From matwood@peruvian.cs.utah.edu  Thu Feb 24 14:25:20 1994
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To: cube-lovers@life.ai.mit.edu
Subject: Book - "Simple Solution To Rubik's Cube"
Date: Thu, 24 Feb 94 12:25:17 MST
From: Mark Atwood <matwood@peruvian.cs.utah.edu>

(I sent my request to be added to this mailing list in to
 cube-lovers-request@ai.ai.mit.edu a few days ago and havnt
 heard back. Hope this works..)

I was going thru my stuff a while back and found my two original
Rubik's Cubes, one of which was given to me several years before
they became wildly popular. The solution I learned was in the
book "The Simple Solution To Rubik's Cube", which was a paperback
of about 20-30 pages.

I remember most of the solution steps outlined in the book, (my
hands remember better than my head does), however, I can't find
a copy of the book anywhere, to refresh my memory.

Anyone got a copy or know where I can get one?

..Mark Atwood

From anandrao@hk.super.net  Fri Feb 25 03:32:04 1994
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Date: Fri, 25 Feb 1994 16:18:56 +0800 (HKT)
From: "Mr. Anand Rao" <anandrao@hk.super.net>
Subject: Re: your mail
To: Jan de Ruiter <jandr@xirion.nl>
Cc: cube-lovers@life.ai.mit.edu
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On Fri, 18 Feb 1994, Jan de Ruiter wrote:
> 
> Sorry about not reporting this earlier, but my search for solutions for
> Rubiks Tangle 10x10 confirms the finding of Don Woods: no solutions!
> 
[snip]
> we could re-define the puzzle as follows:
>   find which four pieces to duplicate in order to find solutions for
>   the 10x10.
> If the number of solutions varies depending on the choice, you could
> even add a restriction:
>   find which four pieces to duplicate in order to find a set which has
>   the minimum number of solutions for the 10x10.
        ^^^^^^^
The kind Mr. Rubik has already done that - the minimum is - ZERO!

The revised problem can be solved fairly easily using your program ( I
don't know, though, how long it takes to run to completion for the 10*10
case) - try to place only 99 tiles  out of the 100 given tiles. You may
have several sub-solutions. It is then easy to determine for each of these
sub-solutions which tile you need to complete the 10*10 mosaic. If this
pattern has already been duplicated, i.e. you need THREE numbers of this
tile to find the complete solution, this sub-solution will not work and so
examine the next sub-solution .... Hopefully you find the solution this way.

After running the program for the 99 tiles, the additional time required
to solve the problem defined by you should not be significant because that
would be  a linear process.

Anand.




From sage@world.std.com  Sun Feb 27 20:08:15 1994
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Date: Sun, 27 Feb 1994 20:08:08 -0500 (EST)
From: Meisha n Thompson <sage@world.std.com>
Subject: Puzzle
To: cube-lovers@life.ai.mit.edu
Cc: Meisha n Thompson <sage@world.std.com>
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Please add me to your mailing list. Thank You Meisha Thompson

From xirion!jandr@relay.nl.net  Mon Mar  7 05:38:41 1994
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From: Jan de Ruiter <jandr@xirion.nl>
Date: Mon, 7 Mar 94 11:37:25 +0100
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X-Organization:  Xirion Unix Software & Consultancy bv
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To: cube-lovers@life.ai.mit.edu

To anandrao@HK.Super.NET
Subject: Re: your mail
Cc: cube-lovers@life.ai.mit.edu
In-Reply-To: <Pine.3.07.9402251654.A20452-b100000@hk.super.net>

>
>On Fri, 18 Feb 1994, Jan de Ruiter wrote:
>> 
>> Sorry about not reporting this earlier, but my search for solutions for
>> Rubiks Tangle 10x10 confirms the finding of Don Woods: no solutions!
>> 
>[snip]
>> we could re-define the puzzle as follows:
>>   find which four pieces to duplicate in order to find solutions for
>>   the 10x10.
>> If the number of solutions varies depending on the choice, you could
>> even add a restriction:
>>   find which four pieces to duplicate in order to find a set which has
>>   the minimum number of solutions for the 10x10.
>        ^^^^^^^
>The kind Mr. Rubik has already done that - the minimum is - ZERO!

Correct, but I think you know what I mean: minimum >= 1

>The revised problem can be solved fairly easily using your program ( I
>don't know, though, how long it takes to run to completion for the 10*10
>case)

More than a week

>      - try to place only 99 tiles  out of the 100 given tiles. You may
>have several sub-solutions. It is then easy to determine for each of these
>sub-solutions which tile you need to complete the 10*10 mosaic.

I am sorry, but I have to disagree on this. It is not that simple.

If you managed to place 99 pieces, you have already placed three or even
all four of the duplicated pieces (depending on which one is left over)
If you placed three, there are tree possibilities for the piece we need:
- it is nonexistent (illegal colour combinations): no solution
- it is one of the duplicated pieces: this means two of the four puzzles
  will be identical which is OK, but not so nice, or
- it is any other piece: we found a good solution

If you placed all four duplicated pieces already, any solution you find
will not satisfy the conditions of the puzzle (i.e. precisely four duplicated
pieces).

And in both cases you have not solved:
   which four pieces to duplicate in order to find solutions for the 10x10.
but:
   which four pieces to duplicate in order to find solutions for the 10x10,
   with the restriction that three of them must be identical to any three
   taken from the set of four duplicates given by Rubik.

Solving the puzzle without this restriction requires a different approach.
I was thinking of starting the program with 5 of each (120 pieces), and after
placing the 5th duplicated piece remove the rest of the duplicates from
the remaining pieces, and see if this leads to a solution. As soon as back-
tracking removes the 5th duplicate, all other duplicates must be made
accessible again.

	Jan

From phiscock@ee.ryerson.ca  Mon Mar 14 14:14:44 1994
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From: phiscock@ee.ryerson.ca (Peter Hiscocks)
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Subject: Anyone solved Rubik's Tangle?
To: Cube-Lovers@ai.mit.edu
Date: Mon, 14 Mar 94 14:09:18 EST
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For those who haven't seen it, Rubik's Tangle is a new puzzle to 
drive us all nuts, break up our families, and divert us from
the things we should be working on.

It consists of 25 tiles, which form a 5x5 pattern. On each
tile is a pattern of coloured ropes, the ends of which must
match the ends of the ropes on the adjacent tiles.

Certain clues are evident: the shape of each rope pattern
is the same, there are equal numbers of each colour,
and each tile given a letter label on the back.

Before I waste my life on this, has anyone solved the
problem?

Peter
-- 
Peter Hiscocks                            Phone: (416) 979-5000 Ext 6109
Department of Electrical Engineering      Fax:   (416) 979-5280
Ryerson Polytechnical University, Toronto, Canada


From Don.Woods@eng.sun.com  Mon Mar 14 16:14:51 1994
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From: Don.Woods@eng.sun.com (Don Woods)
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> Before I waste my life on this, has anyone solved the problem?

Yes, it's been solved, and discussed at some length on this group.
However, I haven't seen anyone who claims to have come up with an
"insightful" solution, i.e. one in which you figure out a general
approach that leads to a solution.  All solutions I've heard of
have been found by exhaustive search, often by computer.

	-- Don.


From anandrao@hk.super.net  Mon Mar 14 21:27:38 1994
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Date: Tue, 15 Mar 1994 10:15:06 +0800 (HKT)
From: "Mr. Anand Rao" <anandrao@hk.super.net>
Subject: Re: Anyone solved Rubik's Tangle?
To: Don Woods <Don.Woods@eng.sun.com>
Cc: phiscock@ee.ryerson.ca, Cube-Lovers@ai.mit.edu
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On Mon, 14 Mar 1994, Don Woods wrote:

> > Before I waste my life on this, has anyone solved the problem?
> 
> Yes, it's been solved, and discussed at some length on this group.

True. However, the 10*10 solution where you use all the four tangle
puzzles  to form a 10*10 pattern with matching edges, has been found to be
impossible( Although the puzzle leaflet says that it is solvable). Once
again there is no 'insightful' solution. Someone has posted that he has
seen an intuitive solution which evades his memory for the time being
but will try to recollect what it was ... reincarnation of Fermat's Last
Problem :). There have been some interesting postings in this group on
this topic in the last few weeks and you should read them .

 > However, I haven't seen anyone who claims to have come up
with an
> "insightful" solution, i.e. one in which you figure out a general
> approach that leads to a solution.  All solutions I've heard of
> have been found by exhaustive search, often by computer.
> 
> 	-- Don.
> 

Anand Rao.




From @mitvma.mit.edu:SHERE@SLACVM.BITNET  Tue Mar 15 14:08:27 1994
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Date: Tue, 15 Mar 1994   11:04 -0800 (PST)
From: SHERE%SLACVM.BITNET@mitvma.mit.edu
To: cube-lovers@life.ai.mit.edu
Subject: Mailing List

Hello, would you please add me.. shere@slac.stanford.edu .. to your
mailing list?  I've forgotten some of my key moves and am trying
to brush up.  I've downloaded your archived mail.  Where might I
find a GZ decompression utility?  Anyway, thanks.  Lee

From @mail.uunet.ca:mark.longridge@canrem.com  Sat Apr  2 21:21:55 1994
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To: cube-lovers@life.ai.mit.edu
Reply-To: CRSO.Cube@canrem.com
Sender: CRSO.Cube@canrem.com
Subject: Invariant Shifting
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.733.5834.0C1993C2@canrem.com>
Date: Sat, 2 Apr 1994 20:12:00 -0500
Organization: CRS Online  (Toronto, Ontario)

Something new to stop the drought of cube posts...

Example of Invariant Shifting
-----------------------------

The resultant position generated by process p8 is invariant under
shifting, specifically 2 X on the Left and Right sides.

P8   2 x ORDER 2:

shift
0                              D2 F2 T2 F2 B2 T2 F2 T2
1                           T2 D2 F2 T2 F2 B2 T2 F2
2                        F2 T2 D2 F2 T2 F2 B2 T2
3                     T2 F2 T2 D2 F2 T2 F2 B2
4                  B2 T2 F2 T2 D2 F2 T2 F2
5               F2 B2 T2 F2 T2 D2 F2 T2
6            T2 F2 B2 T2 F2 T2 D2 F2
7         F2 T2 F2 B2 T2 F2 T2 D2

This is the longest process I've found so far. Certainly this property
is not true of all squares group processes. I suspect there are no
processes in the full group with this property (of any significant
length). Perhaps the fact that the L and R faces never rotate will
give some clue on how to generate processes with this property.

Q: Is this the longest such process?

Further Notes on Antipodes in the Square's Group
------------------------------------------------

I just realized some things about sq group antipodes which
I should have seen before...

The closest 2 antipodes can be is 2 square's group moves.
Take the position produced by p66:

p66  Double 4 corner sw L2 B2 R2 F2 L2 F2 T2 R2 (T2 D2 F2 T2) F2 L2 D2

Any turn will reduce this to a position requiring 14 moves. Undoing
this move will regenerate the antipode. No single move can change
position p66 into another antipode, therefore the closest any 2
antipodes can be is 2 moves.

Futhermore any antipode can not be made into a local maximum
which is 14 moves deep with 1 half turn. I will conclude that
there are no local maxima in the square's group that
neighbour each other closer than 2 moves.

-> Mark <-
Email: mark.longridge@canrem.com

From pbeck@pica.army.mil  Mon Apr  4 08:24:04 1994
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Date:     Mon, 4 Apr 94 8:22:52 EDT
From: Peter Beck (BATDD) <pbeck@pica.army.mil>
To: cube-lovers@life.ai.mit.edu
Cc: pbeck@pica.army.mil
Subject:  puzzle boxes
Message-Id:  <9404040822.aa26168@COR6.PICA.ARMY.MIL>

I found a shop in NYC's chinatown that
stocks Japanese puzzle/trick boxes.

  TING'S GIFT SHOP
  18 DOYERS STREET
  NY, NY 10013
  212-962-1081

-  4-way $25
-  4-way w/music $36
-  10-way $42
-  12-way $46
-  20-way $52

NYC sales tax 8%

good puzzling and good eating

From ishius@ishius.com  Mon Apr  4 14:11:59 1994
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To: cube-lovers@ai.mit.edu
From: ishius@ishius.com (Ishi Press International)
Subject: puzzle boxes

Ishi Press International offers a variety of Japanese Trick Puzzle Boxes,
from 4 moves to 66 moves.  These are handcrafted, wood inlaid Okiyama
trickboxes.  We also have unique puzzle boxes by Kamei.

For a free catalog of our PUZZLES please send us your postal mailing address
and we will mail you one.  Please specify that you are interested in PUZZLES.

Always feel free to write me if you have any questions or comments.

Anton Dovydaitis
Customer Support


========================================================================
Ishi Press International        800/859-2086 voice, 408/944-9110 FAX
76 Bonaventura Drive            ishius@ishius.com           The Americas
San Jose, CA  95134             ishi@cix.compulink.co.uk    Europe



From mmoss@panix.com  Mon Apr  4 17:30:58 1994
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From: Matthew Moss <mmoss@panix.com>
Message-Id: <199404042130.AA12590@panix.com>
Subject: About Rubix's tetrahedron...
To: cube-lovers@life.ai.mit.edu (Cube Mailing List)
Date: Mon, 4 Apr 1994 17:30:46 -0400 (EDT)
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Here's a question for y'all.....
It's about the tetrahedron puzzle from Rubix (I forget the real name).
[the one with legal moves consisting of removing a 4-piece tetrahedron from
 a 10-piece and putting it back on in different orientation]

Anyway, mine is pretty loosed up, and occassionally when I am working on it,
one piece will come loose and go skitting across the floor.

There's no way I can remember the orientation it had on there. Has any study
been done or does someone know if I put that piece back on, will it still be
solveable if it's orientation is wrong (ie, different than it was before it
fell off)?

Do you understand my question? (I hope I'm not too confusing...)

I've been thinking about implementing this via computer, just to test this
out, but I thought I'd ask y'all first.

Thanx.

From @mail.uunet.ca:mark.longridge@canrem.com  Sun Apr 10 23:26:50 1994
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Sender: CRSO.Cube@canrem.com
Subject: More Sq Notes
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.738.5834.0C19A249@canrem.com>
Date: Sun, 10 Apr 1994 17:37:00 -0400
Organization: CRS Online  (Toronto, Ontario)

Additional Notes on Squares Group Patterns
------------------------------------------

Note that p80a, p99a and p108a are 2 DOT patterns,
all of the form U1 (swap edges & corners in U and D faces)  D3
           or   U1 (swap edges & corners in U and D faces)  D1
           or   D1 (swap edges within U and edges within D) D3

P66a alternate method    F2 R2 U2 F2 R2 U3 D3 B2 L2 F2 B2 U1 D1
 (13)
p67a alternate method    F2 R2 F2 U3 D3 L2 B2 D2 L2 B2 U1 D1 B2
 (13)

p80a alternate method    U1 F2 R2 L2 U2 D2 F2 U2 D3
  (9)
p99a alternate method    U1 R2 F2 B2 U2 D2 R2 D1
  (8)

P100a alternate method   F2 U2 D2 F2 R2 L2 D1 F2 R2 L2 B2 U1
 (12)
p108a alternate method   R2 F2 B2 L2 D1 R2 U2 R2 L2 U2 R2 D1
 (12)
p130a alternate method   F2 R2 F2 B2 U1 D1 F2 R2 D2 F2 L2 U3 D3
 (13)
p133a alternate method   R2 U1 F2 R2 L2 U2 D2 F2 U2 D3 R2
 (11)

A) In general, any sequence of half turns which swaps edges and
corners in the U and D faces can be sanwiched between a single quarter
turn of U and a single quarter turn of D. Such a process would lead to
a square's group position.

B) Furthermore, any sequence of half turns which swaps edges within U
and
edges within D can be sanwiched between a single quarter turn of U or D
and a single quarter turn of U or D. Once again, such a process would
lead to a square's group position.

Here is an example of a position which takes over twice as many half
turns as full group moves:

        L2 T2 L2 T2 L2 T2 F2 L2 T2 F2 T2 R2 B2     (13)
        U1 F2 R2 L2 B2 D1                           (6)

As discussed in point A above, sequences which move all elements of
the U face to the D face and also move all elements of the D face
to the U face (excepting the centres naturally) appear as 2 DOT
patterns on the cube. This makes sense, as the initial quarter
turn in process p80a must be balanced by another quarter turn.
Since all of the elements subjected to the quarter turn are now
in the D face, we must turn that face a quarter turn to remain
in the squares group.

