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Philip writes:
> I have also made the square 2 puzzle before it was patented.

Actually it has not been patented, unless you mean the original Square-1. =
Back around 1995 I cut the corners of a Square-1 to make a Square-2 (and na=
med it that) but later when I joined TwistyPuzzles I found two others who h=
ad also on their own made this puzzle (although far later). So I get some =
credit for being the first (so far known, the Square-1 came out in 1993 in =
the US) to build it, but I'm not at all surprised if you also independently=
created it, there are a few of us.
I'd love to see pictures, and get some back story on how and when you made =
it.

Dave
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space; -webkit-line-break: after-white-space; ">Philip writes:

font-family: Georgia; font-size: 13px; line-height: 15px; ">
v>

-size: 13px; line-height: 15px; ">Actually it has not been patented, unless=
you mean the original Square-1. Back around 1995 I cut the corners o=
f a Square-1 to make a Square-2 (and named it that) but later when I joined=
TwistyPuzzles I found two others who had also on their own made this puzzl=
e (although far later). So I get some credit for being the first (so =
far known, the Square-1 came out in 1993 in the US) to build it, but I'm no=
t at all surprised if you also independently created it, there are a few of=
us.

: Georgia; font-size: 13px; line-height: 15px; ">I'd love to see pictures, =
and get some back story on how and when you made it.

class=3D"Apple-style-span" style=3D"font-family: Georgia; font-size: 13px;=
line-height: 15px; ">

" style=3D"font-family: Georgia; font-size: 13px; line-height: 15px; ">Dave=

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From: "schuma" <mananself@gmail.com>
Date: Wed, 20 Mar 2013 05:59:15 -0000
Subject: Re: About me

> What version of the 24 cell would you like me to solve? There are quite a=
few!

In MPUlt, the 24-cell_FT is the shallow-cut puzzle that supports all the mo=
ves. This is THE 24-cell puzzle that got most attention, I think.

> Maybe even a 120 cell pentultimate would be an idea for a future 4d puzzl=
e!

In the latest MPUlt, there is a 120-cell_halfcut. It's the Face turning 120=
-cell where the cutting plane passes the center. I think it can be consider=
ed as the Pentultimate in 4D. I don't think anyone has attempted it. But I'=
m sure you have enough courage to make the move.

> My mind sometimes cannot sleep because of ideas running through my head, =
but it is sooo discouraging that I don't have funds or proper equipment to =
make and sell them...

I also like thinking about puzzles. For me, some ideas end up being simulat=
ors people can play on computers. (see http://nanma80.github.com/) Compared=
to physical puzzles, it's easy to test the concept, easy to share among fr=
iends all over the world, easy to incorporate feedbacks and suggestions and=
iterate to the next version, and of course, go beyond physics, like allowi=
ng reflection moves. About the ideas of physical puzzles, some people go to=
twistypuzzles.com to share some concepts, or, even internal mechanism. Som=
etimes, other builders bring them into reality, and give credit to the orig=
inal contributor. Oskar van Deventer, for example, made a lot of puzzles. B=
ut he acknowledged many people for their original ideas. I think the users =
of this forum have a pretty healthy attitude on recognizing the ideas, than=
ks to good management by David Litwin and others.

About the regular polytopes in high dimensional space, Melinda said,
> Cracking my bible (Regular Polytopes) to chapter and verse 7-8 gives ...
Believe it or not, the first thing after I got home today was also checking=
this book. Before this verse, Coxeter was calculating the dihedral angles.=
I think the argument can be explained using this example.=20

The dihedral angle of a 3D cube is 90 degrees. When you use some cubes to m=
ake a 4D regular polytope, how many cubes can you fit around an edge? It's =
necessary that the sum of dihedral angles is less than 360 degrees to make =
a bounded 4D polytope, or equal to 360 degrees to make an unbounded tessell=
ation. So, two cubes are too few (the outcome is flat). Three cubes around =
an edge are OK (hypercube). Four cubes around an edge will form the tessell=
ation. So three cubes around an edge is the only valid way to make a 4D bou=
nded polytope using cubes as faces, in Euclidean space. The necessary condi=
tion is that the dihedral angle must be small enough. As I understand, equa=
tion (7.77) formalizes this argument in math.

