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Subject: More UR Stuff!
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.799.5834.0C1AF26D@canrem.com>
Date: Mon, 12 Sep 1994 00:28:00 -0400
Organization: CRS Online  (Toronto, Ontario)

                      Notes on the UR Group
                      ---------------------

 Well, some small news about the < U, R > group. Previously I believed
that my 3-cycle of edges:

    UR1 = U3 R3 U2 R1 U1 R3 U1 R1 U1 R1 U2 R3 U3 R1 U3 R3 = 18 q, 16 h

...discovered by hand was minimal. My newly created < U, R > solver
(now being at the point of churning out correct results) as
happily proven me wrong!

    UR2 = U3 R1 U2 R1 U1 R1 U1 R2 U3 R3 U3 R2 U1 = 16 q, 13 h


Also I found 6 "UR-Reflective" processes altogether. This is all
 there are up to and including 12 q turns:

  U3 R1 U1 R1 (U2) R3 U3 R3 U1 = R3 U1 R1 U1 (R2) U3 R3 U3 R1   (10)
  U1 R3 U3 R3 (U2) R1 U1 R1 U3 = R1 U3 R3 U3 (R2) U1 R1 U1 R3   (10)
                 ( U2 R2 ) ^ 3 = ( R2 U2 ) ^ 3                  (12)
    U1 R1 U2 R3 U2 R3 U2 R1 U1 = R1 U1 R2 U3 R2 U3 R2 U1 R1     (12)
            ( U1 R1 U3 R3 ) ^3 = ( R1 U1 R3 U3 ) ^3             (12)
            ( U3 R3 U1 R1 ) ^3 = ( R3 U3 R1 U1 ) ^3             (12)


 The program is still a sluggish beast, but I think with further
refinements it should eventually find other interesting results
like antipodes and pure twists, etc.

             -> Mark <-

 Email: mark.longridge@canrem.com

From BRYAN@wvnvm.wvnet.edu  Mon Sep 12 17:56:28 1994
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Date:      Mon, 12 Sep 1994 15:35:32 EDT
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   God's Algorithm, Q-Moves Through Level 10

Distance   Number   Branching       Number   Branching   Ratio of
from         of      Factor           of      Factor     Cubes to
Start      Cubes                 M-Conjugate             M-Conjugate
                                    Classes               Classes

 0              1                         1
 1             12     12.000              1   1.000        12.000
 2            114      9.500              5   5.000        22.800
 3          1,068      9.368             25   5.000        42.720
 4         10,011      9.374            219   8.760        45.712
 5         93,840      9.374          1,978   9.032        47.442
 6        878,880      9.366         18,395   9.300        47.778
 7      8,221,632      9.355        171,529   9.325        47.931
 8     76,843,595      9.347      1,601,725   9.338        47.976
 9    717,789,576      9.341     14,956,266   9.338        47.993
10  6,701,836,858      9.337    139,629,194   9.336        47.997

Some of you may remember previous results where I calculated equivalence
classes of the form {m'Xmc} for all 48 elements m in M, the set of all
cube rotations and reflections, and for all 24 elements c in C, the set
of all cube rotations.  This is effectively calculating M-conjugate classes
for centerless cubes.  My previous data bases have contained
representative elements Y for each equivalent class {m'Xmc}.  To get
cubes with centers (where rotational orientation makes a difference),
I then calculated Yc for each c in C, forming a matrix indexed by
Y and c.

The previous approach permits a very compact representation of God's
algorithm, and I used it for corners-only cubes and am presently
using it for edges-only cubes.  However, I find that the {m'Xmc}
approach does not work well for whole cubes.  The problem is that
the matrix is extremely sparse close to Start.  With corners-only
or edges-only cube, I can calculate the entire problem.  With the
whole cube, I cannot even come close to calculating the whole problem,
and the matrix representation wastes space rather than saving space.

Hence, for whole cubes, I am calculating equivalence classes (which
are M-conjugate classes) of the form {m'Xm} for all 48 elements m in M.
My data base includes a representative element Z for each M-conjugate
class {m'Xm}.  This reduces the size of the problem by about 48 times,
and lets me calculate about two more levels of the search tree with the
same level of effort as before.

Just to reiterate some obvious points that have appeared before:

   1) X is an arbitrary element of {m'Xm}, but Z is a particular element
      of {m'Xm} chosen with a selection function.

   2) Z is in {m'Xm} and we have {m'Zm} = {m'Xm}.

   3) |Z| = |X| = |m'Xm| = |m'Zm| for all m in M and for all X in
      {m'Xm}.  This trivial equivalence is what makes M-conjugate
      classes a viable approach for brute force calculation of
      God's algorithm.

   4) Most M-conjugate classes of the form {m'Xm} contain 48 elements.
      The size of {m'Xm} can be used as a measure of the symmetry of
      X, with |{m'Xm}|=1 for the most symmetric cubes and |{m'Xm}|=48
      for the least symmetric cubes.  Conversely, Symm(X) is the
      set of all m in M such that m'Xm=X.  |Symm(X)|=48 for the
      most symmetric cubes, |Symm(X)|=1 for the least symmetric cubes,
      and |{m'Xm}| * |Symm(X)| = 48 in all cases.

   5) The ratio of cubes to M-conjugate classes is close to, but not
      exactly equal to, 48.  The reason the equality is inexact is
      symmetry (see item #4 above). The ratio is closer to 48 when
      you get further from Start because the proportion of asymmetric
      cubes is higher when you are further from Start.

I actually calculated (and previously reported) God's Algorithm directly
through level 8.  For levels 9 and 10, I only calculated the number of
M-equivalence classes directly.  I then calculated the size of each
M-equivalence class to obtain the number of cubes.  This particular data
base has 14 bytes for each cube (actually for each representative element
Z).  Hence, 14*139,629,194= 1,954,808,716 bytes are required to store
level 10 (each level is in a separate file).  This is about
2 gigabytes of storage, which is quite large, but which is by no means
outrageous.

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

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

From BRYAN@wvnvm.wvnet.edu  Wed Sep 21 11:32:38 1994
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Date:      Wed, 21 Sep 1994 11:31:42 EDT
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   M-Conjugate Classes as a Group

C is the set of twenty-four rotations of the cube.  After much
bungling (see my notes of 13 Feb 1994, 23 May 1994, and 19 July 1994),
I showed that the left cosets of C, denoted by xC or {Xc}, form
a group, and that the group is isomorphic to a subgroup of G.  I consider
this to be important because I use left cosets of C to model
centerless cubes.

M is the set of forty-eight rotations and reflections of the cube.
I often model the cube with M-conjugate classes of the form {m'Xm}.
Therefore, it seems that I should try to define an operation such that
the M-conjugate classes form a group, and such that the group is
isomorphic to a subgroup of G.

I would like to start by reviewing briefly the results for left cosets.
Two operations were defined:

   1.a.  {Xc} * {Yc} = {(VW)c}
   1.b.     V ** W   = (VW)

where V and W are representative elements of {Xc} and {Yc}, respectively.
Further, the mapping V <--> {Vc} defines an isomorphism between the
set of left cosets and the operation * on the one hand, and the set of
representative elements and the operation ** on the other hand.  Since
the ** operation is simply normal cube multiplication and since the
set of representative elements are a group under **, the set of
representative elements form a subgroup of G.

I tried to define groups without using representative elements and failed.
Not only that, the representative elements had to be selected in a
special way rather than arbitrarily.  For example, we could choose as
the representative element of {Xc} the unique element V such that the
ur cubie is positioned properly.

Positioning the ur cubie properly is not the only selection function for
a representative element which will work, but any selection function must
satisfy two criteria in order to work:

    A. It must select a representative element based on a property which
       is possessed by exactly one element of each coset.

    B. There must be closure in the sense that if V is the representative
       element of {Xc} and W is the representative element of {Yc},
       then (VW) must be the representative element of {(VW)c}.

Criterion #B merits some additional discussion.  First, it is the
criterion that really proves you have a group.  Associativity
for a subset of a group generally follows from the the associativity
of the group.  For a finite group, closure for a subset implies
the identity and the complement for the subset, so closure is the key
factor in demonstrating that a set of cubes is a group.  Second,
criterion #B will bear directly on our attempt to define a group
operation for the M-conjugate classes.

Suppose we choose not to require criterion #B.  We still need to
have closure in order to have a group.  We could obtain closure by
brute force as follows:

   2.a.  {Xc} * {Yc} = {(Repr{(VW)c})c}
   2.b.     V ** W   = Repr{(VW)c}

It is probably a little easier to see what is going on in equation
2.b. than in 2.a., but it is the identical mechanism in both cases.
Suppose we don't have closure.  That is, suppose the selection
function operates in such a way that if V is the representative
element of {Xc} and W is the representative element of {Yc} that
(VW) is not necessarily the representative element of {(VW)c}.
We can still find the representative element of {(VW)c} by simply
applying the selection function, which we have done.

Equations 2.a and 2.b define groups, where the left cosets are
a group under * and the representative elements are a group under
**.  Furthermore, the mapping V <--> {Vc} defines an isomorphism
between the two groups.  But even though the set of representative
elements is a subset of G, and even though they form a group under **,
they are not a subgroup of G.  The problem is that the operation
** as defined by equation 2.b. is not the same operation as
standard cube multiplication as it was in equation 1.b.

Now, let's look at M-conjugate classes.  By analogy with the left
coset case, there are two possibilities to define a group:

   3.a.  {m'Xm} * {m'Ym} = {m'(VW)m}
   3.b.       V ** W     = (VW)

   4.a.  {m'Xm} * {m'Ym} = {m'(Repr{m'(VW)m})m}
   4.b.       V ** W     = Repr{m'(VW)m}

As before, X and Y are any cubes in G, and V and W are the
representative elements of {m'Xm} and {m'Ym}, respectively.

In order to make 3.a. and 3.b. work, we need some characteristic
which can be used by the selection function which possesses the
properties of uniqueness and closure as defined by #A and #B
above.  But I can't think of any such property, and I don't think
such a property exists (see below).

4.a and 4.b certainly work.  That is, they define operations
* and ** under which the set of M-conjugate classes and the set
of representative elements, respectively, form groups, and the
groups are isomorphic under the mapping V <--> {m'Vm}.
However, the groups fail to be subgroups of G for the same reason
elements of left cosets fail to be subgroups of G under equation 2.b.
Namely, the ** operation is not really the same operation as
normal cube multiplication.

As to the question of whether 3.a. and 3.b. can be made to work,
I think we can prove that they cannot.  Suppose the contrary.
That is, suppose that there is some property such that it is
possessed by exactly one element of each M conjugancy class and
such that the normal cube product of two such elements also
possesses the property.  Then, it would be the case that the
set of representative elements would be a subgroup of G.  But the
number of representative elements is the same as the number of
M conjugate classes, and the number of M conjugate classes is
known not to divide the number of cubes in G evenly.  Hence,
the set of representative elements of M-conjugate classes is not
a subgroup of G.  Working backwards contrapositively, the desired
property cannot exist.

So, the final result is that the set of M conjugate classes can be
made into a group, and the set of representative elements of the
M conjugate classes can be made into a group.  But neither group
is a subgroup of G, nor is either group isomorphic to any
subgroup of G.

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

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

From BRYAN@wvnvm.wvnet.edu  Wed Sep 21 18:35:37 1994
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From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <cube-lovers@ai.mit.edu>
Subject:   Re: < U, R> Group
In-Reply-To: Message of 08/09/94 at 01:48:00 from mark.longridge@canrem.com

I wanted to get back to some of the symmetry considerations associated
with the Two-Generator Group.  In particular, I wanted to calculate
God's Algorithm in terms of something analogous to M-conjugates.
I haven't started the calculations, but I wanted to go ahead and
discuss what those calculations will consist of.

First, we can make the obvious observation that the <U,R> group is not
the only Two-Generator group.  Any two adjacent faces can serve
as generators for a Two-Generator Group, and there are twelve such
pairs of adjacent faces.  All twelve groups have identical structures,
and are isomorphic under M-conjugation.

With respect to any particular Two-Generator Group such as <U,R>, the
associated symmetry group is not M, it is a subgroup of M.  Dan Hoey
has determined that M has 98 subgroups, and that the 98 subgroups
may be grouped into 33 classes.  Dan has developed a taxonomy for the
33 classes and 98 subgroups.  I have seen bits and pieces of Dan's
taxonomy posted to the list, and he has sent me a good bit of it
via private E-mail, but I don't think I have ever seen the whole thing
all in one piece.  In any case, let's talk a little bit about the
98 subgroups and 33 classes.

