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Date: Mon, 09 Jun 1997 16:01:06 -0500
To: cube-lovers@ai.mit.edu
From: Tom Magliery
Subject: Re: Designations for the cubes (proposal)
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At 11:31 AM 6/8/97 -0400, Nicholas Bodley unabashedly said:
>
> Peter (Reitan, I think; sorry) (who is not Karen) brought up the
>clumsiness of such designations as "5X5X5". I find these downright clumsy
>to type (although the Caps Lock key helps). In my private world, I simply
>refer to the Pocket Cube as "[the] two", the original Rubik's as "[the]
>three", Revenge as "[the] four", and the biggest available as "[the]
>five".
>
> I think that provided we understand that we are referring to the
>well-known family of true cubes, it should be OK simply to refer to "the
>three", for instance. Granted, these names require more keystrokes, but
>numerals should be OK, as in "the 3".
I have another suggestion, which might be slightly less likely to require
explanation to a newcomer. I know how to *pronounce* it, but I'm not sure
how I would recommend *spelling* it. (Considerations include terseness,
ease of typing -- which is of course not the same thing!, and likeliness to
be mispronounced by a reader.)
The pronunciation is three-bye, four-bye, five-bye, ...
Possible spellings include:
3by, 4by, 5by, ...
3-by, 4-by, 5-by, ...
3x, 4x, 5x, ...
three-by, four-by, five-by, ...
mag
--
.---o Tom Magliery, Research Programmer (217) 333-3198 .---o
`-O-. NCSA, 605 E. Springfield O- mag@ncsa.uiuc.edu `-O-.
o---' Champaign, IL 61820 http://sdg.ncsa.uiuc.edu/~mag/ o---'
From cube-lovers-errors@oolong.camellia.org Mon Jun 9 19:03:57 1997
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Date: Mon, 09 Jun 1997 21:17:47 +0200
From: Herbert Kociemba
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Subject: Re: Detailed explanation of two phase algorithm
References: <970608193131.21411978@iccgcc.cle.ab.com>
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SCHMIDTG@iccgcc.cle.ab.com wrote:
>
> And if we want to show that all depth one nodes will be pruned when
> we are at some search depth d where 1 < d < h[0] we would need to show
> that:
>
> 1.9 1 + h[1] > h[0]
>
Why do you say 1 < d < h[0] and not d = 1? What I wanted to show is,
that unter the assumption
2.0 D < h(0)
all depth-one nodes will be pruned. As you correctly stated before, for
pruning we need
1.8 d + h(d) > D
and in the case d=1 this means
1.9a 1 + h(1) > D,
which is different from 1.9, because of 2.0 .
But 1.9a can be shown easily:
In my last message, I tried to explane that
2.1 |h(n-1)-h(n)| <=1,
I try to explain it once more in other words. A node at depth n is
generated from a node at depth n-1 by applying a single face-turn on it.
And as I told, h is defined by
h(x,y,z):=max{h1(x,y),h2(x,z),h3(y,z)},
where for example h1(x,y) is the length of the shortest maneuver
sequence which transforms (x,y,z) to (x0,y0,z') for any z' (this means
the z-coordinate is ignored). And this length can maximal change by one
when applying a single move. The same holds for h2(x,z) and h3(y,z). For
this reason, h(x,y,z) also can change maximal by one, which implies 2.1
.
In the case n=1, from 2.1 follows
h(0) <= 1 + h(1), and because of 2.0 we have
D < h(0) <= 1 + h(1),
which proves 1.9a .
> (1)
> / \
> (2) (3*) cost = .9
> /
> (4*) cost = .7
>
> Suppose nodes 3 and nodes 4 were both solutions. Even though node 4
> has a lower cost, phase1 would find node 3 to be our first solution
> whereas IDA* wouldn't.
I don't think we are far away from each other. Of course, the phase1 (or
phase2) algorithm does not claim to be an universal IDA* for any sort of
problem. But for a special problem like the cube you can simplify the
general IDA* and the simplified algorithm will be equivalent to the one
I developed for phase1.
Best regards,
Herbert
From cube-lovers-errors@oolong.camellia.org Mon Jun 9 21:30:42 1997
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From: bandecbv@mailhost.rz.ruhr-uni-bochum.de
Message-Id: <199706100041.UAA11455@life.ai.mit.edu>
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To: mgirard@videotron.ca
Date: Tue, 10 Jun 1997 02:38:52 +0000
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Mathieu Girard wrote:
>I am in a quest to buy both 4x4x4 Rubik's Revenge and also 5x5x5
>cube.. but i just can't find any!
Regrettably the 4x4x4 cubes seem to be sold out everywhere in the
world. With the 5x5x5 Magic Cubes (in Japan once sold under the name
Professor's Cube), the situation is much better: They are still
available from me (and as far as I know nowhere else now). Since they
will probably never be produced again (the production cost is too
high and the general interest too low), they will also be sold out
soon. I shall inform the cube-lovers when this is the case.
My price of the 5x5x5 cube is 40 DM or 24 USD plus postage.
I send my free mail order catalog (containing also many other
twisting puzzles like the Magic Dodecahedron , the Skewb, the
Pyraminx, Mickey's Chellenge and several books and details how to
order) to every cube-lover requesting it and providing a postal address.
A few days ago, Joe McGarity complained bitterly about his 5x5x5 cube
which fell apart. Fortunately, I did not encounter this problem
before, and Joe is not in my files so he has probably not bought his cube
from me. On the other hand you will destroy every twisting puzzle by
twisting it with force without sufficient aligning the layers before
every single move. Since the 5x5x5 cube contains 98 visible little cubies
compared to the 26 of the 3x3x3 (and 92 compared to 20 if we only
count the freely floating ones), one should accept that it requires a
little bit more care.
Joe McGarity also mentioned that some orange stickers sometimes do
not behave according to there name. I have to admit that this
sometimes also happens with my 5x5x5 cubes. Furtunately
it happens only to the orange stickers and it can be repaired easily by
warm pressure or - better - some glue.
Christoph
Christoph Bandelow
mailto:Christoph.Bandelow@rz.ruhr-uni-bochum.de
From cube-lovers-errors@oolong.camellia.org Tue Jun 10 15:15:53 1997
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Date: Tue, 10 Jun 1997 08:16:53 -0400 (EDT)
From: Nicholas Bodley
To: Tom Magliery
cc: cube-lovers@ai.mit.edu
Subject: Re: Designations for the cubes (proposal)
In-Reply-To: <3.0.1.32.19970609160106.00ade4a0@sdgmail.ncsa.uiuc.edu>
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I like Tom's version; it is concise, distinctive enough not to be
confused (however, I haven't studied all the various math. notations going
around), and easy to say and type. Of the ones you gave, I prefer such a
form as "3-by". Thanks!
|* Nicholas Bodley *|* Electronic Technician {*} Autodidact & Polymath
|* Waltham, Mass. *|* -----------------------------------------------
|* nbodley@tiac.net *|* When the year 2000 begins, we'll celebrate
|* Amateur musician *|* the 2000th anniversary of the year 1 B.C.E.
--------------------------------------------------------------------------
From cube-lovers-errors@oolong.camellia.org Tue Jun 10 15:16:08 1997
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Date: Tue, 10 Jun 1997 13:07:15 -0400 (Eastern Daylight Time)
From: Jerry Bryan
Subject: Re: Some Face Turn Numbers
In-reply-to:
To: Cube-Lovers
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X-X-Sender: jbryan@PSTCC6.pstcc.cc.tn.us
On Thu, 5 Jun 1997, Jerry Bryan wrote:
> The following table gives the
> best known results for face turns. The results through depth 7 have been
> calculated (my message of 19 July 1994). The rest are based on Dan Hoey's
> recursion formula PH[n] = 6*2*PH[n-1] + 9*2*PH[n-2] for n>2, where PH[n]
> is the number of face turns which are n moves from Start
Rats, here is a little correction. I think my meaning was clear from
the overall context of the note, but Dan's formula is an upper bound, so
it should read PH[n] <= 6*2*PH[n-1] + 9*2*PH[n-2] for n>2. For depth 0
through 7, my table provided exact values. As I hope was clear from the
context, my table included upper bounds rather than exact values for
depths greater than 7. My apologies.
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan) jbryan@pstcc.cc.tn.us
Pellissippi State (423) 539-7198
10915 Hardin Valley Road (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990
From cube-lovers-errors@oolong.camellia.org Tue Jun 10 15:15:37 1997
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Message-ID: <339CB8B7.1B8D@snowcrest.net>
Date: Mon, 09 Jun 1997 19:15:19 -0700
From: Joe McGarity
Reply-To: joemcg3@snowcrest.net
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Subject: 5x cubes
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Yes, the cube I purchased was not from Cristoph Bandelow. He is off the
hook. It was purchased in San Francisco at the Game Gallery. Although
I am considering buying my next one from him. And no, I don't force it
either. I treated it with kindness and love, yet it betrayed me.
From cube-lovers-errors@oolong.camellia.org Wed Jun 11 00:38:42 1997
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From: SCHMIDTG@iccgcc.cle.ab.com
Date: Wed, 11 Jun 1997 0:35:55 -0400 (EDT)
To: cube-lovers@ai.mit.edu
Message-Id: <970611003555.21417ec3@iccgcc.cle.ab.com>
Subject: Re: Detailed explanation of two phase algorithm
Herbert Kociemba wrote:
>SCHMIDTG@iccgcc.cle.ab.com wrote:
>>
>> And if we want to show that all depth one nodes will be pruned when
>> we are at some search depth d where 1 < d < h[0] we would need to show
>> that:
>>
>> 1.9 1 + h[1] > h[0]
>>
>
>Why do you say 1 < d < h[0] and not d = 1?
Oops, I think that should have been 'D' and not 'd'.
>[...slightly different restatement of earlier proof omitted...]
After examining this once again, I have now satisfied myself
that it is correct. It's just that for some reason, I seem to
find the result rather counter-intuitive. But that makes the
result all the more interesting.
So I think this may yet me another case where the phase1 algorithm
differs slightly from IDA*, but the difference is not significant
since, in this case, one is able to prove a special property of the
heuristic that demonstrates that the number of nodes explored by the
two algorithms is comparable. At this point, I think we can wind down
this thread, (I do hope others on this list have found it interesting)
and I will still continue to think of possible ideas for improving
the algorithm.
I do have one last question regarding the pruning tables. While
the three tables used in phase1 are clear, I do not recall reading
a description of the tables that are used in phase2.
I examined Dik Winter's program and he seems to have a few more
"maximum move" (i.e. "mm" tables) than I expected, namely:
phase1
------
mm_twists[]
mm_flips[]
mm_choices[]
/* and the following "mixed" tables */
mm_tf[][] /* twist & flip */
mm_tc[][] /* twist & choice */
mm_fc[][] /* flip & choice */
phase2
------
mm_eperms[] /* edge perms */
mm_cperms[] /* corner perms */
mm_sperms[] /* slice orderliness */
/* "mixed" tables follow */
mm_cs[][] /* corner perms & slice orderliness */
mm_es[][] /* edge perms and slice orderliness */
Are you using the same tables? Or are the "mixed" tables ones that Dik
added to the algorithm? It appears that Dik was able to use them because
he had a machine with more memory at his disposal than your 1MB Atari ST.
His program can be built with or without the "mixed" tables and is 11MB
with them. He also mentions that the small program finds a reasonable
solution in 30 minutes whereas the large program finds it in only a
few seconds.
I have also been studying his code to try to understand how he generates
these tables. He does not seem to be using breadth-first-search to
fill in these tables as Korf does.
I will be interested in looking at your new program when it becomes
available.
Thanks again for your patience.
Best regards,
-- Greg
From cube-lovers-errors@oolong.camellia.org Wed Jun 11 14:05:23 1997
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Date: Wed, 11 Jun 1997 08:39:17 -0700
From: "Jason K. Werner"
Message-Id: <9706110839.ZM926@isdn-rubik.corp.sgi.com>
In-Reply-To: Joe McGarity
"5x cubes" (Jun 9, 19:15)
References: <339CB8B7.1B8D@snowcrest.net>
Reply-to: "Jason K. Werner"
X-Mailer: Z-Mail-SGI (3.2S.3 08feb96 MediaMail)
To: cube-lovers@ai.mit.edu, Mark Pilloff ,
Mathieu Girard , joemcg3@snowcrest.net
Subject: Re: 5x cubes
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Here's the place where I got my 5x cube, which has been extremely solid and
nearly indestructible:
Game Gallery
2855 Stevens Creek Blvd.
Santa Clara, CA 95050
USA
408-241-4263
408-241-5945 FAX
http://www.gamegallery.com
From cube-lovers-errors@oolong.camellia.org Wed Jun 11 14:04:50 1997
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Date: Wed, 11 Jun 1997 08:32:27 -0400 (EDT)
From: der Mouse
Message-Id: <199706111232.IAA00315@Twig.Rodents.Montreal.QC.CA>
To: cube-lovers@ai.mit.edu
Cc: Mathieu Girard
Subject: Re: ...
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> I am in a quest to buy both 4x4x4 Rubik's Revenge and also 5x5x5
> cube.. but i just can't find any!
> I live in Qu=E9bec, Canada,
Where? Montr=E9al, or Qu=E9bec City, or what? I'm in Montr=E9al; I got =
my
5-Cube at Valet de Coeur, on the west side of St-Denis, somewhere a bit
south of Mont-Royal. I don't know whether they still have them; this
_was_ back in 1993 (December 15, according to my records).
> By the way.. if it is not asking too much... could u please give me
> an aproximate of the prices of thoses cubes... in canadian or us
> dollar...
I paid $45.01, including tax, for my 5-Cube. But as I remarked above,
that _was_ three and a half years ago.
der Mouse
mouse@rodents.montreal.qc.ca
7D C8 61 52 5D E7 2D 39 4E F1 31 3E E8 B3 27 4B
From cube-lovers-errors@oolong.camellia.org Wed Jun 11 16:11:56 1997
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To: Cube-Lovers@AI.MIT.EDU
From: Wei-Hwa Huang
Subject: Re: 5x5x5 Stuctural Integrtity
Date: 11 Jun 1997 19:01:30 GMT
Organization: California Institute of Technology, Pasadena
Message-ID: <5nmsma$t63@gap.cco.caltech.edu>
References:
NNTP-Posting-Host: blend.ugcs.caltech.edu
X-Newsreader: NN version 6.5.0 #2 (NOV)
David Litwin writes:
> The orange sticker problem seems to be with all of them. Mine had
>some small documentation with it mentioning that putting a piece of paper
>on the orange side and ironing it a bit will help fix the stickers on the
>cube. It worked well for me and I've not had any problems with them
>anymore.
I must say that the first sticker I lost on my 5x5x5 was red.
(And I don't know where it went! ARGH!!!)
--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
-------------------------------------------------------------------------------
Inspiration strikes suddenly, so be prepared to defend yourself.
From cube-lovers-errors@oolong.camellia.org Wed Jun 11 16:11:24 1997
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Date: Wed, 11 Jun 1997 20:51:44 +0200
From: Herbert Kociemba
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To: cube-lovers@ai.mit.edu
Subject: Re: Detailed explanation of two phase algorithm
References: <970611003555.21417ec3@iccgcc.cle.ab.com>
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SCHMIDTG@iccgcc.cle.ab.com wrote:
> I do have one last question regarding the pruning tables. While
> the three tables used in phase1 are clear, I do not recall reading
> a description of the tables that are used in phase2.
In phase2, the state of the cube also is described by a triple (x,y,z),
in this case 0<=x<8! describes a permutation of the 8 corners, 0<=y<8!
describes a permutation of the 8 UD-slice edges and 0<=z<4! describes a
permutation of the middleslice edges. Because the overall permutation
must be even, only half of the triples belong to physical cubes. We
could correct this, by defining the z coordinate to describe one of the
12 possibilities for the locations of two middleslice edges - the other
two edges will then be corrected automatic. But there are good reasons
not to do so (which I think is not necessary to explain here).
> I have also been studying his code to try to understand how he generates
> these tables. He does not seem to be using breadth-first-search to
> fill in these tables as Korf does.
>
I only use the "mixed" tables. How to generate the tables is quite
obvious and though I don't know how Dik does it I'm sure it is similar:
1. On initialisation set all elements of the table to 0xf (we use four
bits per entry), only the element belonging to (x0,y0,z0) is set to 0.
Set L=0, n_done=1, n_old=1 (n_done denotes the number of valid
tableentries).
2. Check all elements of the table one after the other. If an entry is
0xf, do nothing. If the entry is L, compute the the 18 possible "child
nodes" and check, if the corresponding tableentry is 0xf. Only in this
case set it to L+1 and increment n_done.
3. Check if n_done=n_old. In this case we are ready. Else increment L,
set n_old=n_done and goto 2.
> I will be interested in looking at your new program when it becomes
> available.
I'm writing too much to this mailing list and do not work at my
windows-help! The program itself is ready. I did a two hours run on each
of Rich Korfs 10 random cubes on a Pentium133 with 16MB RAM and the
result were really pleasing: The generated maneuver lenghts were on the
average less than 1 move away from Rich Korfs optimal solutions
(exactly: 9 moves more for the 10 cubes).
Best regards,
Herbert
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From: SCHMIDTG@iccgcc.cle.ab.com
Date: Wed, 11 Jun 1997 20:42:11 -0400 (EDT)
To: cube-lovers@ai.mit.edu
Message-Id: <970611204211.2141971d@iccgcc.cle.ab.com>
Subject: Re: Detailed explanation of two phase algorithm
Herber Kociemba wrote:
[...much additional detailed explanation deleted...]
Thank again, I found this information helpful, especially when my only
other option is to examine code in great detail in order to extract out
the general principles.
>> I will be interested in looking at your new program when it becomes
>> available.
>
>I'm writing too much to this mailing list and do not work at my
>windows-help!
Sorry about that. I'll stop with my questions. In fact, no need
to even answer this response! I'm sure your program will be well
worth the wait :).
> The program itself is ready. I did a two hours run on each
>of Rich Korfs 10 random cubes on a Pentium133 with 16MB RAM and the
>result were really pleasing: The generated maneuver lenghts were on the
>average less than 1 move away from Rich Korfs optimal solutions
>(exactly: 9 moves more for the 10 cubes).
Very impressive. And if you perform some longer runs and find optimal
solutions, please be sure to let us know the run times.
Best regards,
-- Greg
From cube-lovers-errors@oolong.camellia.org Thu Jun 12 13:23:55 1997
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Date: Thu, 12 Jun 1997 01:03:11 -0400 (EDT)
From: Nicholas Bodley
To: Wei-Hwa Huang
cc: Cube-Lovers@ai.mit.edu
Subject: Re: 5x5x5 Structural Integrtity (Stickers)
In-Reply-To: <5nmsma$t63@gap.cco.caltech.edu>
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One cure might be to take a solved cube (more convenient...) and remove
all the red stickers; carefully clean off the adhesive, perhaps with 99%
isopropyl alcohol, and paint the surfaces neatly with the type of paint
used for plastic model kits. I have also thought of removing the stickers,
cleaning all the adhesive off both the stickers and the cubies, and then
reattaching the stickers with a different type of adhesive.
These are just ideas, and I hope no source of trouble.
My best to all,
|* Nicholas Bodley *|* Electronic Technician {*} Autodidact & Polymath
|* Waltham, Mass. *|* -----------------------------------------------
|* nbodley@tiac.net *|* When the year 2000 begins, we'll celebrate
|* Amateur musician *|* the 2000th anniversary of the year 1 B.C.E.
--------------------------------------------------------------------------
From cube-lovers-errors@oolong.camellia.org Wed Jun 18 16:19:15 1997
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To: cube-lovers@ai.mit.edu
Date: Wed, 18 Jun 1997 15:05:54 -0500
Subject: Square One
Message-ID: <19970618.150557.11350.0.shaggy34@juno.com>
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From: Josh D Weaver
Does anyone know how to solve one of those "Square One" puzzles?
Josh
From cube-lovers-errors@oolong.camellia.org Wed Jun 18 18:43:38 1997
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Date: Wed, 18 Jun 1997 22:58:54 +0200
From: Herbert Kociemba
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Subject: Windows95 program now available
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The Windows95 program which implements my algorithm to solve Rubik's
Cube is now availabe at
http://home.t-online.de/home/kociemba/cube.htm
It not only solves Rubik's cube, but also does a few other nice
things...
Herbert
[ Moderator's note: This program is also available in the Cube-Lovers Archive.
See: ftp://ftp.ai.mit.edu/pub/cube-lovers/contrib/cubexp10.zip
- Alan ]
From cube-lovers-errors@oolong.camellia.org Thu Jun 19 01:16:47 1997
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From: Benjamin Wong
To: Josh D Weaver
Date: Thu, 19 Jun 1997 11:37:43 +1000 (EST)
X-Sender: chi@pipe02.orchestra.cse.unsw.EDU.AU
cc: cube-lovers@ai.mit.edu
Subject: Re: Square One
In-Reply-To: <19970618.150557.11350.0.shaggy34@juno.com>
Message-ID:
On Wed, 18 Jun 1997, Josh D Weaver wrote:
._@_.Does anyone know how to solve one of those "Square One" puzzles?
._@_.
http://www.cfar.umd.edu/~arensb/Square1/
is the only page on the net (that i can find)
which describe how to solve square 1
however, either i can not follow instruction,
or error in it's instructuion
i just can not solve it with their algorimthm
I bought square 1 mess them up,
only manage to solve them 2 times.
(beginner luck)
but the page does not help very much.
._@_.Josh
._@_.
o------------------------------------------------------o
|Error: Reality.sys Corrupt? Reboot Universe [Y,N,Q] |
+---------------o--------------------------------------o
| Benjamin Wong | E-mail: chi@cse.unsw.edu.au |
| | or benjaminwong@hotmail.com |
| | http://www.cse.unsw.edu.au/~chi |
o---------------o--------------------------------------o
|=A1u=C2=E5=A5=CD=A1I=BD=D0=B0=DD=A1y=BA=B5=BF=DF=B2=B4=A1z=AA=BA=A6=A8=A6]=ACO=AC=C6=BB=F2=A1H=A1v |
|Quick Quiz: Describe Universe ? Give Three Example. |
o------------------------------------------------------o
From cube-lovers-errors@oolong.camellia.org Thu Jun 19 12:31:32 1997
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Date: Thu, 19 Jun 1997 00:12:04 -0700
From: Joe McGarity
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To: Josh D Weaver
CC: "Mailing List, Rubik's Cube"
Subject: Re: Square One
References: <19970618.150557.11350.0.shaggy34@juno.com>
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After three months of agony and wondering if Square One was actually the
puzzle box from Clive Barker's Hellraiser, I managed to come up with a
solution that covered all the bases, i.e. worked every time. I have
never tried to write it out in a step by step form however. I will try
to cover the basics.
First the puzzle looks like a Rubik's Cube when solved in that it is a
cube with a solid color on each face, but the similarity ends there.
Square One more closely resembles the Orb, Masterball and Smart Alex in
the ways that it moves. If you can solve any of those you will be a
step closer to the Square One. I see the square one as nearly identicle
to the Smart Alex. The shapes of the pieces are different, but they
move as a disk divided into sectors (exactly like the Masterball).
There are six pieces on each face if you count the small sectors as half
pieces. Count the pieces and you will see what I mean. The little ones
are half the the size of the angle of the big ones. The idea then for
me was to get the little ones paired up like the picture in the
instruction booklet. Once they were paired correctly I could solve it
just like the Masterball or Smart Alex making it look like it did when
it was new in the package (remember it came in a slightly scrambled
state with instructions on how to solve it from there in about six
moves). Then I could just follow the booklet for the final part. Like
I said it took three months and scores of note paper to finally get it.
When I did, the walls opened up and the Cenobites took me away, but it
was worth it.
I hope I haven't caused more confusion. It is difficult to describe
without having one in my hands to show you. This is just a sketchy
overview of how I solve it. If I get a chance to document this solution
I will send you a copy, but it probably won't be for a while. I'm sure
that someone has a better solution and I'd be interested in seeing what
others have come up with. My solution takes about twenty minutes to do
and there must be a faster way. The ones I have trouble with are the
Sqewb and the Alexander's Star. Anybody got a good solution for any of
these?
Joe McGarity
From cube-lovers-errors@oolong.camellia.org Thu Jun 19 12:32:15 1997
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Message-ID:
From: "joyner.david"
To: "'Josh D Weaver'"
Cc: "'cube-lovers@ai.mit.edu'"
Subject: RE: Square One
Date: Thu, 19 Jun 1997 08:06:58 -0400
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>----------
>From: Josh D Weaver[SMTP:shaggy34@juno.com]
>Sent: Wednesday, June 18, 1997 4:05 PM
>To: cube-lovers@ai.mit.edu
>Subject: Square One
>
>Does anyone know how to solve one of those "Square One" puzzles?
There's a paper on my web page which indirectly
explains how. (It's actually a math paper written with
a student of mine explaining the group theory of the puzzle.)
What's useful are some of the moves which we give.