To make 5D regular polytopes, we need to pick some 4D regular polytopes to =
be the faces (that's part of the definition of regularity). It turns out, t=
he dihedral angles of all the special 4D polytopes are just too large to fi=
t around an edge. So they cannot be used. So in 5D, we only have {3,3,3,3},=
{4,3,3,3} and {3,3,3,4}.=20

In 6D, the Schlaefli symbol needs to be {p,q,r,s,t}, where {p,q,r,s} may be=
{3,3,3,3}, {4,3,3,3} or {3,3,3,4}, and {q,r,s,t} may be {3,3,3,3}, {4,3,3,=
3} or {3,3,3,4} (this is the definition of regularity). So q,r,s have to be=
3,3,3. And you can't make anything but {3,3,3,3,3}, {4,3,3,3,3} and {3,3,3=
,3,4}. Without fancy building blocks, you just can't make fancy stuff. By t=
he same argument (induction), these polytopes are the only regular ones in =
higher dimensions.

You may argue that the definition of regularity is too strong. If we relax =
that, maybe we get more interesting things.

Nan

--- In 4D_Cubing@yahoogroups.com, Roice Nelson wrote:
>
> In dimension 5 and above, there are only 3 kinds of regular polytopes: th=
e
> n-simplex, the n-cube, and the n-orthoplex.
>=20
> http://en.wikipedia.org/wiki/List_of_regular_polytopes#Five-dimensional_r=
egular_polytopes_and_higher
>=20
> So dimension 4 is very special, having 6 different flavors of regular
> polytopes. Dimension 3 is also special, having 5. If anyone could give
> insight into *why* things change for dimension 5, please do share.
>=20
> Even though there is no "5D dodecahedron", there are 5D polytopes that ar=
e
> at least reminiscent of the dodecahedron. They just aren't regular. For
> instance, you could make a prism based on the 120-cell, aka a {5,3,3}x{}.
> I bet it'd be a horrific puzzle though!
>=20
> Cheers,
> Roice
>=20
>=20
> On Tue, Mar 19, 2013 at 8:18 PM, Philip Strimpel wrote:
>=20
> >
> > Hello Melinda,
> > Many thanks for the kind welcome! :) What version of the 24 cell woul=
d
> > you like me to solve? There are quite a few! Also, I don't know if it i=
s
> > because my computer is two slow or not, but I can't seem to understand =
how
> > to twist anything besides 3^4 and 120 cell. I would really enjoy attemp=
ting
> > some of the 5 and 6d puzzles. My computer is waaay too slow to even try=
the
> > 600 cell though. :( Now THAT would be awesome to solve! Maybe even a 12=
0
> > cell pentultimate would be an idea for a future 4d puzzle! :=95 I am cu=
rious
> > though... How come there can't be a 5d dodecahedron? I know it would
> > probably have hundreds or thousands of cells to it, but does nobody kno=
w of
> > it because it would be too big to comprehend, or is it really virtually=
and
> > physically impossible? Has anybody else thought of this? Just something=
to
> > bring to the table...
> >
> > Best regards,
> > Philip
> >
> >
> > ------------------------------------
> >
> > Yahoo! Groups Links
> >
> >
> >
> >
>

From: "schuma" <mananself@gmail.com>
Date: Wed, 20 Mar 2013 06:20:45 -0000
Subject: Re: About me

--- In 4D_Cubing@yahoogroups.com, "schuma" wrote:
> In the latest MPUlt, there is a 120-cell_halfcut. It's the Face turning 1=
20-cell where the cutting plane passes the center. I think it can be consid=
ered as the Pentultimate in 4D. I don't think anyone has attempted it. But =
I'm sure you have enough courage to make the move.

Let me compare it with the shallow cut 120-cell, which has 7560 stickers, b=
ut only 2520 movable pieces. The 120-cell halfcut has 14400 stickers, all o=
f which are 1C pieces: there are 14400 pieces! The situation is similar to =
the Big Chop (half-cut edge turning dodecahedron). Even the shape of cuts o=
n each dodecahedral cell is like the Big Chop. But this puzzle is still fac=
e turning, so it is a proper analog of Pentultimate.

