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Hello all,
Okay, I'm going to really check what I say here before posting, it's a bit =
abstracted!
But if I have to correct myself, I guess that's okay.=A0 I realized earlier=
I had said that
this theorem applies for any piece, not just corners.=A0 It does, but it ne=
eds to be
modified in one of two ways, depending on whether the pieces are normals or=
wings
(see my Notation page on the wiki for a somewhat detailed explanation of th=
ese
terms), and it would be difficult to describe the difference here.=A0 So, I=
'll stick to
corners, and if anyone wants to know more, feel free to ask.=A0 Permutation=
s are
actually much easier to determine than orientations (so that as long as I
understand the puzzle, I pretty much have a determined procedure for evalua=
ting
the number of reachable configurations!), but there is an additional detail=
in that
multiple families of pieces can interact: when they all have odd permutatio=
ns, we
only divide by two once. (As in the normal 3^3 Rubik's Cube; the permutatio=
ns of
edges and corners bust both be odd or both be even.)
The following is a very generalized theorem regarding the orientations of t=
he corners
of virtually any n-dimensional puzzle for any n.=A0 For some rare cases, it=
will not
work, as the (very undemanding) conditions below need to be met for the the=
orem
to apply.=A0 Only puzzles with special rules will fail.=A0 The Gearcube/Gea=
rcube Extreme
is one example, as it fails because one can only rotate certain faces 180 d=
egrees,
and in fact the orientations of the corners of those puzzles never change.=
=A0 The
Latch cube would meet the conditions.=A0 Any n-dimensional physical puzzle =
with
no restrictions on moves will meet the conditions.=A0 Any MagicTile or Magi=
c
Hyperbolic Tile puzzle imaginable would satisfy the conditions.=A0 In short=
, it applies
to nearly every possible puzzle, in a very practical manner!
We will first state the definitions we will be using for our highly abstrac=
ted corners,
in order to state the conditions that must apply to them.=A0 This discussio=
n will be
somewhat technical, despite the fact that I will not prove the theorem, and=
I
apologize to anyone who finds it less than appealing.
Definitions:
We will unify all of the possible corners of any puzzle into a single conce=
pt.
In order to do this, we remove the corners from the puzzle and imagine them
floating in n-dimensional space.
Definition 1: A corner will be an n-dimensional hypersphere, n >=3D 3, floa=
ting at
a fixed position in n-dimensional space with n stickers on its surface.
Definition 2: A sticker will be a colored 0-dimensional point.
Explanation: We are removing the corners from the puzzle, turning them into
hyperspheres, and imagining the stickers as colored points.=A0 This is all =
done
without loss of generalization, as will be seen.=A0 Any set of corners on a=
ny puzzle
can be imagined to be hyperspheres, with the stickers replaced by points.=
=A0 When
we rotate the corners of a puzzle, we can imagine the hyperspheres moving i=
n the
same way, and the points on the hyperspheres=A0 occupying the appropriate p=
ositions.
In MagicTile, the pieces are 2-dimensional, but in fact pieces that are con=
sidered
to be corners in that program will in fact be isomorphic to 3-dimensional s=
pheres
satisfying the following properties and conditions.=A0 Similarly, in MagicT=
ile, we have
stickers on the inside of the pieces.=A0 This does not matter, it can be ea=
sily seen
that such corners are isomorphic to the 3-dimensional spheres with the poin=
ts on
their surface.
Properties:
The following two properties will describe the method by which our modified=
corners
will function in order to mimic the corners of any Rubik-like puzzle.=A0 Th=
ese properties
are not conditions of the actual puzzle we are considering; but in Conditio=
n 1 we will
see that these properties will have to be satisfied as a whole by our puzzl=
e, of course
as an isomorphic description.=A0 The first will automatically hold for any =
puzzle, the
other can appear to not hold when it actually does (in an isomorphic fashio=
n).
Property 1: Any two corners are a finite, nonzero distance from each
other.
Comments: Obviously!
Property 2: Each corner is the same size as the others, and has the same
number of stickers in the same positions.
Comments: This implies nothing about the sizes of the actual corners of the
puzzle; they could in fact be different sizes (such as in the mirror cube, =
to
which this theorem applies).=A0 This property is a necessity for the follow=
ing
conditions to make sense.
Conditions:
Condition 1: The corners of the puzzle we are applying the theorem to must =
be
isomorphic to the corners of Definition 1, with Properties 1 and 2 holding.
Comments: I cannot even think of a Rubik-like puzzle in which this conditio=
n does
not hold, but need to include it so that the theorem applies to any puzzle =
that
we are examining.
Condition 2: Every corner must have a set of stickers which are all of diff=
erent colors.
Comments: In other words, no corner can have two identical stickers.=A0 The=
re
is no restriction on more than one corner having the same set of stickers, =
for
example as in the 6-color megaminx.
Condition 3: The stickers on a corner must all lie on the vertices of a reg=
ular
simplex, and this simplex cannot lie in an (n-1)-dimensional hyperplane whi=
ch
bisects the corner into two equal halves.
