Odd parity Algorithms (specifically, single edge "flip")

[qqwref]

It isn't a site, it's a program, called ACube (the version I use is JACube, same thing though). This program only does 3x3 but you can add any move constraints you want, so in this case you can simulate the 4x4 middle slices by only allowing R, L, U2, D2, F2, B2 and ignoring the middle slice of the 3x3.

There is a more general program out there called ksolve which can be used to solve pretty much anything (you could even optimally solve 4x4 positions if you had enough computing power/time) but it takes more work to set up and I don't think you would find any good solutions that are not in <r,l,U2,F2,B2,D2>.

[Stefan]

I don't think you would find any good solutions that are not in <r,l,U2,F2,B2,D2>.

Well, here's one of length 28 that I just made up:
Dw' L' U F (r U2 r U2 r U2 r U2 r) F' U' L d L' d' L D L' d L
Ok, so it's a bit longer. But I'm only human.

[Stefan]

here's a 23:
r' D2 l B2 r' U2 r U2 l' B2 U2 D2 r U2 r'

Make that a 21:
r' D2 l B2 r' U2 r U2 l' B2 E2 l U2 l'

[deadalnix]

(Rr)' U R U
[(Rr)' U2] * 3
(Rr)2 U R' U' (Rr)2
U' R' U (Rr)'
This alg is awesome. Finally I have an excuse to use Petrus for speedsolving the 4x4x4.

Is it a good way to scramble the cube, or something I don't get ?

EDIT: OK, say no more, I get it. But how the hell did you find this alg ?

[Stefan]

But how the hell did you find this alg ?

Follow the link.

[Ron]

(Rr)' U R U
[(Rr)' U2] * 3
(Rr)2 U R' U' (Rr)2
U' R' U (Rr)'

R2 U2 R U R2 U2 R2 U R2

[Kenneth]

If you have an alg that is 25 q to solve a case, then it is impossible to have an alg of 24 q to solve the same case...

Parity comes from a odd number of q-turns, if it is even it is not a parity. So, 22 is not possible either.

I have no idea how your 24 q can solve this case, did you ignore a ending U turn?

[Kenneth]

I have no idea how your 24 q can solve this case, did you ignore a ending U turn?

No, I did not. The 24q and moreover, the 23q, are complete algorithms. They correct the "one edge flip" to a completely solved cube state.

Edit:
It is a privilege to hear from you first. I have read your occupation in your public profile. You and I appear to have the same passion (finding algorithms).

Hehe, well, that was like 2 years ago, then I found you can setup the same cases but for corners on a 3x3 and use a solver to do the work for me, then I lost intrest...

If you claim you got an alg of 24 q then I'm sure you do not know what you are talking about and/or don't know how to count your turns.

All algs that solves this case MUST have a uneven number of quarterturns, there are no hidden secrets we do not know about yet. That's why i'm sure you are wrong somehow...

[Stefan]

Kenneth: I'd count R r Rw' as three quarter turns.

[Kenneth]

"If you don't count the following as 10q, then, I am counting wrong: (Rr) B r U L D S E F2."

Really depends, if you count slices as quarter turns, then it is ok but then specify it is in slice metric because I would count that as 13 q-turns (face turns and wide face turns).