将所有边的权重设置为相同的值,然后使用算法找到所有最小生成树。
由于所有生成树都有|V|-1
条边且所有边的权重相等,因此所有生成树都是最小生成树。
我对这个问题产生了兴趣,但还没有找到一个真正令人满意的答案。然而,我找到了一些参考资料:Knuth在TAOCP第4卷第4部分中的算法S和S',Sorensen和Janssens的一篇论文paper by Sorensen and Janssens,以及Knuth的GRAYSPAN、SPSPAN和GRAYSPSPAN。可惜它们都不是我能使用的语言实现...我想我将不得不花些时间编写它们...
from itertools import product
import networkx as nx
import numpy as np
import matplotlib.pyplot as plt
#Optional: Calculate the total number of spanning trees for a given graph (g) using the Matrix Tree Theorem (Kirchhoff's theorem),
def numSpanTrees(g)-> int:
if not nx.is_connected(g):
print("Only for connected graphs")
return 0
if nx.is_directed(g): nx.to_undirected(g)
n = g.number_of_nodes()
laplacian_matrix = nx.laplacian_matrix(g).toarray() #Calculate the Laplacian matrix
cofactor_matrix = laplacian_matrix[1:n, 1:n] # Choose any cofactor by excluding the first row and column
determinant = np.linalg.det(cofactor_matrix) #Calculate the determinant of an cofactor of Laplacian matrix
return int(np.round(abs(determinant)))
#Generate all spanning trees using contraction-deletion algorithm:
def spanTrees(trs, Edg, all_span_trees, k):
if k == 0:
all_span_trees.extend(product(*trs))#Expand parallels edges using the Cartesian product
for i in range(k):
if Edg[k][i] == []: continue
trs.append(Edg[k][i])
Edg[i] = [Edg[i][j] + Edg[k][j] for j in range(i)]#Contraction
spanTrees(trs, Edg, all_span_trees, k-1)
trs.pop()
[Edg[i][j].pop() for j in range(i) for _ in range(len(Edg[k][j]))]#Deletion
#Helper function
def Spanning_Trees_Generator(g, hch = False):
if not nx.is_connected(g):
print("Only for connected graphs")
return 0
if nx.is_directed(g): nx.to_undirected(g)
n = g.number_of_nodes()
edgs = list(g.edges)
Edg = [[[] for _ in range(n)] for _ in range(n)]
mx = len(edgs)
edgNum = dict() #dictionary designed to store labeled edges where the keys are integer labels for the edges, and the values are tuples representing the ordinary edge definitions (in, out)
for ed in edgs:
i, j = sorted(ed)
Edg[j][i] = [mx]
edgNum[mx] = ed
mx -= 1
all_span_trees = []
spanTrees([], Edg, all_span_trees, n-1)
if hch:
return all_span_trees #Generate only the labeled edges (Keys)
else:
return [nx.Graph(edgNum[k] for k in element) for element in all_span_trees] #Generate spanning trees as NetworkX graphics objects
# Example usage
g = nx.complete_graph(5)
print("Number of Spanning Trees : ", numSpanTrees(g)) #Optional and only for undirected graph
spanTreesGraph = Spanning_Trees_Generator(g)
#Plot the first spanning tree
pos = nx.circular_layout(g)
nx.draw(g, pos, with_labels=True, node_color='lightblue', node_size=200, edge_color='b')
nx.draw(spanTreesGraph[0], pos, with_labels=True, node_color='lightgreen', node_size=200, edge_color='r')
plt.show()
https://github.com/tagtog12000/SpanTree/blob/master/Spanning%20Trees%20Generator.py