My task is to calculate the approximate value of pi with an accuracy of at least 10^-6. The Monte Carlo algorithm does not provide the required accuracy. I need to use the calculation only through the volume of the sphere. What do you advise? I would be glad to see examples of code in CUDA or pure C++. Thank you.
To be completely literal about it, you could use a Darboux integral to measure the volume of one octant of the sphere. This code sample measures the area of one quadrant of a circle of radius 2.
#include <iomanip>
#include <iostream>
#include <queue>
#include <vector>
enum class Relationship { kContained, kDisjoint, kIntersecting };
static const double kRadius = 2;
class Rectangle {
public:
Rectangle(const double x0, const double y0, const double x1, const double y1)
: x0_(x0), y0_(y0), x1_(x1), y1_(y1) {}
double Area() const { return (x1_ - x0_) * (y1_ - y0_); }
Relationship RelationshipToCircle() const {
if (x1_ * x1_ + y1_ * y1_ < kRadius * kRadius) {
return Relationship::kContained;
}
if (x0_ * x0_ + y0_ * y0_ > kRadius * kRadius) {
return Relationship::kDisjoint;
}
return Relationship::kIntersecting;
}
std::vector<Rectangle> Subdivide() const {
const double xm = 0.5 * (x0_ + x1_);
const double ym = 0.5 * (y0_ + y1_);
return {
{x0_, y0_, xm, ym},
{x0_, ym, xm, y1_},
{xm, y0_, x1_, ym},
{xm, ym, x1_, y1_},
};
}
private:
double x0_;
double y0_;
double x1_;
double y1_;
};
int main() {
const Rectangle unit_rectangle{0, 0, kRadius, kRadius};
double lower_bound = 0;
double upper_bound = unit_rectangle.Area();
std::queue<Rectangle> rectangles;
rectangles.push(unit_rectangle);
while (upper_bound - lower_bound > 1e-6) {
const Rectangle r = rectangles.front();
rectangles.pop();
switch (r.RelationshipToCircle()) {
case Relationship::kContained:
lower_bound += r.Area();
break;
case Relationship::kDisjoint:
upper_bound -= r.Area();
break;
case Relationship::kIntersecting:
for (const Rectangle s : r.Subdivide()) {
rectangles.push(s);
}
break;
}
}
std::cout << std::setprecision(10) << 0.5 * (lower_bound + upper_bound)
<< '\n';
}