AMC12 2020 B

AMC12 2020 B · Q14

AMC12 2020 B · Q14. It mainly tests Invariants, Games (basic).

Bela and Jenn play the following game on the closed interval $[0, n]$ of the real number line, where $n$ is a fixed integer greater than 4. They take turns playing, with Bela going first. At his first turn, Bela chooses any real number in the interval $[0, n]$. Thereafter, the player whose turn it is chooses a real number that is more than one unit away from all numbers previously chosen by either player. A player unable to choose such a number loses. Using optimal strategy, which player will win the game?

Bela和Jenn在实数轴上的闭区间$[0, n]$上玩以下游戏，其中$n$是大于4的固定整数。他们轮流玩，Bela先手。在他的第一回合，Bela选择区间$[0, n]$内的任意实数。此后，轮到某玩家的回合时，该玩家选择一个与之前任一方玩家选择的所有数字距离大于1的实数。无法选择这样的数的玩家输掉。使用最优策略，谁会赢？

(A) Bela will always win. Bela 总是赢。

(B) Jenn will always win. Jenn 总是赢。

(D) Jenn will win if and only if $n$ is odd. 仅当 $n$ 为奇数时 Jenn 赢。

(E) Jenn will win if and only if $n$ > 8. 仅当 $n$ > 8 时 Jenn 赢。

Answer

Correct choice: (A)

正确答案：（A）

Solution

Answer (A): Bela has a straightforward strategy that will guarantee winning. At his first turn he chooses the number $\frac{n}{2}$. This splits the playing field into two parts—the numbers less than $\frac{n}{2}$ and the numbers greater than $\frac{n}{2}$. Thereafter, whatever legal move Jenn makes, Bela makes the symmetric move in the other half of the playing field. Specifically, if Jenn chooses $x$, then Bela chooses $n-x$. By symmetry, if Jenn’s move is possible, so is Bela’s. Thus the first player to be unable to make a legal move will be Jenn, and she will lose. The game will end after a finite number of moves, because of the restriction that the chosen numbers are at least one unit apart. Note: The game is much more interesting if the chosen numbers are restricted to being integers, in which case it is known as Dawson’s Chess. See Winning Ways for Your Mathematical Plays by Berlekamp, Conway, and Guy for a mathematical theory of games like this.

答案（A）：Bela 有一个简单直接的策略，可以保证获胜。在他第一次行动时，他选择数字 $\frac{n}{2}$。这将游戏区域分成两部分——小于 $\frac{n}{2}$ 的数以及大于 $\frac{n}{2}$ 的数。此后，无论 Jenn 做出什么合法行动，Bela 都在游戏区域的另一半做出对称的行动。具体来说，如果 Jenn 选择 $x$，那么 Bela 选择 $n-x$。由于对称性，如果 Jenn 的行动可行，那么 Bela 的也可行。因此，第一个无法做出合法行动的玩家将是 Jenn，她会输。由于所选数字之间至少相差 1 个单位这一限制，游戏会在有限步内结束。注：如果把可选数字限制为整数，这个游戏会有趣得多，此时它被称为 Dawson’s Chess。关于这类游戏的数学理论，可参见 Berlekamp、Conway 和 Guy 的《Winning Ways for Your Mathematical Plays》。

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