Many methods, if not all, recommend solving it layer by layer, but is it actually possible to solve it face by face? If so, explain how you would do this, if not, prove it impossible.

  • $\begingroup$ I might be wrong, but isn't it clearly impossible to have 5 faces solved and not the sixth?? So face-by-face can not be done if you mean a sequence of face 1, 2, 3, 4, 5, and then 6. $\endgroup$
    – BmyGuest
    Dec 16, 2014 at 23:35
  • $\begingroup$ @BmyGuest not entirely true, what if you solved face#6 during the solving process of face#5? $\endgroup$
    – warspyking
    Dec 16, 2014 at 23:37
  • $\begingroup$ That was basically what i meant by the comment. You have to relax Face-by-face by this condition that 2 faces are solved "at once". $\endgroup$
    – BmyGuest
    Dec 17, 2014 at 8:03

4 Answers 4


The answer is: sort of.

Updated to include pictures. I didn't have a 3x3 handy, so I used my 5x5, only moving the edges.

Unsolved cube:
Unsolved cube

Solve side 1. Same as solving a layer.
Side 1

Solve side 2, adjacent to side 1. (Opposite would mean solving in layers). Again, no real explanation needed.
Side 2a
Side 2b

Solve side 3, adjacent to both sides 1 and 2. While doing this, one of side 1 and 2 will spend a small amount of time in an unsolved state. But not so much as to consider it unsolved. (At least the way I solve it).
Side 3a
Side 3b
Now, because we have 7 corners in the correct position and orientation, the 8th corner is also correct. This means that at most 3 side pieces will be incorrect.

Solving side 4 (any of the remaining 3) means that at most 1 piece will be left incorrect. As this is impossible, once the fourth side is solved, sides 5 and 6 will also be solved.

  • 2
    $\begingroup$ Wow these images are really big :O $\endgroup$ Dec 17, 2014 at 21:47
  • 1
    $\begingroup$ Joel Rondeau, please review my edit and revert as desired. $\endgroup$
    – msh210
    Dec 27, 2020 at 18:50

Here I will show my (simplified) reasoning:

For you to solve one face without the risk of "unsolving" it, you MUST solve the opposite side at the same time.

If you change a square, it isn't solved anymore, even if it is to solve some other face.

This defeats the purpose of solving face-by-face.

Conclusion: you can't, but you can solve in parallel sides.


Yes, it is possible to solve a Rubik's cube side-by-side. It is considered a LOT more difficult than solving it layer-by-layer. Here's a video showing how.


The video does mention that for people who are cubing with the usual layer-based algorithms, it's a LOT harder than solving it normally esp. beyond the first side which is easy, second side is quite difficult and the third side is very challenging, but in that context, I think it might just be about being used to a different reference system.

For example my 10 yr old son has not learnt any algorithms for cubing, but since he started a month or two ago, he has evolved his own way of solving cubes where he's routinely getting to 3 (2 adjacent sides first, then one more side) sides frequently if not every time.

Hope this helps.

  • 1
    $\begingroup$ Could you summarize the important points from the video? We want answers to be valid at any point in time, and that link may rot eventually, making this answer no longer useful. $\endgroup$
    – bobble
    Dec 16, 2020 at 15:40
  • $\begingroup$ sorry, but being a rubik's cube noob, my competence level right now is only at the point where I can look up resources, but I'm not able to (nor my 10 yr. old son can articulate what logic he's following) go further and give a textual summary of what the person in the video could be doing. Maybe someone else here could do that? I just wanted to capture the YT video URL so that at least this unique way of solving the cube can get recorded for my own purpose. $\endgroup$ Dec 27, 2020 at 6:19

This is technically possible but you must remember that if you can only see exactly the one face that you are solving without looking at the other five faces then it would be technically impossible to verify that you have correctly solved at least one side of the cube. This is a proof for why you would definitely need access to the information on the stickers immediately adjacent to the edge and corner stickers of the first face that you are solving.

By this logic then it follows that if you were to solve at least one face of the cube correctly you would also be solving one layer of the cube. That is to say that you would have correctly placed at minimum 4 edge cubies and 4 corner cubies that all occupy the same layer. By induction we can show that solving any additional face requires solving the corner cubies, edges cubies, and center on the layer containing the new solved face. Because this process is possible through valid permutation groups, by induction we have that the entire cube can be solved through valid permutation groups.

It should be noted that this is not possible without a minimum of looking at 4 sides of the cube for each face that is solved. The face that is being solved and 3 additional adjacent sides must be visible to gain enough information to verify that a face is solved. 1 of 4 side faces has redundant information that can be deduced by process of elimination of invalid groups, or conversely by finding the compatible subset of valid permutations that are consistent with the known information such that the group contains isomorphisms restricted to the location and orientation of the layer of cubies that contains the face that is being solved. We know the position and orientation of the last edge cubie and last two corner cubies given everything else.

In some strict sense, you are not technically solving a face ever without solving a layer at the same time. If you did, it would be a new game that does not relate in any way to what is typically considered solved with respect to a Rubik's cube. Like peeling back the stickers to arrive at meaningless maybe even impossible permutation. You are however solving a layer when you are solving a face if you are solving a face.

  • $\begingroup$ Note that you can hide your answers in spoiler tag with >! :) $\endgroup$ Dec 27, 2020 at 21:01

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