--14dae93d94b8d2416f04e39e5164
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: quoted-printable
Hi all,
Check out a new physical model of the exotic {4,4,4} H=C2=B3 honeycomb!
http://shpws.me/oFpu
Each cell is a tiling of squares with an infinite number of facets. All
vertices are ideal (meaning they live at infinity, on the Poincare ball
boundary). Four cells meet at every edge, and an infinite number of cells
meet at every vertex (the vertex figure is a tiling of squares too). This
honeycomb is self-dual.
I printed only half of the Poincare ball in this model, which has multiple
advantages: you can see inside better, and it saves on printing costs. The
view is face-centered, meaning the projection places the center of one
(ideal) 2D polygon at the center of the ball. An edge-centered view is
also possible. Vertex-centered views are impossible since every vertex is
ideal. A view centered on the interior of a cell is possible, but (I
think, given my current understanding) a cell-centered view is also
impossible.
I rendered one tile and all the tiles around it, so only one level of
recursion. I also experimented with deeper recursion, but felt the
resulting density inhibited understanding. Probably best would be to have
two models at different recursion depths side by side to study together. I
had to artificially increase edge widths near the boundary to make things
printable.
These things are totally cool to handle in person, so consider ordering one
or two of the honeycomb models :) As I've heard Henry Segerman comment,
the "bandwidth" of information is really high. You definitely notice
things you wouldn't if only viewing them on the computer screen. The
{3,6,3} and {6,3,6} are very similar to the {4,4,4}, just based on
different Euclidean tilings, so models of those are surely coming as well.
So... whose going to make a puzzle based on this exotic honeycomb? :D
Cheers,
Roice
As a postscript, here are a few thoughts I had about the {4,4,4} while
working on the model...
In a previous thread on the
{4,4,4}
Nan made an insightful comment. He said:
I believe the first step to understand {4,4,4} is to understand {infinity,
> infinity} in the hyperbolic plane.
I can see now they are indeed quite analogous. Wikipedia has some great
pictures of the {=E2=88=9E,=E2=88=9E} tiling and {p,q} tilings that approac=
h it by
increasing p or q. Check out the progression that starts with an {=E2=88=
=9E,3}
tiling and increases q, which is the bottom row of the table here:
http://en.wikipedia.org/wiki/Uniform_tilings_in_hyperbolic_plane#Regular_hy=
perbolic_tilings
The {=E2=88=9E} polygons are inscribed in
horocycles
circle of infinite radius with a unique center point on the disk boundary).
The horocycles increase in size with this progression until, in the limit,
the inscribing circle is* the boundary of the disk itself.* Something
strange about that is an {=E2=88=9E,=E2=88=9E} tile loses its center. A ho=
rocycle has a
single center on the boundary, so the inscribed {=E2=88=9E,q} tiles have a =
clear
center, but because an {=E2=88=9E,=E2=88=9E} tile is inscribed in the entir=
e boundary,
there is no longer a unique center. Tile centers are at infinity for the
whole progression, so you'd think they would also live at infinity in the
limit. At the same time, all vertices have also become ideal in the limit,
and these are the only points of a tile living at infinity. So every
vertex seems equally valid as a tile center. Weird.
This is good warm-up to jumping up a dimension. The {4,4,3} is kind of
like an {=E2=88=9E,q} with finite q. It's cells are inscribed in horospher=
es, and
have finite vertices and a unique center. The {4,4,4} is like the {=E2=88=
=9E,=E2=88=9E}
because cells are inscribed in the boundary of hyperbolic space. They
don't really have a unique center, and every vertex is ideal. Again, each
vertex sort of acts like a center point.
(Perhaps there is a better way to think about this... Maybe when all the
vertices go to infinity, the cell center should be considered to have
snapped back to being finite? Maybe the center is at some average of all
the ideal vertices or at a center of mass? That makes sense for an ideal
tetrahedron, but can it for a cell that is an ideal {4,4} tiling? I don't
know!)
--14dae93d94b8d2416f04e39e5164
Content-Type: text/html; charset=UTF-8
Content-Transfer-Encoding: quoted-printable
ical model of the exotic {4,4,4} H=C2=B3 honeycomb! =C2=A0
div><=
div>http://shpws.me/oFpu
nfinite number of facets. All vertices are ideal=C2=A0(meaning they live at=
infinity, on the Poincare ball boundary). =C2=A0Four cells meet at every e=
dge, and an infinite number of cells meet at every vertex (the vertex figur=
e is a tiling of squares too). =C2=A0This honeycomb is self-dual.
I printed only half of the Poincare ball in this model, which has =
multiple advantages: you can see inside better, and it saves on printing co=
sts. =C2=A0The view is face-centered, meaning the projection places the cen=
ter of one (ideal) 2D polygon at the center of the ball. =C2=A0An edge-cent=
ered view is also possible. =C2=A0Vertex-centered views are impossible sinc=
e every vertex is ideal. =C2=A0A view centered on the interior of=C2=A0a ce=
ll is possible, but (I think, given my current understanding) a cell-center=
ed view is also impossible. =C2=A0
y one level of recursion. =C2=A0I also experimented with deeper recursion, =
but felt the resulting density inhibited understanding. =C2=A0Probably best=
would be to have two models at different recursion depths side by side to =
study together. =C2=A0I had to artificially increase edge widths near the b=
oundary to make things printable.
rson, so consider ordering one or two of the honeycomb models :) =C2=A0As I=
've heard Henry Segerman comment, the "bandwidth" of informat=
ion is really high. =C2=A0You definitely notice things you wouldn't if =
only viewing them on the computer screen. =C2=A0The {3,6,3} and {6,3,6} are=
very similar to the {4,4,4}, just based on different Euclidean tilings, so=
models of those are surely coming as well.
xotic honeycomb? :D
Cheers,
Roice
=
out the {4,4,4} while working on the model... =C2=A0
nsightful comment. =C2=A0He said:"gmail_quote" style=3D"margin:0px 0px 0px 0.8ex;border-left-width:1px;borde=
r-left-color:rgb(204,204,204);border-left-style:solid;padding-left:1ex">
I believe the first step to understand {4,4,4} is to understand {infinity,<=
br>infinity} in the hyperbolic plane.
I can see now they a=
re indeed quite analogous. =C2=A0Wikipedia has some great pictures of the {=
=E2=88=9E,=E2=88=9E} tiling and {p,q} tilings that approach it by increasin=
g p or q. =C2=A0Check out the progression that starts with an {=E2=88=9E,3}=
tiling and increases q, which is the bottom row of the table here:ding:0px">
horocycles=C2=A0(a c=
ircle of infinite radius with a unique center point on the disk boundary). =
=C2=A0The=C2=A0horocycles=C2=A0increase in size with this progression until=
, in the limit, the inscribing circle is the boundary of the disk itself=
.=C2=A0 Something strange about that is an {=E2=88=9E,=E2=88=9E} tile l=
oses its center. =C2=A0A=C2=A0horocycle=C2=A0has a single center on the bou=
ndary, so the inscribed {=E2=88=9E,q} tiles have a clear center, but becaus=
e an {=E2=88=9E,=E2=88=9E} tile is inscribed in the entire boundary, there =
is no longer a unique center. =C2=A0Tile centers are at infinity for the wh=
ole progression, so you'd think they would also live at infinity in the=
limit. =C2=A0At the same time, all vertices have also become ideal in the =
limit, and these are the only points of a tile living at infinity. =C2=A0So=
every vertex seems equally valid as a tile center. =C2=A0Weird.
This is good warm-up to jumping up a dimension. =C2=A0The {4,4,3} is ki=
nd of like an {=E2=88=9E,q} with finite q. =C2=A0It's cells are inscrib=
ed in horospheres, and have finite vertices and a unique center. =C2=A0The =
{4,4,4} is like the {=E2=88=9E,=E2=88=9E} because cells are inscribed in th=
e boundary of hyperbolic space. =C2=A0They don't really have a unique c=
enter, and every vertex is ideal. =C2=A0Again, each vertex sort of acts lik=
e a center point.=C2=A0
aybe when all the vertices go to infinity, the cell center should be consid=
ered to have snapped back to being finite? Maybe the center is at some aver=
age of all the ideal vertices or at a center of mass? =C2=A0That makes sens=
e for an ideal tetrahedron, but can it for a cell that is an ideal {4,4} ti=
ling? =C2=A0I don't know!)
--14dae93d94b8d2416f04e39e5164--
Summary of the long post:
Roice reinvented broccoli.
Remarkable!=20
A thought on the horocycle of {infinity, infinity} is that when the horocyc=
le becomes the boundary of the disk itself, technically the horocycle does =
not exist any more. The boundary is not part of the Poincare disk model. So=
it's natural not to talk about the "center" of a non-existing horocycle.
Nan
--- In 4D_Cubing@yahoogroups.com, Roice Nelson
>
> Hi all,
>=20
> Check out a new physical model of the exotic {4,4,4} H=C2=B3 honeycomb!
>=20
> http://shpws.me/oFpu
>=20
>=20
> Each cell is a tiling of squares with an infinite number of facets. All
> vertices are ideal (meaning they live at infinity, on the Poincare ball
> boundary). Four cells meet at every edge, and an infinite number of cell=
s
> meet at every vertex (the vertex figure is a tiling of squares too). Thi=
s
> honeycomb is self-dual.
>=20
> I printed only half of the Poincare ball in this model, which has multipl=
e
> advantages: you can see inside better, and it saves on printing costs. T=
he
> view is face-centered, meaning the projection places the center of one
> (ideal) 2D polygon at the center of the ball. An edge-centered view is
> also possible. Vertex-centered views are impossible since every vertex i=
s
> ideal. A view centered on the interior of a cell is possible, but (I
> think, given my current understanding) a cell-centered view is also
> impossible.
>=20
> I rendered one tile and all the tiles around it, so only one level of
> recursion. I also experimented with deeper recursion, but felt the
> resulting density inhibited understanding. Probably best would be to hav=
e
> two models at different recursion depths side by side to study together. =
I
> had to artificially increase edge widths near the boundary to make things
> printable.
>=20
> These things are totally cool to handle in person, so consider ordering o=
ne
> or two of the honeycomb models :) As I've heard Henry Segerman comment,
> the "bandwidth" of information is really high. You definitely notice
> things you wouldn't if only viewing them on the computer screen. The
> {3,6,3} and {6,3,6} are very similar to the {4,4,4}, just based on
> different Euclidean tilings, so models of those are surely coming as well=
.
>=20
> So... whose going to make a puzzle based on this exotic honeycomb? :D
>=20
> Cheers,
> Roice
>=20
>=20
> As a postscript, here are a few thoughts I had about the {4,4,4} while
> working on the model...
>=20
> In a previous thread on the
> {4,4,4}
> Nan made an insightful comment. He said:
>=20
> I believe the first step to understand {4,4,4} is to understand {infinity=
,
> > infinity} in the hyperbolic plane.
>=20
>=20
> I can see now they are indeed quite analogous. Wikipedia has some great
> pictures of the {=E2=88=9E,=E2=88=9E} tiling and {p,q} tilings that appro=
ach it by
> increasing p or q. Check out the progression that starts with an {=E2=88=
=9E,3}
> tiling and increases q, which is the bottom row of the table here:
>=20
> http://en.wikipedia.org/wiki/Uniform_tilings_in_hyperbolic_plane#Regular_=
hyperbolic_tilings
>=20
>=20
> The {=E2=88=9E} polygons are inscribed in
> horocycles
> circle of infinite radius with a unique center point on the disk boundary=
).
> The horocycles increase in size with this progression until, in the limi=
t,
> the inscribing circle is* the boundary of the disk itself.* Something
> strange about that is an {=E2=88=9E,=E2=88=9E} tile loses its center. A =
horocycle has a
> single center on the boundary, so the inscribed {=E2=88=9E,q} tiles have =
a clear
> center, but because an {=E2=88=9E,=E2=88=9E} tile is inscribed in the ent=
ire boundary,
> there is no longer a unique center. Tile centers are at infinity for the
> whole progression, so you'd think they would also live at infinity in the
> limit. At the same time, all vertices have also become ideal in the limi=
t,
> and these are the only points of a tile living at infinity. So every
> vertex seems equally valid as a tile center. Weird.
>=20
> This is good warm-up to jumping up a dimension. The {4,4,3} is kind of
> like an {=E2=88=9E,q} with finite q. It's cells are inscribed in horosph=
eres, and
> have finite vertices and a unique center. The {4,4,4} is like the {=E2=
=88=9E,=E2=88=9E}
> because cells are inscribed in the boundary of hyperbolic space. They
> don't really have a unique center, and every vertex is ideal. Again, eac=
h
> vertex sort of acts like a center point.
>=20
> (Perhaps there is a better way to think about this... Maybe when all the
> vertices go to infinity, the cell center should be considered to have
> snapped back to being finite? Maybe the center is at some average of all
> the ideal vertices or at a center of mass? That makes sense for an ideal
> tetrahedron, but can it for a cell that is an ideal {4,4} tiling? I don'=
t
> know!)
>
--------------070702000207070100030901
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Content-Transfer-Encoding: quoted-printable
Lovely, Roice!
This makes me wonder whether it might be possible to add a 3-color=20
{inf,3}=20
to MagicTile something like this:
groups.yahoo.com/group/4D_Cubing/photos/album/1962624577/pic/908182938/view=
/
-Melinda
On 8/10/2013 2:10 PM, Roice Nelson wrote:
>
>
> Hi all,
>
> Check out a new physical model of the exotic {4,4,4} H=C2=B3 honeycomb!
>
> http://shpws.me/oFpu
>
>
> Each cell is a tiling of squares with an infinite number of facets.=20
> All vertices are ideal (meaning they live at infinity, on the Poincare=20
> ball boundary). Four cells meet at every edge, and an infinite number=20
> of cells meet at every vertex (the vertex figure is a tiling of=20
> squares too). This honeycomb is self-dual.
>
> I printed only half of the Poincare ball in this model, which has=20
> multiple advantages: you can see inside better, and it saves on=20
> printing costs. The view is face-centered, meaning the projection=20
> places the center of one (ideal) 2D polygon at the center of the ball.=20
> An edge-centered view is also possible. Vertex-centered views are=20
> impossible since every vertex is ideal. A view centered on the=20
> interior of a cell is possible, but (I think, given my current=20
> understanding) a cell-centered view is also impossible.
>
> I rendered one tile and all the tiles around it, so only one level of=20
> recursion. I also experimented with deeper recursion, but felt the=20
> resulting density inhibited understanding. Probably best would be to=20
> have two models at different recursion depths side by side to study=20
> together. I had to artificially increase edge widths near the=20
> boundary to make things printable.
>
> These things are totally cool to handle in person, so consider=20
> ordering one or two of the honeycomb models :) As I've heard Henry=20
> Segerman comment, the "bandwidth" of information is really high. You=20
> definitely notice things you wouldn't if only viewing them on the=20
> computer screen. The {3,6,3} and {6,3,6} are very similar to the=20
> {4,4,4}, just based on different Euclidean tilings, so models of those=20
> are surely coming as well.
>
> So... whose going to make a puzzle based on this exotic honeycomb? :D
>
> Cheers,
> Roice
>
>
> As a postscript, here are a few thoughts I had about the {4,4,4} while=20
> working on the model...
>
> In a previous thread on the {4,4,4}=20
>
> an insightful comment. He said:
>
> I believe the first step to understand {4,4,4} is to understand
> {infinity,
> infinity} in the hyperbolic plane.=20
>
>
> I can see now they are indeed quite analogous. Wikipedia has some=20
> great pictures of the {=E2=88=9E,=E2=88=9E} tiling and {p,q} tilings that=
approach it=20
> by increasing p or q. Check out the progression that starts with an=20
> {=E2=88=9E,3} tiling and increases q, which is the bottom row of the tabl=
e here:
>
> http://en.wikipedia.org/wiki/Uniform_tilings_in_hyperbolic_plane#Regu=
lar_hyperbolic_tilings
>
>
> The {=E2=88=9E} polygons are inscribed in horocycles=20
>
> with a unique center point on the disk boundary).=20
> The horocycles increase in size with this progression until, in the=20
> limit, the inscribing circle is*the boundary of the disk itself.*=20=20
> Something strange about that is an {=E2=88=9E,=E2=88=9E} tile loses its c=
enter.=20
> A horocycle has a single center on the boundary, so the inscribed=20
> {=E2=88=9E,q} tiles have a clear center, but because an {=E2=88=9E,=E2=88=
=9E} tile is=20
> inscribed in the entire boundary, there is no longer a unique center.=20
> Tile centers are at infinity for the whole progression, so you'd=20
> think they would also live at infinity in the limit. At the same=20
> time, all vertices have also become ideal in the limit, and these are=20
> the only points of a tile living at infinity. So every vertex seems=20
> equally valid as a tile center. Weird.
>
> This is good warm-up to jumping up a dimension. The {4,4,3} is kind=20
> of like an {=E2=88=9E,q} with finite q. It's cells are inscribed in=20
> horospheres, and have finite vertices and a unique center. The=20
> {4,4,4} is like the {=E2=88=9E,=E2=88=9E} because cells are inscribed in =
the boundary=20
> of hyperbolic space. They don't really have a unique center, and=20
> every vertex is ideal. Again, each vertex sort of acts like a center=20
> point.
>
> (Perhaps there is a better way to think about this... Maybe when all=20
> the vertices go to infinity, the cell center should be considered to=20
> have snapped back to being finite? Maybe the center is at some average=20
> of all the ideal vertices or at a center of mass? That makes sense=20
> for an ideal tetrahedron, but can it for a cell that is an ideal {4,4}=20
> tiling? I don't know!)
>
>
>=20
--------------070702000207070100030901
Content-Type: text/html; charset=UTF-8
Content-Transfer-Encoding: quoted-printable
">
Lovely, Roice!
This makes me wonder whether it might be possible to add a
938/view/">3-color
{inf,3} to MagicTile something like this:
groups.yahoo.com/group/4D_Cubing/photos/album/1962624577/pic/908182938/view=
/
-Melinda
wrote:
com"
type=3D"cite">
honeycomb! =C2=A0
40px;border:none;padding:0px">
facets. All vertices are ideal=C2=A0(meaning they live at infinit=
y,
on the Poincare ball boundary). =C2=A0Four cells meet at every
edge, and an infinite number of cells meet at every vertex
(the vertex figure is a tiling of squares too). =C2=A0This
honeycomb is self-dual.
I printed only half of the Poincare ball in this model, which
has multiple advantages: you can see inside better, and it
saves on printing costs. =C2=A0The view is face-centered, meaning
the projection places the center of one (ideal) 2D polygon at
the center of the ball. =C2=A0An edge-centered view is also
possible. =C2=A0Vertex-centered views are impossible since every
vertex is ideal. =C2=A0A view centered on the interior of=C2=A0a =
cell is
possible, but (I think, given my current understanding) a
cell-centered view is also impossible. =C2=A0
one level of recursion. =C2=A0I also experimented with deeper
recursion, but felt the resulting density inhibited
understanding. =C2=A0Probably best would be to have two models at
different recursion depths side by side to study together. =C2=A0=
I
had to artificially increase edge widths near the boundary to
make things printable.
consider ordering one or two of the honeycomb models :)
=C2=A0As I've heard Henry Segerman comment, the "bandwidth" o=
f
information is really high. =C2=A0You definitely notice thing=
s
you wouldn't if only viewing them on the computer screen.
=C2=A0The {3,6,3} and {6,3,6} are very similar to the {4,4,4}=
,
just based on different Euclidean tilings, so models of
those are surely coming as well.
exotic honeycomb? :D
Cheers,
Roice
the {4,4,4} while working on the model... =C2=A0
ge/1226">a
previous thread on the {4,4,4}, Nan=C2=A0made an
insightful comment. =C2=A0He said:
0.8ex;border-left-width:1px;border-left-color:rgb(204,204,204);border-left-=
style:solid;padding-left:1ex">I
believe the first step to understand {4,4,4} is to understand
{infinity,
infinity} in the hyperbolic plane.
