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 '14 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 '14 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 '14 at 8:03

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.
Front: Side 2a Back: 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).
Front: Side 3a Back: 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.

  • $\begingroup$ Nice, could you provide a visual? $\endgroup$ – warspyking Dec 17 '14 at 0:32
  • $\begingroup$ I'll try later tonight. Have to find a good cube program. $\endgroup$ – Joel Rondeau Dec 17 '14 at 0:34
  • $\begingroup$ Okay, can't wait! :D $\endgroup$ – warspyking Dec 17 '14 at 0:41
  • 2
    $\begingroup$ Wow these images are really big :O $\endgroup$ – martijnn2008 Dec 17 '14 at 21:47

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.


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