If you can't print it out (It's a dvi file) I'll mail it to you
if you give me your postal address.
http://www.nadn.navy.mil/MathDept/wdj/rubik.html
The idea, if I remember, is
1. get into a square form,
2. use the special moves we give (moves which permute 3
pieces only and leave the others alone, for example)
to solve the puzzle as one solves the Rubik's cube.
- David Joyner
>
>Josh
>
>
From cube-lovers-errors@oolong.camellia.org Thu Jun 19 12:32:39 1997
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Date: Thu, 19 Jun 1997 09:19:01 -0400
Message-Id: <199706191319.JAA27307@bso.newvision.com>
From: Carl Woolf
To: chi@cse.unsw.edu.au
CC: shaggy34@juno.com, cube-lovers@ai.mit.edu
In-reply-to:
(message from Benjamin Wong on Thu, 19 Jun 1997 11:37:43 +1000 (EST))
Subject: Re: Square One
Square One is a great puzzle!
I think there is an instruction booklet, published in Massachusetts or
thereabouts, and available from Puzzlets (mgreen@puzzletts.com). I
developed a set of techniques that let me solve the thing, but I
haven't worked my notes into a form intelligible by other humans (or
by me on a bad day).
--
-- Carl
-----------------------------------------------------
Business: woolf@newvision.com
Personal: woolf@ccs.neu.edu
http://www.ccs.neu.edu/home/woolf
From cube-lovers-errors@oolong.camellia.org Thu Jun 19 21:55:51 1997
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To: "cube-lovers@ai.mit.edu"
Subject: Square One Solution
Date: Thu, 19 Jun 97 21:53:07 -0500
From: Mike Masonjones
X-Mailer: E-Mail Connection v2.5.03
Cube Lovers,
This is long, but complete solutions are long. I even left quite a bit of
the boring and obvious stuff out, and it still took me 3 hours to write it.
Any errors, let me know. I hope this satisfies.
Apparently I have the least extracurricular life of all of you, since I know
how to solve Square One much quicker than any reports I've seen here. Quite
a dubious honor, I suspect. Anyway, here goes my solution, which with a
little practice, guarantees a solution within 75-100 seconds for someone who
can do The Cube in about 55-60 seconds (assuming there's a correlation in
hand speed from puzzle to puzzle).
The solution is completely my own, as is the notation. Sorry if it offends
anyone.
1. Start by getting to the cube state for the top and bottom faces. Ignore
the middle slice til the very end. This can be most efficiently done by
memorizing a table. A pretty good (but not error-free) one is on the only
square-one web site in existence that I know of. Sorry, you'll have to find
the site yourself with a browser, since I can't look it up right now. I
have a scheme written down somewhere that is quite a bit easier to memorize,
but why should I take the fun away from any of you looking for the solution
yourselves. If there is a big response to this letter, then I will dig out
my easy table.
Tables are difficult to memorize, so I usually just try to get to a six
pointed star on one side and all the little wedges plus the remaining 2 big
wedges on the other. It is easy to get to one of the five possible states
that result, thus requiring memorization of only 5 solutions to get back to
a cube. This method takes about 5-10 seconds longer, an average, than the
table technique.
Using a notation where L represents a large wedgie, and S a small wedgie,
the five possible states can be written as:
1) bottom = LLLLLL, top = LLSSSSSSSS
2) bottom = LLLLLL, top = LSLSSSSSSS
3) bottom = LLLLLL, top = LSSLSSSSSS
4) bottom = LLLLLL, top = LSSSLSSSSS
5) bottom = LLLLLL, top = LSSSSLSSSS
For cases 1),3),5), rotate the top face so that it will be sliced
symmetrically between the two L pieces when the center is flipped. The next
move in each of these cases involves moving the top face one way and the
bottom face the other, when looking from the front. (Front will be the term
used from now on to denote the end nearest you of the central cut through
which flipping occurs (a 180 turn of one half of the cube)). After a flip,
cases 1) and 5) should give two barrel shapes (LLSSLLSS), top and bottom.
You should aim in case 3) for two tomahawk (LLSSLSLS) shapes. Any self-
respecting cubist should be able to get home from here.
Cases 2 and 4 are a little more complicated. For both cases align the top
so that the left half of the top face reads, going clockwise, SLSSS. Flip
right side. Now rotate the bottom so that when you flip with the right hand
, the top will read SlSSSSSLL starting from the front and going clockwise.
Now rotate the top 1/12 turn counterclockwise and the bottom so it reads
LLLLSSL going clockwise from the front and flip again. Now you're in an
easy state to get home from (LLLLSSSS on top and LLSLSSLS on bottom).
2. Now that you're in a cube state top and bottom, get all the wedgies on
their correct side (top and bottom face all the same color, respectively).
This is very straightforward and intuitive. I usually start with one large
wedgie, and sequentially put in one at a time next to it going around a face
until you get down to one S wedgie stuck on the wrong side. Sometimes it is
easier to do LSL on one half of the top, and then do LSL on the other, and
then putting in the second to last S between the groups.
Now position the top face so that the Odd S wedgie (O) is positioned as
LSLOLSLS going clockwise from the front. Put the bottom odd wedgie in front
with the bottom square skewed from the top (bottom should read LSLSLSLO
going clockwise from the front). Now do FT4B1FT-4B-1FT4B1FT-3F, where F =
flip with right hand, Bx = turn bottom face clockwise x/12 of a turn, B-x =
same thing counterclockwise, Tx, T-x mean similar things.
3. Now get L's positioned.
Case 1. No L's are correctly adjacent to each other. Position top and
bottom (top = LSLSLSLS, bottom = SLSLSLSL, each going clockwise from front).
Now go
FB3FT-3B-3FT3F, turn the whole puzzle 180 degrees so that the back of the
central cut is now the front, and repeat the move.
Case 2. Two sets of adjacent pairs are out of whack, one on top, and one on
bottom.
Do the move for case 1 once, with the components of the pairs in question
all nearest the front.
Case 3. Only one adjacent pair correct. Position the top so that the
correctly adjacent pair (denoted as A) is positioned as ASASLSLS, and the
bottom reads LSLSLSLS (same conventions as before). Now do FB-3FB3FB-3FB3F.
Case 4. Only one pair incorrect. Position the top (with the incorrect pair)
so that the correctly adjacent pair is positioned as LSASASLS, and the
bottom reads LSLSLSLS. Do the move in Case 3 twice with a T3 between
instances.
Case 5. One side is OK, the other has no correct adjacent pairs. Bad side =
top. top = LSLSLSLS, bottom = LSLSLSLS. Do the move in Case 3 twice with
T6 between instances.
4. Now check for parity.
With the L's in place it is easy to identify whether you need to change the
parity of the system. It should take an even number of switches to right
the S's at this stage. A cycle of three is even, since it would take two
switches to fix it. A cycle of two or four is odd. If the overall parity
is odd, do the following:
starting with top = LSLSLSLS, bottom = LSLSLSLS, go FT3B3FT1B2FT2B2FT-
2FT2B2FT3B2FT-3B-3T-3B-1FT-2B-2F
This may not be the optimum way, but it preserves the corners, and it's easy
to remember the path. (Try it)
5. Place the S's (they should already be on their correct face). The most
useful moves are the below: All permutations of S's can be solved with
application of a maximum of three of these short moves in sequence, combined
with the appropriate turns in between to set things up.
Move 1. Start with top = LOLSLSLO, bottom = LOLSLSLO, where O = pairs that
will be switched on a given side. Do FT-3FT1B1FT2B-1F. Repeated twice with
a T3 between makes a three-cycle on the top side.
Move 2. top = LSLOLSLO, bottom = LSLOLSLO, O definition same as Move 1. Do
FT1B1FT6FT-1B-1F.
There may be quicker solutions than applying these moves for a 4-cycle/2-
cycle combination or a 4-cycle/4-cycle combination, where you have to apply
3 moves in succession. I'd like to hear about suggestions. I haven't
investigated it too much since these modes come up so rarely.
6. Fix middle slice. If square shaped but wrong, do FT6B6F. Otherwise,
position the bad half on the right, and do FB6FB6F.
Congratulations, you have a solved Square One.
Happy cubing.
Mike Masonjones
From cube-lovers-errors@oolong.camellia.org Thu Jun 19 21:55:37 1997
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Date: Thu, 19 Jun 1997 11:00:50 -0700
To: cube-lovers@ai.mit.edu
From: Mark Pilloff
Subject: Re: Square One
At 12:12 AM 6/19/97 -0700, J.M. wrote:
>The ones I have trouble with are the
>Sqewb and the Alexander's Star. Anybody got a good solution for any of
>these?
>
>Joe McGarity
I finally came up with a solution to the Alexander Star last year. I
haven't ever written out all of the details, but here are some helpful
hints. First of all, the star is almost identical to the Megaminx (aka,
magic dodecahedron, etc.) with all of the corners pieces removed. The only
reason I say "almost" is that on the star, every individual piece is doubly
degenerate. This sometimes leads to a problem wherein using the Megaminx
moves seems to leads to an insoluble position. The trick in this case is
two swap two of the degenerate pieces while disturbing as little of the rest
of the star as possible. This has always worked perfectly for me. As for
the rest of the star, I usually find that I can solve most of the star
(except for the uppermost regions) just by inspection. From there, very
slight modifications of Rubik's cube manipulations are useful. It's
worthwhile to note that locally, the star (and the megaminx) are identical
to the cube (except, perhaps, for the missing corners on the star). Thus,
cube moves which only affect small portions of the cube will often be
successful on the star or megaminx. In any event, I'm not going to write
out explicit moves because I believe solutions to the megaminx are floating
on the net as well, but I hope this is somewhat helpful. If there is really
strong demand for explicit solutions, I'll see what I can do about that.
Good luck,
Mark
************************************
** Mark D. Pilloff **
** mdp1@uclink4.berkeley.edu **
************************************
From cube-lovers-errors@oolong.camellia.org Fri Jun 20 11:36:32 1997
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To: Carl Woolf
From: Georges Helm
Subject: Re: Square One
Cc: cube-lovers@ai.mit.edu
At 09:19 19/06/1997 -0400, you wrote:
>Square One is a great puzzle!
>
>I think there is an instruction booklet, published in Massachusetts or
>thereabouts, and available from Puzzlets (mgreen@puzzletts.com). I
>developed a set of techniques that let me solve the thing, but I
>haven't worked my notes into a form intelligible by other humans (or
>by me on a bad day).
>
There is a solution in Puzzle World. Check out details at
http://ourworld.compuserve.com/homepages/Georges_Helm/cubeold.htm
Georges
geohelm@pt.lu
http://ourworld.compuserve.com/homepages/Georges_Helm
http://www.geocities.com/Athens/2715
From cube-lovers-errors@oolong.camellia.org Fri Jun 20 11:36:18 1997
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To: joemcg3@snowcrest.net
From: Georges Helm
Subject: Re: Square One
Cc: cube-lovers@ai.mit.edu
At 00:12 19/06/1997 -0700, you wrote:
>The ones I have trouble with are the
>Sqewb and the Alexander's Star. Anybody got a good solution for any of
>these?
I have solutions for both.
Check out my list of solutions at
http://ourworld.compuserve.com/homepages/Georges_Helm/cubeold.htm
Alexander's book on his star
Flettermann has a good solution for the skewb (but I realize I don't have
him listed)
I can send copies, though
Georges
geohelm@pt.lu
http://ourworld.compuserve.com/homepages/Georges_Helm
http://www.geocities.com/Athens/2715
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From: Goyra
To: cube-lovers@ai.mit.edu
Subject: Re: Square One
Date: Thu, 19 Jun 1997 19:00:58 +0100
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> >Does anyone know how to solve one of those "Square One" puzzles?
Can anyone tell me what this looks like so I can
put up a Java version?
David Byrden
From cube-lovers-errors@oolong.camellia.org Mon Jun 23 21:05:29 1997
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Date: Mon, 23 Jun 1997 18:33:40 -0400
From: "Jonathan R. Ferro"
Message-Id: <199706232233.SAA51996@knave.ece.cmu.edu>
Organization: Electrical and Computer Engineering, CMU
X-Disclaimer: This disclaimer is not required by Leader Kibo.
This article does not necessarily represent the opinions of Leader Kibo.
Have a nice day!
X-Exclaimer: Yow!
To: cube-lovers@ai.mit.edu
Subject: An art project...
http://www.wunderland.com/EBooks/Window/Pages/SUTW-JD.html
-- Jon
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Date: Tue, 24 Jun 1997 17:30:04 -0700
From: Jin "Time Traveler" Kim
Organization: The Fourth Dimension
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Jonathan R. Ferro wrote:
>
> http://www.wunderland.com/EBooks/Window/Pages/SUTW-JD.html
>
> -- Jon
Very impressive. How many cubes and were they altered in any way except
turning them?
--
Jin "Time Traveler" Kim
chrono@ibm.net
VGL Costa Mesa
From cube-lovers-errors@oolong.camellia.org Wed Jun 25 13:29:10 1997
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Date: Wed, 25 Jun 1997 18:11:31 BST
From: David Singmaster Computing & Maths South Bank Univ
To: cube-lovers@ai.mit.edu
Message-ID: <009B651B.AC285B40.328@vax.sbu.ac.uk>
Subject: Pilloff's query about 5^3; Korf's method.
I've been away and have just seen email for late May and early June.
Pilloff asks about getting the parity of the edges correct on the 5^3.
As he notes, the commutators cannot solve this. Examining the basic moves,
one sees that the rotation of a inner face changes the parity of the edges
while conserving the parity of some pieces, but not of others. However, the
pieces whose parity is not conserved are the pieces next to the centre and
there are four apparently identical copies of these, so one can simulate
an exchange by a 3-cycle with two pieces the same.
Hence one wants to apply a rotation of an inner face. What one does
is to move the two edges into the same inner face. Then rotate the face. Then
make a 3-cycle of the edges. This produces an exchange of edges - and rather
messes up the centre pieces. Then put things back.
The edges go A B C D to D A B C to B A C D.
Because this is relatively easy to do, but messes up the centres, I
normally do the edges and corners first and then put centres in place.
Re: Korf. Someone has said it reminds them of alpha-beta pruning.
It reminds me of branch and bound search. Both are older names for the general
process of using information about the remainder of the problem to estimate
the number of steps for a solution via a partial solution.
Going back to Pilloff's query, I have several methods in my files for
exchanging two edges without moving anything else (I think, but that seems to
contradict what I said about parities??)
DAVID SINGMASTER, Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411; fax: 0171-815 7499;
email: zingmast or David.Singmaster @sbu.ac.uk
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Date: Wed, 25 Jun 1997 09:53:21 -0500
To: cube-lovers@ai.mit.edu
From: Kristin Looney
Subject: Re: An art project...
Cc: jake@wunderland.com
>> http://www.wunderland.com/EBooks/Window/Pages/SUTW-JD.html
>
> Very impressive. How many cubes and were they altered in any
> way except turning them?
100 cubes, scrambled - not altered in any other way.
This is, I believe, the sixth cube sculpture that Jacob has done in
the window of my gameroom. A seventh is currently in progress.
Previous sculptures include: "Rubik's Cube", "Merry Xmas" with a
picture of a Christmas tree, a bizarre (and not too successful)
abstract thingamabob, a pacman with several ghosts, and the Apple
logo. I think the pacman is probably my favorite, it's a hard choice.
They are all very much worth a look... I have pictures of them, and
I will encourage Jake to put them on his web page for all to see.
The one he is currently working on is, believe it or not, TWO SIDED.
Unlike most which he just sorta fiddles with while playing games
at my gameroom table, this one was carefully planned out in advance
with graph paper.
-K.
5th fastest hands in the nation (at least back in 1981)
kristin@wunderland.com
www.wunderland.com/kristin
-------------------------------------------------------------
"I'm really angry that I, a superior human being in every way, have
less money than my neighbor, who's wife I would love to nail, if only
I weren't so busy sleeping and eating pork chops."
-- George "Cannonball" Carlin,
on the 7 deadly sins
From cube-lovers-errors@oolong.camellia.org Wed Jun 25 17:18:25 1997
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Date: Wed, 25 Jun 1997 22:14:32 BST
From: David Singmaster Computing & Maths South Bank Univ
To: cube-lovers@ai.mit.edu
CC: notation@vax.sbu.ac.uk, for@vax.sbu.ac.uk, 4^3@vax.sbu.ac.uk,
and@vax.sbu.ac.uk, 5^3@vax.sbu.ac.uk
Message-ID: <009B653D.9EE57540.331@vax.sbu.ac.uk>
Subject: notation for 4^3 and 5^3
David Barr has described some moves for the 5^3 using his own notation.
In the early 1980s, I extended my 3^3 notation to the 4^3 and this can be used
on the 5^3 as well. I described this in my Cubic Circular but perhaps it would
be useful to give it here.
I will describe the situation for the 4^3. On the 5^3, we don't ever
have to make a slice move of just a central layer. Further, a combination of
4^3 and 3^3 processes will solve the 5^3, so we don't really need to label the
central layers.
Consider the four layers from L to R. I denote the inner layers
by l and r. So the four layers are: LlrR. Similarly we have four layers:
FfbB and UudD. To describe a piece requires more effort than before, but
each piec lies in three layers and we can describe the piece by these layers.
E.g. FUR is a corner piece; FUr is an edge piece, lying on the FU edge -
but there are two of these and they are distinguished as being FUr and FUl
(I usually give the layers in clockwise order, but it is not essential and
there are times when it is more informative to use the other order.); Fur is
a face-centre piece, the one in the upper right corner of the inner four cells
of the F face. If you have been paying attention, you will ask about fur.
This is one of the body-centre cells, invisible to you unless you make a
transparent cube!
Using the standard notation of [F, R] = FRF'R', we find a number
of easy 3-cycles.
[[F,R],L] = (FLU,ULB,RFU)
[[F,R],l] = (FlU,UlB,DfR) (I've copied this from my Circular, but
I wonder if it's right as I thought there'd be some symmetry with the
preceding??)
[[F,r],l] = (Flu,Ulb,Drb)
In theory, these and a careful consideration of parity are sufficient to
solve the 4^3 and the 5^3. However, the parity problem is a bit awkward.
In my original approach, I got the corners in place and then all edges
except leaving the four edges along the FU and BU edges. Examine the
parity of these carefully. If they are in an odd permutation, apply
r^2 D^2 l' D^2 r^2 which 4-cycles these four edges and moves some centres.
Once the parity is corrected, there is little difficulty restoring the
rest of the cube.
For the 5^3, once you have paired up the edges, one can solve the
central edges by treating the 5^3 as a 3^3 with fat slices.
To correct a single pair of edges, one can use the following.
rrDDl'DDrr rrD'RR [[R,U],l'] RRDrr = rrDDl'DR'UR'U'l'URU'lRDrr =
(UBl,UBr) (Ful,Ubl,Bdl,Ufr) (Fdl,Ufl,Bul,Ubr). This messes up some centres,
but they are not too hard to restore. Indeed applying
rrUUr (uurrll)^2 r'UUrr wil correct the F and B centres disturbed by the
above, leaving a 180 degree rotation of the four U centres.
After I had developed the notation and solution, a Peter Lees pointed
out an unexpected feature. The exchange of upper and lower case letters is
a duality. The dual of URF is urf, while the dual of URf is urF.
This gives us a Pricniple of Duality: The dual of a sequence of moves is
the same process on the dual pieces. E.g., we had
[[F,R],l] = (FUl,UBl,DRf), so [[f,r],L] = (fuL,ubL,drF).
This duality allows one to transfer a number of 3^3 processes to
4^3 processes and to solve the invisible interior part of the cube!
By always moving an outer layer with its inner layer, one is
obviously simulating the 2^3. However, if one always moves, say R and l
together, one is also simulating the 2^3 in eight copies! Ooops, one
wants to move R and l' together. If one moves R and l together,
I think you get eight versions of the 2^3, but each is a reflection of
its neighbours!
If you are tired of thinking about God's Algorithm on the 3^3,
try the 4^3. I'm not even sure how to count moves. E.g., to do r, one
normally does Rr and then R', so does r count as one move or two?
Likewise, does Rr count as one move or two?
Enough for now.
DAVID SINGMASTER, Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411; fax: 0171-815 7499;
email: zingmast or David.Singmaster @sbu.ac.uk
From cube-lovers-errors@oolong.camellia.org Wed Jun 25 18:29:12 1997
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Date: Wed, 25 Jun 1997 23:28:16 BST
From: David Singmaster Computing & Maths South Bank Univ
To: cube-lovers@ai.mit.edu
Message-ID: <009B6547.EBD77100.328@vax.sbu.ac.uk>
Subject: 4^3 and 5^3
I've just seen a comment about the 5^3 saying the writer had problems
with the four pieces at distance 1 from the center. My approach treated
both these and the pieces ate distance 2 from the center in the same way.
We know that the commutator of two slice moves on the 3^3 produces two
3-cycles of the centres. Applying this idea to the 4^3, we find that
[f,r] gives two 3-cycles of central pieces, one on each face. By turning
one face and then inverting, we get a 3-cycle of central pieces, two being
in one face. E.g. [[r,b],U] = (Fur,Ubr,Ubl). A similar result holds if
we combine any two inner moves, so we can replace the b above by a central
slice on the 5^3 to obtain a 3-cycle of the pieces at distance 1 from the
central piece, while the process directly gives us a 3-cycle of the pieces at
distance 2 from the central piece. Although tedious, these moves mean that
once one has the corners and edges in place, the rest of the problem is
easy - though very tedious - it generally took me about an hour to do the
5^3, assuming I could get the corners and edges correct without making too
many mistakes.
DAVID SINGMASTER, Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411; fax: 0171-815 7499;
email: zingmast or David.Singmaster @sbu.ac.uk
From cube-lovers-errors@oolong.camellia.org Wed Jun 25 23:51:13 1997
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Date: Thu, 26 Jun 1997 04:44:18 BST
From: David Singmaster Computing & Maths South Bank Univ
To: cube-lovers@ai.mit.edu
Message-ID: <009B6574.123497C0.321@vax.sbu.ac.uk>
Subject: names for cubes
What's wrong with 2^3, 3^3, 4^3, 5^3, pronounced 2 cube,
3 cube, 4 cube, 5 cube. If you have superscripts available, you can use them
instead of the uparrows.
DAVID SINGMASTER, Professor of Mathematics and Metagrobologist
School of Computing, Information Systems and Mathematics
Southbank University, London, SE1 0AA, UK.
Tel: 0171-815 7411; fax: 0171-815 7499;
email: zingmast or David.Singmaster @sbu.ac.uk
From cube-lovers-errors@oolong.camellia.org Thu Jun 26 15:40:18 1997
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Date: Thu, 26 Jun 1997 07:44:43 -0400 (EDT)
From: Nicholas Bodley
To: David Singmaster Computing & Maths South Bank Univ
cc: cube-lovers@ai.mit.edu, notation@vax.sbu.ac.uk, for@vax.sbu.ac.uk,
4^3@vax.sbu.ac.uk, and@vax.sbu.ac.uk, 5^3@vax.sbu.ac.uk
Subject: Hidden cubies; Spaceball
In-Reply-To: <009B653D.9EE57540.331@vax.sbu.ac.uk>
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Reading David Singmaster's recent posts, it seems almost obvious to me
that (for the ambitious, which I'm definitely not!), computer simulations
that are unhindered by real-world mechanical constraints and opacity can
permit manipulation of the normally-hidden (and physically-nonexistent)
internal cubies. It's very likely that a new collection of maneuvers would
need to be developed for this.
I haven't thought much about coloring the internal faces of the outer
cubies...
Incidentally, if this weren't the cube-lovers' List, I would have split
that first sentence into a few shorter ones.
Sorry if "cubie" is not the most-preferred term; should be no great
problem.
On another topic, it seems to me that an ideal device for controlling a
computer-simulated Cube (or other similar puzzle) would be the Spaceball,
a ball that you can grip. It senses torque around all three mutually-
orthogonal axes, as well as "translational" force along those axes. It's
not a consumer item; not sure it's still being made. I'm reasonably sure
of the tradename. It was/is used with workstations. "Spaceball" sounds
much like the name of a puzzle.
(I expect some astute reader to tell me that the MIT Media Lab did just
this thing 5 years ago!)
My best to all,
|* Nicholas Bodley *|* Electronic Technician {*} Autodidact & Polymath
|* Waltham, Mass. *|* -----------------------------------------------
|* nbodley@tiac.net *|* When the year 2000 begins, we'll celebrate
|* Amateur musician *|* the 2000th anniversary of the year 1 B.C.E.
--------------------------------------------------------------------------
From cube-lovers-errors@oolong.camellia.org Thu Jun 26 16:35:00 1997
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Message-ID: <33B2D235.3A77@ibm.net>
Date: Thu, 26 Jun 1997 13:33:57 -0700
From: Jin "Time Traveler" Kim
Organization: The Fourth Dimension
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To: Nicholas Bodley
CC: cube-lovers@ai.mit.edu
Subject: Re: Hidden cubies; Spaceball
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Nicholas Bodley wrote:
>
> On another topic, it seems to me that an ideal device for controlling a
> computer-simulated Cube (or other similar puzzle) would be the Spaceball,
> a ball that you can grip. It senses torque around all three mutually-
> orthogonal axes, as well as "translational" force along those axes. It's
> not a consumer item; not sure it's still being made. I'm reasonably sure
> of the tradename. It was/is used with workstations. "Spaceball" sounds
> much like the name of a puzzle.
