From cube-lovers-errors@mc.lcs.mit.edu Thu Sep 17 17:18:28 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil by mc.lcs.mit.edu (8.8.8/mc) with SMTP
id RAA12950; Thu, 17 Sep 1998 17:18:27 -0400 (EDT)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Date: Mon, 14 Sep 1998 14:14:22 -0400 (Eastern Daylight Time)
From: Jerry Bryan
Subject: Re: Weak Local Maxima, 6f from Start
In-Reply-To: <14Sep1998.111621.Hoey@AIC.NRL.Navy.Mil>
To: Cube Lovers
Message-Id:
> It turns out that 6f is indeed the shortest. There are two such positions
> unique to symmetry which are 6f from Start, the Pons and one other. The
> other one is quite pretty:
>
> L2 R2 D2 U2 B' F (6f*)
>
I didn't notice it originally, but this position is in the slice
subgroup, and is only one slice move from Pons. Half turns such
as L2 can be written equally well as either LL or as L'L',
so we can write (L2 R2) as (L'R)(L'R) and (D2 U2) as (D'U)(D'U).
Thus, the weak local maximum 6f from Start can be written as five
slices, one slice short of Pons.
L'R L'R D'U D'U B'F
All we would have to do to get the Pons would be to add one more
B'F slice.
----------------------
Jerry Bryan
jbryan@pstcc.cc.tn.us
From cube-lovers-errors@mc.lcs.mit.edu Thu Sep 17 20:27:13 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil by mc.lcs.mit.edu (8.8.8/mc) with SMTP
id UAA14346; Thu, 17 Sep 1998 20:27:13 -0400 (EDT)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
From: "Chris and Kori Pelley"
To: "Cube Lovers"
Subject: DOGIC solution
Date: Mon, 14 Sep 1998 18:01:56 -0400
Message-Id: <000a01bde02b$49b5aae0$da460318@CC623255-A.srst1.fl.home.com>
Using Noel Dillabough's PUZZLER program for MS Windows, I was able to verify
the basic moves needed to solve the new DOGIC puzzle.
SPOILER WARNING! If you wish to solve the puzzle yourself, read no further.
What seems fairly obvious is that the DOGIC is essentially a superset of
Impossi-Ball, which is basically the corners of MegaMinx. On DOGIC,
however, these corners have been flattened and could more properly be called
"centers." Using classic 3x3x3 techniques, you can position and orient
these pieces using the following moves:
Center 3-cycle:
(R' U L U') (R U L' U')
Center orientation (pair):
(R' D R) (F D F') U'
(F D' F') (R' D' R) U
Note that these faces refer to the large pentagons and must be "translated"
to fit the dodecahedral nature of DOGIC. R, U, and L form a horseshoe, and
F intersects all three. The D face is not really D, in fact it touches the
U face at one point.
The remaining triangular pieces turn out to be fairly trivial, and any two
can be swapped with the simple sequence:
R u R' u'
In this case, R is a large pentagon and u is any intersecting smaller
pentagon.
A general strategy would be to manually place the top "centers" followed by
their adjacent centers (if you have solved ImpossiBall this should not be
difficult). Then apply the first two moves above to complete the remaining
centers. Finally, place all the smaller triangles with the third move.
Despite having more permutations than most magic puzzles, DOGIC seems to be
fairly easy to solve.
Chris Pelley
ck1@home.com
http://www.chrisandkori.com
From cube-lovers-errors@mc.lcs.mit.edu Fri Sep 18 14:29:06 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil by mc.lcs.mit.edu (8.8.8/mc) with SMTP
id OAA20307; Fri, 18 Sep 1998 14:29:05 -0400 (EDT)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Date: Tue, 15 Sep 1998 12:21:58 -0400 (EDT)
From: der Mouse
Message-Id: <199809151621.MAA19286@Twig.Rodents.Montreal.QC.CA>
To: cube-lovers@ai.mit.edu
Subject: Two-face and three-face subgroups
I've been playing with the two-face subgroup [%] of the 3-Cube and got
to wondering - how much work has been done on the two-face and
three-face subgroups? Certainly the two-face subgroup "feels" like a
much smaller object than even the 2-Cube (though perhaps more tedious
for human solution), perhaps about the size of the Pyraminx.
[%] Okay, strictly speaking there are two different two-face subgroups,
but one of them is not even the least bit interesting.
And what about the three-face subgroups? Certainly the three- and
four-face subgroups are smaller than the whole Cube group, though ISTR
that the five-face (sub)group is actually the whole thing. But how
much smaller, and how difficult of human solution? I'd expect one of
the three-face groups (the one involving two opposite faces - call it
the L-F-R one) to be more tedious but no more difficult than the
two-face group, whereas the other one (involving one face from each
pair of opposite faces - U-F-R, say) should have more interest.
In particular, the two-face subgroup is smaller than the set of all
position that leave unchanged the cubies that the two-face subgroup
never touches. (To put it another way, I'm saying that the subgroup
generated by {R,F} is smaller than the set of positions of the full
group that leaves unmoved the 11 cubies that don't touch either of
those two faces - 7 if you don't count face cubies.) I can see a
factor of 128 smaller, since it's not possible to flip edge cubies in
the two-face group, but I haven't thought about the corners, so it may
be even smaller than that.
What about the three-face subgroups? The L-F-R subgroup is also
smaller, if for no other reason than an inability to flip edge cubies,
like the two-face group. But is the U-F-R subgroup the same as the
subset of the full group that leaves untouched the 7 (4 if you don't
count face centers) cubies in the DBL corner?
What about human solvability? I've taught myself to solve the two-face
group, and with the tools I developed (largely powers, reorientations,
and inverses of F' R' F R) I feel confident I can handle the L-F-R
three-face group or even the L-F-R-B four-face group. Can anyone
comment on how humanly difficult the U-F-R group, or for that matter
the U-F-R-L four-face group, is?
der Mouse
mouse@rodents.montreal.qc.ca
7D C8 61 52 5D E7 2D 39 4E F1 31 3E E8 B3 27 4B
From cube-lovers-errors@mc.lcs.mit.edu Tue Sep 22 16:10:10 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil by mc.lcs.mit.edu (8.8.8/mc) with SMTP
id QAA17882; Tue, 22 Sep 1998 16:10:10 -0400 (EDT)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Date: Fri, 18 Sep 1998 15:52:33 -0400 (EDT)
From: Nicholas Bodley
To: Hana Bizek
Cc: cube-lovers@ai.mit.edu
Subject: Re: Rubik's cube kingdom
In-Reply-To: <35F350B1.626F@ameritech.net>
Message-Id:
My apologies for a delayed reply.
Hana's essay was rather philosophical, and contained some uncommon
points of view; it was appropriate, in my opinion.
One aspect of the Cube (and related puzzles) that seemed to be ignored
is the remarkable ingenuity of their internal mechanisms. I maintain
that the mechanism of the original (i.e., 3^3) Rubik's Cube is one of
the most ingenious ever invented.
I recall being very fatigued, riding the West Side IRT subway in NYC
about 2 AM, perhaps, and catching sight of someone manipulating what
must have been one of the very first Cubes, probably from Hungary*. I
was fairly sure I wasn't hallucinating, but was very troubled that what
I'd seen simply appeared impossible. I've been a somewhat-casual student
of mechanisms all my life.
*This was probably several weeks, or more, before the Scientific
American article, and the later explosion of its popularity.
My regards to all,
|* Nicholas Bodley *|* Electronic Technician {*} Autodidact & Polymath
|* Waltham, Mass. *|* -----------------------------------------------
|* nbodley@tiac.net *|* The personal computer industry will have become
|* Amateur musician *|* mature when crashes become unacceptable.
--------------------------------------------------------------------------
From cube-lovers-errors@mc.lcs.mit.edu Tue Sep 22 18:09:16 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil by mc.lcs.mit.edu (8.8.8/mc) with SMTP
id SAA18211; Tue, 22 Sep 1998 18:09:15 -0400 (EDT)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Date: Thu, 17 Sep 1998 00:03:47 -0400 (EDT)
From: Jerry Bryan
Subject: More on Calculating Weak Local Maxima
To: Cube-Lovers
Message-Id:
In the process of adding the code to my God's Algorithm program to
calculate weak local maxima in the face turn metric, I realized that the
algorithm I posted previously to do so was incomplete in one subtle but
very important respect. This message will provide the missing piece to
the algorithm.
I have posted much of this before, but my program is in general jumping
ahead by more than one move at a time. For example, suppose we can store
all the positions up to five moves from Start. Then, we can determine all
the positions which are eight move s from Start by calculating all the
products xy where x is a position of length five and y is a position of
length three.
Obviously, just because the length of x is five and the length of y is
three does not mean that the length of xy is eight. In fact, the length
of xy could be anywhere from two through eight. To determine the true
length of xy, we compare xy to the stored positions of length two, three,
four, and five. In addition, we compare xy to the calculated positions of
length six and seven, which are calculated in the same manner as is xy.
If xy fails to match all such shorter positions, then its length is indeed
eight.
Next we focus on the quarter turn metric. For some fixed q in Q, the set
of twelve quarter turns, what is the length of xyq if the length of xy is
eight with the length of x equal to five and the length of y equal to
three? First of all, it must be either seven or nine. Second of all, the
length of yq must be either two or four. If the length of yq is two, then
we know that the length of xyq must be seven. But if the length of yq is
four, then we are still not sure. The reason is that there might be some
u not equal to x of length five and some v not equal to y of length three
such that xy=uv, but where the length of vq is two. If so, then the
length of xyq is the same as the length of uvq which is seven.
The basic idea is that if z=xy where the length of x is five and the
length of y is three, then there may be many, many x and y pairs of length
five and three respectively whose product yields z. The length of zq is
nine only if for every such y the length of yq is four. Even if all but
one yq is of length four, it only takes one yq of length two to spoil the
pudding, as it were.
The mechanism which I have posted previously to capture this concept is
the Ends-with function E(z). E(z) is defined to the be set of all moves
with which a minimal maneuver for z can end. So in the case at hand,
since the length of z is eight, the length of zq is nine only if E(z) does
not contain q'. E(z) can be calculated in the case at hand as the union
of E(y) taken over all the y values of length three which can be composed
with an x of length five to create z. Therefore, to say that E(z) does
not contain q' is the same thing as saying that none of the E(y) contain
q'.
So far, so good and there is nothing new here which I haven't posted
before. But let's consider the exact same issue in the face turn metric.
If the length of x is five and the length of y is three, then the length
of xy can be in the range of two through eight as before. And as before,
if we compare xy with all positions of length two through seven without
finding a match, then the length of xy is indeed eight.
But this time we need to consider xyf, where f is some fixed face turn in
the set Q+H of twelve quarter turns and six half turns. What is the
length of xyf? For starters, it is either seven or eight or nine. Also,
the length of yf is two or three or four.
If the length of yf is two, then the length of xyf is guaranteed to be
seven.
If the length of yf is three, then the length of xyf is guaranteed to be
no more than eight. But the length nevertheless might be seven, because
as in the quarter turn case, there may be some u of length five and some v
of length three such that uv=xy, but such that the length of vf is only
two. If so, the length of xyf is the same as the length of uvf which is
guaranteed to be seven.
The definition of Ends-with is the same in the face turn case as in the
quarter turn case, namely E(z) is the set of all face turns with which a
minimal maneuver for z can end. If z=xy then E(z) can be calculated as
the union of E(y) over all the y value s of length three which can be
combined with an x value of length five to form z. To say that the length
of zf is at least eight is the same thing is saying that E(z) does not
contain f' which is the same thing as saying that none of the E(y) contain
f'.
Next, let's suppose that indeed E(z) does not contain f'. We are still
left with the issue of whether the length of z is eight or nine, having
eliminated seven as a possibility. The test is still the length of all
the yf, with a length of two having been eliminated as a possibility. If
all of the yf are of length 4, then xyf is of length nine. But if even so
many as one of the yf are of length three, then xyf is of length 8.
The mechanism I have posted before to capture this concept is the
Ends-with2 function. E2(z) is a little tricky to describe. Informally,
we might say that E2(z) is the set of all f in Q+H with which z can end
without changing it's length. It is probably better to say that E2(z) is
the set of all f in Q+H such that the length of zf' is the same as the
length of z. The technique which I have posted before (and which I must
now correct) to calculate E2(z) is to form the union of E2(y) over all y
values of length three which can be combined with an x value of length
five to form z.
If the length of zf is eight or nine, then this mechanism is fine. But if
the length of zf turns out to be seven, there is a problem. That is,
there may be one y where the length of yf is two and where E(y) contains
f, and there may be another y where the length of yf is is three and where
E2(y) contains f. In such a case, both E(z) and E2(z) would contain f.
Hence, we must always calculate E(z) prior to calculating E2(z), and we
must omit from E2(z) any f values which are already contained in E(z).
With this correction, everything works. A local maximum is a position z
for which |E(z)|+|E2(z)|=18, a strong local maximum is a local maximum z
for which |E(z)|=18 and |E2(z)|=0, and a weak local maximum is a local
maximum z for which |E(z)| < 18 and |E 2(z)| > 0. All my examples have
been specific to y values of length 3 for clarity of exposition, but the
calculation of E(z) and E2(z) is totally general, and is the union of E(y)
and E2(y), respectively, over all y values which can be used to form a z
of the form z=xy, and with any values which are in E(z) omitted from
E2(z).
Finally, my programs also calculate a Starts-with and a Starts-with2
function, which are defined analogously. The same correction must be made
to the Starts-with2 function as was made for the Ends-with2 function.
Equivalently, we can define S(z)=E'(z') and S2(z)=E2'(z'), where E' and
E2' are the set of all inverses of the elements of E and E2, respectively.
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan) jbryan@pstcc.cc.tn.us
Pellissippi State (423) 539-7198
10915 Hardin Valley Road (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990
From cube-lovers-errors@mc.lcs.mit.edu Tue Sep 22 19:05:00 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil by mc.lcs.mit.edu (8.8.8/mc) with SMTP
id TAA18829; Tue, 22 Sep 1998 19:05:00 -0400 (EDT)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Date: Sat, 19 Sep 1998 09:13:58 -0400 (Eastern Daylight Time)
From: Jerry Bryan
Subject: Re: Two-face and three-face subgroups
In-Reply-To: <199809151621.MAA19286@Twig.Rodents.Montreal.QC.CA>
To: der Mouse
Cc: Cube Lovers
Message-Id:
On Tue, 15 Sep 1998 12:21:58 -0400 (EDT) der Mouse
wrote:
> I've been playing with the two-face subgroup [%] of the 3-Cube and got
> to wondering - how much work has been done on the two-face and
> three-face subgroups? Certainly the two-face subgroup "feels" like a
> much smaller object than even the 2-Cube (though perhaps more tedious
> for human solution), perhaps about the size of the Pyraminx.
>
The __ subgroup has been explored fairly thoroughly. For
example, look in the archives 8/31/1994 for a summary of the
first complete God's Algorithm search of this particular
subgroup. There are a number of articles in the archives
thereafter. ____ has been searched in both the quarter turn
metric and the face turn metric, and local maxima have been
investigated as a part of the search.
____ has a very small branching factor and a corresponding
large diameter of 25 in the quarter turn metric, at least I
think it's a large diameter for such a small group. Until Mike
Reid recently showed that the diameter of G in the quarter turn
metric was at least 26, the diameter of ____ was the largest
known for the 3x3x3 cube or any of its subgroups.
Frey and Singmaster's book discusses both two face and three
face subgroups, among other things giving their sizes.
To my knowledge, no God's Algorithm searches have been performed
for the three face subgroups.
We have |____|=73483200, so ____ is slightly smaller than the
corners group at 88179840. The 2x2x2 is 24 times smaller than
the corners group, at 3674160. However, I am of the school
of thought that tends not to equate the size of the group (or
search space, for problems that are not actually groups) with
the difficulty of the problem.
----------------------
Jerry Bryan
jbryan@pstcc.cc.tn.us
From cube-lovers-errors@mc.lcs.mit.edu Tue Sep 22 19:50:07 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil by mc.lcs.mit.edu (8.8.8/mc) with SMTP
id TAA19137; Tue, 22 Sep 1998 19:50:02 -0400 (EDT)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Date: Mon, 21 Sep 1998 16:34:29 -0400 (Eastern Daylight Time)
From: Jerry Bryan
Subject: Local Maxima which Fix the Corners, 12q from Start
To: Cube Lovers
Message-Id:
I am making a run to calculate God's Algorithm out to 12 moves
from Start in the quarter turn metric. It has been running
several weeks, and will probably run several more.
I have made some changes to my program to make it easier to
extract the positions for local maxima, and to checkpoint the
local maxima data. As a part of the checkpointing, I can
actually see the local maxima as they are generated without
having to wait for the program to end.
It is becoming apparent that there are a *lot* of local maxima
12q from Start. It is already known that there are only four
(unique to symmetry) which are 10q from Start (the shortest ones
in the quarter turn metric), and that there are none 11q from
Start. So I am a little surprised that I am seeing so many.
I have looked at quite a few of them, and most of them are not
all that interesting. But the ones which fix the corners are
all quite pretty. Because the positions are being produced in
lexicographic order, and because I am sorting by corners first,
edges second, the positions which fix the corners are the first
ones to appear. There are eight of them as follows.
1. F2 L2 F2 B2 R2 B2
2. F B' U2 D2 F' B R2 L2
3. F B R2 F' B' U D L2 U' D'
4. D' F B' R F R' F' B U F' U' D
5. F B R2 L2 F B U2 D2
6. R L' F2 B2 R L' F2 B2
7. F2 B2 U2 D2 R2 L2
8. R L' U D' F B' R2 L2 U D'
#1 is a 2-H pattern (only four edge cubies are moved).
#2 is a 4-H.
#3 moves four edge cubies, leaving eight of the nine
facelets the same color on four faces, and a solid color on the
other two faces.
#4 moves three edge cubies, leaving eight of the nine facelets
the same color on all six faces.
#5 has 2 H's, 2 checkerboards, and 2 solid faces -- with the
respective H's, checkerboards, and solid faces opposing each
other.
#6 has 4 H's and 2 checkerboards, with the 2 checkerboards
opposing each other.
#7 is the Pons Asinorum, and is included only for completeness
because we already knew that the Pons was a local maximum of
length 12q.
#8 has all six faces being sort of a "three colored
checkerboard".
Some of these positions may have appeared on Cube-Lovers in some
other context, but the only one I recognize for sure is the Pons.
In some ways, #4 is the most interesting to me, because it a
simple 3-cycle on the edges, and who would have thought that
such a position would turn out to be a local maximum? #1 and #3
both consist of two 2-cycles on the edges, and are about as
striking to me as is #4.
----------------------
Jerry Bryan
jbryan@pstcc.cc.tn.us
From cube-lovers-errors@mc.lcs.mit.edu Wed Sep 23 12:37:03 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil by mc.lcs.mit.edu (8.8.8/mc) with SMTP
id MAA22987; Wed, 23 Sep 1998 12:37:03 -0400 (EDT)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Message-Id: <19980922204717.20206.rocketmail@web1.rocketmail.com>
Date: Tue, 22 Sep 1998 13:47:17 -0700 (PDT)
From: "Jorge E. Jaramillo"
Subject: Moves to this pattern
To: Cube-Lovers@ai.mit.edu
Hi I just joined the list!
