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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.)

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.)

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

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.

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'))

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).

Spoiler: Transformation Steps

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'

Only for those who didn't compare those two algorithms before,

Spoiler

"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)

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.