[Unofficial] 4x4 - 59.99

[aronpm]

[youtubehd]Z8VXr3aR9p4[/youtubehd]

Scramble: Uw' Rw F D Fw Uw' F2 L2 U2 F' U B D L R D' Rw2 Fw Uw2 R' U' R2 F2 U' Uw F Uw D2 U2 B L' R Uw F B2 U Rw2 F B Fw'

 

[AustinReed]

WTF parity recog 0___0

 

[tx789]

just sub 1

 

[ottozing]

when you inspected the cube, it looked so much like bld memo

btw nice

 

[Bob]

When I opened this thread, I was expecting to see somebody selling a 4x4 for $59.99 and I was wondering what the hell they did to it to ask for so much.

 

[kirtpro]

inspection was.. cool

was that parity prevention stuff or something

edit: somehow i missed the words at the intro (noob)
anyway... how does that work lol

 

[Weston]

Psh. Not even bld.

 

[Henrik]

anyway... how does that work lol

Count edges --> odd number = parity!

To avoid --> solve two adjacent centers in an odd number of slice moves (Rw/r) (Rw2 counts as even)
This works on the presumption that you use some kind of reduction. (Then if you are not crazy, this should work)

 

[kirtpro]

Count edges --> odd number = parity!

To avoid --> solve two adjacent centers in an odd number of slice moves (Rw/r) (Rw2 counts as even)
This works on the presumption that you use some kind of reduction. (Then if you are not crazy, this should work)

seems kinda complex, i'll choose to stick with normal parity solving lol

 

[Henrik]

seems kinda complex, i'll choose to stick with normal parity solving lol

It is not really.

When you have solved two adjacent centers in an odd number of moves, you can just solve the rest as usual. Because, paring edges takes an even number of slice turns. The same goes for the rest of the the centers.

That is at least how I think Aaron does it. I don't know of other practical ways to do it. I'm sure there are. (Other than "regular" redux methods)

 

[kirtpro]

It is not really.

When you have solved two adjacent centers in an odd number of moves, you can just solve the rest as usual. Because, paring edges takes an even number of slice turns. The same goes for the rest of the the centers.

That is at least how I think Aaron does it. I don't know of other practical ways to do it. I'm sure there are. (Other than "regular" redux methods)

all right then, thank you for the explanation

 

[Brest]

Spoiler: Video

[youtubehd]Z8VXr3aR9p4[/youtubehd]

y u' r F D f u' F2 L2 U2 F' U B D L R D' r2 f u2 R' U' R2 F2 U' u F u D2 U2 B L' R u F B2 U r2 F B f'

x2 // inspection
// centres
U l2' (z) U2' l z' U l' // yellow
x2' F' r U2' r' (z x2') U2' l2' (y' x') U r U2' r' // white
z l' U' r U x F2 r U' x' l' U2' l // red / GR
x2' l' U l U' l' U2 l x' U r2 U2' r2' // green
U r U' r' F x r' U2 l // orange / blue
// edges
z U' R' U 3l u // WR
y' R2 U R2 U' y L' U L u' // WG / YR
y F' R2 u2 y' R U 3l' U R' U' 3l u2 // GO / WO
U' U L U' y L' U L y' R' U R u // YB
y' R U' R' u' // WB
y2 U' y L' U L y' R' U R u // RB
U' L' U L u' // YO
(x y) 3d' L' U L y' u' R U 3l' U R' U' 3l u // YG / OB
// 3x3x3
U x' U' R2' x U L2' x' D // white cross
U2 L U' L' // GO pair
y L' U' L U y L' U' L // GR pair
y U2' R U2' R' U y L' U' L // OB pair
U' R U R' // RG pair
U3 R U2' R2' U' R2 U' R2' U2 R // OLL
U2 R' U2 R U2' R' F R U R' U' R' F' R2 // PLL
View at alg.garron.us

[B]Step	Time	STM	stps	ETM	etps[/B]	
[COLOR="red"]Total	59.99	163	2.72	218	3.63	[/COLOR]
						
Centres	18.46	45	2.44	66	3.58	
Edges	27.61	65	2.35	83	3.01	
[COLOR="blue"]Redux	46.07	110	2.39	149	3.23	[/COLOR]
						
Cross+1	4.34	10	2.30	14	3.23	
F2L	8.67	29	3.34	39	4.50	
LL	5.25	24	4.57	30	5.71	
[COLOR="blue"]3x3x3	13.92	53	3.81	69	4.96	[/COLOR]
						
[B][SIZE="4"]%[/SIZE]						
Step		Time	STM	ETM[/B]		
Centres/Redux	40%	41%	44%			
Edges/Redux	60%	59%	56%			
[COLOR="blue"]Redux/Total	77%	67%	68%	[/COLOR]		
						
Cross+1/F2L	50%	34%	36%			
F2L/3x3x3	62%	55%	57%			
[COLOR="blue"]3x3x3/Total	23%	33%	32%	[/COLOR]

 

[Sebastien]

It is not really.

When you have solved two adjacent centers in an odd number of moves, you can just solve the rest as usual. Because, paring edges takes an even number of slice turns. The same goes for the rest of the the centers.

That is at least how I think Aaron does it. I don't know of other practical ways to do it. I'm sure there are. (Other than "regular" redux methods)

What you say is not completely right. You have to take care until finishing the 3rd center. Only everything after this step has an even number of quarter slice moves.

Also I think that when talking about avoiding parity this only applies to what we know as OLL-Parity.

Anyway, it is amazing to see how Aron is able to do that within preinspection time!

 

[brunovervoort]

My first sub 1 was 59.99

 

[Kirjava]

What you say is not completely right. You have to take care until finishing the 3rd center. Only everything after this step has an even number of quarter slice moves.

You're wrong. He said first two adjacent centres, which is correct.

 

[Sebastien]

Lol, you're completely right. I was just too stupid for not reading adjacent as opposite in the given context. Sorry Henrik.

 

[Mike Hughey]

Sorry I missed this earlier when it came out. This is totally amazing - Aron is amazing! I've been trying it, and it's lots of fun, but I average about a minute inspection time, unfortunately. I've gotta get faster...

 

[MirzaCubing]

Glad to see I'm not the only one who's gotten that time XD