Right now I am trying to get the following distorted position:

On each face only one diagonal is solved and the rest of the cubes are a different colour than the diagonal. For example: If I am looking at the white face, there should exactly be two opposite corner pieces in that face correctly oriented, and no other white piece should be there.

If this is possible, what is an algorithm to get there?

  • $\begingroup$ I feel like you could use the checkerboard algorithm, then 4x rotate three corners algorithm. Note that I don't know what the actual names for these are and I don't have a cube handy to verify. $\endgroup$ – Ian MacDonald Dec 1 '15 at 18:53
  • $\begingroup$ This can be done - please see my post for a photo. $\endgroup$ – Dr Xorile Dec 3 '15 at 21:55
  • $\begingroup$ Did you mean for the stickers other than the diagonals to be all the same color, or just any colors other than the one on the diagonal? $\endgroup$ – user20 Dec 3 '15 at 23:01

This isn't possible to do.

Break the various cubes where all faces have diagonals into two cases:

  1. There does not exist a corner which lies on three diagonals.
  2. There exists a corner which lies on three diagonals.

In each of these cases, I will show why this can't be done.

(I'm using the standard Rubik's Cube color scheme: BOY clockwise, RO/WY/GB opposites format.)

In the first case, where no three diagonals meet, suppose we have two diagonals on adjacent faces. They must meet at a corner, which means (by assumption) the third diagonal does not connect at this adjacent corner.

Suppose (because colors are invariant), that the diagonals point to the DRF corner, and lie on the front and right faces. Because we're in the first case, we know that the top must go from the URF corner to the BLF corner, and the bottom must be rotated.

enter image description here

The only corner that can satisfy this orientation of the blue and yellow corner is the blue-yellow-red corner, which obviously won't work here because the diagonal is red. (The blue-yellow-orange corner is discounted because it would have the wrong positions of blue and yellow.) Thus, in the first case, there is no configuration.

In the second case, suppose without loss of generality that the joining corner is in the front top position, like so:

enter image description here

This is a little more annoying, but we can once again have to split this position into two subcases, since we need to assume one more diagonal to conclude anything about the remaining corners.

Subcase A:

Let's assume the following position:

enter image description here

There's only one corner fitting a couple of these positions, so let's slot them in:

enter image description here

Now we know the rest of the corners, so let's slot those in. (I apparently missed a green corner color here; this is an accident.)

enter image description here

Now, we just slot in the rest of the pieces such that they themselves are a rotation of a Rubik's cube, like so:

enter image description here

(In this specific picture, blank means white. A quirk of the editor I'm using.)

I actually thought this was a legal position when I first looked at it. Indeed, I assembled a cube in this position to try it out. Unfortunately, when I went to solve it, one final corner was rotated, making the state unsolvable.

Since this position is isomorphic to all other positions in of the same type, this case can be ruled out as a viable option.

Subcase B:

Now, we need to take a look at the other rotation of the green side.

enter image description here

Pick an arbitrary edge. By the above assumptions, all other edges are solved with respect to the one you pick. The colors you put on this edge are therefore irrelevant.

I fixed one arbitrary edge, and fixed the remaining twelve edges with respect to it:

enter image description here

(In this picture, the four blank edges are white. Again, a quirk of the editor.)

I've marked two corners. By assumption, the green diagonal points a specific direction. Thus, the other two corners are not on the diagonal, and those squares must be blue.

Therefore, the top back corner in this frame must be a blue-yellow corner, with blue on the green face and yellow on the yellow face. However, this corner isn't available! It's the one being used in the top front of the cube. The only remaining corner that has the right colors to fit is the blue-yellow-red corner, and it has the wrong orientation.

Thus, there is no solution to subcase B.

Since there are no cases where this is possible, the state you're looking for can't be found.

  • 1
    $\begingroup$ I'm not sure what you've proved here, but the OP asked for "On each face only one diagonal is solved and the rest of the cubes are a different colour than the diagonal." This can clearly be done - please see photos of cube in this state in my post. Perhaps I've misunderstood the OP's requirements. $\endgroup$ – Dr Xorile Dec 3 '15 at 21:54
  • $\begingroup$ The OP only asked for the non-diagonal stickers on each face to be a different color from the diagonal, not requiring that they all be the same different color. $\endgroup$ – f'' Dec 3 '15 at 22:30
  • $\begingroup$ @f'' Since the OP accepted this answer, I would imagine this is the pattern they were looking to produce. I decided to answer it this way because the question is ambiguous, and the stickers being of the same color makes more sense than it being irrelevant what their colors are. $\endgroup$ – user20 Dec 3 '15 at 23:03
  • $\begingroup$ @DrXorlie I read it the other way: that the other stickers are "of a [single] color," rather than "[each] of a different color." I've asked the OP to clarify above. $\endgroup$ – user20 Dec 3 '15 at 23:04

This is no problem. Here's a photo of the cube solved in this way:

Orange, green and white stripes Red, yellow and blue stripes

You want all the edges to be wrong, but edges can be twisted in place, provided you do so in pairs. There are 6 pairs of edges, so do that move and all your edges will be "wrong".

Corners can also be twisted in place, provided you do so in pairs, so do the diagonals of one face and the opposite diagonals of the opposing face. So, to be clear, if the cube has side length 1, with corners at 000,001,010,011,100,101,110,111, then twist 000-011 and 101-110.

That will leave all 6 faces with correct diagonals and no other correct square (as seen in the photos).

  • $\begingroup$ A faster way to get all the edges wrong is F2 B2 L2 R2 U2 D2. $\endgroup$ – f'' Dec 2 '15 at 1:32

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