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David, thank you for the excellent email response on the parity discussion
with Levi (and for the fix of my incorrect 4-cycle claim about corners on
MC5D). It did indeed provide more insight for me by laying things out for
general dimensions, and I think I have a good picture of things, with
confidence to look at configurations and determine their possibility. (I'd
like to soon use this new found knowledge gained from Levi and you to
proveto myself that my experimental
conclusions on the possible 4D
checkerboards
right.)
Anyway, I ran across the following small surprise thinking about the parity
discussion (though it isn't related to that), and figured I'd share with
everybody...
A d-dimensional Rubik's cube with n cubies per side has the same number of
cubies as stickers when n is equal to the number of faces (that is, when n =
f = 2d). This occurs when n = 6 on a 3D cube, when n = 8 on a 4D cube, and
so on for the 10^5, 12^6, 14^7, etc. The math showing this is simple. You
just solve for n in the following equation, where the left side represents
the number of cubies and the right the number of stickers.
n^d = 2d * n^(d-1)
when n < 2d, the number of stickers is greater than the number of pieces.
when n > 2d, the number of stickers is less than the number of pieces.
This reversal comes into play since the number of hidden, 0-colored cubies
grows more quickly with increasing n than other piece types, adding to the
piece count but not to the sticker count. 1C cubie numbers are a wash for
the difference of the two counts. Higher colored cubies, which contribute
more stickers than pieces, don't grow in number fast enough to keep up with
the 0Cs, even with all 2C ... dC cubie types combined.
Why do the number of 0-colored pieces grow so much faster than the others,
even taken together? Consider a puzzle with very large n. In the limit,
0-coloreds are the only piece type that are filling up the full dimension of
the cube. 1-coloreds fill up the faces of dimension d-1, 2-coloreds fill up
the d-2 spaces, etc. And higher dimensional spaces are more voluminous, so
it makes sense 0C will win out in the end.
I found n = f neat because a priori, why should the number per side have any
relationship whatsoever to the the number of faces? (maybe this surprise is
just the fact that the number of faces = 2d in disguise.) I also wonder why
the existence of puzzles where the number of stickers and cubies coincide
should even be guaranteed, another fact not a priori obvious to me. A
non-existence conclusion that can be drawn is that no puzzle (of any
dimension) with odd n can have the the same number of stickers and cubies,
since the n = 2d constraint would imply fractional dimension.
Take Care All,
Roice
P.S. I want to defer to the group on the use of m^n verses n^d. In this
email, I wanted to say "n-dimensional" at one point, but that would have
conflicted with my usual labels. It got me second-guessing myself. Any
opinions? Maybe we should make a poll?
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sion with Levi (and for the fix of my incorrect 4-cycle claim about corners=
on MC5D). It did indeed provide more insight for me by laying things=
out for general dimensions, and I think I have a good picture of things, w=
ith confidence to look at configurations and determine their possibility. (=
I'd like to soon use this new found knowledge gained from Levi and you =
to provean> to myself that my ing/message/477">experimental conclusions on the possible 4D checkerboards<=
/a> were right.)
ing about the parity discussion (though it isn't related to that), and =
figured I'd share with everybody...
umber of cubies as stickers when n is equal to the number of faces (that is=
, when n =3D f =3D 2d). This occurs when n =3D 6 on a 3D cube, when n=
=3D 8 on a 4D cube, and so on for the 10^5, 12^6, 14^7, etc. The mat=
h showing this is simple. You just solve for n in the following equat=
ion, where the left side represents the number of cubies and the right the =
number of stickers.
ieces.
when n > 2d, the number of stickers is less than the number of=
pieces.
bies grows more quickly with increasing n than other piece types, adding to=
the piece count but not to the sticker count. 1C cubie numbers are a wash =
for the difference of the two counts. Higher colored cubies, which co=
ntribute more stickers than pieces, don't grow in number fast enough to=
keep up with the 0Cs, even with all 2C ... dC cubie types combined.
ers, even taken together? Consider a puzzle with very large n. =
In the limit, 0-coloreds are the only piece type that are filling up the fu=
ll dimension of the cube. 1-coloreds fill up the faces of dimension d-1, 2-=
coloreds fill up the d-2 spaces, etc. And higher dimensional spaces a=
re more voluminous, so it makes sense 0C will win out in the end.
have any relationship whatsoever to the the number of faces? (maybe this su=
rprise is just the fact that the number of faces =3D 2d in disguise.) =
I also wonder why the existence of puzzles where the number of stickers an=
d cubies coincide should even be guaranteed, another fact not a priori obvi=
ous to me. A non-existence conclusion that can be drawn is that no pu=
zzle (of any dimension) with odd n can have the the same number of stickers=
and cubies, since the n =3D 2d constraint would imply fractional dimension=
.