I was wondering if someone could help me with a
pattern that has been bugging me for a while.
I have been able to solve the cube to this pattern so
i know it is a valid one.
I used one of those on-line solvers entered the
pattern and then reversed the sequence the solver
gave me and it indeed gets the pattern i want but
somehow it seemed too many moves for me for such a
simple pattern.
the pattern I am talking about is: the 4 cubelets
that make the vertex formed by FDR are exchanged with
the vertex from BDL
Can any one give me the set of moves to get to this
pattern from a solved cube? Does this pattern have a
name?
[Here is a] set of moves (they work but I am
sure there is a shorter way):
D2 F B- L2 F- B D- F B- L F- B D2 F B- L2 F- B D F B-
L- F- B F B- L- F- B D2 F B- L- F- B D2 R- D- R D- R-
D2 R D2 L- D- L B D- B- L- D L2 D L- D- F- D- F D B-
D- B D R D R- D F- B D B- F R D- R- T B- T- B- D- B-
This long set of moves reminds me of something else:
There are many Rubik cube annimations that you move
the faces with either the keypad or the mouse. Does
anyone know of one that follows sets of orders you
write?
It would be neat to try this long patterns in one of
those simulators.
Thanks
===
Jorge E Jaramillo
From cube-lovers-errors@mc.lcs.mit.edu Wed Sep 23 19:54:38 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil by mc.lcs.mit.edu (8.8.8/mc) with SMTP
id TAA26229; Wed, 23 Sep 1998 19:54:38 -0400 (EDT)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Message-Id: <19980922213222.29513.rocketmail@web1.rocketmail.com>
Date: Tue, 22 Sep 1998 14:32:21 -0700 (PDT)
From: "Jorge E. Jaramillo"
Subject: Is it only mine?
To: cube
In order to solve the cube faster I developed a
method that would turn a lot the middle faces. I
don't know how you call them in this list I am
talking about the face between Top and Bottom, the
face between Right and Left and even the face between
Front and Back.
Well after twisting it a few times my cube came
undone, fell apart and I thought "Damn made in Taiwan
cubes" (although the ones I can buy here do not say
where are they made). I was going through my second
cube in a short while (I just regained interest in
the cube a short while ago after say 15 years) and it
fell apart again. I took off the plastic color of the
center cubelet that came apart and found a screw and
a spring that keeps the screw tight. I re screwed it
but did not have any glue at hand so I kept on
playing without the color of the center cubelet I was
doing one of these center face moves and saw how the
screw was turning counterclockwise in other words the
way that makes the cube fall apart. Needless to say I
had to re design my method.
This long story is to ask if all the cubes are built
this way or only the ordinary ones I can buy here?
Thanks.
===
Jorge E Jaramillo
From cube-lovers-errors@mc.lcs.mit.edu Fri Sep 25 18:05:43 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil by mc.lcs.mit.edu (8.8.8/mc) with SMTP
id SAA07121; Fri, 25 Sep 1998 18:05:42 -0400 (EDT)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Message-Id: <360964EA.D07A79F8@t-online.de>
Date: Wed, 23 Sep 1998 23:15:22 +0200
Reply-To: Rainer.adS.BERA_GmbH@t-online.de
Organization: BERA Softwaretechnik GmbH
To: "Jorge E. Jaramillo"
Cc: Cube-Lovers@ai.mit.edu
Subject: Re: Moves to this pattern
References: <19980922204717.20206.rocketmail@web1.rocketmail.com>
From: Rainer.adS.BERA_GmbH@t-online.de (Rainer aus dem Spring)
Jorge E. Jaramillo wrote:
> the pattern I am talking about is: the 4 cubelets
> that make the vertex formed by FDR are exchanged with
> the vertex from BDL.
R2 U R2 U2 B2 D L2 U2 L2 B2 D' B2 U2
Rainer adS
PS
If you have a WINTEL system you should download Herbert Kociemba's cube
program.
[ Moderator's note: Jerry Bryan provides another 13f process,
R2 U' B2 R2 U2 R2 U' B2 U2 R2 U' B2 U2, which hasn't been proven optimal.
Steve LoBasso has a longer one. ]
From cube-lovers-errors@mc.lcs.mit.edu Sat Sep 26 00:05:04 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil by mc.lcs.mit.edu (8.8.8/mc) with SMTP
id AAA07745; Sat, 26 Sep 1998 00:05:03 -0400 (EDT)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Date: Fri, 25 Sep 1998 08:36:22 -0400 (Eastern Daylight Time)
From: Jerry Bryan
Subject: Summary of Local Maxima, Face Turn Metric
To: Cube Lovers
Message-Id:
I have posted maneuvers for a number of specific strong and weak
local maximal positions (the strong local maxima at 9f and 10f,
and the weak local maxima at 6f) , but I haven't really posted a
summary of the numbers. Here are the numbers I have so far. In
order to complete the table through 10f from Start for weak
local maxima, I would have to repeat a rather long run. As
might be expected, it appears that the number of weak local
maxima will greatly exceed the number of strong local maxima.
As is the usual case, patterns are M-conjugacy classes (symmetry
classes), and represent the number of positions which are unique
up to symmetry.
Distance Strong Strong Weak Weak
from Lclmax Lclmax Lclmax Lclmax
Start Patterns Positions Patterns Positions
0f 0 0 0 0
1f 0 0 0 0
2f 0 0 0 0
3f 0 0 0 0
4f 0 0 0 0
5f 0 0 0 0
6f 0 0 2 7
7f 0 0 1 6
8f 0 0 37 739
9f 2 32 327 13014
10f 6 107
----------------------
Jerry Bryan
jbryan@pstcc.cc.tn.us
From cube-lovers-errors@mc.lcs.mit.edu Tue Oct 6 15:05:42 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil by mc.lcs.mit.edu (8.8.8/mc) with SMTP
id PAA07299; Tue, 6 Oct 1998 15:05:41 -0400 (EDT)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Date: Sun, 27 Sep 1998 20:35:33 -0400 (EDT)
From: Jerry Bryan
Subject: Corners Only, Ignoring Twist
To: Cube-Lovers
Reply-To: Jerry Bryan
Message-Id:
I have been playing around with the idea of trying to calculate God's
Algorithm all the way to the bitter end for the group which results from
ignoring all twists of the corners and flips of the edges. It's a pretty
big group. The order is |G|/(3^7)/(2^11), which is about 9.7*10^12, call
it about 10^13 to make it a round number. (Another way to calculate it is
8!12!/2.)
This is probably right at the bare edge, maybe even slightly past the bare
edge, of the size of problem I can handle right now, which makes it a
worthy endeavor. Also, it would provide a lower limit on the diameter of
G (although the limit might not be any better than the ones we already
have), which again makes it a worthy endeavor.
Such a result might be suitable as the estimator function required by IDA*
searches. The distance from Start in the no-twist, no-flip group would
certainly be a lower bound for every position where twist and flip *are*
considered. My only hesitation about suggesting this group as a suitable
IDA* estimator function is that there are obvious pathological cases such
as the superflip where this function would be a very poor estimator.
In developing a no-twist, no-flip version of the program, I decided to try
it out on the corners only case. Here are the results.
Distance
from Patterns Positions
Start
0q 1 1
1q 1 12
2q 5 114
3q 24 876
4q 119 4931
5q 301 12972
6q 364 15066
7q 166 6300
8q 3 48
Distance
from Patterns Positions
Start
0f 1 1
1f 2 18
2f 9 243
3f 68 2646
4f 302 12516
5f 418 17624
6f 170 7080
7f 14 192
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan) jbryan@pstcc.cc.tn.us
Pellissippi State (423) 539-7198
10915 Hardin Valley Road (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990
From cube-lovers-errors@mc.lcs.mit.edu Tue Oct 6 16:52:26 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil by mc.lcs.mit.edu (8.8.8/mc) with SMTP
id QAA08831; Tue, 6 Oct 1998 16:52:24 -0400 (EDT)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Date: Fri, 2 Oct 1998 23:18:37 +0200 (METDST)
From: Martin Moller Pedersen
Message-Id: <199810022118.XAA19053@stargazer.daimi.aau.dk>
To: cube-lovers@ai.mit.edu
Subject: cubes at spielmessen in Essen
There will soon be a big gathering for games in Germany - Essen a so-called
Spielmessen.
I am attending this gathering for the first time in three years so I am
looking for companies who will came to the spielmessen and who sells cubes.
The places is big so it would be nice for me to have same names to look for.
and hopefully I will have a real 4x4x4 and 5x5x5 cube to play with in a few weeks :-)
/Martin
From cube-lovers-errors@mc.lcs.mit.edu Tue Oct 6 18:38:32 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil by mc.lcs.mit.edu (8.8.8/mc) with SMTP
id SAA10776; Tue, 6 Oct 1998 18:38:31 -0400 (EDT)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Message-Id:
Date: Sun, 4 Oct 1998 11:12:27 -0600
To: cube-lovers@ai.mit.edu
From: Steve LoBasso
Subject: Cube Solver for Macintosh
I just wrote a Macintosh port to Dik T. Winter's cube solving code.
I put it on my web page at the link below. I haven't had a chance to make
an info page for it so the link below is just the application.
It will run on both 68k and PPC Native.
Be warned the 68k version runs fairly slow and the initialization phase
takes quite a while.
Cube Solver
By Steve LoBasso
slobasso@dtint.com
Written using algorithm code by
Dik T. Winter based on algorithm described by Herbert Kociemba.
http://members.tripod.com/~slobasso/downloads/Cube_Solver.hqx
--
Steve LoBasso mailto:slobasso@dtint.com
Digital Technology International or mailto:slobasso@hotmail.com
500 West 1200 South, Orem, UT, 84058 http://members.tripod.com/~slobasso
(801)226-6142 ext.265 FAX (801)221-9254
From cube-lovers-errors@mc.lcs.mit.edu Tue Oct 6 20:00:34 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil by mc.lcs.mit.edu (8.8.8/mc) with SMTP
id UAA12380; Tue, 6 Oct 1998 20:00:34 -0400 (EDT)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
From: "Chris and Kori Pelley"
To:
Subject: That's Incredible!
Date: Sun, 4 Oct 1998 22:37:28 -0400
Message-Id: <002701bdf009$180cb5e0$da460318@CC623255-A.srst1.fl.home.com>
I recently obtained (courtesy of Peter Beck) the Rubik's Cube-a-Thon video
from the TV show "THAT'S INCREDIBLE" and digitized it in RealVideo format.
The file is rather large (18.1 megabytes) but it's worth a download if
you're into cubic nostalgia. Eleven and a half minutes long, it features
Minh Thai, Jeff Varasano, Kris Wunderlich, and others that may be on this
list.
Here's the URL:
http://www.chrisandkori.com/incredible.htm
It requires the RealPlayer 5.0 or later to view it. Note that you must
download the file, then view it. I do not have a streaming video server.
If anyone would like to host the file on a streaming server, please contact
me.
Chris Pelley
ck1@home.com
From cube-lovers-errors@mc.lcs.mit.edu Thu Oct 8 19:04:05 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil by mc.lcs.mit.edu (8.8.8/mc) with SMTP
id TAA24808; Thu, 8 Oct 1998 19:04:04 -0400 (EDT)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Date: Thu, 8 Oct 1998 15:11:54 +0100
From: David Singmaster
To: cube-lovers@ai.mit.edu
Message-Id: <009CD656.72A81016.35@ice.sbu.ac.uk>
Subject: Davenport's pattern
The pattern given by Jacob Davenport is what I called a cube in a cube
in a cube. I discovered this in 1979 or 1980 and was very pleased with it.
Indeed, I used the cube in a cube as the logo of the late and much lamented
David Singmaster Ltd. in 1980-1982 (approx. dates since I'm not where my
records are). The pattern is in my Notes. There
are various ways to generate the pattern, but the one that I can remember uses
what Roger Penrose called the Y-commutator, which has the form FR'F'R. The
reason this is the Y-commutator is that it affect the three edges adjacent to a
corner and the corner and its three adjacent corners. I.e. the affected pieces
form a Y, while the pieces affected by the ordinary commutator FRF'R' form a
Z. Combining three Y-commutators as follows: FR'F'R RU'R'U UF'U'F gives a
process that twists the corner and the three adjacent edges as a unit and
twists an adjacent corner the opposite direction. NOTE - I'm doing this from
memory and I have a suspicion that the middle group may need to be inverted??
By moving the odd corner to the right place adjacent to the opposite
corner and applying the inverse of the above, one gets the same sort of pattern
at the opposite corner and the odd corner has been restored. Now one 3-cycles
the centers, as is easily done by a commutator of slice moves, and one has the
cube in a cube. Now one can twist the two opposite corners to get the cube in
a cube in a cube, though I find this not as visually dramatic as the cube in a
cube.
Someone - Mike Reid ? - sent me a minimal method for one of these
patterns, but it's not very memorable.
DAVID SINGMASTER, Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411; fax: 0171-815 7499;
email: zingmast or David.Singmaster @sbu.ac.uk
From cube-lovers-errors@mc.lcs.mit.edu Thu Oct 8 19:47:11 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil by mc.lcs.mit.edu (8.8.8/mc) with SMTP
id TAA24896; Thu, 8 Oct 1998 19:47:10 -0400 (EDT)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Date: Thu, 8 Oct 1998 16:45:31 +0100
From: David Singmaster
To: cube-lovers@ai.mit.edu
Message-Id: <009CD663.86DEF2D6.16@ice.sbu.ac.uk>
Subject: Nicholas Bodley's message of 22 Sep 1998.
Nicholas Bodley's message reminds me of when I wrote about the Cube in
1978 or early 1979, I think in the Observer, which seems to have been the first
article outside Hungary. I mentioned that the mechanical problem seemed even
harder than the mathematical problem and this led to about six submissions of
mechanisms from readers. All but one of these were clearly impossible, but the
last was Rubik's mechanism with slight differences - e.g. he had the undersides
of the centers rounded. The submitter of this was a UK patent agent with
obvious mechanical aptitude.
However, one of my students, who had bought a cube from me, told me
that a friend rang her up and asked if she had seen the hoax article about a
cube that moved in all directions. The friend had just proven that such an
object was impossible. My student had to disabuse her.
When the cubes first came out of Hungary, we didn't know what the
mechanism was and they were too precious to fiddle with. Roger Penrose said he
had one face center piece come undone and he carefully wrapped thread around
the exposed part of the screw and worked the screw into place and pulled on the
thread to screw the screw back into the central piece. Sometime in late 1978,
a friend had trouble with his cube and took a screwdriver to it and discovered
the cover plates and the screw heads inside!
Enough for now.
DAVID SINGMASTER, Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411; fax: 0171-815 7499;
email: zingmast or David.Singmaster @sbu.ac.uk
From cube-lovers-errors@mc.lcs.mit.edu Fri Oct 9 18:40:38 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil by mc.lcs.mit.edu (8.8.8/mc) with SMTP
id SAA02313; Fri, 9 Oct 1998 18:40:37 -0400 (EDT)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Message-Id: <361E8D3F.6597@ameritech.net>
Date: Fri, 09 Oct 1998 17:25:03 -0500
From: Hana Bizek
Reply-To: hbizek@ameritech.net
To: cube-lovers@ai.mit.edu
Subject: comments on "davenport's pattern"
David Singmaster claims to have discovered this pattern in 1979 or
1980, so it should be credited to him. In my own book I present a
number of patterns, but I would never dare to claim authorship to any
of them. Singmaster's comments prompted me to look at my own books. In
CUBE GAMES (Taylor and Rylands} this pattern appears on the top of page
37.I have a strong suspicion this pattern could be a combination of the
two cyclicity-three patterns on page 36 therein. One may use this
pattern as a corner in a 3-color design.
Design-construction is a step beyond pattern-construction. My question
has not yet been satisfactorily answered. Has anyone seen construction
of 3-dimensional "sculpture-like" designs? People referred me to
Davenport's creations, but my own designs are quite different.
Hana
From cube-lovers-errors@mc.lcs.mit.edu Fri Oct 30 13:56:08 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil by mc.lcs.mit.edu (8.8.8/mc) with SMTP
id NAA17979; Fri, 30 Oct 1998 13:56:07 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Mail-from: From cube-lovers-request@life.ai.mit.edu Thu Oct 15 11:02:13 1998
Message-Id: <19981015145704.8314.rocketmail@attach1.rocketmail.com>
Date: Thu, 15 Oct 1998 07:57:04 -0700 (PDT)
From: "Jorge E. Jaramillo"
Subject: Moves to this other pattern
To: cube
Please if this is not one of the purposes of this
list someone let me know I don't meant to be rude.
Could someone please tell me the moves to get from a
solved cube to the following pattern:
The top and bottom faces keep their colors.
The 4 columns in the middle of every side face stay
with their color.
The left column on the front face moves to the right
and the right column moves to the left.
The left column on the back face moves to the right
and the right column moves to the left.
Thanks
===
Jorge E Jaramillo
[ Moderator's note: We have a lot of requests for processes for various
and have got a lot of optimal processes. Maybe the hard part is
figuring out how to look them up. This is one of the "4-" patterns,
and probably appears among the quasi-continuous partial isoglyphs. --Dan ]
From cube-lovers-errors@mc.lcs.mit.edu Fri Oct 30 14:17:00 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil by mc.lcs.mit.edu (8.8.8/mc) with SMTP
id OAA18032; Fri, 30 Oct 1998 14:16:59 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Date: Mon, 26 Oct 1998 23:58:27 -0400 (EDT)
From: Jerry Bryan
Subject: 12q From Start
To: Cube-Lovers
Message-Id:
|x| Patterns Lcl Positions Lcl Branching
Max Max Factor
0q 1 0 1 0
1q 1 0 12 0 12
2q 5 0 114 0 9.5
3q 25 0 1068 0 9.368
4q 219 0 10011 0 9.374
5q 1978 0 93840 0 9.374
6q 18395 0 878880 0 9.366
7q 171529 0 8221632 0 9.355
8q 1601725 0 76843595 0 9.347
9q 14956266 0 717789576 0 9.341
10q 139629194 4 6701836858 42 9.337
11q 1303138445 0 62549615248 0 9.333
12q 12157779067 103 583570100997 2913 9.330
The last time a new level was calculated for the quarter turn metric was 4
February 1995.
The cumulative number of positions now identified is 653625391832, or
about 6.5*10^11. This is well past the "geometric halfway point" of
sqrt(|G|), which is about 6.5*10^9. However, it is known that the
diameter of G is at least 26q, strongly indicating that there is a bit of
a tail to the distribution of positions by length.
Of the 103 local maxima of length 12q, 70 of them also have their inverse
as local maxima. For the other 33, the inverse is not a local maximum.
For one of them, the inverse has 11 moves which go closer to Start. For
seven of them, the inverse has 10 moves which go closer to Start. For
eleven of them, the inverse has 8 moves which go closer to Start. For six
of them, the inverse has 6 moves which go closer to Start. For two of
them, the inverse has 4 moves which go closer to Start. And for six of
them, the inverse has only 2 moves which go closer to Start.
