Question

我写了一个用于研究棒渗透的算法（即：在它们之间相交的线段之间的网络）。在我的算法中，在“b”和“h”的矩形框内创建N个（线段），然后逐个算法探索所有线段之间的交集。这是蒙特卡罗模拟，因此“实验”执行多次（不少于100次）。这样写，消耗了很多RAM：

array_x1=uniform.rvs(loc=-b/2, scale=b, size=N)  
array_y1=uniform.rvs(loc=-h/2, scale=h, size=N)
array_x2=uniform.rvs(loc=-b/2, scale=b, size=N)  
array_y2=uniform.rvs(loc=-h/2, scale=h, size=N)  

M = np.zeros([N,N])  

for u in xrange(100):  ----> This '100' is the number of experiments.
    for j in xrange(N):
       if j>0:
         x_A1B1 = array_x2[j]-array_x1[j]
        y_A1B1 = array_y2[j]-array_y1[j]
        x_A1A2 = array_x1[0:j]-array_x1[j]
        y_A1A2 = array_y1[0:j]-array_y1[j]      
        x_A2A1 = -1*x_A1A2
        y_A2A1 = -1*y_A1A2
        x_A2B2 = array_x2[0:j]-array_x1[0:j]
        y_A2B2 = array_y2[0:j]-array_y1[0:j]
        x_A1B2 = array_x2[0:j]-array_x1[j]
        y_A1B2 = array_y2[0:j]-array_y1[j]
        x_A2B1 = array_x2[j]-array_x1[0:j]
        y_A2B1 = array_y2[j]-array_y1[0:j]

        p1 = x_A1B1*y_A1A2 - y_A1B1*x_A1A2
        p2 = x_A1B1*y_A1B2 - y_A1B1*x_A1B2
        p3 = x_A2B2*y_A2B1 - y_A2B2*x_A2B1
        p4 = x_A2B2*y_A2A1 - y_A2B2*x_A2A1

        condition_1=p1*p2
        condition_2=p3*p4                         

        for k in xrange (j):
            if condicion_1[k]<=0 and condicion_2[k]<=0:
                M[j,k]=1

    if j+1<N+4:
        x_A1B1 = array_x2[j]-array_x1[j]
        y_A1B1 = array_y2[j]-array_y1[j]
        x_A1A2 = array_x1[j+1:]-array_x1[j]
        y_A1A2 = array_y1[j+1:]-array_y1[j]     
        x_A2A1 = -1*x_A1A2
        y_A2A1 = -1*y_A1A2
        x_A2B2 = array_x2[j+1:]-array_x1[j+1:]
        y_A2B2 = array_y2[j+1:]-array_y1[j+1:]
        x_A1B2 = array_x2[j+1:]-array_x1[j]
        y_A1B2 = array_y2[j+1:]-array_y1[j]
        x_A2B1 = array_x2[j]-array_x1[j+1:]
        y_A2B1 = array_y2[j]-array_y1[j+1:]

        p1 = x_A1B1*y_A1A2 - y_A1B1*x_A1A2
        p2 = x_A1B1*y_A1B2 - y_A1B1*x_A1B2
        p3 = x_A2B2*y_A2B1 - y_A2B2*x_A2B1
        p4 = x_A2B2*y_A2A1 - y_A2B2*x_A2A1

        condicion_1=p1*p2
        condicion_2=p3*p4                         

        for k in xrange (N-j-1):
            if condicion_1[k]<=0 and condicion_2[k]<=0:
                M[j,k+j+1]=1

这里，如果棒i与棒j相交，则元素Mij = 1，否则Mij = 0。

如何优化算法？在这种情况下，图论是有用的吗？怎么样？

等待你的回答。

非常感谢！

祝你好运

棒（线段）渗流算法 - 图论？

0 个答案: