Difference between revisions of "Talk:Last Four Corners"

From Speedsolving.com Wiki
Line 157: Line 157:
 
|-valign="top"
 
|-valign="top"
 
|
 
|
 +
-Sb is mirror/inverse
 
=== Sa (E) ===
 
=== Sa (E) ===
 
[[File:L4C SE (a).jpg]]
 
[[File:L4C SE (a).jpg]]
  
{{Alg|!}}
+
{{Alg|(y') L U2 B' R B R' U2 L' B' U2 B2 U2 B' }}
  
 
|
 
|
  
 
-Sa is inverse
 
-Sa is inverse
 +
 
=== Sb (E) ===
 
=== Sb (E) ===
 
[[File:L4C SE (b).jpg]]
 
[[File:L4C SE (b).jpg]]

Revision as of 06:51, 21 August 2010

This will be running a bit slower than what has been the later weeks, the rains has gone again so I'm outside daytime now and tired at nights :)

// Kenneth 18:59, 12 August 2010 (UTC)


I think we make a separate page for L3C, it is a stand alone method and it will make this page a little smaller. // Kenneth 06:51, 14 August 2010 (UTC)


I don't know if it is a good order for the 1-twist cases as is, better to group them like I did with the 3-cycles at L3C in quads where 'the same' alg can be used for each group of 4.

I do that later, first I fix the rest of the images... done. // Kenneth 09:37, 18 August 2010 (UTC)


Work area

16 cases, two are E-PLL, leaving 14.


Order 2

Order 2

!

File:!.jpg

Speedsolving Logo tiny.gif Alg !


!

File:!.jpg

Speedsolving Logo tiny.gif Alg !


Order 2

Order 6, but adding y2 makes the inverse case the same as this, in practice a 2-cycle.

Ua (E)

L4C UE (a).jpg

Speedsolving Logo tiny.gif Alg !


Ub (E)

L4C UE (b).jpg

Speedsolving Logo tiny.gif Alg R' F' U' F R B L F U F' L' B'


Order 2

Order 6, but adding y2 makes the inverse case the same as this, in practice a 2-cycle.

Ta (E)

L4C TE (a).jpg

Speedsolving Logo tiny.gif Alg !


Tb (E)

L4C TE (b).jpg

Speedsolving Logo tiny.gif Alg R U2 F' D2 F2 D' F' U F D F2 D2 F U R'


Order 6

Order 6, inverse of previous case

La (E)

L4C LE (a).jpg

Speedsolving Logo tiny.gif Alg (y) F' U2 F2 R' F' R2 B L U2 L' B' R' U2


Lb (E)

L4C LE (b).jpg

Speedsolving Logo tiny.gif Alg U2 R B L U2 L' B' R2 F R F2 U2 F


Order 2 F (Sexy Move)3 F' is a optimal solution (triple fruruf).

Order 2

Ha (E)

L4C HE (a).jpg

Speedsolving Logo tiny.gif Alg F(RUR'U')3F'


Hb (E)

L4C HE (b).jpg

Speedsolving Logo tiny.gif Alg !


Order 2

Order 6, but adding y2 makes the inverse case the same as this, in practice a 2-cycle.

pia (E)

L4C piE (b).jpg

Speedsolving Logo tiny.gif Alg U F U F2 R2 F2 U' F' U' L2 B' D2 B L2


pib (E)

L4C piE (a).jpg

Speedsolving Logo tiny.gif Alg U' R U' R2 D' R U2 R' U' F2 U F2 D R





-Sb is mirror/inverse

Sa (E)

L4C SE (a).jpg

Speedsolving Logo tiny.gif Alg (y') L U2 B' R B R' U2 L' B' U2 B2 U2 B'


-Sa is inverse

Sb (E)

L4C SE (b).jpg

Speedsolving Logo tiny.gif Alg (y') F R F' L B F R' F' R B' L' R'


Sb is inverse

-Sa (E)

L4C aSE (b).jpg

Speedsolving Logo tiny.gif Alg R L B R' F R F' B' L' F R' F' (12f*)


Sa is inverse/mirror

-Sb (E)

L4C aSE (a).jpg

Speedsolving Logo tiny.gif Alg R' U2 B L' B' L U2 R B U2 B2 U2 B