Question

假设您有一个字典，其中包含图形的顶点值，如下所示：

graph = {'A':['B', 'C'], 'B': ['C', 'D'], 'C':['D', 'F'], D:[], F:[E], E: ['F']}

*为简单起见，图表不是双向的。这意味着如果A连接到B，这并不一定意味着B连接到A.

问题：为所有顶点获得高达 N 度的连接度。我知道这是一个非常复杂的问题。那么有什么方法，除了多个for循环来看它吗？也许有些论文你也可以参考我？必须有更好的方法来查看它，因为使用for循环可能达到O（n ^ n）复杂度（如果我没有弄错的话）。

我的代码计算最多 2 连接度是这样的：

def getDegreesOfConnectivity(graph):
    '''
    Based on a dictionary with the graphs vertexes creates a list of dictionaries for degrees of connectivity.
    Args:
        :param graph (dictionary): A dictionary whose keys are the vertexes and the values are the vertexes to which they go 
            **direct** interactions.
    Returns:
        list of dictionaries: A list containing dictionaries. Each dictionary has **3** keys.
            vertex: Starting node in which we start looking at the **path**
            first_degree_of_connectivity: All the vertexes that are directly connected to **initial** vertex
            second_degree_of_connectivity: Multiple **lists** of vertexes connected to each vertex in the **first_degree_of_connectivity**.

    '''
    sys.stdout.write("Calculating degrees of connectivity...\n")
    li = []
    # For easier understanding, assume vertex1 refers to vertex in the 1st level of connectivity.
    # vertex2, to vertex with 2 levels of connectivity, etc...
    for vertex1 in graph:
        dic = {}
        dic['vertex'] = vertex1
        dic['first_degree_of_connectivity'] = graph[vertex1]
        dic['second_degree_of_connectivity'] = []
        # This was the easy part. Now to get the 2nd degree of connectivity...
        #dic['second_degree_of_connectivity']
        for vertex2 in graph[vertex1]:
            # look in to the first degree of connectivity vertexes and compare them with our data. That means
            # go through the graph data again. for every vertex...
            if vertex2 in graph:
                dic['second_degree_of_connectivity'].append(graph[vertex2])

        li.append(dic)
    return li

Answer 1

def findNode(graph,head,target):
    if head == target: return 0
    if not graph[head]: return float('inf') # hackery ... basically means no path
    return 1 + min(findNode(graph,child,target) for  child in graph[head])

可能是我将如何递归地解决这个问题（警告它会破坏非常深的图的递归深度）

Answer 2

PIP存储库中的模块vertex connectivity上存在一个计算python-igraph的python实现。

请参阅文档，网址为

https://igraph.org/python/doc/igraph.Graph-class.html#vertex_disjoint_paths

请注意，它并不以字典作为输入，而是提供了几种实例化图形的方法。

图表中的连接度

2 个答案: