Date: 3 NOV 1980 0945-EST
From: DCP at MIT-MC (David C. Plummer)
Subject: Christman Cross
To: CUBE-HACKERS at MIT-MC
Greetings, to those who are still on this list!!
I have discovered a rather fast method of doing the Christman
Cross. It can be done in 22 quarter twists, and I think it can be
optomized down into 20. When I ask somebody what the current
representation for transforms is, I will send the HOW along.
((NOTE: This means the Plummer Cross can be done in 44 QTwists))
((((REMARK: For those who count Half Twists [[**sigh**]] as one
MOVE, this algorithm is a mere 16 MOVES))))
Date: 4 NOV 1980 0023-EST
From: DCP at MIT-MC (David C. Plummer)
Subject: Christman Cross algorithm described
To: CUBE-HACKERS at MIT-MC
Let the following notation exist:
the faces:
T=Top B=Bottom
F=Front P=Posterior (Rear, Back, but those letters are taken)
R=Right L=Left
The rotations:
+ Clockwise
- Clockwise
++ 180 Rotation (Two moves)
Note that the following algorithm exchanges corners diagonally on
top and bottom: (Parens for major functional components)
(+F+P)(++T++B)(-F-P)(-R-L)(++T++B)(+R+L)(++T++B)
This is now a tool. To do the Christman Cross, we do
-P [(+F+P)(++T++B)(-F-P)(-R-L)(++T++B)(+R+L)(++T++B)] +P
but since transforms associate, and F and P commute, and -P+P = I, we get
(+F)(++T++B)(-F-P)(-R-L)(++T++B)(+R+L)(++T++B) +P
Which is 20 quarter twists.
David Christman thinks he read in Singmaster, Edition 5 that this
cross can be done in less than 20 quarter twists. Can somebody
confirm or deny this recolection.
Date: 4 Nov 1980 09:28 PST
From: McKeeman at PARC-MAXC
Subject: Re: Christman Cross algorithm described
In-reply-to: DCP's message of 4 NOV 1980 0023-EST
To: DCP at MIT-MC (David C. Plummer)
cc: CUBE-HACKERS at MIT-MC
Using Rubik Song,
(+F+P)(++T++B)(-F-P)(-R-L)(++T++B)(+R+L)(++T++B)
becomes
FB UUDD F'B' R'L' UUDD RL UUDD
Date: 12 November 1980 1849-est
From: Paul Schauble
Subject: Cube lubrication
To: cube-lovers @ mit-mc
Pardon me if you've seen this before, I'm not sure it got mailed.
The best material I have found for lubricating a cube is a product
called Tufoil. This is a suspension of fine Teflon particles in a
synthetic base. From the appearance, it is mostly Teflon. The only
drawbacks are that the quantity needed is fairly critical (too much will
make the outside of the cube sticky until it all wipes off) and the
stuff only comes in 8 oz bottles. This should be enough to lubricate
over 1000 cubes. Available in the automotive department at Bradlees in
Boston.
Date: 24 NOV 1980 0451-EST
From: ALAN at MIT-MC (Alan Bawden)
Subject: Old cube mail no longer on-line.
To: CUBE-HACKERS at MIT-MC
There is a klutz amoung us.
This is twice that the file of old cube lovers mail has been deleted
from my directory. I retrieved it last time and put it in
ALAN;CUBE OMAIL and let the new messages continue to accumulate in
ALAN;CUBE MAIL. But somebody just deleted the OMAIL again. I will
retrieve it just once more, and if it goes away AGAIN I will just stop
bothering.
Date: 24 NOV 1980 1743-EST
From: ALAN at MIT-MC (Alan Bawden)
To: CUBE-HACKERS at MIT-MC
ALAN;CUBE OMAIL is back on disk as ALAN;CUBE XXMAIL .
Date: 24 NOV 1980 1926-EST
From: DCP at MIT-MC (David C. Plummer)
Subject: Plumme Cross
To: CUBE-LOVERS at MIT-MC
I have an algorithm to get to the Plummer Cross from solved in
thirty quarter twists. It may get reduced if I notice I can
combine different parts of the algorithm. Description to come in
a few days.
Date: 25 NOV 1980 1308-EST
From: DCP at MIT-MC (David C. Plummer)
Subject: 30 move Plummer Cross algorithm described
To: CUBE-LOVERS at MIT-MC
Using the {Up Down Left Right Front Back} notation with clockwise
normal and ' (single quote) indicating counter-clockwise:
(B D)^7 that's right, 14 moves. this sets the back face
up and does a couple things to the front.
The next two are general tools to play with three corners
and nothing else, and people should play with them for
their utility.
(L D L') U (L D' L') U'
This fixes front upper left, which leaves front
upper-right, lower-right, lower-left. At this point I
would rotate the entire cube CCW about the front-back
axis so that these become front upper-left, upper-right,
and lower-right, in which case the following transform
becomes (in the new coordinates)
U' (R' D' R) U (R' D R)
but in the old coordinates, we get
R' (D' L' D) R (D' L D)
(+ 14. 8. 8.) ==> 30.
I don't think I made a mistake. Have fun.
IF: Plummer == 2*Christman
THEN Christman=14 (must be even)
and also Plummer=28.
Date: Tuesday, 2 December 1980 10:37-EST
From: Jonathan Alan Solomon
To: CUBE-LOVERS at MIT-MC
cc: Solomon at RUTGERS
Subject: Source to CUBE to run it locally (!)
Could someone please point to where the source to the CUBE
program lives? Some users here want to use it and don't
have access to the net. I have the newest MACLISP, so there
is no problem there.
Thanks!
JSol
Date: 2 December 1980 1040-est
From: Bernard S. Greenberg
Subject: Re: Source to CUBE to run it locally (!)
To: SOLOMON at Rutgers
Cc: CUBE-LOVERS at MIT-MC, Solomon at Rutgers
I repeat for all those interested, the :CUBE program, designed
and implemented by yours truly, lives in the BSG directory on
AI. The file .INFO.;CUBE INFO describes, at its end, the organization
of the source modules. I will give anyone any reasonable amount
of help bringing it up elsewhere, and would very much like
to know if you succeed with it. -bsg
Date: 12/02/80 1059-EDT
From: PLUMMER at LL
Subject: Cube lube
To: CUBE-LOVERS at MIT-MC
I have found that talcum powder works. Take apart your cube and
sprinkle generously. For a day or two after some powder will leak
out, but this brushes off easily. --Bill
-------
Date: 3 December 1980 0050-EST
From: James.Saxe at CMU-10A (C410JS30)
To: CUBE-LOVERS at MIT-MC
Subject: 28 qtw Plummer Cross
CC: James.Saxe at CMU-10A
Message-Id: <03Dec80 005058 JS30@CMU-10A>
First,
F (LL RR) F B (LL RR) F.
Now, rotate the whole cube 1/4 turn clockwise about the up-down axis
(so that the top stays on top and the right side becomes the front)
and finish up with
F (LL RR) F B (LL RR) F (UU DD).
Date: 4 DEC 1980 2300-EST
From: DR at MIT-MC (David M. Raitzin)
To: CUBE-LOVERS at MIT-MC
Hi! I'm new to this list, since I just found out about it, so I might
repeat some stuff that has already been said. I guess I want to say
two things. First of all, I saw Greenberg's message saying that some
guy who has written a book on the cube or something does not believe
that solutions in two minutes are possible. Well, even though I can
not do it in two minutes (it takes me just under four minutes to do
it), my roommate does it in a minute and a half (I timed it, so it's
no lie). Second, I've never heard of anyone using the same algorithm
that me and my roommate use. We get the top row first, then we get
the second row, and finally get the third. The only other solution
that I've seen (and/or heard of), and that is from anyone else who has
solved the cube, is getting all the eight corners, and then getting
the middles. (Greenberg's Lisp machine program also solves it in this
manner. In fact, I was quite surprised to see it done in that manner,
but as time went on, I realized that everyone I know of does it in
that manner too.) Is that true? Does anyone have any other algorithm
to the two I've described? Also, I've heard that the most optimum
solution to a randomized cube is at most 41 moves. Is that true? And
if so, what is the algorithm it uses. I immagine that algorithm was
achieved on a computer is that true? As you can see, I have a lot of
questions, but this is my first letter to the mailing list.
Dave
Date: 4 Dec 1980 2118-PST
From: Dave Dyer
Subject: solution sequence
To: cube-lovers at MIT-MC
My original algorithm was:
(1) bottom corners by random twisting about
(2) bottom sides by random twisting about
(3) middle sides mostly by variants of [( r u -r -u )^2 ]^3
with extra "u" operations between the inner and outer loops
(4) top sides mostly by variants of ( r u -r -u )^2 ( -f -u f u)^2
with extra reps of phase 3 to swap positions around
(5) top corner position by variations on ( r u -r -u )^2 ( r u -r -u )z
with extra "b" operations between the major groups
(6) top corners orientation
mostly by repitions of (gasp!)
A = ( r u -r -u )^2
A f A f A f A -f A -f A -f
... which I called the "long walk"
My typical solution times with this methodology was 15-20 minutes. I suspect
If I operated ar warp speeds (like BSG) it would still be 5 minutes.
FORTUNATELY, I have developed better since then.
-------
Date: 5 December 1980 0022-est
From: Ronald B. Harvey
Subject: Your mail of 4 DEC 1980 2300-EST
To: DR at MIT-MC
Cc: CUBE-LOVERS at MIT-MC
Welcome to the Cube-hackers !!
Note that this is my first contribution, although I have read all
previous mail.
First of all, there are other people who solve the cube (I, myself,
average about a minute and three quarters). It is also part of the lore
that there are people (overseas) who do it in 50 seconds or less.
Second, while I personally get eight corners first, and then randomly go
about fixing the edge cubies two at a time or so, David Singmaster, who
publishes some notes in England (availability in previous mail),
included, in his latest edition his algorithm, which proceeds as
follows:
Solve TOP edges
Solve TOP corners
Flip cube so that TOP is BOTTOM from now on
Solve MIDDLE edges
Solve TOP edges
Solve TOP corners
Finally, in regards to the 'most optimum solution' (also known as God's
Algorithm), a person by the name of Morwen B Thistlethwaite (that's from
memory - I left my notes at work) is mentioned all over Singmaster's
notes as contributing algorithms. Yes, he does use a program, and his
algorithm at the time of the 5th Edition took somewhere around 50 moves.
Date: 5 DEC 1980 0132-EST
From: SHL at MIT-MC (Stephen H. Landrum)
Subject: ADDITION
To: CUBE-LOVERS at MIT-MC
COULD I BE ADDED TO THIS LIST?
THANX.
STEPHEN LANDRUM
(SHL@MC)
P.S. I DIDN'T KNOW WHO TO CONTACT SINCE THERE IS NO 'CUBE-LOVERS-REQUEST'
Date: 5 DEC 1980 0135-EST
From: SHL at MIT-MC (Stephen H. Landrum)
To: CUBE-LOVERS at MIT-MC
I ALSO SOLVE THE CUBE TOP ROW, MIDDLE, THEN BOTTOM.
ONE TIME I MANAGED IT IN 58 SEC, BUT I THINK THAT WAS JUST A FREAK CASE.
MY SECOND BEST TIME IS 1 MIN. 42 SEC, I AVERAGE ABOUT THREE MINS.
STEPHEN LANDRUM
Date: 5 DEC 1980 0250-EST
From: ALAN at MIT-MC (Alan Bawden)
To: SLH at MIT-MC
CC: CUBE-LOVERS at MIT-MC
I have added you to the cube-lovers mailing list.
Old mail is archived in MC:ALAN;CUBE MAIL.
I'll Cc this message to CUBE-LOVERS so that anyone else looking
through the (recent) old mail, will find out that I am a person to
mail "add me" requests to.
Date: 5 Dec 1980 00:29 PST
From: Woods.PA at PARC-MAXC
Subject: Re: DR's message of 4 DEC 1980 2300-EST
To: DR at MIT-MC (David M. Raitzin)
cc: CUBE-LOVERS at MIT-MC
I've hardly ever found two people who solve the same way. The most
common method I've found around Stanford is to go for all the edges,
then the corners. My own method is to get the top corners, then the
top edges (via operations like -l r f^2 l -r & similar tweaks) then three of
the middle edges (via things like l^2 u^2 r^x u^2 l^2, where x=1, 2, or -1),
then get the other four corners in the right place (requiring at most one
corner-swap, for which I have a fairly simple macro), then get those four
corners oriented using the "rFUfuR" commutator macro, which also randomly
alters the remaining middle edge, then get the last five edges, then flip
edges as necessary. My typical time is about 5 minutes. "Best" times are
meaningless, since anybody can luck out once or twice; the best measure
of your solving speed (in my opinion) is your WORST solving time over
your most recent ten or so attempts. Of course, for real fun, pick some
pretty pattern and solve to it without going through the normal solved state!
-- Don.
Date: 5 DEC 1980 1240-EST
From: DCP at MIT-MC (David C. Plummer)
Subject: My 6 faces worth
To: CUBE-LOVERS at MIT-MC
The technology exists to solve the cube in any way possible.
I.E., tools exists to do the most primitive of operations to
individual cubies (rotating corners, flipping edges, moving three
corners around, and moveing trhee edges around). Using these, any
method of solving is possible (even solving a corner, then the
edges around that corner, then the corners around those edges,
etc, all the way toward the other corner). Since everybody tends
to develop their own technology, they develop their own favorite
way of solving.
Personally, at present, I solve
the top edges
three of the top corners
three of the middle edges
the last top edge
Up until here the algorithms are quite simple and now I only have
the bottom and one middle edge, which I solve in whatever way
looks most promising. Using this sequence, I am consistently on
the order of three minutes (tending toward 2.5 as I get more
familiar with the method)
Intuitively, I believe that God's algorithm is less than 35 or
so, and I would not be surprised if it is less than 30. Under 25
would really surprise me.
Date: 5 Dec 1980 1038-PST (Friday)
From: Mike at UCLA-SECURITY (Michael Urban)
Subject: Cube approach survey?
To: cube-lovers at mc
Top edges, top corners, middle edges, bottom edges, bottom
corner placing, bottom corner orientation. Takes me about 4
minutes, but I'm out of practice. None of the steps requires
any real unusual tools. I suspect I got in the habit of
solving the cube in that order because the initial challenge
was "get one face". Then I realized that the next step must
surely be "get one face so that it matches the center
faces of the sides".
The beauty of the cube is that, unlike Instant Insanity
or its cousins, the Cube's solutaion can not be stumbled
across by accident. Instead, it's a learning experience,
as you learn to manipulate things, then to manipulate
cubies without messing up what you already have
done. This gets progressively more sophisticated as
more of the cube comes into place. Definitely the
most satisfying puzzle I've ever dealt with. However,
I'd be interested to find out how many as-yet-unsolved
Cubes are sitting even now on the shelves of disgusted customers
who thought they were great puzzlists because they solved Soma,
but weren't patient enough to figure out how to get those
last two corner cubies oriented right.
Mike
-------
Date: 5 December 1980 16:28-EST
From: Alan Bawden
Subject: cube-lovers-request
To: CUBE-LOVERS at MIT-MC
OK, I have created the cube-lovers-request mailing list.
Date: 5 December 1980 1848-est
From: David C. Plummer
Subject: That 28 move Plummer Cross
To: Cube-Lovers@MIT-MC
Something I noticed about that 28 move algorithm to get to the Plummer
Cross. Any twist can be the first one made to take the Cross back to
home. The last instructions are (UU DD) which can obviously be (UU D'D'),
(DD UU), and (DD U'U'), so possible backward applications of the
algorithm can start with U,D,U',D'. Add in the three way rotation
symmetry of the cross and you can get all 12 moves as viable first
moves. So, any move takes you closer to home (by this algorithm), so it
has the chance of being the most distant from home. This requirement
(any moves takes you closer) was mentioned by ALAN some time ago. (I
don't hold that the Plummer Cross is the most distant.)
Date: 5 December 1980 19:10 est
From: Greenberg.Symbolics at MIT-Multics
Subject: Re: That 28 move Plummer Cross
To: Plummer.SIPBADMIN at MIT-Multics (David C. Plummer)
cc: Cube-Lovers at MIT-MC
In-Reply-To: Message of 5 December 1980 18:48 est from David C. Plummer
It is a corollary of what Plummer just said, that
the "maximally distant" configuration, which is God's number of
moves (the maximal length of God's algorithm) would then be
no more than 28 moves from home (bounding God's number thusly).
Date: 5 Dec 1980 1624-PST
From: Dave Dyer
Subject: Re: Re: That 28 move Plummer Cross
To: Greenberg.Symbolics at MIT-MULTICS
cc: cube-lovers at MIT-MC
In-Reply-To: Your message of 5-Dec-80 1910-PST
Not true! It may be just a local maximum.
-------
Date: 5 December 1980 21:15 est
From: Greenberg.Multics at MIT-Multics
Subject: Re: Re: That 28 move Plummer Cross
To: Dave Dyer
cc: cube-lovers at MIT-MC
In-Reply-To: Message of 5 December 1980 19:24 est from Dave Dyer
No. It is DEFINITIONALLY a local maximum, because any move moves it
"closer". DCP was asserting its candidacy for the "ultimate" maximum, if
not its fact.
Date: 6 Dec 1980 13:40 PST
From: McKeeman.PA at PARC-MAXC
Subject: Re: That 28 move Plummer Cross
In-reply-to: Plummer.SIPBADMIN's message of 5 December 1980 1848-est
To: David C. Plummer
cc: Cube-Lovers@MIT-MC, Greenberg.Symbolics at MIT-MULTICS, DDYER at
USC-ISIB, Hofstadter at SU-AI
David,
Interesting observation! Your argument about the Plummer Cross being a local
maximum in QTW metric holds for any completely symmetrical configuration of
the cube, independent of the algorithm used to reach it. There are a lot of them
(including "home").
It raises the question: Can the maximally distant point be proven to be
symmetric? If so, the search for a bound is much simplified.
Bill
Date: 6 Dec 1980 13:57 PST
From: McKeeman.PA at PARC-MAXC
Subject: A Top-Middle-Bottom solution
To: CUBE-LOVERS at MIT-MC
Since I had this writeup in "sendable" form, I thought I would put it out. Most
of you already have this one or better, so just delete it if you are not interested.
---------------------------
Date: 19 Jun. 1980 5:57 pm PST
From: McKeeman
To: McKeeman
File: [Ivy]Rubik.txt
Kertesz' Algorithm for the Rubik Cube. McKeeman -- June 19, 1980
Rubik's Cube is a 3x3x3 cube such that each of its 3x3 planes can rotate freely
about its perpendicular axis. There are 54 visible squares, colored in six sets of
9. There are 26 visible cubelets, each colored uniquely. (Commercially available
cubes are not colored consistently.) The edge cublets have two colors; the corner
cublets have three.
The inventor is a Hungarian architect; my solution comes from Adam Kertesz, a
Hungarian programmer I met in Poland. It may be less precise than you like,
but without pictures it is just too much to give all the details. With a little
struggle this should get you to a solution.
The Problem: From an arbitrary starting position, arrange the cube so that each
3x3 surface is of a uniform color.
Notation: One can limit moves to 90o rotations of the six surfaces without losing
anything. It is convenient to describe the moves in terms of the orientation of
the cube rather than the color. The names of the faces are:
U=up
D=down
R=right
L=left
F=front
B=back.
A move of R, for example, is a 90o clockwise rotation of the right face, RR = R^2
is 180o, and R'=RRR= R^3 is 270o, or 90o counterclockwise. For any move M,
M'M is a no-op.
Macros: There a lot of macro-moves that are useful and interesting. I will use
six here:
X = R'F'UFUF'U'FRUU
Xr = LFU'F'U'FUF'L'UU
Y = RFFDFD'FFR'
Yr = L'FFD'F'DFFL
Z = R'DRFDF'
Z' = FD'F'R'D'R
It is not trivial to use these things. The general idea is to decide on the
orientation of the cube, and then follow the directions mechanically. Note that
the color of the central square of every face is unchanged during a macro. That
is the easiest way to keep from getting messed up. I mumble to myself "Red is
Up, Blue is Front" or some such thing each time just before I dive into a macro.
I find it is even helpful to close my eyes during a macro when doing it from
memory!
Kertesz' Algorithm: Generally speaking the idea is to get the cube right one
layer at a time. They get increasingly hard (tedious) because you must be
careful not to destroy work already done.
----------------
Layer 1: Pick Up color.
Step 1. Do Up edges. The four Up edges must be moved between the two
center squares that have the same colors:
For each edge do
(a) hold target position UF (i.e., up front, nearest you),
(b) get needed cubelet into bottom layer,
(c) rotate it under target position using D^n,
(d) do F^2, in which case you are done or it needs flipping,
(e) in which case F'UL'U' finishes it.
You should now have the top "+" shape correct and matching the side center
squares.
Step 2. Do Up corners. The four Up corners must be moved so that all three of
their colors match their neighbors.
For each corner do
(a) hold target position FUR (i.e., up right, nearest your right hand),
(b) get needed cubelet into bottom layer,
(c) rotate it under target position using D^n,
(d) either DFD'F' or D'R'DR will bring it up,
(e) it may need rotation, in which case use Z for 120o clockwise, Z'
for counterclockwise.
You should now have the top layer correct and matching the side center squares.
----------------
Layer 2: The four remaining edges on layer two are next. Y and Yr leave Up
surface invariant.
For each layer 2 edge target do
(a) get needed cubelet into down layer (using Y or Yr if needed),
(b) rotate it 90o from under target using D^n,
matching color of nearest center square,
(c) hold needed cubelet DF
(d) if it needs F, do Y to bring it into position,
(e) if it needs F', do Yr.
----------------
Layer 3: Turn puzzle over. The macros X and Xr are used over and over
again. The sequence of events is to place the four up corners, place the four up
edges, flip the edges if necessary, and finally rotate the corners if necessary.
Step 1. Place Up corners. One can always be rotated into position with U^n.
Among other things, X will exchange the RFU and RBU corners. At most four
applications of X will place all four corners. It takes a little forethought.
Step 2. Place Up edges. X will also exchange the LU and FU edges. Xr will
exchange the RU and FU edges and the LFU and LBU corners. Used in pairs X
and Xr can exchange edges without changing corners. That is, one use
exchanges corners, the next carefully chosen one restores them.
Step 3. Flip Up edges as necessary. X^2 will flip both LU and FU edges.
Step 4. Rotate Up corners as necessary. The macro ZU^nZ'U^-n will rotate
two Up corners, one 120o and one 240o. Place the one needing 120o in FUR, do
Z, do U^n to bring one needing 240o into FUR, do Z'U^-n. Don't look at the
mess after the first rotation; it will destroy your morale.
-----------------
You should now be done. It took me a couple of days to be able to solve the
thing reliably. Enjoy.
P.S. Real Rubniks will want to write for:
Notes on Rubik's 'Magic Cube', David Singmaster, Mathematical Sciences and
Computing, Polytechnic of the South Bank, London SE1 0AA, England (send $5).
------------------------------------------------------------
Date: 6 Dec 1980 14:16 PST
From: McKeeman.PA at PARC-MAXC
Subject: Re: That 28 move Plummer Cross
In-reply-to: Greenberg's message of 6 December 1980 1644-est
To: Bernard S. Greenberg
cc: McKeeman, Cube-Lovers at MIT-MC, DDYER at at MIT-Multics,
Plummer.SIPBADMIN at MIT-Multics
I do not follow the reasoning. It seems quite possible that there is a
non-symmetric local maximum. In any case, it is not a definition, but rather a
proof that needs doing. It is certainly true that a move from a non-symmetric
configuration will either
a. get closer to home
b. stay the same distance from home
c. get further from home.
Furthermore, it is obvious that there are usually both (a) and (c) cases. What I
don't see is the argument that there must always be a (c) case.
One way of looking at it is that there is an enormous graph connecting all
solutions by one QTW moves. Nearly all nodes are non-symmetric. You argue
that from every non-symmetric node there is a move to a node that is further
from home. I am willing to be convinced, but am not yet, and mere probability
favors the following:
Conjecture: There exists a non-symmetric configuration from which every move
leads to a position that is closer to home.
Bill
Date: 6 Dec 1980 1428-PST
From: Dave Dyer
Subject: Re: That 28 move Plummer Cross
To: McKeeman.PA at PARC-MAXC
In-Reply-To: Your message of 6-Dec-80 1340-PST
Remailed-date: 6 Dec 1980 1429-PST
Remailed-from: Dave Dyer
Remailed-to: cube-lovers at MIT-MC
While I believe it is true that all symmetric positions are local maxima,
It is definitely NOT true that all local maxima are symmetric. As
a counterexample, consider that a repeated permutation is at a local
maximum 1/2 way through its period. So for example, the Idempotent
( r u -r -u )^6 is at a familiar but unsymmetric local maximum
halfway through.
-------
Date: 6 DEC 1980 1745-EST
From: DCP at MIT-MC (David C. Plummer)
Subject: Re: That 28 move Plummer Cross
To: McKeeman.PA at PARC-MAXC
CC: CUBE-LOVERS at MIT-MC
Date: 6 Dec 1980 13:40 PST
From: McKeeman.PA at PARC-MAXC
In-reply-to: Plummer.SIPBADMIN's message of 5 December 1980 1848-est
USC-ISIB, Hofstadter at SU-AI
David,
Interesting observation! Your argument about the Plummer Cross being a local
maximum in QTW metric holds for any completely symmetrical configuration of
the cube, independent of the algorithm used to reach it. There are a lot of them
(including "home").
It raises the question: Can the maximally distant point be proven to be
symmetric? If so, the search for a bound is much simplified.
Bill
I don't know exactly where to start my comments. For one thing,
the Plummer cross is not totally symmetric. What I stated
(actually ALAN, but I seem to be the culprit now):
It is necessary for the maximal state to have the quality
that any quarter twist brings you closer to home.
It is also true that any symmetric state also has this quality.
What I noted was that the 28 move algorithm given shows that the
Plummer Cross also fulfills this. HOWEVER, there may exist a 26
or 24 move algorithm such that only 6 of the 12 possible moves
may be done first in order to fix it.
About your question, even if you could prove the maximal distant
point is symmetric, we still cannot prove how far away a
configuration is away from home. If you could prove that, you
would also God's Algorithm.
Date: 6 DEC 1980 1748-EST
From: DCP at MIT-MC (David C. Plummer)
Subject: food for twisting
To: CUBE-LOVERS at MIT-MC
Here is a piece of food for thought: why can't there be two (in
general more than one) maximally distant point). For example, the
24 Plummer Crosses and the "every edge flipped" configuration.
And the obvious question is: how far apart are these maximally
distant types. Between different types is probably more
interesting than between different "reflections" of the same
type.
Date: 6 DEC 1980 1846-EST
From: DCP at MIT-MC (David C. Plummer)
Subject: Re: That 28 move Plummer Cross
To: McKeeman.PA at PARC-MAXC
CC: CUBE-LOVERS at MIT-MC
Date: 6 Dec 1980 14:16 PST
From: McKeeman.PA at PARC-MAXC
In-reply-to: Greenberg's message of 6 December 1980 1644-est
Plummer.SIPBADMIN at MIT-Multics
I do not follow the reasoning. It seems quite possible that there is a
non-symmetric local maximum. In any case, it is not a definition, but rather a
proof that needs doing. It is certainly true that a move from a non-symmetric
configuration will either
a. get closer to home
b. stay the same distance from home
c. get further from home.
Furthermore, it is obvious that there are usually both (a) and (c) cases. What I
don't see is the argument that there must always be a (c) case.
Bill
Except from solved, there always exists a move taking you closer
to home. Always: There is NEVER (by the QTW metric) a move that
keeps you the same distance, and from the maximally distant state
it is IMPOSSIBLE to get further from home. Notice I have said
nothing about symmetry.
Date: 6 DEC 1980 1841-EST
From: DCP at MIT-MC (David C. Plummer)
Subject: Re: That 28 move Plummer Cross
To: DDYER at MIT-MC
CC: CUBE-LOVERS at MIT-MC
Date: 6 Dec 1980 1428-PST
From: Dave Dyer
In-Reply-To: Your message of 6-Dec-80 1340-PST
Remailed-date: 6 Dec 1980 1429-PST
Remailed-from: Dave Dyer
Remailed-to: cube-lovers at MIT-MC
While I believe it is true that all symmetric positions are local maxima,
It is definitely NOT true that all local maxima are symmetric. As
a counterexample, consider that a repeated permutation is at a local
maximum 1/2 way through its period. So for example, the Idempotent
( r u -r -u )^6 is at a familiar but unsymmetric local maximum
halfway through.
-------
I think what most of mean by LOCAL MAXIMUM is that ANY twist will
bring you closer to home. The proposed counterexample can be
taken farthur away from home (eg (R U -R -U)^3 then D)
Date: 6 Dec 1980 16:30 PST
From: McKeeman.PA at PARC-MAXC
Subject: Re: That 28 move Plummer Cross
In-reply-to: Greenberg's message of 6 December 1980 1836-est
To: Bernard S. Greenberg
cc: McKeeman, Cube-Lovers at MIT-MC, Plummer.SIPBADMIN at MIT-Multics
OK. There may be non-symmetric local maxima. I feel better; thought maybe I
was just so dense I couldn't grok the obvious.
As for a definition of symmetric: For the purposes of local maximum argument,
each of the 6 basic QTW must have the same effect. That is, the configuration
must be isomorphic under the rotation group of the whole cube (of order 24).
Bill
Date: 6 Dec 1980 16:42 PST
From: McKeeman.PA at PARC-MAXC
Subject: Re: That 28 move Plummer Cross
In-reply-to: DCP's message of 6 DEC 1980 1745-EST
To: DCP at MIT-MC (David C. Plummer)
cc: McKeeman, CUBE-LOVERS at MIT-MC
David,
Suppose one could prove local maxima had configurations that were invariant
under the rotation group of the whole cube. (I am not at all sure it is even
true.)
There are a small number of such symmetric configurations, and they could
probably be easily tabulated. One of them would have to be maximally distant
from home. Thus if we had a QTW solution for each of them, the maximum
over that set would bound God's Algorithm.
I see no reason to believe that a QTW cannot take you between two solutions
that are at the same distance. As DPC pointed out, there are a lot of even
identity paths. E.g., (RUR'U')^6. The two furthest points on the path are (by
symmetry) necessarily equally distant, yet connected by a QTW.
Bill
Date: 7 December 1980 00:47-EST
From: Alan Bawden
Subject: Maximally distant states
To: McKeeman.PA at PARC-MAXC
cc: CUBE-LOVERS at MIT-MC
Date: 6 Dec 1980 16:42 PST
From: McKeeman.PA at PARC-MAXC
I see no reason to believe that a QTW cannot take you between two solutions
that are at the same distance. As DPC pointed out, there are a lot of even
identity paths. E.g., (RUR'U')^6. The two furthest points on the path are (by
symmetry) necessarily equally distant, yet connected by a QTW.
I am not sure I understand what you are trying to say here. But I do
know that a single quarter twist can never leave you the same distance
from anything. This is because a single quarter twist is a odd
permutation of the "stickers". Thus if you are N quarter twists away
from something, a single quarter twist will leave you N-1 or N+1
quarter twists away. (And hence the proof that any quarter twist will
bring you closer from a maximally distant state.)
I'm not sure how to apply this to your statement that perhaps a "QTW"
can take you "between two solutions that are at the same distance".
Date: 7 DEC 1980 0108-EST
From: MJA at MIT-MC (Michael J. Aramini)
Subject: maximally distant state
To: CUBE-LOVERS at MIT-MC
well it is possible that to maximally distant states are half twist apart
also if you count half twists as one twist (i dont, but its
still worth thinking about) does that change the set of
maximally distant states?
also it is possible that there exists states for which all directions
lead closer to home (and twist put the cube in a state closer to home)
but the state is not necessarily maximally distant (to use
a continous analogy, think of think of a hill in a funtion of
two variables, that is not necessarily the maximum value of the
function)
Date: 7 DEC 1980 0724-EST
From: DCP at MIT-MC (David C. Plummer)
Subject: maximally distant state, setting the record straight
To: MJA at MIT-MC
CC: CUBE-LOVERS at MIT-MC
Date: 7 DEC 1980 0108-EST
From: MJA at MIT-MC (Michael J. Aramini)
well it is possible that to maximally distant states are half twist apart
WRONG! (I assume you meant "two" for "to" and typo'ed). Read
ALAN's previous message. In the half twist metric, there exist
odd distances away, and there exist even distances away. A QTW
takes the cube from odd to even or from even to odd. The maximally
distant state is the state such that the fewest number of QTW
required to solve it is maximized. This must be odd OR even, and
thus, two states that are maximally distant must be both odd or
both even, which means the distance between them is even, or an
EVEN number of QTW. A single QTW is ODD, and thus cannot separate
maximal states.
also if you count half twists as one twist (i dont, but its
still worth thinking about) does that change the set of
maximally distant states?