From: Melinda Green <melinda@superliminal.com>
Date: Tue, 19 Mar 2013 23:38:37 -0700
Subject: Re: [MC4D] Re: About me

On 3/19/2013 10:59 PM, schuma wrote:
> [...]
> About the regular polytopes in high dimensional space, Melinda said,
>> Cracking my bible (Regular Polytopes) to chapter and verse 7-8 gives ...
> Believe it or not, the first thing after I got home today was also checking this book.

If I ever need to swear on a bible, I will insist on this one.

> Before this verse, Coxeter was calculating the dihedral angles. I think the argument can be explained using this example.
>
> The dihedral angle of a 3D cube is 90 degrees. When you use some cubes to make a 4D regular polytope, how many cubes can you fit around an edge? It's necessary that the sum of dihedral angles is less than 360 degrees to make a bounded 4D polytope, or equal to 360 degrees to make an unbounded tessellation. So, two cubes are too few (the outcome is flat). Three cubes around an edge are OK (hypercube). Four cubes around an edge will form the tessellation. So three cubes around an edge is the only valid way to make a 4D bounded polytope using cubes as faces, in Euclidean space. The necessary condition is that the dihedral angle must be small enough. As I understand, equation (7.77) formalizes this argument in math.
>
> To make 5D regular polytopes, we need to pick some 4D regular polytopes to be the faces (that's part of the definition of regularity). It turns out, the dihedral angles of all the special 4D polytopes are just too large to fit around an edge. So they cannot be used. So in 5D, we only have {3,3,3,3}, {4,3,3,3} and {3,3,3,4}.

Wow, what a great description, Nan. I'm starting to see it! Thanks for
spelling that out.

BTW, the criterion for regularity requires all the faces also be the
same regular polytopes, not just any regular polytopes such as prisms. I
know you know that. I just wanted to clarify it.

>
> In 6D, the Schlaefli symbol needs to be {p,q,r,s,t}, where {p,q,r,s} may be {3,3,3,3}, {4,3,3,3} or {3,3,3,4}, and {q,r,s,t} may be {3,3,3,3}, {4,3,3,3} or {3,3,3,4} (this is the definition of regularity). So q,r,s have to be 3,3,3. And you can't make anything but {3,3,3,3,3}, {4,3,3,3,3} and {3,3,3,3,4}. Without fancy building blocks, you just can't make fancy stuff. By the same argument (induction), these polytopes are the only regular ones in higher dimensions.
>
> You may argue that the definition of regularity is too strong. If we relax that, maybe we get more interesting things.

I feel that when you relax a constraint you *always* get interesting
things. I feel that large parts of pure math are done by branching off
from highly symmetrical roots by carefully relaxing one constraint and
seeing what you find. That's how I got interested in IRPs.

And then there are those really rare cases where someone goes in the
opposite direction by recognizing and describing how two sometimes
seemingly unrelated results share a previously unknown deep relationship.

-Melinda

From: "schuma" <mananself@gmail.com>
Date: Wed, 20 Mar 2013 06:52:21 -0000
Subject: Re: About me

--- In 4D_Cubing@yahoogroups.com, Melinda Green wrote:
> And then there are those really rare cases where someone goes in the=20
> opposite direction by recognizing and describing how two sometimes=20
> seemingly unrelated results share a previously unknown deep relationship.

"Good mathematicians see analogies. Great mathematicians see analogies betw=
een analogies." -- Stefan Banach

Reference:

http://en.wikipedia.org/wiki/Stefan_Banach

From: Melinda Green <melinda@superliminal.com>
Date: Wed, 20 Mar 2013 00:07:49 -0700
Subject: Re: [MC4D] Re: About me

On 3/19/2013 11:52 PM, schuma wrote:
> --- In 4D_Cubing@yahoogroups.com, Melinda Green wrote:
>> And then there are those really rare cases where someone goes in the
>> opposite direction by recognizing and describing how two sometimes
>> seemingly unrelated results share a previously unknown deep relationship.
> "Good mathematicians see analogies. Great mathematicians see analogies between analogies." -- Stefan Banach
>
> Reference:
>
> http://en.wikipedia.org/wiki/Stefan_Banach

Perfect!!

