从飞镖扔到球形飞镖板（Python）的飞镖中估算PI

发布于 2025-02-02 12:02:32 字数 582 浏览 4 评论 0原文

估计PI的熟悉代码如下：

import math,random
random.seed(777)
j = pie = 1
while True:
    pie = pie + math.trunc(random.random()**2 + random.random()**2)
    if j%100000==0: print(4*(1-float(pie)/float(j)))
    j = j + 1
print('done')

模拟飞镖在二维飞镖板上投掷。但是，当您尝试提升到三维飞镖板时，

import math,random
random.seed(777)
j = pie = 1
while True:
    pie = pie + math.trunc(random.random()**2 + random.random()**2 + random.random()**2)
    if j%100000==0: print((3.0/4.0)*8.0*(1-float(pie)/float(j)))
    j = j + 1
print('done')

您会得到PI以外的其他东西。

原文

The familiar code for estimating pi is as follows:

import math,random
random.seed(777)
j = pie = 1
while True:
    pie = pie + math.trunc(random.random()**2 + random.random()**2)
    if j%100000==0: print(4*(1-float(pie)/float(j)))
    j = j + 1
print('done')

which emulates darts being thrown at a two-dimensional dartboard.
But when you try to elevate to a three-dimensional dartboard with:

import math,random
random.seed(777)
j = pie = 1
while True:
    pie = pie + math.trunc(random.random()**2 + random.random()**2 + random.random()**2)
    if j%100000==0: print((3.0/4.0)*8.0*(1-float(pie)/float(j)))
    j = j + 1
print('done')

you get something other than pi.

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我是有多爱你 2025-02-09 12:02:32

通常，我同意@bob__提出的解决方案，因为它更简单。
话虽如此，您的解决方案可以推广到3维dartboard：

import math,random

random.seed(777)
j = pie = 1
pie = 0
while True:
    # pie = pie + math.trunc(random.random()**2 + random.random()**2 + random.random()**2)
    pie = pie + min(1, math.trunc(random.random()**2 + random.random()**2 + random.random()**2))
    if j%100000==0: print((3.0/4.0)*8.0*(1-float(pie)/float(j)))
    j = j + 1
print('done')

要了解问题，我们需要澄清pie> pie变量的真正作用。它包含随机半径长于1.0（超出3D飞镖）的情况。

2D案例之所以有效，是因为MATH.TRUNC始终返回0或1（in或out），但对于3D，它返回0或1或2。将PIE增加2，从而将收敛速度更快。

要解决您的示例，您只需要确保一次pie一次不会增加1个即可。

Generally, I agree with the solution proposed by @Bob__ as it's more straightforward.
Having said that, your solution can be generalized to 3 dimensional dartboard:

import math,random

random.seed(777)
j = pie = 1
pie = 0
while True:
    # pie = pie + math.trunc(random.random()**2 + random.random()**2 + random.random()**2)
    pie = pie + min(1, math.trunc(random.random()**2 + random.random()**2 + random.random()**2))
    if j%100000==0: print((3.0/4.0)*8.0*(1-float(pie)/float(j)))
    j = j + 1
print('done')

To understand the issue we need to clarify what the pie variable really does. It contains the number of cases where a random radius is longer than 1.0 (so beyond the 3D dartboard).

The 2D case works because math.trunc always returns either 0 or 1 (in or out) but for 3D it returns either 0 or 1 or 2. So, in case of 3D, it fails because it sometimes increases pie by 2 and, thus, converges faster.

To fix your example, you just need to make sure that pie is never increased by more than 1 at a time.

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