How to generate random positions with distance between them inside the hexagon? - matlab

Question

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

Answer 1

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: