------=_NextPart_000_0006_01D06EE4.F9C2B3B0
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charset="utf-8"
Content-Transfer-Encoding: quoted-printable
This is not the shortest solution for a specific scrambling but for the wor=
st case scrambling.
If 'gods number for 3^3 got under 20 you try 'under 200' for 3^4 now.
Correct?
Ed
----- Original Message -----=20
From: damienturtle@hotmail.co.uk [4D_Cubing]=20
To: 4D_Cubing@yahoogroups.com=20
Sent: Saturday, April 04, 2015 2:10 PM
Subject: [MC4D] New 3^4 shortest solution! (227 twists)
=20=20=20=20
Hi everyone,
For those newer members, I used to be fairly active here a few years back=
and set the 3^4 fewest moves record with 251 twists, amongst other things.=
I'm not actually returning to being active (I probably won't be taking par=
t in the speedsolving contest being talked about, much as I would like to),=
I just felt like tidying up this solve since I hadn't finished optimising =
the last step before, but when I became busy I just submitted what I had so=
far. Essentially, I had ended with a parity situation which cost moves to =
fix, and I wasn't happy with that. After going to a speedsolving competitio=
n recently (I had a free weekend and had fun even though I was very out of =
practice) and having good luck in FMC with a 28 twist solution (which shoul=
d have been 25 twists, maybe if I had practiced at all I would have spotted=
the really obvious insertion that I missed), I decided to correct the endi=
ng of my 3^4 solution with something much better, thoug h I'm still busy th=
ese days so I could probably have found something even better if I was will=
ing to spend longer on it.
I'll try to explain roughly how my solution works, especially since the e=
nding now isn't very clear now that I'm better at fewest move solving. The =
first 182 twists of the solution are unchanged from before, since I didn't =
really have the time to work through a full solve and because I always cons=
idered my previous submission as an unfinished solve which I wanted to fini=
sh eventually. It is simple blockbuilding mostly, followed by orienting the=
last layer. This means that what is left is essentially a 3^3 solve embedd=
ed into the 3^4 puzzle, and it is this which I revisited. As for the parity=
problem I had, try doing a U2 move on one 3^3 cell, leaving the rest of th=
e puzzle solved, my best solution for this is 22 twists if I remember corre=
ctly.
Optimising the last step is weird. It's based on optimising a 3^3 solve, =
but it's not quite as simple as that. Let's look at that first though, and =
the metric used (as a simplification since the 'correct' metric is too comp=
licated) is quarter-slice turn metric. A quarter turn of any layer is one m=
ove, a half turn of any layer counts as 2 moves. For some orientation I hap=
pened to choose, I looked at what the 3x3 scramble was, and used CubeExplor=
er to find a short scramble to work with:
D L2 F D2 U' F U' F D' U' R' B2 L' B F' U' B U'
Now, due to the 3^4 context and my solution before this point, I needed t=
o perform a net result of an L turn, which in practice means the net result=
of adding the twists of the solution (ignoring which faces are turned) mus=
t be a quarter turn clockwise (so L, R, U2 B', L' F' D', etc. would be vali=
d solutions in this regard). Also, by being able to change the previous twi=
st in the solution, I get a free turn of the front face in terms of movecou=
nt, which I tak e advantage of. My solution looks like this (knowledge of F=
MC techniques required, I can explain in more detail if anyone wants me to)=
:
(F) R' (U' D) L F' //222 block 4/4
U' L2 U L D //223 block, 6/10
(B' L D L D' L' U' L U L2) //Performed on inverse, solves all but 4 corne=
rs + 3 edges, 11/21
Skeleton: (F) R' (U' D) L F' U' L2 * @ U L D ** L2 U' L' U L D L' D' L' B=
//21 moves
Insertions:
* =3D (F B') U' L U L' (B F') //Solves 3 edges + 1 corner, no moves cance=
l, 6/27
@ =3D L U R' U' L' U R U' //Solves 3 corners, 2 moves cancel, 6/33
Solution: (F) R' (U' D) L F' U' L2 (FB') U' L U L' (BF') L U R' U' L' U R=
L D L2 U' L' U L D L' D' L' B //33 moves QSTM
Bonus: insert ** =3D D' (LR') B L2 B' (RL') D L2, to solve 3 edges, which=
cancels 7 moves. Never really checked the corner insertions on this, so it=
might have given a better result actually, but probably not.
Now, I just try to adapt this to the 3^4 solve as efficiently as possible=
. My first serious attempt managed 230 twists, then 229. After a while I fo=
und a 228 twist solution which I nearly kept, but some instinct told me tha=
t I could save one more twist, which I managed to do.=20
The idea is as follows. Say we need to do R U R' U' to solve the 3^3. The=
simple solution is to do R [F] R [F]' R' [F] R' [F]' (8), where [F] means =
to rotate the whole cube as in an F turn, and notice that if we ignore the =
rotations all the moves cancel, which is necessary. I think Ray put some in=
fo on this on the wiki for how this works, if you aren't familiar with it, =
try doing these moves on the 'inside' face in the most obvious way, with on=
e twist in MC4D for each twist and for each rotation. Instead, we could do =
R U [UR] U' R' [UR] (6), where [UR] rotates the cub e around the axis thoug=
h the UR edge and the opposite edge DL, and again all moves cancel once rot=
ations are ignored. With this idea, two moves are saved here, and for full =
solutions the same idea applies but it becomes far more difficult to optimi=
se. With this, my solution turned out to be:
[F] F' M' [F]' R B' U' R2 S R' [UL] R D [DR] D' S' R U [DR] U' L' D' [F]'=
D L R U R [F]' U [UR] U' R' U R D [UFL] D' [RB] R' U' [DR] B
where S is a slice turn which follows F, M is a slice turn which follows =
L, and [UFL] is a clockwise cube rotation about the UFL corner.