From @mail.uunet.ca:mark.longridge@canrem.com  Sun Apr 10 23:27:13 1994
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Subject: More Shifting
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.739.5834.0C19A24A@canrem.com>
Date: Sun, 10 Apr 1994 17:39:00 -0400
Organization: CRS Online  (Toronto, Ontario)

More Notes on Invariant Shifting
--------------------------------

Let us define a process as "Shift Invariant" if it results in the same
displacement even after a series of left or right shifts. That is,
from a process of length N we can generate N-1 processes which
result in the same displacement by shifting the process. Sometimes
the processes generated are not all unique!

e.g. P8   2 x ORDER 2: (symmetry level 3)
     D2 F2 T2 F2 B2 T2 F2 T2
  (8)

Q: Is this the longest such process?

A: No.

The following processes are also shift invariant:

2 Swap                   D2 R2 D2 R2 D2 R2 (6)
(symmetry level 12, SI level 2)

p21  2 H                 L2 R2 B2 L2 R2 F2 (6)
(symmetry level 6, SI level 6)

Amazingly, the process p3 (found using Dik Winter's program) is actually
a series of 20 processes which all result in the same displacement!

p3   12 flip             R1 L1 D2 B3 L2 F2 R2 U3 D1 R3 D2 F3 B3 D3 F2 D3
                            R2 U3 F2 D3 (20)
(symmetry level 1, SI level 20)

Since p3 is shift invariant, we can easily shift the 3 consecutive half
turns to the beginning without fear of altering the end result:
  L2 F2 R2 U3 D1 R3 D2 F3 B3 D3 F2 D3 + R2 U3 F2 D3 R1 L1 D2 B3

From bagleyd@source.asset.com  Thu Apr 14 11:21:48 1994
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From: bagleyd@source.asset.com (David A. Bagley)
Message-Id: <9404141500.AA10500@source.asset.com>
To: cube-lovers@life.ai.mit.edu
Subject: Mailing List


Hi I was wondering if you can add  me to your mailing list.

I put some motif puzzles at ftp.x.org in /contrib/motif_puzzles.  They are:
rubik: a (nxnxn) rubik's cube
pyramid: a (nxnxn) pyraminx with period 2 and period 3 cuts
oct: an (nxnxn) octahedron with period 3 and period 4 cuts
skewb: a diagonal cut rubik's cube
cubes, triangles, & hexagons: sliding block puzzles
 
There are no self-solvers provided with these.
Also you may want to check out the tetris games which only use X.
altetris: polyomino version of tetris
alweltris: polyomino version of welltris
altertris: polyiamond version of tetris (they bounce of the walls)
alhextris: polyhexes version of tetris (again, they bounce off the walls)
These are found in ftp.x.org in /contrib

I recently heard that 10x10 tangle has no solution.  I was trying to solve
that one too.  What a waste! (90 days of compute time, wow thats one
efficient program, fast machine , or both!)

Oh, unfortuately the motif puzzles need motif to compile.  Is there a good
public domain substitute for Motif?
Have fun
David

From bagleyd@source.asset.com  Thu Apr 14 13:21:45 1994
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From: bagleyd@source.asset.com (David A. Bagley)
Message-Id: <9404141655.AA64031@source.asset.com>
To: cube-lovers@ai.mit.edu
Subject: Mailing List


Hi, I was wondering if you can add me to your mailing list. 
  
  I put some motif puzzles at ftp.x.org in /contrib/motif_puzzles.
  They are: 
rubik: a (nxnxn) rubik's cube 
pyramid: a (nxnxn) pyraminx with period 2 and period 3 cuts 
oct: an (nxnxn) octahedron with period 3 and period 4 cuts  
skewb: a diagonal cut rubik's cube 
cubes, triangles, & hexagons: sliding block puzzles 
  There are no self-solvers provided with these. A record keeps track of
how many move it takes you to solve them. A record of 32767 means I never
did it. (I do not follow any standard notation for a move (for example,
on the rubik's cube a move is any 1/4 turn)).
  Unfortuately the motif puzzles need motif to compile.  Is there a good
public domain substitute for Motif?
 
Also you may want to check out the tetris games which only use X. 
altetris: polyomino version of tetris 
alweltris: polyomino version of welltris 
altertris: polyiamond version of tetris (they bounce off the walls) 
alhextris: polyhexes version of tetris (again, they bounce off the walls) 
These are found in ftp.x.org in /contrib 
  
I recently heard that 10x10 tangle has no solution.  I was trying to solve 
that one too.  What a waste! (90 days of compute time, wow thats one 
efficient program, fast machine , or both!)                 
 
Have fun
David

From iavlang@cs.vu.nl  Tue Apr 19 03:52:25 1994
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Date: Tue, 19 Apr 1994 08:01:19 +0200
To: cube-lovers@life.ai.mit.edu
From: Izak van Langevelde <iavlang@cs.vu.nl>
Subject: Cubism for Fun
Message-Id:  <9404190801.aa24220@top.cs.vu.nl>

In the archive of this mailing list I found the following:
>CFF is a newsletter published by the Nederlandse Kubus Club NKC (Dutch
>Cubists Club).  It appears a bit irregular, but a few times a year.
>Yearly membership fee is now NLG 25.- (Dutch Guilders) which amounts to
>approximately $ 15.-.  Institutional membership is also possible.
>Information is available from the editor Gerald Maurice

Unfortunaty, the abovementioned editor didn't respond to my email.
Does anyone know whether CFF still exists? Who is the current editor?

                                                        Thanks,
                                                        Izak van Langevelde




From iavlang@cs.vu.nl  Tue Apr 19 09:32:59 1994
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Date: Tue, 19 Apr 1994 10:10:25 +0200
To: cube-lovers@ai.mit.edu
From: Izak van Langevelde <iavlang@cs.vu.nl>
Subject: Cubism for Fun
Message-Id:  <9404191010.aa25902@top.cs.vu.nl>

In the archive of this mailing list I found the following:
>CFF is a newsletter published by the Nederlandse Kubus Club NKC (Dutch
>Cubists Club).  It appears a bit irregular, but a few times a year.
>Yearly membership fee is now NLG 25.- (Dutch Guilders) which amounts to
>approximately $ 15.-.  Institutional membership is also possible.
>Information is available from the editor Gerald Maurice

Unfortunaty, the abovementioned editor didn't respond to my email.
Does anyone know whether CFF still exists? Who is the current editor?

                                                        Thanks,
                                                        Izak van Langevelde




From dik@cwi.nl  Tue Apr 19 18:17:58 1994
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Date: Wed, 20 Apr 1994 00:15:17 +0200
From: Dik.Winter@cwi.nl
Message-Id: <9404192215.AA05363=dik@boring.cwi.nl>
To: cube-lovers@ai.mit.edu, iavlang@cs.vu.nl
Subject: Re:  Cubism for Fun

 > In the archive of this mailing list I found the following:
 > >CFF is a newsletter published by the Nederlandse Kubus Club NKC (Dutch
 > >Cubists Club).  It appears a bit irregular, but a few times a year.
 > >Yearly membership fee is now NLG 25.- (Dutch Guilders) which amounts to
 > >approximately $ 15.-.  Institutional membership is also possible.
 > >Information is available from the editor Gerald Maurice

 > Unfortunaty, the abovementioned editor didn't respond to my email.
 > Does anyone know whether CFF still exists? Who is the current editor?

It ought to work.  Perhaps mail got lost?  Just today I received CFF 33,
a summary of the contents will be forthcoming.

dik

From dik@cwi.nl  Tue Apr 19 18:47:54 1994
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Date: Wed, 20 Apr 1994 00:46:34 +0200
From: Dik.Winter@cwi.nl
Message-Id: <9404192246.AA05401=dik@boring.cwi.nl>
To: cube-lovers@life.ai.mit.edu
Subject: CFF33

Cubism For Fun number 33.

I just received it.  It is dated February 1994, so it is a bit late ;-).
Here a summary of the contents.

1.  Dr. Dragon's Polycons by Bernhard Wiezorke and Jacques Haubrich
    A new, apparently interesting, puzzle from Japan, a lot like
    polyonimos.  Given a rectangular grid you can make pieces from
    the horizontal and vertical lines connecting the points.  The
    writers coin the term 'monocon' for the piece consisting of a
    single line segment, 'dicon' for the two different pieces that
    consist of two connected line segment (one is angled, the other
    not).  Similarly there are 5 'tricons' and 16 'tetracons'.  The
    puzzle consists of 10 of the 16 'tetracons' that must be put on
    a 5x5 rectangular grid.  The authors also look at extensions of
    the puzzle.
2.  The Hollow Pyramid by Jan de Ruiter.
    In a previous issue there was a puzzle about a hollow pyramid made
    up of balls that must be constructed by the 25 different pieces
    that consist of 4 connected balls.  Jan de Ruiter was the first to
    solve the puzzle (with a computer).  Here he explains the program.
3.  Junior Polycubes by Jacques Haubrich.
    Pieces consist of 1 to 4 connected cubes that must be put on a
    6x6 square.  Not so much a puzzle, more like Tangram: create forms.
4.  Folding Puzzles by Leo Links.
    About puzzles where some intricate folding is needed to solve.
5.  Cross Pattern Piling by Dieter Gebhardt.
    A puzzle where you put counters on a square and its four neighbours.
    The goal is to pile up to a common height for all the squares.  The
    article also discusses a modified version where counting is done mod
    2.  Associated with it comes the 24th CFF contest.
6.  Gouge Packing Puzzle by Gaetan Gouge.
    Description of and some elaborations about a packing puzzle.
7.  Spots Puzzle by Harold Cataquet.
    Elaborations about a puzzle from A.L.Hoffman, Puzzles old and new,
    New York, 1920.
8.  Arrow-Minded by Ivan Moscovich.
    Start with a fully-connected hexagon.  Put random arrows on all edges.
    Next add nodes on all intersections.  This gives 19 nodes in all.  The
    problem is to find a Hamiltonian path along the nodes, minding the
    arrows (not always possible).  An original puzzle by the writer.
9.  Prime Pentacube Pakcings by Frits Gobel.
    Start with the pentacube consisting of a square base of four cubes and
    one cube on top of it in a corner.  Is it possible to pack a 5x5x5
    cube with 25 such pentacubes?  What other figures can be packed?
10. Contest 25 by Ekkehard Kuenzell.
    Pack a figure with all 29 different pentacubes.
11. Rubik's Rabbits by Luc de Smet.
    Discussion of this latest by Rubik.
12. Party Impressions by Gerald Maurice.
    Impressions of the Puzzle Party and Cube Day, Augustus 1993 in the
    Netherlands.
Further results of contest 23, a book review by Mark Peters and the first
of a series of columns by Edward Hordern.

Cubism For Fun is a newsletter published by Nederlandse Kubus Club NKC
(Dutch Cubists Club).  Applications for membership to the treasurer:
   Lucien Matthijsse
   Loenapad 12
   3402 EP IJsselstein
   The Netherlands
Membership fee NLG 25.- (about US$ 15.-; add transaction costs).

dik

From SCHMIDTG@beast.cs.hh.ab.com  Tue Apr 19 19:38:21 1994
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Date: Tue, 19 Apr 1994 19:38:17 -0400 (EDT)
From: SCHMIDTG@beast.cs.hh.ab.com
To: cube-lovers@ai.mit.edu
Message-Id: <940419193817.20407598@iccgcc.cs.hh.ab.com>
Subject: Re:  Cubism for Fun and Games

> > In the archive of this mailing list I found the following:
> > >CFF is a newsletter published by the Nederlandse Kubus Club NKC (Dutch
> > >Cubists Club).  It appears a bit irregular, but a few times a year.
> > >Yearly membership fee is now NLG 25.- (Dutch Guilders) which amounts to
> > >approximately $ 15.-.  Institutional membership is also possible.
> > >Information is available from the editor Gerald Maurice
>
> > Unfortunaty, the abovementioned editor didn't respond to my email.
> > Does anyone know whether CFF still exists? Who is the current editor?
>
>It ought to work.  Perhaps mail got lost?  Just today I received CFF 33,
>a summary of the contents will be forthcoming.
>
>dik

Glad to see someone has had some luck with CFF.  I sent some postage stamps
to the address posted in rec.puzzles a few years ago and never heard back.
Also, obtained no response from my email to Longridge.  Geez, I thought at
least I deserved some sort of reply!

-- Greg Schmidt		schmidtg@iccgcc.decnet.ab.com

From SCHMIDTG@beast.cs.hh.ab.com  Tue Apr 19 20:45:05 1994
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Date: Tue, 19 Apr 1994 20:45:03 -0400 (EDT)
From: SCHMIDTG@beast.cs.hh.ab.com
To: cube-lovers@ai.mit.edu
Message-Id: <940419204503.204073c2@iccgcc.cs.hh.ab.com>
Subject: Re: Cubism for Fun and Games (Correction and Apologies!)

>> > In the archive of this mailing list I found the following:
>> > >CFF is a newsletter published by the Nederlandse Kubus Club NKC (Dutch
>> > >Cubists Club).  It appears a bit irregular, but a few times a year.
>> > >Yearly membership fee is now NLG 25.- (Dutch Guilders) which amounts to
>> > >approximately $ 15.-.  Institutional membership is also possible.
>> > >Information is available from the editor Gerald Maurice
>>
>> > Unfortunaty, the abovementioned editor didn't respond to my email.
>> > Does anyone know whether CFF still exists? Who is the current editor?
>>
>>It ought to work.  Perhaps mail got lost?  Just today I received CFF 33,
>>a summary of the contents will be forthcoming.
>>
>>dik
>
>Glad to see someone has had some luck with CFF.  I sent some postage stamps
>to the address posted in rec.puzzles a few years ago and never heard back.
>Also, obtained no response from my email to Longridge.  Geez, I thought at
>least I deserved some sort of reply!
>

Whoops, I think I got my signals crossed here.  I was actually referring to
DOTC Newsletter (Domain of the Cube) not CFF.  I apologize for any misunder-
standing caused by this.


-- Greg Schmidt		schmidtg@iccgcc.decnet.ab.com

From coxj@rpi.edu  Mon Apr 25 11:13:35 1994
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Date: Mon, 25 Apr 1994 11:12:43 -0400
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From: coxj@rpi.edu (Jeffrey M. Cox)
Subject: Unsubscribe
X-Mailer: <PC Eudora Version 1.4b22>

        I would like to be taken off the mailing list.
|   _____       _   _  |               Jeff "The Master" Cox                 |
|     /  __    / ) / ) |                   coxj@rpi.edu                      |
|  / /  /__)  (_  (_   |                                                     |
|  \/   \__  /   /     |   "There's a fine line between clever and stupid"   |


From ronnie@cisco.com  Mon Apr 25 12:14:29 1994
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To: coxj@rpi.edu (Jeffrey M. Cox)
Cc: cube-lovers@life.ai.mit.edu
Subject: Re: Unsubscribe 
In-Reply-To: Your message of "Mon, 25 Apr 1994 11:12:43 EDT."
             <199404251513.LAA15800@mail1.its.rpi.edu> 
Date: Mon, 25 Apr 1994 09:14:05 -0700
From: "Ronnie B. Kon" <ronnie@cisco.com>

>        I would like to be taken off the mailing list.
>|   _____       _   _  |               Jeff "The Master" Cox                 
>|     /  __    / ) / ) |                   coxj@rpi.edu                      
>|  / /  /__)  (_  (_   |                                                     
>|  \/   \__  /   /     |   "There's a fine line between clever and stupid"   

Would that be the line between cube-lovers and cube-lovers-request?

				Ronnie

From alan@parsley.lcs.mit.edu  Mon Apr 25 21:41:35 1994
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From: Alan Bawden <Cube-Lovers-Request@ai.mit.edu>
Sender: Cube-Lovers-Request@ai.mit.edu
To: coxj@rpi.edu
Cc: cube-lovers@life.ai.mit.edu
In-Reply-To: Jeffrey M. Cox's message of Mon, 25 Apr 1994 11:12:43 -0400 <199404251513.LAA15800@mail1.its.rpi.edu>
Subject: Unsubscribe

   Date: Mon, 25 Apr 1994 11:12:43 -0400
   From: coxj@rpi.edu (Jeffrey M. Cox)
	   I would like to be taken off the mailing list.
   |   _____       _   _  |               Jeff "The Master" Cox
   |     /  __    / ) / ) |                   coxj@rpi.edu
   |  / /  /__)  (_  (_   |
   |  \/   \__  /   /     |   "There's a fine line between clever and stupid"

Despite the fact that you stupidly mailed your administrative request to
the entire mailing list, I have removed you from Cube-Lovers.

I thought I would also take this opportunity to remind the rest of you of
the contents of the greeting message I send all new subscribers.
Old-timers who've seen previous versions of this file can amuse themselves
by noticing that this winter we had some of our heaviest traffic ever.

------- Begin Standard Greeting -------

Don't expect to receive any mail anytime soon.  Cube-Lovers is mostly quiet
these days.  Our addresses are Cube-Lovers@AI.MIT.EDU for submissions and
Cube-Lovers-Request@AI.MIT.EDU for administrivia.