Comments: The first portion of this condition helps ensure that the corner =
behaves
like the corners of a Rubik-like puzzle.=A0 That is, when a corner changes =
orientation
or moves, the stickers occupy positions that were previously occupied by st=
ickers.
The second portion prevents a technical issue in which all of the stickers =
would
lie on a 'great (n-1)-dimensional hypersphere', which could enable the corn=
er to
rotate in ways that would be unanalogous to a Rubik-like puzzle (and the th=
eorem
would not hold).
Condition 4: At regular intervals, the corners will go through a process ca=
lled a
move: Each member of a nonempty subset of them will move to the exact posit=
ion
of a corner that just left its position.=A0 Thus, a move permutes the corne=
rs in any
possible way.=A0 Additionally, each corner that moves must align itself so =
that each
of its stickers occupies a position previously occupied by a sticker.
Comments: This condition emulates a face rotation in a very nonrestrictive =
manner,
compared to what we usually think of as a face rotation moving corners, all=
owing
a huge variety of possible puzzles to satisfy the requirements of this theo=
rem.
We are told that any permutation of the corners is allowed to occur.=A0 The=
second
portion of this condition, along with the previous condition, ensures that =
stickers
always replace other stickers.=A0 Note that this condition does not say tha=
t the
permutation has to be the same at each point in time, which makes sense bec=
ause
in Rubik-like puzzles different face rotations produce different permutatio=
ns.=A0 One
very noteworthy fact to observe is that since the condition places no restr=
ictions
on the nature of the permutations, we can introduce our own restrictions in=
to
a Rubik-like puzzle, and the theorem would still hold.=A0 For example, we c=
ould
use a regular 3^3 Rubik's Cube, but state that the first move must swap two
adjacent corners, followed by swapping two corners along a face diagonal,
followed by swapping two corners on opposite ends of a space diagonal, and
repeat.=A0 As long as the puzzle satisfies the next condition on moves, any
possible restriction is allowable.
Condition 5: The corners must constitute a family; that is, there exists a
permissible sequence of moves such that every corner would occupy the posit=
ion
of every other corner using this sequence.=A0 Also, every corner would be r=
equired
to eventually occupy each position in every possible permutation of its sti=
ckers
given by the rotations of the corner.
Comments: The first portion of this condition means that each corner must
eventually have the freedom to have moved everywhere.=A0 This is naturally =
satisfied
by any 'regular' Rubik-like puzzle I can imagine.=A0 The second portion imp=
lies
that each orientation can occur freely in any position.=A0 This is the cond=
ition
that the Gearcube/Gearcube Extreme fails to meet, as it turns out each corn=
er
can only occupy one of the three possible orientations in each position.
The preparatory work has been accomplished, and we can finally present the
theorem!
Generalized Corner Orientation Theorem:=A0 If a permutation puzzle's corner=
s
satisfy Condition 1 by using k n-dimensional hyperspheres as corners, and
the isomorphic versions of the puzzle's corners and moves satisfy Condition=
s 2
through 5, then the number of orientations the corners of the puzzle can ac=
hieve
is:
((n!/2)^k)/3=A0 if n =3D 3 or 4
and
(n!/2)^k if n >=3D 5.
The proof relies heavily on others' work and thus I cannot really claim it =
as my
own.=A0 But neither the authors of The Rubik Tesseract nor An n-dimensional=
Rubik
Cube applied their results to anything other than the 3^n cube, so I made t=
he
generalization.
It took me over 3.5 hours to write this email (I had no formulation of the =
theorem
in my mind before I started; I had to come up with the entire sphere/point/=
simplex
idea as I wrote, and there were many parts rewritten or rephrased).=A0 Give=
n that,
I hope that this post has been interesting to some of you!=A0 I would appre=
ciate
any comments or questions you have for me.=A0 Thanks to Roice for the sugge=
stion!
All the best,
David
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| top" style=3D"font: inherit;">Hello all, Okay, I'm going to really c= heck what I say here before posting, it's a bit abstracted! But if I hav= e to correct myself, I guess that's okay. I realized earlier I had sa= id that this theorem applies for any piece, not just corners. It d= oes, but it needs to be modified in one of two ways, depending on whethe= r the pieces are normals or wings (see my Notation page on the wiki for = a somewhat detailed explanation of these terms), and it would be difficu= lt to describe the difference here. So, I'll stick to corners, and= if anyone wants to know more, feel free to ask. Permutations are = actually much easier to determine than orientations (so that as long as I uating the number of reachable configurations!), but there is an additio= nal detail in that multiple families of pieces can interact: when they all = have odd permutations, we only divide by two once. (As in the normal 3^3= Rubik's Cube; the permutations of edges and corners bust both be odd or= both be even.) The following is a very generalized theorem regardin= g the orientations of the corners of virtually any n-dimensional puzzle = for any n. For some rare cases, it will not work, as the (very und= emanding) conditions below need to be met for the theorem to apply. = ; Only puzzles with special rules will fail. The Gearcube/Gearcube Ex= treme is one example, as it fails because one can only rotate certain fa= ces 180 degrees, and in fact the orientations of the corners of those pu= zzles never change. The Latch cube would meet the conditions. = ; Any n-dimensional physical puzzle with no restrictions on moves will m= eet the conditions. Any MagicTile or Magic Hyperbolic Tile puzzle imaginable would satisfy the conditions. In short, it applies= to nearly every possible puzzle, in a very practical manner! We = will first state the definitions we will be using for our highly abstracted= corners, in order to state the conditions that must apply to them. = ; This discussion will be somewhat technical, despite the fact that I wi= ll not prove the theorem, and I apologize to anyone who finds it less th= an appealing. Definitions: We will unify all of the possible = corners of any puzzle into a single concept. In order to do this, we rem= ove the corners from the puzzle and imagine them floating in n-dimension= al space. Definition 1: A corner will be an n-dimensional hyperspher= e, n >=3D 3, floating at a fixed position in n-dimensional space with= n stickers on its surface. Definition 2: A sticker will be a colore= d 0-dimensional point. Explanation: We are removing the corners from the puzzle, turning them into hyperspheres, and imagining the stic= kers as colored points. This is all done without loss of generaliz= ation, as will be seen. Any set of corners on any puzzle can be im= agined to be hyperspheres, with the stickers replaced by points. When= we rotate the corners of a puzzle, we can imagine the hyperspheres movi= ng in the same way, and the points on the hyperspheres occupying t= he appropriate positions. In MagicTile, the pieces are 2-dimensional, bu= t in fact pieces that are considered to be corners in that program will = in fact be isomorphic to 3-dimensional spheres satisfying the following = properties and conditions. Similarly, in MagicTile, we have sticke= rs on the inside of the pieces. This does not matter, it can be easil= y seen that such corners are isomorphic to the 3-dimensional spheres wit= h the points on their surface. Properties: The following two properties will describe the method by which our modified co= rners will function in order to mimic the corners of any Rubik-like puzz= le. These properties are not conditions of the actual puzzle we ar= e considering; but in Condition 1 we will see that these properties will= have to be satisfied as a whole by our puzzle, of course as an isomorph= ic description. The first will automatically hold for any puzzle, the= other can appear to not hold when it actually does (in an isomorphic fa= shion). Property 1: Any two corners are a finite, nonzero distance f= rom each other. Comments: Obviously! Property 2: Each corn= er is the same size as the others, and has the same number of stickers i= n the same positions. Comments: This implies nothing about the sizes= of the actual corners of the puzzle; they could in fact be different si= zes (such as in the mirror cube, to which this theorem applies). This property is a necessity for the following conditio= ns to make sense. Conditions: Condition 1: The corners of the= puzzle we are applying the theorem to must be isomorphic to the corners of Definition 1, with Properties 1 and 2 holding.= Comments: I cannot even think of a Rubik-like puzzle in which this conditio= n does not hold, but need to include it so that the theorem applies to any puzzle = that we are examining. Condition 2: Every corner must have a set of stick= ers which are all of different colors. Comments: In other words, no = corner can have two identical stickers. There is no restriction on= more than one corner having the same set of stickers, for example as in= the 6-color megaminx. Condition 3: The stickers on a corner must al= l lie on the vertices of a regular simplex, and this simplex cannot lie = in an (n-1)-dimensional hyperplane which bisects the corner into two equ= al halves. Comments: The first portion of this condition helps ensur= e that the corner behaves like the corners of a Rubik-like puzzle. = That is, when a corner changes orientation or moves, the stickers occup= y positions that were previously occupied by stickers. The second portio= n prevents a technical issue in which all of the stickers would lie on a= 'great (n-1)-dimensional hypersphere', which could enable the corner to rotate in ways that would be unanalogous to a Rubik-like puzz= le (and the theorem would not hold). Condition 4: At regular inte= rvals, the corners will go through a process called a move: Each member = of a nonempty subset of them will move to the exact position of a corner= that just left its position. Thus, a move permutes the