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan) jbryan@pstcc.cc.tn.us
From cube-lovers-errors@mc.lcs.mit.edu Mon Nov 2 09:40:50 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil by mc.lcs.mit.edu (8.8.8/mc) with SMTP
id JAA29402; Mon, 2 Nov 1998 09:40:48 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Message-Id: <363B1799.3534@hrz1.hrz.tu-darmstadt.de>
Date: Sat, 31 Oct 1998 14:58:49 +0100
From: Herbert Kociemba
Reply-To: kociemba@hrz1.hrz.tu-darmstadt.de
To: cube-lovers@ai.mit.edu
Subject: Unauthorized Use of RUBIK'S CUBE and CUBE Design Marks?
About 2 weeks ago I received the following message and it seems to me
that it might be interesting for you too:
> Subject:
> Use of RUBIK'S mark
> Date:
> Sun, 18 Oct 1998 21:05:31 EDT
> From:
> CK4IPLAW@aol.com
> To:
> kociemba@hrz1.hrz.tu-darmstadt.de
>
>
> CLEARY, KOMEN & LEWIS, LLP
> 600 Pennsylvania, Avenue, S.E.
> Suite 200
> Washington, D.C. 20003-4316
> Telephone: 202 675-4700
> Telecopier: 202 675-4716
> E-Mail: ck4iplaw@aol.com
>
>
> October 18, 1998
>
> Via Electronic Mail
>
> Herbert Kociemba
> kociemba@hrz1.hrz.th-darmstadt.de
>
> Re: Unauthorized Use of RUBIK'S CUBE and CUBE Design Marks
>
> Dear Mr. Kociemba:
>
> This firm is intellectual property counsel to Seven Towns Limited ("Seven
> Towns"), the manufacturer and worldwide distributor of the RUBIK'S CUBE three-
> dimensional puzzle and provider of an electronic version of the puzzle via its
> official web site, which is located at http://rubiks.com.
>
> The RUBIK'S CUBE mark is famous throughout the world. The distinctive
> overall appearance of the RUBIK'S CUBE puzzle also is a famous trademark owned
> by Seven Towns. These marks are registered or are the subject of pending
> trademark applications in most of the major countries of the world.
>
> It has come to our attention that your web site features a program under the
> name of Rubik's Cube Explorer. I must advise that your unauthorized use of
> the RUBIK'S CUBE mark owned by Seven Towns constitutes trademark infringement.
> Specifically, the use of this mark in designating the origin of your program
> confuses the public into believing mistakenly that it derives from, is
> associated with, or is endorsed or sponsored by the owner of this commercial
> symbol (i.e., Seven Towns). Moreover, apart from causing consumer confusion,
> your use of the well-known mark dilutes its distinctive value in violation of
> the federal and state anti-dilution laws.
>
> Seven Towns appreciates your interest in the RUBIK'S CUBE puzzle, and it
> certainly does not wish to inhibit legitimate discussion of the puzzle on the
> Internet or in any other medium. However, it also must be vigilant in
> maintaining the value and integrity of its intellectual property. It cannot
> afford to lose control over its commercial reputation, or damage to its
> substantial goodwill, by permitting another party to use its trademarks or
> trade dress in a manner that causes source confusion or otherwise dilutes
> their selling power. Thus, Seven Towns requests that you remove from your web
> site the electronic version of the RUBIK'S CUBE manipulative puzzle, and that
> you discontinue any further use of the term RUBIK'S CUBE or any similar
> designation in the prominent, source-indicating manner of a trademark.
>
> I hope that you are understanding of our client's position, and I thank you
> in advance on behalf of Seven Towns for your prompt attention to this matter.
>
> Sincerely yours,
>
> //sjm//
>
> Scott J. Major
Indeed the headline of my homepage at
http://home.t-online.de/home/kociemba/cube.htm
was "Rubik's Cube Explorer 1.5". So I removed the word "Rubik's" and
added some note at the bottom of the page (...blah blah is not derived
from, is not associated with blah blah...).
I definitely will not remove the program from my homepage. This seems
ridiculous. I know, that other cube fans received similar mail because
they have some similar statements on their homepages now.
What do you think about that?
Herbert Kociemba
From cube-lovers-errors@mc.lcs.mit.edu Mon Nov 2 14:41:00 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil by mc.lcs.mit.edu (8.8.8/mc) with SMTP
id OAA00366; Mon, 2 Nov 1998 14:41:00 -0500 (EST)
Precedence: bulk
Date: Sat, 31 Oct 1998 23:06:34 -0400 (EDT)
From: Jerry Bryan
Subject: All the Local Maxima at 12q
To: Cube-Lovers
Message-Id:
I had originally decided not to post all the 12q local maxima because
there are so many and because not all of them look all that interesting.
But I have been looking at them a bit more, and I think it's worth the
effort. Some of them will be very familiar and some of them not. I would
highlight the following ones.
#9 is a partial isoglyph -- four short T's (or four short U's).
#57 is the well-known four spot.
#68 is one of the more striking of several pseudo two spots.
#80 is worthy of some study. It's the only one where the maximality
of the inverse is 11 -- almost but not quite a local maximum.
#97 along with #98 and #99 are very striking pseudo six spots. #97
and #99 are almost the same pattern, and it took me a while to
see how they differ.
In the table below, the right most column gives the maximality of the
inverse of the pattern, where the maximality is the number of moves which
go closer to Start. The maximality of the inverse gives an indication of
how close the inverse comes to being a local maximum, with 12 indicating a
local maximum. (The maximality of the pattern itself is not given, since
it is 12 by definition.) The table gives the 103 local maxima of length
12q which are unique up to symmetry.
1. F F L L F F B B R R B B 12
2. F B' U U D D F' B R R L L 12
3. F B R R F' B' U D L L U' D' 12
4. U' R' L B L B' R L' D L' U D' 12
5. F B R R L L F B U U D D 12
6. R L' F F B B R L' F F B B 12
7. F F B B U U D D R R L L 12
8. R L' U' D F' B R R L L U' D 12
9. F B U U D D F' B R R L L 12
10. U B' D F D' B' D F' D' B B U' 12
11. L B B R D D R' L B B L L 12
12. U R' U L' U' R' U L U' R R U' 12
13. F F U' D R R U D D B B D' 12
14. F B D D F B' L L U U F F 12
15. U D' F U D R' L' U D F U D' 12
16. U U R L F' B' U D' R R L L 10
17. R R U D F' B U' D F B' U' D 8
18. U U D D F B' R R L L B B 12
19. R' L F' B U U F F B B U' D' 12
20. U D' F F U' D R L F B' U' D 12
21. F B R' L D D F B' R' L U' D 12
22. L L U D' B B R L U' D B B 12
23. F L L F' D D F B' L L B B 12
24. F' R R F' U U F' B L L B B 12
25. F B U U L L B B L L U U 8
26. U' D' R L' F F B B L L U' D 4
27. U D R L' F F B B L L U D' 4
28. F B' U U B B L L B B U U 12
29. F' B U D L L B B R R U D 12
30. F' B U D R R F F L L U D 12
31. F B' U U F F R R F F U U 12
32. F B' U U L L B B L L U U 12
33. F B' U U R R F F R R U U 12
34. L L U D L L F' B R L U U 10
35. F' R R F U U F B' R R B B 12
36. L L F' B D D R R F F B B 12
37. L L F' B D D F F B B L L 12
38. D D F' B U D' F F B B U D' 10
39. F F R' L F B' U' D F F L L 12
40. U R' L F U D' R' L F F B B 8
41. U D R L F B' U D' R' L F F 12
42. U D F F U' D R L F B' U' D 12
43. U D B B U' D R' L' F B' U' D 12
44. F F R L' F B' U D' F F L L 12
45. U D F' B R L U' D B B U' D 12
46. U D F' B R' L' U' D F F U' D 12
47. F B' D R' L' U U R L D B B 8
48. B' U' D R' L B' U' D R R L L 6
49. B' U D' R L' F U D' R R L L 6
50. U D F' B' U' D L L F B' U' D' 8
51. U' F' U' D F B' U' D R R L L 2
52. U' F F B R' L U' D F F B B 8
53. D R R L F' B U' D R R L L 2
54. R U' F' B R L' U D' F F U D' 2
55. L F F B R L' F B' R R L L 2
56. L F F B U D' R' L F F B B 8
57. F B' U U D D F B' R R L L 12
58. F B' R L' U' D F' B' R L' U' D' 12
59. F B R L U D' F B R' L' U D' 12
60. F B' U U D D F' B U U D D 12
61. R L' U' D F' B L L U' D F F 12
62. R L B B U D' R R F B U D' 12
63. F B' U D' R L' F F B B U D' 12
64. F B R R L L U D' R' L' U D' 12
65. F B' U D' R L' B B U D' L L 12
66. F F U' D R R U D F F U U 12
67. F L L B U U F' B L L F F 12
68. B R L F F R' L' B U U L L 12
69. U' F B' L' U D' R L' F F B B 2
70. F B R' L' U D' F' B' R' L' U' D' 12
71. F B R L U D' F B R' L' U' D' 12
72. U' R L' U' D F B' R R L L U' 12
73. F' B R' L U' D B R' L U' D D 12
74. U' R' L F B' U' D F B B U' D 6
75. U R' L F' B U D' B F F U D' 6
76. F B U D R' L F F R' L U' D 12
77. R' L' F' B R R L L U D B B 12
78. R L F B' R R L L U' D' F F 12
79. F F R R F' B' U' D' L L U D 8
80. U U R' L F' B' U' D R R L L 11
81. U U D R L' B U D' R R L L 6
82. U D F' B U' D F B' U' D R R 12
83. F B' U D R L' B B R L' U' D 12
84. U D' F F U D' R' L' F B' U' D 12
85. B R L' D' F B' U D' R R L L 2
86. F' R' L U' F B' R' L U D' B B 12
87. F' R' L U' F' B R' L U' D B B 12
88. R' L F B' U D R R L L D D 12
89. F B R L' F' B R L' U D' F F 12
90. U' D' R L' F F U D' R R L L 12
91. F F R' L F' B' U' D' L L B B 10
92. F U D B B U' D' F R L U D' 10
93. R L U D' F B' R' L' F B' U' D 8
94. F B' U' D L L F B' U' D' F F 10
95. F B' U U D D F B' R' L' U D' 10
96. R' L L F' B U R L' F F B B 6
97. F B' U' D R L F' B U U B B 12
98. F B U' D R' L F' B D D B B 12
99. F B' U' D R L F B' D D F F 12
100. U' D F B' U D' F F B B U D' 12
101. U R L' F' U D' R L' F F B B 8
102. U D' R L' F' B U D' F F B B 12
103. F' B' L L U' D F B U' D R R 8
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan) jbryan@pstcc.cc.tn.us
Pellissippi State (423) 539-7198
10915 Hardin Valley Road (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990
From cube-lovers-errors@mc.lcs.mit.edu Mon Nov 2 18:00:09 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil by mc.lcs.mit.edu (8.8.8/mc) with SMTP
id SAA01157; Mon, 2 Nov 1998 18:00:08 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Message-Id: <3.0.32.19981102164235.0094e100@mail.spc.nl>
Date: Mon, 02 Nov 1998 16:42:36 +0100
To: cube-lovers@ai.mit.edu
From: Christ van Willegen
Subject: Re: Unauthorized Use of RUBIK'S CUBE and CUBE Design Marks?
At 14:58 31-10-1998 +0100, you wrote:
>About 2 weeks ago I received the following message and it seems to me
>that it might be interesting for you too:
[Message deleted for brevity]
>
>Indeed the headline of my homepage at
>
>http://home.t-online.de/home/kociemba/cube.htm
>
>was "Rubik's Cube Explorer 1.5". So I removed the word "Rubik's" and
>added some note at the bottom of the page (...blah blah is not derived
>from, is not associated with blah blah...).
>
>I definitely will not remove the program from my homepage. This seems
>ridiculous. I know, that other cube fans received similar mail because
>they have some similar statements on their homepages now.
>
>What do you think about that?
>
I had the same 'problem' a couple of months ago.
A handheld computer users group I was active in, once published a
program that could be described as 'well, it looks a bit like Tetris
((R), if those lawyers are reading this, as well :-), but it's a long
way off the mark'.
The program was published in our magazine, and was also placed on a
web-page.
A couple of months ago, I received a letter from a Belgian lawyer firm,
addressed to the user's group. This user's group, by the way, died about
5 years (!) ago.
They told us to 'cease and decist' (a couple of things), including publishing
this article on 'our web-page'. Well, since this was someone else's web-
page, there was nothing we could do about it.
We told them (in friendly terms) that the user's group no longer existed,
that we were not affiliated to the user having the article on his web-page,
and that we nover sold the program.
We haven't heard from them (not even a letter stating that they received our
reply!) since.
It's kinda sad: The program was for an HP28S. I would have _loved_ to see
those lawyers type the program in (a couple of kilobytes, via the keyboard!),
only to find out that it was 'almost, but not quite, entirely unlike Tetris'
(Douglas, if you're reading this, quotes are ok, no?).
Just tell them you will take down the name, and they will probably be
off your back.
Christ van Willegen
From cube-lovers-errors@mc.lcs.mit.edu Tue Nov 3 06:37:22 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil by mc.lcs.mit.edu (8.8.8/mc) with SMTP
id GAA02996; Tue, 3 Nov 1998 06:37:21 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Date: Mon, 2 Nov 1998 15:32:29 -0500
From: michael reid
Message-Id: <199811022032.PAA24281@euclid.math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: Re: Unauthorized Use of RUBIK'S CUBE and CUBE Design Marks?
Cc: kociemba@hrz1.hrz.tu-darmstadt.de
> > Re: Unauthorized Use of RUBIK'S CUBE and CUBE Design Marks
> >
> > Dear Mr. Kociemba:
[ ... ]
> What do you think about that?
i think it's not good business strategy to attack the people who
are the biggest fans of their product! to claim ownership of
"the distinctive overall appearance of the RUBIK'S CUBE puzzle"
is ludicrous!
the changes you've described to your web page seem quite
reasonable and appropriate (given the circumstances), without
compromising too much. i hope you have no further trouble with
seven towns. but if you do, please let me know about it.
to make this situation even more ridiculous, i just checked out
their "official" website, which features a java cube
(http://rubiks.com/VRCUBE.html). their applet is stolen
from karl ho"rnell! (http://www.tdb.uu.se/~karl/java/rubik.html)
mike
From cube-lovers-errors@mc.lcs.mit.edu Mon Nov 9 17:39:21 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil by mc.lcs.mit.edu (8.8.8/mc) with SMTP
id RAA01496; Mon, 9 Nov 1998 17:39:20 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Message-Id: <19981104032938.14284.rocketmail@send102.yahoomail.com>
Date: Tue, 3 Nov 1998 19:29:38 -0800 (PST)
From: Han Wen
Subject: Query for Corners-First Method Rubik Solution
To: cube-lovers@ai.mit.edu
Hi,
Does anyone know of any websites that describe the Corners-first
method of the solving the rubik's cube? I know of many layer-first
methods such as Jiri Fridrich's (for which I have spent many hours
learning), but I really haven't seen a comprehensive explanation of
the corners-first method. I'm really curious to understand how anyone
can solve the cube under 30secs by solving the corners first.
-Han-
From cube-lovers-errors@mc.lcs.mit.edu Mon Nov 9 18:35:15 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil by mc.lcs.mit.edu (8.8.8/mc) with SMTP
id SAA01664; Mon, 9 Nov 1998 18:35:13 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Date: Wed, 04 Nov 1998 16:41:26 -0500 (Eastern Standard Time)
From: Jerry Bryan
Subject: Local Maxima Whose Inverses are not Local Maxima
To: Cube Lovers
Message-Id:
On 30 June 1997 I reported that if you could find a local
maximum whose inverse was not a local maximum, then you could
also find a longer local maximum. For example, suppose x is a
local maximum in the quarter turn metric and x' is not. Then,
there exists q in Q such that |x'q| = |x'| + 1 = |x| + 1. But
we know that q'x is a local maximum and we also know that
|q'x| = |x| + 1 because |q'x| is the same as |x'q|.
Because we now have at 12q a good number of local maxima whose
inverses are not local maxima as specimens, I have begun to
wonder if the same process might be able to be repeated several
times to yield progressively longer local maxima. For example,
if x is a local maximum and (q1)x is a local maximum, might also
(q2)(q1)x be a local maximum and also (q3)(q2)(q1)x etc. It
seems to me that good candidates to investigate in this regard
might be those local maxima at 12q whose inverses have a very
small maximality. For example, if x is a local maximum where
the maximality of x' is 2 (and there are several such cases),
then we know that there are 10 local maxima of the form qx.
I am not sure if I have time to investigate this question
further, but I certainly would love to hear from anyone who has
the time and the computing resources to do so.
----------------------
Jerry Bryan
jbryan@pstcc.cc.tn.us
From cube-lovers-errors@mc.lcs.mit.edu Tue Nov 10 06:10:31 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil by mc.lcs.mit.edu (8.8.8/mc) with SMTP
id GAA04442; Tue, 10 Nov 1998 06:10:31 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Sender: bosch@sgi.com
Message-Id: <36477CAE.446B@sgi.com>
Date: Mon, 09 Nov 1998 15:37:18 -0800
From: Derek Bosch
To: Han Wen
Cc: cube-lovers@ai.mit.edu
Subject: Re: Query for Corners-First Method Rubik Solution
References: <19981104032938.14284.rocketmail@send102.yahoomail.com>
Here's the method I use to solve Rubik's Cube in 30 seconds or less.
My notation is as follows:
F = turn front face clockwise 90 degrees.
F' = turn front face counter-clockwise 90 degrees.
F" = turn front face 180 degrees.
U = turn top face clockwise 90 degrees.
R = turn right face " " "
L = turn left face " " "
B = turn back face " " "
D = turn bottom face " " "
^ = move middle slice 90 degrees up.
v = move middle slice 90 degrees down.
OK. Now the order I do things is corners, edges on two OPPOSITE sides (Right
and Left), followed by the middle slice edges.
(1) Corners:
I solve my corners a bit wierdly, but I find it is really fast. I position
any four corners of the same color on a side. I don't care what colors are
on the adjoining faces right now, as I fix them later.
(1a)
Once I have the 4 corners of the same color, I turn the cube so that those
colors are on the down face. Now there are a few combinations that can occur
on the top face:
All corners on top face same color: Goto (1b)
Three corners need to rotate clockwise (position like below o=no rotate)
(+ = needs clockwise rotation)
(- = needs counter-clkwise rot.)
+ + Move: R'U"RUR'UR and goto (1b)
+ o
Three corners need counter-clockwise rotate:
- o Move: RU"R'U'RU'R' and goto (1b)
- -
One corner needs clockwise rotate, One needs counter-clockwise rotate:
3 cases:
+ - Move: RU"RU"RUR" and goto (1b)
o o
- + Move: RUR'U'F'U'F and goto (1b)
o o
o - Move: R'URUBU'B' and goto (1b)
+ o
Two corners need clockwise rotate, Two need counter clockwise:
2 cases:
+ - Move R"U"RU"R" and goto (1b)
- +
- + Move RUR"F'R"UR' and goto (1b)
- +
(1b)
Now, you should have two opposite sides, with the corners of those
two sides the proper color.