I didn't do this on paper, I worked in MC4D, then typed up what my final =
solution was. The strangest part for me was trying (R z' U), which is essen=
tially a double turn with a rotation in the middle, which actually saved a =
rotation overall.
I think I'm unlikely to try beating this, so if anyone wants to take the =
record, they are welcome to it. I just hope I've made it very challenging t=
o do so :), but it is certainly possible. I reckon sub-200 twists is very p=
ossible, so maybe if this isn't achieved for many years, I might try to do =
it myself ...
These days, I'm doing a PhD is mathematics, working on network theory. I =
seen that someone new here recently is also in this area, so I might be abl=
e to have an interesting conversation with them :). I have one paper recent=
ly resubmitted to a journal which will hopefully be accepted soon, and anot=
her related paper nearly ready to be submitted. This is why I don't have mu=
ch time to spend on cubing, but that's life.
I hope as many people take part in the speedsolving contest as possible, =
it was very fun the first time so I recommend having a go. Even though I wo=
n't have time to practice for it, I'll be interesting in the results so ple=
ase make them exciting! Good luck everyone!
Happy hypercubing,
Matt=20
=20=20
------=_NextPart_000_0006_01D06EE4.F9C2B3B0
Content-Type: text/html;
charset="utf-8"
Content-Transfer-Encoding: quoted-printable
=EF=BB=BF
Hi everyone,
For those newer members, I used to be fairly activ=
e=20
here a few years back and set the 3^4 fewest moves record with 251 twists=
,=20
amongst other things. I'm not actually returning to being active (I proba=
bly=20
won't be taking part in the speedsolving contest being talked about, much=
as I=20
would like to), I just felt like tidying up this solve since I hadn't fin=
ished=20
optimising the last step before, but when I became busy I just submitted =
what=20
I had so far. Essentially, I had ended with a parity situation which cost=
=20
moves to fix, and I wasn't happy with that. After going to a speedsolving=
=20
competition recently (I had a free weekend and had fun even though I was =
very=20
out of practice) and having good luck in FMC with a 28 twist solution (wh=
ich=20
should have been 25 twists, maybe if I had practiced at all I would have=
=20
spotted the really obvious insertion that I missed), I decided to correct=
the=20
ending of my 3^4 solution with something much better, thoug h I'm still b=
usy=20
these days so I could probably have found something even better if I was=
=20
willing to spend longer on it.
I'll try to explain roughly how my=
=20
solution works, especially since the ending now isn't very clear now that=
I'm=20
better at fewest move solving. The first 182 twists of the solution are=20
unchanged from before, since I didn't really have the time to work throug=
h a=20
full solve and because I always considered my previous submission as an=20
unfinished solve which I wanted to finish eventually. It is simple=20
blockbuilding mostly, followed by orienting the last layer. This means th=
at=20
what is left is essentially a 3^3 solve embedded into the 3^4 puzzle, and=
it=20
is this which I revisited. As for the parity problem I had, try doing a U=
2=20
move on one 3^3 cell, leaving the rest of the puzzle solved, my best solu=
tion=20
for this is 22 twists if I remember correctly.
Optimising the last=
step=20
is weird. It's based on optimising a 3^3 solve, but it's not quite as sim=
ple=20
as that. Let's look at that first though, and the metric used (as a=20
simplification since the 'correct' metric is too complicated) is quarter-=
slice=20
turn metric. A quarter turn of any layer is one move, a half turn of any =
layer=20
counts as 2 moves. For some orientation I happened to choose, I looked at=
what=20
the 3x3 scramble was, and used CubeExplorer to find a short scramble to w=
ork=20
with:
D L2 F D2 U' F U' F D' U' R' B2 L' B F' U' B U'
Now, =
due=20
to the 3^4 context and my solution before this point, I needed to perform=
a=20
net result of an L turn, which in practice means the net result of adding=
the=20
twists of the solution (ignoring which faces are turned) must be a quarte=
r=20
turn clockwise (so L, R, U2 B', L' F' D', etc. would be valid solutions i=
n=20
this regard). Also, by being able to change the previous twist in the=20
solution, I get a free turn of the front face in terms of movecount, whic=
h I=20
tak e advantage of. My solution looks like this (knowledge of FMC techniq=
ues=20
required, I can explain in more detail if anyone wants me to):
(F)=
R'=20
(U' D) L F' //222 block 4/4
U' L2 U L D //223 block, 6/10
(B' L D L=
D'=20
L' U' L U L2) //Performed on inverse, solves all but 4 corners + 3 edges,=
=20
11/21
Skeleton: (F) R' (U' D) L F' U' L2 * @ U L D ** L2 U' L' U L=
D L'=20
D' L' B //21 moves
Insertions:
* =3D (F B') U' L U L' (B F') //Solv=
es 3=20
edges + 1 corner, no moves cancel, 6/27
@ =3D L U R' U' L' U R U' //So=
lves 3=20
corners, 2 moves cancel, 6/33
Solution: (F) R' (U' D) L F' U' L2 (=
FB')=20
U' L U L' (BF') L U R' U' L' U R L D L2 U' L' U L D L' D' L' B //33 moves=
=20
QSTM
Bonus: insert ** =3D D' (LR') B L2 B' (RL') D L2, to solve 3 =
edges,=20
which cancels 7 moves. Never really checked the corner insertions on this=
, so=20
it might have given a better result actually, but probably not.
No=
w, I=20
just try to adapt this to the 3^4 solve as efficiently as possible. My fi=
rst=20
serious attempt managed 230 twists, then 229. After a while I found a 228=
=20
twist solution which I nearly kept, but some instinct told me that I coul=
d=20
save one more twist, which I managed to do.
