From alan@curry.epilogue.com  Fri May 19 22:01:05 1995
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Date: Fri, 19 May 1995 22:01:05 -0400
Message-Id: <19May1995.194423.Alan@LCS.MIT.EDU>
From: Alan Bawden <Cube-Lovers-Request@ai.mit.edu>
Sender: Cube-Lovers-Request@ai.mit.edu
Reply-To: Cube-Lovers-Request@ai.mit.edu
To: Cube-Lovers@ai.mit.edu
Subject: The Cube-Lovers Archives expands

Occasionally people have approached me and wondered if there was an FTP
archive of Cube-related material (programs, documents, databases, pictures,
whatever).  I have always replied that the only archive I knew of was the
archive of Cube-Lovers mail that I maintain.  Since this question keeps
coming up, there must be a need to be filled, so I propose to expand our
archives to cover any additional Cube-related material that people might
care to submit.

If you would like to submit a contribution to this archive, please send
mail to Cube-Lovers-Request@AI.MIT.EDU (please do -not- send mail to all of
Cube-Lovers) and include:

  o  The location where I can pick up the files you wish to contribute
     (preferably using anonymous FTP).

  o  A brief description of your contribution, to be included in a master
     index file.

I reserve the right to redescribe, repackage, rename, recompress or totally
reject your contribution.

Periodically I will announce new additions to the archive to all of
Cube-Lovers.  Currently the archives contain nothing other than the
electronic mail archives.  (Although I know of at least one potential
contributor who's been waiting in the wings for a couple of months now...)

Some of you will no doubt have forgotten where the archive is: 

Using FTP, connect to FTP.AI.MIT.EDU, login as "anonymous" (any password),
and go to the directory "pub/cube-lovers".  (From the World Wide Web, you
can use the URL: "ftp://ftp.ai.mit.edu/pub/cube-lovers".)

				- Alan

From BRYAN@wvnvm.wvnet.edu  Sun May 21 15:24:47 1995
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Date:      Sun, 21 May 1995 07:18:49 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   AntiSlice Under M-conjugacy (and a problem with slice)

I have had some posts not get through.  The following will serve
to consolidate several of them. Some of this may be a repeat,
but not all, I think.

Mark Longridge's antislice results are as follows:

>                  arrangements           arrangements
>Moves Deep   (2q or anti-slice moves)   (4q or double anti-slice moves)
>
>    0                   1                     1
>    1                   6                     9
>    2                  27                    51
>    3                 120                   265
>    4                 423                   864
>    5               1,098                 1,785
>    6               1,770                 2,017
>    7               1,650                 1,008
>    8                 851                   144
>    9                 198
>                    -----                 -----
>                    6,144                 6,144

We have the following M-conjugacy results for 2q moves.

  Level              Positions             Local
                                          Maxima

   0                    1                    0
   1                    1                    0
   2                    3                    0
   3                   10                    0
   4                   37                    0
   5                   93                    1
   6                  166                    2
   7                  147                    7
   8                   89                   12
   9                   21                   21
                     ----
                      568

The level 5 local maximum is (U'D')(FB)(FB)(UD)(L'R').  The position is
not its own inverse, but we can use as an inverse (U'D')(FB)(FB)(UD)(LR).
Hence, (U'D')(FB)(FB)(UD) forms a nice "middle" of the sequence.  In
fact, the (U'D')(FB)(FB)(UD) position in some ways seems more
interesting than the local maximum itself.  Does it already have
a name?

I have not verified if the length of the local maximum is 10q in G,
nor if it is a local maximum in G.

We have the following M-conjugacy results for 4q moves.  Strong
and weak local maxima are defined according to my preference.
If you prefer Mike Reid's definition, ignore the "weak"
column and read the "total" column as "weak".

      Level       Positions     Strong      Weak        Total
                               Local Max   Local Max   Local Max

        0              1           0           0           0
        1              2           0           0           0
        2              5           0           0           0
        3             25           0           1           1
        4             75           0           2           2
        5            152           0          19          19
        6            184           1          35          36
        7            108           0          46          46
        8             16           0          16          16
                    ----
                     568

Back on the subject of the slice group, we have the following.

Mark Longridge said:

>By the way GAP gives  NumberConjugacyClasses (slice) = 23

>In your calculations of M-conjugacy classes for the slice group the
>total number of classes is 50, but I think GAP does not use
>M-conjugates but C-conjugates instead. The NumberConjugacyClasses
>function always thrashes with any larger groups unfortunately.
>If you could easily tweak your program perhaps you could
>verify my theory.

Recall that in my work with <U,R>, I had to use W3-conjugacy rather
than M-conjugacy.  The simplest explanation is that all M-conjugates
of a position in <U,R> are not in <U,R>, and in particular the
representative element might not be in <U,R>.  W3 is the largest
subgroup of M such that all conjugates of <U,R> are in <U,R>.

I flirted briefly with the notion that I might have the same problem
with the slice group and the antislice group.  But it seems
immediate that M-conjugacy is appropriate for both slice and
antislice.  For example, think of applying M-conjugacy to
all the individual 2q or 4q moves in a slice or antislice process.
Clearly, the result is still in slice and antislice, respectively.

I doubt that Mark's theory about GAP using C-conjugacy for slice
instead of M-conjugacy is correct.  I already have 50 positions
to 23 for GAP, and using C-conjugacy would just make my results
larger.  For example, RL' and R'L are M-conjugate positions,
but not C-conjugate positions.  I don't have a clue why my
results do not match GAP.  I have double and triple checked
my results, and they seem correct.  For example, I can "expand"
my conjugacy classes, and the results then match Mark's exactly.

How does GAP's NumberConjugacyClasses function work?  By that,
I mean how does it know the subgroup with respect to which
you are taking conjugacy classes (if my terminology is correct)?
For example, how does it know to take C or M or whatever
conjugates?

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
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

From mschoene@math.rwth-aachen.de  Mon May 22 05:14:39 1995
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Date: Mon, 22 May 95 11:13 WET DST
From: "Martin Schoenert" <Martin.Schoenert@math.rwth-aachen.de>
To: Cube-Lovers@ai.mit.edu
Cc: BRYAN@wvnvm.wvnet.edu
In-Reply-To: "Jerry Bryan"'s message of Sun, 21 May 1995 07:18:49 -0400 (EDT) <wvmail32.1995may19.084306.bryan@wvnvm.wvnet.edu>
Subject: Re: AntiSlice Under M-conjugacy (and a problem with slice)

Sorry, lately I didn't have any time to follow the discussions on
Cube-Lovers (preparing the second upgrade of GAP 3.4 and working long
hours for GAP 4.0).  But Jerry Bryan's message talks about GAP's
'NumberConjugacyClasses' function.

Jerry wrote

    I doubt that Mark's theory about GAP using C-conjugacy for slice
    instead of M-conjugacy is correct.  I already have 50 positions
    to 23 for GAP, and using C-conjugacy would just make my results
    larger.  For example, RL' and R'L are M-conjugate positions,
    but not C-conjugate positions.  I don't have a clue why my
    results do not match GAP.  I have double and triple checked
    my results, and they seem correct.  For example, I can "expand"
    my conjugacy classes, and the results then match Mark's exactly.

GAP's 'NumberConjugacyClasses' follows the general usage in group theory.
The conjugacy class of an element <g> of <G> is the set of elements <h>
that are G-conjugated to <g> (i.e., there exists an element <x> in <G>,
such that <x>^-1 * <g> * <x> = <h>).

Thus GAP is using <slice>-conjugacy classes.  Since GAP is using a
*larger* group, it is not surprising that GAP finds less conjugacy
classes (if M were a subgroup of <slice>, then this had to be so,
because in this case every M-conjugacy class would be a subset of a
<slice>-conjugacy class).

Jerry continued

    How does GAP's NumberConjugacyClasses function work?  By that,
    I mean how does it know the subgroup with respect to which
    you are taking conjugacy classes (if my terminology is correct)?
    For example, how does it know to take C or M or whatever
    conjugates?

It always takes the whole group itself as the acting group.  With some
related functions (e.g., 'ConjugacyClass' itself) you can specify that
you want another group acting, but not with 'NumberConjugacyClasses'.

Mark Longridge wrote (as cited by Jerry)

    In your calculations of M-conjugacy classes for the slice group the
    total number of classes is 50, but I think GAP does not use
    M-conjugates but C-conjugates instead. The NumberConjugacyClasses
    function always thrashes with any larger groups unfortunately.
    If you could easily tweak your program perhaps you could
    verify my theory.

As I wrote above GAP does not use C-conjugates but <slice>-conjugates.
Conjugacy classes in permutation groups are notoriously difficult.
Computing the conjugacy classes of G (the full cube group) for example
is absolutely impossible (without using some theory anyway).

Martin.

-- .- .-. - .. -.  .-.. --- ...- . ...  .- -. -. .. -.- .-
Martin Sch"onert,   Martin.Schoenert@Math.RWTH-Aachen.DE,   +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany

From mreid@ptc.com  Mon May 22 17:31:26 1995
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From: mreid@ptc.com (michael reid)
Message-Id: <9505222148.AA21637@ducie>
To: cube-lovers@ai.mit.edu
Subject:   AntiSlice Under M-conjugacy (and a problem with slice)
Content-Length: 1713

jerry writes (about the antislice group)

> The level 5 local maximum is (U'D')(FB)(FB)(UD)(L'R').

[ ... ]

> I have not verified if the length of the local maximum is 10q in G,
> nor if it is a local maximum in G.

this is exactly what i tried to explain in my recent posts: the latter
statement follows from the former.  and yes, its length is indeed  10q.
it's pretty easy to find maneuvers which end in each quarter turn:

     (F'B') (UUDD) (FB) (RL)    =
     (U'D') (FFBB) (UD) (R'L')  =
     (R'L') (U'D') (FFBB) (UD)  =
     (R'L') (UD) (FFBB) (U'D')  =
     (RL) (F'B') (UUDD) (FB)    =
     (RL) (FB) (UUDD) (F'B').

> We have the following M-conjugacy results for 4q moves.  Strong
> and weak local maxima are defined according to my preference.

it seems like jerry's terminology is more reasonable, so i'll stop
using mine.  jerry's figures

>       Level       Positions     Strong      Weak        Total
>                                Local Max   Local Max   Local Max
> 
>         0              1           0           0           0
>         1              2           0           0           0
>         2              5           0           0           0
>         3             25           0           1           1
>         4             75           0           2           2
>         5            152           0          19          19
>         6            184           1          35          36
>         7            108           0          46          46
>         8             16           0          16          16
>                     ----
>                      568

beg the obvious question:  what is that strong local maximum,
which is unique up to symmetry?

mike

From hoey@aic.nrl.navy.mil  Tue May 23 13:11:33 1995
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From: hoey@aic.nrl.navy.mil
Message-Id: <9505231711.AA26574@sun13.aic.nrl.navy.mil>
To: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>, cube-lovers@ai.mit.edu
Subject: M-conjugacy vs. C-Conjugacy in the Slice group

Jerry Bryan tried to communicate about this to cube-lovers, but
there's apparently a technical difficulty.

On 20 May, Jerry said:

>I doubt that Mark's theory about GAP using C-conjugacy for slice
>instead of M-conjugacy is correct.  I already have 50 positions
>to 23 for GAP, and using C-conjugacy would just make my results
>larger.  For example, RL' and R'L are M-conjugate positions,
>but not C-conjugate positions.

I emailed him to note to the contrary that RL' and R'L are indeed
C-conjugates, for example under 180 degree rotation around the F-B
axis.  I did wonder, though, whether that meant that there could be
C-conjugate slice positions that were not M-conjugate.

He emailed me:

> We can observe that R and R' are not C-conjugates, nor are L' and L,
> which suckered me into stating that RL' and R'L are not.  But
> rewrite R'L as LR' since opposite face moves commute.  Now, RL'
> and LR' are clearly C-conjugate.

> In fact, I have now verified with a quick search program that
> all M-conjugates in the slice group are also C-conjugates.  Hence,
> there are 50 C-conjugate classes in slice, just as there are
> 50 M-conjugate classes.

> In retrospect, I don't think the search program was necessary....

and continues with an argument that did not convince me, but the
following does:

First, consider the central inversion v, which maps each point of the
cube to its diametric opposite.  Conjugation by v maps each face-turn
(e.g. F) with its diametric opposite in opposite sense (B').  Since
these are the pairs that constitute a slice move, and they commute, we
have:

    v' FB' v = (v' F v)  (v' B' v) = B' F = F B',

and similarly for the other slice moves, showing that each slice move
is its own v-conjugate.  This extends to a proof that each position in
the slice group is its own v-conjugate:

    v' s1 s2 ... sn v = (v' s1 v) (v' s2 v) ... (v' sn v)
                      = s1 s2 ... sn.

