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For the OLL parity (dedge-preserving) algs of this kind, I don't know of any methods. (This is an "unused" subset.) As I stated in the wiki when I found them, they were merely for curiosity (theory). (See footnote 1.) I figured, since OLL parity algs which don't preserve F3L (but preserve the colors of the centers) were found, (probably one of the first well-known was Stefan Pochmann's "semi-2 gen" double parity. Source.), and since algorithms like my (15,13) adjacent double parity were found (which preserves F3L but not the colors of the centers), then the category in question would be the other "combination" to find algorithms for.

After all, even the OLL parity algs which don't preserve F3L (but DO preserve the colors of the centers) such as Bruce Norskog's u m R D R' b 2F m 2D' m' f 2B m' F u are quite useless regarding saving moves. (See footnote 2), because they really only save you two moves (Bruce's (17,13) algs in place of (25,15) algs, for example -- if you are counting the half turn move metric or Bruce's (15,15) algorithms in place of my (and the one and only) (18,18) move alg -- if you are counting the quarter turn move metric). (But cubers do find some of the Petrus algs such as Rw' U R' U2 R U' Rw' U2 Rw' U2 Rw' U' R U2 R' U Rw' to be fast . . . which can be useful if someone is trying to speedsolve with the Petrus method . . . which is kind of strange, but if that was an "event" or competition to see who could solve the 4x4x4 with a Petrus type variant, an alg like that would be most useful!)

But for fewest moves, it's no brainer to detect the permutation of the wings before you begin the solve (I would have said "during inspection", but most --not all-- people can't do this in 15 seconds!) and then complete the first three centers with an odd number of inner layer slice quarter turns if the (starting) permutation of the wings is odd or an even number if the permutation is even -- to save you 15+ moves.

I mean, my Roux-like alg of which I made from studying Bruce's (15,15) alg could be useful for Roux solvers to solve the last 6 dedges in one algorithm, but there are SO many different possible cases which involve 6 dedges (= 12 "edges" (wing edges) that cuBerBruce (Bruce Norskog) and reThinkingTheCube were talking about in those posts!) that it would be infeasible to come up with a subset of such algs for each possible case. I know these and these are useful to Roux solvers (which involve just 3-4 dedges in the M layer), however.

But for the (8,7) 2-cycles (the last alg list in the wiki), I guess you could use them for a hybrid of the Cage Method and Reduction, where you start by pairing dedges with short sequences such as F2 2R2 F2 and 2R U2 2R' (for example, see my Human Optimized 3x3x3 Reduction Method (but the links don't work, as they were do alg.garron.us (which Lucas disabled after his alg.cubing.net), so here are new links to Solution 1 and Solution 2.) and then use one of those (8,7) "2-cycles" to easily pair up the last few dedges. Once all dedges are paired up, solve the "cage" and then complete the centers with commutators. (Just as an example.)

Footnotes:

If you study the (9,9) algs or even better the (10,9) algs, you will see how intuitive they really are (I found all of the ones attributed ONLY to my name by hand). So once you have that knowledge down, then you can use them as a start of a OLL parity fix with preserves the centers but not F3L. Post. (You will have to convert the capital S moves to lowercase when copying the algorithms in the spoiler to alg.cubing.net, but here is the "finished product" linked to alg.cubing.net for convenience. If you find that interesting, you can also check out the process that I explained in this post.

I actually used this (15,15) single parity alg of Bruce's to create a 30q 11 dedge flip, which is indeed the shortest those algorithms come to flip 11 dedges. (So you can think of this as an Application of such algs.) But of course you can see that I made a (30,20) with Herbert Kociemba in the Eleven dedge flip section of the wiki.

But some of the others may be used for bandaged cubes (because of restricted movement). Although not 2-gen, here is an example of algorithms in a restricted move set to "accompany" a specific bandaged 4x4x4. And although it proved to be a failed attempt, I did try to actually use different types of 4x4x4 2-gen for another weird puzzle. (My post is the second in that thread -- of which you can see/read my logic behind it.)

I guess you have seen my 2-gen parity algorithm video series playlist on my YouTube channel? (Youdon't have to watch it, but I guess you know of its existence?) Also, I recently found a <Rw, U> two corner twist for the 5x5x5, just for fun, nothing more.

But some of the others may be used for bandaged cubes (because of restricted movement). Although not 2-gen, here is an example of algorithms in a restricted move set to "accompany" a specific bandaged 4x4x4. And although it proved to be a failed attempt, I did try to actually use different types of 4x4x4 2-gen for another weird puzzle. (My post is the second in that thread -- of which you can see/read my logic behind it.)

I guess you have seen my 2-gen parity algorithm video series playlist on my YouTube channel? (Youdon't have to watch it, but I guess you know of its existence?) Also, I recently found a <Rw, U> two corner twist for the 5x5x5, just for fun, nothing more.

