\frac{2}{8}$ 移动到 $I$，并以概率 $\frac{3}{8}$ 移动到正方形某一边的内部点。类似地，位于 $C$ 的粒子以概率 $\frac{4}{8}$ 移动到 $A$，以概率 $\frac{4}{8}$ 移动到 $I$；位于 $I$ 的粒子以概率 $\frac{2}{8}$ 移动到 $A$，以概率 $\frac{1}{8}$ 移动到 $C$，以概率 $\frac{1}{8}$ 移动到正方形的一个顶点，并以概率 $\frac{4}{8}$ 移动到正方形某一边的内部点。令 $a$、$c$ 和 $i$ 分别表示：当粒子当前位于 $A$、$C$、$I$ 时，它首次在顶点处碰到正方形的概率。上述转移概率导出如下方程组：
$$
\begin{array}{l}
a=\frac{2}{8}a+\frac{1}{8}c+\frac{2}{8}i\\[4pt]
c=\frac{4}{8}a+\frac{4}{8}i\\[4pt]
i=\frac{2}{8}a+\frac{1}{8}c+\frac{1}{8}
\end{array}
$$
用消元法可解得 $a=\frac{1}{14}$，$c=\frac{4}{35}$，$i=\frac{11}{70}$。所求分数是 $c$，其分子与分母之和为 $39$。