>
> (I expect some astute reader to tell me that the MIT Media Lab did just
> this thing 5 years ago!)
>
> My best to all,
>
Actually, the Spaceball that you talk about is still in existence of
sorts. I have three Spaceballs. Actually, they were known as the
Spacetec Spaceball Avengers. Those are no longer produced. They've
been replaced by the newer model, the SpaceOrb 360. Not to get too far
off subject, but the SpaceOrb is used by some people to play Quake. You
can't beat a good Mouse and Keyboard for Quake, but the SpaceOrb's
multiple axes of movement does allow for some interesting
possibilities. Due to its 3D nature, I think the SpaceOrb would be a
natural extension for the solving of 3d puzzles in graphical
environments.
--
Jin "Time Traveler" Kim
chrono@ibm.net
VGL Costa Mesa
From cube-lovers-errors@oolong.camellia.org Thu Jun 26 16:55:53 1997
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Date: Thu, 26 Jun 1997 13:53:47 -0700
From: "Jason K. Werner"
Message-Id: <9706261353.ZM3850@neuhelp.corp.sgi.com>
In-Reply-To: Nicholas Bodley
"Hidden cubies; Spaceball" (Jun 26, 7:44)
References:
Reply-to: "Jason K. Werner"
X-Mailer: Z-Mail-SGI (3.2S.2 10apr95 MediaMail)
To: Nicholas Bodley , cube-lovers@ai.mit.edu
Subject: Re: Hidden cubies; Spaceball
Mime-Version: 1.0
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On Jun 26, 7:44, Nicholas Bodley wrote:
> Subject: Hidden cubies; Spaceball
> On another topic, it seems to me that an ideal device for controlling a
> computer-simulated Cube (or other similar puzzle) would be the Spaceball,
> a ball that you can grip. It senses torque around all three mutually-
> orthogonal axes, as well as "translational" force along those axes. It's
> not a consumer item; not sure it's still being made. I'm reasonably sure
> of the tradename. It was/is used with workstations. "Spaceball" sounds
> much like the name of a puzzle.
In case anyone is interested:
http://www.spacetec.com/
http://www.spaceorb.com/
-Jason
--
Jason K. Werner Email: mrhip@sgi.com
Systems Administrator Phone: 415-933-6397
USFO I/S Technical Services Fax: 415-932-6397
Silicon Graphics, Inc. Pager: 415-317-4084, mrhip_p@sgi.com
"Winning is a habit"-Vince Lombardi;"These go to eleven"-Nigel Tufnel
From cube-lovers-errors@oolong.camellia.org Fri Jun 27 16:39:01 1997
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Date: Fri, 27 Jun 1997 16:28:04 -0400
From: Corey Folkerts
Subject: First corners, then sides
To: Cube-Lovers@ai.mit.edu
Message-ID: <199706271628_MC2-1961-C7B8@compuserve.com>
During the last 5 months or so I have been fiddling around with a
Rubik's Cube that I got for Toys R Us. I have become pretty good at solving
it with the layers method (90 seconds is my record so far.) My question is
this : Is the method of solving the cube corners first and then sides any
faster (once one becomes good at it) then solving it by layers? If so,
could someone please reply with a description of that method. I don't know
the names of any fancy moves, so if possible please use F B U D L R for
the faces during specific moves and expain the other parts of the strategy
clearly. I realize this is asking alot. Thanks in advance to anyone who
replies!
Corey Folkerts
From cube-lovers-errors@oolong.camellia.org Sat Jun 28 14:03:46 1997
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Date: Sat, 28 Jun 1997 05:50:29 -0400 (EDT)
From: Jiri Fridrich
X-Sender: fridrich@bingsun2
To: Corey Folkerts
cc: Cube-Lovers@ai.mit.edu
Subject: Re: First corners, then sides
In-Reply-To: <199706271628_MC2-1961-C7B8@compuserve.com>
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It appears that both systems are good if worked out into sufficient
detail. As far as I know, the fastest speed cubists can achieve an average
of 16-17 seconds irrespective! of whether they use corners-edges or
by-slices systems. You can look at a system which I have developed some
time ago and which enables me to solve the cube in 17 sec. on average. It
is described in detail here: http://ssie.binghamton.edu/~jirif/. I also
recommend the section on speed cubing tips.
Good luck!
Jiri
On Fri, 27 Jun 1997, Corey Folkerts wrote:
>
>
> During the last 5 months or so I have been fiddling around with a
> Rubik's Cube that I got for Toys R Us. I have become pretty good at solving
> it with the layers method (90 seconds is my record so far.) My question is
> this : Is the method of solving the cube corners first and then sides any
> faster (once one becomes good at it) then solving it by layers? If so,
> could someone please reply with a description of that method. I don't know
> the names of any fancy moves, so if possible please use F B U D L R for
> the faces during specific moves and expain the other parts of the strategy
> clearly. I realize this is asking alot. Thanks in advance to anyone who
> replies!
>
> Corey Folkerts
>
>
**********************************************************************
| Jiri FRIDRICH, Research Associate, Dept. of Systems Science and |
| Industrial Engineering, Center for Intelligent Systems, SUNY |
| Binghamton, Binghamton, NY 13902-6000, Tel.: (607) 797-4660, |
| Fax: (607) 777-2577, E-mail: fridrich@binghamton.edu |
| http://ssie.binghamton.edu/~jirif/jiri.html |
**********************************************************************
......................................................................
Remember, the less insight into a problem, the simpler it seems to be!
----------------------------------------------------------------------
From cube-lovers-errors@oolong.camellia.org Sun Jun 29 22:09:51 1997
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Message-Id: <199706292239.CAA27387@telecom.lek.ru>
From: Alex Joukov
To: Cube-Lovers@ai.mit.edu
Subject: Newcomer: Algoritms. How to solv U-slice if D-slice and UD-slice have been solvd yet?
Date: Mon, 30 Jun 1997 03:11:05 +0400
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Dear Cube-Lovers,
How to solv U-slice if D-slice and UD-slice have been solvd yet? I remember
that in about 1994 when I was a child I used 3 or 5 algoritms (with mirrow
variants). But now I don't remember the way.
I read cube-lovers arhives (from #0) and step by step find out some
algoritms. But before I will be able to solve Cube I lose a lot of
interesting in
archive messages. I just can't try a lot not having solved Cube!
Please help me. A need just "1. U2DFTU'RD' 2. UR'D2L'D2 3..." without
comments.
Or, may be somebody have created such type FAQ which is accesible for ftp?
lllykob@telecom.lek.ru
Sasha
From cube-lovers-errors@oolong.camellia.org Mon Jun 30 15:09:03 1997
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From: Jerry Bryan
Subject: Inverses of Local Maxima
To: cube-lovers@ai.mit.edu
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One of the oldest unsolved problems on Cube-Lovers (aside from God's
Algorithm itself) has to do with inverses of local maxima. It seems
obvious that the inverse of a local maximum also ought to be a local
maximum. But is it necessarily so?
In Symmetry and Local Maxima, Jim Saxe and Dan Hoey suggest that it may
not be. Their example is UFF, which can end with F or F' because we can
write it as UF'F'. But the inverse is F'F'U', which can only end with U'
Hence, there are very simple positions where the number of q-turns with
which the position can end is different than the number of q-turns with
with the inverse of the position can end. If the same thing should
happen with a local maximum, then the inverse would not be a local
maximum.
On the other hand, for all known local maxima in G, the inverse is also
a local maximum. What are we to think?
I have some small progress. I can report that for the corners-only
group, there are local maxima for which the inverse is not a local
maximum. The results were obtained with my new Shamir program.
For each position x, we define E(x) to be the set of all quarter-turns
with which a minimal process for the position can end. As an example,
if x=UFF, then E(x)={F,F'}.
E(x) is a subset of Q, the set of twelve quarter-turns, or equivalently
it is an element of P(Q), the power set of Q. As such, it is
conveniently represented in my program as a bit-string of twelve bits.
In this notation, we would say that a position x is a local maximum if
E(x)=Q or if |E(x)|=12.
We also define S(x) to be the set of all quarter-turns with which a
minimal process for a position can start. In this notation, for x=UFF
we would say that |S(x)|=1 and |E(x)|=2. So the general question for
local maxima becomes the following: if |E(x)|=12, does it necessarily
follow that |S(x)|=12?
My program calculates S(x) and E(x) as follows. Any breadth-first
search may be characterized as calculating products of the form z=xy for
suitable choices of x and y. Most typically, x comes from Q[n], the set
of all quarter-turns of length n, and y comes from Q[1], the set of all
quarter-turns of length 1. But in my more general Shamir program, x
comes from Q[m] and y comes from Q[n] to form products of length m+m. In
any case, S(z) is the union of S(x) over all x which can be a part of a
product which produces z, and E(z) is the union of E(y) over all y which
can be a part of a product which produce z. For each q in Q, we
initialize with S(q)=E(q)={q} and go from there.
Here is a portion of a printout from my program.
|x| |E(x)| |S(x)| M-Conjugacy Positions
Classes
0 0 0 1 1
1 1 1 1 12
2 1 1 2 96
2 2 2 3 18
3 1 1 12 576
3 1 2 3 96
3 2 1 3 96
3 2 2 4 96
3 3 3 2 60
As you can see, the effect pointed out by Saxe and Hoey first shows up
three moves from Start, where there are six positions unique up to
M-conjugacy where |S(x)| is not equal to |E(x)|. (Actually, three of the
six positions are just the inverses of the other three.)
The first local maxima are six moves from Start in the corners-only
group.
|x| |E(x)| |S(x)| M-Conjugacy Positions
Classes
6 12 12 8 114
As you can see, there are 114 local maxima of which 8 are unique up to
M-conjugacy. However, for all 8 of them, the inverse is also a local
maximum so we discover nothing new.
The new discovery occurs 7 moves from Start.
|x| |E(x)| |S(x)| M-Conjugacy Positions
Classes
7 12 8 4 120
7 12 10 3 144
7 12 12 14 336
As you can see, there are 21 local maximu unique up to M-conjugacy. For
14 of them, the inverse is also a local maximum. But for the other 7,
the inverse is not a local maximum. In 4 cases, we have |S(x)|=8, and
in 3 cases we have |S(x)|=10.
Here follow summaries for local maximum up to a distance of 11 moves
from Start.
|x| |E(x)| |S(x)| M-Conjugacy Positions
Classes
8 12 6 14 576
8 12 8 12 576
8 12 10 86 4128
8 12 11 13 624
8 12 12 272 12012
9 12 4 26 1152
9 12 6 31 1344
9 12 8 24 1152
9 12 10 14 576
9 12 12 131 5976
10 12 2 14 576
10 12 4 88 4032
10 12 6 218 10368
10 12 8 144 6336
10 12 10 168 8064
10 12 12 140 5664
11 12 4 384 18432
11 12 6 2687 128688
11 12 8 5550 264192
11 12 10 5014 240576
11 12 12 3617 166224
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan) jbryan@pstcc.cc.tn.us
Pellissippi State (423) 539-7198
10915 Hardin Valley Road (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990
From cube-lovers-errors@oolong.camellia.org Mon Jun 30 19:21:04 1997
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Date: Mon, 30 Jun 1997 16:05:31 -0700 (PDT)
From: Bill Webster
Subject: Hi
To: cube-lovers@ai.mit.edu
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Hi,
I first encountered the cube sometime in the early 80s when the fad
hit Australia. Solving it absorbed all my spare time for two weeks
along with several wads of A4 and $15 for a second cube which I used
to develop what I called 'sequences'. My experiments on the second
cube were conducted with great care, as I had not yet discovered that
Satan's Algorithm could be had for the price of a small screwdriver. I
never became a speed-freak, or even employed operators outside my own
meagre discoveries, but could solve the cube comfortably inside
three minutes. I solved corners, then edges because the first
reasonable sequences I discovered were edge-disrupting corner
operators. My operators were short and scant so my method incurred a
lot of short term memory overhead, manipulating faces into susceptible
positions, applying the sequence, then inverting the prior
manipulation.
A friend of mine aquired a cube at about the same time and much to my
chagrin, solved it in less than a day, without paper, without
explicitly developing any operators. In fact, he couldn't give a
satisfactory account of exactly how he'd solved it. He may have just
got *extremely* lucky and stumbled on something close to START, but I
don't think so. I handed him a scrambled cube a couple of weeks later
and he was quite taken aback - he wasn't going through all that again,
he'd done it hadn't he?. I always felt that my own solution was
somewhat contrived after witnessing this feat. Does anyone else have
examples of GestaltCube?
I have coded a C++ class which represents cube states and operators
and which includes methods to manipulate the cube. I would like to
implement overloaded C++ operators in a manner which is consistent
with (and perhaps extends) the appropriate mathematical notation *if
this is feasible*. Is there anyone out there familiar with both
grammars and willing to make suggestions? I am aware and prepared to
accept that the use of some (C++) operators may introduce
inefficiencies in the form of temporary objects created during
expression evaluation - such operators will not be used in time
critical code.
I have been using a freeware ray-tracer, POV-Ray to produce
'photo-realist' cube images. I intend to extend my software to export
animation scripts, so that I can produce (externally rendered)
animated solutions. These take forever to trace, so their value is
aesthetic rather than practical. I have been experimenting with solid
gold cubes inlaid with coloured marble etc., but still prefer the
platonic form. I have the POV source for a static cube if anyone is
interested.
The POV team are true heroes - details...
"The internet home of POV-Ray is reachable on the World Wide Web
via the address http://www.povray.org and via ftp as ftp.povray.org."
"POV-Ray can be used under MS-DOS, Windows 3.x, Windows for
Workgroups 3.11, Windows 95, Windows NT, Apple Macintosh 68k,
Power PC, Commodore Amiga, Linux, UNIX and other platforms."
Regards, Bill Webster
_____________________________________________________________________
Sent by RocketMail. Get your free e-mail at http://www.rocketmail.com
From cube-lovers-errors@oolong.camellia.org Mon Jun 30 21:31:40 1997
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Date: Mon, 30 Jun 1997 20:01:16 -0400 (EDT)
From: Jerry Bryan
Subject: Re: Inverses of Local Maxima
In-reply-to:
To: cube-lovers@ai.mit.edu
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On Mon, 30 Jun 1997, Jerry Bryan wrote:
>
> I have some small progress. I can report that for the corners-only
> group, there are local maxima for which the inverse is not a local
> maximum.
>
There is a minor interesting point that might be added. When we find a
local maximum x for which |S(x)|<12, we can form a new, longer local
maximum qx for suitable q in Q.
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan) jbryan@pstcc.cc.tn.us
Pellissippi State (423) 539-7198
10915 Hardin Valley Road (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990
From cube-lovers-errors@oolong.camellia.org Mon Jun 30 21:57:52 1997
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Date: Mon, 30 Jun 1997 21:37:39 -0400
From: michael reid
To: cube-lovers@ai.mit.edu, jbryan@pstcc.cc.tn.us
Subject: example of a local maximum whose inverse is not a local maximum
jerry bryan asks if the inverse of a local maximum is necessarily a local
maximum. the following example shows that this need not be the case.
the interesting "six-two-one" pattern is produced by the sequence
B U2 F2 R U' R' B' R' U F2 U2 (15q)
this position has six symmetries, generated by the cube rotation C_UFR
and central reflection. therefore we also have the maneuvers
L F2 R2 U F' U' L' U' F R2 F2
D R2 U2 F R' F' D' F' R U2 R2
F' D2 B2 L' D L F L D' B2 D2
R' B2 L2 D' B D R D B' L2 B2
U' L2 D2 B' L B U B L' D2 L2
for the same position. it is not hard to check (by computer) that
these are minimal maneuvers. note that for each quarter turn, we
have a maneuver that ends with that quarter turn. thus, from this
position, any quarter turn brings us closer to start, so our position
is a local maximum.
consider now the inverse position; it is produced by
U2 F2 U' R B R U R' F2 U2 B' (15q)
it is not hard to check (by computer) that applying the quarter turn B'
to this moves us further from start (16q), so this position is not locally
maximal.
note that this is already in the archives; i first reported it on april 20,
1995 in my message "correction and an interesting example"
mike
From cube-lovers-errors@oolong.camellia.org Sat Jul 5 20:21:49 1997
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Date: Sat, 5 Jul 1997 16:58:20 -0400
From: michael reid
To: cube-lovers@ai.mit.edu
Subject: optimal cube solver
the recent work by rich korf on finding optimal solutions has prompted
me to try my hand at writing an optimal cube solving program. so far,
i've done this for the face turn metric. a description of my program
follows.
let H denote the intermediate subgroup __ which
we've seen before. we'll use distances to this intermediate subgroup
for our pruning tables (or "pattern databases"). calculating these
distances involves doing a breadth first search on the coset space
H \ G , and storing these distances in memory. (i've written this as
a right coset space, rather than a left coset space.) this search has
been done several times, by dik winter and by myself.
some review. positions in H are characterized by the following.
corners cannot change orientation; their U or D facelet remains on
the U or D face. similar, edges cannot change orientation.
furthermore, the four U-D slice edges remain in the U-D slice.
therefore, cosets in H \ G are described by triples (c, e, l),
where c denotes corner orientation,
e denotes edge orientation
and l denotes the location of the four U-D slice edges.
there are 3^7 = 2187 possible corner orientations,
2^11 = 2048 possible edge orientations
/ 12 \
and \ 4 / = 495 possible U-D slice edge locations.
all combinations are possible, so there are 2187 * 2048 * 495 =
2217093120 cosets. since this is too many configurations to store
in memory, we use symmetry to to reduce this number.
there are 16 symmetries of the cube that preserve the U-D axis, and
therefore the intermediate subgroup H. rather than store all the
cosets, we'll just store one of each up to symmetry. actually, this
is slightly more complicated than necessary; instead, we could just
divide the corner coordinate by symmetry. this is what i did in my
message of january 7, 1995. however, i encountered a pitfall along
the way. i discovered (very late in the development stage) the need
for very large transformation tables. although i continued with the
same approach at that time, i gave two options for overcoming this
problem:
> i) only use the 8 symmetries that preserve my choice of
> 12 edge facelets.
>
> ii) combine the two coordinates edge and location into a single
> coordinate and divide this coordinate by the 16 symmetries.
of these, clearly the second is the better choice, since it utilizes
more symmetry. this new edge coordinate has 2048 * 495 = 1013760
possibilities. up to symmetry, there are 64430 possibilities. we
need room for 64430 * 2187 = 140908410 cosets in memory. for each
of these, we store its distance to the identity coset. this is an
integer between 0 and 12 (inclusive), so each is stored in half a
byte. thus the whole table requires 67 megabytes.
essentially, what we're doing here is changing coordinates from
(c, e, l) to (c, e', s), where e' is our new edge coordinate,
and s is a symmetry coordinate. some cosets have multiple
coordinates in this new system, but that causes no harm.
a breadth first search of this space takes under 11 minutes. the
increase in speed is partially due to a more powerful computer, and
partially due to switching to "backward searching" (or "bidirectional
search") at the optimal time.
we'll also use distances to the intermediate subgroups
and . we don't need to
store additional coset spaces, since we can derive that information
from our first coset space. note that the cube rotation C_UFR takes
the subgroup ____ to the subgroup
. therefore it transforms the first coset space
into the second coset space. furthermore, it preserves distances, so
the one pattern database suffices for all three applications.
an attractive feature of this approach is that it uses the 16
symmetries to reduce the size of the pattern database, and then uses
the remaining symmetry of the cube in applying it in different
orientations.
these are the only pruning tables my program currently uses. note that
they cannot "see the entire group". specifically, let
H_0 = ____, H_1 = ,
H_2 = , and let T denote the intersection of
these three subgroups. for a given position, the three distances to
these subgroups depend only upon the corresponding coset in T \ G .
thus T might be thought of as a "target subgroup".
this target subgroup T is interesting. it consists of those positions
that "look like" they're in the "square group" ,
i.e. F and B colors mix only with each other, and similarly for
R and L , and also for U and D. however, this is strictly larger
than the square group; it contains the square group as a subgroup of
index 6.
the searching is done in the way that korf describes as "IDA*" (or at
least the "ID" part of that terminology). we traverse the tree of all
sequences of length 1, hoping to find a solution. that generally fails,
so we continue to sequences of length 2, and so forth, until a solution
is found. the "A*" part of the algorithm is to use the pruning tables
to avoid searching large parts of the tree that are guaranteed not to
bear fruit.
in his paper, korf uses the expected value of his heuristic functions to
get an estimate of how effective they are at pruning the search tree.
actually, he should subtract 1 from this expected value, since we must
generate (at least partially) the top node of a subtree that gets pruned.
this is only a rough estimate; getting a more precise figure is a delicate
matter which i won't address here. korf reports an expected value of 8.878.
i generated 10 million random cubes (i did not use the long sequence of
random twists method) and got an expected value of 9.941.
my program generates slightly more than 500000 nodes per second. korf
generates them at 700000 per second, so i've got more overhead per
node. however, it generates many fewer nodes, since it prunes the
search tree more efficiently.
i solved korf's ten random cubes, and found all minimal solutions,
rather than stopping at the first. this entailed one complete search
through length 16f, three through length 17f and six through length 18f.
the position at distance 16f has a unique minimal solution, as do the
three positions at distance 17f. of the six positions at distance 18f,
one has a unique minimal solution, one has 3 minimal solutions, two
have 4 minimal solutions and two have 6 minimal solutions. the total
run time for these was just under 198 hours. korf estimates 4000 hours
for the same search, so on these positions, my program is twenty times
as fast. my computer has a 200 MHz pentium pro processor, and is
configured with 128 megabytes of RAM.
i'd expect a similar increase in performance for most positions, but
not all. for example, positions inside the target subgroup T run
very slowly, as do positions very close to it. hopefully, most of
these are close enough to start, so that searches don't have to go very
deep. i suspect that there are probably also positions that give korf's
program difficulty.
as you can see, i've made only minor modifications to korf's method.
the only differences are:
1. use different pattern databases that allow more efficient pruning.
2. apply the same pattern database in multiple orientations.
3. allow a target subgroup larger than just the identity.
it's clear that more experimentation is needed with different pattern
databases. for any subgroup K of G , we could consider distances
to that subgroup. it seems likely that we want small subgroups, so
that the average distance is large. for this reason, using symmetry
to reduce the size of the database is an important tool. i encourage
others to experiment with different subgroups.
more results to come ...
mike
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Date: Mon, 7 Jul 1997 01:04:35 -0400
From: michael reid
To: cube-lovers@ai.mit.edu
Subject: symmetry reductions for superflip
a simple counting argument shows that some cube positions are at least
18 face turns from start, and thus the diameter of the cube group is
at least 18f. in january 1995, i showed, by exhaustive search,
that the position "superflip" is exactly 20 face turns from start.
therefore the diameter is at least 20f. this gave the first
improvement to the lower bound obtained by the counting argument.
the searching method i used at the time was my version of kociemba's
algorithm. although my symmetry reductions fit together quite well
with kociemba's algorithm, this might not be the most appropriate
searching method to use for this purpose. (i guess i could have
hacked it not to bother looking for solutions longer than 19f.
i don't remember why i didn't do this.)
my new optimal solving program can do an exhaustive search in much
less time. the symmetry reductions are similar, but much simpler.
i will try be more coherent this time with my explanation, hopefully
without being overbearing.
the first thing to note is that dik winter found a maneuver for
superflip in 20f:
F B U2 R F2 R2 B2 U' D F U2 R' L' U B2 D R2 U B2 U (20f)
therefore our concern is with searching for maneuvers of length
at most 19f.
there are three ways to transform a maneuver for superflip to get
another such maneuver, which do not change its length:
1. we may conjugate the maneuver by any symmetry of the cube.
2. we may cyclically shift the maneuver; i.e. replace
sequence_1 sequence_2 by sequence_2 sequence_1
3. we may replace the maneuver by its inverse.
