How to swap the centre squares of a 4x4 Rubik's cube

Question

I'm solving by 4x4 Rubik's cube (aka Rubik's revenge) by reducing it to a 3x3. However, I keep finding that once I've got all the edge pieces paired together I've accidentally moved the centre pieces around, so that they're in places that don't correspond to their positions on a 3x3 cube.

To solve this, I need algorithms to do the following two things:

• swap two opposite centres

• swap two adjacent centres

The first is fairly easy to find online but the second is what I really need. (I'm asking for both so that future visitors will have both in one place.)

I don't need the algorithm to leave the edge or corner pieces invariant, as long as it doesn't split up the edge pieces that I've already paired.

Answer 1

Opposite Centers:

To swap two opposite centers (top and bottom) without moving the corners nor egdes, you can use:

r2 U2 r2 U2 M2 U2 l2 U2 l2

If you don't like Cube notations (like me), here the same algorithm with symbols instead:

Here a shorter alternative with the same result (pure top and bottom):

r2 S2 r2 S2

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Adjacent Centers:

To swap two adjacent centers (this time it will move around some edges and corners), you can use:

r U2 r' l' U2 l

Which will swap the two adjacent centers at the top and front, as well as the FD, FU, FB edges.

I will see if I can find a pure swap as well. Will edit when I do.

Here a pure adjacent center swap, for the front and left faces:

u M2 u' L2 F2 u M2 u' M2 F2 LM2

Or here a shorter alternative adjacent center swap, for the front and right faces:

u M' U2 M u' d' M U2 M' d

Shorter opposite swap and both pure adjacent swaps were not found by me, but by Tall Pawn when I asked for help on the TwistyPuzzles Forum.

Answer 2

You can swap two adjacent center blocks with this algorithm:

u' R2 u d R2 d'

This does not break up any other centers or edge pairs, it just moves the edjes around a bit.

I should add that it is possible to pair up all your edges without moving the centers. I believe that is more efficient in the long run since you don't have to correct previous progress.