为什么我的python程序没有收敛到pi？

2023年3月14日 207次阅读

我写了这个方法：

def approx_pi(n, p):
    """
    Approximates Pi by putting dots in a square and counting
    the dots in the max possible circle in that square.
    :param n: Number of dots
    :param p: Precision used for urandom
    :return: Approximation of Pi
    """
    in_circle = 0
    k = 100
    max_int = (2**(8*p)-1)
    for i in range(n):
        # get two random Numbers between 0 and 1
        x = int.from_bytes(os.urandom(p), byteorder='big')/max_int
        y = int.from_bytes(os.urandom(p), byteorder='big')/max_int

        if x ** 2 + y ** 2 <= 1:
            in_circle += 1

        # Just for debugging
        if (i+1) % k == 0:
            k = int(k*1.01)
            print(i, '\t',4*in_circle/i)
            sys.stdout.flush()

    return 4*in_circle/n

在我的大多数Testruns中,它在3.141处稳定,然后在该值附近发散.
这是urandom的弱点吗？但如果是这样,为什么pi不会向一个方向转移？或者我的代码有问题.

最佳答案 首先.你的代码很混乱.如果您通过名称调用变量,可以更好地了解它们的含义.

我已将代码简化为必要的行.因此,我使用随机库来生成点位置.

这种方法的问题在于收敛速度非常慢.凭借10 ** 8分,我获得了3.14167604.

import random


def approx_pi(points):
    """
    Approximates Pi by putting dots in a square and counting
    the dots in the max possible circle in that square.
    :param points: Number of dots
    :return: Approximation of Pi
    """
    in_circle = 0
    for dummy_i in xrange(points):
        # get two random Numbers between 0 and 1
        x_dot_position = random.random()
        y_dot_position = random.random()

        if x_dot_position ** 2 + y_dot_position ** 2 <= 1:
            in_circle += 1

    return 4.0*in_circle/points

编辑
你对函数random.random()很严格.所以在下一个代码中我使用了random.random_integers,它是[low,high]值之间的均匀分布.

代码是并行化的,我尝试了10 ** 10分获得：

PI = 3.14157765
Tiempo de calculo = 3790秒

import multiprocessing
import time
import numpy as np

starting_point = time.time()

def approx_pi(point):
    """
    Approximates Pi by putting dots in a square and counting
    the dots in the max possible circle in that square.
    :param points: Number of dots
    :return: Approximation of Pi
    """
    # get two random Numbers between 0 and 1
    x_dot_position = float(np.random.random_integers(0,10**10))/10**10
    y_dot_position = float(np.random.random_integers(0,10**10))/10**10

    if x_dot_position ** 2 + y_dot_position ** 2 <= 1:
        return 1
    else: 
        return 0

###########################################################

total_points     = 1*10**10       
paso    = 1*10**8       

in_circle = 0

for in_este_bucle in xrange(0, total_points, paso):
    print "Procesadores disponibles: " + str(multiprocessing.cpu_count())

    pool = multiprocessing.Pool()
    resultado = pool.map(approx_pi, xrange(in_este_bucle, in_este_bucle + paso))
    pool.close()
    pool.join()
    in_circle += sum(resultado)
    del resultado


print 'Total time: ' + str(time.time()-starting_point) +' seconds'
print
print 4.0*in_circle/total_points
点赞