Please note that Cube-Lovers-Request is processed by a human being, not a
computer program (such as LISTSERV or Majordomo).  If your request is not
instantly processed, it is because I don't spend my entire life reading my
electronic mail.  I do know how to interpret many of the commands typically
sent to such programs, but I would prefer it if instead you can remember to
address me in complete sentences.

If you are interested in the archives of the Cube-Lovers mailing list:

Using FTP, connect to FTP.AI.MIT.EDU, login as "anonymous" (any password),
and in the directory "pub/cube-lovers" you will find the twelve (compressed)
files "cube-mail-0.gz" through "cube-mail-11.gz".

Archive vital statistics (when uncompressed):

	   File		   From		   To	     Size (bytes)
	   ----		   ----		   --	     ------------
	cube-mail-0	12 Jul 80	23 Oct 80	185037
	cube-mail-1	 3 Nov 80	 9 Jan 81	135719
	cube-mail-2	10 Jan 81	 3 Aug 81	138566
	cube-mail-3	 3 Aug 81	 3 May 82	137753
	cube-mail-4	 4 May 81	11 Dec 82	139660
	cube-mail-5	11 Dec 82	 6 Jan 87	173364
	cube-mail-6	10 Jan 87	13 Apr 90	216733
	cube-mail-7	12 Oct 90	 9 Sep 91	137508
	cube-mail-8	 1 Nov 91	25 May 92	171205
	cube-mail-9	28 May 92	 7 Jan 93	155755
	cube-mail-10	20 Mar 93	 6 Dec 93	171881
	cube-mail-11	 6 Dec 93	18 Feb 94	349772

In addition, the file "recent-mail" contains a copy of the currently active
section of the archive.  (Unfortunately, due to the way mail works here at
the AI Lab, it is not possible to have new mail accumulate directly into
this file, so there may be some delay before a new message arrives here.)
Finally, the file "README" contains the information you are currently reading.

				- Alan

------- End Standard Greeting -------

From 70410.1050@compuserve.com  Fri Apr 29 13:28:11 1994
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Date: 29 Apr 94 13:24:23 EDT
From: Jerry Slocum <70410.1050@compuserve.com>
To: "Cube Lovers @ MIT" <Cube-Lovers@ai.mit.edu>
Subject: Mechanical Puzzles
Message-Id: <940429172422_70410.1050_CHV84-1@CompuServe.COM>

Announcements

An exhibition of "MAZES and PUZZLES" opens May 27 and closes Sept.5 at the 
Museum of Science and Industry in Chicago. It includes a "people maze", 80 
hands-on mechanical puzzles of many types to challenge visitors and 640 
mechanical puzzles of all types and ages that are displayed in 21 cases.
I will send a flyer with details to anyone upon request.

A Directory of Puzzle Collectors (232), Mail Order puzzle sellers (96), Puzzle
periodicals (6), and Retail puzzle stores (147), worldwide, has just been 
published by the non-profit Slocum Puzzle Foundation. Is is available for $10.
postpaid.



                The Slocum Puzzle Foundation

The Slocum Puzzle Foundation was established on August 10, 1993 as a nonprofit
public benefit Corporation. It has been approved by the State of California 
and the U.S. Government as a charitable and educational Foundation.

The purpose of the Foundation is to educate the public on puzzles, their 
history, development, and use in various cultures of the world. The 
Foundation will actively support the use of puzzles for education.

The Foundation will educate the public on puzzles through:

Puzzle exhibitions at museums, libraries, universities, and primary and 
secondary schools, with emphasis on interactive, hands-on puzzles.

Publication of books, compendiums, and research papers on puzzles, and a 
Directory of puzzle collectors.

Supporting and encouraging study and research of the history, development, and
use of puzzles in various cultures of the world.

Supporting and encouraging communication among puzzle experts, educators, 
historians, and the public.

Building and maintaining a collection of puzzles and a library to support 
these activities and to be available for puzzle exhibitions, education, 
research and study.

The first project of the Foundation is to support a Maze and Puzzle exhibition
at the Museum of Science and Industry in Chicago, Illinois.

The Directory is the first publication of the Slocum Puzzle Foundation.

We are interested in suggestions of projects to support exhibitions, 
publications, research and educational activities. 

We would welcome volunteers to help with the activities of the Foundation.

We would welcome donations of puzzles and puzzle literature and financial 
support.

All gifts are deductible from California and Federal income taxes.


Jerry Slocum
Internet:70410.1050@compuserve.com
Address: 257 South Palm Drive, Beverly Hills, CA 90212
Phone:310-273-2270
Fax:310-274-3644

From e0f2m2wm@credit.erin.utoronto.ca  Sat Apr 30 23:59:34 1994
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From: Do Anh Vu <e0f2m2wm@credit.erin.utoronto.ca>
To: cube-lovers@life.ai.mit.edu
Subject: Request to unsubscribe from Cube-Lovers list
Message-Id: <94May1.000231edt.34069@credit.erin.utoronto.ca>
Date: 	Sun, 1 May 1994 00:02:25 -0400

Hi, I would like to be removed from the Cube-Lovers list.
...I'm staying away from my mail for now.

Thanks in advance.

Do Anh

From bagleyd@source.asset.com  Sun May  1 00:00:57 1994
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From: bagleyd@source.asset.com (David A. Bagley)
Message-Id: <9405010338.AA38537@source.asset.com>
To: 70410.1050@compuserve.com, Cube-Lovers@ai.mit.edu
Subject: Re:  Mechanical Puzzles


Mr Jerry Slocum
  I just got your Directory, great job!

  The last order seems mixed up.  I ordered:
     Jug w/ <>'s
     1989 Puzzle Calendar
  I only received Jug w/ Diamonds.  (The directory was from    a previous
order).  I included $29 , I believe.

  Also , thanks for the info on        cube-lovers.
  If this is garbled its because my modem is not working so well.
David Bagley
58 Winsor Place
Glen Ridge NJ O7028

From pbeck@pica.army.mil  Thu May 19 09:09:47 1994
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Date:     Thu, 19 May 94 9:09:30 EDT
From: Peter Beck (BATDD) <pbeck@pica.army.mil>
To: cube-lovers@life.ai.mit.edu
Subject:  rubik's 94 items
Message-Id:  <9405190909.aa08091@COR6.PICA.ARMY.MIL>

NEWer RUBIK'S PUZZLES

There is a line of Rubik's puzzles, currently available that are
distributed by Western Publishing in their line of GOLDEN GAMES.  This
line was to have two new items for 1994.  They are available in europe
but Western has decided not to make them available in the USA.

1 - Rubik's Maze:  six connected cubes that lay in a plane and have
lines drawn on them.  The object is turn the blocks until a continuous
path is constructed.
2 - Rubik's Rabbits:  This looks a magician's top hat.  Looking down
at the hat it is divided into 8 wedges.  The layers(5) are turned
until a rabbit appears in each wedge or in no wedges at all.

PRE-1994 ITEMS
1 - Rubik's Cube 4: standard cube with rubik's likeness on a center
sticker.
2 - RUBIK's Fifteen:  a plunger type sequential motion puzzle
3 - Rubik's Dice:  a hollow cube with holes where the die spots go.
 Internally there are colored sheets of plastic that by flipping the
cube can be made to cover up the holes
4 - Rubik's Tangle:  comes in 4 versions, discussed at length on cube
lovers

MY PRICES:
 - Rubik's Cube 4:     $10
 - RUBIK's Fifteen:   $10
 - Rubik's Dice:        $12
 - Rubik's Tangle:     $8 each or $30 for a set of all 4

CONTINENTAL USA POSTAGE:  $2 for 1 item, $3 for 2 or more
outside of CONTINENTAL USA 20% surface , 40% 1st class.

PETER BECK
54 RICHWOOD PLACE
DENVILLE , NJ 07834

DAYS  201-724-4812


From pbeck@pica.army.mil  Thu May 19 09:12:09 1994
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Date:     Thu, 19 May 94 9:11:55 EDT
From: Peter Beck (BATDD) <pbeck@pica.army.mil>
To: cube-lovers@life.ai.mit.edu
Subject:  PIONEER PUZZLES
Message-Id:  <9405190911.aa08959@COR6.PICA.ARMY.MIL>


I was wondering if anybody has bought puzzles from:

PIONEER PUZZLES
POB 183
CHEROKEE, TEXAS 76832
1800-441-1796.

They make wire disentanglement puzzles.

thanks 

pete

From @wvnvm.wvnet.edu:BRYAN@WVNVM.WVNET.EDU  Mon May 23 16:51:56 1994
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Date:      Mon, 23 May 1994 11:00:24 EDT
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Modelling Centerless Cubes

On 13 Feb 1994, I proposed a way to model centerless cubes which would
(in Dan Hoey's words) retain the symmetrical nature of the problem.
I need to post a partial correction/retraction.

The conventional model for centerless cubes loses the symmetrical
nature of the problem.  For example, for a corners-only cube, seven
cubies are modeled rather than eight, and for an edges-only cube,
eleven cubies are modeled rather than twelve.

My proposal in February was to use cosets of the form xC to model
centerless cubes, where x is a cube and where C is the set of
twenty-four whole cube rotations.  This proposal in turn requires
an interpretation of C such that C is a subset of G, the entire cube
group.

C is a group, but normally it is not considered to be a subset
of G, hence it is not normally considered to be a subgroup of G.
That is, C moves the centers of the faces, but G does not.  The
required interpretation is obtained by removing the centers of each
face, and defining rotational orientation by convention so that the
cube is solved only when the Up color is Up, the Front Color is
Front, and so forth.

Under this interpretation, C is indeed a subset (and hence a subgroup)
of G.  More correctly, C[even] is a subset of G, C is a subset of
GC (the corners only cube), C is a subset of GE (the edges only cube),
and C is a subset of GS, where GS=<Q,S> (Q is the set of quarter turns
and S is the set of slice moves).  That is, when you start
talking about C as a subset of G, you have to worry about odd and
even permutations.  Hence, you have to say C is a subset of GS or
C[even] is a subset of G in order not to violate parity rules.

All of the above was posted in February, and I am still comfortable
with it.  However, I went on to say that GS/C, G/C[even], GC/C, and
GE/C were all groups under the operation xC * yC = (xy)C.  I find
that I must retract this claim.

In my note in February, I did not give a proof, but rather appealed
to a proof in Frey and Singmaster's _Handbook of Cubik Math_.  I now
find that I mis-applied their proof.  In order to show the nature of
the problem, I find it useful to go through an attempted proof and
determine the point at which it fails.

Note that the proposed group elements are not cubes, they are cosets.
We proceed as follows:

  1. Associativity:  (xC * yC) * zC =
                     (xy)C * zC =
                     ((xy)z)C =
                     (x(yz))C =
                     xC * (yz)C
                     xC * (yC * zC)

     Note that the associativity of the proposed group G/C derives
     directly from the associativity of G.

  2. Identity: we propose that the identity is iC

                     iC * xC = (ix)C = xC
                     xC * iC = (xi)C = xC

     Note that the identity of the proposed group G/C derives
     directly from the identity i of G.  Further note that the
     identity iC of the proposed group G/C is C, which is
     precisely what you would want for the identity of a centerless
     cube.

  3. Inverse: we propose that (xC)'=x'C

                     xC  * x'C = (xx')C = iC
                     x'C * xC  = (x'x)C = iC

     Note that the inverse of xC in the proposed group G/C derives
     from the inverse of x in G.

  4. Closure: This is where we have our problem.  We require that
     if xC * yC = (xy)C, then (xy)C must be a coset of C.  But the
     representation of xC and yC is not unique.  That is, xC=(xd)C,
     where d is in C, and yC=(ye)C where e is in C.  It is the
     case that (x(ye))C = (xy)C, but in general it is not the case
     that ((xd)y)C = (xy)C.  Hence, we can have xC=(xd)C, but have
     it be the case that xC * yC is not equal (xd)C * yC.  Hence,
     we do not have closure.

     Strictly speaking, this same problem afflicts our "proof" for
     the inverse, but I deferred discussing the problem until I got
     to closure.  If the problem is repaired for closure, it is also
     repaired for inverses (see the next paragraph for a discussion
     of normal subgroups).

Cosets of a subgroup H are said to be normal if xH = Hx for all x.
I was implicitly and incorrectly assuming that C is a normal
subgroup of G, but it is not.  For normal subgroups, closure of
coset multiplication is readily proven.  Frey and Singmaster's proof
is for normal subgroups only, and I was applying it to C, which is
not normal.

It is instructive to consider briefly what xC vs. Cx means for cubes.
We can interpret the left coset xC as simply holding a cube in your
hands and rotating it any way you wish in space without performing
any twists.

The right coset Cx is a little trickier.  The cube x
must be pre-multiplied by each c in C to form Cx.  If you have cube
x in your hands, there is no obvious thing you can do to form Cx.
The thing that is most intuitive to me personally is to think in
terms of "rotation by color", which is the way I described
pre-multiplication when I first posted some of the results of my
work back in December.  That is, think of holding the cube still, but
recoloring it by permuting the colors.  The elements of the coset Cx
look the "same" but with the colors permuted.  It is not possible to
perform this operation on a real cube (short of pulling off the stickers
and putting them back on), but the operation can be readily performed
on a computer model.

Having said all this, I keep thinking that there must be a way to define
an operation on the cosets xC so that they form a group.  However, I
have been unsuccessful in doing so.  I would welcome any advice from
the net.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From ishius@ishius.com  Thu May 26 14:32:42 1994
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Date: Thu, 26 May 1994 11:28:21 -0800
To: cube-lovers@ai.mit.edu
From: ishius@ishius.com (ishius@holonet.net)
Subject: ISHI PRESS MAILING LIST!

ISHI PRESS MAILING LIST!

Earlier, I have intruded on this fine discussion to solicit e-mail addresses
for Ishi Press's retail puzzle e-mail list.  Due to some unforseen hardware
problems, however, we were cut off from the net for a couple weeks, and
have lost all of our e-mail lists.

If you would like to receive e-mailings of Ishi Press's impressive line
of mechanical puzzles, send us your e-mail address.  If you would like
a full color catalog, including puzzle reference guide, send us your
postal address as well.

PLEASE INDICATE THAT YOU ARE INTERESTED IN PUZZLES AND/OR GO (an ancient
oriental strategy game).

Retail E-mailings are sent about once per month, so we won't be stuffing
your e-mail box with more junk to slog through.  While some e-mailings
duplicate our paper sales literature, there are often descriptions, reviews,
and offers that we would never include in a general mailing (such as damaged
or second merchandise, unique items, out of print books).

Always feel free to write me if you have any questions or comments.

Anton Dovydaitis
Customer Support

===========================================================================
Ishi Press International                408/944-9900 vc, 408/944--9110 FAX
76 Bonaventura Drive                    800/859-2086 Toll Free Order Line
San Jose, CA  95134                     ishius@ishius.com (or @holonet.net)



From bagleyd@source.asset.com  Fri May 27 13:46:56 1994
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Date: Fri, 27 May 1994 13:42:15 -0400
From: bagleyd@source.asset.com (David A. Bagley)
Message-Id: <9405271742.AA64537@source.asset.com>
To: cube-lovers@ai.mit.edu
Subject: xrubik


Hi
  I just finished up "xrubik", a UNIX-X Rubik's cube.  It has been tested
on Linux, SunOS, and HP-UX.  It currently resides on ftp.x.org at
/contrib/games/puzzles.  Here's the blurb from the README in that 
directory:
-------------------------------------------------------------
xrubik has been converted from xmrubik.
  New features:
        hold down control key to move whole cube
        letters that represent colors can now be changed in mono-mode
  Bug fix:
        when xmrubik did not recognize when the cube was solved nontrivially
        (i.e. the number of cubes on an edge > 1  OOPS.)
 
The /R5contrib/xmpuzzles:
xmpyramid       xmoct           xmskewb
xmcubes         xmtriangles     xmhexagons
are currently being changed to exclude Motif dependencies.
The Motif versions will no longer be maintained.
 
The proposed collection includes:
SLIDING BLOCK PUZZLES
xcubes:         expanded 15 puzzle
xtriangles:     same complexity as 15 puzzle
xhexagons:      2 modes: one ridiculously easy, one harder than 15 puzzle
 
ROTATIONAL 3D PUZZLES
xrubik:         a nxnxn rubik's cube
xpyramid:       a nxnxn tetrahedron (a nxnxn pyraminx)
                with Period 2, Period 3, and Combined cut modes
xoct:           a nxnxn octahedron with Period 3, Period 4, and Combined cut
                modes
xskewb:         a cube with diagonal cuts
The rest of the platonic solids (the dodecahedron and the icosahedron) seem
too hard for me.
 