Suppose that we have two M-conjugate positions X, Y in the slice
group.  So X = m' Y m for some m in M.  If m is in C, then X and Y are
C-conjugate and we are done.  Otherwise take the central inversion v;
we know that mv is in C.  We also know that

    X = v' X v = v' m' Y m v = (mv)' Y (mv).

So X and Y are C-conjugate in this case as well.  QED.

Note: "Being its own v-conjugate" might as well be called "being
v-symmetric".

Dan
Hoey@AIC.NRL.Navy.Mil

From BRYAN@wvnvm.wvnet.edu  Tue May 23 14:39:03 1995
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Message-Id: <wvmail32.1995may22.150528.bryan@wvnvm.wvnet.edu>
Date:      Tue, 23 May 1995 14:04:03 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@lcs.mit.edu>
Subject:   M-conjugacy vs. C-Conjugacy in Slice and Antislice

On 20 May, I said the following:

>I doubt that Mark's theory about GAP using C-conjugacy for slice
>instead of M-conjugacy is correct.  I already have 50 positions
>to 23 for GAP, and using C-conjugacy would just make my results
>larger.  For example, RL' and R'L are M-conjugate positions,
>but not C-conjugate positions.

We already know from Martin Schoenert that GAP is using neither
M-conjugacy nor C-conjugacy, but <slice>-conjugacy.  But my
statement about C-conjugacy vs. M-conjugacy was completely incorrect
in any case.

Dan Hoey pointed out to me that RL' and R'L in fact *are* C-conjugates
under 180 degree rotation around the U-D axis.

We can observe that R and R' are not C-conjugates, nor are L' and L,
which suckered me into stating that RL' and R'L are not.  But
rewrite R'L as LR' since opposite face moves commute.  Now, RL'
and LR' are clearly C-conjugate.

In fact, I have now verified with a quick search program that
all M-conjugates in the slice group are also C-conjugates.  Hence,
there are 50 C-conjugate classes in slice, just as there are
50 M-conjugate classes.

In retrospect, I don't think the search program was necessary.
Suppose X and Y are M-conjugates in the slice group.  Then,
they can be written as M-conjugate sequences.  That is, they
can be written so that the individual slice moves are respective
M-conjugates for some fixed m in M.  (The fact that it might be
possible also to write them so that the individual slice moves
are not respective M-conjugates for some fixed m in M is
irrelevant.)  Furthermore, write the sequence for X so that the
clockwise half of each slice is written prior to the
counter-clockwise half of the slice.  The sequence for
Y with individual slice moves being respective M conjugates of
X may or not have this property.  But if not, then simply reorder
the halves of the slices of Y to put the clockwise half first, and Y
will still be piecewise M-conjugate with X.  Then, the piecewise
M-conjugate slices are also C-conjugate, and therefore the X and
Y positions are C-conjugate.

Antislice is totally different.  For example, RL is M-conjugate to
R'L', but it not C-conjugate.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
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

From BRYAN@wvnvm.wvnet.edu  Tue May 23 14:43:48 1995
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Message-Id: <wvmail32.1995may22.150528.bryan@wvnvm.wvnet.edu>
Date:      Tue, 23 May 1995 08:50:40 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   M-conjugacy vs. C-Conjugacy in Slice and Antislice

On 20 May, I said the following:

>I doubt that Mark's theory about GAP using C-conjugacy for slice
>instead of M-conjugacy is correct.  I already have 50 positions
>to 23 for GAP, and using C-conjugacy would just make my results
>larger.  For example, RL' and R'L are M-conjugate positions,
>but not C-conjugate positions.

We already know from Martin Schoenert that GAP is using neither
M-conjugacy nor C-conjugacy, but <slice>-conjugacy.  But my
statement about C-conjugacy vs. M-conjugacy was completely incorrect
in any case.

Dan Hoey pointed out to me that RL' and R'L in fact *are* C-conjugates
under 180 degree rotation around the U-D axis.

We can observe that R and R' are not C-conjugates, nor are L' and L,
which suckered me into stating that RL' and R'L are not.  But
rewrite R'L as LR' since opposite face moves commute.  Now, RL'
and LR' are clearly C-conjugate.

In fact, I have now verified with a quick search program that
all M-conjugates in the slice group are also C-conjugates.  Hence,
there are 50 C-conjugate classes in slice, just as there are
50 M-conjugate classes.

In retrospect, I don't think the search program was necessary.
Suppose X and Y are M-conjugates in the slice group.  Then,
they can be written as M-conjugate sequences.  That is, they
can be written so that the individual slice moves are respective
M-conjugates for some fixed m in M.  (The fact that it might be
possible also to write them so that the individual slice moves
are not respective M-conjugates for some fixed m in M is
irrelevant.)  Furthermore, write the sequence for X so that the
clockwise half of each slice is written prior to the
counter-clockwise half of the slice.  The sequence for
Y with individual slice moves being respective M conjugates of
X may or not have this property.  But if not, then simply reorder
the halves of the slices of Y to put the clockwise half first, and Y
will still be piecewise M-conjugate with X.  Then, the piecewise
M-conjugate slices are also C-conjugate, and therefore the X and
Y positions are C-conjugate.

Antislice is totally different.  For example, RL is M-conjugate to
R'L', but it not C-conjugate.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
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

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Date:      Wed, 24 May 1995 15:46:45 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <cube-lovers@lcs.mit.edu>
Subject:   Re: AntiSlice Under M-conjugacy (and a problem with slice)
In-Reply-To: Message of 05/22/95 at 17:48:42 from mreid@ptc.com

On 05/22/95 at 17:48:42 mreid@ptc.com said:

>>       Level       Positions     Strong      Weak        Total
>>                                Local Max   Local Max   Local Max
>>
>>         6            184           1          35          36

>beg the obvious question:  what is that strong local maximum,
>which is unique up to symmetry?

I haven't verified in the antislice data base, but it *has* to be
Pons Asinorum, accomplshed as six antislices.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
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

From BRYAN@wvnvm.wvnet.edu  Wed May 24 17:14:08 1995
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Date:      Wed, 24 May 1995 16:36:30 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@lcs.mit.edu>
Subject:   Re: AntiSlice Under M-conjugacy (and a problem with slice)
In-Reply-To: Message of 05/22/95 at 11:13:00 from ,
           Martin.Schoenert@math.rwth-aachen.de

On 05/22/95 at 11:13:00 Martin Schoenert said:

>GAP's 'NumberConjugacyClasses' follows the general usage in group theory.
>The conjugacy class of an element <g> of <G> is the set of elements <h>
>that are G-conjugated to <g> (i.e., there exists an element <x> in <G>,
>such that <x>^-1 * <g> * <x> = <h>).

Just to give an example that I am familiar with, suppose the group
in question were M itself.  Then, NumberConjugacyClasses should yield
10, because the 48 elements in M yield 10 conjugacy classes under
M-conjugation.  If anybody who has GAP also has defined M, you
might give it a try.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
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

From mschoene%math.rwth-aachen.de@samson.math.rwth-aachen.de  Thu May 25 09:39:22 1995
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From: "Martin Schoenert" <Martin.Schoenert@math.rwth-aachen.de>
To: Cube-Lovers@lcs.mit.edu
Cc: BRYAN@wvnvm.wvnet.edu
In-Reply-To: "Jerry Bryan"'s message of Wed, 24 May 1995 16:36:30 -0400 (EDT) <wvmail32.1995may24.155259.bryan@wvnvm.wvnet.edu>
Subject: Re: Re: AntiSlice Under M-conjugacy (and a problem with slice)

Jerry Bryan wrote

    Just to give an example that I am familiar with, suppose the group
    in question were M itself.  Then, NumberConjugacyClasses should yield
    10, because the 48 elements in M yield 10 conjugacy classes under
    M-conjugation.  If anybody who has GAP also has defined M, you
    might give it a try.

GAP also thinks that M has 10 conjugacy classes (under M-conjugation).

Martin.

-- .- .-. - .. -.  .-.. --- ...- . ...  .- -. -. .. -.- .-
Martin Sch"onert,   Martin.Schoenert@Math.RWTH-Aachen.DE,   +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany

From BRYAN@wvnvm.wvnet.edu  Thu May 25 12:01:36 1995
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Message-Id: <wvmail32.1995may23.232649.bryan@wvnvm.wvnet.edu>
Date:      Thu, 25 May 1995 11:50:55 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Dan Hoey" <hoey@aic.nrl.navy.mil>,
        "Cube Lovers List" <cube-lovers@lcs.mit.edu>
Subject:   Re: M-conjugacy vs. C-Conjugacy in the Slice group
In-Reply-To: Message of 05/23/95 at 13:11:27 from hoey@AIC.NRL.Navy.Mil

I said:

>> In fact, I have now verified with a quick search program that
>> all M-conjugates in the slice group are also C-conjugates.  Hence,
>> there are 50 C-conjugate classes in slice, just as there are
>> 50 M-conjugate classes.

>> In retrospect, I don't think the search program was necessary....

On 05/23/95 at 13:11:27 hoey@AIC.NRL.Navy.Mil said:

>and (Jerry) continues with an argument that did not convince me, but the
>following does:

I think I can both greatly simplify and greatly strengthen the
argument that did not convince Dan.  My argument is based on the idea
(copied from _Symmetry and Local Maxima_) that M-conjugation can be
viewed as a permutation on Q, the set of twelve quarter turns.

Call the six clockwise quarter turns Q+ and the six counter-clockwise
quarter turns Q-.  We can observe that the 24 rotations in M all
map Q+ to Q+ and map Q- to Q-, and that the 24 reflections in M all
map Q+ to Q- and map Q- to Q+.  We also note that in particular,
the central inversion v is a reflection.

Suppose X and Y are M-conjugates in <slice> with Y=m'Xm for some fixed
m in M.  Write X as pairs of quarter turns (each pair is a slice), and
write Y as pairs of quarter turns which are respective M-conjugates
(via the fixed permutation m) of the quarter turns in X.  If the
respective quarter turns have been mapped Q+ to Q+ and Q- to Q-, then
m is a rotation and we are done.

Otherwise, commute the halves of each slice in Y.  We first note that
so commuting is the identity on Y.  We also note that so commuting is
equivalent to performing the permutation operation v on Q, and is
therefore equivalent to performing v-conjugation on Y. (In passing,
we see that this effectively proves Dan's first point, namely that
X=v'Xv for all X in <slice>.  Given that, I would shorten the rest of
Dan's argument by saying Y=m'Xm=v'('m'Xm)v=v'm'Xmv, and noting that
either m or mv is a rotation).

But having started with the "commuting the halves of slices" argument,
I would continue as follows.  Having commuted the halves of the slices,
we still have an M-conjugate (and still the same M-conjugate)
because commuting is equivalent to v-conjugation, v is in M, and
v-conjugation is the identity in <slice>.  Finally, having commuted
the halves of the slices, we are now mapping Q+ to Q+ and Q- to Q-,
so we have a rotation.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
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

From @mail.uunet.ca:mark.longridge@canrem.com  Thu May 25 19:56:07 1995
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Subject: GAP notes
From: mark.longridge@canrem.com (Mark Longridge)
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Date: Thu, 25 May 1995 18:41:00 -0400
Organization: CRS Online  (Toronto, Ontario)

On 05/22/95 at 11:13:00 Martin Schoenert said:

>GAP's 'NumberConjugacyClasses' follows the general usage in
> group theory.
>The conjugacy class of an element <g> of <G> is the set of elements <h>
>that are G-conjugated to <g> (i.e., there exists an element <x> in <G>,
>such that <x>^-1 * <g> * <x> = <h>).

On  05-24-95 (18:16) Jerry Bryan said:

>Just to give an example that I am familiar with, suppose the group
>in question were M itself.  Then, NumberConjugacyClasses should yield
>10, because the 48 elements in M yield 10 conjugacy classes under
>M-conjugation.  If anybody who has GAP also has defined M, you
>might give it a try.

