Alice 能保证必胜吗

把这些数都写成 $2^k$：集合为 $\{2^{-4},2^{-3},\ldots,2^4\}$。

三数乘积为 1 等价于指数和为 0：

$$2^{k_1}2^{k_2}2^{k_3}=1\iff k_1+k_2+k_3=0.$$

将每个指数平移 $+5$（把 $\{-4,-3,\ldots,4\}$ 映射到 $\{1,2,\ldots,9\}$），条件变为“三个数之和为 15”。这正是经典的 15 游戏，与井字棋同构。

结论：在双方最优对弈下，先手 不能保证必胜（最优结果为不输/逼平；对手失误时才可赢）。

实用策略：先手优先拿 $1$（指数 0，对应井字棋中心格），随后每步优先阻止对手形成任意三数乘积为 1 的组合，并尝试制造“双威胁”（下一步有两种不同方式可凑出乘积为 1）。

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英文解析

Write all these numbers as $2^k$: the set is $\{2^{-4}, 2^{-3}, \ldots, 2^4\}$.

The product of three numbers equals 1 if and only if the sum of their exponents equals 0:

$$2^{k_1}2^{k_2}2^{k_3}=1\iff k_1+k_2+k_3=0.$$

Shift each exponent by $+5$ (mapping $\{-4,-3,\ldots,4\}$ to $\{1,2,\ldots,9\}$), and the condition becomes "the sum of three numbers equals 15". This is exactly the classic 15 puzz \leq, which is isomorphic to Tic-Tac-Toe.

Conclusion: Under optimal play by both sides, the first player cannot guarantee a win (the optimal result is a draw; the first player can only win if the opponent makes a mistake).

Practical strategy: The first player should prioritize taking $1$ (exponent 0, corresponding to the center square in Tic-Tac-Toe), and then in each subsequent move, prioritize blocking the opponent from forming any combination of three numbers whose product equals 1, while attempting to create a "double threat" (two different ways to form a product of 1 in the next move).