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I need good V perm (TH&OH)
For TH I use: R' U R' d' R' F' R2 U' R' U R' F R F
For OH I use: R U2' R' D R U' R U' R U R2 D R' U' R D2
OH one is kinda good, if I only know good way of fingertricking it might be good for TH.
Thanks!
 

TDM

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I need good V perm (TH&OH)
For TH I use: R' U R' d' R' F' R2 U' R' U R' F R F
For OH I use: R U2' R' D R U' R U' R U R2 D R' U' R D2
OH one is kinda good, if I only know good way of fingertricking it might be good for TH.
Thanks!
Don't know about OH, but if you like RUD for 2H, then here are a few:
R' U R' U' R D' R' D R' [U D'] R2 U' R2 D R2 // slight regrip before the R' [U D'], but this can be done during the D move and doesn't waste much time. Aside from this part, I find it much easier getting a good TPS on this alg than the second.
y R U' R U R' D R D' R U' D R2 U R2 D' R2 // F/B mirror of above. Last part is a bit slower, but the whole alg is regripless.
z D' R2 D R2' U R' D' R U' R U R' D R U' // Has a rotation at the start, so AUF is a bit harder, but aside from that probably the fastest of these three.
 
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For your specific case, I get 19 single slice turns: z2 R 3U' F 3U' F' 3U' F' 3U' F 3U' B U2 B' 3U B U2 B' 3U' R'

However, we can cycle 4 fixed centers with only 9 single slice turns: 3L' U' 3L' U 3L' U 3L' U' 3L'.
Interesting, thanks!

I don't know how I keep painting myself into a corner when I do solves.

So, in general, does such a 4-cycle of the centers require at least two medges (or whatever they're called) to be out of place? Can the 4 centers be rotated within a completed cage?
 
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Interesting, thanks!

I don't know how I keep painting myself into a corner when I do solves.

So, in general, does such a 4-cycle of the centers require at least two medges (or whatever they're called) to be out of place? Can the 4 centers be rotated within a completed cage?
The minimum number of midges that need to be unsolved when four fixed centers are "cycled" (that is, in a 4-cycle/odd permutation) like in the above algs is indeed two, as you have guessed. (The two midges of course need to be swapped, not just merely unoriented.)

Yes, we can "cycle" four fixed centers without disturbing a completed cage if and only if it's the even permutation 2 2-cycle which can be done in 8 single slice turn moves: 3U 3F 3D 3B2 3D' 3F' 3U' 3B2

If by chance you really did mean "rotated" (as in the supercube), yes, we can also rotate four fixed centers within their own locations in very few moves.
 
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For some reason, I missed the part "Because inner-layer moves suck" when I first saw your post, but looking at it again since unsolved replied,

Standard algorithms have 5 inner layer slice turns, because, for example, instead of having 7 inner slice turns with the "old standard" alg:
3R2 B2 U2 3L U2 3R' U2 3R U2 F2 3R F2 3L' B2 3R2
we can instead make the first and last inner slice turns into wide turns and achieve the same result:
3r2 B2 U2 3L U2 3R' U2 3R U2 F2 3R F2 3L' B2 3r2
[HR][/HR]It seems like you want algorithms which handle this "pure" case which have fewer than 5 inner layer slice turns.

I will list algorithms that I found by hand which conform this this idea.

Here are two very short algorithms (the first I named "cmowlaparity") which only contain 3 inner layer slice turns. (Note that you may convert the inner slice turns of the first alg into wide turns for traditional speedsolving use, but not in the second one.)
x' 3r2 U2 3l' U2 3R U2 3r U2 x' U 3R U' F2 U 3R' U 3r2 x
x' 3r2 U2 3l' U2 3R U2 3r U2 x' U' 3R U F2 U' 3R' U' 3r2 x

If you want algorithms which only contain 3 inner layer slice turns that only contain face turns of U, here is one such alg (longer):
3r U2 3r2 U L' U 3R U' L U2 L' U 3R' U' L U 3r' U2 3R U2 3r' U2 3r'
[HR][/HR]If you want algorithms which only contain 2 inner layer slice turns, here are two such algs. (Even longer).
3r U2 3r U 3r U2 3r U2 3r' U2 3r' U 3R' U' 3r U2 3r U2 3r' U2 3r' U 3R2 U2 3r' U2 3r'
3r U2 3r U' 3r U2 3r U2 3r' U2 3r' U' 3R' U 3r U2 3r U2 3r' U2 3r' U' 3R2 U2 3r' U2 3r'

[HR][/HR]I'm pretty sure algorithms which only contain 1 inner slice turn for this case do not exist/are impossible, despite that 0 inner layer slices is possible for the 4x4x4 case and the 5x5x5 case.

(Note, however, that 1 inner slice turn is sufficient for the adjacent double parity case (I also found this by hand), for example.)

Note that I did not find these two linked 0 inner layer slice algs by hand: they were found using k-solve--the 4x4x4 alg by Bruce Norskog and the 5x5x5 alg by Ben Whitmore.

I do note, Lucas Garron as my witness, that I was the first to publicly post such an alg for the 4x4x4 case (which I did find by hand) which is nearly 200 moves long (my first was about 1100 moves long, but then I decreased it to about 200 moves before the optimal algorithms were found with k-solve). You can find my explanation for how I found this in the "Derivation" spoiler within the "2-Gen 2-Cycles" spoiler in this post.

