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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asked Aug 31, 2021 at 6:42
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1$\begingroup$ This is a nice idea! What made you think of this puzzle? $\endgroup$– BmyGuestAug 31, 2021 at 6:59
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$\begingroup$ @BmyGuest Well I just got my first ever Rubik's cube and I like polyominoes, so I just put the two together. $\endgroup$– Dmitry KamenetskyAug 31, 2021 at 7:04
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1$\begingroup$ 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. $\endgroup$– Daniel MathiasAug 31, 2021 at 23:31
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2 Answers
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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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5$\begingroup$ 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. $\endgroup$ Aug 31, 2021 at 15:35
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$\begingroup$ Oh wow, you are right, I completely misunderstood the question. $\endgroup$– aphyerAug 31, 2021 at 15:55
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1$\begingroup$ 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 $\endgroup$ Aug 31, 2021 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.
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...
(This solution works for an Australian Rubik's Cube)
answered Aug 31, 2021 at 23:09
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2$\begingroup$ 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, 2021 at 23:26
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$\begingroup$ 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, 2021 at 23:39
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3$\begingroup$ 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, 2021 at 0:23
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$\begingroup$ 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, 2021 at 0:31