I can see now they are indeed quite analogous. =C2=A0Wikipedia has
some great pictures of the {=E2=88=9E,=E2=88=9E} tiling and {p,q} t=
ilings that
approach it by increasing p or q. =C2=A0Check out the progression
that starts with an {=E2=88=9E,3} tiling and increases q, which is =
the
bottom row of the table here:
40px;border:none;padding:0px">
href=3D"http://en.wikipedia.org/wiki/Horocycle">horocycles>=C2=A0(a
circle of infinite radius with a unique center point on the
disk boundary). =C2=A0The=C2=A0horocycles=C2=A0increase in size=
with this
progression until, in the limit, the inscribing circle is
the boundary of the disk itself.=C2=A0 Something strange
about that is an {=E2=88=9E,=E2=88=9E} tile loses its center.
=C2=A0A=C2=A0horocycle=C2=A0has a single center on the boundary=
, so the
inscribed {=E2=88=9E,q} tiles have a clear center, but because =
an
{=E2=88=9E,=E2=88=9E} tile is inscribed in the entire boundary,=
there is no
longer a unique center. =C2=A0Tile centers are at infinity for
the whole progression, so you'd think they would also live
at infinity in the limit. =C2=A0At the same time, all vertices
have also become ideal in the limit, and these are the only
points of a tile living at infinity. =C2=A0So every vertex seem=
s
equally valid as a tile center. =C2=A0Weird.
This is good warm-up to jumping up a dimension. =C2=A0The {4,4,=
3}
is kind of like an {=E2=88=9E,q} with finite q. =C2=A0It's cell=
s are
inscribed in horospheres, and have finite vertices and a
unique center. =C2=A0The {4,4,4} is like the {=E2=88=9E,=E2=88=
=9E} because cells
are inscribed in the boundary of hyperbolic space. =C2=A0They
don't really have a unique center, and every vertex is
ideal. =C2=A0Again, each vertex sort of acts like a center
point.=C2=A0
Maybe when all the vertices go to infinity, the cell center
should be considered to have snapped back to being finite?
Maybe the center is at some average of all the ideal
vertices or at a center of mass? =C2=A0That makes sense for an
ideal tetrahedron, but can it for a cell that is an ideal
{4,4} tiling? =C2=A0I don't know!)
=20=20=20=20=20=20
--------------070702000207070100030901--
--------------010400020305010508080402
Content-Type: text/plain; charset=UTF-8; format=flowed
Content-Transfer-Encoding: quoted-printable
Here's a slightly less awful sketch:
http://groups.yahoo.com/group/4D_Cubing/photos/album/1962624577/pic/4087515=
2/view/
On 8/11/2013 4:34 PM, Melinda Green wrote:
>
>
> Lovely, Roice!
>
> This makes me wonder whether it might be possible to add a 3-color=20
> {inf,3}=20
>
> to MagicTile something like this:
> groups.yahoo.com/group/4D_Cubing/photos/album/1962624577/pic/908182938/vi=
ew/
>
> -Melinda
>
> On 8/10/2013 2:10 PM, Roice Nelson wrote:
>> Hi all,
>>
>> Check out a new physical model of the exotic {4,4,4} H=C2=B3 honeycomb!
>>
>> http://shpws.me/oFpu
>>
>>
>> Each cell is a tiling of squares with an infinite number of facets.=20
>> All vertices are ideal (meaning they live at infinity, on the=20
>> Poincare ball boundary). Four cells meet at every edge, and an=20
>> infinite number of cells meet at every vertex (the vertex figure is a=20
>> tiling of squares too). This honeycomb is self-dual.
>>
>> I printed only half of the Poincare ball in this model, which has=20
>> multiple advantages: you can see inside better, and it saves on=20
>> printing costs. The view is face-centered, meaning the projection=20
>> places the center of one (ideal) 2D polygon at the center of the=20
>> ball. An edge-centered view is also possible. Vertex-centered views=20
>> are impossible since every vertex is ideal. A view centered on the=20
>> interior of a cell is possible, but (I think, given my current=20
>> understanding) a cell-centered view is also impossible.
>>
>> I rendered one tile and all the tiles around it, so only one level of=20
>> recursion. I also experimented with deeper recursion, but felt the=20
>> resulting density inhibited understanding. Probably best would be to=20
>> have two models at different recursion depths side by side to study=20
>> together. I had to artificially increase edge widths near the=20
>> boundary to make things printable.
>>
>> These things are totally cool to handle in person, so consider=20
>> ordering one or two of the honeycomb models :) As I've heard Henry=20
>> Segerman comment, the "bandwidth" of information is really high. You=20
>> definitely notice things you wouldn't if only viewing them on the=20
>> computer screen. The {3,6,3} and {6,3,6} are very similar to the=20
>> {4,4,4}, just based on different Euclidean tilings, so models of=20
>> those are surely coming as well.
>>
>> So... whose going to make a puzzle based on this exotic honeycomb? :D
>>
>> Cheers,
>> Roice
>>
>>
>> As a postscript, here are a few thoughts I had about the {4,4,4}=20
>> while working on the model...
>>
>> In a previous thread on the {4,4,4}=20
>>
>> Nan made an insightful comment. He said:
>>
>> I believe the first step to understand {4,4,4} is to understand
>> {infinity,
>> infinity} in the hyperbolic plane.=20
>>
>>
>> I can see now they are indeed quite analogous. Wikipedia has some=20
>> great pictures of the {=E2=88=9E,=E2=88=9E} tiling and {p,q} tilings tha=
t approach it=20
>> by increasing p or q. Check out the progression that starts with an=20
>> {=E2=88=9E,3} tiling and increases q, which is the bottom row of the tab=
le here:
>>
>> http://en.wikipedia.org/wiki/Uniform_tilings_in_hyperbolic_plane#Reg=
ular_hyperbolic_tilings
>>
>>
>> The {=E2=88=9E} polygons are inscribed in horocycles=20
>>
>> with a unique center point on the disk boundary).=20
>> The horocycles increase in size with this progression until, in the=20
>> limit, the inscribing circle is*the boundary of the disk itself.*=20
>> Something strange about that is an {=E2=88=9E,=E2=88=9E} tile loses its =
center.=20
>> A horocycle has a single center on the boundary, so the inscribed=20
>> {=E2=88=9E,q} tiles have a clear center, but because an {=E2=88=9E,=E2=
=88=9E} tile is=20
>> inscribed in the entire boundary, there is no longer a unique center.=20
>> Tile centers are at infinity for the whole progression, so you'd=20
>> think they would also live at infinity in the limit. At the same=20
>> time, all vertices have also become ideal in the limit, and these are=20
>> the only points of a tile living at infinity. So every vertex seems=20
>> equally valid as a tile center. Weird.
>>
>> This is good warm-up to jumping up a dimension. The {4,4,3} is kind=20
>> of like an {=E2=88=9E,q} with finite q. It's cells are inscribed in=20
>> horospheres, and have finite vertices and a unique center. The=20
>> {4,4,4} is like the {=E2=88=9E,=E2=88=9E} because cells are inscribed in=
the boundary=20
>> of hyperbolic space. They don't really have a unique center, and=20
>> every vertex is ideal. Again, each vertex sort of acts like a center=20
>> point.
>>
>> (Perhaps there is a better way to think about this... Maybe when all=20
>> the vertices go to infinity, the cell center should be considered to=20
>> have snapped back to being finite? Maybe the center is at some=20
>> average of all the ideal vertices or at a center of mass? That makes=20
>> sense for an ideal tetrahedron, but can it for a cell that is an=20
>> ideal {4,4} tiling? I don't know!)
>
>
>
>=20
--------------010400020305010508080402
Content-Type: text/html; charset=UTF-8
Content-Transfer-Encoding: quoted-printable
">
Here's a slightly less awful sketch:
/40875152/view/">http://groups.yahoo.com/group/4D_Cubing/photos/album/19626=
24577/pic/40875152/view/
wrote:
">
pe">
Lovely, Roice!
This makes me wonder whether it might be possible to add a
href=3D"groups.yahoo.com/group/4D_Cubing/photos/album/1962624577/pic/908182=
938/view/">3-color
{inf,3} to MagicTile something like this:
groups.yahoo.com/group/4D_Cubing/photos/album/1962624577/pic/908182938/view=
/
-Melinda
wrote:
com"
type=3D"cite">
honeycomb! =C2=A0
40px;border:none;padding:0px">
of facets. All vertices are ideal=C2=A0(meaning they live at
infinity, on the Poincare ball boundary). =C2=A0Four cells meet
at every edge, and an infinite number of cells meet at every
vertex (the vertex figure is a tiling of squares too). =C2=A0Th=
is
honeycomb is self-dual.
I printed only half of the Poincare ball in this model,
which has multiple advantages: you can see inside better,
and it saves on printing costs. =C2=A0The view is face-centered=
,
meaning the projection places the center of one (ideal) 2D
polygon at the center of the ball. =C2=A0An edge-centered view =
is
also possible. =C2=A0Vertex-centered views are impossible since
every vertex is ideal. =C2=A0A view centered on the interior of=
=C2=A0a
cell is possible, but (I think, given my current
understanding) a cell-centered view is also impossible. =C2=A0<=
/div>
one level of recursion. =C2=A0I also experimented with deeper
recursion, but felt the resulting density inhibited
understanding. =C2=A0Probably best would be to have two models =
at
different recursion depths side by side to study together.
=C2=A0I had to artificially increase edge widths near the
boundary to make things printable.
consider ordering one or two of the honeycomb models :)
=C2=A0As I've heard Henry Segerman comment, the "bandwidth"
of information is really high. =C2=A0You definitely notice
things you wouldn't if only viewing them on the computer
screen. =C2=A0The {3,6,3} and {6,3,6} are very similar to t=
he
{4,4,4}, just based on different Euclidean tilings, so
models of those are surely coming as well.
exotic honeycomb? :D
Cheers,
Roice
the {4,4,4} while working on the model... =C2=A0
sage/1226">a
previous thread on the {4,4,4}, Nan=C2=A0made an
insightful comment. =C2=A0He said:
0.8ex;border-left-width:1px;border-left-color:rgb(204,204,204);border-left-=
style:solid;padding-left:1ex">I
believe the first step to understand {4,4,4} is to
understand {infinity,
infinity} in the hyperbolic plane.
I can see now they are indeed quite analogous. =C2=A0Wikipedia ha=
s
some great pictures of the {=E2=88=9E,=E2=88=9E} tiling and {p,q}=
tilings that
approach it by increasing p or q. =C2=A0Check out the progression
that starts with an {=E2=88=9E,3} tiling and increases q, which i=
s the
bottom row of the table here:
40px;border:none;padding:0px">
href=3D"http://en.wikipedia.org/wiki/Horocycle">horocycles<=
/a>=C2=A0(a
circle of infinite radius with a unique center point on
the disk boundary). =C2=A0The=C2=A0horocycles=C2=A0increase i=
n size with
this progression until, in the limit, the inscribing
circle is the boundary of the disk itself.=C2=A0
Something strange about that is an {=E2=88=9E,=E2=88=9E} tile=
loses its
center. =C2=A0A=C2=A0horocycle=C2=A0has a single center on th=
e boundary,
so the inscribed {=E2=88=9E,q} tiles have a clear center, but
because an {=E2=88=9E,=E2=88=9E} tile is inscribed in the ent=
ire boundary,
there is no longer a unique center. =C2=A0Tile centers are at
infinity for the whole progression, so you'd think they
would also live at infinity in the limit. =C2=A0At the same
time, all vertices have also become ideal in the limit,
and these are the only points of a tile living at
infinity. =C2=A0So every vertex seems equally valid as a tile
center. =C2=A0Weird.
This is good warm-up to jumping up a dimension. =C2=A0The
{4,4,3} is kind of like an {=E2=88=9E,q} with finite q. =C2=
=A0It's
cells are inscribed in horospheres, and have finite
vertices and a unique center. =C2=A0The {4,4,4} is like the
{=E2=88=9E,=E2=88=9E} because cells are inscribed in the boun=
dary of
hyperbolic space. =C2=A0They don't really have a unique cente=
r,
and every vertex is ideal. =C2=A0Again, each vertex sort of
acts like a center point.=C2=A0
Maybe when all the vertices go to infinity, the cell
center should be considered to have snapped back to being
finite? Maybe the center is at some average of all the
ideal vertices or at a center of mass? =C2=A0That makes sense
for an ideal tetrahedron, but can it for a cell that is an
ideal {4,4} tiling? =C2=A0I don't know!)
=20=20=20=20=20=20
--------------010400020305010508080402--
From: Roice Nelson <roice3@gmail.com>
Date: Sun, 11 Aug 2013 22:06:17 -0500
Subject: Re: [MC4D] The exotic {4,4,4}
--089e011770db70448304e3b7689b
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: quoted-printable
The puzzle in your pictures *needs* to be made!
It feels like the current MagicTile engine will fall woefully short for
this task, though maybe I am overestimating the difficulty. Off the cuff,
an approach could be to try to allow building up puzzles using fundamental
domain triangles rather than entire tiles, because it will be necessary to
only show portions of these infinite-faceted tiles. (In the past, I've
wondered if that enhancement is going to be necessary for uniform tilings.)
It does seem like a big piece of work, and it might even be easier to
write some special-case code for this puzzle rather than attempting to fit
it into the engine.
I bet there is an infinite set of coloring possibilities for this tiling
too.
On Sun, Aug 11, 2013 at 7:19 PM, Melinda Green
ote:
>
>
> Here's a slightly less awful sketch:
>
> http://groups.yahoo.com/group/4D_Cubing/photos/album/1962624577/pic/40875=
152/view/
>
>
> On 8/11/2013 4:34 PM, Melinda Green wrote:
>
> Lovely, Roice!
>
> This makes me wonder whether it might be possible to add a 3-color {inf,3=
}
>
> groups.yahoo.com/group/4D_Cubing/photos/album/1962624577/pic/908182938/vi=
ew/
>
> -Melinda
>
> On 8/10/2013 2:10 PM, Roice Nelson wrote:
>
> Hi all,
>
> Check out a new physical model of the exotic {4,4,4} H=C2=B3 honeycomb!
>
> http://shpws.me/oFpu
>
>
> Each cell is a tiling of squares with an infinite number of facets. All
> vertices are ideal (meaning they live at infinity, on the Poincare ball
> boundary). Four cells meet at every edge, and an infinite number of cell=
s
> meet at every vertex (the vertex figure is a tiling of squares too). Thi=
s
> honeycomb is self-dual.
>
> I printed only half of the Poincare ball in this model, which has multipl=
e
> advantages: you can see inside better, and it saves on printing costs. T=
he
> view is face-centered, meaning the projection places the center of one
> (ideal) 2D polygon at the center of the ball. An edge-centered view is
> also possible. Vertex-centered views are impossible since every vertex i=
s
> ideal. A view centered on the interior of a cell is possible, but (I
> think, given my current understanding) a cell-centered view is also
> impossible.
>
> I rendered one tile and all the tiles around it, so only one level of
> recursion. I also experimented with deeper recursion, but felt the
> resulting density inhibited understanding. Probably best would be to hav=
e
> two models at different recursion depths side by side to study together. =
I
> had to artificially increase edge widths near the boundary to make things
> printable.
>
> These things are totally cool to handle in person, so consider ordering
> one or two of the honeycomb models :) As I've heard Henry Segerman
> comment, the "bandwidth" of information is really high. You definitely
> notice things you wouldn't if only viewing them on the computer screen.
> The {3,6,3} and {6,3,6} are very similar to the {4,4,4}, just based on
> different Euclidean tilings, so models of those are surely coming as well=
.
>
> So... whose going to make a puzzle based on this exotic honeycomb? :D
>
> Cheers,
> Roice
>
>
> As a postscript, here are a few thoughts I had about the {4,4,4} while
> working on the model...
>
> In a previous thread on the {4,4,4}
> Nan made an insightful comment. He said:
>
> I believe the first step to understand {4,4,4} is to understand {infinity=
,
>> infinity} in the hyperbolic plane.
>
>
> I can see now they are indeed quite analogous. Wikipedia has some great
> pictures of the {=E2=88=9E,=E2=88=9E} tiling and {p,q} tilings that appro=
ach it by
> increasing p or q. Check out the progression that starts with an {=E2=88=
=9E,3}
> tiling and increases q, which is the bottom row of the table here:
>
>
> http://en.wikipedia.org/wiki/Uniform_tilings_in_hyperbolic_plane#Regular_=
hyperbolic_tilings
>
>
> The {=E2=88=9E} polygons are inscribed in horocycles
> circle of infinite radius with a unique center point on the disk boundary=
).
> The horocycles increase in size with this progression until, in the limi=
t,
> the inscribing circle is* the boundary of the disk itself.* Something
> strange about that is an {=E2=88=9E,=E2=88=9E} tile loses its center. A =
horocycle has a
> single center on the boundary, so the inscribed {=E2=88=9E,q} tiles have =
a clear
> center, but because an {=E2=88=9E,=E2=88=9E} tile is inscribed in the ent=
ire boundary,
> there is no longer a unique center. Tile centers are at infinity for the
> whole progression, so you'd think they would also live at infinity in the
> limit. At the same time, all vertices have also become ideal in the limi=
t,
> and these are the only points of a tile living at infinity. So every
> vertex seems equally valid as a tile center. Weird.
>
> This is good warm-up to jumping up a dimension. The {4,4,3} is kind of
> like an {=E2=88=9E,q} with finite q. It's cells are inscribed in horosph=
eres, and
> have finite vertices and a unique center. The {4,4,4} is like the {=E2=
=88=9E,=E2=88=9E}
> because cells are inscribed in the boundary of hyperbolic space. They
> don't really have a unique center, and every vertex is ideal. Again, eac=
h
> vertex sort of acts like a center point.
>
> (Perhaps there is a better way to think about this... Maybe when all the
> vertices go to infinity, the cell center should be considered to have
> snapped back to being finite? Maybe the center is at some average of all
> the ideal vertices or at a center of mass? That makes sense for an ideal
> tetrahedron, but can it for a cell that is an ideal {4,4} tiling? I don'=
t
> know!)
>
>
>
>
>
>=20
>
--089e011770db70448304e3b7689b
Content-Type: text/html; charset=UTF-8
Content-Transfer-Encoding: quoted-printable
div>
t for this task, though maybe I am overestimating the difficulty. =C2=A0Off=
the cuff, an approach could be to try to allow building up puzzles using f=
undamental domain triangles rather than entire tiles, because it will be ne=
cessary to only show portions of these infinite-faceted tiles. =C2=A0(In th=
e past, I've wondered if that=C2=A0enhancement is going to be necessary=
for uniform tilings.) =C2=A0It does seem like a big piece of work, and it =
might even be easier to write some special-case code for this puzzle rather=
than attempting to fit it into the engine.
s for this tiling too.
a Green <arget=3D"_blank">melinda@superliminal.com> wrote:x #ccc solid;padding-left:1ex">
=20=20=20=20=20=20=20=20
=20=20
=20=20=20=20
=20=20
Here's a slightly less awful sketch:
577/pic/40875152/view/" target=3D"_blank">http://groups.yahoo.com/group/4D_=
Cubing/photos/album/1962624577/pic/40875152/view/
wrote:
=20=20=20=20=20=20
=20=20=20=20=20=20
Lovely, Roice!
This makes me wonder whether it might be possible to add a "http://groups.yahoo.com/group/4D_Cubing/photos/album/1962624577/pic/908182=
938/view/" target=3D"_blank">3-color
{inf,3} to MagicTile something like this:
pic/908182938/view/" target=3D"_blank">groups.yahoo.com/group/4D_Cubing/pho=
tos/album/1962624577/pic/908182938/view/
-Melinda
wrote:
=20=20=20=20=20=20=20=20
=20=20=20=20=20=20=20=20
honeycomb! =C2=A0
0px">
of facets. All vertices are ideal=C2=A0(meaning they live at
infinity, on the Poincare ball boundary). =C2=A0Four cells meet
at every edge, and an infinite number of cells meet at every
vertex (the vertex figure is a tiling of squares too). =C2=A0Th=
is
honeycomb is self-dual.