corners in an= y possible way. Additionally, each corner that moves must align it= self so that each of its stickers occupies a position previously occupie= d by a sticker. Comments: This condition emulates a face rotation in= a very nonrestrictive manner, compared to what we usually think of as a= face rotation moving corners, allowing a huge variety of possible puzzl= es to satisfy the requirements of this theorem. We are told that any per= mutation of the corners is allowed to occur. The second portion of= this condition, along with the previous condition, ensures that stickers always replace other stickers. Note that this condition = does not say that the permutation has to be the same at each point in ti= me, which makes sense because in Rubik-like puzzles different face rotat= ions produce different permutations. One very noteworthy fact to o= bserve is that since the condition places no restrictions on the nature = of the permutations, we can introduce our own restrictions into a Rubik-= like puzzle, and the theorem would still hold. For example, we could<= br>use a regular 3^3 Rubik's Cube, but state that the first move must swap = two adjacent corners, followed by swapping two corners along a face diag= onal, followed by swapping two corners on opposite ends of a space diago= nal, and repeat. As long as the puzzle satisfies the next conditio= n on moves, any possible restriction is allowable. Condition 5: T= he corners must constitute a family; that is, there exists a permissible sequence of moves such that every corner would occupy the= position of every other corner using this sequence. Also, every c= orner would be required to eventually occupy each position in every poss= ible permutation of its stickers given by the rotations of the corner. Comments: The first portion of this condition means that each corner = must eventually have the freedom to have moved everywhere. This is= naturally satisfied by any 'regular' Rubik-like puzzle I can imagine.&n= bsp; The second portion implies that each orientation can occur freely i= n any position. This is the condition that the Gearcube/Gearcube E= xtreme fails to meet, as it turns out each corner can only occupy one of= the three possible orientations in each position. The preparatory w= ork has been accomplished, and we can finally present the theorem! puzzle's corners satisfy Condition 1 by using k n-dimensional hypersphe= res as corners, and the isomorphic versions of the puzzle's corners and = moves satisfy Conditions 2 through 5, then the number of orientations th= e corners of the puzzle can achieve is: ((n!/2)^k)/3 if n = =3D 3 or 4 and (n!/2)^k if n >=3D 5. The proof reli= es heavily on others' work and thus I cannot really claim it as my own.&= nbsp; But neither the authors of The Rubik Tesseract nor An n-dimensional R= ubik Cube applied their results to anything other than the 3^n cube, so = I made the generalization. It took me over 3.5 hours to write thi= s email (I had no formulation of the theorem in my mind before I started= ; I had to come up with the entire sphere/point/simplex idea as I wrote,= and there were many parts rewritten or rephrased). Given that, I = hope that this post has been interesting to some of you! I would appreciate any comments or questions you have for me. Thanks to R= oice for the suggestion! All the best, David |
| top" style=3D"font: inherit;">Hi Brandon, No apologies necessary!&nb= sp; I just had the time available, and made use of it. It appears th= at you have found some extra puzzles I had not considered! However, I be= lieve that when the corner orientation cannot be handled by the theorem,= it is usually simple to figure out; i.e. all orientations or no orienta= tions. But I could be wrong. My theorem should apply to any ve= rtex or edge-twisting puzzle (with an important exception you made me aw= are of, see below). I can't see why it doesn't apply to the puzzle= in the first link you posted, the 1.2.2. I'll note that if it is = possible to rotate a single corner, then of course it is easy to calcula= te the number of orientations; it will be the maximum with no restrictio= ns. You said, > violates your first theorem because the permutation+orientation > formula for it is ((20!/2)^3) I'm not sure what you mean by "fi= rst theorem". Also you mention that the "permutation+ orientation = formula" for it is (20!/2)^3. I'm not certain if you're just talki= ng about the corners, and why you added the word 'permutation'. Where= did you get this formula? Upon analyzing it, it appears that you = meant, '(20!/2)^30', because there are 30 edges. I can tell you th= is: From looking at the image alone, and assuming that the only ty= pe of twist one can perform is the one displayed, then it appears that t= he corners can have 20!/2 permutations, and the corners can have 3^20/3 = orientations. > Also, the Helicopter cube has the /3 orientation = restriction but does > not have the /2 permutation restriction. " This seems perfectly fine and consistent with my formula. Y= our image here was very insightful: http://users.skynet.be/gelatinbr= ain/Applets/Magic%20Polyhedra/icosa_v4.gif I simply hadn't considere= d the (most likely obvious) fact that corners can have more stickers tha= n the number of dimensions! I have to admit, I made some claims ab= out my theorem holding for almost all puzzles, but I was just considering riants! In the case of a corner having at least n stickers if it's c= orner sphere is n-dimensional, I need to make the following changes: = Definition 1: A corner will be an n-dimensional hypersphere, n >=3D = 3, floating at a fixed position in n-dimensional space with =3D"font-weight: bold;">at least n stickers on its surface. C= ondition 3: The stickers on a corner must all lie on the vertices of a regular polytope, and this ensional hyperplane which bisects the corner into two equal halves. <= br>Generalized Corner Orientation Theorem: If a permutation puzzle's corners satisfy Condition 1 by using k n-dimensional hypersphe= res as corners, each font-weight: bold;">having s stickers,span> and the isomorphic versions of the puzzle's corners and moves sati= sfy Conditions 2 through 5, then the number of orientations the corners = of the puzzle can achieve is: ((o= rder of rotation group of the regular (n-1)-dimensional polytope with> s vert= ices)^k)/3 if n =3D 3 or 4 and weight: bold;"> (order of rotation gr= oup of the regular (n-1)-dimensional polytope with weight: bold;"> s vertices)^k if n >=3D 5. The complex phras= e, "order of rotation group..." could be simplified, as there are only s= o many regular polytopes of each dimension. For 3-dimensional puzz= les, it would equal s, which corresponds to n!