The idea is as follow=
s.=20
Say we need to do R U R' U' to solve the 3^3. The simple solution is to d=
o R=20
[F] R [F]' R' [F] R' [F]' (8), where [F] means to rotate the whole cube a=
s in=20
an F turn, and notice that if we ignore the rotations all the moves cance=
l,=20
which is necessary. I think Ray put some info on this on the wiki for how=
this=20
works, if you aren't familiar with it, try doing these moves on the 'insi=
de'=20
face in the most obvious way, with one twist in MC4D for each twist and f=
or=20
each rotation. Instead, we could do R U [UR] U' R' [UR] (6), where [UR]=20
rotates the cub e around the axis though the UR edge and the opposite edg=
e DL,=20
and again all moves cancel once rotations are ignored. With this idea, tw=
o=20
moves are saved here, and for full solutions the same idea applies but it=
=20
becomes far more difficult to optimise. With this, my solution turned out=
to=20
be:
[F] F' M' [F]' R B' U' R2 S R' [UL] R D [DR] D' S' R U [DR] U'=
L'=20
D' [F]' D L R U R [F]' U [UR] U' R' U R D [UFL] D' [RB] R' U' [DR]=20
B
where S is a slice turn which follows F, M is a slice turn which=
=20
follows L, and [UFL] is a clockwise cube rotation about the UFL=20
corner.
I didn't do this on paper, I worked in MC4D, then typed up=
what=20
my final solution was. The strangest part for me was trying (R z' U), whi=
ch is=20
essentially a double turn with a rotation in the middle, which actually s=
aved=20
a rotation overall.
I think I'm unlikely to try beating this, so i=
f=20
anyone wants to take the record, they are welcome to it. I just hope I've=
made=20
it very challenging to do so :), but it is certainly possible. I reckon=20
sub-200 twists is very possible, so maybe if this isn't achieved for many=
=20
years, I might try to do it myself ...
These days, I'm doing a=
PhD=20
is mathematics, working on network theory. I seen that someone new here=20
recently is also in this area, so I might be able to have an interesting=
=20
conversation with them :). I have one paper recently resubmitted to a jou=
rnal=20
which will hopefully be accepted soon, and another related paper nearly r=
eady=20
to be submitted. This is why I don't have much time to spend on cubing, b=
ut=20
that's life.
I hope as many people take part in the speedsolving=20
contest as possible, it was very fun the first time so I recommend having=
a=20
go. Even though I won't have time to practice for it, I'll be interesting=
in=20
the results so please make them exciting! Good luck everyone!
Happ=
y=20
hypercubing,
Matt
Hi everyone,
--------------020900090900040706050907
Content-Type: text/plain; charset=utf-8; format=flowed
Content-Transfer-Encoding: quoted-printable
I was wondering if anyone would mind that I update the record. I suppose=20
it's valid because we allow people to save solution states and revert to=20
them as needed. I guess I'll plan to update the solution date as well=20
unless anyone objects. BTW, a solution time measured in years is=20
unlikely to get a speedsolving prize unless it's for the *slowest* solution=
!
-Melinda
On 4/4/2015 5:59 AM, damienturtle@hotmail.co.uk [4D_Cubing] wrote:
>
>
> This was just for a random scramble generated by MC4D, I never tried=20
> to get an unusually good or bad scramble, I took the first one=20
> generated. 200 is just a nice milestone, in my own opinion, so there=20
> is no real significance to the 200 moves I mentioned as far as I know.
>
> I sent the solution to Melinda last weekend (it's taken me this long=20
> to type up the post) but it doesn't seem to have been updated in the=20
> HOF yet so I realise the log file isn't available to look at.
>
> Matt
>
>
> ---In 4D_Cubing@yahoogroups.com,
>
> =EF=BB=BF
> This is not the shortest solution for a specific scrambling but for=20
> the worst case scrambling.
> If 'gods number for 3^3 got under 20 you try 'under 200' for 3^4 now.
> Correct?
> Ed
>
> ----- Original Message -----
> *From:* damienturtle@... [4D_Cubing]
>
> *To:* 4D_Cubing@yahoogroups.com
> *Sent:* Saturday, April 04, 2015 2:10 PM
> *Subject:* [MC4D] New 3^4 shortest solution! (227 twists)
>
> Hi everyone,
>
> For those newer members, I used to be fairly active here a few
> years back and set the 3^4 fewest moves record with 251
> twists, amongst other things. I'm not actually returning to
> being active (I probably won't be taking part in the
> speedsolving contest being talked about, much as I would like
> to), I just felt like tidying up this solve since I hadn't
> finished optimising the last step before, but when I became
> busy I just submitted what I had so far. Essentially, I had
> ended with a parity situation which cost moves to fix, and I
> wasn't happy with that. After going to a speedsolving
> competition recently (I had a free weekend and had fun even
> though I was very out of practice) and having good luck in FMC
> with a 28 twist solution (which should have been 25 twists,
> maybe if I had practiced at all I would have spotted the
> really obvious insertion that I missed), I decided to correct
> the ending of my 3^4 solution with something much better,
> thoug h I'm still busy these days so I could probably have
> found something even better if I was willing to spend longer
> on it.
>
> I'll try to explain roughly how my solution works, especially
> since the ending now isn't very clear now that I'm better at
> fewest move solving. The first 182 twists of the solution are
> unchanged from before, since I didn't really have the time to
> work through a full solve and because I always considered my
> previous submission as an unfinished solve which I wanted to
> finish eventually. It is simple blockbuilding mostly, followed
> by orienting the last layer. This means that what is left is
> essentially a 3^3 solve embedded into the 3^4 puzzle, and it
> is this which I revisited. As for the parity problem I had,
> try doing a U2 move on one 3^3 cell, leaving the rest of the
> puzzle solved, my best solution for this is 22 twists if I
> remember correctly.
