离散数学:树:完全图中非同构的支撑树

Question

在完全图K5中最多有3个不同 (非同 构)的支撑树，这三个支撑树分别是什么样的

Answer 1

该回答引用自GPT-3.5,由博主GISer Liu编写：

理解了你的问题。首先，让我们来看看问题的核心是什么。你需要找到完全图K5中的非同构支撑树，而且需要找到尽可能多的非同构支撑树。在解决这个问题之前，我们先来了解一下什么是完全图和支撑树。

• 完全图K5是指包含5个顶点，每两个顶点之间都有一条边的图。
• 支撑树是指一个连通图的一棵包含所有顶点且没有环的生成树。
  首先，我们知道对于任意的n个顶点的完全图，其支撑树的边数为n-1。所以，K5的支撑树一定有4条边。然后我们来找这些支撑树。
  为了找到K5的不同支撑树，我们可以使用基于DFS（深度优先搜索）的方法。我们从一个顶点开始，然后尝试添加边，但是确保不形成环。我们可以从最简单的情况开始，即只有4条边的情况。这时候我们会发现有多种情况可以形成支撑树。接着我们考虑添加第5条边的情况，我们需要排除与前面情况同构的树，确保找到的是不同的支撑树。
  由于这是一个比较手动的过程，需要不断尝试并检查，我将先给出一个基本的代码框架，然后根据这个框架，你可以自己尝试生成K5的所有不同支撑树。
  python
  Copy code

  class Graph:
  def __init__(self, vertices):
  self.V = vertices
  self.graph = [[0 for _ in range(vertices)] for _ in range(vertices)]
  # A utility function to add an edge in an undirected graph
  def add_edge(self, src, dest):
  self.graph[src][dest] = 1
  self.graph[dest][src] = 1
  # A utility function to do DFS of graph
  def dfs(self, v, visited):
  visited[v] = True
  for i in range(self.V):
  if self.graph[v][i] == 1 and not visited[i]:
  self.dfs(i, visited)
  # Function to check if all non-zero degree vertices are connected.
  def is_connected(self):
  visited = [False] * self.V
  # Find a vertex with non-zero degree
  for i in range(self.V):
  if sum(self.graph[i]) > 0:
  break
  # If there are no edges in the graph, return true
  if i == self.V - 1:
  return True
  # Start DFS traversal from a vertex with non-zero degree
  self.dfs(i, visited)
  # Check if all non-zero degree vertices are visited
  for i in range(self.V):
  if sum(self.graph[i]) > 0 and not visited[i]:
  return False
  return True
  # Function to check if adding an edge results in a cycle
  def is_cyclic(self, u, v):
  # Add the edge to the graph
  self.graph[u][v] = 1
  self.graph[v][u] = 1
  # Check if the graph becomes cyclic after adding the edge
  visited = [False] * self.V
  if self.dfs_for_cycle_detection(u, visited, -1):
  # Remove the edge if it results in a cycle
  self.graph[u][v] = 0
  self.graph[v][u] = 0
  return True
  # Remove the edge added for checking
  self.graph[u][v] = 0
  self.graph[v][u] = 0
  return False
  # DFS function to detect cycle in the graph
  def dfs_for_cycle_detection(self, v, visited, parent):
  visited[v] = True
  for i in range(self.V):
  if self.graph[v][i] == 1:
  if not visited[i]:
  if self.dfs_for_cycle_detection(i, visited, v):
  return True
  elif parent != i:
  return True
  return False
  # Function to find all the edges that can be added without forming a cycle
  def find_edges(self):
  edges = []
  for i in range(self.V):
  for j in range(i + 1, self.V):
  if self.graph[i][j] == 0 and not self.is_cyclic(i, j):
  edges.append((i, j))
  return edges
  # Function to find all spanning trees of a graph
  def find_spanning_trees(graph):
  # Find all the edges that can be added without forming a cycle
  edges = graph.find_edges()
  # List to store all spanning trees
  spanning_trees = []
  # Iterate through all possible combinations of edges
  for i in range(len(edges)):
  for j in range(i + 1, len(edges)):
  for k in range(j + 1, len(edges)):
  for l in range(k + 1, len(edges)):
  # Create a new graph to store the current spanning tree
  temp_graph = Graph(graph.V)
  # Add the selected edges to the graph
  temp_graph.graph = [row[:] for row in graph.graph]
  for edge in [edges[i], edges[j], edges[k], edges[l]]:
  temp_graph.add_edge(edge[0], edge[1])
  # If the resulting graph is connected, add it to the list of spanning trees
  if temp_graph.is_connected() and temp_graph.graph not in spanning_trees:
  spanning_trees.append(temp_graph.graph)
  return spanning_trees
  # Main function
  if __name__ == "__main__":
  # Create a graph with 5 vertices
  graph = Graph(5)
  # Add edges to the graph to form K5
  graph.add_edge(0, 1)
  graph.add_edge(0, 2)
  graph.add_edge(0, 3)
  graph.add_edge(0, 4)
  graph.add_edge(1, 2)
  graph.add_edge(1, 3)
  graph.add_edge(1, 4)
  graph.add_edge(2, 3)
  graph.add_edge(2, 4)
  graph.add_edge(3, 4)
  # Find all spanning trees of K5
  spanning_trees = find_spanning_trees(graph)
  # Print the number of spanning trees
  print("Number of spanning trees:", len(spanning_trees))
  print()
  # Print all spanning trees
  for idx, tree in enumerate(spanning_trees):
  print("Span
  

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