/2 if n =3D 3 and s =3D 3. > ((n!/2)^k)/3 if n =3D 3 or 4 >=20 > and >=20 > (n!/2)^k if n >=3D 5. > Hmm, did you swap n and k in these? > That is, k is the number of corners and they are restricted to even > permutations and if they are 3 or 4 dimensional than the last corner's > orientation is uniquely defined by the others. > To me that would be: > ((k!/2)^n)/3 if n =3D 3 or 4 > and > (k!/2)^n if n >=3D 5. > If this isn't the case they maybe I don't follow you after all... No, I have it correctly. I think you are just making a simple= oversight, as I had been doing with the Klein's Quartic counts! :)&nbs= p; I believe you're making the mistake of thinking in terms of permutati= ons, hence your reasoning for the term, "k!/2". My formula concerns only= orientations. It needs to be multiplied by the permutation count = (which is simple, but as I said before depends upon the other piece familie= s due to parity interaction) to finish the calculation for that piece.&n= bsp; The n!/2 term comes from the number of ways one can rotate an n-dim= ensional simplex. We raise it to the power of k because there are = k corners. Really, the only two outstanding features of this theor= em are that for dimensions greater than 4, there is no division by three= (no corner restriction), and that it can be generalized so far as I hav= e done. > On the topic of vertex-twisting puzzles, what about duals? For > example, Gelatinbrain's 2.2.6 > (gelatinbrain/Applets/Magic%20Polyhedra/icosa_v4.gif">http://users.skynet.be= /gelatinbrain/Applets/Magic%20Polyhedra/icosa_v4.gif) > is the dual of the Penultimate (1.1.7 / nbrain/Applets/Magic%20Polyhedra/dodeca_f6.gif">> http://users.skynet.be= /gelatinbrain/Applets/Magic%20Polyhedra/dodeca_f6.gif) > It is nice to be able to think of centers with super-stickers as > corners and corners and centers. Using the 1.1.7 <-> 2.2.6 dual > relationship is troublesome though because the number of orientations > of a 1.1.7 center / 2.2.6 corner is 5 and isn't tied to the > dimensionality of the puzzle. Perhaps this is how you've avoided > vertex-twisting puzzles by using definition 1 where the number of > stickers matches the number of dimensions? That doesn't handle the > Helicopter cube though. My theorem should apply to vertex-twisting and edge-twisting puzzle= s, but it is important to keep in mind that it only works for corners, i= .e. pieces with at least as many stickers as dimensions. It should= n't matter if we are twisting about the vertex, as the other surrounding= vertices are moved by the rotation. For duals, we would have to r= ecalculate the corner orientation; we can't interchange corners for cent= ers and centers for corners. It's a good idea however. I ho= pe this has been helpful. Thanks again for welcoming me! It's g= reat to be back. All the best, David --- On Tue, 5/3= /11, Brandon Enright <bmenrigh@ucsd.edu> wrote: dding-left: 5px;"> From: Brandon Enright <bmenrigh@ucsd.edu> Su= bject: Re: [MC4D] Generalized corner orientation theorem To: 4D_Cubing@yahoogroups.com Cc: djs314djs314@yahoo.com, bmenrigh@ucsd.edu= Date: Tuesday, May 3, 2011, 8:33 PM =20=20=20=20=20=20 =20=20=20=20=20=20 -----BEGIN PGP SIGNED MESSAGE----- =20=20=20=20=20 =20 |
Hi David,
It looks like your result works for hypercubes, dodecahedron and 120-cell a=
nd for the most 2D Magic Tile with fixed centers (for which we use n=3D3, n=
ot 2). But there are puzzles with 2D and 3D stickers where all corner orien=
tations are possible, or where the factor is different from 3.
1) Everybody knows that you can twist single corner in pyraminx :)
2) For face-turn octahedron number of possible orientations is only 1/2 of =
"order^k". But it doesnt's satisfy conditions: either you should paint only=
4 its faces - and corners will be 2-colored, or not all permutations of st=
icker colors will be reachable (each 4-colored corner may be only on two or=
ientations).
3) in "hemidodecahedron" from Magic Tile you also can twist one corner. Num=
ber of corner orientations there will be 6^10/2 (puzzle is non-orientable s=
o corner piece has 6 possible orientations)
4) in {6,3,3} puzzle with 8 colors I saw a corner sticker twisted by 3-cycl=
e - so all 12^28 corner orientations are possible.
5) It looks like for the face-turn icosahedron number of corner orientation=
s is 5^12/5 (I never tried it, but you can see it just by the picture - sel=
ect 4 faces without common corners and count sum of corner rotations during=
twists - it always will be zero).
I afraid that there are too many possible permutation groups even for corne=
r stickers to have some common theorem about them. May be, something like "=
you always can change orientation of one corner to any possible state so th=
at only two corners will be twisted (this and one more) - and other k-2 cor=
ners will save their orientation" will work? I'm not sure even in this.
Regards,
Andrey
=C2=A0
=20=20
=20=20=20=20
=20=20=20=20=20=20
=20=20=20=20=20=20
Hi David,
It looks like your result works for hypercubes, dodecahedron and 120-cell a=
nd for the most 2D Magic Tile with fixed centers (for which we use n=3D3, n=
ot 2). But there are puzzles with 2D and 3D stickers where all corner orien=
tations are possible, or where the factor is different from 3.
1) Everybody knows that you can twist single corner in pyraminx :)
2) For face-turn octahedron number of possible orientations is only 1/2 of =
"order^k". But it doesnt's satisfy conditions: either you should paint only=
4 its faces - and corners will be 2-colored, or not all permutations of st=
icker colors will be reachable (each 4-colored corner may be only on two or=
ientations).
