How to obtain a 3D matrix from Graphics3D?

Question

I have a capillary system and I want to convert this into a numeric matrix. My aim is to create a matrix with NxNxN dimensions of 1's and 0's where the 0's represent the capillary and the 1's represent the transparent region into the box (similar to 2d's case questioned and answered).

The capillary was built manually; it is consists of tubes with different diameters/radius

(*trunk*)
t0 = {{4, 4, 0}, {4, 4, 2}};
r0 = 0.7;
(*first branches*)
t11 = {{4, 4, 2}, {2, 4, 4}};
t12 = {{4, 4, 2}, {6, 4, 4}};
t13 = {{4, 4, 2}, {4, 2, 4}};
t14 = {{4, 4, 2}, {4, 6, 4}};
r1 = 0.4;
(*first branche's secondary branches*)
t21 = {{3, 4, 3}, {2, 5, 5}};
t22 = {{3, 4, 3}, {2, 3, 5}};
t23 = {{5, 4, 3}, {6, 5, 5}};
t24 = {{5, 4, 3}, {6, 3, 5}};
t25 = {{4, 3, 3}, {3, 2, 5}};
t26 = {{4, 3, 3}, {5, 2, 5}};
t27 = {{4, 5, 3}, {5, 6, 5}};
t28 = {{4, 5, 3}, {3, 6, 5}};
r2 = 0.2;
(*second branches*)
t31 = {{4, 2, 4}, {1, 0, 5}};
t32 = {{4, 2, 4}, {7, 0, 5}};
t33 = {{4, 6, 4}, {1, 8, 5}};
t34 = {{4, 6, 4}, {7, 8, 5}};
t35 = {{2, 4, 4}, {0, 7, 5}};
t36 = {{2, 4, 4}, {0, 1, 5}};
t37 = {{6, 4, 4}, {8, 7, 5}};
t38 = {{6, 4, 4}, {8, 1, 5}};
r3 = 0.3;
(* branches to center*)
t41 = {{4, 2, 4}, {4, 3, 6}};
t42 = {{4, 6, 4}, {4, 5, 6}};
t43 = {{2, 4, 4}, {3, 4, 6}};
t44 = {{6, 4, 4}, {5, 4, 6}};
r4 = 0.1;
(*branches to center (vertical)*)
t51 = {{4, 3, 6}, {4, 3, 8}};
t52 = {{4, 5, 6}, {4, 5, 8}};
t53 = {{3, 4, 6}, {3, 4, 8}};
t54 = {{5, 4, 6}, {5, 4, 8}};
r5 = 0.1;
(*secondary branches conection*)
t61 = {{3, 2, 5}, {1, 1, 8}};
t62 = {{5, 2, 5}, {7, 1, 8}};
t63 = {{5, 6, 5}, {7, 7, 8}};
t64 = {{3, 6, 5}, {1, 7, 8}};
t65 = {{2, 5, 5}, {1, 7, 8}};
t66 = {{6, 5, 5}, {7, 7, 8}};
t67 = {{6, 3, 5}, {7, 1, 8}};
r6 = 0.1;

To generate the tubes with different diameters, I joined the radius to the branches

