Tomographic Reconstruction of a Convex Polyhedron from its Silhouettes

Question

So, I can construct a random polyhedron and find its 3 silhouettes onto the 3 standard planes.

For example,

    polyhedron = PolyhedronCoordinates[RandomPolyhedron[{"ConvexHull", 10}]];
    
    (* Projection Functions *)
    projections = polyhedron[[All, #]] & /@ {{1, 2}, {1, 3}, {2, 3}};
    
    (* Find Convex Hulls and Create Graphics *)
    graphics = Graphics[Polygon[MeshCoordinates[#][[MeshCells[#, 2][[1, 1]]]]]] & /@ (ConvexHullMesh /@ projections);

    (* Labeled Graphics *)
    titles = {"XY Projection", "XZ Projection", "YZ Projection"};
    labeledGraphics = MapThread[Labeled[#1, #2, Bottom] &, {graphics, titles}];
    
    (* Display *)
    GraphicsGrid[{labeledGraphics}]

An example polyhedron and its projections are shown below:

which gives the three orthographic projections:

The question is, given the three standard orthographic projections, how do I find the original convex polyhedron knowing that the polyhedron is convex?

UPDATE:

So, cvgmt's answer is correct, it was not unique with how the question was originally posed, and there is a lot of value in that answer. But I feel that it would be good to extend the question to include solutions that would reconstruct the original convex polyhedron using minimal information. I am moving the goalpost, that is why I approved cvgmt's answer, and am setting a bounty on this. I think there are multiple ways to do this. For example, looking at a different set of silhouettes. Or looking at a small continuous set of silhouettes instead of discrete.

One way I am thinking is something like this, but not quite sure how to code it exactly, :

Shadows in a Bounding Box

Imagine the Z-axis goes in and out of the board, positive Z out of the board, the X-axis is left to right, and the Y-axis is up and down. So, since we have the projections onto the bounding box, let's say we use the XY+ projection, the XY- projection, the YZ+ projection, and the YZ- projection. The XY+ projection and the XY- projection are the exact same, except that XY- has the same Z-coordinate as the smallest Z from the original polyhedron and XY+ has the largest Z coordinate as the largest Z from the original polyhedron. The YZ+ projection and the YZ- projection are the exact same, except that YZ- has the same X-coordinate as the smallest X from the original polyhedron and YZ+ has the largest X coordinate as the largest X from the original polyhedron. The respected projections, XY+ to YZ+ to XY- to YZ- form a 'ring' around the the original bounding box. Note how we don't need the XZ projections. The 'cylinder' is then from XZ- to XZ+, lines are drawn to connect the line segments with the same Y-value. Then find the convex hull. I am not sure if that would work or not.

Cylinder is a poor word choice, but I cannot think of anything better.

Answer 1

• We only need the data of the boundary points of the projections.

Clear[data];
polyhedron = 
  PolyhedronCoordinates[RandomPolyhedron[{"ConvexHull", 10}]];
projections = polyhedron[[All, #]] & /@ {{1, 2}, {1, 3}, {2, 3}};
data = MeshCoordinates /@ ConvexHullMesh /@ projections

• Not unique,but we can construct the maximun one of such polyhedron from the boundary vertexs of the three projections,do not assume that we know the interior points of the projections.

regs = {reg12, reg13, reg23} = data // Map@ConvexHullMesh;

{{{x1, x2}, {y1, y2}}, {{x1, x2}, {z1, z2}}, {{y1, y2}, {z1, z2}}} = 
  RegionBounds /@ regs;

Reg12 = BoundaryMesh[RegionProduct[reg12, Line[{{z1}, {z2}}]], 
   ViewPoint -> {0, 0, 1}, ViewProjection -> "Orthographic"];

Reg13 = RegionProduct[reg13, Line[{{y1}, {y2}}]] // BoundaryMesh;
Reg13 = BoundaryMeshRegion[MeshCoordinates[Reg13][[;; , {1, 3, 2}]], 
   MeshCells[Reg13, 2], ViewPoint -> {0, -1, 0}, 
   ViewProjection -> "Orthographic"];

Reg23 = RegionProduct[reg23, Line[{{x1}, {x2}}]] // BoundaryMesh;
Reg23 = BoundaryMeshRegion[MeshCoordinates[Reg23][[;; , {3, 1, 2}]], 
   MeshCells[Reg23, 2], ViewPoint -> {1, 0, 0}, 
   ViewProjection -> "Orthographic"];

result = RegionIntersection[Reg12, Reg13, Reg23]

Grid[{{reg12, reg13, reg23}, {Reg12, Reg13, Reg23}, 
  Show[result, ViewPoint -> #, 
     ViewProjection -> "Orthographic"] & /@ {{0, 0, 1}, {0, -1, 
     0}, {1, 0, 0}}}]

• Such polyhedron is not unique. We can calculus their RegionDifference and relate volume and draw the solids with difference colors to show this.

