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Void Parity Resolution/Algs

Lucas Garron

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The "Void Parity" is the issue that occurs when solving anything equivalent to a 3x3x3, particularly a void cube, where you have odd total parity with your current assignment of centers.
Layer methods hit this (i.e. you can recognize the parity) at PLL, and I think with CF or Roux, you can try to avoid it.

It's well-known that a reassigning centers with a quarter-slice twist + resolving pieces will resolve this. My approach used to be something like an M-fix with a 4-flip and two 3-cycles.
But for speedsolving, you'd want something efficient. Tomas once asked us for a fix in #rubik, and so I got kinda interested.

The most practical is to find a fix that preserves OLL (which can be applied before PLL). It's not hard to imagine that the best fixes are URrML-ish and preserve CPLL, which is also good for CF and Roux, depending on how you do it (and what kind of mistakes you make).

After a while, I found l U r' U M U' M' r U' R' U M U'.
Then, qqwref found the very nice M' U M' U M' U' M' U' M' U2' M' U' M'.

So, does anyone have any other algs, or better approaches?
 
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blah

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Y'know what? I used to have a number of these algs, back when I first proposed my "new" 5x5x5 BLD method, but then I never went back to solving a 5x5x5 BLD again, so I think I may have deleted that file (can't find it right now) :(
 

cuBerBruce

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I've been using the following to swap UF and UR: (M' U')4 (R2 d' M' d2 M' d' R2) (F2 M' F2 M2).
Essentially a 4-flip, an Allan (U-Perm), and an M-layer 3-cycle + M move. But qqwref's alg (which swaps UR and UB after adding a missing U' on the end :) ) is shorter.
For swapping UFL and UBR, I've been using (R' U L' U2 R U' L)2 U' F R F' (M' U')4 F R' F' B2 M B2 M2.
Not great algs, but I didn't have to memorize anything new.
 

Meep

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My approach to that was to simply do M', then insert the two cross pieces again via the M layer, then orient any edges that were misoriented from doing that.

Ex. M' + U' M' U M U M U2 (M' + M <-- These cancel out) U M U M U2 M' U M' U M' U'

A really inefficient intuitive approach. =\
 

cuBerBruce

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M' U M' U' M U' M U2 M' R U R U' M' U R U' r'
edit: joey again
joey, your alg seems to move a lot of pieces. Did you make a mistake?

BTW, there is this old thread that is somewhat related to this one.

http://www.speedsolving.com/forum/showthread.php?t=2516

(And joey, your alg in that thread doesn't work for me either. Are there people who actually post algs without using an alg checker first? ;) )
 

Lucas Garron

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joey's alg is M' U M' U' M U' M U2 (M' R) U R' U' M' U R U' r'. Missed a '.

(And thanks, I know about that thread. I made this thread specifically to discuss approaches to centerless speedsolving, though. :))
 
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Ethan Rosen

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On the last layer make it so all you have is two edges opposite each other on the layer that need to be swapped. Then do cube rotations so those two pieces are BR and FR. Do E' L2 E L2 E'. The problem with this is that it then leaves you with four edges to orient.
 

Stefan

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M' (D' R2 D) (U M' U' M) (D' R2 D)
The M' changes parity, the (conjugated) 5-edges-cycle repairs the damage.

Now with ACube and this input: FD ? -? ? BD DR -? DL FR FL BR BL ? ? ? ? DRF DFL DLB DBR
13 qtm: M' U' B D B U S' U' F' R' B'
9 stm, 12 htm: M' U B2 U2 M' U M U B2
 
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Stefan

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I came up with a quite elegant approach now, not sure whether your algs use it, but at least you didn't explicitly talk about it. With one alg implementing this approach I got a 1.66 execution last night after some practice. What best times did you get with what algs?

I'll show my approach/alg later, but because I find it so elegant, I want to give you guys the opportunity to discover it as well (or maybe even another one). I'm sure if you see it, you'll wonder why you didn't think of it immediately.
 

Kenneth

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So, does anyone have any other algs, or better approaches?

Solve all but the last edges, if the case is a parity do M or M' and parity is gone and edges are solveable.

EDIT: Use Roux, look for parity before edge permutation and do as I wrote above...

Hmmm, on a second, then you have to redo orientation and orient four edges...

To do it before orientation makes recognition much harder but after some testing I found it seems possible to find out about parity counting how many edges that are at uneven positions = if an edge is adjacent to its place (U) it is uneven, if it is opposite (M = U + U = U2) it is even and if it is M2 = even = M + M, a M + U is uneven and so on...

If it is a even number of uneven edges then it is parity. Using that is a little slow =/ (but wery useful for void cube FMC =)

Like to try a void but don't know where to get one???

Remove the centre stickers :p
 
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Stefan

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To clarify and not be confused with Kenneth's suggestion: What I meant was still for the CFOP method, applying an alg between OLL and PLL. With "approach" I meant a certain way to build algs, i.e., algs based on a certain idea.
 

Kenneth

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EDIT: Use Roux, look for parity before edge permutation and do as I wrote above...

Hmmm, on a second, then you have to redo orientation and orient four edges...

To do it before orientation makes recognition much harder but after some testing I found it seems possible to find out about parity counting how many edges that are at uneven positions = if an edge is adjacent to its place (U) it is uneven, if it is opposite (M = U + U = U2) it is even and if it is M2 = even = M + M, a M + U is uneven and so on...

If it is a even number of uneven edges then it is parity. Using that is a little slow =/ (but wery useful for void cube FMC =)

That was not useful...

The first approach works, the parity alg to use is:

M U' B2 U2 M' U' M U' B2 ... first M solves parity, the rest orients the four edges of the M-slice.

