Thread: "Calculating the number of permutation of 2by2by2by2by2 (2^5)"

I think I can provide a more clear-cut answer to Charlie's
question: "Should the 31 be 32?"

My answer:

Only if you wish to regard as being distinct arrangements
which differ only by a reorientation of the whole pile.

Personally, I do not think one should wish that. Indeed, the
way the arrangements are counted in the odd order cases takes
care of this by not counting permutations of the facet-center
cubies - i.e., those cubies that are in center slices on all
axes but one. In the odd order cases, those provide a
convenient reference for the orientation of the pile. So we
are only counting arrangements for which the pile is in a
standard orientation as defined by the stickers on those
cubies. The factor left out by using 31 is 5!*2^4 which is
the number of ways in which you can arrange the facet-centers
(or orient a 5-cube).

When the order is even, there is not such a convenient
reference for a standard orientation. However, one way to do
it is to pick one of the corner cubies and say that the pile
is in standard orientation when that one cubie has all its
stickers facing in the same direction as in the initial
state. Again, there 5!*2^4 ways of positioning that one
corner. Now saying that the pile is in standard orientation
is equivalent to saying that there is no choice in the
placement of that reference corner; so there are only 31
corners which we are free to place arbitrarily.

In the odd order situation, you really only need one of the
facet-center stickers on each axis for a reference. Thus
there are 5 (or the dimension of the puzzle) stickers for an
orientation reference in either case. To be consistent I
tend to favor the stickers facing in negative direction on the
facet-center cubies (for the odd order cases) and the
stickers on the corner cubie whose position coordinates are
all negative (for even order). It would actually be even
more consistent to base standard orientation on a corner
cubie for all orders, in which case you would count the
arrangements of facet-center cubies when they are present.
You get the same count either way.

It may be noted that, in Smith's formulae for odd order
cases, he does use 32 and not 31 for the corners.

Regards,
David V.



----- Original Message -----
From: Andrew Gould
To: 4D_Cubing@yahoogroups.com
Sent: Sunday, May 06, 2012 4:35 PM
Subject: RE: [MC4D] Calculating the number of permutation of
2by2by2by2by2 (2^5)

The choice between 31 and 32 comes down to how you define the
locations of pieces. If you define all their locations
relative to one of the pieces it's 31, but if you define what
moves and what doesn't for each twist you can make it 32. I
note that for 32, it would be tricky to say that rotating the
entire puzzle doesn't change the state.



--

Andy



----- Original Message -----
From: 4D_Cubing@yahoogroups.com
[mailto:4D_Cubing@yahoogroups.com] On Behalf Of
charliemckiz@rocketmail.com
Sent: Wednesday, May 02, 2012 12:46
To: 4D_Cubing@yahoogroups.com
Subject: [MC4D] Calculating the number of permutation of
2by2by2by2by2 (2^5)


2x2x2x2x2

(31!/2)(60^31) =

545356551753081970586352633891109632137647267774464
00000000000000000000000000000000000000

approx. = 5.4 x 10^88

Should the 31 be 32?

Mine is

32!/2*(5!/2)^(32) =

104708457936591738352579705707093049370428275412697088
000000000000000000000000000000000000000

approx. = 1.0 x 10^92

I did not download Melinda's message until after I posted my
own preceding. Thus mine was not written as a response to
Melinda. Nevertheless, there is a sense in which it does
respond:

I would argue that the way one should identify colors is by
the direction in which the corresponding stickers face in
the initial state. Actual color, as in red vs. yellow, is
really just a rendering issue and does not bear on the
nature of the puzzle. So, in this sense, changing the color
used for a given direction is not changing anything at all.

Now you could take it that reassigning the colors
corresponds to redefining the standard orientation of the
pile; and I am guessing that Melinda may have been thinking
of it in this sense. However, you would have to be careful
about that, as the pair of colors facing in opposite
directions on a given axis would have to remain the same on
any axis you might use for the pair in initial state.
Otherwise, it would be difficult to give 'physical' meaning
to a rearrangement of sticker colors. With that
restriction, we come back to the n!*2^(n-1) expression,
where n is the dimension. The 2n colors have to retain
their pairings, so that the 2^n factor corresponds to
choosing which of each pair faces in positive direction on
their axis. The n! factor comes from ways of associating
the color pairs with axes. (The factor of 1/2 is to count
only orientations and to exclude ways that create mirror
images.)