We have to correct the 4 remaining sides to get corners in the right
place, before we can move onto edges. To do this, count the number
of sides that have the upper pair of corners the same color. Also
counter the number of sides that have the lower pair of corners the
same color.
All four sides (upper and lower) corner pairs match. Goto (2)
No sides' corner pairs match. Do Move R"F"R". Goto (2)
One Bottom corner pair matches.
Move that corner pair to the Down-Left position.
Move R"UR"U'R"UR"U'R. Goto (2)
One Top corner pair matches. Turn Cube over, and do previous moves.
One Top and one Bottom pair matches.
Move both corner pairs to the front face.
Move R"UR"U"F"UF". Goto (2)
All Bottom pairs match.
Move R"UR"U"F"UF"U"L"UL". Goto (2)
All Top pairs match. Turn Cube over, and do previous move.
All Bottom pairs match. One Top pair match.
Move Top Pair to Left-Upper position.
Move R"UR"U'R"F"U'F"UF". Goto(2)
All Top pairs match. One Bottom pair match. Turn Cube over, and do prev.
(2) Solving two Opposite Sides.
Now, all the corners should be solved. You should move the center of
each cube to its respective corners, to get an X on each side (at least
on two opposite sides). From now on orient the cube so that the two
opposite sides are right and left.
(2a)
Solve three edges on the left face with the following moves.
U'^U - moves the edge piece in the Front-Down position to the Up-Left
position.
UvU' - moves the edge piece in the Back-Down position to the Up-Left
position.
This is easier done with a cube in your hand, and try and see how this
works. This will mess up the centers and edges in the middle slice, as well
as the Up-Right edge. Don't worry about this. As long as you keep this
orientation, and rotate the left face to get ready for a new edge to be moved
you can solve three out of four of the edges on the left face.
(2b)
Solve four edges on the right face:
First, rotate the left face, so that the unsolved edge is in the Up-Left
position.
Then, using the following moves, solve all four edges (similarly to step 2a).
U^U' - moves the edge piece in the Front-Down position to the Up-Right
position.
U'vU - moves the edge piece in the Back-Down position to the Up-Right
position.
(2c)
Solve remaining edge on left face:
2 cases (other than already solved):
edge in place, needs to be flipped: Use U'vUvUvU'
otherwise, move edge to Down-Front position, using v or ^.
if the front color (of the DF edge) is the same as the left face color,
Use U'vU"vU'
else Use vU^U"^U
(3)
Solve middle slice edges.
First use ^ or v to position middle slice centers in proper faces.
(3a) Position edges:
3 cases (other than all in place):
only three edges out of place:
position cube such that DF needs to go to UB and UB needs to go to UF.
Use ^U"vU".
all four edges need to move:
if UF needs to go to DB, Use ^F"B"vF"B".
otherwise, position so that UF needs to go to UB, Use U'^^U'^^.
(3b) Flip required edges:
3 cases (other than no flips needed).
all four need flipping, use FR'F'^U^U^U^UFRF'
two edges need flipping, both on same face. Turn cube so that these
edges are the UF and UB edges. use ^U^U^U"vUvUvU"
otherwise, turn cube so UB and DF need flipping, use F"^U^U^U"vUvUvU"F"
That should do it. I apologize for the roughness of this solution. I think
diagrams would help it a lot. If you have any criticism or ideas that could
help this solution become more readable, let me know.
Note, this solution is very close to Jeff Verasano (sp) and
Minh Thai's methods...
D
--
Derek Bosch "A little nonsense now and then
(650) 933-2115 is relished by the wisest men"... W.Wonka
bosch@sgi.com
From cube-lovers-errors@mc.lcs.mit.edu Tue Nov 10 07:33:39 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil by mc.lcs.mit.edu (8.8.8/mc) with SMTP
id HAA04496; Tue, 10 Nov 1998 07:33:37 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Date: Mon, 9 Nov 1998 19:15:58 -0500 (EST)
From: Alchemist Matt
Reply-To: Alchemist Matt
To: Han Wen
Cc: cube-lovers@ai.mit.edu
Subject: Re: Query for Corners-First Method Rubik Solution
In-Reply-To: <19981104032938.14284.rocketmail@send102.yahoomail.com>
Message-Id:
My page at http://www.unc.edu/~monroem/rubik.html describes a
method that is sort of "corners first". Although, in my first step I say
to solve the first layer before going on, one could effectively simply
place only the four corners in the top layer, then move on to the four
corners in the bottom layer (specified in steps 2 and 3), then begin
filling in the gaps on the top and bottom layers (steps 4 and 5), and
lastly finish the middle layer.
In fact, a chemistry professor at my current school, Holden Thorp,
competed in one of the Rubik's cube playoff contests that was aired on the
TV show That's Incredible. Someone posted the video of it about a month
ago (and mentioned it in this discussion list), and he saw it here at my
school after I downloaded it. He then looked at my page and mentioned
that the winner of the contest actually used the solution shown on my page
(probably modified slightly). I can only solve a well-scrambled cube in 2
to 3 minutes using the solution, but I'm sure someone quite adept, nimble,
and fast could push it to under one minute. (Please note this isn't "my"
solution; I simply learned it from a book many years ago. Further, I have
never been in a cube solving competition).
Matt
-----------------------------------------------------------------------
Matthew Monroe Monroem@UNC.Edu
Analytical Chemistry http://www.unc.edu/~monroem/
UNC - Chapel Hill, NC This tagline is umop apisdn
From cube-lovers-errors@mc.lcs.mit.edu Thu Nov 12 14:18:47 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil by mc.lcs.mit.edu (8.8.8/mc) with SMTP
id OAA24676; Thu, 12 Nov 1998 14:18:45 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Message-Id: <19981111170341.14375.rocketmail@send105.yahoomail.com>
Date: Wed, 11 Nov 1998 09:03:41 -0800 (PST)
From: Han Wen
Subject: RE: Query for Corners-First Method Rubik Solution
To: Noel Dillabough
Cc: cube-lovers@ai.mit.edu
Hi,
Thanks for the link to your Puzzler program.
You're not going to believe this, but you can still purchase the
Professor's Cube (5x5x5) and the Megaminx! Since it's difficult... no,
impossible to find anyone that sell these puzzles, I think it's worth
mentioning. You can get them from Meffert's site:
http://ue.net/mefferts-puzzles/
Your Puzzler program is a tremendously useful tool to develop moves.
I've got 11/12 sides of the Megaminx solved. But for the last side, I
need to figure out corner/edge twisting/permuting moves. You're
Puzzler program's great for that. I'm surprised how many of my
Rubik's cube moves can be applied with minor modifications to the
Megaminx.
-Han-
---Noel Dillabough wrote:
> I actually solve all the cubes this way (or at least centers ->
> corners -> edges for larger cubes) I just find it more logical and
> easier to memorize than other methods.
> You can check out my solution at
> http://www.mud.ca/puzzler/puzzler.html. Its in the puzzler help
> file under "solving the cube". I will be adding other solutions
> soon that are clearer, let me know if you would like them I could
> mail them to you.
> -Noel.
From cube-lovers-errors@mc.lcs.mit.edu Thu Nov 12 15:00:46 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil by mc.lcs.mit.edu (8.8.8/mc) with SMTP
id PAA26362; Thu, 12 Nov 1998 15:00:45 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Date: Thu, 12 Nov 1998 15:01:40 +0000
From: David Singmaster
To: cube-lovers@ai.mit.edu
Message-Id: <009CF1D5.D1348312.50@ice.sbu.ac.uk>
Subject: Use of the name Rubik's Cube
The lawyers are being obsessively zealous as the name is
certainly well on its way to becoming a common noun. It was included
in the Oxford English Dictionary in the mid-1980s. Other examples are
Kleenex and Aspirin, which were both originally tradenames and their
owners fought to retain them but eventually lost. Xerox is fighting a
rear-guard action on its name. If you don't want to get involved in
legal hassle, I suggest that you use the name Magic Cube which was
the original name and is such a common term that they can't claim it
is a trademark.
DAVID SINGMASTER, Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411; fax: 0171-815 7499;
email: zingmast or David.Singmaster @sbu.ac.uk
[ Moderator's note: I am still dropping messages that consist mainly of
generic comments on intellectual property issues. There a great variety
of individualistic and contentious debate on these topics that you may
follow in dedicated fora such as the Usenet group misc.int-property. I
am not yet persnickety enough to elide the third and fourth sentences
from the above, but they are on the edge.
I will also note that the term "Magic Cube" is also used to refer to a
cubical array of natural numbers whose orthogonal and diagonal rows sum
to the same number, as a generalization of "Magic Square", so it is
advisable to include context such as "The geometrical puzzle originally
known as the Hungarian Magic Cube." ]
From cube-lovers-errors@mc.lcs.mit.edu Wed Nov 18 12:52:47 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil (sun28.aic.nrl.navy.mil [132.250.84.38])
by mc.lcs.mit.edu (8.9.1a/8.9.1-mod) with SMTP id MAA28044
for ; Wed, 18 Nov 1998 12:52:47 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Date: Thu, 12 Nov 1998 17:05:37 -0500 (EST)
From: Nicholas Bodley
To: Han Wen
Cc: Noel Dillabough , cube-lovers@ai.mit.edu
Subject: Solutions (Was: RE: Query for Corners-First Method Rubik Solution)
In-Reply-To: <19981111170341.14375.rocketmail@send105.yahoomail.com>
Message-Id:
On Wed, 11 Nov 1998, Han Wen wrote:
{snips}
}mentioning. You can get them from Meffert's site:
}http://ue.net/mefferts-puzzles/
It might be of interest to mention that Meffert has solutions to many
puzzles at his Web site. The Contributors section gives generous credit
to a number of experts; they wrote the solutions.
Best,
|* Nicholas Bodley *|* Electronic Technician {*} Autodidact & Polymath
|* Waltham, Mass. *|* -----------------------------------------------
|* nbodley@tiac.net *|* The personal computer industry will have become
|* Amateur musician *|* mature when crashes become unacceptable.
--------------------------------------------------------------------------
From cube-lovers-errors@mc.lcs.mit.edu Wed Nov 18 13:27:34 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil (sun28.aic.nrl.navy.mil [132.250.84.38])
by mc.lcs.mit.edu (8.9.1a/8.9.1-mod) with SMTP id NAA28426
for ; Wed, 18 Nov 1998 13:27:29 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Message-Id: <3.0.32.19981113091853.00948790@mail.spc.nl>
Date: Fri, 13 Nov 1998 09:18:55 +0100
To: cube-lovers@ai.mit.edu
From: Christ van Willegen
Subject: MegaMinx/5^3 (Was: RE: Query for Corners-First Method Rubik Solution)
At 09:03 11-11-1998 -0800, you wrote:
>Hi,
>
>Thanks for the link to your Puzzler program.
>
>You're not going to believe this, but you can still purchase the
>Professor's Cube (5x5x5) and the Megaminx! Since it's difficult... no,
>impossible to find anyone that sell these puzzles, I think it's worth
>mentioning. You can get them from Meffert's site:
>http://ue.net/mefferts-puzzles/
Also, a store in The Netherlands sells these! Last time I was there
(last saturday), they had;
- a couple of 3^3's
- a couple of 5^3's
- skewb
I can't recall if they had a Megaminx at that time.
The store is based in Eindhoven. If anybody wants some, I can buy
them and send them out.
5^3 costs F. 50 (about $25). 3^3 costs F. 10 (about $5).
Before anybody gets this wrong:
- I do not work for them, I'm a happy customer
- I don't get paid to do this
- I make no money out of this
You can call them at: +31-40-2461376
Business hours are 0900 to 1800. The Netherlands is at CET
(differs +6 hours with NY, +9 with CA)
Christ van Willegen
From cube-lovers-errors@mc.lcs.mit.edu Wed Nov 18 17:23:08 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil (sun28.aic.nrl.navy.mil [132.250.84.38])
by mc.lcs.mit.edu (8.9.1a/8.9.1-mod) with SMTP id RAA29198
for ; Wed, 18 Nov 1998 17:23:06 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Message-Id: <19981114070437.9789.rocketmail@attach1.rocketmail.com>
Date: Fri, 13 Nov 1998 23:04:37 -0800 (PST)
From: "Jorge E. Jaramillo"
Subject: RE: Moves to this other pattern
To: David Singmaster ,
Maybe (although I don't think so since some people
already answered what I was asking) I made a mistake
when describing the position I wanted to accomplish.
What I wanted was:
L B R
L B R
L B R
F F F T T T F F F D D D
L L L T T T R R R D D D
B B B T T T B B B D D D
L F R
L F R
L F R
And until now the best solution is:
F B L F2 T D- L2 B- T- D- F2 R- T- D L- R D
===
Jorge E Jaramillo
From cube-lovers-errors@mc.lcs.mit.edu Wed Nov 18 18:05:42 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil (sun28.aic.nrl.navy.mil [132.250.84.38])
by mc.lcs.mit.edu (8.9.1a/8.9.1-mod) with SMTP id SAA29431
for ; Wed, 18 Nov 1998 18:05:41 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Message-Id: <365115AC.22369936@erco.com>
Date: Tue, 17 Nov 1998 07:20:28 +0100
From: "michael ehrt"
Reply-To: m.ehrt@erco.org
To: Cube Lovers Mail
Subject: Getting 2x2x2 cubes
If anyone is interested in getting 2x2x2 cubes, during my holiday in the
UK two weeks ago I found a shop in Sheffield which has them in stock.
It's called "The Puzzle Shop" and in situated in Meadowhall shopping
centre. The cubes are GBP 5 each, and they have a few other things like
keyring 3x3x3s etc.
Michael
From cube-lovers-errors@mc.lcs.mit.edu Thu Nov 19 13:41:00 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil (sun28.aic.nrl.navy.mil [132.250.84.38])
by mc.lcs.mit.edu (8.9.1a/8.9.1-mod) with SMTP id NAA03513
for ; Thu, 19 Nov 1998 13:40:59 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Message-Id: <19981117105414.18936.rocketmail@attach1.rocketmail.com>
Date: Tue, 17 Nov 1998 02:54:14 -0800 (PST)
From: "Jorge E. Jaramillo"
Subject: The Cylinder
To: cube
I was checking the Rubik official website and I was
surprised not to find one product that I seem to find
here (I live in Colombia South America) fairly
easily. I am talking about the cylinder.
When I first saw it I bought it and thought it was
going to be some amazing and tricky to solve puzzle,
it ended up being a 3x3 cube with the corners cut, so
corner cubelets only have 2 colors and there are two
types of borders, 8 borders with the usual two colors
and 4 with only one.
Does it mean that this cube was "invented" by some
manufacturer other than Mr Rubik and that is not so
common?
===
Jorge E Jaramillo
[Moderator's note: I own such a puzzle, but I would call its shape
an octagonal prism, rather than a cylinder. On mine, the solved
position is not an octagonal prism because one beveled face is
rotated 90 degrees, forming a decahedron whose faces are six
rectangles and four irregular hexagons. I don't remember whether
it was originally manufactured this way or whether I altered the
color tabs. ]
From cube-lovers-errors@mc.lcs.mit.edu Thu Nov 19 17:36:55 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil (sun28.aic.nrl.navy.mil [132.250.84.38])
by mc.lcs.mit.edu (8.9.1a/8.9.1-mod) with SMTP id RAA04360
for ; Thu, 19 Nov 1998 17:36:54 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Date: Thu, 19 Nov 1998 15:00:06 -0500 (Eastern Standard Time)
From: Jerry Bryan
Subject: Re : The Cylinder
In-Reply-To: <19981117105414.18936.rocketmail@attach1.rocketmail.com>
To: "Jorge E. Jaramillo"
Cc: cube
Message-Id:
Jorge E. Jaramillo's message of 17 Nov 1998 included:
>[Moderator's note: I own such a puzzle, but I would call its shape
> an octagonal prism, rather than a cylinder. On mine, the solved
> position is not an octagonal prism because one beveled face is
> rotated 90 degrees, forming a decahedron whose faces are six
> rectangles and four irregular hexagons. I don't remember whether
> it was originally manufactured this way or whether I altered the
> color tabs. ]
I also own such a puzzle, although I have never seen one in
a store. I got mine at a garage sale for $0.25.
I haven't played with it in a long time. But my best
recollection is that it can be solved basically the same
way as a 3x3x3 cube, except that *I think* (don't remember
for sure) that the color scheme permits invisible swaps of
identically colored pieces which can make the puzzle seem
"impossible" to solve unless you realize that the
identically colored pieces must be swapped. It is also my
best recollection that such a puzzle is mentioned briefly
and is pictured in one of Douglas Hofstadters's cube
articles in Scientific American back in the early 80's. So
I don't think it is any kind of new invention.
----------------------------------------
Jerry Bryan
jbryan@pstcc.cc.tn.us
Pellissippi State Technical Community College
From cube-lovers-errors@mc.lcs.mit.edu Fri Nov 20 11:11:16 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil (sun28.aic.nrl.navy.mil [132.250.84.38])
by mc.lcs.mit.edu (8.9.1a/8.9.1-mod) with SMTP id LAA07158
for ; Fri, 20 Nov 1998 11:11:15 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Message-Id: <19981117070139.11901.rocketmail@send103.yahoomail.com>
Date: Mon, 16 Nov 1998 23:01:39 -0800 (PST)
From: Han Wen
Subject: Fwd: Re: HOW TO SOLVE THE PROBLEM OF THE PROFESSOR CUBE
To: cube-lovers@ai.mit.edu
I thought this would be of value to Rubik fans out there with
Professor Cubes (5x5x5)...
note: forwarded msgs attached.
_________________________________________________________
---Uwe Meffert wrote: [reprinted here with his permission]
> Dear Mr. Wen
> Thank you for your interest in my puzzles, I am sorry to hear that
> you are having problems with the Prof. Cube.
> You received one of the last ones from the last production batch and
> the next production is not for another month.
> What unfortunately happened is that when gluing the center small
> caps excess glue fixed the screw to the plastic centre piece that it
> should turn in. So when you turn these sections it will tighten /
> loosen that one screw.
> If you are skilful enough you can try and carefully remove the
> centre label and then pry open and remove the centre cap of the blue
> and orange side. Then try to remove the excess glue from around the
> screw with a sharp object and try turning the screw with a
> screwdriver firmly holding the plastic piece so you can break the
> glue bond. Once the screw can freely turn inside the plastic part,
> re-tighten it to the same tension as it was originally, so as to
> allow smooth turning without any pieces falling out during play.
> Then carefully using only very little glue fix the centre cap back
> into place and re-attach the color label.
> Good Luck and Happy Puzzeling.
> Please let me know the outcome of this recommended procedure.
> With warm regards
> Uwe Meffert
________________________________________________________________
Date: Mon, 16 Nov 1998 22:57:20 -0800 (PST)
From: Han Wen
Subject: Re: HOW TO SOLVE THE PROBLEM OF THE PROFESSOR CUBE
To: Uwe Meffert
Cc: Jing Meffert
Hi,
My cube is all fixed.