>
> Optimising the last step is weird. It's based on optimising a
> 3^3 solve, but it's not quite as simple as that. Let's look at
> that first though, and the metric used (as a simplification
> since the 'correct' metric is too complicated) is
> quarter-slice turn metric. A quarter turn of any layer is one
> move, a half turn of any layer counts as 2 moves. For some
> orientation I happened to choose, I looked at what the 3x3
> scramble was, and used CubeExplorer to find a short scramble
> to work with:
>
> D L2 F D2 U' F U' F D' U' R' B2 L' B F' U' B U'
>
> Now, due to the 3^4 context and my solution before this point,
> I needed to perform a net result of an L turn, which in
> practice means the net result of adding the twists of the
> solution (ignoring which faces are turned) must be a quarter
> turn clockwise (so L, R, U2 B', L' F' D', etc. would be valid
> solutions in this regard). Also, by being able to change the
> previous twist in the solution, I get a free turn of the front
> face in terms of movecount, which I tak e advantage of. My
> solution looks like this (knowledge of FMC techniques
> required, I can explain in more detail if anyone wants me to):
>
> (F) R' (U' D) L F' //222 block 4/4
> U' L2 U L D //223 block, 6/10
> (B' L D L D' L' U' L U L2) //Performed on inverse, solves all
> but 4 corners + 3 edges, 11/21
>
> Skeleton: (F) R' (U' D) L F' U' L2 * @ U L D ** L2 U' L' U L D
> L' D' L' B //21 moves
> Insertions:
> * =3D (F B') U' L U L' (B F') //Solves 3 edges + 1 corner, no
> moves cancel, 6/27
> @ =3D L U R' U' L' U R U' //Solves 3 corners, 2 moves cancel, 6/3=
3
>
> Solution: (F) R' (U' D) L F' U' L2 (FB') U' L U L' (BF') L U
> R' U' L' U R L D L2 U' L' U L D L' D' L' B //33 moves QSTM
>
> Bonus: insert ** =3D D' (LR') B L2 B' (RL') D L2, to solve 3
> edges, which cancels 7 moves. Never really checked the corner
> insertions on this, so it might have given a better result
> actually, but probably not.
>
> Now, I just try to adapt this to the 3^4 solve as efficiently
> as possible. My first serious attempt managed 230 twists, then
> 229. After a while I found a 228 twist solution which I nearly
> kept, but some instinct told me that I could save one more
> twist, which I managed to do.
>
> The idea is as follows. Say we need to do R U R' U' to solve
> the 3^3. The simple solution is to do R [F] R [F]' R' [F] R'
> [F]' (8), where [F] means to rotate the whole cube as in an F
> turn, and notice that if we ignore the rotations all the moves
> cancel, which is necessary. I think Ray put some info on this
> on the wiki for how this works, if you aren't familiar with
> it, try doing these moves on the 'inside' face in the most
> obvious way, with one twist in MC4D for each twist and for
> each rotation. Instead, we could do R U [UR] U' R' [UR] (6),
> where [UR] rotates the cub e around the axis though the UR
> edge and the opposite edge DL, and again all moves cancel once
> rotations are ignored. With this idea, two moves are saved
> here, and for full solutions the same idea applies but it
> becomes far more difficult to optimise. With this, my solution
> turned out to be:
>
> [F] F' M' [F]' R B' U' R2 S R' [UL] R D [DR] D' S' R U [DR] U'
> L' D' [F]' D L R U R [F]' U [UR] U' R' U R D [UFL] D' [RB] R'
> U' [DR] B
>
> where S is a slice turn which follows F, M is a slice turn
> which follows L, and [UFL] is a clockwise cube rotation about
> the UFL corner.
>
> I didn't do this on paper, I worked in MC4D, then typed up
> what my final solution was. The strangest part for me was
> trying (R z' U), which is essentially a double turn with a
> rotation in the middle, which actually saved a rotation overall.
>
> I think I'm unlikely to try beating this, so if anyone wants
> to take the record, they are welcome to it. I just hope I've
> made it very challenging to do so :), but it is certainly
> possible. I reckon sub-200 twists is very possible, so maybe
> if this isn't achieved for many years, I might try to do it
> myself ...
>
>
> These days, I'm doing a PhD is mathematics, working on network
> theory. I seen that someone new here recently is also in this
> area, so I might be able to have an interesting conversation
> with them :). I have one paper recently resubmitted to a
> journal which will hopefully be accepted soon, and another
> related paper nearly ready to be submitted. This is why I
> don't have much time to spend on cubing, but that's life.
>
> I hope as many people take part in the speedsolving contest as
> possible, it was very fun the first time so I recommend having
> a go. Even though I won't have time to practice for it, I'll
> be interesting in the results so please make them exciting!
> Good luck everyone!
>
> Happy hypercubing,
> Matt
>
>
>
>=20
--------------020900090900040706050907
Content-Type: text/html; charset=utf-8
Content-Transfer-Encoding: quoted-printable
">
I was wondering if anyone would mind that I update the record. I
suppose it's valid because we allow people to save solution states
and revert to them as needed. I guess I'll plan to update the
solution date as well unless anyone objects. BTW, a solution time
measured in years is unlikely to get a speedsolving prize unless
it's for the *slowest* solution!
-Melinda
This was just for a random scramble generated by MC4D, I never
tried to get an unusually good or bad scramble, I took the first
one generated. 200 is just a nice milestone, in my own opinion, so
there is no real significance to the 200 moves I mentioned as far
as I know.