In this case, where the issue is to count all possible
distinct arrangements; I think the approach of counting only
arrangements when the pile has a standard orientation and
identifying the colors by the directions they face in
initial state is all that's needed.

For really gory details, check out this article from way
back in '94:
http://www.math.rwth-aachen.de/~Martin.Schoenert/Cube-Lovers/Dan_Hoey__The_real_size_of_cube_space.html
or http://preview.tinyurl.com/ys3pf2
which is referenced, taking a simplified point of view, from
here: http://www.cube20.org/.

Though the above is just for the regular 3D puzzle, the
basic idea remains the same. In their approach, they
admitted not just orientations of the pile (of which there
are 24 in the 3D case) but also more general symmetry
transformations - i.e., including transformations which
create mirror images - for a total of 48 (the full 3!*2^3).
From a computational point of view that makes perfectly good
sense. From a point of view of manipulating the puzzles, it
does not make sense unless one's implementation allows
reflections of the pile.

Regards,
David V.


----- Original Message -----
From: Melinda Green
To: 4D_Cubing@yahoogroups.com
Sent: Sunday, May 06, 2012 6:39 PM
Subject: Re: [MC4D] Calculating the number of permutation of
2by2by2by2by2 (2^5)


I may have answered my own question which is that to account
for color symmetry we can simply divide by (n - 1)! where n
is the number of colors. For a 2-colored puzzle there is
only one color permutation whereas for 3 color puzzles there
are two because we can fix one color and either swap or not
swap the other two. (n - 1)! gives the number of unique
color swapping patterns. Is that right? I usually don't
expect to fully understand the equations here.

-Melinda

On 5/6/2012 4:18 PM, Melinda Green wrote:

I feel that it's not just tricky but it is wrong in most
conceptualizations of the idea of puzzle state
spaces. Taking this natural idea one step further, I
would argue that states that have identical patterns of
stickers should be thought of as the same state. For
example, if you scramble any twisty puzzle and then swap
all red and green stickers, then I feel that you still
have the same state in terms of permutations since
anything you can say about one version also applies to
the other. For example, twist one face of a Rubik's
cube. For our purposes, it doesn't matter which face was
twisted. When talking about that state with each other
we will never think to ask about the particular colors.

Would anyone like to attempt to find the formula for the 3D
and 4D cubes with this extra "color symmetry" constraint?

-Melinda

On 5/6/2012 2:35 PM, Andrew Gould wrote:

The choice between 31 and 32 comes down to how you
define the locations of pieces. If you define all
their locations relative to one of the pieces it's
31, but if you define what moves and what doesn't
for each twist you can make it 32. I note that for
32, it would be tricky to say that rotating the
entire puzzle doesn't change the state.

Still out of phase. Significant redundancy between my last
post and Roice's. :-(

It dawns on me that the problem the cube20.org folks were
addressing is deeper than the one David Smith was addressing
(and which I was defending) and more like what Melinda was
probably driving at. I.e., given two arrangements that are
visually different when the pile is viewed in standard
orientation and based on the sticker IDs inferred from
initial state as in Smith's calculations, when can those two
still be regarded as the 'same' based on a symmetry of the
pile? This is where conjugation by a symmetry enters the
picture, and that can be viewed in a sense which remaps axis
IDs (along with the corresponding sticker 'color' pairs).
As may be seen from the '94 Cube Lovers article, the precise
issue gets very messy and David Smith was not trying to
address this more complex version. What Smith was
calculating is much more straightforward, is still
meaningful, and is more easily understood. The deeper view
reduces the "real size" by a factor less than n!*2^n (but
not quite n!*2^n), which is actually rather minor compared
to the sizes of the numbers Smith was computing. In my
view, the main point of his efforts is not the precise
values (which are relatively useless) but to get some feel
for just how immense these numbers are. They are
incomprehensibly large with or without the factor of n!*2^n.
The little ol' order 3 3D problem was close enough to being
tractible that a factor of nearly 48 actually did make a big
difference in the ultimate analysis there.

One thing that should be pointed out is that, if you admit
mirroring transformations, then it does create problems for
the purpose of counting distinct arrangements. The
problem is that there exist (a small minority of)
arrangements which possess mirror symmetry, so that the very
same arrangement would occur for two different symmetry
transformations of the pile. Accounting for the number of
distinct positions possessing such symmetry is not at all
easy; so the basic counting problem is much easier if you do
not admit reflecting transformations for achieving standard
orientation. But the cube20.org folks were not interested
so much in _counting_ arrangements as in analyzing relations
between explicit instances of state, so the extra complexity
that gained another factor of 2 was well worth it for them.

