Find a scramble which will produce a Rubik's cube fulfilling the following criteria which uses as few turns of faces as possible (Half-Turn-Metric):

  • Each of the six faces show exactly three colours
  • No face shows its opposite colour (the colour of the centre of the opposite face)
  • Each of the eight collections of three faces which share a corner show all six colours.

As an example of a conforming state here is a seventeen turn scramble (which is, of course, too long):

R' B F' R U2 F D' U R2 B F' U B2 L' D U' L

Which, when applied to a solved cube, produces:


Note that: there is no orange on the top, no green on the front, etc...; that each side shows exactly three colours; and looking directly at any corner one will see all six colours.

  • $\begingroup$ Are E, E', M, M', S, S' and x,y,z allowed ? $\endgroup$
    – 19aksh
    Commented Jun 9, 2019 at 17:46
  • $\begingroup$ @Ak19 The measure I have chosen for the puzzle is HTM so x,y, and z are effectively free, while the slices would cost the face turns required. I'd also suggest writing out your solution using just face turns for clarity. $\endgroup$ Commented Jun 9, 2019 at 17:58

1 Answer 1


I worked through several approaches to solve this and really enjoy the challenge. Given the number of permutations that would satisfy the given conditions, proving that a solution is optimal may require brute force. I decided to do it the hard way, trying combinations of patterns both in Cube Explorer and my own cube.

At first I started with leaving as much red as possible on the red face assuming that replacing fewer pieces would mean less moves. This means only replacing two pieces with different colors. Intuitively, I think the solution would need to have symmetry for the pattern to hold true from all sides. So replacing any two pieces somewhat dictates where the rest of the pieces should go in turn. This approach ended up not working out too well with the few permutations that I tried.

I discovered that if all 4 corners are the same color on one face, they have to be in a normal permutation relative to each other. Moving pairs of pieces together reduces move count.

Using these principles, here is the best solution I've come up with so far:

B' F L R' D U' B' F D2 U2 - 10 turns HTM! If slices were allowed this is only 5 moves: E2 S' M' S' M'

enter image description here

  • $\begingroup$ One of my all time favourite patterns! $\endgroup$ Commented Jun 13, 2019 at 5:21
  • $\begingroup$ @JonathanAllan Thanks. Are you waiting for a proof or to see if anyone else has more solutions? $\endgroup$ Commented Jun 18, 2019 at 15:53
  • $\begingroup$ I know yours is not the shortest. A formal proof of such a solution would of course be great, I personally know only by exhaustion. (Note I did not add the no-computers since I think even as a programming based puzzle people may enjoy this one) $\endgroup$ Commented Jun 18, 2019 at 17:41
  • 1
    $\begingroup$ OK, that's fair. Given the upper bound I discovered, even a brute force approach is feasible. Especially since this is color-neutral, there are only 2 possible first moves that have to be considered. $\endgroup$ Commented Jun 20, 2019 at 16:34

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