Happy solving! :)
[HR][/HR]EDIT (hopefully my final one):

Just in case you're curious, based on the 0 inner slice turn algs I linked to above, I did find a relatively brief pure 3 dedge flip algorithm by hand which contains 0 inner layer slices.
r U2 r U2 r U R U r U2 r' U' R' U' r' R' U R U r U2 r' U' R' U' r U2 r' U2 r'

In addition, if you solve the last layer of the 4x4x4 by solving corners first and dedges last, here is a pretty fast 3 flip double parity I found by hand using a similar idea to the above which only contains only 2 inner layer slices.
r' U2 2R U2 r' x' U2 2R' U' R' U' r' U2 r U R U' r R U2 x
 
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I modified my 5x5x5 program to accept a series of positions to solve, one after another, and just run automated until no more positions were in its task list. I asked it to find the quickest solutions for every possible way to exchange 1, 2, or 3 centers, from top to front or top to bottom.

All 1 center exchanges had 8 move solutions.
Most 2-center exchanges had 10 move solutions, so I naturally thought all of them could be done in 10 moves.

But here is one that took 12 moves for some reason:

https://alg.cubing.net/?setup=U_2R_...back&alg=2L-_U2_3R_U-_2R_U_3R-_U2_2L_U_2R-_U-

Can this be done in 10?

The closest position like it can be formed with:

3R U 2R U' 2L' 3R' U 2R' U' 2L
 

CAL

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For some reason, I missed the part "Because inner-layer moves suck" when I first saw your post, but looking at it again since unsolved replied,

Standard algorithms have 5 inner layer slice turns, because, for example, instead of having 7 inner slice turns with the "old standard" alg:
3R2 B2 U2 3L U2 3R' U2 3R U2 F2 3R F2 3L' B2 3R2
we can instead make the first and last inner slice turns into wide turns and achieve the same result:
3r2 B2 U2 3L U2 3R' U2 3R U2 F2 3R F2 3L' B2 3r2
[HR][/HR]It seems like you want algorithms which handle this "pure" case which have fewer than 5 inner layer slice turns.

I will list algorithms that I found by hand which conform this this idea.

Here are two very short algorithms (the first I named "cmowlaparity") which only contain 3 inner layer slice turns. (Note that you may convert the inner slice turns of the first alg into wide turns for traditional speedsolving use, but not in the second one.)
x' 3r2 U2 3l' U2 3R U2 3r U2 x' U 3R U' F2 U 3R' U 3r2 x
x' 3r2 U2 3l' U2 3R U2 3r U2 x' U' 3R U F2 U' 3R' U' 3r2 x

If you want algorithms which only contain 3 inner layer slice turns that only contain face turns of U, here is one such alg (longer):
3r U2 3r2 U L' U 3R U' L U2 L' U 3R' U' L U 3r' U2 3R U2 3r' U2 3r'
[HR][/HR]If you want algorithms which only contain 2 inner layer slice turns, here are two such algs. (Even longer).
3r U2 3r U 3r U2 3r U2 3r' U2 3r' U 3R' U' 3r U2 3r U2 3r' U2 3r' U 3R2 U2 3r' U2 3r'
3r U2 3r U' 3r U2 3r U2 3r' U2 3r' U' 3R' U 3r U2 3r U2 3r' U2 3r' U' 3R2 U2 3r' U2 3r'

[HR][/HR]I'm pretty sure algorithms which only contain 1 inner slice turn for this case do not exist/are impossible, despite that 0 inner layer slices is possible for the 4x4x4 case and the 5x5x5 case.

(Note, however, that 1 inner slice turn is sufficient for the adjacent double parity case (I also found this by hand), for example.)

Note that I did not find these two linked 0 inner layer slice algs by hand: they were found using k-solve--the 4x4x4 alg by Bruce Norskog and the 5x5x5 alg by Ben Whitmore.

I do note, Lucas Garron as my witness, that I was the first to publicly post such an alg for the 4x4x4 case (which I did find by hand) which is nearly 200 moves long (my first was about 1100 moves long, but then I decreased it to about 200 moves before the optimal algorithms were found with k-solve). You can find my explanation for how I found this in the "Derivation" spoiler within the "2-Gen 2-Cycles" spoiler in this post.

Happy solving! :)
[HR][/HR]EDIT (hopefully my final one):

Just in case you're curious, based on the 0 inner slice turn algs I linked to above, I did find a relatively brief pure 3 dedge flip algorithm by hand which contains 0 inner layer slices.
r U2 r U2 r U R U r U2 r' U' R' U' r' R' U R U r U2 r' U' R' U' r U2 r' U2 r'

In addition, if you solve the last layer of the 4x4x4 by solving corners first and dedges last, here is a pretty fast 3 flip double parity I found by hand using a similar idea to the above which only contains only 2 inner layer slices.
r' U2 2R U2 r' x' U2 2R' U' R' U' r' U2 r U R U' r R U2 x
Wow :O
Huge Thanks
So now I have to find the Most fingertricky-one for me.
 
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N-perm

Here are the ones that I have now

na perm= (r' D' r U2) x5
nb perm=(r D' r' U2) x5

I am looking for better ones, maybe some with less than 20 moves
 

Kudz

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I was using this Nb perm (for Roux):
(R' U L' U2 R U' L)2 U

Short but not that fast, I'm going to replace it with this one:
R' U R U' R' F' U' F R U R' U' R d' R U R'
(or R' U R U' R' F' U' F R U R' F R' F' R U' R)
Actually thanks, I need to replace it too. What do you use for Na?
 
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