Edit: hehe, same as Stefans... promise I did not look at the algs above until now I just used setups and the usual Roux "OLL" = M' U M ("OLL" to avoid Stefans "It is not OLL" else I had written only OLL :) )

Like to try a void but don't know where to get one???

Remove the centre stickers :p

Made one, it works fine... first I thought the Rubik's logo on one of the centres would ruin it but it makes the same, you can even leave the stickers on that and the opposite side.

Edit: 25.34, but no parity = boring record, I want to do it with parity =)

Used columns first (solve pairs + CLL), place three D edges, orient, parity, permute.
 
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Stefan

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Ok, here's my idea (the other reason I didn't post it earlier was lack of time).

Observations:
- We want to "change permutation parity", most obvious way is to do an inner slice turn.
- We allow permutation of LL edges.

Idea: Combine those two observations. Setup the four edges in one inner slice (with proper orientation), turn it, undo the setup.

Obvious one resulting in nice pattern on normal 3x3 (btw I think a riddle from Per at RWC2005):
(F L B F R F') E (F R' F' B' L' F')

Here's one in 12htm/14qtm: (R' F D2 r U) M' (U' r' D2 F' R)

After that, I thought about going backwards. So instead of setting the four edges up into the M slice, I set them up there and tried to find a nice way to "solve the cross" from there. Because I have more experience with that. I found (U' R2' F r U') and built this alg from it which is the one with the 1.66 seconds mentioned above:
(B r' U' x') (R2 U R' r U' R2' F r U' x)
 
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Kenneth

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I wanna change the name for the puzzle, "Level 1 Menger Sponge" is much cooler :p

Started to try to do parity after I solved all but 5 edges, before orientation. then I simply AUF and do a quick "BLD memo" to see if it is parity, it works but is a bit slow. With practice it may be better.

Probably also a wery nice way to practice edge memo... The void is good for such things, like F2L practice, it is much harder to see where to put the pair without the cross as a guide. Try it boys and gals you are only six stickers away from a Menger :)
 
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Kenneth

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M2 U M U' M2 ... solves parity and orient four LL edges :)
M U2 M' U M' U M U' M ... parity + opposite EOLL
M' U M U' M U' M U M' ... p + adjacent EOLL

These algs is a start for a 2xELL that always solves LL edges and parity, it is a total of 59 cases, probably the best approach you can have, no extra step, just more algs.

2xPLL works the same, learn it and you can use it for 4x4x4 = no more PLL paritys =)
 
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Kenneth

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Like to solve parity on this?

229.jpg


It's a Level 3 Menger Sponge, exactly 8,000 pieces :p

I use the image for my profile page at Svekub. Made it in a program I wrote myself, including the 3D image generator (use a separate Z table for the shadows as it was the lights view of the scene and do a 2-phase render, one for shadows, then the image, this one I cheated and made two exactly the same but diffrent direction of light and then I merged the images in PS).

Maybe not on topic but I like to show off :D
 
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Ron

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Cube Explorer can easily find optimal solutions for this.
On a solved cube in Cube Explorer do:
1) moves U + D'
2) color the edge pieces in middle layer correctly (after this cube is solved except for 4 centers)
3) make all stickers in upper layer yellow with stripes (using CTRL+click)
4) click add and solve, error message will come and after removing the error message the outside of upper layer has become gray
5) click add and solve again

Here are some optimal solutions between OLL and PLL:
R D B' L' D2 F' U' R F D2 L B' (12f*)
R B' D2 L' F' U' D R F D2 L B' (12f*)
R B' L' D2 F' U' R F D2 L D B' (12f*)
R B' L' D2 F' U' D R D2 F L B' (12f*)
R' F D2 L B U' D L' F' D2 R' B (12f*)
R' F D2 L B U' L' D2 F' R' D B (12f*)
R' F L D2 B U' D L' D2 F' R' B (12f*)
R' D F D2 L B U' L' D2 F' R' B (12f*)
F R' D2 B' L' U' D F L D2 B R' (12f*)
F R' B' D2 L' U' F L D2 B D R' (12f*)
F R' B' D2 L' U' D F D2 L B R' (12f*)
F D R' B' D2 L' U' F L D2 B R' (12f*)
F' D L D2 B R U' B' D2 L' F' R (12f*)
F' L D2 B R U' D B' L' D2 F' R (12f*)
F' L D2 B R U' B' D2 L' F' D R (12f*)
F' L B D2 R U' D B' D2 L' F' R (12f*)
L F' R' D2 B' U' D L D2 B R F' (12f*)
L F' R' D2 B' U' L B D2 R D F' (12f*)
L F' D2 R' B' U' D L B D2 R F' (12f*)
L D F' R' D2 B' U' L B D2 R F' (12f*)
L' D B D2 R F U' R' D2 B' L' F (12f*)
L' B R D2 F U' D R' D2 B' L' F (12f*)
L' B D2 R F U' R' D2 B' L' D F (12f*)
L' B D2 R F U' D R' B' D2 L' F (12f*)
B D L' F' D2 R' U' B R D2 F L' (12f*)
B L' F' D2 R' U' D B D2 R F L' (12f*)
B L' F' D2 R' U' B R D2 F D L' (12f*)
B L' D2 F' R' U' D B R D2 F L' (12f*)
B' R F D2 L U' D F' D2 R' B' L (12f*)
B' R D2 F L U' F' D2 R' B' D L (12f*)
B' R D2 F L U' D F' R' D2 B' L (12f*)
B' D R D2 F L U' F' D2 R' B' L (12f*)

You may want to try other center positions. Maybe there are shorter solutions for those cases.

Have fun,

Ron
 
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