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I have an old Rubik's Cube, the kind with stickers. The stickers tend to fall off.

That got me wondering: How many stickers can you remove from a Rubik's Cube at most, while preserving the property that the solution is unique?

Another way to look at it is: How many squares can you hide with opaque tape and still be able to solve the cube correctly (i.e. when you remove the tape it reveals a solved cube)?

The cube should be solvable without prior knowledge of the original color pattern, i.e. which color must be adjacent to which color on the solved cube. (Without this condition it could be possible to solve the cube to a different color pattern).

I don't know the answer yet. I don't think it is easy. I will validate the most convincing answer.

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I posted the problem without knowing the answer.

In the meantime I have been thinking about it and here is the best I could do.

Here is my cube missing 23 stickers.

cube with stickers missing

I think of the left half as the "front", and the right side as the "back".

And here is why it is unique

Let's start with the corners (here we see the exposed faces):

8 corners

The first 3 corners imply that blue, red and yellow are adjacent around a corner. There are 2 possible orientations, but one of them would place all 3 corners in the same place. So without loss of generality let's place blue, yellow and red on the front (left half) and place the 3 corners accordingly.

The next 2 corners have a yellow sticker. They must occupy the 2 missing yellow corners. One also has an orange sticker, it cannot be on the front, it has to go to the back. The other one goes to the front center.

The yellow+orange corner forces the orange face to be opposite the red one. Then notice that the center cross has a yellow and green sticker, placing these faces side by side. Anyway, we also have a yellow+green edge. This places the green face adjacent to the yellow and opposite the blue face. Then the white face goes to the bottom, opposite the yellow face.

The two stickers on the center cross let us orient it correctly.

Now that we know where the colors are, the red and blue corners must occupy the remaining corner of their respective faces.

The all-black corner occupies the last corner at the back. Its orientation is invisible. But because of the "parity" of the corners orientation, when all other corners are in place, the black corner must be oriented correctly.

So far we have placed all the corners and the center cross. At the right the placement of the faces.

partial solution with corners and centers

Second part:

Now let's place the edges.

The first 6 edges have 2 stickers, they can be placed easily.

6 more corners

partial solution with 6 pieces missing

For the remaining edges I will use Jaap Scherphuis's clever trick.

enter image description here

The 6 missing edges form a cycle around all 6 colors. The available edges all have a single sticker, allowing exactly 2 placement for the group, these placements being a rotation of each other. But a rotation of 6 edges is an odd permutation and we know a Rubik's cube can only have even permutation of the edges (at least once all the corners are in place). This means that only one of the 2 placements can be reached with legal moves.

From this it results that despite the missing stickers, only one placement of the pieces solves the cube.

completed cube, the same as the first picture

I'll leave the question open since I haven't proven the solution is optimal.

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Here is a way to remove 20 stickers.

enter image description here

To explain why the pattern works, consider the following cube with 18 stickers removed:

enter image description here

The cube has all its centre colours, so the 6 edges and 6 corners with 2 stickers have unique solved locations. The remaining 6 edges each have 1 sticker. There are exactly 2 ways they could be solved - the pattern above, or the following pattern:

enter image description here

Those edges have been moved in a 6-cycle, which is an odd permutation. Odd permutations are not possible on the Rubik's cube, so the two corners without stickers must also be swapped for this 6-cycle to happen. By putting back one sticker on the front corner, the corner swap is no longer possible, and this therefore forces the location of all the pieces. This added sticker also forces the orientation of that front corner, and the orientation of the stickerless corner is forced too because it is impossible to twist a single corner in isolation on a Rubik's cube.

Lastly, we can remove three centre stickers, in particular the blue, orange, and yellow ones. There are white-blue and orange-white corner pieces. This means the missing blue and orange centres must be adjacent to the white face, putting the missing yellow centre to the face opposite the white. Those two corner pieces also force the order of the orange and blue centres - if the colours were swapped both corners would have to fulfil the role of the orange-white-blue corner piece.

I don't think it is possible to remove a fourth centre sticker, but am not completely sure.


This reminded me of a sticker variation of the Rubik's cube that I came up with, that I called the Minimalist Cube. It came out of a similar idea, namely to use as few stickers as possible to still make an interesting cube puzzle. I ended up with this:

enter image description here

It uses only 1 sticker of each colour, but those stickers are each cut into quarters and placed on 4 pieces. Each piece has at most 1 quarter sticker. This puzzle does have multiple solutions, but also some non-solutions where a single centre is twisted 90 degrees.

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  • $\begingroup$ You can peel the yellow centre; there are corner pieces placing yellow adjacent to both red and green. It seems plausible that even a third centre might be peelable, but I haven't found a way to do that yet. $\endgroup$
    – Bass
    Mar 26, 2022 at 8:13
  • $\begingroup$ I checked it and validate your 18-sticker solution. But there are better solutions. $\endgroup$
    – Florian F
    Mar 26, 2022 at 9:27
  • $\begingroup$ @FlorianF I have now removed two more centre stickers, making it 20 peeled stickers. $\endgroup$ Mar 26, 2022 at 10:04
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    $\begingroup$ I validate the new 20-stickers-off solution. But I believe the optimum to be 23 or 24. $\endgroup$
    – Florian F
    Mar 26, 2022 at 11:36
  • $\begingroup$ @FlorianF I think it may be possible to take two more stickers off the corners in my solution, but 3 or 4 looks a bit too ambitious for now. I'll leave it or now, and maybe someone else will have a go. $\endgroup$ Mar 26, 2022 at 12:39
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The position of the colors are fixed, so all 8 corners have 3 colors such as if you hide one of them, if you still have the other 2 you can predict which one is there.

The 6 middle squares are unique so if you hide one of them you can predict which one it is by comparing with the other 5.

Finally you can lose one of the two stickers for at most one of each color from the 12 middle-corner parts because if you remove more than that some parts will become ambiguous. For example: if from the blue-white part you remove the white one, you can't remove the red square from the blue-red part, and you can't remove any other white squares from the other 3 white middle-corners.

That rounds up to a total of 15 specific stickers (not random) you can remove and still complete the Rubik's cube as if the stickers were still there.

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    $\begingroup$ I think the assumption in your first sentence is contradicted in the question by "The cube should be solvable without prior knowledge of the original color pattern" $\endgroup$
    – bobble
    Mar 26, 2022 at 0:58
  • $\begingroup$ @bobble I thought that meant the current settings of the cube, and assumed the cube has the original design $\endgroup$ Mar 28, 2022 at 14:19
  • $\begingroup$ That is clarified right after: "The cube should be solvable without prior knowledge of the original color pattern, i.e. which color must be adjacent to which color on the solved cube", and has been since the first version of the question $\endgroup$
    – bobble
    Mar 28, 2022 at 14:51
  • $\begingroup$ I see. In which case my solution is reduced from 15 to 7 stickers $\endgroup$ Mar 29, 2022 at 20:35

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