I printed only half of the Poincare ball in this model,
which has multiple advantages: you can see inside better,
and it saves on printing costs. =C2=A0The view is face-centered=
,
meaning the projection places the center of one (ideal) 2D
polygon at the center of the ball. =C2=A0An edge-centered view =
is
also possible. =C2=A0Vertex-centered views are impossible since
every vertex is ideal. =C2=A0A view centered on the interior of=
=C2=A0a
cell is possible, but (I think, given my current
understanding) a cell-centered view is also impossible. =C2=A0<=
/div>
one level of recursion. =C2=A0I also experimented with deeper
recursion, but felt the resulting density inhibited
understanding. =C2=A0Probably best would be to have two models =
at
different recursion depths side by side to study together.
=C2=A0I had to artificially increase edge widths near the
boundary to make things printable.
consider ordering one or two of the honeycomb models :)
=C2=A0As I've heard Henry Segerman comment, the "b=
andwidth"
of information is really high. =C2=A0You definitely notice
things you wouldn't if only viewing them on the compute=
r
screen. =C2=A0The {3,6,3} and {6,3,6} are very similar to t=
he
{4,4,4}, just based on different Euclidean tilings, so
models of those are surely coming as well.
exotic honeycomb? :D
Cheers,
Roice
the {4,4,4} while working on the model... =C2=A0
previous thread on the {4,4,4}, Nan=C2=A0made an
insightful comment. =C2=A0He said:
ex;border-left-width:1px;border-left-color:rgb(204,204,204);border-left-sty=
le:solid;padding-left:1ex">I
believe the first step to understand {4,4,4} is to
understand {infinity,
infinity} in the hyperbolic plane.
I can see now they are indeed quite analogous. =C2=A0Wikipedia ha=
s
some great pictures of the {=E2=88=9E,=E2=88=9E} tiling and {p,q}=
tilings that
approach it by increasing p or q. =C2=A0Check out the progression
that starts with an {=E2=88=9E,3} tiling and increases q, which i=
s the
bottom row of the table here:
g:0px">
circle of infinite radius with a unique center point on
the disk boundary). =C2=A0The=C2=A0horocycles=C2=A0increase i=
n size with
this progression until, in the limit, the inscribing
circle is the boundary of the disk itself.=C2=A0
Something strange about that is an {=E2=88=9E,=E2=88=9E} tile=
loses its
center. =C2=A0A=C2=A0horocycle=C2=A0has a single center on th=
e boundary,
so the inscribed {=E2=88=9E,q} tiles have a clear center, but
because an {=E2=88=9E,=E2=88=9E} tile is inscribed in the ent=
ire boundary,
there is no longer a unique center. =C2=A0Tile centers are at
infinity for the whole progression, so you'd think they
would also live at infinity in the limit. =C2=A0At the same
time, all vertices have also become ideal in the limit,
and these are the only points of a tile living at
infinity. =C2=A0So every vertex seems equally valid as a tile
center. =C2=A0Weird.
This is good warm-up to jumping up a dimension. =C2=A0The
{4,4,3} is kind of like an {=E2=88=9E,q} with finite q. =C2=
=A0It's
cells are inscribed in horospheres, and have finite
vertices and a unique center. =C2=A0The {4,4,4} is like the
{=E2=88=9E,=E2=88=9E} because cells are inscribed in the boun=
dary of
hyperbolic space. =C2=A0They don't really have a unique c=
enter,
and every vertex is ideal. =C2=A0Again, each vertex sort of
acts like a center point.=C2=A0
Maybe when all the vertices go to infinity, the cell
center should be considered to have snapped back to being
finite? Maybe the center is at some average of all the
ideal vertices or at a center of mass? =C2=A0That makes sense
for an ideal tetrahedron, but can it for a cell that is an
ideal {4,4} tiling? =C2=A0I don't know!)
=20=20=20=20=20=20
=20=20
=20=20=20=20
=20=20=20=20
--089e011770db70448304e3b7689b--
From: Melinda Green <melinda@superliminal.com>
Date: Sun, 11 Aug 2013 20:32:30 -0700
Subject: Re: [MC4D] The exotic {4,4,4}
--------------070308070208030708040901
Content-Type: text/plain; charset=UTF-8; format=flowed
Content-Transfer-Encoding: quoted-printable
Hello Roice,
I'm glad that you think that this puzzle makes sense. Also, I like your=20
idea of using fundamental domain triangles. As for other colorings (and=20
topologies), I would first hope to see the simplest one(s) first. This=20
3-coloring seems about as simple as possible though perhaps one could=20
remove an edge or two by torturing the topology a bit. As for=20
incorporating into MT versus creating a stand-alone puzzle, I have a=20
feeling that there might be some clever ways to incorporate it. One way=20
might be to implement it as a {N,3} for some large N. If a user were to=20
pan far enough to see the ragged edge, so be it. If it must be a=20
stand-alone puzzle, it might allow for your alternate colorings and=20
perhaps other interesting variants that would otherwise be too difficult.
-Melinda
On 8/11/2013 8:06 PM, Roice Nelson wrote:
>
>
> The puzzle in your pictures *needs* to be made!
>
> It feels like the current MagicTile engine will fall woefully short=20
> for this task, though maybe I am overestimating the difficulty. Off=20
> the cuff, an approach could be to try to allow building up puzzles=20
> using fundamental domain triangles rather than entire tiles, because=20
> it will be necessary to only show portions of these infinite-faceted=20
> tiles. (In the past, I've wondered if that enhancement is going to be=20
> necessary for uniform tilings.) It does seem like a big piece of=20
> work, and it might even be easier to write some special-case code for=20
> this puzzle rather than attempting to fit it into the engine.
>
> I bet there is an infinite set of coloring possibilities for this=20
> tiling too.
>
>
>
> On Sun, Aug 11, 2013 at 7:19 PM, Melinda Green=20
>
>
>
>
> Here's a slightly less awful sketch:
> http://groups.yahoo.com/group/4D_Cubing/photos/album/1962624577/pic/4=
0875152/view/
>
>
>
> On 8/11/2013 4:34 PM, Melinda Green wrote:
>> Lovely, Roice!
>>
>> This makes me wonder whether it might be possible to add a
>> 3-color {inf,3}
>>
>> to MagicTile something like this:
>> groups.yahoo.com/group/4D_Cubing/photos/album/1962624577/pic/9081829=
38/view/
>>
>>
>> -Melinda
>>
>> On 8/10/2013 2:10 PM, Roice Nelson wrote:
>>> Hi all,
>>>
>>> Check out a new physical model of the exotic {4,4,4} H=C2=B3 honeyc=
omb!
>>>
>>> http://shpws.me/oFpu
>>>
>>>
>>> Each cell is a tiling of squares with an infinite number of
>>> facets. All vertices are ideal (meaning they live at infinity,
>>> on the Poincare ball boundary). Four cells meet at every edge,
>>> and an infinite number of cells meet at every vertex (the vertex
>>> figure is a tiling of squares too). This honeycomb is self-dual.
>>>
>>> I printed only half of the Poincare ball in this model, which
>>> has multiple advantages: you can see inside better, and it saves
>>> on printing costs. The view is face-centered, meaning the
>>> projection places the center of one (ideal) 2D polygon at the
>>> center of the ball. An edge-centered view is also possible.
>>> Vertex-centered views are impossible since every vertex is
>>> ideal. A view centered on the interior of a cell is possible,
>>> but (I think, given my current understanding) a cell-centered
>>> view is also impossible.
>>>
>>> I rendered one tile and all the tiles around it, so only one
>>> level of recursion. I also experimented with deeper recursion,
>>> but felt the resulting density inhibited understanding.
>>> Probably best would be to have two models at different
>>> recursion depths side by side to study together. I had to
>>> artificially increase edge widths near the boundary to make
>>> things printable.
>>>
>>> These things are totally cool to handle in person, so consider
>>> ordering one or two of the honeycomb models :) As I've heard
>>> Henry Segerman comment, the "bandwidth" of information is really
>>> high. You definitely notice things you wouldn't if only viewing
>>> them on the computer screen. The {3,6,3} and {6,3,6} are very
>>> similar to the {4,4,4}, just based on different Euclidean
>>> tilings, so models of those are surely coming as well.
>>>
>>> So... whose going to make a puzzle based on this exotic
>>> honeycomb? :D
>>>
>>> Cheers,
>>> Roice
>>>
>>>
>>> As a postscript, here are a few thoughts I had about the {4,4,4}
>>> while working on the model...
>>>
>>> In a previous thread on the {4,4,4}
>>>
>>> Nan made an insightful comment. He said:
>>>
>>> I believe the first step to understand {4,4,4} is to
>>> understand {infinity,
>>> infinity} in the hyperbolic plane.=20
>>>
>>>
>>> I can see now they are indeed quite analogous. Wikipedia has
>>> some great pictures of the {=E2=88=9E,=E2=88=9E} tiling and {p,q} t=
ilings that
>>> approach it by increasing p or q. Check out the progression
>>> that starts with an {=E2=88=9E,3} tiling and increases q, which is =
the
>>> bottom row of the table here:
>>>
>>> http://en.wikipedia.org/wiki/Uniform_tilings_in_hyperbolic_plan=
e#Regular_hyperbolic_tilings
>>>
>>>
>>> The {=E2=88=9E} polygons are inscribed in horocycles
>>>
>>> radius with a unique center point on the disk boundary).
>>> The horocycles increase in size with this progression until, in
>>> the limit, the inscribing circle is*the boundary of the disk
>>> itself.* Something strange about that is an {=E2=88=9E,=E2=88=9E} =
tile loses
>>> its center. A horocycle has a single center on the boundary, so
>>> the inscribed {=E2=88=9E,q} tiles have a clear center, but because =
an
>>> {=E2=88=9E,=E2=88=9E} tile is inscribed in the entire boundary, the=
re is no
>>> longer a unique center. Tile centers are at infinity for the
>>> whole progression, so you'd think they would also live at
>>> infinity in the limit. At the same time, all vertices have also
>>> become ideal in the limit, and these are the only points of a
>>> tile living at infinity. So every vertex seems equally valid as
>>> a tile center. Weird.
>>>
>>> This is good warm-up to jumping up a dimension. The {4,4,3} is
>>> kind of like an {=E2=88=9E,q} with finite q. It's cells are inscri=
bed
>>> in horospheres, and have finite vertices and a unique center.
>>> The {4,4,4} is like the {=E2=88=9E,=E2=88=9E} because cells are in=
scribed in
>>> the boundary of hyperbolic space. They don't really have a
>>> unique center, and every vertex is ideal. Again, each vertex
>>> sort of acts like a center point.
>>>
>>> (Perhaps there is a better way to think about this... Maybe when
>>> all the vertices go to infinity, the cell center should be
>>> considered to have snapped back to being finite? Maybe the
>>> center is at some average of all the ideal vertices or at a
>>> center of mass? That makes sense for an ideal tetrahedron, but
>>> can it for a cell that is an ideal {4,4} tiling? I don't know!)
>>
>
>
>
>
>
>
>=20
--------------070308070208030708040901
Content-Type: text/html; charset=UTF-8
Content-Transfer-Encoding: quoted-printable
">
Hello Roice,
I'm glad that you think that this puzzle makes sense. Also, I like
your idea of using fundamental domain triangles. As for other
colorings (and topologies), I would first hope to see the simplest
one(s) first. This 3-coloring seems about as simple as possible
though perhaps one could remove an edge or two by torturing the
topology a bit. As for incorporating into MT versus creating a
stand-alone puzzle, I have a feeling that there might be some clever
ways to incorporate it. One way might be to implement it as a {N,3}=C2=
=A0
for some large N. If a user were to pan far enough to see the ragged
edge, so be it. If it must be a stand-alone puzzle, it might allow
for your alternate colorings and perhaps other interesting variants
that would otherwise be too difficult.
-Melinda
wrote:
com"
type=3D"cite">
woefully short for this task, though maybe I am overestimating
the difficulty. =C2=A0Off the cuff, an approach could be to try t=
o
allow building up puzzles using fundamental domain triangles
rather than entire tiles, because it will be necessary to only
show portions of these infinite-faceted tiles. =C2=A0(In the past=
,
I've wondered if that=C2=A0enhancement is going to be necessary f=
or
uniform tilings.) =C2=A0It does seem like a big piece of work, an=
d
it might even be easier to write some special-case code for
this puzzle rather than attempting to fit it into the engine.v>
for this tiling too.
Melinda Green < href=3D"mailto:melinda@superliminal.com" target=3D"_blank">me=
linda@superliminal.com>
wrote:
.8ex;border-left:1px #ccc solid;padding-left:1ex">
Here's a slightly less awful sketch:
href=3D"http://groups.yahoo.com/group/4D_Cubing/photos/album/1962624577/pic=
/40875152/view/"
target=3D"_blank">http://groups.yahoo.com/group/4D_Cubing/p=
hotos/album/1962624577/pic/40875152/view/
Lovely, Roice!
This makes me wonder whether it might be possible to
add a href=3D"http://groups.yahoo.com/group/4D_Cubing/photos/album/1962624577/pic=
/908182938/view/"
target=3D"_blank">3-color {inf,3} to MagicTile
something like this:
href=3D"http://groups.yahoo.com/group/4D_Cubing/photos/album/1962624577/pic=
/908182938/view/"
target=3D"_blank">groups.yahoo.com/group/4D_Cubing/ph=
otos/album/1962624577/pic/908182938/view/
-Melinda
exotic {4,4,4} H=C2=B3 honeycomb! =C2=A0
40px;border:none;padding:0px">
target=3D"_blank">http://shpws.me/oFpuiv>
infinite number of facets. All vertices are
ideal=C2=A0(meaning they live at infinity, on the
Poincare ball boundary). =C2=A0Four cells meet at
every edge, and an infinite number of cells
meet at every vertex (the vertex figure is a
tiling of squares too). =C2=A0This honeycomb is
self-dual.
I printed only half of the Poincare ball in
this model, which has multiple advantages: you
can see inside better, and it saves on
printing costs. =C2=A0The view is face-centered,
meaning the projection places the center of
one (ideal) 2D polygon at the center of the
ball. =C2=A0An edge-centered view is also possibl=
e.
=C2=A0Vertex-centered views are impossible since
every vertex is ideal. =C2=A0A view centered on t=
he
interior of=C2=A0a cell is possible, but (I think=
,
given my current understanding) a
cell-centered view is also impossible. =C2=A0v>
around it, so only one level of recursion. =C2=A0=
I
also experimented with deeper recursion, but
felt the resulting density inhibited
understanding. =C2=A0Probably best would be to ha=
ve
two models at different recursion depths side
by side to study together. =C2=A0I had to
artificially increase edge widths near the
boundary to make things printable.
in person, so consider ordering one or two
of the honeycomb models :) =C2=A0As I've hear=
d
Henry Segerman comment, the "bandwidth" of
information is really high. =C2=A0You
definitely notice things you wouldn't if
only viewing them on the computer screen.
=C2=A0The {3,6,3} and {6,3,6} are very simila=
r
to the {4,4,4}, just based on different
Euclidean tilings, so models of those are
surely coming as well.
based on this exotic honeycomb? :D
Cheers,
Roice
thoughts I had about the {4,4,4} while
working on the model... =C2=A0
/4D_Cubing/message/1226"
target=3D"_blank">a previous thread on the
{4,4,4}, Nan=C2=A0made an insightful
comment. =C2=A0He said:
style=3D"margin:0px 0px 0px
0.8ex;border-left-width:1px;border-left-color:rgb(204,204,204);border-left-=
style:solid;padding-left:1ex">I
believe the first step to understand {4,4,4}
is to understand {infinity,
infinity} in the hyperbolic plane.
I can see now they are indeed quite analogous.
=C2=A0Wikipedia has some great pictures of the {=E2=
=88=9E,=E2=88=9E}
tiling and {p,q} tilings that approach it by
increasing p or q. =C2=A0Check out the progression
that starts with an {=E2=88=9E,3} tiling and increa=
ses
q, which is the bottom row of the table here:
40px;border:none;padding:0px">
gular_hyperbolic_tilings"
target=3D"_blank">http://en.wikipedia.org/w=
iki/Uniform_tilings_in_hyperbolic_plane#Regular_hyperbolic_tilings>
href=3D"http://en.wikipedia.org/wiki/Horocycl=
e"
target=3D"_blank">horocycles=C2=A0(a circ=
le
of infinite radius with a unique center
point on the disk boundary).
=C2=A0The=C2=A0horocycles=C2=A0increase in size=
with this
progression until, in the limit, the
inscribing circle is the boundary of the
disk itself.=C2=A0 Something strange abou=
t
that is an {=E2=88=9E,=E2=88=9E} tile loses its=
center.
=C2=A0A=C2=A0horocycle=C2=A0has a single center=
on the
boundary, so the inscribed {=E2=88=9E,q} tiles =
have
a clear center, but because an {=E2=88=9E,=E2=
=88=9E} tile is
inscribed in the entire boundary, there is
no longer a unique center. =C2=A0Tile centers a=
re
at infinity for the whole progression, so
you'd think they would also live at infinity
in the limit. =C2=A0At the same time, all
vertices have also become ideal in the
limit, and these are the only points of a
tile living at infinity. =C2=A0So every vertex
seems equally valid as a tile center.
=C2=A0Weird.
This is good warm-up to jumping up a
dimension. =C2=A0The {4,4,3} is kind of like an
{=E2=88=9E,q} with finite q. =C2=A0It's cells a=
re
inscribed in horospheres, and have finite
vertices and a unique center. =C2=A0The {4,4,4}
is like the {=E2=88=9E,=E2=88=9E} because cells=
are
inscribed in the boundary of hyperbolic
space. =C2=A0They don't really have a unique
center, and every vertex is ideal. =C2=A0Again,
each vertex sort of acts like a center
point.=C2=A0
about this... Maybe when all the vertices go
to infinity, the cell center should be
considered to have snapped back to being
finite? Maybe the center is at some average
of all the ideal vertices or at a center of
mass? =C2=A0That makes sense for an ideal
tetrahedron, but can it for a cell that is
an ideal {4,4} tiling? =C2=A0I don't know!)v>
=20=20=20=20=20=20
--------------070308070208030708040901--
From: Roice Nelson <roice3@gmail.com>
Date: Mon, 12 Aug 2013 19:37:40 -0500
Subject: Re: [MC4D] The exotic {4,4,4}
--089e014940eac9cd2204e3c9723d
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: quoted-printable
Hi Melinda,
I liked your idea to do large N puzzles, so I configured some biggish ones
and added them to the download :) They are in the tree at "Hyperbolic ->
Large Polygons". They take a bit longer to build and the textures get a
little pixelated, but things work reasonably well. Solving the {32,3} 3C
will effectively be the same experience as an {inf,3} 3C, though I would
still like to see the infinite puzzle someday too. One strange thing about
{inf,3} will be that no matter how much you hyperbolic pan, you won't be
able to separate tiles from the disk boundary, whereas in these puzzles you
can drag a tile across the disk center and to the other side.
Download link:
http://www.gravitation3d.com/magictile/downloads/MagicTile_v2.zip
And some pictures:
http://groups.yahoo.com/group/4D_Cubing/photos/album/1694853720/pic/4354759=
20/view
http://groups.yahoo.com/group/4D_Cubing/photos/album/1694853720/pic/8107180=
37/view
http://groups.yahoo.com/group/4D_Cubing/photos/album/1694853720/pic/6004978=
13/view
seeya,
Roice
On Sun, Aug 11, 2013 at 10:32 PM, Melinda Green
rote:
>
>
> Hello Roice,
>
> I'm glad that you think that this puzzle makes sense. Also, I like your
> idea of using fundamental domain triangles. As for other colorings (and
> topologies), I would first hope to see the simplest one(s) first. This
> 3-coloring seems about as simple as possible though perhaps one could
> remove an edge or two by torturing the topology a bit. As for incorporati=
ng
> into MT versus creating a stand-alone puzzle, I have a feeling that there
> might be some clever ways to incorporate it. One way might be to implemen=
t
> it as a {N,3} for some large N. If a user were to pan far enough to see
> the ragged edge, so be it. If it must be a stand-alone puzzle, it might
> allow for your alternate colorings and perhaps other interesting variants
> that would otherwise be too difficult.