Frey and Singmaster use script characters to describe whole cube
rotations.  For the purposes of this note, I will use lower
case letters.  For example, r will be used to describe grabbing
the right face and turning the whole cube 90 degrees clockwise
(to be distinguished from R, which means to turn only the right
face 90 degrees clockwise).  In this notation, <r> = {i,r,rr,rrr}
is one of the 98 subgroups of M.  (Note that <l> is the same
group as <r>.)  Similarly, <u> = {i,u,uu,uuu} and
<f> = {i,f,ff,fff} are subgroups of M.  The groups <r>, <u>, and
<f> have identical structures (isomorphic under rotation), and the
collection (<r>, <u>, <f>) is one of Dan's 33 classes.

Similarly, (<r2>, <u2>, <f2>) is another of Dan's 33 classes.  <r2>
is a subgroup of <r>, <u2> is a subgroup of <u>, and <f2> is a
subgroup of <f>.  Dan's taxonomy includes a complete description
of group-subgroup relationships within the subgroups of M, or maybe
I should say that it is a complete description of class-subclass
relationships.

The Frey-Singmaster script notation is adequate for rotations, but it
is not adequate for reflections.  Instead, I will use a notation which
I believe originated with Dan Hoey (e.g., 28 Dec 1993).  For example,
you could write r=(FUBD), where the upper case letters describe movements
of whole faces (*not* quarter-turns in this context!).  (FUBD) means Front
goes to Up goes to Back goes to Down goes to Front, which is what happens
when you perform r.  In the same notation, a Front-Back reflection
would be (FB), etc.

With that all said, the symmetry group for <U,R> is

  <(UR)(DL),(FB)> = {I, (FB), (UR)(DL), (FB)(UR)(DL)}

In Dan's taxonomy, this group is a member of the W class, and there
are six such groups  -- W1 through W6.  I am not sure which one this one
is (I only have a list of the classes, not of the groups), but let's
say for the sake of the argument it is W3.  Then for <U,R>, I will
be calculating W3-conjugate classes of the form {w'Xw} for all w in
W3.  The size of the problem will be reduced by about four times,
compared to a reduction of about forty-eight times for whole cube
problems where M-conjugation can be used.

I was initially surprised that there are twelve groups M-conjugate with
<U,R>, but only six corresponding symmetry groups in M.  This arises,
for example, because the <U,R> and <D,L> (diagonally opposed) groups share
the same symmetry group.  I really shouldn't have been surprised.
After all, we know how many subgroups of G there are, namely
"over three beelion" (Dan Hoey, 20 Aug 1992), and we know how
many subgroups of M there are, namely 98.  Since "over three beelion"
is a lot more than 98, there must be many, many subgroups of G
which share the same symmetry properties in the sense of sharing
a subgroup of M.

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

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

From BRYAN@wvnvm.wvnet.edu  Tue Sep 27 01:22:13 1994
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From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <cube-lovers@ai.mit.edu>
Subject:   Re: < U, R> Group (W-conjugate results)
In-Reply-To: Message of 08/09/94 at 01:48:00 from mark.longridge@canrem.com

I have completed the God's Algorithm calculations for the <U,R> group
in terms of W-conjugate classes (or really, in terms of representative
elements of W-conjugate classes), with the results below.  In general,
the use of W-conjugates reduces the size of the problem by about
four times.  However, I was surprised to see that for levels 1, 3, 5,
7, 9, and 11 the number of cubes was exactly four times larger than the
number of W-conjugate classes.  My interpretation is that all cubes
in <U,R> at these levels are completely "asymmetric" with respect to
W.  (They are somewhat symmetric with respect to M, of course.)  However,
when level 13 turned out not to have a ratio of exactly 4 between cubes
and W-conjugate classes, I was rescued from the task of explaining why
all cubes at an odd distance from Start were asymmetric.


Level   W-Conjugate  Branching      Total   Branching   Ratio
           Classes     Factor       Cubes     Factor      of
                                                        Cubes to
                                                        Classes

  0             1                        1              1
  1             1      1                 4     4        4
  2             3      3                10     2.5      3.333
  3             6      2                24     2.4      4
  4            15      2.5              58     2.416    3.866
  5            35      2.333           140     2.413    4
  6            85      2.429           338     2.414    3.976
  7           204      2.4             816     2.414    4
  8           493      2.417         1,970     2.414    3.996
  9         1,189      2.412         4,756     2.414    4
 10         2,863      2.408        11,448     2.407    3.999
 11         6,862      2.397        27,448     2.398    4
 12        16,324      2.379        65,260     2.378    3.998
 13        38,550      2.362       154,192     2.363    3.9997
 14        90,192      2.340       360,692     2.339    3.9992
 15       206,898      2.294       827,540     2.294    3.9997
 16       462,893      2.237     1,851,345     2.237    3.9996
 17       992,268      2.144     3,968,840     2.144    3.9998
 18     1,973,209      1.989     7,891,990     1.988    3.9996
 19     3,415,314      1.731    13,659,821     1.755    3.9996
 20     4,618,491      1.352    18,471,682     1.352    3.9995
 21     4,147,448      0.898    16,586,822     0.898    3.9993
 22     2,010,449      0.485     8,039,455     0.485    3.9988
 23       378,110      0.118     1,511,110     0.188    3.9965
 24        11,894      0.031        47,351     0.031    3.9811
 25            27      0.002            87     0.002    3.222

Total  18,373,824               73,483,200              3.9993

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

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

From bagleyd@source.asset.com  Thu Sep 29 14:03:19 1994
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Date: Thu, 29 Sep 1994 13:34:37 -0400
From: bagleyd@source.asset.com (David A. Bagley)
Message-Id: <9409291734.AA25973@source.asset.com>
To: Cube-Lovers@ai.mit.edu
Subject: X-Puzzles

HI

  I've been busy updating X puzzles on ftp.x.org in /contrib/games/puzzles.     
Here's the blurb from the xpuzzle.README file there. 
-------------------------------
What's new?:                   
xrubik now has a undo, save, and recall
  as well as self-solver (computer solves cube) up to 3x3x3.
  Currently it is the only one in this collection with a self-solver,  
  undo, save, and recall.
xmball recently added, atan2 problem on Suns fixed
xmlink recently added, initialize error fixed                      
xhexagons minor update  
                        
The collection includes:          
SLIDING BLOCK PUZZLES                       
xcubes:         expanded 15 puzzle                                       
xtriangles:     same complexity as 15 puzzle                             
xhexagons:      2 modes: one ridiculously easy, one harder than 15 puzzle
                                          
ROTATIONAL 3D PUZZLES                                      
  hold down control key to move whole cube                 
  letters that represent colors can be changed in mono-mode  

xrubik:         a nxnxn Erno Rubik's Cube(tm) (or Magic Cube)
                self-solves 2x2x2 and 3x3x3 (non-orient mode).
xpyraminx:      a nxnxn Uwe Meffert's Pyraminx(tm) (and Senior Pyraminx),
                a tetrahedron with Period 2, Period 3, and Combined cut modes
                and it also a sticky mode to simulate a Halpern's Tetrahedron
                or a Pyraminx Tetrahedron
xoct:           a nxnxn Uwe Meffert's Magic Octahedron (or Star Puzzler) and
                Trajber's Octahedron with Period 3, Period 4, and Combined
                cut modes and it also includes a sticky mode
xskewb:         a Meffert's Skewb (or Pyraminx Cube), a cube with diagonal cuts
xmball:         a variable cut Masterball(tm), variable number of latitudinal
                and longitudinal cuts on a sphere, where the longitudinal cuts
                permit only 180 degree turns.

COMBINATION ROTATIONAL AND SLIDING 3D PUZZLES
xmlink:         a nxm Erno Rubik's Missing Link(tm)

Future directions:
  Sorry about the lack of self-solvers, but I would rather write the puzzle
  than the tedious solution.
  The ability to take back moves, record moves, and start with a
  entered position to other puzzles besides xrubik should be done.
  Currently the saved file for xrubik is cryptic (not intentionally).
  Also xmlink and xmball need better algorithms for drawing sectors than just
  a series of arcs.
  A Billion Barrel would be nice but only with a self-solver (the puzzle is
  too hard (I confess, I never solved it)).

Newbies (especially DOS users 8-) ):
  DOS/Windows & Mac users, sorry no port currently available.
  What you need:
    80386 or better, or Risc, etc.
    UNIX: Linux and FreeBSD are freely available (it may work on VMS).
    X: XFree86 is freely available on Linux and FreeBSD distributions.
    gunzip: freely available from GNU and the above distributions.
    tar: freely available from GNU also.
  What you do:
    After transfering the PUZZLE file to your machine
        (DOS users may want to rename the file PUZZLE.tar.gz to PUZZLE.tgz)
      gunzip PUZZLE.tar.gz  (or gunzip PUZZLE.tgz)
      tar xvf PUZZLE.tar
        (tar xvzf PUZZLE.tar.gz or tar xvzf PUZZLE.tgz may work as a short cut)
    Then read the README generated by the above command.

----------
I hope you enjoy
David

From ishius@ishius.com  Fri Sep 30 18:16:33 1994
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Subject: COMMERCIAL: Penrose, Cutler, Masterball

COMMERCIAL POSTING

September 30, 1994

Dear Puzzle Enthusiast:

        Sometimes the pieces of a puzzle just seem to fall into place.
For the Fall season, we have several new puzzles whose pieces may easily
fall into place for you. Or perhaps they'll prove more challenging than
watching autumn leaves.

         Chief among our Fall puzzles is our new plastic puzzle, Sneaky
Squares.  It's hard to talk about this puzzle without saying that it is
truly a great puzzle. A great puzzle should be simple in concept and design,
but require a right angle turn of the mind in order to arrive at the
solution. Sneaky Squares was invented by veteran puzzle designer Bill
Cutler. He calls it his finest achievement. It consists of just 4 blocks
that must be fitted into a box. What could be simpler you say? Yet it
stumps 99% of the people we show it to. But, once you have figured out
the solution, you can demonstrate it to your friends in seconds. Because
of its simple elegance, this is an excellent gift for people who might
be intimidated by a complex or esoteric puzzle.

        At just $15, Sneaky Squares is a must for your collection, and a
great gift for your friends!

        Birds of an entirely different feather comprise the Perplexing Poultry
series of puzzles. These intriguing puzzles, from England, are based on
the tiling theories of mathematician and cosmologist Roger Penrose.
Penrose became interested in the shapes of tiles that will cover a plane.
Some regular shapes (such as squares) do this, but Penrose came up with
a number of irregular tile sets that could cover a plane. These tiles
produce patterns that are non-periodic (that is, the patterns do not
repeat). They are called 'quasi-periodic' since the pattern appears
to repeat regularly until you examine it closely. (Scientists have
since found some real-world crystals that form in a quasi-periodic way.)

        The quasi-periodic tile sets make interesting puzzles. To decide the
position of the next tile you place, you must take into account more
than just the neighbor tiles -- you have to think about the 'whole-board
position.'

        Choose either a color or black and white version of the Perplexing
Poultry.
The Black & White is reminiscent of the work of M.C. Escher. (Many of
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Additionally, four of the tile sets have been made into jigsaw puzzles.
The jigsaws are unusual and colorful. 500 die-cut pieces build to a 19"
diameter circle in each of them. Although these are conventional jigsaws,
they are quite difficult because of the quasi-periodicity of the pattern.
See the enclosed flyer for illustrations of each jigsaw and for more
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         As a special offer, to get your fingers moving as fast as your brain,
we'll throw in free shipping on orders of $50 or more, if you place them
by October 15! So hurry and dig into these new brainteasers.

        Happy puzzling,

        James W. Connelley
        President
        Ishi Press International

P.S. We picked up a limited number of Circusmaster and Dragonmaster
Masterballs at a special price. Masterball is a rotational puzzle
with the sphere divided along lines of latitude and longitude.
Masterballs usually sell for $19.95. While they last, we are offering
these at the special price of just $11.70 each. Because they came
from a European shop, the packaging is not in perfect condition, but
the puzzles are just fine. Take advantage of this special offer before
we run out!

Dear Ishi - Please send me the following puzzles to brighten my Fall days!