I sent the solution to Melinda last weekend (it's taken me this
long to type up the post) but it doesn't seem to have been updated
in the HOF yet so I realise the log file isn't available to look
at.
Matt
---In g@yahoogroups.com">4D_Cubing@yahoogroups.com, rfc2396E" href=3D"mailto:ed.baumann@..."><ed.baumann@...> wrote :=
solution for a specific
scrambling but for the worst case scrambling.
under 20 you try 'under
200' for 3^4 now.
solid;PADDING-LEFT:5px;PADDING-RIGHT:0px;MARGIN-LEFT:5px;MARGIN-RIGHT:0px;"=
>
-----
title=3D"4D_Cubing@yahoogroups.com" target=3D"_blank"
href=3D"mailto:4D_Cubing@yahoogroups.com">4D_Cubing@yah=
oogroups.com
April 04, 2015 2:10
PM
3^4 shortest
solution! (227 twists)
=C2=A0
For those newer members, I used to be fairly active
here a few years back and set the 3^4 fewest moves
record with 251 twists,
amongst other things. I'm not actually returning to
being active (I probably
won't be taking part in the speedsolving contest
being talked about, much as I
would like to), I just felt like tidying up this
solve since I hadn't finished
optimising the last step before, but when I became
busy I just submitted what
I had so far. Essentially, I had ended with a parity
situation which cost
moves to fix, and I wasn't happy with that. After
going to a speedsolving
competition recently (I had a free weekend and had
fun even though I was very
out of practice) and having good luck in FMC with a
28 twist solution (which
should have been 25 twists, maybe if I had practiced
at all I would have
spotted the really obvious insertion that I missed),
I decided to correct the
ending of my 3^4 solution with something much
better, thoug h I'm still busy
these days so I could probably have found something
even better if I was
willing to spend longer on it.
I'll try to explain roughly how my
solution works, especially since the ending now
isn't very clear now that I'm
better at fewest move solving. The first 182 twists
of the solution are
unchanged from before, since I didn't really have
the time to work through a
full solve and because I always considered my
previous submission as an
unfinished solve which I wanted to finish
eventually. It is simple
blockbuilding mostly, followed by orienting the last
layer. This means that
what is left is essentially a 3^3 solve embedded
into the 3^4 puzzle, and it
is this which I revisited. As for the parity problem
I had, try doing a U2
move on one 3^3 cell, leaving the rest of the puzzle
solved, my best solution
for this is 22 twists if I remember correctly.
Optimising the last step
is weird. It's based on optimising a 3^3 solve, but
it's not quite as simple
as that. Let's look at that first though, and the
metric used (as a
simplification since the 'correct' metric is too
complicated) is quarter-slice
turn metric. A quarter turn of any layer is one
move, a half turn of any layer
counts as 2 moves. For some orientation I happened
to choose, I looked at what
the 3x3 scramble was, and used CubeExplorer to find
a short scramble to work
with:
D L2 F D2 U' F U' F D' U' R' B2 L' B F' U' B U'
Now, due
to the 3^4 context and my solution before this
point, I needed to perform a
net result of an L turn, which in practice means the
net result of adding the
twists of the solution (ignoring which faces are
turned) must be a quarter
turn clockwise (so L, R, U2 B', L' F' D', etc. would
be valid solutions in
this regard). Also, by being able to change the
previous twist in the
solution, I get a free turn of the front face in
terms of movecount, which I
tak e advantage of. My solution looks like this
(knowledge of FMC techniques
required, I can explain in more detail if anyone
wants me to):
(F) R'
(U' D) L F' //222 block 4/4
U' L2 U L D //223 block, 6/10
(B' L D L D'
L' U' L U L2) //Performed on inverse, solves all but
4 corners + 3 edges,
11/21
Skeleton: (F) R' (U' D) L F' U' L2 * @ U L D ** L2
U' L' U L D L'
D' L' B //21 moves
Insertions:
* =3D (F B') U' L U L' (B F') //Solves 3
edges + 1 corner, no moves cancel, 6/27
@ =3D L U R' U' L' U R U' //Solves 3
corners, 2 moves cancel, 6/33
Solution: (F) R' (U' D) L F' U' L2 (FB')
U' L U L' (BF') L U R' U' L' U R L D L2 U' L' U L D
L' D' L' B //33 moves
QSTM
Bonus: insert ** =3D D' (LR') B L2 B' (RL') D L2, to
solve 3 edges,
which cancels 7 moves. Never really checked the
corner insertions on this, so
it might have given a better result actually, but
probably not.
Now, I
just try to adapt this to the 3^4 solve as
efficiently as possible. My first
serious attempt managed 230 twists, then 229. After
a while I found a 228
twist solution which I nearly kept, but some
instinct told me that I could
save one more twist, which I managed to do.
The idea is as follows.
Say we need to do R U R' U' to solve the 3^3. The
simple solution is to do R
[F] R [F]' R' [F] R' [F]' (8), where [F] means to
rotate the whole cube as in
an F turn, and notice that if we ignore the
rotations all the moves cancel,
which is necessary. I think Ray put some info on
this on the wiki for how this
works, if you aren't familiar with it, try doing
these moves on the 'inside'
face in the most obvious way, with one twist in MC4D
for each twist and for
each rotation. Instead, we could do R U [UR] U' R'
[UR] (6), where [UR]
rotates the cub e around the axis though the UR edge
and the opposite edge DL,
and again all moves cancel once rotations are
ignored. With this idea, two
moves are saved here, and for full solutions the
same idea applies but it
becomes far more difficult to optimise. With this,
my solution turned out to
be:
[F] F' M' [F]' R B' U' R2 S R' [UL] R D [DR] D' S' R
U [DR] U' L'
D' [F]' D L R U R [F]' U [UR] U' R' U R D [UFL] D'
[RB] R' U' [DR]
B
where S is a slice turn which follows F, M is a
slice turn which
follows L, and [UFL] is a clockwise cube rotation
about the UFL
corner.