Maybe it does, maybe it doesn't. It is much harder to tell because
counting half twists has no analog to the QTW odd/even property of
distance, and this is one reason several of us don't count half twists.
For example, (R L R) and (L [RR]) are equivalent manipulations, but in
half twist counting, one is three and the other is two moves. (assume []
means grouping two moves into one.)
also it is possible that there exists states for which all directions
lead closer to home (and twist put the cube in a state closer to home)
but the state is not necessarily maximally distant (to use
a continous analogy, think of think of a hill in a funtion of
two variables, that is not necessarily the maximum value of the
function)
We have been saying this all along! Simple example: (RRLL UUDD FFBB)
is a local max (any twist takes you closer), and it is definitely not
absolute max (abs max must be at least 21 from combinatoric arguments).
Date: 6 December 1980 1836-est
From: Bernard S. Greenberg
Subject: Re: That 28 move Plummer Cross
To: McKeeman.PA at PARC-Maxc
Cc: Cube-Lovers at MIT-MC, Plummer.SIPBADMIN at MIT-Multics
I confess to not only having no proof to this proposition, but no
longer believing it. The false equivalence of "symmetric" and
"locally maximal" seemed to me visually obvious, given that all symmetric
positions are in fact locally maximal. If one believes this
subconsciously, it is but a short step to the conscious "proof" that
all asymmetric positions are but "on the way" to a symmetric, maximal
one; this of course, is bogus. Now a good definition of symmetric
is also needed here; I assume that what we have been meaning is that
all rotations of the cube are Isomorphic. Note that the Plummer and
Christman crosses don't qualify under this definition. What is a good
definition?
Date: 6 December 1980 1644-est
From: Bernard S. Greenberg
Subject: Re: That 28 move Plummer Cross
To: McKeeman.PA at PARC-Maxc
Cc: Cube-Lovers at MIT-MC, DDYER at at MIT-Multics,
Plummer.SIPBADMIN at MIT-Multics
The maximally distant point is definitionally symmetric. It was
Bawden (ALAN) who first stated this; were it NOT symmetric, there
would be some twist you could make that would make it WORSE
as opposed to the ones that would make it better. Were this
the case, it would then not be a maximum.
Date: 7 December 1980 1218-est
From: Bernard S. Greenberg
Subject: Symmetry of configurations
To: McKeeman.PA at PARC-MAXC
Cc: CUBE-LOVERS at MIT-MC
The definition of symmetry proposed by you does not seem adequate.
Plummer's Cross is NOT symmetrical by that definition. Given
Plummer's Cross in any orientation, there are plenty of elements
of the 24-move whole-cube rotation group which will shift it to
something NOT isomorphic through a substitution of colors to the
original. Yet, we know, intuitively, that the CP is "highly
symmetric", and it is a local maximum. It is as though I were
saying that this equivalence of all twists "out" of this position
almost defined symmetry.....
Date: 7 December 1980 1401-est
From: Bernard S. Greenberg
Subject: Re: That 28 move Plummer Cross
To: DDYER at MIT-Multics
Cc: McKeeman.PA at PARC-Maxc, CUBE-LOVERS at MIT-MC
I do not believe that (r u r' u')^3 is a local maximum.
An move other than r, u, r', u', (e.g., l) seems to pervert
it further than any of the four which bring it back.
Date: 7 December 1980 1620-est
From: Bernard S. Greenberg
Subject: Re: Maximally distant states
To: McKeeman.PA at PARC-Maxc
Cc: CUBE-LOVERS at MIT-MC, ALAN at MIT-MC
King of bogosity! Fencepost errors abound. Here's an "even"
identity transform for you: (R R)^2. Halfway through
this transform is ONE state. There are no "two furthest points
on the path", there is one. There ARE, however, two ways out of it.
And similar to all even identity transforms.
Date: 7 December 1980 16:58-EST
From: Alan Bawden
To: CUBE-LOVERS at MIT-MC, Greenberg at MIT-MULTICS
cc: ELLEN at MIT-MC
To sum up everything we know:
1) A Symetric position is a local maximum.
2) A maximally distant state is a local maximum.
3) If Plummer's cross is 28 Qs away from home, then it is a local maximum.
All the other implications that have been flying around are unproven
as far as I know. I welcome PROOFS of anything else that I might not
be aware of. The volume of air-headed speculation on this subject has
reached the point where the MC mailer is so overloaded with gubbish
that a "CUBE-LOVERS Digest" is being considered. Now that is just
plain silly given that NORMALLY we only get one or two pieces of mail
a week through this list. If you don't understand why one of the
three facts above are true, DON'T send mail to the whole list! Look
through the old mail (they are all explained there) or send JUST ME a
question, and I will try and answer it.
Date: 8 December 1980 1309-EST (Monday)
From: Guy.Steele at CMU-10A
To: McKeeman.PA at PARC-MAXC
Subject: Re: That 28 move Plummer Cross
CC: cube-lovers at MIT-MC
In-Reply-To: McKeeman.PA@PARC-MAXC's message of 6 Dec 80 19:42-EST
Message-Id: <08Dec80 130910 GS70@CMU-10A>
A QTW cannot take you between two positions of equal distance, I believe
-- is there not some parity quantity obeyed by QTW's (net twist of
the six center cubies)? If so, then there cannot be two positions of
equal distance separated by a QTW, for then there would be an odd
length identity transformation, which would violate parity.
(In the example you gave, there are *not* two positions of
equal distance, but only one -- the halfway point.)
Date: 8 December 1980 1117-EST (Monday)
From: Guy.Steele at CMU-10A
To: ddyer at USC-ISIB
Subject: Re: That 28 move Plummer Cross
CC: cube-lovers at MIT-MC
In-Reply-To: Dave Dyer's message of 6 Dec 80 17:28-EST
Message-Id: <08Dec80 111740 GS70@CMU-10A>
The position halfway through the cycle of a permutation need not
be a local maximum, since moves other than the permutation cycle
might take it further away. It is more like a saddle point (most
positions probably are in that class: at least two moves take
it closer, and at least two take it further).
There is one symmetric position which is not a local maximum:
the solved state!
Date: 8 Dec 1980 17:03 PST
From: McKeeman.PA at PARC-MAXC
Subject: A Proposed Definition of Symmetry
To: cube-lovers at MIT-MC
The discussion of local maxima for the Q measure of distance led to an informal
use of symmetry. It is not clear to me just what symmetry is needed to carry
through the maxima argument but I suggest the following is sufficient (although
perhaps too restrictive).
Let C by the rotation group of the cube (closure of IJK: order 24)
Let G be Rubik's group (closure of UDRLFB: order 10^19 or so)
Both groups can be represented as a permutation group on [0, 1, ...53] for some
arbitrary numbering of the 54 faces. We can also use the names UDRLFB for the
six colors; where the association is made once and for all for any given physical
puzzle. Like U=red, F=blue, etc.).
The elements of g are 1-1 with the observable configurations of the standard
cube; and in fact are the recipes to reach the configurations from "home". g' is
the "solution" that returns g to home.
The elements of G*C are also 1-1 with the observable configurations except now
the correspondence must also take into account the observed orientation of the
cube.
Each g in G is represented by a permutation of the cubelet faces. Each face in g
is a fixed color.
For color X, let X[g] be the set of faces of g colored X. |X[g]| = 9.
Let Coloring[g] = {U[g], D[g], R[g], L[g], F[g], B[g]}.
Then g is totally symmetric if for all c in C, Coloring[gc] = Coloring[g].
----
It is true that "home" and UUDDRRLLFFBB are totally symmetric by this
definition. "home" is a minimum (special case). UUDDRRLLFFBB is a local
maximum.
Questions:
Is there a simpler equivalent definition?
How many totally symmetric configurations are there?
Is there a less restrictive definition that guarantees local maxima?
Date: Mon, 8 Dec 1980 1956-CST
Message-id: <345171394.17@DTI>
From: aramini at DTI (Michael Aramini)
To: cube-lovers@mc
Subject: "notes on rubikics cube" booklet
can someone please tell me the address of the seller of this booklet in
England (I know a can find it in the archive file, but i prefer not
to try to edit it at a mere 300 baud, without the benefit of being
able to do an i-search (local host traps ^s's)
also, is it better to order it from this english address, or from the one
in virginia?
also, what is the current price for the booklet from each of these 2 sources?
-----
Date: 9 December 1980 1638-EST (Tuesday)
From: Dan Hoey at CMU-10A
To: McKeeman.PA at PARC-MAXC
Subject: Re: A Proposed Definition of Symmetry
CC: Cube-Lovers at MIT-MC
In-Reply-To: McKeeman.PA@PARC-MAXC's message of 8 Dec 80 20:03-EST
Message-Id: <09Dec80 163821 DH51@CMU-10A>
Given a subroup G of permutations, we may define a cube position P to be
G-symmetric if every permutation in G preserves P up to color
relabelling. Explicitly, if x and y are two facelets which have the
same color in position P, then g(x) and g(y) also have the same color
in P.
Consider, for instance, the group R of whole-cube moves. The Pons
Asinorum is R-symmetric. Any whole-cube move moves the set of blue
facelets to positions occupied (before the move) by facelets of some
single other color. In fact, even if the cube were reflected, this
would be true. This can be verified by looking at one blue facelet of
the cube and realizing that you know from that alone where all the
other blue facelets are. Letting the M denote the group composed of R
augmented by the set of whole-cube mirror reflections. The Pons
Asinorum is also M-symmetric.
McKeeman has stated that any R-symmetric position (with the exception
of the solved state) is a local maximum. I do not know if this is true,
but I can show that any M-symmetric position (with the same exception)
is a local maximum.
Consider a robot cubenik which knows how to do whole-cube moves, but
can only QTW the "up" face. Clearly, this robot has no problems,
because it can move any face to the top, perform U or U', and move it
back. The robot could get along without U' by doing U^3 instead, but if
we're counting QTWs it's not going to win any races. Solution? Build a
transdimensional robot which can perform whole-cube reflections. This
robot performs U' by reflecting the cube, performing U, and reflecting
it back. The point of this parable is that for any two QTWs x and y,
there is an element m in M for which y = (m x m'). Note also that for
any m in M, the permutation (m x m') is a QTW.
Let P be an M-symmetric position, and let (x1 x2 ... xn) = P' be the
shortest solution of P in QTWs. Assume that P is not the solved
position, so n > 0. For any QTW y, I will demonstrate an n-step
solution of P which begins with y. Write y = (m x1 m'). Since P is
M-symmetric, (P m) is a relabelling of P. This implies that (P m P')
and therefore (P m P' m') are relabellings of I. Therefore the sequence
((m x1 m') (m x2 m') ... (m xn m')) = (m P' m') will essentially solve
P, up to a whole-cube move. The existance of an n-step solution
starting with an arbitrary QTW y implies that P is a local maximum.
There is a twelve-element subgroup T of M which will suffice instead of
M for this argument. Representing elements as permutations of faces, T
is generated by the permutations (represented as cycles):
(F L U)(R D B) -- Rotating the cube about the FLU-RBD axis
(F B)(U R)(L D) -- Rotation exchanging corners FLU and RBD
(L U)(R B) -- Reflection in the LU-RB plane
Question: Does there exist a position other than the solved position
and the Pons Asinorum which is T-symmetric or R-symmetric?
Date: 9 December 1980 23:57-EST
From: Alan Bawden
Subject: Re: A Proposed Definition of Symmetry
To: Hoey at CMU-10A
cc: CUBE-LOVERS at MIT-MC
Date: 9 December 1980 1638-EST (Tuesday)
From: Dan Hoey at CMU-10A
There is a twelve-element subgroup T of M which will suffice
instead of M for this argument. Representing elements as
permutations of faces, T is generated by the permutations
(represented as cycles):
(F L U)(R D B) -- Rotating the cube about the FLU-RBD axis
(F B)(U R)(L D) -- Rotation exchanging corners FLU and RBD
(L U)(R B) -- Reflection in the LU-RB plane
AH! Excellent! (I believe you mean that last permutation to be
(L U)(R D).) It took me a while to realize that this is the subgroup
of M that leaves the FLU-RBD "diagonal" fixed.
Question: Does there exist a position other than the solved
position and the Pons Asinorum which is T-symmetric or
R-symmetric?
Hmm. I hadn't realized that we don't really know that many symmetric
positions. I have another favorite pattern that happens to be fully
M-symmetric. It is the pattern obtained by "flipping" all of the edge
cubies:
U B U
L U R
U F U
L U L F U F R U R B U B
B L F L F R F R B R B L
L D L F D F R D R B D B
D F D
L D R
D B D
This pattern has another interesting property, it is the only other
permutation besides the identity that commutes with every other
element of the cube group! I have often thought that this position is
a good candidate for maximality. Dave Plummer has shown that this
position can be also be reached in 28 moves...
Date: 10 DEC 1980 0157-EST
From: DCP at MIT-MC (David C. Plummer)
Subject: The 28 QTW all-edge-flipper mentioned by ALAN
To: CUBE-LOVERS at MIT-MC
Indeed it can be done in 28.
Tool: flip the top edges and the bottom edges:
(R L) (U D) (F B) (R L) (U D) (F B)
/\
||
To do the other four, insert the following things at the indicated
place
(u b') (u' D b' u' d l' u' d f' u' d r') (b u')
There may be another place to put this to find a shorter path.
Those with time are free to try and find one.
Date: 10 DEC 1980 0834-EST
From: DCP at MIT-MC (David C. Plummer)
Subject: A configuration symmetric wrt the first QTW
To: CUBE-LOVERS at MIT-MC
I believe the following is symmetric in the sense that any QTW
will bring you closer to home:
All corners are rotate, and letting + indicate clockwise
and - counterclockwise, each face has the following
corner configuration
+-
-+
(and all the edges are intact)
a total map of the cube corners might look like
+-
-+
+- -+ +- -+
-+ +- -+ +-
+-
-+
Each face is essentially the same: edges OK and all corners
rotated so that opposite corners are rotated in the same
direction. It is rather intuitive to me that rotating a face
clockwise is the same as rotating the face counterclockwise. This
fulfills the condition needed for maximality, but what flavor of
symmetric is it (if the symmetry is easily describable). Also,
does anybody have a 28 QTW algorithm (OR LESS!!) to go between
solved and this position.
Date: 14 December 1980 1916-EST (Sunday)
From: Dan Hoey, Jim Saxe
To: Cube-lovers at MIT-MC
Subject: Symmetry and Local Maxima (long message)
Reply-To: Dan Hoey at CMU-10A
Message-Id: <14Dec80 191649 DH51@CMU-10A>
Symmetry and Local Maxima
-- Dan Hoey
Jim Saxe
1. Introduction
===============
In this note, we attempt to give a uniform treatment of the
issues raised in the recent discussions of symmetry and local
maxima. We have attempted to restate and justify the correct
observations on these subjects that have been made in mail to
cube-lovers in recent days and also to refute a number of incorrect
ones. We include a description of 71 local maxima, which we believe
to be all of the local maxima that can be proven using known
techniques other than exhaustive search.
We let G denote the "Rubik group", consisting of all
transformations of the cube which can be achieved by twisting
faces. G does not include transformations which require movements
of the whole cube. Also, G does not take account of the
orientations of the face centers. We will defer discussion of the
Supergroup, in which face center orientations are significant (but
whole-cube motions still excluded), to Section 5 of this message.
We will sometimes (particularly towards the end of this message)
take the liberty of identifying a transformation with the position
reached by applying that transformation to SOLVED.
We let Q = {U, D, L, R, F, B, U', D', L', R', F', B'} be
the set of all possible quarter-twists. Q is a subset, but not a
subgroup, of G. The set {U, D, L, R, F, B} of all clockwise
quarter-twists is called Q+, and the set {UU, DD, LL, RR, FF, BB}
of all half-twists is called Q2. The "length" of an element s of G
(denoted |s|) is the length of the shortest sequence of quarter-
twists whose product is s. The "distance" between two elements, s
and t, of G is the length of the shortest r such that t = s r. Note
that the length of s is same as the distance between s and the
identity permutation. Note also that we are measuring distance in
the "quarter-twist metric." We defer discussion of the "half-twist
metric" to Section 5.
Two elements, s and t, of G are "neighbors" if there is
some q in Q such that t = s q (i.e., if the distance between s and
t is 1). An element, s, of G is said to be a "local maximum" if no
neighbor of s is longer than s. It is a consequence of parity
considerations that all neighbors of any local maximum, s, have the
same length, namely |s| - 1. Conversely, any element s of G (except
for the identity permutation) whose neighbors are all equally long
must be a local maximum. [Anyone who *still* doesn't understand why
neighbors cannot be equally long in the quarter-twist metric should
either send mail to one of the authors, or learn about even and odd
permutations from a book on group theory and think about how a
quarter-twist permutes the positions of the corner cubies.]
2. Symmetry
===========
It has been asserted that any "symmetric" element of G must
have all its neighbors equally long and must therefore be a local
maximum (or the identity). The first occurrence of this assertion
in cube-lovers mail was by Alan Bawden in his message of 10 Sep
1980, 11:09 pm PDT. The recent spate of messages on this subject
has made it clear that Bawden's notion of symmetry was not clearly
defined. In what follows, we make the notion of symmetry more
precise and categorize those kinds of symmetry for which the above
assertion is correct.
We let M denote the group generated by all rotations and
reflections of the whole cube and C denote the subgroup of M which
contains only the rotations. C has 24 elements (any of the six
faces can be put on top, after which any of the four adjacent faces
can be put in front, uniquely determining the positions of the
remaining faces). M has 48 elements (six choices for U, then four
for F, then two for L). [Our use of G and C agrees with that in
McKeeman's note of 8 Dec 1980, 17:03 PST. Hoey's note of 9 Dec
1980, 16:38 EST used the letter R rather than C for the latter
group, a practice which we hereby retract.]
Let G+M be the group of all transformations achievable by
any sequence of face twists and/or whole cube moves, including
reflections. Note that G is a subgroup of G+M and that the elements
of G are precisely those elements of G+M which leave the positions
of the face centers fixed (to forestall possible confusion, we
remark, at certain risk of belaboring the obvious, that the phrases
"face center positions" and "face center orientations" have
different meanings). We say that two elements, s and t, of G+M are
"M-conjugates" of each other if there exists some m in M such that
t = m' s m. We assert the following results without proof because
they are obvious.
Fact 1: Any M-conjugate of an element of Q is an element of Q,
and any M-conjugate of an element of G is an element of G.
Fact 2: If two elements of G are M-conjugates, they are equally
long.
[Anyone who does not consider the preceding facts obvious is urged
to direct further inquiries to one of us rather than bothering
everyone on the mailing list. The same goes for anyone who believes
that any other assertions made in this message are in error; this
procedure will help to reduce either duplication of erratum notices
or proliferation of false counterexamples.]
Let W be a subgroup of M. Then an element s of G is said to
be W-symmetric iff s w = w s (or, equivalently, w s w' = s) for
every w in W. [This definition is equivalent to Hoey's definition
(9 Dec 1980, 16:38 EST) as we will now show. By a "recoloring" of
the cube, we intuitively mean an operation which "consistently"
changes the colorings of all facelets of the cube. Note that the
permutation performed by a recoloring depends not only on the
chosen mapping of colors to colors, but also on the configuration
the cube is in when we start recoloring. Thus, a particular mapping
of colors to colors doesn't appear to correspond to any fixed
element of G+M. This is not a particularly satisfying situation. We
can rectify this situation by always doing the recolorings when the
cube is in the SOLVED state, that is, by thinking of recoloring a
configuration as pre-multiplication of the permutation that
achieves that configuration (from SOLVED) by a recoloring of SOLVED.
More succinctly, two elements, s and t, of G+M are equivalent up to
recoloring iff there is some m in M such that t = m s. Thus, by
Hoey's definition, an element s of G is W-symmetric iff for every w
in W there is some m in M such that
s w = m s.
But s doesn't move the face centers. So the only way m can be chosen
so that m s will leave the face-centers in the same places as s w is
to pick m = w.]
3. Transitivity
===============
For which choices of W can we guarantee that W-symmetric
patterns are local maxima? To approach this question, we introduce
the notion of Q-transitivity. Recall that Q is the set (not a
group) of all quarter-twists. We define a subgroup W of M to be
"Q-transitive" iff for each two elements p and q of Q there is some
w in W such that q = w' p w. (Q+-transitivity and Q2-transitivity
are defined analogously) We now come to our principal result:
Theorem 1: Let W be any Q-transitive subgroup of M and let s be
any W-symmetric element of G. Then any two neighbors of s
are m-conjugates of each other.
Proof: Let p and q be any two elements of Q. We must show that
sq is an M-conjugate of sp. Since W is Q-transitive, produce
some w in W such that q = w' p w. Thus,
s q = s w' p w
= w' (s p) w.
Since W is a subgroup of M, w must be an element of M. So
the definition of M-conjugacy is satisfied.
Q.E.D.
Corollary: Let W be any Q-transitive subgroup of M and let s be
any W-symmetric element of G other than the identity. Then
s is a local maximum.
If an element s of G is both V-symmetric and W-symmetric,
where V and W are subgroups of M, then it follows that s is
V+W-symmetric, where V+W is the closure of the union of V and W
(that is, V+W is the group of all elements of M which can be
expressed as the product of a sequence of elements of the union of
V and W). Thus we may unambiguously define the "symmetry group" of
any element s of G as the largest subgroup W of M such that s is
W-symmetric. The elements of this group will be precisely those
elements of M which commute with s. To see that this set of
elements forms a group, simply note that for any elements v and w
of M that commute with s,
1. (v w) s = v (w s) = v (s w) = (v s) w = (s v) w = s (v w),
and
2. v' s = v' s v v' = v' v s v' = s v'.
By the corollary to Theorem 1, any element of G (except the
identity) whose symmetry group is Q-transitive is a local maximum.
In order to find local maxima, we will first find the Q-transitive
subgroups of M. The search for Q-transitive subgroups is simplified
by realizing that the number of elements of any Q-transitive
subgroup W must be a multiple of twelve. To show this, it will be
useful to introduce another way of looking at the group M, namely
as a group of permutations on the set Q. We associate each element
m of M with a 1-1 function from Q to Q defined by the rule
(q) = m' q m for all q in Q.
These functions form a group under the operation of functional
composition, which we will write in the left-to-right manner so
that (*)(q) is definitionally equivalent to ((q)). Call
this group . The function <> which maps each element m of M to
the corresponding element of is an isomorphism as may be
seen by noting that for any two elements m and n of M and for any
element q of Q,
(*)(q) = ((q))
= n' (m' q m) n
= (m n)' q (m n)
= (q)
The isomorphism <> maps each subgroup of W of M into a subgroup
of , while is in turn a subgroup of the group of all
permutations of Q. If W is Q-transitive, then is transitive on
Q in the usual group-theoretic sense that for any two elements p
and q of Q there is some element of such that (p) = q.
This may be taken as motivation/"justification" for our use of the
term "transitive" in the former context. It is a well-known result
of group theory that every transitive group of permutations on a
set X has a number of elements that is a multiple of the number of
elements of X. This proves our claim that each Q-transitive
subgroup of M has a multiple of twelve elements. [For those not
familiar with the group-theoretic result mentioned above, the proof
for this specific case goes like this. Let be a transitive
subgroup of . We must show that has a multiple of twelve
elements. Let q be any element of Q and let V be the subgroup of
containing all elements of such that (q) = q. For
any element of , the right coset V of V is the set
{* | is in V}
Since (*)(q) = ((q)) is equal to (q) iff (q) = q,
V consists of precisely those elements of which map q to
(q). All cosets must have the same size, so the number of
elements of must be a multiple of the number of cosets, which
is the number of distinct values of (q). Since is transitive
on Q, the set of such values is all of Q, which has twelve
elements.]
We generated the 98 subgroups of M by computer, and found
that only 9 of them had a multiple of 12 elements. We examine these
subgroups below. In the preceding proof, we found it useful to
think of elements of M as permutations on the set Q. In what
follows, we will sometimes find it useful to think of elements of M
as permutations of the set of face center positions. An element of
M will be called "even" or "odd" according as it induces an even or
odd permutation on the six face centers. Also, we refer to elements
of M which are not rotations of the cube as "reflections". This
applies even to permutations which are not simple geometric
reflections but must be expressed as the composition of a
reflection and a rotation. An example is the element of M which
permutes face centers in the cycle (U,B,R,D,F,L).
The largest subgroup of M is, of course, M itself (48
elements). M has three subgroups of size 24: the group C of
rotations of the cube, the group of all even elements of M, which
we call AM (for alternating M), and the group of elements which are
in either both or neither of C and AM. We will call this third
group H. Thus the group H contains the even rotations and the odd
reflections. M has five subgroups of size 12. One of these is the
group of all even rotations, which we call AC. The other four are
of the kind called "T" by Hoey in his message of 9 Dec 1980, 16:38
EST (corrected by Bawden, same date, 23:57 EST). Let Z be any of
the four long diagonals of the cube (e.g., UFL-DRB). Then T(Z) is
the contains all elements of M which map the corner cubies at the
ends of Z either to themselves or to each other. Of these nine
groups, all except C and AC are Q-transitive.
4. A Catalog of Local Maxima
============================
In this section, we examine the seven Q-transitive subgroups of M,
and describe the 72 corresponding symmetric positions. In general,
given a geometric interpretation of a subgroup W of M, verifying
that a position is W-symmetric is immediate, and no proof will be
given. To prove that our catalog contains all W-symmetric positions
is more difficult, and we defer this to a later message.
There are four M-symmetric positions: SOLVED, the Pons
Asinorum (reached by RRLL UUDD FFBB), SOLVED with all edges flipped
(Bawden, 9 Dec 1980, 23:57 EST), and the Pons Asinorum with all
edges flipped. The Pons Asinorum is interesting in that it is our
only example of a PROVEN local maximum which has been PROVEN NOT to
be a global maximum (it is known to have a length of at most 12,
while the global maximum must be longer because |G| > 12^12). This
was pointed out by Plummer (7 Dec 1980, 07:24 EST).
The only AM-symmetric elements of G are those which are
also M-symmetric. Since the symmetry group of a position is the
largest subgroup W of M such that the position is W-symmetric,
there is no position which has AM as its symmetry group.
Plummer (10 Dec 1980, 23:27 EST) has already presented an
example of an H-symmetric position which is not M-symmetric. The
position is SOLVED with adjacent corners rotated in opposite
directions. Another position, whose H-symmetry leads to our choice
of nomenclature, is shown below.
U U U
D U D
U U U
L L L F B F R R R B F B
R L R F F F L R L B B B
L L L F B F R R R B F B
D D D
U D U
D D D
There are two such "six-H" positions; composing the two yields the
Pons Asinorum. This gives four possibilities for edge cubie
positions. The corners of an H-symmetric position may be in any of
three orientations, all home, Plummer's configuration, or Plummer's
configuration with the twists reversed. In any position, the edges
may all be flipped or not. Composing the choices yields twenty-four
H-symmetric positions, twenty of which are not M-symmetric.
There are four groups of the form T(Z). To make our
presentation more specific, we will fix Z as the (UFL-DRB)
diagonal. We define the "girdle" of the cube as the set containing
all the corner cubies other than UFL and DRB and all edge cubies
which are NOT adjacent to either UFL or DRB. Thus,
Girdle = {ULB, LB, DBL, DL, DLF, DF,
DFR, RF, URF, UR, UBR, UB}
A position which is T-symmetric but not M-symmetric (T has no
proper supergroups in M except for M itself) may be obtained by
flipping all edges on the girdle, as shown.
U B U
U U R
U U U
L L L F F F R U R B U B
B L L F F R F R R B B L
L D L F D F R R R B B B
D F D
L D D
D D D
Also, each edge on the girdle may be swapped with the diametrically
opposite edge, provided that the corners on the girdle are swapped
with their opposites as well.
R D D
U U D
U U B
D L L F F D L L F L F F
R L L F F B L R R B B F
R R B R B B U R R B B U
U U L
U D D
F D D
These positions may be composed with each other and with the four
M-symmetric positions to yield sixteen T-symmetric positions,
twelve of which are not M-symmetric. Counting the positions
symmetric with respect to the four different T groups yields 48
positions whose symmetry groups are T groups.
This completes the catalog of positions with Q-transitive
symmetry groups. Summarizing the numbers of positions of each kind,
we have
M-symmetric 4
AC-symmetric but not M-symmetric 4-4 = 0
H-symmetric but not M-symmetric 24-4 = 20
T-symmetric but not M-symmetric 4*(16-4) = 48
for a total of 72 positions, one of which is the identity and 71 of
which are local maxima.
5. Generalizations
===================
The group of whole cube rotations, C, is Q+-transitive, but
not Q-transitive, because U = c U' c' has no solution for c in C.
This means that McKeeman's suggestion (8 Dec 1980, 17:03 PST) that
C-symmetry was a sufficient condition for being a local maximum is
not an immediate corollary of Theorem 1. However, it happens that
all C-symmetric positions are also M-symmetric and they are
therefore local maxima with the exception of the identity. Thus
McKeeman's claim turns out to be true "by accident". However, the
case for the Supergroup is a different story.
Analysis of the Supergroup, in which the orientations of
the face centers are significant, is trivial given the analysis for
G. The only operations on the face centers which yield Q-transitive
symmetry groups are to leave them all alone or two rotate them all
by 180 degrees. Thus there are a total of 72 * 2 = 144 elements of
the Supergroup which have Q-transitive symmetry groups. One of
these is the identity and the other 143 are local maxima.
Considering face orientations as significant also allows us to
construct a position which is C-symmetric but not M-symmetric,
namely Big Ben, the position reached from SOLVED by turning all the
face centers 90 degrees clockwise. Big Ben is a good candidate for
a counterexample (in the Supergroup) to McKeeman's (8 Dec 1980,
17:03 PST) suggestion that C-symmetric positions are local maxima.
Possibly Big Ben is a local maximum, but it sure isn't obvious to
us that, say, U and U' will lead to positions equally near to
SOLVED).
Those who are interested in counting half-twists as single
moves may be pleased to hear that all 71 (143 in the Supergroup)
positions described above are also local maxima in the half-twist
metric. To see this, first note that every Q-transitive subgroup of
M is also Q2-transitive. This means that for any position p among
those 71 (or 143), all positions reached by quarter-twists from p
are M-conjugate (and thus equally far from SOLVED) and all positions
reached by half-twists are also M-conjugate. The positions in one
of these two sets must all be one step closer to SOLVED (in this
metric) than p. The positions in the other set cannot be further
from SOLVED than p since they are only one move away from positions
in the first set. Note that this proof depends on BOTH
Q-transitivity and Q2-transitivity. We do NOT make the claim that
any position whose symmetry group is Q2-transitive must be a local
maximum in the half-twist metric (in fact, we suspect that the
six-spot pattern mentioned below is a counterexample).
6. On Conjectures
==================
The point of this section is not to make conjectures, but
to examine conjectures which have recently appeared in the light of
our results. As an example, we will first discuss a conjecture that
has not been made, but which would likely have been baldly stated
as fact had anyone thought to do so.
"Of course, the inverse of a local maximum is also a local
maximum." Easily said, but is it true? All local maxima we know
about have Q-transitive symmetry groups, and the symmetry groups of
an element and its inverse are equal. But suppose the local maximum
were not symmetric. Consider the position reached from SOLVED by
performing the twists (U F F). From this position, either F or F'
will move the cube closer to SOLVED. From its inverse, (F' F' U'),
only U will move closer to SOLVED. Is it not conceivable that from
some position, any of the twelve twists would move closer to SOLVED,
yet only eleven or fewer would move its inverse closer? Such a
position would be a counterexample to the first statement of this
paragraph. Of course, the example we provide is not a local
maximum, and indeed there may exist no local maxima except the
(symmetric) ones we have found. But there is also no reason to
believe they can't exist, and there is no reason to believe that
their inverses are local maxima. Of course, the inverse of a global
maximum is also a global maximum.
The symmetry group of a Plummer cross has six elements and
is the intersection of H with T(Z) for an appropriate choice of
diagonal Z. This group is not Q-transitive, but is Q2-transitive.
Consequently if the algorithm presented by Saxe in his message of 3
Dec 1980, 00:50 EST, is optimal, then the Plummer cross is a local
maximum. The reason for this is that Saxe's algorithm ends with a
half-twist. This means that, if the algorithm is optimal, then, by
virtue of the Q2-transitivity of the symmetry group, performing any
half-twist on a Plummer cross brings you two qtw closer to SOLVED.
This implies that performing any quarter-twist on a Plummer cross
would bring you one qtw closer to SOLVED, since the quarter-twist
could be continued into a half-twist for a total gain of two qtw.
This observation (Saxe's algorithm optimal => Plummer cross a local
maximum) was first made by David Plummer (5 Dec 1980, 20:29 EST),
who offered a slightly different, but correct, proof. We emphasize
that this is all based on the purely speculative conjecture that
Saxe's algorithm is optimal. The Plummer cross is NOT known to be a
local maximum merely by virtue of its symmetry, Greenberg's (bogus)
statement of 7 Dec 1980, 12:18 EST ("Yet, we know, intuitively,
that CP is 'highly symmetric', and it is a local maximum.")
notwithstanding. To drive this point home, consider the six-spot
configuration (Pavelle, 16 Jul 1980, 20:51 EDT) produced by moving
(L R' F B' U' D L R'). This position has exactly the same symmetry
group as the Plummer cross, but is not a local maximum. Any of the
six quarter-twists L', R, F', B, U, or D' will bring you closer to
SOLVED (obvious), any of the other six quarter-twists will take you
further away (based on exhaustive search by computer). An even more
symmetric position is the twelve-L's [not to be confused with
Singmaster's less symmetric but visually similar 6-2L, obtained
from SOLVED by (F B U D R' L' F B)]:
L R R
L U R
L L R
B F F U U U F B B U U U
B L F U F D F R B U B D
B B F D D D F F B D D D
L R R
L D R
L L R
The symmetry group of this position is AC, the group of even
rotations. AC is Q+-symmetric, so all clockwise twists have the
same effect, and all counterclockwise twists have the same effect.
If the two sets of neighbors should happen to have the same
lengths, then this position would be a local maximum. Need we say
that there is no reason to believe this to be the case?
Michael Aramini (7 Dec 1980, 01:08 EST) mentions the
possibility that two maxima might be one half-twist apart in the
half-twist metric. This was claimed impossible by Plummer (7 Dec
1980, 07:24 EST). We do not follow the reasoning, and we conjecture
that he misread "half" as quarter and then mistyped "quarter" as
half. We also do not see anything to prevent two (local or global)
maxima in the half-twist metric from being a quarter-twist apart or
two maxima in the quarter-twist metric from being a half-twist
apart. Parity considerations do not stand in the way of such
occurrences and, while none of the known (symmetric) local maxima
are so close to each other, we have no proof that either local or
global maxima must be symmetric. We have no proof that such closely
neighboring maxima (or any non-symmetric maxima at all) *do* exist,
either.
While we know 71 local maxima, we know only 25 distinct
ones up to M-conjugacy (3 having symmetry group H, 12 having
symmetry group T, and 10 [yes, 10] having symmetry group H).
McKeeman (6 Dec 1980, 16:42 PST) has correctly (provided we
substitute our corrected definition of symmetry) noted that, if we
could show that maxima had to be symmetric, then the maximum of the
best known solutions to these configurations would bound the length
of the global maximum. Unfortunately, we have no proof of this
conjecture, nor any strong reason to think it true.
Dan Hoey (Hoey @ CMU-10A)
Jim Saxe (Saxe @ CMU-10A)
Date: 15 DEC 1980 1851-EST
From: DCP at MIT-MC (David C. Plummer)
Subject: Administrata and an Algorithm
To: CUBE-LOVERS at MIT-MC
CC: ELLEN at MIT-MC, DUFFEY at MIT-MC
First the Administrata. CUBE-LOVERS came close to a crisis this
weekend. There were thoughts of having it digested, but DUFFEY
came through and made some very valuable suggestions. What
happened is that the Hoey-Saxe message was long enough to confuse
the mailer, and it ended up sending it twice. It took at least an
hour and a half to send, so the mailer response was terrible.
Things have been reorganized to help make things more efficient,
and we really thank DUFFEY for doing this. In the future, please
limit your messages to one to two thousand characters. If you
must send long messages, please break it up into sections and
send a piece every hour or so. Please keep this mailing list
winning. Thanks.
Now for the algorithm. Hoey and Saxe mentioned an H pattern. By a
simple method of doing it, it can be done as follows:
FF (LLRR) BB (LLRR) RR (UUDD) LL (UUDD) DD (FFBB) UU (FFBB)
and after removing the NOPs
==> FF (LLRR) BB (LL) (UUDD) LL (UU) (FFBB) UU (FFBB)
which is another 28 mover. But I have found another way to do IT:
(UD LLRR FFBB UD) (FFBB LR FFBB L'R') ==> 24 QTW
Hoey and Saxe said that the Pons Asinorum is the only local
maximum that is PROVEN not to be a global maximum. It still is,
but if somebody can prove that Global Max must be larger than 24
(it is currently at 22), then this would be another example. The
other possibility is to find a faster algorithm to this pattern.
I have an intuitive sense that the global maximum is an even
distance away. I cannot prove it. Can anybody?
Date: 16 December 1980 1841-EST
From: James.Saxe at CMU-10A (C410JS30)
To: Cube-Lovers at MIT-MC
Subject: 16 qtw algs for CC and H patterns
CC: James.Saxe at CMU-10A
Message-Id: <16Dec80 184127 JS30@CMU-10A>
In his note of 4 Nov 1980, 00:23 EST, David Plummer gives a
20 qtw algorithm for the Christman cross based on the following tool
for producing a 4-cross pattern:
4+ = FB UUDD F'B' R'L' UUDD RL UUDD
This can be reduced to 16 moves as follows:
4+ = FB UD LLRR UD FB UUDD
Consequently, Plummer's 16 qtw Christman cross algorithm, conceptually
B' 4+ B, can be reduced to
B' [FB UD LLRR UD FB UUDD] B
= F UD LLRR UD FB UUDD B (16 qtw).
[Note: There is another 4-cross pattern besides the above, namely
LLRR F LLRR FB LLRR B' LLRR F'B'.]
The H pattern which Dan Hoey and I described in our earlier
message (14 Dec 1980, 19:16 EST, Sec. 4) can be achieved in 16 qtw
as follows:
FF LL DD FF BB DD RR FF
This makes it the second proven example of a local maximum which is
not a global maximum. Of course this applies equally to the second
H pattern which is Pons Asinorum away from the above. I count these
two as only one example since they are M-conjugates.
--Jim Saxe
Date: 30 DEC 1980 0109-EST
From: DCP at MIT-MC (David C. Plummer)
Subject: Hackery (92 line message) (first of three)
To: CUBE-LOVERS at MIT-MC
Greetings...I'm back from vacation. I took a copy the old mail
home and read it. It was an interesting experience.
Existence proofs:
Saxe and Hoey described (14 December 1980, 19:16 EST)
several local maximums that were previously not described. I have
algorithms for a couple that are 28 moves long, showing that
THESE PARTICULAR local maximums are no further than 28 away from
SOLVED.
==========
The edges flipped along the girdle (hereafter refered to as
GIRDLE EDGES FLIPPED [until somebody else thinks of a better
name]) can be done as follows:
(F D'U L D'U B D'U R D'U ) (U'L'F)
(L'R F' L'R U' L'R B' L'R D') (F'L U)
[This is a Sprat Wrench, get-ready, another Sprat Wrench,
un-get-ready.] At first glance this is 30 moves, but notice that
the U and U' near the end of the first line cancel, making it 28.
==========
The edges not along the girdle flipped (hereafter refered to as
OFF GIRDLE EDGES FLIPPED) (NOTE: This is GIRDLE EDGES FLIPPED
compounded with ALL EDGES FLIPPED), can be done as follows:
(L B'F U B'F R B'F D B'F ) (F F U)
(R'L U' R'L F' R'L D' R'L B') (U'F F)
[Again, Sprat-mung-Sprat-unmung] And again, 30 at first glance,
but this time the FFF near the end of the first line becomes F',
so 30-2 is again 28.
==========
==========
I noticed something peculiar about the second kind of girdle
configuration the mentioned. Hoey and Saxe describe it as "each
edge on the girdle may be swapped with the diametrically opposite
edge, provided that the corners on the girdle are swapped with
their opposites as well" and they give a picture. It is indeed
reachable, but the most interesting thing (in my opinion) is that
it is an odd distance away!! (Of course so are the related
configurations obtained by performing on it: PONS, ALL EDGES
FLIPPED, and PONS WITH ALL EDGES FLIPPED.) ODD!! Most of (if not
all) the previously described "nifty patterns" were even. And on
top of that, this odd permutation is a local max. Most of the
tools people seem to use are even. Several of them fall along
the lines of MUNG/DO SOMETHING/UNMUNG, and DO SOMETHING is often
even, and two MUNGS is even. Now the question is: HOW FAR AWAY
ARE THESE, AND HOW DO YOU (THE HUMAN) TRY TO FIND THE ALGORITHMS
TO DO THEM? My suggestion: Do something, Mung, Do something
(maybe different). I also suggest that Do somethingbe of length
12, and that mung be 3 or 5. I would really apprecitate it if it
were less than 28 -- I happen to like 28 as the maximal length.
==========
The local max which I described (Plummer 10 December 1980, 08:34
EST) which is all corners rotated such that adjacent corners are
rotated in opposite directions (shall we call this ALL CORNERS
ROTATED? I know it isn't explicit, but it is descriptive enough
for our purposes) can be done in 40 by a Christman Cross (16,
thanks to Saxe 10 December 1980, 18:41 EST), two half twists (4),
another Christman Cross (16), and undoing the two half twists
(4). I have been trying for the better part of a week to get this
down to 28 (my favorite number), but have not yet succeeded. What
I want is a 14 move algorithm to exchange all corners with their
diametrically opposite corner in such a way that the top forms a
cross of the top as the cross and the bottom at the corners. Two
of these with a whole cube manipulation in between will do the
trick. Alternately, if somebody could come up with a 12 mover to
move FDL to FUR (and do this so that it sould look this way if
viewed from F,L,R or B), you could do U U' and the
opposite corners would be swapped as desired. I have an almost
algorithm, but it falls short in that the four edges on the
center horizontal slice are swapped with their diametric
opposites. The algorithm is
R'L' F'B' R'L' FB RL FB
If anybody can find the desired algorithms, please publish.
==========
Reading cube mail, I saw several things about "this Sprat Wrench
is best executed by a right handed person." The way I do the
Wrench does not descriminate, and I think it is worth it to
describe it:
(0 QTW) Holding the right face in the right hand,
(0 QTW) holding the left face in the left hand,
(2 QTW) rotate the center verticle slice toward you,
(1 QTW) rotate the NEW front face clockwise (either hand will do)
Do this a total of four times, giving 4*3=12 QTW. Try it, and
also try turning the NEW front face counterclockwise.
==========
That's all I can think of on this subject. Coming soon:
Design of the higher order cubes
Thoughts on manipulating the higher order cubes.
Date: 30 DEC 1980 0144-EST
From: DCP at MIT-MC (David C. Plummer)
Subject: The higher order cubes (just the 4x4x4 for now) [93 lines]
To: CUBE-LOVERS at MIT-MC
And you thought the 3x3x3 was a complicated beastie...
I have plans for the 4x4x4 which I think can work. A first order
approximation is to take the 3x3x3 and split the edges in two.
This doesn't take care of the centers or axis, but those are
perhaps the most complicated anyway, and will get considerable
thought.
WARNING: If you really want to see this, get out the graph paper
and correct the aspect ratio that the characters here will have.
The way I have labeled the cubies is as follows:
Corners are labelled A (there are 8 of these)
Edges are labeled both B and D (this defines the
orientation) (there are 24 of these)
Centers are labeled C (there are 24 of these)
So a face would look like:
ABDA
DCCB
BCCD
ADBA
Take a slice down the center of one of the planes and open it up.
It should look something like:
.....&&&&&&&&++++++++
.....&&&&&&&&++++++++ where & and @ are pieces of
..xxxxxx&&&++++++++++ the C type of cubie
..xxxxxx&&&++++++++++ + is a B/D type
....xx&&&&&++++++++++ x is the central axis
....xx&&&&&++++++++++
....xx&&&&&++++++++++
....xx&&&&&++++++++++
....xx&&&&&++++++++@@
....xx&&&&&++++++++@@
....xx&&&&/@@@@@@@@@@
\...xx&&&/@@@@@@@@@@@
.\..xx&&/@@@@@@@@@@@@
..\.xx&/@@@@@@@@@xx@@
...\xx/@@@@@@@@@@xx@@
xxxxxxxxxxxxxxxxxxx@@
xxxxxxxxxxxxxxxxxxx..
.../xx\..........xx..
../.xx.\.........xx..
./..xx..\............
Gross, isn't it? There are a few things going on here that are
hard to show. That central cross must be able to rotate, so the
innermost parts of C and B/D are carved in somewhat. There is
another hairy constraint: the central axis MUST BE RIGIDLY
CONNECTED TO ONE AND ONLY ONE OF THE C TYPE CUBIES. The reason is
hard to describe. Hint: suppose you rotate a half cube portion.
It is possible that when the rotation is finished, the central
axis is misaligned. Connecting it to one face cubie forces it to
win. This may also make the central cross somewhat more fragile.
Forgetting about the central axis and the corner cubies (they
shouldn't be too able to work into the picture) we have the
following diagrams for the B/D and C type cubies. The numbers
indicate elevations:
B/D C
44444444 4444
44444444 ACTUALLY THEY 4444
4443333333 ARE CURVED, 4443333333
4443333333 BUT CHARS 4443333333.
4433333333 HAVE POOR 4433333333.2
4433333333 RESOLUTION 4433333333.2.
4433333333 4433333333.2.1
4433333333 4433333333.2.1.
I seriously think these will work. I hope to find somebody in
a material science type of lab that can make plastic molds so I
can actually try and build one of these. I hope to use a soft
plastic that is machinable, carvable, sandpaperable, etc, and
when I have a good one, make several pieces out of harder plastic
and see what happens. Perhaps I'll have something by the end of
January.
I think that the 5x5x5 cube (though the tolerances might be
tighter) may actually be easier to construct. It may also be a
more interesting cube to work with.
Next: Notes on transforms on the 4x4x4 and 5x5x5 cubes.
PS: A mind blower: a DODECAHEDRON frob (can't call it a cube). I
thought of this one coming back on the bus. I haven't put
anything down on paper, but my minds eye tells me it has a
chance.
Comments welcome.
Date: 30 DEC 1980 0152-EST
From: DCP at MIT-MC (David C. Plummer)
Subject: Notes on transforms on the 4x4x4 and 5x5x5 cubes. (65 lines)
To: CUBE-LOVERS at MIT-MC
First we have to define what a twist is. I propose that a twist
is a twisting of face or faces about an axis such that there is
only one plane on which sliding friction happens. This means that
slice turns are not single turns, but rather multiple turns. This
is consistent with the 3x3x3 definition, and it is generalizable
to nxnxn. Also, I think it is wisest to consider quarter twists
as the only single twist "move" (don't count half twists as one!!)
