What is the complexity of the following code? Is it O(n^3log(n))?

Question

What is the complexity of the following code? Is it O(n^3log(n))?

#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

The complexity of constructing a minimum_spanning_tree is O(n^2). What is the complexity of finding a shortest path in a minimum_spanning_tree? Is it O(nlog(n))?

The code is based on Madhav-99's code, see:

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

Answer 1

This is the code with comments added about each section's complexity in O:

#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

given the complexity of MST = nx.minimum_spanning_tree(G) is O(n²).

Now adding up the complexity for the loops:

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

Now we have three complexities:

• O(n²) for initializing the MST
• O(n²) for initializing the matrix
• O(n³ log n) for the loop

As O(n³ log n) > O(n²) (read: O(n³ log n) dominates O(n²)), the overall complexity of cal_minimax_path_matrix(n) is O(n³ log n).

_{References:
[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}