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7x7x7 Last Edges Parity

Tesseract

Member
Joined
Oct 9, 2009
Messages
19
Location
Ukraine, Kiev
// first of all, sorry for my bad english.

Hi, all.
I have a question (and my own suggestion) to cubers, who likes big cubes.
How are you solving Parities of last two Edges on big cudes, like 6x6, 7x7 etc?

Clearly, that:
- using formulas of 5x5x5;
- thats formulas can be corrected by moving not only two sides together (like Rw), but three also;
- too many cases, when you have to use several formulas one by one;
- so too much time spends for these operations.
 

Tesseract

Member
Joined
Oct 9, 2009
Messages
19
Location
Ukraine, Kiev
So, I have my invention for these situations.
(I`d like to listen yours responses for the one)

At the begining of the solving first 8 edges - I pile up its on the up and down sides of the cube.
During, I looking for such edge - solution of the one will be easy to do, using one formula alone (Or make such edge by myself).
And put this "unsolve" edge to the up side, remember it as like not completly solved.


see picture (1):
stockpiled edge situated on the up side
(black-blue, with one "unsolved" red-yellow element)

mirror plased behind.
(click to enlarge image)


When all of 8 edges are solved and putted to the up and down sides (so my "unsolved" edge too), -
as usual, I solve last two of four edges, in the working space.

And now come in handy my "unsolved" edge - I get it into working space.
I conclude three edges, with not breaking a main part of "unsolved" edge (in current example: it is black-blue block of the one).

Eventually I get harvested parity of last two edges, for using known simple formula!

Watch the result:

see picture (2):
In the end you`ll get known parity, simultaneously of adjacent colors
(in current example: they are black-blue and red-yellow)

mirror plased behind.
(click to enlarge image)


This way is good enough as:
- "unsolved" edge you can to provide any time, during pilling up first 8 edges to the up and down sides;
- actually, you can to make this edge by yourself, with your favorite formula for solveing parity;
- You know the parity beforehand where and when.

That is my idea.
 
Last edited:

Novriil

Member
Joined
Mar 2, 2009
Messages
738
Location
Estonia
WCA
2009KRUU01
YouTube
ukx14
first: they are algorithms (or just algs.. )

second 7x7 are just the same algs than you have on 5x5.. if you want to change those then just do the parity alg (Ll' U2 Ll' U2 F2 Ll' F2 Rr U2 Rr' U2 Ll2)

just when you have the inner 3 edges being paired up then you use 3L' U2 3L' and so on but when the outer like you have on the pic then you use 2L' U2 2L' and so on.
 

Tesseract

Member
Joined
Oct 9, 2009
Messages
19
Location
Ukraine, Kiev
second 7x7 are just the same algs than you have on 5x5.
That is right, I know. However, you have to use not one alg - but several, one by one, until complete two edges parity both.

Also I know about 3L' U2 3L' or 2L' U2 2L' and so on... But again: you have to solve last two edges with SEVERAL algorithms. So my idea produce decision of using ONLY ONE alg for solving two_last_edges_parity.
 

Jake Gouldon

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Aug 30, 2009
Messages
566
Location
New Jersey
YouTube
Jakegouldon
Whoops. I meant avoid the two wings flipped pairity, not the two wings swapped parity. Just set up a cube with both of those parities, and use AVG. You'll see what I mean.
 
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