Question

我尝试在6个维度中实现简单的蒙特卡罗集成。积分仅覆盖两个3D球体，在下文中球体的坐标标记为r1和r2。当使用笛卡尔坐标并忽略球体之外的任何东西时，积分工作正常。当被积函数取决于r1和r2之间的角度时，使用球面坐标失败。

似乎r1和r2可能指向同一个方向，而我希望对齐是完全随机的。

用于生成球面坐标的变换可在此处找到：http://de.mathworks.com/help/matlab/math/numbers-placed-randomly-within-volume-of-sphere.html

以下代码使用两个不同的积分和基于笛卡尔坐标和球坐标的蒙特卡罗积分来说明这一点：

import numpy as np
from math import *

N=1e5
R=2
INVERT = False # if this is set to True, the results are correct
INTEGRAND = None


def spherical2cartesian(r, cos_theta, cos_phi):
    sin_theta = sqrt(1-cos_theta**2)
    sin_phi = sqrt(1-cos_phi**2)
    return np.array([r*sin_theta*cos_phi, r*sin_theta*sin_phi, r*cos_theta])

# Integrands (cartesian)

def integrand_sum_sqr(r1,r2):
    r1=np.linalg.norm(r1)
    r2=np.linalg.norm(r2)
    if r1>R or r2>R:
        return 0.0;
    return r1**2 + r2**2

def integrand_diff_sqr(r1,r2):
    delta = r1-r2
    r1=np.linalg.norm(r1)
    r2=np.linalg.norm(r2)
    if r1>R or r2>R:
        return 0.0;
    return np.linalg.norm(delta)**2

# integrand for spherical integration (calls cartesian version)

def integrand_spherical(r1,r2):
    # see the following link for the transformation used: http://de.mathworks.com/help/matlab/math/numbers-placed-randomly-within-volume-of-sphere.html
    r1 = spherical2cartesian((3*r1[0])**(1.0/3.0), r1[1], cos(r1[2]))
    r2 = spherical2cartesian((3*r2[0])**(1.0/3.0), r2[1], cos(r2[2]))
    if INVERT:
        s = (-1)**np.random.choice(2,6)
        r1=s[0:3]*r1
        r2=s[3:6]*r2
    return INTEGRAND(r1,r2)

# monte carlo integration routines

def monte_carlo(name,func,samples,V):
    results=np.array([func(x[0:3],x[3:6]) for x in samples])
    avg = results.mean()*V
    std = results.std(ddof=1)*V/sqrt(len(samples))
    print name,": ",avg," +- ",std

def monte_carlo_cartesian():
    V=(2*R)**6
    samples = np.random.rand(N,6)
    samples = R*(2*samples-1)
    monte_carlo('cartesian',INTEGRAND,samples,V)

def monte_carlo_spherical():
    V=(4.0/3.0*R**3*pi)**2
    samples = np.random.rand(6,N)
    samples = np.array([R**3*samples[0]/3.0, 2*samples[1]-1, 2*pi*samples[2], R**3*samples[3]/3.0, 2*samples[4]-1, 2*pi*samples[5]])
    samples = samples.T
    monte_carlo('spherical',integrand_spherical,samples,V)

# run all functions with all different monte carlo versions
print "Integrating sum_sqr, expected:",32.0/15.0*R**8*pi**2
INTEGRAND=integrand_sum_sqr
monte_carlo_cartesian()
monte_carlo_spherical()

print "Integrating diff_sqr, expected:",32.0/15.0*R**8*pi**2
INTEGRAND=integrand_diff_sqr
monte_carlo_cartesian()
monte_carlo_spherical()

典型输出：

Integrating sum_sqr, expected: 5390.11995025
cartesian :  5406.6226422  +-  29.5405030567
spherical :  5392.72811794  +-  5.23871574928
Integrating diff_sqr, expected: 5390.11995025
cartesian :  5360.20055643  +-  34.8851044924
spherical :  4141.88319573  +-  9.69351527901

最后一个积分显然是错误的。为什么r1和r2相关？我该如何解决这个问题？

Answer 1

上述代码的问题在以下一行

sin_phi = sqrt(1-cos_phi**2)

这只会产生正面结果，而sin(phi)会产生phi > pi

的负面结果

球面坐标中的蒙特卡罗积分

1 个答案: