move P clockwise as seen from the front =
~~ move S clockwise as seen from the right side =
Move combinations are two-letter names beginning with C, E or A;
C1-C9 and (CA-CZ, Ca-Cz if needed) describe corner-combinations;
E1-E9 and EA-EZ (Ea-Ez if needed) describe edge-combinations;
A1-A9, AA-AZ & Aa-Az describe combinations that mess with both.
I also use V,W,X,Y,Z and v,w,x,y,z as temporary macros
Permutations = { list of cubies -> list of new spots }. The permutation
described puts the first list into the new LOCATIONS described by the
second list, moving each face of each cubie in the order specified.
E.g., does {flu fu fur fr frd fd fdl -> fur fr frd fd fdl fu fdl}
Useful basic corner combinations:
C1: <6(RurFU)> :: {fdl flu bul -> dlf luf ulb} twist corners clockwise
C2: <6(rFRuf)> :: {fdl flu bul -> lfd ufl lbu} " " ccw (lf2 + bul)
C3: <3(RUru)> :: {frd fur bul bru -> rfu rdf ubr ulb}
switch those pairs of corners, twisting each
C4: <3(fuFU)> :: {frd fur flu bul -> rfu rdf bul flu} (ditto)
C5: <3(FrfR)> :: {frd fur flu bru -> rfu rdf bru flu} (ditto)
compound ones:
C6: = :: {frd fur fdl flu -> rfu rdf luf lfd}
switch those pairs of corners all in the front face
C7: = <6(RurFU)M6(rFRuf)m> :: {fdl bld -> dlf dbl}
twist fdl cw and bld ccw.
C8: = <3(FrfR)3(RUru)> :: {flu bul bru -> bru ufl ulb}
cycle three top corners (left two & bru) "cw"
C9: = <3(FrfR)3(fuFU)> :: {flu bul bru -> bul bru flu}
cycle three top corners (left two & bru) "ccw"
Useful basic edge (side) combinations:
E1: <4(FH)4(HF)> :: {fl fr -> lf rf} twist two sides in place
E2: <2(ssdd)> :: {fu bu fd bd -> bu fu bd fd}
"doubleswap" swaps front for back sides on top and bottom
E3: <3(FRRF)> :: {fl fr ru rd -> fr fl rd ru}
E4: <3(ffrr)> :: {fu fd ru rd -> fd fu rd ru}
E5: <3(URRU)> :: {ul ur rf rb -> ur ul rb rf}
E6: <3(ffuu)> :: {ul ur fl fr -> ur ul fr fl}
E7: <3(FLLF)> :: {lu ld fl fr -> ld lu fr fl}
E8: <3(ULLU)> :: {lf lb ul ur -> lb lf ur ul}
E9: = <3(FRRF)3(ffrr)> :: {fl fr fu fd -> fr fl fd fu}
EA: = :: {fu ul ur -> ul ur fu}
cycle top (left right and front) sidies clockwise
EB: = :: {fu ul ur -> ur ul fu}
cycle top (left right and front) sidies counterclockwise
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Pons Asinorum: HHSSPP (180 degrees on each centerslice)
Laughter: <4(lrfb)> = <6(lrK)> (but in the latter the cube is KK'd)...
Another laugh resolves the cube; a K first produces the 4 crosses.
Crux Christmani: or, call E2 (the doubleswap) `Z',
. This is 6 crosses in 3 pairs: colors go top with
bottom, front with right side and back with left side.
Crux Plummeri (the fancy one): ; or (again):
or if Y := , it's simply !
This is 6 crosses in 2 sets of three each: Front, Right and Top each with
each other; and Back, Left and Bottom each with each other.
Greenberg's "well-known center-cubie rotation algorithm" has 12 varieties:
Pick two centerslices (say X and Y) and a direction (cw or ccw):
does ccw and does cw. Details:
(xyz cw) means the center (face) cubies of the faces x y and z are
rotate clockwise; ... ,ccw means counterclockwise.
fur cw :: or or , ccw :: or or
frd cw :: or or , ccw :: or or
fdl cw :: or or , ccw :: or or
flu cw :: or or , ccw :: or or
Outstanding questions mentioned in the note about Singmaster's talk:
(I.e., I ain't seen these yet ... )
1. Conway's Monoswop
2. Rubik's Duotwist
(ignore the two lines above beginning with Pick ... and ending with
... Details: they're wrong. I'd go edit them but my Unix is out
of space on /tmp and on my filesystem and I only have 300-baud access to
it anyway).
Have fun
DalGorf
Date: 27 August 1980 1:58 pm PDT (Wednesday)
From: Woods at PARC-MAXC
Subject: Lexicon
To: Cube-Lovers@MIT-MC
We need a term to refer to the fear that someone will randomise your cube. Any suggestions?
-- Don.
ACW@MIT-AI 08/29/80 13:24:04
To: Cube-hackers at MIT-MC
Date: 29 AUG 1980 1316-EDT
From: ACW at MIT-AI (Allan C. Wechsler)
On this coast we have a colorful slang term for randomizing the
cube. Hint: the command to randomize the cube in Bernie's
:CUBE program is "F".
---Wechsler
Date: 29 Aug 1980 10:55 PDT
From: Lynn.ES at PARC-MAXC
Subject: Re: Lexicon
In-reply-to: Woods's message of 27 August 1980 1:58 pm PDT (Wednesday)
To: Woods.PA
cc: Cube-Lovers@MIT-MC
Incongruphobia seems nice, but may imply more of a fear of lack of order, rather
than of the change from order to lack thereof. It doesn't specify the objects that
are disordered either. Cubic discombobulaphobia might be a little better in these
respects.
/Don Lynn
MJA@MIT-AI 08/29/80 21:45:41 Re: Knowledge-based Rubik's Cube solver?
To: cube-hackers at MIT-MC
Please excuse the previous badly munged versions of this message i accidently
sent.
I am taking an artificial intelligence course at the University of
Illinois at Urbana-Champaign and I need to do a term project for the
course that involves developing and/or using a knowledge based system.
Since I am interested in the Rubik's cube and since it seems like
there are human expert cube solvers (as is evident from msgs on this mailing
list) that use heuristic knowledge (for example the various macro's (as I like
to call them, they are more properly called composite elements of the
group, i guess) that people have come up with to do some particular operations
on the cube), it seems natural to me that a knowledge based system would
be just right for this task. I would like to hear:
(1) Pros/cons on why this can/can't be done and how effective
such a system would be.
(2) Considering that I have but one semester to work on this project
(while concurently taking 5 courses, incl. this AI course) is it reasonable to
attempt this for a term project? (The actual system would not have to be
100% complete at the end of the term, but I would at least like to have
something to show for a semester's worth of work.)
(3) If this is too ambitious for a term project for a course, what
about the use of this as a topic for a master's thesis. (I would have until
the summer of '81 to finish since thats how long my company will wait for me
to get a master's degree if im going to continue working for them.)
(4) Of course in addition to all of the above trivia, I would like
some ideas for such a system (even if it isnt feasable to do in this limited
amount for time, im still interested in at least thinking about such a system
with the idea in mind that someday it can be implemented).
I dont know how interested the majority of the people on this mailing
list are in knowledge based systems, so we might consider a seperate mailing
list for knowledge based cube solvers to spare those on the current mailing
list from mail they dont want.
/ MJA
ZILCH@MIT-AI 08/30/80 01:39:38
To: CUBE-LOVERS at MIT-MC
Subjects covered:random notes on the cube and cubing
1> To enable new and old cube-lovers to communicate on an equal basis
I propose that a file be established that describes FBLRUD
(or whatever),the I,J,K and centerslice move and any other concepts
that might be usefull in describing transformations.(I can not do this as I know nothing about establishing files or editting them, and as a tourist I am
unsure of my status as file-creater.)
2> Today having nothing better to do, I fiddled with the 3x3x2 version
of the cube (actually I just didn't allow certain moves on my
3x3x3). At this point I have two transforms which would enable me
to solve the 3x3x2 version if and when I ever see it. I derived the number of orbits and the reasons behind them, but will not describe them here because
the 3x3x2 is a novel idea and I don't want to ruin all the fun.
Also I came up with the idea of a 2x2x3 version which basically operates on
the same principles as the 3x3x2 , but it looks different.
solving the 3x3x2 can give a slight amount of insight about
Thistlewaite's algorithim ,described by
McKeeman on 12 Aug. @15:10 PDT.
3> On July 15 @1413 EDT, ALAN asked if anyone had learned to solve the extended cube problem independently. I had known the faces could assume different
orientations when the cube was solved but hadn't done any work on this
and didn't know what the possible transforms might be. In a later note
cmb described what his transforms accomplished and from that idea
I recently derived similar (if not equal) transforms.With a little foresight
I find that only one of these transforms might be needed.
This evening a friend and I came up with a modification to the cube
construction that would facilitate the extended problem. this is simply affixing another cubie to each face cubie and coloring appropriatly.
4> As suggested by CSD.VANDERSCHEL on 9 Aug. @1610 pdt has anyone given any
thoughts to 3-faced cubing (2 types) or 4-faced cubing (again 2
types). As far as I can tell 5-faced cubing is not profitable.
Also does CSD.VANDERSCHEL know a good way (at this time) to show
why only 1/6 of the 6! possible permutations of the corners are possible?
There is also an extended problem for the 2-faced problem
involving face orientations.
5>An addendum to WOOD's message of 23 JUL. @5:23pm
regarding repetitive sequences to get to the identity. As it turns out,
by my calculations the number 1260 is sufficient even including the prblem
where the faces are permitted to move. The only subcycles of the faces
themselves have lengths 4,3,2,and of course 1.
These subccles may immediatly be generated by I (4 times), I^2
(2 times),^2 (also 2 times)
and finally (3 times).
(I do not use IJK notation on these last two because I have noticed a little
disagreement on exactly what these mean.)
These subcycles make no difference in the total number because subcycles of
lengths 4,3,and 2 have already been taken into consideration in the computation
of 1260.
I have no idea whether th e face cubies might change their orientations in a
sequence of length 1260,when the faces are allowed to move,
but I know it does nothe number above 1260 when the faces are not allowed to
move.
6> Singmaster's solution from his version 5, as reported by McLure
on 16 Aug. @1053 PDT sounds almost exactly like my solution
except that I keep the first face completed in the up poition at
all times. He reported that his method takes less than 200
(of his) moves. My transformations yield a solution in a maximum of 190
quarter-twists. My actual solving length (from when I bothered counting)
averages about 125 qtws , and my time (when I bother) is almost
consistantly 2.5 minutes, occasionaly under 2, with worst case about 3 min. 10sec when I messed
up. Usually my fast times do not use all of my algorithms techniques ,
because they are new to me and I don't know them by heart.
Breakdown of moves:
1. Top edges in proper position and orientation 20 (3+6+6+5)
2. " corners " " " " " 36 (9+9+9+9)
3> 3 middle edges " " " " " 45 (15+15+15)
4. 4th middle edge in proper position and orientation
and bottom edges in proper orientation 23
5. bottom edges in proper position 18
6.bottom corners in proper position 20
7. " " " " " 28
Total 190
Note: in step 4 proper orientation means that say if the down face is
white , after step 4 all bottom edges will be showing white on their
down sides.
This algorhitim has not been optimized much and uses lookahead only in step
4. Step 2 gives the worst case for any corner, but if only worst cases are
present then they cancel each other out somewhat.
Replys, questions, and comments are welcome.
Chris
DMM@MIT-ML 08/30/80 18:41:24 Re: DAL@ucla-security's msg of 27-Aug
To: cube-lovers at MIT-MC
Genereally, I have found themacros supplied by DAL to be quite useful,
but t|ere are a few bugs i've noticed. First of all, oC8&C9,
I believe DAL has the cw & ccw notations reversed.
Regarding C7: What is M? I have not seen it referenced anywhere.
and last, in maco EB: the last move is f, not F.
(I`guess it was the 5 beers and 2 Drambuies.)
Date: 3 September 1980 2108-EDT
From: James.Saxe at CMU-10A (C410JS30)
Subject: Orbit classification revisited
To: cube-lovers at MIT-MC
Message-ID: <03Sep80 210846 JS30@CMU-10A>
Hi, folks.
Having read Vanderschel's msg of Aug. 6, it appears to me that the
explanations of the orientation parities are unnecessarily complex,
though the material on permutation parities is presented in a way
that should be immediately convincing to anyone familiar with the
notion of even and odd permutations from elementary group theory (and
anyone who isn't should be!). Here's my attempt at a more elegant
demonstration.
In what follows, I will use the term "facelet" to denote any (visible)
face of a cubie. Thus, each edge cubie has two facelets and each
corner cubie has three facelets.
I will address Edge Orientation Parity (EOP) first. Consider the
following diagram:
+-------+
| 1 |
|0 U 0|
| 1 |
+-------+-------+-------+-------+
| 1 | 0 | 1 | 0 |
|0 L 0|1 F 1|0 R 0|1 B 1|
| 1 | 0 | 1 | 0 |
+-------+-------+-------+-------+
| 1 |
|0 D 0|
| 1 |
+-------+
The numbers label absolute positions, not facelets, and therefore
remain in the same configuration when the cube is manipulated.
Imagine that a mark is placed on each facelet that occupies a
position labeled "0" when the cube is in the solved configuration.
Thus, each edge cubie will have one marked and one unmarked facelet.
(Unlike the numbers, the marks are attached to the facelets and will
move as the cube is manipulated). The parity of an edge cubie in an
arbitrary configuration is defined as the number labelling the
position occupied by its marked face, and the EOP is defined as the
sum of the parities of all edge cubies modulo 2. A quarter turn of
any face reverses the parity of 4 edge cubies, thus leaving the EOP
fixed. By induction, no sequence of manipulations starting from the
solved configuration can produce a configuration EOP = 1. So much
for EOP.
[Just for grins, here's a cute way to determine the parity of an edge
cubie without consulting (or reconstructing) the diagram: Assign
each of the cubie's two facelets a number in {0,1,2} according as it
is oriented parallel to its home position ("Self" -> 0), parallel
to the other facelet's home position ("Other" -> 2) or perpendicular
to both home positions ("Neither" -> 1); add these two numbers and
reduce modulo 3 (yes, three!) giving the parity of the cubie. The
sum can never be 2 because that would imply that the two facelets
were parallel to each other. Here's an addition table with
double-ended arrows showing the possible transitions:
(S) 0 (N) 1 (O) 2
+-------+-------+-------+
| | |no |
(S) 0 | 0 <-+-> 1 | + |
| ^ | ^ | way|
+---+---+---+---+-------+
| v |no | | |
(N) 1 | 1 <-+---+---+-> 0 |
| | |way| ^ |
+-------+---+---+---+---+
|no | v | v |
(O) 2 | + | 0 <-+-> 1 |
| way| | |
+-------+-------+-------+
]
Now for Corner Orientation Parity (COP). Consider the diagram below:
+-------+
|0 0|
| U |
|0 0|
+-------+-------+-------+-------+
|2 1|2 1|2 1|2 1|
| L | F | R | B |
|1 2|1 2|1 2|1 2|
+-------+-------+-------+-------+
|0 0|
| D |
|0 0|
+-------+
Once again, imagine that we mark all (corner) facelets which occupy
positions labeled "0" when the cube is in the solved configuration.
The parity of a corner cubie in an arbitrary configuration is defined
as the number which labels the position of the cubie's marked face.
The COP is the sum of all the parities of all corner cubies modulo 3.
Inspection of the diagram will reveal that the twists U, u, D, and d
leave the parities of all corner cubies unchanged. Any of the other
possible quarter twists will increment (modulo 3) the parities of two
corner cubies and decrement the parities of two others, thereby
leaving the COP unchanged.
[As Vanderschel pointed out, one way to compute the parity of a
corner cubie by looking at it is to note the number of clockwise (as
viewed from outside the cube) 120 degree twists of the cubie that it
would take to bring the marked facelet parallel to its home position.
Note that the parity of a corner cubie, unlike that of an edge cubie,
depends on the selection of a particular pair of opposing colors for
the marked facelets. While this lack of symmetry may be considered
unfortunate, it is an inevitable result of the fact that four is not
divisible by three. It is easy to show that the COP as a whole is
independent of the choice of distinguished faces.]
Vanderschel also mentions the extended problem, but does not quite
make it clear that the FOP changes every time a qtw is done. This
constrains all three of {FOP, EPP, CPP} to be the same, so that only
two of the eight plausible states of the vector are
actually achievable by twisting.
Jim
Date: 3 SEP 1980 2123-EDT
From: DCP at MIT-MC (David C. Plummer)
Subject: New problem
To: CUBE-HACKERS at MIT-MC
This was inspired by Tanya Sienko (if that means anything!!)
It is possible for each of the six faces to have a capital "T" on them.
That is, each face looks like
X X X
O X O
O X O
QUESTIONS:
1) How many classes are there?
2) How many members are in each class?
3) How easy is it to get to them from solved?
My answers:
[in a few days, I don't want to spoil the fun]
Good luck, and happy hacking.
- DCP
Date: 3 September 1980 2149-EDT
From: James.Saxe at CMU-10A (C410JS30)
Subject: Re: Lexicon
To: Woods at Parc-Maxc
CC: Cube-Lovers at MIT-MC, James.Saxe at CMU-10A
Message-ID: <03Sep80 214930 JS30@CMU-10A>
My suggested term for the fear that someone will randomize one's cube
is "nobility," since that is what sufferers of this phobia typically
have to solve their cubes.
Jim
Date: 3 September 1980 2328-EDT (Wednesday)
From: Dan Hoey at CMU-10A
To: Cube-Lovers (and Hackers) at MIT-MC
Subject: Addictiveness/Disassembly/Taxonomy/Lubrication/Spoilers
Message-Id: <03Sep80 232808 DH51@CMU-10A>
Hello. I saw the notice announcing the formation of this list a couple
months ago, and decided that it was one of those things I could forgo
-- until I got my hands on a physical cube. It was an immediate
necessity to own one. I bought an Ideal brand cube, which, I
understand, is of the species C. Americanus in spite of its "Made in
Hungary" label.
I had owned the cube less than ten minutes before a facie cover fell
off, without the aid of chemical additives. This was not very
destructive; just about any gummy material (I used gluestick) suffices
to hold it on. However, the screw head revealed by this unusual
transformation leads to a new method for disassembly. Unscrewing does
not stress the cube as does prying, and probably avoids the deleterious
side-effects observed by Greenberg (16 Aug 1453). This method is not
without its hazards, however. It is EXTREMELY easy to strip the threads
on the plastic X that holds the cube together. I have paper shims in
the two threads I stripped and they seem to suffice. Still, it is
probably better just to loosen the screw until the cube comes apart
with gentle prying.
There is at least one good reason for taking a screwdriver to the cube.
Mine had been assembled with several of the screws so tight that the
springs were completely compressed. Due to mfg inaccuracies in the
cubies, this made the cube difficult to twist. By prying each of the
facies off with a fingernail I was able to correct the tension.
Jim Saxe's cube, putatively of the same species but purchased in a
different store at 80% the price, has facies which seem impervious to
this prying even after disassembly a la Greenberg (17 Jul 2118). Jim
and I exhausted our fingernails to no avail, and careful prying with a
knife was unsuccessful. Additionally, this cube has a strong tendency
to jam, due either to its uncorrectable looseness or to its edge
cubies, which have oversized tongues with extremely sharp edges. The
differences lead us to believe that our cubes may belong to different
species in spite of their outward similarity.
I am amazed that anyone would put molybdenum disulfide on their cube.
Isn't that stuff poisonous? Graphite works well but is messy if you
overdo it. Silicone lubricant was mentioned by Zimmerman (25 Aug 0907)
-- has anyone any experience with this? Merrick Furst recommends soap.
For anyone who cubes in public, the only word for LACK of fear that
someone will F your cube is Cubemeistership. I made a mistake in taking
the cube to one session of a recent conference. The sequence
4(Borrow)4(Borrow') appeared to have an entropic effect on the cube and
a negative effect on the transfer of information.
SPOILER WARNING: I have one transform which I haven't seen here, and
which I find useful: an 8-qtw move to permute three corners of a face.
Specifically, for {fdl flu fur -> ufl rfu lfd} do .
Mnemonically, you move a socket back and forth between flu/fdl with the
f/F transforms, alternating with moving one of the three pointies
(cornies?) to be permuted into the socket. Why it works is a mystery to
me, but it's useful.
Another, which should be obvious but improves Landauer's (27 Aug 0128)
C7, is the cw monotwist {flu -> luf in u-face} =. Then
{flu fur -> luf rfu} is = taking 16qtw
instead of 60 (assuming =~~*, =**).
All for now. -- Dan
Date: 4 SEP 1980 0852-EDT
From: JURGEN at MIT-MC (Jon David Callas)
Subject: Lexicon...
To: woods at PARC-MAXC
CC: CUBE-LOVERS at MIT-MC
Worse yet is the fear that someone will put your cube in an
unsolvable orbit......
Happy phobias,
Jurgen at mit-mc
Date: 6 Sep 1980 0021-PDT (Saturday)
From: Dal at UCLA-SECURITY (Doug Landauer)
Subject: errata
To: cube-lovers at mit-mc
Attn: DMM@mit-ml:
No, unless my message got garbled somewhere in the midwest, it was
correct about c8 and c9, the three corner-cubie movers. I.e., c8
( <3(FrfR)3(RUru)> ) cycles them cw, and c9 ( <3(FrfR)3(fuFU)> ) ccw.
(Much shorter ways to do similar things exist, e.g. to
move around three front corners (left two and fur) cw).
In these descreptions, the `cw' and `ccw' do not refer
to individual cubies but rather to the triangle of three cubies involved,
that's why I call it "cycle" == move cubies around, rather than
"twist" == re-orient a corner cubie in place, or "flip" == do the
same for an edge (side) cubie.
You're right about EB.
Attn: Dan Hoey at CMU-10A
Regarding C7: When I first made this list up, I used m,n and o to
represent i,j and k because it was unclear which of i,j and k was
which. m is the same as i, and M == I. My cube-turning moves are:
I,J,K=front-top-right clockwise
** move cube's front to top ** or ** top to front
move top to right side or right side to top
move right side to front or front to right side
thank you...
... dalgorf
-------
Date: 8 SEP 1980 0853-EDT
From: ZIM at MIT-MC (Mark Zimmermann)
Subject: Singmaster 5th; 2x3x3 "domino"
To: CUBE-HACKERS at MIT-MC
I received the 5th edition of Singmaster's booklet a few days ago, from Logical
Games, Inc.--presumably they have some $4 copies left (they sent mine 1st
class mail)....Bela Szalai (LGI) loaned me a 2x3x3 to play with. Conceptually,
of course, it's the same as the 3x3x3 with moves restricted to *__...physically, however, it was MUCH less aesthetic to play with...
turned poorly, and looked ugly. Bela had a couple of them, both of which
were rather more "delicate" than the 3x3x3 (in that they tended to get jammed
or come apart). -- Mark
Date: 10 Sep 1980 1500-PDT
From: Alan R. Katz
Subject: randomizing
To: cube-lovers at MIT-MC
Here's a question for all of you. How do you maximally randomize a cube?
In other words, suppose you were going to have a cube solving compitition
and wanted to make it as hard as possible to solve a cube, what precautions
would you take?
One thing I would do would be to make sure the corners are not correct in
relation to each other. Anyone have other ideas??
Alan (Katz@isif)
-------
Date: 10 September 1980 18:12 edt
From: Greenberg at MIT-Multics (Bernard S. Greenberg)
Subject: Re: randomizing
Sender: Greenberg.Multics at MIT-Multics
To: Alan R. Katz
cc: cube-lovers at MIT-MC
In-Reply-To: Message of 10 September 1980 18:00 edt from Alan R. Katz
My personal theory on making a cube as hard to solve as possible
is as follows: Put the cube in
some very complex pretty, regular, pattern, such as the Plummer's
Cross with centers trebly rotated, or the Pons Asinorum on that
which looks like the simple Pons: the challenged cubemeister
will INVARIABLY say, "Oh,
this, this is just one of those and solved like this" and go through
a long hairy procedure, hopefully the wrong one,
or make an error, or try hard at understanding it, and so forth.
In short, s/he will take MORE time trying to undo what s/he perceives
as a "simple" hairy hack than a straightforward application of
solve-from-random technology.
Date: Wed, 10 Sep 1980 1854-CDT
Message-id: <337474464.2@DTI>
From: aramini at DTI (Michael Aramini)
To: cube-hackers@mit-mc
Subject: randomizing
There are two types of aproaches to what can be meant by total
(or maximal) randomizing. One is to consider states of the cube that
are maximally far away from being solved (that is the minimal number of
quarter twists needed to solve from a given state is maximallized.
the other, which may be more useful for making the cube more difficult
to solve, is based on maximallizing the amount of time it would take
a person (or a program, for that matter) to solve the cube. Of course,
since this amount of time is both dependent on the state of the cube, and
how the person (or program) goes about solving it. Thus the set of most
randomized positions is dependent on who is solving the cube.
the advantage of the first definition is that it seems to be of more
theorectical significance (the number of quarter turns needed to
solve from this position gives you the diameter of the state graph).
The second approach seems more kludgy since it much less well defined
since the amount of time it takes to solve the cube if a function of
many variable besides simply the initial state.
This distinction is much like differing strategies for writing chess
playing programs, you could write it assuming the oponent to be a
"perfect" player (as if a complete look ahead to the leaves of the tree
appraoach were being used by the oponent) or by considering how the
oponent is likely to play (have the program try to confuse the oponent
by taking advantage of something that the oponent wont recognise).
the second method thus will take into account the knowledge-base
available to the solver (for example trying to trick the solver into
thinking that the cube is in one of the easy to solve classes but really
isnt, thus leading the solver down a blind alley)
This gives me an idea for a game where oponents are each trying to bring
a randomized cube into two different final states (each of which
is hopefully equally far away from the initial state, just to make the
game fair) by alternately taking turns making quarter turns. of course
one must make up some rules so that that the game terminates (ei going
around cycles in the state graph an indefinite number of times is
disallowed), and these rules might not be very obvious,although
they would prob be similar to some of the stalemates rules of chess.
-----
Date: 10 September 1980 2303-edt
From: Bernard S. Greenberg
Subject: Randomizing
To: CUBE-LOVERS at MIT-MC
I enter the following tidbit into the transcript of this society,
apropos to the current discussion:
RMS at AI (Richard Stallman) has suggested the following paradigm
for cubing tournaments:
(Assuming physically standard cubes are used) The two participants
acquire (or create) solved cubes. Each one randomizes it his/her
favorite way, and hands it to the other to solve. The resources
expended by the solver (either qtw or real-time) are added to his/her
resource expenditure in solving the other's cube. Minimum resource
expenditure wins.
Date: 11 SEP 1980 0016-EDT
From: ALAN at MIT-MC (Alan Bawden)
Subject: How do you maximally randomize a cube?
To: CUBE-HACKERS at MIT-MC
I am interested in maximally distant states of the cube. I have often
wondered just what a maximally distant state would look like. I also
wonder HOW MANY of them there are!
Interesting fact (offered without proof (it's not hard)):
Assuming we are counting quarter-twists. If I hand you a cube in a
maximally distant state, and ask you to solve it in as few twists as
possible, you don't have to think at all in order to know what to do
first! ANY first twist will bring it closer to home (after that it
gets harder).
Call a state with this property a "local maximum". Any maximally
distant state is also a local maximum. Also, any symmetric state is a
local maximum. This doesn't mean that a maximally distant state is
symmetric, but it does get you thinking along those lines!
Date: 10 Sep 1980 9:37 pm PDT (Wednesday)
From: Woods at PARC-MAXC
Subject: Re: How do you maximally randomize a cube?
In-reply-to: ALAN's message of 11 SEP 1980 0016-EDT
To: ALAN at MIT-MC (Alan Bawden)
cc: CUBE-HACKERS at MIT-MC
It is not obvious to me that any twist in a maximally distant state
brings you closer to the home position. How do you know there
aren't two equally distant states that are a qtw apart? There is
probably a parity argument that proves this can't be so, but if you
count half-twists as single operations I'd be much surprised if you
could find a simple proof.
-- Don.
Date: 11 September 1980 01:04-EDT
From: Alan Bawden
Subject: Re:"Assuming we are counting quarter-twists."
To: Woods at PARC-MAXC
cc: CUBE-HACKERS at MIT-MC
I said:
"Assuming we are counting quarter-twists."
I realize that there are people out there who like to count
half-twists as a single twist. I don't. I'm not trying to force my
way of counting twists on anyone. I always try to be carefull to make
this assumption explicit out of courtesy to those who might want to
count otherwise. (Sort of like the Axiom of Choice.)
In fact I DON'T know if that result is true if you count half-twists
as well. I suspect it is, but I don't have a proof.
You needn't jump up and down and point every time one of us
quarter-twisters reveals himself. This one, at least, is well aware
of his assumptions.
Date: 10 Sep 1980 11:09 pm PDT (Wednesday)
From: Woods at PARC-MAXC
Subject: Re: "Assuming we are counting quarter-twists."
In-reply-to: ALAN's message of 11 September 1980 01:04-EDT
To: (Alan Bawden and the rest of) CUBE-HACKERS at MIT-MC
If you'll reread my message, you'll note that I claimed the parity
argument wasn't obvious to me even in the case of a qtw metric;
my reference to half-twists was intended along the lines of "this
is even less obvious". I'd be interested in seeing your "obvious"
proof in the qtw case, if you'd care to send it to me (no need to
bother the whole mailing list with it).
-- Don.
Date: 11 September 1980 22:06 edt
From: Greenberg.Multics at MIT-Multics
Subject: Singmaster v.5
To: cube-lovers at MIT-MC
(To MIT area cubists only:)
I have received "Notes on Rubik's Magic Cube" version 5
from Singmaster, if anybody wants to see it. It
is 75 pages long and AWESOME.
Date: 15 Sep 1980 1842-PDT
From: Alan R. Katz
Subject: number of reachable states
To: cube-lovers at MIT-MC
I have seen the number 4.3 * 10^19 for the number of reachable states
for the cube, can anyone tell me how you calculate it? This may have
been answered before in this list, but I couldn't find it.
Also, someone mentioned that one can make a checkerboard pattern from
the Pons Asinorum by trebly rotating the centers by a simple
transformation. Can anyone tell me this transformation? (again I may
have missed reading it)
Reply to either me or the list.
Alan
-------
Date: 15 SEP 1980 2156-EDT
From: ALAN at MIT-MC (Alan Bawden)
Subject: Where to find old mail.
To: CUBE-HACKERS at MIT-MC
Just a reminder to you all that ALL of the old cube-lovers mail is
archived in the file ALAN;CUBE MAIL on MIT-MC.
Date: 16 SEP 1980 0746-EDT
From: RP at MIT-MC (Richard Pavelle)
Subject: number of reachable states
To: KATZ at USC-ISIF
CC: RP at MIT-MC, CUBE-LOVERS at MIT-MC
Date: 15 Sep 1980 1842-PDT
From: Alan R. Katz
I have seen the number 4.3 * 10^19 for the number of reachable states
for the cube, can anyone tell me how you calculate it? This may have
been answered before in this list, but I couldn't find it.
The number is (12! * 2^12 * 8! * 3^8)/12. This comes from the following.
There are 8 corners and there are 3 positions- hence 8!*3^8. There are
12 edges with 2 positions hence 12!*2^12. Finally, the /12 comes from
parity considerations. Only 1/4 of the positions in the flippling of
two edges are possible while 1/3 of the toppling of two edges are possible.
Also, someone mentioned that one can make a checkerboard pattern from
the Pons Asinorum by trebly rotating the centers by a simple
transformation. Can anyone tell me this transformation? (again I may
have missed reading it)
The moving of centers is easy- 4 moves of the center slice while rotating
the cube 90 degrees in your hand between moves. With the transformation
in hand you can move the centers easily to possible positions.
Date: 16 SEP 1980 0946-EDT
From: DCP at MIT-MC (David C. Plummer)
Subject: number of reachable states
To: KATZ at USC-ISIF
CC: CUBE-HACKERS at MIT-MC
Date: 15 Sep 1980 1842-PDT
From: Alan R. Katz
I have seen the number 4.3 * 10^19 for the number of reachable states
for the cube, can anyone tell me how you calculate it? This may have
been answered before in this list, but I couldn't find it.