(in fact, we won't use 3 here, but it might be helpful elsewhere.)
our first result is
proposition 1. any maneuver for superflip in 19f contains a 180 degree
face turn.
proof. if the proposition were false, then superflip would be an
odd number of quarter turns from start, contradiction. qed.
the relevance of this proposition is
proposition 2. suppose that a maneuver for superflip contains a 180 degree
face turn. then it can be transformed, using the above
tranformations, into a maneuver that begins with U R2.
proof. we first claim that the maneuver has two consecutive "syllables"
such that the first contains a 90 degree face turn and the second
contains a 180 degree face turn. a "syllable" is a sequence of
one or two face turns along the same axis; e.g. U D2. by
hypothesis, the maneuver has a syllable that contains a half turn.
if the claim is false, then the preceding syllable contains no
90 degree turns, and therefore consists only of half turns. but
then the syllable before that contains only half turns, by the
same reasoning. continuing in this way, we see that every syllable
consists only of half turns. therefore we have a maneuver for
superflip consisting only of half turns. this is a contradiction,
so the claim is true.
now, since the individual face turns within a syllable
commute, we may suppose that the maneuver has a 90 degree face
turn followed by a 180 degree face turn, which are along
different axes, and thus are adjacent faces. now we may
conjugate by an appropriate symmetry of the cube to suppose that
these turns are U R2. finally, we may cyclically shift the
maneuver so that these are the first two turns. qed.
proposition 3. suppose that superflip is exactly 19 face turns from
start. then applying the sequence U R2 to it brings
us 2 face turns closer to start, i.e. 17f from start.
proof. apply proposition 1 and proposition 2. qed.
we now know how to handle the case that superflip's distance from start
is exactly 19f. if the distance is less than 19f, we use the following
proposition 4. under any circumstances, applying the sequence U R2
to superflip brings us at least 1f closer to start.
proof. a minimal maneuver for superflip must contain a 90 degree twist,
and we may suppose that the next face turned is an adjacent one.
by cyclically shifting the maneuver, we may bring these two
turns to the beginning. furthermore, by symmetry, we may
suppose that the first turn is U and the second is some twist
of the R face. now by applying U to superflip, we've moved
1f closer to start, and applying R2 to this doesn't move us
any further from start, since it either combines with, or cancels
the next turn in the minimal maneuver. qed.
putting this all together, we get our desired result.
proposition 5. suppose that superflip is within 19f of start. then the
position superflip U R2 is within 17f of start.
proof. this is just combining props 3 and 4. qed.
i don't claim that these are the best reductions possible. they
suffice for our purposes.
i tested the position superflip U R2 (i.e. the position obtained by
first doing superflip, and then doing the sequence U R2) with my
optimal solver. my program took 2 hours and 40 minutes to exhaustively
search this position through 17 face turns (not including about 11
minutes to generate all the lookup tables). there were no solutions.
thus superflip is exactly 20 face turns from start. when i did the
search in january 1995, the run time was 6 days. so we see quite a bit
of improvement.
mike
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Date: Tue, 8 Jul 1997 00:18:18 -0400
From: michael reid
To: cube-lovers@ai.mit.edu
Subject: superfliptwist requires 20 face turns
i can now show that the pattern "superfliptwist" is exactly 20 face
turns from start. this position was proposed as a likely antipode
of start by cubologist christoph bandelow. in the german edition
of his book "einfuerung in die cubologie" he offered a prize for the
shortest maneuver for this pattern. the prize was collected by rainer
aus dem spring, who found a maneuver in 22 face turns. much later,
a maneuver of length 20f was found by herbert kociemba:
D F2 U' B2 R2 B2 R2 L B' D' F D2 F B2 U F' L R U2 F' (20f)
as one of the first applications of his ingenious searching algorithm.
i'll try not to be so verbose with my symmetry reductions this time.
first note that "superfliptwist" does not describe a unique position
of the cube; there are two possible orientations. in this context,
i use the term "position" to refer to one of the 43252003274489856000
possible configurations, and the term "pattern" to refer to an
equivalence class of positions under symmetries of the cube.
(this concept has been discussed by dan hoey and jerry bryan as
the "real size of cube space" i.e. the number of patterns.)
the following two facts are easily verified:
* superfliptwist commutes with the square of each face turn.
* it does not commute with 90 degree slices (e.g. U D') or 90 degree
antislices (e.g. U D), however, if A is a 90 degree slice or
antislice, then
A superfliptwist A^(-1)
is also superfliptwist, but in the other orientation.
these facts lead to the importance of the following
proposition. superfliptwist is not in the subgroup generated by slices
and antislices. (note that this group contains all squares
of face turns.)
proof. we may ignore the corners and just show that all edges cannot be
flipped in this subgroup. to do this, we choose dominant facelets
on the 12 edges as follows: choose the U or D facelet of the edges
in the R-L slice, the R or L facelet of the edges in the F-B slice
and the F or B facelet of the edges in the U-D slice. now we may
define the flip of an edge that is not in its correct location.
all edges start in the correct orientation. a 90 degree slice or
antislice along the U-D axis changes the orientation of all eight
edges in the F-B slices and R-L slices. similarly, a 90 degree
slice or antislice along the F-B or R-L axis flips all edges in
two different slices. within this subgroup, either all edges in
a given slice are flipped, or none are flipped, and furthermore,
the number of the three slices with flipped edges is even, i.e.
0 or 2. however, superfliptwist has all three slices with
flipped edges, so it is not in this subgroup. qed.
now consider the first syllable of a minimal maneuver for superfliptwist.
("syllable" was defined in my previous message.) if this is a single
180 degree turn, then we may cyclically shift this to the end of the
maneuver. similarly, a slice squared may also be shifted to the end
of the maneuver. furthermore, 90 degree slices and or antislices may
also be shifted to the end of the maneuver, with only the mild effect
of changing which orientation of superfliptwist we're doing. from the
proposition, we eventually find a syllable which is not of these types,
and is therefore of type U or D2 U. in the case of D2 U , we may
shift the D2 to the back of the maneuver, so we may suppose that the
first face turn is U . furthermore, by conjugating by the cube rotation
C_U, if necessary, we may suppose that our maneuver solves our preferred
orientation of superflip. the second face turn is in a different syllable,
so it is an adjacent face. conjugating by C_U2, if necessary, brings this
face to either R or F. therefore we may suppose that the first two
face turns are one of the six sequences
U R , U R2 , U R', U F , U F2 or U F' .
to show that superfliptwist is not within 19f of start, i tested the
six patterns obtained by applying these sequences to it. it took my
program 7.5 hours to exhaustively search all of these through 17f.
(these positions ran a bit faster than most of the others i've tested.
this is partly because superfliptwist is 15 face turns from my "target"
subgroup, so larger parts of the search tree are pruned.) no solutions
were found, so superfliptwist requires 20 face turns.
i also let the first situation run partially through depth 18f. in about
4 and a half hours, it found a solution which yields
U R F' B U' D' F U' D F L F' L' U R D F U R L (20f, 20q)
this is automatically minimal in the quarter turn metric!
mike
From cube-lovers-errors@oolong.camellia.org Tue Jul 8 17:33:24 1997
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Date: Tue, 8 Jul 1997 16:43:50 -0400
From: michael reid
To: cube-lovers@ai.mit.edu
Subject: composition of superflip and pons asinorum
the position which is the composition of superflip and pons asinorum is
exactly 19 face turns from start. some time ago, jerry bryan found that
this position is exactly 20 quarter turns from start, and he gave all
minimal maneuvers, up to symmetry. one of these is 19 face turns long:
B' D' L' F' D' F' B U F' B R2 L U D' F L U R D (19f, 20q)
symmetry reductions for this position are much simpler (but not nearly
as good) as for superflip and/or superfliptwist. if the first face turn
is a 90 degree turn, then by symmetry, we may suppose it is U . if the
first face turn is a 180 degree turn, then we may suppose it is U2 .
i tested the two positions obtained by applying these possible initial
turns. my program took about 6 and a half hours to exhaustively search
these through 17 face turns. no solutions were found, and therefore the
original position is more than 18 face turns from start.
i realize that this is not nearly as satisfying as obtaining all minimal
maneuvers. that will take about 13 times as long, but is feasible with
my current program.
mike
From cube-lovers-errors@oolong.camellia.org Thu Jul 10 00:56:16 1997
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Date: Thu, 10 Jul 1997 00:54:58 -0400
From: michael reid
To: cube-lovers@ai.mit.edu
Subject: all minimal maneuvers for superflip (face turn metric)
i can now give all minimal maneuvers for superflip in the face turn
metric. recall that there are three operations we may apply to any
maneuver for superflip which give another maneuver of the same length:
1. we may conjugate the maneuver by any symmetry of the cube.
2. we may cyclically shift the maneuver; i.e. replace
sequence_1 sequence_2 by sequence_2 sequence_1
3. we may replace the maneuver by its inverse.
my original search (january 1995) for superflip in 19 face turns was
divided into 16 cases. since i used my (unhacked) version of kociemba's
algorithm, the search through each case produced maneuvers for superflip,
and 8 of these cases found maneuvers of length 20f. i previously reported
that these were each equivalent to dik winter's maneuver, using the three
operations above. however, i was mistaken about this; there were two
different maneuvers which differ only very slightly.
to facilitate an exhaustive search through 20f, i'll use a result of a
previous search.
proposition. any maneuver for superflip in 20f contains a 180 degree
face turn.
proof. otherwise the maneuver would be 20 quarter turns long. however, i
did an exhaustive search through 20q and found no maneuvers. qed.
(in fact, this quarter turn result was later improved by jerry bryan, who
showed that superflip is not within 22q of start, and therefore is exactly
24q from start.)
now the symmetry reductions show that we may take the first two face turns
to be U R2 . my program exhaustively searched the position superflip U R2
through 18f. it took 35 hours, and found 30 maneuvers, which came in two
different types:
U R2 F B R B2 R U2 L B2 R U' D' R2 F R' L B2 U2 F2 (20f)
U R2 F B R B2 R U2 L B2 R U' D' R2 F D2 B2 U2 R' L (20f)
note that these are the same except for the last 5 face turns. (this gives
the relation R' L B2 U2 F2 R L' U2 B2 D2 = identity; alternatively, the
same sequence produces (f++)(d++) in the supergroup.)
from this, we can count the exact number of 20f sequences for superflip.
both of the above may be cyclically shifted in 23 different ways. we get
23 different ways, instead of 20, because there are three separate pairs
of consecutive twists of opposite faces. we'd consider
sequence_1 U D sequence_2 and sequence_1 D U sequence_2
to be the same, but we wouldn't consider
U sequence D and D sequence U
to be the same. yet cyclic shifting of these last two produces the same
maneuver.
we can also conjugate by any of the 48 symmetries of the cube, and we can
also invert any of the maneuvers. all these operations produce different
maneuvers, so we get a total of 2 * 23 * 48 * 2 = 4416 different maneuvers.
by counting, the number of different sequences of length <= 19f is about
82 times as many positions the cube has. thus a position has, on average,
82 maneuvers of length <= 19f, although superflip has 0. the number of
different sequences of length 20f is about 1016 times the number of
positions, so a position has, on average, 1016 different maneuvers of
length 20f. superflip has more than 4 times that many.
here are the 30 solutions my program found for superflip U R2. hopefully
i haven't made any mistakes this time. they should all be equivalent to
one of the two listed above.
U R2 F U2 F2 D2 R' L U R2 F' B' R D2 L F2 R D2 R D (20f)
U R2 F B R B2 R U2 L B2 R U' D' R2 F R' L B2 U2 F2 (20f)
U R2 F B R B2 R U2 L B2 R U' D' R2 F D2 B2 U2 R' L (20f)
U R2 F R' L F2 D2 B2 U R2 F' B' R D2 L F2 R D2 R D (20f)
U R2 F2 L2 F D2 R L D R2 D F2 U R2 D F' B' D2 L D' (20f)
U R2 F2 L2 F' U2 R' L' U' R2 U' F2 D' R2 U' F B U2 L' D' (20f)
U R2 F2 L2 B D2 R' L' D F2 U R2 D F2 D F B D2 R D' (20f)
U R2 F2 L2 B' U2 R L U' F2 D' R2 U' F2 U' F' B' U2 R' D' (20f)
U R2 F' U2 B2 D2 R L' D' R2 F' B' R' B2 R' D2 L' B2 R' D (20f)
U R2 F' B' R D2 L F2 R D2 R U D R2 F U2 F2 D2 R' L (20f)
U R2 F' B' R D2 L F2 R D2 R U D R2 F R' L F2 D2 B2 (20f)
U R2 F' R L' B2 D2 F2 D' R2 F' B' R' B2 R' D2 L' B2 R' D (20f)
U R2 U B2 D R2 U F' B' U2 L F2 R2 B2 U' D B U2 R L (20f)
U R2 U B2 D R2 U F' B' U2 L U' D R2 B2 L2 B U2 R L (20f)
U R2 U R L U2 F L2 F2 R2 U' D L U2 F' B' U R2 D F2 (20f)
U R2 U R L U2 F U' D F2 R2 B2 L U2 F' B' U R2 D F2 (20f)
U R2 U2 L2 F' B R F2 U' D' F L2 B U2 F L2 F R L F2 (20f)
U R2 B R' L B2 U2 F2 D R2 F' B' R U2 L B2 R U2 R D (20f)
U R2 B D2 B2 U2 R' L D R2 F' B' R U2 L B2 R U2 R D (20f)
U R2 B' R L' F2 U2 B2 U' R2 F' B' R' F2 R' U2 L' F2 R' D (20f)
U R2 B' D2 F2 U2 R L' U' R2 F' B' R' F2 R' U2 L' F2 R' D (20f)
U R2 D F2 U R2 U R L U2 F L2 F2 R2 U' D L U2 F' B' (20f)
U R2 D F2 U R2 U R L U2 F U' D F2 R2 B2 L U2 F' B' (20f)
U R2 D F2 U R' L' U2 B L2 F2 R2 U' D R U2 F B U F2 (20f)
U R2 D F2 U R' L' U2 B U' D F2 R2 B2 R U2 F B U F2 (20f)
U R2 D F2 D F B D2 R U D' R2 F2 L2 B D2 R' L' D F2 (20f)
U R2 D F2 D F B D2 R B2 R2 F2 U D' B D2 R' L' D F2 (20f)
U R2 D F' B' D2 L U D' R2 F2 L2 F D2 R L D R2 D F2 (20f)
U R2 D F' B' D2 L B2 R2 F2 U D' F D2 R L D R2 D F2 (20f)
U R2 D2 L2 F B' L B2 U D B D2 B L2 F D2 B R' L' B2 (20f)
mike
From cube-lovers-errors@oolong.camellia.org Sun Jul 13 19:35:33 1997
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Date: Sun, 13 Jul 1997 19:37:58 -0400
From: michael reid
To: cube-lovers@ai.mit.edu
Subject: minimal maneuvers for composition of superflip and pons asinorum
i've finished calculating all minimal maneuvers (in the face turn metric)
for the composition of superflip and pons asinorum. in my search for
maneuvers of length <= 18f for this position, i used symmetry to show
that we may suppose that the first face turn is either U or U2 .
in fact, there is more symmetry available, and this time i will use it.
in the case beginning with U , there are four symmetries, generated
by the cube rotation C_U . using these, we may suppose that the second
face turn is one of D , D2 , D' , R , R2 or R' .
in the case beginning with U2 , there are eight symmetries. these are
generated by the cube rotation C_U and reflection through the left-right
plane. using these, we may assume that the second face turn is one of
D , D2 , R or R2 .
we can reduce these cases somewhat further. the cases beginning with
U D2 and with U2 D are equivalent, so only one needs to be seacrhed.
the cases beginning with U D' and with U2 D2 can also be eliminated.
both U D' and U2 D2 commute with both pons asinorum and with superflip,
so we may cyclically shift these turns to the end of the maneuver. our
position cannot be achieved only using these "slice" turns, so we'll
always be able to cyclically shift until we do not begin with a slice turn.
(alternatively, note that any maneuver of length 19f , or any odd length
cannot consist only of slice turns!)
that leaves seven cases to search. my program took just less than one
day to search all through 17f. it found 26 maneuvers, 16 for the case
beginning with U D . however, this case has 8 symmetries, so there are
just 2 different maneuvers, each in 8 different orientations.
this leaves 12 different maneuvers, which come in 6 pairs of inverses.
they are:
U R F D R U' D L' U' D F' B2 R L' D' F' L' B' R' (19f)
U D F R L' F B' L D2 R L F' B' U' L2 F B' U2 L' (19f)
U D F' B' L' U2 F' B L2 U' R' L' F' U' D F' B D' L2 (19f)
U2 R F U F B' L' D' F B' L B R L' U D2 B' R' U2 (19f)
U2 R F U2 D' R' L F' L' F B' U L F B' D' B' R' U2 (19f)
U2 R U2 D2 R U' L' U B R F2 U' D B' R' F' D B' L2 (19f)
and their inverses. the first of these is the maneuver found by
jerry bryan. it's also the only of these that is 20 quarter turns
long, which is consistent with his findings.
mike
From cube-lovers-errors@oolong.camellia.org Mon Jul 14 12:37:17 1997
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Date: Mon, 14 Jul 1997 11:39:21 -0400
To: cube-lovers@ai.mit.edu
From: karen angelli
Subject: hockey puck puzzle
Mime-Version: 1.0
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I recently purchased a hockey puck puzzle from Whole Systems Design, who
also brought us the Masterball. Since I received it, I've played with for
only a couple of hours without beginning any systematic attempt at solving
it. My initial impression is that, although it looks like is should be
very easy, it may actually be pretty hare. Or, it's one of those puzzles
that is so easy that I'm thinking too hard and can't see the answer. I've
never heard anyone in the cube-lovers discuss the puzzle, and I wondered if
any of you had any impressions of it.
The puzzle has the shape, size and feel of a regulation hockey puck, and is
divided into twelve wedges. It's basically a flattened masterball in which
only the lines of longitude twist, and the lines of latitude do not. There
are several designs, with various degrees of difficulty and redundancy.
For example, on some, there are printed hockey players, Maple leafs, or
American flags. On my version, the wedges are numbered consecutively from
one through twelve. The pristine version is with the numbers lined up in
consecutive order.
I'm not really interested in learning a solution from anybody, but I would
be interested in comments about whether you think the puzzle is harder or
easier than it looks.
YOu can see pictures of the hockey puck puzzle or order them at
www.wsd.com/HockeyPuck/home.
'e-ya later,
Pete.
From cube-lovers-errors@oolong.camellia.org Mon Jul 14 14:02:22 1997
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Date: Mon, 14 Jul 1997 13:59:16 -0400 (EDT)
Message-Id: <199707141759.NAA00043@spork.bbn.com>
From: Allan Wechsler
To: karen angelli
Cc: cube-lovers@ai.mit.edu
Subject: hockey puck puzzle
In-Reply-To: <3.0.1.32.19970714113921.006aa4a4@pop.tiac.net>
References: <3.0.1.32.19970714113921.006aa4a4@pop.tiac.net>
Please give a clearer description of the puck. The photo gives only a
slight clue about how many pieces there are, and how they are
arranged. Are there twelve pieces, or twenty-four, or more?
-A
From cube-lovers-errors@oolong.camellia.org Tue Jul 15 13:18:50 1997
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Date: Tue, 15 Jul 1997 17:57:11 +0200
From: Rob Hegge
Subject: Description of hockey puck puzzle
To: cube-lovers@ai.mit.edu
Message-id: <9707151557.AA06449@sumatra.mp.tudelft.nl>
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The hockey puck puzzle is a flat disk with a diameter of about 9 cm
or 3.5 inches and a thickness of about 2.5 cm or 1 inch. It basically
consists of a circle (the hart) and a "ring" surrounding the circle.
The circle is cut into two equal halves like "(|)". The two halves are
connected so that you can turn one half upside down, while holding the
other half. The ring is cut (from front to back) into 12 equal wedges,
each of which is attached to the circle by a dovetail so that the ring
with the wedges can be moved around the circle. One can also flip six
wedges including one half of the circle around so that afterwards those
6 wedges and the half circle face backwards. Thus the puzzle is similar
to a puzzle called saturn (which has only 8 wedges ?). The type of moves
reminds me of moves possible on square-1.
In the puzzle I own the 12 wedges on the front are numbered from 1 to 12
and on the back with the letters of "hockeypuzzle", while the left half
circle contains the letters "pu" and right half circle the letters "ck"
as shown below. I do not have it here so this was straight from memory.
front: back:
12 1 c k
11 2 o e
10 | 3 h | y
| pu|ck
9 | 4 p | e
8 5 u l
7 6 z z
The three "|" denote the cut through the circle.
A flip as described above would give for instance
12 k c 1
11 e o 2
10 | y h | 3
|ck pu|
9 | e p | 4
8 l u 5
7 z z 6
while then a clockwise turn of the ring for one wedge would give:
11 12 1 2
10 k c 3
9 | e o | 4
|ck pu|
8 | y h | 5
7 e p 6
z l u z
For a rotational puzzle it is not that difficult.
Rob
From cube-lovers-errors@oolong.camellia.org Tue Jul 15 13:18:23 1997
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Date: Tue, 15 Jul 1997 09:41:28 -0400 (EDT)
Message-Id: <199707151341.JAA00944@spork.bbn.com>
From: Allan Wechsler
To: jandr@xirion.nl
Cc: Allan Wechsler , cube-lovers@ai.mit.edu
Subject: Re: hockey puck puzzle
In-Reply-To: <9707151143.AA27610@la1.apd.dec.com>
References: <199707141759.NAA00043@spork.bbn.com>
<9707151143.AA27610@la1.apd.dec.com>
[Jan de Ruiter:]
The puzzle contains edge pieces and center pieces, all with the same
thickness as the puck.
There are always two center pieces which together form the inner
circle.
In this case there are 12 edge pieces which together form the outer
ring. Simpeler pucks might contain less than 12 edge pieces, but always
an even number.
The possible moves are:
1. rotate one center piece with half of the edge pieces 180 degrees
relative to the other center piece and the other half of the edge
pieces.
2. rotate the edge pieces around the center pieces, always multiples of
30 degrees, or 1/12 of a circle (or more if there are less edge
pieces)
I know the puck with 6 edge pieces is near trivial to solve.
I haven't tried the other ones yet.
Jan de Ruiter
Thanks for the description -- Pete ("Karen Angelli") provided an
identical one in a private reply. But this is still incomplete. Are
the obverse and reverse of the individual pieces distinguishable?
Suppose I manage to flip every other edge piece over in place (not
sure this is possible). Does it then look solved? Or do the two
sides have different colors or a distinguishing mark or something?
I haven't tried it, but I can't imagine that this puzzle would be very
difficult.
-A
From cube-lovers-errors@oolong.camellia.org Tue Jul 15 13:17:34 1997
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Subject: Re: hockey puck puzzle
To: Allan Wechsler
Date: Tue, 15 Jul 1997 13:42:54 +0200 (CETS)
From: Jan de Ruiter
Cc: cube-lovers@ai.mit.edu
In-Reply-To: <199707141759.NAA00043@spork.bbn.com> from "Allan Wechsler" at Jul 14, 97 01:59:16 pm
Reply-To: jandr@xirion.nl
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>
> Please give a clearer description of the puck. The photo gives only a
> slight clue about how many pieces there are, and how they are
> arranged. Are there twelve pieces, or twenty-four, or more?
>
> -A
>
The puzzle contains edge pieces and center pieces, all with the same
thickness as the puck.
There are always two center pieces which together form the inner
circle.
In this case there are 12 edge pieces which together form the outer
ring. Simpeler pucks might contain less than 12 edge pieces, but always
an even number.
The possible moves are:
1. rotate one center piece with half of the edge pieces 180 degrees
relative to the other center piece and the other half of the edge
pieces.
2. rotate the edge pieces around the center pieces, always multiples of
30 degrees, or 1/12 of a circle (or more if there are less edge
pieces)
I know the puck with 6 edge pieces is near trivial to solve.
I haven't tried the other ones yet.
Jan de Ruiter
From cube-lovers-errors@oolong.camellia.org Wed Jul 16 09:37:39 1997
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Message-ID:
From: "joyner.david"
To: "'cube-lovers@ai.mit.edu'"
Subject: RE: hockey puck puzzle
Date: Wed, 16 Jul 1997 07:53:21 -0400
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>----------
>From: Jan de Ruiter[SMTP:jandr@apd.dec.com]
>Sent: Tuesday, July 15, 1997 7:42 AM
>To: Allan Wechsler
>Cc: cube-lovers@ai.mit.edu
>Subject: Re: hockey puck puzzle
>
>>
>> Please give a clearer description of the puck. The photo gives only a
>> slight clue about how many pieces there are, and how they are
>> arranged. Are there twelve pieces, or twenty-four, or more?
Spencer Robinson, a former student, and I have written a WWW
page which sketches an easy but rather inefficient solution of the 12
piece
hockey puck puzzle:
http://www.nadn.navy.mil/MathDept/wdj/mball/puck.htm
Have fun! - David Joyner
>>
>> -A
>>
>
From cube-lovers-errors@oolong.camellia.org Wed Jul 16 10:18:34 1997
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From: Jerry Bryan
Subject: No Local Maxima 11q from Start
To: cube-lovers@ai.mit.edu
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My new Shamir program has now generated the entire search tree for the
standard cube group G up to 11q from Start. This search was
accomplished once before using my old tape spinning programs, so there
is limited new information.
One good result is that all the numbers match between the two programs.
The matching results were obtained using different programs,
implementing different algorithms, in different programming languages,
on different hardware platforms, and under a different operating
systems. So I feel pretty good about the numbers. They have been
posted before, so I won't post them again. With problems this big, it
is always good to have some sort of independent verification because it
is impossible to verify anything by hand.
Another interesting result is in fact new. The old program was only
able to determine local maxima up to 10q from Start while calculating
the 11q tree. The new program is able to determine local maxima up to
the same distance from Start it is searching. There are no local maxima
11q from Start. I find this result somewhat surprising, since there are
four local maxima (unique up to M-conjugacy) 10q from Start. The new
program did confirm the previously known 10q local maxima, but failed to
find any 11q local maxima.
In its search for local maxima, for each position x the program
calculates the set E(x) of quarter turns with which a minimal process
for the position can end. We call |E(x)| the maximality of x, and a
position is a local maximum if its maximality is 12. At a distance of
11q from Start, there exist positions with maximality values for every
number in 1..11. This is the first time we have found any positions
with a maximality of 9 or 11. (See my note of 16 June 1995, "10q Local
Maxima Search Matrix".) There seem to be more positions with even
maximality values than odd, and a maximality of 11 is especially
interesting because such a position is "almost" a local maximum.
I am disappointed in the speed of my program. For this run, it
identified about 1100 patterns (representative elements of M-conjugacy
classes) per second. This corresponds to about 50,000 positions per
second (about 48 times 1100). The program is running on a Pentium P166
with 16MB memory under Windows/95. My concern is that I have worked so
hard to make the program run in small amounts of memory that it is
running too slow. I am now going to take out a few of the memory saving
techniques to see if I can speed it up a bit.
The program is actually about 20MB, and runs successfully on a 16MB
machine due to the good graces of virtual memory. In fact, I can
calculate out to 11q from Start even on an 8MB machine. But trying to
calculate out to 12q from Start fails on the 16MB machine (the program
is the same size for 11q from Start and for 12q from Start because I am
storing all positions up to 6q from Start. The program would only have
to be made larger if I were to try calculating 13q from Start or 14q
from Start.)
When I say the program fails at 12q from Start, I mean that the virtual
memory thrashes unmercifully, and therein lies an interesting tale. Why
should the program be able to calculate 11q without thrashing, but
thrash so badly at 12q? It has to do with the Shamir algorithm itself.
Recall that we are producing products of the form ST in lexicographic
order. To be specific, we are producing products of the from St in
lexicographic order for all t in T and merging the results. S itself is
already in lexicographic order. Think of processing a dictionary, and
thing of processing S in lexicographic order. We essentially process
all the A's, followed by all the B's, then all the C's, etc. There is
very good locality of reference as far as the virtual memory goes.
Moving up to St, we might first process all the N's, then all the E's,
then all the Z's, etc, but there is still very good locality of
reference. There is an occasional big jump in where we are referencing
memory, but most of the time we reference elements of the set S which
are very close together in memory.
When we calculate 11q from Start, S is the set Q[6] of positions which
are 6q from Start, and T is the set Q[5]. Because Q[5] is only about
1/9 as big as Q[6], the real memory working set to calculate Q[6]Q[5] is
only about 10% of the total virtual memory of the program, maybe about
2MB. But when we move up to calculating 12q, we move up to Q[6]Q[6] and
the real memory working set becomes the whole 20M program. This simply
doesn't work on a 16MB machine.
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan) jbryan@pstcc.cc.tn.us
Pellissippi State (423) 539-7198
10915 Hardin Valley Road (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990
From cube-lovers-errors@oolong.camellia.org Thu Jul 17 14:17:15 1997
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Date: Wed, 16 Jul 1997 22:41:08 -0400 (EDT)
From: Jerry Bryan
Subject: Re: No Local Maxima 11q from Start
In-reply-to:
To: cube-lovers@ai.mit.edu
Reply-to: Jerry Bryan
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On Wed, 16 Jul 1997, Jerry Bryan wrote:
> (See my note of 16 June 1995, "10q Local
> Maxima Search Matrix".)
> There seem to be more positions with even
> maximality values than odd, ...
This statement is bogus, which is clear if you look at my chart from 1995.
Most of the positions have a maximality of 1 this close to Start. I was
thinking of another situation. That is, when I looked at local maxima in
the corners only group, the inverses of the local maxima tended to have
even maximality.
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan) jbryan@pstcc.cc.tn.us
Pellissippi State (423) 539-7198
10915 Hardin Valley Road (423) 694-6435 (fax)
P.O. Box 22990
Knoxville, TN 37933-0990
From cube-lovers-errors@oolong.camellia.org Wed Jul 23 16:50:10 1997
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Date: Wed, 23 Jul 1997 06:42:55 -0400
From: Edwin Saesen
Subject: Where can I get...?
To: CUBE
Message-ID: <199707230643_MC2-1B6E-D8FD@compuserve.com>
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Hi everyone,
sorry if this is a boring question for most of you (and if it is, please
reply by private mail instead of to the list), but I'm new here, and my
most important question at the moment is where to get any of the followin=
g
puzzles:
Rubik's Wahn (5x5x5) (maybe also called Professor's cube, Ultimate or
Master Revenge)
Rubik's Revenge (4x4x4)
Rubik's Domino (3x3x2)
Rubik's Pocket Cube (2x2x2)
Pyraminx
Pyraminx Octahedron
Megaminx
I'd prefer sources in germany, although I doubt it that I can find any of=
them here...
Michael
From cube-lovers-errors@oolong.camellia.org Wed Jul 23 17:35:00 1997
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Message-ID: <33D6768B.707E@ibm.net>
Date: Wed, 23 Jul 1997 14:24:27 -0700
From: Jin "Time Traveler" Kim
Reply-To: chrono@ibm.net
Organization: The Fourth Dimension
X-Mailer: Mozilla 3.01Gold (Win95; I)
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To: CUBE
Subject: Re: Where can I get...?
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Edwin Saesen wrote:
>
> Rubik's Wahn (5x5x5) (maybe also called Professor's cube, Ultimate or
> Master Revenge)
> Rubik's Revenge (4x4x4)
> Rubik's Domino (3x3x2)
> Rubik's Pocket Cube (2x2x2)
> Pyraminx
> Pyraminx Octahedron
> Megaminx
>
> I'd prefer sources in germany, although I doubt it that I can find any of
> them here...
>
> Michael
You are quite in luck. There is a great source of puzzles in Germany.
Christoph.Bandelow@rz.ruhr-uni-bochum.de
bandecbv@rz.ruhr-uni-bochum.de
I'm not sure which is correct, or maybe they both work, but Dr.
Christoph Bandelow has a catalog available for people to buy various
puzzles.
He should have all of the above puzzles available for purchase, except
the 4x4x4, 3x3x2, and the Megaminx. All of them are rather hard to
locate. Especially tough is the 4x4x4 because of its high demand. I've
been looking for one as well for a number of years. By many accounts,
there are no more for sale anywhere except maybe private collections.
And if you own one, you're not likely to part with it anyway. So good
luck.
--
Jin "Time Traveler" Kim
chrono@ibm.net
VGL Costa Mesa
http://www.geocities.com/timessquare/alley/9895
http://www.slamsite.com
From cube-lovers-errors@oolong.camellia.org Wed Jul 23 22:49:31 1997
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From: Christoph Bandelow
To: cube-lovers@ai.mit.edu
Date: Thu, 24 Jul 1997 04:04:23 +0000
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Subject: Where can I get...?
Reply-to: Christoph.Bandelow@ruhr-uni-bochum.de
Priority: normal
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Edwin Saesen asked for a source of "Rubik's Wahn" (5x5x5 magic cube)
and some other rotational puzzles.
As our busy "Time Traveler" Jin Kim already kindly remarked, the
5x5x5 Magic Cube and many other rotational puzzles are still
available from me. This refers also to the Megaminx, both the one
made in Hong Kong - sold as Megaminx - and the slightly better one
made in Hungary - sold as Super Nova or as Magic Dodecahedron.
By the way, I do now sell the 5x5x5 cube under the name "Giant Magic
Cube" (In Germany: "Riesen-Zauberwuerfel", in France: "Cube Magique
Geant") , and I hope this sounds nice and doesn't create too much new
confusion. At least the price is still the same: 40 DM or 24 USD.
Just email me your postal address to receive your free copy of my
mail order catalog.
Cube-Lovers: Please notice my new slightly simplified email address.
Christoph
Christoph Bandelow
mailto:Christoph.Bandelow@ruhr-uni-bochum.de
From cube-lovers-errors@oolong.camellia.org Wed Jul 23 22:49:09 1997
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From: Tim Mirabile
To: CUBE
Subject: Re: Where can I get...?
Date: Wed, 23 Jul 1997 23:16:49 GMT
Organization: http://www.webcom.com/timm/
Message-ID: <33d68e99.988371@mail.htp.com>
References: <199707230643_MC2-1B6E-D8FD@compuserve.com> <33D6768B.707E@ibm.net>
In-Reply-To: <33D6768B.707E@ibm.net>
On Wed, 23 Jul 1997 14:24:27 -0700, Jin "Time Traveler" Kim
wrote:
>He should have all of the above puzzles available for purchase, except
>the 4x4x4, 3x3x2, and the Megaminx. All of them are rather hard to
>locate. Especially tough is the 4x4x4 because of its high demand. I've
>been looking for one as well for a number of years. By many accounts,
>there are no more for sale anywhere except maybe private collections.
>And if you own one, you're not likely to part with it anyway. So good
>luck.
Hmmm. I would imagine a large number of the 4x4x4 cubes sold in the U.S. as
"Rubik's Revenge" were sold to non-cube-lovers and are now tucked away in attics
and basements with all the other toys the kids have outgrown. For this reason
I've considered checking out local garage and yard sales. (Before it became
widely known what the value of old baseball cards could be, these sales were an
excellent way for collectors to pick up large boxes of old cards for only a few
dollars).
Anyway, I hope Christoph still has the Megaminx because I just sent him a check
for some items including this. :) There was no indication he was out of them in
the catalog I just got.
--
For USCF & FIDE rated chess on Long Island -> http://www.webcom.com/timm/
TimM on the Free Internet Chess Server - telnet://fics.onenet.net:5000/
Webmaster, tech support - ICD/Your Move Chess & Games: http://www.icdchess.com/
The opinions of my employers are not necessarily mine, and vice versa.
From cube-lovers-errors@mc.lcs.mit.edu Fri Jul 25 12:21:37 1997
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Mail-from: From R.F.Hegge@MP.TUDelft.NL Thu Jul 24 04:40:13 1997
Date: Thu, 24 Jul 1997 10:36:30 +0200
From: R.F.Hegge@MP.TUDelft.NL (Rob Hegge)
Subject: Re: Where can I get...?
To: CUBE-LOVERS@ai.mit.edu
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> Rubik's Wahn (5x5x5) (maybe also called Professor's cube, Ultimate or
> Master Revenge)
> Rubik's Revenge (4x4x4)
> Rubik's Domino (3x3x2)
> Rubik's Pocket Cube (2x2x2)
> Pyraminx
> Pyraminx Octahedron
> Megaminx
>
> I'd prefer sources in germany, although I doubt it that I can find any of
> them here...
Most of them can indeed be bought from C.Bandelow.
The only source for a 4x4x4 that I know of is Puzzletts, where I bought mine
a year ago for about 50 US $. You can see their online mailorder catalog at
www.puzzletts.com. They are based in Seattle, USA.
I myself am still looking for Rubik's Domino (3x3x2).
Rob,
r.f.hegge@ctg.tudelft.nl
PS
If you or anyone else is interested in puzzles like this. There still exist
a club called "Nederlandse Kubus Club" (NKC) or Dutch Cubist Club.
About half of its members are not Dutch, despite its name, but American,
German, Japanese etc. They publish a magazine called Cubism For Fun (CFF)
three times a year, which deals with all kinds of (mechanical) puzzles, like
polyform puzzles, sliding puzzles, rotational puzzles, burrs etc. They also
organise a Cube Day once a year in the Netherlands for members with talks
about puzzles and where new and SECOND HAND puzzles can be traded, bought,
admired etc.
From cube-lovers-errors@mc.lcs.mit.edu Fri Jul 25 12:39:18 1997
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Mail-from: From whuang@ugcs.caltech.edu Thu Jul 24 15:30:43 1997
To: mlist-cube-lovers@nntp-server.caltech.edu
From: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
Subject: Re: Where can I get...?
Date: 24 Jul 1997 19:27:15 GMT
Organization: California Institute of Technology, Pasadena
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Tim Mirabile writes:
>Hmmm. I would imagine a large number of the 4x4x4 cubes sold in the
>U.S. as "Rubik's Revenge" were sold to non-cube-lovers and are now
>tucked away in attics and basements with all the other toys the kids
>have outgrown. For this reason I've considered checking out local
>garage and yard sales.
I suggest any cube-lover who is considering going to local garage
and yard sales to not do so! This way, *I* can get all of the
old cubes and other puzzles!! In fact, I've acquired TWO 4x4x4's
by this method in the last five years, as well as one of Nob
Yoshigahara's Pineapple Puzzles for only fifty cents! The old
lady who sold it to me even cautioned me not to eat it!
(Hmm... I'm not doing a very good job of convincing you guys to
avoid these sales, am I? ;-) )
--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
finger me for /etc/passwd
From cube-lovers-errors@mc.lcs.mit.edu Fri Jul 25 13:54:12 1997
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Mail-from: From pdwyer@ecn.net.au Fri Jul 25 05:16:15 1997
Message-Id: <199707250905.AA10662@ecn.net.au>
From: "Peter Dwyer"
To: "CUBE"
Subject: Re: Where can I get...?
Date: Fri, 25 Jul 1997 18:57:16 +1000
> Hmmm. I would imagine a large number of the 4x4x4 cubes sold in the
> U.S. as "Rubik's Revenge" were sold to non-cube-lovers and are now
> tucked away in attics and basements with all the other toys the kids
> have outgrown. For this reason I've considered checking out local
> garage and yard sales. (Before it became widely known what the
> value of old baseball cards could be, these sales were an excellent
> way for collectors to pick up large boxes of old cards for only a
> few dollars).
I got my 4x4x4 from a fleamarket for $1. I was so exicited when I saw
it I told the lady I would pay anything and she said $1 :-)
Donna
From cube-lovers-errors@mc.lcs.mit.edu Fri Jul 25 20:58:24 1997
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Mail-from: From chrono@ibm.net Fri Jul 25 17:00:49 1997
Message-Id: <33D8EF4F.69C5@ibm.net>
Date: Fri, 25 Jul 1997 11:24:15 -0700
From: "Jin \"Time Traveler\" Kim"
Reply-To: chrono@ibm.net
Organization: The Fourth Dimension
To: CUBE
Subject: Re: Where can I get...?
References: <199707250905.AA10662@ecn.net.au>
Peter Dwyer wrote:
>
> > Hmmm. I would imagine a large number of the 4x4x4 cubes sold in the
> > U.S. as "Rubik's Revenge" were sold to non-cube-lovers and are now
> > tucked away in attics and basements with all the other toys the kids
> > have outgrown. For this reason I've considered checking out local
> > garage and yard sales. (Before it became widely known what the
> > value of old baseball cards could be, these sales were an excellent
> > way for collectors to pick up large boxes of old cards for only a
> > few dollars).
>
> I got my 4x4x4 from a fleamarket for $1. I was so exicited when I saw
> it I told the lady I would pay anything and she said $1 :-)
>
> Donna
>
Indeed, while we are sharing 4x4x4 stories, I got mine (on an "extended"
borrow) while digging around in a friend's garage. (Just hunting for
junk) I found the Rubik's Revenge in a plastic bucket behind some old
paint cans with all teh stickers peeled off. The cube is in great
shape, but the stickers are pretty trashed (had to Krazy Glue them back
on). But at least it works well.
--
Jin "Time Traveler" Kim
chrono@ibm.net
VGL Costa Mesa
http://www.geocities.com/timessquare/alley/9895
http://www.slamsite.com
From cube-lovers-errors@mc.lcs.mit.edu Fri Jul 25 21:02:50 1997
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Mail-from: From tim@mail.htp.com Fri Jul 25 19:32:56 1997
From: tim@mail.htp.com (Tim Mirabile)
To: CUBE-LOVERS@ai.mit.edu
Subject: Re: Where can I get...?
Date: Fri, 25 Jul 1997 23:29:20 GMT
Organization: http://www.webcom.com/timm/
Message-Id: <33dc3688.524425@mail.htp.com>
References: <9707240836.AA12191@sumatra.mp.tudelft.nl>
In-Reply-To: <9707240836.AA12191@sumatra.mp.tudelft.nl>
On Thu, 24 Jul 1997 10:36:30 +0200, R.F.Hegge@MP.TUDelft.NL
(Rob Hegge) wrote:
>The only source for a 4x4x4 that I know of is Puzzletts, where I
>bought mine a year ago for about 50 US $. You can see their online
>mailorder catalog at www.puzzletts.com. They are based in Seattle,
>USA.
I tried to order from them first, in the beginning of June. There was
no response to my order or the followup email I sent.
--
TimM on the Free Internet Chess Server - telnet://fics.onenet.net:5000/
The opinions of my employers are not necessarily mine, and vice versa.
From cube-lovers-errors@mc.lcs.mit.edu Sat Jul 26 12:33:57 1997
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Mail-from: From ERCO@compuserve.com Sat Jul 26 05:11:01 1997
Date: Sat, 26 Jul 1997 05:06:24 -0400
From: Edwin Saesen
Subject: Re: Re: Where Can I get...
To: CUBE
Message-Id: <199707260506_MC2-1B9B-F7B3@compuserve.com>
Hi everyone,
thanks for all of the responses for my original inquiry about those various
cubes.
Rob wrote:
>There still exist
>a club called "Nederlandse Kubus Club" (NKC) or Dutch Cubist Club.
This sounds interesting. Can you tell me how to contact them? And, does
anyone know if there's a similar organization in germany?
Wei-Hwa Huang wrote:
>I suggest any cube-lover who is considering going to local garage
>and yard sales to not do so! This way, *I* can get all of the
>old cubes and other puzzles!!
Well, it seem s to me that all those people in the USA are much more lucky
than I am in germany. I've been going to flea markets for years (not only
for cubes, mainly for buying records), and I've never ever seen a 4x4x4 for
sale there, only lots of 3x3x3s.
Concerning Puzzletts:
>The only source for a 4x4x4 that I know of is Puzzletts,
>I tried to order from them first, in the beginning of June. There was
>no response to my order or the followup email I sent.
I asked a friend of mine in the USA to order one for me, so I simply *HOPE*
that the above described situation isn't their regular way of doing
business. What might help is the fact that one of their retailers is in the
city where my friend lives, so he might have a chance of getting one there.
Concerning my old 4x4x4: Is there any way to fix broken center pieces?
That's the reason why I need a new one, I still have all the pieces but two
of them are broken and I have no idea if there's a way to fix them (I doubt
it, though).
Furthermore, does anyone of you maybe have a broken 4x4x4 and is willing to
sell that one? Maybe I can fix my old one with some of those pieces then.
(But probably they wouldn't fit together if they were done by different
companies).
I also could do with someone willing to sell a working 4x4x4 by the way :-)
Sorry for the length of this
Michael
From cube-lovers-errors@mc.lcs.mit.edu Sun Jul 27 21:30:37 1997
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Mail-from: From nbodley@tiac.net Sun Jul 27 11:09:16 1997
Date: Sun, 27 Jul 1997 11:05:58 -0400 (EDT)
From: Nicholas Bodley
To: Edwin Saesen
Cc: CUBE
Subject: Broken 4^3s; advice on repairs to plastic (medium length)
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Edwin's mention of broken center pieces reminds me of a period roughly 15
years ago when I was quite active disassembling various cubes and their
relatives; I'm fairly sure that it was these center pieces that impressed
me as being fragile. Even the wondrous innards of the 5^3 are not as
delicate, although fascinating.
I'm trying to remember without much success whether I actually broke a
piece. I think I did. Some plastics, I'm just about sure, can't be
dissolved by any readily-available liquids, and i think the 4^3s were made
of such material. (Acrylics can be "solvent-welded" very successfully.)
Plastic welding is possible with probably any thermoplastic (i.e., one
that can be melted after being molded), but is a skill just as is metal
welding, and needs expensive hot-air tools (others also?) designed for the
purpose. I would not recommend it for something so precious as a 4^3,
unless you can find an expert.
Finally, adhesives are worth considering. After all, the colored stickers
are retained by adhesive. Not at all sure, but I think I did have success
with cyanoacrylate (CA) (famous in the USA for its tradename "Krazy
Glue"). This is a strange substance that seems not to be well understood,
and it might be a good idea to learn more about it (even if you already
think you know) before trying a critical repair. At least, if your repair
is unsuccessful it will come apart, with probably little damage (no
promises!) to the surfaces. The adhesive can then be scraped off. (Of
course, you'll let it cure before reassembly...)
As with almost any adhesive, clean surfaces are quite important. 99%
isopropyl alcohol (no longer costly; try a good drugstore) is a worthy
cleaner; rather few, uncommon contemporary plastics are attacked by it
(but the clear printhead drive rack in an Oki printer disintegrated in a
very few minutes, maybe 8 years ago). I doubt that any plastics used for
these sorts of puzzles would be sensitive to alcohol.
Not sure of my information, but I believe that curing of CA is triggered
by absence of oxygen combined with an imperceptibly thin film of water on
the surfaces (don't try to wet them!). Very low humidity might inhibit
curing.
For those who are really serious, the model-builders' magazines have ads
for different versions of CA adhesives; after all, models have been made
of plastics since WW II. Look into (i.e., catalog pages (Web?)) the
products of the Loctite Corp., which makes a variety of industrial
adhesives. There are other companies like Loctite, but Loctite has the
most-successful marketers IMHO. (Note that there's no "k" in "Loctite"!)
By the way, the Pocket Cube (2^3) is a bear to disassemble and even worse
to reassemble. If it weren't for the really-good-quality polymer chosen
for it, it (more than likely) could not be manufactured. The difficulty is
in that the cubies have to be distorted ("sprung") to disassemble it.
Whether this plastic retains its ability over many years to be bent out of
shape but not crack, I don't know!
All of this is offered with the best intentions; if I'm wrong about some
particular, and you're reasonably sure that I am, by all means please let
me know!
The brittle plastics used before WW II, in general, are a different
matter; alcohol is probably a bad idea in general. The modest amount I do
know is really off-topic.
I hope that this isn't too lengthy; this List seems to be willing to
carry long messages at times.
|* Nicholas Bodley *|* Electronic Technician {*} Autodidact & Polymath
|* Waltham, Mass. *|* -----------------------------------------------
|* nbodley@tiac.net *|* When the year 2000 begins, we'll celebrate
|* Amateur musician *|* the 2000th anniversary of the year 1 B.C.E.
--------------------------------------------------------------------------
From cube-lovers-errors@mc.lcs.mit.edu Sun Jul 27 21:32:43 1997
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Mail-from: From nbodley@tiac.net Sun Jul 27 11:32:09 1997
Date: Sun, 27 Jul 1997 11:28:40 -0400 (EDT)
From: Nicholas Bodley
Reply-To: Nicholas Bodley
To: Edwin Saesen
Cc: Cube Mailing List ,
"Dr. Christoph Bandelow"
Subject: Fit of 4^3 pieces; 5^3 query
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(Sorry to have forgotten to include this in the previous message about
repairing plastic.)
Others (such as Drs. Christoph Bandelow or David Singmaster) might know
better than I, but I strongly suspect that there was only one specific
maker, and probably one set of molding dies, for the 4^3. If so, all
pieces regardless of by what path they reached the owner, should fit. A
good, close look, perhaps under a magnifier, would give an initial
judgment about whether a given piece should be trial-assembled. If it has
about the same amount of friction as the others, it really ought to be OK.
The 3^3 most definitely has been made by several different companies from
their own molding dies; interchangeability is by no means assured!
Could someone enlighten me about the 5^3? I got mine from Dr. Uwe
Meffert; I wonder whether Dr. Bandelow's came from the same set of dies?
|* Nicholas Bodley *|* Electronic Technician {*} Autodidact & Polymath
|* Waltham, Mass. *|* -----------------------------------------------
|* nbodley@tiac.net *|* When the year 2000 begins, we'll celebrate
|* Amateur musician *|* the 2000th anniversary of the year 1 B.C.E.
--------------------------------------------------------------------------
From cube-lovers-errors@mc.lcs.mit.edu Sun Jul 27 21:34:49 1997
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Mail-from: From nbodley@tiac.net Sun Jul 27 12:09:36 1997
Date: Sun, 27 Jul 1997 12:06:18 -0400 (EDT)
From: Nicholas Bodley
To: Cube Mailing List
Subject: 4^3 innards: There's a ball in there, but which way does it point?
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Uniquely to all cubes from the Pocket Cube to the 5^3, only the 4^3
(Rubik's Revenge) has a ball inside it. (I do hope that I don't cause
curious people to break their fragile center pieces trying to open up
theirs!) This is really a repetition of some old information, but with the
current interest in 4^3s, it seems sufficiently interesting to repeat it.
However, afaik, the questions about the ball's orientation are probably
new.
The ball (made of at least 9 pieces plus eight screws, as I recall)
consists of a center piece, essentially a sphere (possibly two
hemispheres) with "octants" fastened to it; these have the geometry of
90-90-90-degree right spherical triangles, although in practical detail
they differ. Gaps between these octants define three circumferential
grooves that correspond to the Earth's equator and two orthogonal
meridians of longitude. (All 3 grooves are orthogonal.) (Sorry for the
redundancies; trying to be clear to everybody).
