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cubeone

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This is a j-perm that I found in a cubing book that I have but is not in the ss wiki.

F2' L' U' r U2' l' U R' U' R2 (x2)


Hopefully there aren't any copyright issues because of the fact that it was from a book. (By the way, in case you were wondering, the book is Speedsolving The Cube by Dan Harris.)
 

Christopher Mowla

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This is a j-perm that I found in a cubing book that I have but is not in the ss wiki.

F2' L' U' r U2' l' U R' U' R2 (x2)


Hopefully there aren't any copyright issues because of the fact that it was from a book. (By the way, in case you were wondering, the book is Speedsolving The Cube by Dan Harris.)
It's in the wiki, just not in that exact form:

(y x') R2 u' R' u R2 (x' y') R' U R' U' R2 (x2)
and
(y2 x) U2 r' U' r U2 l' U R' U' l2

are equivalents.

All are equal to the decomposition: [B2 R': (U')] [L': [D, L'] ] .
 

Robert-Y

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Yu Sajima's OLL parity algorithm as seen from here http://www.youtube.com/watch?v=OkqaXYabgW0

Original:
r U2' r' U2' r U2' r U2' l' U2' r U2' r' U2' x' U2' r2

Mirrors and inverses:
r2 U2 x U2 r U2 r' U2 l U2 r' U2 r' U2 r U2 r'
r' U2 r U2 r' U2 r' U2 l U2 r' U2 r U2 x U2 r2
r2 U2 x' U2 r' U2 r U2 l' U2 r U2 r U2 r' U2 r

Lefty algs:
l' U2 l U2 l' U2 l' U2 r U2 l' U2 l U2 x' U2 l2
l2 U2 x U2 l' U2 l U2 r' U2 l U2 l U2 l' U2 l
l U2 l' U2 l U2 l U2 r' U2 l U2 l' U2 x U2 l2
l2 U2 x' U2 l U2 l' U2 r U2 l' U2 l' U2 l U2 l'

Just in case anyone is wondering, he does it in 2.6 seconds in that video with a 0.2 sec lockup
 
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r' F R F' r U' R' U2 R' F R F'

Is this new? Can't find it anywhere, it's way better than what I used to do for this case (C(M)LL). Inverse isn't as pretty as far as I can tell, might find something nice for it though. For CMLL, you can do R' instead of the first move.
 

lachose

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The reverse is quite nice and actually I was looking for something for this case. Thanks.
(But for the one you posted, I like (FRUR'U'F') (RUR'U'R'FRF'))
 

Christopher Mowla

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R perm
R U R' U2' R' U2 L R2 U' R' U L' R' U2' R U2'

RUL. I think I am going to use it for OH.
Found it a long time ago but only thought it was worth posting now
Nice alg. It's already in the wiki though. "Alg.5" under R Permutation b on the PLL page.


Anyway, here's a transformation of my M layer double parity algorithm: r U2 r' U2 r' D2 r D2 r' B2 r B2 r' (19,13) which I posted on 1-9-2010 (that post no longer exists though).
r U2 r' U2 l2 r U2 r U2 l' U2 r U2 l' (20,14)

This might have been made in the past, but I didn't see it posted anywhere in public. So I hope I'm not posting an old (version) of my (19,13).

r U2 r' U2 r' D2 r D2 r' B2 r B2 r'
= r U2 r' U2 r' x2 U2 r U2 r' x2 x' U2 r U2 r' x
= r U2 r' U2 r' x2 U2 r U2 r' x U2 r U2 r' x
= r U2 r' U2 r' (L2 l2 r2 R2) U2 r U2 r' (L' l' r R) U2 r U2 r' (L' l' r R)
= r U2 r' U2 l2 r (L2 R2) U2 r U2 l' (L' R) U2 r U2 l' (L' R)

Omitting the L and R outer layer turns to obtain the transformation,
= r U2 r' U2 l2 r U2 r U2 l' U2 r U2 l'
---------------------------------------------------------------------------------------------------------------------
Also, for those who participated in the WANTED: New Dedge Flip Algorithm! thread, everyone was telling reThinking the Cube that his "reParity" algorithm Rw U2 Lw' U2 x' Rw' U2 Lw U2 Rw' U2 Lw U2 Lw' U2 y' M2 y' was identical to my 4-cycle double parity algorithm r U2 r' F2 l' B2 l B2 l' D2 l D2 l' F2 U2. Well, it is, as others have said for the inner layer turn portion of the algorithms.