        Puzzle                           Price                 s/h

o Sneaky Squares                         $15.00                 1
o Puzzling Poultry, B&W (PX01)           $69.00                 6
o Puzzling Poultry, Color (PX02)         $99.00                 6
o Perplexing Poultry Jigsaw (PX05)       $15.00                 2
o Cat Amongst the Pigeons (PX06)         $15.00                 2
o Perplexing Pisces (PX07)               $15.00                 2
o Pentaplex (PX08)                       $15.00                 2
o Circusmaster Masterball                $11.70                 2
o Dragonmaster Masterball                $11.70                 2

                                Total   $_______              ____

FREE shipping on orders of $50
or more received by Oct. 15, 1994!

Please send these puzzles to:
________________________________________
________________________________________
________________________________________
________________________________________

MC/VISA___________________________ exp:__________

California residents please include 8.25% sales tax.

Toll Free order line: (800) 859-2086


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

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



From @mail.uunet.ca:mark.longridge@canrem.com  Mon Oct  3 05:48:56 1994
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Sender: CRSO.Cube@canrem.com
Subject: < U, R > Processes
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.808.5834.0C1B2945@canrem.com>
Date: Mon, 3 Oct 1994 01:13:00 -0400
Organization: CRS Online  (Toronto, Ontario)

Alas, no antipodes yet, but some interesting results nonetheless.
Process UR8 improves on the best known process for a certain
quad-twist in the U layer at 20 q turns. Table 3 in Winning Ways
gives a 22 q turn process.

The following results should be particularly interesting to
the physical cube solver as it is easier to execute a sequence
of 2 adjacent sides compared to a sequence using 3 or more
sides, which may require some re-orienting of the cube. I will
measure the "face index" of a process by the number of different
sides used in a certain cube sequence. Such a measure could be
used to evaluate the relative elegance of two equally long
processes with respect to their face indices.


Jerry Bryan mentions:
>                         Also, the global maxima are of length 25.
> Does this tell us anything about the Q-turn length of the global
> maxima for the full cube group?

Well, that reminds me of one of the hardest patterns that Dik Winter
tried to find an optimal sequence for:

p141a alternate method   F1 R1 L2 U3 R2 L3 U3 D2 R2 F1 D1 B1 D1 F2 U3
of Superfliptwist + 6 X        R3 D3 F2 D2 L2
**This process was one of the hardest ever to reduce to 20 moves,
requiring  over 19 hours on an SGI R4K Indigo, 28 q turns**

My own $.02 worth is that an antipode for the full group of the 3x3x3
cube is probably deeper than an antipode for the < U, R > group.


        Optimal Sequences for < U, R > group elements (positions)
        ---------------------------------------------------------

Edge 3-cycle
UR1 = U3 R1 U2 (R1 U1)^2 R2 U3 R3 U3 R2 U1                 (16 q, 13 h)

Double adjacent edge swap
UR2 = U3 R3 U3 R2 U1 R1 U1 R3 U3 R1 U1 (R1 U3)^2 R3 U3     (18 q, 17 h)

Diagonal Corner twist
UR3 = U1 R1 U3 R1 U3 R2 U1 R1 U1 R3 (U3 R1)^2 U2 R3 U3 R3  (20 q, 18 h)

Double opposite edge swap, also in sq group 24 q, 12 h
UR4 = R2 U2 R3 (U2 R2)^2 U2 R3 U2 R2                       (20 q, 11 h)

Edge 7-cycle, equivalent to (U1 R1)^15
UR5 = U3 R1 U3 R3 U3 R1 U2 R3 U1 R3 U2 R1 U3 R3 (U3 R1)^2  (20 q, 18 h)

Corner Tri-Twist
UR6 = (U3 R3)^2 U1 R1 U3 R3 U3 R2 U1 R2 U3 R3 U3 R1 U1 R3  (20 q, 18 h)

Corner Quad-Twist, Flat style
UR7 = R1 U3 (R1 U1)^2 (R3 U3)^2 R2 U3 R1 U1 R3 U3 R1 U3 R3 (20 q, 19 h)

Corner Quad-Twist, Arms & Legs style                       (20 q, 20 h)
UR8 = R1 U1 R3 U1 R3 U3 R1 U1 R1 (U3 R3)^2 (U1 R1)^2 U3 R3 U3

ML Doodle Position
UR9 = (U2 R2)^2 U2 R3 U1 R2 (U3 R2)^2 U1 R1                (22 q, 14 h)
Same position found by hand: (a non-optimal 24 q, 15 h)
      (U2 R2)^3 U1 R1 (U2 R3)^2 U2 R1 U1

4 Opp Corner Swap, also in sq group at 26 q, 13 h
UR10 = U3 R3 (U1 R1)^2 U2 R3 U1 R1 (U2 R2)^2 U1 R3 U1      (22 q, 17 h)

        Other Subgroups within reach
        ----------------------------

11. |<U, R2, L2>|              = 2^12 3^4 5^2 7   =             58060800
12. |<U2, R, L2>|              = 2^12 3^4 5^2 7   =             58060800
17. |<U, R2, F2>|              = 2^8 3^5 5^2 7    =             10886400
21. |<U, R2, L2, D2>|          = 2^13 3^4 5^2 7   =            116121600
22. |<U, R2, L2, D>|           = 2^15 3^4 5^2 7^2 =           3251404800

I welcome any proposed < U, R > group antipodes. I haven't really
looked for anything exotic like < U, R > positions which are
shift invariant, or even if such a beast is possible!

Of course I already mentioned that...

        U2 R2 U2 R2 U2 R2 = R2 U2 R2 U2 R2 U2

...but aside from that nothing comes to mind.

Generally when there are elements which occur in both the square's
group AND the < U, R > group the latter is the shorter in q turns.

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

From BRYAN@wvnvm.wvnet.edu  Fri Oct  7 14:48:43 1994
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Date:      Fri, 7 Oct 1994 10:52:58 EDT
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <cube-lovers@ai.mit.edu>
Subject:   Re: < U, R> Group  -- Q+H
In-Reply-To: Message of 08/09/94 at 01:48:00 from mark.longridge@canrem.com

Distance    Number     Branching       Number    Branching     Ratio of
 from         of        Factor           of       Factor       Cubes to
 Start   W-Conjugate                    Cubes                  W-Conjugate
           Classes                                              Classes

      0          1                            1                  1
      1          2          2                 6     6            3
      2          5          2.5              18     3            3.6
      3         14          2.8              54     3            3.857
      4         41          2.929           162     3            3.951
      5        122          2.976           486     3            3.984
      6        365          2.992         1,457     2.998        3.992
      7      1,091          2.989         4,360     2.992        3.996
      8      3,256          2.984        13,016     2.985        3.998
      9      9,627          2.957        38,482     2.957        3.997
     10     28,282          2.938       113,094     2.939        3.9987
     11     82,243          2.908       328,920     2.908        3.9994
     12    235,611          2.865       942,351     2.865        3.9996
     13    654,297          2.777     2,616,973     2.777        3.9997
     14  1,693,858          2.589     6,774,848     2.589        3.9997
     15  3,776,718          2.230    15,105,592     2.230        3.9997
     16  6,058,483          1.604    24,231,019     1.604        3.9995
     17  4,856,334          0.802    19,421,274     0.802        3.9992
     18    961,504          0.198     3,843,568     0.198        3.997
     19     11,954          0.012        47,465     0.012        3.971
     20         16          0.001            54     0.002        3.375

  Total 18,373,824                   73,483,200                  3.9993

Notice that using Q+H turns instead of Q turns reduces the maximum
distance from Start from 25 down to 20.

When I first calculated God's Algorithm for <U,R> for Q turns, I calculated
it for cubes first, then for W-conjugate classes.  In this case, I
really did it only for W-conjugate classes (problem is four times
smaller).  The "Number of Cubes" column is then derived by calculating
the size of each W-conjugate class; no real search is needed to
obtain the number of cubes if the W-conjugate classes are already
in hand.

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

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

From dik@cwi.nl  Wed Oct 12 21:55:07 1994
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From: Dik.Winter@cwi.nl
Message-Id: <9410130155.AA21510=dik@boring.cwi.nl>
To: cube-lovers@life.ai.mit.edu
Subject: CFF 34, summary of contents

CFF #34 came out, it ought to have been in June but is a bit late.
Still, the editors expect the next issue in December.

Summary of contents.

Leo Links: On Folding Puzzles.
  A discussion about folding puzzles made from paper or cardboard
  that display a particular figure or text when folded.

Frits Goebel and Bernhard Wiezorke: Problems for Einstein.
  They discuss a puzzle consisting of 8 octonimo's which is marketed
  as IQ CREATOR, MAGIC BLOCK or EINSTEIN PUZZLE.  They show a few
  pretty patterns that can be formed with the 8 pieces.

Vic Stok: Skyline Tetracubes.
  This discusses figures that can be formed from the 8 different
  pieces consisting of 4 cubes glued together.

Jacques Haubrich and Nanco Bordewijk: Cube Chains.
  This discusses a number of puzzles.  Each consists of 27 cubes
  connected to each other by an elastic string.  The objective is
  to form a 3x3x3 cube.

Bernhard Wiezorke: On Nob's L-Puzzle.
  This duscusses Nob's puzzle.  It consists of 10 L shaped pieces,
  3 squares high, 2 squares wide; all in the same orientation, one
  such piece 4 squares high (also the same orientation) and one
  3 square high piece in different orientation (i.e. turned over).
  The objective is to fill a 7x7 square, turnover of the pieces is
  not permitted.

Jacques Haubrich: Pantactic Patterns and Puzzles.
  This discusses an extension of the memory wheel.  On the wheel
  the digits 0 and 1 are written such that when you look at 3
  consecutive digits, all 8 different can be created.  This can
  be generalized to n consecutive digits.  It is well known (since
  N.G. de Bruijn) that 2^n digits are needed.  An 2-dimensional
  extension was made by B. Astle who had a 5x5 square with a
  black-white pattern such that when you look at the 16 different
  2x2 subsquares you will find all 16 different configurations.
  C.J. Bouwkamp made this into a puzzle (in the early 70's) as
  follows: You have 16 2x2 squares with all possible patterns.
  The puzzle is to put them together in a larger square such that
  the borders match.  Rotation is *not* permitted.

Torsten Sillke: Three 3x3 Matching-Puzzles.
  A discussion about three puzzles consisting of 9 squares that
  have to be put in a 3x3 square where some form of marking has
  to match.

Jacques Haubrich: Cube 216.
  The puzzle Gemini consists of 10 pieces where each piece is
  made by joining two 1x2x2 blocks together.  This is done in
  all possible ways.  It is known that there are 50 possible
  ways to pack them in a 4x4x5 block.  Yoshikatsu Hara extended
  this with 22 pieces that form all possible ways to join two
  1x3x3 blocks together.  One result is that there are (at
  least) 11 selections of 12 of these pieces so that they can
  be packed in a 6x6x6 cube in an unique way.  There are more
  results and the author asks also for input.

Chris Roothart: Polylambdas.
  Polylambdas are formed from the 30/60/90 degree triangle.
  Lambdas can be joined at corresponding edges.  Joining
  along the hypothenusa is not allowed.  There are 4 dilambdas,
  4 trilambdas, 11 tetralambdas en 12 pentalambdas.  These 31
  pieces can fill a parallelogram of 4 by 31 units (the short
  leg is the unit).  Many other problems are stated.

Columns:

Mark Peters: Books and Magazines (book reviews)
Edward Hordern: What's Up? (details some new puzzles)
------
CFF (Cubism For Fun) is the newsletter published by the
Nederlands Kubus Club NKC (Dutch Cubists Club).

Membership fee is NLG 25 individual, NLG 80 institutional.
(USD 1 ~ NLG 1.70).  Applications for membership to the
treasurer:
   Lucien Matthijsse
   Loenapad 12
   3402 PE  IJsselstein
   The Netherlands

If you write, please add an international reply coupon
(can be obtained at your post office).


From f94dk@efd.lth.se  Fri Oct 14 15:24:33 1994
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Date: Fri, 14 Oct 1994 16:43:57 MET
From: David Kaspar <f94dk@efd.lth.se>
To: CUBE-LOVERS@life.ai.mit.edu
Subject: You are my only hope...




Hi !!

My name is David. Long time ago I was able to solve Rubik's Cube but I ha=
ve
unfortunatly (Ooops the spelling) forget it now. Can you please help me??=


I would be very grateful, because you are my only hope.


					Many thankx, David

				email: f94dk@efd.lth.se =


From ma2gapen@lucano.uco.es  Tue Oct 18 08:13:23 1994
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Date: Tue, 18 Oct 94 13:12:54 +0100
From: ma2gapen@lucano.uco.es (Nicolas G. Pedrajas)
Message-Id: <9410181212.AA29888@lucano.uco.es>
To: cube-lovers@life.ai.mit.edu
Subject: help!


hello,
	I used to know how to resolve rubik's cube, but
i've forgotten it. Can anybody here help me?
	Thanks in advance for any help.

