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yahoo论坛四阶盲拧的例子 [复制链接]

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发表于 2006-4-9 00:31:36 |只看该作者 |倒序浏览

有朋友在研究高阶盲拧的吗？转发yahoo的一个4阶盲拧例子。

From: "cmhardw" <foozman17@...>
Date: Sat Nov 26, 2005 3:07 pm
Subject: An example 4x4x4 BLD solve (by request) cmhardw
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Hey everyone,

I finally got another 4x4 BLD solve successfully. It's actually my
new record at 24 minutes 10.61 seconds. I decided to go a bit
slower than I had been going on memorization, but still go full
speed on the solving.

Anyway, I had gotten some requests for a sample 4x4x4 BLD solve, and
I wanted to wait until I got a successful one so I could talk about
it practically, rather than in theory.

Here's the scramble:
F R' b2 L2 r' B2 R' L' F' U' b d f2 d' B' R' u R2 b u2 F' l L f B' D
U' r2 F2 L u2 b r2 u2 B' R d' f' B' d

The first thing I do is to get as many centers solved as I can. My
color scheme that I solve to is yellow on top and green on front.
Do the rotation z right now (rotate the whole cube as if you're
doing the move F).

This gets 7 centers solved right off the bat. Ok now I have to
memorize the centers. The cycle I used is

(lbU -> ruF -> rbU -> buL -> lfU -> brD -> dbL -> ufR -> blD -> frU -
> frD -> ubR -> fdL -> lfD -> ruB -> dlF -> luB)

The notation means the only piece on the intersection of the three
slices named. The piece lbU is the only piece on the intersection
of "l" "b" and "U".

I have names for each spot, using the name my list is

(Larry -> airplane -> Batman -> Giraffe -> the Terminator -> Book ->
Alligator -> the South Pole -> sword -> Spiderman -> rubber duckie -
> Classroom/Lecture Hall -> hamster -> chair -> skydiving -> the
back to the future III train time machine -> mountaineering)

And the story I made up is:

LARRY is the pilot of an AIRPLANE and accidently flies it into
BATMAN, who was in freefall in the air. Meanwhile, below the plane
accident, a GIRAFFE is hunting the TERMINATOR, who is reading a BOOK
with a picture of an ALLIGATOR on it. The Giraffe kicks the
Terminator to the SOUTH POLE. Once he gets his bearings straight
the Terminator draws a SWORD to do battle with SPIDERMAN, who
unfortunately is only armed with a RUBBER DUCKIE. This pans out to
a CLASSROOM where the students were watching this on a discovery
channel episode. Suddenly a giant HAMSTER rips through the roof and
starts ripping apart the classroom and CHAIRs are flying
everywhere. Meanwhile, above, a SKYDIVER in midair falls into the
wood storage car of the BACK TO THE FUTURE TRAIN TIME MACHINE. Just
when he realizes where he is the time machine flies into a MOUNTAIN
and explodes.

I know the story is really dumb and silly, but that is exactly why
it is easy to remember. I read a lot about memory, and one of the
techniques is not only to make a story silly, but also to try to
make it shocking. That's why I have the train explode, and the
hamster destroy the classroom with students in it. It's sound
weird, but the technique says that whatever makes you laugh, or
shocks you, is easier to remember.

Ok, so those are the centers. The edges cycle as:

(lUB -> fUR -> uFL -> dFR -> uBL -> dBL -> fDL -> rFU -> uBR -> bDL -
> rFD -> rBU -> dBR -> lBD -> dFL -> lFD -> fUL -> bDR -> bUR ->
rBD -> fDR)

For the edges I just memorize where they go. I try to picture it in
my head as one piece that moves through each of these locations, and
I picture the intermittent moves sometimes as well, whatever helps.

The corners cycle as (UBL -> DFL -> UBR) and (UFL -> DBR -> UFR) and
(DBL -> DFR)

and they flip as UBL needs to flip clockwise and DFL needs to flip
counterclockwise.

Ok here's the solve:
-----------------------------------

First I solve corner orientation, since it's the last thing I
memorize and I can just take a quick snapshot of it if I solve it
first.