Thank you for your prompt reply. You were right, the glue used to fix
the caps also fixed the spindle screw! Actually, your instructions
gave me the perfect excuse to take your cube apart. I was dying to
find out how the heck all these pieces are held together.
Anyways, I popped of the caps carefully using a razor blade, scraped
off all the excess glue, greased the screw head and before screwing it
back together, I took all the pieces for one face out just to see and
understand the engineering holding all the pieces together. Wow, what
an amazing bit of engineering. It's like a cube spindle inside
another cube spindle! Amazing.
Actually, the center caps don't really need glue. They fit nice and
snug, and it also leaves me the option to adjust the screw again in
case it becomes loose.
Now that I understand the mechanism, I've decided to only rotate faces
clockwise to minimize the possibility that a counter-clockwise
rotation will actually loosen one of the spindle screws.
I still haven't messed the faces up though. I'm so close to finishing
the Megaminx. I just have two edge pieces to swap on the last face!
The other 11 sides were fairly straightforward to solve. I also got
the corners of the last face fairly quickly by using Sune's move to
twist corner pieces (a standard Rubik's cube move). However, getting
those edge pieces was a different story. I had to develop quite a few
moves to rotate and twist the edge pieces around. I'm close... so
close..! :)
I hope you keep inventing and making new puzzles. I eagerly click on
your new releases on your web page quite regularly, hoping to find a
worthy successor to the Megaminx or the Professor's Cube. Maybe a
7x7x7?!! Or a Buckyball? One can only imagine...
-Han-
From cube-lovers-errors@mc.lcs.mit.edu Fri Nov 20 14:44:20 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil (sun28.aic.nrl.navy.mil [132.250.84.38])
by mc.lcs.mit.edu (8.9.1a/8.9.1-mod) with SMTP id OAA08578
for ; Fri, 20 Nov 1998 14:44:17 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Message-Id: <3654ED01.E6D38EE8@hurstlinks.com>
Date: Thu, 19 Nov 1998 23:16:01 -0500
From: "Guy N. Hurst"
Organization: HurstLinks Sites On the Internet
To: "Jorge E. Jaramillo"
Cc: cube
Subject: Re: The Cylinder
References: <19981117105414.18936.rocketmail@attach1.rocketmail.com>
I have seen one or both of these puzzles, and they were very different
from each other. The cylinder, or prism, was actually the first cube
I learned to solve, when my cousin from Luxembourg visited back in
1981. I have pleasant memories of it, because it was very well made
and pleasing to view. It is harder to solve than the cube since the
four of the edges are "cut", so it is impossible to match edges to
centers - leaving the possibility of having to backtrack later and
figure out which "corner" (and matching edge) is in the wrong place!
But I had it down and could quickly readjust (usually had to swap
corners diagonally in the top two layers if I found a single flipped
edge left in the bottom layer when almost done solving it, if I
remember).
I liked it so much, I requested and obtained 4 more after my cousin
returned to Europe! I would take them to school, one at a time, until
(unfortunately) they all eventually disappeared. At least two were
stolen out of my (locked) locker on different occasions. Someone else
liked them, too.
Anyway, I never found that puzzle in the US, and could only get it
from my cousin in Europe. (Who I think may have gotten them from
England).
But the other puzzle, as described by the moderator, was available in
the US back then, I think in the following year or so after my cousin
visited, since one of my friends had one. But it wasn't as nice
looking or well made. So I didn't care for it. It was more like the
cube with its corners cut, forming rectangles and triangles in a
spherical symmetry, as opposed to the prism from Europe which has four
of its 3-piece-edges cut, forming only rectangles and having a
cylindrical symmetry.
Guy N. Hurst
From cube-lovers-errors@mc.lcs.mit.edu Fri Nov 20 15:16:19 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil (sun28.aic.nrl.navy.mil [132.250.84.38])
by mc.lcs.mit.edu (8.9.1a/8.9.1-mod) with SMTP id PAA08664
for ; Fri, 20 Nov 1998 15:16:18 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Message-Id:
Date: Fri, 20 Nov 1998 08:46:30 +0100 (CET)
From: Bas de Bakker
To: Cube-Lovers@ai.mit.edu
In-Reply-To: (message from Jerry
Bryan on Thu, 19 Nov 1998 15:00:06 -0500 (Eastern Standard Time))
Subject: Re: The Cylinder
References:
>>>>> "Jerry" == Jerry Bryan writes:
[About the octagonal "cube"]
Jerry> I haven't played with it in a long time. But my best
Jerry> recollection is that it can be solved basically the same
Jerry> way as a 3x3x3 cube, except that *I think* (don't remember
Jerry> for sure) that the color scheme permits invisible swaps of
Jerry> identically colored pieces which can make the puzzle seem
Jerry> "impossible" to solve unless you realize that the
Jerry> identically colored pieces must be swapped.
Your recollection is not exact. There are no identically colored
pieces to swap, but you can swap complete columns consisting of two
"corners" (what would have been corners on the cube) and one "edge"
without noticing.
In fact, if you create an even permutation of those columns, there is
no problem. But if you create an odd permutation, it will become
impossible to solve the upper layer.
Presuming you solve cubes in layers, the easiest way out is to not
start at one of the octagonal layers (which seems the most natural
way), but to start with a "side" layer. If you do it this way, it
will always be possible to solve the last layer.
I hope I'm making myself at least somewhat clear,
Bas.
From cube-lovers-errors@mc.lcs.mit.edu Fri Nov 20 17:41:06 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil (sun28.aic.nrl.navy.mil [132.250.84.38])
by mc.lcs.mit.edu (8.9.1a/8.9.1-mod) with SMTP id RAA09316
for ; Fri, 20 Nov 1998 17:41:05 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Message-Id: <3.0.1.32.19981120105340.009ac040@icex5.cc.ic.ac.uk>
Date: Fri, 20 Nov 1998 10:53:40 +0000
To: jbryan@pstcc.cc.tn.us
From: "Andrew R. Southern"
Subject: Space Shuttle.
Cc: Cube-Lovers@ai.mit.edu, kingeorge@rocketmail.com
I got a similar puzzle in the same way. Mine was called the "Space
Shuttle". The lack of the name Rubik probably meant it wasn't from Rubiks.
But since they only have a patent in Hungary, and everywhere else they are
protected by copyrights on the name and the external appearence, this is
probably legally legitimate. The colour scheme allowed the puzzle to be
solved in more ways than the cube and so I reckon its easier.
Each of the chamfered sides was coloured in a colour which did not relate
to the rest of the puzzle, and so these were (within the boundaries of a
2-swap) possible to position ("correctly") in a few different positions.
The mid-edges of the chamfered edges were rotationally symmetrical every
180 degreees and so, since they were all one colour, it was possible to
have one (or three) of them rotated, and hence one of the other mid edges
rotated and it would look majorly FUBAR'd. Once you'd realised what
happened, the puzzle was easier than the cube though.
-Andy Southern.
From cube-lovers-errors@mc.lcs.mit.edu Mon Nov 23 13:45:28 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil (sun28.aic.nrl.navy.mil [132.250.84.38])
by mc.lcs.mit.edu (8.9.1a/8.9.1-mod) with SMTP id NAA18663
for ; Mon, 23 Nov 1998 13:45:28 -0500 (EST)
Message-Id: <199811231845.NAA18663@mc.lcs.mit.edu>
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Mail-from: From cube-lovers-request@life.ai.mit.edu Fri Nov 20 22:15:27 1998
Date: Fri, 20 Nov 1998 22:13:38 -0500 (EST)
From: Nicholas Bodley
To: Jerry Bryan
Cc: "Jorge E. Jaramillo" ,
cube
Subject: Re: Re : The Cylinder
In-Reply-To:
Uwe Meffert had a printed color catalogue around 1985 that showed some
very interesting moving-piece "group theory" puzzles (is that a proper
term?). Although I have a copy safely stashed somewhere, I don't know
where.
I'm just about sure that one was a cylinder, possibly in three layers
like layer cake; it also, iirc, had maybe three more "cutting planes"
that were spherical sectors bounded by the cylinder. Rotating the pieces
would exchange top and bottom.
|* Nicholas Bodley *|* Electronic Technician {*} Autodidact & Polymath
|* Waltham, Mass. *|* -----------------------------------------------
|* nbodley@tiac.net *|* The personal computer industry will have become
|* Amateur musician *|* mature when crashes become unacceptable.
--------------------------------------------------------------------------
From cube-lovers-errors@mc.lcs.mit.edu Mon Nov 23 16:35:08 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil (sun28.aic.nrl.navy.mil [132.250.84.38])
by mc.lcs.mit.edu (8.9.1a/8.9.1-mod) with SMTP id QAA19314
for ; Mon, 23 Nov 1998 16:35:06 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Message-Id: <001e01be155b$9a456940$6bc4b0c2@home.icl.web>
From: roger.broadie@iclweb.com (Roger Broadie)
To: "cube"
Cc: "Jorge E. Jaramillo"
Subject: Re: The Cylinder
Date: Sat, 21 Nov 1998 14:29:22 -0000
I was given a cylinder here in England in 1981. I no longer have the
packaging, but I suspect it was Taiwanese, unless the Hungarians made
this variant. It was my first cube puzzle, and its shape was so
unappealing when disturbed that I put it on one side and got a genuine
cube to learn on - well, almost genuine: it came from a street trader
in Regent Street.
The apparently impossible state is a monoflip of a top or bottom edge
piece. There will be a matching flip of a middle-layer edge piece,
but that will be invisible, since the piece has only one face.
I wondered if it would be possible to get the puzzle into the solved
shape and then restore the positions of the pieces without losing the
shape, that is, only allowing turns from the group ____, where S and A are slice and anti-slice moves of the middle
layers (I needed them). In fact it is not. There may always be a
hidden flip in the middle layer and you can't correct that without
moving the piece out of the middle layer, which needs a turn like F,
and that destroys the shape. But if you cheat a little and make sure
the flips are got right before the shape is finally restored, then it
can be done.
Andy Southern has already made the point about the flips. He also
pointed out that the configuration is not unique because columns
corresponding to the vertical edges on a normal cube can be swapped.
As a rider to that point, the pretty pattern stripes on the normal
cube is not distinguishable on the octagonal prism, because it's
striped already.
It should be possible to work out whether our moderator's puzzle came
in the form he now has by counting stickers. The octagonal prism has
2 sets of 9 (the top and bottom) and 8 of 3 (the side columns). If
I've grasped his configuration correctly, if it came in that form
originally it should have 4 sets of 6 and 6 of 3.
Roger Broadie
[ Yes, my decahedron's stickers are incompatible with an octagonal
prism solution. I just can't remember whether I replaced some of
the stickers to make this new shape. --Dan ]
From cube-lovers-errors@mc.lcs.mit.edu Mon Nov 23 18:13:16 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil (sun28.aic.nrl.navy.mil [132.250.84.38])
by mc.lcs.mit.edu (8.9.1a/8.9.1-mod) with SMTP id SAA19716
for ; Mon, 23 Nov 1998 18:13:15 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Message-Id: <19981123071931.2655.rocketmail@send105.yahoomail.com>
Date: Sun, 22 Nov 1998 23:19:31 -0800 (PST)
From: Han Wen
Subject: Method for Solving the Megaminx
To: Cube-Lovers@ai.mit.edu
Hi,
Well, it took me a 2-3 days, but I finally solved the Megaminx. Whew.
I know, big deal. It's been done.. many many times... over a decade
ago. But, I thought some of the moves I found may of interest to some
of the Megaminx aficionados out there. So, here it is, I apologize
for its length:
Solving the Megaminx faces 1-11 are fairly straightforward.
Ironically, the larger number of faces makes it easier to solve than
the Rubik's cube, because they provide a lot more "free lanes" to move
pieces around. There's actually just one move you need to remember to
solve these faces. It's the same move when solving the middle layer
of the Rubik's cube, when you want to move edge pieces from the bottom
layer to their respective position in the middle layer. Namely,
D'R'DRF'RFR'
Solving the last face, however, is another matter. The general
strategy I followed is the same as some of the standard methods for
solving the bottom layer of the Rubik's cube. Namely, I first solve
the 5 corners, then I solve the 5 edge pieces. To solve the corners,
I simply used Sune's move applied with slight modification to the
Megaminx. For the remaining edge pieces, I had to develop moves that
only moved the edge pieces around, while leaving the corners
unchanged. Noel Dillabough's Puzzler program was an invaluable tool
for helping me experiment with various edge moves. Anyways, the
following are my notes describing some of the more useful moves I've
found. I'm pretty sure they're not the most efficient method for
solving the Megaminx, but they're the best I could come up with.
___________________________________________
Notation for Solving the Last Face corner pieces:
F=Front Face, D=Lower Face, L=Left Face, R=Right Face
The F and D faces are adjacent
The last layer containing the corners you need to flip/permute should
be positioned at the D-face
____________________________________________
Move for Solving the Last Face corner pieces:
Name: Sune's Double-Swap
Description: Sune's Rubik's Cube move applied to the Megaminx
Number of pairs of corners swapped: 2
Number of corners twisted counterclockwise: 3
Move: R'D'RD'R'D'3R
____________________________________________
Strategy for Solving the Last Face corner pieces:
- Position the corners
- Twist the corners in place by applying Sune's Double-Swap
move twice
============================================
Notation for Solving the Last Face edge pieces:
F=Front Face, U=Upper Face, L=Left Face, R=Right Face
The F and U faces are adjacent
X= L'R U2 LR' F2
X'=L'R U'2 LR' F'2
X2= X X = (L'R U2 LR' F2) (L'R U2 LR' F2)
Xa= L'R U2 LR' F'2
Y= LR' F2 L'R U2
The last layer containing the edges you need to flip/permute should be
positioned as the F-face or the U-face depending on the move described
below:
_____________________________________________
Moves for Solving the Last Face edge pieces:
Name: F Tricycle 1
Description: Permutes 3 adjacent edges clockwise on the lower left of
the F-face
No. Edges permuted: 3
No. Edges flipped: 2
Move: (Xa3 X'2)^2
Name: F Tricycle 2
Description: Permutes 3 adjacent edges clockwise on the upper half of
the F-face
No. Edges permuted: 3
No. Edges flipped: 2
Move: (Xa3 X2)^2
Name: U Tricycle 1
Description: Permutes 3 edges clockwise on the U-face
No. Edges permuted: 3
No. Edges flipped: 2
Move: F' X2 Y'2 F
Name: U Tricycle 2
Description: Permutes 3 edges counterclockwise on the U-face
No. Edges permuted: 3
No. Edges flipped: 2
Move: F X'2 Y2 F'
Name: Cross-country Tricycle
Description: Permutes 3 edges across the U and F faces
No. Edges permuted: 3
No. Edges flipped: 1
Move: (X2 X'2)^4
Name: U Bi-Flip 1
Description: Flips two opposite edges on the U-face
No. Edges permuted: 0
No. Edges flipped: 2
Move: (Xa3 X'2)^3
Name: U Bi-Flip 2
Description: Flips two adjacent edges on the U-face
No. Edges permuted: 0
No. Edges flipped: 2
Move: (X2 X2 X'2)^5
Name: Cross-country Bi-Flip
Description: Flips two edges, one on the U-face, one on the F-face
No. Edges permuted: 0
No. Edges flipped: 2
Move: (Xa3 X2)^3
Name: "W"-Cycle
Description: Permutes all edges on the F-face in a "W" pattern
No. Edges permuted: 5
No. Edges flipped: 2
Move: (X2 X2 X'2)^2
Name: "Figure 8"-Cycle
Description: Permutes all edges on the F-face in a "Figure 8" pattern
No. Edges permuted: 5
No. Edges flipped: 4
Move: (X2 X2 X'2)^4
____________________________________________
Strategy for Solving the Last Face edge pieces:
- You should only need to use F Tricycle and the Bi-Flip moves to
completely solve the edges. The F Tricycle move usually needs
to be applied twice.
If anything is vague/unclear please feel free to request clarification.
-Han Wen-
From cube-lovers-errors@mc.lcs.mit.edu Tue Nov 24 16:09:26 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil (sun28.aic.nrl.navy.mil [132.250.84.38])
by mc.lcs.mit.edu (8.9.1a/8.9.1-mod) with SMTP id QAA24075
for ; Tue, 24 Nov 1998 16:09:26 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
To: cube-lovers@ai.mit.edu
From: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
Subject: Re: The Cylinder
Date: 24 Nov 1998 18:53:56 GMT
Organization: California Institute of Technology, Pasadena
Message-Id: <73evc4$cq0@gap.cco.caltech.edu>
References:
roger.broadie@iclweb.com (Roger Broadie) writes:
>I was given a cylinder here in England in 1981. I no longer have the
>packaging, but I suspect it was Taiwanese, unless the Hungarians made
>this variant. It was my first cube puzzle, and its shape was so
>unappealing when disturbed that I put it on one side and got a genuine
>cube to learn on - well, almost genuine: it came from a street trader
>in Regent Street.
There is a Taiwanese manufacture of the octagonal prism. I have
part of one in my collection. (Got it when I was 10, and many
cubies have disappeared since then.)
I also have one of the "truncated cubes" mentioned earlier in this
thread.
I find the discussion on these two quite strange, since I always
thought of these as cubes with weird cubies -- no more special than,
say, that spherical "cube" they had a few years back.
--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
StethoPHONE, not stethoSCOPE. What do doctors SEE in those things anyway?
From cube-lovers-errors@mc.lcs.mit.edu Tue Dec 1 14:27:08 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil (sun28.aic.nrl.navy.mil [132.250.84.38])
by mc.lcs.mit.edu (8.9.1a/8.9.1-mod) with SMTP id OAA18757
for ; Tue, 1 Dec 1998 14:27:07 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
To: Cube-Lovers@ai.mit.edu
From: "Andrew R. Southern"
Subject: Uwe Meffert's Re-issueing of Prof. Cube
Message-Id:
Date: Mon, 30 Nov 1998 22:48:11 +0000
Dear Cube Lovers,
I have written a website for Uwe Meffert (with input from both W. David
Joyner and David Byrden) that can be found at:
http://www.ue.net/mefferts-puzzles/
and was speaking with him earlier today.
Uwe is going to make another batch of Professor Cubes (5x5x5) in the next
week or so, and is taking orders through his site.
This is a subject that is often raised on the newsgroup, and I hope people
don't think of this as taking too much of a liberty.
The website contains a credit card order page, information about the puzzles
(including a solution to all of his popular puzzles) and multiple links to
other pages. Whilst I am not involved with the day to day running of the
website, if people would like their pages added to the links, please forward
the URL to this address WITH A SHORT SUMMARY that will appear with the link.
Puzzle available include some of the more recent ones (Orbix, Pyramorphix,
Megaminx, Prof Cube).
I am told that orders *may* still be on time for delivery before Chirstmas.