These programs do not have self-solvers like "magiccube" (Motif) or
"puzzle" (X).
-----------------------------
Have fun
David (the newbie)

From @wvnvm.wvnet.edu:BRYAN@WVNVM.WVNET.EDU  Mon May 30 22:48:07 1994
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Date:      Mon, 30 May 1994 21:36:07 EDT
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Branching Factors and God's Algorithm Search Trees

At various times, there have been discussions about what the maximum
distance from Start might be in God's algorithm.  One argument is made
with respect to worst/best case branching factors.  For example,
a simple calculation is that the first move has at most twelve
possibilities and that each subsequent move has eleven possibilities,
when dealing with Q-turns only.  For Q-turns plus H-turns, the same
argument would be eighteen possibilities for the first move and seventeen
possibilities for each subsequent move.

My experience is that search trees tend to develop relatively
constant branching factors after some sort of variable startup.
I expect Rubik's cube to be no different.  I just wonder if anyone
has calculated some number of levels for the full Rubik's cube,
enough levels for the hypothesized steady state branching factor
to be achieved.  I have not done so, but if anyone has, it might
shed considerable light on the question of the maximum distance
from Start.

Subsets of the cube such as corners only and edges only have been
calculated.  It is suggestive to examine branching factors for the
cases which have already been calculated.  The question of "average
branching factor" is subject to interpretation because it is not
necessarily clear when the distribution has achieved its steady
state.  I am including a number of tables giving branching factors
for the cases which have been calculated already.  I will preface
the tables with the following comments:

   1. The distributions for edges-only cubes have a variable branching
      factor during a startup phase, then have a relatively constant
      branching factor for several levels. and finally the distribution
      has sort of a tail.

   2. The distributions for corners-only cubes have a variable branching
      factor during a startup phase, and almost immediately the
      distribution has a tail.  The number of cases simply is not
      large enough to support an extended constant branching factor
      in the middle of the distribution.  It's sort of like a very
      short airplane flight where it is time to descend about the
      time the ascent is completed.

   3. I would expect the distributions for a full cube to have an
      even longer period with a constant branching factor than
      the distributions for edges-only cubes because the number
      of cases is so much larger.  There should be plenty of time
      for a plateau between the startup phase and any tail of the
      distribution.

   4. There are an equal number of odd and even permutations.  For
      the cases where you restrict yourself to Q-turns, there are
      therefore equal numbers of states an even distance from Start
      and an odd distance from Start.  Hence, the distribution tends
      either to have two adjacent levels with approximately equal
      numbers of states, or else tends to have one dominant level with
      a level on each side of the dominant level with about half
      the number of states in the dominant level.

   5. For the cases where you allow both Q-turns and H-turns, there
      tends to be one dominant level which contains most of the
      of the states.

   6. Those of you who followed all the traffic on this list in
      December and January will recall that my work with God's
      algorithm exploits symmetric conjugates in order to reduce
      the size of the problem.  It turns out that using conjugates
      does not change the average branching factor once you get
      past the startup portion of the distribution.  This effect
      can be a bit hard to see for corners-only cubes because the
      steady state portion of the distribution is so short, but
      the effect is very striking for edges-only cubes.  I would
      expect the effect to be very striking, as well, for the
      case of the full cube.

      ------------------------------------------------------------------

              2x2x2 Cube using Q-turns and H-turns

    Distance    Number of  Branching    Number of  Branching   Ratio of
        from        Cubes     Factor       M          Factor   Cubes to
       Start                           Conjugates            Conjugates

           0            1                       1                  1.00
           1            9       9.00            2       2.00       4.50
           2           54       6.00            5       2.50      10.80
           3          321       5.94           19       3.80      16.89
           4         1847       5.75           68       3.58      27.16
           5         9992       5.41          271       3.99      36.87
           6        50136       5.02         1148       4.24      43.67
           7       227536       4.54         4915       4.28      46.29
           8       870072       3.82        18364       3.74      47.38
           9      1887748       2.17        39707       2.16      47.54
          10       623800       0.33        13225       0.33      47.17
          11         2644       0.00           77       0.01      34.34

    Total/Avg     3674160     ? 4.83        77802     ? 3.54      47.22

      ------------------------------------------------------------------

              2x2x2 Cube using Q-turns

    Distance    Number of  Branching    Number of  Branching   Ratio of
        from        Cubes     Factor       M          Factor   Cubes to
       Start                           Conjugates            Conjugates

           0            1                       1                  1.00
           1            6       6.00            1       1.00       6.00
           2           27       4.50            3       3.00       9.00
           3          120       4.44            6       2.00      20.00
           4          534       4.45           17       2.83      31.41
           5         2256       4.22           59       3.47      38.24
           6         8969       3.98          217       3.68      41.33
           7        33058       3.69          738       3.40      44.79
           8       114149       3.45         2465       3.34      46.31
           9       360508       3.16         7646       3.10      47.15
          10       930588       2.58        19641       2.57      47.38
          11      1350852       1.45        28475       1.45      47.44
          12       782536       0.58        16547       0.58      47.29
          13        90280       0.12         1976       0.12      45.69
          14          276       0.00           10       0.01      27.60

    Total/Avg     3674160     ? 3.05        77802     ? 2.92      47.22

      ------------------------------------------------------------------

              Corners of 3x3x3 Cube using Q-turns and H-turns

    Distance    Number of  Branching    Number of  Branching   Ratio of
        from        Cubes     Factor       M          Factor   Cubes to
       Start                           Conjugates            Conjugates

           0            1                       1                  1.00
           1           18      18.00            2       2.00       9.00
           2          243      13.50            9       4.50      27.00
           3         2874      11.83           71       7.89      40.48
           4        28000       9.74          637       8.97      43.96
           5       205416       7.34         4449       6.98      46.17
           6      1168516       5.69        24629       5.54      47.44
           7      5402628       4.62       113049       4.59      47.79
           8     20776176       3.85       433611       3.84      47.91
           9     45391616       2.18       947208       2.18      47.92
          10     15139616       0.33       316823       0.33      47.79
          11        64736       0.00         1481       0.00      43.71

    Total/Avg    88179840     ? 4.74      1841970     ? 4.63      47.87

      ------------------------------------------------------------------

              Corners of 3x3x3 Cube using Q-turns

    Distance    Number of  Branching    Number of  Branching   Ratio of
        from        Cubes     Factor       M          Factor   Cubes to
       Start                           Conjugates            Conjugates

           0            1                       1                  1.00
           1           12      12.00            1       1.00      12.00
           2          114       9.50            5       5.00      22.80
           3          924       8.11           24       4.80      38.50
           4         6539       7.08          149       6.21      43.89
           5        39528       6.04          850       5.70      46.50
           6       199926       5.06         4257       5.01      46.96
           7       806136       4.03        16937       3.98      47.60
           8      2761740       3.43        57800       3.41      47.78
           9      8656152       3.13       180639       3.13      47.92
          10     22334112       2.58       466052       2.58      47.92
          11     32420448       1.45       676790       1.45      47.90
          12     18780864       0.58       392558       0.58      47.84
          13      2166720       0.12        45744       0.12      47.37
          14         6624       0.00          163       0.00      40.64

    Total/Avg    88179840     ? 4.48      1841970     ? 4.29      47.87

      ------------------------------------------------------------------

         Edges of 3x3x3 Cube Without Centers using Q-turns and H-Turns

    Distance    Number of  Branching    Number of  Branching   Ratio of
        from        Cubes     Factor       M          Factor   Cubes to
       Start                           Conjugates            Conjugates

           0            1                       1                  1.00
           1           18      18.00            2       2.00       9.00
           2          243      13.50            9       4.50      27.00
           3         3240      13.33           75       8.33      43.20
           4        42359      13.07          919      12.25      46.09
           5       538034      12.70        11344      12.34      47.43
           6      6666501      12.39       139325      12.28      47.85
           7     79820832      11.97      1664347      11.95      47.96
           8    888915100      11.14     18524022      11.13      47.99
           9   8056929021       9.06    167864679       9.06      48.00
          10  27958086888       3.47    582489607       3.47      48.00
          11   3883792136       0.14     80930364       0.14      47.99
          12         8827       0.00          314       0.00      28.11

    Total/Avg 40874803200    ? 12.26    851625008    ? 11.99      48.00


      ------------------------------------------------------------------

              Edges of 3x3x3 Cube Without Centers Using Q-turns

    Distance    Number of  Branching    Number of  Branching   Ratio of
        from        Cubes     Factor       M          Factor   Cubes to
       Start                           Conjugates            Conjugates

           0            1                       1                  1.00
           1           12      12.00            1       1.00      12.00
           2          114       9.50            5       5.00      22.80
           3         1068       9.37           25       5.00      42.72
           4         9759       9.14          215       8.60      45.39
           5        88144       9.03         1860       8.65      47.39
           6       786500       8.92        16481       8.86      47.72
           7      6916192       8.79       144334       8.76      47.92
           8     59623239       8.62      1242992       8.61      47.97
           9    495496593       8.31     10324847       8.31      47.99
          10   3695351994       7.46     76993295       7.46      48.00
          11  17853871137       4.83    371975385       4.83      48.00
          12  18367613703       1.03    382690120       1.03      48.00
          13    395043663       0.02      8235392       0.02      47.97
          14         1080       0.00           54       0.00      20.00
          15            1       0.00            1       0.02       1.00

    Total/Avg 40874803200     ? 8.80    851625008     ? 8.63      48.00

      ------------------------------------------------------------------

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From Joel.Franklin@altosax.reed.edu  Thu Jun  9 15:09:41 1994
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Date: 09 Jun 94 12:07:54 PDT
From: Joel.Franklin@altosax.reed.edu (Joel Franklin)
Subject: 
To: CUBE-LOVERS@life.ai.mit.edu

How do I subscribe to this list?

From rprakash@cdotp.ernet.in  Wed Jun 22 08:37:49 1994
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Date: Wed, 22 Jun 94 17:14:48+050
From: rprakash@cdotp.ernet.in (PRAKASH R.)
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To: cube-lovers@life.ai.mit.edu

Dear cube-lovers-request,
	please send me some information & problems about the cube. i haven't
been able to solve the cube fully yet. i can get about 60-70% . i need some
suggestions about how to solve the cube also.
	thanking you,
						-love
						prakash r.


From @wvnvm.wvnet.edu:BRYAN@WVNVM.WVNET.EDU  Wed Jun 29 14:11:23 1994
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Date:      Wed, 29 Jun 1994 13:45:42 EDT
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Comments on Cube Lengths (Long, 1 of 2)

As you all know, the length of a cube X is defined as the shortest
process P such that XP = I, and is denoted as |X|.  One definition
of God's Algorithm is simply that God's algorithm is the knowledge
of |X| for all cubes.  I wish to make some observations about
|X| as related to various models of the cubes and their symmetries.

The set C is the set of 24 rotations of the cube. The set M is the
set of 48 rotations and reflections of the cube, where half of M is
C and the other half of M is C reflected.

The first (obvious) observation is that |X| = |m'Xm| for all m in M.
That is, the set of all M-conjugates of a cube have the same
length.  Another way to say the same thing is that |m'Xm| = |n'Xn|
for all m and n in M.

Actually, there is an even more obvious observation that probably
should be made.  The set G is the set of all cubes, where
G is generated as G=<Q>, where Q is the set of 12 quarter-turns
of the cube.  The *really* obvious observation is that if X
is in G, then m'Xm is in G for all m in M.  Furthermore, if GC
is the set of corners-only cubes, then X in GC implies m'Xm in GC,
and if GE is the set of edges-only cubes, then X in GE implies
m'Xm in GE.  Finally, the observation that |X| = |m'Xm| remains
true whether X is in G, in GC, or in GE.

The process of forming M-conjugates in G (or in GC or in GE)
induces a partition which is an equivalence relation.  Hence, the
set {m'Xm} for all m in M is an equivalence class.  Since,
|m'Xm| = |n'Xn| for all m and n in M, it is meaningful to speak
of the length of {m'Xm}, namely |{m'Xm}| = |Y|, where Y is
any element of {m'Xm}.

Now, consider cubes of the form Xc where X is in G and c is in C.
We first observe that Xc is in G if and only if c is even.   Half the
elements in C are even, and half are odd.  An odd permutation in C is
even on the corners but is odd on the edges; hence, Xc is not in G when
c is odd.  On the other hand, Xc is in GC for all X in GC, and Xc is in
GE for all X in GE.

The fact that Xc is in G only for even elements of C is why I thought
it was important to make the "really obvious" observation that
m'Xm is in G for all X in G and all m in M.  The two cases m'Xm and
Xc are similar on the surface, but different when you dig a
little bit deeper.

With respect to lengths, we can observe that |Xc| >= |X| whenever
Xc is well-defined (that is, whenever c is even for G, or for all c
for GC and GE).

The process of performing rotations in G (even rotations in G, or any
rotation in GC or in GE) induces a partition which is an equivalence
relation.  Hence, the set {Xc} for all (or even, as appropriate) c
in C is an equivalence class.  The collection of all sets of the form
{Xc} can serve as a model for cubes without centers.

However, it is not the case that |Xc| = |Xd| for all c and d in C.
Nonetheless, it is meaningful to speak of |{Xc}|.  Namely,
|{Xc}| = min{|Xd|} for all (or even) d in C.  Hence, we have
|{Xc}| <= |Xd| for all (or even) d in C.

The definition |{Xc}| = min{|Xd|} probably requires a bit of
justification.  For a cube without centers, the solved or Start
state is {Ic} for all (or even) c in C.  Hence, Start is C (or
C[even]), and we need the shortest process P such that XP is in C
in order to calculate |{Xc}|.

Consider the set {P[1], P[2], ... P[24]} where P[n] is the shortest
process for which (Xc[n])P[n] = I.  Observe, that XP[n] is in C for
all n in 1..24.  This immediately gives us |P| <= |P[n]| for all
n in 1..24.

Conversely, if XP is in C, then there exists some c[n] in C such that
Xc[n]P = I.  This gives us |P[n]| <= |P| for some n in 1..24.
Therefore, |P| = min{|P[n]|} for n in 1..24.

Note that we have |{Xc}| <= |X| <= |Xd|.  On its face, this may seem
somewhat paradoxical, but I believe that it is entirely correct.
There is a huge difference is speaking of |{Xc}| as opposed to
speaking of |Xd|.  Xd is an (atomic) element of G;  {Xc} is a set.
Elements of {Xc} are also in G, but the *set* {Xc} is not in G.

My model for cubes without centers is really {m'Xmc} rather than
{Xc}.  However, the results from above are readily combined.  That is,
we can speak of |{m'Xmc}|, namely |{m'Xmc}| = min{|(m'Xm)d|} for all
(or even) d in C.  As before, we have |{m'Xmc}| <= |m'Xm| <= |m'Xmd|.
Note that in the middle of this last string of inequalities we could
insert any of |X| = |m'Xm| = |{m'Xm}|.

In my God's algorithm data base for cubes without centers, I store
ordered pairs of the form (Y,|{m'Xmc}|), where Y is a representative
element of the set {m'Xmc}.  Note that Y is in G (or GC or GE, as
appropriate).  It is a picky point, but the data base does *not*
consist of ordered pairs of the form (Y,|Y|).  Remember that
|Y| >= |{m'Xmc}|.

My God's algorithm data base for cubes with centers nominally consists
of ordered pairs of the form (Z,|{m'Xm}|), where Z is a representative
element of the set {m'Xm}.  Unlike the case without centers, we do
have |Z| = |{m'Xm}|, so we could also say the data base elements are
of the form (Z,|Z|).

However, the data representation is really a bit different to take
advantage of the relationship between sets of the form {m'Xmc}
and sets of the form {m'Xm}.  A set of the form {m'Xmc} can be
partitioned into (up to) twenty-four sets of the form {m'Xm},
where the (up to) twenty-four sets are indexed by C.

Let Y=Repr({m'Xmc}).  Then, the data base is ordered pairs of the form
(Yc[i],|Yc[i]|) for i in 1..24.  Note that Yc[i] is in G, but can be
said to be a representative element for sets of the form {m'(Yc[i])m},
which in turn is a set of the form {m'Xm} for some X in G.

Finally, there is no real need to store the Yc[i];  it is only
necessary to store the lengths.  Hence, a data base
element for cubes with centers is really, really of the form:

  (Y,|{m'Xmc}|,|Yc[1]|,|Yc[2]|, .... |Yc[24]|),

where Y is a representative element of {m'Xmc}.

Note that this is a very compressed format.  The representative element
Y is stored only once for the 24 different values for c.  Note also that
the solution for cubes without centers is stored in the same data base
as the solution for cubes with centers.   Finally, since
|m'Yc[i]m| = |Yc[i]|, we have stored the length of all cubes by storing
the length of only one cube for each M-conjugancy class.

It is not really necessary explicitly to store the solution for cubes
without centers to have the solution for cubes without centers in the same
data base.  That is, |{m'Xmc}|=min(|Yc[i]|) for i in 1..24.  But it takes
very little space to do so and is convenient for certain calculations.