Ok... let's define C in the context of GAP:

c := Group(
    ( 1, 3, 8, 6)( 2, 5, 7, 4)( 9,33,25,17)(10,34,26,18)(11,35,27,19)
    (20,12,36,28)(21,13,37,29)
    (46,48,43,41)(44,47,45,42)(38,30,22,14)(39,31,23,15)(40,32,24,16),
    ( 9,11,16,14)(10,13,15,12)( 1,17,41,40)( 4,20,44,37)( 6,22,46,35)
    (2,18,42,39)(7,23,47,34)
    (30,32,27,25)(28,31,29,26)(19,43,38,3) (21,45,36,5) (24,48,33, 8)
);;

M is the same as C but with the central reflection:

m := Group(
    ( 1, 3, 8, 6)( 2, 5, 7, 4)( 9,33,25,17)(10,34,26,18)(11,35,27,19)
    (20,12,36,28)(21,13,37,29)
    (46,48,43,41)(44,47,45,42)(38,30,22,14)(39,31,23,15)(40,32,24,16),
    ( 9,11,16,14)(10,13,15,12)( 1,17,41,40)( 4,20,44,37)( 6,22,46,35)
    (2,18,42,39)(7,23,47,34)
    (30,32,27,25)(28,31,29,26)(19,43,38,3) (21,45,36,5) (24,48,33, 8),
    (1,8)(3,6)(2,7)(4,5)
    (17,24)(19,22)(18,23)(20,21)
    (9,16)(11,14)(10,15)(12,13)
    (25,32)(27,30)(26,31)(28,29)
    (33,40)(35,38)(34,39)(36,37)
    (41,48)(43,46)(42,47)(44,45)
);;

Then we have

Size (c) = 24
NumberConjugacyClasses (c) = 5

Size (m) = 48
NumberConjugacyClasses (m) = 10

These results concur with Dan's message from Tue, 28 Dec 93 18:40:52 EST
from the archives.

We can also use GAP to calculate the size of the M-conjugacy class of
a given element:

Size (ConjugacyClass (m, cross4)) = 3

Here we see there are three possible 4 Cross order 2 patterns.

I've tried dabbling in some GAP programming. Say we are looking for an
element in the slice group with 4 variants under M-conjugacy....

a := 0;
x := 0;
z := Elements (slice);

repeat
  a := a+1
  x := Size (ConjugacyClass (m, Random (slice)));
until a = 768 or x = 4

This short program found no elements of size 4 in the slice group.

-> Mark <-

From BRYAN@wvnvm.wvnet.edu  Fri May 26 19:37:17 1995
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Date:      Fri, 26 May 1995 19:33:50 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <mreid@ptc.com>, "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Re: AntiSlice Under M-conjugacy (and a problem with slice)
In-Reply-To: Message of 05/24/95 at 17:14:09 from mreid@ptc.com

On 05/24/95 at 17:14:09 mreid@ptc.com said:
>hi jerry,

>you said

>> On 05/22/95 at 17:48:42 mreid@ptc.com said:
>>
>> >>       Level       Positions     Strong      Weak        Total
>> >>                                Local Max   Local Max   Local Max
>> >>
>> >>         6            184           1          35          36
>>
>> >beg the obvious question:  what is that strong local maximum,
>> >which is unique up to symmetry?
>>
>> I haven't verified in the antislice data base, but it *has* to be
>> Pons Asinorum, accomplshed as six antislices.

>no it hasn't.  this is the equivalent of the face turn metric,
>i.e. your "level" is half the face turn distance.  (RL  and  RRLL
>each count as 1, which is half their face turn count.)

>then pons asinorum is at level 3, and gives the weak local maximum
>at that level.

Mike is correct.

Try instead, (L2R2)(U2D2)(F'B')(U2D2)(L2R2)(F'B').

This is a very pretty pattern which may well have a name, but I
don't know what the name is.  Also, it is its own inverse.

Is the length 12h in <Q,H>?  Is it a local maximum (strong or otherwise)
in <Q,H>?   Is the length 20q in <Q>?  Is it a local maximum in <Q>?

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
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

From mreid@ptc.com  Tue May 30 15:39:31 1995
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From: mreid@ptc.com (michael reid)
Message-Id: <9505301629.AA15624@ducie>
To: cube-lovers@ai.mit.edu
Subject:   Re: AntiSlice Under M-conjugacy (and a problem with slice)
Content-Length: 1649

jerry writes

[ ... ]

> >> >>       Level       Positions     Strong      Weak        Total
> >> >>                                Local Max   Local Max   Local Max
> >> >>
> >> >>         6            184           1          35          36
> >>
> >> >beg the obvious question:  what is that strong local maximum,
> >> >which is unique up to symmetry?

[ ... ]

> Try instead, (L2R2)(U2D2)(F'B')(U2D2)(L2R2)(F'B').
> 
> This is a very pretty pattern which may well have a name, but I
> don't know what the name is.  Also, it is its own inverse.
> 
> Is the length 12h in <Q,H>?  Is it a local maximum (strong or otherwise)
> in <Q,H>?   Is the length 20q in <Q>?  Is it a local maximum in <Q>?

no, yes (otherwise), no, and yes, respectively.

we have seen this pattern several times recently.  this is one of 
those positions with 16 symmetries.  i called it "four pluses" in my 
message of may 11 (although i gave it in a different orientation)

)     four pluses                 ( R2 F2 R2 U'D F2 R2 F2 UD' )

in fact, this maneuver is minimal in both the quarter turn and the
face turn metrics, so its length is  16q, 10f.  it is a weak local
maximum in the face turn metric; one can check that no minimal
maneuver ends with the face turn R.  however, using the 16 symmetries
which preserve the  U-D axis, and inversion, we can give minimal
maneuvers which end with turns of any of the six faces.  this shows
that it's a weak local maximum in the face turn metric.  local
maximality in the quarter turn metric follows in a similar manner.

also, mark pointed out on april 16 that this position lies in the
center of the antislice group.

mike

From @mail.uunet.ca:mark.longridge@canrem.com  Sat Jun  3 03:39:09 1995
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Sender: CRSO.Cube@canrem.com
Subject: Super Groups
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.1148.5834.0C1E504D@canrem.com>
Date: Sat, 3 Jun 1995 04:14:00 -0400
Organization: CRS Online  (Toronto, Ontario)

Notes on the various Super-Groups
---------------------------------

I have calculated the size of the super-groups for various subgroups
of the cube. I have suffixed the standard group names with the
letter c to show that the centre orientations are significant.

The groups are (ranked smallest to largest):

Size (slice)       =         768
Size (slicec)      =      24,576     Size (slicec) / Size (slice)   = 32

The following reference confirms this calculation and expounds further
on the nature of the slice group...

The Slice Group in Rubik's Cube,
by David Hecker, Ranan Banerji
Mathematics Magazine, Vol. 58 No. 4 Sept 1985

Size (antisl)      =       6,144
Size (antislc)     =      49,152     Size (antislc) / Size (antisl) =  8

Size (sq)          =     663,552
Size (sqc)         =   5,308,416     Size (sqc) / Size (sq)         =  8

Size (ur)          =  73,483,200
Size (urc)         = 587,865,600     Size (urc) / Size (ur)         =  8

Size (cube)  =     43,252,003,274,489,856,000
Size (cubec) = 88,580,102,706,155,225,088,000

Size (cubec) / Size (cube) = 2,048

The case of the super squares group (sqc) is interesting. It is only
possible to rotate opposite centres 180 degrees. There are actually
8 centres in the super square's group:

(1 way)    Identity
(1 way)    All 6 centres rotated 180 degrees
(3 ways)   2 opposite centres rotated 180 degrees
(3 ways)   2 pairs of opposite centres rotated 180 degrees

-> Mark <-

From @mail.uunet.ca:mark.longridge@canrem.com  Sun Jun  4 03:28:58 1995
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Subject: Super Squares Group
From: mark.longridge@canrem.com (Mark Longridge)
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Date: Sun, 4 Jun 1995 04:21:00 -0400
Organization: CRS Online  (Toronto, Ontario)

Way back on Thu Aug 20 20:10:13 1992 Mike wrote:

>  R2 F2 B2 L2 U2 L2 F2 B2 R2  ~  D2 ,
>            so that  <U2, F2, R2, L2, B2> = <U2, F2, R2, L2, B2, D2>.

I bet he didn't realize at the time he was finding a minimal
sequence to rotate 2 centres in the super square's group!

If we tack D2 on the end we get the sequence:

R2 F2 B2 L2 U2 L2 F2 B2 R2 D2 =  turn U & D centres 180 degrees
 in 10 face turns.

-> Mark <-

From bagleyd@source.asset.com  Wed Jun  7 17:31:08 1995
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From: bagleyd@source.asset.com (David A. Bagley)
Message-Id: <9506072107.AA36858@source.asset.com>
To: Cube-Lovers@ai.mit.edu
Subject: Dinosaur Rubik's Cube

Hi
  I just just added new modes to my xdino puzzle.  It is the Rubik's
Dinosaur Puzzle recently mentioned.  (A cube with diagonal X cuts.)
In addition to the Period 3 movement it now has a Period 2 movement with
the faces cut up like:
   ___
  |\ /|
  | X |
  |/ \|
   ---
as opposed to just
   \ /
    X
   / \
In the Period 3 movement the cube turns around a corner while in Period 2
movement it turns around the center of an edge.
 
Of course if you want to make it harder there is a "Both" mode where
you can have both turning modes at once.
 
I would like to thank Derek Bosch <bosch@sgi.com> for suggesting the
Period 2 movement -> Bosch's Cube :)
 
I spent a good 10 minutes trying to solve Bosch's Cube and did
not get anywhere.  The Period 3 seems easier.  Be the first in the
Universe to solve it. :)
 
All those who have X should be able to run it.  There are many other
puzzles in the collection as well.
 
Any problems with the compilation or bugs ... let me know.
 
Cheers,
      --__---------------------------------------------------------------
     /  \ \   /           David A. Bagley                                \
    |    \ \ /            bagleyd@source.asset.com                        |
    |     \//\            Some days are better than other days.           |
    |     / \ \                -- A short lived character of Blake's 7    |
     \   /   \_\puzzles   Available at: ftp.x.org/contrib/games/puzzles  /
      -------------------------------------------------------------------

From norgomez@itecs5.telecom-co.net  Mon Jun 12 22:31:21 1995
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Date: Mon, 12 Jun 1995 21:28:16 -0400
Message-Id: <95061221281669@itecs5.telecom-co.net>
From: norgomez@itecs5.telecom-co.net (ORLANDO GOMEZ CAMACHO, BOGOTA-COLOMBIA.)
To: JCCANNEK@ukcc.uky.edu, MTY.ITESM.MX@ukcc.uky.edu,
        commune-list@stealth.acf.nyu.edu, COMP-CEN%UCCVMA.BITNET@vm1.nodak.edu,
        CONFOCAL%UBVM.BITNET@vm1.nodak.edu,
        CW-MAIL%TECMTYVM.BITNET@vm1.nodak.edu,
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        CRTNET%PSUVM.BITNET@cunyvm.cuny.edu,
        CUMREC-L%NDSUVM1.BITNET@cunyvm.cuny.edu,
        CVNET%YORKVM1.BITNET@cunyvm.cuny.edu,
        Cyber-L%Bitnic.BITNET@cunyvm.cuny.edu,
        DANCE-L%HEARN.BITNET@cunyvm.cuny.edu, Comp-Soc@limbo.intuitive.com,
        concrete-blonde@ferkel.ucsb.edu,
        CONS-L%MCGILL1.BITNET@cornellc.cit.cornell.edu,
        441495@acadvm1.uottawa.ca, com-priv@psi.com, cjr2@cornell.edu,
        CORPORA-REQUEST@x400.hd.uib.no, CORRYFEE@hasara11.telecom-co.net,
        CPAE@catfish.valdosta.peachnet.edu, CPE-LIST@uncvm1.oit.unc.edu,
        CREA-CPS@nic.surfnet.nl, CREWRT-L@mizzou1.missouri.edu,
        CROMED-L@aearn.bitnet, CSNET-FORUM@sh.cs.net
Subject: Oportunidades de Negocios
X-Vms-To: @LISTA1.DIS

======================================================================

                      OPORTUNIDADES DE NEGOCIOS

         SOLO PARA PERSONAS RESIDENTES EN LOS ESTADOS UNIDOS,
         CON BUENOS CONTACTOS EN EMPRESAS DE ALTA TECNOLOGIA
           EN INFORMATICA, OFIMATICA Y TELECOMUNICACIONES
-----------------------------------------------------------------------

Apreciados amigos de la lista:

Nuestra empresa ha sido contratada por una firma que edita una publicaciOn 
especializada en las Areas de  Computadoras, Telecomunicaciones, OfimAtica 
y Servicios relacionados que  circula  en Venezuela, Ecuador, Peru, Brazil 
y Colombia,  constituyEndose en un medio Unico  de informaciOn sobre estos 
tOpicos en las naciones descritas.

