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I have 4x4x4 with a pattern. It is possible to get a situation where one centre block needs to be rotated 90 degrees.

I have seen only 2 other cases online, but neither have documented a concrete algorithm. My only way to solve currently is to scramble and hope that it resolves itself.

To be clear, I am not talking about rotating 2 centres. The 5 other sides are correctly orientated, just one centre 2x2 block is rotated 90 degrees. enter image description here

Here's a link to a Reddit thread with a similar problem.

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  • $\begingroup$ I've seen this on patterned 3x3x3s as well, but I don't necessarily remember a algorithm for it. $\endgroup$
    – Chipster
    Jun 22, 2020 at 15:39
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    $\begingroup$ On the 3x3x3 it will always be two centres that are off. $\endgroup$
    – PassKit
    Jun 22, 2020 at 15:40
  • $\begingroup$ Oh, you're right. Sorry, I'm being silly. $\endgroup$
    – Chipster
    Jun 22, 2020 at 15:42
  • $\begingroup$ All 6 sides are solved with the centres aligned, except for orange pictured above. $\endgroup$
    – PassKit
    Jun 22, 2020 at 15:59

2 Answers 2

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Speedsolving.com user rjohnson_8ball has figured out a way to reorient 2 of the centre pieces on a single side (while reordering them a bit, which shouldn't be an issue here), so running his algorithm twice (properly orienting the cube each time, of course) should do the trick:

https://www.speedsolving.com/threads/4x4-picture-cube.5181/post-68020

I'm copying the algorithm here in case the source article ever disappears:

(rf'r'f) F (f'rfr') F (rf'r'f) F2 (f'rfr')

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  • $\begingroup$ How would this work when there is an odd number of misaligned centres? $\endgroup$
    – PassKit
    Jun 22, 2020 at 18:05
  • $\begingroup$ @PassKit Assuming you mean "odd number of misaligned pieces among the 4 centre pieces on a single side", I don't think you can get into that kind of position without taking apart the cube. You can use the given method to reduce all odd numbers into 1, and in general, you can't do anything to a single piece only. $\endgroup$
    – Bass
    Jun 22, 2020 at 18:17
  • $\begingroup$ Yes you can on a 4x4x4. See the refit comment. There is also a YouTube that explains it explicitly but I wasn’t able to find it as it is so obscure. Order one of these cubes from Amazon and you can experience yourself $\endgroup$
    – PassKit
    Jun 22, 2020 at 18:19
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    $\begingroup$ If you say so. I have absolutely no idea what you mean by "the refit comment" though. $\endgroup$
    – Bass
    Jun 22, 2020 at 18:27
  • $\begingroup$ Sorry - the comment I added to my post $\endgroup$
    – PassKit
    Jun 22, 2020 at 18:35
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You are wrong in assuming you need to do a 1/4 turn of the center.

Turning the 4 pieces of the center by 90° is impossible. It amounts to an odd permutation of the centers, which is impossible to do because all moves on an 4x4x4 do even permutations on the centers.

But you will notice that the centers are not all different. On the picture you see that the 2 opposing squares of the center are identical. They are interchangeable. This allows for other permutations that are even and are visually equivalent to a rotation.

What you need to do is to swap the 2 upper centers and swap the 2 lower centers.

That you can do with (problem face facing up): r'u'r U r'ur U' r'u'r U' r'ur U' r'u'r U r'ur U.
(note how the r'u'r and the r'ur alternate).

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