   (to be continued)

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From @wvnvm.wvnet.edu:BRYAN@WVNVM.WVNET.EDU  Wed Jun 29 15:00:05 1994
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Date:      Wed, 29 Jun 1994 13:56:02 EDT
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Comments on Cube Lengths (Long, 2 of 2)

   (continuation)

I have described this data base structure once before (a little
less formally before), but I wanted to describe it again because
there is some interesting (to me, at least) analysis that can be
derived from the data base, over and above God's algorithm.

First, it is interesting to compare |{m'Xmc}| to the various |Yc[i]|.
Recall that |Yc[i]| >= |{m'Xmc}| for all i in 1..24.  Also, by the
definition of |{m'Xmc}|, there is at least one i in 1..24 such that
|Yc[i]| = |{m'Xmc}|.

I posted a comparison of |{m'Xmc}| to |Yc[i]| for corners-only cubes
on 4 December 1993.  (I have the "without centers" part of the edges-only
data base done, but it will take many more months to
complete the "with centers" part.  So corners-only is the only
complete data base we have to work with.)

At the time, I received a couple of comments to the effect
that people didn't understand what I was comparing to what.  I hope
this note clarifies the situation.  For example (and referring to my
previous note with respect to corners-only cubes), there is only one
element of the form {m'Xmc} for which |{m'Xmc}| = 0.  The only element
for which the length is 0 is Start, but in a corners only cube without
centers, any rotation of Start is still at Start and still has length 0.

Here is a brief excerpt from my note of 4 December 1993.

              Corresponding Distances from Start
                     Using Only q-turns

        Without           With
        Centers         Centers         Number
     Distance from    Distance from    of Nodes
         Start         Start

           0
                           0                1
                           2                1
                           4                2
                           6                1

What this means is as follows.  First, we have |{m'Imc}| = 0.  Let
Y=Repr({m'Imc}).  Then, there is 1 element of the form Yc for which
|Yc|=0, 1 element of the form Yc for which |Yc|=2, 2 elements of the
form Yc for which |Yc|=4, and 1 element of the form Yc for which
|Yc|=6.

In words, suppose you have a corners-only cube (peel off the edge
tabs, but keep the corners and centers).  Then, suppose the
corners look "solved" if you ignore the centers.  The corners will
be rotated relative to the centers.  In all, there are 24 different
ways they can be rotated, including the identity, where they
are not rotated.

But under M-conjugancy, some of the 24 rotations are equivalent.
Under M-conjugancy, there is one way they can be 0 moves from Start,
one way they can be two moves from Start (RL' is equivalent to DU',
for example ), two ways they can be four moves from Start, and one way
they can be six moves from Start.  Among other things, this says that any
rotation of the corners (ignoring the edges) can be accomplished
in no more than six quarter turns.

This example illustrates why a set of the form {m'Xmc} may be
partitioned into "up to" twenty-four elements of the form
{m'Xm}, rather than "exactly" twenty-four elements.  Normally,
a set of the form  {m'Xmc} contains 1152 elements,
where 1152=24*48.  It can in turn be partitioned into twenty-four
elements of the form {m'Xm} which contain forty-eight elements each.
But cubes which are "symmetric" reduce the number because
various M-conjugates are equivalent.

I normally think of the God's algorithm data base as a matrix, with
the rows indexed by the representative elements Y, and the columns
indexed by C (or more simply, by 1..24).  Because of M-conjugate
symmetry, there are always a few empty cells in the matrix.

M-conjugate symmetry did not cause me any computational difficulty
when I was working with cubes without centers.  That is,
suppose {m'(X1)mc} and {m'(X2)mc} are the same set for X1 not equal
X2.  My "representative element calculator" would calculate the
same representative element Y in both cases.  But in the case of
cubes with centers, the "representative element calculator" had to
calculate both a representative element Y and an associated rotation
index Cind in 1..24.

When a set {m'Xmc} had exactly 1152 elements (most of the time), the
calculation of Cind was correct.  But when a set {m'Xmc} had fewer
than 1152 elements, I would get a different Cind depending on which
element of the set I started with.  That is, the loops in the program
actually calculate 1152 elements in any case, but if the set really has
less than 1152 elements, then some of the elements are generated
multiple times.  (The loops have no way of knowing ahead of time how
many elements are going to be in the set.)  The generation of the same
set elements multiple times severely messed up the calculation of Cind
until I figured out what was going on.

I want to finish by getting back to what I started with, the lengths
of cubes.  As I said, the God's algorithm results for edges without
centers are complete (posted to the list back in December), but the
God's algorithm calculations for edges with centers are still work
in progress.  However, I noticed something striking about the
partial edges with centers results when I compared them with the
completed edges without centers results.  For example, here is a
table which compares the results when using q-turns only.


    Distance    Number of  Branching    Number of  Branching
        from   M-Conjugate   Factor    M-Conjugate   Factor
       Start     Classes                 Classes
                Without                  With
                Centers                 Centers

           0           1                     1
           1           1       1.00          1         1.00
           2           5       5.00          5         5.00
           3          25       5.00         25         5.00
           4         215       8.60        215         8.60
           5        1860       8.65       1886         8.77
           6       16481       8.86      16902         8.96
           7      144334       8.76     150442         8.90
           8     1242992       8.61    1326326         8.81
           9    10324847       8.31   11505339         8.67
          10    76993295       7.46   96755918         8.40
          11   371975385       4.83  750089528         7.75
          12   382690120       1.03      ....
          13     8235392       0.02      work
          14          54       0.00       in
          15           1       0.02    progress

    Total      851625008

As you can see, with or without centers, there are the same number
of cubes (actually, equivalence classes) at each distance from
Start from level 0 through level 4.  From level 5 on, there are more
cubes with centers than without.  Why is the number the same
through level 4, and what happens at level 5 to make the numbers
different?  Actually, overall there are about twenty-four times more
cubes with centers than without, so it is not surprising to find
more cubes with centers than without at fairly low levels in the
search tree.  So fundamentally, the question is, why does the
divergence occur at level 5?

Well, I can't explain why it is level 5 exactly, but I can explain
what is going on.  Consider level 0.  There is one row in the data
base where |{m'Xmc}|=0.  There are twenty-four cells in the same row
for |Yc[i]|, corresponding to the twenty-four rotations of the
representative element Y.  For exactly one of these cells, we have
|Yc[i]|=0.  The remainder of the cells are either undefined (meaning
the cell represents a rotation which is M-conjugate equivalent with
another rotation), or else we have |Yc[i]|>=5.  Hence, any
rotation of the edges of the cube requires at least 5 q-turns to
accomplish.  After the data base is complete, we can determine
exactly how many q-turns are required to accomplish each rotation of
the edges, just as we can already do with the corners.

Similar comments apply to level 1 through 4.  There is exactly one
rotation of the representative element that has the same length as
representative element.  All the other rotations of the representative
element are either M-conjugate equivalent to the representative
element, or else have a length greater than or equal to 5.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From jkato@tmastb.eec.toshiba.co.jp  Tue Jul  5 20:56:33 1994
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Date: Wed, 6 Jul 94 09:56:14 JST
From: jkato@tmastb.eec.toshiba.co.jp (Toshi Kato)
Return-Path: <jkato@tmastb.eec.toshiba.co.jp>
Message-Id: <9407060056.AA03691@tmastb.eec.toshiba.co.jp>
To: cube-lovers@life.ai.mit.edu
Subject: Cube-Lovers ML

I am a member of NKC(CFF) and Puzzle KONWAKAI(Academy of Recreational 
Mathematics, Japan).
I knew your Mailing List from Jerry's Puzzle Collectors Directory.
I would like to join your ML.

In this summer I am going to 14th International Puzzle collectors Party 
in Seattle.
------
                         Thank you,
                         Toshi(Junk) Kato
                         2-14-60 Hishinuma, Chigasaki 253 Japan
                         Tel/Fax: +81-467-52-1447
                         E-mail: jkato@tmastb.eec.toshiba.co.jp
   






From jkato@tmastb.eec.toshiba.co.jp  Wed Jul  6 07:00:52 1994
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Date: Wed, 6 Jul 94 20:00:14 JST
From: jkato@tmastb.eec.toshiba.co.jp (Toshi Kato)
Return-Path: <jkato@tmastb.eec.toshiba.co.jp>
Message-Id: <9407061100.AA05730@tmastb.eec.toshiba.co.jp>
To: Cube-Lovers@ai.mit.edu
Subject: SBP "Magic sQ"

                                   Sliding Block Puzzle "Magic sQ"
Fig.1 is incomplete.                      +---+---+---+
Can you complete a magic square           | 2 | 9 | 4 | 
with minimum sliding steps?               +---+---+---+
                                          | 7 | 5 | 3 |
You, very easy or not?                    +---+---+---+---+
                                          | 1 | 6 | 8 |   |  Fig.1
                                          +---+---+---+---+
------
                        Toshi(Junk) Kato from Japan
                        E-mail: jkato@tmastb.eec.toshiba.co.jp
                        Tel/Fax: +81-467-52-1447






From mouse@collatz.mcrcim.mcgill.edu  Fri Jul  8 15:24:40 1994
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Date: Fri, 8 Jul 1994 15:24:30 -0400
From: der Mouse <mouse@collatz.mcrcim.mcgill.edu>
Message-Id: <199407081924.PAA08873@Collatz.McRCIM.McGill.EDU>
To: Cube-Lovers@ai.mit.edu
Subject: Re:  SBP "Magic sQ"
Cc: jkato@tmastb.eec.toshiba.co.jp

>                                    Sliding Block Puzzle "Magic sQ"
> Fig.1 is incomplete.                      +---+---+---+
> Can you complete a magic square           | 2 | 9 | 4 |
> with minimum sliding steps?               +---+---+---+
>                                           | 7 | 5 | 3 |
> You, very easy or not?                    +---+---+---+---+
>                                           | 1 | 6 | 8 |   |  Fig.1
>                                           +---+---+---+---+

Not hard, but cutely deceptive.  The figure as supplied is a magic
square with the 1 and 6 switched.  It is not possible to switch two
adjacent tiles in a quadrilateral sliding-block puzzle of this sort
(there's an easy induction proof that only even permutations are
possible).  Thus, either it's not possible or the solution involves
some other magic square.

Since the 8 must clearly be in the lower right corner of the resulting
magic square, there are only two squares possible.  One is the magic
square that is almost present already; the other is its reflection:

        2 9 4        2 7 6
        7 5 3        9 5 1
        6 1 8        4 3 8

Fortunately, the "other" magic square is an even permutation from the
provided start position.  It's then just a straightforward tree search
to find a solution.  A simple brute-force "meet in the middle"
breadth-first search finds a solution easily.  Move the 8 aside, then
move as follows (* represents the blank space):

2 9 4    2 9 4    2 9 4    2 9 4    2 9 4    2 9 4    2 9 4    2 9 4
7 5 3 -> 7 5 3 -> 7 5 3 -> * 5 3 -> 5 * 3 -> 5 3 * -> 5 3 6 -> 5 3 6 ->
1 6 *    1 * 6    * 1 6    7 1 6    7 1 6    7 1 6    7 1 *    7 * 1
                                                    +-------+
2 9 4    2 * 4    2 4 *    2 4 6    2 4 6    2 4 6  | 2 4 6 |  2 4 6
5 * 6 -> 5 9 6 -> 5 9 6 -> 5 9 * -> 5 9 1 -> 5 9 1 -> 5 9 1 -> * 9 1 ->
7 3 1    7 3 1    7 3 1    7 3 1    7 3 *    7 * 3  | * 7 3 |  5 7 3
                                                    +-------+
2 4 6    2 * 6    * 2 6    9 2 6    9 2 6    9 2 6    9 2 6    9 2 6
9 * 1 -> 9 4 1 -> 9 4 1 -> * 4 1 -> 4 * 1 -> 4 7 1 -> 4 7 1 -> * 7 1 ->
5 7 3    5 7 3    5 7 3    5 7 3    5 7 3    5 * 3    * 5 3    4 5 3

* 2 6    2 * 6    2 7 6    2 7 6    2 7 6
9 7 1 -> 9 7 1 -> 9 * 1 -> 9 5 1 -> 9 5 1
4 5 3    4 5 3    4 5 3    4 * 3    4 3 *

Then put the 8 back, and you're done, in a total of 30 moves (28 shown,
plus the two moves of the 8).  For those who care about such things,
the boxed position is the midpoint at which the two searches met.
(This is fairly obvious - since the number of moves is even, the
configuration on which the searches met must be the middle one.)

This solution exhibits curious near-symmetries in portions of the path
taken by the blank space.  Anyone have any thoughts on this?  Perhaps I
should modify the program so it reports _all_ solutions of this length;
there may be something interesting lurking here.

					der Mouse

			    mouse@collatz.mcrcim.mcgill.edu

From jkato@tmastb.eec.toshiba.co.jp  Mon Jul 11 00:08:01 1994
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Date: Mon, 11 Jul 94 13:06:53 JST
From: jkato@tmastb.eec.toshiba.co.jp (Toshi Kato)
Return-Path: <jkato@tmastb.eec.toshiba.co.jp>
Message-Id: <9407110406.AA17550@tmastb.eec.toshiba.co.jp>
To: mouse@collatz.mcrcim.mcgill.edu
Cc: Cube-Lovers@ai.mit.edu
In-Reply-To: der Mouse's message of Fri, 8 Jul 1994 15:24:30 -0400 <199407081924.PAA08873@Collatz.McRCIM.McGill.EDU>
Subject: Re: SBP "Magic sQ"

Many thanks. 
Especially to Dan Hoey and der Mouse <mouse@Collatz.McRCIM.McGill.EDU>.

I have recieved E-mail individually from Mr.Dan Hoey too,
 
  Date: Thu, 7 Jul 94 16:43:02 EDT
  From: hoey@AIC.NRL.Navy.Mil
  To: jkato@tmastb.eec.toshiba.co.jp (Toshi Kato)
  Subject: Re: SBP "Magic sQ"

  A little hack tells me you can solve that by moving
  1,7,3,6,1,9,4,1,7,5,9,4,2,9,4,7,5,9,2,5,8.  That's 21 moves, or 30 if
  you count by the tile.  It's optimal in both metrics.

  Do you know anything about Rubik's Cube?

  Dan Hoey
  Hoey@AIC.NRL.Navy.Mil

I was surprised to recieve solutions both so quickly.
Cubes-Lover is very smart team,$@!!(JI think.

Thanks again,
Junk Kato
jkato@tmastb.eec.toshiba.co.jp



From @mail.uunet.ca:mark.longridge@canrem.com  Fri Jul 15 03:19:03 1994
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Sender: CRSO.Cube@canrem.com
Subject: DOTC 1.4 is done
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.770.5834.0C1A658F@canrem.com>
Date: Fri, 15 Jul 1994 01:32:00 -0400
Organization: CRS Online  (Toronto, Ontario)

        Domain of the Cube 1.4 is finally done!
        ---------------------------------------

   I've finally finished the new issue of the DOTC newsletter, and I'm
basically happy with it. I believe I owe my fellow cubists an apology
for taking so long, especially Greg Schmidt and Dan Hoey.

 I've enjoyed using the cube since 1981 and I wish I had more time and
energy to put into it. I was also rather ill earlier this year, and
things at work seemed to always interrupt. Nevertheless, the first
20 copies are finally ready to mail. Despite the fact these initial
copies are slightly flawed I am no longer willing to wait. This time
the issues have beige covers and are stapled like a booklet, much
the same as David Singmaster's Cubic Circular.

   I'm pleased with the printer's results, and I am mailing out the
first issues tomorrow. I have considerable work done on issue 1.5
and I expect the next issue to be ready relatively soon.

 - Mark

New Technique for Pattern Finding:

Cycle a process until you find the identity, e.g. (F1 B1 R1 D1)^24 = I
 then bisect the process if the order is even,
   ( F1 B1 R1 D1 ) ^ 12 = Pattern, naturally this process is order 2.

---------------------------------------------------------------------

Hmmmm, actually I have some questions that have been bugging me for
some time. I while back a guy was watching me use my cube program
and I explained that the reason I like studying group theory is
because it provided greater insights into the cube. He then asked
me:  "What are other uses of group theory?" and "What are the
practical uses of group theory" to which I haltingly replied (somewhat
vaguely) that it helped show relationships between geometry and
algebra. I felt this explanation unsatisfactory. I also mumbled
about symmetry and architecture. I'm sure there is a better
answer than that!

Also why is it in math that  |-11| means absolute value and can also
be the order of G, e.g. Let G be a Group, and |G| means the order
of G.

