Cubing Forever
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What are the uses of parity algs that don't preserve F3L or centers ?
In short, are there any methods that use these kinds of algs ?
In short, are there any methods that use these kinds of algs ?
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.What are the uses of parity algs that don't preserve F3L or centers ?
In short, are there any methods that use these kinds of algs ?
I really don't think any of them are used anywhere (for the most part). But <r, U> (where r is just the inner layer slice = 2R in SiGN notation) was used to set the upperbound for solving the 4x4x4 LL wing edges (for the K4 Method) (by Bruce Norskog).Another question: Where are type 5 2 gen parity (<Rw,3Uw> I think?) algs used ?
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 questionI really don't think any of them are used anywhere (for the most part). But <r, U> (where r is just the inner layer slice = 2R in SiGN notation) was used to set the upperbound for solving the 4x4x4 LL wing edges (for the K4 Method) (by Bruce Norskog).
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? (You don'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.
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.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?
So long that shortest known alg is 98 quarter turnsAll 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.)Is setting up to the (9, 9) or (10,9) parity algs in the wiki and undoing them a good idea?
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 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?
Thank you!!!Hope this helps
WIN_20210330_09_27_00_Pro.mp4
drive.google.com
no, not if you don't consider regular PLL parity to be a parity. It is just pll parity and t permFor 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?