>
> -Melinda
>
>
> On 8/11/2013 8:06 PM, Roice Nelson wrote:
>
> The puzzle in your pictures *needs* to be made!
>
> It feels like the current MagicTile engine will fall woefully short for
> this task, though maybe I am overestimating the difficulty. Off the cuff=
,
> an approach could be to try to allow building up puzzles using fundamenta=
l
> domain triangles rather than entire tiles, because it will be necessary t=
o
> only show portions of these infinite-faceted tiles. (In the past, I've
> wondered if that enhancement is going to be necessary for uniform tilings=
.)
> It does seem like a big piece of work, and it might even be easier to
> write some special-case code for this puzzle rather than attempting to fi=
t
> it into the engine.
>
> I bet there is an infinite set of coloring possibilities for this tiling
> too.
>
>
>
> On Sun, Aug 11, 2013 at 7:19 PM, Melinda Green
wrote:
>
>>
>>
>> Here's a slightly less awful sketch:
>>
>> http://groups.yahoo.com/group/4D_Cubing/photos/album/1962624577/pic/4087=
5152/view/
>>
>>
>> On 8/11/2013 4:34 PM, Melinda Green wrote:
>>
>> Lovely, Roice!
>>
>> This makes me wonder whether it might be possible to add a 3-color
>> {inf,3}
>>
>> groups.yahoo.com/group/4D_Cubing/photos/album/1962624577/pic/908182938/v=
iew/
>>
>> -Melinda
>>
>> On 8/10/2013 2:10 PM, Roice Nelson wrote:
>>
>> Hi all,
>>
>> Check out a new physical model of the exotic {4,4,4} H=C2=B3 honeycomb!
>>
>> http://shpws.me/oFpu
>>
>>
>> Each cell is a tiling of squares with an infinite number of facets. All
>> vertices are ideal (meaning they live at infinity, on the Poincare ball
>> boundary). Four cells meet at every edge, and an infinite number of cel=
ls
>> meet at every vertex (the vertex figure is a tiling of squares too). Th=
is
>> honeycomb is self-dual.
>>
>> I printed only half of the Poincare ball in this model, which has
>> multiple advantages: you can see inside better, and it saves on printing
>> costs. The view is face-centered, meaning the projection places the cen=
ter
>> of one (ideal) 2D polygon at the center of the ball. An edge-centered v=
iew
>> is also possible. Vertex-centered views are impossible since every vert=
ex
>> is ideal. A view centered on the interior of a cell is possible, but (I
>> think, given my current understanding) a cell-centered view is also
>> impossible.
>>
>> I rendered one tile and all the tiles around it, so only one level of
>> recursion. I also experimented with deeper recursion, but felt the
>> resulting density inhibited understanding. Probably best would be to ha=
ve
>> two models at different recursion depths side by side to study together.=
I
>> had to artificially increase edge widths near the boundary to make thing=
s
>> printable.
>>
>> These things are totally cool to handle in person, so consider ordering
>> one or two of the honeycomb models :) As I've heard Henry Segerman
>> comment, the "bandwidth" of information is really high. You definitely
>> notice things you wouldn't if only viewing them on the computer screen.
>> The {3,6,3} and {6,3,6} are very similar to the {4,4,4}, just based on
>> different Euclidean tilings, so models of those are surely coming as wel=
l.
>>
>> So... whose going to make a puzzle based on this exotic honeycomb? :D
>>
>> Cheers,
>> Roice
>>
>>
>> As a postscript, here are a few thoughts I had about the {4,4,4} while
>> working on the model...
>>
>> In a previous thread on the {4,4,4}
>> Nan made an insightful comment. He said:
>>
>> I believe the first step to understand {4,4,4} is to understand {infinit=
y,
>>> infinity} in the hyperbolic plane.
>>
>>
>> I can see now they are indeed quite analogous. Wikipedia has some great
>> pictures of the {=E2=88=9E,=E2=88=9E} tiling and {p,q} tilings that appr=
oach it by
>> increasing p or q. Check out the progression that starts with an {=E2=
=88=9E,3}
>> tiling and increases q, which is the bottom row of the table here:
>>
>>
>> http://en.wikipedia.org/wiki/Uniform_tilings_in_hyperbolic_plane#Regular=
_hyperbolic_tilings
>>
>>
>> The {=E2=88=9E} polygons are inscribed in horocycles
>> circle of infinite radius with a unique center point on the disk boundar=
y).
>> The horocycles increase in size with this progression until, in the lim=
it,
>> the inscribing circle is* the boundary of the disk itself.* Something
>> strange about that is an {=E2=88=9E,=E2=88=9E} tile loses its center. A=
horocycle has a
>> single center on the boundary, so the inscribed {=E2=88=9E,q} tiles have=
a clear
>> center, but because an {=E2=88=9E,=E2=88=9E} tile is inscribed in the en=
tire boundary,
>> there is no longer a unique center. Tile centers are at infinity for th=
e
>> whole progression, so you'd think they would also live at infinity in th=
e
>> limit. At the same time, all vertices have also become ideal in the lim=
it,
>> and these are the only points of a tile living at infinity. So every
>> vertex seems equally valid as a tile center. Weird.
>>
>> This is good warm-up to jumping up a dimension. The {4,4,3} is kind of
>> like an {=E2=88=9E,q} with finite q. It's cells are inscribed in horosp=
heres, and
>> have finite vertices and a unique center. The {4,4,4} is like the {=E2=
=88=9E,=E2=88=9E}
>> because cells are inscribed in the boundary of hyperbolic space. They
>> don't really have a unique center, and every vertex is ideal. Again, ea=
ch
>> vertex sort of acts like a center point.
>>
>> (Perhaps there is a better way to think about this... Maybe when all
>> the vertices go to infinity, the cell center should be considered to hav=
e
>> snapped back to being finite? Maybe the center is at some average of all
>> the ideal vertices or at a center of mass? That makes sense for an idea=
l
>> tetrahedron, but can it for a cell that is an ideal {4,4} tiling? I don=
't
>> know!)
>>
>>
>>
>>
>>
>>
>
>
>
>=20
>
--089e014940eac9cd2204e3c9723d
Content-Type: text/html; charset=UTF-8
Content-Transfer-Encoding: quoted-printable
do large N puzzles, so I configured some biggish ones and added them to the=
download :) =C2=A0They are in the tree at "Hyperbolic -> Large Pol=
ygons". =C2=A0They take a bit longer to build and the textures get a l=
ittle pixelated, but things work reasonably well. =C2=A0Solving the {32,3} =
3C will effectively be the same experience as an {inf,3} 3C, though I would=
still like to see the infinite puzzle someday too. =C2=A0One strange thing=
about {inf,3} will be that no matter how much you hyperbolic pan, you won&=
#39;t be able to separate tiles from the disk boundary, whereas in these pu=
zzles you can drag a tile across the disk center and to the other side.
tures:
ya,x #ccc solid;padding-left:1ex">
=20=20=20=20=20=20=20=20
=20=20
=20=20=20=20
=20=20
Hello Roice,
I'm glad that you think that this puzzle makes sense. Also, I like
your idea of using fundamental domain triangles. As for other
colorings (and topologies), I would first hope to see the simplest
one(s) first. This 3-coloring seems about as simple as possible
though perhaps one could remove an edge or two by torturing the
topology a bit. As for incorporating into MT versus creating a
stand-alone puzzle, I have a feeling that there might be some clever
ways to incorporate it. One way might be to implement it as a {N,3}=C2=
=A0
for some large N. If a user were to pan far enough to see the ragged
edge, so be it. If it must be a stand-alone puzzle, it might allow
for your alternate colorings and perhaps other interesting variants
that would otherwise be too difficult.r=3D"#888888">
-Melinda
wrote:
=20=20=20=20=20=20
=20=20=20=20=20=20
woefully short for this task, though maybe I am overestimating
the difficulty. =C2=A0Off the cuff, an approach could be to try t=
o
allow building up puzzles using fundamental domain triangles
rather than entire tiles, because it will be necessary to only
show portions of these infinite-faceted tiles. =C2=A0(In the past=
,
I've wondered if that=C2=A0enhancement is going to be necessa=
ry for
uniform tilings.) =C2=A0It does seem like a big piece of work, an=
d
it might even be easier to write some special-case code for
this puzzle rather than attempting to fit it into the engine.v>
for this tiling too.
Melinda Green <erliminal.com" target=3D"_blank">melinda@superliminal.com>
wrote:
er-left:1px #ccc solid;padding-left:1ex">
Here's a slightly less awful sketch:
um/1962624577/pic/40875152/view/" target=3D"_blank">http://groups.yahoo.com=
/group/4D_Cubing/photos/album/1962624577/pic/40875152/view/
Lovely, Roice!
This makes me wonder whether it might be possible to
add a g/photos/album/1962624577/pic/908182938/view/" target=3D"_blank">3-color {i=
nf,3} to MagicTile
something like this:
os/album/1962624577/pic/908182938/view/" target=3D"_blank">groups.yahoo.com=
/group/4D_Cubing/photos/album/1962624577/pic/908182938/view/
-Melinda
exotic {4,4,4} H=C2=B3 honeycomb! =C2=A0
:none;padding:0px">
infinite number of facets. All vertices are
ideal=C2=A0(meaning they live at infinity, on the
Poincare ball boundary). =C2=A0Four cells meet at
every edge, and an infinite number of cells
meet at every vertex (the vertex figure is a
tiling of squares too). =C2=A0This honeycomb is
self-dual.
I printed only half of the Poincare ball in
this model, which has multiple advantages: you
can see inside better, and it saves on
printing costs. =C2=A0The view is face-centered,
meaning the projection places the center of
one (ideal) 2D polygon at the center of the
ball. =C2=A0An edge-centered view is also possibl=
e.
=C2=A0Vertex-centered views are impossible since
every vertex is ideal. =C2=A0A view centered on t=
he
interior of=C2=A0a cell is possible, but (I think=
,
given my current understanding) a
cell-centered view is also impossible. =C2=A0v>
around it, so only one level of recursion. =C2=A0=
I
also experimented with deeper recursion, but
felt the resulting density inhibited
understanding. =C2=A0Probably best would be to ha=
ve
two models at different recursion depths side
by side to study together. =C2=A0I had to
artificially increase edge widths near the
boundary to make things printable.
in person, so consider ordering one or two
of the honeycomb models :) =C2=A0As I've =
heard
Henry Segerman comment, the "bandwidth&q=
uot; of
information is really high. =C2=A0You
definitely notice things you wouldn't if
only viewing them on the computer screen.
=C2=A0The {3,6,3} and {6,3,6} are very simila=
r
to the {4,4,4}, just based on different
Euclidean tilings, so models of those are
surely coming as well.
based on this exotic honeycomb? :D
Cheers,
Roice
thoughts I had about the {4,4,4} while
working on the model... =C2=A0
{4,4,4}, Nan=C2=A0made an insightful
comment. =C2=A0He said:
px 0px 0px 0.8ex;border-left-width:1px;border-left-color:rgb(204,204,204);b=
order-left-style:solid;padding-left:1ex">I
believe the first step to understand {4,4,4}
is to understand {infinity,
infinity} in the hyperbolic plane.
I can see now they are indeed quite analogous.
=C2=A0Wikipedia has some great pictures of the {=E2=
=88=9E,=E2=88=9E}
tiling and {p,q} tilings that approach it by
increasing p or q. =C2=A0Check out the progression
that starts with an {=E2=88=9E,3} tiling and increa=
ses
q, which is the bottom row of the table here:
er:none;padding:0px">
es=C2=A0(a circle
of infinite radius with a unique center
point on the disk boundary).
=C2=A0The=C2=A0horocycles=C2=A0increase in size=
with this
progression until, in the limit, the
inscribing circle is the boundary of the
disk itself.=C2=A0 Something strange abou=
t
that is an {=E2=88=9E,=E2=88=9E} tile loses its=
center.
=C2=A0A=C2=A0horocycle=C2=A0has a single center=
on the
boundary, so the inscribed {=E2=88=9E,q} tiles =
have
a clear center, but because an {=E2=88=9E,=E2=
=88=9E} tile is
inscribed in the entire boundary, there is
no longer a unique center. =C2=A0Tile centers a=
re
at infinity for the whole progression, so
you'd think they would also live at infinit=
y
in the limit. =C2=A0At the same time, all
vertices have also become ideal in the
limit, and these are the only points of a
tile living at infinity. =C2=A0So every vertex
seems equally valid as a tile center.
=C2=A0Weird.
This is good warm-up to jumping up a
dimension. =C2=A0The {4,4,3} is kind of like an
{=E2=88=9E,q} with finite q. =C2=A0It's cel=
ls are
inscribed in horospheres, and have finite
vertices and a unique center. =C2=A0The {4,4,4}
is like the {=E2=88=9E,=E2=88=9E} because cells=
are
inscribed in the boundary of hyperbolic
space. =C2=A0They don't really have a uniqu=
e
center, and every vertex is ideal. =C2=A0Again,
each vertex sort of acts like a center
point.=C2=A0
about this... Maybe when all the vertices go
to infinity, the cell center should be
considered to have snapped back to being
finite? Maybe the center is at some average
of all the ideal vertices or at a center of
mass? =C2=A0That makes sense for an ideal
tetrahedron, but can it for a cell that is
an ideal {4,4} tiling? =C2=A0I don't know!)=
=20=20=20=20=20=20
=20=20
=20=20=20=20
=20=20=20=20
--089e014940eac9cd2204e3c9723d--
From: "schuma" <mananself@gmail.com>
Date: Tue, 13 Aug 2013 01:23:52 -0000
Subject: Re: The exotic {4,4,4}
Neat. Just solved {32,3}3C with two moves... That's a good appetizer for di=
nner.
Nan
--- In 4D_Cubing@yahoogroups.com, Roice Nelson
>
> Hi Melinda,
>=20
> I liked your idea to do large N puzzles, so I configured some biggish one=
s
> and added them to the download :) They are in the tree at "Hyperbolic ->
> Large Polygons". They take a bit longer to build and the textures get a
> little pixelated, but things work reasonably well. Solving the {32,3} 3C
> will effectively be the same experience as an {inf,3} 3C, though I would
> still like to see the infinite puzzle someday too. One strange thing abo=
ut
> {inf,3} will be that no matter how much you hyperbolic pan, you won't be
> able to separate tiles from the disk boundary, whereas in these puzzles y=
ou
> can drag a tile across the disk center and to the other side.
>=20
> Download link:
> http://www.gravitation3d.com/magictile/downloads/MagicTile_v2.zip
>=20
> And some pictures:
> http://groups.yahoo.com/group/4D_Cubing/photos/album/1694853720/pic/43547=
5920/view
> http://groups.yahoo.com/group/4D_Cubing/photos/album/1694853720/pic/81071=
8037/view
> http://groups.yahoo.com/group/4D_Cubing/photos/album/1694853720/pic/60049=
7813/view
>=20
> seeya,
> Roice
>=20
>=20
>=20
> On Sun, Aug 11, 2013 at 10:32 PM, Melinda Green
>=20
> >
> >
> > Hello Roice,
> >
> > I'm glad that you think that this puzzle makes sense. Also, I like your
> > idea of using fundamental domain triangles. As for other colorings (and
> > topologies), I would first hope to see the simplest one(s) first. This
> > 3-coloring seems about as simple as possible though perhaps one could
> > remove an edge or two by torturing the topology a bit. As for incorpora=
ting
> > into MT versus creating a stand-alone puzzle, I have a feeling that the=
re
> > might be some clever ways to incorporate it. One way might be to implem=
ent
> > it as a {N,3} for some large N. If a user were to pan far enough to se=
e
> > the ragged edge, so be it. If it must be a stand-alone puzzle, it might
> > allow for your alternate colorings and perhaps other interesting varian=
ts
> > that would otherwise be too difficult.
> >
> > -Melinda
> >
> >
> > On 8/11/2013 8:06 PM, Roice Nelson wrote:
> >
> > The puzzle in your pictures *needs* to be made!
> >
> > It feels like the current MagicTile engine will fall woefully short fo=
r
> > this task, though maybe I am overestimating the difficulty. Off the cu=
ff,
> > an approach could be to try to allow building up puzzles using fundamen=
tal
> > domain triangles rather than entire tiles, because it will be necessary=
to
> > only show portions of these infinite-faceted tiles. (In the past, I've
> > wondered if that enhancement is going to be necessary for uniform tilin=
gs.)
> > It does seem like a big piece of work, and it might even be easier to
> > write some special-case code for this puzzle rather than attempting to =
fit
> > it into the engine.
> >
> > I bet there is an infinite set of coloring possibilities for this tili=
ng
> > too.
> >
> >
> >
> > On Sun, Aug 11, 2013 at 7:19 PM, Melinda Green
> >
> >>
> >>
> >> Here's a slightly less awful sketch:
> >>
> >> http://groups.yahoo.com/group/4D_Cubing/photos/album/1962624577/pic/40=
875152/view/
> >>
> >>
> >> On 8/11/2013 4:34 PM, Melinda Green wrote:
> >>
> >> Lovely, Roice!
> >>
> >> This makes me wonder whether it might be possible to add a 3-color
> >> {inf,3}
> >>
> >> groups.yahoo.com/group/4D_Cubing/photos/album/1962624577/pic/908182938=
/view/
> >>
> >> -Melinda
> >>
> >> On 8/10/2013 2:10 PM, Roice Nelson wrote:
> >>
> >> Hi all,
> >>
> >> Check out a new physical model of the exotic {4,4,4} H=C2=B3 honeycom=
b!
> >>
> >> http://shpws.me/oFpu
> >>
> >>
> >> Each cell is a tiling of squares with an infinite number of facets. A=
ll
> >> vertices are ideal (meaning they live at infinity, on the Poincare bal=
l
> >> boundary). Four cells meet at every edge, and an infinite number of c=
ells
> >> meet at every vertex (the vertex figure is a tiling of squares too). =
This
> >> honeycomb is self-dual.
> >>
> >> I printed only half of the Poincare ball in this model, which has
> >> multiple advantages: you can see inside better, and it saves on printi=
ng
> >> costs. The view is face-centered, meaning the projection places the c=
enter
> >> of one (ideal) 2D polygon at the center of the ball. An edge-centered=
view
> >> is also possible. Vertex-centered views are impossible since every ve=
rtex
> >> is ideal. A view centered on the interior of a cell is possible, but =
(I
> >> think, given my current understanding) a cell-centered view is also
> >> impossible.
> >>
> >> I rendered one tile and all the tiles around it, so only one level of
> >> recursion. I also experimented with deeper recursion, but felt the
> >> resulting density inhibited understanding. Probably best would be to =
have
> >> two models at different recursion depths side by side to study togethe=
r. I
> >> had to artificially increase edge widths near the boundary to make thi=
ngs
> >> printable.
> >>
> >> These things are totally cool to handle in person, so consider orderi=
ng
> >> one or two of the honeycomb models :) As I've heard Henry Segerman
> >> comment, the "bandwidth" of information is really high. You definitel=
y
> >> notice things you wouldn't if only viewing them on the computer screen=
.
> >> The {3,6,3} and {6,3,6} are very similar to the {4,4,4}, just based o=
n
> >> different Euclidean tilings, so models of those are surely coming as w=
ell.
> >>
> >> So... whose going to make a puzzle based on this exotic honeycomb? :D
> >>
> >> Cheers,
> >> Roice
> >>
> >>
> >> As a postscript, here are a few thoughts I had about the {4,4,4} whil=
e
> >> working on the model...
> >>
> >> In a previous thread on the {4,4,4}
> >> Nan made an insightful comment. He said:
> >>
> >> I believe the first step to understand {4,4,4} is to understand {infin=
ity,
> >>> infinity} in the hyperbolic plane.
