非重叠随机定位的圆圈

Question

我需要随机生成固定数量的非重叠圆圈。我可以显示圆圈，在本例中为 20，使用这段代码随机定位，

  for i =1:20
  x=0 + (5+5)*rand(1)
  y=0 + (5+5)*rand(1)
  r=0.5
  circle3(x,y,r)
  hold on 
  end

但是圆圈重叠，我想避免这种情况。这是以前使用 Mathematica https://mathematica.stackexchange.com/questions/69649/generate-nonoverlapping-random-circles 的用户实现的，但我使用的是 MATLAB，我想坚持下去。

为了重现性，这是我用来画圆圈的函数 circle3

function h = circle3(x,y,r)
 d = r*2;
 px = x-r;
 py = y-r;
 h = rectangle('Position',[px py d d],'Curvature',[1,1]);
 daspect([1,1,1])

谢谢。

Answer 1

您可以保存所有先前绘制的圆圈的列表。后 随机化一个新圆检查它是否与之前绘制的圆相交。

代码示例：

nCircles = 20;
circles = zeros(nCircles ,2);
r = 0.5;

for i=1:nCircles
    %Flag which holds true whenever a new circle was found
    newCircleFound = false;
    
    %loop iteration which runs until finding a circle which doesnt intersect with previous ones
    while ~newCircleFound
        x = 0 + (5+5)*rand(1);
        y = 0 + (5+5)*rand(1);
        
        %calculates distances from previous drawn circles
        prevCirclesY = circles(1:i-1,1);
        prevCirclesX = circles(1:i-1,2);
        distFromPrevCircles = ((prevCirclesX-x).^2+(prevCirclesY-y).^2).^0.5;
        
        %if the distance is not to small - adds the new circle to the list
        if i==1 || sum(distFromPrevCircles<=2*r)==0
            newCircleFound = true;
            circles(i,:) = [y x];
            circle3(x,y,r)
        end
    
    end
    hold on
end

*请注意，如果圆的数量相对于绘制 x 和 y 坐标的范围而言太大，则循环可能会无限运行。 为了避免它 - 相应地定义这个范围（它可以定义为 nCircles 的函数）。

Answer 2

如果您对暴力破解感到满意，请考虑以下解决方案：

N = 60;                        % number of circles
r = 0.5;                       % radius
newpt = @() rand([1,2]) * 10;  % function to generate a new candidate point

xy = newpt();  % matrix to store XY coordinates
fails = 0;     % to avoid looping forever
while size(xy,1) < N
    % generate new point and test distance
    pt = newpt();
    if all(pdist2(xy, pt) > 2*r)
        xy = [xy; pt];  % add it
        fails = 0;      % reset failure counter
    else
        % increase failure counter,
        fails = fails + 1;
        % give up if exceeded some threshold
        if fails > 1000
            error('this is taking too long...');
        end
    end
end

% plot
plot(xy(:,1), xy(:,2), 'x'), hold on
for i=1:size(xy,1)
    circle3(xy(i,1), xy(i,2), r);
end
hold off

Answer 3

稍微修改了代码@drorco，以确保绘制出我想要的确切数量的圆圈

nCircles = 20;
circles = zeros(nCircles ,2);
r = 0.5;
c=0; 

for i=1:nCircles
%Flag which holds true whenever a new circle was found
newCircleFound = false;

%loop iteration which runs until finding a circle which doesnt intersect with     previous ones
while ~newCircleFound & c<=nCircles
    x = 0 + (5+5)*rand(1);
    y = 0 + (5+5)*rand(1);

    %calculates distances from previous drawn circles
    prevCirclesY = circles(1:i-1,1);
    prevCirclesX = circles(1:i-1,2);
    distFromPrevCircles = ((prevCirclesX-x).^2+(prevCirclesY-y).^2).^0.5;

    %if the distance is not to small - adds the new circle to the list
    if i==1 || sum(distFromPrevCircles<=2*r)==0
        newCircleFound = true;
        c=c+1
        circles(i,:) = [y x];
        circle3(x,y,r)
    end

end
hold on

结束

Answer 4

虽然这是一篇旧帖子，并且因为我在分享我的解决方案之前遇到了同样的问题，它使用匿名函数：https://github.com/davidnsousa/mcsd/blob/master/mcsd/cells.m。此代码允许从用户定义的单元半径分布创建 1、2 或 3-D 单元环境。目的是为生物组织中扩散的蒙特卡罗模拟创建一个复杂的环境：https://www.mathworks.com/matlabcentral/fileexchange/67903-davidnsousa-mcsd

此代码的更简单但不太灵活的版本是二维环境的简单情况。下面创建N 随机定位且不重叠的圆的空间分布，其半径为R，与其他单元格的最小距离为D。全部打包在一个长度为S的方形区域中。

function C = cells(N, R, D, S)
  C = @(x, y, r) 0;
  for n=1:N
    o = randi(S-R,1,2);
    while C(o(1),o(2),2 * R + D) ~= 0
      o = randi(S-R,1,2);
    end
    f = @(x, y) sqrt ((x - o(1)) ^ 2 + (y - o(2)) ^ 2);
    c = @(x, y, r) f(x, y) .* (f(x, y) < r);
    C = @(x, y, r) + C(x, y, r) + c(x, y, r);
  end
  C = @(x, y) + C(x, y, R);
end

其中返回C 是所有圈子的组合匿名函数。我相信，尽管它是一种蛮力解决方案，但它既快速又优雅。