Regards,
David V.


----- Original Message -----
From: Roice Nelson
To: 4D_Cubing@yahoogroups.com
Sent: Sunday, May 06, 2012 8:00 PM
Subject: Re: [MC4D] Calculating the number of permutation of
2by2by2by2by2 (2^5)

Hi Melinda,


To help solve the "God's Number" problem, the cube20 guys
used the trick of considering states that only differ by a
symmetry of the cube to be the same. I think this is
precisely what you are describing. Their site links to a
cube lovers post titled "The real size of cube space", which
writes:

...

Melinda wrote:
>David: Regarding your comments, I didn't completely follow
>your description but I do not think we are talking about the
>same thing but I could easily misunderstand. Here is a way
>that I think might clarify what I am saying: Imagine that
>you toss me a scrambled Ribik's cube. I then pick two
>sticker colors, peal them off the cube and swap them. I may
>repeat that process several times and then toss it back to
>you. I claim that I've not changed your puzzle state at
>all.

I agree with that. That was the point I was making about
the specific colors just being a rendering issue. The
assignment of colors to directions in initial state (or a
subsequent state resulting from some specific twist
sequence) does not change the nature of the puzzle one whit.

>You said:

>"you could take it that reassigning the colors corresponds
>to redefining the standard orientation of the pile."

>This is not what I am saying.

Sorry. I did not claim that you were. The sentence is in
subjunctive mood. I was searching for a meaningful
something that you might have meant. Nevertheless, whether
you want to agree yet or not, I do still believe that, for
what you are trying to do to work right, it _does_ need to
correspond to a reorientation of the pile.

>If for example I take a pristine Rubik's cube and swap the
>colors of adjacent faces I will get a coloration that you
>cannot duplicate by reorienting the puzzle.

I don't see the relevance of this sort of recoloring, as it
does not change the puzzle at all. Not only do there exist
such recolorings that cannot be achieved by reorientation
(or even twisting), there is no motivation for wishing to be
able to do so.

The relevant issue would be to consider two puzzles which
started with the same color scheme and consider different
arrangements of the these two puzzles. Now you may ask, "Is
there a way to remap the sticker colors on one of them
(still maintaining, relative to initial state, the same
opposing pairs for any given axis), so that the resulting
pattern of sticker colors is identical?" If so, we can
regard the two patterns, which initially appeared to be
different based on the particular sticker colors as
presented, as being the same for all practical purposes. I
think this is the sort of equivalence you seek.

It is possible to explain the above approach in terms of
conjugation, without having to worry about specific colors
(which I claim are irrelevant). The way to do it is to
regard a state as being that which results from a particular
permutation of the stickers relative to their positions in
the space (no matter what colors). Now a symmetry
transformation of the cube, when applied to the whole pile,
can also be regarded as inducing a permutation of the
stickers. So consider two states A and B resulting from
permutations Pa and Pb. If there is a permutation Ps which
results from a symmetry transformation of the n-cube such
that Pa = Ps*Pb*Ps' (' for inverse, * for composition), then
we say that the states A and B are equivalent. Indeed, they
are equivalent under conjugation by a symmetry.

One way to understand the above is as follows. Suppose I
have a twist sequence that generates state B starting from
initial state, and I represent that sequence in a fixed
system of space coordinates. Now suppose, starting from
initial state, that I reorient the whole puzzle using the
symmetry transform that produces permutation Ps, and then
apply my twist sequence with the (otherwise initialized)
pile in that different orientation. Then I put the pile
back in its original orientation by applying the inverse of
the symmetry transformation to the whole pile. The
resulting permutation of stickers is now Pa, resulting in
state A. There is a strong sense in which the very same
twist sequence produces both states A and B.

Coming back to Melinda's sticker color point of view, one
can perceive the reorientation of the puzzle before applying
the twist sequence as a reassignment of the colors; as this
is indeed the effect it has on the resulting state in the
sense in which I think she wishes to compare them. (But do
note that you cannot violate the opposing color pair matchup
by simply reorienting the pile.) In dimensions higher than
3, the possibilities for reorienting the pile become much
more complex.

>Your permutations are therefore only a subset of it's color
>symmetry. Agreement or not, all three of us seem to come
>up with different numbers. Can someone set us straight?