(*trunk*)
J0 = {{{4, 4, 0}, {4, 4, 2}}, r0};(*start and end point*)
(*first branches*)
J11 = {{{4, 4, 2}, {2, 4, 4}}, r1};
J12 = {{{4, 4, 2}, {6, 4, 4}}, r1};
J13 = {{{4, 4, 2}, {4, 2, 4}}, r1};
J14 = {{{4, 4, 2}, {4, 6, 4}}, r1};
(*first branche's secondary branches*)
J21 = {{{3, 4, 3}, {2, 5, 5}}, r2};
J22 = {{{3, 4, 3}, {2, 3, 5}}, r2};
J23 = {{{5, 4, 3}, {6, 5, 5}}, r2};
J24 = {{{5, 4, 3}, {6, 3, 5}}, r2};
J25 = {{{4, 3, 3}, {3, 2, 5}}, r2};
J26 = {{{4, 3, 3}, {5, 2, 5}}, r2};
J27 = {{{4, 5, 3}, {5, 6, 5}}, r2};
J28 = {{{4, 5, 3}, {3, 6, 5}}, r2};
(*second branches*)
J31 = {{{4, 2, 4}, {1, 0, 5}}, r3};
J32 = {{{4, 2, 4}, {7, 0, 5}}, r3};
J33 = {{{4, 6, 4}, {1, 8, 5}}, r3};
J34 = {{{4, 6, 4}, {7, 8, 5}}, r3};
J35 = {{{2, 4, 4}, {0, 7, 5}}, r3};
J36 = {{{2, 4, 4}, {0, 1, 5}}, r3};
J37 = {{{6, 4, 4}, {8, 7, 5}}, r3};
J38 = {{{6, 4, 4}, {8, 1, 5}}, r3};
(* branches to center*)
J41 = {{{4, 2, 4}, {4, 3, 6}}, r4};
J42 = {{{4, 6, 4}, {4, 5, 6}}, r4};
J43 = {{{2, 4, 4}, {3, 4, 6}}, r4};
J44 = {{{6, 4, 4}, {5, 4, 6}}, r4};
(*branches to center (vertical)*)
J51 = {{{4, 3, 6}, {4, 3, 8}}, r5};
J52 = {{{4, 5, 6}, {4, 5, 8}}, r5};
J53 = {{{3, 4, 6}, {3, 4, 8}}, r5};
J54 = {{{5, 4, 6}, {5, 4, 8}}, r5};
(*secondary branches conection*)
J61 = {{{3, 2, 5}, {1, 1, 8}}, r6};
J62 = {{{5, 2, 5}, {7, 1, 8}}, r6};
J63 = {{{5, 6, 5}, {7, 7, 8}}, r6};
J64 = {{{3, 6, 5}, {1, 7, 8}}, r6};
J65 = {{{2, 5, 5}, {1, 7, 8}}, r6};
J66 = {{{6, 5, 5}, {7, 7, 8}}, r6};
J67 = {{{6, 3, 5}, {7, 1, 8}}, r6};
J = {J0,J11,J12,J13,J14,J21,J22,J23,J24,J25,J26,J27,J28,J31,J32,J33,J34,J35,J36,J37,J38,J41,J42, J43,J44,J51,J52,J53,J54,J61,J62,J63,J64,J65,J66,J67};
Graphics3D[Tube[#[[1]], #[[2]]] & /@ J]

The figure generated by this branching process is

How to obtain a 3d matrix (with dimension 128x128x128)?

Is it possible to save this image in 3D (pixel)? If yes, so could import it a list of images as a 3D volume:

n = 128;
image = Import["image", "Image3D"]
image = ImageResize[image, {n,n,n}];
data = ImageData[MorphologicalBinarize[image]];
MatrixPlot[data]
MatrixPlot[1 - data]

Or alternatively, would it be possible some alternative way to get this 3D matrix?

n = 128; (*dimension of matrix*)
f[i_, j_, k_] := (*rule to generated the tube*)
s = SparseArray[{{i_, j_, k_} -> f[i, j]}, {n, n, n}]; (*matrix*)
MatrixForm[s];

Can anybody help me?

Thank you in advance.

Answer 1

You can apply FillingTransform directly to the Image3D produced using RegionImage to fill the hollow tubes:

si = Graphics3D[Tube[#[[1]], #[[2]]] & /@ J];
si2 = RegionImage[DiscretizeGraphics[si]];
si3 = si2 // FillingTransform // Binarize;
Image3DSlices[si3]

Answer 2

Another way is use CapsuleShape instead of Tube.

reg = Region[RegionUnion[CapsuleShape[#[[1]], #[[2]]] & /@ J]]
ri = RegionImage[DiscretizeRegion[reg, MaxCellMeasure -> 0.01]]
result = Image3DSlices[ri]
GraphicsGrid[Partition[result, UpTo[12]]]

• MaxCellMeasure -> 0.001

Answer 3

Actually, the Tube is not a filled 3D-polygon. This is just a pipe. However, the question about how to obtain the cross-section of a region is a really interesting. The simplest way is:

Rasterize[
 Graphics3D[{
   DiscretizeRegion[
    MeshRegion@
     DiscretizeGraphics[si, MaxCellMeasure -> 0.00001],
     {{0, 8}, {0,8}, {4.995, 5}}, MaxCellMeasure -> 0.000001]}, 
  ViewPoint -> Above, Boxed -> False],
 ImageSize -> 800,
 RasterSize -> 1920]

It gives you:

As one can see, choosing the third range (z-value range) in DiscretizeRegion you can take the position of desired cross-section. But these are still just points not a curve.. Playing with z-range, RasterSize and ImageSize one can adjust the image quality. Or you can look on this site for a post-processing of raster to make curves.