Para comercializar ESPACIOS PUBLICITARIOS de esta publicaciOn, se requiere 
de una compan~Ia o persona que maneje presupuesto de negocios, preferible-
mente  con capacidad econOmica propia e idoneidad para contratar. Es nece-
sario que la persona o empresa resida en los Estados Unidos de AmErica.

A continuaciOn se presentan las caracterIsticas principales de la publica-
ciOn:

    * CirculaciOn efectiva en Brazil, Colombia, Ecuador, Peru y Venezuela.
    * A los anunciadores  se les proveen  en medio magnEtico, las bases de
      datos con informaciOn importante  de estos paises.  A las Empresas y 
      Usuarios de cada naciOn tambiEn se les harA entrega de este material 
      que incluirA informaciOn sobre proveedores de bienes  y servicios en
      los sectores   de Telecomunicaciones, Software, Equipos  de  Oficina,
      Computadores, y Suministros.
    * En  la publicaciOn figuran  empresas  Norteamericanas  que fabrican,
      representan y proveen equipos, programas y servicios en las areas de
      telecomunicaciones, sector electronico, de comunicacion, computacion
      perifericos, componentesm accesorios, suministros, servicios y areas
      afines
    * El tiraje es de Cuarenta Mil (40,000) ejemplares para ser distribui-
      dos en forma gratuita.
    * El perfil de los anunciadores corresponde a compan~Ias Estadouniden-
      ses de estos sectores que se encuentren en ampliar y consolidar su
      operacionalidad en estos paises.

La Empresa OFRECE:

    * Exclusividad
    * Participacion atractiva sobre los negocios
    * Excelente calidad editorial y material informativo y util para generar
      nuevas y mejores oportunidades de negocios en forma proactiva.

==============================================================================
A las personas interesadas, se les ruega contactarnos por esta misma via,
indicando los siguientes datos:
Nombre   :
e-mail   :
Direccion:
Ciudad   :
Telefono :
Fax      :
==============================================================================
Mayores informes:

COMUNICACIONES INTERACTIVAS
Contacto  : Orlando Gomez Camacho
e-mail    : norgomez@itecs5.telecom-co.net
Voice Mail: (+571) 5002072
------------------------------------------------------------------------------ 

From MULL4195@splava.cc.plattsburgh.edu  Wed Jun 14 16:17:44 1995
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Date: Wed, 14 Jun 1995 16:17:58 -0500 (EST)
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signoff cube-lovers

From comnetlu!georges.helm@eo.net  Thu Jun 15 15:38:16 1995
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To: cube-lovers@ai.mit.edu
Subject: Andras Mezei's book
From: georges.helm@comnet.eo.lu (GEORGES HELM)
Message-Id: <8AB54D5.0063000243.uuout@comnet.eo.lu>
Date: Thu, 15 Jun 95 20:37:00 +1
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Reply-To: georges.helm@comnet.eo.lu (GEORGES HELM)
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Mark Longridge wrote

> Does anyone on Cube-Lovers have that book?

Yes, I do.

The illustrations cover:
Rubik himself, Polytoys, Konsumex, cubes, books, cube-related items
(stickers, pencils, lp's, earrings, t-shirts...)
There is an article on a competition in 1982 in Budapest.
And many more articles I don't understand a word of.

Georges <cubemeister>

From comnetlu!georges.helm@eo.net  Fri Jun 16 03:20:00 1995
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Subject: Andras Mezei's book
From: georges.helm@comnet.eo.lu (GEORGES HELM)
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I have the book. - Georges

From BRYAN@wvnvm.wvnet.edu  Fri Jun 16 14:10:52 1995
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Date:      Fri, 16 Jun 1995 14:11:02 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   10q Local Maxima

There are four of them unique up to M-conjugacy (more
information to come).

 Distance  Cubes  Branch Lcl     {m'Xm}   Branch   Ratio   Local
 from             Factor Max   (M-Conj.   Factor    of       Max
 Start                          Classes)          Cubes to
                                                  Classes
 0              1                       1                     0
 1             12 12.000  0             1   1.000  12.000     0
 2            114  9.500  0             5   5.000  22.800     0
 3          1,068  9.368  0            25   5.000  42.720     0
 4         10,011  9.374  0           219   8.760  45.712     0
 5         93,840  9.374  0         1,978   9.032  47.442     0
 6        878,880  9.366  0        18,395   9.300  47.778     0
 7      8,221,632  9.355  0       171,529   9.325  47.931     0
 8     76,843,595  9.347  0     1,601,725   9.338  47.976     0
 9    717,789,576  9.341  0    14,956,266   9.338  47.993     0
10  6,701,836,858  9.337 42   139,629,194   9.336  47.997     4
11 62,549,615,248  9.333    1,303,138,445   9.333  47.9992

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
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

From BRYAN@wvnvm.wvnet.edu  Fri Jun 16 14:17:43 1995
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Date:      Fri, 16 Jun 1995 14:17:51 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   10q Local Maxima Search Matrix

The individual cells in this chart give numbers of M-conjugacy
classes.  The local maxima are in column 12.  Still more
information to come.

                         Number of Moves Which Go Closer to Start

Level  Total  0         1        2       3    4    5   6  7  8 9 1 1 1
      Classes                                                    0 1 2

 0         1  1         0        0       0     0   0   0  0  0 0 0 0 0
 1         1  0         1        0       0     0   0   0  0  0 0 0 0 0
 2         5  0         2        3       0     0   0   0  0  0 0 0 0 0
 3        25  0        20        4       1     0   0   0  0  0 0 0 0 0
 4       219  0       182       34       2     1   0   0  0  0 0 0 0 0
 5      1978  0      1677      280      20     1   0   0  0  0 0 0 0 0
 6     18395  0     15642     2561     184     8   0   0  0  0 0 0 0 0
 7    171529  0    145974    23773    1721    61   0   0  0  0 0 0 0 0
 8   1601725  0   1362579   222235   16241   663   1   3  0  3 0 0 0 0
 9  14956266  0  12719643  2077549  153026  5954  74  15  2  3 0 0 0 0
10 139629194  0 118711701 19418503 1438825 58862 925 318 11 37 0 8 0 4

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
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

From BRYAN@wvnvm.wvnet.edu  Sat Jun 17 00:35:23 1995
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Date:      Sat, 17 Jun 1995 00:35:31 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   10q Local Maxima Positions

1. (UU)  (F'B')(RL)(RL)(FB)
2. (UD') (F'B')(RL)(RL)(FB)
3. (UD)  (F'B')(RL)(RL)(FB)

4. (UD')(FB')(LR')(FB')(FB')

#4 is the one Mike Reid already found in the slice group.  #3 is the
one I already found in the antislice group.  #1, #2, and #3 are
obviously closely related.  #1 and #2 appear not to be in either
slice or antislice, but I have been fooled before by alternative
sequences which yield the same position.

#1, #2, and #3 all have the property that |{m'Xm}|=6 and
|Symm(X)|=8.  As has already been discussed, #4 has the
property that |{m'Xm}|=24 and |Symm(X)|=2.

The symmetry groups for #1, #2, and #3 are of a type Dan Hoey's
taxonomy calls class P, class S, and class AX, respectively.
These particular classes are hard to describe succinctly without
introducing a lot of notation.  But in all three cases, the symmetry
groups (subgroups of M such that X=m'Xm} consist of four rotations
and four reflections, and have as an axis of symmetry one of
the three major axes of the cube (U-D, F-B, or R-L).  There
are three groups P1, P2, P3 with axis of symmetry U-D, F-B,
and R-L, respectively, and similarly for S1, S2, and S3, and
for AX1, AX2, and AX3.

For #4, we have Symm(X)=HV in Dan's taxonomy, where HV={i,v}, and
where i is the identity in M and v is the central inversion in M.
If proper typography were available, the i and the v would be
upper case script letters to follow Frey and Singmaster.

There are relatively few positions in all of cube space
for which Symm(X)=Pi or Symm(X)=Si or Symm(X)=AXi (i in 1..3).
There are only 10 P positions through level 10 in the search tree (of which
just one is a local maximum).  There is only one S position through
level 10, and only one AX position through level 10, both of
which are of course local maxima.  The positions
are not Q-transitive, but the positions look "symmetric", and they
fulfill the (incorrect) intuition that "symmetric" positions must
be local maxima.  We have no reason to say that other P or S or AX
positions further down the search will be local maxima.

I find position #4 extremely intriguing.  In general, HV is not
very strong symmetry, and there are relatively speaking, quite
a few HV positions in cube space.

We could create an HV position as follows.
Put any edge cubie anywhere (say UF in RD).  Put the "opposite"
cubie in the "opposite" cubicle (DB in LU in this case).  Continue
for the remaining edge cubies, and then do the same thing for the
corners, remembering only to make sure the edges and corners have
the same parity.  You can easily make an HV position that looks
quite "random" to the casual glance, and in fact most HV positions
don't look very "symmetric".

But Mike's position looks very "symmetric" at a casual glance, as
if its symmetry must be much stronger than HV.  I certainly would
not have expected to find an HV position as a local maximum close
to Start.  I think the "look" of Mike's position as "symmetric",
and the fact that it is a local maximum close to Start are
related.  Without getting too long winded, I think the reasons
are two-fold.  First, the corners and edges have much stronger
symmetry separately than they do collectively.  Second, the
symmetry looks much stronger if you ignore the centers (i.e.,
if you ignore the rotational positioning of the cubies), perhaps
in the sense of Dan's CSymm function.  For example, the corners
are properly positioned with respect to each other, even though
they are not properly positioned with respect to the fixed face
centers.

In the next few days, I intend to calculate Symm(X)
for the corners and edges separately for Mike's position, as
well as calculating CSymm(X) for the corners and edges
separately and combined.  I think the results will be
enlightening.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
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

From @mail.uunet.ca:mark.longridge@canrem.com  Sat Jun 17 22:00:06 1995
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To: cube-lovers@life.ai.mit.edu
Reply-To: CRSO.Cube@canrem.com
Sender: CRSO.Cube@canrem.com
Subject: Crazy Corner Pattern
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.1161.5834.0C1E7553@canrem.com>
Date: Sat, 17 Jun 1995 05:38:00 -0400
Organization: CRS Online  (Toronto, Ontario)

On Mon, 17 Jan 1994 09:06:59 EST, Jerry wrote:
>
>     Counting M-conjugacy classes of the corners of Rubik's cube
>     -----------------------------------------------------------
>
>                   M-Class     Number     Number
>                     Size        of         of
>                              Classes    Elements
>
>                         1        1  =        1     Start
>                         2        1  =        2     +4 -4 Twist
>                         3        3  =        9
>                         4        1  =        4
>                         6       34  =      204
>                         8       33  =      264
>                        12      301  =     3612
>                        16      104  =     1664
>                        24     9064  =   217536
>                        48  1832428  = 87956544
>
>                     Total  1841970    88179840

I'm trying to find the 1 pattern with M-class size of 4 of the
corners group.
The only pattern that I can find is 4 alternate corners twisted
clockwise which is in the twist orbit.

It does not seem to be any pattern with just corners twisted in
place.

Jerry, if you could *please* identify this pattern before I go
nuts...... !

-> Mark <-

From BRYAN@wvnvm.wvnet.edu  Sun Jun 18 00:35:58 1995
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Date:      Sat, 17 Jun 1995 17:18:00 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   A Third Way to Calculate the Real Size of Cube Space?

We define the real size of cube space to be the number of M-conjugate
classes {m'Ym} for m in M, set of 48 rotations and reflections
of the cube, and for Y in G.

Dan Hoey has calculated the real size of cube space using the
Polya-Burnside theorem.

Dan and I (mostly Dan) have also calculated the same result using
exhaustive computer search.  The computer search is much less
elegant than the Polya-Burnside results, but the search does
provide additional information, such as the number of positions
associated with each symmetry group.  The results from the
computer search have not yet been posted to the list, but a draft
paper is in progress.

In the meantime, it occurs to me that perhaps  --  but only perhaps --
there is a third way to calculate the real size of cube space.
The third way would not require (much) computer searching, but would
provide the same level of detail about number of positions per
symmetry group as does the full blown search.

The idea is based on a posting from Mike Reid.  Mike calculated the
number of positions in G whose symmetry preserves the U-D axis.
Such positions have a symmetry group which is called X1
in Dan's taxonomy.  For these positions, we say Symm(Y)=X1,
where in general for Y in G we have Symm(Y) is the set
(and group) of all m in M such that Y=m'Ym.