Melinda, I don't know about "different numbers", but I think
I have clarified the issue of what it means for arrangements
to be equivalent under conjugation by a symmetry transform.
Furthermore, I do think that it is the same basic idea that
you are trying to get at with your concept of rearranging
sticker colors. Not only is the conjugation approach more
amenable to mathematical treatment, but, as far as I am
concerned, the conjugation approach is much clearer because
there is no need to talk about specific colors. They can be
whatever they are as determined by the direction they face
in initial state. Your reassignment corresponds to what
results from the reorientation by the symmetry transform.
It is better to view that as primarily a permutation of
coordinate axes with an associated direction sign on each
axis, so as to prevent violating the opposing sticker color
pair matchup. I do believe that the process is meaningless
if you do violate the opposing pair matchups, as there is no
corresponding 'physical' process achievable with the
puzzles. (E.g., as you observed, there is no linear
transformation of n-space that would swap the stickers on
just two adjacent facets of a puzzle.)

I do not think there is any disparity between what I am
saying and what Roice was saying. I am hoping that you can
understand it so that all three of us can be in agreement.
If you really believe that you are driving at some sort of
equivalence that goes beyond conjugation by a symmetry, then
I believe the burden is on you to spell it out in much
greater detail, identify its utility, and make it meaningful
in terms of the 'mechanics' of the puzzles.

Regards,
David V.

Brandon wrote:
> You can think of re-orienting a puzzle as permuting the
> colors in some way. For example, on a cube if you rotate
> the whole cube about the U face you've done the equivalent
> of a 4-cycle of the face colors in the equator between the
> U and D faces. In that sense there are 24 ways to
> "permute" the colors on a Rubik's cube. The way the
> calculation is done though avoids counting this extra
> factor of 24. On the Rubik's cube it is easy to avoid
> this extra factor of 24 because it has the centers.

Thank you, Brandon, for echoing exactly what I was
originally trying to explain to Melinda. However, you have
overlooked the same issue that I did on first pass. What
she is driving at is that there is a sense in which certain
arrangements, which appear to be distinct based on the
particular arrangement of sticker colors (even when
correcting for pile orientation), may be very closely
related if you are allowed to reassign the colors. The
relevant sense of similarity that Roice and I have tried to
point her to is conjugation by a symmetry. This is not the
same thing as reorientation of the pile, as it does produce
equivalent arrangements which cannot be made to match up
(sticker-color-wise) by straight reorientation. Please
refer to my last post, which attempts to explain this
distinction in greater detail.

Regards,
David V.


----- Original Message -----
From: "Brandon Enright"
To: <4D_Cubing@yahoogroups.com>
Cc: ;
Sent: Sunday, May 06, 2012 9:45 PM
Subject: Re: [MC4D] Calculating the number of permutation of
2by2by2by2by2 (2^5)


On Sun, 06 May 2012 19:13:13 -0700
Melinda Green wrote:

> I think that your interpretation of my post is correct
> though I wonder about the quantities. For example, I would
> factor out (6-1)! = 120 from the 3D cube due to color
> symmetry, and (8-1)! = 5,040 for the 4D cube but then I've
> not yet read your citations.

Hi Melinda,

Regarding the factoring out the color permutations, it would
only be correct to do so in cases where it is possible to
actually "swap" two colors.

For example, on the 3x3x3 the centers of each face are fixed
relative to each other. You can never solve the puzzle such
that the white face is adjacent to the yellow face. Even if
you took away the stickers on the centers (void cube) you
still can't because the 2-color edges and 3-color corners
define how the colors are related to each other.

You can think of re-orienting a puzzle as permuting the
colors in some way. For example, on a cube if you rotate
the whole cube about the U face you've done the equivalent
of a 4-cycle of the face colors in the equator between the U
and D faces. In that sense there are 24 ways to "permute"
the colors on a Rubik's cube. The way the calculation is
done though avoids counting this extra factor of 24. On the
Rubik's cube it is easy to avoid this extra factor of 24
because it has the centers.

For the 2x2x2 cube, the color scheme is still fully defined
by the corners but it doesn't have centers. There are two
basic ways to calculate the number of distinct states for
puzzles like the 2x2x2. The first is to over count and then
divide by the total orientations, and the second is to fix
one of the pieces in space, call it solved, and then count
how all the rest of the puzzle can be permuted about this
piece.

