2021 AMC 10B Spring 第 24 题

先试着解答 2021 AMC 10B Spring 第 24 题,然后核对你的答案与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2021 AMC 10B Spring 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

24.

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 44 和 22 的一组墙,经过一步可以变成以下任意一种:(3,2)(3,2),(2,1,2)(2,1,2),(4)(4),(4,1)(4,1),(2,2)(2,2),或 (1,1,2)(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 44 and 22 can be changed into any of the following by one move: (3,2),(3,2), (2,1,2),(2,1,2), (4),(4), (4,1),(4,1), (2,2),(2,2), or (1,1,2).(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)(6,1,1)
(6,2,1)(6,2,1)
(6,2,2)(6,2,2)
(6,3,1)(6,3,1)
(6,3,2)(6,3,2)
答案:B
知识点:组合游戏
小提示:
计算每个相关单独墙长的游戏值。
Compute the game value for a single wall of each relevant size
大提示:
当所有墙的游戏值按位异或为 00 时,轮到行动的玩家处于必败状态。
A position is losing for the player to move exactly when the xor of the wall values is 00
解答:
对长度为 nn 的单独一堵墙,根据所有可能操作计算它的 Sprague-Grundy 值。这里用到的各个墙长的值为
g(1)=1,g(2)=2,g(3)=3,g(4)=1,g(5)=4,g(6)=3。 \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}\text{。}
多堵墙的局面在这些值按位异或为 00 时,正好是轮到行动者的必败局面。逐一计算选项:
(6,1,1):3⊕1⊕1=3,(6,1,1): 3\oplus1\oplus1=3\text{,}
(6,2,1):3⊕2⊕1=0,(6,2,1): 3\oplus2\oplus1=0\text{,}
(6,2,2):3⊕2⊕2=3,(6,2,2): 3\oplus2\oplus2=3\text{,}
(6,3,1):3⊕3⊕1=1,(6,3,1): 3\oplus3\oplus1=1\text{,}
(6,3,2):3⊕3⊕2=2。(6,3,2): 3\oplus3\oplus2=2\text{。}
只有 (6,2,1)(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,n, compute its Sprague-Grundy value from the possible moves. For the wall lengths needed here, the values are
g(1)=1,g(2)=2,g(3)=3,g(4)=1,g(5)=4,g(6)=3. \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 0.0. Evaluating the choices gives
(6,1,1):3⊕1⊕1=3,(6,1,1): 3\oplus1\oplus1=3,
(6,2,1):3⊕2⊕1=0,(6,2,1): 3\oplus2\oplus1=0,
(6,2,2):3⊕2⊕2=3,(6,2,2): 3\oplus2\oplus2=3,
(6,3,1):3⊕3⊕1=1,(6,3,1): 3\oplus3\oplus1=1,
(6,3,2):3⊕3⊕2=2.(6,3,2): 3\oplus3\oplus2=2.
Only (6,2,1)(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 .
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