I have a case on my square-1 that I can't solve where only two edges of one layer are swapped. I am not looking for an algorithm as much as for an explenation how that is possible. I can 3-cycle edges pure and also corners can be 3-cycled just like on a 3x3x2. Since edges and corners behave exactly the same on the puzzle (8 "identical" pieces) how can such a two-swap parity occur on only one part of pieces? If I do a 90° turn I get rid of the edge parity but create a corner parity.
What's happening there?

Hi alaskajoe,


have a look to this
http://www.twistypuzzles.com/forum/view ... t=square+1
perhaps this explains it. Specifically the parity.


Cheers,
Andrea

As you have noticed, when the puzzle is in the cubic shape, a dihedral turn swaps two corners and two edges and a 2 swaps for the corners in even and 2 swaps for the edges is even. As long as you stay in a cubic state you can't find a move that violates this.

When you move out of a cubic state though you can arrange different number of swaps for the edges and corners. For example, what Jaap calls the "Scallop" pattern:

If you arrange both the top and bottom into this pattern then if you do a dihedral turn with 3 corners swapping with 3 other corners you have changed the parity of the corners independent of the edges. This is just one example of many. I haven't reverse engineered any "parity fix" routines to see what swaps there are doing but i suspect they aren't using the Scallop pattern.

Thanks to you both. I thought once it's as far as I had it, it can be solved without breaking the cubic shape again.