Bilbirothawroids (D4.3 to D4.9)

Discussion of known convex regular-faced polytopes, including the Johnson solids in 3D, and higher dimensions; and the discovery of new ones.

Re: Johnsonian Polytopes

Postby Marek14 » Sat Feb 08, 2014 10:12 am

quickfur wrote:Now look at something like the snub square antiprism. There are 4 layers of vertices, the top and bottom are o4x and x4o. To fix the points in the middle two layers, you need to simultaneously solve for the heights of the x4o (resp. o4x's) and the height of the middle layers, with unit edge lengths constraining the height between the middle layers and between the middle layer to the outer layer. There is no direct constraint that lets you solve for one height first, because the circumradius of the middle layers is dependent on the height. Almost all of the parts are interdependent, so I'm expecting that you need to solve at least a cubic equation in order to get everything fixed. So in a sense, the snub square antiprism is somehow "inherently" more complex than the bilunabirotunda or the triangular hebesphenorotunda.


I actually tried to compute its coordinates in Mathematica and you're right: the set of 3 quadratic equations needed to set constraints for it leads to a cubic equation.

I set coordinates of the bottom layer to (0,+-1,+-1), second layer to (x,+-y,0) and (x,0,+-y), third layer to (z,+-y*Sqrt(2)/2,+-y*Sqrt(2)/2) and fourth layer to (x+z,+-Sqrt2/2,0) and (x+z,0,+-Sqrt2/2) and tried to solve by setting all distances in triangle (0,1,1)-(x,y,0)-(z,y*Sqrt(2)/2,y*Sqrt(2)/2) equal to 2. I got extremely complicated cubic expression as solution, with numerical values (if all 3 are positive) of x = 0.759133, y = 1.65094 and z = 1.39426. With this values, the triangle is equilateral.

For snub disphenoid, I can take vertices (0,0,+-1), (x,+-y,0), (z,0,+-y) and (x+z,1,0) and search for equilateral triangle (0,0,1)-(x,y,0)-(z,0,y). Unlike snub square antiprism, this one is actually quadratic, though with very complicated expressions.

{x -> 1/2 (2 Sqrt[-1 + Sqrt[7 - 2 Sqrt[2]] + Sqrt[
2 (7 - 2 Sqrt[2])]] - Sqrt[
2 (-1 + Sqrt[7 - 2 Sqrt[2]] + Sqrt[2 (7 - 2 Sqrt[2])])]),
z -> Sqrt[-1/2 + Sqrt[1/2 (7 - 2 Sqrt[2])] +
1/2 Sqrt[7 - 2 Sqrt[2]]],
y -> 1/2 (1 - Sqrt[2] + Sqrt[7 - 2 Sqrt[2]])}

or numerically

{x -> 0.580704, z -> 1.40194, y -> 0.814115}
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Re: Johnsonian Polytopes

Postby student91 » Sat Feb 08, 2014 5:30 pm

quickfur wrote:[...]
I know what you're getting at, but sometimes even "nice" things have ugly coordinates. Take the snub cube, for example. It requires solving a bunch of degree-3 polynomials, and the resulting coordinates have nested cube roots. Yet the thing is uniform! So we cannot say that just because those crazy nested cube roots are coming out, the polyhedron isn't going to close up. Similarly, the snub disphenoid has some pretty crazy coordinates that I haven't even been able to derive myself, I just copied somebody else's work. :P

To be honest, I think the snubs are the most ugly uniforms among the convex ones. Besides, 4D-snubs don't work out very well, so
maybe "hard" crown jewels might not work out very well in 4D as well, although they do in 3D. I still don't know why exactly the snubs don't work out, but wendy's last post made that much clearer to me. (to be honest, I just copied the content of that post) Has it been proven that all snubs in dimension >3 except s3s4o3o don't work out? If so, maybe a similar proof can be done for the crown jewels. Of course, such a proof is much harder without vertex-transitivity.
EDIT: I should've reread wendy's post, It just gives such a proof. unfortunately it doesn't extend to non-uniforms very well.

The idea of determening a polytope's complexity (or CVP) appeals to me as well. It's just what we need.

About the rings, when was thinking about this, I realized that (only sure with parallel faces though) if you want a regular n-ring, the dichoral angle between the cells of the ring is equal to the angle of an n-gon. this means the dihedral angle between two things is fixed, limiting the amound of polyhedra that can be inserted. You could of course make less-regular rings, but still the dichoral angles are limited to have a total of 180n.
Last edited by student91 on Sun Feb 09, 2014 9:19 pm, edited 1 time in total.
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Re: Johnsonian Polytopes

Postby quickfur » Sat Feb 08, 2014 6:43 pm

Interesting. So the snub disphenoid is only order 2 after all. Maybe that means it will be relatively easy to find a CRF containing snub disphenoid cells? :D
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Re: Johnsonian Polytopes

Postby Marek14 » Sat Feb 08, 2014 6:50 pm

quickfur wrote:Interesting. So the snub disphenoid is only order 2 after all. Maybe that means it will be relatively easy to find a CRF containing snub disphenoid cells? :D


For starters you might try to start starting from an edge that has only snub disphenoids around it. The dihedral angle for the "top" edge (edge connecting two degree 4-vertices) is 96.1983, so something might appear when you fold three snub disphenoids around such an edge.
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Re: Johnsonian Polytopes

