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What is the most number of distinct free polyominoes you can form on the faces of a standard 3x3x3 Rubik's cube? Here a polyomino is considered as a set of orthogonally-adjacent cells of the same colour lying on the same face of the cube. Two free polyominoes are considered distinct if they are not a rigid transformation (translation, rotation, reflection, glide-reflection) of each other.

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    $\begingroup$ This is a nice idea! What made you think of this puzzle? $\endgroup$
    – BmyGuest
    Aug 31 at 6:59
  • $\begingroup$ @BmyGuest Well I just got my first ever Rubik's cube and I like polyominoes, so I just put the two together. $\endgroup$ Aug 31 at 7:04
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    $\begingroup$ Confinement of each polyomino to a single face is forced by the design of the cube. A monochrome polyomino could never wrap around from one face to another. $\endgroup$ Aug 31 at 23:31
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No pictures right now, sorry.

We have 9x6=54 squares to work with.

The maximum we can fit is:

1 monomino = 1

1 domino = 2 (total 3)

2 trominos = 6 (total 9)

4 tetrominos (line cannot fit) = 16 (total 25)

This leaves 29 squares left, which can fit at most 5 pentominoes or bigger.

So our theoretical maximum is 13.

We can do this with e.g.

U pentomino + T tetromino

L pentomino + square tetromino

L tetromino + extended square pentomino

W pentomino + L tromino + monomino

S tetromino + line tronimo

Domino + anything else that fits in the 7 remaining squares.

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    $\begingroup$ While this shows what polyominos could be packed onto the 6 faces of a rubiks cube, it doesn't demonstrate that it would actually be possible to configure a cube like this, which from my understanding is part of the requirements of the question. $\endgroup$
    – StephenTG
    Aug 31 at 15:35
  • $\begingroup$ Oh wow, you are right, I completely misunderstood the question. $\endgroup$
    – aphyer
    Aug 31 at 15:55
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    $\begingroup$ Still valuable showing the upper bound, and if your proposed arrangement actually ends up being possible all you'd need to add is a picture of a cube or the like $\endgroup$
    – StephenTG
    Aug 31 at 15:57
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Here is an example of the six faces tiled with the maximum number of distinct free polyominoes, ie 13.

Method: I tiled a 3x3 with all possible combinations of polyominoes of size 1 through 7. Then I searched for all disjoint cliques and discarded several with only 12 pieces, and rendered just one of them.

Update: I re-ran the whole thing with octominoes included to get a count of all possible solutions. There are 78 ways of selecting 6 sets of pieces for the faces. 14 of them use 13 polyominoes, the rest 12. If you allow one of the faces to be a 3x3 there are 325 sets of polyominoes in total.

enter image description here

I'll leave the original answer here as an example of tiling the faces of a cube 'with wrap': Here is a 3-cube tiled with all polyominoes to size four plus five pentominoes. I tiled a net of a cube, you just have to fold it up in your imagination. 14 polyominoes total.

Method: I ran my tiler on this shape and told it to use up to five of the 12 pentominoes plus all of the smaller polyominoes. It's a zillion lines of hieroglyphic code designed for all sorts of other things so I won't post it here...

enter image description here

(This solution works for an Australian Rubik's Cube) enter image description here

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    $\begingroup$ Each polyomino must be confined to a single face of the cube. This is forced by the coloring of the pieces of the cube. $\endgroup$ Aug 31 at 23:26
  • $\begingroup$ Ah thanks, I missed the "Here a polyomino is considered as a set of orthogonally-adjacent cells of the same colour lying on the same face of the cube.". I'll try again... $\endgroup$ Aug 31 at 23:39
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    $\begingroup$ Even if the author hadn't updated to say "lying on the same face of the cube", having a polyomino extend over an edge to another face is impossible. That would require a cubelet with two facets of the same color, which does not occur on a Rubik's cube. $\endgroup$ Sep 1 at 0:23
  • $\begingroup$ Maybe... All the cubes around my house live in an 'unsolved' state. There are no resident competent solvers. But I don't really want to try to fit my solution to a valid unsolved cube so I just fixed my answer to what was intended....ah... I read your comment more carefully... your are correct of course. $\endgroup$ Sep 1 at 0:31

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