Roice
to the group on the use of m^n verses n^d. In this email, I wanted to=
say "n-dimensional" at one point, but that would have conflicted=
with my usual labels. It got me second-guessing myself. Any op=
inions? Maybe we should make a poll?
--0016360e3bd6c941d10461fcb602--
That's a delightful finding. If n=3D2d, then stickers=3Dcubies. I've=20
always thought the 6^3 had a sort of hidden beauty to it.
As for the m^n or n^d. My preference for m^n stems from my use of=20
single letters to label my algoriths, from a to g. But n^d uses more=20
descriptive variable names. I think developing a standard is=20
essential for simplifying discussions.=20
-Levi
--- In 4D_Cubing@yahoogroups.com, Roice Nelson
>
> David, thank you for the excellent email response on the parity=20
discussion
> with Levi (and for the fix of my incorrect 4-cycle claim about=20
corners on
> MC5D). It did indeed provide more insight for me by laying things=20
out for
> general dimensions, and I think I have a good picture of things,=20
with
> confidence to look at configurations and determine their=20
possibility. (I'd
> like to soon use this new found knowledge gained from Levi and you=20
to
> proveto myself that my experimental
> conclusions on the possible 4D
>=20
checkerboards
> right.)
>=20
> Anyway, I ran across the following small surprise thinking about=20
the parity
> discussion (though it isn't related to that), and figured I'd share=20
with
> everybody...
>=20
> A d-dimensional Rubik's cube with n cubies per side has the same=20
number of
> cubies as stickers when n is equal to the number of faces (that is,=20
when n =3D
> f =3D 2d). This occurs when n =3D 6 on a 3D cube, when n =3D 8 on a 4D=20
cube, and
> so on for the 10^5, 12^6, 14^7, etc. The math showing this is=20
simple. You
> just solve for n in the following equation, where the left side=20
represents
> the number of cubies and the right the number of stickers.
>=20
> n^d =3D 2d * n^(d-1)
>=20
> when n < 2d, the number of stickers is greater than the number of=20
pieces.
> when n > 2d, the number of stickers is less than the number of=20
pieces.
>=20
> This reversal comes into play since the number of hidden, 0-colored=20
cubies
> grows more quickly with increasing n than other piece types, adding=20
to the
> piece count but not to the sticker count. 1C cubie numbers are a=20
wash for
> the difference of the two counts. Higher colored cubies, which=20
contribute
> more stickers than pieces, don't grow in number fast enough to keep=20
up with
> the 0Cs, even with all 2C ... dC cubie types combined.
>=20
> Why do the number of 0-colored pieces grow so much faster than the=20
others,
> even taken together? Consider a puzzle with very large n. In the=20
limit,
> 0-coloreds are the only piece type that are filling up the full=20
dimension of
> the cube. 1-coloreds fill up the faces of dimension d-1, 2-coloreds=20
fill up
> the d-2 spaces, etc. And higher dimensional spaces are more=20
voluminous, so
> it makes sense 0C will win out in the end.
>=20
> I found n =3D f neat because a priori, why should the number per side=20
have any
> relationship whatsoever to the the number of faces? (maybe this=20
surprise is
> just the fact that the number of faces =3D 2d in disguise.) I also=20
wonder why
> the existence of puzzles where the number of stickers and cubies=20
coincide
> should even be guaranteed, another fact not a priori obvious to=20
me. A
> non-existence conclusion that can be drawn is that no puzzle (of any
> dimension) with odd n can have the the same number of stickers and=20
cubies,
> since the n =3D 2d constraint would imply fractional dimension.
>=20
> Take Care All,
> Roice
>=20
> P.S. I want to defer to the group on the use of m^n verses n^d. In=20
this
> email, I wanted to say "n-dimensional" at one point, but that would=20
have
> conflicted with my usual labels. It got me second-guessing=20
myself. Any
> opinions? Maybe we should make a poll?
