feat(graph/tree-centroid.md)：增加对重心若干性质的证明 #5431
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purple-vine
commented
Feb 25, 2024
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purple-vine
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- 我已认真阅读贡献指南 (contributing guidelines) 和社区公约 (code of conduct)，并遵循了如何参与页及格式手册页的相应规范。
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请认真阅读贡献指南 (contributing guidelines) 和社区公约 (code of conduct)，并在之后勾选选择框，否则您的 PR 将无法通过 checks。 |
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可以的 |
Tiphereth-A
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Mar 26, 2024
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| 左推右：考虑证明逆否命题，若存在某个 $v$ 使得 $siz(u, v) > n/2$，则 $u$ 不是重心。 | ||
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| 将根节点换为 $v$，$siz(v, u) = n-siz(u, v) < siz(u, v)$，又 $\displaystyle \sum_{(w, v) \in E, w \neq u} siz(v, w) = siz(u, v) - 1 \Longrightarrow \max_{(w, v) \in E, w \neq u} \{siz(v, w)\} < siz(u, v)$，故以 $v$ 为根的最大子树大小严格小于以 $u$ 为根的最大子树大小，与重心的定义矛盾。 | ||
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| 右推左：设 $u$ 满足 $\forall (u, v) \in E, siz(u, v) \leq n/2$ 且不为重心，重心为 $v$，考虑 $u,v$ 之间的路径，注意到 $siz(v, u) = n - siz(u, v) \geq n-n/2$ 即 $n-n/2 \leq siz(u, v) \leq n/2$。若 $n$ 不为偶数或者 $siz(u, v) \neq n/2$，矛盾。否则 $u, v$ 为根的最大子树大小相等，均为 $n/2$，与 $u$ 不是重心矛盾。 |
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明明证充分性和必要性都是用的反证法，为什么要用两种不同的说法 🤔
另外，应该用一些比较明显的方式来标识证明内容的范围，比如用 Collapsible Blocks 包起来（参考 https://oi-wiki.org/math/number-theory/lift-the-exponent/）或者单列一小节（参考 https://oi-wiki.org/math/number-theory/inverse/#%E8%AF%81%E6%98%8E）。
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| 下面证明，树中所有点到某个点的距离和中，到重心的距离和是最小的；如果有两个重心，那么到它们的距离和一样。 | ||
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| 考虑 $sum(u) = \sum_v dis(u, v)$，其中 $dis(u, v)$ 是 $u$ 到 $v$ 的唯一简单路径的长度。若 $\min_x\{sum(x)\}=sum(u)$，$\forall (u, v) \in E, sum(v)-sum(u) = siz(v, u) - siz(u, v) = n-2siz(u, v) \geq 0$，因此 $u$ 是重心。 |
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这里 u 和 v 的取值范围是什么，现在的写法说的是树中任意节点都是重心 🤔
HeRaNO
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你好 👋 这个 Pull Request 已经 30 天未响应。 感谢理解与支持。 |
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