Also, someone mentioned that one can make a checkerboard pattern from
the Pons Asinorum by trebly rotating the centers by a simple
transformation. Can anyone tell me this transformation? (again I may
have missed reading it)
Reply to either me or the list.
Alan
-------
Consider the corners. There are 8 of them, and they can go
anyplace. This leads to 8 factorial permutations. Each corner can
take on three orientations, so this is another factor of 3^8. But
the corners have three possible states (trarity [three way
parity]) so divide by 3. Now do the same with the edges. 12 edges
gives 12 factorial arrangements, times 2^12 oreintations. But the
edges have two parities involved, so divide by four (thus giving
rise to the 12 states of the cube, one of which has the solved
configuration as a member). So if you evaluate
8 12
8!*3 *12!*2
-----------
3*4
you will get 4.3 * 10^19.
DPC@MIT-AI 09/17/80 17:02:00 Re: cube mode on lisp machines
To: cube-lovers at MIT-MC
cube mode has been fixed (at least temporarily) to work in
the new window system on the lisp machines. to invoke it:
(load "bsg;cubpkg")
(cube)
it is self explanitory once you get in. have fun.
-dpc
Date: Wed, 17 Sep 1980 2047-CDT
Message-id: <338086051.13@DTI>
From: aramini at DTI (Michael Aramini)
To: cube-hackers@mc
Cc: boken@mc
Subject: cubing hazordous to your health?
I saw someone cubing near the digital computer lab at the Univ. of
Ill. today. Considering how much traffic there is on springfeild av
(the street in front of said bldg.) i though it might be a good idea
to be crossing the street while intently solving a cube (i didnt see the
person try crossing the street, but he was at least walking without looking
where he was going).
-----
Date: 20 September 1980 2144-edt
From: Bernard S. Greenberg
Subject: Cubesys/LISPM fixed
To: CUBE-LOVERS at MIT-MC
:cube's Lisp Machine instantiation has now been fixed to work in color
on Color LISPM's again.
To invoke, (load "bsg;cubpkg"), wait till it loads, and (cube).
Should be self-documenting.
Date: 23 SEP 1980 0851-EDT
From: ZIM at MIT-MC (Mark Zimmermann)
Subject: report on DC area cube contest
To: CUBE-HACKERS at MIT-MC
Hecht's (a local dept. store chain) offerred $100 gift certificate to
any one who could get the cube done in 5 minutes (and T-shirts promoting
the cube to anyone getting 1 face right in that time), for a 4-hour period
last Saturday...I was the 7th to get the cube done (there were a bunch of
mathematicians from the University of Maryland there before me)...the store
took everyone's name & address & social security numbers...haven't heard
anything from them yet, though...there may have been more solvers after
I left. --Mark
Date: 25 SEP 1980 0859-EDT
From: ZIM at MIT-MC (Mark Zimmermann)
Subject: search-from-both-ends ultimate cube algorithm
To: CUBE-HACKERS at MIT-MC
There are some standard algorithms involving time-memory tradeoffs for solving
problems like Knapsack in N^(1/2) steps instead of N...Hellman had a paper
in some IEEE journal about an application to breaking cryptosystems. The same
could be applied to solving the cube "exhaustively", given several hundred
billion memory locations storage and a willingness to go through a similar
number of table-lookups. Just make a table of all the positions that can
be reached within 10 meoves or less (with the route to that position recorded
too!)...then in order to solve an arbitrary cube set-up, begin working out
from that set-up, looking for a position included in the table. If Singmaster
is right and any position can be reached in 20 moves or less, one will
always succeed within 10 moves from the arbitrary start....
The several hundred billion entries in the table are a bit too large for my
home computer, but perhaps a smaller sub-group of the cube (slice, or
anti-slice, or such) could be done this way in a reasonable amount of time....
Hellman seems to think that solving the DES (which has 2^56 keys, I think) is
not impractical, given enough money and a few years.
How about a competition to find the shortest ways to get to the Crux
Christmanni & Plummeri(?)...a simpleminded corner-shifting I tried took 84
moves, I think, to get whichever one has two cycles of 3 colors...that leaves
lots of room for improvement! --Mark
Date: 23 Oct 1980 2155-PDT
From: KATZ at USC-ISIF
Subject: whats happening?
To: cube-lovers at MC
cc: katz
There hasn't been much activity on this list for long time, whats happening?
Maybe we are all cubed out?? Anyway, heres a question for the LA people.
Does anyone know where I can get a Hungarian version of Rupic's cube
(NOT the ideal version) in the L A area??
Alan
-------
__