# Can a Rubik's cube be mapped knowing only two sides?

Is it possible to know the entire configuration of a Rubik's cube looking at only two sides and not rotating the cube? In other words: what is the minimum information required to create a two-dimensional map of a $3\times 3\times 3$ Rubik's cube?

• Two sides are definitely not sufficient; one 'minimal' move is a flip of two edges. This implies that at least four sides are necessary. Mar 24, 2014 at 15:01
• Answer: NO! Why: my first method of solving was layer by layer, I'd do white first, the the middle layer, then finally the last. If I do that white bit and once face adjacent to white I have 2 sides solved and known, but I can permute three centres that do not affect that adjacent side nor the whites and have a different layout. Mar 24, 2014 at 15:40
• There's some lovely algebra to the cube BTW! I see a proof by contradiction though. Mar 24, 2014 at 15:40
• I've disassembled one and I can tell that each 1x1x1 cube is not connected physically to any other except for the centers. Mar 25, 2014 at 4:01

Take a solved cube and flip all four edges with a red face. Now you can permute those edges freely without the effect being visible on any other side than the one with the red center. So even seeing five sides of the cube is not enough to reconstruct the last one.

(And by "freely" I mean that any even permutation of the edges can be reached by legal cube moves, of course).

• This finishes it out perfectly! Mar 24, 2014 at 15:09
• I haven't spent much time with Rubik's cubes and am having a hard time picturing what you mean by "flip all four edges with a red face"; could you please clarify this? Mar 24, 2014 at 15:39
• @KyleStrand: There is a sequence of moves that turns each of these four edges 180° about their axis of symmetry. It's a bit long if you don't want anything else to happen, but since we don't really care about the corners or other edges, you can just hold the cube with the red face up and do $FLBRF$ -- that will leave all top edges still in the top layer but now with the red face pointing out instead of up. From this state, a permutation of these edges will affect only how the "up" side looks. Mar 24, 2014 at 15:52
• Thank you! I didn't know such a sequence existed. Mar 24, 2014 at 16:06
• Oops, I meant $FRBLF$, of course. Mar 24, 2014 at 16:17

If you look at two opposite sides and they are solid colors, you can't distinguish rotations of the other sides.

• Perhaps this is better off as a comment :) Mar 24, 2014 at 15:03
• @Shaun: It answers the question by showing that two sides are not sufficient. Mar 24, 2014 at 15:09
• Yeah, that's fair enough; sorry :) Mar 24, 2014 at 15:22
• Well, strictly speaking it shows that knowing two opposite sides does not suffice. "Looking at only two sides" seems to imply that the sides are choosen by the solver, so he might choose adjacent sides (Ok, we now know that this case is also closed). Mar 24, 2014 at 16:23
• @Drunix: Yes, it depends on how you interpret the question. But the other answer has it in any case. Mar 24, 2014 at 17:33

Since you can see two sides, you can see, in total, 6 out of the 8 corners. By twisting the two you can't see in opposite directions, you can change the orientation of these corners while having a cube that can still be solvable.

• ... assuming the two sides you can see are adjacent. If they are opposite, you get Charles's case. Feb 11 at 9:41
• Good point. I’m back almost 10 years later and I can see that my past self only came up with a single counterexample and didn’t think to consider any further than that. Mar 1 at 1:42