Here is another tidbit for the cube archives:

Rare 11-cycle of edges: ( L2 B1 R1 D3 L3 ) ^ 7   (35)
alternately:  F2 R3 U1 D3 B3 D1 L3 U3 D1 B1 L1 D1 B2 U2 D2 R2 B2 D1 (18)


-> Mark <-

From jkato@tmastb.eec.toshiba.co.jp  Mon Jul 18 06:11:51 1994
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Date: Mon, 18 Jul 94 19:11:24 JST
From: jkato@tmastb.eec.toshiba.co.jp (Toshi Kato)
Return-Path: <jkato@tmastb.eec.toshiba.co.jp>
Message-Id: <9407181011.AA16075@tmastb.eec.toshiba.co.jp>
To: Cube-Lovers@ai.mit.edu
Subject: A real robot solve the Rubik's Cube but...

A real robot which had artificial eyes and arms and computer brain was 
manufactured at Kawasaki Heavy Industry Co.,Ltd in Japan last year. 
He can solve the real commercial Rubik's Cube.

As he has not so intelligent, it takes about 12 minutes and 120 steps
average between starting and finishing the Cube to solve it.

Are there any other live robot like him over the world?  Do you know?

And you can help him more clever with your solving algolithm, can't you?
------
                        Toshi(Junk) Kato from Japan
                        E-mail: jkato@tmastb.eec.toshiba.co.jp
                        Tel/Fax: +81-467-52-1447









From @wvnvm.wvnet.edu:BRYAN@WVNVM.WVNET.EDU  Mon Jul 18 10:44:30 1994
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X-Acknowledge-To: <BRYAN@WVNVM.WVNET.EDU>
Date:      Mon, 18 Jul 1994 10:41:49 EDT
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   God's Algorithm, Additional Level

The following is for Q-turns only, whole cubes (both corners and
edges), and not considering M-conjugates.  I think it would be
possible to squeeze out another couple of levels by considering
M-conjugates.  The best previous result I have found in the
archives was through level 7 (reported on 7 December 1981, and again
on 3 August 1992).

            Distance              Number     Branching
              from                  of         Factor
             Start                Cubes


                 0                      1
                 1                     12     12.00
                 2                    114      9.50
                 3                  1,068      9.37
                 4                 10,011      9.37
                 5                 93,840      9.37
                 6                878,880      9.37
                 7              8,221,632      9.35
                 8             76,843,595      9.35   (new)


 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From newfield@vsl.ist.ucf.edu  Mon Jul 18 11:56:37 1994
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From: newfield@vsl.ist.ucf.edu (Dale Newfield)
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To: Cube-Lovers@ai.mit.edu
Subject: Re: A reaal robot solve the Rubik's Cube but...


Sorry to pick nits, but if it is autonomous(sp?), which I think you implied,
wouldn't it be an android, instead of a robot?
-Dale

Dale Newfield							   I'd rather
newfield@vsl.ist.ucf.edu					   be playing
dn1l@{cs,andrew}.cmu.edu					       xlife.


From jkato@tmastb.eec.toshiba.co.jp  Mon Jul 18 22:40:44 1994
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Date: Tue, 19 Jul 94 11:40:15 JST
From: jkato@tmastb.eec.toshiba.co.jp (Toshi Kato)
Return-Path: <jkato@tmastb.eec.toshiba.co.jp>
Message-Id: <9407190240.AA17895@tmastb.eec.toshiba.co.jp>
To: Cube-Lovers@ai.mit.edu
Subject: Re: A real robot solve the Rubik's Cube but...

Dale Newfield said:
 <Sorry to pick nits, but if it is autonomous(sp?), which I think you implied,
 <wouldn't it be an android, instead of a robot?

Pardon me. I wonder if I shoudn't use such words "real" and "live". 
-----
                                  Toshi(Junk) Kato
                                  jkato@tmastb.eec.toshiba.co.jp








From diamond@jrdv04.enet.dec.com  Mon Jul 18 23:11:47 1994
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From: Norman Diamond  19-Jul-1994 1206 <diamond@jrdv04.enet.dec.com>
To: cube-lovers@ai.mit.edu
Apparently-To: cube-lovers@ai.mit.edu
Subject: Re: A real robot solve the Rubik's Cube but...
Mime-Version: 1.0
Content-Type: TEXT/PLAIN; charset=ISO-2022-JP

Toshi Kato says:

>Dale Newfield said:
> <Sorry to pick nits, but if it is autonomous(sp?), which I think you implied,
> <wouldn't it be an android, instead of a robot?
>
>Pardon me. I wonder if I shoudn't use such words "real" and "live".

Half right.  It was a real dead robot :-)

-- Norman Diamond       diamond@jrdv04.enet.dec.com
[Digital did not write this.]

From jkato@tmastb.eec.toshiba.co.jp  Tue Jul 19 00:24:58 1994
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Date: Tue, 19 Jul 94 13:24:38 JST
From: jkato@tmastb.eec.toshiba.co.jp (Toshi Kato)
Return-Path: <jkato@tmastb.eec.toshiba.co.jp>
Message-Id: <9407190424.AA18202@tmastb.eec.toshiba.co.jp>
To: cube-lovers@ai.mit.edu
Subject: Re: A real robot solve the Rubik's Cube but...

Norman Diamond$@!!(Jsaid:
> >Pardon me. I wonder if I shouldn't use such words "real" and "live".
>
>Half right.  It was a real dead robot :-)

Thanx. I think the robotic machine isn't alive now too. 
-----
                                 Toshi(Junk) Kato
                                 E-mail:jkato@tmastb.eec.toshiba.co.jp




From @wvnvm.wvnet.edu:BRYAN@WVNVM.WVNET.EDU  Tue Jul 19 10:55:43 1994
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From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   More on Centerless Cubes

On 13 Feb 1994, I proposed a model for centerless cubes which
I claimed met two criteria:  1) it was a group, and 2) it
maintained the symmetrical nature of the problem.  On
23 May 1994, I retracted the claim that the proposed model
was a group.

I am now of the opinion that it is impossible to satisfy
both criteria simultaneously.  I can make a very small modification
to the proposed model to make it a group, but the small
modification costs the model its cubic symmetry.

G is the full cube group, GC is the corners only cube group,
and GE is the edges only cube group.  The proposed model
for centerless cubes consisted of partitioning any of G,
GC, or GE into sets of the form {Xc} for all c in C, where
C is the set of twenty-four rotations of the cube and X is
a cube.  The sets are the elements of the proposed group.
The sets are called cosets and can also be denoted as
xC.  The partitions are denoted as G/C, GC/C, and GE/C,
respectively.

Originally, the proposed group operator was {Xc} * {Yc} = {XYc}.
This operator fails to maintain closure, and hence fails to
define a group.

In order to illustrate the slight modification which will define
a group, we will start by restricting ourselves to GC.  An
operator which works to define GC/C as a group is
{Xc} * {Yc} = {VWc}, where V is the unique element of {Xc} such
that the urf cubie is properly positioned in the urf cubicle,
and W is the unique element of {Yc} such that the urf cubie is
properly positioned in the urf cubicle.

Any other corner could have been used instead of urf, but once
you choose a corner the problem loses its symmetric nature.
Well, I guess it still has symmetry, but it is not the uniform
symmetry of the cube any more, because there is a preferred
orientation.

I have found only limited discussion in the archives, but
previous investigators have modeled a corners only, centerless
cube by leaving one corner fixed.  Such a model is clearly
a group.  For example, if we leave the urf corner fixed,
we can generate the group JC as JC=<L,L',D,D',B,B'>, where
we omit all twists of the U, R, and F faces from the set
of generators.

It is easy to find an isomorphism between GC/C and JC.  I would
express it as something like {Xc} = {Wc} <--> W, where W is
defined as before.  W is an element of JC, and as well is an
element of {Xc} = {Wc}.  {Xc} = {Wc} is an element of GC/C.
But W is a particular element of {Xc} = {Wc}, whereas X is
an arbitrary element.  Also, X is in GC, but X is not in JC
unless X = W.  The mapping {Wc} <--> W is clearly one-to-one and
onto in both directions.

For the edges GE, we need to keep one edge cubie fixed, so the
centerless cube could be generated by something like
JE=<D,D',L,L',R,R',B,B'>, where we keep the uf cubie fixed by
omitting all twists of the U and F faces from the set of
generators.  The isomorphism between GE/C and JE is expressed
as {Xc} = {Wc} <--> W, where X is an arbitrary element of
GE, and W is the unique element of {Xc} such that the uf
cube is properly placed in the uf cubicle.  As before, any
edge cube would do as well, but once chosen, it is no longer
arbitrary.

For the whole cube G, at first blush it appears we could model
centerless cubes either by keeping a corner cubie fixed, or by
keeping an edge cubie fixed.  But if we keep a corner cubie
fixed, the three immediately adjacent edge cubies are never moved
by any Q-turns.  We could solve the difficulty by admitting
slice turns.  But slice quarter-turns are odd on edges and
even on corners, so we have to restrict ourselves to slice
half-turns.  I find this ugly, plus I would prefer to generate
G with Q-turns only.  Hence, I would prefer to model a
centerless full cube as J=<D,D',L,L',R,R',B,B'>, where it
is an edge cubie which is held fixed rather than a corner cubie.

I said at the beginning that I thought it was impossible for a
model of centerless cubes both to be a group and also to
maintain cubic symmetry.  The reason is as follows:  it seems
to me that for any model which is a group, it should be
possible to find an isomorphism between the model and
J (or JC or JE, as appropriate).  But J and JC and JE
do not have cubic symmetry because there is a preferred
orientation.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From @wvnvm.wvnet.edu:BRYAN@WVNVM.WVNET.EDU  Tue Jul 19 19:09:07 1994
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From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   God's Algorithm, Minor Progress, Q+H

Surprisingly, there seems not to be anything in the archives for
God's Algorithm for Q+H moves for the whole cube past level 3.
Here are some updated results:

                Distance           Number     Branching
                 from                 of       Factor
                Start               Cubes

                   0                    1
                   1                   18     18.000
                   2                  243     13.500
                   3                3,240     13.333
                   4               43,239     13.345  (new)
                   5              574,908     13.296  (new)
                   6            7,618,438     13.252  (new)
                   7          100,803,036     13.231  (new)

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From jkato@tmastb.eec.toshiba.co.jp  Sun Aug  7 22:18:28 1994
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Date: Fri, 5 Aug 94 19:30:13 JST
From: jkato@tmastb.eec.toshiba.co.jp (Toshi Kato)
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Message-Id: <9408051030.AA03406@tmastb.eec.toshiba.co.jp>
To: Cube-Lovers@ai.mit.edu
Subject: HIKIMI Wooden Puzzle Competition

	To promote puzzles throughout the world
	and to convey the warmth of wood to as many people as possible
 THE 6th HIKIMI WOODEN PUZZLE COMPETITION
	will be held in October 1994.
	We invite puzzlers from around the world to enter.

 APPLICATION GUIDE
 1.Conditions
  1)Puzzle must be made of wood
  2)Puzzle must be original
  3)Puzzle has never been sold commercially
  4)Puzzle can be easily mass produced
  5)Do not submit puzzles that emphasize artistic design
  6)Puzzle may have two or more inventors
  7)Entry with more than one puzzle permitted
 
 2.Points for consideration
  1)Entry into the competition is free of charge, but each contestant must bear
    the expense for the sending the puzzle to Hikimi.
  2)We may request to purchase the puzzle within the constraints of our budget.
  3)Copyright of the puzzles entered belongs to the inventor, but  
    Hikimi Chamber of Commerce reserves the right to first negotiation.
  4)Deadline: Puzzle and application must be received by October 14,1994.

 3.For an application form, write to:

   Hikimi-cho Shokokai,Hikimi-cho,Mino-gun,Shimane Prefecture 698-12 Japan
 
    Applications in Japanese are prefered. However, since this may be a 
   difficult requirement for non-Japanese entrants, you may send your 
   application in English.

 4.Judging
   Judging will be held sometime during October,1994. All applicants will be
  notified directly of the results. The commendation ceremony will be held on
  November 12,1994 in Hikimi Town.

 5.Judges
   Chief Judge: Saburo Oguro
   Judge      : Nob Yoshigahara
   Judge      : Shigeo Takagi
   Judge      : A. Yamashita
   Judge      : T. Ohhata

 6.Prizes
   Grand Prize (one person): \500,000(about 5,000 US$)
   2nd Prize  (two persons): \300,000(about 3,000 US$)each
   3rd Prize(three persons): \100,000(about 1,000 US$)each
   Runner-ups  (several)   : \ 50,000(about   500 US$)each

 Sponsored by: Hikimicho Shokokai(Hikimi Chamber of Commerce)
                Tel:+81-856-56-0310  Fax:+81-856-56-0753

APPLICATION FORM FOR THE 6TH HIKIMI WOODEN PUZZLE COMPETITION
                                                            Date:
+--------------------------------------+--------------------------------------+
|Applicant's name:              Age(  )|Co-inventors of puzzle                |
|                                      |       Name         Age   Occupation  |
+--------------------------------------+-------------------+---+--------------+
|Address:                              |                   |   |              |
|                                      |                   |   |              |
|Tel:                Fax:              |                   |   |              |
|Occupation:                           |                   |   |              |
+--------------------------------------+-------------------+---+--------------+
|Name of company where you're employed:|Rating by Judges                     *|
|                                      +--------------------------------------+
|Address:                              |Application number                   *|
|                                      +--------------------------------------+
|Tel:                Fax:              |Remarks                              *|
+--------------------------------------+                                      |
|Name of puzzle:                       |                                      |
|                                      |                                      |
|Number of pieces in the puzzle:       |                                      |
+--------------------------------------+--------------------------------------+
|Object of puzzle:                     |Solution:                             |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
+--------------------------------------+                                      |
|Applicant's comments on the puzzle:   |                                      |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
|                                      |                                      |
+--------------------------------------+--------------------------------------+
[Note] 1)Please type or write in English block letters. For multiple entries, 
make copies of this form and submit a separate one for each puzzle entered.
2)If there are any co-inventors of the puzzle entered, be sure to write each
of their names. 3)Do not write in spaces marked(*).
 
   















From ybanezs%geds@mhsgate.salem.ge.com  Tue Aug  9 11:18:04 1994
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Date: Tue, 9 Aug 94 09:42:55 EST
From: Ybanez Sheldon <ybanezs%geds@mhsgate.salem.ge.com>
To: cube-lovers@ai.mit.edu
Subject: Cube Availability?
Return-Receipt-To: <ybanezs%geds@mhsgate.salem.ge.com>
X-Mailer: XGATE 2.12 MHS/SMTP Gateway


Fellow Cubers:

Does anyone know if the original 3x3 Rubik's Cube is still available?  My 
cube finally fell apart without any hope of repairing....  I still have 
my 'Revenge (which is also not too far from its last days) but I would 
still like to be able to play with the original... and keep my fingers 
nimble....  If the original cube is still available could you please post 
an address or something so that I can try to acquire one....

thanks,
                                                         ,,,
______________________________________________________  (o o)  _________
+----------------------------------------------------ooO-(_)-Ooo-------+
| Sheldon Ybanez  [ybanez-s@salem.ge.com]  GE Drive Systems  Salem, VA |
|     Always "Remember. No matter where you go, there you are." 88     |
+======================================================================+



From @mail.uunet.ca:mark.longridge@canrem.com  Tue Aug  9 15:17:23 1994
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Subject: < U, R> Group
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.783.5834.0C1A9A39@canrem.com>
Date: Tue, 9 Aug 1994 01:48:00 -0400
Organization: CRS Online  (Toronto, Ontario)

Well I decided to pull a "Jerry Byran" and take another look at
 some cube results, plus take a look at some new groups.


        Analysis of the 3x3x3 squares group
        -----------------------------------

                  (h only)      branching
Moves Deep       arrangements    factor      loc max (h only)

  0                    1           --             0
  1                    6           6              0
  2                   27           4.5            0
  3                  120           4.444          0
  4                  519           4.325          0
  5                1,932           3.722          0
  6                6,484           3.356          1  (6 X pattern)
  7               20,310           3.132          0
  8               55,034           2.709         65
  9              113,892           2.069      1,482
 10              178,495           1.567      7,379
 11              179,196           1.004     25,980
 12               89,728           0.501     50,320
 13               16,176           0.180     11,328
 14                1,488           0.092        912
 15                  144           0.096        144
                 -------                     ------
                 663,552                     97,611

        Analysis of the 3x3x3 <U, R> group
        ----------------------------------

                                          branching
Moves Deep       arrangements (q only)     factor

  0                    1                      --
  1                    4                       4
  2                   10                       2.5
  3                   24                       2.4
  4                   58                       2.416
  5                  140                       2.413
  6                  338                       2.414
  7                  816                       2.414
  8                1,970                       2.414
 program starts to really bog down after this...

  I leave it to Jerry or Dan to check my results. I checked up to 2
moves deep by hand and verified 10 different positions. What I don't
understand is how Jerry manages to look at so many cube positions:

On full 3x3x3 cube,   7     100,803,036     13.231  (new)

  Using 10 bytes to store a single cube position would still
need over 1 billion bytes, or am I missing something?

