Question

Arjun 和 Beth 玩一个游戏：他们轮流从若干堵砖墙中的一堵移除一块砖，或移除相邻的两块砖；移除后产生的空隙可能把一堵墙分成新的墙。每堵墙都只有一块砖高。例如，大小为 $4$ 和 $2$ 的一组墙，经过一步可以变成以下任意一种： $(3,2),(2,1,2),(4),(4,1),(2,2)$ ，或 $(1,1,2)$ 。

Arjun 先手，移除最后一块砖的玩家获胜。对于哪一种初始配置，Beth 有必胜策略？

Arjun and Beth play a game in which they take turns removing one brick or two adjacent bricks from one "wall" among a set of several walls of bricks, with gaps possibly creating new walls. The walls are one brick tall. For example, a set of walls of sizes $4$ and $2$ can be changed into any of the following by one move: $(3,2),(2,1,2),(4),(4,1),(2,2),$ or $(1,1,2).$

Arjun plays first, and the player who removes the last brick wins. For which starting configuration is there a strategy that guarantees a win for Beth?

$(6,1,1)$

$(6,2,1)$

$(6,2,2)$

$(6,3,1)$

$(6,3,2)$

答案：B

知识点：组合游戏

难度评级：2390

解答：

对长度为 $n$ 的单独一堵墙，根据所有可能操作计算它的 Sprague-Grundy 值。

$\begin{aligned} &g(1)=1,\quad g(2)=2, \\ &g(3)=3,\quad g(4)=1, \\ &g(5)=4,\quad g(6)=3. \end{aligned}$

多堵墙的局面在这些值按位异或为 $0$ 时，正好是轮到行动者的必败局面。逐一计算选项：

$(6,1,1): 3\oplus1\oplus1=3,$

$(6,2,1): 3\oplus2\oplus1=0,$

$(6,2,2): 3\oplus2\oplus2=3,$

$(6,3,1): 3\oplus3\oplus1=1,$

$(6,3,2): 3\oplus3\oplus2=2.$

只有 $(6,2,1)$ 是轮到行动者的必败局面，所以 Beth 恰好在这个初始配置下有必胜策略。

所以答案是 B。

For a single wall of length $n,$ compute its Sprague-Grundy value from the possible moves. For the wall lengths needed here, the values are

$\begin{aligned} &g(1)=1,\quad g(2)=2, \\ &g(3)=3,\quad g(4)=1, \\ &g(5)=4,\quad g(6)=3. \end{aligned}$

For several walls, the position is losing for the player to move exactly when the xor of the wall values is $0.$ Evaluating the choices gives

$(6,1,1): 3\oplus1\oplus1=3,$

$(6,2,1): 3\oplus2\oplus1=0,$

$(6,2,2): 3\oplus2\oplus2=3,$

$(6,3,1): 3\oplus3\oplus1=1,$

$(6,3,2): 3\oplus3\oplus2=2.$

Only $(6,2,1)$ is losing for the player to move, so Beth has a guaranteed win exactly for that starting configuration.

Thus, the answer is B .

2021 AMC 10B Spring 第 24 题

答案：B

难度评级：2390

解答：

其他年份的第 24 题