Postby quickfur » Sat Feb 08, 2014 6:51 pm

student91 wrote:
quickfur wrote:[...]
I know what you're getting at, but sometimes even "nice" things have ugly coordinates. Take the snub cube, for example. It requires solving a bunch of degree-3 polynomials, and the resulting coordinates have nested cube roots. Yet the thing is uniform! So we cannot say that just because those crazy nested cube roots are coming out, the polyhedron isn't going to close up. Similarly, the snub disphenoid has some pretty crazy coordinates that I haven't even been able to derive myself, I just copied somebody else's work. :P

To be honest, I think the snubs are the most ugly uniforms among the convex ones. Besides, 4D-snubs don't work out very well, so
maybe "hard" crown jewels might not work out very well in 4D as well, although they do in 3D. I still don't know why exactly the snubs don't work out, but wendy's last post made that much clearer to me. (to be honest, I just copied the content of that post) Has it been proven that all snubs in dimension >3 except s3s4o3o don't work out? If so, maybe a similar proof can be done for the crown jewels. Of course, such a proof is much harder without vertex-transitivity.

Well, as Klitzing has said, all snubs "work" in the sense of being actual polytopes, the problem is whether they can be made uniform. Most of them can't.

The idea of determening a polytope's complexity (or CVP) appeals to me as well. It's just what we need.

It's not so easy to prove the CVP order of a polytope, though. Just because we cannot find an order 2 polynomial to express a polytope, doesn't mean one doesn't exist. Any polytope can always be expressed as a higher-order polynomial, albeit not a minimal one. Proving minimality is not always straightforward, as Marek's post proves by finding a closed-form quadratic for the snub disphenoid that I couldn't find by hand.

About the rings, when was thinking about this, I realized that (only sure with parallel faces though) if you want a regular n-ring, the dichoral angle between the cells of the ring is equal to the angle of an n-gon. this means the dihedral angle between two things is fixed, limiting the amound of polyhedra that can be inserted. You could of course make less-regular rings, but still the dichoral angles are limited to have a total of 180n.

Yes, that was what I said in my post. :)
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Re: Johnsonian Polytopes

Postby Keiji » Sat Feb 08, 2014 7:19 pm

quickfur wrote:
The idea of determening a polytope's complexity (or CVP) appeals to me as well. It's just what we need.

It's not so easy to prove the CVP order of a polytope, though. Just because we cannot find an order 2 polynomial to express a polytope, doesn't mean one doesn't exist. Any polytope can always be expressed as a higher-order polynomial, albeit not a minimal one. Proving minimality is not always straightforward, as Marek's post proves by finding a closed-form quadratic for the snub disphenoid that I couldn't find by hand.


Hang on - what are we going on here? Marek's post has roots inside roots inside roots. Doesn't that make it an eighth root? Or are we literally going on the maximum root written, which would leave only primes as possible maximum roots, since an expression with, for example, a fourth root could be expressed as a square root inside a square root?
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Re: Johnsonian Polytopes

Postby Marek14 » Sat Feb 08, 2014 7:22 pm

I just realized that I used wrong values for the previous computations (was accidentally searching for edge length Sqrt(2) instead of 2). Let me redo the calculations...

So. Snub disphenoid now actually gives much simpler symbolic solution {y -> Sqrt[-5 + 4 Sqrt[3]], x -> -Sqrt[2] + Sqrt[6], z -> Sqrt[2]} with numerical values {y -> 1.3886, x -> 1.03528, z -> 1.41421}.

As for snub square antiprism, it still gives cubic equation with numerical solution {x -> 0.982523, z -> 1.72495, y -> 2.42641}

I tried to compute next Johnson solid, sphenocorona. My form came out with 5 different variables:

Let's call the two endpoints of the single square-square edge (0,0,-1) and (0,0,1).
Then the next layer contains four vertices: (x,y,-1), (x,y,1), (x,-y,-1) and (x,-y,1).
Third layer contains (z,0,-w) and (z,0,w).
And the fourth layer contains (v,-1,0) and (v,1,0).