 Adios.


From @mail.uunet.ca:mark.longridge@canrem.com  Sun Oct 23 03:41:06 1994
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Sender: CRSO.Cube@canrem.com
Subject: Cross and X's
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.814.5834.0C1B6751@canrem.com>
Date: Sun, 23 Oct 1994 02:36:00 -0400
Organization: CRS Online  (Toronto, Ontario)

-----------------------------
Possible Legal Cross Patterns
-----------------------------

Plummer Cross   (6 Cross order 3) =  8 patterns
Christman Cross (6 Cross order 2) =  6 patterns
4 Cross order 2 (sq group)        =  3 patterns
4 Cross order 4                   =  6 patterns
                                    -----------
      14 Six Cross + 9 Four Cross = 23 total legal Cross patterns

There are 0 total cross patterns in the swap orbit

-------------------------
Possible Legal X Patterns
-------------------------

6 X order 3                       =  8 patterns
6 X order 6                       =  8 patterns
6 X order 2 (sq group)            =  1 pattern
4 X order 2 (sq group)            =  3 patterns
2 X order 2 (sq group)            =  3 patterns
                                    -----------
    17 Six X + 3 Four X + 3 Two X = 23 total legal X patterns

For a while I thought that [6 x order 3] combined with the [2 x pattern]
would make a new sort of [6 x order 6], but combining [6 x order 3] with
the [2 x pattern] is essentially the same as combining 6 x order 3 with
the pons asinorum or 6 x order 2.

------------------------------
Possible Swap-Orbit X Patterns
------------------------------

6 X order 2                       =  6 patterns
6 X order 4                       =  6 patterns
4 X order 2                       =  6 patterns
4 X order 4                       =  6 patterns
                                    -----------

             12 Six X + 12 Four X = 24 total swap-orbit X patterns

Some description of the swap-orbit patterns is in order.
The 6 X order 2 pattern has a 2-cycle of opposite edges and
2 sets of 2-cycles of adjacent edges.
The 6 X order 4 pattern has a 2-cycle of opposite edges and
a 4-cycle of edges of adjacent faces.
The 4 X order 2 has 2 sets of 2-cycles of adjacent edges.
The 4 X order 4 has a 4-cycle of edges of adjacent faces.

To make any of these swap-orbit patterns one would have to first
exchange any 2 edge cubies.

Interestingly, a thin line 6 X order 3 is possible on the 5x5x5
cube. No process as yet....

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

From mschoene@math.rwth-aachen.de  Mon Oct 24 16:59:39 1994
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Date: Mon, 24 Oct 94 21:58 PST
From: Martin.Schoenert@math.rwth-aachen.de
To: cube-lovers@life.ai.mit.edu
Subject: Shift invariant processes

Mark Longridge wrote in is e-mail message of 1994/04/02

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

    P8   2 x ORDER 2:

    shift
    0                              D2 F2 T2 F2 B2 T2 F2 T2
    1                           T2 D2 F2 T2 F2 B2 T2 F2
    ...
    7         F2 T2 F2 B2 T2 F2 T2 D2

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

I have classified all such shift invariant processes, using a little bit
of group theory and the computer algebra system GAP.

Let me first repeat the definition.
A *process* g_1 g_2 ... g_n, where the letters g_i come from the set
{U,U2,U3,D,D2,D3,...,B,B2,B3}, is called *shift invariant* if each of
the processes g_1 g_2 ... g_n, g_2 ... g_n g_1, ..., g_n g_1 ... g_{n-1}
effects the same element g in the cube group G.

In the following I will be a bit sloppy and neither distinguish between
letters and the corresponding generators of the cube group nor between
processes and the elements of the cube group they effect.
With this terminology a shift invariant processes would be one where
g_1 g_2 ... g_n = g_2 ... g_n g_1 = g_n g_1 ... g_{n-1}.

So lets assume that g = g_1 g_2 ... g_n is a shift invariant process.
Then for every letter g_i in g we have
g_i' g = g_i' (g_i g_{i+1} ... g_{i-1}) = (g_{i+1} ... g_{i-1})
       = (g_{i+1} ... g_{i-1} g_i) g_i' = g g_i'.
That means that g commutes with (the inverses) of each of its letters.

Because g commutes with each of its letters, it also commutes with all
elements of the subgroup H = < g_1, g_2, ..., g_n > generated by its
letters.  The set of those elements of H which commute with all elements
of H is called the centre of H.  Thus g lies in the centre of H.

Obviously the other direction is also true.  If g lies in the centre of
H = < g_1, g_2, ..., g_n >, i.e., if it commutes with every element of H,
then it especially commutes with its letters, and so the corresponding
process is shift invariant.

This says that if we have an element g in the centre of a subgroup
H = < g_1, g_2, ..., g_n >, then *every* process that effects g and
uses the letters g_1, g_2, ..., g_n will be a shift invariant process.

So there are finitely many such elements (after all there are only
finitely many elements in the entire cube group), but each gives
rise to infinitely many different shift invariant processes.
In particular there is *no* longest shift invariant process.

So the task is to search for subgroups H generated by subsets of
{U,U2,U3,D,D2,D3,...,B,B2,B3} that have non-trivial centres.
There are 729 = 3^6 such subgroups.  Of course we are only interested in
representatives under the operation of M (the subgroup of symmetries of
the entire cube), which leaves us with 56 subgroups.

Of those 21 have a non-trivial centre (for this computation I used GAP).
The centres are all very small and contain mostly the same elements,
i.e., the same element lies in the centre of different such subgroups.
I do not want to bore you with the details.  Allow me to jump to the
discussion of the results.

There are 5 different types of elements that give rise to shift invariant
processes.

1)  The ``trivial'' element.
    The identity element lies in the centre of every subgroup H.
    Thus every process that effects the identity is shift invariant.
    There is exactely one such element in the entire group.

2)  The ``central'' element.
    The superflip, which flips all edges, is in the centre of G.
    Thus every process that effects the superflip is shift invariant.
    There is exactely one such element in the entire group.

3)  The ``abelian'' elements.
    The subgroups < U > and < U, D > (and their conjugates under M)
    are abelian, and are therefore equal to their centre.
    Therefore every element in < U > and < U, D > is shift invariant.
    There are 45 such elements in the entire group.

4)  The ``odd'' element.
    The element (UR)^140 lies in the centre of the subgroup <U,R>.
    It is the only shift invariant element of odd order (hence the name).
    Thus this process and its inverse are shift invariant.
    There are 24 such elements in the entire group (two for each edge).

5)  The ``square'' elements.
    The following elements live in the ``square ring'' group <U2,D2,R2,L2>
    (though some of them are central in proper supergroups of it).

5a) The ``single square'' elements.
    The element (U2 R2)^3 lies in the centre of <U2,D,R2,L>.
    Thus this process is shift invariant.
    There are 12 such elements in the entire group (one for each edge).

5b) The ``edge square'' elements.
    The element (U2 R2)^3 (U2 L2)^3 = (D2 R2)^3 (D2 L2)^3
    lies in the centre of <U,D,R2,L2>.
    Thus this process is shift invariant.
    There are 6 such elements in the entire group (two for each axis).

5c) The ``diagonal square'' elements.
    The element (U2 R2)^3 (D2 L2)^3 = (U2 L2)^3 (D2 R2)^3
    lies in the centre of <U2,D2,R2,L2>.
    Thus this process is shift invariant.
    There are 3 such elements in the entire group (one for each axis).

For me the most amazing elements were the ``odd'' element and the
``diagonal square'' element.

They are special in the sense that the smallest subgroup in which they
lie and the largest subgroup in which they are central are equal.
That means that you have no choice which letters to choose to write
them (you have lots of choices how arrange those letters and how often
to repeat them of course).  You cannot use less, because they do not
lie in a smaller group, and you cannot use more, because they are not
central in a larger group.

Let me return to Mark's e-mail and discuss it in the light of the above.

Mark writes

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

    P8   2 x ORDER 2:

    shift
    0                              D2 F2 T2 F2 B2 T2 F2 T2
    ...

Believe it or not, this is a process for the ``diagonal square'' element.

Mark writes

    This is the longest process I've found so far.

How about (UR)^140 or (UR)^1400?  As mentioned above, you can make the
processes as long as you wish.

Mark writes

    Certainly this property is not true of all squares group processes.

No, only for processes that effect one of the 21 elements mentioned above
(31 if you want to count the ``trivial'' and ``abelian'' elements in the
square group).

Mark writes

    I suspect there are no processes in the full group with this property
    (of any significant length).

Not true.  The interesting ones are the processes effecting the
``central'' element and the ``odd'' elements.

Mark writes

    Perhaps the fact that the L and R faces never rotate will give some
    clue on how to generate processes with this property.

Now this remark makes me very suspicious.  Did Mark know the full story?
The squares subroup has trivial centre (containing only the identity),
you have to leave out at least two generators belonging to opposite
faces, to get a subgroup with non-trivial centre.

Mark writes in another e-mail message of 1994/04/10

    The following processes are also shift invariant:

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

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

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

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

The first is obviously a ``single square'' element, the second is a
``edge square'' element, and 'p3' is the ``central'' element.

Thus at this time all non-trivial such elements had been found, except
for the ``odd'' element.

Have a nice day.

Martin.

PS: GAP is really a nice program to analyze the cube group from the
    group-theoretical side, though I would not use it to enumerate
    positions in search for god's algorithm.

PPS: Of course I am biased, because I am one of the main authors of GAP ;-)

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

From dbisel@adrian.adrian.edu  Wed Oct 26 10:26:30 1994
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Date: Wed, 26 Oct 1994 10:25:43 -0400
Message-Id: <94102610254293@adrian.adrian.edu>
From: dbisel@adrian.adrian.edu
To: cube-lovers@life.ai.mit.edu
X-Vms-To: smtp%"cube-lovers@ai.ai.mit.edu"

Does the "mit" in the address stand for Michigan Tech University? 
I am a student at Adrian College.

What is the record in minutes (or seconds) for the time to solve a rubik's
cube?

Do you happen to have any brain teasers or hypothetical questions?

I would love to hear from you.

Diana Bisel

From MALONEY9146@a12t.cc.fredonia.edu  Wed Oct 26 11:03:59 1994
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Date:         Wed, 26 Oct 1994 11:02 am EDT (15:02:37 UT)
From: "Daniel P. Maloney" <MALONEY9146@fredonia.edu>
Organization: State University of New York - College at Fredonia
To: dbisel@adrian.adrian.edu
Cc: cube-lovers@life.ai.mit.edu
In-Reply-To:  Your message of 26 Oct 1994 10:25:43 -
Message-Id: <35483102694110235@FREDONIA>

I'm not sure what the record is, but I used to have a book
called "Jeff Conquers The Cube In 45 Seconds (And So Can You!)".
Needless to say, I never got cloise to that.

Dan

BTW MIT is probably Massachusetts Institute of Technology
  (a big, expensive school)

Dan

From nivek@frc2.frc.ri.cmu.edu  Wed Oct 26 11:19:52 1994
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Date: Wed, 26 Oct 94 11:18:26 EDT
From: Kevin Dowling <nivek@frc2.frc.ri.cmu.edu>
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In-Reply-To: <94102610254293@adrian.adrian.edu> (dbisel@adrian.adrian.edu)
Subject: MIT
Reply-To: nivek@cmu.edu (Kevin Dowling)

No sorry, it stands for Massachusetts Institute of Technology, a
little technical school located in Cambridge, MA, on the Charles River
near Boston.

CMU doesn't stand for Central Michigan University either.


				nivek

aka:    Kevin Dowling         
tel:    412.268.8830          Carnegie Mellon University
fax:    412.682.1793          The Robotics Institute
net:    <nivek@cmu.edu>       5000 Forbes Avenue
                              Pittsburgh, PA 15213




From bosch@smiteo.esd.sgi.com  Wed Oct 26 11:28:44 1994
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From: "Derek Bosch" <bosch@smiteo.esd.sgi.com>
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Date: Wed, 26 Oct 1994 08:27:59 -0700
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        "" (Oct 26, 11:02am)
References: <35483102694110235@FREDONIA>
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To: "Daniel P. Maloney" <MALONEY9146@fredonia.edu>, dbisel@adrian.adrian.edu
Cc: cube-lovers@life.ai.mit.edu
Subject: Fast cubing
Content-Type: text/plain; charset=us-ascii
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I too, have read the book, Jeff Conquers The Cube in 45 seconds, as well as Minh
Thai's book on the cube (he's the world record holder, with 22 seconds as an
official world record.  I used to compete back in the cubing days, and could
regularly get under 25 seconds, using a strategy of solving the corners,
solving the edges on two opposite sides, followed by the middle slice.