One thing to note: Even order cubes should have a STANDARD
coloring. This is nice to have since there is now visible axis to
align the corners on. I know it isn't necessary, but if I own a
cube and you own a cube, and they are colored differently, and we
swap for a day, both of us will probably have a hard time trying
to get used to a different color arrangement than what each is
used to. May I suggest
Front Right Up Back Left Down
RED WHITE BLUE ORANGE YELLOW GREEN
The general PONS can be described as follows (for cubes, not
necessarily dodecahedra):
Pick an axis. Make 180 twists along all twistable planes.
Pick another axis. Do the same.
Do the same on the last axis.
This will produce a checkerboard pattern on all faces using
complementary (from the opposite side of the cube) colors. On the
4x4x4 (and in general on even order cubes), to corner cubies do
not move (but then again it is hard to tell without a fixed
reference).
I have done more thinking about the 5x5x5 than the 4x4x4 since
the 5x5x5 offers several immediate advantages. First of all, from
solved we can do things like: do a transform as we would on a
3x3x3 and consider the 3x3 set of cubies in the center of the
faces as the center of our favorite 3x3x3 Hungarian. Another way
to view this is to do normal Hungarian rotations using only one
level deep of cubies. Then run the algorithm backwards doing
twists using two levels deep (considering theface center as a
center and 2x2 at the corners as corners, and 1x2 around the
edges as edges. (Bad explanation, but it is 1:30 in the morning).
I believe doing the traditional SIX-DOTS pattern forward then
backward using this method will produce something that looks
like:
+++++ +++++
+@@@+ +@@@+
+@+@+ +@x@+
+@@@+ +@@@+
+++++ +++++
The one on the right is what happens if you do it again in the
same direction. I think the Plummer's Cross will look like:
+@+@+
@@+@@
+++++
@@+@@
+@+@+
And if I am not mistaken, the approriate EXTENDED-SIX-DOTS
mentioned above performed on this will produce a 3-cycle
checkerboard. And the PONS on top of that will produce a 6-cycle
checkerboard. Of course, I could be wrong, but we wont know until
(a) somebody carfully works it out on paper (b) somebody writes a
computer program to simulate the crazy thing, or (c) somebody
actually builds one.
That's it for tonight folks, aren't you glad!!
Date: 29 Dec 1980 23:54 PST
From: McKeeman at PARC-MAXC
Subject: Re: The higher order cubes (just the 4x4x4 for now) [93 lines]
In-reply-to: DCP's message of 30 DEC 1980 0144-EST
To: DCP at MIT-MC (David C. Plummer)
cc: CUBE-LOVERS at MIT-MC
David,
Interesting thoughts.
It seems to me that the icosahedron is a more attractive physical target than the
dodecahedron. It has 12 shear planes. A move would be a 72o rotation of a
five-triangle subsection. My intuition is that it is buildable more or less along
the line of the Rubik cube, except that the fixed axes point at vertices instead of
center faces.
The dual of the 2x2x2 cube seems to be an octahedron with shear planes along its
edges.
More generally slice a sphere a buncha times. The cuts are shear planes. Don't
let it fall apart (there is the trick!). Now paint the surface pieces. Now twist to
mix.
The actual shape does not seem to matter much. It is the interaction of the shear
planes that makes the puzzle.
Happy sphereing,
Bill
Date: 31 DEC 1980 0330-EST
From: ACW at MIT-MC (Allan C. Wechsler)
Subject: Groups and Cayley Graphs
To: CUBE-LOVERS at MIT-MC
If there is any popular response I will write
a short introduction to Group Theory for Cubists.
= - = - = - = - = - = - = - = - = - = - = - = - =
Problem: the cube is known to have "small mountains",
states that are locally maximally distant from the
home state but not globally so. The Pons Asinorum
is an example: it has been proved (a refreshing relief
from unmitigated conjecture) to be 12 qtw from home
and all its neighbors are closer.
The cube's Cayley graph in its six generators is vertex-
transitive (all the states look the same) like all Cayley
Graphs. In addition, because all the generators are
conjugate in the big group (quarter-twists plus whole-
cube motions) the Cayley graph is edge-transitive
(all transitions between adjacent states look the same).
Can anyone find a small example of an edge-transitive
graph that has local maxima that are not global?
---Wechsler
Date: 31 DEC 1980 1115-EST
From: DCP at MIT-MC (David C. Plummer)
Subject: the higher order SPHERE slices
To: McKeeman at PARC-MAXC
CC: CUBE-LOVERS at MIT-MC
Sigh, my goemetry needs to be zapped back into working order. I
really did mean icosahedron instead of dodecahedron (they both
have 12 faces). I was originally thinking of the fixed axes
pointing at the center faces, but pointing at the vertices is an
interesting idea. I shall have to think about it a little.
About the general case: "slice a sphere a buncha times." Holding
it together is REALLY the tricky part. Consider the 7x7x7 CUBE.
Do a 45 degree clockwise rotation on the top face (half a quarter
twist). The problem is that the ENTIRE FRU corner cubie
COMPLETELY hangs over the front face. This means that the 3-D
jigsaw-puzzle idea that keeps the 3x3x3 cube together will not
work on order 7 and higher order CUBES.
Date: 31 Dec 1980 0729-EST
From: ZILCH at MIT-DMS (Chris C. Worrell )
To: CUBE-LOVERS at MIT-MC
Subject: relatives of the plummer cross
Message-id: <[MIT-DMS].177333>
Two relatives of the plummer cross which people may find interesting
are given here:
The first is called a baseball.
lL U L
L U U
L L L
F F F U U U B R B D B D
L L F U F F R R B D B B
F L F U F U B B B D D D
R D R
D D R
R R R
I do this one in 50 qtw though I know how to shorten it to 42.
The second is called a cube in a cube.
L L L
U U L
U U L
F L L F F U B B B D D D
F L L F F U B R R B B D
F F F U U U B R R B B D
R R R
R D D
R D D
I do this one in 70 qtw, though I could reduce it to 62.
These configurations both have two minor variants
which leave the essential nature of the configuration the same.
these other patterns are obtained by twisting the FUL AND BDR
corners. Both of these patterns were devised by people at Caltech.
Chris Worrell (ZILCH @MIT-AI)
Date: 31 DEC 1980 1210-EST
From: DCP at MIT-MC (David C. Plummer)
Subject: the 5x5x5 [133 lines]
To: CUBE-LOVERS at MIT-MC
OK, folks, I'm considering going further than 4x4x4 and entering
the realm of the 5x5x5.
Cubies:
C := Corner
X := aXis (center)
E := Edge (outside center)
L := Left (external edge)
R := Right (external edge)
D := Diagonal (internal [to the face] corner)
A := Adjacent (to the center, thanks to WER)
(internal [to the face] edge)
A 3-D view would look like this
z=-5 +---+---+---+---+---+
/ / / / / /|
/ C / L / E / R / C / |
+---+---+---+---+---+ |
/ / / / / /|C +
/ R / D / A / D / L / | /|
+---+---+---+---+---+ |/ |
z=0 / / / / / /|R + |
/ E / A / X / A / E / | /|L +
+---+---+---+---+---+ |/ | /|
/ / / / / /|E + |/ |
/ L / D / A / D / R / | /|D + |
+---+---+---+---+---+ |/ | /|E +
/ / / / / /|L + |/ | /|
/ C / R / E / D / C / | /|A + |/ |
y,z=5 +---+---+---+---+---+ |/ | /|A + |
| | | | | |C + |/ | /|R +
| C | L | E | R | C | /|D + |/ | /|
| | | | | |/ | /|X + |/ |
y=3 +---+---+---+---+---+ |/ | /|D + |
| | | | | |R + |/ | /|C +
| R | D | A | D | L | /|A + |/ | /
| | | | | |/ | /|A + |/
y=1 +---+---+---+---+---+ |/ | /|L +
| | | | | |E + |/ | /
y=0 | E | A | X | A | E | /|D + |/
| | | | | |/ | /|E +
y=-1 +---+---+---+---+---+ |/ | /
| | | | | |L + |/
| L | D | A | D | R | /|R +
| | | | | |/ | /
y=-3 +---+---+---+---+---+ |/
| | | | | |C +
| C | R | E | L | C | /
| | | | | |/
y=-5 +---+---+---+---+---+
x=-5 -3 -1 1 3 5
LOVE THAT ASPECT RATIO !!!!
All in all there are
6 aXis faces
8 Corners
12 Edges
24 Left/Right type edges
24 Diagonals
24 Adjacents
--
98 = 5^3 - 3^3 = 125-27 visible cubies
Computation (inaccurate, but within a couple orders of magnitude)
of the number of reachable positions:
Axes: lets not hack the extended problem yet -> 1
Corners:8 of them anywhere -> 8!
each can take 3 orientations -> 3^8
parity of the corner -> 1/3
Edges: 12 of them anywhere -> 12!
each can take 2 orientations -> 2^12
position/orientation restriction -> 1/4
L/R: 24 of them anywhere -> 24!
orientation defined (they cannot flip) -> 1
parity (cannot swap only two) -> 1/2 (I think)
Adjac: 24 of them anywhere: -> 24!
one edge always touches a face center -> 1
parity -> 1/2 (at least)
Diags: 24 of them anywhere -> 24!
one corner always touches a face center -> 1
parity -> 1/2 (at least)
It may not be accurate, but this computation gives
1.291318 * 10^90
A slice through the center (z=0) probably looks something like
y=5\
/ ..XXXXXXXXXX++++++++++EEEEEEEEEE
..XXXXXXXXXX++++++++++EEEEEEEEEE
.....XXXX++++++++EEEEEEEEEEEEEEE
.....XXXX++++++++EEEEEEEEEEEEEEE X is an axis cubie
y=4\ .....XXXX++++++++EEEEEEEEEEEEEEE E is an edge cubie
/ .....XXXX++++++++EEEEEEEEEEEEEEE + is one adjacent cubie
.....XXXX++++++++EEEEEEEEEEEEEEE ~ is another adjacent
.....XXXX++++++++EEEEEEEEEEEEEEE
.....XXXX++++++++EEEEEEEEEEEEEEE
y=3\ .....XXXX++++++++EEEEEEEEEEEEEEE
/ .....XXXX++++++++EEEEEEEEEEEEE~~
.....XXXX++++++++EEEEEEEEEEEEE~~
.....XXXX++++++++EEEEEEEEEEEEE~~
.....XXXX++++++++EEEEEEEEEEEEE~~
y=2\ .....XXXX++++++++EEEEEEEEEEEEE~~
/ .....XXXX+++++++/~~~~~~~~~~~~~~~
.....XXXX++++++/~~~~~~~~~~~~~~~~
.....XXXX+++++/~~~~~~~~~~~~~~~~~
\....XXXX++++/~~~~~~~~~~~~~~~~~~
y=1\ .\...XXXX+++/~~~~~~~~~~~~~~~~~~~
/ ..\..XXXX++/~~~~~~~~~~~~~~~~~~XX
...\.XXXX+/~~~~~~~~~~~~~~~~~~~XX
....\XXXX/~~~~~~~~~~~~~~~~~~~~XX
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
y=0\ XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
/ XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
..../XXXX\....................XX
.../.XXXX.\...................XX
y=-1\ ../..XXXX..\..................XX
/
/\ /\ /\ /\ /\
x=-1 0 1 3 5
This time the central axis is rigid in the sense that it does
form a cross, but each of the six spokes can rotate as in the
3x3x3 cube. The curvature and tolerances of some of the pieces
gets a little hairy, but I'm working with graph paper and looking
at the other slices through the cube. Wish me luck -- I have
thoughts of construction.
Date: 31 December 1980 1511-EST (Wednesday)
From: Guy.Steele at CMU-10A
To: DCP at MIT-MC (David C. Plummer)
Subject: Icosahedron
CC: cube-lovers at MIT-MC
In-Reply-To: DCP@MIT-MC's message of 31 Dec 80 11:15-EST
Message-Id: <31Dec80 151120 GS70@CMU-10A>
The icosahedron has twenty faces, not twelve. It and the
dodecahedron are duals: each has the same number of vertices
as the other has faces. If you join the centers of the faces
of one with lines, you get the other. They have the same
number of edges.
Date: 31 DEC 1980 1230-PST
From: WOODS at PARC-MAXC
Subject: Re: relatives of the plummer cross
To: ZILCH at MIT-DMS, CUBE-LOVERS at MIT-MC
In response to the message sent 31 Dec 1980 0729-EST from ZILCH@MIT-DMS
Your "baseball" is listed in Singmaster's booklet as "the Worm" (though
I must admit I like your name for it), with a 28-qtw solution (23 if
you count half-twists=1). I prefer the pattern named the "Snake":
DUD
DUD
DUD
BBB RFR FFF LBL
BLL FFR FRR BBL
BLB RRR FRF LLL
UUU
DDD
UUU
which is known to be soluble in 20 qtw (16 if htw=1). Meanwhile, your
cube-in-a-cube has a 24-qtw (20-htw/qtw) solution in Singmaster, though
I generally find it easy enough to make in 34 qtw using the commutator
operation [spoiler warning!] nicknamed "rFUfuR": rFRfUfuFuRUr.
-- Don.
-------
Date: 31 Dec 1980 1545-PST
From: Dave Dyer
Subject: bigger&better cubes
To: cube-lovers at MIT-MC
Clearly, we must not be deterred by mere mechanical limitations!
I'm sure a color bit-mapped raster could do a dandy job of displaying
cube-like objects in any size and geometry you can imagine. If you
must have a physical cube, how about a box with LCD's on all faces
(in six colors?), pressure sensors to detect what ytou want to twist,
and a uP inside to do all the work.
-------
Date: 31 Dec 1980 1634-PST
From: Dave Dyer
Subject: Singmasters book
To: cube-lovers at MIT-MC
Does anyone know of a domestic source for it?
-------
Date: 31 Dec 1980 1651-PST
From: MERRITT at USC-ISIB
Subject: Bigger & better cubes
To: cube-lovers at MIT-MC
In light of all the talk of cubes greater than 3 x 3 x 3, such as
4 x 4 x 4, or 5 x 5 x 5, perhaps you would like to comtemplate a
cube that is 3 x 3 x 3 x 3, or 4 x 4 x 4 x 4. The idea of adding
dimensions is really reasonable, since it needn't be mechanically
possible. This Rubics hypercube could probably be represented by
a video display in some form.
This is at least a matter for further contemplation.
-IHM
-------
Date: 31 DEC 1980 1740-PST
From: WOODS at PARC-MAXC
Subject: Re: Bigger & better cubes
To: MERRITT at USC-ISIB, cube-lovers at MIT-MC
In response to the message sent 31 Dec 1980 1651-PST from MERRITT@USC-ISIB
And then of course there's the megacube, formed by taking a standard 3x3x3
cube and replacing each cubie with a 3x3x3 Rubik's cube. The idea would be
that each megaface could be twisted as usual, but the megafacies (the faces
of the smaller 3x3x3 cubes) could be twisted only if they were visible.
(Otherwise all you really have is a collection of independent 3x3x3 cubes.)
The orientation of the colors on all the microcubes is the same -- that
is, the beast starts with all 81 microfacies on a given side being the same
color.
-- Don.
-------
Date: 31 December 1980 22:35-EST
From: Ed Schwalenberg
Subject: bigger&better cubes
To: DDYER at USC-ISIB
cc: cube-lovers at MIT-MC
Obviously the age of mechanical polyhedra is long past. I propose
that the center cubies (polyhedries?) of the electronic marvels
be equipped with fixed arrows so that those who are hacking the
higher-order problem can be happy. (I really like this idea because
it closely resembles one of my favorite ideas: a keyboard which has
small character displays in each keytop in lieu of engraving; when you
hit SHIFT the legends change from qwertyiop to QWERTYUIOP and when you
hit CONTROL+META they change to ETAIONSHRDLU, etc.)
The all-electronic cube has many other potential features: instant
resettability, a stack of saved states, subroutines.... Given
extensible, customizible, self-documenting editors, I don't see why
we should settle for polyhedra that are any less featureful.
Speaking of higher-order polyhedra, I hereby nominate the tetrahedron
as being appropriate to my own level of expertise.
Date: 1 JAN 1981 0221-EST
From: ACW at MIT-MC (Allan C. Wechsler)
Subject: How much intellect does it take to solve the cube?
To: CUBE-LOVERS at MIT-MC
Well, happy new year.
I solved it while not exactly sober.