> >>
> >>
> >> I can see now they are indeed quite analogous. Wikipedia has some gre=
at
> >> pictures of the {=E2=88=9E,=E2=88=9E} tiling and {p,q} tilings that ap=
proach it by
> >> increasing p or q. Check out the progression that starts with an {=E2=
=88=9E,3}
> >> tiling and increases q, which is the bottom row of the table here:
> >>
> >>
> >> http://en.wikipedia.org/wiki/Uniform_tilings_in_hyperbolic_plane#Regul=
ar_hyperbolic_tilings
> >>
> >>
> >> The {=E2=88=9E} polygons are inscribed in horocycles
> >> circle of infinite radius with a unique center point on the disk bound=
ary).
> >> The horocycles increase in size with this progression until, in the l=
imit,
> >> the inscribing circle is* the boundary of the disk itself.* Something
> >> strange about that is an {=E2=88=9E,=E2=88=9E} tile loses its center. =
A horocycle has a
> >> single center on the boundary, so the inscribed {=E2=88=9E,q} tiles ha=
ve a clear
> >> center, but because an {=E2=88=9E,=E2=88=9E} tile is inscribed in the =
entire boundary,
> >> there is no longer a unique center. Tile centers are at infinity for =
the
> >> whole progression, so you'd think they would also live at infinity in =
the
> >> limit. At the same time, all vertices have also become ideal in the l=
imit,
> >> and these are the only points of a tile living at infinity. So every
> >> vertex seems equally valid as a tile center. Weird.
> >>
> >> This is good warm-up to jumping up a dimension. The {4,4,3} is kind o=
f
> >> like an {=E2=88=9E,q} with finite q. It's cells are inscribed in horo=
spheres, and
> >> have finite vertices and a unique center. The {4,4,4} is like the {=
=E2=88=9E,=E2=88=9E}
> >> because cells are inscribed in the boundary of hyperbolic space. They
> >> don't really have a unique center, and every vertex is ideal. Again, =
each
> >> vertex sort of acts like a center point.
> >>
> >> (Perhaps there is a better way to think about this... Maybe when all
> >> the vertices go to infinity, the cell center should be considered to h=
ave
> >> snapped back to being finite? Maybe the center is at some average of a=
ll
> >> the ideal vertices or at a center of mass? That makes sense for an id=
eal
> >> tetrahedron, but can it for a cell that is an ideal {4,4} tiling? I d=
on't
> >> know!)
> >>
> >>
> >>
> >>
> >>
> >>
> >
> >
> >
> >=20
> >
>
From: Melinda Green <melinda@superliminal.com>
Date: Mon, 12 Aug 2013 18:30:03 -0700
Subject: Re: [MC4D] The exotic {4,4,4}
--------------040301010507020902060707
Content-Type: text/plain; charset=UTF-8; format=flowed
Content-Transfer-Encoding: quoted-printable
Very nice, Roice!
I was pretty sure the {inf,3} wouldn't be very difficult but I didn't=20
expect it to be this easy It seems like god's number for it can be=20
counted on one hand! One nit: Scrambling it with 1000 twists rings the=20
"solved" bell a whole bunch of times as it accidentally solves it self=20
many times. Silencing the solved sound during scrambling will be=20
helpful, but then you should probably also discard all the twists that=20
led to it since it just becomes unneeded log file baggage.
The experience of scrolling around in the {32,3} is better than I=20
imagined. Somehow I expected to see only the fundamental polygons. With=20
N >=3D 100 you can probably just mask off the outermost few pixels of the=20
limit making it indistinguishable from the inf version. You might also=20
not center a face in the disk to disguise its finiteness. Users can=20
still scroll a face to the center but they'd almost need to be trying to=20
do that, and the larger the N, the harder that will be.
The ragged borders are indeed unsightly. Normally that's not a problem=20
for solvers but it does go against the wonderful amount of polish you've=20
applied to MT in general. At the very least it shows us what the=20
experience can be which will be important in finding out how interesting=20
these puzzles are compared with other potential puzzles.
As for the time needed to initialize these puzzles, perhaps you can=20
cache all the build data for all puzzles so that you never pay more than=20
once for each? It might also be nice to ship with the build data for=20
whichever puzzle you make the default. One last minor suggestion: If=20
it's not tricky, would you please see if you can make the expanding=20
circles animation spawn new circles centered on the mouse pointer when=20
it's in the frame? That would provide a nice distraction while waiting.
Really nice work, Roice. Thanks a lot!
-Melinda
On 8/12/2013 5:37 PM, Roice Nelson wrote:
>
>
> Hi Melinda,
>
> I liked your idea to do large N puzzles, so I configured some biggish=20
> ones and added them to the download :) They are in the tree at=20
> "Hyperbolic -> Large Polygons". They take a bit longer to build and=20
> the textures get a little pixelated, but things work reasonably well.=20
> Solving the {32,3} 3C will effectively be the same experience as an=20
> {inf,3} 3C, though I would still like to see the infinite puzzle=20
> someday too. One strange thing about {inf,3} will be that no matter=20
> how much you hyperbolic pan, you won't be able to separate tiles from=20
> the disk boundary, whereas in these puzzles you can drag a tile across=20
> the disk center and to the other side.
>
> Download link:
> http://www.gravitation3d.com/magictile/downloads/MagicTile_v2.zip
>
> And some pictures:
> http://groups.yahoo.com/group/4D_Cubing/photos/album/1694853720/pic/43547=
5920/view
> http://groups.yahoo.com/group/4D_Cubing/photos/album/1694853720/pic/81071=
8037/view
> http://groups.yahoo.com/group/4D_Cubing/photos/album/1694853720/pic/60049=
7813/view
>
> seeya,
> Roice
>
>
>
> On Sun, Aug 11, 2013 at 10:32 PM, Melinda Green=20
>
>
>
>
> Hello Roice,
>
> I'm glad that you think that this puzzle makes sense. Also, I like
> your idea of using fundamental domain triangles. As for other
> colorings (and topologies), I would first hope to see the simplest
> one(s) first. This 3-coloring seems about as simple as possible
> though perhaps one could remove an edge or two by torturing the
> topology a bit. As for incorporating into MT versus creating a
> stand-alone puzzle, I have a feeling that there might be some
> clever ways to incorporate it. One way might be to implement it as
> a {N,3} for some large N. If a user were to pan far enough to see
> the ragged edge, so be it. If it must be a stand-alone puzzle, it
> might allow for your alternate colorings and perhaps other
> interesting variants that would otherwise be too difficult.
>
> -Melinda
>
>
> On 8/11/2013 8:06 PM, Roice Nelson wrote:
>> The puzzle in your pictures *needs* to be made!
>>
>> It feels like the current MagicTile engine will fall woefully
>> short for this task, though maybe I am overestimating the
>> difficulty. Off the cuff, an approach could be to try to allow
>> building up puzzles using fundamental domain triangles rather
>> than entire tiles, because it will be necessary to only show
>> portions of these infinite-faceted tiles. (In the past, I've
>> wondered if that enhancement is going to be necessary for uniform
>> tilings.) It does seem like a big piece of work, and it might
>> even be easier to write some special-case code for this puzzle
>> rather than attempting to fit it into the engine.
>>
>> I bet there is an infinite set of coloring possibilities for this
>> tiling too.
>>
>>
>>
>> On Sun, Aug 11, 2013 at 7:19 PM, Melinda Green
>>
>>
>>
>>
>> Here's a slightly less awful sketch:
>> http://groups.yahoo.com/group/4D_Cubing/photos/album/1962624577/=
pic/40875152/view/
>>
>>
>>
>> On 8/11/2013 4:34 PM, Melinda Green wrote:
>>> Lovely, Roice!
>>>
>>> This makes me wonder whether it might be possible to add a
>>> 3-color {inf,3}
>>>
>>> to MagicTile something like this:
>>> groups.yahoo.com/group/4D_Cubing/photos/album/1962624577/pic/90=
8182938/view/
>>>
>>>
>>> -Melinda
>>>
>>> On 8/10/2013 2:10 PM, Roice Nelson wrote:
>>>> Hi all,
>>>>
>>>> Check out a new physical model of the exotic {4,4,4} H=C2=B3
>>>> honeycomb!
>>>>
>>>> http://shpws.me/oFpu
>>>>
>>>>
>>>> Each cell is a tiling of squares with an infinite number of
>>>> facets. All vertices are ideal (meaning they live at
>>>> infinity, on the Poincare ball boundary). Four cells meet
>>>> at every edge, and an infinite number of cells meet at
>>>> every vertex (the vertex figure is a tiling of squares
>>>> too). This honeycomb is self-dual.
>>>>
>>>> I printed only half of the Poincare ball in this model,
>>>> which has multiple advantages: you can see inside better,
>>>> and it saves on printing costs. The view is face-centered,
>>>> meaning the projection places the center of one (ideal) 2D
>>>> polygon at the center of the ball. An edge-centered view
>>>> is also possible. Vertex-centered views are impossible
>>>> since every vertex is ideal. A view centered on the
>>>> interior of a cell is possible, but (I think, given my
>>>> current understanding) a cell-centered view is also
>>>> impossible.
>>>>
>>>> I rendered one tile and all the tiles around it, so only
>>>> one level of recursion. I also experimented with deeper
>>>> recursion, but felt the resulting density inhibited
>>>> understanding. Probably best would be to have two models
>>>> at different recursion depths side by side to study
>>>> together. I had to artificially increase edge widths near
>>>> the boundary to make things printable.
>>>>
>>>> These things are totally cool to handle in person, so
>>>> consider ordering one or two of the honeycomb models :) As
>>>> I've heard Henry Segerman comment, the "bandwidth" of
>>>> information is really high. You definitely notice things
>>>> you wouldn't if only viewing them on the computer screen.
>>>> The {3,6,3} and {6,3,6} are very similar to the {4,4,4},
>>>> just based on different Euclidean tilings, so models of
>>>> those are surely coming as well.
>>>>
>>>> So... whose going to make a puzzle based on this exotic
>>>> honeycomb? :D
>>>>
>>>> Cheers,
>>>> Roice
>>>>
>>>>
>>>> As a postscript, here are a few thoughts I had about the
>>>> {4,4,4} while working on the model...
>>>>
>>>> In a previous thread on the {4,4,4}
>>>>
>>>> Nan made an insightful comment. He said:
>>>>
>>>> I believe the first step to understand {4,4,4} is to
>>>> understand {infinity,
>>>> infinity} in the hyperbolic plane.=20
>>>>
>>>>
>>>> I can see now they are indeed quite analogous. Wikipedia
>>>> has some great pictures of the {=E2=88=9E,=E2=88=9E} tiling an=
d {p,q}
>>>> tilings that approach it by increasing p or q. Check out
>>>> the progression that starts with an {=E2=88=9E,3} tiling and
>>>> increases q, which is the bottom row of the table here:
>>>>
>>>> http://en.wikipedia.org/wiki/Uniform_tilings_in_hyperbolic=
_plane#Regular_hyperbolic_tilings
>>>>
>>>>
>>>> The {=E2=88=9E} polygons are inscribed in horocycles
>>>>
>>>> infinite radius with a unique center point on the disk
>>>> boundary). The horocycles increase in size with this
>>>> progression until, in the limit, the inscribing circle
>>>> is*the boundary of the disk itself.* Something strange
>>>> about that is an {=E2=88=9E,=E2=88=9E} tile loses its center.
>>>> A horocycle has a single center on the boundary, so the
>>>> inscribed {=E2=88=9E,q} tiles have a clear center, but because=
an
>>>> {=E2=88=9E,=E2=88=9E} tile is inscribed in the entire boundary=
, there is no
>>>> longer a unique center. Tile centers are at infinity for
>>>> the whole progression, so you'd think they would also live
>>>> at infinity in the limit. At the same time, all vertices
>>>> have also become ideal in the limit, and these are the only
>>>> points of a tile living at infinity. So every vertex seems
>>>> equally valid as a tile center. Weird.
>>>>
>>>> This is good warm-up to jumping up a dimension. The
>>>> {4,4,3} is kind of like an {=E2=88=9E,q} with finite q. It's =
cells
>>>> are inscribed in horospheres, and have finite vertices and
>>>> a unique center. The {4,4,4} is like the {=E2=88=9E,=E2=88=9E=
} because
>>>> cells are inscribed in the boundary of hyperbolic space.
>>>> They don't really have a unique center, and every vertex
>>>> is ideal. Again, each vertex sort of acts like a center
>>>> point.
>>>>
>>>> (Perhaps there is a better way to think about this... Maybe
>>>> when all the vertices go to infinity, the cell center
>>>> should be considered to have snapped back to being finite?
>>>> Maybe the center is at some average of all the ideal
>>>> vertices or at a center of mass? That makes sense for an
>>>> ideal tetrahedron, but can it for a cell that is an ideal
>>>> {4,4} tiling? I don't know!)
>>>
>>
>>
>>
>>
>
>
>
>
>
>
>=20
--------------040301010507020902060707
Content-Type: text/html; charset=UTF-8
Content-Transfer-Encoding: quoted-printable
">
Very nice, Roice!
I was pretty sure the {inf,3} wouldn't be very difficult but I
didn't expect it to be this easy It seems like god's number for it
can be counted on one hand! One nit: Scrambling it with 1000 twists
rings the "solved" bell a whole bunch of times as it accidentally
solves it self many times. Silencing the solved sound during
scrambling will be helpful, but then you should probably also
discard all the twists that led to it since it just becomes unneeded
log file baggage.
The experience of scrolling around in the {32,3} is better than I
imagined. Somehow I expected to see only the fundamental polygons.
With N >=3D 100 you can probably just mask off the outermost few
pixels of the limit making it indistinguishable from the inf
version. You might also not center a face in the disk to disguise
its finiteness. Users can still scroll a face to the center but
they'd almost need to be trying to do that, and the larger the N,
the harder that will be.
The ragged borders are indeed unsightly. Normally that's not a
problem for solvers but it does go against the wonderful amount of
polish you've applied to MT in general. At the very least it shows
us what the experience can be which will be important in finding out
how interesting these puzzles are compared with other potential
puzzles.
As for the time needed to initialize these puzzles, perhaps you can
cache all the build data for all puzzles so that you never pay more
than once for each? It might also be nice to ship with the build
data for whichever puzzle you make the default. One last minor
suggestion: If it's not tricky, would you please see if you can make
the expanding circles animation spawn new circles centered on the
mouse pointer when it's in the frame? That would provide a nice
distraction while waiting.
Really nice work, Roice. Thanks a lot!
-Melinda
wrote:
com"
type=3D"cite">
I liked your idea to do large N puzzles, so I configured some
biggish ones and added them to the download :) =C2=A0They are in th=
e
tree at "Hyperbolic -> Large Polygons". =C2=A0They take a bit
longer to build and the textures get a little pixelated, but
things work reasonably well. =C2=A0Solving the {32,3} 3C will
effectively be the same experience as an {inf,3} 3C, though I
would still like to see the infinite puzzle someday too. =C2=A0One
strange thing about {inf,3} will be that no matter how much you
hyperbolic pan, you won't be able to separate tiles from the
disk boundary, whereas in these puzzles you can drag a tile
across the disk center and to the other side.
Melinda Green < href=3D"mailto:melinda@superliminal.com" target=3D"_blank">me=
linda@superliminal.com>
wrote:
.8ex;border-left:1px #ccc solid;padding-left:1ex">
Hello Roice,
I'm glad that you think that this puzzle makes sense.
Also, I like your idea of using fundamental domain
triangles. As for other colorings (and topologies), I
would first hope to see the simplest one(s) first. This
3-coloring seems about as simple as possible though
perhaps one could remove an edge or two by torturing the
topology a bit. As for incorporating into MT versus
creating a stand-alone puzzle, I have a feeling that there
might be some clever ways to incorporate it. One way might
be to implement it as a {N,3}=C2=A0 for some large N. If a us=
er
were to pan far enough to see the ragged edge, so be it.
If it must be a stand-alone puzzle, it might allow for
your alternate colorings and perhaps other interesting
variants that would otherwise be too difficult.
-Melinda
to be made!
will fall woefully short for this task, though
maybe I am overestimating the difficulty. =C2=A0Off
the cuff, an approach could be to try to allow
building up puzzles using fundamental domain
triangles rather than entire tiles, because it
will be necessary to only show portions of these
infinite-faceted tiles. =C2=A0(In the past, I've
wondered if that=C2=A0enhancement is going to be
necessary for uniform tilings.) =C2=A0It does seem
like a big piece of work, and it might even be
easier to write some special-case code for this
puzzle rather than attempting to fit it into the
engine.
possibilities for this tiling too.
7:19 PM, Melinda Green <
href=3D"mailto:melinda@superliminal.com"
target=3D"_blank">melinda@superliminal.com&=
gt;
wrote:
0 0 .8ex;border-left:1px #ccc
solid;padding-left:1ex">
Here's a slightly less awful sketch:
href=3D"http://groups.yahoo.com/group/4D_Cubing/photos/album/1962624577/pic=
/40875152/view/"
target=3D"_blank">http://groups.yahoo.com/gro=
up/4D_Cubing/photos/album/1962624577/pic/40875152/view/
wrote:
Lovely, Roice!
This makes me wonder whether it might
be possible to add a
href=3D"http://groups.yahoo.com/group/4D_Cubing/photos/album/1962624577/pic=
/908182938/view/"
target=3D"_blank">3-color {inf,3}
to MagicTile something like this:
href=3D"http://groups.yahoo.com/group/4D_Cubing/photos/album/1962624577/pic=
/908182938/view/"
target=3D"_blank">groups.yahoo.com/grou=
p/4D_Cubing/photos/album/1962624577/pic/908182938/view/
-Melinda
Nelson wrote:
model of the exotic {4,4,4} H=C2=B3
honeycomb! =C2=A0
0px
40px;border:none;padding:0px">
squares with an infinite number
of facets. All vertices are
ideal=C2=A0(meaning they live at
infinity, on the Poincare ball
boundary). =C2=A0Four cells meet at
every edge, and an infinite
number of cells meet at every
vertex (the vertex figure is a
tiling of squares too). =C2=A0This
honeycomb is self-dual.
I printed only half of the
Poincare ball in this model,
which has multiple advantages:
you can see inside better, and
it saves on printing costs. =C2=A0T=
he
view is face-centered, meaning
the projection places the center
of one (ideal) 2D polygon at the
center of the ball. =C2=A0An
edge-centered view is also
possible. =C2=A0Vertex-centered vie=
ws
are impossible since every
vertex is ideal. =C2=A0A view
centered on the interior of=C2=A0a
cell is possible, but (I think,
given my current understanding)
a cell-centered view is also
impossible. =C2=A0
the tiles around it, so only one
level of recursion. =C2=A0I also
experimented with deeper
recursion, but felt the
resulting density inhibited
understanding. =C2=A0Probably best
would be to have two models at
different recursion depths side
by side to study together. =C2=A0I
had to artificially increase
edge widths near the boundary to
make things printable.
cool to handle in person, so
consider ordering one or two
of the honeycomb models :)
=C2=A0As I've heard Henry
Segerman comment, the
"bandwidth" of information
is really high. =C2=A0You
definitely notice things you
wouldn't if only viewing
them on the computer screen.
=C2=A0The {3,6,3} and {6,3,6} a=
re
very similar to the {4,4,4},
just based on different
Euclidean tilings, so models
of those are surely coming
as well.
make a puzzle based on
this exotic honeycomb? :D
Cheers,
Roice
a few thoughts I had about
the {4,4,4} while working on
the model... =C2=A0
href=3D"http://games.groups.y=
ahoo.com/group/4D_Cubing/message/1226"
target=3D"_blank">a previous
thread on the {4,4,4},
Nan=C2=A0made an insightful
comment. =C2=A0He said:
style=3D"margin:0px 0px 0px
0.8ex;border-left-width:1px;border-left-color:rgb(204,204,204);border-left-=
style:solid;padding-left:1ex">I
believe the first step to
understand {4,4,4} is to
understand {infinity,
infinity} in the hyperbolic
plane.
