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Mike Mathcook
Asked: 2024-07-06 10:05:57 +0800 CST

以下代码的复杂度是多少？是 O(n^3log(n)) 吗？

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以下代码的复杂度是多少？是 O(n^3log(n)) 吗？

#G is an undirected dense graph, which has N vertices.
import networkx as nx
def cal_minimax_path_matrix(G):
    MST = nx.minimum_spanning_tree(G)
    minimax_matrix = np.zeros((N, N))

    for i in range(N):
        for j in range(N):
            if j > i:
                max_weight = -1 
                path = nx.shortest_path(MST, source=i, target=j)
                for k in range(len(path)-1):
                    if( MST.edges[path[k],path[k+1]]['weight'] > max_weight):
                        max_weight = MST.edges[path[k],path[k+1]]['weight']
                minimax_matrix[i,j] = minimax_matrix[j,i] = max_weight
                
    return minimax_matrix

构建最小生成树的复杂度为 O(n^2)。在最小生成树中查找最短路径的复杂度是多少？是 O(nlog(n)) 吗？

该代码基于 Madhav-99 的代码，请参阅：

https://github.com/Madhav-99/Minimax-Distance

graph-theory

1 个回答

  1. Best Answer

    这是添加了关于 O 中每个部分的复杂性的注释的代码：

    #G is an undirected dense graph, which has N vertices.
import networkx as nx
def cal_minimax_path_matrix(G):
    MST = nx.minimum_spanning_tree(G) # O(n²) (given)
    minimax_matrix = np.zeros((N, N)) # O(n²) (Initializes a NxN matrix, takes NxN time)

    for i in range(N): # O(n) (loop)
        for j in range(N): # O(n) (loops combined: O(n)*O(n)=O(n²))
            if j > i:
                max_weight = -1 # O(1) (constant)
                path = nx.shortest_path(MST, source=i, target=j) # Uses Dijkstra algorithm by default [1], which is O(ElogV) [2], applied to a minimum spanning tree this will be O(V-1 log V) [3] -> O((n-1) log n) -> O(n log n)
                for k in range(len(path)-1): # O(n) as the path can be at most n long
                    if( MST.edges[path[k],path[k+1]]['weight'] > max_weight):
                        max_weight = MST.edges[path[k],path[k+1]]['weight']
                minimax_matrix[i,j] = minimax_matrix[j,i] = max_weight
                
    return minimax_matrix

    假设的复杂度MST = nx.minimum_spanning_tree(G)为O(n²)。

    现在增加循环的复杂性：

    O(n) (inner loop)
+ O(n log n) (`nx.shortest_path`)
= O(n log n)
* O(n²) (outer loops)
= O(n³ log n)

    现在我们面临三个复杂情况：

    • 初始化 MST 所需的时间为 O(n²)
    • 初始化矩阵的时间为 O(n²)
    • 循环时间为 O(n³ log n)

    由于 O(n³ log n) > O(n²)（读取：O(n³ log n) 大于 O(n²)），cal_minimax_path_matrix(n) 的整体复杂度为 O(n³ log n)。

    参考文献：
    [1] https://networkx.org/documentation/stable/reference/algorithms/generated/networkx.algorithms.shortest_paths.generic.shortest_path.html
    [2] https://stackoverflow.com/a/26548129/8517948
    [3] https://math.stackexchange.com/questions/1524709/minimum-spanning-tree-edge-count

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