网络 X 无法计算代数连接

Question

我可以在几分之一秒内计算 20 个顶点上完整图的代数连通性

import networkx
D = {}
for i in range(20):
    D[i] = [j for j in range(20)]
G = networkx.Graph(D)
networkx.algebraic_connectivity(G)

然而，在一个过程中，我生成了一个图(在 20 个节点上(，我要求 networkx 计算它的代数连接，它永远运行，没有错误。这是图表：

import networkx
D = {0: [32, 33, 19, 5, 21, 37, 6, 38, 39, 41, 26, 42, 11, 43, 28, 44, 15, 31], 5: [32, 0, 33, 19, 37, 21, 6, 22, 38, 39, 41, 26, 42, 11, 43, 44, 28, 15, 31], 6: [0, 32, 33, 19, 5, 37, 21, 22, 38, 39, 41, 26, 42, 11, 43, 28, 44, 15, 31], 11: [32, 0, 33, 19, 21, 37, 5, 6, 22, 38, 39, 41, 26, 42, 43, 28, 44, 15, 31], 15: [0, 32, 33, 19, 5, 21, 37, 6, 22, 38, 39, 41, 26, 42, 11, 43, 28, 44, 31], 19: [0, 32, 33, 5, 21, 37, 6, 22, 38, 39, 41, 26, 42, 11, 43, 28, 44, 15, 31], 21: [32, 0, 33, 19, 37, 5, 6, 22, 38, 39, 41, 26, 42, 11, 43, 28, 44, 15, 31], 22: [32, 33, 19, 5, 21, 37, 6, 38, 39, 41, 26, 42, 11, 43, 28, 44, 15, 31], 26: [0, 32, 33, 19, 5, 21, 37, 6, 22, 38, 39, 41, 42, 11, 43, 28, 44, 15, 31], 28: [32, 0, 33, 19, 21, 37, 5, 6, 22, 38, 39, 41, 26, 42, 11, 43, 44, 15, 31], 31: [32, 0, 33, 19, 5, 21, 37, 6, 22, 38, 39, 41, 26, 42, 11, 43, 28, 44, 15], 32: [0, 33, 19, 5, 21, 37, 6, 22, 38, 39, 41, 26, 42, 11, 43, 28, 44, 31, 15], 33: [0, 32, 19, 5, 21, 37, 6, 22, 38, 39, 41, 26, 42, 11, 43, 28, 44, 15, 31], 37: [32, 0, 33, 19, 5, 21, 6, 22, 38, 39, 41, 26, 42, 11, 43, 28, 44, 31, 15], 38: [32, 0, 33, 19, 21, 37, 5, 6, 22, 39, 41, 26, 42, 11, 43, 28, 44, 15, 31], 39: [0, 32, 33, 19, 5, 21, 37, 6, 22, 38, 41, 26, 42, 11, 43, 28, 44, 15, 31], 41: [32, 0, 33, 19, 21, 37, 5, 38, 6, 22, 39, 26, 42, 11, 43, 28, 44, 15, 31],  42: [32, 0, 33, 19, 21, 37, 5, 6, 22, 38, 39, 41, 26, 11, 43, 28, 44, 15, 31], 43: [32, 0, 33, 19, 21, 37, 5, 6, 22, 38, 39, 41, 26, 42, 11, 28, 44, 15, 31], 44: [32, 0, 33, 19, 5, 21, 37, 38, 6, 22, 39, 41, 42, 26, 11, 43, 28, 15, 31]}
G = networkx.Graph(D)
networkx.algebraic_connectivity(G)

为什么会这样，以及如何解决它？

Answer 1

tracemin 方法中似乎有一个错误，它是 networkx.algebraic_connectivity(( 的默认值。 尝试使用

networkx.algebraic_connectivity(G, method='lanczos')