京都大学 情報学研究科 数理工学専攻 2018年8月実施 グラフ理論

标签：

• 京都大学
• 最短路径问题

Author

祭音Myyura

Description

大学公表の原題

日本語版

$G = (V, E)$ を節点集合 $V$、枝集合 $E$ から成る単純強連結有向グラフ、$N = [G, w]$ を $G$ の各枝 $e \in E$ に実数値の重み $w(e)$ を与えて得られるネットワークとする。 節点 $u$ から節点 $v$ への有向枝は $(u, v)$ と書き、その枝重みは $w(u,v)$ とも書く。 節点 $u$ から節点 $v$ への距離 $\text{dist}(u,v)$ を $N$ における $u$ から $v$ への単純路上の枝重みの和の最小値と定める。 枝重み和が負である有向閉路を負閉路と呼ぶ。以下の問いに答えよ。

(i) 次の条件を満たす節点の実数値重み $p(v)$, $v \in V$ が存在するとき、$N$ に負閉路が存在しないことを証明せよ。

$$w(u, v) + p(u) - p(v) \geqq 0, \quad \forall (u, v) \in E.$$

(ii) 次を満たす節点 $s \in V$ と枝 $(u,v) \in E$ の組が存在するとき、$N$ に負閉路が存在することを証明せよ。

$$\text{dist}(s, u) + w(u, v) < \text{dist}(s, v)$$

(iii) 各枝の重みが非負であると仮定する。 ある部分集合 $S \subseteq V$ と節点 $s \in S$ に対して、$S$ から $V \setminus S$ へ向かう枝 $(u, v) \in E$ の中で $\text{dist}(s, u) + w(u,v)$ の値が最小とする枝を $(u^*, v^*)$ とする。 このとき、$\text{dist}(s, v^*) = \text{dist}(s, u^*) + w(u^*, v^*)$ が成り立つことを証明せよ。

English Version

Let $G = (V, E)$ denote a simple, strongly connected digraph with a vertex set $V$ and an edge set $E$, and let $N= [G, w]$ denote a network obtained from $G$ by assigning a real value $w(e)$ ot each edge $e \in E$ as its weight. A directed edge from a vertex $u$ to a vertex $v$ is denoted by $(u, v)$ and its weight is writen as $w(u, v)$. Define the distance $\text{dist}(u, v)$ from a vertex $u$ to a vertex $v$ to be the minimum summation of weights of edges in a simple path from $u$ to $v$ in $N$. A directed cycle is called a negative cycle if the sum of edge weights in the cycle is negative. Answer the following questions.

(i) Prove that $N$ has no negative cycle if there is a set of real weights $p(v), v \in V$ such that

$$w(u, v) + p(u) - p(v) \geqq 0, \quad \forall (u, v) \in E.$$

(ii) Prove that $N$ has a negative cycle if there is a pair of a vertex $s \in V$ and an edge $(u, v) \in E$ such that

$$\text{dist}(s, u) + w(u, v) < \text{dist}(s, v)$$

(iii) Assume that the weight of each edge is non-negative. For a subset $S \subseteq V$ and a vertex $s \in S$, let $(u^*, v^*)$ be an edge that minimizes $\text{dist}(s, u) + w(u,v)$ among all edges $(u, v) \in E$ directed from $S$ to $V \setminus S$. Prove that $\text{dist}(s, v^*) = \text{dist}(s, u^*) + w(u^*, v^*)$.

题目描述

设 $G=(V,E)$ 为简单强连通有向图，网络 $N=[G,w]$ 由给每条有向边 $e\in E$ 赋实权 $w(e)$ 得到。用 $(u,v)$ 表示从 $u$ 到 $v$ 的边，权重记为 $w(u,v)$；$\operatorname{dist}(u,v)$ 是 $N$ 中从 $u$ 到 $v$ 的简单有向路的最小边权和。总权重为负的有向回路称为负环。回答：

