假设我有半径为r的n个圆,我想将它们随机放置在大小为AxA的矩形内。
保证它们可以放下。我们可以假设所有圆的面积总和约为矩形面积的60%。
我可以通过回溯法尝试放置,不断尝试,但应该有更好的方法来解决这个问题。
保证它们可以放下。我们可以假设所有圆的面积总和约为矩形面积的60%。
我可以通过回溯法尝试放置,不断尝试,但应该有更好的方法来解决这个问题。
std::complex
类型表示。
请注意,我使用了srand
和rand
进行测试目的。根据您的约束条件,您可以使用更好的随机算法。
根据我进行的测试,对于密度为60%,收敛似乎是有保证的。我还进行了一些密度为70%的测试:有时收敛,有时不收敛。
复杂度为O(n^2 n_iter)
,其中n
是圆的数量,n_iter
是迭代次数。对于密度为60%,n_iter
通常在100到300之间。通过放宽收敛标准,可以将其减少。
与评论中的其他提案相比,它可能看起来非常复杂。实际上,对于 n = 15
,在我的电脑上可以在不到30毫秒内完成工作。这个时间可能很长或足够快,具体取决于上下文。我已经包含了一张图来说明算法。
#include <cstdlib>
#include <iostream>
#include <fstream>
#include <vector>
#include <ctime>
#include <complex>
#include <cmath>
#include <tuple>
#include <ios>
#include <iomanip>
using dcomplex = std::complex<double>;
void print (const std::vector<dcomplex>& centers) {
std::cout << std::setprecision (9);
std::cout << "\ncenters:\n";
for (auto& z: centers) {
std::cout << real(z) << ", " << imag(z) << "\n";
}
}
std::tuple<bool, int, double> process (double A, double R, std::vector<dcomplex>& centers, int n_iter_max = 100) {
bool check = true;
int n = centers.size();
std::vector<dcomplex> moves (n, 0.0);
double acceleration = 1.0001; // to accelerate the convergence, if density not too large
// could be made dependent of the iteration index
double dmin;
auto limit = [&] (dcomplex& z) {
double zx = real(z);
double zi = imag(z);
if (zx < R) zx = R;
if (zx > A-R) zx = A-R;
if (zi < R) zi = R;
if (zi > A-R) zi = A-R;
return dcomplex(zx, zi);
};
int iter;
for (iter = 0; iter < n_iter_max; ++iter) {
for (int i = 0; i < n; ++i) moves[i] = 0.0;
dmin = A;
for (int i = 0; i < n; ++i) {
for (int j = i+1; j < n; ++j) {
auto vect = centers[i] - centers[j];
double dist = std::abs(vect);
if (dist < dmin) dmin = dist;
double x = std::max (0.0, 2*R*acceleration - dist) / 2.0;
double coef = x / (dist + R/10000);
moves[i] += coef * vect;
moves[j] -= coef * vect;
}
}
std::cout << "iteration " << iter << " dmin = " << dmin << "\n";
if (dmin/R >= 2.0 - 1.0e-6) break;
for (int i = 0; i < n; ++i) {
centers[i] += moves[i];
centers[i] = limit (centers[i]);
}
}
dmin = A;
for (int i = 0; i < n; ++i) {
for (int j = i+1; j < n; ++j) {
auto vect = centers[i] - centers[j];
double dist = std::abs(vect);
if (dist < dmin) dmin = dist;
}
}
std::cout << "Final: dmin/R = " << dmin/R << "\n";
check = dmin/R >= 2.0 - 1.0e-6;
return {check, iter, dmin};
}
int main() {
int n = 15; // number of circles
double R = 1.0; // ray of each circle
double density = 0.6; // area of all circles over total area A*A
double A; // side of the square
int n_iter = 1000;
A = sqrt (n*M_PI*R*R/density);
std::cout << "number of circles = " << n << "\n";
std::cout << "density = " << density << "\n";
std::cout << "A = " << A << std::endl;
std::vector<dcomplex> centers (n);
std::srand(std::time(0));
for (int i = 0; i < n; ++i) {
double x = R + (A - 2*R) * (double) std::rand()/RAND_MAX;
double y = R + (A - 2*R) * (double) std::rand()/RAND_MAX;
centers[i] = {x, y};
}
auto [check, n_iter_eff, dmin] = process (A, R, centers, n_iter);
std::cout << "check = " << check << "\n";
std::cout << "Relative min distance = " << std::setprecision (9) << dmin/R << "\n";
std::cout << "nb iterations = " << n_iter_eff << "\n";
print (centers);
return 0;
}
n = 15
解决方案。虽然不如维基百科的76.2%解决方案好,但考虑到算法的简单性(天真?),这并不荒谬。 - Damien
n = 15, r = 1 and A = 4 + sqrt(2) + sqrt(6)
,你必须能够生成这个:https://upload.wikimedia.org/wikipedia/commons/thumb/e/ee/15_circles_in_a_square.svg/512px-15_circles_in_a_square.svg.png - orlp