The solution in symbolic form is: {v -> (9675/2 (34432/5625 + (128 Sqrt[6])/625)^(3/2) Sqrt[
2/75 + (24 Sqrt[6])/25 +
1/2 Sqrt[34432/5625 + (128 Sqrt[6])/625]] -
7668 (34432/5625 + (128 Sqrt[6])/625)^(3/2) Sqrt[
6 (2/75 + (24 Sqrt[6])/25 +
1/2 Sqrt[34432/5625 + (128 Sqrt[6])/625])] +
84581824/
625 Sqrt[(34432/5625 + (128 Sqrt[6])/625) (2/75 + (24 Sqrt[6])/
25 + 1/2 Sqrt[34432/5625 + (128 Sqrt[6])/625])] +
12516864/625 Sqrt[
6 (34432/5625 + (128 Sqrt[6])/625) (2/75 + (24 Sqrt[6])/25 +
1/2 Sqrt[34432/5625 + (128 Sqrt[6])/625])])/(-1022976/5 + (
497664 Sqrt[6])/5 + 51840 Sqrt[34432/5625 + (128 Sqrt[6])/625]),
y -> 1/103680(103680 +
84581824/625 Sqrt[34432/5625 + (128 Sqrt[6])/625] +
9675/2 (34432/5625 + (128 Sqrt[6])/625)^(3/2) -
7668 Sqrt[6] (34432/5625 + (128 Sqrt[6])/625)^(3/2) +
12516864/625 Sqrt[6 (34432/5625 + (128 Sqrt[6])/625)] -
647626752/(
25 (-18944/25 + (9216 Sqrt[6])/25 +
192 Sqrt[34432/5625 + (128 Sqrt[6])/625])) - (
399900672 Sqrt[6])/(
25 (-18944/25 + (9216 Sqrt[6])/25 +
192 Sqrt[34432/5625 + (128 Sqrt[6])/625])) + (
8045641728 Sqrt[34432/5625 + (128 Sqrt[6])/625])/(
125 (-18944/25 + (9216 Sqrt[6])/25 +
192 Sqrt[34432/5625 + (128 Sqrt[6])/625])) + (
2592000 (34432/5625 + (128 Sqrt[6])/625)^(3/2))/(-18944/25 + (
9216 Sqrt[6])/25 + 192 Sqrt[34432/5625 + (128 Sqrt[6])/625]) - (
3732480 Sqrt[6] (34432/5625 + (128 Sqrt[6])/625)^(
3/2))/(-18944/25 + (9216 Sqrt[6])/25 +
192 Sqrt[34432/5625 + (128 Sqrt[6])/625]) - (
1855291392 Sqrt[6 (34432/5625 + (128 Sqrt[6])/625)])/(
125 (-18944/25 + (9216 Sqrt[6])/25 +
192 Sqrt[34432/5625 + (128 Sqrt[6])/625]))),
w -> (31232/125 + (2541824 Sqrt[2/3])/15625 + (1563392 Sqrt[6])/
15625 - (3492032 Sqrt[34432/5625 + (128 Sqrt[6])/625])/5625 -
25 (34432/5625 + (128 Sqrt[6])/625)^(3/2) +
36 Sqrt[6] (34432/5625 + (128 Sqrt[6])/625)^(3/2) +
89472/625 Sqrt[6 (34432/5625 + (128 Sqrt[6])/625)])/(-18944/
25 + (9216 Sqrt[6])/25 +
192 Sqrt[34432/5625 + (128 Sqrt[6])/625]),
x -> 1/3600(48 Sqrt[
2/75 + (24 Sqrt[6])/25 +
1/2 Sqrt[34432/5625 + (128 Sqrt[6])/625]] -
225/4 (34432/5625 + (128 Sqrt[6])/625)^(3/2) Sqrt[
2/75 + (24 Sqrt[6])/25 +
1/2 Sqrt[34432/5625 + (128 Sqrt[6])/625]] +
1728 Sqrt[
6 (2/75 + (24 Sqrt[6])/25 +
1/2 Sqrt[34432/5625 + (128 Sqrt[6])/625])] +
108 (34432/5625 + (128 Sqrt[6])/625)^(3/2) Sqrt[
6 (2/75 + (24 Sqrt[6])/25 +
1/2 Sqrt[34432/5625 + (128 Sqrt[6])/625])] -
430244/625 Sqrt[(34432/5625 + (128 Sqrt[6])/625) (2/75 + (
24 Sqrt[6])/25 + 1/2 Sqrt[34432/5625 + (128 Sqrt[6])/625])] -
405984/625 Sqrt[
6 (34432/5625 + (128 Sqrt[6])/625) (2/75 + (24 Sqrt[6])/25 +
1/2 Sqrt[34432/5625 + (128 Sqrt[6])/625])]),
z -> Sqrt[
2/75 + (24 Sqrt[6])/25 + 1/2 Sqrt[34432/5625 + (128 Sqrt[6])/625]]}

And in numerical form:
{v -> 2.62659, y -> 1.70545, w -> 1.57886, x -> 1.04471, z -> 1.9144}

So despite slightly wrong numbers the result stands: snub square antiprism has CVP 3 (which should mean it's not actually constructible by Euclidean means) while snub disphenoid and sphenocorona have CVP only 2.
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Re: Johnsonian Polytopes

Postby Marek14 » Sat Feb 08, 2014 7:29 pm

Keiji wrote:
quickfur wrote:
The idea of determening a polytope's complexity (or CVP) appeals to me as well. It's just what we need.

It's not so easy to prove the CVP order of a polytope, though. Just because we cannot find an order 2 polynomial to express a polytope, doesn't mean one doesn't exist. Any polytope can always be expressed as a higher-order polynomial, albeit not a minimal one. Proving minimality is not always straightforward, as Marek's post proves by finding a closed-form quadratic for the snub disphenoid that I couldn't find by hand.


Hang on - what are we going on here? Marek's post has roots inside roots inside roots. Doesn't that make it an eighth root? Or are we literally going on the maximum root written, which would leave only primes as possible maximum roots, since an expression with, for example, a fourth root could be expressed as a square root inside a square root?


It's not about maximum roots, it's about the polynoms. Square roots, no matter how nested and how deep, will still lead to Euclidean construction with straightedge and compass. The moment you get a single CUBE root anywhere, this chance is gone. For this reason the incredibly complex symbolic solution to snub square antiprism

