# [Help Thread]4x4 Supercube: Swapping two center cubies diagonally

#### Galva101

##### Member
Hello everyone, this is my first post here.

I have encountered the following situation in the image while solving a 4x4 axis cube, but let's just treat it as a super cube.
I would like to ask whether anyone knows of an algorithm that swaps two diagonally opposite cubies on the last center of a 4x4 super cube.
I know the following ones already:
$f'\;U'\;(r\;f\;r')\;U'\;(r\;f'\;r')\;U2\;f\;U'$ (Which swaps 2 cubies on the Front face: Bottom Right to Top Right and reversed)
$(r\;f'\;r'\;f)\;F\;(f'\;r\;f\;r')\;F\;(r\;f'\;r'\;f)\;F2\;(f'\;r\;f\;r')$ (Which swaps 3 cubies on the Top face: Bottom Right to Top Right, Top Right to Top Left and
Top Left to Bottom Right)

I know one could also use both of these algorithms after eachother, but if there is one to do it directly, I would be very thankful.

Thank you very much
Roman

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#### Christopher Mowla

Hi Roman, welcome to the forum!

First of all I think you meant to write the second algorithm as:
$(r\;f'\;r'\;f)\;U\;(f'\;r\;f\;r')\;U\;(r\;f'\;r'\;f)\;U2\;(f'\;r\;f\;r')$

Second, I'm not sure what the moveset constraint is, as you are applying these algorithms to a puzzle other than the 4x4x4 supercube.

It's very easy to come up with a short algorithm which solves/generates your case directly.
$f'\;L'\;f\;r2\;f'\;L\;f\;D\;f2\;D'\;f2\;D'\;r2$
= Fw' L' Fw Rw2 Fw' L Fw D Fw2 D' Fw2 D' Rw2

or its mirror,
$f\;R\;f'\;l2\;f\;R'\;f'\;D'\;f2\;D\;f2\;D\;l2$
= Fw R Fw' Lw2 Fw R' Fw' D' Fw2 D Fw2 D Lw2

But is there a constraint that we can only have moves <U,r,f>, where r and f must be quarter turns?

Because if not, if you don't like the above, then just a 3 move conjugation of my 12 move adjacent algorithm that you are using is:

$f\;U\;f'\;f'\;U'\;r\;f\;r'\;U'\;r\;f'\;r'\;U2\;f\;U'\;f\;U'\;f'$
= Fw U Fw' Fw' U' Rw Fw Rw' U' Rw Fw' Rw' U2 Fw U' Fw U' Fw'

But the bold moves become Fw2, which may violate a constraint that may be in place.

So, are:
• Half turns (other than U2) allowed?
• Other turns such as D and L allowed?

EDIT:

Alternatively, I guess the following four move conjugation of the adjacent algorithm you are using will fit the above constraints. So never mind. But I'm still curious to know if there are any constraints.

Rw' U' Rw U'
Fw' U' Rw Fw Rw' U' Rw Fw' Rw' U2 Fw U'
U Rw' U Rw

= Rw' U' Rw U' Fw' U' Rw Fw Rw' U' Rw Fw' Rw' U2 Fw Rw' U Rw

=
$r'\;U'\;r\;U'\;f'\;U'\;r\;f\;r'\;U'\;r\;f'\;r'\;U2\;f\;r'\;U\;r$

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#### Galva101

##### Member
Thank you very much Christopher, those were exactly the algorithms I was looking for, they work great.
The 4x4 axis cube has no constraints at all, it is basically just a shape mod of a 4x4 with oriented centerpieces, so Half turns (other than U2) are allowed, as well as other turns such as D and L.

#### Galva101

##### Member
I have now come up with 3 very simple to remember algorithms, that should be able to solve all cases of a 4x4 Supercube center. They all use the common "large cube centercubie algorithm" that on larger cubes swaps two centerpieces, but is actually a 3 Cycle:
Small letters mean inner faces.
$l\;F\;r’\;F’\;l’\;F\;r\;F’$
I will call this sequence "Cycle" below for simplicity.
1. Cycle the Top Face’s Centercubies counter clockwise (Top Left to Bottom Right, and then to
Top Right): $(Cycle)\;U’\;(Cycle)\;U$
2. Swap the Top Face’s Bottom Centercubies across (Bottom Left to Bottom right and vice
versa). The last U is to restore the edges in case they are already solved, else it can be omitted: $(Cycle)\;U2\;(Cycle)\;U\;(Cycle)\;(U)$
3. Swap the Top Face’s Centercubies diagonally (Top Right to Bottom Left and vice versa):$(Cycle)\;U’\;(Cycle)\;(Cycle)\;U’\;(Cycle)$
The most important thing is, that these algorithms do not change anything else about the Cube, so they are absolutely Supercube Safe.

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