{result, ConvexHullMesh@polyhedron, 
  RegionDifference[result, ConvexHullMesh@polyhedron]} // Volume
Graphics3D[{{Opacity[.2], Yellow, result}, {Opacity[.9], Green, 
     ConvexHullMesh@polyhedron}, {Red, Opacity[.1], 
     RegionDifference[result, ConvexHullMesh@polyhedron]}}, 
   ViewProjection -> "Orthographic", Boxed -> False, 
   ViewPoint -> #] & /@ {{0, 0, 1}, {0, -1, 0}, {1, 0, 0}, {1, 1, 1}}

• We test non-convex region, for example, the regions construct by some fonts.
• Since RegionIntersection not always work for non-convex region, we use OpenCascadeLink instead.

Clear["Global`*"];
reg12 = BoundaryDiscretizeGraphics[Text["G"], _Text];
reg23 = BoundaryDiscretizeGraphics[Text["E"], _Text];
reg13 = BoundaryDiscretizeGraphics[Text["B"], _Text];
{{x1, x2}, {y1, y2}} = RegionBounds[reg12];
reg23 = RegionResize[reg23, {y2 - y1, Automatic}];
{{y1, y2}, {z1, z2}} = RegionBounds[reg23];
reg13 = RegionResize[reg13, {{x1, x2}, {z1, z2}}];
Reg12 = BoundaryMesh[RegionProduct[reg12, Line[{{z1}, {z2}}]], 
   ViewPoint -> {0, 0, 1}, ViewProjection -> "Orthographic"];
Reg13 = RegionProduct[reg13, Line[{{y1}, {y2}}]] // BoundaryMesh;
Reg13 = BoundaryMeshRegion[MeshCoordinates[Reg13][[;; , {1, 3, 2}]], 
   MeshCells[Reg13, 2], ViewPoint -> {0, -1, 0}, 
   ViewProjection -> "Orthographic"];
Reg23 = RegionProduct[reg23, Line[{{x1}, {x2}}]] // BoundaryMesh;
Reg23 = BoundaryMeshRegion[MeshCoordinates[Reg23][[;; , {3, 1, 2}]], 
   MeshCells[Reg23, 2], ViewPoint -> {1, 0, 0}, 
   ViewProjection -> "Orthographic"];

Needs["OpenCascadeLink`"];
Needs["NDSolve`FEM`"];
{shape12, shape13, shape23} = 
  OpenCascadeShape[ToBoundaryMesh@#] & /@ {Reg12, Reg13, Reg23};
shape = OpenCascadeShapeIntersection[shape12, shape13, shape23];
bm = OpenCascadeShapeSurfaceMeshToBoundaryMesh[shape];
reg = MeshRegion@bm;
Grid[{{reg12, reg13, reg23}, {Reg12, Reg13, Reg23}, 
  RegionPlot3D[reg, ViewPoint -> #, ViewProjection -> "Orthographic", 
     ColorFunction -> Hue] & /@ {{0, 0, 1}, {0, -1, 0}, {1, 0, 
     0}, {-1, -1, 3}}}]

Answer 2

Try

polyhedron = PolyhedronCoordinates[RandomPolyhedron[{"ConvexHull", 10}]];

(* Projection Functions *)
projections = polyhedron[[All, #]] & /@ {{1, 2}, {1, 3}, {2, 3}};

define symbollically unknown polyhedron and its projections

ph = Table[{x[k], y[k], z[k]}, {k, Length[polyhedron]}]
phproj = ph[[All, #]] & /@ {{1, 2}, {1, 3}, {2, 3}}

fit the coordinates of ph

null = phproj - projections // Flatten;
mini=NMinimize[null . null, Flatten[ph]]
(*{-1.77636*10^-15, {x[1] -> 0.91212, y[1] -> 0.067305, 
z[1] -> 0.404641, x[2] -> 0.774921, y[2] -> 0.0863487, 
z[2] -> 0.0197736, x[3] -> 0.30895, y[3] -> 0.218595, 
z[3] -> 0.976788, x[4] -> 0.396847, y[4] -> 0.177632, 
z[4] -> 0.781611, x[5] -> 0.0989693, y[5] -> 0.437153, 
z[5] -> 0.328092, x[6] -> 0.978541, y[6] -> 0.330944, 
z[6] -> 0.328021, x[7] -> 0.374619, y[7] -> 0.547744, 
z[7] -> 0.0331691, x[8] -> 0.330898, y[8] -> 0.867104, 
z[8] -> 0.0107966, x[9] -> 0.776468, y[9] -> 0.599847, 
z[9] -> 0.167013}}*)