Parity on Square-1

[hawkmp4]

So...I know how to correct parity on square-1, that's not what this thread's about...
How exactly does this occur, where two pieces are swapped? I know, for example, that on 4x4 parity occurs when a scramble has an odd/even number of slice turns, and when solving, one reduces it with an even/odd number of slice turns. What's happening with square-1? I wish I could phrase this better, I know its a vague question...

 

[shelley]

After you are in the square shape, every turn you make that preserves the square shape swaps even pairs (every time you do a / you are swapping two pairs of edges and two pairs of corners). However, while the puzzle is not in square shape, it is possible to swap an odd number of pairs (e.g. swapping three pairs of corners).

 

[hawkmp4]

That makes perfect sense, exactly what I was looking for. Thanks shelley!

 

[Kenneth]

Easiest is to go to star shape, there it is only corners at one of the sides so you can 6-cycle there. 50% of the positions of the star makes parity and 50% does not.

 

[fanwuq]

Easiest is to go to star shape, there it is only corners at one of the sides so you can 6-cycle there. 50% of the positions of the star makes parity and 50% does not.

So are you saying that if you get to star shape, there is a way to correct parity there?

 

[hawkmp4]

Yes, I remember seeing something about how if you're able to recognize parity when all you have done is separation of U/D pieces, you can apply a much shorter algorithm to correct parity. Wish I could tell you where I found it >.<

 

[masterofthebass]

If you want to recognize parity before CP (lars method) then you recognize if there's parity on each of the U/D faces, just like a 4x4. If both faces are the same (parity/parity, none/none) then you don't have a parity issue. If they are different, then you can apply this alg:

/(-3,0)/(-3,0)/(-5,0)/(-2,0)/(4,0)/(-4,0)/(-2,0)/(5,0)/(-3,0)/

(thanks andrew)

This is an easier, 2gen alg for fixing parity that also messes up CP. Personally, I'm not fast enough at recognizing each face to apply it to my solves, but it's a viable option.

 

[cuBerBruce]

Easiest is to go to star shape, there it is only corners at one of the sides so you can 6-cycle there. 50% of the positions of the star makes parity and 50% does not.

So are you saying that if you get to star shape, there is a way to correct parity there?

Consider having 6 corners on one layer. Turn that layer 60 degrees and obviously you are performing a 6-cycle of corners, while keeping the other corners and all the edges fixed. The 6-cycle obviously changes the parity of the corners, while the parity of the edges doesn't change (since they remained fixed). (Note that one even-length cycle or any odd number of even-length cycles changes parity.)

Now you can go from cube shape to having a layer with 6 corners in four twists. Think of that as a setup move. Do the 60-degree turn of the layer with all corners. Then undo the setup move to get back to cube shape. (Be sure to do the exact reverse of the setup move.) Obviously the corners will now be 6-cycled from the cube shape position you started from, while the edges will be back to where they started from. Hence, the parity has been changed. Of course, this scrambles up the corners a bit!

When the corners are all in the same layer (which I think of as "snowflake" shape), you could also do a 180 turn of the layer, as that performs three 2-cycles, which is also an odd permutation.

 

[blade740]

mmm, 2gen parity alg. It looks almost as good written out as it feels to execute.

My method involves several different parity fixes, one for each possible corner permutation. That is one of them.

 

[TMOY]

Consider having 6 corners on one layer. Turn that layer 60 degrees and obviously you are performing a 6-cycle of corners, while keeping the other corners and all the edges fixed. The 6-cycle obviously changes the parity of the corners, while the parity of the edges doesn't change (since they remained fixed). (Note that one even-length cycle or any odd number of even-length cycles changes parity.)

Now you can go from cube shape to having a layer with 6 corners in four twists. Think of that as a setup move. Do the 60-degree turn of the layer with all corners. Then undo the setup move to get back to cube shape. (Be sure to do the exact reverse of the setup move.) Obviously the corners will now be 6-cycled from the cube shape position you started from, while the edges will be back to where they started from. Hence, the parity has been changed. Of course, this scrambles up the corners a bit!

That's precisely the alg I am using
After performing it, just redo the corners orientation step. Then everything will be back in place except for the edges of one layer which will be 4-cycled.

 

[Kenneth]

Easiest is to go to star shape, there it is only corners at one of the sides so you can 6-cycle there. 50% of the positions of the star makes parity and 50% does not.

So are you saying that if you get to star shape, there is a way to correct parity there?

Yes, that's the only alg I know

Cube -> Star (turn star one step) Star -> Cube

As TMOY writes, you only need to insert a few turns to get it to a 4-cycle edges in one layer instead of the 6-cycle corners.
------
It is possible to do parity in only one turn before you go from star to cube in the beginning of the solve, but how to do recognition is a mystery to me

EDIT: I once had an idéa to do the whole solve in star shape and only go to cube in the wery end.

Discuss: is this possible?

 

[Swordsman Kirby]

Easiest is to go to star shape, there it is only corners at one of the sides so you can 6-cycle there. 50% of the positions of the star makes parity and 50% does not.

So are you saying that if you get to star shape, there is a way to correct parity there?

Yes, that's the only alg I know

Cube -> Star (turn star one step) Star -> Cube

As TMOY writes, you only need to insert a few turns to get it to a 4-cycle edges in one layer instead of the 6-cycle corners.
------
It is possible to do parity in only one turn before you go from star to cube in the beginning of the solve, but how to do recognition is a mystery to me

EDIT: I once had an idéa to do the whole solve in star shape and only go to cube in the wery end.

Discuss: is this possible?

Do you know of a good way to swap individual edges? That's the main problem I see.

 

[Kenneth]

Not if you learn to recognise parity before you solve permutation, if one layer has a non valid PLL an the other has a OK PLL then it is parity. If both are bad or if both are good, then it is not parity.

If it is parity, then you fix the bad side by doing a 4-cycle edges to it.

 

[Swordsman Kirby]

Not if you learn to recognise parity before you solve permutation, if one layer has a non valid PLL an the other has a OK PLL then it is parity. If both are bad or if both are good, then it is not parity.

If it is parity, then you fix the bad side by doing a 4-cycle edges to it.

I meant if you had actual algorithms for cycling individual edges.

 

[Kenneth]

I understand that but... If I had one I would have posted it

I'm not sure but the problem is not that easy to solve. The case is both a two edge swap but also Z-PLL + a 4-cycle corners after AUF (if it is opposite edges, adjacent makes U-PLL + 4-cycle corners). The case with a 4-cycle edges can also be edges solved and a 4-cycle corners after AUF (or a H-PLL + 4 corners). The cases are diffrent, you will need something else than my 6-cycle corners to 4-edges variation, something that also does Z-PLL.

Hmmm... poor mans in the end of the first variation, I see if I find it

 

[Kenneth]

Got one but it is not short...

(-1,0) / (1,1) / (0,-1) / (-3,-3) / (-1,-2) / (2,2) / (2,0) / (-2,-2) / (-2,-1) / (-3,-3) / (-1,-3) / (-3,-3) / (0,-3) / (1,1) / (0,2)

The first three starts the Z-PLL the rest of it is in the end and in the middle my usual parity fix.

Hope I got it right, I'm not used to this puzzle, I actually had to relearn my poor mans solution to even get it solved so I was able to look for the alg, took half an hour or so