I hope this has been of use to you guys,
Andy Southern.
a.southern@ic.ac.uk
From cube-lovers-errors@mc.lcs.mit.edu Tue Dec 1 15:45:22 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil (sun28.aic.nrl.navy.mil [132.250.84.38])
by mc.lcs.mit.edu (8.9.1a/8.9.1-mod) with SMTP id PAA19069
for ; Tue, 1 Dec 1998 15:45:21 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Message-Id: <19981201060918.6975.rocketmail@send104.yahoomail.com>
Date: Mon, 30 Nov 1998 22:09:18 -0800 (PST)
From: Han Wen
Subject: Method for Solving the Professor's Cube (5x5x5)
To: Cube Lovers
Cc: Charles Lin , Keith Miller
Hi,
Okay, another so what, big deal. I finally solved the Professor's
Cube. For those who may not be familiar, the Professor's Cube is a
5x5x5 Rubik's cube. Whew that was hard. It took me a good 4 days to
figure out all the moves. Gees, it made the Megaminx seem like
child's play in comparison. Once again, Noel Dillabough's Puzzler
program was an invaluable tool to visualize and experiment with
various moves. Thanks Noel!
For those brave souls who would like to conquer this beast, the
following solution may provide some enlightenment. It's a layers
solution, in contrast to the corners-first solution that I have seen
posted on various web sites. Good luck to you. The Professor's Cube
is a truly challenging puzzle.
______________________________________________________
Method for Solving the Professor's Cube (5x5x5)
I will use Noel Dillabough's system for referring to various slices or
layers, as described in his Puzzler's F1 help.
________________________________________
Notation:
U - The upper slice
u - One slice away from the upper slice
e - The equator slice
d - One slice away from the lower slice
D - The lower slice
L - The leftmost slice
l - One slice away from the leftmost slice
m - The middle slice
r - One slice away from the rightmost slice
R - The rightmost slice
F - The facing slice
f - Once slice away from the facing slice
M - The facing middle slice
b - One slice away from the back face
B - The back slice
I will use the words "slice" and "layer" synonymously. A "face" is one
of the six outer slices; namely, U, D, L, R, F or B. Rotations of the
middle slices e, m or M will be in the same direction as the U, R and
F faces, respectively.
Let y denote one of the slices.
y - represents a clockwise 1/4 turn of the y-slice
y' - represents a counterclockwise 1/4 turn of the y-slice
y2 - represents a clockwise 1/2 turn of the y-slice
(For example, Rrm represents clockwise 1/4 turns using the RIGHT-hand
of the R, r and m slices. Ll represents clockwise 1/4 turns using the
LEFT-hand of the L and l slices.)
Finally, let's consider the pieces or cubelets on any given face.
There are four types of cubelets: corners, edges, centrals and a
center. For a given face, there are 4 corners, 12 edges, 8 centrals
and 1 center. With these four types and the intersection of any two
slices using Dillabough's notation, we can specify the location of any
cubelet. For example, consider the F-face:
LU-corner: the corner cubelet on the upper left-hand corner of the
F-face
re-central: the central cubelet adjacent and to the right of the
center cubelet of the F-face
_______________________________________
First Layer (U slice):
Solving the first layer is fairly straightforward. Basically the same
as solving the Rubik's cube. The central pieces are the only thing
really different.
_______________________________________
Second Layer (u slice):
1. First, solve for the mid-central pieces (F-face mu, B-face mu,
L-face Mu, R-face Mu). Get one of the mid-central piece on the same
color face, and then rotate it into position by using the "free lane"
from the opposite face. For example, let's say we want have a
mid-central piece at the re position of the F-face. Use the D-"free
lane" of the B face to position the mid-central piece without
affecting your newly completed U slice, by moving: B2 U2 F' U2 B2.
2. Now, solve for the left and right central pieces (F-face lu, ru,
L-face bu, fu, etc). Here's where we'll use a genuinely new move.
Position one of the left/right central pieces on the D-face so that it
and the position you want to move the cubelet into lie in the save
vertical slice. For example, let's say we want to move the left
central cubelet into the F-face lu position. Position the left
central cubelet at the D-face lb position and perform the following
u-layer DF Swing move:
>From the D-face lb position: l d' l' d' l d2 l'
See how that works? The corresponding move at the D-face rb position
is:
>From the D-face rb position: r' d r d r' d2 r
This same concept is used to move the left/right central pieces into
position for both the Second (u-slice) and Fourth (d-slice) layers.
"Hey, what if my left/right central piece is on the F face? How do I
move the piece to the D face so that I can apply this move?" Good
question. Position the piece on the F-face ld or rd position and
apply the corresponding move described above. That should move the
cubelet to the D face where you can then apply the move again to move
it into the correct left/right central position.
3. Finally, solve for the left and right edges (F-face and B-face Lu,
Ru). Use the classic Rubik's cube move to rotate an D-edge piece into
one of the middle layer edge positions. Namely, if the cubelet is at
the F-face rD or lD position and the destination position is F-face Ru
or Lu then perform the following:
F-Edge Swing Moves:
Destination position F-face Ru: D' R' D R F' R F R'
Destination position F-face Lu: D L D' L' F L' F' L
_______________________________________
Third Layer (e slice):
1. Solve for the left/right central pieces (F-face le, re, L-face be,
fe, etc). You'll notice that the DF Swing moves will not work here.
Darn. Instead, we'll use the F-Edge Swing move adapted for the l and
r slices. Position the cubelet at the F-face md position then perform
the following:
F-Central Swing Moves:
Destination position F-face Re: d'r'dD rR f'F' r fF r'R'
Destination position F-face Le: d l d'D' l'L' fF l' f'F' lL
"Hey, what if my left/right central piece is on the D face? How do I
move the piece to the F face so that I can apply this move?" Same
problem. Position the cubelet at the D-face rM position then apply
the Re F-Central Swing move.
2. Solve for the left and right edges (F-face and B-face Le, Re).
Again, a slight variation of the F-Edge Swing move will do. Position
the edge piece on the F-face mD position and perform the following:
e-Layer F-edge Swing Moves:
Destination position F-face Re: D' R' D rR F' R F r'R'
Destination position F-face Le: D L D' L'l' F L' F' Ll
______________________________________
Fourth Layer (d slice):
1. First, solve for the mid-central pieces (F-face md, B-face md,
L-face Md, R-face Md). This is one of the most difficult steps. The
mid-central pieces will be on either the d-slice or on the D-face. To
move them into there correct positions, you'll need to use a few
modified Rubik's cube moves:
Place the D-face as the U-face when applying these moves:
The following sets of cubelets are affected by these moves:
cL = (central L-face Lu, edge U-face LM and central U-face lM)
cR = (central R-face Ru, edge U-face RM and central U-face rM)
cF = (central F-face mu, edge U-face mF and central U-face mf)
Mid-central Tricycle:
move: T2(U) = F2 f2 Uu Ll r'R' F2 f2 L'l' rR Uu F2 f2
action: Permutes the three sets of cubelets (cL, cR, cF) clockwise:
Mid-central Bi-Flip Tricycle:
move: S2(B) = L'l' rR bB Ll r'R' U2u2 L'l' rR Bb Ll r'R'
action: Permutes the three sets of cubelets (cL, cR, cF) clockwise and
flips the cR and cF sets. Let's clarify "flipping". Let's say for
the cR set you have the colors: blue, (blue, yellow), yellow
corresponding to the three cubelets. After flipping the cR set you'll
have the colors: yellow, (yellow, blue), blue.
Use these two moves to position all the mid-central pieces for the
Fourth Layer. Now, if you're lucky, and Murphy's Law says that you
will be, you may end up in a configuration where you'll have three of
the mid-central pieces positioned properly, but the fourth mid-central
position will be on the D-face. Okay, now we're going to start having
fun. Position the central cubelet at the D-face lM position (i.e. on
the left-hand side). Place the D-face as the U-face and then apply
the following sequence of moves:
S2(B) T2(U') U2 T2(U) S2(B') U' S2(B')
Yes, all that trouble just to move one mid-central cubelet from the
U-face to the F-face.
2. Whew, congratulate yourself if you've made it this far. Now, solve
for the left/right central cubelets, (F-face ld, rd, L-face bd, fd,
etc).
Position the left central cubelet at the D-face lf or rf position and
perform the following d-layer DF Swing move:
>From the D-face lf position: l d l' d l d2 l'
>From the D-face rf position: r' d' r d' r' d2 r
3. Solve for the left and right edges (F-face and B-face Ld, Rd).
Again, a slight variation of the F-Edge Swing move will do. Position
the edge piece on the F-face lD or rD position and perform the
following:
d-Layer F-edge Swing Moves:
Destination position F-face Rd: D' R' D mrR F' R F m'r'R'
Destination position F-face Ld: D L D' L'l'm F L' F' Llm'
______________________________________
Fifth Layer (D slice):
1. Solve for the corner cubelets using standard Rubik's cube moves.
First, position the corners in their correct locations using the usual
corner swappers:
Adjacent corners swap: R' D' R F D F' R' D R D2
Diagonal corners swap: R' D' R F D2 F' R' D R D
And then rotate or twist the corners in position using Sune's move:
Sune's 3-corner twister: : R' D' R D' R' D2 R D2
2. Solve for the mid-edges (mF, RM, mB, LM) using a slight
modification to the Tricycle moves.
Place the D-face as the U-face when applying these moves:
Mid-edge Tricycle:
move: F2 U Ll r'R' F2 L'l' rR U F2
action: Permutes the three edges (LM, RM, mF) clockwise:
Mid-edge Bi-Flip Tricycle:
move: L'l' rR B Ll r'R' U2 L'l' rR B Ll r'R'
action: Permutes the three edges (LM, RM, mF) clockwise and flips the
RM and mF.
3. Solve for the left/right edges (lF,rF, Rf, Rb, lB, rB, Lf, Lb).
Now, we're going to have some serious fun. The hardest part of this
step is not getting lost while performing the long sequence of moves.
Also while spinning all these slices, another difficulty is preventing
the cube from exploding and keeping the central pieces from twisting
around.
Again, place the D-face as the U-face with applying these collection
of moves:
LR-edge Tricycle:
move: F2 U Lm'R' F2 L'mR U F2
action: Permutes the three pairs of edges ((Lf,Lb), (Rf,Rb), (lF,rF))
clockwise:
LR-edge Bi-Flip Tricycle:
move: L'mR B Lm'R' U2 L'mR B Lm'R'
action: Permutes the three pairs of edges ((Lf,Lb), (Rf,Rb), (lF,rF))
clockwise and flips the Rf, Rb, lF and rF.
To get those last remaining cubelets in place, a few more exotic moves
are necessary:
Definitions:
T(x) = F2 U x F2 x' U F2
x1 = L r' R'
x2 = L l R'
x3 = L m' R'
(T(x) is a generalized form of the Mid-edge Tricycle)
X1 = T(x1) T(x1) T(x1)
X2 = T(x2) T(x2) T(x2)
X3 = T(x3)
Name: Double pair F swap
Description: Swap two pairs of edges: (lF - Lb) and (rF - Rb)
Move: X2 X1
Name: Double pair F cross swap
Description: Swap two pairs of edges: (lF -Lf ) and (rF -Rf )
Move: X1 X2
Name: Double pair R swap
Description: Swap two pairs of edges: (Rb - Lf) and (Rf - lF)
Move: X2 X1 X3
Name: Double pair R cross swap
Decription: Swap two pairs of edges: (Rb - rF) and (Rf - Lb)
Move: X1 X2 X3
Name: Double pair L swap
Description: Swap two pairs of edges: (Lb - Rf) and (Lf - rF)
Move: X3 X2 X1
Name: Double pair L cross swap
Description: Swap two pairs of edges: (Lb - lF) and (Lf - Rb)
Move: X3 X1 X2
Name: LRL-edge Bi-Flip Tricycle
Description: Permutes (lF, Lf, Rf) edges clockwise and flip lF and Lf
edges
Move: X3 X1
Name: LLR-edge Bi-Flip Tricycle
Description: Permutes (lF, Lb, Rb) edges clockwise and flip Lb and Rb
edges
Move: X1 X3
Name: RRL-edge Bi-Flip Tricycle
Description: Permutes (rF, Lf, Rf) edges clockwise and flip Lf and Rf
edges
Move: X2 X3
Name: RLR-edge Bi-Flip Tricycle
Description: Permutes (rF, Lb, Rb) edges clockwise and flip rF and Lb
edges
Move: X3 X2
With these collection of moves, you should be able to finish off the
Professor's Cube! *Sigh*
-Han-
P.S. Thanks "Professor" Meffert. For those folks like myself who have
wrestled and completed your 5x5x5 cube, we can only ask and plead,
"What's Next?!!" :)
From cube-lovers-errors@mc.lcs.mit.edu Tue Dec 1 19:18:47 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil (sun28.aic.nrl.navy.mil [132.250.84.38])
by mc.lcs.mit.edu (8.9.1a/8.9.1-mod) with SMTP id TAA20686
for ; Tue, 1 Dec 1998 19:18:47 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Date: Mon, 30 Nov 1998 23:13:51 -0500
From: michael reid
Message-Id: <199812010413.XAA11878@euclid.math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: new types of cyclic shifters
a few months ago, i introduced the position superflip composed with
four spot and showed that it had a new type of cyclic shifting
property. i've now found some new ways to generalize cyclic shifting,
and this in turn suggests some new positions to consider.
first, some brief review.
the position superflip is central, so it commutes with all turns.
therefore, if x y produces superflip, so does y x . we can shift
one turn at a time, removing the first turn of the sequence and
shifting it to the end. in other words, if m produces superflip,
then x m = m x for all turns x . clearly, any position m with
the property that x m = m x for all turns x , is central, so the
only such positions are superflip and start.
i showed in an earlier message, that if x is any turn and m is
superflip composed with four spot, then x m = m y , where y is
x conjugated by the cube rotation C_U2 . more generally, we can
ask for positions m such that for any turn x , there is another
turn y satisfying x m = m y . for such a position, we can
cyclically shift any maneuver, one turn at a time, by replacing
the turn x^(-1) at the beginning with the corresponding y^(-1)
at the end. some other positions with this property are: four
spot, six spot, six spot composed with superflip.
a new way to generalize this is to consider positions m such that
for any turn x , we have x m = n x , where n is the same
pattern as m , but perhaps in a different orientation. for such
a position, we can cyclically shift any maneuver, by shifting the
first turn to the end, and then conjugating by the appropriate cube
symmetry. for example, consider the position in which the UFR corner
is twisted clockwise, and the other seven corners are twisted
counterclockwise. (i'll call this "1-7-twist" for now, but this
pattern needs a better name.) this position is created by
U F2 B' U B U D2 R2 U2 B U' D2 B U F2 B L2 B2
now, cyclically shift the U at the beginning to the end to get
F2 B' U B U D2 R2 U2 B U' D2 B U F2 B L2 B2 U
which produces a different orientation of the same position; this
time, the ULF corner is twisted clockwise. now conjugate this
maneuver by C_U to get
R2 L' U L U D2 B2 U2 L U' D2 L U R2 L F2 L2 U
which produces the original position, in its original orientation.
actually, there are 3 cube symmetries by which one could conjugate,
since the position has 3-fold symmetry. another position with
this type of cyclic shifting property is 1-7-twist composed with
superflip.
we can combine both types of generalizations, and ask for positions
m that have the property that for any turn x , we have x m = n y ,
where y is another turn, and n is the same pattern as m , but
perhaps in a different orientation. for such positions, we can
cyclically shift any maneuver by replacing x^(-1) at the beginning
of the maneuver by the corresponding y^(-1) at the end, and then
conjugating the whole maneuver by the appropriate cube symmetry.
two such positions are: 1-7-twist composed with four spot, and
1-7-twist composed with four spot composed with superflip.
here's all the examples of cyclic shifters that i know, along with
minimal maneuvers:
1. central positions
start (0q*, 0f*)
superflip
R' U2 B L' F U' B D F U D' L D2 F' R B' D F' U' B' U D' (24q*, 22f)
F' B' D2 L' B2 L2 F2 U' D B' D2 R L D' F2 U' L2 D' F2 D' (20f*, 28q)
2.
four spot F2 B2 U D' R2 L2 U D' (12q*, 8f*)
six spot F B' U D' R L' F B' (8q*, 8f*)
four spot composed with superflip
U2 D2 L F2 U' D R2 B U' D' R L F2 R U D' R' L U F' B' (26q*, 21f)
F U2 R L D F2 U R2 D F2 D F' B' U2 L F2 R2 B2 U' D (20f*, 28q)
six spot composed with superflip
R' U D R' U F' D R' B U' L' U' F' D F' B' D' R' F D F D' R2
(24q*, 23f)
U2 F B' R F L2 F2 D B2 D2 R2 B' L2 F' D2 R2 D' B R B2 (20f*, 30q)
3.
1-7-twist
F R' U' L' F' U' B' L' U' R2 F L' D' R' F' D' B' R' D' L2 (22q*, 20f)
F R2 L' F L F B2 U2 F2 L F' B2 L F R2 L D2 L2 (18f*, 26q)
1-7-twist composed with superflip
F R F' R U B L D B D' L' D L F B' R B D F R' (20q*, 20f)
F R L' B L D F' L2 B D2 B' L' F2 B' D B U L B (19f*, 22q)
4.
1-7-twist composed with four spot
F B2 L' U' B' L U' D L2 U R' U' R F' U' L' F D B' U' (22q*, 20f*)
1-7-twist composed with four spot composed with superflip
F U' R' L F' U R L' U2 D' B R L F' R D' R' F2 L U' (22q*, 20f*)
as usual, i give a maneuver which is minimal in both metrics whenever
this is possible. i don't claim that i've found all positions in
these categories, but these are all that i know. if you find any others,
they'd be good candidates for positions far from start.
mike
From cube-lovers-errors@mc.lcs.mit.edu Wed Dec 2 14:51:54 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil (sun28.aic.nrl.navy.mil [132.250.84.38])
by mc.lcs.mit.edu (8.9.1a/8.9.1-mod) with SMTP id OAA24883
for ; Wed, 2 Dec 1998 14:51:53 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Date: Mon, 30 Nov 1998 23:39:25 -0500
From: michael reid
Message-Id: <199812010439.XAA11918@euclid.math.brown.edu>
To: cube-lovers@ai.mit.edu
Subject: asymmetric local maxima
based on the previous analysis, i can now give 2 asymmetric positions
that are local maxima, namely
1-7-twist composed with four spot, and
1-7-twist composed with four spot composed with superflip.
all previous examples of local maxima had some symmetry (although
jerry bryan recently gave a bunch of new local maxima at distance 12q;
perhaps these contain some other examples.)
to show that a position is locally maximal, i must give a minimal
maneuver that ends with each possible quarter turn. bear in mind
that F2 = F F = F' F'.
