Python 蒙特卡罗模拟无法识别连通图

我正在写一个蒙特卡罗模拟，计算使n顶点图连接所需的最小有向边数的期望值。它从一个全0邻接矩阵开始，添加一条有向边，然后测试该矩阵，看它是否表示一个连通图。这个过程是循环的，直到构建了一个连通图，然后迭代的次数就是单个试验的样本量。该程序对于小型图似乎是精确的，但一旦图超过10个顶点，就越来越清楚地看到，它会在已构造连通图之后继续添加顶点。本质上，该算法在大型图上停止得不够早

似乎可能的罪魁祸首是断开连接的函数，但我不能完全确定。如有任何建议，将不胜感激

import numpy as np
import math
import random

# Randomly adds an edge to the graph by
# choosing a random 0 from the adjecency
# matrix and changing it to a 1.
# @param mat - the list type matrix
# @param k   - the dimension of the matrix
def addEdge(mat, k):

    flag = False

    while flag == False:

        colNum = random.randint(0, k-1)
        rowNum = random.randint(0, k-1)

        if mat[rowNum][colNum] == 0:
            mat[rowNum][colNum] = 1
            flag = True 

# Runs singleTrial multiple times and finds
# the average of their sample sizes
def getExpectedValue(size, trials):

    sampleSum = 0.0

    flag = True

    for i in range(trials):
        sample = singleTrial(size)
        sampleSum += sample

    expectedValue = float(sampleSum/trials)

    return expectedValue


# Adds edges to an initially edgeless
# graph UNTIL the graph becomes connected.
def singleTrial(size):

    # Create all zero matrix
    mat = np.random.randint(0,1,size=(size,size))

    sample = 0

    flag = True

    while flag:
        # Uncomment this code to see each matrix that is 
        # generated in a single trial. Upon viewing it
        # is clear that this while loop does not terminate
        # early enough.
        #print mat
        #print "\n"
        addEdge(mat, size)
        sample += 1
        if isConnected(mat, size):
            print mat
            flag = False

    print sample
    return sample 


# Checks if a given matrix is connected by
# calculating the sum of the number of 1 step
# paths though the number of k step paths
# for every pair of verticies. If this number
# is zero for any pair, then the graph is 
# not connected.
def isConnected(A, k):

    B = A

    for i in range(2, k-1):
        B = B + np.linalg.matrix_power(A, i)
    if np.prod(B) != 0:
        return True
    else:
        return False


if __name__ == '__main__' :


    # Perform 1 trial on an 11 vertex graph
    print getExpectedValue(11, 1)

---

我解决了这个问题。我使用函数numpy.prod（mat）检查矩阵是否包含零，因为它返回矩阵中所有项的乘积。我没有考虑过溢出的可能性。我假设如果numpy.prod的值变得太大（它肯定会这么做），它将重置为零。这会造成许多错误的否定。

为什么要使用np.prod（B）而不是（B==0）。any（）？这是一个python学校的项目，我的老师更喜欢数学，他建议使用prod（）。虽然我已经引起了他的注意，而且已经解决了。我刚开始学习python，所以我不知道any（）函数。谢谢你的提示。