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笼中狮子

lions in a cage

专题
Brainteaser / 脑筋急转弯
难度
L2

题目详情

笼子里有 nn 头狮子。有人扔进来一块肉。每头狮子要么把整块肉吃掉，要么完全不吃。

如果某头狮子吃了肉，它会立刻睡着，并在其他狮子眼里变成“一块肉”（可以被吃掉）。

问：最后会发生什么？

英文原题

There are nn lions in a cage. A piece of meat is thrown into the cage. A lion can either eat the whole piece of meat or none of it. If it eats the meat, it falls asleep and becomes to the other lions a piece of meat. What happens?

解析

假设狮子都足够理性：只有在“吃了不会让自己被其他狮子吃掉”时才会吃。

用归纳：

  • n=1n=1：唯一的狮子吃掉肉，安全。
  • n=2n=2：若其中一头吃肉就会睡着，另一头会把它吃掉，因此没人敢吃。
  • 假设对 n−1n-1 成立。

当有 nn 头狮子时，若某头狮子吃肉睡着，那么笼中还剩 n−1n-1 头清醒狮子面对“一块肉”（睡着的那头）。根据归纳假设：

  • 若 n−1n-1 为奇数（即 nn 为偶数），会有狮子去吃这块“肉”，于是最初吃肉那头会被吃掉，因此没有狮子敢先吃；
  • 若 n−1n-1 为偶数（即 nn 为奇数），不会有狮子去吃这块“肉”，因此最初吃肉那头是安全的，于是会有人吃。

结论：n 为奇数时会有狮子吃；n 为偶数时无人吃。\boxed{n\text{ 为奇数时会有狮子吃；}n\text{ 为偶数时无人吃。}}

英文解析

Assuming the lions are sufficiently rational: they will eat only if eating will not leave them vulnerable to being eaten by another lion.

Using induction:

-n=1n=1: The single lion eats the meat and remains safe.
-n=2n=2: If one lion eats the meat, it will fall asleep; the other lion will then eat it, so no lion dares to eat.

  • Assume the statement holds for n−1n-1.

When there are nn lions, if one lion eats the meat and falls asleep, the cage is left with n−1n-1 awake lions facing "a piece of meat" (the sleeping one). According to the inductive hypothesis:

  • If n−1n-1 is odd (i.e., nn is even), a lion will eat this "piece of meat," so the lion that initially ate will be eaten; therefore, no lion dares to eat first.
  • If n−1n-1 is even (i.e., nn is odd), no lion will eat this "piece of meat," so the lion that initially ate remains safe; therefore, someone will eat.

Conclusion: Lions will eat when n is odd; no one will eat when n is even.\boxed{\text{Lions will eat when } n \text{ is odd; no one will eat when } n \text{ is even.}}