Several people on this mailing list have done serious analysis trying to reach
"God's Algorithm", which isn't terribly useful to me.  The operators that these
analyses generate are really slow to crank out on the cube.  I prefer slightly
longer ones, that are more optimized for speed (hand positions, etc).

Derek

-- 
Derek Bosch                   "Time flies like an arrow,
(415) 390-2115                 but fruit flies like bananas"
bosch@sgi.com                  J. Blaylock


From HOWSER@lua6.lu.edu  Wed Oct 26 11:59:14 1994
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From: HOWSER@lua6.lu.edu
Date: 26 OCT 94 11:03   
To: <cube-lovers@life.ai.mit.edu>
Subject: Record times for the cube
Comments: Automatic Return Receipt Requested
Message-Id: <CM3N02353B9ECC@LUA6>

Ouch!  I fat-fingered it, sorry.
-------  Forwarded message -------

Date: Wed, 26 Oct 1994 10:23 am CDT (15:23:26 UT)
From: Gerry Howser <HOWSER @ LUA6>
To: <cube-lobers @ LIFE.AI.MIT.EDU>
bcc: Gerry Howser <HOWSER @ LUA6>
Subject: Record times for the cube
Comments: Automatic Return Receipt Requested
Message-ID: <CM3M23241CBE63@LUA6>

I recall a demo on Johnny Carson where a guy solved seven cubes in seven minutes
and I think he had the current world record of around 21 seconds.  I have solved
a cube in 39 seconds but it was luck more than anything else.  When I was play-
ing around with making modifications to my solution to the cube I could solve
any cube in about a minute and a half, which was fast enough to earn me a few
drinks in bars.  I think that a legitimate record would be around 30-45 seconds
and would have to be an average for multiple cubes.

------------------------------------------------------------------------
    Gerry Howser
    INTERNET:  howser@lua6.lu.edu
               Postmaster@lua6.lul.edu
               howser@penny.lu.edu (Alternate)

    VOICE:  (314) 681-5400
    FAX:    (314) 681-5566
------------------------------------------------------------------------

------- End of forwarded message(s) -------

------------------------------------------------------------------------
    Gerry Howser
    INTERNET:  howser@lua6.lu.edu
               Postmaster@lua6.lul.edu
               howser@penny.lu.edu (Alternate)

    VOICE:  (314) 681-5400
    FAX:    (314) 681-5566
------------------------------------------------------------------------


From diamond@jrdv04.enet.dec-j.co.jp  Wed Oct 26 20:35:05 1994
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From: Norman Diamond  27-Oct-1994 0932 <diamond@jrdv04.enet.dec-j.co.jp>
To: cube-lovers@life.ai.mit.edu
Apparently-To: cube-lovers@life.ai.mit.edu
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>Ouch!  I fat-fingered it, sorry.
>From: Gerry Howser <HOWSER @ LUA6>
>To: <cube-lobers @ LIFE.AI.MIT.EDU>
>I could solve any cube in about a minute and a half,

Nonsense.  Anyone who can't hit the "v" key on their keyboard is surely
incapable of manipulating the right cubie :-)

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

From dbisel@adrian.adrian.edu  Thu Oct 27 15:47:08 1994
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From: dbisel@adrian.adrian.edu
To: cube-lovers@life.ai.mit.edu
Subject: other games
X-Vms-To: smtp%"cube-lovers@ai.ai.mit.edu"

What other games are you interested in besides the Rubik's Cube? Do you know
of any other addresses where I can get fun, cool information at?
Diana Bisel



From ybanezs%geds@mhsgate.salem.ge.com  Thu Oct 27 16:22:05 1994
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Date: Thu, 27 Oct 94 16:20:08 EST
From: Ybanez Sheldon <ybanezs%geds@mhsgate.salem.ge.com>
To: cube-lovers@ai.mit.edu
Subject: Solution..
X-Mailer: XGATE 2.12 MHS/SMTP Gateway

I have been able to solve the Cube in under a minute... but that was 
years ago when my reflexes and memory was better in Junior High... now I 
pull the old cube out for limbering the fingers... and seeing how much I 
remember the solutions... now they are so ingrained in me... I no longer 
remember them as separate moves.... but a conglomeration of twists and 
turns... 

the book --the title I can't remember-- I originally learned from, showed 
the solution as a TOP to BOTTOM approach... doing the first top layer... 
then the center edges... and then completing the last layer...  I noticed 
then Mihn used the top, bottom, middle approach... when he won the 
World's Cube solving championship on the show 'THAT"S INCREDIBLE', which 
was also advocated by the solution book that was available from the 
address that was included with the original cubes...

so then I was able to solve it either way... finding the latter approach 
a little faster...  now the Revenge I can solve in about 5 minutes... 
maybe quicker, but I never really bothered to accurately time myself... I 
only learned the one way to solve the 4x4x4 cube... from Mihn's book.

Now since I joined this mailing list I have been inundated with all these 
algorithms.... how do I translate them?  Being a neophyte to cube 
'theory' its pretty frustrating trying to figure out what all the letters 
and numbers mean... and what they are trying to achieve....

can anyone help?

thanks in advance...


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



From BRYAN@wvnvm.wvnet.edu  Thu Oct 27 17:47:30 1994
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From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Ybanez Sheldon" <ybanezs%geds@mhsgate.salem.ge.com>,
        "Cube Lovers List" <cube-lovers@ai.mit.edu>
Subject:   Re: Solution..
In-Reply-To: Message of 10/27/94 at 16:20:08 from ,
           ybanezs%geds@mhsgate.salem.ge.com

On 10/27/94 at 16:20:08 Ybanez Sheldon said:

>Now since I joined this mailing list I have been inundated with all these
>algorithms.... how do I translate them?  Being a neophyte to cube
>'theory' its pretty frustrating trying to figure out what all the letters
>and numbers mean... and what they are trying to achieve....

>can anyone help?

I was thinking of suggesting a few references, but then it occurs that
perhaps there are not very many references currently in print.  Here
is a little Cube Theory 101.

In the "Standard Model" (or maybe the "Singmaster Model") of the 3x3x3
cube, the cube is not rotated in space.  The only thing you can do is
twist one of the six faces.  Singmaster designates the faces as
Up, Down, Right, Left, Front, and Back.  The names are chosen so that
no two of the faces start with the same letter.  There have been some
latter day efforts to rename Up as Top so that all the faces have names
beginning with consonants.

Twists are designated by the first letter of their name  --  U, D, R, L,
F, and B for clockwise quarter-turns; U', D', R', L', F', and B' for
counter-clockwise quarter-turns; U2, D2, R2, L2, F2, and B2 for half-
turns (180 degrees).  In proper typography, the "2" in "U2" is
written as a superscript.  Sometimes U3, D3, etc. are used to denote
counter-clockwise quarter-turns because three clockwise quarter-turns
produce the same result as one counter-clockwise quarter-turn.

A sequence of twists is written left-to-right  -- e.g., FRU'LLR.

The complement notation which is used to convert clockwise quarter-turns
into counter-clockwise quarter-turns may also be applied to a group
of twists  --  e.g., (FRU')' is equal to UR'F' (twisting in the opposite
order and in the opposite direction).

The same sort of notation is used to describe cubies  -- the up-right
cubie is ur.  Singmaster distinguishes between cubies and cubicles
via italic and Roman text, but that is a bit hard to do via E-mail.

Things get a bit more complicated when you consider slice moves,
cubes larger than 3x3x3, and rotations of the whole cube.  Note that
most people solving "real cubes" (as opposed to mathematical
models of cubes) do indeed rotate the whole cube, for example they
move the Bottom face to the Up (or Top), to make it easier to twist.
However, the "Standard Model" does not rotate the whole cube because
mathematically it is just as easy to twist one face as any other.

Hope this helps.


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

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

From BRYAN@wvnvm.wvnet.edu  Fri Oct 28 08:54:11 1994
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Date:      Fri, 28 Oct 1994 08:53:52 EDT
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Ybanez Sheldon" <ybanezs%geds@mhsgate.salem.ge.com>,
        "Cube Lovers List" <cube-lovers@ai.mit.edu>
Subject:   Re: Solution..
In-Reply-To: Message of 10/27/94 at 17:00:30 from BRYAN@wvnvm.wvnet.edu

On 10/27/94 at 17:00:30 Jerry Bryan said:
>On 10/27/94 at 16:20:08 Ybanez Sheldon said:

>The same sort of notation is used to describe cubies  -- the up-right
>cubie is ur.  Singmaster distinguishes between cubies and cubicles
>via italic and Roman text, but that is a bit hard to do via E-mail.

Ooops.  I just pulled out my Frey and Singmaster.  Cubies and cubicles
are both italics, and the distinction is one of upper case italics
vs. lower case italics (still hard to do on E-mail).  Face twists are
Roman (block) letters.  Whole cube rotations are script letters.
Also, in proper typography, a complement (as in R') would normally
be a superscript "-1" rather than an apostrophe.

(By the way, even with a word processor or text processor, I have trouble
with script letters.  I can't get Word Perfect to do script letters, nor
Waterloo Script.  I used to use TeX, and I don't think it could do script
letters.  I haven't tried desk top publishing of the Pagemaker ilk.
Does anybody have any suggestions?  If so, I suspect this is the sort of
thing where private E-mail might be more appropriate than broadcasting
to the entire list.  Thank in advance.)

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

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

From hoey@aic.nrl.navy.mil  Fri Oct 28 11:38:16 1994
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From: hoey@aic.nrl.navy.mil (Dan Hoey)
Message-Id: <9410281538.AA17190@Sun0.AIC.NRL.Navy.Mil>
To: "Cube Lovers List" <cube-lovers@ai.mit.edu>
Cc: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
Subject: Cube colors and face names
Keywords: Rubiksong, Varga, Colors, Humor

> Singmaster designates the faces as Up, Down, Right, Left, Front, and
> Back.  The names are chosen so that no two of the faces start with
> the same letter.  There have been some latter day efforts to rename
> Up as Top so that all the faces have names beginning with
> consonants.

Yes, this is the main reason for using Top, because of the Rubiksong
introduced by Varga that I described (unfortunately with many typos)
on 22 Feb 90 (<ftp:://ftp.ai.mit.edu/pub/cube-lovers/cube-mail-6.gz>
is a URL that I hope works--anyone who is actually able to point and
click on this, please let me know).

But there's another reason.  Remember the annoying feature that the
color assignments to faces were never standardized?  The first cube I
bought had red opposite yellow, blue opposite white, and orange
opposite green (I think).  Even though in later days most cubes are
manufactured with opposite faces ``differing by yellow''--red opposite
orange, blue opposite green, and yellow opposite white--there does not
seem to be a standard for the handedness of the coloring.  This has
long been a problem on cube-lovers, where everyone starts out asking
``I've got my cube solved except a blue sticker on the white face, a
white sticker on the green face, and a green sticker on the blue
face,'' and the puzzle becomes trying to figure out where those faces
are.  (This was fixed in Square 1, where they printed a full-color
instruction book coordinated with the puzzle).

My modest proposal is to define the Standard Earth-Tone Cube, which
has the faces in standard and easily remembered places.  The colors
are taupe, dun, fawn, beige, loam, and roan.  This supports the use of
Top over Up, because ``taupe'' is so much more evocative than
``umber''.  ``Dun'' is also a major win, and I wish I had better names
for the other faces.

I have yet not tried painting such a cube, because I can't figure out
which color is which.

Dan
Hoey@AIC.NRL.Navy.Mil

From @mail.uunet.ca:mark.longridge@canrem.com  Fri Oct 28 11:50:15 1994
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Sender: CRSO.Cube@canrem.com
Subject: Shift Invariant Part 2
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.825.5834.0C1B76FC@canrem.com>
Date: Thu, 27 Oct 1994 21:56:00 -0400
Organization: CRS Online  (Toronto, Ontario)

Continuing the previous discussion on shift invariance...

Mark writes:
>>    This is the longest process I've found so far.

Martin writes:
>How about (UR)^140 or (UR)^1400?  As mentioned above, you can make the
>processes as long as you wish.

...or (U1 R1)^35 ? And indeed, (U1 R1)^(35 * 40) is shift invariant.
I meant to say (and should have said):

 "This is the longest optimal process I've found so far."