3) in "hemidodecahedron" from Magic Tile you also can twist one corner. Num=
ber of corner orientations there will be 6^10/2 (puzzle is non-orientable s=
o corner piece has 6 possible orientations)
4) in {6,3,3} puzzle with 8 colors I saw a corner sticker twisted by 3-cycl=
e - so all 12^28 corner orientations are possible.
5) It looks like for the face-turn icosahedron number of corner orientation=
s is 5^12/5 (I never tried it, but you can see it just by the picture - sel=
ect 4 faces without common corners and count sum of corner rotations during=
twists - it always will be zero).
I afraid that there are too many possible permutation groups even for corne=
r stickers to have some common theorem about them. May be, something like "=
you always can change orientation of one corner to any possible state so th=
at only two corners will be twisted (this and one more) - and other k-2 cor=
ners will save their orientation" will work? I'm not sure even in this.
Regards,
Andrey
=20=20=20=20
=20=20=20=20=20
=20=20=20=20
=20=20=20=20
=20
=20=20
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| top" style=3D"font: inherit;">Hi Andrey, You are absolutely right, m= y theorem does not apply for as many puzzles as I claimed. Thank y= ou for pointing this out! But it is correct as written, and is very be modified to handle different cases such as the ones you mentioned, bu= t I haven't included all of the variations. It is related to a muc= h stronger property of corner orientations in general, as I mentioned in= the beginning of my post, and I'm sure that I can modify it as needed t= o apply it to any puzzle. It is also useful to check if the condition= s apply, and if not then one knows how to fix it. Of course, when= you can rotate a single corner, the theorem does not apply. :) But thos= e cases are easy to handle, so the practicality of the theorem is not affected. I find your examples of the factor being something other t= han 3 to be very interesting; I haven't worked with them too much and wi= ll study them later to see why the theorem does not apply as-is, and how= to modify it to work. Also, this statement was obviously in error:<= br> ((order of rotation group of the = regular (n-1)-dimensional polytope with d;">s vertices)^k)/3 if n = =3D 3 or 4 with the division by 3. I'll fix this later when I = get home. An easy way to see if my theorem applies as-is is first to= make sure that the number of stickers is equal to the number of dimensi= ons. That is the first condition that more unique puzzles are like= ly to fail. But the most restrictive condition is this: eve= ry corner would be required to eventually occupy each position in every possible permutation of its stickers given by the rotations of the corn= er. This is basically the key condition that Rubik-like puzzles are = likely to fail to qualify for the theorem, and studying why it fails can= often provide the insight as to how to modify the theorem to make it wo= rk. An easy way to check if this condition holds is to rotate two = adjacent faces, and see if you can get a corner to achieve all of its or= ientations on it home square. If it can, then the above requiremen= t would be satisfied, by using conjugates. I thank you very much for= pointing out the inaccuracy in my statement that my theorem works for a= lmost every puzzle! With modifications, I believe it can handle th= em all. Later I will study the examples you presented, and also give<= br>a brief overview of my general orientation-counting technique. Th= anks again, and best wishes! All the best, David --- On Wed, 5/4/11, Andrey <andreyastrelin@= yahoo.com> wrote: rgb(16, 16, 255); margin-left: 5px; padding-left: 5px;"> |
Hi David,
=C2=A0
=20=20
=20=20=20=20
=20=20=20=20=20=20
=20=20=20=20=20=20
Hi Andrey,
You are absolutely right, my theorem does not apply for as many puzzles as =
I
claimed.=C2=A0 Thank you for pointing this out!=C2=A0 But it is correct as =
written, and is very
useful in determining orientations.=C2=A0 For practical purposes, It can be=
modified to
handle different cases such as the ones you mentioned, but I haven't includ=
ed
all of the variations.=C2=A0 It is related to a much stronger property of c=
orner orientations
in general, as I mentioned in the beginning of my post, and I'm sure that I=
can modify
it as needed to apply it to any puzzle.=C2=A0 It is also useful to check if=
the conditions
apply, and if not then one knows how to fix it.
Of course, when you can rotate a single corner, the theorem does not apply.=
:)
But those cases are easy to handle, so the practicality of the theorem is
not
affected.=C2=A0 I find your examples of the factor being something other t=
han 3 to
be very interesting; I haven't worked with them too much and will study the=
m
later to see why the theorem does not apply as-is, and how to modify it to =
work.
Also, this statement was obviously in error:
((order of rotation group of the regular (n-1)-dimensional polytope with
s vertices)^k)/3
if n =3D 3 or 4
with the division by 3.=C2=A0 I'll fix this later when I get home.
An easy way to see if my theorem applies as-is is first to make sure that t=
he
number of stickers is equal to the number of dimensions.=C2=A0 That is the =
first
condition that more unique puzzles are likely to fail.=C2=A0 But the most r=
estrictive
condition is this:
every corner would be required
to eventually occupy each position in every
possible permutation of its stickers
given by the rotations of the corner.