I can see now they are indeed
quite analogous. =C2=A0Wikipedia has
some great pictures of the {=E2=88=9E=
,=E2=88=9E}
tiling and {p,q} tilings that
approach it by increasing p or q.
=C2=A0Check out the progression that
starts with an {=E2=88=9E,3} tiling a=
nd
increases q, which is the bottom
row of the table here:
0px 0px
40px;border:none;padding:0px">
inscribed in
href=3D"http://en.wikipedia.org=
/wiki/Horocycle"
target=3D"_blank">horocycles>=C2=A0(a
circle of infinite radius with
a unique center point on the
disk boundary).
=C2=A0The=C2=A0horocycles=C2=A0in=
crease in
size with this progression
until, in the limit, the
inscribing circle is the
boundary of the disk itself.>=C2=A0
Something strange about that
is an {=E2=88=9E,=E2=88=9E} tile =
loses its
center. =C2=A0A=C2=A0horocycle=C2=
=A0has a
single center on the boundary,
so the inscribed {=E2=88=9E,q} ti=
les
have a clear center, but
because an {=E2=88=9E,=E2=88=9E} =
tile is
inscribed in the entire
boundary, there is no longer a
unique center. =C2=A0Tile centers
are at infinity for the whole
progression, so you'd think
they would also live at
infinity in the limit. =C2=A0At t=
he
same time, all vertices have
also become ideal in the
limit, and these are the only
points of a tile living at
infinity. =C2=A0So every vertex
seems equally valid as a tile
center. =C2=A0Weird.
This is good warm-up to
jumping up a dimension. =C2=A0The
{4,4,3} is kind of like an
{=E2=88=9E,q} with finite q. =C2=
=A0It's
cells are inscribed in
horospheres, and have finite
vertices and a unique center.
=C2=A0The {4,4,4} is like the {=
=E2=88=9E,=E2=88=9E}
because cells are inscribed in
the boundary of hyperbolic
space. =C2=A0They don't really ha=
ve
a unique center, and every
vertex is ideal. =C2=A0Again, eac=
h
vertex sort of acts like a
center point.=C2=A0
way to think about this...
Maybe when all the vertices go
to infinity, the cell center
should be considered to have
snapped back to being finite?
Maybe the center is at some
average of all the ideal
vertices or at a center of
mass? =C2=A0That makes sense for =
an
ideal tetrahedron, but can it
for a cell that is an ideal
{4,4} tiling? =C2=A0I don't know!=
)
=20=20=20=20=20=20
--------------040301010507020902060707--
From: Roice Nelson <roice3@gmail.com>
Date: Tue, 13 Aug 2013 11:05:00 -0500
Subject: Re: [MC4D] The exotic {4,4,4}
--001a11c238e83755c504e3d6678f
Content-Type: text/plain; charset=UTF-8
Content-Transfer-Encoding: quoted-printable
I want to recall that Andrey pointed out all the 3C puzzles are
topologically the same, and behave like the {4,3} 3C (hemicube). The state
space does feel too small to really be enjoyable...even as an appetizer :)
Thanks so much for the suggestions Melinda. I especially like the idea to
cache the puzzle build data. Don't know why I never thought of that
before! I added your thoughts to my running trello list.
Cheers,
Roice
On Mon, Aug 12, 2013 at 8:30 PM, Melinda Green
ote:
>
>
> Very nice, Roice!
>
> I was pretty sure the {inf,3} wouldn't be very difficult but I didn't
> expect it to be this easy It seems like god's number for it can be counte=
d
> on one hand! One nit: Scrambling it with 1000 twists rings the "solved"
> bell a whole bunch of times as it accidentally solves it self many times.
> Silencing the solved sound during scrambling will be helpful, but then yo=
u
> should probably also discard all the twists that led to it since it just
> becomes unneeded log file baggage.
>
> The experience of scrolling around in the {32,3} is better than I
> imagined. Somehow I expected to see only the fundamental polygons. With N
> >=3D 100 you can probably just mask off the outermost few pixels of the l=
imit
> making it indistinguishable from the inf version. You might also not cent=
er
> a face in the disk to disguise its finiteness. Users can still scroll a
> face to the center but they'd almost need to be trying to do that, and th=
e
> larger the N, the harder that will be.
>
> The ragged borders are indeed unsightly. Normally that's not a problem fo=
r
> solvers but it does go against the wonderful amount of polish you've
> applied to MT in general. At the very least it shows us what the experien=
ce
> can be which will be important in finding out how interesting these puzzl=
es
> are compared with other potential puzzles.
>
> As for the time needed to initialize these puzzles, perhaps you can cache
> all the build data for all puzzles so that you never pay more than once f=
or
> each? It might also be nice to ship with the build data for whichever
> puzzle you make the default. One last minor suggestion: If it's not trick=
y,
> would you please see if you can make the expanding circles animation spaw=
n
> new circles centered on the mouse pointer when it's in the frame? That
> would provide a nice distraction while waiting.
>
> Really nice work, Roice. Thanks a lot!
> -Melinda
>
>
> On 8/12/2013 5:37 PM, Roice Nelson wrote:
>
> Hi Melinda,
>
> I liked your idea to do large N puzzles, so I configured some biggish
> ones and added them to the download :) They are in the tree at "Hyperbol=
ic
> -> Large Polygons". They take a bit longer to build and the textures get=
a
> little pixelated, but things work reasonably well. Solving the {32,3} 3C
> will effectively be the same experience as an {inf,3} 3C, though I would
> still like to see the infinite puzzle someday too. One strange thing abo=
ut
> {inf,3} will be that no matter how much you hyperbolic pan, you won't be
> able to separate tiles from the disk boundary, whereas in these puzzles y=
ou
> can drag a tile across the disk center and to the other side.
>
> Download link:
> http://www.gravitation3d.com/magictile/downloads/MagicTile_v2.zip
>
> And some pictures:
>
> http://groups.yahoo.com/group/4D_Cubing/photos/album/1694853720/pic/43547=
5920/view
>
> http://groups.yahoo.com/group/4D_Cubing/photos/album/1694853720/pic/81071=
8037/view
>
> http://groups.yahoo.com/group/4D_Cubing/photos/album/1694853720/pic/60049=
7813/view
>
> seeya,
> Roice
>
>
>
> On Sun, Aug 11, 2013 at 10:32 PM, Melinda Green
>
>>
>>
>> Hello Roice,
>>
>> I'm glad that you think that this puzzle makes sense. Also, I like your
>> idea of using fundamental domain triangles. As for other colorings (and
>> topologies), I would first hope to see the simplest one(s) first. This
>> 3-coloring seems about as simple as possible though perhaps one could
>> remove an edge or two by torturing the topology a bit. As for incorporat=
ing
>> into MT versus creating a stand-alone puzzle, I have a feeling that ther=
e
>> might be some clever ways to incorporate it. One way might be to impleme=
nt
>> it as a {N,3} for some large N. If a user were to pan far enough to see
>> the ragged edge, so be it. If it must be a stand-alone puzzle, it might
>> allow for your alternate colorings and perhaps other interesting variant=
s
>> that would otherwise be too difficult.
>>
>> -Melinda
>>
>>
>> On 8/11/2013 8:06 PM, Roice Nelson wrote:
>>
>> The puzzle in your pictures *needs* to be made!
>>
>> It feels like the current MagicTile engine will fall woefully short for
>> this task, though maybe I am overestimating the difficulty. Off the cuf=
f,
>> an approach could be to try to allow building up puzzles using fundament=
al
>> domain triangles rather than entire tiles, because it will be necessary =
to
>> only show portions of these infinite-faceted tiles. (In the past, I've
>> wondered if that enhancement is going to be necessary for uniform tiling=
s.)
>> It does seem like a big piece of work, and it might even be easier to
>> write some special-case code for this puzzle rather than attempting to f=
it
>> it into the engine.
>>
>> I bet there is an infinite set of coloring possibilities for this
>> tiling too.
>>
>>
>>
>> On Sun, Aug 11, 2013 at 7:19 PM, Melinda Green
>>
>>>
>>>
>>> Here's a slightly less awful sketch:
>>>
>>> http://groups.yahoo.com/group/4D_Cubing/photos/album/1962624577/pic/408=
75152/view/
>>>
>>>
>>> On 8/11/2013 4:34 PM, Melinda Green wrote:
>>>
>>> Lovely, Roice!
>>>
>>> This makes me wonder whether it might be possible to add a 3-color
>>> {inf,3}
>>>
>>> groups.yahoo.com/group/4D_Cubing/photos/album/1962624577/pic/908182938/=
view/
>>>
>>> -Melinda
>>>
>>> On 8/10/2013 2:10 PM, Roice Nelson wrote:
>>>
>>> Hi all,
>>>
>>> Check out a new physical model of the exotic {4,4,4} H=C2=B3 honeycomb=
!
>>>
>>> http://shpws.me/oFpu
>>>
>>>
>>> Each cell is a tiling of squares with an infinite number of facets.
>>> All vertices are ideal (meaning they live at infinity, on the Poincare =
ball
>>> boundary). Four cells meet at every edge, and an infinite number of ce=
lls
>>> meet at every vertex (the vertex figure is a tiling of squares too). T=
his
>>> honeycomb is self-dual.
>>>
>>> I printed only half of the Poincare ball in this model, which has
>>> multiple advantages: you can see inside better, and it saves on printin=
g
>>> costs. The view is face-centered, meaning the projection places the ce=
nter
>>> of one (ideal) 2D polygon at the center of the ball. An edge-centered =
view
>>> is also possible. Vertex-centered views are impossible since every ver=
tex
>>> is ideal. A view centered on the interior of a cell is possible, but (=
I
>>> think, given my current understanding) a cell-centered view is also
>>> impossible.
>>>
>>> I rendered one tile and all the tiles around it, so only one level of
>>> recursion. I also experimented with deeper recursion, but felt the
>>> resulting density inhibited understanding. Probably best would be to h=
ave
>>> two models at different recursion depths side by side to study together=
. I
>>> had to artificially increase edge widths near the boundary to make thin=
gs
>>> printable.
>>>
>>> These things are totally cool to handle in person, so consider
>>> ordering one or two of the honeycomb models :) As I've heard Henry
>>> Segerman comment, the "bandwidth" of information is really high. You
>>> definitely notice things you wouldn't if only viewing them on the compu=
ter
>>> screen. The {3,6,3} and {6,3,6} are very similar to the {4,4,4}, just
>>> based on different Euclidean tilings, so models of those are surely com=
ing
>>> as well.
>>>
>>> So... whose going to make a puzzle based on this exotic honeycomb? :D
>>>
>>> Cheers,
>>> Roice
>>>
>>>
>>> As a postscript, here are a few thoughts I had about the {4,4,4} while
>>> working on the model...
>>>
>>> In a previous thread on the {4,4,4}
>>> Nan made an insightful comment. He said:
>>>
>>> I believe the first step to understand {4,4,4} is to understand
>>>> {infinity,
>>>> infinity} in the hyperbolic plane.
>>>
>>>
>>> I can see now they are indeed quite analogous. Wikipedia has some grea=
t
>>> pictures of the {=E2=88=9E,=E2=88=9E} tiling and {p,q} tilings that app=
roach it by
>>> increasing p or q. Check out the progression that starts with an {=E2=
=88=9E,3}
>>> tiling and increases q, which is the bottom row of the table here:
>>>
>>>
>>> http://en.wikipedia.org/wiki/Uniform_tilings_in_hyperbolic_plane#Regula=
r_hyperbolic_tilings
>>>
>>>
>>> The {=E2=88=9E} polygons are inscribed in horocycles
>>> circle of infinite radius with a unique center point on the disk bounda=
ry).
>>> The horocycles increase in size with this progression until, in the li=
mit,
>>> the inscribing circle is* the boundary of the disk itself.* Something
>>> strange about that is an {=E2=88=9E,=E2=88=9E} tile loses its center. =
A horocycle has a
>>> single center on the boundary, so the inscribed {=E2=88=9E,q} tiles hav=
e a clear
>>> center, but because an {=E2=88=9E,=E2=88=9E} tile is inscribed in the e=
ntire boundary,
>>> there is no longer a unique center. Tile centers are at infinity for t=
he
>>> whole progression, so you'd think they would also live at infinity in t=
he
>>> limit. At the same time, all vertices have also become ideal in the li=
mit,
>>> and these are the only points of a tile living at infinity. So every
>>> vertex seems equally valid as a tile center. Weird.
>>>
>>> This is good warm-up to jumping up a dimension. The {4,4,3} is kind of
>>> like an {=E2=88=9E,q} with finite q. It's cells are inscribed in horos=
pheres, and
>>> have finite vertices and a unique center. The {4,4,4} is like the {=E2=
=88=9E,=E2=88=9E}
>>> because cells are inscribed in the boundary of hyperbolic space. They
>>> don't really have a unique center, and every vertex is ideal. Again, e=
ach
>>> vertex sort of acts like a center point.
>>>
>>> (Perhaps there is a better way to think about this... Maybe when all
>>> the vertices go to infinity, the cell center should be considered to ha=
ve
>>> snapped back to being finite? Maybe the center is at some average of al=
l
>>> the ideal vertices or at a center of mass? That makes sense for an ide=
al
>>> tetrahedron, but can it for a cell that is an ideal {4,4} tiling? I do=
n't
>>> know!)
>>>
>>>
>>>
>>>
>>>
>>>
>>
>>
>>
>>
>
>
>
>=20
>
--001a11c238e83755c504e3d6678f
Content-Type: text/html; charset=UTF-8
Content-Transfer-Encoding: quoted-printable
uzzles are topologically the same, and behave like the {4,3} 3C (hemicube).=
=C2=A0The state space does feel too small to really be enjoyable...even as=
an appetizer :)
ly like the idea to cache the puzzle build data. =C2=A0Don't know why I=
never thought of that before! =C2=A0I added your thoughts to my running tr=
ello list.
8:30 PM, Melinda Green <erliminal.com" target=3D"_blank">melinda@superliminal.com> wr=
ote:x #ccc solid;padding-left:1ex">
=20=20=20=20=20=20=20=20
=20=20
=20=20=20=20
=20=20
Very nice, Roice!
I was pretty sure the {inf,3} wouldn't be very difficult but I
didn't expect it to be this easy It seems like god's number for=
it
can be counted on one hand! One nit: Scrambling it with 1000 twists
rings the "solved" bell a whole bunch of times as it accident=
ally
solves it self many times. Silencing the solved sound during
scrambling will be helpful, but then you should probably also
discard all the twists that led to it since it just becomes unneeded
log file baggage.
The experience of scrolling around in the {32,3} is better than I
imagined. Somehow I expected to see only the fundamental polygons.
With N >=3D 100 you can probably just mask off the outermost few
pixels of the limit making it indistinguishable from the inf
version. You might also not center a face in the disk to disguise
its finiteness. Users can still scroll a face to the center but
they'd almost need to be trying to do that, and the larger the N,
the harder that will be.
The ragged borders are indeed unsightly. Normally that's not a
problem for solvers but it does go against the wonderful amount of
polish you've applied to MT in general. At the very least it shows
us what the experience can be which will be important in finding out
how interesting these puzzles are compared with other potential
puzzles.
As for the time needed to initialize these puzzles, perhaps you can
cache all the build data for all puzzles so that you never pay more
than once for each? It might also be nice to ship with the build
data for whichever puzzle you make the default. One last minor
suggestion: If it's not tricky, would you please see if you can mak=
e
the expanding circles animation spawn new circles centered on the
mouse pointer when it's in the frame? That would provide a nice
distraction while waiting.
Really nice work, Roice. Thanks a lot!r=3D"#888888">
-Melinda
wrote:
=20=20=20=20=20=20
=20=20=20=20=20=20
I liked your idea to do large N puzzles, so I configured some
biggish ones and added them to the download :) =C2=A0They are in th=
e
tree at "Hyperbolic -> Large Polygons". =C2=A0They tak=
e a bit
longer to build and the textures get a little pixelated, but
things work reasonably well. =C2=A0Solving the {32,3} 3C will
effectively be the same experience as an {inf,3} 3C, though I
would still like to see the infinite puzzle someday too. =C2=A0One
strange thing about {inf,3} will be that no matter how much you
hyperbolic pan, you won't be able to separate tiles from the
disk boundary, whereas in these puzzles you can drag a tile
across the disk center and to the other side.
wnloads/MagicTile_v2.zip
Melinda Green <erliminal.com" target=3D"_blank">melinda@superliminal.com>
wrote:
er-left:1px #ccc solid;padding-left:1ex">
Hello Roice,
I'm glad that you think that this puzzle makes sense.
Also, I like your idea of using fundamental domain
triangles. As for other colorings (and topologies), I
would first hope to see the simplest one(s) first. This
3-coloring seems about as simple as possible though
perhaps one could remove an edge or two by torturing the
topology a bit. As for incorporating into MT versus
creating a stand-alone puzzle, I have a feeling that there
might be some clever ways to incorporate it. One way might
be to implement it as a {N,3}=C2=A0 for some large N. If a us=
er
were to pan far enough to see the ragged edge, so be it.
If it must be a stand-alone puzzle, it might allow for
your alternate colorings and perhaps other interesting
variants that would otherwise be too difficult.lor=3D"#888888">
-Melinda
to be made!
will fall woefully short for this task, though
maybe I am overestimating the difficulty. =C2=A0Off
the cuff, an approach could be to try to allow
building up puzzles using fundamental domain
triangles rather than entire tiles, because it
will be necessary to only show portions of these
infinite-faceted tiles. =C2=A0(In the past, I'v=
e
wondered if that=C2=A0enhancement is going to be
necessary for uniform tilings.) =C2=A0It does seem
like a big piece of work, and it might even be
easier to write some special-case code for this
puzzle rather than attempting to fit it into the
engine.
possibilities for this tiling too.
7:19 PM, Melinda Green <f=3D"mailto:melinda@superliminal.com" target=3D"_blank">melinda@superlimina=
l.com>
wrote:
0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">
Here's a slightly less awful sketch:
ing/photos/album/1962624577/pic/40875152/view/" target=3D"_blank">http://gr=
oups.yahoo.com/group/4D_Cubing/photos/album/1962624577/pic/40875152/view/a>
wrote:
Lovely, Roice!
This makes me wonder whether it might
be possible to add a oups.yahoo.com/group/4D_Cubing/photos/album/1962624577/pic/908182938/view/"=
target=3D"_blank">3-color {inf,3}
to MagicTile something like this:
4D_Cubing/photos/album/1962624577/pic/908182938/view/" target=3D"_blank">gr=
oups.yahoo.com/group/4D_Cubing/photos/album/1962624577/pic/908182938/view/<=
/a>
-Melinda
Nelson wrote:
model of the exotic {4,4,4} H=C2=B3
honeycomb! =C2=A0
px 40px;border:none;padding:0px">
squares with an infinite number
of facets. All vertices are
ideal=C2=A0(meaning they live at
infinity, on the Poincare ball
boundary). =C2=A0Four cells meet at
every edge, and an infinite
number of cells meet at every
vertex (the vertex figure is a
tiling of squares too). =C2=A0This
honeycomb is self-dual.
I printed only half of the
Poincare ball in this model,
which has multiple advantages:
you can see inside better, and
it saves on printing costs. =C2=A0T=
he
view is face-centered, meaning
the projection places the center
of one (ideal) 2D polygon at the
center of the ball. =C2=A0An
edge-centered view is also
possible. =C2=A0Vertex-centered vie=
ws
are impossible since every
vertex is ideal. =C2=A0A view
centered on the interior of=C2=A0a
cell is possible, but (I think,
given my current understanding)
a cell-centered view is also
impossible. =C2=A0
the tiles around it, so only one
level of recursion. =C2=A0I also
experimented with deeper
recursion, but felt the
resulting density inhibited
understanding. =C2=A0Probably best
would be to have two models at
different recursion depths side
by side to study together. =C2=A0I
had to artificially increase
edge widths near the boundary to
make things printable.
cool to handle in person, so
consider ordering one or two
of the honeycomb models :)
=C2=A0As I've heard Henry
Segerman comment, the
"bandwidth" of inform=
ation
is really high. =C2=A0You
definitely notice things you
wouldn't if only viewing
them on the computer screen.
