V Cube 7 Last two centers help please

[Icedece]

hi does anyone know a website that has an in depth tutorial for solving the last two centers of the v cube 7? note: please don't send links to videos as my computer's recently gone haywire and can't load any type of video. Thanks

 

[Marvolo]

Build a 3x3 in center, next connect a "edge" of center and solve like 5x5

 

[MichaelErskine]

hi does anyone know a website that has an in depth tutorial for solving the last two centers of the v cube 7? note: please don't send links to videos as my computer's recently gone haywire and can't load any type of video. Thanks

Hmmm, in my opinion this sort of thing is best learned by demonstration: it's a lot easier to show somebody what to do than it is to write about it. Maybe that's why there's so few pages about it.

I solve the last two centres opposite each other and use what I call a "Pendulum" movement (maybe it has a better name) to swap individual pairs of corresponding opposite pieces that are out of place -- kind of hard to describe but should be recognised by those who use it: I tend to hold the cube with the final centres in F & B and rotate these two opposing centres so they each have a piece to swap (of the opposite side's colour) in the same location in the upper left area of F and equivalent lower left of B (i.e. in the same place when when you do x2). You don't need to focus on the upper-left but I teach this convention as I find it helps me keep track of the movement. Then I send the vertical slice that contains the swapping piece (i.e. with the B colour) to the other side of the cube with an l2 move. The opposite swapping piece (with the F colour) has now moved to the front face, along with the entire slice. The top and bottom faces also have a slice of their own opposite colour but don't worry as we're going to fix that soon! Next we take this swapping piece out of its column and send it back to the far side: an F move puts the swapping piece in the upper-right of F (or centre if the swapping pieces were in the middle row) and then an r2 move sends the vertical column to the back. Now we observe the top and bottom sides have two slices of their opposing colour. We fix the one on the left by doing F' (to move the remainder of the slice from the other side back into its vertical column) and the appropriate l2 slice move to send it home. Now we fix the other slice with F and another r2 slice. The two swapping pieces have now been swapped!

So to summarise: l2 F r2 F' l2 F r2 whilst choosing the appropriate slice to take. Swapping pieces in the centre rows and columns can cause some confusion when the slices are next to each other but this becomes easier with use. You can also swap larger blocks on opposing faces with this move and after a while it becomes intuitive to swap pieces, slices and blocks on opposing and adjacent sides using moves in any direction and being able to recognise the "cleanup" moves to repair the other faces.



...I told you it was best learned by demonstration

 

[masterofthebass]

That's how I do them... I think it's pretty fast.

 

[trying-to-speedcube...]

The way I do it is first solving the middle 3x3, then solving "pseudo-edges" and then solve the big 3x3.

This question is posted in the wrong section.

(If I have time I will give a better explanation)

 

[MichaelErskine]

That's how I do them... I think it's pretty fast.

"Niklas-type commutators"? Is that the correct term for the sequence I described?

Aha! http://www.speedsolving.com/wiki/index.php/Niklas - very useful.

 

[kickinwing2112]

The Dan Cohen way is the fastest in my opinion.

 

[MichaelErskine]

The Dan Cohen way is the fastest in my opinion.

Well that isn't so surprising considering Dan Cohen's achievements, however, I couldn't find any references for what Dan Cohen's way of solving the last two centres of a 7x7x7 actually is! Care to enlighten us?

 

[ShadenSmith]

The Dan Cohen way is the fastest in my opinion.

Well that isn't so surprising considering Dan Cohen's achievements, however, I couldn't find any references for what Dan Cohen's way of solving the last two centres of a 7x7x7 actually is! Care to enlighten us?

Watch the video a few posts up

 

[rjohnson_8ball]

@ msemtd, your "pendulum" movement seems equivalent to the Niklas commutator, but the 180 degree turns might be slower.

@ Icedece, the most important thing you need is the Niklas commutator to swap a piece on U with a piece on F (plus an invisible side effect of swapping 2 cubies of the same color on F). Find a bad color center cubie on U. Rotate U so the cubie is in the lower right area on U. Now put the corresponding bad cubie on F in the exact same lower right area. Now do R(x) U L'(y) U' R'(x) U L(y) U'. Here, I do not mean the x or y to be cube rotations. Instead I mean the x,y to be like Cartesian coordinates. By R(x) I mean perform R on the single slice containing the bad cubies. By L'(y) I mean perform L' on the single slice that contains the new location of the bad cubie.

By the way, my old way of swapping 2 cubies was more difficult than the Niklas commutator.

 

[MichaelErskine]

Watch the video a few posts up

I did! So that was Dan Cohen -- I couldn't see his face! (I really need to spend more time on my people recognition - I'm still working out who's who here!)

EDIT: I've now (after much anguish) changed my last two centres technique from solving opposite centres to solving adjacent centres.