There are some puzzles that can be solved into their mirror
color scheme (such as the Dino Cube) and some puzzles that
can be solved into any permutation of their colors such as
the Big Chop (and Little Chop). In the case where a puzzle
can be solved into the mirror state you'd divide by 2 and in
the case where all permutations of colors are available it
would be #colors!. I'm sure there are puzzles that can only
be solved into an even permutation of the colors so those
would be #colors! / 2.

Brandon

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On Sun, May 6, 2012 at 9:45 PM, Brandon Enright wrote:

> There are some puzzles that can be solved into their mirror color
> scheme (such as the Dino Cube) and some puzzles that can be solved into
> any permutation of their colors such as the Big Chop (and Little
> Chop). In the case where a puzzle can be solved into the mirror state
> you'd divide by 2 and in the case where all permutations of colors are
> available it would be #colors!. I'm sure there are puzzles that can
> only be solved into an even permutation of the colors so those would be
> #colors! / 2.
>


cool! I didn't know that, so thanks for sharing :)

In dividing by these numbers, are you calling these different pristine
positions "the same", just like we call a solution to the Rubik's Cube the
same when you reorient the whole puzzle by 90 degrees? To me, a mirroring
(or other permutation) of the coloring scheme feels like a fundamentally
different state, because you must do some twisting to get from one of these
pristine states to another.

Intriguing stuff!
Roice

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Melinda wrote:
>It sounds like we are getting closer to closure.

Does this mean that you now understand why you must maintain
the pairs of opposing colors when reassigning sticker
colors? With all my effort at trying to explain it for you,
I was hoping for an "I see now!" I was also hoping that you
would realize that it is easier to talk about conjugation by
a symmetry than to get involved in talking about remapping
sticker colors, which are actually somewhat tangential to
the real problem.

>The main MC4D page claims to give the exact counts for the
>3D and 4D puzzles. Is that not true? If not, then I need to
>change it.

It is true. Also the counts that David Smith came up with
for 5D are intended to be exact. (I am not ruling out the
possibility of an error.) As I said, the problem that Eric
and David were solving is much more straightforward and
admits a precise answer. The problem only arises when you
start trying to get the number of equivalence classes of the
distinct states they counted that are related through
conjugation by a symmetry, which is the complication that it
would seem you were trying to inject. That would result in
a reduction of Eric's and David's numbers by a factor close
to (but less than) n!*2^n. But it is not practical to do
that in an exact sense.

>Please help me to understand the part that is "not
>quite". Here is what I am hearing: The main thing that makes
>it tricky to be exact seems to involve mirror symmetry. In
>particular, we would need to account for some relatively
>rare cases of states that happen to have mirror symmetry,
>and that counting those special cases appears to be very
>difficult.

That is my understanding; but I must confess that I have not
myself fully absorbed all the gory details of the "real
size" article. There may be other relevant symmetries
besides mirroring.

>If it can't be done then it can't be done. If that is the
>case, then what is our best estimate of the truly unique
>positions when accounting for color and mirror symmetries,
>and in particular, what is the term that we need to divide
>by and how is it derived?

I say don't try. Stay with the distinct arrangement counts
we have. They are meaningful. Since you have an interest
in the equivalence classes of these arrangements under
conjugation by a symmetry, you might want to add the comment
that the number of such equivalence classes will be less by
a factor close to n!*2^n. Regarding the nature of the
issue, you could just link to the "real size" article,
saying that the analogous problem exists for any dimension.

You have already worked out a simple example giving
representative members of the equivalence classes for the
case of the 2D puzzle with mirroring twists allowed. I
would hope you would remember the previous occasion on which
some folks on the list tried to explain the relationship of
your equivalence classes to conjugation by a symmetry:
http://games.groups.yahoo.com/group/4D_Cubing/message/1843
Perhaps that previous discussion will make better sense to
you now.

Regards,
David V.

Melinda wrote:
> I'm sorry that I have exasperated you. I really don't mean to
> argue
> anything and I believe that I get what you are saying. I think
> the
> tension is due to the fact that each of us is interested in
> related but
> different concepts here. In the 2D case you see 24 distinct
> states in a
> way that is both mathematically clean, meaningful and
> interesting to
> you. I on the other hand see only 8 states that are interesting
> to me,
> however messy they may be to count.