X1 contains sixteen elements (eight rotations and eight reflections),
and preserves the U-D axis.  X2 and X3 are conjugate subgroups of X1
and similarly preserve the F-B and the R-L axes, respectively. If Y is
X1-symmetric, then we have {m'Ym}={Y1,Y2,Y3}.  One of the Yi is Y
and is X1-symmetric, one of the Yi is X2-symmetric, and the other
Yi is X3-symmetric.

Mike determined (without computer search) that there are 128
X1-symmetric positions.  Since four of the positions are also
M-symmetric, we have 124 positions Y for which Symm(Y)=X1.
Similar results hold for X2 and X3.  Hence, there are 124
M-conjugacy classes containing cubes for which Symm(Y)=Xi, or
perhaps we might say for which SymmClass(Y)=X.  The important
fact here is that we have determined that there are
124 M-conjugacy classes for symmetry class X without having
to do a computer search.

If we could similarly determine the number of K-symmetric positions
for each of the 98 subgroups K of M without computer search, then
we could calculate the real size of cube space.  You really only
have to determine the size of 33 subgroups.  Just as the solution
for X1 also gave us the solution for X2 and X3, similarly the
solution for any subgroup provides the solution for all conjugate
subgroups, and there are 33 classes of conjugate subgroups.

I usually get myself in trouble when I delve too much into things
I don't understand, but let's try a few examples.  The subgroup
HV={i,v} is easy to understand, where v is the central inversion.
For the edges, the number of HV-symmetric positions should be
24*20*16*12*8*4.  That is, put the first cubie anywhere (24
possibilities) which dictates the location of the respective
"opposite" cubie.  There are then 20 possibilities for the
location of the third cubie which again dictates the position
of the respective "opposite" cubie, and so forth.  In the same
manner, the number of HV-symmetric corner positions is
24*18*12*6.  The number of HV-symmetric positions is then
(24*20*16*12*8*4)*(24*18*12*6)/2 to take parity into account.

Now we have the rub.  In order to calculate the positions for
which Symm(Y)=HV, we must subtract out the HV-symmetric
positions which have stronger symmetry than HV, just as we
subtracted out the M-symmetric positions in Mike's X1 case.
But to do so, we cannot take the subgroups of M in isolation.
We have to do them all, starting with M and working our way
down.  (And HV is pretty far down the food chain.)

Some of the subgroups I can do pretty easily, and for others I have
not a clue.  Recall that A is the subgroup of M consisting of the
24 even rotations and reflections and that C is the subgroup of
M consisting of the 24 rotations (12 even and 12 odd).  As long
ago as _Symmetry and Local Maxima_, Dan Hoey and Jim Saxe
determined that there are only four A-symmetric positions and
only four C-symmetric positions, namely the four that are also
M-symmetric.  Hence, there are no positions for which Symm(Y)=A
nor for which Symm(Y)=C.  But I haven't a clue how they knew,
nor how to go about constructing an A-symmetric or a C-symmetric
position from scratch.  You can't get very far with my proposal
unless you can figure out how to construct K-symmetric positions
for any K.

For one more example, consider H, the set of 12 even rotations and
12 odd reflections.  I know from computer search and also from
_Symmetry and Local Maxima_ that there are 24 H-symmetric positions,
of which 4 are M-symmetric and 20 are H-symmetric without also being
M-symmetric.  The 20 H-symmetric but not M-symmetric positions form
10 M-conjugacy classes for which we would say SymmClass(Y)=H.  It
ought to be easy to derive this result without a computer search, but
again I confess I haven't a clue as to go about constructing the 24
H-symmetric positions from scratch.

Well, I could cheat and look up the Class H positions
in _Symmetry and Local Maxima_, but what about the classes
that haven't been figured out yet?  Also, I could cheat and
use the results from computer search, but that's hardly the
point.

One final point: just as Mike's 128 X1-symmetric positions formed
a group, similarly the set of K-symmetric positions form a group
for all 98 possible values of K.  We have to be a little careful
with our terminology.  The X1-symmetric positions form a group,
as do the X2-symmetric and the X3-symmetric positions.  But if
we want to talk about the X-symmetric positions, we no longer
have a group.  For example, we do not in general have closure
when forming the composition of X1-symmetric positions with
X2-symmetric positions.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
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

From BRYAN@wvnvm.wvnet.edu  Sun Jun 18 08:48:57 1995
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Message-Id: <wvmail32.1995jun18.081858.bryan@wvnvm.wvnet.edu>
Date:      Sun, 18 Jun 1995 08:25:49 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <cube-lovers@life.ai.mit.edu>
Subject:   Re: Crazy Corner Pattern
In-Reply-To: Message of 06/17/95 at 05:38:00 from mark.longridge@canrem.com

On 06/17/95 at 05:38:00 mark.longridge@canrem.com said:
>On Mon, 17 Jan 1994 09:06:59 EST, Jerry wrote:
>>
>>     Counting M-conjugacy classes of the corners of Rubik's cube
>>     -----------------------------------------------------------
>>
>>                   M-Class     Number     Number
>>                     Size        of         of
>>                              Classes    Elements
>>
>>                         1        1  =        1     Start
>>                         2        1  =        2     +4 -4 Twist
>>                         3        3  =        9
>>                         4        1  =        4
>>                         6       34  =      204
>>                         8       33  =      264
>>                        12      301  =     3612
>>                        16      104  =     1664
>>                        24     9064  =   217536
>>                        48  1832428  = 87956544
>>
>>                     Total  1841970    88179840

>I'm trying to find the 1 pattern with M-class size of 4 of the
>corners group.
>The only pattern that I can find is 4 alternate corners twisted
>clockwise which is in the twist orbit.

>It does not seem to be any pattern with just corners twisted in
>place.

>Jerry, if you could *please* identify this pattern before I go
>nuts...... !

The position is called T-symmetric in Dan's taxonomy (actually,
there are four T subgroups of M, T1 through T4).  The symmetry
is related to opposite corners, e.g., the UFL-DRB axis,
which is why there are four T subgroups.

Also, the position is Q-transitive, so you can check it out
in _Symmetry and Local Maxima_.  I quote ".... each edge on the
girdle may be swapped with the diametrically opposite edge,
provided that the corners on the girdle are swapped with
their opposites as well."   Here, you would fix the edges and
pay attention only to the swapping of the corners.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
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

From BRYAN@wvnvm.wvnet.edu  Sun Jun 18 15:55:14 1995
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Date:      Sun, 18 Jun 1995 15:55:22 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Re: A Third Way to Calculate the Real Size of Cube Space?
In-Reply-To: Message of 06/17/95 at 17:18:00 from BRYAN@wvnvm.wvnet.edu

I should have said "a fourth way", I think.  Martin Schoernert
performed the same calculation with GAP.  Hence, we have three
ways in hand: 1) Dan's Polya-Burnside method, 2) Martin's GAP
calculations, and 3) brute force computer search.  My new
proposal would then be a fourth way.

Here is a question for Martin: is there any way with GAP to calculate
the number of M-conjugacy classes associated with each symmetry class?
It is this additional information about the "real cube space" which
*is* available via computer search, and for which I am proposing
an alternative which does not involve computer search.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
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

From @mail.uunet.ca:mark.longridge@canrem.com  Sun Jun 18 23:54:16 1995
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Sender: CRSO.Cube@canrem.com
Subject: Re: Crazy Corner Pattern
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.1167.5834.0C1E77E1@canrem.com>
Date: Mon, 19 Jun 1995 00:36:00 -0400
Organization: CRS Online  (Toronto, Ontario)

>>I'm trying to find the 1 pattern with M-class size of 4 of the
>>corners group.
>>The only pattern that I can find is 4 alternate corners twisted
>>clockwise which is in the twist orbit.

>>It does not seem to be any pattern with just corners twisted in
>>place.

>>Jerry, if you could *please* identify this pattern before I go
>>nuts...... !

>The position is called T-symmetric in Dan's taxonomy (actually,
>there are four T subgroups of M, T1 through T4).  The symmetry
>is related to opposite corners, e.g., the UFL-DRB axis,
>which is why there are four T subgroups.

>Also, the position is Q-transitive, so you can check it out
>in _Symmetry and Local Maxima_.  I quote ".... each edge on the
>girdle may be swapped with the diametrically opposite edge,
>provided that the corners on the girdle are swapped with
>their opposites as well."   Here, you would fix the edges and
>pay attention only to the swapping of the corners.

Hmmmmm, that's very interesting.

Below is my interpretation of "...corners on the girdle are swapped
with their opposites as well." Let's call it pattern X.

        D U D
        U U U
        D U D
F L B   R F L   B R F   L B R
L L L   F F F   R R R   B B B
F L B   R F L   B R F   L B R
        U D U
        D D D
        U D U

If I understand the terminology correctly, then for this pattern
X = |{m'Xm}|=3 and |Symm(X)|=16, same as the 4 spot.
Also X = |{c'Xc}|=3.

But perhaps this is not the same pattern....

Let's call this next cube arrangement pattern Y.

        F U U
        U U U
        D U L
R L B   R F B   D R R   B B D
L L L   F F F   R R R   B B B
U L L   F F U   L R F   L B F
        D D B
        D D D
        R D U

Y = |{m'Ym}|=4 and |Symm(Y)|=12, and Y=|{c'Yc}| = 4

This pattern was created by swapping 3 pairs of diametrically opposite
corners, which is in the *swap* orbit on a normal cube, but since we
are dealing with corners of Rubik's cube and ignoring edges we can
realize permutations with an odd number of pairs of corners swapped.

-> Mark <-

From BRYAN@wvnvm.wvnet.edu  Mon Jun 19 09:57:45 1995
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Date:      Mon, 19 Jun 1995 09:57:53 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <cube-lovers@life.ai.mit.edu>
Subject:   Re: Crazy Corner Pattern
In-Reply-To: Message of 06/19/95 at 00:36:00 from mark.longridge@canrem.com

Perhaps I should have quoted a little more from _Symmetry and
Local Maxima_.  Here is the T-symmetric position given by
Saxe and Hoey.  In their position, the UFL and RBD corners are
in place, and the other three pairs are swapped.  The "girdle"
includes the three pairs that are swapped.  Hence, there
is an axis of symmetry along the UFL-RBD axis.  The odd number of
swaps is compensated by an odd number in the edges.  The compensation
is not required for corners only.

        R D D
        U U D
        U U B

 D L L  F F D  L L F  L F F
 R L L  F F B  L R R  B B F
 R R B  R B B  U R R  B B U

        U U L
        U D D
        F D D

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
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

From BRYAN@wvnvm.wvnet.edu  Mon Jun 19 14:04:49 1995
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Date:      Mon, 19 Jun 1995 13:41:51 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   3x3x3 Cubes for Sale

For the first time in years, I have seen 3x3x3 cubes for sale in a
store.  (For American readers, the store is Hills, which is a regional
chain which competes against WalMart and Kmart.  The price is $8.97 --
U.S. dollars.)  The box has a note signed by Erno Rubik, and gives
the proper size of the cube group.  Also, and I couldn't see inside
the box to verify this, the Face centers seemed to be marked in such
a way as to support the Supergroup.  Rubik's note about the size of
the  problem says it is 4^4 times bigger than the regular problem.
The manufacturer (or maybe distributor?) is Matchbox Toys (or is it
Matchbook Toys?).

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
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

From news@nntp-server.caltech.edu  Mon Jun 19 18:36:29 1995
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From: whuang@cco.caltech.edu (Wei-Hwa Huang)
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Subject: Re: 3x3x3 Cubes for Sale
Date: 19 Jun 1995 22:36:17 GMT
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"Jerry Bryan" <BRYAN@wvnvm.wvnet.edu> writes:

>For the first time in years, I have seen 3x3x3 cubes for sale in a
>store.  (For American readers, the store is Hills, which is a regional
>chain which competes against WalMart and Kmart.  The price is $8.97 --
>U.S. dollars.)  The box has a note signed by Erno Rubik, and gives
>the proper size of the cube group.  Also, and I couldn't see inside
>the box to verify this, the Face centers seemed to be marked in such
>a way as to support the Supergroup.  Rubik's note about the size of
>the  problem says it is 4^4 times bigger than the regular problem.
>The manufacturer (or maybe distributor?) is Matchbox Toys (or is it
>Matchbook Toys?).