  I also used GAP (quite a good program) to calculate the size of
  < U1, R1 > on the magic dodecahedron: 7,999,675,084,800.

  Once again, I welcome any verification.

  -> Mark <-

From BRYAN@wvnvm.wvnet.edu  Wed Aug 10 08:38:50 1994
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Date:      Tue, 9 Aug 1994 23:18:06 EDT
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <Cube-Lovers@ai.mit.edu>
Subject:   Re: < U, R> Group
In-Reply-To: Message of 08/09/94 at 01:48:00 from mark.longridge@canrem.com

On 08/09/94 at 01:48:00 mark.longridge@canrem.com said:

>  I leave it to Jerry or Dan to check my results. I checked up to 2
>moves deep by hand and verified 10 different positions. What I don't
>understand is how Jerry manages to look at so many cube positions:

>On full 3x3x3 cube,   7     100,803,036     13.231  (new)

>  Using 10 bytes to store a single cube position would still
>need over 1 billion bytes, or am I missing something?

Well, when I deal with the big problems I want to solve to the
bitter end, I use the M-conjugate and centerless cube tricks
I have described at much too great length in the past.  This one
is a quick and dirty program using no conjugate tricks.  The
only real "trick" is that I externalize the data.  I decided
long ago that the problems I wanted to solve were too big to
keep in memory.  Hence, I keep everything in simple (but large)
flat files and sort and merge the files like crazy.  In this quick
and dirty program, the cube itself is 13 bytes and the level is
1 byte, for a total of 14 bytes per cube.  I guess that makes
the file size about 1.4 gigabytes (10^9).  I am
leery of using the word "billion" on E-mail forums because
E-mail is international and "billion" means 10^12 in some
countries.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From BRYAN@wvnvm.wvnet.edu  Sat Aug 13 17:17:43 1994
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Date:      Sat, 13 Aug 1994 17:14:53 EDT
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   GC Local Maxima, Additional Information

In all my God's algorithm searches, I have never calculated local
maxima.  My search technique and data representation do not lend
themselves very well to calculating local maxima.  However, I thought
I would give it a try anyway.  I decided to start by calculating local
maxima for GC, the corners only group, since local maxima for this group
have been calculated before and I could check my answers.  I have
come up with some surprising results.

For a given cube X, there are twelve (not necessarily distinct)
neighbors of the form Xq, where q is in Q, the set of twelve
quarter turns.  Each Xq is either one move closer to Start or one
move further from Start than X.  I decided to determine for each cube
X, how many of the Xq were closer to Start and how many were further
from Start.  This is a superset of the local maxima problem.  Those
X for which I determine that all twelve of the Xq are closer to Start
are the local maxima, but I also determine for how many of the X
there are eleven neighbors Xq closer to Start, for how many of the X
there are ten neighbors Xq closer to Start, etc.  This is where the
surprising results come in.

The following table summarizes the results.  The table is a little hard
to read.  The rows (from 0 to 14) are the distances from Start.  The
columns (from 0 to 12) are the number of qturns which take you closer
to Start.  For example, row 4 column 3 contains 480.  This means that
there 480 cubes that are 4 moves from Start and for which 3 of the 12
qturns will take you closer to Start.  The table is too wide for a
computer screen, so I split it.  Columns 0 through 6 appear first, and
then columns 7 through 12 are below.

Column 12 represents the local maxima.  The "Total" column is simply
the total number of cubes at each level. The "Total" column and the
local maxima column appear several times in the archives, and my
numbers match the archives.  The first occurrence I have found is
Dan Hoey's note of 20 August 1984.

                             Number of Twists which Go Towards Start

     Level    Total  0      1        2     3        4     5        6
              Cubes

        0         1  1      0        0     0        0     0        0
        1        12  0     12        0     0        0     0        0
        2       114  0     96       18     0        0     0        0
        3       924  0    672      192    60        0     0        0
        4      6539  0   4032     1920   480       51     0       56
        5     39528  0  19104    14904  3792      984   216      384
        6    199926  0  71184    90984 16656    13212  1872     3936
        7    806136  0 123360   478008 42768   117576  7824    16656
        8   2761740  0  23328  1911312  9024   643536  2736   121872
        9   8656152  0      0  5573376     0  2327616     0   558336
       10  22334112  0      0 11167488     0  7057440     0  2818176
       11  32420448  0      0  4661568     0  8314272     0  8893248
       12  18780864  0      0    19008     0   123840     0   591744
       13   2166720  0      0        0     0        0     0        0
       14      6624  0      0        0     0        0     0        0

     Total 88179840  1 241788 23918778 72780 18598527 12648 13004408


                            7        8     9       10    11       12

        0                   0        0     0        0     0        0
        1                   0        0     0        0     0        0
        2                   0        0     0        0     0        0
        3                   0        0     0        0     0        0
        4                   0        0     0        0     0        0
        5                 144        0     0        0     0        0
        6                 528     1344     0       96     0      114
        7                1680     9096  1536     6552   480      600
        8                 384    23232    96     7584   720    17916
        9                   0   167616     0    19008     0    10200
       10                   0  1020384     0   235584     0    35040
       11                   0  6746688     0  2986560     0   818112
       12                   0  2202912     0  6189120     0  9654240
       13                   0      288     0    39168     0  2127264
       14                   0        0     0        0     0     6624

                         2736 10171560  1632  9483672  1200 12670110

The first surprising thing I noticed is the diagonals, especially
close to Start.  I am not quite sure why there should be diagonals.
I would describe the diagonals as a weak feature of the chart, but
they are surely there.  I think the diagonals mean something as follows.
Pick a cube X which is N moves from Start, and for which M qturns
will take you closer to Start.  Move to a cube Xq which is N+1 moves
from Start.  Then there is a *tendency* (not a certainty!) for there
to be M+1 moves which will take Xq closer to Start.  I can't think
of any reason for this to be so, but the chart has diagonals.

The second surprising thing is that the chart contains a preponderance
of even numbers.  There are only
two odd numbers in the whole chart.  Row 0 column 0
contains a 1, and row 4 column 4 contains a 51.  All the other cells
in the chart contain an even number.  Furthermore, by comparison to
each other, the even columns are densely populated and the odd columns
are sparsely populated.  Finally, below level 8, the
odd columns are completely empty.

It is often the case that when there are an even number of objects, there
is some natural pairing that can be performed on the objects.  In this
case, I think the pairing that can be performed to explain the even numbers
is twists of opposite faces.  R can be paired with L, R' can be paired
with L', U can be paired with D, etc.

The corners-only group GC is "almost" the same as the corners-only-
without-centers group GC/C.  (GC/C is also called a 2x2x2 cube or a
pocket cube).  In GC/C, the pairing between moves of opposite faces
is absolute.  You can always choose either of two opposite faces with
equivalent results.  In GC, the pairing is relative.  You can "almost"
solve GC the same way as GC/C, but sometimes you have to be sensitive
to which of two opposite faces you twist in order to rotate the
corners properly.  Dan's 20 August 1984 note explains this phenomenon
much better than I can:

>The alert reader will notice that rows 10 through 14 contain values
>exactly 24 times as large as those for the pocket cube.  This is not
>surprising, given that the groups are identical except for the position
>of the entire assembly in space, and each generator of the corner cube
>is identical to the inverse of the corresponding generator for the
>opposite face except for the whole-cube position.  Thus when solving a
>corner-cube position at 10 qtw or more from solved, it can be solved as
>a pocket cube, making the choice between opposite faces in such a way
>that the whole-cube position comes out right with no extra moves.

I can't fully explain why Dan's results are for rows 10 through 14,
whereas in my chart the odd columns are empty for rows 9 through 14.
Also, any rotation of the corners can be accomplished in no more than
6 qturns (see my note of 4 December 1993 concerning the corners of
the 3x3x3).  I think that the explanation is something to the effect
that for rows 10 through 14, if (for example) R will take you closer
to Start, then so too will L, and vice versa.  I don't think you have
to start choosing between (for example) R and L to accomplish the proper
rotation until you get below level 10.

Perhaps Dan can fully explain the mystery:  why is it
that a rotation of the corners only requires 6 qturns, full pairing
of opposite face turns kicks in at level 9, and GC becomes
exactly 24 times GC/C at level 10?  What is happening between level 6
and level 10?  Why don't all three phenomena kick in at the same level?

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From BRYAN@wvnvm.wvnet.edu  Fri Aug 19 16:26:59 1994
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From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Updated Upper Limits, Q-turns

On 9 January 1981, Dan Hoey provided a recursive procedure which
gives the best known upper bound on the number of
cubes at each distance from Start.  With Dan's recursive procedure,
the upper bound for any level is a function of the known value or
upper bound for the immediately preceding four levels.  Dan's procedure
takes into account identities of the form XX'=I and RL=LR.

At the time Dan performed his calculations, only level 0 through level 4
were known for sure.  We now have 8 levels, so Dan's calculations can be
updated.  I am going to give the new calculations, and I am also
going to include Dan's original calculations for comparison purposes.
In both tables, P[n] is the number of cubes which are n moves from
Start.

Dan's recursion formula is:

>   P[n]  <= 4*2*P[n-1] + 6*2*P[n-2] + 4*2*P[n-3] + 1*2*P[n-4]


Dan's calculations:

> P[0]  =           1   P[9]  <  724,477,008     P[18] <  4.048*10^17
> P[1]  =          12   P[10] <  6.792*10^9      P[19] <  3.795*10^18
> P[2]  =         114   P[11] <  6.366*10^10     P[20] <  3.557*10^19
> P[3]  =       1,068   P[12] <  5.967*10^11     P[21] <  3.334*10^20
> P[4]  =      10,011   P[13] <  5.594*10^12     P[22] <  3.125*10^21
> P[5]  <=     93,840   P[14] <  5.243*10^13     P[23] <  2.930*10^22
> P[6]  <     879,624   P[15] <  4.915*10^14     P[24] <  2.746*10^23
> P[7]  <   8,245,296   P[16] <  4.607*10^15     P[25] <  2.574*10^24
> P[8]  <  77,288,598   P[17] <  4.319*10^16


New Calculations:

  P[0]  =           1   P[9]  <= 720,627,064     P[18] <= 4.026*10^17
  P[1]  =          12   P[10] <= 6.755*10^09     P[19] <= 3.774*10^18
  P[2]  =         114   P[11] <= 6.332*10^10     P[20] <= 3.538*10^19
  P[3]  =       1,068   P[12] <= 5.935*10^11     P[21] <= 3.316*10^20
  P[4]  =      10,011   P[13] <= 5.563*10^12     P[22] <= 3.108*10^21
  P[5]  =      93,840   P[14] <= 5.215*10^13     P[23] <= 2.914*10^22
  P[6]  =     878,880   P[15] <= 4.888*10^14     P[24] <= 2.731*10^23
  P[7]  =   8,221,632   P[16] <= 4.582*10^15     P[25] <= 2.560*10^24
  P[8]  =  76,843,595   P[17] <= 4.295*10^16

I think that the two most interesting things about the new calculations
are: 1) they are nearly the same as the old calculations, and 2) they
are not exactly the same as the old calculations.  In both cases, the
question is "why?".

My interpretation is that Dan's analysis not only puts an upper
bound on the number cubes at each level, it also puts an upper bound
on the branching factor.  We trivially have an absolute upper limit
on the branching factor of 12.  After level 1, we trivially have an
upper limit on the branching factor of 11 (i.e., "don't undo the move
you just made", or "don't have a sequence of the form XX'").  As before,
moves of opposite faces commute.  Taking commutations of opposite faces
into account, the branching factor is reduced (empirically ) to an upper
limit of about 9.37.

This empirical analysis is starting with a high branching factor and
subtracting out the cubes we should not count, so that we are
dealing with identities of the form XX' and commutations of the form
RL=LR separately.  Dan's analysis deals with cubes we *should* count,
and he thereby deals with identities of the form XX' and commutations
of the form RL=LR in one fell swoop.

But Dan's analysis does not yield exact figures, only limits.  It seems
therefore that there must be other cases our empirical approach must
choose not to count.  What might those other cases be?  It seems that
there must be cases where a sequence X1 X2 ... Xn
is equal to a sequence Y1 Y2 ... Ym, but where there is no obvious
way to characterize the relationship between two sequences (e.g., they
are not simple commutations of each other), and where we cannot even
find the sequences without some sort of exhaustive search.

I would interpret that fact that the new upper limits do not equal the
old upper limits as meaning that such "duplicate" sequences do exist
close to Start.  I would interpret the fact that the new upper limits
are close to the old upper limits as meaning that there are not very
many such "duplicate" sequences close to Start.

But consider another quote from Dan in the same article:

>The recurrence on which this bound relies is due to the
>relations F^4 = FBF'B' = I (and their M-conjugates.)  It may be
>possible to improve the recurrence by recognizing other short
>relations.  Exhaustive search has shown that there are none of
>length less than 10.

I am afraid I need Dan to explain this further.  Dan's logic seems
impeccable.  But on the other hand
there must be cases where X1 X2 ... Xn = Y1 Y2 ... Ym, where
the sum of the length of the sequences is less than 10, and where
the equality is not explained by the relations F^4 = FBF'B' = I.
Otherwise, Dan's calculations would yield exact values rather than
upper limits close to Start, and the "new calculations" for upper
limits would equal the "old calculations".

Let me think out loud just for a second.  Consider relations such as
LRLRLRLR = I or RR'RR'RR'RR' = I.  These are *sequences* of length 8
but *cubes* of length 0.  Is it possible that such sequences are being
counted too many or not enough times when the recursion is four
levels deep?

Finally, I have argued on purely empirical grounds that the branching
factor will remain relatively constant from about level 3 to some
unknown level (maybe about level 18 or 19 or 20?), where the branching
factor will decay rapidly because you run out of cubes.  Well, I think
I want to argue further that during this "relatively constant" portion
of the distribution the branching factor *will* decay.  It might not
decay very much, and I don't see any easy way to calculate how much
it will decay.  The argument is very simple.  Any time a "duplicate
sequence" occurs, it reduces the branching factor at that level, but
also at subsequent levels.  That is, longer sequences can contain the
"duplicate sequence" as a sub-sequence.  Hence, any decay in the branching
factor at one level is propagated to all subsequent levels.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From mouse@collatz.mcrcim.mcgill.edu  Sun Aug 21 06:33:13 1994
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Date: Sun, 21 Aug 1994 06:33:02 -0400
From: der Mouse <mouse@collatz.mcrcim.mcgill.edu>
Message-Id: <199408211033.GAA06700@Collatz.McRCIM.McGill.EDU>
To: Cube-Lovers@ai.mit.edu
Subject: Re: Updated Upper Limits, Q-turns

> Dan's recursion formula is:

>>   P[n]  <= 4*2*P[n-1] + 6*2*P[n-2] + 4*2*P[n-3] + 1*2*P[n-4]

> Dan's calculations:

>> P[0]  =           1
>> P[1]  =          12
>> P[2]  =         114
>> P[3]  =       1,068
>> P[4]  =      10,011

Ummm.  4*2*P[4-1] + 6*2*P[4-2] + 4*2*P[4-3] + 1*2*P[4-4] =
       4*2*1068 + 6*2*114 + 4*2*12 + 1*2*1 = 10010 < P[4].

What have I missed?  Is Dan's formula not valid until n=5 or something?

					der Mouse

			    mouse@collatz.mcrcim.mcgill.edu

From BRYAN@wvnvm.wvnet.edu  Sun Aug 21 09:08:01 1994
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Date:      Sun, 21 Aug 1994 08:18:29 EDT
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "der Mouse" <mouse@collatz.mcrcim.mcgill.edu>,
        "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Re: Updated Upper Limits, Q-turns
In-Reply-To: Message of 08/21/94 at 06:33:02 from ,
           mouse@collatz.mcrcim.mcgill.edu

On 08/21/94 at 06:33:02 der Mouse said:
>> Dan's recursion formula is:

>>>   P[n]  <= 4*2*P[n-1] + 6*2*P[n-2] + 4*2*P[n-3] + 1*2*P[n-4]

>> Dan's calculations:

>>> P[0]  =           1
>>> P[1]  =          12
>>> P[2]  =         114
>>> P[3]  =       1,068
>>> P[4]  =      10,011

>Ummm.  4*2*P[4-1] + 6*2*P[4-2] + 4*2*P[4-3] + 1*2*P[4-4] =
>       4*2*1068 + 6*2*114 + 4*2*12 + 1*2*1 = 10010 < P[4].

>What have I missed?  Is Dan's formula not valid until n=5 or something?

>                                        der Mouse

I had just noticed the same thing, and intended to investigate.  I don't
know what happens to Dan's formula for n=4.

At the time Dan's chart was first published, P[n] was only known
for sure for n = 0..4.  Dan showed strict equality for these
levels, and I assume P[4] came from the known values rather than
from the formula.  It still does not explain why the formula
yields a value which is too low for P[4]  --  I could easier
understand why it yielded a value which is too high, but it seems
to me that it ought to yield the exact value that close to Start.

