Thread: "new hemi-cube and hemi-dodecahedron puzzles"

; PADDING-LEFT: 1ex" class=3D"gmail_quote">I just got time to solve the hem=
i-dodecahedron (length-3). It's a beautiful puzzle. During the solve I =
ran into the situation that I need to twist a corner by +120 deg and its an=
tipode -120 deg. It's a surprise for me and it took me a while to figur=
e out why it happened. It's a nice twist!

; PADDING-LEFT: 1ex" class=3D"gmail_quote">For me this puzzle is a bandaged=
Megaminx, that is, a Megaminx painted with six colors, and with the modifi=
cation that opposite faces are bandaged so that they move together. I don&#=
39;t know an internal physical structure to make such a bandage, but logica=
lly it's like that. Because of this bandage, when you look downward at =
the top of Megaminx and turn the top face CW, the bottom face is also turne=
d CW viewed from top, that is, CCW viewed from bottom. Therefore when the M=
egaminx is projected to a plane, the two faces of the same color always tur=
n in opposite directions.


Since I view this puzzle as a bandaged version of the familiar Megaminx=
, I don't think in the fancy way that it's a non-orientable puzzle =
on a projective plane.

; PADDING-LEFT: 1ex" class=3D"gmail_quote">Another natural way to construct=
a hemi-cube or hemi-dodecahedron is that, opposite faces turn clockwise si=
multaneously or counter-clockwise simultaneously. The physical meaning is t=
hat opposite faces are not bandaged but connected using differential gears.=
Such a hemi-cube is basically a gear cube/caution cube<//www.youtube.com/watch?v=3DUDVb9NExsA8" target=3D"_blank">http://www.youtu=
be.com/watch?v=3DUDVb9NExsA8>, where L and R always go together if y=
ou take the middle slice as the reference. I think the hemi-cube of this ki=
nd is more non-trivial than the current hemi-cube (I'm only talking abo=
ut length-3).

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>
>
>> Another natural way to construct a hemi-cube or hemi-dodecahedron is that,
>> opposite faces turn clockwise simultaneously or counter-clockwise
>> simultaneously. The physical meaning is that opposite faces are not bandaged
>> but connected using differential gears. Such a hemi-cube is basically a gear
>> cube/caution cube, where L
>> and R always go together if you take the middle slice as the reference. I
>> think the hemi-cube of this kind is more non-trivial than the current
>> hemi-cube (I'm only talking about length-3).
>>
>>
> This does sound interesting, but note that in this case we are no longer
> talking about hemi-puzzles. A puzzle where opposite faces would be coupled
> like you are describing would necessarily be 6-faced for the cube and
> 12-faced for the dodecahedron. You are no longer able to identify opposite
> faces as one and the same after a twist. You could color opposite faces the
> same, but they would still behave differently. A hemi-cube is an abstract
> polytope with 3 faces, and a hemi-dodecahedron one with 6 faces.
>

I was wrong on the hemi-cube. It does still work there if you twist as you
described (which is why we were already getting the hemi-cube behavior with
the other 3-colored orientable puzzles). But for the hemi-dodecahedron, I
think what I wrote is correct, and that the new puzzle would have have 12
faces.

On Mon, Mar 7, 2011 at 3:09 PM, Andrey wrote:

>
> If you twist opposite sides in same direction (relative to sphee surface),
> you'll get orientable puzzle. I don't know, if there are sphere paintings
> that are invariant to this transformation (i.e. order of adjacent colors of
> all intances of one face is the same). There is 5-color painting if
> icosahedron (all faces of outscribed tetrahedra have the same color), but
> I'm not sure that it will work.