What you saw was "Rubik's Cube 4th Dimension."  This was released some
time in the early 90s.  The only difference between this and the typical 3x3x3
is that four faces have pictures on their center squares, requiring 
the solver to orient them correctly.  

I belive last year there was a European Rubik Puzzle series that rereleased
the cube, along with other rubik puzzles.

-- 
   -- Wei-Hwa Huang (whuang@cco.caltech.edu)
Homepage (under construction): http://www.ugcs.caltech.edu/~whuang/
I have a proof that NP = P; however, it requires exponential time to write down.

From pbeck@pica.army.mil  Tue Jun 20 08:00:48 1995
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Date:     Tue, 20 Jun 95 8:00:16 EDT
From: Peter Beck (FSAC) <pbeck@pica.army.mil>
To: cube-lovers@ai.mit.edu
Subject:  cube availability
Message-Id:  <9506200800.aa03200@COR6.PICA.ARMY.MIL>


In Feb at the NY international toy show the KOOSH people
told me that they had bought the marketing rights for
RUBIK items from Western Publishing (the Matchbox LABEL).
They had new packaging and were changing the items to
be sold (eg, they said they were going to sell Rubuik's
amgic).

So the items seen in Hills with Matchbox packaging is
old stock that will soon no longer be around.

THE FUTURE IS PUZZLING,
BUT CUBING IS FOREVER !!!

pete beck

From mschoene@math.rwth-aachen.de  Tue Jun 20 08:40:27 1995
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Date: Tue, 20 Jun 95 14:39 WET DST
From: "Martin Schoenert" <Martin.Schoenert@math.rwth-aachen.de>
To: Cube-Lovers@ai.mit.edu
Cc: BRYAN@wvnvm.wvnet.edu
In-Reply-To: "Jerry Bryan"'s message of Sun, 18 Jun 1995 15:55:22 -0400 (EDT) <wvmail32.1995jun18.154620.bryan@wvnvm.wvnet.edu>
Subject: Re: Re: A Third Way to Calculate the Real Size of Cube Space?

Jerry wrote in his message of 1995/06/17

    We define the real size of cube space to be the number of M-conjugate
    classes {m'Ym} for m in M, set of 48 rotations and reflections
    of the cube, and for Y in G.

    Dan Hoey has calculated the real size of cube space using the
    Polya-Burnside theorem.

    Dan and I (mostly Dan) have also calculated the same result using
    exhaustive computer search.  The computer search is much less
    elegant than the Polya-Burnside results, but the search does
    provide additional information, such as the number of positions
    associated with each symmetry group.  The results from the
    computer search have not yet been posted to the list, but a draft
    paper is in progress.

    In the meantime, it occurs to me that perhaps  --  but only perhaps --
    there is a third way to calculate the real size of cube space.
    The third way would not require (much) computer searching, but would
    provide the same level of detail about number of positions per
    symmetry group as does the full blown search.

    ... detailed description of the method deleted ...

And he continued in his message of 1995/06/18

    I should have said "a fourth way", I think.  Martin Schoernert
    performed the same calculation with GAP.  Hence, we have three
    ways in hand: 1) Dan's Polya-Burnside method, 2) Martin's GAP
    calculations, and 3) brute force computer search.  My new
    proposal would then be a fourth way.

Well, I don't know about *four* ways.

Dan used the Polya-Burnside theorem.  That is, he computed the
number of M-conjugacy classes as the average number of fixed points
of the elements of M w.r.t. to their action on G.  He computed
the number of fixed points of an element m using clever arguments
about the cycle structure of elements of G that m would fix.

I simply observed that the number of fixed points of an element m is
the size of the centralizer of in G, and then used GAP to compute
those.  So I don't think it is correct to call this a way of its own.

The method you propose is indeed different from Dan's method that
uses Polya-Burnside.

I can't figure out how the brute force computer search works.
So I can't tell whether it is really different from the other methods
(and if indeed it is a method to compute the real size of the cube ;-).
Jerry, could you say a little bit more about this computation?

It appears to me that Dan and Jim Saxe must have realized all the
important pieces for your new method when they wrote their seminal
``Symmetry and Local Maxima (long message)'' message of 1980/12/14.
As Jerry points out, they did calculate the important values for
9 of the 33 conjugacy classes of subgroups of M (those whose sizes
are a multiple of 12).  It is neither clear from their message how
they found those 9 classes (in fact they apparently found all 98
subgroups of M), nor how they computed the numbers of elements of
G that have a specific subgroup of M as symmetry group.
Perhaps Dan can say a little bit more about this?

Jerry continued in his message of 1995/06/18

    Here is a question for Martin: is there any way with GAP to calculate
    the number of M-conjugacy classes associated with each symmetry class?
    It is this additional information about the "real cube space" which
    *is* available via computer search, and for which I am proposing
    an alternative which does not involve computer search.

Of course there is ;-).

Given a subgroup H of M, the centralizer of H in G is the set of all
those elements of G that commute with all elements of H.  But this is
of course simply the set of those elements of G that have either H or
a larger group as their symmetry group.

So we can compute the numbers of elements of G with symmetry group H
by computing the size of the centralizer of H in G, and then subtracting
the numbers of those elements that have a symmetry group that is a
proper supergroup of H.  This is easy if we compute those numbers
for all subgroups of M, from the larger subgroups down to the smaller.
Of course it is not neccessarry to do this for all 98 subgroups
of M, but only for one subgroup for each of the 33 conjugacy classes.

Then if we simply divide the number of elements with symmetry group H by
the index of H in M, we obtain the number of M-conjugacy classes into
which those elements fall.

As a GAP program this looks as follows

    # compute the conjugacy classes of subgroups of M
    classes := ConjugacyClassesSubgroups( M );
    numbers := [];

    # for all conjugacy classes of subgroups of M
    for i  in [Length(classes),Length(classes)-1..1]  do

        # select a representative for this conjugacy class
        rep := Representative( classes[i] );

        # compute how many elements have at least this symmetry group
        number := Size( Centralizer( G, rep ) );

        # subtract the number of elements that have a larger symmetry group
        for k  in [Length(classes),Length(classes)-1..i+1]  do
            for sub  in Elements( classes[k] )  do
                if IsSubgroup( sub, rep )  then
                    number := number - numbers[k];
                fi;
            od;
        od;

        # store the number
        numbers[i] := number;

        # print the number of the class
        Print( i, ":\t" );

        # the size of the subgroups in the class
        Print( Size(rep), "\t" );

        # the number of subgroups in the class
        Print( Size(classes[i]), "\t" );

        # the number of elements whose symmetry group lies in the class
        Print( Size(classes[i]), " * ", number, "\t" );

        # and the number of M-conjugacy classes of those elements
        Print( Size(classes[i]), " * ", number, " / ", Index(M,rep), "\n" );

    od;

*Do not try this with GAP 3.4.2 (our latest release).*
GAP 3.4.2 contains several naive functions for permutation groups,
that cause this computation to take a very long time.
But with GAP 3.5 (our current development version),
this produces in about a minute the following table.

CLASS   SIZE   LENGHT NUMBER                    REAL		NAME
33:     48     1      1 * 4                     1 * 4 / 1       (M)
32:     24     1      1 * 0                     1 * 0 / 2       (C)
31:     24     1      1 * 0                     1 * 0 / 2       (AM)
30:     24     1      1 * 20                    1 * 20 / 2      (H)
29:     16     3      3 * 124                   3 * 124 / 3     (X1,X2,X3)
28:     12     4      4 * 12                    4 * 12 / 4      (T1,T2,T3)
27:     12     1      1 * 48                    1 * 48 / 4
26:      8     3      3 * 384                   3 * 384 / 6
25:      8     3      3 * 1408                  3 * 1408 / 6
24:      8     3      3 * 2944                  3 * 2944 / 6
23:      8     3      3 * 1920                  3 * 1920 / 6
22:      8     3      3 * 384                   3 * 384 / 6
21:      8     3      3 * 896                   3 * 896 / 6
20:      8     1      1 * 11892                 1 * 11892 / 6
19:      6     4      4 * 416                   4 * 416 / 8
18:      6     4      4 * 32                    4 * 32 / 8
17:      6     4      4 * 7740                  4 * 7740 / 8
16:      4     6      6 * 96232                 6 * 96232 / 12
15:      4     6      6 * 96256                 6 * 96256 / 12
14:      4     3      3 * 92928                 3 * 92928 / 12
13:      4     3      3 * 437504                3 * 437504 / 12
12:      4     3      3 * 574208                3 * 574208 / 12
11:      4     3      3 * 1163520               3 * 1163520 / 12
10:      4     3      3 * 144640                3 * 144640 / 12
9:       4     3      3 * 62208                 3 * 62208 / 12
8:       4     1      1 * 280272                1 * 280272 / 12
7:       3     4      4 * 3770864               4 * 3770864 / 16
6:       2     6      6 * 424415168             6 * 424415168 / 24
5:       2     6      6 * 2547748032            6 * 2547748032 / 24
4:       2     3      3 * 15285460992           3 * 15285460992 / 24
3:       2     3      3 * 18342768640           3 * 18342768640 / 24
2:       2     1      1 * 45862360944           1 * 45862360944 / 24
1:       1     1      1 * 43252003109885814336  1 * 43252003109885814336 / 48

As expected the numbers in the fourth column add to the size of G.
And the numbers in the fifth column add to 901083404981813616,
the real size of the cube (|M\MG/M|).

For those classes that I could identify I have added their names.
If somebody could describe Dan's taxonomy, I will name the other
classes as well.

Have a nice day.

Martin.

-- .- .-. - .. -.  .-.. --- ...- . ...  .- -. -. .. -.- .-
Martin Sch"onert,   Martin.Schoenert@Math.RWTH-Aachen.DE,   +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany

From mschoene@math.rwth-aachen.de  Tue Jun 20 09:29:59 1995
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Date: Tue, 20 Jun 95 14:40 WET DST
From: "Martin Schoenert" <Martin.Schoenert@math.rwth-aachen.de>
To: Cube-Lovers@ai.mit.edu
Cc: BRYAN@wvnvm.wvnet.edu
In-Reply-To: "Jerry Bryan"'s message of Sun, 18 Jun 1995 15:55:22 -0400 (EDT) <wvmail32.1995jun18.154620.bryan@wvnvm.wvnet.edu>
Subject: Re: Re: A Third Way to Calculate the Real Size of Cube Space?

Looking through the old messages about the real size of the cube group,
it appeared to me that no one has shown a proof for the Polya-Burnside
theorem.  Since it is not difficult to prove, I decided to write one up.

In the following I will use TeX notation for formulae, i.e., formulae
are included in '$' signs, '{}' are used to group terms, '^' is used
for superscripts, and '_' for subscripts.

If $g \in G$, then I denote the set of elements that are really
equivalent to $g$ by $g^M$.  Jerry denotes this set by {m'gm},
but $g^M$ is the more common notation in group theory.

The sum $\sum_{g \in h^M}{1/|g^M|}$ is simply 1, since it is the sum over
all elements in one M-conjugacy class (h^M) of 1 over the length of
that M-conjugacy class.  Thus the sum $\sum_{g \in G}{1/|g^M|}$ is the
number of M-conjugacy classes.

Now we need a standard lemma from group theory, which tells us that the
length of a class $g^M$ of an element $g$ under the action of a group $M$
is equal to the size of the group $M$ divided by the size of the subgroup
of those elements of $M$ that fix $g$ (more precisely the index of that
subgroup in $M$, since the lemma is true, even if $M$ is infinite).

So using Jerry's notation this lemma gives $1/|g^M| = |Symm(g)|/|M|$.
Applying that to the above formula we see that the number of
M-conjugacy classes in $G$ is $\sum_{g \in G} {|Symm(g)|/|M|}$
Or, after a trivial change, $1/|M| \sum_{g \in G} {|Symm(g)|}$.

Assume that $(g^m == g)$ is 1 if $g^m$ is equal to $g$ and 0 otherwise.
Then we have $|Symm(g)| = \sum_{m \in M}{(g^m == g)}$.
Thus the number of M-conjugacy classes is
$1/|M| \sum_{g \in G} \sum_{m \in M} {(g^m == g)}$.

Now we can simply change the order of the two summations, so we get
$1/|M| \sum_{m \in M} \sum_{g \in G} {(g^m == g)}$.

But of course $\sum_{g \in G} {(g^m == g)}$ is obviously the number of
fixpoints of $m$.  So we obtain the Polya-Burnside lemma: ``The number of
M-conjugacy classes is the average number of fixpoints of the elements
of $M$ w.r.t. their operation on $G$''.