---Wechsler
Date: 1 JAN 1981 1315-EST
From: DCP at MIT-MC (David C. Plummer)
Subject: several subjects
To: CUBE-LOVERS at MIT-MC
One last try!! What I meant was the 12 sided frob built out of
pentagons. And after refering to better and better dictionaries I
discovered this thing is called a pentagonal dodecahedron, and I
meant the faces to be the points of rotation. Perhaps McKeeman
thought I meant the rhombic dodecahedron, and subsequent messages
got me confused and I jumped the gun without thinking very
heavily.
Woods: Could you please send the manipulations for "baseball,"
"snake," and "cube-in-cube" for the benifit of those who do not
have Singmaster. Please use prime notation (R' instead of (lower
case) r) for counterclockwise twists since that seems to be the
notation currently in use in this list.
In general, my opinion is that it would be nice if people would
send along the short algorithms that are known. ZILCH's 50 and 70
qtw algotithms are a little too long, but anything under 36
should probably be sent. I know it may be a spoiler, but (1)
there seem to be several configurations mentioned and perhaps
some people don't have time to find nice fast ways to get there,
(2) it reduces needless duplication of effort, (3) parts of the
algorithm (or the algorithm itself) might serve as a subroutine
for other algorithms under development.
On the concept of the higher order "cubes:"
N dimensions
has a reasonable geometric interpretation (maybe it
doesn't have to have this condition?)
built out of some number of "cubies" of dimension N
Each "cubie" is in turn a "cube," perhaps of a different
order than the larger cube (eg, a 3x3x3 cube
whose cubies are 5x5x5)
Each "cubie" of the "cubie" is a "cube," ad infinitum as
desired
In addition to all this, each faclet is a "cube" of
dimension N-1, ad infinitum (at least until the
dimensions run out!!)
The thing I am doing is trying to PHYSICALLY construct a higher
order (flavor: dimension 3, cubical, order 5, cubies are "atomic"
(ie, not cubes in themselves), faclets are atomic). Personally, I
think half the fun is being able to hold one of the beasties and
mung it by twisting.
ACW: I think it would be instructive to have an short intro to
Group Theory for Cubist. This would benefit newcomers to the
mailing list, and people who hack the cube and want to know some
of the theory behind the cube. (5-10K characters if you can keep
it down to that. If not, send it when machines are generally
lightly loaded.) I vote: plase do.
Date: 3 JAN 1981 0248-EST
From: ZILCH at MIT-MC (Chris C. Worrell )
Subject: How to play with the corners of your cube (long message)
To: CUBE-LOVERS at MIT-MC
At this point I have found sufficient Algorithms such that given a
cube with everything correct except for possibly the four
corner cubies on one side, I can (with a little thought and
reference to my notes) solve the cube in 24qtws.
[SPOILER WARNING]
If all the cubes are in the right place, but possibly oriented
wrong, the following transforms are used to TWIST the corners to
the proper orientations:
T1: F' (R' D' R D' F D F' D)^2 F 18qtws
(FDL,RDF,BDR,LDB) => (DLF,FRD,RBD,DBL)
T2: L D (D L' D' L)^2 D' L' R' D' (D' R D R')^2 D R 24qtws
(FDL,RDF,BDR,LDB) => (LFD,DFR,RBD,DBL)
T3: (D' L' D R D' L D R') (B' L D^2 L' B L B' D^2 B L')
(FDL,RDF,BDR) => (DLF,DFR,DRB) 20qtws
T4: (R D' L' D R' D' L D) (L B' D^2 B L' B' L D^2 L' B)
(FDL,RDF,BDR) => (LFD,FRD,RBD) 20qtws
Note: T3 and T4 are inverses based on the same components, which
happen to commute. (see C1 and C2 below)
T5: (L' U L F U F') D' (F U' F' L' U' L) D 14qtws
(FDL,RDF) => (DLF,FRD)
T6: L' F' D' L' D R D' L D F L F' R' F 14qtws
(FDL,RDF) => (LFD,DRB)
Along the same line as T5 and T6, but not usefull in the
present discussion, shown to me in a private message from
Dan.Hoey@CMU-10A.
T7: F' R' D' R U R' D R F D F' U' F D' 14qtws
(FDL,BUR) => (LFD,RBU)
If all of the corner cubies are not in the proper positions
it is more profitable to execute several corner moving transforms
rather then one corner moving one then one corner twisting one.
As presented here all transforms cycle the cubies in the same
manner (clockwise) , though twisting the cubies in all possible
ways. Their inverses (counter-clockwise) should also be kept
in mind.
C1: D' L' D R D' L D R' 8qtws
(FDL,RDF,BDR) => (FRD,RBD,LFD) (twist all clockwise)
C2: L B' D^2 B L' B' L D^2 L' B 12qtws
(FDL,RDF,BDR) => (DFR,DRB,DLF) (twist all clockwise)
Note: make reference in c1 (twist all counter-clockwise)
C3: F L^2 D' R' D L' D' R D L' F' 12qtws
(FDL,RDF,BDR) => (RDF,BDR,FDL) (don't twist at all)
C4: F' R' B' R F R' B R 8qtws
(FDL,RDF,BDR) => (FRD,BDR,DLF)
C5: F L F' R F L' F' R' 8qtws
(FDL,RDF,BDR) => (RDF,RBD,DLF)
Note: C4 and c5 have been adopted from those presented by DCP in
his message of 25 nov. 1308-EST
C6: L F L' D^2 L F' L' F D^2 F' 12qtws
(FDL,RDF,BDR) => (FRD,DRB,FDL)
C7: R' D^2 R B' R' B D^2 B' R B 12qtws
(FDL,RDF,BDR) => (DFR,RBD,FDL)
C8: (C5)' (C1)'= 16qtws
(R F L F' R' F L' F') (R D' L' D R' D' L D)
(FDL,RDF,BDR) => (RDF,DRB,LFD)
C9: (C1)' (C4)'= 16qtws
(R D' L' D R' D' L D) (R' B' R F' R' B R F) (FDL,RDF,BDR) => (DFR,BDR,LFD)
These nine transforms are the only possible legal ones (along with
their inverses) which exchange three corners on a face (with the
possibility of twists0, though I can't guarentee minimum lengths for
any of them.
If all of the corners are not in their proper positions then there
are three possibilities:
1) One of the corners is in the right position and has the correct
orientation.
to fix: do the appropriate transform, or its inverse from the
list given above.
max length=16qtws
2) None of the corners is in the proper position
to fix: using C1,C4, or C5 (the shortest ones) move one of the
corners to the proper position and orientation, then
continue as in case 1.
max length=8+16=24qtws
3) One of the corners is in the correct position, but is in the wrong
orientation.
to fix: Preferably using C1,C4, or C5 move theproperly
positioned corner out of its spot and at the same time move
another corner into its proper position and orientation, then
procede as in case 1.
If none of C1, C4, or C5 will do the proper thing then
a combination of C2 and C3 must be used, C2 first, to orient
the corners correctly (with respect to the bottom) then use C3 to
position the corners correctly.
max length=8+16=12+12=24qtws
These algorithms may be usefull to someone making a sides first,
corners second cube solving algorithm.
If anyone has any shorter algorithms for any of these transforms, please
send them to the list.
Unfortunatly I probably won't be able to answer any questions about this
method as I am going back to school (Caltech) tommorrow (today?)(Sat. 3rd)
and I don't have decent net access from there.
Chris Worrell (ZILCH@MIT-MC)
p.s.: sorry about the length of this message.
Date: 3 JAN 1981 0433-EST
From: ZILCH at MIT-MC (Chris C. Worrell )
Subject: oops....
To: CUBE-LOVERS at MIT-MC
Sorry folks, I missed one possibility.
In case 2,my algorithm does not account for the configuration reached by
L R D^2 R' L' F' B' D^2 F B D^2 14qtws
(FDL,RDF,BDR,LDB) => (BDR,LDB,FDL,RDF)
Either add this transform to the list given, or when you get to
this particular configuration execute C3 twice 24qtws, but still inmy
range and still along the general lines of the rest of the possibilities.
Chris Worrell (ZILCH@MIT-MC)
ACW@MIT-AI 01/05/81 00:01:05 Re: Group theory and Cayley graphs.
To: CUBE-LOVERS at MIT-AI
I have some other short term resposibilities,
but will get around to this in a couple of
days.
---Wechsler
From: Don Woods
Subject: excerpts from Singmaster
To: CUBE-LOVERS at MIT-MC
[In reply to message from Plummer sent 1 JAN 1981 1315-EST.]
** SPOILER WARNING!! SPOILER WARNING!! **
This message gives the shortest solutions known to Singmaster (as of
the fifth edition of his booklet) for three pretty and moderately
complex patterns, recently referred to as "baseball" (or "worm"),
"snake", and "cube-in-a-cube". People wishing to investigate these
patterns (as described in earlier messages) may not wish to read
further.
Notes: Singmaster uses the half-twist measure. His notation also
includes special representation for the slice (center-twist) and
antislice (such as L+R as opposed to L'+R) operations, which he counts
as two twists; I have expanded this to strict "befuddler" (BFUDLR)
notation in this message.
BASEBALL or WORM
RUFFD'RL'FB'D'F'R'FFRUUFRRF'R'U'F'UUFR
SNAKE
BRL'D'RRDR'LB'RR + UBBU'DBBRLUUR'L'BBD'
It's not too hard to get to the snake, if you don't mind wasting few
twists. Start with DLLRRD' and you're most of the way there. This
assumes you know simple macros for swapping two pairs of edge cubies
and for flipping two edge cubies.
CUBE IN A CUBE
BL'DDLDF'DDFD'B' + F'RUUR'U'BUUB'UF
Again, the cube-in-a-cube is not hard to generate using two instances
of the commutator macro -- R'FRF' UF'U'F U'RUR' -- plus a few simple
extra twists. This is left as an exercise to the reader.
-- Don.
Date: 7 January 1981 12:59 est
From: Greenberg.Symbolics at MIT-Multics
Subject: Lisp Machine Color cubesys
To: cube-lovers at MIT-MC
Lisp machine color cubesys has been fixed.
Date: 7 January 1981 1352-EST (Wednesday)
From: Dan Hoey at CMU-10A
To: Cube-Lovers at MIT-MC
Subject: Pons Asinorum -- Part 1: Optimality
Message-Id: <07Jan81 135220 DH51@CMU-10A>
The Pons Asinorum (obtained by UUDDFFBBLLRR, and known as
6-X in Singmaster) is well-known to readers of this list. It is
perhaps surprising that this so well-known position has anything
more to teach us.
The first surprise is that the 12-move sequence given above
is provably optimal in the quarter-twist metric. Proofs were sent
to me by David C. Plummer, who attributed it to Alan Wechsler, and
by Chris C. Worrell. While it is well-known (See Alan Bawden's
messages of 31 July 1980 13:06-EDT and 31 JUL 1980 2159-EDT) that
some positions require at least 21 moves, the longest sequence
which has previously been proven optimal is LR'FB'DU'LR' for the
six-spot configuration. It is good to see a 12-move sequence
proven optimal -- and in a way not dependent on the vagaries of
programming errors and cosmic rays.
The proof of optimality relies on the "Oriented Distance
from Home" (ODH), used by Vanderschel (6 Aug 1980 1909-PDT) in his
proof of edge orientation parity conservation. The ODH of an edge
cubie (in some position and orientation) is defined to be the
minimum number of quarter-twists required to move that cubie to its
home position and orientation. A table of possible ODH values of
the UF cubie is given below, indexed by the position of that
cubie's F tab.
+ 3 +
2 U 2
+ 3 +
+ 1 + + 0 + + 1 + + 2 +
2 L 2 1 F 1 2 R 2 3 B 3
+ 3 + + 2 + + 3 + + 4 +
+ 3 +
2 D 2
+ 3 +
The Pons Asinorum moves every edge cubie to a position and
orientation which has an ODH of 4. To move all twelve cubies in
this way requires a total of 48 edge moves, and only four edge
moves can be accomplished by each quarter-twist. Thus the Pons
Asinorum requires twelve quarter-twists.
Unfortunately, this seems to be the only really impressive
result to be derived from counting ODH. All edges flipped
(Singmaster's 12-Flip) can be shown to require at least ten
quarter-twists, but this is a far cry from the 28-qtw process which
is the best known (Plummer, 10 Dec 1980 0157-EST). A brute force
technique for deriving such results is of course possible, but to
show a twelve-move lower bound seems to require sorting and merging
two lists of over one hundred thousand positions each, an act which
is viewed as unsociable (or, more usually, impossible) on the
systems to which I have access. Anyone who has a gigabit to spare
should get in touch -- there are several good problems for which
brute force seems attractive if there is enough of it.
Surprise number two -- Pons Asinorum in the Supergroup -- in
an hour or two.
Dan
Date: 7 January 1981 1615-EST (Wednesday)
From: Dan Hoey at CMU-10A
To: Cube-lovers at MIT-MC
Subject: Pons Asinorum -- Part 2: Pons in the Supergroup
Message-Id: <07Jan81 161515 DH51@CMU-10A>
The second observation I would like to make regarding the
Pons Asinorum involves the Supergroup (also known as "the extended
problem") in which the orientation of face centers is considered.
The process UUDDFFBBLLRR turns each of the face centers 180
degrees, so Pons Asinorum is symmetric in the Supergroup as well.
(Turning each face center 180 degrees is the M-symmetric position
Big Ben Squared, which I will call Noon.) There is another optimal
way of making a (pseudo-) Pons Asinorum, (UD'FB')^3, which differs
from the true Pons only in the face center orientations. According
to an exhaustive search I carried out by hand, this is the only
pseudo-Pons (up to M-conjugacy) that can be obtained with six slice
moves. I would be very interested in hearing about any other
twelve-qtw positions which differ from the Pons Asinorum only in
the Supergroup.
I have found a 16-qtw process for Pons Asinorum composed
with Noon, (UD FB FB UD)(FB UD UD FB), which looks like Pons
Asinorum, but does not rotate the face centers. This in turn gives
a 20-qtw process for Noon itself:
LLRR UUDD (UD FB FB UD) (FB UD UD FB) FFBB
= LLRR (U'D' FB FB UD) (FB UD UD F'B').
Of course, there's no assurance of optimality here.
It occurs to me that many readers of this list may find
details of the Supergroup uninteresting. I have more on this
subject, so if you would or wouldn't like to know more about the
Supergroup, send a vote to Hoey@CMU-10A and we'll see what to do.
Dan
Date: 8 JAN 1981 1515-EST
From: DR at MIT-MC (David M. Raitzin)
To: CUBE-LOVERS at MIT-MC
I've been reading the old mail from this mailing list, and I've
noticed some discussion on the definition of what is a single twist.
As far as I can see, the definition of a twist should be the change in
position of all the cubies in a single plane performed by a single
motion. In other words that should include a 90 degree twist, a 180
degree twist, and a twist of the middle slice (the last one can be
thought of as holding the U and D parts of the cube in place with one
hand, while moving the middle slice with the other).
Date: 8 JAN 1981 1523-EST
From: DR at MIT-MC (David M. Raitzin)
To: CUBE-LOVERS at MIT-MC
The other day, having nothing to do, I started optimizing my
transformations. Counting 90-degree twist, 180-degree twist and a
twist of the middle slice as a single move, I reached the following
numbers: for Plummer cross: 24 moves, and for what I call inverting
the cube (that means flipping every non-corner cubie) 29 moves. Now
realizing that neither of those comes close to the optimum # of moves,
I'd like to know how far of the so far achieved optimum am I?
Date: 8 January 1981 20:06 cst
From: VaughanW.REFLECS at HI-Multics (Bill Vaughan)
Subject: Befuddled by BFUDLR
To: CUBE-LOVERS at MIT-MC
cc: VaughanW.REFLECS at HI-Multics
This is a plea for another notation.
BFUDLR is sufficient to describe anything. So what? It's about as
readable as a LISP s-expression, as rich as the average grad
student, and (my particular gripe) it's impossible to express an
elegant sequence in it and keep any of the elegance.
I want to keep this short, so I'll only give a few examples. The
first is the Sprat Wrench. BFUDLR calls it: RL'URL'BRL'DRL'F. But
the way most everyone does it, it's: (XU)*4, where X means "move
the LR slice clockwise as viewed from the left."