I am interested both in the number of distinct states as
determined by
the positions and orientations of the cubies as well as the
equivalence classes of those states under conjugation by a
symmetry.
Your "interest" in the equivalence classes under conjugation (or
whatever you have in mind) seems to go so far that you would
actually
exclude the distinct states on which the published counts are
based - to the extent that you expressed concern that the
published
numbers are in error. I remain exasperated that you do not seem
to be
willing to embrace both the permutations themselves as well as
their
equivalence classes related by conjugation. BOTH are worth
studying,
and it _starts_ with the distinct permutations from which the
equivalence classes can be derived. To take the attitude that
our
apparent disagreement stems from some personal focuses of
interest is
absurd. If you are so focussed on certain equivalence classes,
it
would behoove you to try to understand the issues in a manner
better
than than your confused writings have indicated so far.

It occurred to me earlier today that it may well be that you lack
adequate grasp of the group theory involved - as I cannot imagine
that, if you really understood the concept of conjugation in the
context of permutations of a finite set, you would remain as
confused
as you apparently have. I urge you to study a little group
theory as
it relates to permutations. In particular, learn about
representing a
permutation in terms of its disjoint cycles, what cycle structure
is,
and why permutations with the same cycle structure must be
conjugates
of one another. This is not deep stuff. It is basic group
theory.
You should not be sticking your head in the sand every time
somebody
says "conjugation"! And please note that there are definitely
others
besides myself who have tried to point you in the direction of
conjugation. I had merely been a little more patient about it -
up to
now.

> Regarding recolorings, please don't think that I care about the
> particular colors. I only care about the patterns that they
> create.
> Perhaps we can clear this up by talking about those color pairs
> that are
> important to you. Imagine a pristine cube. If you swap two
> opposite sets
> of stickers, then you feel that nothing has changed even though
> you
> can't rotate the whole cube to match the original colors.

The symmetry in that case is reflection in the zero plane of the
axis
on which you swapped the sticker colors.

>(Perhaps you mean to allow only an even number of pair swaps.)

No. Swapping just one opposing pair is OK.

>For the things that I care about, definitely nothing has changed
>when
>you do that because the puzzle is still in what I consider to be
>*the* solved state which I define as the one in which all sides
>have
>a single different color.

You seem to be losing sight of the fact that you want to use
color
remapping on different piles in permuted states to make them
match up.
I agree that doing it on a single pile that is in initial state
does
not mean much. But, when you do it on piles, which started out
with
the same color scheme but which have been permuted in distinct
ways,
with the purpose of relating those permutations, you have to go
about
it in such a way that you do not create two different puzzles.
I.e.,
both piles must still contain the same set of cubies as
determined by
their sticker colors. Going back to the basic Rubik example, if
you
exchanged the blue and yellow colors without also swapping white
and
green, then the edge cubie that had been white/blue would become
white/yellow, and the puzzle is not supposed to contain any edge
cubies colored that way. It is no longer meaningful to compare
permutations between puzzles that differ in such a profound way.

>For that matter I am also perfectly happy swapping the stickers
>of
>two adjacent faces because that gives the same result as when
>swapping opposite sides. The puzzle remains solved. Unless I've
>just
>said something whacky, I think we can leave it at that.

Unfortunately, you have indeed said something wacky because what
you
have described does not relate in any valid way to the method
which I
think you propose to define your notion of distinct states. (I
have
to be vague about this because you have never actually spelled
your
method out unambiguously. I have been trying to help you make it
meaningful by urging you to add the opposing pair restriction,
which
addition you resist. Nevertheless, adding that restriction makes
it
equivalent to conjugation by a symmetry, which is on a sound
basis and
does not even require talking about colors.) It's wacky either
because it is trivial (and not worth saying in the first place)
or
because it is meaningless for whatever purpose you would seek to
employ it.

Earlier in this thread I wrote:

If you really believe that you are driving at some sort of
equivalence that goes beyond conjugation by a symmetry, then
I
believe the burden is on you to spell it out in much greater
detail, identify its utility, and make it meaningful in terms
of
the 'mechanics' of the puzzles.

You have not made any credible attempt to do so. Furthermore -
and to
put it bluntly - I don't believe that you can.

Regards,
David V.

I accidentally composed my last message in this thread
with a line length that was too long for the way I have
set up my email program for plain text. Thus it really
trashed the formatting to an extent that I imagine a lot
of folks would not want to read. Thus I am reposting
the message with a line length that will render
properly. Sorry about the first mess.