However, here the operation is special, so we can simplify even further.
$g^m$ here means $m^{-1} g m$, so $(g^m == g)$ means $(m^{-1} g m == g)$,
which is equivalent to $(m == g^{-1} m g)$ (multiply the equation first
by $m$ and then by $g^{-1}$ from the left), which is $(m == m^g)$.

So the number of M-conjugacy classes is
$1/|M| \sum_{m \in M} \sum_{g \in G} {(m == m^g)}$.

But $\sum_{g \in G} {(m == m^g)}$ is simply the size of the subgroup of
those elements in $G$ that fix $m$.  This is the centralizer of
$m$ in $G$.  So the number of M-conjugacy classes is finally
$1/|M| \sum_{m \in M} |Centralizer(G,m)|$.

This is the formulation that I used to compute the real size of the cube
group with GAP.

Have a nice day.

Martin.

-- .- .-. - .. -.  .-.. --- ...- . ...  .- -. -. .. -.- .-
Martin Sch"onert,   Martin.Schoenert@Math.RWTH-Aachen.DE,   +49 241 804551
Lehrstuhl D f"ur Mathematik, Templergraben 64, RWTH, 52056 Aachen, Germany

From BRYAN@wvnvm.wvnet.edu  Tue Jun 20 18:04:35 1995
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Date:      Tue, 20 Jun 1995 13:48:51 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Re: Re: A Third Way to Calculate the Real Size of Cube Space?
In-Reply-To: Message of 06/20/95 at 14:39:00 from ,
           Martin.Schoenert@Math.RWTH-Aachen.DE

On 06/20/95 at 14:39:00 Martin Schoenert said:

>I can't figure out how the brute force computer search works.
>So I can't tell whether it is really different from the other methods
>(and if indeed it is a method to compute the real size of the cube ;-).
>Jerry, could you say a little bit more about this computation?

I look at corners and edges separately, and then combine the results.

I can't speak for how Dan did his corner search and his edge search,
but I can describe mine.  Our results do match, which is always nice.

Dan did most of the figuring out of "combining the results" for corners
and edges.

Conceptually, I look at every single corner position X and calculate
Symm(X), and I look at every single edge position Y and calculate
Symm(Y).  In practice, there are some important shortcuts.

I have a data base containing "every" corner position and a second
data base containing "every" edge position with "every" defined in
the following sense.  The set of all positions is partitioned into
equivalence classes of the form {m'Xmc} for the corners and {m'Ymc}
for the edges, for m in M (48 rotations and reflections) and c in C
(24 rotations).  The data bases contain a representative element
from each equivalence class.

For all the cases where |{m'Xmc}|=1152 and |{m'Ymc}|=1152, no searching
is required.  For these cases, we know *a priori* that there are 24
M-conjugacy classes containing 48 elements each, and that for each
of the 1152 positions we have Symm(X)=I and Symm(Y)=I.  Fortunately,
for the vast majority of the cases (over 97% for corners and well
over 99% for edges), the so-called B-class size is 1152.

Suppose the representative for {m'Xmc} is V and for {m'Ymc} is W.
For the remaining cases, the idea is to use Vc and Wc for
each fixed c in C as a base or representative element for an
M-conjugacy class of the form {m'(Vc)m} and {m'(Wc)m}.  The tricky
part here is that while Vc and Vd are distinct when c and d in C
are not equal, nonetheless Vc and Vd may be M-conjugate.  Hence,
we use each Vc and Wc which are distinct up to M-conjugacy as
a representative element for an M-conjugacy class.

Conceptually, we calculate Symm(T) for each T in {m'(Vc)m} for each
fixed c (and for Vc distinct up to M-conjugacy) and summarize the
results.  In practice, it is sufficient to calculate Symm(Vc).

Here comes another tricky part (at least it was until I figured it out).
Initially, I assumed that Symm(T) was the same for all T in {m'(Vc)m}.
However, such is not the case.  Rather, if you calculate each Symm(T), you
will discover that each one will be a group from a class of conjugate
groups, and that each group in the class of conjugate groups will appear
an equal number of times.

Given that you have picked an arbitrary representative from the M-conjugacy
class, you don't know which conjugate group you are going to get when
you calculate Symm(m'(Vc)m).  I got around this issue originally by
calculating a representative group (the 98 subgroups of M then have
33 representative groups, one for each of the 33 symmetry classes).
In the work I am doing now, I map directly from each of the 98 subgroups
to their respective symmetry class.

The exhaustive search then consists of performing the above calculations
for each B-class of the form {m'Xmc} and {m'Ymc} and summarizing the
results (that is, counting all the M-conjugacy classes).  In addition
to an overall total, you summarize by symmetry class, which gives you
the additional information I have talked about, over and above Dan's
Polya-Burnside results.

At this point, the summary results give you the real size of the corners
only problem and the real size of the edges only problem.  While you are
at it, you have to count the M-conjugacy classes for even positions and odd
positions separately in order to put corners and edges together
properly.

Finally, you put the corners and edges together in all possible ways.
The putting together of the corners and edges is described in the
draft I have mentioned, so I will just wait until the draft is
ready before posting the rest.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
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

From hoey@aic.nrl.navy.mil  Thu Jun 22 03:44:23 1995
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Date: Thu, 22 Jun 95 03:43:59 EDT
From: hoey@aic.nrl.navy.mil (Dan Hoey)
Message-Id: <9506220743.AA25013@Sun0.AIC.NRL.Navy.Mil>
To: Cube-Lovers@ai.mit.edu,
        "Martin Schoenert" <Martin.Schoenert@math.rwth-aachen.de>
Subject: Ways to Calculate the Real Size of Cube Space?
In-Reply-To: <m0sO2aQ-00026zC@hobbes.math.rwth-aachen.de>

Martin Schoenert says:
> I can't figure out how the brute force computer search works.

and while Jerry Bryan gives one answer, I have another.  For you see,
I also ran a brute force computer search for the symmetry classes,
too.  And while it agrees with Jerry's answers, mine was a
significantly different algorithm, which I discuss a little a the end
of this message.

> It appears to me that Dan and Jim Saxe must have realized all the
> important pieces for your new method when they wrote their seminal
> ``Symmetry and Local Maxima (long message)'' message of 1980/12/14.
> As Jerry points out, they did calculate the important values for
> 9 of the 33 conjugacy classes of subgroups of M (those whose sizes
> are a multiple of 12).  It is neither clear from their message how
> they found those 9 classes (in fact they apparently found all 98
> subgroups of M), nor how they computed the numbers of elements of
> G that have a specific subgroup of M as symmetry group.
> Perhaps Dan can say a little bit more about this?

First, to find all the subgoups of M.  I represented the elements of M
as a list of permutations on faces, found easily enough by finding the
closure of the generators.  Then I did a depth-first search for
subgroups, branching by computing the closure of the current subgroup
with each possible element not in that subgroup, and cutting off the
search on previously-seen subgroups.  It's quick, it's dirty, it
works.  I found the nine subgroups of order a multiple of 12, as shown
in the Hasse subgroup diagram:

    order 48 . . . M_
                  /|\\_ 
                 / | \ \_ 
                /  |  \  \_
    order 24 . A . H . C   \_
                \  |  /      \_
                 \ | /         \_
                  \|/            \
    order 12 . . . E . . . . .T[1..4]

(except we called E "AC" back then).  The trick then was to find all
the E-symmetric positions and all the T1-symmetric positions; the
tasks of finding the full symmetry group of such positions and
counting the positions for T2, T3, and T4 were straightforward.

The best tool we had for figuring out symmetric positions was
essentially the one I wrote about in ``The real size of cube space''
on 4 Nov 94.  For a subgroup J of M, if a position g is J-symmetric,
then g must commute with each operation m in J.  Recall:

The fundamental principle we use in finding whether g commutes with
m can be found by examining the cycles of m.  Suppose m permutes a
cycle (c1,c2,...,c[k-1],ck), so that
    c2=m(c1), c3=m(c2), ..., ck=m(c[k-1]), and c1=m(ck).
For g to commute with m, we have
    g(c2)=m(g(c1)), g(c3)=m(g(c2)), ..., g(ck)=m(g(c[k-1])),
    and g(c1)=m(g(ck)).
So (g(c1),g(c2),...,g(ck)) is also a cycle of m.  Thus g must map each
k-cycle of m to another k-cycle of m, and in the same order.

The orientation question was a lot more difficult, so we ran through a
bunch of little results.  The following is a cleaned-up sample of the
sort of arguments, as I remember them.  In it FRD is an unoriented
corner cubie/cubicle and FRD.D is its down-facing color-tab/facicle.
================================================================
    Lemma 1: Suppose X and Y are corners, and m is in C, m(X)=Y.
             Suppose g(X)=X and g(Y)=Y, and g commutes with m,
             Then g applies the same twist to corners X and Y.
    Proof: Let TX,TY be the clockwise 120-degree rotation of corners
           cubies X and Y, respectively.  Then m(TX(.))=TY(m(.)), as
           can be seen by the fact that we could apply a twist to X in
           place (TX) before moving it to Y with m, or we can perform
           the same twist on Y (TY) after moving it.  So if
           g(X)=TX^k(X), then
           g(Y)=g(m(X))=m(g(X))=m(TX^k(X))=TY^k(m(X))=TY^k(Y).
           performing the same twist on Y as on X,               QED.

    Lemma 2: If g is E-symmetric, then each corner cubie remains in
             its home cubicle (not considering orientation).
    Proof: Supposing otherwise, take a moved cubie (without loss of
           generality) to be FRD, and suppose (w.l.o.g.) it moves to
           one of locations FDL, FLT, or BTL.
        Case 1.  If g(FRD)=FDL, consider operation m to be the
                 120-degree rotation about FRD.
                 m(FRD)=FRD, m(FDL)=FTR.  So
                 g(FRD)=g(m(FRD))=m(g(FRD))=m(FDL)=FTR, contradicting
                 g(FRD)=FDL.
        Case 2.  If g(FRD)=FLT, then take m as in case 1; m(FLT)=BRT,
                 so g(FRD)=g(m(FRD))=m(g(FRD))=m(FLT)=BRT,
                 contradicting g(FRD)=FDL.
        Case 3.  If g(FRD)=BTL, then g(FRD.F) is BTL.B, BTL.T, or
                 BTL.L.
        Case 3a. If g(FRD.F)=BTL.B, then g(FRD.R)=BTL.T, by clockwise
                 adjacency.  But m(FRD.F)=FRD.R and m(BTL.B)=BTL.L,
                 and g(FRD.R)=g(m(FRD.F))=m(g(FRD.F))=m(BTL.B)=BTL.L
                 contradicts G(FRD.F)=BTL.B.
        Cases 3b and 3c work the same way.
           The contradictions establish that FRD does not move,  QED.

    Lemma 3: If g is E-symmetric, then the twists of the four corner
             cubies FRD, FLT, BLD, and BRT agree with each other, and
             and the other four also agree with each other.
    Proof: For any two of the four corners (e.g. FRD, FLT), there is
           a 120-degree rotation in E taking one to the other
           (e.g. the rotation about FTR).  Lemma 1 applies
           immediately to show the twists agree,                 QED.

    Lemma 4. If g is E-symmetric, then the corner cubies are all
             solved, or are rotated alternately in opposite
             directions.
    Proof: From Lemma 2, all the cubies are in their home cubicles.
           If one of the sets from Lemma 3 is twisted, their total
           twist is the twist of a single cubie (since there are four
           of them) and so must be counteracted by having the other
           set twisted in the opposite direction, which is alternate
           corners twisted oppositely; otherwise all corner are
           solved,                                               QED.

    Lemma 5. Let T1 refer to the group that fixes the FRD-BTL axis.
             Then any T1-symmetric position g must keep the FRD and
             BTL cubies in their solved position, and rotated by the
             same amount.
    Proof: Let m be the 120-degree rotation about FRD, which is in T1.
           Since FRD and BTL are the only 1-orbits of m, they are kept
           in place or swapped.  From the proof of Lemma 2 (which uses
           the same m) they cannot be swapped.  Otherwise, the two
           cubies are kept in place, and 180 degree rotation about the
           FL-BR axis, also in T1 fulfills the requirements of Lemma 1
           to show they will both be rotated the same amount,    QED.
================================================================
That's about all I feel like remembering and formalizing right now.
As you can see, it's long, mechanical, and boring.  That's why we
never got around to writing it all down.