=C2=A0The {3,6,3} and {6,3,6} a=
re
very similar to the {4,4,4},
just based on different
Euclidean tilings, so models
of those are surely coming
as well.
make a puzzle based on
this exotic honeycomb? :D
Cheers,
Roice
a few thoughts I had about
the {4,4,4} while working on
the model... =C2=A0
thread on the {4,4,4},
Nan=C2=A0made an insightful
comment. =C2=A0He said:
le=3D"margin:0px 0px 0px 0.8ex;border-left-width:1px;border-left-color:rgb(=
204,204,204);border-left-style:solid;padding-left:1ex">I
believe the first step to
understand {4,4,4} is to
understand {infinity,
infinity} in the hyperbolic
plane.
I can see now they are indeed
quite analogous. =C2=A0Wikipedia has
some great pictures of the {=E2=88=9E=
,=E2=88=9E}
tiling and {p,q} tilings that
approach it by increasing p or q.
=C2=A0Check out the progression that
starts with an {=E2=88=9E,3} tiling a=
nd
increases q, which is the bottom
row of the table here:
0px 40px;border:none;padding:0px">
inscribed in .wikipedia.org/wiki/Horocycle" target=3D"_blank">horocycles=C2=A0(a
circle of infinite radius with
a unique center point on the
disk boundary).
=C2=A0The=C2=A0horocycles=C2=A0in=
crease in
size with this progression
until, in the limit, the
inscribing circle is the
boundary of the disk itself.>=C2=A0
Something strange about that
is an {=E2=88=9E,=E2=88=9E} tile =
loses its
center. =C2=A0A=C2=A0horocycle=C2=
=A0has a
single center on the boundary,
so the inscribed {=E2=88=9E,q} ti=
les
have a clear center, but
because an {=E2=88=9E,=E2=88=9E} =
tile is
inscribed in the entire
boundary, there is no longer a
unique center. =C2=A0Tile centers
are at infinity for the whole
progression, so you'd think
they would also live at
infinity in the limit. =C2=A0At t=
he
same time, all vertices have
also become ideal in the
limit, and these are the only
points of a tile living at
infinity. =C2=A0So every vertex
seems equally valid as a tile
center. =C2=A0Weird.
This is good warm-up to
jumping up a dimension. =C2=A0The
{4,4,3} is kind of like an
{=E2=88=9E,q} with finite q. =C2=
=A0It's
cells are inscribed in
horospheres, and have finite
vertices and a unique center.
=C2=A0The {4,4,4} is like the {=
=E2=88=9E,=E2=88=9E}
because cells are inscribed in
the boundary of hyperbolic
space. =C2=A0They don't reall=
y have
a unique center, and every
vertex is ideal. =C2=A0Again, eac=
h
vertex sort of acts like a
center point.=C2=A0
way to think about this...
Maybe when all the vertices go
to infinity, the cell center
should be considered to have
snapped back to being finite?
Maybe the center is at some
average of all the ideal
vertices or at a center of
mass? =C2=A0That makes sense for =
an
ideal tetrahedron, but can it
for a cell that is an ideal
{4,4} tiling? =C2=A0I don't k=
now!)
=20=20=20=20=20=20
=20=20
=20=20=20=20
=20=20=20=20
--001a11c238e83755c504e3d6678f--
From: Melinda Green <melinda@superliminal.com>
Date: Tue, 13 Aug 2013 16:19:18 -0700
Subject: Re: [MC4D] The exotic {4,4,4}
--------------090909010406070802060503
Content-Type: text/plain; charset=UTF-8; format=flowed
Content-Transfer-Encoding: quoted-printable
You mean this message?:=20
http://games.groups.yahoo.com/group/4D_Cubing/message/1492 Perhaps I=20
shouldn't be surprised but I am. I've spent a little time with the other=20
infinite puzzles without much progress so far. Has anybody else tried=20
them? If so, do you feel that there is any difference between these and=20
the rest of MT's bestiary?
Best,
-Melinda
On 8/13/2013 9:05 AM, Roice Nelson wrote:
>
>
> I want to recall that Andrey pointed out all the 3C puzzles are=20
> topologically the same, and behave like the {4,3} 3C (hemicube). The=20
> state space does feel too small to really be enjoyable...even as an=20
> appetizer :)
>
> Thanks so much for the suggestions Melinda. I especially like the=20
> idea to cache the puzzle build data. Don't know why I never thought=20
> of that before! I added your thoughts to my running trello list.
>
> Cheers,
> Roice
>
>
>
> On Mon, Aug 12, 2013 at 8:30 PM, Melinda Green=20
>
>
>
>
> Very nice, Roice!
>
> I was pretty sure the {inf,3} wouldn't be very difficult but I
> didn't expect it to be this easy It seems like god's number for it
> can be counted on one hand! One nit: Scrambling it with 1000
> twists rings the "solved" bell a whole bunch of times as it
> accidentally solves it self many times. Silencing the solved sound
> during scrambling will be helpful, but then you should probably
> also discard all the twists that led to it since it just becomes
> unneeded log file baggage.
>
> The experience of scrolling around in the {32,3} is better than I
> imagined. Somehow I expected to see only the fundamental polygons.
> With N >=3D 100 you can probably just mask off the outermost few
> pixels of the limit making it indistinguishable from the inf
> version. You might also not center a face in the disk to disguise
> its finiteness. Users can still scroll a face to the center but
> they'd almost need to be trying to do that, and the larger the N,
> the harder that will be.
>
> The ragged borders are indeed unsightly. Normally that's not a
> problem for solvers but it does go against the wonderful amount of
> polish you've applied to MT in general. At the very least it shows
> us what the experience can be which will be important in finding
> out how interesting these puzzles are compared with other
> potential puzzles.
>
> As for the time needed to initialize these puzzles, perhaps you
> can cache all the build data for all puzzles so that you never pay
> more than once for each? It might also be nice to ship with the
> build data for whichever puzzle you make the default. One last
> minor suggestion: If it's not tricky, would you please see if you
> can make the expanding circles animation spawn new circles
> centered on the mouse pointer when it's in the frame? That would
> provide a nice distraction while waiting.
>
> Really nice work, Roice. Thanks a lot!
> -Melinda
>
>
> On 8/12/2013 5:37 PM, Roice Nelson wrote:
>> Hi Melinda,
>>
>> I liked your idea to do large N puzzles, so I configured some
>> biggish ones and added them to the download :) They are in the
>> tree at "Hyperbolic -> Large Polygons". They take a bit longer
>> to build and the textures get a little pixelated, but things work
>> reasonably well. Solving the {32,3} 3C will effectively be the
>> same experience as an {inf,3} 3C, though I would still like to
>> see the infinite puzzle someday too. One strange thing about
>> {inf,3} will be that no matter how much you hyperbolic pan, you
>> won't be able to separate tiles from the disk boundary, whereas
>> in these puzzles you can drag a tile across the disk center and
>> to the other side.
>>
>> Download link:
>> http://www.gravitation3d.com/magictile/downloads/MagicTile_v2.zip
>>
>> And some pictures:
>> http://groups.yahoo.com/group/4D_Cubing/photos/album/1694853720/pic/=
435475920/view
>> http://groups.yahoo.com/group/4D_Cubing/photos/album/1694853720/pic/=
810718037/view
>> http://groups.yahoo.com/group/4D_Cubing/photos/album/1694853720/pic/=
600497813/view
>>
>> seeya,
>> Roice
>>
>>
>>
>> On Sun, Aug 11, 2013 at 10:32 PM, Melinda Green
>>
>>
>>
>>
>> Hello Roice,
>>
>> I'm glad that you think that this puzzle makes sense. Also, I
>> like your idea of using fundamental domain triangles. As for
>> other colorings (and topologies), I would first hope to see
>> the simplest one(s) first. This 3-coloring seems about as
>> simple as possible though perhaps one could remove an edge or
>> two by torturing the topology a bit. As for incorporating
>> into MT versus creating a stand-alone puzzle, I have a
>> feeling that there might be some clever ways to incorporate
>> it. One way might be to implement it as a {N,3} for some
>> large N. If a user were to pan far enough to see the ragged
>> edge, so be it. If it must be a stand-alone puzzle, it might
>> allow for your alternate colorings and perhaps other
>> interesting variants that would otherwise be too difficult.
>>
>> -Melinda
>>
>>
>> On 8/11/2013 8:06 PM, Roice Nelson wrote:
>>> The puzzle in your pictures *needs* to be made!
>>>
>>> It feels like the current MagicTile engine will fall
>>> woefully short for this task, though maybe I am
>>> overestimating the difficulty. Off the cuff, an approach
>>> could be to try to allow building up puzzles using
>>> fundamental domain triangles rather than entire tiles,
>>> because it will be necessary to only show portions of these
>>> infinite-faceted tiles. (In the past, I've wondered if
>>> that enhancement is going to be necessary for uniform
>>> tilings.) It does seem like a big piece of work, and it
>>> might even be easier to write some special-case code for
>>> this puzzle rather than attempting to fit it into the engine.
>>>
>>> I bet there is an infinite set of coloring possibilities for
>>> this tiling too.
>>>
>>>
>>>
>>> On Sun, Aug 11, 2013 at 7:19 PM, Melinda Green
>>>
>>> wrote:
>>>
>>>
>>>
>>> Here's a slightly less awful sketch:
>>> http://groups.yahoo.com/group/4D_Cubing/photos/album/196262=
4577/pic/40875152/view/
>>>
>>>
>>>
>>> On 8/11/2013 4:34 PM, Melinda Green wrote:
>>>> Lovely, Roice!
>>>>
>>>> This makes me wonder whether it might be possible to
>>>> add a 3-color {inf,3}
>>>>
>>>> to MagicTile something like this:
>>>> groups.yahoo.com/group/4D_Cubing/photos/album/1962624577/p=
ic/908182938/view/
>>>>
>>>>
>>>> -Melinda
>>>>
>>>> On 8/10/2013 2:10 PM, Roice Nelson wrote:
>>>>> Hi all,
>>>>>
>>>>> Check out a new physical model of the exotic {4,4,4}
>>>>> H=C2=B3 honeycomb!
>>>>>
>>>>> http://shpws.me/oFpu
>>>>>
>>>>>
>>>>> Each cell is a tiling of squares with an infinite
>>>>> number of facets. All vertices are ideal (meaning they
>>>>> live at infinity, on the Poincare ball boundary).
>>>>> Four cells meet at every edge, and an infinite number
>>>>> of cells meet at every vertex (the vertex figure is a
>>>>> tiling of squares too). This honeycomb is self-dual.
>>>>>
>>>>> I printed only half of the Poincare ball in this
>>>>> model, which has multiple advantages: you can see
>>>>> inside better, and it saves on printing costs. The
>>>>> view is face-centered, meaning the projection places
>>>>> the center of one (ideal) 2D polygon at the center of
>>>>> the ball. An edge-centered view is also possible.
>>>>> Vertex-centered views are impossible since every
>>>>> vertex is ideal. A view centered on the interior of a
>>>>> cell is possible, but (I think, given my current
>>>>> understanding) a cell-centered view is also impossible.
>>>>>
>>>>> I rendered one tile and all the tiles around it, so
>>>>> only one level of recursion. I also experimented with
>>>>> deeper recursion, but felt the resulting density
>>>>> inhibited understanding. Probably best would be to
>>>>> have two models at different recursion depths side by
>>>>> side to study together. I had to artificially
>>>>> increase edge widths near the boundary to make things
>>>>> printable.
>>>>>
>>>>> These things are totally cool to handle in person, so
>>>>> consider ordering one or two of the honeycomb models
>>>>> :) As I've heard Henry Segerman comment, the
>>>>> "bandwidth" of information is really high. You
>>>>> definitely notice things you wouldn't if only viewing
>>>>> them on the computer screen. The {3,6,3} and {6,3,6}
>>>>> are very similar to the {4,4,4}, just based on
>>>>> different Euclidean tilings, so models of those are
>>>>> surely coming as well.
>>>>>
>>>>> So... whose going to make a puzzle based on this
>>>>> exotic honeycomb? :D
>>>>>
>>>>> Cheers,
>>>>> Roice
>>>>>
>>>>>
>>>>> As a postscript, here are a few thoughts I had about
>>>>> the {4,4,4} while working on the model...
>>>>>
>>>>> In a previous thread on the {4,4,4}
>>>>>
>>>>> Nan made an insightful comment. He said:
>>>>>
>>>>> I believe the first step to understand {4,4,4} is
>>>>> to understand {infinity,
>>>>> infinity} in the hyperbolic plane.=20
>>>>>
>>>>>
>>>>> I can see now they are indeed quite analogous.
>>>>> Wikipedia has some great pictures of the {=E2=88=9E,=E2=
=88=9E} tiling
>>>>> and {p,q} tilings that approach it by increasing p or
>>>>> q. Check out the progression that starts with an
>>>>> {=E2=88=9E,3} tiling and increases q, which is the bottom=
row
>>>>> of the table here:
>>>>>
>>>>> http://en.wikipedia.org/wiki/Uniform_tilings_in_hyper=
bolic_plane#Regular_hyperbolic_tilings
>>>>>
>>>>>
>>>>> The {=E2=88=9E} polygons are inscribed in horocycles
>>>>>
>>>>> infinite radius with a unique center point on the disk
>>>>> boundary). The horocycles increase in size with this
>>>>> progression until, in the limit, the inscribing circle
>>>>> is*the boundary of the disk itself.* Something strange
>>>>> about that is an {=E2=88=9E,=E2=88=9E} tile loses its cen=
ter.
>>>>> A horocycle has a single center on the boundary, so
>>>>> the inscribed {=E2=88=9E,q} tiles have a clear center, bu=
t
>>>>> because an {=E2=88=9E,=E2=88=9E} tile is inscribed in the=
entire
>>>>> boundary, there is no longer a unique center. Tile
>>>>> centers are at infinity for the whole progression, so
>>>>> you'd think they would also live at infinity in the
>>>>> limit. At the same time, all vertices have also
>>>>> become ideal in the limit, and these are the only
>>>>> points of a tile living at infinity. So every vertex
>>>>> seems equally valid as a tile center. Weird.
>>>>>
>>>>> This is good warm-up to jumping up a dimension. The
>>>>> {4,4,3} is kind of like an {=E2=88=9E,q} with finite q. =
It's
>>>>> cells are inscribed in horospheres, and have finite
>>>>> vertices and a unique center. The {4,4,4} is like the
>>>>> {=E2=88=9E,=E2=88=9E} because cells are inscribed in the =
boundary of
>>>>> hyperbolic space. They don't really have a unique
>>>>> center, and every vertex is ideal. Again, each vertex
>>>>> sort of acts like a center point.
>>>>>
>>>>> (Perhaps there is a better way to think about this...
>>>>> Maybe when all the vertices go to infinity, the cell
>>>>> center should be considered to have snapped back to
>>>>> being finite? Maybe the center is at some average of
>>>>> all the ideal vertices or at a center of mass? That
>>>>> makes sense for an ideal tetrahedron, but can it for a
>>>>> cell that is an ideal {4,4} tiling? I don't know!)
>>>>
>>>
>>>
>>>
>>>
>>
>>
>>
>>
>
>
>
>
>
>
>=20
--------------090909010406070802060503
Content-Type: text/html; charset=UTF-8
Content-Transfer-Encoding: quoted-printable
">
You mean this message?:
m/group/4D_Cubing/message/1492">http://games.groups.yahoo.com/group/4D_Cubi=
ng/message/1492 Perhaps I
shouldn't be surprised but I am. I've spent a little time with the
other infinite puzzles without much progress so far. Has anybody
else tried them? If so, do you feel that there is any difference
between these and the rest of MT's bestiary?
Best,
-Melinda
wrote:
l.com"
type=3D"cite">
are topologically the same, and behave like the {4,3} 3C
(hemicube). =C2=A0The state space does feel too small to really b=
e
enjoyable...even as an appetizer :)
Thanks so much for the suggestions Melinda. =C2=A0I especially like
the idea to cache the puzzle build data. =C2=A0Don't know why I nev=
er
thought of that before! =C2=A0I added your thoughts to my running
trello list.
Melinda Green < href=3D"mailto:melinda@superliminal.com" target=3D"_blank">me=
linda@superliminal.com>
wrote:
.8ex;border-left:1px #ccc solid;padding-left:1ex">
Very nice, Roice!
I was pretty sure the {inf,3} wouldn't be very difficult
but I didn't expect it to be this easy It seems like god's
number for it can be counted on one hand! One nit:
Scrambling it with 1000 twists rings the "solved" bell a
whole bunch of times as it accidentally solves it self
many times. Silencing the solved sound during scrambling
will be helpful, but then you should probably also discard
all the twists that led to it since it just becomes
unneeded log file baggage.
The experience of scrolling around in the {32,3} is better
than I imagined. Somehow I expected to see only the
fundamental polygons. With N >=3D 100 you can probably
just mask off the outermost few pixels of the limit making
it indistinguishable from the inf version. You might also
not center a face in the disk to disguise its finiteness.
Users can still scroll a face to the center but they'd
almost need to be trying to do that, and the larger the N,
the harder that will be.
The ragged borders are indeed unsightly. Normally that's
not a problem for solvers but it does go against the
wonderful amount of polish you've applied to MT in
general. At the very least it shows us what the experience
can be which will be important in finding out how
interesting these puzzles are compared with other
potential puzzles.
As for the time needed to initialize these puzzles,
perhaps you can cache all the build data for all puzzles
so that you never pay more than once for each? It might
also be nice to ship with the build data for whichever
puzzle you make the default. One last minor suggestion: If
it's not tricky, would you please see if you can make the
expanding circles animation spawn new circles centered on
the mouse pointer when it's in the frame? That would
provide a nice distraction while waiting.
Really nice work, Roice. Thanks a lot!=
-Melinda
I liked your idea to do large N puzzles, so I
configured some biggish ones and added them to the
download :) =C2=A0They are in the tree at "Hyperbolic
-> Large Polygons". =C2=A0They take a bit longer t=
o
build and the textures get a little pixelated, but
things work reasonably well. =C2=A0Solving the {32,3}
3C will effectively be the same experience as an
{inf,3} 3C, though I would still like to see the
infinite puzzle someday too. =C2=A0One strange thing
about {inf,3} will be that no matter how much you
hyperbolic pan, you won't be able to separate
tiles from the disk boundary, whereas in these
puzzles you can drag a tile across the disk center
and to the other side.
10:32 PM, Melinda Green <
href=3D"mailto:melinda@superliminal.com"
target=3D"_blank">melinda@superliminal.com&=
gt;
wrote:
0 0 .8ex;border-left:1px #ccc
solid;padding-left:1ex">
Hello Roice,
I'm glad that you think that this puzzle
makes sense. Also, I like your idea of using
fundamental domain triangles. As for other
colorings (and topologies), I would first
hope to see the simplest one(s) first. This
3-coloring seems about as simple as possible
though perhaps one could remove an edge or
two by torturing the topology a bit. As for
incorporating into MT versus creating a
stand-alone puzzle, I have a feeling that
there might be some clever ways to
incorporate it. One way might be to
implement it as a {N,3}=C2=A0 for some large N.
If a user were to pan far enough to see the
ragged edge, so be it. If it must be a
stand-alone puzzle, it might allow for your
alternate colorings and perhaps other
interesting variants that would otherwise be
too difficult.
-Melinda
wrote:
pictures *needs* to be made!
MagicTile engine will fall
woefully short for this task,
though maybe I am overestimating
the difficulty. =C2=A0Off the cuff, a=
n
approach could be to try to allow
building up puzzles using
fundamental domain triangles
rather than entire tiles, because
it will be necessary to only show
portions of these infinite-faceted
tiles. =C2=A0(In the past, I've
wondered if that=C2=A0enhancement is
going to be necessary for uniform
tilings.) =C2=A0It does seem like a b=
ig
piece of work, and it might even
be easier to write some
special-case code for this puzzle
rather than attempting to fit it
into the engine.
of coloring possibilities for this
tiling too.
