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Dunno. =C2=A0Still kinda don't get how to link the 4D cube back to the 3D c=
ube, and I can't visualise how the F2L pieces move around on a 4D cube. =C2=
=A0I basically only used Fridrich, but for the last layer, I suggest using =
commutators (Beyer-Hardwick method could be helpful). =C2=A0btw, my times f=
or 3^3 cube have dropped down a few seconds, and I got a new pb on hi-games=
.net=C2=A0hi-games.net/cube-3x3x3/watch?u=3D3139.
Good luck with the last layer,
Brandon
________________________________
From: Ray Zhao
To: 4D_Cubing@yahoogroups.com=20
Sent: Friday, June 1, 2012 10:29 AM
Subject: [MC4D] MC4D: CFOP method test [1 Attachment]
=20
=C2=A0=20
[Attachment(s) from Ray Zhao included below]
Results for CFOP/Friedrich method:
Complete first two layers (2/3 of 3^4) in 377 moves, not bad.
BUT
last layer #moves=3Df2l #moves (total around 650 moves), therefore the last=
layer method still needs work.
Also, no shortcuts (e.g. recorded macros, edge turns, vertex turns) were us=
ed.
The edge/corner "parity" problems are inefficient with face-only turns, so =
I left them until last, which isn't a good idea.
Identifying the pieces often took a while and positioning 3c/4c pairs too.
Well, I don't know any other methods like roux or petrus so you can try sol=
ving the 3^4 using those methods.
--=20
while(true)
Console.Writeline("HI!")
=C2=A0
=20
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ll kinda don't get how to link the 4D cube back to the 3D cube, and I can't=
visualise how the F2L pieces move around on a 4D cube. I basically o=
nly used Fridrich, but for the last layer, I suggest using commutators (Bey=
er-Hardwick method could be helpful). btw, my times for 3^3 cube have=
dropped down a few seconds, and I got a new pb on hi-games.net hi-gam=
es.net/cube-3x3x3/watch?u=3D3139.
he last layer,
es, serif; ">
=3D"1"> From: Ray Zhao
<thermostatico@gmail.com>
=
To: 4D_Cubing@yahoogroups.com
: bold;">Sent: Friday, June 1, 2012 10:29 AM
=3D"font-weight: bold;">Subject: [MC4D] MC4D: CFOP method test [=
1 Attachment]
=20=20=20=20=20=20
=20=20=20=20=20=20=20=20
=20=20=20=20=20=20
Complete first two layers =
(2/3 of 3^4) in 377 moves, not bad.
BUT
last layer #moves=3Df2l #move=
s (total around 650 moves), therefore the last layer method still needs =
work.
Also, no shortcuts (e.g. recorded macros, edge turns, vertex turns) wer=
e used.
The edge/corner "parity" problems are inefficient with face-only=
turns, so I left them until last, which isn't a good idea.
">
Identifying the pieces often took a while and positioning 3c/4c pairs too.<=
br>
Well, I don't know any other methods like roux or petrus so you can =
try solving the 3^4 using those methods.
--
div>
=20=20=20=20=20
<=
/div>
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