计算半径R和尺寸D的球体内的整数点
我正在尝试编写一种有效的算法，该算法计算半径R和尺寸D的球体内的点数。球始终位于原点。假设我们有一个半径为5的尺寸为2（圆形）的球面。
我的策略是在第一个象限内生成所有可能的点，因此对于上面的示例，我们知道（1,2）在圆中，因此该点的所有+/-组合都必须是简单的尺寸平方。因此，对于在n维球体的单个象限中找到的每个点，我们将2
^维添加到总数中。
我不确定是否有更有效的解决方案来解决这个问题，但这是我到目前为止的实现方式。
int count_lattice_points(const double radius, const int dimension) {int R = static_cast<int>(radius);
int count = 0;
std::vector<int> points;
std::vector<int> point;
for(int i = 0; i <= R; i++)
points.push_back(i);
do {
for(int i = 0; i < dimension - 1; i++)
point.push_back(points.at(i));
if(isPointWithinSphere(point, radius)) count += std::pow(2,dimension);
point.clear();
}while(std::next_permutation(points.begin(), points.end()));
return count + 3;
}
在这种情况下我可以解决或改善什么？
回答：
我在这里介绍了我的2D算法（带有一些源代码和一个难看但方便的插图）
它比MBo在四分之一之一的圆的原点和边缘之间的计数点快3.4倍。
您只是想像一个刻有正方形的正方形，并且只计算该圆圈内该正方形之外的东西的八分之一。
public static int gaussCircleProblem(int radius) { int allPoints=0; //holds the sum of points
double y=0; //will hold the precise y coordinate of a point on the circle edge for a given x coordinate.
long inscribedSquare=(long) Math.sqrt(radius*radius/2); //the length of the side of an inscribed square in the upper right quarter of the circle
int x=(int)inscribedSquare; //will hold x coordinate - starts on the edge of the inscribed square
while(x<=radius){
allPoints+=(long) y; //returns floor of y, which is initially 0
x++; //because we need to start behind the inscribed square and move outwards from there
y=Math.sqrt(radius*radius-x*x); // Pythagorean equation - returns how many points there are vertically between the X axis and the edge of the circle for given x
}
allPoints*=8; //because we were counting points in the right half of the upper right corner of that circle, so we had just one-eightth
allPoints+=(4*inscribedSquare*inscribedSquare); //how many points there are in the inscribed square
allPoints+=(4*radius+1); //the loop and the inscribed square calculations did not touch the points on the axis and in the center
return allPoints;
}
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