problem with convhulln degenerate

Hi my name is noé alvarado I want to find de convex hull of the next point:

xyzw = [1 -1 -1.375 0;-0.5 -0.8 -0.2830 0;-1.3 0.7 0.217 0;1.5 1.3 1.74 0;0.6 0 0 0]
xyzw =
    1.0000   -1.0000   -1.3750         0
   -0.5000   -0.8000   -0.2830         0
   -1.3000    0.7000    0.2170         0
    1.5000    1.3000    1.7400         0
    0.6000         0         0         0
    tess = convhulln(xyzw)

however i have the error:

Error using ==> qhullmx The data is degenerate in at least one dimension - ND set of points lying in (N+1)D space.

Could yoyu help me?

thanks

Answer 1

[Nshine]

xyzw = [1 -1 -1.375 0;-0.5 -0.8 -0.2830 0;-1.3 0.7 0.217 0;1.5 1.3 1.74 0;0.6 0 0 0]
convhulln(xyzw)

The error occurs because the last column is all zeros. If you delete the last column and try again:

xyzw(:,4) = [];
convhulln(xyzw)

you get the answer

ans =

     1     4     3
     4     2     3
     2     1     3
     1     2     4
     

ie: facet 1 is made by the triangle connecting vertex 1, 4 and 3, etc.

Answer 2

[Image Analyst]

I'm not sure I understand. You have a 2D matrix. Why not use the 2D version of convex hull, convhull()?

And what is "the next point"? What were the prior points for that matter?

You can't have the convex hull of just a point - you need at least 3 points to give a meaningfull convex hull.

[Nshine]

This is an old thread, but the the argument for convhulln, X is an m-by-n array (ie: 2D array) representing m points in n-D space, as can be read in the documentation:

help convhulln

Answer 3

[ilPlus30]

Hi everybody,

even keeping the last column , you could try this way:

xyzw = [1 -1 -1.375 0;-0.5 -0.8 -0.2830 0;-1.3 0.7 0.217 0;1.5 1.3 1.74 0;0.6 0 0 0]
convhulln (xyzw, {'QJ'})

[Walter Roberson]

The last column of xyzw is all 0.

[ilPlus30]

Sorry you're right, however in that way the last column could go on, for example it could be important not to delete it.

Answer 4

[Shannu]

what id degeneracy of matrix

A=[-1 0 0 0
    0 -1 0 0 
    0 0 -1 0
    0 0 0 -1];

[Walter Roberson]

You have 4 points in a 4 dimensional space. In order to generate a convex hull, you need at least 5 points for a 4 dimensional space.