>
=20=20=20=20
David, thank you for the excellent email response on the parity=
discussion with Levi (and for the fix of my incorrect 4-cycle claim about =
corners on MC5D). =A0It did indeed provide more insight for me by laying th=
ings out for general dimensions, and I think I have a good picture of thing=
s, with confidence to look at configurations and determine their possibilit=
y. (I'd like to soon use this new found knowledge gained from Levi and you =
to prove to myself that my experimental conclusions on the possible 4D chec=
kerboards were right.)
Anyway, I ran across the following small surprise thinking about the parity=
discussion (though it isn't related to that), and figured I'd share with e=
verybody...
=A0
A d-dimensional Rubik's cube with n cubies per side has the same number of =
cubies as stickers when n is equal to the number of faces (that is, when n =
=3D f =3D 2d).=A0 This occurs when n =3D 6 on a 3D cube, when n =3D 8 on a =
4D cube, and so on for the 10^5, 12^6, 14^7, etc.=A0 The math showing this =
is simple.=A0 You just solve for n in the following equation, where the lef=
t side represents the number of cubies and the right the number of stickers=
.
=A0
n^d =3D 2d * n^(d-1)
=A0
when n < 2d, the number of stickers is greater than the number of pieces.
when n > 2d, the number of stickers is less than the number of pieces.
=A0
This reversal comes into play since the number of hidden, 0-colored cubies =
grows more quickly with increasing n than other piece types, adding to the =
piece count but not to the sticker count. 1C cubie numbers are a wash for t=
he difference of the two counts.=A0 Higher colored cubies, which contribute=
more stickers than pieces, don't grow in number fast enough to keep up wit=
h the 0Cs, even with all 2C ... dC cubie types combined.
=A0
Why do the number of 0-colored pieces grow so much faster than the others, =
even taken together?=A0 Consider a puzzle with very large n.=A0 In the limi=
t, 0-coloreds are the only piece type that are filling up the full dimensio=
n of the cube. 1-coloreds fill up the faces of dimension d-1, 2-coloreds fi=
ll up the d-2 spaces, etc.=A0 And higher dimensional spaces are more volumi=
nous, so it makes sense 0C will win out in the end.
=A0
I found n =3D f neat because a priori, why should the number per side have =
any relationship whatsoever to the the number of faces? (maybe this surpris=
e is just the fact that the number of faces =3D 2d in disguise.)=A0 I also =
wonder why the existence of puzzles where the number of stickers and cubies=
coincide should even be guaranteed, another fact not a priori obvious to m=
e.=A0 A non-existence conclusion that can be drawn is that no puzzle (of an=
y dimension) with odd n can have the the same number of stickers and cubies=
, since the n =3D 2d constraint would imply fractional dimension.
=A0
Take Care All,
Roice
P.S. I want to defer to the group on the use of m^n verses n^d. =A0In this =
email, I wanted to say "n-dimensional" at one point, but that would have co=
nflicted with my usual labels. =A0It got me second-guessing myself. =A0Any =
opinions? =A0Maybe we should make a poll?
=20=20=20=20=20=20
=20=20=20=20
=20=20=20=20
=09
=09=20
=09
=09
=09
=09
=09
=20=20=20=20=20=20
--0-251679478-1233690763=:54546
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| You're welcome, Roice! I'm glad my observations helped, and great to hear you are off to solve another problem. As for your observation, check another unique insight off the list for Roice! ;) I thought of a cool way to visualize why your fact must be true, and it could even be used to intuitively prove the result without solving the equation you gave. When the number of cubies per edge equals the number faces, this of course also means that the number of faces equals the number of layers ((n-1)-dimensional slices of an n-dimensional cube). Now just mentally reposition each sticker on a particular face to the inside of a corresponding cubie on a particular layer of the cube. Do this for each face, and we see that each sticker rests within each cubie, establishing a 1-to-1 correspondence of cubies to stickers. This is a lot easier to visualize than to describe in words. This "proof without words" can also be used to establish the inequalities you gave. As for the terminology of higher-dimensional Rubik's cubes, I prefer n^d, which is what I use in my formulas. (At least I will use d eventually! ;)) Strangely however, I also use the term n-dimensional, as you can see above. I believe the deference to n as the dimension (perhaps unconsciously) is because this is used almost everywhere for dimensions in mathematics. As an example, consider the standard term, "Euclidian n-space" for the set of all n-tuples of real numbers, denoted by R^n. Anyway, I hope this helps establish the a priori justification you were looking for, and good luck with your checkerboard problem! All the best, David --- On Tue, 2/3/09, Roice Nelson <roice3@gmail.com> wrote: From: Roice Nelson <roice3@gmail.com> |
On Monday, February 02, "Roice Nelson"
>Why do the number of 0-colored pieces grow so much
>faster than the others, even taken together?