These grooves are "undercut" on one side, so they have an inverted
L-shaped cross-section. The cubies (center ones only, as I recall) have
feet that tuck under the extended edges of the octants' grooves; this
keeps them in place. Machinists know well of the T-slots that work with
clamps to hold work in place on machine tools; these are similar, but the
cubies are free to slide in the L-slots.
If I'm thinking clearly (not too sure!), the ball has a 120-degree
rotational symmetry about an unique diameter. One octant has no undercuts;
its opposite, I think, has all three edges undercut. Other octants have
some edges undercut.
Practical details dictate that when one half of the 4^3 is rotated, one
half is definitely locked to the internal ball. However, you probably
don't know which half it is! (The mathematical folk here might find it fun
to predict which half is the locked half for any given configuration; this
might even not be a trivial problem. (As well, does the locked part always
end up in the same place when the Cube is solved? I suspect so, but am not
sure.) (Anybody for a translucent-cubie 4^3?)
Is it possible to maneuver the internal ball so that it has
effectively revolved by a half turn (or quarter turn?) about any given
axis, while preserving the exterior configuration?
My best regards to all,
|* Nicholas Bodley *|* Electronic Technician {*} Autodidact & Polymath
|* Waltham, Mass. *|* -----------------------------------------------
|* nbodley@tiac.net *|* When the year 2000 begins, we'll celebrate
|* Amateur musician *|* the 2000th anniversary of the year 1 B.C.E.
--------------------------------------------------------------------------
From cube-lovers-errors@mc.lcs.mit.edu Sun Jul 27 21:38:21 1997
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Mail-from: From nbodley@tiac.net Sun Jul 27 20:25:05 1997
Date: Sun, 27 Jul 1997 20:21:46 -0400 (EDT)
From: Nicholas Bodley
Reply-To: Nicholas Bodley
To: Cube Mailing List
Cc: Javier Susaeta ,
Mark Glusker
Subject: Making parts for puzzles (somewhat off-topic) (medium length)
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For reasons I'm not certain of, but probably having to do with a case of
brain fade caused by not eating a proper breakfast (I live alone!), I
posted a short article to the wrong mailing list, no less. It was to an
unmoderated list that goes to people interested in mechanical calculators,
so the few replies I have had so far found it interesting.
I do hope this is not too far off-topic; I have included replies that
contain material that would be helpful to anyone who has any ideas for
making new or replacement parts, and has access to the technologies.
The length is not hopeless...
Subject: Rapid Prototyping and conceivable Cube reissues
(Posted 09:53 AM 27/07/97 -0700 with follow-up (by gracious permission)
from a couple of people who >do< know what they're talking about.)
This is close to being peripheral to the topics of the List, but it might
be worth pointing out that it is now possible by several methods to
"print" solid objects in 3 dimensions from digital data in the proper
format. I'm fairly sure it's even possible to make molds.
These processes are sometimes known as Rapid Prototyping, and (betraying
linguistic ignorance or lack of concern) something like 3-D Lithography.
("Lithography" has roots meaning writing on stone.) (I suggest Solid
Object Synthesis...) The field is still evolving and improving.
The data files for these processes are ordinarily created by CAD
(Computer-Aided Design/Drafting) programs. The programs create the data
for numerous thin slices of an object.
The data for a slice directs the machine to make a solid counterpart, and
the object is created progressively by creating a bonded stack of the
requisite number of slices. (I shouldn't go into how, but one early method
uses a pool of photopolymer and a flat "stage" that progressively moves
deeper into the pool. A bright UV light source positioned by the data
triggers polymerization (solidifies).)
My point is that it is now considerably less expensive and much quicker
than it used to be, to make a prototype of a puzzle design, although the
materials used for this process (afaik) don't have the requisite
durability yet. With copyrights taken care of, it might be possible to
make limited numbers of given designs. It would (at present) be too costly
to make individual puzzles by these processes, but the economics could
change in a decade.
Whether it is possible to pour melted plastic into molds to make the
pieces of a Cube-like puzzle is at present very doubtful, but it might be,
in the future. (Rattlebacks, also known as celts, look just like
solidified poured plastic.)
It's unfortunate that the economics that make possible the affordable
production of large numbers of such puzzles as the 4^3 also mean that
starting another production run has to be economically justified.
Let's hope that the tooling for manufacturing such miracles as the 5^3
and the Magic Dodecahedron is preserved!
Failing that, at least any CAD files for doing numerical machining of the
molding dies really ought to be kept and backed up.
Hope I'm not too far off-topic.
{"afaik" = "as far as I know", pointed out as a courtesy to non-native
users of English}
{END of the text I posted to the wrong List -nb}
* * *
{The following replies are posted with the gracious permission of their
respective authors. }
Here's a reply from Javier Susaeta:
Yes, I have read a bit about rapid prototyping and believe that it has a
tremendous potential. I cannot tell you where I read it, but I remember a
case where a complete intake manifold was designed by CAD/CAM and then a
single copy was (slowly) built, layer by layer, with one of the just-born
rapid-prototyping machines. The material used was aluminium powder plus a
plastic agglomerant.
Once the manifold was "finished" it was baked in an oven in order to
eliminate the agglomerant and weld together the metal particles. The result
was not so solid as a cast aluminium manifold, but nevertheless perfectly
usable.
(Referring to more details, Javier said:)
I remember having read it in internet, a few months ago. The intake
manifold was for a specially-built big diesel of a bulldozer or a similar
machine. I am not sure, but perhaps it was from Caterpillar or a similar
US company. They resorted to rapid prototyping because of costs. It was a
very special engine, and they needed 1 (yes, one) intake manifold. The
fixed costs for a single casting were enormous, and a manifold is so
convoluted that machining it out of a block of metal was impossible. So
they resorted to this new technique.
Regards
Javier Susaeta
Here's some more on the topic, from Mark Glusker:
Date: Sun, 27 Jul 1997 14:33:36 -0700 (PDT)
From: Mark Glusker
This process has improved dramatically over the past few years. The
materials are now quite durable. It is not necessarily cheaper to use
stereo lithography: it is best for small parts with lots of detail-per-
cubic-inch. Simple parts or large parts are best fabricated using
conventional machining methods. If you are making a part to be used as a
master for a mold, don't forget to enlarge the original by several percent
(depending on the final material) to account for molding shrinkage.
Despite the improvements in this field, the best stereo lithography part
is still not nearly as good as a well machined part. It's very much
like the difference between a well printed photograph and a scanned image
printed on a good laserwriter. Used appropriately, it is a great tool
but Bridgeport is not about to go the way of Friden or Monroe.
{Bridgeport is a very-famous maker of milling machines; Friden and Monroe
were once-vital makers of mechanical desktop calculators; they are now
history. -nb}
[Mark also wrote:]
--
Mark Glusker, glusk@sgi.com
>From glusk@mechcad3.engr.sgi.com Sun Jul 27 19:13:05 1997
Date: Sun, 27 Jul 1997 16:04:04 -0700 (PDT)
From: Mark Glusker
I just received some stereo lithography parts last week that were made of
polycarbonate. It is finally possible to use that process for more than
just verifying the shape of objects. However, if you wanted to replicate
a cube (I assume you mean a Rubik-like mechanism) you would need to do
lots of post-finishing to remove the "raster" ridges from the parts so
they will move smoothly against one another. This post-finishing is done
by hand and will certainly affect your final tolerances.
I use ProEngineer (on a Silicon Graphics workstation, naturally) which
can automatically generate an STL file, the standard data format used by
the stereo lithography vendors. There are lots of vendors around here,
and several have relationships with prototype die casters for limited
production runs of parts (5 to 100 pieces, typically).
{At this point, Mark offered help; however, since the help was a
personal offer to me, I don't want to post his comments directly. -nb}
{snip}
... I could go on for quite some time on this subject!
There are similar digital processes for replicating flat metal parts from
a CAD file, with similar economic tradeoffs, in this case related to
perimeter-per-area of part. That would be a great way to replicate a
missing piece of a mechanical calculator,... {Here, Mark's helpful
comments were welcome, but off-topic for this List. -nb}
Regards,
Mark
|* Nicholas Bodley *|* Electronic Technician {*} Autodidact & Polymath
|* Waltham, Mass. *|* -----------------------------------------------
|* nbodley@tiac.net *|* When the year 2000 begins, we'll celebrate
|* Amateur musician *|* the 2000th anniversary of the year 1 B.C.E.
--------------------------------------------------------------------------
From cube-lovers-errors@mc.lcs.mit.edu Sun Jul 27 22:51:19 1997
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Mail-from: From Hoey@AIC.NRL.Navy.Mil Sun Jul 27 22:45:48 1997
Date: Sun, 27 Jul 1997 22:45:37 -0400
Message-Id: <199707280245.WAA12048@sun30.aic.nrl.navy.mil>
From: Dan Hoey
To: cube-lovers@ai.mit.edu
Cc: Nicholas Bodley
Subject: Administrata and a reference
The administrata is that Alan Bawden has asked me to take over
cube-lovers-request for a few weeks, while he is recovering from
surgery. After seventeen years of running cube-lovers, he deserves a
vacation, but I would rather he find someplace more pleasant to spend
it than a hospital. I'm sure we all wish him a speedy recovery.
I will be adding and removing addresses and filtering out abuse, but I
will not be updating the archives or the collection of reader
contributions. As always, send to cube-lovers-request@ai.mit.edu for
administrative services.
The reference is for Nicholas Bodley, who in one of his very
informative messages on Rubik's Revenge raised questions of the
possible orientations achievable by the internal sphere without
changing the exterior. I answered that question in my message on
"Invisible Revenge" on 9 August 1982. The sphere can be placed in any
of 24 orientations, and I showed how to do so. If we consider the
sphere modulo its functional symmetry (fixing one corner of the cube)
we will distinguish only 8 of these orientations. I also mentioned
how to determine which of these 8 orientations the interior sphere is
in on a physical cube, without disassembly. See
ftp://ftp.ai.mit.edu/pub/cube-lovers/cube-mail-4
for details.
Dan Hoey
Hoey@AIC.NRL.Navy.Mil
From cube-lovers-errors@mc.lcs.mit.edu Mon Jul 28 12:11:59 1997
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Mail-from: From CFolkerts@compuserve.com Mon Jul 28 05:22:17 1997
Date: Mon, 28 Jul 1997 05:18:01 -0400
From: Corey Folkerts
Subject: 2^3 Reassembly
To:
Message-Id: <199707280518_MC2-1BB6-48C8@compuserve.com>
My 2^3 burst into pieces while I was playing around with it a while
back. I was amazed and intrigued by the number of internal pieces it
contained; many more than the 3^3. Anyway, after a couple minutes I
got it all put back together, and started playing with it again. One
problem: when I attempted to rotate the cube on one of the axes, it
gave me a lot of resistance. If I continued to force it, the whole
thing burst and was reduced once again to a pile of little black
plastic pieces. After a few more random tests, I examined the pieces
and noticed, as I'm sure many have, that some of the small internal
pieces are slightly different than the others. This fact leads me to
believe that the 'special' pieces need to be oriented correctly with
respect to each other in order for the cube to work correctly. I
would be most appreciative if someone could please inform the manner
in which they need to be placed.
Thanks in advance,
Corey Folkerts
From cube-lovers-errors@mc.lcs.mit.edu Mon Jul 28 12:14:45 1997
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Mail-from: From R.F.Hegge@MP.TUDelft.NL Mon Jul 28 05:30:18 1997
Date: Mon, 28 Jul 1997 11:26:33 +0200
From: R.F.Hegge@MP.TUDelft.NL (Rob Hegge)
Subject: Re: Where Can I get...
To: CUBE-LOVERS@ai.mit.edu
Message-Id: <9707280926.AA23236@sumatra.mp.tudelft.nl>
Sorry,
but I have just been informed that Puzzletts has
been out of 4x4x4's for several years now. I probably
received one of their last.
>I tried to order from them first, in the beginning of June. There was
>no response to my order or the followup email I sent.
Maybe they put your order on their 'wish list' and will contact you
after they obtained one. But given the fact the 4x4x4 are quite rare
now it does not seem likely.
> >There still exist
> >a club called "Nederlandse Kubus Club" (NKC) or Dutch Cubist Club.
> This sounds interesting. Can you tell me how to contact them? And, does
> anyone know if there's a similar organization in germany?
I will look up the address etc and send the info to the list asap.
For germany I would not know.
Rob
From cube-lovers-errors@mc.lcs.mit.edu Mon Jul 28 12:56:42 1997
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Mail-from: From hart@netprofitlc.com Mon Jul 28 11:16:15 1997
Date: Mon, 28 Jul 1997 09:01:14 -0600 (MDT)
From: Paul Hart
To: "Jin \"Time Traveler\" Kim"
Cc: CUBE
Subject: Re: Where can I get...?
In-Reply-To: <33D8EF4F.69C5@ibm.net>
Message-Id:
On Fri, 25 Jul 1997, Jin "Time Traveler" Kim wrote:
> Indeed, while we are sharing 4x4x4 stories, I got mine (on an "extended"
> borrow) while digging around in a friend's garage.
I've got another interesting story, too, about my two 4x4x4 cubes.
Earlier this year, by sheer coincidence, I happened across two Rubik's
Revenge cubes at a local hobby store. Each of the cubes was authentic,
and still in the (unopened) original box. I easily slapped down my $12.95
(USD) for each cube, needlessly to say. :-)
Paul Hart
From cube-lovers-errors@mc.lcs.mit.edu Mon Jul 28 15:20:07 1997
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Mail-from: From chrono@ibm.net Mon Jul 28 14:58:51 1997
Message-Id: <33DCE54F.5463@ibm.net>
Date: Mon, 28 Jul 1997 11:30:39 -0700
From: "Jin \"Time Traveler\" Kim"
Reply-To: chrono@ibm.net
Organization: The Fourth Dimension
To: Cube-Lovers@ai.mit.edu
Subject: Re: 2^3 Reassembly
References: <199707280518_MC2-1BB6-48C8@compuserve.com>
Corey Folkerts wrote:
>
> My 2^3 burst into pieces while I was playing around with it a while
> back. I was amazed and intrigued by the number of internal pieces it
> contained; many more than the 3^3. Anyway, after a couple minutes I
> got it all put back together, and started playing with it again. One
> problem: when I attempted to rotate the cube on one of the axes, it
> gave me a lot of resistance. If I continued to force it, the whole
> thing burst and was reduced once again to a pile of little black
> plastic pieces. After a few more random tests, I examined the pieces
> and noticed, as I'm sure many have, that some of the small internal
> pieces are slightly different than the others. This fact leads me to
> believe that the 'special' pieces need to be oriented correctly with
> respect to each other in order for the cube to work correctly. I
> would be most appreciative if someone could please inform the manner
> in which they need to be placed.
>
> Thanks in advance,
> Corey Folkerts
You have experienced a problem which has led me to purchase a total of
FOUR 2x2x2 cubes. Not even of my own undoing either. In two cases,
friends attempted to play with the cube and disassembled them, and were
unable to properly reassemble them. I have pieces of each in separate
boxes, minus several pieces each. In one case I dropped it, it flew
open, and I DID manage to reassemble it properly. But it ALSO fell
victim to a careless reassembly by a friend who also carelessly
disassembled it. That one I reassembled AGAIN and gave it to someone.
They later told me they "broke" it, which means it's in pieces and since
they live 450 miles away, I can't exactly help them. I have a fourth,
still in its bag, untouched by human hands. Oh yes, and I bought a
fifth one (actually, it was the fourth, so the untouched one is
technically the fifth) but I had to return it and get another one (the
fifth) because it had been disassembled before and reassembled
incorrectly (the pieces only turned on one axis).
--
Jin "Time Traveler" Kim
chrono@ibm.net
VGL Costa Mesa
http://www.geocities.com/timessquare/alley/9895
http://www.slamsite.com
From cube-lovers-errors@mc.lcs.mit.edu Tue Jul 29 10:14:22 1997
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Mail-from: From ERCO@compuserve.com Tue Jul 29 02:39:59 1997
Date: Tue, 29 Jul 1997 02:22:43 -0400
From: Edwin Saesen
Subject: Re: Where can I get...?
To: CUBE
Message-Id: <199707290223_MC2-1BD0-779D@compuserve.com>
Paul wrote:
>I've got another interesting story, too, about my two 4x4x4 cubes.
>I easily slapped down my $12.95
>(USD) for each cube, needlessly to say. :-)
>From all of these stories of almost everyone in the USA either finding two
copies or a single one for around US$1, I take it that they were MUCH more
common in the USA than they were in germany :-(
Rob wrote:
>but I have just been informed that Puzzletts has
>been out of 4x4x4's for several years now. I probably
>received one of their last.
:-((( Ok then, I'm desperate now. Is there ANYONE willing to sell a spare
4x4x4 one? I'll be willing to pay the $50 puzzletts were charging for them
(as I tried to order one from them...).
(Hey Paul, this is *THE* chance for you to get your money back, and have
your own copy of the 4x4x4 for free plus having about $24 leftover to buy
24 copies of the 4x4x4 on your local flea markets...).
But honest, if anyone ever sees one of those for sale somewhere, please get
it for me. I think my chances of ever finding one here in germany are
virtually zero, as I haven't been able to replace my own copy for six years
or so...
Michael
From cube-lovers-errors@mc.lcs.mit.edu Tue Jul 29 10:52:34 1997
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Mail-from: From nbodley@tiac.net Tue Jul 29 03:22:14 1997
Date: Tue, 29 Jul 1997 03:10:32 -0400 (EDT)
From: Nicholas Bodley
To: Corey Folkerts
Cc: Cube-Lovers@ai.mit.edu
Subject: Re: 2^3 Reassembly
In-Reply-To: <199707280518_MC2-1BB6-48C8@compuserve.com>
Message-Id:
My, it's been a while since I opened up mine. I hope I remember. What
perplexes me (I was about to say, "puzzles me") is that I don't recall how
it could be assembled wrong; surely there aren't two internal mechanisms
that differ in some details?
Perhaps it would help if I describe the internal structure (from memory).
The basic structure that I remember is built on a "jack"; it's basically
mutually-orthogonal extensions from a center, so to speak; like a physical
embodiment of the axes of 3-D Cartesian coordinates. What I recall is that
three of these, all adjacent, have a square cross-section; the other three
have a circular cross-section, with a diameter significantly smaller than
a side of the square.
Each of three round projections fits into a hole through a rotating
long, thin square prism. (In the vernacular, square sticks with round
holes through their centers.) In mine, I am just about certain that all of
these had the same length.
Each cubie (all were identical internally) is hollow, but cut away with
concave arcs that allow them to turn with respect to their neighbors.
The cubies are kept together by 12 "clips". These fit into the cutout
arcs; when you assemble the Cube, you put two cubies next to each other
(they touch) and fit this "clip" so that it keep s them together. To
install it, you move the clip away from the imaginary geometrical center
of the whole puzzle.
As I recall them, the "clips" are essentially quadrants (1/4 circles).
They consist of two parallel planes with a gap between them; the sides of
the cubies fit into this gap. The parallel planes are joined at the inner
edges.
When the whole Cube is assembled, the square extensions of the "jack", as
well as the square sticks that turn on the other ends of the jack serve to
keep the clips from moving toward the center of the whole Cube.
This is a structure that could not be either assembled or disassembled if
it were made of rigid materials. It's only because the cubies (at least!)
are made of a strong plastic that has good mechanical spring properties
and can be harmlessly deformed (within limits), that it is possible to
make this structure.
>>> It's conceivable that the cube was misassembled so that one or more
"clips" didn't actually straddle both of its cubies. When assembled, there
should not be any gaps between the cubies, and all movement should be
reasonably free of friction.
I'd love to know how these were assembled in the first place; did the
mfr. have special tools to temporarily deform the parts? Did the
assemblers develop very strong hand muscles?
Btw, it's a challenge to describe the innards in words.
I hope this helps!
|* Nicholas Bodley *|* Electronic Technician {*} Autodidact & Polymath
|* Waltham, Mass. *|* -----------------------------------------------
|* nbodley@tiac.net *|* When the year 2000 begins, we'll celebrate
|* Amateur musician *|* the 2000th anniversary of the year 1 B.C.E.
--------------------------------------------------------------------------
From cube-lovers-errors@mc.lcs.mit.edu Tue Jul 29 14:33:28 1997
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Mail-from: From chrono@ibm.net Tue Jul 29 13:43:28 1997
Message-Id: <33DE2A54.62F2@ibm.net>
Date: Tue, 29 Jul 1997 10:37:24 -0700
From: "Jin \"Time Traveler\" Kim"
Reply-To: chrono@ibm.net
Organization: The Fourth Dimension
To: Cube-Lovers@ai.mit.edu
Subject: Re: 2^3 Reassembly
References:
Nicholas Bodley wrote:
> The cubies are kept together by 12 "clips". These fit into the cutout
> arcs; when you assemble the Cube, you put two cubies next to each other
> (they touch) and fit this "clip" so that it keep s them together. To
> install it, you move the clip away from the imaginary geometrical center
> of the whole puzzle.
>
> As I recall them, the "clips" are essentially quadrants (1/4 circles).
> They consist of two parallel planes with a gap between them; the sides of
> the cubies fit into this gap. The parallel planes are joined at the inner
> edges.
Part of the problem is that the clips weren't all identically shaped.
If all of the clips were shaped the same, then reassembly wouldn't be a
problem. But because they ARE shape differently, there is a question of
whether the position of one is important relative to the position of the
others.
--
Jin "Time Traveler" Kim
chrono@ibm.net
VGL Costa Mesa
http://www.geocities.com/timessquare/alley/9895
http://www.slamsite.com
From cube-lovers-errors@mc.lcs.mit.edu Tue Jul 29 15:10:58 1997
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Mail-from: From CFolkerts@compuserve.com Tue Jul 29 14:09:01 1997
Date: Tue, 29 Jul 1997 13:58:09 -0400
From: Corey Folkerts
Subject: Re: 2^3 Reassembly
To: Cube-Lovers
Message-Id: <199707291358_MC2-1BDB-4183@compuserve.com>
Nicholas Bodley writes:
> The cubies are kept together by 12 "clips". These fit into the cutout
>arcs; when you assemble the Cube, you put two cubies next to each other
>(they touch) and fit this "clip" so that it keep s them together. To
>install it, you move the clip away from the imaginary geometrical center
>of the whole puzzle.
My 2^3 has these 12 clips as I discovered when it first burst. However, I
would like to confirm something. Nine of my clips are identical, the 1/4
circle shape. However, its the other three that are causing me trouble. One
of them is identical to the other nine except that on one of the two planes
it has a very small notch cut out of it. The notch is an arc and I'm
guessing it is probably about 1 mm deep. The other two have one of the 1/4
planes identical to the first nine, but the second plane extends far
beyond, doubling the "height" of the clip. If viewed from the side which
has the extended plane it is a diamond instead of a 1/4 circle. All of the
other internal pieces are identical to your description
I would like to know if everyone else has these altered clips in
their 2^3s.
Corey Folkerts
From cube-lovers-errors@mc.lcs.mit.edu Tue Jul 29 16:24:18 1997
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Mail-from: From whuang@ugcs.caltech.edu Tue Jul 29 14:32:02 1997
To: Cube-Lovers@AI.MIT.Edu
From: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
Subject: Re: Broken 4^3s; advice on repairs to plastic (medium length)
Date: 29 Jul 1997 18:28:24 GMT
Organization: California Institute of Technology, Pasadena
Message-Id: <5rlco8$em3@gap.cco.caltech.edu>
References:
Nicholas Bodley writes:
> By the way, the Pocket Cube (2^3) is a bear to disassemble and even worse
>to reassemble. If it weren't for the really-good-quality polymer chosen
>for it, it (more than likely) could not be manufactured. The difficulty is
>in that the cubies have to be distorted ("sprung") to disassemble it.
>Whether this plastic retains its ability over many years to be bent out of
>shape but not crack, I don't know!
Does anyone know whether the mechanisms for the old 2^3's are the same
as the recent new 2^3 releases?
--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
I hate formication. It should be abolished entirely.
From cube-lovers-errors@mc.lcs.mit.edu Wed Jul 30 10:35:07 1997
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Mail-from: From nbodley@tiac.net Wed Jul 30 08:15:13 1997
Date: Wed, 30 Jul 1997 08:11:44 -0400 (EDT)
From: Nicholas Bodley
To: "Jin \"Time Traveler\" Kim"
Cc: Cube-Lovers@ai.mit.edu
Subject: Re: 2^3 Reassembly
In-Reply-To: <33DE2A54.62F2@ibm.net>
Message-Id:
On Tue, 29 Jul 1997, Jin "Time Traveler" Kim wrote:
}Nicholas Bodley wrote:
}
}> The cubies are kept together by 12 "clips". These fit into the cutout
{Snips}
}Part of the problem is that the clips weren't all identically shaped.
}If all of the clips were shaped the same, then reassembly wouldn't be a
}problem. But because they ARE shaped differently, there is a question of
}whether the position of one is important relative to the position of the
}others.