1-7-twist composed with four spot:
F L' U' B' L U' D L2 U R' U' R F' U' L' F D B' U' F2 (22q*)
U' R' D L F' U' L' B U' B' U F2 U' D F R' U' F' L R2 (22q*)
D' L B' D' L' F B2 D' F' U B R' D' B' L D' L' D R2 U (22q*)
F B2 L' U' B' L U' D L2 U R' U' R F' U' L' F D B' U' (22q*)
R' U L' F B D' F L' D B' L F' L' F D' B D' B L' U R' B (22q*)
U' F L' U' B' L D R' U' R L2 F' U' L' F U' D F2 U B' (22q*)
U' R' D L F' U' L' B U' B' U F2 U' D F R' U' F' R2 L (22q*)
U' F' L U' L' U R2 U' D R B' U' R' F B2 U' F' D B L' (22q*)
F' U' F B2 L' U' B' L U' D L2 U R' U' R F' U' L' F D (22q*)
L B' D' L' F B2 D' F' U B R' D' B' L D' L' D R2 U D' (22q*)
1-7-twist composed with four spot composed with superflip:
R' D' R B' R L F U2 D' R L' U B' R' L U' B U' R F2 (22q*)
B U' R U' F B' R' U F' B U2 D' L F B R' B D' B' R2 (22q*)
B D R B R L' U F' B' L' F L D' B2 U R' B L B D' U (22q*)
F U' R' L F' U R L' U2 D' B R L F' R D' R' F2 L U' (22q*)
U' D2 L F B R' F U' F' R2 B D' L D' F B' L' D F' B (22q*)
U' F L' F B R U' D2 F B' D L' F' B D' L D' F R2 B' (22q*)
U' F L2 B' D' B L' F B R U2 D' F' B U L' F B' U' L (22q*)
B' D R L' U' D2 F R L B' R U' R' B2 L D' F D' R L' (22q*)
R L' D F' R' L D' F D' R B2 L' U' L B' R L F U' D2 (22q*)
these positions are also strong local maxima in the face turn metric.
1-7-twist composed with four spot:
U D' B2 D F' D' F R' D' B' R U L' D' R2 L F' D' R' F (20f*)
F L' U' B' L U' D L2 U R' U' R F' U' L' F D B' U' F2 (20f*)
U' R' B D F' U' F B2 L' U' B' L U' D L2 U R' U' R F' (20f*)
F' U' L' B U' B' U F2 U' D F R' U' F' R2 L U' L' D R (20f*)
D R' D' R B' D' L' B U F' D' F B2 R' D' B' R U D' R2 (20f*)
F B2 D' F' U B R' D' B' L D' L' D R2 U D' R F' D' R' (20f*)
D' L B' D' L' F B2 D' F' U B R' D' B' L D' L' D R2 U (20f*)
D' B D F2 B R2 B2 D F2 B' D B D' R2 U2 B2 U2 D2 F' U2 (20f*)
F B2 L' U' B' L U' D L2 U R' U' R F' U' L' F D B' U' (20f*)
L2 U R2 B D2 B L2 U' F' B2 D F D' R2 L2 D2 B2 L2 D' B (20f*)
B L2 U F' L2 U' F' B U F D' L2 U F2 D F2 D' R2 U2 B2 (20f*)
U' F L' U' B' L D R' U' R L2 F' U' L' F U' D F2 U B' (20f*)
U' R' D L F' U' L' B U' B' U F2 U' D F R' U' F' R2 L (20f*)
D2 F2 U' L2 U L2 D B2 U' L D R L' D' B2 L' D B2 R L2 (20f*)
U' F' L U' L' U R2 U' D R B' U' R' F B2 U' F' D B L' (20f*)
F' U' F B2 L' U' B' L U' D L2 U R' U' R F' U' L' F D (20f*)
F2 B R2 D F' R2 D' F' B D F U' R2 D F2 U F2 U' L2 D2 (20f*)
L B' D' L' F B2 D' F' U B R' D' B' L D' L' D R2 U D' (20f*)
1-7-twist composed with four spot composed with superflip:
R L F' L U' L' F2 R D' B D' R' L B' D R L' U' D2 F (20f*)
R' D' R B' R L F U2 D' R L' U B' R' L U' B U' R F2 (20f*)
R L F U' D2 R' L D B' R L' D' B D' L F2 R' U' R F' (20f*)
R F' L U' L' F2 R D' B D' R' L B' D R L' U' D2 F R (20f*)
B U' R U' F B' R' U F' B U2 D' L F B R' B D' B' R2 (20f*)
F' B D' L D' F R2 B' U' B R' F B L U' D2 F' B D R' (20f*)
F D' B' R2 D' F R U B' L D2 B R2 U2 B2 U' R L2 B U (20f*)
R2 U' F B2 R U R D' L' B2 D' R B U L' F D2 L B2 U2 (20f*)
F U' R' L F' U R L' U2 D' B R L F' R D' R' F2 L U' (20f*)
U' D2 L F B R' F U' F' R2 B D' L D' F B' L' D F' B (20f*)
B L U L D' R' F2 D' L F U R' B D2 R F2 U2 R2 U' B2 (20f*)
U' F L' F B R U' D2 F B' D L' F' B D' L D' F R2 B' (20f*)
U' F L2 B' D' B L' F B R U2 D' F' B U L' F B' U' L (20f*)
D' F2 B R D R U' L' F2 U' R F D L' B U2 L F2 D2 L2 (20f*)
B' D R L' U' D2 F R L B' R U' R' B2 L D' F D' R L' (20f*)
R L' U' D2 F R L B' R U' R' B2 L D' F D' R L' F' D (20f*)
R L' D F' R' L D' F D' R B2 L' U' L B' R L F U' D2 (20f*)
F D' R B2 L' U' L B' R L F U' D2 R' L D B' R L' D' (20f*)
mike
From cube-lovers-errors@mc.lcs.mit.edu Wed Dec 2 20:31:57 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil (sun28.aic.nrl.navy.mil [132.250.84.38])
by mc.lcs.mit.edu (8.9.1a/8.9.1-mod) with SMTP id UAA27103
for ; Wed, 2 Dec 1998 20:31:57 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Date: Wed, 2 Dec 1998 00:49:45 -0500
Message-Id: <0008F2BF.C22092@scudder.com>
From: Jacob_Davenport@scudder.com (Jacob Davenport)
Subject: Re: Method for Solving the Professor's Cube (5x5x5)
To: Cube Lovers
It is difficult. I also took a long time to solve it, but I used a very
different solution. Check out www.wunderland.com/WTS/Jake/5x5x5.html for
my solution. Perhaps you can combine the best moves of both solutions to
find a way of solving this interesting puzzle that you find most pleasing.
If you do use my solution and have any comments about how I can make it
better, either in my writing or my moves, please let me know.
-Jacob
From cube-lovers-errors@mc.lcs.mit.edu Thu Dec 3 15:12:43 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil (sun28.aic.nrl.navy.mil [132.250.84.38])
by mc.lcs.mit.edu (8.9.1a/8.9.1-mod) with SMTP id PAA02280
for ; Thu, 3 Dec 1998 15:12:42 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
To: cube-lovers@ai.mit.edu
From: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
Subject: Re: Method for Solving the Professor's Cube (5x5x5)
Date: 2 Dec 1998 18:21:22 GMT
Organization: California Institute of Technology, Pasadena
Message-Id: <7440f2$q5v@gap.cco.caltech.edu>
References:
Han Wen writes:
>For those brave souls who would like to conquer this beast, the
>following solution may provide some enlightenment. It's a layers
>solution, in contrast to the corners-first solution that I have seen
>posted on various web sites. Good luck to you. The Professor's Cube
>is a truly challenging puzzle.
>______________________________________________________
>Method for Solving the Professor's Cube (5x5x5)
[snip]
Well, since we're sharing solutions, here's my solution to the 5x5x5:
First, a preliminary exercise that
should be mastered before the solution is attempted:
Let's ignore all the corners and all the cubies adjacent or
diagonally adjacent to the corners. (In other words, ignore the
"supercorners," where "super" is a prefix meaning "two layers deep.")
Ignore the centers, too. Paint all of them black, if you want. :-)
Now, all we have left are the 12 "superedges." Each superedge is
composed of a normal edge piece and two attached edge centers. Or,
in other words, each of the 24 edge centers are attached to
an edge piece face. In a normal messed-up cube, these edge
centers will not match their edge piece faces. Our goal in this
exercise will be to match all the edge centers with their edge piece
faces.
Note that superface turns never destroy an edge center pairing.
Now, consider the following sequence of moves:
1. Rotate any face (NOT superface) 180 degrees.
2. Turn any center slice (as much as you want).
3. Rotate the same face in step 1 180 degrees.
(i.e., perform the inverse move of step 1.)
Now, if you chose the center slice to be parallel to the face,
obviously this sequence doesn't do anything. Ditto for when you
turned the center slice some multiple of 360 degrees.
In all other cases, this will essentially perform two swaps of
edge centers. Step 1 swaps two pairs of edge centers all around the
face, but one of those swaps gets undone by Step 3. Step 2
moves the other pair out of the way and puts another pair in its
place to be swapped again. So, if you choose wisely, you can
increase the number of correctly matched edge center pairs by this
move.
Many of these moves, interspersed with superface turns, will
allow you to match all the edge center pairs. Practice this on
your cube.
Thus ends the preliminary exercise. Note that all the moves in
the exercise do not disturb the individual supercorners (well, one
move does for a bit, but then it undoes the damage) but does change
their orientation with respect to each other.
Now, the solution!
Step 1. Ignore the superedges and the centers. You now have what
is equivalent to a 4x4x4. Solve it.
Step 2. Match the edge centers with the edges as detailed in the
preliminary exercise.
Step 3. You now have a cube with correct supercorners (as done in
step 1) and correct superedges (as done is step 2). This means
that your cube is equivalent to a 3x3x3, using only superface
turns. Solve it.
Step 4. Tada! Your 5x5x5 is now solved.
--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
O*e T*o: "Thre* *our fi*e s*x; se*en *ight *ine, *en!"
From cube-lovers-errors@mc.lcs.mit.edu Thu Dec 3 16:32:04 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil (sun28.aic.nrl.navy.mil [132.250.84.38])
by mc.lcs.mit.edu (8.9.1a/8.9.1-mod) with SMTP id QAA02610
for ; Thu, 3 Dec 1998 16:32:04 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Message-Id:
In-Reply-To: <872566CD.007745CF.00@notes.dtint.com>
Date: Wed, 2 Dec 1998 23:15:57 -0700
To: Cube Lovers
From: Steve LoBasso
Subject: Re: Method for Solving the Professor's Cube (5x5x5)
Although I use a different method, centrals first, edge combinations, edge
parity corrections, finish using 3x3x3 solution.
I was playing with a layered solution last week also, amazing coincidence.
Much of my solution is the same as Han's.
Most of the differences are in the 4th and 5th layers.
To move the 4th layer mid centrals into place:
central D-face bm to central F-face dm: F l D l' D' F'
If there are no central pieces in the bottom central area, simply move a
bottom central up causing another central to go down.
To move 4th layer edges into place:
edge L-face Db to edge F-Rd: R' D' r D R D' r'
edge R-face Db to edge F-Ld: L D l' D' L' D l
My 5th layer edge moves are a bit different but I haven't had time to write
them down with this terminology.
--
Steve LoBasso mailto:slobasso@dtint.com
Digital Technology International or mailto:slobasso@hotmail.com
500 West 1200 South, Orem, UT, 84058 http://members.tripod.com/~slobasso
(801)226-6142 ext.265 FAX (801)221-9254
From cube-lovers-errors@mc.lcs.mit.edu Thu Dec 3 20:49:15 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil (sun28.aic.nrl.navy.mil [132.250.84.38])
by mc.lcs.mit.edu (8.9.1a/8.9.1-mod) with SMTP id UAA03518
for ; Thu, 3 Dec 1998 20:49:14 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Date: Thu, 3 Dec 1998 17:47:33 -0500
Message-Id: <00096300.C22092@scudder.com>
From: Jacob_Davenport@scudder.com (Jacob Davenport)
Subject: (5x5x5) edge parity corrections
To: Cube Lovers
I don't like the edge parity correction move that I use in my solution, and
I'm hoping that someone can give me a better one.
The parity problem is found in 5x5x5 cubes (and 4x4x4 cubes, I understand)
when two of the edges right next to the corners (which I call "wings") are
switched. Some fairly simple moves can get all three edges in line with
each other, but half the time two wings need to be switched. By the time I
figure this out when doing a 5x5x5 cube, I've solved most of it, and my
parity fixing move messes up many of the edges I've been working on.
How do other people fix this problem?
-Jacob
From cube-lovers-errors@mc.lcs.mit.edu Fri Dec 4 11:30:15 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil (sun28.aic.nrl.navy.mil [132.250.84.38])
by mc.lcs.mit.edu (8.9.1a/8.9.1-mod) with SMTP id LAA05048
for ; Fri, 4 Dec 1998 11:30:14 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Message-Id: <19981204043241.26555.rocketmail@send106.yahoomail.com>
Date: Thu, 3 Dec 1998 20:32:41 -0800 (PST)
From: Han Wen
Subject: Re: Method for Solving the Professor's Cube (5x5x5)
To: Steve LoBasso
Cc: Cube-Lovers@ai.mit.edu
Hi,
> To move the 4th layer mid centrals into place:
>
> central D-face bm to central F-face dm: F l D l' D' F'
>
This is a stunningly elegant move. You've reduce the difficulty in
solving the 4th layer by an order of magnitude. I tried this move
out, actually it swaps two centrals:
F-face md <-> D-face lM (mid central swap)
F-face rd <-> D-face lf (right central swap)
Beautiful move.
There is one particular move that I haven't figured out yet. It pops
up occasionally when I solve the edges of the D-layer. Sometimes I
end up with every cubie in place except for two right centrals on
adjacent faces. For example: F-face rD and L-face Lf. The two pieces
only need to be swapped. No flipping is needed. Does anyone know how
to perform this move? I've been beating the Puzzler program for a
while, but I have been unsuccessful so far.
______________________________________________
Han Wen
Applied Materials
3050 Bowers Ave, MS 1145
Santa Clara, CA 95054
e-mail: Han_Wen@amat.com / hansker@yahoo.com
From cube-lovers-errors@mc.lcs.mit.edu Fri Dec 4 13:17:06 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil (sun28.aic.nrl.navy.mil [132.250.84.38])
by mc.lcs.mit.edu (8.9.1a/8.9.1-mod) with SMTP id NAA05373
for ; Fri, 4 Dec 1998 13:17:05 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Date: Thu, 3 Dec 1998 23:58:19 -0500 (EST)
From: der Mouse
Message-Id: <199812040458.XAA26116@Twig.Rodents.Montreal.QC.CA>
To: Cube-Lovers@ai.mit.edu
Subject: Re: (5x5x5) edge parity corrections
> The parity problem is found in 5x5x5 cubes (and 4x4x4 cubes, I
> understand) when two of the edges right next to the corners (which I
> call "wings") are switched.
Yes, it does occur equally on the 4-Cube. Though I have never seen
one, I feel certain that similar parity problems will occur on all
higher-order Cubes as well, though above order 5 there will be multiple
distinct types of "wings", each of which will have its own comparable
potential problem.
Note that the problem goes away entirely if cube faces are marked such
that symmetrically placed face cubies are not visually
indistinguishable, because the parity problem in question always occurs
in conjunction with a similar parity problem on face cubies, but the
latter is invisible on most cubes.
> [...] half the time two wings need to be switched.
> How do other people fix this problem?
Most briefly, how I do it is to make a single quarter-turn of a slice
containing one of the wing pieces involved, then fix up the damage by
moving wings back into place using commutators rather than slice moves.
der Mouse
mouse@rodents.montreal.qc.ca
7D C8 61 52 5D E7 2D 39 4E F1 31 3E E8 B3 27 4B
From cube-lovers-errors@mc.lcs.mit.edu Fri Dec 4 14:26:23 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil (sun28.aic.nrl.navy.mil [132.250.84.38])
by mc.lcs.mit.edu (8.9.1a/8.9.1-mod) with SMTP id OAA05692
for ; Fri, 4 Dec 1998 14:26:23 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
To: cube-lovers@ai.mit.edu
From: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
Subject: Method for Solving the 4x4x4
Date: 4 Dec 1998 17:26:00 GMT
Organization: California Institute of Technology, Pasadena
Message-Id: <7495v8$3nv@gap.cco.caltech.edu>
References:
I have been told that my solution for the 5x5x5 includes
knowing how to solve the 4x4x4, which is of course not
trivial. With the post asking about the parity problem, I
thought I might as well post my solution to the 4x4x4.
Yes, the biggest barrier is the parity problem where two
adjacent edge cubies are flipped. My earliest attempt at a 4x4x4
solution was the following:
1. Match all the centers.
2. Match all the edges.
3. You now have a 3x3x3. Solve.
Unfortunately, with the parity problem you can often
end up with an unsolvable 3x3x3 by the time you get to step 3.
Any simple moves that fix the parity problem tend to mess up the
rest of the cube quite badly -- I wrestled with this problem
a long time until I realized one thing:
Most solutions of the 3x3x3 treat the centers as static, using them
as "anchors" for the entire cube. But this is entirely unnecessary!
If you solve the 3x3x3 while IGNORING the centers, you will eventually
get a solved cube where the centers are either in the "6 dots" or
"4 dots" situation well known to cubists -- and these have rather
simple solutions, essentially consisting of a slice turn conjugated
with another slice turn.
So, my most favorite 4x4x4 solution is now:
1. Match all the edges.
2. Solve the parity problem, if necessary (postpone until after
step 3 if desired).
3. Ignore the centers and treat the cube as a 3x3x3. Solve.
4. Solve the centers.
Okay. Now to qualify the solution. Part 1 is simple and can be
done anyway you wish (the move rF2r'F2 will be rather useful in
the later stages). Part 3 is simple, with the caveat that you
may be treating the "centers" in the wrong manner!
Part 2 stems from the fact that the cube apparently has an "even"
permutation (a 2-cycle involving two edge pieces), an apparent
paradox since 2-cycles should not exist (e.g., on the 3x3x3 it is
impossible to swap exactly two edges). The reason this is only
an "apparent" paradox, however, is because of the misassumption that
the centers of the 4x4x4 are static, which they certainly are not!
In fact, just rotate one slice incident on your 2-cycle, and you
have magically turned the 2-cycle into a 5-cycle, which is perfectly
solvable!
Personally, I solve the 5-cycle by two or more 3-cycles, which generally
take on the form:
FR'F' r FRF' r'
This move performs a cycle on the three edges fUR, FUr, and FDr,
without disturbing the corners, but doing rather annoying things to
the centers. (This move is an extension of the perhaps-not-so-well-known
sequence for the 3x3x3: FR'F'LR'DRD'L'R that rotates 3 edges.)
For FR'F' you may substitute any sequence of moves that brings
your desired edge piece (in this case fUR) to the FUr position,
as long as it does not disturb any other edges on the r slice (specifically,
the edges FDr, BUr, and BDr). You will also have to substitute
the inverse of your sequence for FRF'. As an example, the move
F'L2F r F'L2F r'
cycles BdL, FUr, and FDr. You may also use r2 instead of r and r', which
means that BDr is affected instead of FDr.
And finally, step 4: the centers. This is solved by a
generalization of the "6-dots" rule.