So I think the above answers the question of whether their are sphere
paintings which work in an orientable fashion. Yes. But the reason it
works in the hemi-cube case is that each face has both left-handed and
right-handed piece versions. Maybe that is always necessary - I'm not sure.

seeya,
Roice

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Roice,=20

I think you are definitely right. I think of the alternative hemi-dodecahed=
ron because my interpretation is essentially 12-face but not 6-face. The he=
mi-cube works in either way because the consequence of a CW turn is always =
as same as that of a CCW turn.=20

Nan

--- In 4D_Cubing@yahoogroups.com, Roice Nelson wrote:
>
> >
> >
> >> Another natural way to construct a hemi-cube or hemi-dodecahedron is t=
hat,
> >> opposite faces turn clockwise simultaneously or counter-clockwise
> >> simultaneously. The physical meaning is that opposite faces are not ba=
ndaged
> >> but connected using differential gears. Such a hemi-cube is basically =
a gear
> >> cube/caution cube, where=
L
> >> and R always go together if you take the middle slice as the reference=
. I
> >> think the hemi-cube of this kind is more non-trivial than the current
> >> hemi-cube (I'm only talking about length-3).
> >>
> >>
> > This does sound interesting, but note that in this case we are no longe=
r
> > talking about hemi-puzzles. A puzzle where opposite faces would be cou=
pled
> > like you are describing would necessarily be 6-faced for the cube and
> > 12-faced for the dodecahedron. You are no longer able to identify oppo=
site
> > faces as one and the same after a twist. You could color opposite face=
s the
> > same, but they would still behave differently. A hemi-cube is an abstr=
act
> > polytope with 3 faces, and a hemi-dodecahedron one with 6 faces.
> >
>=20
> I was wrong on the hemi-cube. It does still work there if you twist as y=
ou
> described (which is why we were already getting the hemi-cube behavior wi=
th
> the other 3-colored orientable puzzles). But for the hemi-dodecahedron, =
I
> think what I wrote is correct, and that the new puzzle would have have 12
> faces.
>=20
> On Mon, Mar 7, 2011 at 3:09 PM, Andrey wrote:
>=20
> >
> > If you twist opposite sides in same direction (relative to sphee surfac=
e),
> > you'll get orientable puzzle. I don't know, if there are sphere paintin=
gs
> > that are invariant to this transformation (i.e. order of adjacent color=
s of
> > all intances of one face is the same). There is 5-color painting if
> > icosahedron (all faces of outscribed tetrahedra have the same color), b=
ut
> > I'm not sure that it will work.
>=20
>=20
> So I think the above answers the question of whether their are sphere
> paintings which work in an orientable fashion. Yes. But the reason it
> works in the hemi-cube case is that each face has both left-handed and
> right-handed piece versions. Maybe that is always necessary - I'm not su=
re.
>=20
> seeya,
> Roice
>

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>
> On Mon, Mar 7, 2011 at 3:09 PM, Andrey wrote:
>
>> If you twist opposite sides in same direction (relative to sphee surface),
>> you'll get orientable puzzle. I don't know, if there are sphere paintings
>> that are invariant to this transformation (i.e. order of adjacent colors of
>> all intances of one face is the same). There is 5-color painting if
>> icosahedron (all faces of outscribed tetrahedra have the same color), but
>> I'm not sure that it will work.
>
>
Here is something else interesting I realized about the
"orientable hemi-cube" we've been talking about... In order for twists to
move in the same direction relative to the surface, and for it to still be
interpreted as a hemi-cube puzzle, the identification of 2C/3C piece pairs
must change after every twist! Without allowing this kind of wackiness,
which honestly doesn't feel legit at all, I think the answer to Andrey's
question about the existence of orientable periodic paintings for the sphere
is provably 'no' for any tiling on the sphere (regular, uniform, whatever).

Here is how the logic showing that could go... If an orientable painting was
possible, we are talking about choosing a subset of tiles, which when
repeated would cover the sphere (the "universal cover" of the chosen subset
would be the sphere). The topology of the surface represented by this
subset of tiles must admit elliptical geometry, but there are only two
topologies which do so, the sphere and the projective plane (see chapter 12
of Jeff Weeks' "The Shape of
Space").
The latter is non-orientable, so the topology of our subset must be the
sphere. However, the universal cover of the sphere is the sphere itself
(see here ), so the the
only set of tiles which will work is the full set of tiles, which already
covers the entire sphere. So there is no periodic painting possible.

