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.

  • 1
    $\begingroup$ 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
  • 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. $\endgroup$ Aug 31 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.

  • 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. $\endgroup$
    – StephenTG
    Aug 31 at 15:35
  • $\begingroup$ Oh wow, you are right, I completely misunderstood the question. $\endgroup$
    – aphyer
    Aug 31 at 15:55
  • 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 $\endgroup$
    – StephenTG
    Aug 31 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.

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

  • 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 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
  • 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 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.