CBMS-2016-10

题目来源：[[文字版题库/CBMS/2016#Problem 10|2016#Problem 10]] 日期：2024-07-31 题目主题：CS-图论-欧拉路径与欧拉回路

解题思路

这道题目主要涉及以下几个方面：

1. 根据给定的字符串构建有向图
2. 在构建的有向图中寻找欧拉路径和欧拉回路
3. 证明欧拉回路与图的连通性和平衡性之间的关系

我们需要先理解如何从字符串构建有向图，然后分析这些图的特性来找出欧拉路径和欧拉回路。对于证明部分，我们将使用欧拉回路的性质来论证图的连通性和平衡性，然后反过来证明连通且平衡的图一定存在欧拉回路。

Solution

1. Eulerian paths and cycles in $G_{L,2}$ for $L=ACACA$

Given $L=ACACA$, we construct $G_{L,2}$ as follows:

Vertex set: $V=\{AC,CA\}$

Labeled arc set: $E=\{(AC,CA,1),(CA,AC,2),(AC,CA,3)\}$

To find all Eulerian paths and cycles, we need to traverse all arcs exactly once.

Eulerian paths:

1. $(AC,CA,1)\to (CA,AC,2)\to (AC,CA,3)$
2. $(CA,AC,2)\to (AC,CA,3)\to (AC,CA,1)$
3. $(AC,CA,3)\to (CA,AC,2)\to (AC,CA,3)$

There are no Eulerian cycles in $G_{L,2}$.

2. Eulerian paths and cycles in $G_{L,3}$ and $G_{L,4}$ for $L=GCGCGCAGCG$

For $G_{L,3}$

Vertex set: $V=\{GCG,CGC,GCA,CAG,AGC\}$

Labeled arc set:

$E=\{(GCG,CGC,1),(CGC,GCG,2),(GCG,CGC,3),(CGC,GCA,4),(GCA,CAG,5),(CAG,AGC,6),(AGC,GCG,7)\}$

Eulerian cycle (also as Eulerian path):

$GCG\to (GCG,CGC,1)\to (CGC,GCG,2)\to (GCG,CGC,3)\to (CGC,GCA,4)\to (GCA,CAG,5)\to (CAG,AGC,6)\to (AGC,GCG,7)\text{rightarro}GCG$

For $G_{L,4}$

Vertex set: $V=\{GCGC,CGCG,GCGC,CGCA,GCAG,CAGC,AGCG\}$

Labeled arc set: $E=\{(GCGC,CGCG,1),(CGCG,GCGC,2),(GCGC,CGCA,3),(CGCA,GCAG,4),(GCAG,CAGC,5),(CAGC,AGCG,6)\}$

Eulerian path:

$(GCGC,CGCG,1)\to (CGCG,GCGC,2)\to (GCGC,CGCA,3)\to (CGCA,GCAG,4)\to (GCAG,CAGC,5)\to (CAGC,AGCG,6)$

There are no Eulerian cycles in $G_{L,4}$.

3. Proof: A directed graph with an Eulerian cycle is connected and balanced

Let $G=(V,E)$ be a directed graph with an Eulerian cycle $C$.

Note: We assume that $G$ does not contain any isolated vertices (vertices with both in-degree and out-degree equal to 0), or if such vertices exist, we consider them as irrelevant to the Eulerian cycle and the properties we are about to prove.

Connectedness

Suppose $G$ is not connected. Then there exist vertices $u,v\in V$ such that there is no path from $u$ to $v$. However, the Eulerian cycle $C$ visits every edge in $G$ exactly once, which means it visits every non-isolated vertex at least once. Therefore, $C$ provides a path from $u$ to $v$, contradicting our assumption. Thus, $G$ must be connected.

Balance

Let $v\in V$ be any non-isolated vertex in $G$. Every time the Eulerian cycle $C$ enters $v$, it must also leave $v$, including the starting/ending vertex which is entered in the last step of the cycle. Therefore, the number of edges entering $v$ is equal to the number of edges leaving $v$. Since this is true for all non-isolated vertices, $G$ is balanced.

4. Proof: A connected and balanced directed graph has an Eulerian cycle

Let $G=(V,E)$ be a connected and balanced directed graph. We will prove this by constructing an Eulerian cycle.

1. Start at any vertex $v_0\in V$.
2. Choose any outgoing arc and follow it to the next vertex.
3. Continue this process, each time choosing an unused arc, until we return to $v_0$.

This process must terminate because:

a) The graph is balanced, so we can always leave a vertex we’ve entered (except possibly $v_0$).

b) The number of arcs is finite, so we must eventually return to $v_0$.

Let’s call the cycle we’ve constructed $C$. If $C$ includes all arcs in $G$, we’re done. If not, consider the subgraph $G^{\prime }=(V^{\prime },E^{\prime })$ where $V^{\prime }$ is the set of vertices incident to unused arcs, and $E^{\prime }$ is the set of unused arcs.

1. $G^{\prime }$ is balanced: The balance of vertices in $G^{\prime }$ is unaffected by removing the equal number of incoming and outgoing arcs in $C$.
2. $G^{\prime }$ is connected to $C$: If it weren’t, $G$ would not be connected.

Choose a vertex $v^{\prime }$ in $G^{\prime }$ that’s also in $C$. Start a new cycle $C^{\prime }$ from $v^{\prime }$ in $G^{\prime }$ using the same process as before. We can then combine $C$ and $C^{\prime }$ into a larger cycle.

Repeat this process until all arcs are used. The result is an Eulerian cycle in $G$.

知识点

图论欧拉路径有向图连通性平衡性

难点思路

1. 构建从字符串到有向图的映射关系
2. 在给定的图中找出所有的欧拉路径和欧拉回路
3. 理解并
4. 证明欧拉回路与图的连通性和平衡性之间的关系

解题技巧和信息

1. 欧拉路径和欧拉回路的定义：欧拉路径是遍历图中每条边恰好一次的路径，欧拉回路是起点和终点相同的欧拉路径。
2. 构建有向图：注意字符串到图的映射关系，特别是顶点和边的定义。
3. 寻找欧拉路径和回路：从任意顶点开始，每次选择一条未使用的边，直到无法继续为止。
4. 证明技巧：使用反证法和构造法来证明定理。
5. 图的性质：理解连通性和平衡性对欧拉回路存在性的影响。

重点词汇

1. Eulerian path 欧拉路径
2. Eulerian cycle 欧拉回路
3. Directed graph 有向图
4. Connected graph 连通图
5. Balanced graph 平衡图
6. Vertex 顶点
7. Arc 弧（有向边）
8. Cycle 环

参考资料

1. Bondy, J.A. and Murty, U.S.R. (2008). Graph Theory. Springer. Chapter 5: Eulerian Graphs.
2. Diestel, R. (2017). Graph Theory (5th ed.). Springer. Chapter 1: The Basics.
3. Cormen, T.H., Leiserson, C.E., Rivest, R.L., and Stein, C. (2009). Introduction to Algorithms (3rd ed.). MIT Press. Chapter 22: Elementary Graph Algorithms.