Code: Select all
{x -> 1/8 (-2 \[Sqrt](-20/3 + (8 Sqrt[2])/3 - (1138 2^(1/3))/(
         9 (-28 +
            25 Sqrt[2] + \[ImaginaryI] Sqrt[32616 - 21816 Sqrt[2]])^(
          2/3)) + (224 2^(5/6))/(
         3 (-28 +
            25 Sqrt[2] + \[ImaginaryI] Sqrt[32616 - 21816 Sqrt[2]])^(
          2/3)) + (104 2^(1/6))/(
         9 (-28 +
            25 Sqrt[2] + \[ImaginaryI] Sqrt[32616 - 21816 Sqrt[2]])^(
          1/3)) + (20 2^(2/3))/(
         9 (-28 +
            25 Sqrt[2] + \[ImaginaryI] Sqrt[32616 - 21816 Sqrt[2]])^(
          1/3)) + 4/9 2^(
          5/6) (-28 +
            25 Sqrt[2] + \[ImaginaryI] Sqrt[32616 - 21816 Sqrt[2]])^(
          1/3) + 4/
          9 (2 (-28 +
              25 Sqrt[2] + \[ImaginaryI] Sqrt[
               32616 - 21816 Sqrt[2]]))^(1/3) -
         1/9 (2 (-28 +
              25 Sqrt[2] + \[ImaginaryI] Sqrt[
               32616 - 21816 Sqrt[2]]))^(2/3)) - (206 2^(
        1/3) \[Sqrt](-20/3 + (8 Sqrt[2])/3 - (1138 2^(1/3))/(
           9 (-28 +
              25 Sqrt[2] + \[ImaginaryI] Sqrt[
               32616 - 21816 Sqrt[2]])^(2/3)) + (224 2^(5/6))/(
           3 (-28 +
              25 Sqrt[2] + \[ImaginaryI] Sqrt[
               32616 - 21816 Sqrt[2]])^(2/3)) + (104 2^(1/6))/(
           9 (-28 +
              25 Sqrt[2] + \[ImaginaryI] Sqrt[
               32616 - 21816 Sqrt[2]])^(1/3)) + (20 2^(2/3))/(
           9 (-28 +
              25 Sqrt[2] + \[ImaginaryI] Sqrt[
               32616 - 21816 Sqrt[2]])^(1/3)) +
           4/9 2^(5/
             6) (-28 +
              25 Sqrt[2] + \[ImaginaryI] Sqrt[
               32616 - 21816 Sqrt[2]])^(1/3) +
           4/9 (2 (-28 +
                25 Sqrt[2] + \[ImaginaryI] Sqrt[
                 32616 - 21816 Sqrt[2]]))^(1/3) -
           1/9 (2 (-28 +
                25 Sqrt[2] + \[ImaginaryI] Sqrt[
                 32616 - 21816 Sqrt[2]]))^(2/3)))/(9 (-28 +
          25 Sqrt[2] + \[ImaginaryI] Sqrt[32616 - 21816 Sqrt[2]])^(
        2/3)) + (466 2^(
        5/6) \[Sqrt](-20/3 + (8 Sqrt[2])/3 - (1138 2^(1/3))/(
           9 (-28 +
              25 Sqrt[2] + \[ImaginaryI] Sqrt[
               32616 - 21816 Sqrt[2]])^(2/3)) + (224 2^(5/6))/(
           3 (-28 +
              25 Sqrt[2] + \[ImaginaryI] Sqrt[
               32616 - 21816 Sqrt[2]])^(2/3)) + (104 2^(1/6))/(
           9 (-28 +
              25 Sqrt[2] + \[ImaginaryI] Sqrt[
               32616 - 21816 Sqrt[2]])^(1/3)) + (20 2^(2/3))/(
           9 (-28 +
              25 Sqrt[2] + \[ImaginaryI] Sqrt[
               32616 - 21816 Sqrt[2]])^(1/3)) +
           4/9 2^(5/
             6) (-28 +
              25 Sqrt[2] + \[ImaginaryI] Sqrt[
               32616 - 21816 Sqrt[2]])^(1/3) +
           