For P[5], Dan's original chart showed "=<".  Subsequent computer
search changed this to strict equality, which is a great victory
for Dans' formula.  The first term for which
Dan's chart is too high is P[6].  I had therefore intended to
start my investigations at that point until I discovered the
discrepancy at P[4].

Just as a reality check, let me mention some trivial points.  Suppose
it is discovered that (X1 X2 ... Xn) = (Y1 Y2 ... Ym).  Define
X = (X1 X2 ... Xn) and Y = (Y1 Y2 ... Ym).  Since X = Y, it is
immediate that XY' = Y'X = X'Y = YX' = I.  Conversely, a sequence
(X1 X2 ... Xn) = I can be decomposed into X = (X1 X2 ... Xk) and
Y = (X[k+1] ... Xn).  Then, XY = I and hence X and Y' (and also
X' and Y) are what I have called "duplicate sequences", that is
different sequences which yield the same cube.  This is why
identities are so important for bounding P[n].

I seem to do everything backwards, so I would just look for the duplicate
sequences.  However, it is probably more elegant to look for the identities.
Dan's original note said that computer
search has shown that there are not any identities other than the ones we
already know about up through length 10.  It looks to me like Dan's
formula takes care of the identities we already know about.  So
as usual, I am probably missing something obvious.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From BRYAN@wvnvm.wvnet.edu  Sun Aug 21 09:57:44 1994
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From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Analysis of Turns Towards Start for Whole Cube

The following chart for level 0 through level 8 of the whole
cube using Q-turns gives the number of Q-turns for each cube which will
move towards Start.  (I recently gave the same analysis for corners only.)

Columns 9 through 12 are omitted from the chart since they contain
all zeros.  Any local maxima would appear in column 12.

This adds one new level known not to contain any local maxima.  It would
be extremely interesting to be able to extend the chart at least to
level 12 because level 12 is known to include a local maximum.

As with the corners-only case, the chart contains almost all even numbers.
However, unlike the corners-only case, the numbers do not cluster in
the even columns.  Rather, they cluster towards column 1.  This means that
(close to Start, at least) most cubes have only one "good" move that
takes you closer to Start.  It also serves to illustrate why "random"
moves so quickly scramble the cube.

                           Number of Q-turns which Move
                                Closer to Start

     Level  Total  0         1         2      3      4   5    6   7   8
            Cubes

       0        1  1         0         0      0      0   0    0   0   0
       1       12  0        12         0      0      0   0    0   0   0
       2      114  0        96        18      0      0   0    0   0   0
       3     1068  0       912       144     12      0   0    0   0   0
       4    10011  0      8544      1368     96      3   0    0   0   0
       5    93840  0     80088     12816    912     24   0    0   0   0
       6   878880  0    749376    120612   8640    252   0    0   0   0
       7  8221632  0   7001712   1135104  82152   2664   0    0   0   0
       8 76843595  0  65391504  10645824 777936  28200  48   56   0  27

  Total  86049153  1  73232244  11915886 869748  31143  48   56   0  27

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From BRYAN@wvnvm.wvnet.edu  Wed Aug 24 23:07:21 1994
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Date:      Wed, 24 Aug 1994 23:06:52 EDT
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "der Mouse" <mouse@collatz.mcrcim.mcgill.edu>,
        "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Re: Updated Upper Limits, Q-turns
In-Reply-To: Message of 08/21/94 at 06:33:02 from ,
           mouse@collatz.mcrcim.mcgill.edu

On 08/21/94 at 06:33:02 der Mouse said:
>> Dan's recursion formula is:

>>>   P[n]  <= 4*2*P[n-1] + 6*2*P[n-2] + 4*2*P[n-3] + 1*2*P[n-4]

>> Dan's calculations:

>>> P[0]  =           1
>>> P[1]  =          12
>>> P[2]  =         114
>>> P[3]  =       1,068
>>> P[4]  =      10,011

>Ummm.  4*2*P[4-1] + 6*2*P[4-2] + 4*2*P[4-3] + 1*2*P[4-4] =
>       4*2*1068 + 6*2*114 + 4*2*12 + 1*2*1 = 10010 < P[4].

>What have I missed?  Is Dan's formula not valid until n=5 or something?

I just rechecked Dan's original note of 9 January 1981.  He specifically
says the formula is good for n > 4.  Mea culpa.  However, I still do
not fully understand *why* this should be the case.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From hoey@aic.nrl.navy.mil  Thu Aug 25 14:43:40 1994
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Date: Thu, 25 Aug 94 14:43:13 EDT
From: hoey@aic.nrl.navy.mil (Dan Hoey)
Message-Id: <9408251843.AA03267@Sun0.AIC.NRL.Navy.Mil>
To: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>, <mouse@collatz.mcrcim.mcgill.edu>
Cc: <Cube-Lovers@ai.mit.edu>
Subject: Re: Updated Upper Limits, Q-turns

Jerry Bryan was looking at some formulas I had in the archives:

    >> Dan's recursion formula is:
    >>>   P[n]  <= 4*2*P[n-1] + 6*2*P[n-2] + 4*2*P[n-3] + 1*2*P[n-4]

    ...

    I just rechecked Dan's original note of 9 January 1981.  He
    specifically says the formula is good for n > 4.  Mea culpa.
    However, I still do not fully understand *why* this should be the
    case.

I thought I put the bound in....  I've been meaning to look that up
and explain it, but time's been short.

The analysis is done by breaking up the minimal processes into
"syllables", where a syllable is a maximal sequence of commuting turns.
So for each pair (x,y) in {(F,B),(T,D),(L,R)} there are
    four syllables of length 1: x, x', y, and y';
    six syllables of length 2: xx, xy, xy', x'y, x'y', and yy;
    four syllables of length 3: xxy, xxy', xyy, and x'yy;
    one syllable of length 4: xxyy.

(It's not really a coincidence that this is most of the fifth line of
Pascal's triangle.)

Now for the first syllable, we can pick any of the three pairs for
(x,y).  But for succeeding syllables, we must pick a pair that is not
equal to the preceding pair.  So each term in the recurrence refers to
the length of the last syllable:

Length of last syllable=1            2            3            4

  n=1    P[n] = 4*3 P[n-1];
  n=2    P[n] = 4*2 P[n-1] + 6*3 P[n-2]
  n=3    P[n] = 4*2 P[n-1] + 6*2 P[n-2] + 4*3 P[n-3]
  n=4    P[n] = 4*2 P[n-1] + 6*2 P[n-2] + 4*2 P[n-3] + 1*3 P[n-4]
  n>4    P[n] = 4*2 P[n-1] + 6*2 P[n-2] + 4*2 P[n-3] + 1*2 P[n-4]

The second part of each coefficient is 2, except that when the length
of the last syllable is equal to n (so that we are counting the first
syllable), the second part of the coefficient is 3.

In response to my description:
> >The recurrence on which this bound relies is due to the
> >relations F^4 = FBF'B' = I (and their M-conjugates.)  It may be
> >possible to improve the recurrence by recognizing other short
> >relations.  Exhaustive search has shown that there are none of
> >length less than 10.

Jerry continues:
> ... there must be cases where X1 X2 ... Xn = Y1 Y2 ... Ym, where
> the sum of the length of the sequences is less than 10, and where
> the equality is not explained by the relations F^4 = FBF'B' = I.
> Otherwise, Dan's calculations would yield exact values rather than
> upper limits close to Start, and the "new calculations" for upper
> limits would equal the "old calculations".

No.  The bounds fail to be exact when we have a relation r=s with
|r|=|s|=n.  This corresponds to a relation r s'=I of length 2n.  The
shortest relations of length >4 are the ones of length 12 (as I
reported on 22 March 1981) so my bounds become inexact at length 6.

Chris Worrell listed the length-12 relations on 08/02/81, and I
reported that his list was complete on 14 August 1981 0111-EDT.  The
12-qtw identities (up to M-conjugacy) are:

I12-1	FR' F'R    UF' U'F    RU' R'U
I12-2	FR' F'R   UF'   F'L FL'   U'F
I12-3	FR' F'R   UF'   UL' U'L   FU'

As Allan C. Wechsler noted on 17 August 1981, any two of them can be
combined to form the third.

> Consider relations such as LRLRLRLR = I or RR'RR'RR'RR' = I.

The first is a consequence of the relations L^4=R^4=LRL'R'=I.  The
second is a consequence of group theory; no relations are needed.  The
recurrence deals with these: it models the freeest group specified by
the given relations.

I have tried unsuccessfully to create a recurrence that will deal with
the 12-qtw identities, but it's complicated.  For instance, repeatedly
putting I12-1 in the center of another I12-1 yields identities of the
form:

     F (R'F' RU)^n F'U' (FR U'R')^n U

There are a bunch of other cases, too.

Dan
Hoey@AIC.NRL.Navy.Mil

From BRYAN@wvnvm.wvnet.edu  Wed Aug 31 17:17:08 1994
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Date:      Wed, 31 Aug 1994 15:23:47 EDT
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <cube-lovers@ai.mit.edu>
Subject:   Re: < U, R> Group
In-Reply-To: Message of 08/09/94 at 01:48:00 from mark.longridge@canrem.com

On 08/09/94 at 01:48:00 mark.longridge@canrem.com said:

>        Analysis of the 3x3x3 <U, R> group
>        ----------------------------------

>                                          branching
>Moves Deep       arrangements (q only)     factor

>  0                    1                      --
>  1                    4                       4
>  2                   10                       2.5
>  3                   24                       2.4
>  4                   58                       2.416
>  5                  140                       2.413
>  6                  338                       2.414
>  7                  816                       2.414
>  8                1,970                       2.414
> program starts to really bog down after this...

>  I leave it to Jerry or Dan to check my results. I checked up to 2
>moves deep by hand and verified 10 different positions.

Ok, here it is.  This search is narrower and deeper than any
I have ever done before.  Frey and Singmaster give <U,R>
a good bit of attention in their book, pointing out that it
is trickier than it might first appear.

It is called the Two-Generator Group.  The size of the group
can be calculated as (7!5!/2)(3^6/3) = 73,483,200.  The
3^6 factor accounts for twisting the corners, but there is no
2^n factor as the edges cannot be flipped.

These results are in terms of Q turns without any conjugate
class checking.  I would regard the following as open problems:
local maxima, results with Q+H turns, and results in terms of
conjugate classes.  In this particular case, it would not be
M-conjugates.  I would have to look at Dan Hoey's 98 subgroups
of M to see which subgroup applies to <U,R>.

  0                    1
  1                    4                       4
  2                   10                       2.5
  3                   24                       2.4
  4                   58                       2.416
  5                  140                       2.413
  6                  338                       2.414
  7                  816                       2.414
  8                1,970                       2.414
  9                4,756                       2.414
 10               11,448                       2.407
 11               27,448                       2.398
 12               65,260                       2.378
 13              154,192                       2.363
 14              360,692                       2.339
 15              827,540                       2.294
 16            1,851,345                       2.237
 17            3,968,840                       2.144
 18            7,891,990                       1.988
 19           13,659,821                       1.755
 20           18,471,682                       1.352
 21           16,586,822                       0.898
 22            8,039,455                       0.485
 23            1,511,110                       0.188
 24               47,351                       0.031
 25                   87                       0.002

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

From @mail.uunet.ca:mark.longridge@canrem.com  Fri Sep  2 00:08:42 1994
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Sender: CRSO.Cube@canrem.com
Subject:  < U, R > revisited
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.796.5834.0C1AD333@canrem.com>
Date: Thu, 1 Sep 1994 22:56:00 -0400
Organization: CRS Online  (Toronto, Ontario)

        Analysis of the 3x3x3 <U, R> group (continued)
        ----------------------------------

                                          branching
Moves Deep       arrangements (q only)     factor

  0                    1        1             --
  1                    4        5              4
  2                   10       15              2.5
  3                   24       39              2.4
  4                   58       97              2.416
  5                  140      237              2.413
  6                  338      575              2.414
  7                  816    1,391              2.414
  8                1,970    3,361              2.414
  9                4,756    8,117              2.414
 10               11,448   19,565              2.407
 11               27,448   47,013              2.401
ML's Conjecture: The < U, R > group is >=20 turns deep in qt metric

UR Reflective processes: (in the q metric)
A different sort of symmetry which I started to investigate, having
 been inspired by my friend who solves his cube 2 adjacent faces
 last!

These are the only UR reflective processes at 10 q turns:
     U3 R1 U1 R1 (U2) R3 U3 R3 U1 = R3 U1 R1 U1 (R2) U3 R3 U3 R1   (10)
     U1 R3 U3 R3 (U2) R1 U1 R1 U3 = R1 U3 R3 U3 (R2) U1 R1 U1 R3   (10)

Here is the obvious one we all know:
                    ( U2 R2 ) ^ 3 = ( R2 U2 ) ^ 3                  (12)

I liked this pattern in particular...
       U1 R1 U2 R3 U2 R3 U2 R1 U1 = R1 U1 R2 U3 R2 U3 R2 U1 R1     (12)

I hope to have an algorithm to plumb the depths of the < U, R > group
 soon. Amusingly my friend complained about not been able solve
 the cube completely as he was stuck in a position with 2 flipped
 edges. After watching him squirm for a few weeks I did tell him
 you can't flip edges in the < U, R > group!     ;->

Congrats to Dan Hoey, Dik Winter, Jerry Bryan and Ludwig Plutonium
 for making it into the 1994 Internet White Pages! I'm in good
 company.

        -> Mark <-
        Email: mark.longridge@canrem.com

P.S. I just read the last J.B. post and see I've been somewhat
     overshadowed. Ok let's see some antipodes! At least our
     results are the same though. So, ummmm I guess ML's
     conjecture is correct!

From BRYAN@wvnvm.wvnet.edu  Mon Sep  5 09:25:29 1994
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Date:      Mon, 5 Sep 1994 09:08:14 EDT
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <cube-lovers@ai.mit.edu>
Subject:   Re: < U, R > revisited (with Local Maxima)
In-Reply-To: Message of 09/01/94 at 22:56:00 from mark.longridge@canrem.com

I performed a Local Maxima analysis.  (This is for Q-turns only.)  The Local
Maxima are in column 4.  The shortest local maxima (six of them) are of
length 12.  Interestingly, this is the same length which is suspected of
being the shortest Q-turn length for a local maximum in the full cube group.
Is there any connection?  Also, the global maxima are of length 25.  Does
this tell us anything about the Q-turn length of the global maxima for the
full cube group?  Finally, pick any cube X in <U,R>.  We know
|X| in G <= |X| in <U,R>.  Can anybody find a cube X such that
|X| in G < |X| in <U,R>?  Alternatively, can anybody prove
|X| in G = |X| in <U,R> for all X in <U,R>?  (This assumes Q-turns only in
all cases.  The questions would all have to be asked again for Q+H-turns.)

                         Number of Moves Which Go Closer to Start

     Level       Total   0           1           2           3          4
                 Cubes

         0           1   1           0           0           0          0
         1           4   0           4           0           0          0
         2          10   0           8           2           0          0
         3          24   0          20           4           0          0
         4          58   0          48          10           0          0
         5         140   0         116          24           0          0
         6         338   0         280          58           0          0
         7         816   0         676         140           0          0
         8       1,970   0       1,632         338           0          0
         9       4,756   0       3,940         816           0          0
        10      11,448   0       9,448       1,996           4          0
        11      27,448   0      22,584       4,836          28          0
        12      65,260   0      53,236      11,862         156          6
        13     154,192   0     125,196      28,616         360         20
        14     360,692   0     289,908      69,196       1,472        116
        15     827,540   0     652,792     168,008       6,180        560
        16   1,851,345   0   1,428,560     398,634      21,860      2,291
        17   3,968,840   0   2,938,808     934,908      84,312     10,812
        18   7,891,990   0   5,422,844   2,109,480     309,916     49,750
        19  13,659,821   0   8,065,268   4,288,796   1,068,480    237,277
        20  18,471,682   0   7,948,748   6,625,644   2,947,320    949,970
        21  16,586,822   0   3,485,748   5,507,066   4,831,060  2,762,948
        22   8,039,455   0     286,176   1,165,888   2,665,080  3,922,311
        23   1,511,110   0         740      15,202     156,432  1,338,736
        24      47,351   0           0           8         332     47,011
        25          87   0           0           0           0         87

            73,483,200   1  30,736,780  21,331,532  12,092,992  9,321,895

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
Robert G. Bryan (Jerry Bryan)              (304) 293-5192
Associate Director, WVNET                  (304) 293-5540 fax
837 Chestnut Ridge Road                     BRYAN@WVNVM
Morgantown, WV 26505                        BRYAN@WVNVM.WVNET.EDU

If you don't have time to do it right today, what makes you think you are
going to have time to do it over again tomorrow?