I didn't do this on paper, I worked in MC4D, then
typed up what
my final solution was. The strangest part for me was
trying (R z' U), which is
essentially a double turn with a rotation in the
middle, which actually saved
a rotation overall.
I think I'm unlikely to try beating this, so if
anyone wants to take the record, they are welcome to
it. I just hope I've made
it very challenging to do so :), but it is certainly
possible. I reckon
sub-200 twists is very possible, so maybe if this
isn't achieved for many
years, I might try to do it myself ...
These days, I'm doing a PhD
is mathematics, working on network theory. I seen
that someone new here
recently is also in this area, so I might be able to
have an interesting
conversation with them :). I have one paper recently
resubmitted to a journal
which will hopefully be accepted soon, and another
related paper nearly ready
to be submitted. This is why I don't have much time
to spend on cubing, but
that's life.
I hope as many people take part in the speedsolving
contest as possible, it was very fun the first time
so I recommend having a
go. Even though I won't have time to practice for
it, I'll be interesting in
the results so please make them exciting! Good luck
everyone!
Happy
hypercubing,
Matt</ p>
=20=20=20=20=20=20
--------------020900090900040706050907--
Hi everyone,
--------------020505090105070407070904
Content-Type: text/plain; charset=utf-8; format=flowed
Content-Transfer-Encoding: quoted-printable
I've updated the HOF with Matthew's new shortest solution and link to=20
his log file
like how it took 5 years to shave 24 twists! That brings it to under the=20
250 twist milestone and within sight of sub-200.
Nice going, Matthew!
On 4/4/2015 5:59 AM, damienturtle@hotmail.co.uk [4D_Cubing] wrote:
>
>
> This was just for a random scramble generated by MC4D, I never tried=20
> to get an unusually good or bad scramble, I took the first one=20
> generated. 200 is just a nice milestone, in my own opinion, so there=20
> is no real significance to the 200 moves I mentioned as far as I know.
>
> I sent the solution to Melinda last weekend (it's taken me this long=20
> to type up the post) but it doesn't seem to have been updated in the=20
> HOF yet so I realise the log file isn't available to look at.
>
> Matt
>
>
> ---In 4D_Cubing@yahoogroups.com,
>
> =EF=BB=BF
> This is not the shortest solution for a specific scrambling but for=20
> the worst case scrambling.
> If 'gods number for 3^3 got under 20 you try 'under 200' for 3^4 now.
> Correct?
> Ed
>
> ----- Original Message -----
> *From:* damienturtle@... [4D_Cubing]
>
> *To:* 4D_Cubing@yahoogroups.com
> *Sent:* Saturday, April 04, 2015 2:10 PM
> *Subject:* [MC4D] New 3^4 shortest solution! (227 twists)
>
> Hi everyone,
>
> For those newer members, I used to be fairly active here a few
> years back and set the 3^4 fewest moves record with 251
> twists, amongst other things. I'm not actually returning to
> being active (I probably won't be taking part in the
> speedsolving contest being talked about, much as I would like
> to), I just felt like tidying up this solve since I hadn't
> finished optimising the last step before, but when I became
> busy I just submitted what I had so far. Essentially, I had
> ended with a parity situation which cost moves to fix, and I
> wasn't happy with that. After going to a speedsolving
> competition recently (I had a free weekend and had fun even
> though I was very out of practice) and having good luck in FMC
> with a 28 twist solution (which should have been 25 twists,
> maybe if I had practiced at all I would have spotted the
> really obvious insertion that I missed), I decided to correct
> the ending of my 3^4 solution with something much better,
> thoug h I'm still busy these days so I could probably have
> found something even better if I was willing to spend longer
> on it.
>
> I'll try to explain roughly how my solution works, especially
> since the ending now isn't very clear now that I'm better at
> fewest move solving. The first 182 twists of the solution are
> unchanged from before, since I didn't really have the time to
> work through a full solve and because I always considered my
> previous submission as an unfinished solve which I wanted to
> finish eventually. It is simple blockbuilding mostly, followed
> by orienting the last layer. This means that what is left is
> essentially a 3^3 solve embedded into the 3^4 puzzle, and it
> is this which I revisited. As for the parity problem I had,
> try doing a U2 move on one 3^3 cell, leaving the rest of the
> puzzle solved, my best solution for this is 22 twists if I
> remember correctly.
>
> Optimising the last step is weird. It's based on optimising a
> 3^3 solve, but it's not quite as simple as that. Let's look at
> that first though, and the metric used (as a simplification
> since the 'correct' metric is too complicated) is
> quarter-slice turn metric. A quarter turn of any layer is one
> move, a half turn of any layer counts as 2 moves. For some
> orientation I happened to choose, I looked at what the 3x3
> scramble was, and used CubeExplorer to find a short scramble
> to work with:
>
> D L2 F D2 U' F U' F D' U' R' B2 L' B F' U' B U'
>
> Now, due to the 3^4 context and my solution before this point,
> I needed to perform a net result of an L turn, which in
> practice means the net result of adding the twists of the
> solution (ignoring which faces are turned) must be a quarter
> turn clockwise (so L, R, U2 B', L' F' D', etc. would be valid
> solutions in this regard). Also, by being able to change the
> previous twist in the solution, I get a free turn of the front
> face in terms of movecount, which I tak e advantage of. My
> solution looks like this (knowledge of FMC techniques
> required, I can explain in more detail if anyone wants me to):
>
> (F) R' (U' D) L F' //222 block 4/4
> U' L2 U L D //223 block, 6/10
> (B' L D L D' L' U' L U L2) //Performed on inverse, solves all
> but 4 corners + 3 edges, 11/21
>
> Skeleton: (F) R' (U' D) L F' U' L2 * @ U L D ** L2 U' L' U L D
> L' D' L' B //21 moves
> Insertions:
> * =3D (F B') U' L U L' (B F') //Solves 3 edges + 1 corner, no
> moves cancel, 6/27
> @ =3D L U R' U' L' U R U' //Solves 3 corners, 2 moves cancel, 6/3=
3
>
> Solution: (F) R' (U' D) L F' U' L2 (FB') U' L U L' (BF') L U
> R' U' L' U R L D L2 U' L' U L D L' D' L' B //33 moves QSTM
>
> Bonus: insert ** =3D D' (LR') B L2 B' (RL') D L2, to solve 3
> edges, which cancels 7 moves. Never really checked the corner
> insertions on this, so it might have given a better result
> actually, but probably not.
