需要帮助修复近似pi的算法 [英] Need help fixing an algorithm that approximates pi
问题描述
我正在尝试为近似 pi 的算法编写C代码.它应该得到一个立方体的体积和那个立方体内部的球体的体积(球体的半径是立方体的侧面的1/2).然后我应该将立方体的体积除以球体的体积，然后乘以6得到pi.
I'm trying to write the C code for an algorithm that approximates pi. It's supposed to get the volume of a cube and the volume of a sphere inside that cube (the sphere's radius is 1/2 of the cube's side). Then I am supposed to divide the cube's volume by the sphere's and multiply by 6 to get pi.
它正在工作，但是在应该获得体积的部分中却做了一些奇怪的事情.我认为这与我为近似值选择的增量有关. 有了方4的立方体，而不是给我64的体积，它给了我6400.使用球体而不是33，给了我3334.
It's working but it's doing something weird in the part that is supposed to get the volumes. I figure it's something to do the with delta I chose for the approximations. With a cube of side 4 instead of giving me a volume of 64 it's giving me 6400. With the sphere instead of 33 it's giving me 3334. something.
有人能弄清楚吗?这是代码(我对相关部分进行了注释):
Can someone figure it out? Here is the code (I commented the relevant parts):
#include <stdio.h>
int in_esfera(double x, double y, double z, double r_esfera){
double dist = (x-r_esfera)*(x-r_esfera) + (y-r_esfera)*(y-r_esfera) + (z-r_esfera)*(z-r_esfera);
return dist <= (r_esfera)*(r_esfera) ? 1 : 0;
}
double get_pi(double l_cubo){
double r_esfera = l_cubo/2;
double total = 0;
double esfera = 0;
//this is delta, for the precision. If I set it to 1E anything less than -1 the program continues endlessly. Is this normal?
double delta = (1E-1);
for(double x = 0; x < l_cubo; x+=delta){
printf("x => %f; delta => %.6f\n",x,delta);
for(double y = 0; y <l_cubo; y+=delta){
printf("y => %f; delta => %.6f\n",y,delta);
for(double z = 0; z < l_cubo; z+=delta){
printf("z => %f; delta => %.6f\n",z,delta);
total+=delta;
if(in_esfera(x,y,z,r_esfera))
esfera+=delta;
}
}
}
//attempt at fixing this
//esfera/=delta;
//total/=delta;
//
//This printf displays the volumes. Notice how the place of the point is off. If delta isn't a power of 10 the values are completely wrong.
printf("v_sphere = %.8f; v_cube = %.8f\n",esfera,total);
return (esfera)/(total)*6;
}
void teste_pi(){
double l_cubo = 4;
double pi = get_pi(l_cubo);
printf("%.8f\n",pi);
}
int main(){
teste_pi();
}
推荐答案
事情是，像a * b * c这样的整数上乘以与将1 + 1 + 1 + 1 + ... + 1 a * b * c乘以相同，对吗?
The thing is that multiplication over integers like a * b * c is the same as adding 1 + 1 + 1 + 1 + ... + 1 a * b * c times, right?
您要添加delta + delta + ... (x / delta) * (y / delta) * (z / delta)次.换句话说，就是(x * y * z) / (delta ** 3)次.
You're adding delta + delta + ... (x / delta) * (y / delta) * (z / delta) times. Or, in other words, (x * y * z) / (delta ** 3) times.
现在，delta s的总和与此相同:
Now, that sum of deltas is the same as this:
delta * (1 + 1 + 1 + 1 + ...)
^^^^^^^^^^^^^^^^^^^^ (x * y * z) / (delta**3) times
因此，如果delta是10的幂，则(x * y * z) / (delta**3)将是一个整数，并且将等于括号中1的总和(因为它与 product 相同) > x * y * (z / (delta**3))，其中最后一项是整数-请参阅此答案的第一句).因此，您的结果将如下所示:
So, if delta is a power of 10, (x * y * z) / (delta**3) will be an integer, and it'll be equal to the sum of 1's in parentheses (because it's the same as the product x * y * (z / (delta**3)), where the last term is an integer - see the very first sentence of this answer). Thus, your result will be the following:
delta * ( (x * y * z) / (delta ** 3) ) == (x * y * z) / (delta**2)
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ the sum of ones
这就是您最终计算乘积除以delta平方的方式.
That's how you ended up calculating the product divided by delta squared.
要解决此问题，请将所有体积乘以delta * delta.
To solve this, multiply all volumes by delta * delta.
但是，我认为不可能将这种逻辑用于功率不为10的delta.而且，实际上，代码会为delta == 0.21和l_cubo == 2带来各种麻烦，例如:您将得到9.261000000000061而不是8.
However, I don't think it's possible to use this logic for deltas that aren't a power of 10. And indeed, the code will go all kinds of haywire for delta == 0.21 and l_cubo == 2, for example: you'll get 9.261000000000061 instead of 8.
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