Melinda wrote:
> I'm sorry that I have exasperated you. I really don't
> mean to argue anything and I believe that I get what you
> are saying. I think the tension is due to the fact that
> each of us is interested in related but different
> concepts here. In the 2D case you see 24 distinct states
> in a way that is both mathematically clean, meaningful
> and interesting to you. I on the other hand see only 8
> states that are interesting to me, however messy they may
> be to count.

I am interested both in the number of distinct states as
determined by the positions and orientations of the cubies
as well as the equivalence classes of those states under
conjugation by a symmetry. Your "interest" in the
equivalence classes under conjugation (or whatever you have
in mind) seems to go so far that you would actually exclude
the distinct states on which the published counts are based
- to the extent that you expressed concern that the
published numbers are in error. I remain exasperated that
you do not seem to be willing to embrace both the
permutations themselves as well as their equivalence
classes related by conjugation. BOTH are worth studying,
and it _starts_ with the distinct permutations from which
the equivalence classes can be derived. To take the
attitude that our apparent disagreement stems from some
personal focuses of interest is absurd. If you are so
focussed on certain equivalence classes, it would behoove
you to try to understand the issues in a manner better than
than your confused writings have indicated so far.

It occurred to me earlier today that it may well be that
you lack adequate grasp of the group theory involved - as I
cannot imagine that, if you really understood the concept
of conjugation in the context of permutations of a finite
set, you would remain as confused as you apparently have.
I urge you to study a little group theory as it relates to
permutations. In particular, learn about representing a
permutation in terms of its disjoint cycles, what cycle
structure is, and why permutations with the same cycle
structure must be conjugates of one another. This is not
deep stuff. It is basic group theory. You should not be
sticking your head in the sand every time somebody says
"conjugation"! And please note that there are definitely
others besides myself who have tried to point you in the
direction of conjugation. I had merely been a little more
patient about it - up to now.

> Regarding recolorings, please don't think that I care
> about the particular colors. I only care about the
> patterns that they create. Perhaps we can clear this up
> by talking about those color pairs that are important to
> you. Imagine a pristine cube. If you swap two opposite
> sets of stickers, then you feel that nothing has changed
> even though you can't rotate the whole cube to match the
> original colors.

The symmetry in that case is reflection in the zero plane
of the axis on which you swapped the sticker colors.

>(Perhaps you mean to allow only an even number of pair
>swaps.)

No. Swapping just one opposing pair is OK.

>For the things that I care about, definitely nothing has
>changed when you do that because the puzzle is still in
>what I consider to be *the* solved state which I define as
>the one in which all sides have a single different color.

You seem to be losing sight of the fact that you want to
use color remapping on different piles in permuted states
to make them match up. I agree that doing it on a single
pile that is in initial state does not mean much. But,
when you do it on piles, which started out with the same
color scheme but which have been permuted in distinct ways,
with the purpose of relating those permutations, you have
to go about it in such a way that you do not create two
different puzzles. I.e., both piles must still contain the
same set of cubies as determined by their sticker colors.
Going back to the basic Rubik example, if you exchanged the
blue and yellow colors without also swapping white and
green, then the edge cubie that had been white/blue would
become white/yellow, and the puzzle is not supposed to
contain any edge cubies colored that way. It is no longer
meaningful to compare permutations between puzzles that
differ in such a profound way.

>For that matter I am also perfectly happy swapping the
>stickers of two adjacent faces because that gives the same
>result as when swapping opposite sides. The puzzle remains
>solved. Unless I've just said something whacky, I think
>we can leave it at that.

Unfortunately, you have indeed said something wacky because
what you have described does not relate in any valid way to
the method which I think you propose to define your notion
of distinct states. (I have to be vague about this because
you have never actually spelled your method out
unambiguously. I have been trying to help you make it
meaningful by urging you to add the opposing pair
restriction, which addition you resist. Nevertheless,
adding that restriction makes it equivalent to conjugation
by a symmetry, which is on a sound basis and does not even
require talking about colors.) It's wacky either because
it is trivial (and not worth saying in the first place) or
because it is meaningless for whatever purpose you would
seek to employ it.

Earlier in this thread I wrote:

If you really believe that you are driving at some sort
of equivalence that goes beyond conjugation by a
symmetry, then I believe the burden is on you to spell
it out in much greater detail, identify its utility,
and make it meaningful in terms of the 'mechanics' of
the puzzles.

You have not made any credible attempt to do so.
Furthermore - and to put it bluntly - I don't believe that
you can.

Regards,
David V.