To solve I did U' x R' U' R U' R' U2 R L U L' U L U2 L' x' U
(x is the rotation that means to turn your whole cube as if you were
doing the move R, x' is like R')

Now I do corner permutation
(UBL -> DFL -> UBR)
F2 R2 B2 R F R' B2 R F' R F2

(UFL -> DBR -> UFR)
B2 F2 L2 F' R' F L2 F' R F' B2

(DBL -> DFR)
For this I can't solve them yet, since any alg that moves just two
corners must also perform an odd cycle somewhere on the centers
(think 4x4x4 supercube). Instead I just want to get them ready to
solve later.

F L' F' L F L' F' L F L' F' L - F' L F L' F' L F L' F' L F L' - R2
B2 R F R' B2 R F' R

I'm sure there is a better way to do this, but with the clock
ticking and me wanting to use center safe algs this is how I did
it. I end up with the corner cycle (UBL <-> UBR).

EDGES:
------------------------------------
Here all I use are commutators, I have absolutely no memorized algs
for any sort of cycles.

(lUB -> fUR -> uFL -> dFR -> uBL -> dBL -> fDL -> rFU -> uBR -> bDL -
> rFD -> rBU -> dBR -> lBD -> dFL -> lFD -> fUL -> bDR -> bUR ->
rBD -> fDR)

Here are the commutators I used. I will only write them as [X,Y]
which means to do the permutation X Y X' Y'

I'll write setup moves around the commutator notation as well.

(lUB -> fUR -> uFL) = f' [B' u2 B, U] f
Ok, so to make sure things are clear X=(B' u2 B) and Y=(U) so f'
[B' u2 B, U] f = f' B' u2 B U B' u2 B U' f

(lUB -> dFR -> uBL) = L [U, B d2 B'] L'
again just to help clarify L [U, B d2 B'] L' = L U B d2 B' U' B d2
B' L'

(lUB -> dBL -> fDL) = [L', D' l' D]

(lUB -> rFU -> uBR) = [U2, F' u2 F]

(lUB -> bDL -> rFD) = L2 D2 [B d B', U] D2 L2

(lUB -> rBU -> dBR) = r' [U2, F d' F'] r

(lUB -> lBD -> dFL) = L' [U, B d B'] L

(lUB -> lFD -> fUL) = F' [L d' L', U'] F

(lUB -> bDR -> bUR) = [R d R', U]

(lUB -> rBD -> fDR) = U' L [D', R' d2 R] L' U

And now the edges are solved.

CENTERS:
-------------------------------------
Same as the edges I have no memorized algs for this. I solve this
entirely using commutators. I'll use the same [X,Y] notation here
as above.

(lbU -> ruF -> rbU -> buL -> lfU -> brD -> dbL -> ufR -> blD -> frU -
> frD -> ubR -> fdL -> lfD -> ruB -> dlF -> luB)

(lbU -> ruF -> rbU) = [U', b' u' b]

(lbU -> buL -> lfU) = [U, l u' l']

(lbU -> brD -> dbL) = r2 [U2, f d2 f'] r2

(lbU -> ufR -> blD) = D2 f' [l u l', U] f D2

(lbU -> frU -> frD) = r F' r' [U2, r u2 r'] r F r'

(lbU -> ubR -> fdL) = f [l' u' l, U'] f'

(lbU -> lfD -> ruB) = D2 r' [U2, l u2 l'] r D2

(lbU -> dlF -> luB) = F2 r [U, r' u2 r] r' F2

------------------------

FINISHING:
Ok now we have to finish off our corners. To do this swap two
edges, then end like a 3x3x3.

(Uu)2 (Ll)2 U2 l2 U2 (Ll)2 (Uu)2 y' L' U' L U L F' L2 U L U L' U' L F

And now the cube is solved.

I realize that the centers can be done more efficiently, but I am
taking Dror's advice that he gave me at RWC2005 to try to put the
centers into one long cycle to help with memorization. I may try to
optimize them later, but for now I like making one long cycle to
ease the memorization.

Anyway there is an example solve, let me know if you have any
questions. This solve is not theory, but is actually a
reconstruction of my current record solve.

Hope this helps,
Chris

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