From: Roice Nelson <roice3@gmail.com>
Date: Wed, 20 Mar 2013 11:33:13 -0500
Subject: Re: [MC4D] About me

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On Tue, Mar 19, 2013 at 10:49 PM, Melinda Green wrote:
>
> Also, I don't know if it is because my computer is two slow or not, but =
I can't seem to understand how to twist anything besides 3^4 and 120 cell.
>
>
> Most stickers in most of those puzzles are not controls. We call them
> "grips" in the code. Just poke at different shaped stickers to see which
> will twist and how they work.
>
>
Magic Puzzle Ultimate has some more complex twisting modes too (for
situations where single clicks aren't enough to specify a twist). I was
going to search through the archives to find Andrey's posts about this, but
then noticed he has instructions on his site:

http://astr73.narod.ru/MPUlt/instr.html

4D is definitely the only one in the rich habitable zone and is where all
> the action is. 4D is also one of the few spaces where spheres pack
> perfectly, along with 2D where 6 pennies can exactly surround a 7th. I
> wonder if this has anything to do with the fact that the volume of the un=
it
> spheres is greatest between dimensions 4 and 5. I've attached some code I
> wrote to calculate that maximum. It's interesting because that maximum is
> not on a integer dimension though it is closer to 5 than to 4.
>
>
I think you'd find the following interesting. If you vary the n-ball
radius, the dimension where the max volume occurs can be tuned to anything.

http://www.johndcook.com/blog/2012/10/23/dimension-5-isnt-so-special/

> Cracking my bible (Regular Polytopes) to chapter and verse 7-8 gives
>
> *When n=3D4 we have a Schl=C3=A4fli symbol {p,q,r}, where both {p,q} and =
{q,r}
> must occur among the Platonic solids.... Since the only regular polytope =
in
> five dimensions are =CE=B15, =CE=B25, =CE=B35, it follows by induction th=
at in more than
> five dimensions the ony regular polytopes are =CE=B1n, =CE=B2n, =CE=B3n.*
>
> I still don't get it but I bet Roice will figure it out and explain it to
> us. :-)
>

Luckily Nan did while I slept :D His description definitely helped me
understand things more, so thanks Nan!

seeya,
Roice

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On Tue, Mar 19, 2013 at 10:49 PM, Melinda Gr=
een=C2=A0wrote:

;border-left:1px #ccc solid;padding-left:1ex">
=3D"#FFFFFF">
Also, I don't know if it is because my computer is two slow =
or not, but I can't seem to understand how to twist anything besides 3^=
4 and 120 cell. 
Most stickers in most of those puzzles are not controls. We call
them "grips" in the code. Just poke at different shaped stick=
ers to
see which will twist and how they work.

Magic Puzzle Ultimate has so=
me more complex twisting modes too (for situations where single clicks aren=
't enough to specify a twist). =C2=A0I was going to search through the =
archives to find Andrey's posts about this, but then noticed he has ins=
tructions on his site:

htt=
p://astr73.narod.ru/MPUlt/instr.html

=C2=A0

iv>

:1px #ccc solid;padding-left:1ex">

4D is definitely the only one in the rich habitable zone and is
where all the action is. 4D is also one of the few spaces where
spheres pack perfectly, along with 2D where 6 pennies can exactly
surround a 7th. I wonder if this has anything to do with the fact
that the volume of the unit spheres is greatest between dimensions 4
and 5. I've attached some code I wrote to calculate that maximum.
It's interesting because that maximum is not on a integer dimension
though it is closer to 5 than to 4.

I think you'd find the f=
ollowing interesting. =C2=A0If you vary the n-ball radius, the dimension wh=
ere the max volume occurs can be tuned to anything.

v>
cial/">http://www.johndcook.com/blog/2012/10/23/dimension-5-isnt-so-special=
/

=C2=A0

" style=3D"margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">

Cracking my bible (Regular Polytopes) to chapter and verse 7-8 gives>

When n=3D4 we have a Schl=C3=A4fli symbol {p,q,r}, where
both {p,q} and {q,r} must occur among the Platonic solids....
Since the only regular polytope in five dimensions are =CE=B15, =CE=
=B25,
=CE=B35, it follows by induction that in more than five dimensions
the ony regular polytopes are =CE=B1n, =CE=B2n, =CE=B3n.

I still don't get it but I bet Roice will figure it out and explain
it to us.=C2=A0 :-)

Luckily Nan did while I slept :D =C2=A0His description definitely hel=
ped me understand things more, so thanks Nan!

seey=
a,

Roice

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