京都大学 情報学研究科 数理工学専攻 2018年8月実施 アルゴリズム基礎

标签：

• Kyoto-University
• Articulation-Point
• Depth-First-Search

Author

祭音Myyura

Description

大学公表の原題 Let $G=(V,E)$ be a connected simple undirected graph with a set $V$ of $n\ge 2$ vertices and a set $E$ of edges, let $T=(V,F)$ be a spanning tree of $G$ rooted at a vertex $s\in V$, and let $\ell:V\to \{1,2,\ldots,n\}$ be a numbering on $V$, where we assume that the following conditions (a) and (b) hold.

(a) For each edge $uv\in E$, vertex $u$ is either an ancestor or a descendant of $v$ in $T$.

(b) For each vertex $v\in V\setminus\{s\}$ and the parent $u$ of $v$ in $T$, $\ell(u)<\ell(v)$.

Let $L$ denote the set of leaves in $T$. For each vertex $v\in V$, let $N(v)$ denote the set of neighbors of $v$ in $G$, and let $D(v)$ denote the set consisting of vertex $v$ and the descendants of $v$ in $T$. Define a function

$$\operatorname{lowpt}: V\setminus (L\cup\{s\}) \to \{1,2,\ldots,n\}$$

such that

$$\operatorname{lowpt}(v) = \min\left\{ \ell(y)\mid y\in \bigcup_{x\in D(v)}N(x) \right\}, \ v\in V\setminus (L\cup\{s\}).$$

Answer the following questions.

(i) Prove that no leaf $u\in L$ is a cut-vertex in $G$.

(ii) Prove that a necessary and sufficient condition for the root $s$ to be a cut-vertex in $G$ is that $s$ has at least two children in $T$.

(iii) Prove that a necessary and sufficient condition for a vertex $u\in V\setminus(L\cup\{s\})$ to be a cut-vertex in $G$ is that $u$ has a child $v$ in $T$ such that

$$\operatorname{lowpt}(v)\ge \ell(u).$$

题目描述

设 $G=(V,E)$ 是含 $n\ge2$ 个顶点的连通简单无向图，$T=(V,F)$ 是以 $s\in V$ 为根的生成树，$\ell:V\to\{1,2,\ldots,n\}$ 是顶点编号，并满足：

1. 对每条边 $uv\in E$，$u$ 在 $T$ 中是 $v$ 的祖先或后代；
2. 对每个非根顶点 $v$ 及其父顶点 $u$，有 $\ell(u)<\ell(v)$。

令 $L$ 为 $T$ 的叶集，$N(v)$ 为 $v$ 在 $G$ 中的邻接点集合，$D(v)$ 为 $v$ 及其在 $T$ 中所有后代组成的集合。对每个非根顶点 $v\in V\setminus(L\cup\{s\})$ 定义

$$\operatorname{lowpt}(v)= \min\left\{\ell(y)\ \middle| y\in\bigcup_{x\in D(v)}N(x)\right\}.$$

回答：

1. 证明任一叶顶点 $u\in L$ 都不是 $G$ 的割点。
2. 证明根 $s$ 是割点的充要条件是 $s$ 在 $T$ 中至少有两个子顶点。
3. 证明非叶、非根顶点 $u\in V\setminus(L\cup\{s\})$ 是割点的充要条件是：$u$ 在 $T$ 中存在子顶点 $v$ 满足 $\operatorname{lowpt}(v)\ge\ell(u)$。

Kai

以下では lowpt を同じ式で全ての非根頂点に定義する。葉 $v$ でも親が隣接するので最小値は存在します。

(1)

Let $u \in L$, i.e., $u$ is a leaf of $T$.

Sinc $T$ is a spanning tree of $G$, $T$ is connected.

After removing the vertex $u$ from $T$, the remaining graph $T - u$ is still connected and spans all the vertices $V \setminus \{u\}$.

Note that $T - u$ is a subgraph of $G - u$, hence $G - u$ is connected, which implies that $u$ is not a cut-vertex.

(2)

(Necessity) Assume that $s$ is a cut-vertex.

If $s$ has only one child, then $T-s$ is precisely the subtree rooted at that child and is connected. Hence $G-s$ is connected.

This contradicts the assumption that $s$ is a cut-vertex.

(Sufficiency) Assume that $s$ has at least two children in $T$. Let two of them be $v_1$ and $v_2$.

Consider the two subtrees $D(v_1)$ and $D(v_2)$. Since $v_1$ and $v_2$ are different children of the root $s$, the vertices in $D(v_1)$ and $D(v_2)$ are not in an ancestor-descendant relationship with each other.

By condition (a), there cannot be any edge between a vertex in $D(v_1)$ and a vertex in $D(v_2)$.

After deleting $s$, the two subtrees $D(v_1)$ and $D(v_2)$ cannot be connected to each other. Hence $G - s$ is disconnected, which implies that $s$ is a cut-vertex.

(3)

(Necessity) Assume that $u$ is a cut-vertex.

We prove that there must exist a child $v$ of $u$ such that $\text{lowpt}(v) \ge \ell(u)$ by contradiction.

Suppose that every child $v$ of $u$ satisfies $\text{lowpt}(v) < \ell(u)$. Then for each child $v$ of $u$, there exist some vertex $x \in D(v)$ and some neighbor $y \in N(x)$ such that $\ell(y) < \ell(u)$.

Since $x \in D(v)$, the vertex $x$ is a descendant of $u$. By condition (b), every descendant of $u$ has numbering greater than $\ell(u)$.

Therefore, the vertex $y$ of $\ell(y) < \ell(u)$ is not in $D(v)$.

By condition (a), since $x$ is a descendant of $u$, and $y$ has numbering smaller than $\ell(u)$, the vertex $y$ must be a proper ancestor of $u$.

Thus every child subtree $D(v)$ of $u$ has an edge to some proper ancestor of $u$, i.e., after deleting $u$, every child subtree $D(v)$ can still connect to the part above $u$ through an edge to a proper ancestor of $u$.

Therefore, $G - u$ remains connected, which contradicts the assumption that $u$ is a cut-vertex.

(Sufficiency) Assume that $u$ has a child $v$ satisfying $\text{lowpt}(v) \ge \ell(u)$.

Note that $v$ is a neighbor of $u$, i.e., $uv \in E$, when computing $\text{lowpt}(v)$, the vertex $u$ can be reached from $v$ by one graph edge.

Hence $\text{lowpt}(v) \le \ell(u)$, together with the assumption we have

$$\text{lowpt}(v) = \ell(u)$$

This means that the subtree $D(v)$ cannot reach any proper ancestor of $u$ through a graph edge. The smallest numbered vertex reachable from $D(v)$ is exactly $u$.

Equivalently, $D(v)$ has no edge to any proper ancestor of $u$.

After deleting $u$, the subtree $D(v)$ is disconnected from the part of the graph above $u$. Therefore $G - u$ is disconnected.

Hence $u$ is a cut-vertex.