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TheEpicCuber

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It probably cannot get any easier than something like these:
Uw'
m F
U' 2R' U F2
U' 2R' U F2
U' 2R' U F2
U' 2R' U F2
U' 2R' U F2
F m'
Uw
[Link]

Uw L' R
L' U' 2R U L U2
L' U' 2R U L U2
L' U' 2R U L U2
L' U' 2R U L U2
L' U' 2R U L U2
U2
L R' Uw'
[Link]

Uw' L' R
R2 U' 2R U R2 U2
R2 U' 2R U R2 U2
R2 U' 2R U R2 U2
R2 U' 2R U R2 U2
R2 U' 2R U R2 U2
U2
L R' Uw
[Link]

They are just (2R U2)4 2R with 1-2 setup moves around the 2R move.
(2R U2)4 2R
=
2R U2
2R U2
2R U2
2R U2
2R
=
2R U2
2R U2
2R U2
2R U2
2R U2
U2

So if you add, say, the setup moves R2 U' before every 2R and undo with U R2, you get:
R2 U' 2R U R2 U2
R2 U' 2R U R2 U2
R2 U' 2R U R2 U2
R2 U' 2R U R2 U2
R2 U' 2R U R2 U2
U2

And then do a few conjugates to make a single dedge flip:
Uw' L' R
R2 U' 2R U R2 U2
R2 U' 2R U R2 U2
R2 U' 2R U R2 U2
R2 U' 2R U R2 U2
R2 U' 2R U R2 U2
U2
L R' Uw

It's possible that the entire sequence is what's repeated except for one move, but it's nastier.
(2L' B2 2L B2 U2 F2 D2 2L D2 F2)4 2R'

And it's possible that all of the moves of the alg are the result of a repeated sequence, but those can be very long:
(3Uw' Rw2 3Uw2 Rw2 3Uw2 Rw 3Uw2 Rw' 3Uw2 Rw)15

And even longer if the sequence that is repeated is shorter!
(3Uw' Rw2 3Uw Rw 3Uw' Rw' 3Uw2 Rw)585

_________________________________________________________
EDIT:

If you happen to know the PLL parity algorithm: 2R2 F2 U2 2R2 U2 F2 2R2

Then if you make the middle move a quarter turn, a very easy alg to remember can be made from it also:
  1. Do the first bit: 2R2 F2 U2 2R U2 F2 2R2

  2. Do the move F2: 2R2 F2 U2 2R U2 F2 2R2 (F2)

  3. Do the sequence to swap two adjacent centers (like in reduction):
    2R2 F2 U2 2R U2 F2 2R2 (F2)(2L' U2 2L 2R U2 2R')

  4. Do the obvious finishing move (F2) and then add two setup moves (the same setup moves as those in first algs listed in this post).
    (Rw' s) 2R2 F2 U2 2R U2 F2 2R2 (F2)(2L' U2 2L 2R U2 2R')(F2) (s' Rw)

But that's all I've got.
Also, taking a look at these algorithms, I don’t think they are any easier than drew/Lucas parity algorithms.
 

Christopher Mowla

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Also, taking a look at these algorithms, I don’t think they are any easier than drew/Lucas parity algorithms.
Even if that's the case, what does that matter to him? He's the one who asked the question. He's the one who gets to decide what's best for him. If the well-known algorithms are easy for you to learn, good for you. That doesn't mean that the alternative algorithms I provided are not exactly what he's looking for. Not everyone's mind works the same, and you would be arrogant if you think for a second that everyone ought to "toughen up and think like you".

Algorithms that you already know are easy. Because you know them already. And you can't just say that because you (now) can memorize a 15 move algorithm with ease, the ease you now feel is in part due to your experience with memorizing algorithms. He has not had that experience. And if algorithms were easy for you to learn "from the very beginning", good for you. This is not the case for him. That's why he's asking for help.

We can sit here and teach him tricks to learn/view the well-known algorithms differently, but at the end of the day, that's not what he asked for.

P.S.

You keep mentioning Lucas Parity. That's the wide turn version. That does not just flip a single dedge, clearly. So you are yet again not answering his question. If you can get away with that, then can you honestly say that Lucas parity is easier to learn, than, say Rw' (F2 U' Lw' U)5 Rw?