1. 若存在顶点实数势函数 $p(v)$ 满足

   $$w(u,v)+p(u)-p(v)\geqq0,\qquad\forall (u,v)\in E,$$

   证明 $N$ 中不存在负环。

2. 若存在 $s\in V$ 与 $(u,v)\in E$ 满足

   $$\operatorname{dist}(s,u)+w(u,v)<\operatorname{dist}(s,v),$$

   证明 $N$ 中存在负环。

3. 假设所有边权非负。对 $S\subseteq V$、$s\in S$，在所有从 $S$ 指向 $V\setminus S$ 的边中，令 $(u^*,v^*)$ 使 $\operatorname{dist}(s,u)+w(u,v)$ 最小。证明

   $$\operatorname{dist}(s,v^*)= \operatorname{dist}(s,u^*)+w(u^*,v^*).$$

Kai

(i)

Please refer to 京都大学 情報学研究科 数理工学専攻 2023年8月実施 グラフ理論.

For any directed cycle $C$,

$$\sum_{(u,v)\in C}w(u,v) =\sum_{(u,v)\in C}\bigl(w(u,v)+p(u)-p(v)\bigr)\geq0,$$

because the potential terms telescope. Hence no directed cycle has negative weight.

(ii)

Let $P_{s,u}$ denote a simple path from $s$ to $u$ of distance $\text{dist}(s, u)$. Let $P'_{s, v}$ denote the path by concatenating $P_{s,u}$ and edge $(u,v)$.

If $\text{dist}(s, u) + w(u, v) < \text{dist}(s, v)$, then by the definition of "distance" we know that $P'_{s, v}$ is not a simple path otherwise $\text{dist}(s, u) + w(u, v) \geq \text{dist}(s, v)$.

Since $P_{s,u}$ is a simple path but $P'_{s, v}$ is not, we have $v \in V(P_{s,u})$. Hence there exists a cycle $C$ in $P'_{s, v}$ which can be written as $(v, \ldots, u, v)$.

Let $P_{s,u}^v$ denote the subpath of $P_{s,u}$ ends at $v$ (note that $P_{s,u}^v$ is also a simple path). If $C$ is not a negative cycle, i.e., $w(C) \geq 0$, then we have

$$\begin{aligned} w(P_{s,u}^v) &= w(P_{s,u}) - w(C) + w(u,v) \\ &= \text{dist}(s,u) - w(C) + w(u,v) \\ &< \text{dist}(s,v) \end{aligned}$$

which contradicts the definition of $\text{dist}(s,v)$. Therefore, $C$ is a negative cycle.

(iii)

(Note: this question is actually asking a proof of correctness of Dijkstra's algorithm)

Since the weight of each edge is non-negative, there does not exist negative cycle.

By the contrapositive of Question (2), since there is no negative cycle, for every edge $(u,v)$

$$\text{dist}(s,v) \leq \text{dist}(s,u) + w(u,v)$$

Hence

$$\text{dist}(s,v^*) \leq \text{dist}(s,u^*) + w(u^*,v^*)$$

Any path from $s$ to $v^*$ must go through an edge from $S$ to $V \setminus S$. Hence, let $P_{s, v^*} = \{s, \ldots, u', v', \ldots, v^*\}$ denote a shortest simple path from $s$ to $v^*$. Let $(u',v')$ be the first edge on $P_{s,v^*}$ that crosses from $S$ to $V \setminus S$ and let $P_1 = \{s, \ldots, u'\}$ and $P_2 = \{v', \ldots, v^*\}$ denote the subpaths of $P_{s, v^*}$. Then,

$$\begin{aligned} \text{dist}(s,v^*) &= w(P_{s, v^*}) = w(P_1) + w(u'v') + w(P_2) \\ &\geq w(P_1) + w(u'v') \ \ (\text{the weight of each edge is non-negative}) \\ &\geq \text{dist}(s,u') + w(u'v') \\ &\geq \text{dist}(s,u^*) + w(u^*,v^*) \ \ (\text{definition of }(u^*,v^*)) \end{aligned}$$

Therefore,

$$\text{dist}(s,v^*) = \text{dist}(s,u^*) + w(u^*,v^*)$$