How can I visualize 6 square matrices as a cube?

Question

I have six square matrices of the same dimensions in a list, for example:

n = 4; (*Matrix dimension*)
cubedata = Table[ConstantArray[i, {n, n}], {i, 1, 6}] (*Matrices 1 to 6*)

I would like to make a visualization of the cube that is formed when you put said matrices as its faces. This should be the orientation of the faces.

I want to see the numbers and, if possible, the cube colored (like a Rubik's cube).

I don't even know where to start. I think I should use Graphics3D but I'm not familiar with it, at all.

I would greatly appreciate it if you could help me.

Answer 1

So you need two thing. Six matrices and and a cube with each side painted with a matrix. In such wall painting Texture is the most handy option. Let me show an example.

n = 3; (*Matrix Dimension*)
color = {Red, Blue, Green, Yellow, Orange, White};
Table[mat[k] = 
  Grid[Table[RandomInteger[{1, n^2}], {i, 1, n}, {j, 1, n}], 
    Frame -> All, 
    ItemStyle -> Directive[FontSize -> 20, Bold], 
    Background -> Table[RandomChoice[color], {i, 1, n}, {j, 1, n}]],
  {k, 1, 6}];

You can choose your matrix and style as you wish. Now I am going to make a cube using Polygon and I will use these matrices as wallpaper.

vtc = {{0, 0}, {1, 0}, {1, 1}, {0, 1}}; (*VertexTextureCoordinates*)
coords = {{{0, 0, 0}, {0, 1, 0}, {1, 1, 0}, {1, 0, 0}}, 
         {{0, 0,0}, {1, 0, 0}, {1, 0,1},{0, 0, 1}}, 
         {{1, 0, 0}, {1, 1, 0}, {1,1, 1}, {1, 0, 1}}, 
         {{1, 1, 0}, {0, 1, 0}, {0, 1, 1},{1, 1,1}}, 
         {{0, 1, 0}, {0, 0, 0}, {0, 0, 1}, {0, 1, 1}}, 
         {{0, 0,1}, {1, 0, 1}, {1, 1, 1},{0, 1, 1}}};

Graphics3D[Table[{Texture[mat[k]],Polygon[coords[[k]], VertexTextureCoordinates -> vtc]},{k, 1, 6}], Boxed -> False]

Tada, your cube is served.

Answer for modified question

You just make the job simpler.

n = 3;(*Matrix Dimension*)
color = {Red, Blue, Green, Yellow, Orange, Black};
Table[mat[k] = Grid[Table[RandomInteger[{1, n^2}], {i, 1, n}, {j, 1, n}], ItemStyle ->     Table[{FontSize -> 20, Bold, RandomChoice[color]}, {i, 1, n}, {j,1, n}], Frame -> All], {k, 1, 6}];
(*For 2D visual*)   
Grid[{{, mat[6], ,}, {mat[4], mat[1], mat[2], mat[3]}, {, mat[5], ,}},Frame -> None, Spacings -> {0.1, 0.1}]
(*For 3D visual*)
Graphics3D[Table[{Texture[mat[k]],Polygon[coords[[k]], VertexTextureCoordinates -> vtc]},{k, 1, 6}], Boxed -> False]
(*use the same coords and vtc as before*)

This time instead of using Graphics3D with that large coordinate set, I'm simply using a Grid to create the output in a order you suggested.

Anyway I put both the outputs (just in case you want to play in higher dimension :) ).

Answer 2

My first version had to be repaired because I didn't think about the orientation of the faces, and I cheated...

faces = First@Normal[PolyhedronData["Cube", "Faces"]];
grids = ImagePad[
     Rasterize[#, ImageSize -> 400],
     10,
     Padding -> White] & /@ Table[
    Grid[
     ConstantArray[i, {4, 4}],
     ItemStyle -> {Automatic, 
       Automatic, {{1, 1} -> Red, {3, 3} -> Green}}, 
     Background -> 
      RandomChoice[{LightRed, LightBlue, LightYellow, LightGreen}]],
    {i, {5, 1, 4, 3, 6, 2}}];
textures = MapThread[ImageRotate[#1, #2] &,
   { grids, {Right -> Top, 0, Left -> Top, Left -> Top, Right -> Left,
      Left -> Top}}];
cube = Graphics3D[
   Table[{Texture[textures[[n]]],
     faces[[n]]}, {n, 1, 6}],
   Lighting -> "Neutral",
   Background -> None,
   Boxed -> False,
   ViewPoint -> {2, -2, -1},
   ViewAngle -> .6,
   ViewVertical -> {0, 0, -1}] /. Polygon[l_] :>
   Polygon[l,
    VertexTextureCoordinates -> {{0, 0, 0}, {1, 0, 0}, {1, 1, 0}, 
     {0, 1, 0}}]

By the way, you'll find a complete Rubik's cube implementation on the Wolfram demonstration site.

I don't think this page can afford any more animated spinning business, because Kuba has pursued that to a logical - if not absurd - conclusion. But, to check that the back face of the cube is correct, hold it up to a mirror:

reflection = Graphics3D[
    Table[{Texture[textures[[n]]],
      faces[[n]]}, {n, 1, 6}],
    Lighting -> "Neutral",
    Background -> Gray,
    Boxed -> False, 
    ViewVertical -> {0, 0, -1},
    ViewAngle -> .8,
    ViewPoint -> {-1.5, 1.5, -1}] /. Polygon[l_] :>
    Polygon[l,
     VertexTextureCoordinates -> {{0, 0, 0}, {1, 0, 0}, {1, 1, 0}, 
       {0, 1, 0}}];
Show[
 ImageAdjust[
  ImagePad[
   Blur@ImageResize[
     ImageReflect[reflection, Right -> Left], 450],
   {{-30, -30}, {30, 30}}, 
   Padding -> Gray],
   {-.2, -.2}],
  Image[cube]
 ]

I'm starting to regret writing a quick answer to this question this morning... :)

Answer 3

Do not go further if You have epileptic-seizures/migraines related health problems.

I know You have meant what my followers showed but this is fun:

n = 6;
cubedata = Table[i + j + k, {k, 6}, {i, n}, {j, n}] ;
da = Table[{
             cubedata[[ k, i, j]], 
             Insert[1/(2 n) + {(i - 1)/n, (j - 1)/n}, If[OddQ@k, 0, 1], Ceiling[k/2]]
           }
         , {k, 6}, {i, n}, {j, n}];
Graphics3D[Text[Style[#1, Bold], Scaled[#2]] & @@@ Flatten[da, 2]]

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---

And one clearer version:

cubedata = Table[i + j + k, {k, 6}, {i, n}, {j, n}] ;
da =Table[{cubedata[[ k, i, j]],Insert[1/(2 n) + {(i - 1)/n,(j - 1)/n},
           If[OddQ@k, 0, 1], 
           Ceiling[k/2]], k/6}, {k, 6}, {i, n}, {j, n}];
Graphics3D[{Orange, Cuboid[], 
Inset[Style[#1, Bold, 18, Hue@#3], Scaled[#2]] & @@@ Flatten[da, 2]}]

Notice the difference if Inset->Text. Then Text can be seen through walls :)