Divide a 3D shape into equal-volume subregions?

I have a 3D shape that can be approximated with Delauny triangularization or a convex hull. I would like to divide the shape into a certain number of equal-volume subregions; ideally each subregion would have nearly the same surface area. Here are a couple places I looked but did not find a solution,

• Object functions that you can use on a Delauny triangularization object, a polyshape object, or an alphashape object
• File exchange polygon2voxel

Below is an example conical 3D shape. How can I divide the volume enclosed by the shape into 100 equal-volume subregions? The subregions can all be entirely enclosed by the hull, leaving some of the shape's volume unenclosed by subregions, as if you were to fill a conical container with ice cubes.

Is the best way just to grid the region in xyz and adjust the grid spacing until I get 100 cubes?

load('seamount.mat', 'x', 'y', 'z'); % Generic data that comes with Matlab; contains 3D points
DT = delaunayTriangulation(x,y,z); % 3D triangulization of points
[C, volHull] = convexHull(DT); % Compute convex hull, which is cone-shaped
figure('Name', 'Convex Hull');  
hPatch = trisurf(C,DT.Points(:,1),DT.Points(:,2),DT.Points(:,3), ...
    'FaceAlpha', 0.1);
% Now how to find 100 subregions of equal volume within this shape

[darova]

Please attach the data

[darova]

Is this correcT?

[KAE]

Yes! The subregions do not have to be cubes, they can be any shape as long as they all have the same volume, and have reasonably similar surface area (i.e. not long rods, for example).

[darova]

Did you try to serach in fileexchange? Honestly, i don't think such program exists

[KAE]

Yes, I only could find this, but I am new to 3D discretization and may not have been using the right terms.

[John D'Errico]

I sincerely doubt it will be easy to do for a completely general shape, especially not if you may have a shape that need not even be convex or one that lacks some simple regularity.

That said, it is trivial to do, IF you will allow the shape to be sliced into very flat sections. But you seem to want to force the pieces to be themselves nicely regular, that will now tile the domain. Shiver.

I can only offer luck in your quest. I would strongly suggest that you formulate the problem more carefully and completely. For example, you are not very clear about the necessary shape of the pieces. But vagueness is bad when you are dealing with mathematics.

[KAE]

Perhaps I should take my cue from the fact that voxels are usually shown as cubes, and stick to cube shapes. Thanks for the suggestion to define the problem more clearly: I am trying to come up with regular locations inside a 3D shape for a sampling problem. So I will divide the shape up into 3D cubes of equal size, requiring that the cubes fit fully inside. I put a draft of how to do this in my answer.

Answer 1

[KAE]

Based on the comments, my original question was too open ended, so I am answering this to bring it to a close. A simple 3D division of the space inside the volume into cubes is good enough, something like,

load('seamount.mat', 'x', 'y', 'z'); % Generic data that comes with Matlab; contains 3D points
% Make the example shape less flat!
x = x*5000; 
y = y*8000; 
DT = delaunayTriangulation(x,y,z); % 3D triangulization of points
[C, volHull] = convexHull(DT); % Compute convex hull, which is cone-shaped
figure('Name', 'Convex Hull');  
hPatch = trisurf(C,DT.Points(:,1),DT.Points(:,2),DT.Points(:,3), ...
    'FaceAlpha', 0.1);
nCubesDesired = 100; 
cubeLength = range(z)/round(nCubesDesired^(1/3)); 
xGrid = min(x):cubeLength:max(x); 
yGrid = min(y):cubeLength:max(y);
zGrid = min(z):cubeLength:max(z);
% Loop through each cube and check if it is inside the volume using 
% in_polyhedreon from File Exchange
% https://www.mathworks.com/matlabcentral/fileexchange/48041-in_polyhedron
% If it is not inside the volume, throw the cube out.
set(gca, 'xtick', xGrid, 'ytick', yGrid, 'ztick', zGrid);
grid on;
% Iterate on cubeLength until you get the desired number of cubes

[darova]

Are you talking about griddata?

[KAE]

Sorry, I was not clear. I just meant I would slice up the 3D space's volume into equal-sized cubes. I moved the draft of how to do this from my comment to the answer.