And, again, may I ask where you saw the other alg is being called "drew"? (Out of curiosity?)
 

Christopher Mowla

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but does the same job as Lucas parity
Just so that people (who would get confused by this) don't get confused,

Lucas Parity: r U2 x r U2 r U2 r' U2 l U2 r' U2 r U2 r' U2 r'
The old standard alg: 2R2 B2 U2 2L U2 2R' U2 2R U2 F2 2R F2 2L' B2 2R2

He was asking for an algorithm which just flips the dedge. These are not the same regarding that specific objective (which I believe is definitely relevant given that we replied to his question).

Sure, you can convert all of the turns of the old standard to wide turns to achieve the same result as Lucas Parity: r2 B2 U2 l U2 r' U2 r U2 F2 r F2 l' B2 r2, but the reverse is not true. If you convert Lucas Parity into single slice turns, you get 2R U2 x 2R U2 2R U2 2R' U2 2L U2 2R' U2 2R U2 2R' U2 2R', which doesn't just flip a single dedge.

Yes, it's possible to properly convert Lucas parity to single slice turns:
  1. Move the cube rotation inwards to where the r' move is:
    r U2 r F2 r F2 x r' U2 l U2 r' U2 r U2 r' U2 r'

  2. Since x r' = l' (on the 4x4x4, the cube of focus in this demonstration),
    = (by wide turn equivalences)
    r U2 r F2 r F2 l' U2 l U2 r' U2 r U2 r' U2 r'

  3. Now convert to single slice turns:
    2R U2 2R F2 2R F2 2L' U2 2L U2 2R' U2 2R U2 2R' U2 2R'
    (Single slice turn version of Lucas parity. But "Lucas Parity" is not referring to this alg. This alg has not been named.)
 

Christopher Mowla

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Okay, so maybe you can just use the slice moves to flip just the edge. I really didn't understand the question. I thought he wanted a standard parity alg, not an alg to swap the two edges.

Can someone give me the easiest oll parity for when it’s just one edge is flipped? I’m old and apparently stupid so I don need the fastest, just the easiest.

EDIT:
But I guess you can interpret this in two different ways. (People who think of "just one edge flipped" when thinking of last layer yellow cross edges after F3L.) But I took the wording of his question literally.
 

TheEpicCuber

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EDIT:
But I guess you can interpret this in two different ways. (People who think of "just one edge flipped" when thinking of last layer yellow cross edges after F3L.) But I took the wording of his question literally.
Yeah. I thought he just wanted a parity alg. So I supplied a parity alg. From what I understand, you gave him a direct way to solve for 1 flipped/swapped pair of edge(s).
 
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I use an inefficient reduction method I figured out for 4x4

what should I learn if I want to solve under 1 minute?

i have a sub2 single but I don't time myself often, wanna practice more seriously
 

CubableYT

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While I think Hoya is equal to Yau on 5x5 (if not better) I personally think Yau is the best for 4x4.
Think about it. Same average move count, better ergonomics, the cross is solved when you start edges, you have have 4 options for first center instead of 2, and 3-2-3 is the same.
 
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Think about it. Same average move count, better ergonomics, the cross is solved when you start edges, you have have 4 options for first center instead of 2, and 3-2-3 is the same.

I think for big cube (specifically Yau and Hoya) the method you use doesn't really matter and they're pretty equal but at least Yau has very good results and is more than proven to be extremely good whereas Hoya, which can probably get the same times, has not been proven as much.

You can't make the same arguments for large cubes as you can for 3x3 because it's mostly pure personal preference.
 

CubableYT

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I think for big cube (specifically Yau and Hoya) the method you use doesn't really matter and they're pretty equal but at least Yau has very good results and is more than proven to be extremely good whereas Hoya, which can probably get the same times, has not been proven as much.

You can't make the same arguments for large cubes as you can for 3x3 because it's mostly pure personal preference.
I agree. My main point is that they are at least equal, and if more people don't start using it, it will always be a little known method.
 

CubableYT

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More top solvers use Yau than Hoya simply because Hoya is not as sufficient on 4x4 as Yau is.
Not necessarily. Yau is just better known, therefore more people use it. It's kind of like the CFOP-Roux thing, where they are very simalir, but because cfop was more widespread and invented first.
 
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