>Consider a puzzle with very large n. In the limit,
>0-coloreds are the only piece type that are filling
>up the full dimension of the cube. 1-coloreds fill up
>the faces of dimension d-1, 2-coloreds fill up the
>d-2 spaces, etc. And higher dimensional spaces are
>more voluminous, so it makes sense 0C will win out in
>the end.
As I see it, the phenomenon here is just a quantized
scale effect. The cubies with any stickers at all
(without regard to how many stickers) are all on the
'surface'. The size of that surface (in cubie count)
is proportional (asymptotically) to n^(d-1).
Meanwhile, the others fill a volume the size of which
(again in cubie count) is proportional
(asymptotically) to n^d.
>I found n = f neat because a priori, why should the
>number per side have any relationship whatsoever to
>the the number of faces? (maybe this surprise is just
>the fact that the number of faces = 2d in disguise.)
Roice, I understand your surprise. It is similar to
my surprise about reorientations of a 3-cube: Every
one of the 23 possible reorientations is achievable by
rotation about one of the three types of rotation axis
(through the origin and through a corner, the middle
of an edge, or the middle of a face) and that axis is
uniquely determined for any reorienting
transformation.
I suspect there may be deeper insights which are
eluding us and which would make these relationships
appear more plausible.
>I also wonder why the existence of puzzles where the
>number of stickers and cubies coincide should even be
>guaranteed, another fact not a priori obvious to me.
Indeed.
>P.S. I want to defer to the group on the use of m^n
>verses n^d. In this email, I wanted to say
>"n-dimensional" at one point, but that would have
>conflicted with my usual labels. It got me
>second-guessing myself. Any opinions? Maybe we
>should make a poll?
I prefer m^n. The "nth dimension" is too well
ingrained in me. I, too, often wish to say
"n-dimensional". I also would tend to say something
like, "an order-5 4-puzzle", to imply 5^4.
Regards,
David V.
> I prefer m^n. The "nth dimension" is too well
> ingrained in me. I, too, often wish to say
> "n-dimensional". I also would tend to say something
> like, "an order-5 4-puzzle", to imply 5^4.
>=20
> Regards,
> David V.
>
Order-m n-puzzle, I've been trying to think of that terminology for a=20
while. Thank you! That'll simplify most of my posts in the future! I=20
think I prefer order-m n-cube, but n-puzzle is a little more=20
descriptive. However, I think with the existence of the magic 120-cell,=20
which is a 4-puzzle, we shouldn't assume the puzzle is an n-cube.
-Levi=20
On Tuesday, February 03, "rev_16_4"
>Order-m n-puzzle, I've been trying to think of that
>terminology for a while. Thank you! That'll simplify
>most of my posts in the future! I think I prefer
>order-m n-cube, but n-puzzle is a little more
>descriptive. However, I think with the existence of
>the magic 120-cell, which is a 4-puzzle, we shouldn't
>assume the puzzle is an n-cube.
I can debate that. I don't think that there is all
that much terminology from higher dimensional Rubik
analogues that applies directly to other twisty
puzzles. One can push generalization so far that you
wind up having to qualify the special cases too much.
I do not like to call an instance of a puzzle a "cube"
because it really isn't _one_. It is a pile of little
cubes with some constraints on what arrangements are
possible. Admittedly, the shape of the pile is
cubical; but it is its transformability that is of
real interest.
Regards,
David V.
On Tuesday, February 03, "David Vanderschel"
>On Monday, February 02, "Roice Nelson"
>>I found n = f neat because a priori, why should the
>>number per side have any relationship whatsoever to
>>the the number of faces? (maybe this surprise is just
>>the fact that the number of faces = 2d in disguise.)
>Roice, I understand your surprise. ...
>I suspect there may be deeper insights which are
>eluding us and which would make these relationships
>appear more plausible.