           4/9 (2 (-28 +
                25 Sqrt[2] + \[ImaginaryI] Sqrt[
                 32616 - 21816 Sqrt[2]]))^(1/3) -
           1/9 (2 (-28 +
                25 Sqrt[2] + \[ImaginaryI] Sqrt[
                 32616 - 21816 Sqrt[2]]))^(2/3)))/(9 (-28 +
          25 Sqrt[2] + \[ImaginaryI] Sqrt[32616 - 21816 Sqrt[2]])^(
        2/3)) - (28 2^(
        1/6) \[Sqrt](-20/3 + (8 Sqrt[2])/3 - (1138 2^(1/3))/(
           9 (-28 +
              25 Sqrt[2] + \[ImaginaryI] Sqrt[
               32616 - 21816 Sqrt[2]])^(2/3)) + (224 2^(5/6))/(
           3 (-28 +
              25 Sqrt[2] + \[ImaginaryI] Sqrt[
               32616 - 21816 Sqrt[2]])^(2/3)) + (104 2^(1/6))/(
           9 (-28 +
              25 Sqrt[2] + \[ImaginaryI] Sqrt[
               32616 - 21816 Sqrt[2]])^(1/3)) + (20 2^(2/3))/(
           9 (-28 +
              25 Sqrt[2] + \[ImaginaryI] Sqrt[
               32616 - 21816 Sqrt[2]])^(1/3)) +
           4/9 2^(5/
             6) (-28 +
              25 Sqrt[2] + \[ImaginaryI] Sqrt[
               32616 - 21816 Sqrt[2]])^(1/3) +
           4/9 (2 (-28 +
                25 Sqrt[2] + \[ImaginaryI] Sqrt[
                 32616 - 21816 Sqrt[2]]))^(1/3) -
           1/9 (2 (-28 +
                25 Sqrt[2] + \[ImaginaryI] Sqrt[
                 32616 - 21816 Sqrt[2]]))^(2/3)))/(3 (-28 +
          25 Sqrt[2] + \[ImaginaryI] Sqrt[32616 - 21816 Sqrt[2]])^(
        1/3)) + (32 2^(
        2/3) \[Sqrt](-20/3 + (8 Sqrt[2])/3 - (1138 2^(1/3))/(
           9 (-28 +
              25 Sqrt[2] + \[ImaginaryI] Sqrt[
               32616 - 21816 Sqrt[2]])^(2/3)) + (224 2^(5/6))/(
           3 (-28 +
              25 Sqrt[2] + \[ImaginaryI] Sqrt[
               32616 - 21816 Sqrt[2]])^(2/3)) + (104 2^(1/6))/(
           9 (-28 +
              25 Sqrt[2] + \[ImaginaryI] Sqrt[
               32616 - 21816 Sqrt[2]])^(1/3)) + (20 2^(2/3))/(
           9 (-28 +
              25 Sqrt[2] + \[ImaginaryI] Sqrt[
               32616 - 21816 Sqrt[2]])^(1/3)) +
           4/9 2^(5/
             6) (-28 +
              25 Sqrt[2] + \[ImaginaryI] Sqrt[
               32616 - 21816 Sqrt[2]])^(1/3) +
           4/9 (2 (-28 +
                25 Sqrt[2] + \[ImaginaryI] Sqrt[
                 32616 - 21816 Sqrt[2]]))^(1/3) -
           1/9 (2 (-28 +
                25 Sqrt[2] + \[ImaginaryI] Sqrt[
                 32616 - 21816 Sqrt[2]]))^(2/3)))/(9 (-28 +
          25 Sqrt[2] + \[ImaginaryI] Sqrt[32616 - 21816 Sqrt[2]])^(
        1/3)) - 2/9 2^(
      5/6) (-28 +
        25 Sqrt[2] + \[ImaginaryI] Sqrt[32616 - 21816 Sqrt[2]])^(
      1/3) \[Sqrt](-20/3 + (8 Sqrt[2])/3 - (1138 2^(1/3))/(
         9 (-28 +
            25 Sqrt[2] + \[ImaginaryI] Sqrt[32616 - 21816 Sqrt[2]])^(
          2/3)) + (224 2^(5/6))/(
         3 (-28 +
            25 Sqrt[2] + \[ImaginaryI] Sqrt[32616 - 21816 Sqrt[2]])^(
          2/3)) + (104 2^(1/6))/(
         9 (-28 +
            25 Sqrt[2] + \[ImaginaryI] Sqrt[32616 - 21816 Sqrt[2]])^(
          1/3)) + (20 2^(2/3))/(
         9 (-28 +
            25 Sqrt[2] + \[ImaginaryI] Sqrt[32616 - 21816 Sqrt[2]])^(
          1/3)) + 4/9 2^(
          5/6) (-28 +
            25 Sqrt[2] + \[ImaginaryI] Sqrt[32616 - 21816 Sqrt[2]])^(
          1/3) + 4/
          9 (2 (-28 +
              25 Sqrt[2] + \[ImaginaryI] Sqrt[
               32616 - 21816 Sqrt[2]]))^(1/3) -
         1/9 (2 (-28 +
              25 Sqrt[2] + \[ImaginaryI] Sqrt[
               32616 - 21816 Sqrt[2]]))^(2/3)) +
     2/9 2^(1/
       6) (-28 +
        25 Sqrt[2] + \[ImaginaryI] Sqrt[32616 - 21816 Sqrt[2]])^(
      2/3) \[Sqrt](-20/3 + (8 Sqrt[2])/3 - (1138 2^(1/3))/(
         9 (-28 +
            25 Sqrt[2] + \[ImaginaryI] Sqrt[32616 - 21816 Sqrt[2]])^(
          2/3)) + (224 2^(5/6))/(
         3 (-28 +
            25 Sqrt[2] + \[ImaginaryI] Sqrt[32616 - 21816 Sqrt[2]])^(
          2/3)) + (104 2^(1/6))/(
         9 (-28 +
            25 Sqrt[2] + \[ImaginaryI] Sqrt[32616 - 21816 Sqrt[2]])^(
          1/3)) + (20 2^(2/3))/(
         9 (-28 +
            25 Sqrt[2] + \[ImaginaryI] Sqrt[32616 - 21816 Sqrt[2]])^(
          1/3)) + 4/9 2^(
          5/6) (-28 +
            25 Sqrt[2] + \[ImaginaryI] Sqrt[32616 - 21816 Sqrt[2]])^(
          1/3) + 4/
          9 (2 (-28 +
              25 Sqrt[2] + \[ImaginaryI] Sqrt[
               32616 - 21816 Sqrt[2]]))^(1/3) -
         1/9 (2 (-28 +
              25 Sqrt[2] + \[ImaginaryI] Sqrt[
               32616 - 21816 Sqrt[2]]))^(2/3)) +
     1/9 (2 (-28 +
          25 Sqrt[2] + \[ImaginaryI] Sqrt[32616 - 21816 Sqrt[2]]))^(
      2/3) \[Sqrt](-20/3 + (8 Sqrt[2])/3 - (1138 2^(1/3))/(
         9 (-28 +
            25 Sqrt[2] + \[ImaginaryI] Sqrt[32616 - 21816 Sqrt[2]])^(
          2/3)) + (224 2^(5/6))/(
         3 (-28 +
            25 Sqrt[2] + \[ImaginaryI] Sqrt[32616 - 21816 Sqrt[2]])^(
          2/3)) + (104 2^(1/6))/(
         9 (-28 +
            25 Sqrt[2] + \[ImaginaryI] Sqrt[32616 - 21816 Sqrt[2]])^(
          1/3)) + (20 2^(2/3))/(
         9 (-28 +
            25 Sqrt[2] + \[ImaginaryI] Sqrt[32616 - 21816 Sqrt[2]])^(
          1/3)) + 4/9 2^(
          5/6) (-28 +
            25 Sqrt[2] + \[ImaginaryI] Sqrt[32616 - 21816 Sqrt[2]])^(
          1/3) + 4/
          9 (2 (-28 +
              25 Sqrt[2] + \[ImaginaryI] Sqrt[
               32616 - 21816 Sqrt[2]]))^(1/3) -
         1/9 (2 (-28 +
              25 Sqrt[2] + \[ImaginaryI] Sqrt[
               32616 - 21816 Sqrt[2]]))^(2/3)) +
     2 \[Sqrt](2 (-20/3 + (8 Sqrt[2])/3 - (1138 2^(1/3))/(
           9 (-28 +
              25 Sqrt[2] + \[ImaginaryI] Sqrt[
               32616 - 21816 Sqrt[2]])^(2/3)) + (224 2^(5/6))/(
           3 (-28 +
              25 Sqrt[2] + \[ImaginaryI] Sqrt[
               32616 - 21816 Sqrt[2]])^(2/3)) + (104 2^(1/6))/(
           9 (-28 +
              25 Sqrt[2] + \[ImaginaryI] Sqrt[
               32616 - 21816 Sqrt[2]])^(1/3)) + (20 2^(2/3))/(
           9 (-28 +
              25 Sqrt[2] + \[ImaginaryI] Sqrt[
               32616 - 21816 Sqrt[2]])^(1/3)) +
           4/9 2^(5/
             6) (-28 +
              25 Sqrt[2] + \[ImaginaryI] Sqrt[
               32616 - 21816 Sqrt[2]])^(1/3) +
           4/9 (2 (-28 +
                25 Sqrt[2] + \[ImaginaryI] Sqrt[
                 32616 - 21816 Sqrt[2]]))^(1/3) -
           1/9 (2 (-28 +
                25 Sqrt[2] + \[ImaginaryI] Sqrt[
                 32616 - 21816 Sqrt[2]]))^(2/3)))),
z -> \[Sqrt](-20/3 + (8 Sqrt[2])/3 - (1138 2^(1/3))/(
     9 (-28 +
        25 Sqrt[2] + \[ImaginaryI] Sqrt[32616 - 21816 Sqrt[2]])^(
      2/3)) + (224 2^(5/6))/(
     3 (-28 +
        25 Sqrt[2] + \[ImaginaryI] Sqrt[32616 - 21816 Sqrt[2]])^(
      2/3)) + (104 2^(1/6))/(
     9 (-28 +
        25 Sqrt[2] + \[ImaginaryI] Sqrt[32616 - 21816 Sqrt[2]])^(
      1/3)) + (20 2^(2/3))/(
     9 (-28 +
        25 Sqrt[2] + \[ImaginaryI] Sqrt[32616 - 21816 Sqrt[2]])^(
      1/3)) + 4/9 2^(
      5/6) (-28 +
        25 Sqrt[2] + \[ImaginaryI] Sqrt[32616 - 21816 Sqrt[2]])^(
      1/3) + 4/
      9 (2 (-28 +
          25 Sqrt[2] + \[ImaginaryI] Sqrt[32616 - 21816 Sqrt[2]]))^(
      1/3) - 1/
      9 (2 (-28 +
          25 Sqrt[2] + \[ImaginaryI] Sqrt[32616 - 21816 Sqrt[2]]))^(
      2/3)), y ->
  1/3 (-2 + Sqrt[2]) - (2^(2/3) (-21 + 8 Sqrt[2]))/(
   3 (-28 + 25 Sqrt[2] + \[ImaginaryI] Sqrt[32616 - 21816 Sqrt[2]])^(
    1/3)) + 1/
    3 (2 (-28 +
        25 Sqrt[2] + \[ImaginaryI] Sqrt[32616 - 21816 Sqrt[2]]))^(1/3)
    }