}
}--
}Jin "Time Traveler" Kim
}chrono@ibm.net
It's not likely that there are two or more designs of the 2^3; sorry if I
misled anyone. I might have been lucky; however, I don't recall needing to
sort the clips.
This >is< something to look out for if you disassemble a 2^3.
It's likely that the clips for the swiveling long blocks would be
different from those for the rigid extensions of the "jack". Less likely
is that the different cavities used to mold several clips at once had
different shapes where the differences had no effect on operation, but
anyone familiar with such mechanisms would be able to tell. The
mechanical engineer on this project has the answers!
Sure hope this isn't memory fade; I'm going on 62...
My best regards to all,
|* Nicholas Bodley *|* Electronic Technician {*} Autodidact & Polymath
|* Waltham, Mass. *|* -----------------------------------------------
|* nbodley@tiac.net *|* When the year 2000 begins, we'll celebrate
|* Amateur musician *|* the 2000th anniversary of the year 1 B.C.E.
--------------------------------------------------------------------------
From cube-lovers-errors@mc.lcs.mit.edu Wed Jul 30 12:26:38 1997
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Mail-from: From nbodley@tiac.net Wed Jul 30 08:21:49 1997
Date: Wed, 30 Jul 1997 08:18:30 -0400 (EDT)
From: Nicholas Bodley
To: Corey Folkerts
Cc: Cube-Lovers
Subject: Re: 2^3 Reassembly
In-Reply-To: <199707291358_MC2-1BDB-4183@compuserve.com>
Message-Id:
As to the notches Corey mentions, they might not matter, but all three
pieces I would guess might of a later design that would simplify assembly;
they are probably the last three pieces to be assembled.
It sounds as though one needs to very careful and make sketches when
disassembling a 2^3! All of mine were bought probably around 1985 or so,
or earlier; they are old.
Please don't think that simply because I'm partly informed that I'm an
expert! I'm not.
Good luck to all,
NB
On Tue, 29 Jul 1997, Corey Folkerts wrote:
}Nicholas Bodley writes:
}
}> The cubies are kept together by 12 "clips". These fit into the cutout
}>arcs; when you assemble the Cube, you put two cubies next to each other
}>(they touch) and fit this "clip" so that it keep s them together. To
}>install it, you move the clip away from the imaginary geometrical center
}>of the whole puzzle.
}
}My 2^3 has these 12 clips as I discovered when it first burst. However, I
}would like to confirm something. Nine of my clips are identical, the 1/4
}circle shape. However, its the other three that are causing me trouble. One
}of them is identical to the other nine except that on one of the two planes
}it has a very small notch cut out of it. The notch is an arc and I'm
}guessing it is probably about 1 mm deep. The other two have one of the 1/4
}planes identical to the first nine, but the second plane extends far
}beyond, doubling the "height" of the clip. If viewed from the side which
}has the extended plane it is a diamond instead of a 1/4 circle. All of the
}other internal pieces are identical to your description
}
} I would like to know if everyone else has these altered clips in
}their 2^3s.
}
} Corey Folkerts
}
}
|* Nicholas Bodley *|* Electronic Technician {*} Autodidact & Polymath
|* Waltham, Mass. *|* -----------------------------------------------
|* nbodley@tiac.net *|* When the year 2000 begins, we'll celebrate
|* Amateur musician *|* the 2000th anniversary of the year 1 B.C.E.
--------------------------------------------------------------------------
From cube-lovers-errors@mc.lcs.mit.edu Wed Jul 30 15:23:06 1997
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Mail-from: From whuang@ugcs.caltech.edu Wed Jul 30 14:13:18 1997
To: Cube-Lovers@AI.MIT.Edu
From: whuang@ugcs.caltech.edu (Wei-Hwa Huang)
Subject: Re: 2^3 Reassembly
Date: 30 Jul 1997 18:09:49 GMT
Organization: California Institute of Technology, Pasadena
Message-Id: <5ro01d$as6@gap.cco.caltech.edu>
References:
Corey Folkerts writes:
>Nicholas Bodley writes:
>> The cubies are kept together by 12 "clips". These fit into the cutout
>>arcs; when you assemble the Cube, you put two cubies next to each other
>>(they touch) and fit this "clip" so that it keep s them together. To
>>install it, you move the clip away from the imaginary geometrical center
>>of the whole puzzle.
>My 2^3 has these 12 clips as I discovered when it first burst. However, I
>would like to confirm something. Nine of my clips are identical, the 1/4
>circle shape. However, its the other three that are causing me trouble. One
>of them is identical to the other nine except that on one of the two planes
>it has a very small notch cut out of it. The notch is an arc and I'm
>guessing it is probably about 1 mm deep. The other two have one of the 1/4
>planes identical to the first nine, but the second plane extends far
>beyond, doubling the "height" of the clip. If viewed from the side which
>has the extended plane it is a diamond instead of a 1/4 circle. All of the
>other internal pieces are identical to your description
> I would like to know if everyone else has these altered clips in
>their 2^3s.
I believe so. Those "special" serve the same purpose as the protruding
octants on the 4^3 internal ball -- to anchor one of the blocks in
each plane. Otherwise, one of the planes of "clips" may be offset
45 degrees (not obvious from the outside), and the other planes
become unturnable. Make sure that one weird clip is in each
plane before assembly.
--
Wei-Hwa Huang, whuang@ugcs.caltech.edu, http://www.ugcs.caltech.edu/~whuang/
---------------------------------------------------------------------------
I hate formication. It should be abolished entirely.
From cube-lovers-errors@mc.lcs.mit.edu Thu Jul 31 17:38:54 1997
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Mail-from: From ponder@austin.ibm.com Thu Jul 31 16:55:58 1997
Date: Thu, 31 Jul 1997 15:52:23 -0500
From: ponder@austin.ibm.com (Ponder)
Message-Id: <9707312052.AA16660@roosevelt.austin.ibm.com>
To: Cube-Lovers@ai.mit.edu
Subject: Rubik's octahedron's etc.
Cc: ponder@austin.ibm.com
I have a Rubik's Cube and a Megaminx Dodecahedron. There
are some tetrahedral puzzles available but they do not
correspond precisely to the Rubik's cube, in that they do
not have well-defined center pieces and the corners are
freely rotating. As far as I can tell, nobody has an octa-
hedron or an icosahedron that works on these principles
either.
Its hard to expect the puzzle companies to come out with
anything like these since they're in it for a profit. I
heard that Meffert's company closed down before they could
produce most of the puzzles they intended. Does anyone
have designs for puzzles like these that could be built in
a machine-shop? (Preferably that you've already patented,
to eliminate any legal concerns!!). I imagine I could try
to hack something together, but it would take an awful lot
of trial-and-error especially since some internal designs
would hold together better than others.
I'm publishing a paper in the Journal of Recreational
Mathematics on solving these other puzzles, but it would
be real nice to have demo models, even if it takes some
work. The Octahedron is particularly interesting because
it forbids edge-flips and it would be more convincing if
I do more than show it on paper.
Thanks,
Carl Ponder
ponder@austin.ibm.com
From cube-lovers-errors@mc.lcs.mit.edu Thu Jul 31 18:52:22 1997
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Mail-from: From ponder@austin.ibm.com Thu Jul 31 17:16:55 1997
Date: Thu, 31 Jul 1997 16:13:34 -0500
From: ponder@austin.ibm.com (Ponder)
Message-Id: <9707312113.AA33486@roosevelt.austin.ibm.com>
To: Cube-Lovers@ai.mit.edu
Subject: puzzle to be simulated
Cc: ponder@austin.ibm.com
I've seen a number of Rubik's Cube simulations on the web,
and was wondering if any of you would be interested in
implementing the following puzzle that I call "the hell-hole".
It has 16 faces that each work like the faces of a rubik's
cube, but:
1] The faces are layed out on a 4x4 grid. Each
"corner" joins four surrounding faces instead
of three.
2] The grid is rolled into a cylinder and then
joined at both ends to forma torus.
However, the torus is given a "twist" when you join the two
ends together, as follows:
1 2 3 4
_ _ _ _
a|_|_|_|_|b
b|_|_|_|_|c
c|_|_|_|_|d
d|_|_|_|_|a
1 2 3 4
First join the 1-2-3-4 sides together to form the cylinder,
then the a-b-c-d ends together to get the torus. Each of the
squares is a 3x3 face like a Rubik's Cube.
The combinatorics get a *lot* messier because of the twist.
Without it, you can't "flip" the edge pieces. With it, you
can, but only by moving the edge-piece in a full-circle around
the torus.
No way to build it, either, since the pieces would need to
flex between convex and concave. It could be simulated on a
computer, however. I have a paper coming out in the Journal of
Recreational Mathematics on how to solve these kinds of things,
and it is pretty messy.
Thanks,
Carl Ponder
ponder@austin.ibm.com
From cube-lovers-errors@mc.lcs.mit.edu Thu Jul 31 21:36:12 1997
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Date: Thu, 31 Jul 1997 21:38:18 -0400
Message-Id: <199708010138.VAA24279@sun30.aic.nrl.navy.mil>
From: Dan Hoey
To: Cube-Lovers@ai.mit.edu, ponder@austin.ibm.com
In-Reply-To: <9707312113.AA33486@roosevelt.austin.ibm.com>
Subject: Re: puzzle to be simulated
Carl Ponder describes his "hell-hole" puzzle that (if I undestand
correctly) has 36 facets laid out like
m : n n n : p p p : r r r : s s s : a
..:.......:.......:.......:.......:..
s : a a a : b b b : c c c : d d d : e
s : a a a : b b b : c c c : d d d : e
s : a a a : b b b : c c c : d d d : e
..:.......:.......:.......:.......:..
d : e e e : f f f : g g g : h h h : j
d : e e e : f f f : g g g : h h h : j
d : e e e : f f f : g g g : h h h : j
..:.......:.......:.......:.......:..
h : j j j : k k k : l l l : m m m : n
h : j j j : k k k : l l l : m m m : n
h : j j j : k k k : l l l : m m m : n
..:.......:.......:.......:.......:..
m : n n n : p p p : r r r : s s s : a
m : n n n : p p p : r r r : s s s : a
m : n n n : p p p : r r r : s s s : a
..:.......:.......:.......:.......:..
s : a a a : b b b : c c c : d d d : e
with opposite boundaries identified so that the letters match up. A
turn rotates 25 facets--one of the 3x3 "faces" marked with dots--and
the sixteen neighboring facets from the neighboring faces.
> The combinatorics get a *lot* messier because of the twist.
> Without it, you can't "flip" the edge pieces. With it, you
> can, but only by moving the edge-piece in a full-circle around
> the torus.
If I've got the puzzle right, you could get edge flippability just by
using a 3x3 array of faces instead of 4x4. Or perhaps 3x5. Another
nice idea that uses a "square" torus is
k : l l l : m m m : n n n : d
..:.......:.......:.......:..
h : a a a : b b b : c c c : j
h : a a a : b b b : c c c : j
h : a a a : b b b : c c c : j j j : k k k : l
..:.......:.......:.......:.......:.......:..
n : d d d : e e e : f f f : g g g : h h h : a
n : d d d : e e e : f f f : g g g : h h h : a
n : d d d : e e e : f f f : g g g : h h h : a
..:.......:.......:.......:.......:.......:..
c : j j j : k k k : l l l : m m m : n n n : d
c : j j j : k k k : l l l : m m m : n n n : d
c : j j j : k k k : l l l : m m m : n n n : d
..:.......:.......:.......:.......:.......:..
f : g g g : h h h : a a a : b b b : c c c : j
This corresponds to a square of side sqrt(13) with opposite edges
identified, cut on the bias into 13 square faces.
............................
:.' a .' `.. m .' `.:
: `.. .' `..' n .':
: `..' b .' `.. .' :
: d .' `.. .' `..' d:
: .' `..' c .' `..:
:`..' e .' `.. .' :
:.' `.. .' `..' j :
:' `..' f .' `.. .:
: k .' `.. .' `..':
:.. .' `..' g .' `:
: `..' l .' `.. .' :
:h .' `.. .' `..' h:
: .' a `..' m .'`.. :
:.'.......'b`.......'.n..`.:
after which each face is cut up into nine facets.
Dan Hoey
Hoey@AIC.NRL.Navy.Mil
From cube-lovers-errors@mc.lcs.mit.edu Fri Aug 1 09:34:41 1997
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Mail-from: From nbodley@tiac.net Thu Jul 31 22:03:44 1997
Date: Thu, 31 Jul 1997 22:00:20 -0400 (EDT)
From: Nicholas Bodley
To: Ponder
Cc: Cube-Lovers@ai.mit.edu
Subject: Re: Rubik's octahedrons etc.
In-Reply-To: <9707312052.AA16660@roosevelt.austin.ibm.com>
Message-Id:
If you were lucky, you could use a good CAD program to define the shapes,
and NC machine tools to produce them; also possible is Rapid Prototyping.
A graphic computer simulation would also be a substitute for a physical
puzzle, although holding one in your hand beats just about any graphics.
Good luck!
|* Nicholas Bodley *|* Electronic Technician {*} Autodidact & Polymath
|* Waltham, Mass. *|* -----------------------------------------------
|* nbodley@tiac.net *|* When the year 2000 begins, we'll celebrate
|* Amateur musician *|* the 2000th anniversary of the year 1 B.C.E.
--------------------------------------------------------------------------
From cube-lovers-errors@mc.lcs.mit.edu Fri Aug 1 10:47:57 1997
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Mail-from: From Goyra@iol.ie Fri Aug 1 09:14:33 1997
Message-Id: <199708011310.OAA00278@mail.iol.ie>
From: "David Byrden"
To:
Subject: Re: Rubik's octahedrons, etc.
Date: Fri, 1 Aug 1997 14:09:19 +0100
> From: Ponder
> I'm publishing a paper in the Journal of Recreational
> Mathematics on solving these other puzzles, but it would
> be real nice to have demo models, even if it takes some
> work. The Octahedron is particularly interesting because
> it forbids edge-flips
Just for you, I have put up a new Java Octahedron at the
Rubik Gallery
http://www.iol.ie/~goyra/Rubik.html
The new one is in the Cousteau Collection and has corner-centred
slices.There is another one in the Plato Collection with face-centred
slices. If the one you are thinking about is deeper or it twists in a
different way, drop me a line and I can probably brew it up for you.
David
From cube-lovers-errors@mc.lcs.mit.edu Fri Aug 1 20:34:49 1997
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Mail-from: From Hoey@AIC.NRL.Navy.Mil Fri Aug 1 20:32:48 1997
Date: Fri, 1 Aug 1997 20:31:56 -0400
Message-Id: <199708020031.UAA28142@sun30.aic.nrl.navy.mil>
From: Dan Hoey
To: Cube-Lovers@ai.mit.edu, ponder@austin.ibm.com
In-Reply-To: <199708010138.VAA24279@sun30.aic.nrl.navy.mil> (message from Dan
Hoey on Thu, 31 Jul 1997 21:38:18 -0400)
Subject: Re: puzzle to be simulated
Continuing on my quest for Ponderesque cube-planes made of a square
torus, possibly cut on the bias, I notice that any square torus will
have k=a^2+b^2 faces for some a>b; if b=0 there is no bias. Carl
Ponder suggests (if I understand it correctly) cutting each face into
nine facets, and permuting the puzzle by turning a face together with
the 16 neighboring facets. As with Rubik's cube, the edge facets move
in pairs (called edge cubies). The corner facets move in quadruplets
(called corner cubies).
There are k corner cubies that can apparently achieve any of four
orientations (twist) each, and 2k edge cubies that can apparently
achieve any of two orientations (flip) each. The face center cubies
can achieve any of four orientations (twist) each but never move.
So the "constructible" group size--the size of the group before we
consider parity and orientation constraints--is k! (2k)! 16^k, or
k! (2k)! 64^k for the supergroup.
But if a+b is even, we can shade the faces in a checkerboard, and the
shades never change when we turn the faces. So in this case, the
edges never flip, and the corners have only two orientations. The
checkerboard-constructible group size is then k! (2k)! 2^k, or
k! (2k)! 8^k for the supergroup.
Everyone who knows Rubik's cube will suspect (and everyone who
understands orientation theory will know!) that the corner
orientations must sum to zero (mod 4) and the edge orientations must
sum to zero (mod 2). If a+b is even, there is only a corner
orientation constraint (mod 2). [See my article of 23 September 1982
for a sketch of orientation theory. Essentially, if the orientation
group of a kind of piece is Abelian then there is an orientation
constraint of the order of that orientation group.]
The permutation parity constraint is also familiar to anyone who knows
the cube. The edge permutation parity must equal the corner
permutation parity, and in the supergroup the parity of the face
center twist must also be equal (mod 180 degrees).
So we should find groups of size k! (2k)! 2^f(k), where
f(k)=4k - 4 a+b odd, or
f(k)= k - 2 a+b even
for the permutation group, and
f(k)=6k - 5 a+b odd, or
f(k)=3k - 3 a+b even
for the supergroup.
I've used GAP to find the group sizes for
(a,b) = (2,0), (2,1), (3,0), (3,1), (3,2), (4,0), (4,1),
and the group sizes agree except for (2,0) and (2,1). The group
is smaller than expected by a factor of
5040 for the (2,0) permutation group,
20160 for the (2,0) supergroup,
6 for the (2,1) permutation group, and
12 for the (2,1) supergroup.
I'm not too surprised about the (2,0) groups (for instance, all four
corner cubies move cyclically on every turn!) but I don't see why
(2,1) is pathological. Maybe it's one of those special group things
that happen for just one permutation group.
By the way, I suggest that a simulation of these should not try to
map them to a curved torus, but to a toroidal tesselation of the
plane. Then when you turn one piece, you see a lattice of other
pieces turning in synchrony.
Dan Hoey
Hoey@AIC.NRL.Navy.Mil
From cube-lovers-errors@mc.lcs.mit.edu Mon Aug 4 10:22:51 1997
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Mail-from: From R.F.Hegge@MP.TUDelft.NL Mon Aug 4 09:38:48 1997
Date: Mon, 04 Aug 1997 15:34:47 +0200
From: R.F.Hegge@MP.TUDelft.NL (Rob Hegge)
Subject: CFF was Re: Where Can I get...
To: Cube-Lovers@ai.mit.edu
Message-Id: <9708041334.AA09170@sumatra.mp.tudelft.nl>
Edwin wrote:
> Rob wrote:
> >There still exist
> >a club called "Nederlandse Kubus Club" (NKC) or Dutch Cubist Club.
> This sounds interesting. Can you tell me how to contact them? And, does
> anyone know if there's a similar organization in germany?
Sorry I do not know of anything similar to NKC in Germany.
I put most of the information about CFF/NKC including some tables of
contents on
http://wwwtg.mp.tudelft.nl/~rob/cff.html
Please don't hesitate to email me if you still have questions.
After finally finding an original Rubik's Cube I have some 3x3x3's left.
What are the most interesting bandaged cubes or other puzzles one can
make of them ?
Rob
From cube-lovers-errors@mc.lcs.mit.edu Mon Aug 4 19:43:16 1997
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Mail-from: From Hoey@AIC.NRL.Navy.Mil Mon Aug 4 19:41:34 1997
Date: Mon, 4 Aug 1997 19:41:24 -0400
Message-Id: <199708042341.TAA09295@sun30.aic.nrl.navy.mil>
From: Dan Hoey
To: Cube-Lovers@ai.mit.edu, ponder@austin.ibm.com
In-Reply-To: <199708020031.UAA28142@sun30.aic.nrl.navy.mil> (message from Dan
Hoey on Fri, 1 Aug 1997 20:31:56 -0400)
Subject: Re: puzzle to be simulated
I've found out why the cube-plane groups related to the 1^2+2^2 square
torus are 1/6 the size we would expect. It's the corners. The group
has five corners {1,2,3,4,5} and five generators {A,B,C,D,E} that
operate on corners as
5..CC/DDD`5 A: (1,2,4,3)
EEE`1..DD/E B: (2,3,5,4)
.EE/AAA`2.. C: (3,4,1,5)
B`3..AA/BBB D: (4,5,2,1)
B/CCC`4..BB E: (5,1,3,2)
5..CC/DDD`5
These generators do not generate the 120-element group S5, rather they
generate a 20-element subgroup known to GAP as
5:4 = A split extension of C5 by C4 or equivalently
H(2^2,5) = .
Neither of these tells me a lot, except that the fact that this group
has index 6 in S5 means that there are six "orbits" of corner
permutations.
Dan
Hoey@AIC.NRL.Navy.Mil
From cube-lovers-errors@mc.lcs.mit.edu Mon Aug 4 21:26:36 1997
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Mail-from: From cubeman@idirect.com Mon Aug 4 20:14:53 1997
Message-Id: <33E6709C.4277@idirect.com>
Date: Mon, 04 Aug 1997 20:15:25 -0400
From: Mark Longridge
To: cube lovers
Subject: Megaminx a.k.a. Supernova
Has anyone ever written a simulation of the Megaminx for the PC?
I'm thinking about Megaminx moves and I'm on the verge of writing
a simulation from scratch (starting with my file for GAP), to
help myself to compose sequences for patterns.
I am particularly interested in processes for the 10-spot and
12-spot.
Ultimately I should write one, something with coarse face movement,
algebraic move entries like ( F+ B- )^12 with whole megaminx
rotations.
This is interesting to me as it lies outside of recorded cube
literature.
Mark
Web Page At: http://web.idirect.com/~cubeman
I've also managed to compile Mike Reid's ANSI C cube solver for
MS DOS using DJGPP. It makes good use of DPMI.
From cube-lovers-errors@mc.lcs.mit.edu Tue Aug 5 10:16:49 1997
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Mail-from: From Goyra@iol.ie Tue Aug 5 06:34:57 1997
Message-Id: <199708051031.LAA16445@mail.iol.ie>
From: "David Byrden"
To: " From: Mark Longridge
> Has anyone ever written a simulation of the Megaminx for the PC?
Mark:
There is a virtual Megaminx at my Rubik Gallery
http://www.iol.ie/~goyra/Rubik.html
It's not written for the PC: it's in Java, so it works on alll the major
kinds of computer. In fact, you don't need to download or install
anything, just use a Java browser and you can play with the
puzzles immediately.
Control is via the mouse. The puzzles are not self-solving
but if anyone wants to write a solver and make use of my graphical
representation of the puzzles, get in touch.
> I am particularly interested in processes for the 10-spot and
> 12-spot.
What exactly are these, I may want to put them in the
Gallery.
David
From cube-lovers-errors@mc.lcs.mit.edu Tue Aug 5 11:42:18 1997
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Mail-from: From Goyra@iol.ie Tue Aug 5 06:34:57 1997
Message-Id:
From: "David Byrden"
To: Cube-Lovers@ai.mit.edu
Subject: Re: Megaminx a.k.a. Supernova
Date: Tue, 5 Aug 1997 11:30:44 +0100
> From: Mark Longridge
> Has anyone ever written a simulation of the Megaminx for the PC?
Mark:
There is a virtual Megaminx at my Rubik Gallery
http://www.iol.ie/~goyra/Rubik.html
It's not written for the PC: it's in Java, so it works on alll the major
kinds of computer. In fact, you don't need to download or install
anything, just use a Java browser and you can play with the
puzzles immediately.
Control is via the mouse. The puzzles are not self-solving
but if anyone wants to write a solver and make use of my graphical
representation of the puzzles, get in touch.
> I am particularly interested in processes for the 10-spot and
> 12-spot.
What exactly are these, I may want to put them in the
Gallery.
David
[ Moderator's note: Sorry for those of you who get this Cube-Lovers
message twice. I accidentally sent a version with mangled headers,
which several mailer daemons refused to process. -- Dan]
From cube-lovers-errors@mc.lcs.mit.edu Tue Aug 5 12:36:40 1997
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Mail-from: From Hoey@AIC.NRL.Navy.Mil Tue Aug 5 12:04:25 1997
Date: Tue, 5 Aug 1997 12:04:08 -0400
Message-Id: <199708051604.MAA13056@sun30.aic.nrl.navy.mil>
From: Dan Hoey
To: Goyra@iol.ie
Cc: cube-lovers@ai.mit.edu
In-Reply-To: <199708051031.LAA16445@mail.iol.ie> (Goyra@iol.ie)
Subject: Re: Megaminx a.k.a. Supernova
Mark Longridge wrote:
> I am particularly interested in processes for the 10-spot and
> 12-spot.
"David Byrden" asked for clarification.
Spot patterns are those in which all the corner and edge cubies agree
with each other, but not necessarily with all the face centers. They
are so named because the non-matching face centers show up as
contrasting spots.
Mark reported some analysis on them on 31 Oct 95, apparently from GAP.
As he reported, there are five conjugacy classes:
0. The identity,
1. The 72-degree twelve-spot,
2. The 144-degree twelve-spot,
3. The 120-degree ten-spot,
4. The 180-degree ten-spot.
The angle given is the displacement of the corners-and-edges ensemble
from the face-centers ensemble. In cases 1 and 2, the rotation is
about an axis through two opposite face centers; in case 3, through
opposite corners; in case 4, through opposite edges.
Of course, there's no reason to expect optimal processes for these
patterns to the same length. Interestingly, while the square of the
72-degree is the 144-degree, it is also the case that the square of
the 144-degree is the 72-degree (up to conjugacy).