This is basically the key condition that Rubik-like puzzles are likely to f=
ail to
qualify for the theorem, and studying why it fails can often provide the in=
sight
as to how to modify the theorem to make it work.=C2=A0 An easy way to check
if this condition holds is to rotate two adjacent faces, and see if you can=
get
a corner to achieve all of its orientations on it home square.=C2=A0 If it =
can, then
the above requirement would be satisfied, by using conjugates.
I thank you very much for pointing out the inaccuracy in my statement that
my theorem works for almost every puzzle!=C2=A0 With modifications, I belie=
ve it can
handle them all.=C2=A0 Later I will study the examples you presented, and a=
lso give
a brief overview of my general orientation-counting technique.
Thanks again, and best wishes!
All the
best,
David
--- On Wed, 5/4/11, Andrey
From: Andrey
Subject: Re: [MC4D] Generalized corner orientation theorem
To: 4D_Cubing@yahoogroups.com
Date: Wednesday, May 4, 2011, 12:32 AM
=C2=A0
=20=20=20=20
=20=20=20=20=20=20
=20=20=20=20=20=20
Hi David,
It looks like your result works for hypercubes, dodecahedron and 120-cell a=
nd for the most 2D Magic Tile with fixed centers (for which we use n=3D3, n=
ot 2). But there are puzzles with 2D and 3D stickers where all corner orien=
tations are possible, or where the factor is different from 3.
1) Everybody knows that you can twist single corner in pyraminx :)
2) For face-turn octahedron number of possible orientations is only 1/2 of =
"order^k". But it doesnt's satisfy conditions: either you should paint only=
4 its faces - and corners will be 2-colored, or not all permutations of st=
icker colors will be reachable (each 4-colored corner may be only on two or=
ientations).
3) in "hemidodecahedron" from Magic Tile you also can twist one corner. Num=
ber of corner orientations there will be 6^10/2 (puzzle is non-orientable s=
o corner piece has 6 possible orientations)
4) in {6,3,3} puzzle with 8 colors I saw a corner sticker twisted by 3-cycl=
e - so all 12^28 corner orientations are possible.
5) It looks like for the face-turn icosahedron number of corner orientation=
s is 5^12/5 (I never tried it, but you can see it just by the picture - sel=
ect 4 faces without common corners and count sum of corner rotations during=
twists - it always will be zero).
I afraid that there are too many possible permutation groups even for corne=
r stickers to have some common theorem about them. May be, something like "=
you always can change orientation of one corner to any possible state so th=
at only two corners will be twisted (this and one more) - and other k-2 cor=
ners will save their orientation" will work? I'm not sure even in this.
Regards,
Andrey
=20=20=20=20
=20=20=20=20=20
=20
=20=20=20=20
=20=20=20=20=20
=20=20=20=20
=20=20=20=20
=20
=20=20
--0-242416848-1304613427=:20947
Content-Type: text/html; charset=utf-8
Content-Transfer-Encoding: quoted-printabletop" style=3D"font: inherit;">Hi everyone,
I would like to thank And=
rey again for his very insightful observations. Boy, was I
ignoran=
t of how many puzzles there were! Thank you Andrey, for informing me!=
! :)
In particular, he said:
> I afraid that there are too=
many possible permutation groups even for=20
corner stickers
> to have some common theorem about them.
He m=
ay be very correct here, and if I waste a lot of time searching for one wit=
hout
success, I'll wish I had taken his advice! But, I'm not going=
to give up quite just yet.
I'll describe how I'm going to restrict myse=
lf to much more practical puzzles, rather
than spheres in space, shortly=
.
> May be,=20
something like "you always can change orientation of one corner to any
&=
gt; possible state so that only two corners will be twisted (this and one=20
more) - and
> other k-2 corners will save their orientation" will wor=
k?=20
I'm not sure even in this.
Yes, this sounds a bit too simple. =
I appreciate Andrey's help very much though!
Well, things are gettin=
g much more complex than I anticipated. There are far too
many var=
iations to consider. I'm going to consider only the regular polytopes=
,
and account for vertex, edge, face, hyperplane, etc. based rotations. =
(I believe that
if vertex-turning puzzles are cut deep enough, not all o=
rientations might be possible.
Also, permutations of other pieces and or=
ientations of the corners would interact.)
My goal is to come up with a =
general permutation and orientation counting technique
for the complete =
puzzle for any possible polytope. I'll be considering mathematical
mpletely
account for all of the physical aspects. Here are some of=
them:
- The impossibility to actually build a puzzle that
would not fall apart.
- In some puzzles, such as the edge-turning o=
ctahedron, there are unintentional
and possibly unavoidable corner twist=
s possible.
- Many puzzles have the ability to jumble, or shape-shif=
t. (There is a technical
difference between the two.) This involve=
s turning a face part of the way, and
then having the ability to turn an=
other face. This of course hugely complicates
finding the number o=
f reachable positions. In fact, it is entirely possible that
if a =
MagicCube4D 4.0 puzzle were to be physically built in 4-space, it might hav=
e
the ability to jumble or shape-shift and change the number of permutat=
ions.
