如何在六边形内生成它们之间有距离的随机位置? [英] How to generate random positions with distance between them inside the hexagon?

问题描述

我正在尝试在 500 米六边形内创建给定距离的 N 个随机点对(N = 50).通过使用(dmax-dmin).* rand(N，1)+ dmin 创建的距离D，其中 dmin = 10 和 dmax = 100 在Matlab中.我确信第一个点必须生成一组点([x1 y1])，它们与六边形主边界的距离至少为 D 与第一组具有精确距离 D 的点([[x2 y2])的数量.但是有时我会遇到六角形之外的第二个点的问题，因为如果六角形边界上的第一个位置加上 D 距离，那么第二个位置是六角形之外的(我是说我想在hexagol内生成随机对位置).有人可以帮助我产生这种情况并解决问题吗?谢谢.

例如

  R = 500;己二醇半径％N = 50；数字对位置百分比d_min = 10;最小距离百分比d_max = 100;最大距离百分比D =(d_max-d_min).* rand(N，1)+ d_min;％随机距离X = [0,0];己醇中心％j = 0;而j<ñj = j + 1;theta(j)= 2 * pi * rand(1,1);u = rand()+ rand();如果u<1个r(j)= R * u;别的r(j)= R *(2-u);结尾％创建第一个位置x1(j)= r(j)* cos(theta(j))+ X(1,1);前x个排名的百分比y1(j)= r(j)* sin(theta(j))+ X(1,2);前首位％结尾％创建第二个位置x2(j)= x1(j)+ D(j);％秒x排名y2(j)= y1(j)+ D(j);％第二年y位置 

解决方案

这很像您的

极坐标空间中六边形的公式与外切圆非常相似，不同之处在于，六边形的半径在每个 t (角度)上均不同.我们将此变化半径称为 r2 .因此，如果我们找到对所有 t s返回 r2 的函数 R2 ，那么我们可以为六边形写极坐标公式:

此图演示了问题的参数:

此处的关键参数是α.现在我们需要一个函数 Alpha ，它为所有 t s返回α:

现在，所有点都在极坐标空间的六边形的边界上:

  r = 500;T = linspace(0，2 * pi，181);Alpha = @(t)pi/2-abs(rem(t，pi/3)-(pi/6));R2 = @(t)r * cos(pi/6)./sin(Alpha(t));X = R2(T).* cos(T);Y = R2(T).* sin(T);坚持，稍等情节(X，Y，'.b');图((r).* cos(T)，(r).* sin(T)，'.r') 

正多边形的极性公式:

在我继续之前，我想概括一下 Alpha 和 R2 函数，以涵盖所有

答案:

现在，我们可以像处理圆弧问题一样生成成对的点:

  r = 500;n = 6；a = 10;b = 50；N = 100；D =(b-a).* rand(N，1)+ a;Alpha = @(t)pi/2-abs(rem(t，2 * pi/(n))-(pi/(n)));R2 = @(t)r * cos(pi/n)./sin(Alpha(t));T1 = rand(N，1)* 2 * pi;RT1 = rand(N，1).*(R2(T1)-D);X1 = RT1.* cos(T1);Y1 = RT1.* sin(T1);T2 = rand(N，1)* 2 * pi;X2 = X1 + D. * cos(T2);Y2 = Y1 + D. * sin(T2); 

旋转多边形:

要旋转多边形，我们只需要更新 Alpha 函数:

  t0 = pi/8;Alpha = @(t)pi/2-abs(rem(t + t0，2 * pi/(n))-(pi/(n))); 

这是对 n = 7 ， N = 50000 和 t0 = pi/10 的测试:

I am trying to create N random pairs of points (N = 50) of a given distances, inside a 500 meters hexagon. The distance D created by using (dmax - dmin).*rand(N,1) + dmin, with dmin = 10 and dmax = 100 in Matlab. I understant that the first I have to generate a set of points ([x1 y1]) that have at least distance D from the main hexagon border, then generate the second set of points ([x2 y2]) that have exact distance D from the first set. But sometime I got the problem with the second point outside of hexagon, because if the first position on the hexagol border and plus Ddisance, then the second position is outside of hexagon (I mean that I want to generate random pair position inside of hexagol). Could anybody help me in generating this kind of scenario and fix the problem? Thanks.

