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Divide a 3D shape into equal-volume subregions?

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KAE
KAE am 24 Feb. 2020
Bearbeitet: KAE am 26 Feb. 2020
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,
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
  7 Kommentare
John D'Errico
John D'Errico am 25 Feb. 2020
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
KAE am 25 Feb. 2020
Bearbeitet: KAE am 26 Feb. 2020
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.

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KAE
KAE am 26 Feb. 2020
Bearbeitet: KAE am 26 Feb. 2020
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
  2 Kommentare
darova
darova am 26 Feb. 2020
Are you talking about griddata?
KAE
KAE am 26 Feb. 2020
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.

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