This move creates "6 dots":
u'r'ur
This permutes bUr, Fur, and buR, as well as their "opposites"
fDl, Bdl, and fdL, while affecting no other cubies. This is two 3-cycles
on 6 faces, which is rather unwieldy, so I conjugate (is that the
right word?) it with a simple face turn to get
u'r'ur F r'u'ru F'
which permutes Fur, Fdr, and buR in a simple 3-cycle. Both Fur and Fdr
are on the same face, which makes this move rather easy to deal with.
Especially, if one of those is the right color already, it can be
involved in the 3-cycle without "increasing error."
I think I may have extended my personal jargon a bit more into this
post -- if you wish to understand anything in this post or, conversely,
would like to teach me more "Standard" jargon, please e-mail me.
--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
O*e T*o: "Thre* *our fi*e s*x; se*en *ight *ine, *en!"
From cube-lovers-errors@mc.lcs.mit.edu Fri Dec 4 15:28:52 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil (sun28.aic.nrl.navy.mil [132.250.84.38])
by mc.lcs.mit.edu (8.9.1a/8.9.1-mod) with SMTP id PAA05995
for ; Fri, 4 Dec 1998 15:28:52 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Message-Id:
From: Noel Dillabough
To: "'Cube Lovers'"
Cc: "'Jacob_Davenport@scudder.com'" ,
"'noel@mud.ca'"
Subject: RE: (5x5x5) edge parity corrections
Date: Fri, 4 Dec 1998 12:44:07 -0500
The parity problem can be solved on a 4x4x4 or 5x5x5 by using the
following move (can be pasted into puzzler's move macro):
r2D2l1D2l1D1l3r3d2l1r1D3l3r3d2B2r1B2l3B2l1B2r2
For the 4x4x4, this is all that is needed, but for the 5x5x5, two
crosses (centre edges) are swapped. So you'll need to use the following
to solve the crosses:
First, get the crosses across from each other with:
F2l3F2e1l2e3l2F2l1F2
Now swap the opposite crosses with:
R2e1l2e3l2R2e1l2e3l2
Parity problem solved...
If anyone has a better solution to this rather long one, let me know,
I'm sure some moves could be shaved off.
From cube-lovers-errors@mc.lcs.mit.edu Fri Dec 4 16:11:19 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil (sun28.aic.nrl.navy.mil [132.250.84.38])
by mc.lcs.mit.edu (8.9.1a/8.9.1-mod) with SMTP id QAA06137
for ; Fri, 4 Dec 1998 16:11:18 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Message-Id: <199812041806.NAA21024@pike.sover.net>
Date: Fri, 04 Dec 1998 13:07:36 -0500
To: Jacob Davenport
From: Nichael Lynn Cramer
Subject: Re: (5x5x5) edge parity corrections
Cc: Cube Lovers
In-Reply-To: <00096300.C22092@scudder.com>
Jacob Davenport wrote:
>I don't like the edge parity correction move that I use in my solution, and
>I'm hoping that someone can give me a better one.
>
>The parity problem is found in 5x5x5 cubes (and 4x4x4 cubes, I understand)
>when two of the edges right next to the corners (which I call "wings") are
>switched. Some fairly simple moves can get all three edges in line with
>each other, but half the time two wings need to be switched. By the time I
>figure this out when doing a 5x5x5 cube, I've solved most of it, and my
>parity fixing move messes up many of the edges I've been working on.
>
>How do other people fix this problem?
>
>-Jacob
Hi Jacob
In both cases (4X and 5X) I solve this problem in the following way:
1] I solve the rest of the cube, leaving me with the two "switched wings"
(in your terminology).
2] I then arrange things so both "wings" are on the same
"off-center-slice". (Also it will always be the case that both of these
winds are now on the same face.)
This will be easy to do using the 3-wing swapping operators.
3] At this point I now rotate the "off-center-slice" containing the
"switched wings" by a quarter turn.
As a result of this move it will be the case that that the
"off-center-slice" now has one of the previously "switched wings" in its
"correct cubicle". The other three "wings" will be now be in a cyclic
permutation.
4] Since --from your note above-- I assume you understand how to cycle
three "wings", all you have to do now is put the "wings" in the right place
and replace the damage to the off-center central faces that were messed up
during that initial quarter-turn above. (And since they are in "paired"
clusters, this should be pretty straightforward.)
(In short, the quarter-turn of the non-central slice puts the cube back in
the proper "orbit" for finishing up.)
Now clearly this is far from maximal. And it's certainly not terribly
fast. But I find it a very simple, and an easy (and easy-to-remember [and
easy-to-explain]) way to clean up this potentially messy situation.
Hope this helps
Nichael
--
Nichael Cramer
nichael@sover.net deep autumn--
http://www.sover.net/~nichael/ my neighbor what does she do
From cube-lovers-errors@mc.lcs.mit.edu Fri Dec 4 16:49:43 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil (sun28.aic.nrl.navy.mil [132.250.84.38])
by mc.lcs.mit.edu (8.9.1a/8.9.1-mod) with SMTP id QAA06282
for ; Fri, 4 Dec 1998 16:49:43 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Message-Id:
In-Reply-To: <872566D0.000F48EF.00@notes.dtint.com>
Date: Fri, 4 Dec 1998 11:26:10 -0700
To: Jacob_Davenport@scudder.com (Jacob Davenport)
From: Steve LoBasso
Subject: Re: (5x5x5) edge parity corrections
Cc: Cube-Lovers@ai.mit.edu
This should solve the edge parity problem by swapping the
edge F-Ru and edge F-Rd pieces.
R2 d L2 d L2 d' R2 u' F2 u2 F2 u' F2 L2 F l' F' L2 F l F
This move swaps only these two pieces and some centrals, but only within
their face. A variant of this move should be scalable to solve parity
issues in any NxNxN cube.
The only way I can think of to not have the parity problem, or at least not
require such a long series, is to solve centrals last. Another other idea
would be to spot the parity problem much earlier by counting edge flips.
Not very easy for a person to do, but I have seen it done in software for
normal 3x3x3 cubes. If it were were known very early in either the centers
first or layered solution, it would be trivial to fix.
>I don't like the edge parity correction move that I use in my solution, and
>I'm hoping that someone can give me a better one.
>
>The parity problem is found in 5x5x5 cubes (and 4x4x4 cubes, I understand)
>when two of the edges right next to the corners (which I call "wings") are
>switched. Some fairly simple moves can get all three edges in line with
>each other, but half the time two wings need to be switched. By the time I
>figure this out when doing a 5x5x5 cube, I've solved most of it, and my
>parity fixing move messes up many of the edges I've been working on.
>
>How do other people fix this problem?
>
>-Jacob
--
Steve LoBasso
Digital Technology International mailto:slobasso@dtint.com
500 West 1200 South or mailto:slobasso@hotmail.com
Orem, UT 84058 http://members.tripod.com/~slobasso
(801)226-6142 ext.265 FAX (801)221-9254
From cube-lovers-errors@mc.lcs.mit.edu Fri Dec 4 17:13:44 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil (sun28.aic.nrl.navy.mil [132.250.84.38])
by mc.lcs.mit.edu (8.9.1a/8.9.1-mod) with SMTP id RAA06366
for ; Fri, 4 Dec 1998 17:13:43 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Message-Id: <36683B51.A50833E4@switchview.com>
Date: Fri, 04 Dec 1998 14:43:13 -0500
From: Michael Swart
Organization: Switchview
To: Cube-Lovers@ai.mit.edu
Subject: Re: (5x5x5) edge parity corrections
References: <199812040458.XAA26116@Twig.Rodents.Montreal.QC.CA>
I got this from the archives, it may be relevant to repost it. It's a
way of solving the parity problem:
r2 U2 r l' U2 r' U2 r U2 r l U2 l U2 r U2 l r2 U2
I'm confident you can't do too much better than this.
Mike
From cube-lovers-errors@mc.lcs.mit.edu Tue Dec 8 11:22:45 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil (sun28.aic.nrl.navy.mil [132.250.84.38])
by mc.lcs.mit.edu (8.9.1a/8.9.1-mod) with SMTP id LAA18601
for ; Tue, 8 Dec 1998 11:22:43 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Message-Id: <199812042200.RAA02240@pike.sover.net>
Date: Fri, 04 Dec 1998 16:40:51 -0500
To: der Mouse
From: Nichael Lynn Cramer
Subject: Re: (5x5x5) edge parity corrections
Cc: Cube-Lovers@ai.mit.edu
In-Reply-To: <199812040458.XAA26116@Twig.Rodents.Montreal.QC.CA>
der Mouse wrote:
>> The parity problem is found in 5x5x5 cubes (and 4x4x4 cubes, I
>> understand) when two of the edges right next to the corners (which I
>> call "wings") are switched.
>
>Yes, it does occur equally on the 4-Cube. [...]
The appearance is particularly striking on the 4X cube. Especially in the
situation where the two out-of-place "wings" are side-by-side.
It looks very similar to a solved 3X cube with a single edge-cubie flipped.
This is an interesting state to leave your cube in, when it is just lying
around your office, for visitors to find.
N
From cube-lovers-errors@mc.lcs.mit.edu Tue Dec 8 12:38:33 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil (sun28.aic.nrl.navy.mil [132.250.84.38])
by mc.lcs.mit.edu (8.9.1a/8.9.1-mod) with SMTP id MAA18817
for ; Tue, 8 Dec 1998 12:38:32 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Message-Id: <199812051241.VAA08521@soda3.bekkoame.ne.jp>
Date: Sat, 5 Dec 1998 21:47:39 +0900
To: Cube-Lovers@ai.mit.edu
From: Ishihama Yoshiaki
Subject: 4DRubik Cube
I have simulated 4DRubikCube for Macintosh.
It is madeup of 2x2x2x2 hypercubes.
It is on my HomePage.
//----------------------------------------//
Ishihama Yoshiaki
Tokyo Chofu
E-mail: ishmnn@cap.bekkoame.or.jp (Until 1999/3/31)
ishmnn@cap.bekkoame.ne.jp ( This is correct address)
HomePage : http://www.asahi-net.or.jp/~hq8y-ishm/
//--------------------------------------//
From cube-lovers-errors@mc.lcs.mit.edu Tue Dec 8 13:31:28 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil (sun28.aic.nrl.navy.mil [132.250.84.38])
by mc.lcs.mit.edu (8.9.1a/8.9.1-mod) with SMTP id NAA19371
for ; Tue, 8 Dec 1998 13:31:28 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Message-Id: <366C1ED9.C11@hrz1.hrz.tu-darmstadt.de>
Date: Mon, 07 Dec 1998 19:30:49 +0100
From: Herbert Kociemba
Reply-To: kociemba@hrz1.hrz.tu-darmstadt.de
To: cube-lovers@ai.mit.edu
Subject: Optimal Cube Solver
New Optimal Cube Solver
I wrote an optimal Cube Solver and experimented with coordinates
different of those I use in my Cube Explorer program or of those in Mike
Reid's Optimal Cube Solver. Its pruning tables are not very large (about
25MB), so the performance is relatively low (at least in comparison with
Mike's program), but I think it is worth to give you some information
about it.
Some general considerations on the use of "coordinates" in cube solving
algorithms first. Instead of representing a state of the cube by the
positions of corners or edges, the use of coordinates not only increases
the speed of computing a face-turn but also serves as an index for the
pruning tables.
If we have an arbitrary subgroup H of the Cube Group G, we map the right
cosets Ha to natural numbers from 0 to ord(G)/ord(H)-1). A face-turn T
(which also is an element from G) now induces a map on these numbers,
which can be implemented as a simple lookup-table. For this to work we
have to ensure that if x=h1*a and y=h2*a are in the same coset Ha, then
x*T and y*T are in the same coset Hb. But this is true because
(x*T)*(y*T)^-1 = (h1*a*T)*(h2*a*T)^-1 = h1*h2^-1 is in H.
If we take for example H1={all g from G with corner orientations 0,
corner permutations and edges arbitrary} the resulting coordinate
(0<=x<2187) represents the orientation of the corners.
It also should be possible to reduce the size of the coordinates by the
48 symmetries of the cube (or at least by a subgroup of the symmetry
group M). This is done by defining equivalence classes on the cosets.
Two cosets Ha and Hb are called equivalent, if there is an m from M with
Hb = m*Ha*m^-1. But to make this definition work we have to ensure, that
the elements of a coset Ha are really all mapped to the same coset Hb by
the conjugation with m. This only is true, if
(1) mHm^-1=H
The subgroup H1 from above for example does have this property only for
symmetries which do not change the UD-axis in the way the orientations
of the corners are usually defined. So the corner orientation coordinate
can only be reduced by 16 symmetries. Is it possible to define the
corner orientations in another way, so that (1) holds for all 48
symmetries? I do not believe it, but I do not know how to prove this.
For the analogous case of the edge orientations there is a possibility
to define the orientations in a way (different to the way usually used)
which allows reduction by all 48 symmetries: every quarter turn changes
the orientation of any involved edge.
In my program I use 3 coordinates. The first (let's call it the
X2-coordinate) is defined by the subgroup, where the edges are arbitrary
and the corners are generated by . There are 918540
different cosets. Because (1) holds for all m, they can be reduced by
all 48 symmetries and we get 19926 equivalence classes.
The second coordinate is the edge orientation defined by the subgroup
{all g from G with edge orientations 0, edge permutations and corners
arbitrary}. There are 2048 cosets. I do not reduce them by symmetries
because the number is relative small.
The third coordinate describes the edge permutation. Because there are
12! coordinate values, even reduction by 48 symmetries still gives too
many coordinate values. So for use in a turntable we define two edge
permutations a and b equivalent, if a=m1*b*m2, were m1 and m2 are in M.
In this way we get 208816 equivalence classes c. If now m1*c*m2 is a
(not necessarily unique) representation of an edge permutation applying
a faceturn T is done like that:
(m1*c*m2)*T = m1*c*[m2*T*m2^-1]*m2 = m1*[c*T']*m2=
[m1*m1']*c'*[m2'*m2]=m1''*c'*m2''
The operations in square brackets are done by table lookups:
[m2*T*m2^-1]:=T', [c*T']:=m1'*c'*m2', [m1*m1']:=m1'' and [m2'*m2]:=
m2''.
A cube, which has all three coordinates zero, is in a subgroup with 96
elements, were the edges are in place and the corner orientations are
correct. To find such states, I use two pruning-tables. The first
combines the X2-coordinate and the edge-orientation coordinate which
takes 19926*2048/2=20404224Bytes of memory (we only need 4 bit per
entry). The maximal table entry is 12, with an average of about 9.5. The
second is a pruning table for the edge-permutation. It takes
208816*48/2=5011584Bytes, the maximal table entry is 10 (so it takes not
more then 10 faceturns to position all edges ignoring the orientations).
The program produces about 1 million nodes per second on a P350 and a
depth 15 search is done in about 4 minutes (depending on the situation).
So a complete depth 18 search will need a few days which of course is
not very satisfying. A possible improvement could be to use the subgroup
instead of for the first
coordinate. The subgroup has only 4 elements, so the coset-space has 24
times the size. The pruning table will need about 480MB instead of 20MB
which is above that what is possible for me in the moment. But a
complete depth 18 search should be done in about 1/24 of the time which
will be a few hours then.
Herbert
From cube-lovers-errors@mc.lcs.mit.edu Tue Dec 8 14:34:56 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil (sun28.aic.nrl.navy.mil [132.250.84.38])
by mc.lcs.mit.edu (8.9.1a/8.9.1-mod) with SMTP id OAA19972
for ; Tue, 8 Dec 1998 14:34:56 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
From: jmb184@frontiernet.net (John Bailey)
To: ishmnn@cap.bekkoame.or.jp (Ishihama Yoshiaki)
Cc: Submissions Cube-Lovers
Subject: Re: 4DRubikCube
Date: Sat, 05 Dec 1998 13:10:10 GMT
Message-Id: <36692f00.213266943@mail.frontiernet.net>
References:
On Sat, 05 Dec 1998 18:57:03 +0900, in rec.puzzles you wrote:
>I have created 4Dimension Rubik Cube for Macintosh.
>URL: http://www.asahi-net.or.jp/~hq8y-ishm/
I went there the instant I read your post. Unfortunately, I am
running a Pentium based machine. Could you put a gif image of your
cube on the page? Maybe even a screen copy bitmap of the user
interface. We IBM-PC types can stare and drool.
If you haven't checked out my 2x2x2x2 cube, it's at
http://www.ggw.org/donorware/4D_Rubik
John
http://www.frontiernet.net/~jmb184
From cube-lovers-errors@mc.lcs.mit.edu Tue Dec 8 16:03:32 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil (sun28.aic.nrl.navy.mil [132.250.84.38])
by mc.lcs.mit.edu (8.9.1a/8.9.1-mod) with SMTP id QAA20365
for ; Tue, 8 Dec 1998 16:03:31 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Message-Id: <199812042200.RAA02263@pike.sover.net>
Date: Fri, 04 Dec 1998 17:00:42 -0500
To: cube-lovers@ai.mit.edu
From: Nichael Lynn Cramer
Subject: Re: Method for Solving the Professor's Cube (5x5x5)
In-Reply-To: <7440f2$q5v@gap.cco.caltech.edu>
References:
>>Method for Solving the Professor's Cube (5x5x5)
>
>[snip]
This is not a formal solution, but --say when I want to kill some time-- I
often find it entertaining to solve the 5X cube in "ascending spirals".
By which I mean: Start with the center face on a particular color (I
always start with blue). Next solve the non-center face cubies, one by
one, in order moving clockwise around the "loop". When that loop is done,
then solve one of the blue-faced corners and then solve the remaining
blue-sided edge cubies (in order). Then move up, solving each
parallel-to-the-blue-face internal slice in order; and so on.
Needless to say, this is hardly an optimal solution (in either time or
number of moves). But think of it as a way to "practice scales" (Or as I
say, just a good way to kill some time. ;-)
There are obvious variations on this. For example, solve the individual
faces in "ascending spirals" like the above, but instead of starting on a
center face cubie, start on a corner cubie and work your way diagonally, in
slices, across the cube toward the opposite corner.
Or, for the truly masochistic, solve the cube --again a cubie at a time--
in a checkboard pattern (i.e. the result of putting the 5X cube through the
Pons Asinorum transformation) doing first the half of the cubies in the
first "phase" and then the cubies in the other.
--
Nichael Cramer
work: ncramer@bbn.com
home: nichael@sover.net
http://www.sover.net/~nichael/
From cube-lovers-errors@mc.lcs.mit.edu Tue Dec 8 19:08:13 1998
Return-Path:
Received: from sun28.aic.nrl.navy.mil (sun28.aic.nrl.navy.mil [132.250.84.38])
by mc.lcs.mit.edu (8.9.1a/8.9.1-mod) with SMTP id TAA22642
for ; Tue, 8 Dec 1998 19:08:13 -0500 (EST)
Precedence: bulk
Errors-To: cube-lovers-errors@mc.lcs.mit.edu
Date: Tue, 8 Dec 1998 18:37:02 -0500 (EST)
From: Alchemist Matt
Reply-To: Alchemist Matt
To: Herbert Kociemba
Cc: cube-lovers@ai.mit.edu
Subject: Re: Optimal Cube Solver
In-Reply-To: <366C1ED9.C11@hrz1.hrz.tu-darmstadt.de>
Message-Id: __