is still considered cubic.

This is also probably reason why solving quartic equations is, in general, no harder than solving cubic ones: fourth power doesn't actually bring anything new. However, equations of degree 5 and more generally don't have solutions that can be written in terms of any simpler numbers than themselves (no combination of integers, with roots of any kinds, will work). So if a polytope would lead to equation of degree higher than 4, the degree of that equation is the only possible definition of CVP.
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Re: Johnsonian Polytopes

Postby quickfur » Sat Feb 08, 2014 7:50 pm

Hold on, when you have nested square roots, the CVP is no longer 2, because those values cannot be expressed as the root of a single quadratic polynomial.

Though you do have a point that cubics have no straightedge-compass constructions. Hmm. So cubics have a higher complexity than even nested square roots?
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Re: Johnsonian Polytopes

Postby Keiji » Sat Feb 08, 2014 8:06 pm

In that case, and after re-reading, CVP order isn't a good name at all, as I was trying to refer to something completely different. :oops:

You are talking about the maximum polynomial order needed to determine the coordinates, when split into possibly multiple steps. The CVP order could well end up less than the maximum polynomial order, even for a figure that is just as hard to solve for.

So no, we should not be using CVP order, rather, maximum polynomial order.
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Re: Johnsonian Polytopes

Postby Marek14 » Sat Feb 08, 2014 8:07 pm

quickfur wrote:Hold on, when you have nested square roots, the CVP is no longer 2, because those values cannot be expressed as the root of a single quadratic polynomial.

