# Most polyominoes on a Rubik's cube

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.

• This is a nice idea! What made you think of this puzzle? Aug 31, 2021 at 6:59
• @BmyGuest Well I just got my first ever Rubik's cube and I like polyominoes, so I just put the two together. Aug 31, 2021 at 7:04
• 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. Aug 31, 2021 at 23:31

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.

• 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. Aug 31, 2021 at 15:35
• Oh wow, you are right, I completely misunderstood the question. Aug 31, 2021 at 15:55
• 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 Aug 31, 2021 at 15:57

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)

• Each polyomino must be confined to a single face of the cube. This is forced by the coloring of the pieces of the cube. Aug 31, 2021 at 23:26
• 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... Aug 31, 2021 at 23:39
• 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. Sep 1, 2021 at 0:23
• 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. Sep 1, 2021 at 0:31