 Although I was inspecting (U1 R1)^N patterns in the quest for shift
invariance, (U1 R1)^35 = (R1 U1)^35 escaped me. In fact it was my
mistaken belief that the < U , R > group had no shift invariant
processes.

 I did not realize the connection between the centre of a group and
shift invariance until Martin's message of Mon Oct 24 17:10:27 1994.
I actually did use GAP on the < U, R > group but I couldn't resolve
the resulting position (can GAP use letters? I should have used
letters).

The missing insight was realizing that, although the full group had
a unique centre, other subgroups have different centres.

So without further adieu:

6 Counterclockwise Twist,
Equivalent to (U1 R1)^35= (R1 U1)^35 & Shift Invariant
UR11 = U2 R1 U1 R1 U1 R3 U1 R3 U1 R3 U2 R1 U1 R1 U1 R3 U1 R3 U1 R3
       (22 q  or  20 h  moves)

(U3 R3)^35 would execute a 6 clockwise twist.

Martin writes:
> 4)  The ``odd'' element.
>  The element (UR)^140 lies in the centre of the subgroup <U,R>.
>  It is the only shift invariant element of odd order (hence the name).
>  Thus this process and its inverse are shift invariant.
>  There are 24 such elements in the entire group (two for each edge).

Is this odd due to ( U1 R1 )^35?
Actually everything about the above description appears even.
It is an even number of quarter turns...

Martin writes:
> For me the most amazing elements were the ``odd'' element and the
> ``diagonal square'' element.

 I concur completely, although the all-commuting 12-flip is definitely
interesting too. I was surprised to see the process was shift invariant.

Martin writes:
> Thus at this time all non-trivial such elements had been found, except
> for the ``odd'' element.

For which I refer to the process UR11, 22 q turns.

 Martin, you will be pleased to hear that I like GAP, but I need a
bigger hard drive for that beast!

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

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Sender: CRSO.Cube@canrem.com
Subject: Speed Cubing
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.826.5834.0C1B76FD@canrem.com>
Date: Thu, 27 Oct 1994 21:57:00 -0400
Organization: CRS Online  (Toronto, Ontario)

Derek Bosch writes:
> I too, have read the book, Jeff Conquers The Cube in 45 seconds, as
> well as Minh Thai's book on the cube (he's the world record holder,
> with 22 seconds as an official world record.  I used to compete back
> in the cubing days, and could regularly get under 25 seconds, using
> a strategy of solving the corners, solving the edges on two opposite
> sides, followed by the middle slice.


The "official" world record was set by Minh Thai at the 1982 World
Championships in Budapest Hungary, with a time of 22.95 seconds.

Keep in mind mathematicians provided standardized dislocation patterns
for the cubes to be randomized as much as possible.

I think the Guiness Book of Records dropped the entry in the 1985
edition due to the fact that the contests all dried up.

Interestingly David Allen, the #2 cubist in the United States, also
uses the Jeff Varasano method. I met him in Buffalo NY in the
a regional American Cube-a-thon on Sept 18, 1982. (Yes, that long
ago!)

Did you enter any of the tournaments Derek?

Derek continues:
> Several people on this mailing list have done serious analysis
> trying to reach "God's Algorithm", which isn't terribly useful to me.
> The operators that these analyses generate are really slow to crank
> out on the cube.  I prefer slightly longer ones, that are more
> optimized for speed (hand positions, etc).

I can't agree entirely. I use computer generated sequences for a lot
of patterns and I find them quite useable in some cases. Also the
< U, R > group processes only use 2 sides, and those I can do
without moving the cube in space. Usually I rotate them in
space first.

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

From BRYAN@wvnvm.wvnet.edu  Fri Oct 28 12:55:14 1994
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Date:      Fri, 28 Oct 1994 12:54:56 EDT
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <cube-lovers@life.ai.mit.edu>
Subject:   Re: Speed Cubing
In-Reply-To: Message of 10/27/94 at 21:57:00 from mark.longridge@canrem.com

Has any analysis of speed cubing been performed in the sense of how
many twists were performed?  How many twists does somebody accomplish
in under 45 seconds or in 22.95 seconds?  For example, you might
video tape somebody and replay it in slow motion.  It would still be
hard to get an accurate count, I think.  You would have to ignore
whole cube rotations, and it might be hard to distinguish between
half and quarter turns, plus somebody might be using slice moves.
But if such an analysis *could* be done, it would be interesting to
to compare the results to what is known mathematically about
God's 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

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

From ybanezs%geds@mhsgate.salem.ge.com  Fri Oct 28 15:57:15 1994
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Date: Fri, 28 Oct 94 15:56:21 EST
From: Ybanez Sheldon <ybanezs%geds@mhsgate.salem.ge.com>
To: cube-lovers@ai.mit.edu, bryan@wvnvm.wvnet.edu
Subject: Re: Speed Cubing
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>Has any analysis of speed cubing been performed in the sense of how
>many twists were performed?  How many twists does somebody accomplish
>in under 45 seconds or in 22.95 seconds?



I too had wondered this...  and in what way was the solved cube 
'scrambled' for most of you cubists know that all scrambling is not 
equal...  I have found some of my quickest times involved arriving at the 
solution much sooner than expected by not needing to perform some 
auxiliary, but essential and long routines....   Could these world record 
times have been accomplished with random scrambling..... I have always 
pondered how fast I would have completed the cube if I was handed the 
'same' cube that Minh 'flew' on.... A move count would be very 
interesting indeed.... 

With the right equipment and a good copy of the world record video 
tape... it may be conceivable to actually count the moves.... 

In the Hey day of cubing, during a younger version myself, I also found 
by making a good guess at which way to solve the cube (ie.. Top to bottom 
or
top, bottom, then middle) I could easily cut a few seconds off my S.T. 
(solution time)...  but I can't for the life of me remember the 
criteria..


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



From mschoene@math.rwth-aachen.de  Fri Oct 28 17:54:41 1994
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Date: Fri, 28 Oct 94 22:13 PST
From: Martin.Schoenert@math.rwth-aachen.de
To: cube-lovers@life.ai.mit.edu
In-Reply-To: Mark Longridge's message of Thu, 27 Oct 1994 21:56:00 -0400 <60.825.5834.0C1B76FC@canrem.com>
Subject: Re: Shift Invariant Part 2

Mark Longridge writes in his e-mail message of 1994/01/27

    ...or (U1 R1)^35 ? And indeed, (U1 R1)^(35 * 40) is shift invariant.

Mark  kindly points out, that my process (UR)^140 for the ``odd'' element
is a strange choice, given that (UR)^140 = (UR)^35.

I can't recall how I arrived at this process.  Somehow I simply missed
that (UR)^140 = (UR)^35, which is especially strange since I know that
(UR) has order 105 since 1982.

Mark continues

    Equivalent to (U1 R1)^35= (R1 U1)^35 & Shift Invariant
    UR11 = U2 R1 U1 R1 U1 R3 U1 R3 U1 R3 U2 R1 U1 R1 U1 R3 U1 R3 U1 R3
           (22 q  or  20 h  moves)

Is UR11 the shortest process effecting the ``odd'' element in <U,R>?

Mark continues

    Is this odd due to ( U1 R1 )^35?
    Actually everything about the above description appears even.
    It is an even number of quarter turns...

The ``odd'' element o has odd order as element of the cube group,
i.e., o^3 = id.  All other shift invariant elements e have even order,
i.e., either e^2 = id or e^4 = id (for some ``abelian'' elements).

Mark continues

    I actually did use GAP on the < U, R > group but I couldn't resolve
    the resulting position (can GAP use letters? I should have used
    letters).

I assume you wonder whether GAP can find a process for a given element.
In fact GAP can do this (you define a homomorphism from the free group
on U,D,L,R,F,B to the cube group and then ask for a preimage of the
element).  But the process is usually extremly long, e.g., for
the ``central'' element GAP finds a process that has length > 2*10^6
(don't try this at home ;-).

There is an improved algorithm by Philip Osterlund, which is a lot
better, but still not good enough to help in the quest for god's
algorithm.  For example it finds a process for the ``central''
element of length 228.

Mark continues

    Martin, you will be pleased to hear that I like GAP,
    but I need a bigger hard drive for that beast!

Look at it this way:
The system costs you $200, and you even get a hard drive for free!

Seriously, you don't need the full distribution (32 MByte),
but only the executable and the library (5 MByte).
However, you should have enough real memory;
8 Mbyte is the minimum, 16 MByte is better,
and the 64 MByte that I have in my workstation don't hurt.

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 brett@math.toronto.edu  Mon Oct 31 14:36:30 1994
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Subject: 
To: cube-lovers@ai.mit.edu (cube)
Date: Mon, 31 Oct 1994 14:35:34 -0500 (EST)
From: "Brett Stevens" <brett@math.toronto.edu>
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I joined this list in the middle of an ongoing discussion
on shift invariance.  can someone fill me in with the important
definitions and what has already been discussed  (if it is not 
too much work)

brett stevens

plus I am interested, scince seeing a part of Jerry Slokam's
collection at the exhibit in chicago in august if anyone
had a list or reference to a list of all made rubiks cube
type puzzles  ie   external shape and intenal, rotaional structure
I wouls very much like to know where I cvan get the following

1  a 2X2X2  cube
2  a the various puzzles with the pyramids internal structure but
   and also the cubes internal structue  (3X3X3) but different
   external structurs.
3  conglomeration cubes--Ideal made one of these that was two
   3X3X3  cubes sharing three cubies in a row  I have made
   one of these myself by surgery on two cubes but I know
   that there are other "conglomerates out there"
4  kitsch-cubes   (my name)   but things like royal4  kitsch-cubes   (my name)   but things like royal4  kitsch-cubes   (my name)   but things like royal4  kitsch-cubes   (my name)   but things like royal
wedding cubes,  mount rushmore etc.

thanks 

brett stevens

brett@math.toronto.edu
   

From brett@math.toronto.edu  Mon Oct 31 14:43:38 1994
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Subject: diameter
To: cube-lovers@ai.mit.edu (cube)
Date: Mon, 31 Oct 1994 14:42:43 -0500 (EST)
From: "Brett Stevens" <brett@math.toronto.edu>
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  what is the diameter of the group of 3X3X3  rubiks cube rotations?
  ie. the longest cyclic subgroup.  what is the shortest path
  to solved?

  also I thought that red-orange   blue-white   yellow-green
  was standard.  all the ideal manufacturede c ubes were this way.

  and the two orientations available with the above colouring
  are not only an inconvienience  --Dr. Hana Bizek at
  ARgonne NAtional LAbs has used these two parities to do 
  veery intersesting cube sculptures or designs as she calls
   them

   brett stevens

   brett@math.toronto.edu


From brett@math.toronto.edu  Mon Oct 31 14:45:33 1994
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Subject: diameter
To: cube-lovers@ai.mit.edu (cube)
Date: Mon, 31 Oct 1994 14:42:43 -0500 (EST)
From: "Brett Stevens" <brett@math.toronto.edu>
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  what is the diameter of the group of 3X3X3  rubiks cube rotations?
  ie. the longest cyclic subgroup.  what is the shortest path
  to solved?

  also I thought that red-orange   blue-white   yellow-green
  was standard.  all the ideal manufacturede c ubes were this way.

  and the two orientations available with the above colouring
  are not only an inconvienience  --Dr. Hana Bizek at
  ARgonne NAtional LAbs has used these two parities to do 
  veery intersesting cube sculptures or designs as she calls
   them

   brett stevens

   brett@math.toronto.edu


From alan@curry.epilogue.com  Mon Oct 31 16:31:07 1994
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Date: Mon, 31 Oct 1994 16:34:39 -0500
Message-Id: <31Oct1994.155118.Alan@LCS.MIT.EDU>
From: Alan Bawden <Cube-Lovers-Request@ai.mit.edu>
Sender: Cube-Lovers-Request@ai.mit.edu
To: brett@math.toronto.edu, cube-lovers@ai.mit.edu
In-Reply-To: Brett Stevens's message of Mon, 31 Oct 1994 14:35:34 -0500 (EST) <9410311936.AA07023@life.ai.mit.edu>
Subject: Administrivia

   Date: Mon, 31 Oct 1994 14:35:34 -0500 (EST)
   From: Brett Stevens <brett@math.toronto.edu>
   I joined this list in the middle of an ongoing discussion
   on shift invariance.  can someone fill me in with the important
   definitions and what has already been discussed  (if it is not 
   too much work)

This is what the archives are for!