Though you do have a point that cubics have no straightedge-compass constructions. Hmm. So cubics have a higher complexity than even nested square roots?


Maybe CVP 2 should not be thought as being root of single quadratic polynomial but rather as root of ONLY quadratic polynomials, i.e. never during search for these values are we forced to solve polynomial of degree 3 or higher.

In other words, CVP is the highest degree of a polynomial you have to solve to find coordinates for the given polytope, no matter how many polynomials of that type you have to solve.

EDIT: I basically say the same thing as Keiji, just in a bit different words.
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Re: Johnsonian Polytopes

Postby student91 » Sat Feb 08, 2014 9:46 pm

If you take the minimal polynominal for a number, and write down the prime factors of the powers, doesn't the biggest prime number give what we want? I mean, if there is a factor 3 in the powers somewhere, you have to take the cubic root to get a solution, and if there are only factors of 2, you get only nested square roots.
Furthermore, I thought of a definition of the CVP of a polytope as follows: Take a vertex figure of every vertex. calculate the minimal polynominal of the distances between every vertex inside this vertex figure. Get the biggest prime factor of the exponents. Do this for every distance and take the biggest number.
So basically you compute the biggest prime factor of the exponent for every two vertices that connect to the same vertex.
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Re: Johnsonian Polytopes

Postby Marek14 » Sat Feb 08, 2014 9:58 pm

student91 wrote:If you take the minimal polynominal for a number, and write down the prime factors of the powers, doesn't the biggest prime number give what we want? I mean, if there is a factor 3 in the powers somewhere, you have to take the cubic root to get a solution, and if there are only factors of 2, you get only nested square roots.
Furthermore, I thought of a definition of the CVP of a polytope as follows: Take a vertex figure of every vertex. calculate the minimal polynominal of the distances between every vertex inside this vertex figure. Get the biggest prime factor of the exponents. Do this for every distance and take the biggest number.
So basically you compute the biggest prime factor of the exponent for every two vertices that connect to the same vertex.


Biggest prime number, yes, but please note that I'm not entirely sure just what IS the biggest prime number here. I don't even know how it is for polygons. I did some calculations some years ago to see that regular heptagon requires solving a cubic equation and thus has CVP of 3, but I don't actually know if there even ARE polygons or polyhedra that would require solving a degree 5 equation. It's possible that above-cubic polygons don't exist or that they are all big enough that the polyhedra based on them (prism/antiprism) are not used in any CRF constructions, and we won't have to consider anything more than cubic equations while working.
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Re: Johnsonian Polytopes

Postby student91 » Sat Feb 08, 2014 10:14 pm

Marek14 wrote:Biggest prime number, yes, but please note that I'm not entirely sure just what IS the biggest prime number here. I don't even know how it is for polygons. I did some calculations some years ago to see that regular heptagon requires solving a cubic equation and thus has CVP of 3, but I don't actually know if there even ARE polygons or polyhedra that would require solving a degree 5 equation. It's possible that above-cubic polygons don't exist or that they are all big enough that the polyhedra based on them (prism/antiprism) are not used in any CRF constructions, and we won't have to consider anything more than cubic equations while working.

I thought as well that the actual number of the CVP (or whatever we'll call it) when defined this way doesn't really matter if CVP>3, it's just too big to be worked with nicely
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Re: Johnsonian Polytopes

Postby Marek14 » Sat Feb 08, 2014 10:20 pm

student91 wrote:
Marek14 wrote:Biggest prime number, yes, but please note that I'm not entirely sure just what IS the biggest prime number here. I don't even know how it is for polygons. I did some calculations some years ago to see that regular heptagon requires solving a cubic equation and thus has CVP of 3, but I don't actually know if there even ARE polygons or polyhedra that would require solving a degree 5 equation. It's possible that above-cubic polygons don't exist or that they are all big enough that the polyhedra based on them (prism/antiprism) are not used in any CRF constructions, and we won't have to consider anything more than cubic equations while working.

I thought as well that the actual number of the CVP (or whatever we'll call it) when defined this way doesn't really matter if CVP>3, it's just too big to be worked with nicely


So far we have a working hypothesis that bigger CVP means lesser chance to find CRF. Just a hypothesis, mind you.
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Re: Johnsonian Polytopes

Postby quickfur » Sat Feb 08, 2014 11:01 pm

Marek14 wrote:
student91 wrote:If you take the minimal polynominal for a number, and write down the prime factors of the powers, doesn't the biggest prime number give what we want? I mean, if there is a factor 3 in the powers somewhere, you have to take the cubic root to get a solution, and if there are only factors of 2, you get only nested square roots.
Furthermore, I thought of a definition of the CVP of a polytope as follows: Take a vertex figure of every vertex. calculate the minimal polynominal of the distances between every vertex inside this vertex figure. Get the biggest prime factor of the exponents. Do this for every distance and take the biggest number.
So basically you compute the biggest prime factor of the exponent for every two vertices that connect to the same vertex.


Biggest prime number, yes, but please note that I'm not entirely sure just what IS the biggest prime number here. I don't even know how it is for polygons. I did some calculations some years ago to see that regular heptagon requires solving a cubic equation and thus has CVP of 3, but I don't actually know if there even ARE polygons or polyhedra that would require solving a degree 5 equation. It's possible that above-cubic polygons don't exist or that they are all big enough that the polyhedra based on them (prism/antiprism) are not used in any CRF constructions, and we won't have to consider anything more than cubic equations while working.

