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cube explorer for 5x5x5?

x-colo-x

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Hi,
I would know if there's a program as cube explorer for the 5x5x5
or if there isn't, someone can find for me an algorithm similar to the parity algo of M2 but that rotates only the middle layer and exchanges the three edges of UB with the three of UL ?
Thanks
 
Last edited:

qqwref

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Nope, there is nothing like CubeExplorer for the 5x5, and there won't be at any time in the next decade or two. There are just too many positions to search through, even if you break the cube into two steps like CubeExplorer does. Keep in mind, the 5x5 has about as many positions as 3.8 3x3s, so you'd need something like 7 steps to be able to solve each one as easily as in CubeExplorer (which has two steps for solving the 3x3). But the Two-Phase Algorithm wouldn't work easily with three steps, let alone seven...
 

Shack

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I do: m2 F2 Rw2 F2 U2 r2 U2 F2 Rw2 F2 y T-perm y' F2 Rw2 F2 U2 r2 U2 F2 Rw2 F2 and the J-perm on the back
 

Christopher Mowla

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F' U x u' 2U' m2' x' U2 2f U2 m2' U2 2f U2 2f2 U2 F' U' F (SiGN notation)

Not sure if this qualifies, but it is the best I can come up with.

Edit:
4r u' 2U' m2' x' U2 2f U2 m2' U2 2f U2 2f2 U2 F' L'
(He didn't specify whether or not the tredges were to be oriented or not, so this algorithm is just as valid as the other).

Edit:
Using different set-up moves and optimizing,
L'2 U' 3u2' y' m'2 x' U2 2b U2' m'2 y U2 2r U2 2r2 U2 L U2 z y2
which can be translated to the 4X4X4 like so: L'2 U' u2' y' m'2 x' U2 b U2' m'2 y U2 r U2 r2 U2 L U2 z y2
 
Last edited:

Cubepark

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Thanks Stefan
I thought of a way to solve the 5x5 Blind, is pretty easy, 80% is how to solve the 4x4. I missed only the algorithm for the parity of the centers. It 's a slow method. I solve 4x4 blind on 24/26 minutes minutes, I am old, I take it easy. I think x-colo-x could stay on 14 minutes. After the Italian Open will try to write a guide ...
GOD Bless The BLD
 
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Stefan

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Last edited:

Cubepark

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What can I add Stefan? thanks a lot, I think it's perfect! with U2 solve the centers , as in 4 x 4 bld. For centers that form the cross used m2.
to the edges using the same algorithm of 4x4, central to the edges using M2, is quite simple, a bit long, but to me it works! As Woody Allen says,Whatever Works!
 
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