I take it back. This one is easy to see. I think I
can explain it in a way that is more obvious than what
David Smith wrote:
Given an order-m n-puzzle, we note that there is a 1-1
correspondence between the set of m^(n-1) stickers
which lie in a face and the m^(n-1) n-cubies which
make up the corresponding external slice. Consider
cloning that set of cubies and set them aside. Do
this for each of the 2n faces and pile up the cloned
slices. The height of the cloned-cubie pile is 2n and
the number of cubies in it is the same as the number
of stickers on the puzzle. The slices of the puzzle
are the same size ( m^(n-1) n-cubies ) as those in the
clone pile and its height is m cubies. So now
comparing the number of stickers to the number of
cubies is just a matter of comparing the heights (2n
and m) of the two piles.
It really does come down to the fact that the number
of cloned slices is the number of faces.
Regards,
David V.
--- In 4D_Cubing@yahoogroups.com, David Vanderschel
>
> On Tuesday, February 03, "rev_16_4"
> >Order-m n-puzzle, I've been trying to think of that
> >terminology for a while. Thank you! That'll simplify
> >most of my posts in the future! I think I prefer
> >order-m n-cube, but n-puzzle is a little more
> >descriptive. However, I think with the existence of
> >the magic 120-cell, which is a 4-puzzle, we shouldn't
> >assume the puzzle is an n-cube.
>=20
> I can debate that. I don't think that there is all
> that much terminology from higher dimensional Rubik
> analogues that applies directly to other twisty
> puzzles. One can push generalization so far that you
> wind up having to qualify the special cases too much.
>=20
> I do not like to call an instance of a puzzle a "cube"
> because it really isn't _one_. It is a pile of little
> cubes with some constraints on what arrangements are
> possible. Admittedly, the shape of the pile is
> cubical; but it is its transformability that is of
> real interest.
>=20
> Regards,
> David V.
>
David V.-
I would actually make the opposite argument. Attempting to make a=20
general formula for all possible permutations of every type puzzle in=20
every dimension would be nearly impossible (however David S. has made=20
great strides with the cubical variety). There are some things that=20
do generalize considerably better. In fact I think the=20
generalizations are what some of us find so interesting about these n-
puzzles.
For example, I have a strong feeling that the 120-cell is solvable=20
using the same caging method I use to solve order-m n-cubes, with=20
only slight modifications to the algorithms. This 4-puzzle has even=20
parity, which increases my confidence further.=20
I agree that the "cube" isn't a single cube, but in fact a composite=20
of many smaller pieces put together to make a cube. That's why I=20
think saying "order-m" is so powerful and perfect. We have m pieces=20
extending out along each of the edges to make up the whole cubical=20
puzzle. That the individual cubies just happen to be cubes is a=20
coincidence. This can be seen in the magic dodecahedron/120-cell.=20
Some are rhombic, some are triangular, and some are pentagonal. I see=20
the opposing argument like "It's not a beach, it's trillions of=20
grains of sand."
I think we are generalizing by saying n-puzzle, when it would be just=20
as easy to say n-cube, which is the ultimate shape of these puzzles=20
(MC4D & MC5D). The frequency with which the generalized terminology=20
is required makes this point trivial. You can safely pull the meaning=20
of n-puzzle out of the context from which it's taken.=20
Finally, while during a discussion of each class of n-puzzle, we may=20
not generalize between classes too frequently, if ever. However this=20
discussion itself shows the need for a terminology for the general n-
puzzles.
-Levi
On Wednesday, February 04, "rev_16_4"
>I think we are generalizing by saying n-puzzle, when
>it would be just as easy to say n-cube, which is the
>ultimate shape of these puzzles (MC4D & MC5D).
The problem with that is that we do need to be able to
talk about a simple n-cube also when we are talking
about an n-puzzle (not necessarily the same n's).
>The frequency with which the generalized terminology
>is required makes this point trivial. You can safely
>pull the meaning of n-puzzle out of the context from
>which it's taken.
Indeed. And for me the context is Rubik analogues, in
which context I do not include non-cubical puzzles.
>Finally, while during a discussion of each class of
>n-puzzle, we may not generalize between classes too
>frequently, if ever. However this discussion itself
>shows the need for a terminology for the general n-
>puzzles.
The problem is hard enough for the Rubik analogues
alone. Indeed, I agree about the need for some
agreement on terminology. Furthermore, I have a
proposal for some aspects of the terminology and
notation issues for Rubik analogues. You might want
to consider the extent to which something of the sort
could be extended to other puzzle types.
I am going to start a new thread for my notation and
terminology proposal as it is really a new subject and
not directly related to the current thread. However,
it is responsive to concerns expressed by both Levi
and Roice about the need to get together on language
and notation.
Regards,
David V.