Can anyone explain me the logic behind a J perm(or similar)?

Question

I learned to solve a Rubik's cube recently by intuition and some basic cube theory (not by myself). At the last step, I use corner 3-cycles and corner rotations to reach the final solved state. I understood corner rotation this way: If a corner is rotated CW, then another corner can be rotated CCW without disturbing any other part of cube. Similarly I understood a corner 3-cycle this way (using commutator notations):

• A - replace a corner C1 from left face from right face C2. [U'RU]
• B - Rotate left face so that a new corner from left face C3 gets the place where C1 was. [L]
• A' - place the corner C3 from where C2 was taken. [U'R'U]
• B' - Rotate left face the other way. [L']

Move: ABA'B'
Result: A corner 3-cycle.

I can't apply the same logic to do a 2 corner swap + 2 edge swap.

Can any one explain how J perm (or any other PLL which swaps 2 corners and 2 edges simultaneously) works?

P.S : I learned to solve the cube from this video.

Answer 1

A J-perm swaps two adjacent corners and two adjacent edges in the last layer.

If you combine this with a further quarter turn of the last layer, then one of those swapped corners and one swapped edge returns to its original position. What is left is a 3-cycle of the other corners and a 3-cycle of the other edges.

The technique you describe is commutation, and ABA'B' is called a commutator. Since A and B are both undone in A' and B', a commutator move sequence contains an even number of quarter turns. Every quarter turn changes the permutation parity of the corner from even to odd or vice versa (and the same for the edge permutation parity), so a commutator does not change the permutation parities at all. Unfortunately the effect of a J-perm is an odd permutation on corners and on edges (a single swap is an odd permutation). It is therefore not possible to do a J-perm using commutators only. It needs an odd number of quarter turns to change parity.

So to do a J-perm you need to make an odd number of quarter turns, and you can most easily take care of that by doing a single quarter turn of the last layer. What is left is then an even permutation which can be done by commutators. In particular, you have two 3-cycles, which you already understand how to do by commutation.

The actual J-perm move sequences that speedsolvers use are generally incomprehensible, like most of their move sequences. They are optimised for speed and were often found by computer. They usually don't have as much structure as slightly longer sequences that were found by hand without computer assistance.