Early last year I wrote a computer program to find *all* the symmetry
groups of *all* the positions.  The first part did the corners and
edges separately, counting the number of positions for each symmetry
group and permutation parity.  For each of the 8! or 12! permutations,
I checked to see if the permutation commuted with some nontrivial
operation of M; if not, I just counted the appropriate number of
I-symmetric positions.  Otherwise I applied orientations and counted
up the symmetry groups of each possible orientation.  (It could
probably have been made to go faster, by cutting off partial
permutation or orientation generation as soon as all the non-trivial
operations were ruled out.)

In the above counts, I also kept track each time of whether the
permutation was even or odd.  Then after I had the count of even and
odd permutations for corners and edges in each symmetry group, I had a
program that intersected the symmetry groups, and for each pair of
subgroups J,K, and each parity P, I added
    Corners[J,P] * Edges[K,P]  to  Whole[J intersect K, P], 
with some fancy footwork so I only needed to deal with conjugacy
classes for J and K.  Then for each K, the number of whole K-symmetric
positions was Whole[K,Even]+Whole[K,Odd].

The program finished about the time I realized the application of the
Polya-Burnside theorem.  Then for most of the year, I put off writing
it all up.  I have Jerry to thank for reminding me to get with it, and
for useful comments and discussions on the early drafts.

Dan

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Date:      Tue, 27 Jun 1995 23:22:50 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Constructing K-symmetric Cubes

This is a followup on several recent messages concerning the
question of K-symmetric cubes, where K is one of the
98 subgroups of M.

We recall that a permutation is a special kind of function,
namely a one-to-one and onto function on a set.  A very
common technique used with functions is to restrict the
domain to a (usually proper) subset of the original domain.
In the paradigm of a function as a general rule, the general rule
is applied to the subset of the domain to obtain the restriction
of the function.  In the paradigm of a function as a set of ordered
pairs, the restriction is simply a (usually proper) subset of the
set of ordered pairs.  A restriction of a permutation is usually
not a permutation (certainly not a permutation on the original
domain), but it is still a function.

We will treat a permutation on the cube as a set of restrictions
(functions) of the several cubicles.  We take as our first example
the function UFL->UFL.  Unlike cycle notation, it is not
assumed that the other cubicles are fixed;  rather, the other
function values are undefined (e.g., URF->?).

Let X be any function (not necessarily a permutation) whose domain
is some subset of the cubicles.  We define Symm(X) in the standard
fashion  --  Symm(X) is the set of all m in M such that
m'Xm=X.  If X is the function UFL->UFL, we have Symm(X)=AT4, not
Symm(X)=M as you might expect.  AT4 is a subgroup in Dan's taxonomy
containing six elements, and which has an axis of symmetry along
the UFL-DBR axis.

This definition of Symm(X) perhaps requires a minor bit of
justification.  In a function composition such as FG (left-to-
right notation) or G(F(x)) (right-to-left "calculus" notation),
it is sometimes taken as a convention that the range of F must
match the domain of G.  But we can also take the restriction of
G to the intersection of the range of F with the domain of G,
and we do so.  Having done so, Symm(X) is well defined.

We wish to build an M-symmetric permutation containing the
function UFL->UFL, but Symm(UFL->UFL)=AT4 is not a very
auspicious start.  Rather, we define the conditions under which
a function is K-symmetric is follows.  A function X is K-symmetric
if the union of the K-conjugates k'Xk is a function.

Given this definition, the only M-symmetric function on UFL is
in fact UFL->UFL, so we really have made a good start.  Furthermore,
any M-symmetric permutation that contains UFL->UFL must also
contain all the M-conjugates of UFL->UFL, and the union of the
M-conjugates is simply the identity permutation on the corners.

The fact that Symm(UFL->UFL)=AT4 can be of some benefit in our
investigations.  In particular, the fact that |AT4|=6 means that
there are 8 M-conjugates, so taking all the M-conjugates of
UFL->UFL means that all 8 corners are specified.

Our next example will be UFL->LUF (a twist of the corner).  In
this case, we have Symm(X)=ET4.  ET4 is the subgroup of M in
Dan's taxonomy which contains 3 elements including the identity
plus the 1/3 and 2/3 rotations around the UFL-DBR axis.

Of more import, UFL->LUF is not M-symmetric.  However, it is both
C-symmetric (C is the set of 24 rotations) and H-symmetric
(H is the set of 12 even rotations and 12 odd reflections).  As
an aside, we note that it is not A-symmetric, where A is the set
of 24 even rotations and reflections.  But since it is both
C-symmetric and H-symmetric, there is not a unique largest
subgroup K for which we can say it is K-symmetric.

It is easy to see that UFL->LUF is not M-symmetric.  The set of
M-conjugates contains both UFL->LUF and UFL->FLU, so the union
of the M-conjugates is not a function.  We can see the same thing
from the fact that |ET4|=3.  Since |ET4|=3, there are 16 M-conjugates,
but there are only 8 corners to represent the 16 M-conjugates.

Since UFL->LUF is C-symmetric, let's see if we can build a
C-symmetric permutation.  There are 8 C-conjugates (a good
start!), and the 8 C-conjugates twist each of the 8 corners
by 1/3 in the same direction.  Hence, this is a C-symmetric
but not M-symmetric permutation.  Of course, it is an "illegal"
position in the sense that it is not in the same orbit
as Start.

In the same manner, we can build a K-symmetric permutation for
any K.  We start with a K-symmetric function on a single
cubicle. (A function which is K-symmetric is L-symmetric for
any L which is a subgroup of K).  We include all
K-conjugates.  If the cube is completely specified, we stop.
Otherwise, we choose another K-symmetric function for any
previously unspecified cubicle, add in the new K-conjugates,
and so forth, repeating until the entire permutation is
specified.

Needless to say, this construction process suffers from not
preserving orbit.  Additional steps must be taken to assure that
the constructed position is in the desired orbit (usually, the
Start orbit).  And some orbits do not have representatives from
some subgroups, for example it is well known that there are
no C-symmetric but not M-symmetric permutations in the Start
orbit.

To use Dan's adjectives, this process very quickly can become
long, mechanical, and boring.  But I now see how to build a
K-symmetric permutation for any K.

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
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

From BRYAN@wvnvm.wvnet.edu  Wed Jun 28 12:52:18 1995
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Date:      Wed, 28 Jun 1995 10:09:51 -0400 (EDT)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Re: Re: A Third Way to Calculate the Real Size of Cube Space?
In-Reply-To: Message of 06/20/95 at 14:39:00 from ,
           Martin.Schoenert@Math.RWTH-Aachen.DE

I guess I am going to have to break down and get a copy of GAP.  It
is truly impressive how much GAP can do so easily.

My interpretation of Martin's GAP program is that it implements the
general outline of the algorithm I described, except that GAP was
able to calculate the number of K-symmetric permutations in a very
simple and direct way, whereas I was going to have to puzzle each one
out by hand.

The heart of Martin's program appears to be the following, and I have
a couple of questions.


>        # compute how many elements have at least this symmetry group
>        number := Size( Centralizer( G, rep ) );

The first question is:  how does the Size function work?  As a simpler
example than the one above, what if you simply say Size(G)?  I am
naively assuming that G is specified to GAP in terms of generators
only, and that it makes no attempt to actually represent each element
of G (too big!).  And I have seen snippets of GAP libraries for G
posted by Mark Longridge, and they look like generators.  I have been
in several group theory books lately, and I don't recall seeing
a general algorithm presented for determining the size of a finite
group based on its generators.

The second question is like unto the first:  how does the
Centralizer function work?  In this particular case, we don't
really need the Centralizer, we only need the Size of the Centralizer,
but the question remains in either case.  Surely, GAP does not
literally try each element of G and each element of rep to see
which elements commute (too big again).  So what is the general
algorithm?

 = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
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

From rodrigo@lsi.usp.br  Wed Jun 28 17:15:45 1995
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Date: Wed, 28 Jun 1995 14:12:53 +48000
From: Rodrigo de Almeida Siqueira <rodrigo@lsi.usp.br>
To: steinark@ifi.uio.no, bagleyd@source.asset.com,
        Reinaldo Augusto da Costa Bianchi <rbianchi@mozart.lsi.usp.br>
Cc: cube-lovers@life.ai.mit.edu
Subject: Rubik Cube... for Windows ?
In-Reply-To: <199506281753.OAA19377@jaguar.lsi.usp.br>
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On Wed, 28 Jun 1995, steinark@ifi.uio.no wrote:

> id: Email: steinark@ifi.uio.no
> comments:
> Just thought I would mention a Windows version Rubic Cube
> program. It's located at several ftp sites and the file 
> to look for is called 'cubic.zip'. Has a far better
> look-and-feel than XRubic I think. It has its own solving
> algorithm, but this can be rather slow...
> Check it out. If you don't find it I can probably send it
> to you some how.
> 
> Steinar

Hello Steinar,

Would you tell me where (ftp site) can I find cubic.zip ?
I would like to try it and put a link in the "Robot can play with the 
Cube Web homepage" (http://www.lsi.usp.br/~daia/celula/cubo/)

Thank you,

Rodrigo Siqueira
rodrigo@lsi.usp.br
(http://www.lsi.usp.br/usp/rod/rod.html)

From @mail.uunet.ca:mark.longridge@canrem.com  Mon Jul  3 14:15:37 1995
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Subject: Antislice Patterns
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.1181.5834.0C1E9C05@canrem.com>
Date: Mon, 3 Jul 1995 14:53:00 -0400
Organization: CRS Online  (Toronto, Ontario)

Patterns in the Anti-Slice Group
--------------------------------

p4   8 flip (Op sides)  (R1 L1 U1 D1 F1 B1) ^2                    (12)
p10a pons asinorum      (L3 R1 U3 D1)^3                           (12)
p16a 4 cross order 2     F1 B1 U1 D1 L2 R2 U1 D1 F1 B1 U2 D2      (12)
p17  4 diagonal         (F1 B1 R1 L1) ^3                          (12)
p18a 4 diagonal,2 cross (F1 B1 R3 L3) ^3                          (12)
p22  2 DOT, 2 Stripe     R1 L1 U2 D2 R3 L3                         (6)
p64a 4 Z                 F1 B1 L3 R3 F1 B1 L1 R1 F3 B3 L1 R1      (12)
p143 Pinwheels           F1 B1 L1 R1 F3 B3 U3 D3 L1 R1 U1 D1      (12)
p175a 6 H order 2        U3 D3 L3 R3 F2 B2 U2 D2 L3 R3 U1 D1      (12)
p198a 2 X, 4 Diag no C   L1 R1 F1 B1 L3 R3 F3 B3 L1 R1 F1 B1      (12)
p201 Pinwheels + Pons    L1 R1 F3 B3 L1 R1 U3 D3 F1 B1 U3 D3      (12)

p201 is a quite interesting position.
The square's group equivalent is no shorter in q turns:

p175 6 H order 2 type 2  U2 B2 L2 U2 D2 L2 F2 U2                   (8)

Note that p201 = |{m'Xm}|=2 and |Symm(X)|=24.

From @mail.uunet.ca:mark.longridge@canrem.com  Mon Jul  3 14:15:29 1995
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Subject: Crazy Corner Pattern Revisited
From: mark.longridge@canrem.com (Mark Longridge)
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Date: Mon, 3 Jul 1995 14:59:00 -0400
Organization: CRS Online  (Toronto, Ontario)

Jerry writes:

>                In their position, the UFL and RBD corners are
>in place, and the other three pairs are swapped.  The "girdle"
>includes the three pairs that are swapped.  Hence, there
>is an axis of symmetry along the UFL-RBD axis.  The odd number of
>swaps is compensated by an odd number in the edges.  The compensation
>is not required for corners only.
>
>        R D D
>        U U D
>        U U B
>
> D L L  F F D  L L F  L F F
> R L L  F F B  L R R  B B F
> R R B  R B B  U R R  B B U
>
>        U U L
>        U D D
>        F D D

Much is explained! (And this is all the way back in file "cube01" in
the archives, Date: 14 December 1980 1916-EST).

What does the 1916- mean in front of EST???

If we put the edges in place in Jerry's (and Dan's) position
then we have a position M-conjugate to the one I posted.
(re-posted below).

        F U U
        U U U
        D U L
R L B   R F B   D R R   B B D
L L L   F F F   R R R   B B B
U L L   F F U   L R F   L B F
        D D B
        D D D
        R D U

Jerry Continues:

>...there is an axis of symmetry along the UFL-RBD axis.  The odd
>number of swaps is compensated by an odd number in the edges.
>The compensation is not required for corners only.

Great! 