Some of you old-timers may have forgotten where the archives are, and it's
been a while (several years) since I reminded everybody about the existence
of Cube-Lovers-Request, so I have included the standard greeting message I
send to all new subscribers below.

Some quick administrative observations while I have your attention:

Today are 151 entries on the main mailing list.  Some of those entries are
local redistribution lists.  I estimate there are about 160 of us.

We celebrated our 14 birthday last July.  I'd bet we are among the top ten
oldest active mailing lists on the Internet.

I periodically get requests for FTP archives of cube-related material other
than our mailing list archives (simulators and other programs, tables of
results, catalogs of merchandise, etc.)  I am not aware of any such
centralized collection of Cubist stuff.  If someone knows of such a
collection, or would like to organize one, or simply has a list of cube
related resources on the network, I'd like to hear from them.

My policy on advertising: Since Cube-Lovers is an unmoderated mailing list,
I really have no control over what is sent here, but I do complain to
people who send advertising that isn't obviously Cube related.  And I
reserve the right to start complaining about -all- advertising should it
ever get out of hand.

------- Begin Greeting Message -------

Our addresses are Cube-Lovers@AI.MIT.EDU for submissions and
Cube-Lovers-Request@AI.MIT.EDU for administrivia.

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

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

Using FTP, connect to FTP.AI.MIT.EDU, login as "anonymous" (any password),
and in the directory "pub/cube-lovers" you will find the thirteen (compressed)
files "cube-mail-0.gz" through "cube-mail-12.gz".

Archive vital statistics (when uncompressed):

	   File		   From		   To	     Size (bytes)
	   ----		   ----		   --	     ------------
	cube-mail-0	12 Jul 80	23 Oct 80	185037
	cube-mail-1	 3 Nov 80	 9 Jan 81	135719
	cube-mail-2	10 Jan 81	 3 Aug 81	138566
	cube-mail-3	 3 Aug 81	 3 May 82	137753
	cube-mail-4	 4 May 81	11 Dec 82	139660
	cube-mail-5	11 Dec 82	 6 Jan 87	173364
	cube-mail-6	10 Jan 87	13 Apr 90	216733
	cube-mail-7	12 Oct 90	 9 Sep 91	137508
	cube-mail-8	 1 Nov 91	25 May 92	171205
	cube-mail-9	28 May 92	 7 Jan 93	155755
	cube-mail-10	20 Mar 93	 6 Dec 93	171881
	cube-mail-11	 6 Dec 93	18 Feb 94	349772
	cube-mail-12	24 Feb 94	 5 Sep 94	181193

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

				- Alan

------- End Greeting Message -------

From BRYAN@wvnvm.wvnet.edu  Mon Oct 31 16:48:57 1994
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Date:      Mon, 31 Oct 1994 15:39:04 -0500 (EST)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: "Cube Lovers List" <Cube-Lovers@ai.mit.edu>
Subject:   Speed Cubing Path Lengths

I have received several private E-mail messages indicating that
the algorithms used by speed cubists solve the cube in 50 or
60 moves.  On the one hand, that seems astonishingly good to me,
being fairly close to the solutions from early Thistlethwaite
programs.  On the other hand, it is roughly double (depending, I
suppose on whether H-turns are counted or not) what is probably
the true God's Algorithm.  Hence, it doesn't tell us much about
God's Algorithm except that the speed cubists are very, very
good.

On another subject, my Cube Theory 101 article said that the
apostrophe was used in E-mail to denote complements, when of
course it is used to denote inverses  --  not the same thing
as complements at all.

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

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

From dik@cwi.nl  Mon Oct 31 18:14:47 1994
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Date: Tue, 1 Nov 1994 00:14:45 +0100
From: Dik.Winter@cwi.nl
Message-Id: <9410312314.AA06573=dik@boring.cwi.nl>
To: Cube-Lovers@ai.mit.edu
Subject: Re:  Speed Cubing Path Lengths

 > I have received several private E-mail messages indicating that
 > the algorithms used by speed cubists solve the cube in 50 or
 > 60 moves.  On the one hand, that seems astonishingly good to me,
 > being fairly close to the solutions from early Thistlethwaite
 > programs.  On the other hand, it is roughly double (depending, I
 > suppose on whether H-turns are counted or not) what is probably
 > the true God's Algorithm.  Hence, it doesn't tell us much about
 > God's Algorithm except that the speed cubists are very, very
 > good.

The best current algorithm has a proven upperbound of 37 turns (q
and h).  God's Algorithm is probably much shorter.  In fact the
program that implements Kociemba's algorithms has not yet found
a configuration (out of many thousands random configurations
tested) that could not be solved in 20 turns or less.  If we look
at distributions for similar puzzles it is expected that more than
one in three configurations requires the maximum number of turns
minus 1 or 2.  So I expect God's Algorithm to be at most 22 turns.
Still a long way to go.

dik

From devo@vnet.ibm.com  Tue Nov  1 13:20:25 1994
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   Tue, 01 Nov 94 12:33:01 EST
Date: Tue, 1 Nov 94 12:33:54 EST
From: "Dave Eaton" <devo@vnet.ibm.com>
To: cube-lovers@life.ai.mit.edu
Subject: Is there a symbolic cube program?

Is there a program that allows you to type in Singmaster-style
moves and then prints out the resultant state, something like
this (not actual results):

INPUT:   (R U2 R3 U2)2
OUTPUT:  (fur,drb,rdf) (fr,dr)

This is what I tried to write long ago, but I never had all
the tricks needed to get the program to work.

If no program like this exists, is there something similar?
I guess I would be looking for nicely-portable C or a DOS
binary.  Thanks to you all for sharing cube information.

......Dave Eaton, N2NOQ, Owego NY, devo@vnet.ibm.com


From MALONEY9146@a12t.cc.fredonia.edu  Wed Nov  2 12:56:09 1994
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Date:         Wed, 2 Nov 1994 12:35 pm EST (17:35:07 UT)
From: "Daniel P. Maloney" <MALONEY9146@fredonia.edu>
Organization: State University of New York - College at Fredonia
To: cube-lovers@ai.mit.edu
Subject:      Help ma, please!
Message-Id: <28373110294123506@FREDONIA>

I hate posting this message to the entire list, but how do you
unsubscribe from this list?

Dan

From alan@curry.epilogue.com  Wed Nov  2 14:57:30 1994
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Date: Wed, 2 Nov 1994 14:59:04 -0500
Message-Id: <2Nov1994.145313.Alan@LCS.MIT.EDU>
From: Alan Bawden <Cube-Lovers-Request@ai.mit.edu>
Sender: Cube-Lovers-Request@ai.mit.edu
To: MALONEY9146@fredonia.edu
Cc: cube-lovers@ai.mit.edu
In-Reply-To: "Daniel P. Maloney"'s message of Wed, 2 Nov 1994 12:35 pm EST (17:35:07 UT) <28373110294123506@FREDONIA>
Subject: Help ma, please!

   Date:         Wed, 2 Nov 1994 12:35 pm EST (17:35:07 UT)
   From: "Daniel P. Maloney" <MALONEY9146@fredonia.edu>

   I hate posting this message to the entire list, but how do you
   unsubscribe from this list?

   Dan

As I reminded everybody just last weekend, 
so you can't say you didn't see it:

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

Your request to unsubscribe was sent to me less than 24 hours ago.
I'm sorry I didn't drop everything just to deal with it.  How about if I do
it later on this evening?  Can you wait that long?

From hoey@aic.nrl.navy.mil  Fri Nov  4 11:46:53 1994
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From: hoey@aic.nrl.navy.mil
Message-Id: <9411041646.AA21659@sun13.aic.nrl.navy.mil>
To: Cube-Lovers@life.ai.mit.edu
Subject: The real size of cube space

In January of this year, Jerry Bryan and I wrote of counting the
number of M-conjugacy classes of Rubik's cube.  In the sense that (for
instance) there is really only one position 1 QT from start, even
though that QT may be applied in twelve different ways, this task
amounts to counting the true number of positions of the cube.  The
earlier discussion centered on calculations involving computer
analysis of large numbers of positions.  However, a look in Paul B.
Yale's book _Geometry and Symmetry_ gave me a clue: the Polya-Burnside
theorem is a tool that allows us to perform this calculation by hand.

The Polya-Burnside theorem describes a relation between a finite group
J and a _representation_ of the group.  For our purposes, a represen-
tation is a homomorphism of J into a permutation group, R: J -> S[X].
Here, S[X] refers to the group of all permutations of a set X; the
image of J, called R(J), need not be the whole of S[X], but R(J) will
be a subgroup of S[X].  The _orbits_ of R(J) are the equivalence
classes of X under the relation x~y, said to be true if there is some
permutation p in R(J) for which p(x)=y.  The _fixed points_ of a
permutation p in R(J) are the elements x of X for which p(x)=x.  The
Polya-Burnside theorem states that the average number of fixed points
of permutations in R(J) is equal to the number of orbits of R(J).
That is,
         |R(J)| |Orbits(R(J))| = Sum[p in R(J)] |FixedPoints(p)|.
The average may also be taken over J:
            |J| |Orbits(R(J))| = Sum[j in J] |FixedPoints(R(j))|,
a nontrivial distinction, since R may not be one-to-one (though it is
for our application).  The Polya-Burnside theorem is not very
inaccessible nor hard to prove, but I will not prove it here.

For our purpose, we take the group J to be M, the 48-element group of
symmetries of the cube.  X will be the set of all cube positions,
which we usually call Gx (for GE, GC, or G, depending on whether we
consider edges, corners, or both; we are considering the positions
relative to fixed face centers in all three cases).  And the repre-
sentation R is the operation of M-conjugation: (R(m))(g) = m' g m.
Verifying that R is a homomorphism is an exercise in associativity
that Jim Saxe and I carried out in the Symmetry and Local Maxima
paper, in the archives [cube-mail-1, 14 December 1980].

R has been so chosen because we wish to calculate the number of
M-conjugacy classes of Gx, |Gx\Conj(M)|, which is be the number of
orbits of R(M).  To apply the Polya-Burnside theorem for this, we need
to calculate, for each element of m of M, the number of fixed points
of R(m).  That is the number of elements g of Gx for which m' g m = g.
Multiplying by m, this becomes g m = m g: the fixed points we wish to
count are just those elements g of Gx that commute with m.

There are several tools to make the counting easier.  First, I'll note
that just as there are M-conjugacy classes of Gx, there are
M-conjugacy classes of M itself.  The number of fixed points of R(m)
is the same for any m in a given conjugacy class.  So to calculate the
total number of fixed points over R(M), we need only calculate the
number of g in Gx commuting with each of the ten classes of cube
symmetry and multiply by the size of the class.

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,...,ck), so that c2=m(c1), c3=m(c2),...,ck=m(c[k-1]),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.  Conversely, if g acts thus on
cycles, then g will commute with m, and so g is a fixed point of R(m).

Suppose that m has j different k-cycles of cubies.  There are j! k^j
possibilities for g's action on the cubies in those k-cycles: j!
permutations of cycles, and for each g:(c1,c2,...,ck)->(d1,d2,...,dk),
k choices for g(c1) among {d1,...,dk}.  It turns out to be a fairly
easy exercise to show that half of those possibilities are even
permutations and half odd, though the partition by parity is
surprisingly different depending on whether k is even or odd.  This
will allow us to combine the results for GE and GC simply by
multiplying together and dividing by two.

Now consider orientation of cubies.  This is similar to the case of
permutation, in that the orientation that g imposes on a cubie is a
constant for all cubies in a cycle.  I will first discuss the edge
orientation, which is fairly straightforward, and continue to corner
orientation, which has some surprising features.

For edge orientation, if all the cycles have even length, then g's
orientation parity is zero over each cycle, and so zero over the
entire cube.  So we can choose the orientation of imposed by c1->g(c1)
for each cycle (c1,...,ck) in 2^j ways.  If there are odd-length
cycles, then half of the orientations will have nonzero orientation
parity, and only 2^(j-1) possible orientations can be achieved.