But this of course could probably never be accounted for in a prog=
ram (I'm not
even certain how one would describe the resulting shapes fr=
om such jumbling
of higher-dimensional puzzles mathematically, and to de=
termine how such puzzles
could jumble - it would be extremely,
extremely complex, and probably impossible).
We'll all see how long=
I work on this before giving up. :) I might end up just
consideri=
ng ordinary face rotations.
Thanks to everyone for your support!
=
All the best,
David
--- On Wed, 5/4/11, David Smith <=
;djs314djs314@yahoo.com> wrote:eft: 2px solid rgb(16, 16, 255); margin-left: 5px; padding-left: 5px;">
=
From: David Smith <djs314djs314@yahoo.com>
Subject: Re: [MC4D] Gen=
eralized corner orientation theorem
To: 4D_Cubing@yahoogroups.com
Dat=
e: Wednesday, May 4, 2011, 9:11 AM
=20=20=20=20=20=20
=20=20=20=20=20=20
Hi Andrey,
You are abso=
lutely right, my theorem does not apply for as many puzzles as I
claimed=
. Thank you for pointing this out! But it is correct as written=
, and is very
useful in determining orientations. For practical pu=
rposes, It can be modified to
handle different cases such as the ones yo=
u mentioned, but I haven't included
all of the variations. It is r=
elated to a much stronger property of corner orientations
in general, as=
I mentioned in the beginning of my post, and I'm sure that I can modify
f the conditions
apply, and if not then one knows how to fix it.
=
Of course, when you can rotate a single corner, the theorem does not apply.=
:)
But those cases are easy to handle, so the practicality of the theor=
em
is
not
affected. I find your examples of the factor being something other t=
han 3 to
be very interesting; I haven't worked with them too much and wi=
ll study them
later to see why the theorem does not apply as-is, and how=
to modify it to work.
Also, this statement was obviously in error:<=
br>
((order of rotation group of the =
regular (n-1)-dimensional polytope with
d;">s vertices)^k)/3
if n =
=3D 3 or 4
with the division by 3. I'll fix this later when I =
get home.
An easy way to see if my theorem applies as-is is first to=
make sure that the
number of stickers is equal to the number of dimensi=
ons. That is the first
condition that more unique puzzles are like=
ly to fail. But the most restrictive
condition is this:
eve=
ry corner would be required
to eventually occupy each position in every
possible permutation of its stickers
given by the rotations of the corn=
er.
This is basically the key condition that Rubik-like puzzles are =
likely to fail to
qualify for the theorem, and studying why it fails can=
often provide the insight
as to how to modify the theorem to make it wo=
rk. An easy way to check
if this condition holds is to rotate two =
adjacent faces, and see if you can get
a corner to achieve all of its or=
ientations on it home square. If it can, then
the above requiremen=
t would be satisfied, by using conjugates.
I thank you very much for=
pointing out the inaccuracy in my statement that
my theorem works for a=
lmost every puzzle! With modifications, I believe it can
handle th=
em all. Later I will study the examples you presented, and also give<=
br>a brief overview of my general orientation-counting technique.
Th=
anks again, and best wishes!
All the
best,
David
--- On Wed, 5/4/11, Andrey <andreyastrelin@=
yahoo.com> wrote: rgb(16, 16, 255);">
From: Andrey <andreyastrelin@yahoo.com>
Su=
bject: Re: [MC4D] Generalized corner orientation theorem
To: 4D_Cubing@y=
ahoogroups.com
Date: Wednesday, May 4, 2011, 12:32 AM
=20=20=20=20=20=20
=20=20=20=20=20=20
It looks like your result works for hypercubes, dodecahedron and 120-cell a=
nd for the most 2D Magic Tile with fixed centers (for which we use n=3D3, n=
ot 2). But there are puzzles with 2D and 3D stickers where all corner orien=
tations are possible, or where the factor is different from 3.
1) Everybody knows that you can twist single corner in pyraminx :)
2) For face-turn octahedron number of possible orientations is only 1/2 of =
"order^k". But it doesnt's satisfy conditions: either you should paint only=
4 its faces - and corners will be 2-colored, or not all permutations of st=
icker colors will be reachable (each 4-colored corner may be only on two or=
ientations).
3) in "hemidodecahedron" from Magic Tile you also can twist one corner. Num=
ber of corner orientations there will be 6^10/2 (puzzle is non-orientable s=
o corner piece has 6 possible orientations)
4) in {6,3,3} puzzle with 8 colors I saw a corner sticker twisted by 3-cycl=
e - so all 12^28 corner orientations are possible.
5) It looks like for the face-turn icosahedron number of corner orientation=
s is 5^12/5 (I never tried it, but you can see it just by the picture - sel=
ect 4 faces without common corners and count sum of corner rotations during=
twists - it always will be zero).
I afraid that there are too many possible permutation groups even for corne=
r stickers to have some common theorem about them. May be, something like "=
you always can change orientation of one corner to any possible state so th=
at only two corners will be twisted (this and one more) - and other k-2 cor=
ners will save their orientation" will work? I'm not sure even in this.
Regards,
Andrey
=20=20=20=20=20
=20
=20=20=20=20=20
=20
--0-242416848-1304613427=:20947--