2021 AMC 12B Spring 第 22 题

先试着解答 2021 AMC 12B Spring 第 22 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 12B 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。

22.

Arjun 和 Beth 玩一个游戏:他们轮流从若干堵砖墙中的一堵移走一块砖,或移走两块相邻的砖;移走后产生的空隙可能形成新的砖墙。每堵砖墙都只有一块砖高。例如,大小为 44 和 22 的一组砖墙可以通过一步变成以下任意一种:(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,移走最后一块砖的玩家获胜。对于哪一个初始配置,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:组合游戏不变量
小提示:
计算长度为 nn 的单堵砖墙的 Grundy 值,其中一步可以移走 11 块或 22 块相邻砖,并可能把砖墙分裂
Compute the Grundy value of a single wall of length n,n, where a move removes 11 or 22 adjacent bricks and may split the wall
大提示:
Beth(后手)获胜当且仅当各堵砖墙 Grundy 值的异或为 00。
Beth (the second player) wins exactly when the XOR of the walls’ Grundy values is 00
解答:
把每堵砖墙看作一个类似 Nim 的堆,并赋予 Grundy 值。一步移走 11 块或 22 块相邻砖,可能把一堵墙分成长度为 a,ba,b 的两堵。因此,g(n)g(n) 是所有满足 a+b=n−1a+b=n-1 或 a+b=n−2a+b=n-2 的 g(a)⊕g(b)g(a)\oplus g(b) 的最小未出现非负整数。
从 g(0)=0g(0)=0 开始,这个递推给出 g(1),g(2),…,g(6)g(1),g(2),\ldots,g(6) 依次为 1,2,3,1,4,31,2,3,1,4,3。
后手 Beth 获胜当且仅当各墙 Grundy 值的异或为 00。检查每个选项,只有 (6,2,1)(6,2,1) 给出 g(6)⊕g(2)⊕g(1)g(6)\oplus g(2)\oplus g(1) =3⊕2⊕1=0=3\oplus 2\oplus 1=0。
所以正确答案是 B。
Treat each wall as a Nim-like heap with a Grundy value. A move removes 11 or 22 adjacent bricks, possibly splitting a wall into lengths a,b.a,b. Thus g(n)g(n) is the mex of g(a)⊕g(b)g(a)\oplus g(b) over a+b=n−1a+b=n-1 or a+b=n−2.a+b=n-2.
Starting with g(0)=0,g(0)=0, this recurrence gives g(1),g(2),…,g(6)g(1),g(2),\ldots,g(6) equal to 1,2,3,1,4,3,1,2,3,1,4,3, respectively.
The second player Beth wins exactly when the XOR of the walls’ Grundy values is 0.0. Checking each option, only (6,2,1)(6,2,1) gives g(6)⊕g(2)⊕g(1)g(6)\oplus g(2)\oplus g(1) =3⊕2⊕1=0.=3\oplus 2\oplus 1=0.
Thus, the correct answer is B.
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