For example as

R               = 500; % hexagol radius
N               = 50;  % number pair positions
d_min           = 10;  % minimum distance          
d_max           = 100; % maximum distance       
D               = (d_max - d_min).*rand(N,1) + d_min; % randomly distance
X               = [0,0]; % hexagol center
j=0;
while j < N 
    j=j+1;
    theta(j)=2*pi*rand(1,1);
    u= rand()+ rand();
    if u < 1
       r(j) = R * u;
    else
       r(j) = R * (2 - u);
    end
% to create the first position
    x1(j)=r(j)*cos(theta(j)) + X(1,1); % first x positions
    y1(j)=r(j)*sin(theta(j)) + X(1,2); % first y positions
end
% to create the second position
x2(j) = x1(j) + D(j); % second x positions
y2(j) = y1(j) + D(j); % second y positions

解决方案

This is quite like your other question and its solution is almost the same, but it needs a little more math. Let’s focus on one pair of points. There still are two steps:

Step 1: Find a random point that is inside the hexagon, and has distance d from its border.

Step 2: Find another point that has distance d from first point.

Main problem is step 1. We can say that a points that has distance d form a hexagon with radius r, is actually inside a hexagon with radius r-d. Then we just need to find a random point that lays on a hexagon!

Polar Formula of Hexagons:

I want to solve this problem in polar space, so I have to formulate hexagons in this space. Remember circle formula in polar space:

The formula of a hexagon in polar space is pretty much like its circumscribe circle, except that the radius of the hexagon differs at every t (angle). Let’s call this changing radius r2. So, if we find the function R2 that returns r2 for all ts then we can write polar formula for hexagon:

This image demonstrates parameters of the problem:

The key parameter here is α. Now we need a function Alpha that returns α for all ts:

Now we have all points on border of the hexagon in polar space:

r = 500;
T = linspace(0, 2*pi, 181);
Alpha = @(t) pi/2-abs(rem(t, pi/3)-(pi/6));
R2 = @(t) r*cos(pi/6)./sin(Alpha(t));
X = R2(T).*cos(T); 
Y = R2(T).*sin(T);

hold on
plot(X, Y, '.b');
plot((r).*cos(T), (r).*sin(T), '.r')

Polar Formula of a Regular Polygon:

Before I go on I’d like to generalize Alpha and R2 functions to cover all regular polygons:

Alpha = @(t) pi/2-abs(rem(t, 2*pi/(n))-(pi/(n)));
R2 = @(t) r*cos(pi/n)./sin(Alpha(t));

Where n is the number of edges of the polygon.

Answer:

Now we can generate pairs of points just like what we did for the circle problem:

r = 500; n = 6;
a = 10; b = 50;
N = 100;
D = (b - a).*rand(N,1) + a;

Alpha = @(t) pi/2-abs(rem(t, 2*pi/(n))-(pi/(n)));
R2 = @(t) r*cos(pi/n)./sin(Alpha(t));

T1 = rand(N, 1) * 2 * pi;
RT1 = rand(N, 1) .* (R2(T1)-D);
X1 = RT1.*cos(T1);
Y1 = RT1.*sin(T1);

T2 = rand(N, 1) * 2 * pi;
X2 = X1+D.*cos(T2);
Y2 = Y1+D.*sin(T2);

Rotating the polygon:

For rotating the polygon we just need to update the Alpha function:

t0 = pi/8;
Alpha = @(t) pi/2-abs(rem(t+t0, 2*pi/(n))-(pi/(n)));

This is a test for n=7, N=50000 and t0=pi/10:

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