Also, I have fixed the issue Nan found with the even-length hemi-cube
puzzles, and uploaded a new version. Thank you again Nan!

Roice

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Okay. Take regular tetrahedron and group of it's movements (A4). Its fundam=
ental area is 1/3 of tetrahedron face, like in this pucture: http://groups.=
yahoo.com/group/4D_Cubing/photos/album/1184359940/pic/2087318906/view
May be we can do something from set of tiles that are invariant to A4? Of c=
ourse, verices should not be included in movement (or they may be centers o=
f special faces), but otherwise I don't see any problems...

Andrey



--- In 4D_Cubing@yahoogroups.com, Roice Nelson wrote:
>
> >
> > On Mon, Mar 7, 2011 at 3:09 PM, Andrey wrote:
> >
> >> If you twist opposite sides in same direction (relative to sphee surfa=
ce),
> >> you'll get orientable puzzle. I don't know, if there are sphere painti=
ngs
> >> that are invariant to this transformation (i.e. order of adjacent colo=
rs of
> >> all intances of one face is the same). There is 5-color painting if
> >> icosahedron (all faces of outscribed tetrahedra have the same color), =
but
> >> I'm not sure that it will work.
> >
> >
> Here is something else interesting I realized about the
> "orientable hemi-cube" we've been talking about... In order for twists t=
o
> move in the same direction relative to the surface, and for it to still b=
e
> interpreted as a hemi-cube puzzle, the identification of 2C/3C piece pair=
s
> must change after every twist! Without allowing this kind of wackiness,
> which honestly doesn't feel legit at all, I think the answer to Andrey's
> question about the existence of orientable periodic paintings for the sph=
ere
> is provably 'no' for any tiling on the sphere (regular, uniform, whatever=
).
>=20
> Here is how the logic showing that could go... If an orientable painting =
was
> possible, we are talking about choosing a subset of tiles, which when
> repeated would cover the sphere (the "universal cover" of the chosen subs=
et
> would be the sphere). The topology of the surface represented by this
> subset of tiles must admit elliptical geometry, but there are only two
> topologies which do so, the sphere and the projective plane (see chapter =
12
> of Jeff Weeks' "The Shape of
> Space20&linkCode=3Das2&camp=3D1789&creative=3D390957&creativeASIN=3D0824707095>"=
).
> The latter is non-orientable, so the topology of our subset must be the
> sphere. However, the universal cover of the sphere is the sphere itself
> (see here ), so the the
> only set of tiles which will work is the full set of tiles, which already
> covers the entire sphere. So there is no periodic painting possible.
>=20
> Also, I have fixed the issue Nan found with the even-length hemi-cube
> puzzles, and uploaded a new version. Thank you again Nan!
>=20
> Roice
>

--- In 4D_Cubing@yahoogroups.com, Roice Nelson wrote:
>=20
> Also, I have fixed the issue Nan found with the even-length hemi-cube
> puzzles, and uploaded a new version. Thank you again Nan!
>=20

The previous version worked better. New version is the same as 3-layer :(

I have started to solve KleinQuartic of MagicTile. I find the edge solved K=
leinQuartic is more beautiful than the finished one.
I thought that I can write sequences in the log file and let them execute (=
in order to replace the missing macro possibility). I was wrong. MagicTile =
saves the endposition in a packed array in the beginning of the log file (I=
think packed are the actual colors of the stickers places) and allows only=
Undo and Redo from there. So I'm forced to to the rather lengthy vertices =
solving "by hand". Perhaps I will finish it nevertheless because the puzzle=
is very pleasent.