>
> Now, I just try to adapt this to the 3^4 solve as efficiently
> as possible. My first serious attempt managed 230 twists, then
> 229. After a while I found a 228 twist solution which I nearly
> kept, but some instinct told me that I could save one more
> twist, which I managed to do.
>
> The idea is as follows. Say we need to do R U R' U' to solve
> the 3^3. The simple solution is to do R [F] R [F]' R' [F] R'
> [F]' (8), where [F] means to rotate the whole cube as in an F
> turn, and notice that if we ignore the rotations all the moves
> cancel, which is necessary. I think Ray put some info on this
> on the wiki for how this works, if you aren't familiar with
> it, try doing these moves on the 'inside' face in the most
> obvious way, with one twist in MC4D for each twist and for
> each rotation. Instead, we could do R U [UR] U' R' [UR] (6),
> where [UR] rotates the cub e around the axis though the UR
> edge and the opposite edge DL, and again all moves cancel once
> rotations are ignored. With this idea, two moves are saved
> here, and for full solutions the same idea applies but it
> becomes far more difficult to optimise. With this, my solution
> turned out to be:
>
> [F] F' M' [F]' R B' U' R2 S R' [UL] R D [DR] D' S' R U [DR] U'
> L' D' [F]' D L R U R [F]' U [UR] U' R' U R D [UFL] D' [RB] R'
> U' [DR] B
>
> where S is a slice turn which follows F, M is a slice turn
> which follows L, and [UFL] is a clockwise cube rotation about
> the UFL corner.
>
> I didn't do this on paper, I worked in MC4D, then typed up
> what my final solution was. The strangest part for me was
> trying (R z' U), which is essentially a double turn with a
> rotation in the middle, which actually saved a rotation overall.
>
> I think I'm unlikely to try beating this, so if anyone wants
> to take the record, they are welcome to it. I just hope I've
> made it very challenging to do so :), but it is certainly
> possible. I reckon sub-200 twists is very possible, so maybe
> if this isn't achieved for many years, I might try to do it
> myself ...
>
>
> These days, I'm doing a PhD is mathematics, working on network
> theory. I seen that someone new here recently is also in this
> area, so I might be able to have an interesting conversation
> with them :). I have one paper recently resubmitted to a
> journal which will hopefully be accepted soon, and another
> related paper nearly ready to be submitted. This is why I
> don't have much time to spend on cubing, but that's life.
>
> I hope as many people take part in the speedsolving contest as
> possible, it was very fun the first time so I recommend having
> a go. Even though I won't have time to practice for it, I'll
> be interesting in the results so please make them exciting!
> Good luck everyone!
>
> Happy hypercubing,
> Matt
>
>
>
>=20
--------------020505090105070407070904
Content-Type: text/html; charset=utf-8
Content-Transfer-Encoding: quoted-printable
">
I've updated the HOF with Matthew's new shortest solution and link
to his
file. I like how it took 5 years to shave 24 twists! That
brings it to under the 250 twist milestone and within sight of
sub-200.
Nice going, Matthew!
This was just for a random scramble generated by MC4D, I never
tried to get an unusually good or bad scramble, I took the first
one generated. 200 is just a nice milestone, in my own opinion, so
there is no real significance to the 200 moves I mentioned as far
as I know.
I sent the solution to Melinda last weekend (it's taken me this
long to type up the post) but it doesn't seem to have been updated
in the HOF yet so I realise the log file isn't available to look
at.
Matt
---In g@yahoogroups.com">4D_Cubing@yahoogroups.com, rfc2396E" href=3D"mailto:ed.baumann@..."><ed.baumann@...> wrote :=
solution for a specific
scrambling but for the worst case scrambling.
under 20 you try 'under
200' for 3^4 now.
solid;PADDING-LEFT:5px;PADDING-RIGHT:0px;MARGIN-LEFT:5px;MARGIN-RIGHT:0px;"=
>
-----
title=3D"4D_Cubing@yahoogroups.com" target=3D"_blank"
href=3D"mailto:4D_Cubing@yahoogroups.com">4D_Cubing@yah=
oogroups.com
April 04, 2015 2:10
PM
3^4 shortest
solution! (227 twists)
=C2=A0
For those newer members, I used to be fairly active
here a few years back and set the 3^4 fewest moves
record with 251 twists,
amongst other things. I'm not actually returning to
being active (I probably
won't be taking part in the speedsolving contest
being talked about, much as I
would like to), I just felt like tidying up this
solve since I hadn't finished
optimising the last step before, but when I became
busy I just submitted what
I had so far. Essentially, I had ended with a parity
situation which cost
moves to fix, and I wasn't happy with that. After
going to a speedsolving
competition recently (I had a free weekend and had
fun even though I was very
out of practice) and having good luck in FMC with a
28 twist solution (which
should have been 25 twists, maybe if I had practiced
at all I would have
spotted the really obvious insertion that I missed),
I decided to correct the
ending of my 3^4 solution with something much
better, thoug h I'm still busy
these days so I could probably have found something
even better if I was
willing to spend longer on it.
