MATLAB Answers

selecting a triangle in an array of delaunay points

3 views (last 30 days)
luc
luc on 16 Dec 2014
Answered: luc on 18 Dec 2014
The code explains it all.
Try to run it, it might give you a good result, but that would be random.
This code is supposed to display a point, and color the nearest triangle green. It does not do this, and gives "nan" errors on seemingly random iterations of the loop.
I would appreciate some help :)
clc;
clear all;
close all;
P = [ 2.5 8.0
6.5 8.0
2.5 5.0
6.5 5.0
1.0 6.5
8.0 6.5];
dt=delaunayTriangulation(P);
triplot(dt);
hold on
%%example functions
dt.Points
dt.ConnectivityList
%abs vals x
a=min(P(:,1))
b=max(P(:,1))
%abs vals y
c=min(P(:,2))
d=max(P(:,2))
for n=1:5
random_x=a + (b-a).*rand(1);
random_y=c + (d-c).*rand(1);
triangleId=pointLocation(dt, random_x,random_y) %select the triangle ID of the triangle closest to the point generated by the random generator
scatter(random_x,random_y) %display the points
%%now... the part that does not work
%%selecting the right triangle and turning it a different color
tri = dt(triangleId, [1:end 1]);
patch(P(tri,1), P(tri,2), 'r', 'LineWidth',1, 'FaceColor','g')
end

  6 Comments

Show 3 older comments
luc
luc on 16 Dec 2014
Ill check tommorrow if that is the problem, if it is, thanks :). If it aint, I'll post here.
John D'Errico
John D'Errico on 17 Dec 2014
My point is, pointLocation will return NaN in that case. SURPRISE!
How often will that event happen? As it turns out, that should happen roughly 21.429% of the time that a point falls outside of your triangulated region, but inside the rectangle that you use to generate the random points.
(1 - (8-2.5)*3/(3*(8-1)))*100
ans =
21.429
luc
luc on 17 Dec 2014
Thanks! This solved my problem. I think an addition to the source code of this function would be to still pick the nearest triangle to the point, accompanied by an error message saying that the point itself is not inside a triangle.

Sign in to comment.

Accepted Answer

More Answers (0)


Translated by