Only for those who didn't compare those two algorithms before,
"reParity"
= Rw U2 Lw' U2 x' Rw' U2 Lw U2 Rw' U2 Lw U2 Lw' U2 S2 y2

In single slice turns,
= r U2 l' U2 x' (L' R) r' U2 l U2 r' U2 l U2 l' U2 S2 y2
= r U2 l' U2 (L l r' R') (L' R) r' U2 l U2 r' U2 l U2 l' U2 S2 y2
= r U2 l' U2 l r2 U2 l U2 r' U2 l U2 l' U2 S2 y2

Note from this point, when I say "reParity", I am referring to the single slice turn version above).

To affect only the M-Layer,
= r U2 l' U2 l r2 U2 l U2 r' U2 l U2 l' U2 S2 y2 + y2 S2 U2 (l2 r2)
= r U2 l' U2 l r2 U2 l U2 r' U2 l U2 l r2
Let's call this algorithm (*).

Now, is (*) a transformation of my alg (let's just temporarily call it "cmowla") (which existed before reParity)?

Is (*) = T(cmowla)?

cmowla
= r U2 r' F2 l' B2 l B2 l' D2 l D2 l' F2 U2
= r U2 r' x U2 x' l' x' U2 l U2 l' x x2 U2 l U2 x2 l' x U2 x' U2
= r U2 r' (x) U2 l' (x2) U2 l U2 l' (x') U2 l U2 l' (x') U2 (x') U2
= r U2 r' (L' l' r R) U2 l' (L2 l2 r2 R2) U2 l U2 l' (L l r' R') U2 l U2 l' (L l r' R') U2 (L l r' R') U2
= r U2 l' (L' R) U2 l r2 (L2 R2) U2 l U2 r' (L R') U2 l U2 r' (L R') U2 l r' (L R') U2

Removing the L and R outer layer turns to create the transformation,
T(cmowla) = r U2 l' U2 l r2 U2 l U2 r' U2 l U2 r' U2 l r' U2

And (*) + (l' r) U2 (l r') U2
= r U2 l' U2 l r2 U2 l U2 r' U2 l U2 l r2 + (l' r) U2 (l r') U2
= r U2 l' U2 l r2 U2 l U2 r' U2 l U2 (l r') r' (l' r) U2 (l r') U2
= r U2 l' U2 l r2 U2 l U2 r' U2 l U2 r' U2 l r' U2 = T(cmowla)

So (*)
= T(cmowla) - (l' r) U2 (l r') U2
= T(cmowla) + ((l' r) U2 (l r') U2)'
= T(cmowla) + U2 (l' r) U2 (l r')

Since (*) = reparity + y2 S2 U2 (l2 r2), then reparity
= (*) - y2 S2 U2 (l2 r2)
= (*) + (l2 r2) U2 S2 y2

Therefore, (reParity)
= (*) + (l2 r2) U2 S2 y2
= [T(cmowla) + U2 (l' r) U2 (l r')] + (l2 r2) U2 S2 y2
= T(cmowla) + U2 (l' r) U2 (l' r) U2 S2 y2

and it turns out that T(reParity)
= cmowla + U2 (l' r) U2 (l' r) U2 S2 y2
= r U2 r' F2 l' B2 l B2 l' D2 l D2 l' F2 U2 + U2 (l' r) U2 (l' r) U2 S2 y2
= r U2 r' F2 l' B2 l B2 l' D2 l D2 l' F2 (l' r) U2 (l' r) U2 S2 y2
However, the first 13 turns of "reParity" are completely identical to r U2 r' U2 l2 r U2 r U2 l' U2 r U2 l' (20,14) (the transformation of my (19,13), which I posted even earlier than r U2 r' F2 l' B2 l B2 l' D2 l D2 l' F2 U2...a total of 6 days before reThinking the Cube posted his "reParity").

Just add cube rotations to the beginning and end of the algorithm in wide turns.
(y2 x') Rw U2 Rw' U2 x Lw' U2 Rw U2 Lw' U2 Rw U2 Lw' (x' y2)

Just simplify:
y2 Lw U2 Rw' U2 x Lw' U2 Rw U2 Lw' U2 Rw U2 Rw' y2
Rw U2 Lw' U2 x' Rw' U2 Lw U2 Rw' U2 Lw U2 Lw'
Versus:
Rw U2 Lw' U2 x' Rw' U2 Lw U2 Rw' U2 Lw U2 Lw' U2 y' M2 y'

I can't believe no one compared his algorithm to my (19,13) M layer double parity algorithm. They're identical!

**Note to mods**
I didn't want to bump that old thread back up again to just mention this. However, if you all want me to move the second part of this post into a new post in that thread, just let me know.
 
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