It's also interesting to consider star patterns, in which the edges
agree with the face centers, and the corners agree with each other.
These come in the same classes as the spots.
A third type of pattern is a distorted checkerboard, in which the
corners and face centers agree with each other, and the edges agree
with each other. These come in the same classes as well. I had hoped
to find some in which the edges were apparently reflected with respect
to the face centers (as in the Pons Asinorum and the order 6 6-X
patterns on the cube) but they seem to be in the wrong orbit for the
Megaminx.
Dan
Hoey@AIC.NRL.Navy.Mil
From cube-lovers-errors@mc.lcs.mit.edu Tue Aug 5 15:30:18 1997
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Mail-from: From reid@math.brown.edu Tue Aug 5 14:31:16 1997
Message-Id: <199708051827.OAA07000@life.ai.mit.edu>
Date: Tue, 5 Aug 1997 14:33:36 -0400
From: michael reid
To: cube-lovers@ai.mit.edu
Subject: 20f maneuvers for superfliptwist
because of the historical interest in the pattern superfliptwist, i
decided to find all 20f maneuvers for it. this took about 114 hours
of searching, using the symmetry reductions i described in an earlier
message.
by "all" maneuvers, i mean that any 20 face turn sequence for
superfliptwist can be transformed in to one on my list by conjugating
by one of the 24 symmetries that fix superfliptwist, or by inverting
the sequence and conjugating by the cube rotation C_U , and perhaps
again by one of the 24 symmetries. some extra processing by hand
was required to eliminate inverse pairs. hopefully i haven't made
any errors here. the list of maneuvers is
U R F' B U' D' F U' D F L F' L' U R D F U R L (20f, 20q)
U F D L U R' F' R F U' D F U' D' F' B L U R L (20f, 20q)
U R' F' U' F' R' B' D2 R' D R L' B R F B2 R' U' B D' (20f, 22q)
U F L D L F R U2 F U' F' B R' F' R2 L' F D R' D' (20f, 22q)
U R' U' D F' U' F2 B' U L F' R L' U B L B U F R2 (20f, 22q)
U R' U' D F' U' B' R' F' R U F2 R U B L B U F R2 (20f, 22q)
U F L D L F R U2 F U' F' B R' F' L' U' R' U F R2 (20f, 22q)
U F R' F' L' U' R' U' D F' U F2 R U L B L U F R2 (20f, 22q)
U F D L D F R L' U' L F U2 D' F' U' F' B R' F R2 (20f, 22q)
U F D L D F R U2 F R U' R' D' F' U' F' B R' F R2 (20f, 22q)
U R B D2 L B R' D' R' B L' D2 L B' D2 R' F B2 D' R (20f, 24q)
U R' B L2 U' L2 U' B U' L2 D R B D F U2 R' L' B' R' (20f, 24q)
U F B' R' U2 L U' R2 B' L' F2 U' R' D' L2 U D B D' B (20f, 24q)
U R L2 F U F U F L D2 L' D' L U' D F2 B L' F R2 (20f, 24q)
U R L' B' R' F R' F' B' D2 F U B L2 D R U2 B D' B2 (20f, 24q)
U R' U2 D F B' R F' R' F2 R U B U B U R2 L B L2 (20f, 24q)
U R' B R U' L' U2 B' R2 L' D B2 L U' B R F U B L2 (20f, 24q)
U R' D2 B' U' F2 R' D' L' U2 R L B L' B R F B' D' R2 (20f, 24q)
U F D L U F' R U2 B R' L2 U' F2 R' F' L U L' F R2 (20f, 24q)
U F2 R' U' F' R' L2 U B U L' F B' U2 D L U' D' B R2 (20f, 24q)
U F' B' L F B2 U' D L' B U B R' L2 D' B' R' D2 B R2 (20f, 24q)
U R2 B L' U2 B' R' L F2 D F L2 D R' F2 D' R L' U2 B (20f, 26q)
U F R' L D B R2 U2 L2 D' R' D' R L2 U' F L D2 B R2 (20f, 26q)
U F2 L D B' R L2 F' R' F' L2 B2 R2 U F R' L D B R2 (20f, 26q)
U F2 R' L' U F' U' D2 B2 U' B D R' L2 D2 L2 D2 F U2 B' (20f, 28q)
the maneuver that herbert kociemba found is equivalent to the 28q
maneuver.
the two maneuvers that are 20q long can be obtained from one another
by inverting the first, then cyclically shifting the antislice to the
end of the maneuver, and then reorienting.
the first of the 26q maneuvers is quite interesting. it can also be
written as
(U R2 B L' U2 B' R' L F2 D C_UF)^2
where C_UF is a cube rotation about the UF - DB edge axis (as in
bandelow's book).
mike
From cube-lovers-errors@mc.lcs.mit.edu Tue Aug 5 19:12:26 1997
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Mail-from: From Hoey@AIC.NRL.Navy.Mil Tue Aug 5 19:11:22 1997
Date: Tue, 5 Aug 1997 19:11:12 -0400
Message-Id: <199708052311.TAA13218@sun30.aic.nrl.navy.mil>
From: Dan Hoey
To: Cube-lovers@ai.mit.edu
Subject: Glyph patterns
Years ago I thought for a while on a taxonomy of some of the pretty
patterns. Mark's bringing some of them up on the Megaminx has
reminded me of them.
My favorite class of pretty patterns is the "glyph" patterns. These
are the patterns on which each face of the cube has facets of at most
two colors. They include most of the pretty patterns we've discussed
on the list. The "glyphs" here are the partition of the nine facets
into two colors, where we aren't concerned with which colors but with
the partition. I call the part of the glyph that includes the face
center the "figure" and its complement the "ground".
There are only 51 glyphs up to the symmetry of the square, or 70 if we
distinguish chiral pairs. Some of the common ones we have discussed
are blank, X, plus, dot, bar, T, slash, and H. I recall seeing a
cubing book that assigns 26 of the glyphs to letters of the alphabet,
where you try to place all the letters of your favorite six-letter
word on the cube, or something like that.
Classification and analysis of glyph patterns is often simplified by
separating out the corner-glyph from the edge-glyph. There are only 6
each of these subglyphs (up to symmetry), mostly determined by how
many "figure" facets of each type there are.
Name 0 1 2 D 3 4
+-----+-----+-----+-----+-----+-----+
|. .|X .|X X|X .|X X|X X|
Corner | . | . | . | . | . | . |
|. .|. .|. .|. X|. X|X X|
+-----+-----+-----+-----+-----+-----+
| . | X | X | X | X | X |
Edge |. . .|. . .|X . .|. . .|X . X|X . X|
| . | . | . | X | . | X |
+-----+-----+-----+-----+-----+-----+
So a type-2D glyph would have the corner-glyph 2 and the edge-glyph D.
There are two type-2D glyphs, called T and U.
An important subclass of the glyph patterns are the "isoglyphs", which
have the same glyph on all six faces. We've talked about the 6-T,
6-plus, 6-X, 6-H, and 6-spot patterns. Recall that you can twist just
two opposite corners of the cube--I think Hofstadter called this a
boson or something. I was amused to find that there is just one other
6-corner isoglyph of the cube.
Another subclass are the "continuous" glyph patterns, in which the
glyphs on neighboring faces match along the edge. That is to say, a
facet of an edge cubie and an adjacent facet of a corner cubie have
the same color if and only if the other facet of the edge cubie and
the adjacent facet of the corner cubie have the same color. This
matching condition gives the 6-plus patterns much of their charm.
When every cubie of a continuous glyph pattern has either all "figure"
facets or all "ground" facets, we call the pattern a "reassembled"
glyph pattern. In this case, we can envision the cube having been cut
apart into figure and ground cubies and put back together in a
different orientation. Note that the reorientation may include a
reflection, as we see in the Pons Asinorum.
Some of the prettiest reassembled glyph patterns have corner type 4 on
all faces--I call them "path patterns", because you can consider them
a road map going around the cube. In 1981 Dave Ackley found one he
called the "four-way street", which is the unique continuous type-41
isoglyph. If you can find it, you know what I'm talking about.
I've been considering writing a program (or perhaps sparking someone
else's interest in writing a program) to count and classify all the
glyph patterns, possibly by using corner-edge reduction. It might be
interesting to see if there is a set of nine cubes that exhibits all
51 glyphs, or if not what the smallest panglyphic set is.
Dan Hoey
Hoey@AIC.NRL.Navy.Mil
From cube-lovers-errors@mc.lcs.mit.edu Wed Aug 6 11:16:57 1997
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Mail-from: From reid@math.brown.edu Tue Aug 5 22:42:50 1997
Message-Id: <199708060239.WAA23888@life.ai.mit.edu>
Date: Tue, 5 Aug 1997 22:45:08 -0400
From: michael reid
To: cube-lovers@ai.mit.edu
Subject: more patterns at distance 20f
i can also give three more patterns that are exactly 20 face turns from
start. all three are very symmetric; they have 24 symmetries. the
symmetry group is "H" in dan hoey's taxonomy. in particular, all are
local maxima (in the quarter turn metric). see hoey and saxe's
"symmetry and local maxima" (december 14, 1980) for more info about this.
the first pattern is the composition of superfliptwist with pons asinorum.
you may recall that i suggested this pattern to dik winter when he was
looking for positions that couldn't be solved in 20 face turns or less.
he did succeed in solving it in 20f, using kociemba's algorithm, but it
took much longer than most other positions did.
the other two are inverses of one another. they can both be described
as the composition of superfliptwist with 6 H's. however, the patterns
"6 H's" and "superfliptwist" each come in two orientations. therefore,
fix your favorite orientation of 6 H's; now there are two different
orientations of superfliptwist which may be composed. this gives two
distinct patterns, and the positions are inverses.
by symmetry, we may assume that the first face turn of a maneuver for
any of these positions is either U or U2. to confirm that the pattern
superfliptwist . pons asinorum is not within 19f of start, we need to
search the positions
superfliptwist . pons asinorum . U
and superfliptwist . pons asinorum . U2
completely through depth 18f. similarly, for the second pattern,
two complete searches through depth 18f were required. the third
pattern is the same distance from start as is its inverse, so this one
doesn't require further testing.
my optimal solver did not find the minimal maneuvers for these, although
it certainly would have, if i'd let it continue searching long enough.
however, one can find 20f maneuvers using kociemba's algorithm:
superfliptwist . pons asinorum:
D' R' B' L2 U' L B' D' R' D' B2 D2 B' U D2 R2 F2 D' L' B' (20f)
superfliptwist . 6 H's:
B' L2 D B2 R' D2 F' L2 U' L' F' B U' R D' R2 F2 R2 U' D2 (20f)
it would be nice to find a position that was not within 20f of start.
of course, we don't know if any such positions exist. my guess is that
they do, but that's only a hunch. dik winter examined 9000 random
positions and found that they were all within 20f of start. therefore
the positions we're looking for are extremely scarce. i think that
looking at positions with a lot of symmetry seems to be the right way
to approach this. i've tested some of the most symmetric positions,
but each that i examined was solved in 20f or less.
mike
From cube-lovers-errors@mc.lcs.mit.edu Wed Aug 6 12:37:59 1997
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Mail-from: From saxonia@imn.htwk-leipzig.de Wed Aug 6 11:44:11 1997
Date: Wed, 6 Aug 1997 17:40:38 +0200
From: saxonia@imn.htwk-leipzig.de (Ralf Laue)
Message-Id: <9708061540.AA24642@imn.htwk-leipzig.de>
To: Cube-Lovers@ai.mit.edu
Subject: Rubik's Cube World Records
RUBIK'S CUBE WORLD RECORD LIST
------------------------------
I have created a list of Rubik's Cube world records in the WWW with the URL:
http://www.imn.htwk-leipzig.de/~saxonia/list/rubik.html
It is about speed cubing world records and other funny stuff (cube marathon
record etc.)
I would be glad about comments and corrections to this list. (New record
categories are very welcome: Particularly I am interested in the 4x4x4
cube speed solving world record!)
If you do not have access to the WWW, just send me an e-mail, and I will send
you the list by e-mail.
If you have your own WWW site about Rubik's Cube, I would be glad if you
would create a link to my URL.
(The list is a part of my WWW information about unusual world records at
http://www.imn.htwk-leipzig.de/~saxonia/homepage.html )
Very Best Wishes,
Ralf Laue
___________________________________________________________________________
Please excuse me for a delay in replying to your e-mail.
I cannot read my incoming mail daily.
-----------------------------------------------------------------------------
Ralf Laue e-mail: saxonia@imn.htwk-leipzig.de
P. O. Box 80 Read my Homepage about unusual world records:
04181 Leipzig http://www.imn.htwk-leipzig.de/~saxonia/homepage.html
Germany
-----------------------------------------------------------------------------
From cube-lovers-errors@mc.lcs.mit.edu Wed Aug 6 19:25:21 1997
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Mail-from: From Hoey@AIC.NRL.Navy.Mil Wed Aug 6 19:23:53 1997
Date: Wed, 6 Aug 1997 19:10:30 -0400
Message-Id: <199708062310.TAA17135@sun30.aic.nrl.navy.mil>
From: Dan Hoey
To: Cube-lovers@ai.mit.edu
Subject: Reassembled patterns (was Glyph patterns)
I wrote:
> When every cubie of a continuous glyph pattern has either all "figure"
> facets or all "ground" facets, we call the pattern a "reassembled"
> glyph pattern. In this case, we can envision the cube having been cut
> apart into figure and ground cubies and put back together in a
> different orientation....
On second thought, I prefer the definition that a (2-part) reassembled
pattern is one that can be partitioned into two sets of cubies, where
the cubies of each set are in agreement with each other. This
definition differs from the previous in two ways. Reassembled
patterns need not be continuous--"laughter" is a noncontinuous glyph
pattern. And not all continous glyph patterns with figure/ground
cubies meet this definition--e.g. flip the LF and RD edge cubies.
We may also speak of 3-part reassembled patterns, though they are not
necessarily glyph patterns. Are there any particularly nice ones?
Cube-in-a-cube-in-a-cube comes to mind.
Call an "N-part" pattern one that requires cutting into at least N
parts for reassembly. Surely every position can be reassembled from
at most 21 parts, since that's all the pieces there are. Is this
achievable? We could restrict the reorientation of the parts to C,
but in some cases (e.g. pons asinorum) we can manage with fewer parts
if we allow reorienting some of the edges by M. Is there a 20-part
pattern that would require 21 parts if the orientations were
restricted to C?
In the supergroup, can we manage a 24-part position? A 23-part
position that requires 24 parts for C-reorientation?
Dan Hoey
Hoey@AIC.NRL.Navy.Mil
From cube-lovers-errors@mc.lcs.mit.edu Wed Aug 6 22:13:35 1997
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Mail-from: From reid@math.brown.edu Wed Aug 6 22:12:07 1997
Message-Id: <199708070208.WAA08375@life.ai.mit.edu>
Date: Wed, 6 Aug 1997 22:14:21 -0400
From: michael reid
To: cube-lovers@ai.mit.edu
Subject: optimal solver for quarter turns
i have my optimal cube solver working for quarter turns. it seems to
be as effective as the face turn version. some minimal maneuvers
it has found are
cube in a cube in a cube
U' L' U' F' R2 B' R F U B2 U B' L U' F U R F' (20q)
six X's, order 6
F U' L2 F' L' D R U' D L' B U2 F' L' D' F D R (20q)
ron's cube within the cube
F D' F' R D F' R' D R D L' F L D R' F D' (17q)
and it has also confirmed minimality of known maneuvers for several
other patterns.
mike
From cube-lovers-errors@mc.lcs.mit.edu Thu Aug 7 10:59:18 1997
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Mail-from: From reid@math.brown.edu Wed Aug 6 23:51:22 1997
Message-Id: <199708070348.XAA10924@life.ai.mit.edu>
Date: Wed, 6 Aug 1997 23:53:38 -0400
From: michael reid
To: cube-lovers@ai.mit.edu
Subject: superflip requires 24 quarter turns
with my optimal solver, i can show that superflip is exactly
24 quarter turns from start. this was already shown by jerry
bryan, so this confirms his result.
first some history. david plummer gave a 28q maneuver for
superflip on december 10, 1980. apparently there was no
improvement to this until january 1995, when i implemented
kociemba's algorithm for quarter turns. after a lot of searching,
where i specified the initial sequence R' U2 , it found
R' U2 B L' F U' B D F U D' L D2 F' R B' D F' U' B' U D' (24q)
mark longridge notes that this sequence has a remarkable symmetry,
namely that it may be written as
(R' U2 B L' F U' B D F U D' C_-1)^2 ,
where C_-1 denotes central reflection.
later in january 1995, i completed an exhaustive search for superflip
in 20 quarter turns, without finding any maneuvers. i used my quarter
turn version of kociemba's algorithm, which took 29 cpu hours. this
improved the lower bound of the diameter of the cube group to 22q.
the previous lower bound was 21q, obtained by a counting argument.
in february 1995, jerry bryan improved this result to show that superflip
is not within 22 quarter turns, and thus is exactly 24 quarter turns
from start. this also improved the lower bound for the diameter to 24q.
we'll use symmetry to reduce the size of the search space dramatically.
consider three cases for a minimal maneuver for superflip.
1) it contains a half turn (i.e. two consecutive quarter turns of the
same face).
2) it does not contain a half turn, but contains two consecutive turns
of opposite faces.
3) otherwise.
in case 1, as in the face turn situation, we may suppose that the first
three quarter turns are U R2 .
in case 2, by cyclically shifting, we may suppose these two turns are
the first two. if they form a slice (U D') then we may take the
first three quarter turns to be U D' R . if they form an antislice,
then we may take the first three quarter turns to be either U D R or
U D R' .
in case 3, i claim that we may find three consecutive turns of mutually
adjacent faces. otherwise, if the first two faces turned were U and R,
then we'd only be turning U , R , D and L . however, edges cannot
change orientation when only these faces are turned. thus the claim holds,
and by cyclically shifting, we may suppose that these three faces are
U , R and F . by symmetry, we may suppose that they're turned in that
order. now we have eight cases:
U R F U R F' U R' F U R' F'
U' R F U' R F' U' R' F U' R' F'
we can eliminate two of these by using inversion. inverting the case
U' R F gives F' R' U . conjugating this by the appropriate cube
reflection gives U R F' , and these three turns can be cyclically
shifted to the beginning of the maneuver. similarly, the case U' R' F
can be transformed to the case U' R F .
thus ten cases remain. to show that superflip is not within 22q of start,
these cases must be searched through 19q. my program took 22 hours to
searched these completely, and no maneuvers were found.
iw would be nice to know all the minimal maneuvers for superflip. the
branching factor is about 9.37, so an exhaustive search would take about
22 * (9.37)^2 hours, which is about 80 days. this is feasible, but is
definitely a long term project. i've already searched the first case,
(beginning with U R2) which would seem to be the most likely, through 21q.
this took about 147 hours. i expected it to find a lot of maneuvers, but
it only found 4, in two inverse pairs. the first is equivalent to the
maneuver above, and the new one is
U R2 F' R D' L B' R U' R U' D F' U F' U' D' B L' F' B' D' L' (24q)
mike
From cube-lovers-errors@mc.lcs.mit.edu Thu Aug 7 15:46:48 1997
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Mail-from: From reid@math.brown.edu Thu Aug 7 15:46:42 1997
Message-Id: <199708071943.PAA07912@life.ai.mit.edu>
Date: Thu, 7 Aug 1997 15:48:49 -0400
From: michael reid
To: cube-lovers@ai.mit.edu
Subject: composition of superflip and pons asinorum
my optimal cube solver has also found all minimal maneuvers for the
composition of superflip and pons asinorum. this was previously done
by jerry bryan, so the purpose here is to confirm his results.
up to symmetry, there are 10 maneuvers of length 20q, which occur in
5 inverse pairs. they are
U R F D R U' D L' U' D F' B2 R L' D' F' L' B' R' (20q)
U R U F U F B' L' F B' R L' B' R L' U F' U' R' U' (20q)
U R U F D R L' B' R L' F B' L' F B' D F' U' R' U' (20q)
U R D B U R L' F' R L' F' B L' F' B U B' D' R' U' (20q)
U R D B D F' B L' F' B R L' F' R L' D B' D' R' U' (20q)
this agrees exactly with jerry bryan's results.
mike
From cube-lovers-errors@mc.lcs.mit.edu Fri Aug 8 11:22:01 1997
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Mail-from: From kociemba@hrz1.hrz.th-darmstadt.de Fri Aug 8 09:15:32 1997
Message-Id: <33EB1ABE.1DC8@hrz1.hrz.th-darmstadt.de>
Date: Fri, 08 Aug 1997 15:10:22 +0200
From: Herbert Kociemba
To: Dan Hoey , cube-lovers@ai.mit.edu
Subject: Continuous isoglyph patterns
References: <199708052311.TAA13218@sun30.aic.nrl.navy.mil>
Dan Hoey made some proposals concerning 2-colored cube patterns. The
"coninuous"-condition i find especially interesting. I added this
feature to my Cube Explorer program and found exactly 34 continuous
isoglyphs (plus the trivial solution). I don't know if there are any
among them, which have not been described somewhere else before.
Here are generators for the patterns (I searched only about 1 minute for
most generators of the patterns, so there is no claim for the maneuvers
to be optimal):
D F2 D' . R B2 R' D F2 D' R B2 R' (12)
U . L' D' U B2 D U' L' U' (9)
U2 L' B2 . F R' D' R2 D B2 F' U L (12)
D' U . L' R B' F D' U (8)
R2 D L2 U' B2 D' U2 . R' F' U R B' L' D' F L2 B2 R U' (19)
F2 L2 U L2 U' F2 D . B L R' D' U' F' D2 F' R2 F R (18)
B2 U' B2 D' B2 D . L' B2 F' U R D2 R' D' U' F U (17)
L2 D2 B D2 B' D2 B L2 . D B R D2 F L2 D F' R' (17)
B2 U' B2 L2 B2 U2 B2 U' D2 . R U R' D' L U F U' D' L (19)
U R2 D . F' L D2 U2 R' D2 U2 F D' R2 U' (14)
U B2 . L B F' L2 R' B' F D U2 L' B2 U' (14)
D' U . F' U L' R B' U F D' U R' (12)
U' B2 F2 L2 U B2 U' L2 F2 . B' U R' F D' R2 D2 R' F' (18)
F2 R2 D R2 D U F2 D' . R' D' F L2 F' D R U' (16)
D U2 R2 D' U' . R D B2 R2 B2 R2 D B2 D2 R U' (16)
D U2 L2 U R2 U' L2 U . R' B2 L2 F' L2 B' R' F' L D U' (19)
D2 U F2 D' L2 U R2 B2 . R B2 R2 U2 B' L2 D2 R2 D R' U' (19)
D2 R B2 R . F L B' F U' R L' U' F' D2 F' L2 (16)
D' B2 F2 D' U L2 . F' L R' F' D U' R D B2 R (16)
U' R2 F2 U2 . L' D2 B' L2 U' L2 D2 L U2 F' U2 (15)
L2 U2 R' . B' D U' B2 D' R' D L D2 F D U2 L2 (16)
U' F2 U . R U2 R2 U2 R' F' R2 F U' F2 U (14)
U2 R2 F2 U B2 D' . L' F L' F L' F D B2 U (15)
D2 R F2 L' D2 R . B D2 F' L2 U' R' D L F D L' D L' (19)
D' L2 F2 L2 B2 R2 U F2 U2 . L' F R B D R U' L F' U2 F (20)
B2 L2 R2 U B2 R2 D F2 U' . B F U2 R' B2 L2 D U' B' L' R' (20)
U B2 U2 L2 U F2 R2 B2 U' L2 D2 F2 U' . B L2 R2 D2 U2 F' (19)
L2 . R' B2 F2 D2 B2 F2 L2 R2 U2 R' (11)
D U L2 B2 D U' . F' U F' R F2 R' F D' B2 L2 D' U' (18)
L2 U' B2 F2 D . R D F' U' R2 B2 U' B D2 B' F' L U' (18)
D F2 R2 F2 R2 U F2 . R F2 R D2 U' F L' F' L D (17)
B2 R2 F' U2 D2 L2 R2 B . U' L R B' F U D B2 F2 R' F2 (19)
D' L2 R2 D2 B2 F2 U' . R' B' F D' U L R' F2 D2 U2 F' (18)
B2 F2 L2 R2 D2 U2 (6)
If you copy and paste the maneuvers from this message to a text file,
you can load them into Cube Explorer and directly watch the results.
The response to my Cube Explorer 1.0 program showed me, that the
userinterface and the terminology of the program are confusing (if not
to say bad) and some features are missing which should be there.
I almost completed Version 1.5 now. When I put it to my homepage in a
few days, I will send another message.
--Herbert
From cube-lovers-errors@mc.lcs.mit.edu Fri Aug 8 12:47:55 1997
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Mail-from: From ERCO@compuserve.com Fri Aug 8 03:45:46 1997
Date: Fri, 8 Aug 1997 00:52:32 -0400
From: Edwin Saesen __