11, 2013 at 7:19 PM, Melinda Green
<
href=3D"mailto:melinda@superlimin=
al.com"
target=3D"_blank">melinda@superli=
minal.com>
wrote:
style=3D"margin:0 0 0
.8ex;border-left:1px #ccc
solid;padding-left:1ex">
Here's a slightly less awful
sketch:
href=3D"http://groups.yahoo.com/group/4D_Cubing/photos/album/1962624577/pic=
/40875152/view/"
target=3D"_blank">http://groups=
.yahoo.com/group/4D_Cubing/photos/album/1962624577/pic/40875152/view/
Melinda Green wrote:
Lovely, Roice!
This makes me wonder
whether it might be
possible to add a
href=3D"http://groups.yahoo.com/group/4D_Cubing/photos/album/1962624577/pic=
/908182938/view/"
target=3D"_blank">3-color
{inf,3} to
MagicTile something like
this:
href=3D"http://groups.yahoo.com/group/4D_Cubing/photos/album/1962624577/pic=
/908182938/view/"
target=3D"_blank">groups.=
yahoo.com/group/4D_Cubing/photos/album/1962624577/pic/908182938/view/
-Melinda
PM, Roice Nelson
wrote:
physical model of
the exotic {4,4,4}
H=C2=B3 honeycomb! =
=C2=A0
0px 0px
40px;border:none;padd=
ing:0px">
=3D"true"
href=3D"http://shpws.me/oFpu" target=3D"_blank">http://shpws.me/oFpuiv>
tiling of squares
with an infinite
number of facets.
All vertices are
ideal=C2=A0(meaning
they live at
infinity, on the
Poincare ball
boundary). =C2=A0Four
cells meet at
every edge, and an
infinite number of
cells meet at
every vertex (the
vertex figure is a
tiling of squares
too). =C2=A0This
honeycomb is
self-dual.
I printed only
half of the
Poincare ball in
this model, which
has multiple
advantages: you
can see inside
better, and it
saves on printing
costs. =C2=A0The view
is face-centered,
meaning the
projection places
the center of one
(ideal) 2D polygon
at the center of
the ball. =C2=A0An
edge-centered view
is also possible.
=C2=A0Vertex-centered
views are
impossible since
every vertex is
ideal. =C2=A0A view
centered on the
interior of=C2=A0a ce=
ll
is possible, but
(I think, given my
current
understanding) a
cell-centered view
is also
impossible. =C2=A0iv>
tile and all the
tiles around it,
so only one level
of recursion. =C2=A0I
also experimented
with deeper
recursion, but
felt the resulting
density inhibited
understanding.
=C2=A0Probably best
would be to have
two models at
different
recursion depths
side by side to
study together. =C2=
=A0I
had to
artificially
increase edge
widths near the
boundary to make
things printable.v>
things are
totally cool
to handle in
person, so
consider
ordering one
or two of the
honeycomb
models :) =C2=A0A=
s
I've heard
Henry Segerman
comment, the
"bandwidth" of
information is
really high.
=C2=A0You
definitely
notice things
you wouldn't
if only
viewing them
on the
computer
screen. =C2=A0The
{3,6,3} and
{6,3,6} are
very similar
to the
{4,4,4}, just
based on
different
Euclidean
tilings, so
models of
those are
surely coming
as well.
whose going to
make a puzzle
based on this
exotic
honeycomb? :D
Cheers,
Roice
postscript,
here are a few
thoughts I had
about the
{4,4,4} while
working on the
model... =C2=A0div>
=3D"true"
href=3D"http://games.groups.yahoo.com/group/4D_Cubing/message/1226"
target=3D"_blank"=
>a
previous
thread on the
{4,4,4},
Nan=C2=A0made an
insightful
comment. =C2=A0He
said:
style=3D"margin:0px
0px 0px
0.8ex;border-left-width:1px;border-left-color:rgb(204,204,204);border-left-=
style:solid;padding-left:1ex">I
believe the first
step to understand
{4,4,4} is to
understand
{infinity,
infinity} in the
hyperbolic plane.
I can see now they
are indeed quite
analogous.
=C2=A0Wikipedia has som=
e
great pictures of
the {=E2=88=9E,=E2=88=
=9E} tiling and
{p,q} tilings that
approach it by
increasing p or q.
=C2=A0Check out the
progression that
starts with an {=E2=88=
=9E,3}
tiling and increases
q, which is the
bottom row of the
table here:
0px 0px
40px;border:none;pa=
dding:0px">
=3D"true"
href=3D"http://en.wikipedia.org/wiki/Uniform_tilings_in_hyperbolic_plane#Re=
gular_hyperbolic_tilings"
target=3D"_blank">http://en.wikipedia.org/wiki/Uniform_tilings_in_hyperboli=
c_plane#Regular_hyperbolic_tilings
polygons are
inscribed in
target=3D"_blank"=
>horocycles=C2=A0(a
circle of
infinite radius
with a unique
center point on
the disk
boundary).
=C2=A0The=C2=A0horo=
cycles=C2=A0increase
in size with
this progression
until, in the
limit, the
inscribing
circle is the
boundary of
the disk
itself.=C2=A0
Something
strange about
that is an {=E2=88=
=9E,=E2=88=9E}
tile loses its
center.
=C2=A0A=C2=A0horocy=
cle=C2=A0has
a single center
on the boundary,
so the inscribed
{=E2=88=9E,q} tiles=
have
a clear center,
but because an
{=E2=88=9E,=E2=88=
=9E} tile is
inscribed in the
entire boundary,
there is no
longer a unique
center. =C2=A0Tile
centers are at
infinity for the
whole
progression, so
you'd think they
would also live
at infinity in
the limit. =C2=A0At
the same time,
all vertices
have also become
ideal in the
limit, and these
are the only
points of a tile
living at
infinity. =C2=A0So
every vertex
seems equally
valid as a tile
center. =C2=A0Weird=
.
This is good
warm-up to
jumping up a
dimension. =C2=A0Th=
e
{4,4,3} is kind
of like an {=E2=88=
=9E,q}
with finite q.
=C2=A0It's cells ar=
e
inscribed in
horospheres, and
have finite
vertices and a
unique center.
=C2=A0The {4,4,4} i=
s
like the {=E2=88=9E=
,=E2=88=9E}
because cells
are inscribed in
the boundary of
hyperbolic
space. =C2=A0They
don't really
have a unique
center, and
every vertex is
ideal. =C2=A0Again,
each vertex sort
of acts like a
center point.=C2=A0=
there is a
better way to
think about
this... Maybe
when all the
vertices go to
infinity, the
cell center
should be
considered to
have snapped
back to being
finite? Maybe
the center is at
some average of
all the ideal
vertices or at a
center of mass?
=C2=A0That makes
sense for an
ideal
tetrahedron, but
can it for a
cell that is an
ideal {4,4}
tiling? =C2=A0I don=
't
know!)
=20=20=20=20=20=20
--------------090909010406070802060503--
From: "Andrey" <andreyastrelin@yahoo.com>
Date: Thu, 22 Aug 2013 14:52:07 -0000
Subject: Re: [MC4D] The exotic {4,4,4}
Hi all,
I've tried 4 and 6 colors. As I expected, they are just versions of our f=
amiliar tetrahedron and cube. Unfortunately, I almost forgot how to solve t=
hem, so results are not very good: 60 and 142 twists...
I'll think about coloring of {infinity,3} - it should be not difficult.
Andrey
--- In 4D_Cubing@yahoogroups.com, Melinda Green
>
> You mean this message?:=20
> http://games.groups.yahoo.com/group/4D_Cubing/message/1492 Perhaps I=20
> shouldn't be surprised but I am. I've spent a little time with the other=
=20
> infinite puzzles without much progress so far. Has anybody else tried=20
> them? If so, do you feel that there is any difference between these and=20
> the rest of MT's bestiary?
>=20
> Best,
> -Melinda
>=20
> On 8/13/2013 9:05 AM, Roice Nelson wrote:
> >
> >
> > I want to recall that Andrey pointed out all the 3C puzzles are=20
> > topologically the same, and behave like the {4,3} 3C (hemicube). The=20
> > state space does feel too small to really be enjoyable...even as an=20
> > appetizer :)
> >
> > Thanks so much for the suggestions Melinda. I especially like the=20
> > idea to cache the puzzle build data. Don't know why I never thought=20
> > of that before! I added your thoughts to my running trello list.
> >
> > Cheers,
> > Roice
> >
> >
> >
> > On Mon, Aug 12, 2013 at 8:30 PM, Melinda Green=20
> >
> >
> >
> >
> > Very nice, Roice!
> >
> > I was pretty sure the {inf,3} wouldn't be very difficult but I
> > didn't expect it to be this easy It seems like god's number for it
> > can be counted on one hand! One nit: Scrambling it with 1000
> > twists rings the "solved" bell a whole bunch of times as it
> > accidentally solves it self many times. Silencing the solved sound
> > during scrambling will be helpful, but then you should probably
> > also discard all the twists that led to it since it just becomes
> > unneeded log file baggage.
> >
> > The experience of scrolling around in the {32,3} is better than I
> > imagined. Somehow I expected to see only the fundamental polygons.
> > With N >=3D 100 you can probably just mask off the outermost few
> > pixels of the limit making it indistinguishable from the inf
> > version. You might also not center a face in the disk to disguise
> > its finiteness. Users can still scroll a face to the center but
> > they'd almost need to be trying to do that, and the larger the N,
> > the harder that will be.
> >
> > The ragged borders are indeed unsightly. Normally that's not a
> > problem for solvers but it does go against the wonderful amount of
> > polish you've applied to MT in general. At the very least it shows
> > us what the experience can be which will be important in finding
> > out how interesting these puzzles are compared with other
> > potential puzzles.
> >
> > As for the time needed to initialize these puzzles, perhaps you
> > can cache all the build data for all puzzles so that you never pay
> > more than once for each? It might also be nice to ship with the
> > build data for whichever puzzle you make the default. One last
> > minor suggestion: If it's not tricky, would you please see if you
> > can make the expanding circles animation spawn new circles
> > centered on the mouse pointer when it's in the frame? That would
> > provide a nice distraction while waiting.
> >
> > Really nice work, Roice. Thanks a lot!
> > -Melinda
> >
> >
> > On 8/12/2013 5:37 PM, Roice Nelson wrote:
> >> Hi Melinda,
> >>
> >> I liked your idea to do large N puzzles, so I configured some
> >> biggish ones and added them to the download :) They are in the
> >> tree at "Hyperbolic -> Large Polygons". They take a bit longer
> >> to build and the textures get a little pixelated, but things work
> >> reasonably well. Solving the {32,3} 3C will effectively be the
> >> same experience as an {inf,3} 3C, though I would still like to
> >> see the infinite puzzle someday too. One strange thing about
> >> {inf,3} will be that no matter how much you hyperbolic pan, you
> >> won't be able to separate tiles from the disk boundary, whereas
> >> in these puzzles you can drag a tile across the disk center and
> >> to the other side.
> >>
> >> Download link:
> >> http://www.gravitation3d.com/magictile/downloads/MagicTile_v2.zip
> >>
> >> And some pictures:
> >> http://groups.yahoo.com/group/4D_Cubing/photos/album/1694853720/pi=
c/435475920/view
> >> http://groups.yahoo.com/group/4D_Cubing/photos/album/1694853720/pi=
c/810718037/view
> >> http://groups.yahoo.com/group/4D_Cubing/photos/album/1694853720/pi=
c/600497813/view
> >>
> >> seeya,
> >> Roice
> >>
> >>
> >>
> >> On Sun, Aug 11, 2013 at 10:32 PM, Melinda Green
> >>
> >>
> >>
> >>
> >> Hello Roice,
> >>
> >> I'm glad that you think that this puzzle makes sense. Also, I
> >> like your idea of using fundamental domain triangles. As for
> >> other colorings (and topologies), I would first hope to see
> >> the simplest one(s) first. This 3-coloring seems about as
> >> simple as possible though perhaps one could remove an edge or
> >> two by torturing the topology a bit. As for incorporating
> >> into MT versus creating a stand-alone puzzle, I have a
> >> feeling that there might be some clever ways to incorporate
> >> it. One way might be to implement it as a {N,3} for some
> >> large N. If a user were to pan far enough to see the ragged
> >> edge, so be it. If it must be a stand-alone puzzle, it might
> >> allow for your alternate colorings and perhaps other
> >> interesting variants that would otherwise be too difficult.
> >>
> >> -Melinda
> >>
> >>
> >> On 8/11/2013 8:06 PM, Roice Nelson wrote:
> >>> The puzzle in your pictures *needs* to be made!
> >>>
> >>> It feels like the current MagicTile engine will fall
> >>> woefully short for this task, though maybe I am
> >>> overestimating the difficulty. Off the cuff, an approach
> >>> could be to try to allow building up puzzles using
> >>> fundamental domain triangles rather than entire tiles,
> >>> because it will be necessary to only show portions of these
> >>> infinite-faceted tiles. (In the past, I've wondered if
> >>> that enhancement is going to be necessary for uniform
> >>> tilings.) It does seem like a big piece of work, and it
> >>> might even be easier to write some special-case code for
> >>> this puzzle rather than attempting to fit it into the engine.
> >>>
> >>> I bet there is an infinite set of coloring possibilities for
> >>> this tiling too.
> >>>
> >>>
> >>>
> >>> On Sun, Aug 11, 2013 at 7:19 PM, Melinda Green
> >>>
> >>> wrote:
> >>>
> >>>
> >>>
> >>> Here's a slightly less awful sketch:
> >>> http://groups.yahoo.com/group/4D_Cubing/photos/album/1962=
624577/pic/40875152/view/
> >>>
> >>>
> >>>
> >>> On 8/11/2013 4:34 PM, Melinda Green wrote:
> >>>> Lovely, Roice!
> >>>>
> >>>> This makes me wonder whether it might be possible to
> >>>> add a 3-color {inf,3}
> >>>>
> >>>> to MagicTile something like this:
> >>>> groups.yahoo.com/group/4D_Cubing/photos/album/1962624577=
/pic/908182938/view/
> >>>>
> >>>>
> >>>> -Melinda
> >>>>
> >>>> On 8/10/2013 2:10 PM, Roice Nelson wrote:
> >>>>> Hi all,
> >>>>>
> >>>>> Check out a new physical model of the exotic {4,4,4}
> >>>>> H=C2=B3 honeycomb!
> >>>>>
> >>>>> http://shpws.me/oFpu
> >>>>>
> >>>>>
> >>>>> Each cell is a tiling of squares with an infinite
> >>>>> number of facets. All vertices are ideal (meaning they
> >>>>> live at infinity, on the Poincare ball boundary).
> >>>>> Four cells meet at every edge, and an infinite number
> >>>>> of cells meet at every vertex (the vertex figure is a
> >>>>> tiling of squares too). This honeycomb is self-dual.
> >>>>>
> >>>>> I printed only half of the Poincare ball in this
> >>>>> model, which has multiple advantages: you can see
> >>>>> inside better, and it saves on printing costs. The
> >>>>> view is face-centered, meaning the projection places
> >>>>> the center of one (ideal) 2D polygon at the center of
> >>>>> the ball. An edge-centered view is also possible.
> >>>>> Vertex-centered views are impossible since every
> >>>>> vertex is ideal. A view centered on the interior of a
> >>>>> cell is possible, but (I think, given my current
> >>>>> understanding) a cell-centered view is also impossible.
> >>>>>
> >>>>> I rendered one tile and all the tiles around it, so
> >>>>> only one level of recursion. I also experimented with
> >>>>> deeper recursion, but felt the resulting density
> >>>>> inhibited understanding. Probably best would be to
> >>>>> have two models at different recursion depths side by
> >>>>> side to study together. I had to artificially
> >>>>> increase edge widths near the boundary to make things
> >>>>> printable.
> >>>>>
> >>>>> These things are totally cool to handle in person, so
> >>>>> consider ordering one or two of the honeycomb models
> >>>>> :) As I've heard Henry Segerman comment, the
> >>>>> "bandwidth" of information is really high. You
> >>>>> definitely notice things you wouldn't if only viewing
> >>>>> them on the computer screen. The {3,6,3} and {6,3,6}
> >>>>> are very similar to the {4,4,4}, just based on
> >>>>> different Euclidean tilings, so models of those are
> >>>>> surely coming as well.
> >>>>>
> >>>>> So... whose going to make a puzzle based on this
> >>>>> exotic honeycomb? :D
> >>>>>
> >>>>> Cheers,
> >>>>> Roice
> >>>>>
> >>>>>
> >>>>> As a postscript, here are a few thoughts I had about
> >>>>> the {4,4,4} while working on the model...
> >>>>>
> >>>>> In a previous thread on the {4,4,4}
> >>>>>
> >>>>> Nan made an insightful comment. He said:
> >>>>>
> >>>>> I believe the first step to understand {4,4,4} is
> >>>>> to understand {infinity,
> >>>>> infinity} in the hyperbolic plane.=20
> >>>>>
> >>>>>
> >>>>> I can see now they are indeed quite analogous.
> >>>>> Wikipedia has some great pictures of the {=E2=88=9E,=
=E2=88=9E} tiling
> >>>>> and {p,q} tilings that approach it by increasing p or
> >>>>> q. Check out the progression that starts with an
> >>>>> {=E2=88=9E,3} tiling and increases q, which is the bott=
om row
> >>>>> of the table here:
> >>>>>
> >>>>> http://en.wikipedia.org/wiki/Uniform_tilings_in_hyp=
erbolic_plane#Regular_hyperbolic_tilings
> >>>>>
> >>>>>
> >>>>> The {=E2=88=9E} polygons are inscribed in horocycles
> >>>>>
> >>>>> infinite radius with a unique center point on the disk
> >>>>> boundary). The horocycles increase in size with this
> >>>>> progression until, in the limit, the inscribing circle
> >>>>> is*the boundary of the disk itself.* Something strange
> >>>>> about that is an {=E2=88=9E,=E2=88=9E} tile loses its c=
enter.
> >>>>> A horocycle has a single center on the boundary, so
> >>>>> the inscribed {=E2=88=9E,q} tiles have a clear center, =
but
> >>>>> because an {=E2=88=9E,=E2=88=9E} tile is inscribed in t=
he entire
> >>>>> boundary, there is no longer a unique center. Tile
> >>>>> centers are at infinity for the whole progression, so
> >>>>> you'd think they would also live at infinity in the
> >>>>> limit. At the same time, all vertices have also
> >>>>> become ideal in the limit, and these are the only
> >>>>> points of a tile living at infinity. So every vertex
> >>>>> seems equally valid as a tile center. Weird.
> >>>>>
> >>>>> This is good warm-up to jumping up a dimension. The
> >>>>> {4,4,3} is kind of like an {=E2=88=9E,q} with finite q.=
It's
> >>>>> cells are inscribed in horospheres, and have finite
> >>>>> vertices and a unique center. The {4,4,4} is like the
> >>>>> {=E2=88=9E,=E2=88=9E} because cells are inscribed in th=
e boundary of
> >>>>> hyperbolic space. They don't really have a unique
> >>>>> center, and every vertex is ideal. Again, each vertex
> >>>>> sort of acts like a center point.
> >>>>>
> >>>>> (Perhaps there is a better way to think about this...
> >>>>> Maybe when all the vertices go to infinity, the cell
> >>>>> center should be considered to have snapped back to
> >>>>> being finite? Maybe the center is at some average of
> >>>>> all the ideal vertices or at a center of mass? That
> >>>>> makes sense for an ideal tetrahedron, but can it for a
> >>>>> cell that is an ideal {4,4} tiling? I don't know!)
> >>>>
> >>>
> >>>
> >>>
> >>>
> >>
> >>
> >>
> >>
> >
> >
> >
> >
> >
> >
> >
>