Create/choose a rondom point from an existed data matrix?

I have data matrix (3000 * 3) and then I created from these data delaunay triangles with vertices P0, P1 and P2. I want to establish rondom points Pr which come originally from the data matrix (3000 * 3) and these random points are located inside the generated delaunay triangles vertices P0, P1 and P2. Could you please help me to achieve this ?

Answer 1

[Roger Stafford]

I will give an answer in terms of a single triangle with vertices at P0, P1, and P2, (either in 2D or 3D.) Define

   r1 = rand;
   r2 = sqrt(rand); % <-- The square root operation is important here

Then the value P

   P = (1-r2)*P0+r2*(r1*P1+(1-r1)*P2);

will be a random point within the triangle P0P1P2, with a statistically uniform probability distribution therein.

I will leave the problem of doing this for the entire triangulation to you. It can easily be vectorized.

[Stephen23]

Susan Arnold's "Answer" moved here:

The method above worked unfortunately not :(

[Susan Arnold]

Thank you Mr.Roger Stafford. It worked now, but I need please some mathematical explanation for the above equation, if you please do not mind.

[Susan Arnold]

It worked without taking the square root for the second matrix, which I do not want to do it, but when I calculate for example the areas of the generated Delaunay triangles I got unbalanced results. Triangle A for example has the majority area, let's say 12. And the other two triangles B and C have both 0.5.

[Susan Arnold]

I tried to apply inpolygon function, but I do not know how to apply it for three vertices. Please I need your help.

[Roger Stafford]

Removing the square root operation will still generate points within the triangle, but they will not be statistically distributed uniformly area-wise within the triangle. They will tend to concentrate near the P0 vertex, which I assume you do not want.

You can verify this using the following example, both with and without the 'sqrt' operation in m2:

P0 = randn(1,2); % Random triangle
P1 = randn(1,2);
P2 = randn(1,2);
N = 4000; % Number of random points generated within triangle
m1 = rand(N,1);
m2 = sqrt(rand(N,1));
P = (1-m2)*P0+(m2.*m1)*P1+(m2.*(1-m1))*P2;
plot(P(:,1),P(:,2),'y.',... % Yellow dots within red triangle
     [P0(1),P1(1),P2(1),P0(1)],[P0(2),P1(2),P2(2),P0(2)],'r-')
axis equal

For values of 0<=m1<=1 and 0<=m2<=r, the area covered in the triangle is another smaller similar triangle whose area is proportional to the square of r. This fact is what justifies the use of the square root above in conjunction with the 'rand' function. See the discussion in:

https://en.wikipedia.org/wiki/Triangular_distribution

in the "Generating Triangular-distributed random variates" section, which is closely related to this, or perhaps the 2012 discussion at:

http://www.mathworks.com/matlabcentral/newsreader/view_thread/323426