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On Tue, Mar 8, 2011 at 12:57 AM, Andrey wrote:

>
> --- In 4D_Cubing@yahoogroups.com, Roice Nelson wrote:
> >
> > Also, I have fixed the issue Nan found with the even-length hemi-cube
> > puzzles, and uploaded a new version. Thank you again Nan!
> >
>
> The previous version worked better. New version is the same as 3-layer :(


The previous version was broken and resulted in inconsistent states (the
even puzzles would not have been solvable). Yes, the even hemi-cubes behave
the same as odd puzzles of smaller order, but this is the nature of the
hemi-cube. To do something different, they would need to be bandaged
puzzles like Nan described, but that will require further coding of the
concepts required for those. Internally right now in the code, there are 3
"master" faces and 3 "slave" faces which mimic the masters, rather than 6
faces with configured attachments. Sorry, all I can say for now is that I
added the feature request of bandaged puzzles to my list. If implemented,
I would put the puzzle you are wanting in a different section of the menu.

seeya,
Roice

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Roice,
I understand about hemi-cubes, question is about hemi-dodecahedra. If you=
take kilominx, youu can do 2 possible bandaged rotations (1+5, 2+4) that l=
eave layer 3 unmoved. In hemi-dodecahendron-4 rotation of layer 1 works as =
expected, but "layer 2" rotates layers 2,3,4 from kilominx, not only 2 and =
4 (like that was in broken version).

Andrey.

>=20
> The previous version was broken and resulted in inconsistent states (the
> even puzzles would not have been solvable). Yes, the even hemi-cubes beh=
ave
> the same as odd puzzles of smaller order, but this is the nature of the
> hemi-cube. To do something different, they would need to be bandaged
> puzzles like Nan described, but that will require further coding of the
> concepts required for those. Internally right now in the code, there are=
3
> "master" faces and 3 "slave" faces which mimic the masters, rather than 6
> faces with configured attachments. Sorry, all I can say for now is that =
I
> added the feature request of bandaged puzzles to my list. If implemente=
d,
> I would put the puzzle you are wanting in a different section of the menu=
.
>=20
> seeya,
> Roice
>

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Thank you for the clarification Andrey, and sorry I missed what your email
intended. The hemi-Kilominx behavior is now reverted to what it was before.
The hemi-Impossiball should work now as well (it hadn't yet).

www.gravitation3d.com/magictile

I appreciate the testing help. It can be difficult to get every case right.

seeya,
Roice


On Tue, Mar 8, 2011 at 10:58 AM, Andrey wrote:

> Roice,
> I understand about hemi-cubes, question is about hemi-dodecahedra. If you
> take kilominx, youu can do 2 possible bandaged rotations (1+5, 2+4) that
> leave layer 3 unmoved. In hemi-dodecahendron-4 rotation of layer 1 works as
> expected, but "layer 2" rotates layers 2,3,4 from kilominx, not only 2 and 4
> (like that was in broken version).
>
> Andrey.
>

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oops, I meant "hemi-Master Kilominx" :S In any case, both the length-2 and
length-4 versions should work now...


On Tue, Mar 8, 2011 at 10:05 PM, Roice Nelson wrote:

> Thank you for the clarification Andrey, and sorry I missed what your email
> intended. The hemi-Kilominx behavior is now reverted to what it was before.
> The hemi-Impossiball should work now as well (it hadn't yet).
>
> www.gravitation3d.com/magictile
>
> I appreciate the testing help. It can be difficult to get every case
> right.
>
> seeya,
> Roice
>
>
> On Tue, Mar 8, 2011 at 10:58 AM, Andrey wrote:
>
>> Roice,
>> I understand about hemi-cubes, question is about hemi-dodecahedra. If you
>> take kilominx, youu can do 2 possible bandaged rotations (1+5, 2+4) that
>> leave layer 3 unmoved. In hemi-dodecahendron-4 rotation of layer 1 works as
>> expected, but "layer 2" rotates layers 2,3,4 from kilominx, not only 2 and 4
>> (like that was in broken version).
>>
>> Andrey.
>>
>

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oops, I meant "hemi-Master Kilominx" :S =A0In any case, both the =
length-2 and length-4 versions should work now...