IS CS-2024S-03

题目来源：Problem 3 日期：2024-08-17 题目主题：CS-图论-连通性与生成树

解题思路

在这一题中，我们将处理连通图的连通性检查、生成树的构建、生成树的替换边操作以及寻找割点的算法。这些问题主要涉及图论中的基本算法，如深度优先搜索（DFS）以及图的性质分析。

Solution

Question 1: Describe an algorithm to check whether or not $G$ is connected. Estimate the time complexity of the algorithm

Algorithm

To check whether the graph $G=(V,E)$ is connected, you can perform a Depth-First Search (DFS) or Breadth-First Search (BFS) starting from any vertex $v\in V$. The algorithm proceeds as follows:

1. Start from an arbitrary vertex $v_0$ and perform DFS or BFS.
2. Mark all visited vertices during the search.
3. After the search is complete, check if all vertices have been visited.

If all vertices are visited, then the graph is connected; otherwise, it is not.

Time Complexity

• The time complexity of both DFS and BFS is $O(n+m)$, where $n=|V|$ and $m=|E|$. This is because, in the worst case, you will visit every vertex and every edge exactly once.

Question 2: Describe a $O(m)$ time algorithm to find a spanning tree $T$, given a connected graph $G$

Algorithm

To find a spanning tree $T$ for a connected graph $G$, you can use a Depth-First Search (DFS) or Breadth-First Search (BFS):

1. Start from an arbitrary vertex $v_0$ and initialize $T$ as an empty set of edges.
2. Perform DFS or BFS, adding each edge traversed to $T$ until all vertices are visited.
3. The resulting set of edges $T$ forms a spanning tree.

Time Complexity

• The time complexity of this algorithm is $O(m)$ because each edge is considered exactly once.

Question 3: Prove that one can obtain another spanning tree of $G$ from $T$ by replacing $e$ with another edge $f\ne e$ of $G$

Proof

Let $T$ be a spanning tree of $G$ and $v$ be a non-leaf node in $T$. Suppose $v$ is not a cut vertex of $G$ and $e$ is an edge incident to $v$ in $T$.

1. Since $v$ is not a cut vertex, removing $v$ from $G$ does not disconnect the graph. Therefore, there exists another path in $G\setminus \{v\}$ that connects the components formed by the removal of $v$.
2. Let $f$ be an edge in $G$ that connects two components of $T\setminus \{e\}$.
3. Adding edge $f$ to $T$ will create a cycle because $T$ is a spanning tree.
4. Remove $e$ from the cycle, and you will obtain a new spanning tree $T^{\prime }$. The edge $f$ replaces $e$, forming $T^{\prime }$, which is also a spanning tree.

Thus, replacing edge $e$ with $f$ gives another spanning tree.

Question 4: Describe a $o(mn)$ time algorithm, given a connected graph $G$, to find all cut vertices of $G$

Algorithm

To find all cut vertices of $G$ efficiently, we can use a Depth-First Search (DFS) based algorithm, known as Tarjan’s algorithm:

1. Perform a DFS traversal of $G$, numbering the vertices in the order they are visited.
2. For each vertex $v$, maintain two values:
   • DFS number: The order in which the vertex was visited.
   • Low number: The lowest DFS number reachable from $v$ using back edges.
3. A vertex $v$ is a cut vertex if:
   • It is the root of the DFS tree and has more than one child.
   • It is not the root, and there is a child $u$ such that no vertex in the subtree rooted at $u$ can reach a vertex higher up in the DFS tree than $v$.

Time Complexity

• The time complexity of Tarjan’s algorithm is $O(n+m)$, which is much more efficient than $o(mn)$.

知识点

DFS图论连通性生成树割点切点

解题技巧和信息

1. 图的连通性检查: DFS 和 BFS 是检查图的连通性的基本工具。
2. 生成树构造: DFS 或 BFS 都可以用来构造图的生成树，时间复杂度为 $O(m)$。
3. 割点的寻找: Tarjan 算法是寻找割点的经典算法，其时间复杂度为 $O(n+m)$，适合大规模图的处理。

重点词汇

• Cut vertex: 割点
• Spanning tree: 生成树
• Depth-First Search (DFS): 深度优先搜索
• Connected graph: 连通图

参考资料

1. Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. “Introduction to Algorithms.” MIT Press, Chapter 22, “Elementary Graph Algorithms”.
2. Robert Tarjan, “Depth-First Search and Linear Graph Algorithms”, SIAM Journal on Computing, 1972.