Yes I watched your series on 2 gen parity and I even saw the algs in the 2 gen section and the shortest type 5 alg there was 87 BHTM I think. That's why I asked this question

BTW conclusion is that <Rw,3Uw> (type 5)and other 2 gen parities (except a few) are useful only for bandaged cubes and K4 right?

All 5 can be useful useful for bandaged cubes (if applicable . . . I doubt <Rw, 3Uw> can really be useful for anything, since its move sequences are SO LONG, LOL), but not K4. Only <r, U> for K4.

EDIT:

For <r, U>, you can see my LL 4-cycle PDF. (Several algs are <r, U> or <l, U> (same difference).)

All 5 can be useful useful for bandaged cubes (if applicable . . . I doubt <Rw, 3Uw> can really be useful for anything, since its move sequences are SO LONG, LOL), but not K4. Only <r, U> for K4

I'm not sure what you mean, but maybe take a look at this? (Basically, you tag on a face-turn only sequence at the end of the alg which is the inverse of all face-turn portions of the parity alg.)

That's the only thing that comes to my mind which may somehow answer your question.
OR
If you were talking simply about the generator (called "Moves" on alg.cubing.net) or the solver (called "Algorithm" on alg.cubing.net), yeah, I think it's much better to view the (9,9) and (10,9) algs in "Moves" (generator) mode instead of "Algorithm" (solver) mode to understand how they work . . . because I made all of them from playing with a solved virtual cube (CubeTwister).

And if this is the case, I don't know how to tweak Template:Alg5 to be in "Moves" (generator) mode for certain algorithms. I recall alg.garron.us was in generator mode by default with links from the wiki, but Lucas Garron decided to switch it (probably for 3x3x3 last layer cases).

I solve a 4x4 by solving the centers, solving all of the edges, and then solving it like a 3x3. But I want to start learning 3, 2, 3 edge paring, because it does not look too hard, and the instructions say that you have to solve the cross first. So I am guessing you would solve the centers, solve the 4 cross edges 1 by 1, and then do 3, 2, 3 edge paring? Is that how it works?

I solve a 4x4 by solving the centers, solving all of the edges, and then solving it like a 3x3. But I want to start learning 3, 2, 3 edge paring, because it does not look too hard, and the instructions say that you have to solve the cross first. So I am guessing you would solve the centers, solve the 4 cross edges 1 by 1, and then do 3, 2, 3 edge paring? Is that how it works?

Almost. Solve two opposite centers (one of which is your cross color), then solve 3 cross edges, then solve the remaining 4 centers, then the last cross edge. THEN do 3-2-3 edge pairing.

I am almost certain this is an easy fix for this community but I have very little experience with Rubiks cubes. My son 7 got this cube and was playing around with it and mixed it up like this (see pictures). Can anyone help me solve it. Any help would be appreciated as he wants to display it in the solved state. Both the red and orange sides are solid, but the 2 middle pieces are swapped on opposite sides (white to yellow and blue to green.)

You need to learn a second OLL parity alg to do that, and even then, the PLL parity tracking process is a bit more involved (you need to trace cycles or do something to that effect).

Spoiler: explanation

The standard OLL parity algs flip one edge pair, while also swapping a pair of corners and a pair of edge pairs. This means that the permutation parity of the corners and the edge pairs (considered together) is unaffected, regardless of which edge you "do the OLL parity alg to". In other words, it cannot affect whether you'll get PLL parity afterwards.

There's another type of OLL parity alg that flips one whole 4×1×1 bar, which is the same thing as flipping one edge pair while swapping two corners. These algs will always change the permutation parity of the corners and edge pairs, again regardless of which edge you flip using the algs. Again, it cannot affect whether you'll get PLL parity afterwards.

(This also applies to reduction/Yau/Hoya into CFOP, by the way. It's not method-specific.)

If you want to avoid PLL parity, you have to choose which of these two types of OLL parity algs to use, based on the permutation parity of the corners and edge pairs. Unless you're super good at 3BLD, it's impossible to determine this faster than just doing the PLL parity alg, so avoiding PLL parity in this way is rarely worth it. It's not as bad in CFOP since OLL parity is done during OLL, when all the pieces left to solve are on the top face (and there are only 8 pieces left, rather than 10 in Roux after solving F2B), but it's still bad enough that it's almost never useful.

For 4 by 4, PLL parity is considered not to be real parity, since two wings are needed to be swapped. However for the diagonal-c case, where two corners are swapped and everything else is solved, is that considered parity?

For 4 by 4, PLL parity is considered not to be real parity, since two wings are needed to be swapped. However for the diagonal-c case, where two corners are swapped and everything else is solved, is that considered parity?