For corners, we might expect there to be 3^(j-1) orientations, except
3^j for cycles of length a multiple of three, and this is often so.
But there are two important exceptions.  First, if m is a reflection
(i.e., not a proper rotation in C) then alternate cubies in each cycle
must be given the opposite orientation by g.  If the cycle has even
length, this conserves orientation, so there will be 3^j possibili-
ties.  If the cycle has odd length, this implies that the orientation
of each cubie must be its own opposite (i.e., zero twist).  Thus,
there there is only one possible orientation of the 1-cycles in the
diagonal reflections.  The second exception, an even bigger surprise,
occurs when m is either the 120-degree rotation or the 60-degree in-
verted rotation.  It turns out that the orientation constraint forbids
any permutation that exchanges the two 1-cycles in these positions.
(This constraint on permutations would throw off the equality between
even and odd permutations, except that these classes of m have other
corner cycles that restore the balance.)  The impossibility of m
commuting with an exchange of the two corners can be verified by
examining the possible orientations, but I haven't got any good way of
characterizing when it would be be a problem in general.  In fact, I
did not notice it until I investigated discrepancies with the
exhaustive computer analysis.

Using the above analysis, we may carry out the calculation as in the
three tables below.  The first two tables count the number of fixed
points of R(m) for an element m of each class, multiply by the class
size, and divide by |J|=48 to get the number of orbits as in the
Polya-Burnside theorem.  The third table calculates the number of
fixed points by combining the results of the first two tables, divided
by the class size (which was multiplied in both for edges and for
corners), and divided by 2 (because only half the combined positions
have matching permutation parity).

     Counting M-conjugacy classes of the edges of Rubik's cube.

M class            Cycles                     Total F.P.      Numeric
  (class size)      of m     Perms   Oris     in class       Total/48
==============  ===========  ======  ======   ==========  ===========
Identity   (1)  12 1-cycles  12!     2^12/2   12! 2^11    20437401600

Axis Rot/2 (3)   6 2-cycles  6! 2^6  2^6      6! 3 2^12        184320

Rot/3      (8)   4 3-cycles  4! 3^4  2^4/2    4! 3^4 2^6         2592

Diag Rot/2 (6)   5 2-cycles  5! 2^5  2^5
                 2 1-cycles  2       2^2/2    5! 3 2^13         61440

Rot/4      (6)   3 4-cycles  3! 4^3  2^3      3! 3 2^10           384

Inv Rot/4  (6)   3 4-cycles  3! 4^3  2^3      3! 3 2^10           384

Diag Ref   (6)   5 2-cycles  5! 2^5  2^5
                 2 1-cycles  2       2^2/2    5! 3 2^13         61440

Inv Rot/6  (8)   2 6-cycles  2! 6^2  2^2      2! 3^2 2^7           48

Axis Ref   (3)   4 2-cycles  4! 2^4  2^4  
                 4 1-cycles  4!      2^4/2    4! 3^2 2^14       73728

Inversion  (1)   6 2-cycles  6! 2^6  2^6      6! 2^12           61440
                                                          -----------
                                        | GE\Conj(M) | =  20437847376


     Counting M-conjugacy classes of the corners of Rubik's cube.

M class            Cycles                   Total F.P.    Numeric
  (class size)      of m     Perms   Oris   in class     Total/48
===============  ==========  ======  =====  ===========   =======
Identity   (1)   8 1-cycles  8!      3^8/3  8! 3^7        1837080

Axis Rot/2 (3)   4 2-cycles  4! 2^4  3^4/3  4! 3^4 2^4        648

Rot/3      (8)   2 3-cycles  2! 3^2  3^2  
                 2 1-cycles  1       3^2/3  3^5 2^4            81

Diag Rot/2 (6)   4 2-cycles  4! 2^4  3^4/3  4! 3^4 2^5       1296

Rot/4      (6)   2 4-cycles  2! 4^2  3^2/3  3^2 2^6            12

Inv Rot/4  (6)   2 4-cycles  2! 4^2  3^2    3^3 2^6            36

Diag Ref   (6)   2 2-cycles  2! 2^2  3^2
                 4 1-cycles  4!      1      4! 3^3 2^4        216

Inv Rot/6  (8)   1 6-cycle   6       3
                 1 2-cycle   1       3      3^3 2^4             9

Axis Ref   (3)   4 2-cycles  4! 2^4  3^4    4! 3^5 2^4       1944

Inversion  (1)   4 2-cycles  4! 2^4  3^4    4! 3^4 2^4        648
                                                          -------
                                        | GC\Conj(M) | =  1841970


    Counting M-conjugacy classes of the entire Rubik's cube

M class             Edge         Corner       Corner times edge
  (class size)      F.P.          F.P.             / (96*class size)
===============  ==========     =========    =======================
Identity   (1)   12! 2^11       8! 3^7       901,083,401,551,872,000

Axis Rot/2 (3)    6! 3 2^12     4! 3^4 2^4               955,514,880

Rot/3      (8)    4! 3^4 2^6    3^5 2^4                      629,856

Diag Rot/2 (6)    5! 3 2^13     4! 3^4 2^5               318,504,960

Rot/4      (6)    3! 3 2^10     3^2 2^6                       18,432

Inv Rot/4  (6)    3! 3 2^10     3^3 2^6                       55,296

Diag Ref   (6)    5! 3 2^13     4! 3^3 2^4                53,084,160

Inv Rot/6  (8)    2! 3^2 2^7    3^3 2^4                        1,296

Axis Ref   (3)    4! 3^2 2^14   4! 3^5 2^4             1,146,617,856

Inversion  (1)    6! 2^12       4! 3^4 2^4               955,514,880
                                             -----------------------
                             | G\Conj(M) | = 901,083,404,981,813,616

These results have been corroborated and expanded by use of
combinatorial computer programs, to be described in a later message.

Dan Hoey
Hoey@AIC.NRL.Navy.Mil

From @mail.uunet.ca:mark.longridge@canrem.com  Sat Nov  5 23:49:54 1994
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Sender: CRSO.Cube@canrem.com
Subject: Shifty Invariance
From: mark.longridge@canrem.com (Mark Longridge)
Message-Id: <60.846.5834.0C1BB9FA@canrem.com>
Date: Sat, 5 Nov 1994 22:16:00 -0500
Organization: CRS Online  (Toronto, Ontario)

----------------------------------------
Even more thoughts on "Shift Invariance"
----------------------------------------

>>Mark continues
>>
>>    Equivalent to (U1 R1)^35= (R1 U1)^35 & Shift Invariant
>>    UR11 = U2 R1 U1 R1 U1 R3 U1 R3 U1 R3 U2 R1 U1 R1 U1 R3 U1 R3 U1 R3
>>           (22 q  or  20 h  moves)
>>
Martin asks:
>Is UR11 the shortest process effecting the ``odd'' element in <U,R>?

After a bit of computer cubing I found:

p183 6 Twist      R1 U3 R2 U3 R1 D3 U3 R1 U3 R3 D2 R3 U3 R1 D3 U3
                  (18 q  or  16 h  moves)

 This requires using the larger group of <U1, R1, D1>, although I
expected a 16 turn process. Note the fact this larger group has face
index 3 (rather than 2). But now the process is NOT shift invariant
and we see the route itself can determine whether it will be
shift invariant!

I welcome any mathematical explanation!

With even more contemplation I noticed that the process for
 the edge 3-cycle

UR1 = U3 R1 U2 (R1 U1)^2 R2 U3 R3 U3 R2 U1     (16 q, 13 h)

...was reducible to

UR1a= F1 U2 (F1 U1)^2 F2 U3 F3 U3 F2           (14 q, 11 h)]

Of course, now we are using <U1, F1> rather than <U1, R1>.

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

From BRYAN@wvnvm.wvnet.edu  Sun Nov  6 09:15:57 1994
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Date:      Sun, 6 Nov 1994 09:15:37 -0500 (EST)
From: "Jerry Bryan" <BRYAN@wvnvm.wvnet.edu>
To: <cube-lovers@life.ai.mit.edu>
Subject:   Re: Shifty Invariance
In-Reply-To: Message of 11/05/94 at 22:16:00 from mark.longridge@canrem.com

On 11/05/94 at 22:16:00 mark.longridge@canrem.com said:

>p183 6 Twist      R1 U3 R2 U3 R1 D3 U3 R1 U3 R3 D2 R3 U3 R1 D3 U3
>                  (18 q  or  16 h  moves)
                    ^^^^^^^^^^^^^^^^^^^^^

This is not a shift invariance question, but rather two
questions about your searches.  One question is, do you perform separate
searches for q-turns and h-turns, or only for h-turns?  The reason I
ask is the obvious fact that optimal processes in q-turns need not
contain h-turns.  The second question is, how on earth do you keep track
of all those processes in your searches?  I have been asked how I search
so many positions.  I have answered the question before, but I guess
another part of the answer that I haven't mentioned is that I don't
keep up with processes at all, only positions.  If I am asked to provide
processes, I can do so, but it is a very painful task.  I have thought
about keeping up with processes, but I am quite sure that if I did
so it would reduce the number of positions I could 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

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

From mschoene@math.rwth-aachen.de  Sun Nov  6 17:31:30 1994
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Date: Sun, 6 Nov 94 23:29 PST
From: Martin.Schoenert@math.rwth-aachen.de
To: cube-lovers@life.ai.mit.edu
Cc: CRSO.Cube@canrem.com
In-Reply-To: Mark Longridge's message of Sat, 5 Nov 1994 22:16:00 -0500 <60.846.5834.0C1BB9FA@canrem.com>
Subject: Re: Shifty Invariance

Mark writes in his e-mail message of 1994/11/05

    After a bit of computer cubing I found:

    p183 6 Twist      R1 U3 R2 U3 R1 D3 U3 R1 U3 R3 D2 R3 U3 R1 D3 U3
                      (18 q  or  16 h  moves)

     This requires using the larger group of <U1, R1, D1>, although I
    expected a 16 turn process. Note the fact this larger group has face
    index 3 (rather than 2). But now the process is NOT shift invariant
    and we see the route itself can determine whether it will be
    shift invariant!

    I welcome any mathematical explanation!

As I tried to explain in my first e-mail message, a shift invariant
process is a process in a subgroup X of G corresponding to an element
x in the centre *of this subgroup*.

The ``odd'' element is an element in the centre of the subgroup < U, R >.
Thus any process effecting this element written in U and R is a shift
invariant process.  UR11 is one such process.

However, the ``odd'' element does not lie in the centre of the subgroup
< U, R, D > (in fact this subgroup has trivial centre).  Thus a process
effecting this element *involving D*, will *not* be shift invariant.

Some shift invariant processes are in fact in the centre of multiple
subgroups.  For example the square elements, except for the ``diagonal
square'' element, have this property.

For such elements one has some choice which generators to use.  For
example the ``single square'' elements (U2 R2)^3 lies in the centre of
< U2, R2 > and < U2, D, R2, L > (and all subgroups inbetween), so every
process effecting this element involving any subset of U2, D, D2, R2, L,
and L2, will be a shift invariant process.

For the ``odd'' element, one has now choice.  It lies in the centre of
< U, R >, but not in the centre of any larger group.  Thus a shift
invariant process effecting the ``odd'' element must involve U and R,
and cannot involve more generators.

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  Mon Nov  7 19:20:36 1994
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Date: Tue, 8 Nov 94 01:18 PST
From: Martin.Schoenert@math.rwth-aachen.de
To: Cube-Lovers@life.ai.mit.edu
Cc: hoey@aic.nrl.navy.mil
In-Reply-To: hoey@aic.nrl.navy.mil's message of Fri, 4 Nov 94 11:46:50 EST <9411041646.AA21659@sun13.aic.nrl.navy.mil>
Subject: Re: The real size of cube space

Dan Hoey writes in his e-mail message of 1994/11/04

    In January of this year, Jerry Bryan and I wrote of counting the
    number of M-conjugacy classes of Rubik's cube.  In the sense that (for
    instance) there is really only one position 1 QT from start, even
    though that QT may be applied in twelve different ways, this task
    amounts to counting the true number of positions of the cube.  The
    earlier discussion centered on calculations involving computer
    analysis of large numbers of positions.  However, a look in Paul B.
    Yale's book _Geometry and Symmetry_ gave me a clue: the Polya-Burnside
    theorem is a tool that allows us to perform this calculation by hand.

    ...a very nice application of the Polya-Burnside theorem,
       to compute the number of M-conjugacy classes in G...

Yes, a little bit of group theory can answer many questions arising from
the cube.  In fact I have noticed that quite a few of well known results
in group theory have been rediscovered in this forum.  Note that I don't
think this is a bad thing.  At least for me results that I ``knew'' are
now, that they have been demonstrated for the cube, much easier to grasp
than they were before (grasp is certainly an appropriate term in
connection with the cube).

Dan continues

    For our purpose, we take the group J to be M, the 48-element group of
    symmetries of the cube.  X will be the set of all cube positions,
    which we usually call Gx (for GE, GC, or G, depending 