I'll try to explain roughly how my
solution works, especially since the ending now
isn't very clear now that I'm
better at fewest move solving. The first 182 twists
of the solution are
unchanged from before, since I didn't really have
the time to work through a
full solve and because I always considered my
previous submission as an
unfinished solve which I wanted to finish
eventually. It is simple
blockbuilding mostly, followed by orienting the last
layer. This means that
what is left is essentially a 3^3 solve embedded
into the 3^4 puzzle, and it
is this which I revisited. As for the parity problem
I had, try doing a U2
move on one 3^3 cell, leaving the rest of the puzzle
solved, my best solution
for this is 22 twists if I remember correctly.
Optimising the last step
is weird. It's based on optimising a 3^3 solve, but
it's not quite as simple
as that. Let's look at that first though, and the
metric used (as a
simplification since the 'correct' metric is too
complicated) is quarter-slice
turn metric. A quarter turn of any layer is one
move, a half turn of any layer
counts as 2 moves. For some orientation I happened
to choose, I looked at what
the 3x3 scramble was, and used CubeExplorer to find
a short scramble to work
with:
D L2 F D2 U' F U' F D' U' R' B2 L' B F' U' B U'
Now, due
to the 3^4 context and my solution before this
point, I needed to perform a
net result of an L turn, which in practice means the
net result of adding the
twists of the solution (ignoring which faces are
turned) must be a quarter
turn clockwise (so L, R, U2 B', L' F' D', etc. would
be valid solutions in
this regard). Also, by being able to change the
previous twist in the
solution, I get a free turn of the front face in
terms of movecount, which I
tak e advantage of. My solution looks like this
(knowledge of FMC techniques
required, I can explain in more detail if anyone
wants me to):
(F) R'
(U' D) L F' //222 block 4/4
U' L2 U L D //223 block, 6/10
(B' L D L D'
L' U' L U L2) //Performed on inverse, solves all but
4 corners + 3 edges,
11/21
Skeleton: (F) R' (U' D) L F' U' L2 * @ U L D ** L2
U' L' U L D L'
D' L' B //21 moves
Insertions:
* =3D (F B') U' L U L' (B F') //Solves 3
edges + 1 corner, no moves cancel, 6/27
@ =3D L U R' U' L' U R U' //Solves 3
corners, 2 moves cancel, 6/33
Solution: (F) R' (U' D) L F' U' L2 (FB')
U' L U L' (BF') L U R' U' L' U R L D L2 U' L' U L D
L' D' L' B //33 moves
QSTM
Bonus: insert ** =3D D' (LR') B L2 B' (RL') D L2, to
solve 3 edges,
which cancels 7 moves. Never really checked the
corner insertions on this, so
it might have given a better result actually, but
probably not.
Now, I
just try to adapt this to the 3^4 solve as
efficiently as possible. My first
serious attempt managed 230 twists, then 229. After
a while I found a 228
twist solution which I nearly kept, but some
instinct told me that I could
save one more twist, which I managed to do.
The idea is as follows.
Say we need to do R U R' U' to solve the 3^3. The
simple solution is to do R
[F] R [F]' R' [F] R' [F]' (8), where [F] means to
rotate the whole cube as in
an F turn, and notice that if we ignore the
rotations all the moves cancel,
which is necessary. I think Ray put some info on
this on the wiki for how this
works, if you aren't familiar with it, try doing
these moves on the 'inside'
face in the most obvious way, with one twist in MC4D
for each twist and for
each rotation. Instead, we could do R U [UR] U' R'
[UR] (6), where [UR]
rotates the cub e around the axis though the UR edge
and the opposite edge DL,
and again all moves cancel once rotations are
ignored. With this idea, two
moves are saved here, and for full solutions the
same idea applies but it
becomes far more difficult to optimise. With this,
my solution turned out to
be:
[F] F' M' [F]' R B' U' R2 S R' [UL] R D [DR] D' S' R
U [DR] U' L'
D' [F]' D L R U R [F]' U [UR] U' R' U R D [UFL] D'
[RB] R' U' [DR]
B
where S is a slice turn which follows F, M is a
slice turn which
follows L, and [UFL] is a clockwise cube rotation
about the UFL
corner.
I didn't do this on paper, I worked in MC4D, then
typed up what
my final solution was. The strangest part for me was
trying (R z' U), which is
essentially a double turn with a rotation in the
middle, which actually saved
a rotation overall.
I think I'm unlikely to try beating this, so if
anyone wants to take the record, they are welcome to
it. I just hope I've made
it very challenging to do so :), but it is certainly
possible. I reckon
sub-200 twists is very possible, so maybe if this
isn't achieved for many
years, I might try to do it myself ...
These days, I'm doing a PhD
is mathematics, working on network theory. I seen
that someone new here
recently is also in this area, so I might be able to
have an interesting
conversation with them :). I have one paper recently
resubmitted to a journal
which will hopefully be accepted soon, and another
related paper nearly ready
to be submitted. This is why I don't have much time
to spend on cubing, but
that's life.
I hope as many people take part in the speedsolving
contest as possible, it was very fun the first time
so I recommend having a
go. Even though I won't have time to practice for
it, I'll be interesting in
the results so please make them exciting! Good luck
everyone!
Happy
hypercubing,
Matt</ p>
=20=20=20=20=20=20
--------------020505090105070407070904--