The next example (I don't know its name) flips all top and bottom
edges. BFUDLR calls it: LRUDFBLRUDFB. Interesting, but this is
nicer: (R-B-)*3, where "foo-" means "foo antislice" and is done by
twisting foo and its adjacent slice clockwise 1qtw, then twisting
foo another qtw.
A move yielding (RF, BL, RB) has been published (Don Woods 6 Jan
81) in BFUDLR as: BRL'D'RRDR'LB'RR. Now where's the symmetry in
that? But annotate the same move BXB'RR.BX'B'RR (X as in first
example) and you can see how pretty it is. And it's a lot easier
to remember.
The key is that the fixed orientation of the center cubies
shouldn't be a sacred cow. Often, keeping a corner cubie as a
fixed point will yield a far more natural notation.
The commonest compound moves: slice, antislice, and possibly
Singmaster's Y and Z commutators, should have specialized
notations. A move that I use commonly in solving the cube is a
monotwist: Y(f,r)*2.L.Y'(f,r)*2.L'. That's a lot harder to
understand as: FR'F'RFR'F'RLR'FRF'R'FRF'L'.
I don't have good notations to offer for slice and antislice. What
I do with paper and pencil involves overstrikes that my CRT can't
handle. Something nice and linear is needed, with all the
characters in ASCII. Any suggestions, anyone?
Date: 8 January 1981 2235-est
From: Bernard S. Greenberg
Subject: Cubesys and FLUBRD
To: CUBE-LOVERS at MIT-MC
A new hack for interpreting transforms expressed in FLUBRD
notation has been added to ITS and LISPM cubesys.
(As I am no longer with Honeywell, I can no longer modify
Multics cubesys, but I will give instructions below for
how to use this on Multics).
Transforms are defined in Lisp, with the "defxform" macro.
"defxform" is followed by a name to assign to the transform,
and one to any number of "cube operations". A cube operation
is either
1) a flubrd syllable
2) (apply ), meaning do that transform
3) (inverse ), meaning "undo" that transform
4) any other list, which is interpreted as a list of cube-operations
A flubrd syllable is an atomic symbol whose name is a character string
consisting of one to any number of flubrdniks.
A flubrdnik is either
1) F L U B R D f l u b r d (case doesnt count)
2) F* L* U* .... etc, meaning counterclockwise turn
3) F2 L2 U2 .... etc, meaning 180 degree turn.
Here is the provided, automatically-loaded library of
transforms:
(defxform monotwist-op
(ld l*d*) ldl*)
(defxform monotwist
(apply monotwist-op) u (inverse monotwist-op) u*)
(defxform quark
r2 (apply monotwist) r2)
(defxform pons f2 b2 r2 l2 u2 d2)
(defxform christman-cross ;Saxe 16 dec 1980
f ud llrr ud fb uudd b)
(defxform plummer-cross ;Saxe 3 dec 1980
f (ll rr) f b (ll rr) f
l (bb ff) l r (bb ff) l (uu dd))
To load new transforms...
ITS: ^X break, load a file full of defxforms, ^G back.
Lispm: , load the file,^Z
(oh yes, Lispm cubesys now has a ^Z handler)
Multics: ESC X loadfile the file.
To run a transform:
ITS: Use the X command. the transform name will be prompted for.
Lispm. Use the X command. A MENU OF KNOWN TRANSFORMS WILL POP UP!
Mouse at it.
Multics: ESC X run-xform
On Multics, you will have to load the new package before doing
any of this, with esc-x loadfile. Currently, it is
>udd>sym>bsg>cxfrm .... it may move or go away. Multicians, contact
me if you have any interest in it.
Date: 9 January 1981 0551-EST (Friday)
From: Dan Hoey at CMU-10A
To: Cube-lovers at MIT-MC
Subject: The Supergroup -- Part 1: Physical reality
Message-Id: <09Jan81 055144 DH51@CMU-10A>
Well, whoever doesn't like the Supergroup didn't send me
any messages, and several who do did, so here goes. This message
is part one of three, separated for the benefit of MIT's
notoriously fragile mailer.
In case anyone has managed to miss it, the Supergroup is
the group underlying the cube when face center orientation is taken
into account. By the "orientation" of a face center, we refer to
the number of 90o twists of that face center from the position
designated "solved." To visualize this, Singmaster suggests
replacing the solid colors on the sides of the cube by some
nine-piece pictures, so that the centers must be restored to their
initial (untwisted) state to solve the cube. He reports that this
was done (on two sides only) by a company in England, which printed
its logo on cubes for a promotion. The term "Supergroup" is also
due to Singmaster, and I adopt it in favor of the term "the
extended problem," which has appeared in Cube-lovers.
To make the face center orientation visible on my cube, I
first used magic marker, which rubbed off, then paint, which
attacked the colortabs and looked and felt awful. [Then I went to a
stationery store and got plastic tape and replaced my colortabs --
no orange, so my cube now has a tan face.] I marked the face center
orientation by cutting out circles from the plastic colortabs to
let the black plastic show through. I like it, though some people
think it looks like the cube has been used for target practice.
Each cutout circle has a diameter of about 3/8" (1/2 the side of a
cubie) and is centered at the corner of a face center, overlapping
two edge cubies and one corner cubie. The orientation is then
visible if either the corners or the edges are in the home
position. It doesn't particularly matter which corners of the face
centers are used; I chose the pattern which has the same symmetry
group as Plummer's cross (unique up to M-conjugacy).
There will be a short intermission while we change reels.
Date: 9 January 1981 0629-EST (Friday)
From: Dan Hoey at CMU-10A
To: Cube-lovers at MIT-MC
Subject: The Supergroup -- Part 2: At least 25 qtw, and why
Message-Id: <09Jan81 062915 DH51@CMU-10A>
Alan Bawden (31 JUL 1980 2159-EDT) calculated that it must
take at least 21 quarter-twists to solve an ordinary cube, and 24
qtw to solve a cube in the Supergroup. This message explains how
the first bound can be obtained, improves the second, and points
toward a technique for possible further improvements.
Express any (optimal) sequence of twists as a sequence of segments,
where each segment is a sequence of twists on two opposite faces,
and no two adjacent segments operate on the same pair of faces.
Because the quarter-twist has period four, and opposite faces
commute, a segment operating on faces X and Y has one of the forms
X, X', Y, Y' (1 qtw -- 4 ways)
XX, YY, XY, YX, X'Y, Y'X, XY', Y'X (2 qtw -- 6 ways)
XXY, XXY', XYY, X'YY (3 qtw -- 4 ways)
XXYY (4 qtw -- 1 way).
There are 3 ways of choosing X and Y for the first segment, and two
ways of choosing them for every succeeding segment. Let P[n] be the
number of positions that are exactly n qtw from SOLVED. Then
bounding P[n] by the number of n-qtw sequences,
P[0] = 1
P[1] <= 4*3*P[0]
P[2] <= 4*2*P[1] + 6*3*P[0]
P[3] <= 4*2*P[2] + 6*2*P[1] + 4*3*P[0]
P[4] <= 4*2*P[3] + 6*2*P[2] + 4*2*P[1] + 1*3*P[0]
P[n] <= 4*2*P[n-1] + 6*2*P[n-2] + 4*2*P[n-3] + 1*2*P[n-4]
for n > 4.
Evaluating this recurrence, substituting strict (in)equality where
I have personally verified it, and rounding truthfully yields:
P[0] = 1 P[9] < 724,477,008 P[18] < 4.048*10^17
P[1] = 12 P[10] < 6.792*10^9 P[19] < 3.795*10^18
P[2] = 114 P[11] < 6.366*10^10 P[20] < 3.557*10^19
P[3] = 1,068 P[12] < 5.967*10^11 P[21] < 3.334*10^20
P[4] = 10,011 P[13] < 5.594*10^12 P[22] < 3.125*10^21
P[5] <= 93,840 P[14] < 5.243*10^13 P[23] < 2.930*10^22
P[6] < 879,624 P[15] < 4.915*10^14 P[24] < 2.746*10^23
P[7] < 8,245,296 P[16] < 4.607*10^15 P[25] < 2.574*10^24
P[8] < 77,288,598 P[17] < 4.319*10^16
There are 4.325*10^19 positions in the standard cube; since
P[0]+P[1]+...+P[20] < 3.982*10^19, there must be a position at
least 21 qtw from SOLVED (The number 22 has appeared in Cube-lovers
recently, but it was an error). There are 8.858*10^22 positions in
the Supergroup; since P[0]+P[1]+...+P[23] < 3.280*10^22, there must
be a position at least 24 qtw from SOLVED. But this can be
improved: half of the positions in the Supergroup are an odd number
of qtw from SOLVED, and since P[1]+P[3]+...+P[23] < 2.963*10^22 is
less than half the Supergroup, there must be some odd-length
elements of the Supergroup at least 25 qtw from SOLVED. QED. (If
you think there's something fishy here, mail to Hoey@CMU-10A for
clarification. I am responsible for any cruft that has crept into
the original, elegant, formulation due to Jim Saxe.)
The recurrence on which this bound relies is due to the
relations F^4 = FBF'B' = I (and their M-conjugates.) It may be
possible to improve the recurrence by recognizing other short
relations. Exhaustive search has shown that there are none of
length less than 10. The most promising ones I know of come from
Singmaster:
I = FR'F'R UF'U'F RU'R'U (12 qtw),
I = (FFBB RRLL)^2 (16 qtw),
and a 14-qtw relation which holds only in the standard group, since
it twists a face center 180o (see part 3). Unfortunately, the
number of intermediate terms grows too large to be comfortably
hand-computable, and there are a few conceptual problems to hacking
it out. If you can improve this, or know of any other relations
shorter than 24qtw, I'd like to hear about it.
Coming up next: SPOILERS
Date: 9 January 1981 0756-EST (Friday)
From: Dan Hoey at CMU-10A
To: Cube-lovers at MIT-MC
Subject: The Supergroup -- Part 3: A Super-H and Spoilers
Message-Id: <09Jan81 075610 DH51@CMU-10A>
IMPROVEMENT
Jim Saxe (16 December 1980 1841-EST) gave a 16-qtw process
for the H position: FF LL DD FF BB DD RR FF. Unfortunately, the
position obtained by this process is not H-symmetric in the
Supergroup: the F, L, B, and R face centers are rotated 180
degrees, but the U and D face centers are in the home orientation.
I have a 16-qtw process which leaves all face centers in the home
orientation: (FB UD)^2 (UD FB)^2. This may look familiar --
conjugation by FBUD yields the 16-qtw "Bridge at Midday" I sent day
before yesterday.
SPOILERS
I developed a solution method on my own, but it was a long
one. The following good methods, which affect only face centers,
are from Singmaster. It seems simplest to solve the cube before
applying them, since some of the most popular processes (e.g., the
Spratt wrench (but not mono-ops!)) change face center orientation.
The first method can be used to perform any multiple of
four face-center quarter-twists on the faces in a centerslice. At
most two applications are necessary to accumulate any remaining
twist in one face center. The method is given in Singmaster as two
examples, but he doesn't explain how they work. Hopefully the
following discussion will make them easy to use.
Choose a face center X that needs to be twisted, and a
centerslice containing X and other face centers to be twisted (call
these FCT's). Place the cube with X up and with the FCT's in the FB
slice (i.e., among R, D, and L). The basic move has much of the
flavor of a mono-op:
1: Move the LR centerslice toward you. (Move a face
center from U to F. X and the FCT's are now in
the UD centerslice.
2: Select an FCT, and move it to the F face with UD
centerslice moves.
3: Move the LR centerslice away from you. (Undo step 1.
The U face now contains the selected FCT and all the
U edge and corner cubies.
4: Twist the U face the amount required by the selected
FCT.
Repeat the basic move until all FCTs have been selected. Then
perform the move one last time, but select X instead of an FCT
in step 2 and twist the U face to fix the cube in step 4.
[After sending the previous part, I realized that in the case of
one FCT, this is another 14-qtw relation in the standard cube:
RL'FB'UD' R DU'BF'LR' U'
But the one alluded to is in the next paragraph.]
Since the total number of 90o face center twists must be
even (see Vanderschel's message of 6 Aug 1980 1909-PDT), the
preceding will solve the cube up to a 180o twist. The process
(URLUUR'L')^2 twists the U face center 180o. This is the only short
transform given in Singmaster for twisting face centers by a
nonmultiple of four; I'd be interested in any others you know.
That's it for now. Happy Supercubing!
Dan
Date: 9 Jan 1981 14:59 PST
From: McKeeman at PARC-MAXC
Subject: Re: Befuddled by BFUDLR
In-reply-to: VaughanW.REFLECS's message of 8 January 1981 20:06 cst
To: VaughanW.REFLECS at HI-Multics (Bill Vaughan)
cc: CUBE-LOVERS at MIT-MC
Bill,
Good points. I am not sure how to solve the problem either. But here are some
comments. RubikSong is simply assembly language to drive the human
computer. That has both its advantages and disadvantages. The original
suggestions also included parameterless subroutines and whole cube moves.
Your point it, I think, that there are some easy manipulations that are obscure, at
best, in FLUBRD + IJK. One could take the position that definitions need not be
mnemonic since in steady state their names will be used with a higher level of
semantics. Then we get
Slice = L' R J'
SpratWrench = (Slice U)*4
That seems tolerable but still not well matched to the human at the lowest level.
A nice property the notation could have would be to be concise where the
human manipulation is concise, without introducing a large number of
unmnemonic primitives. For example, we might use Xnnn for multiple parallel
twists.
"R antislice" = L012 = L' R' J
Then we have, for example,
LRUDFBLRUDFB = (L012 U012)*3,
and
R' = L001
R = L001'
Slice = L010.
Commutators are pretty common (i.e., X Y X'). Somebody suggested X[Y] for
them. The idea is that X is a function to apply to Y. The main problem is that
the human machine isn't very good at doing complicated inverses without some
special practice. But, assuming the practice, then
BRL'D'RRDR'LB'RR
= B[Slice] (RRB)[Slice']
= (B[Slice])[RR] RR
= (B[Slice]RR)[]
Feedback appreciated...
Bill
Date: 9 Jan 1981 1648-PST
From: Dave Dyer
Subject: nomenclature
To: cube-lovers at MIT-MC
The real problem with RLUDFB is that it is nearly impossible for
the casual observer to determine if two "transformations" are the
same or not. Transformations should be described such that there
is ONE description of each transformation. We can't really get all
the way to the ideal, because there are discinct sequences that lead
to the same termination. But we can eliminate a lot of the confusion.
I Propose that all transformations be described in terms of twists
done by both hands without changing the overall cube orientation.
my proposed primitives are
1 "forward" twist of left or right face
2 1/2 twist of left or right face
3 "backward" of left or right face
URF orient the cube to work on named face,
with your right hand. Names are lambda-bound from the
start of the transformation. To instantiating a transformation
one can simply substitute the color for the orientation label
notation is <(optional) left hand>
For example, a sprat wrench:
UDLRFB calls it: RL' U RL' B RL' D RL' F.
I call it: R11 U1 R11 F03 R11 U03 R11 F1
-------
Date: 9 January 1981 19:40 cst
From: VaughanW.REFLECS at HI-Multics (Bill Vaughan)
Subject: Re: nomenclature
To: Dave Dyer
cc: CUBE-LOVERS at MIT-MC
In-Reply-To: Msg of 01/09/81 18:48 from Dave Dyer
It would seem that there are two needs immediately visible.
One need is for a notation to communicate a metamorphosis of the cube
from one cube-lover to another hexahedrophile. This notation should
capture the spirit of the thing being described; it should be rich and
chunky (like soup?).
The other need is for an archival notation, for reference and
cataloguing. It must be spare and canonical. There must be one and only
one way to describe a sequence of moves. (Implication: a way to get
reflections, inversions etc. out of the notation? how?) No nourishment
for the spirit there, but when you need to look something up...
Well, clearly they can't be the same notation. (Though even the archival
notation could invoke functions.) Or can they? Food for thought...
Anyway, plain ole BFUDLR can't do either job decently. Not rich or
expressive enough for the first need, too expressive for the second
(gee, I wonder how many ways there are to annotate PONS?)
Well, other people can muddle this out - I'm going home to get some
dinner.