A 7,7-duoprism, according to what you said, would require solving a degree-3 equation, so it would have CVP 3. Yet duoprisms are "easy" to construct in some sense. So maybe this method of measurement is not that useful after all?

OTOH, what I had in mind really wasn't specifically the degree of a polynomial; that was just the most convenient approximation I could think of at the time. What I had in mind was how many degrees of freedom you have to simultaneously solve before you can obtain coordinates. For a 7,7-duoprism, since you already know what the angles must be, it really doesn't leave any degree of freedom at all, and that corresponds with our intuition that duoprisms are "easy" to construct. Whereas if you take a snub square antiprism, you can't determine the angles one at a time; you have to simultaneously balance several different things in order for everything to work out. That's what I was trying to get at -- and perhaps the polynomial degree isn't a good measure of that. So in that sense the snub square antiprism is "harder" than a 7,7-duoprism. I thought polynomial degree was a good way to measure this, but I guess I was wrong.
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Re: Johnsonian Polytopes

Postby Marek14 » Sat Feb 08, 2014 11:18 pm

quickfur wrote:A 7,7-duoprism, according to what you said, would require solving a degree-3 equation, so it would have CVP 3. Yet duoprisms are "easy" to construct in some sense. So maybe this method of measurement is not that useful after all?

OTOH, what I had in mind really wasn't specifically the degree of a polynomial; that was just the most convenient approximation I could think of at the time. What I had in mind was how many degrees of freedom you have to simultaneously solve before you can obtain coordinates. For a 7,7-duoprism, since you already know what the angles must be, it really doesn't leave any degree of freedom at all, and that corresponds with our intuition that duoprisms are "easy" to construct. Whereas if you take a snub square antiprism, you can't determine the angles one at a time; you have to simultaneously balance several different things in order for everything to work out. That's what I was trying to get at -- and perhaps the polynomial degree isn't a good measure of that. So in that sense the snub square antiprism is "harder" than a 7,7-duoprism. I thought polynomial degree was a good way to measure this, but I guess I was wrong.


Well, 7-7 duoprism is easy to construct in abstract sense, but if you search for its coordinates, you'd have to use cubic equation to determine them. I guess it's easy because it belongs to easily-generalized group that contains some quite easy things.

Well, for today I'd suggest looking into the "edge surrounded by 3 snub disphenoids" configuration :)
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Re: Johnsonian Polytopes

Postby student91 » Sat Feb 08, 2014 11:23 pm

quickfur wrote:
[...]

A 7,7-duoprism, according to what you said, would require solving a degree-3 equation, so it would have CVP 3. Yet duoprisms are "easy" to construct in some sense. So maybe this method of measurement is not that useful after all?

OTOH, what I had in mind really wasn't specifically the degree of a polynomial; that was just the most convenient approximation I could think of at the time. What I had in mind was how many degrees of freedom you have to simultaneously solve before you can obtain coordinates. For a 7,7-duoprism, since you already know what the angles must be, it really doesn't leave any degree of freedom at all, and that corresponds with our intuition that duoprisms are "easy" to construct. Whereas if you take a snub square antiprism, you can't determine the angles one at a time; you have to simultaneously balance several different things in order for everything to work out. That's what I was trying to get at -- and perhaps the polynomial degree isn't a good measure of that. So in that sense the snub square antiprism is "harder" than a 7,7-duoprism. I thought polynomial degree was a good way to measure this, but I guess I was wrong.
(hadn't read mareks post yet)
Maybe the CVP as we have it doesn't tell you how easy it was to construct a polytope, but I think this algebraic twist to polytopes might be a way to say what polytopes can be used in CRF's. After all, it is quite perculiar that things with CVP=3 seemesly only occur in cartesian products, so I think, although it's not what you wanted, the CVP is a very potential invention.
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Re: Johnsonian Polytopes

Postby Klitzing » Sun Feb 09, 2014 12:16 am

As up to the name of quickfurs J92 beasty, my working title would go along of the Bowers style namings, generally. His acronym for J92 is thawro = trigonal hebe spheno (=w) rotunda. And your polychoron uses 4 such thingies in a 60 degree rhomb. That is, up to some better name, I'd call it frankly a "thawro rhomb". - But we might ask HedronDude directly about his usually keen sense for namings ...

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Re: Johnsonian Polytopes

Postby Klitzing » Sun Feb 09, 2014 12:28 am

Finally mapped my mind around quickfur's find, that "thawro rhomb".

Wrt. "rhomb" just cf. his already given lace city display:
Code: Select all
               x3o
          f3x       f3x
     o3F                 o3F
               x3F
x3x                           x3x
               F3x
     F3o                 F3o
          x3f       x3f
               o3x


That one then helped me to fiddle out its incidence matrix:
Code: Select all
12  *  * *  * |  2  2  2  0  0  0  0  0  0  0  0 0 0 0 | 1  2  2  1  2  2  0  0  0  0  0 0  0  0  0  0  0  0 0 | 2  2  2 0  0 0 0  0  x3x (of l.c.) ones
 * 12  * *  * |  0  2  0  2  2  2  0  0  0  0  0 0 0 0 | 0  1  0  2  2  2  1  1  2  2  1 0  0  0  0  0  0  0 0 | 1  1  4 1  1 0 0  0  o3F (of l.c.) ones
 *  * 24 *  * |  0  0  1  1  1  0  1  1  1  1  1 0 0 0 | 0  0  1  0  1  1  1  1  1  1  0 1  1  1  1  1  1  1 0 | 1  1  1 1  1 1 1  1  f3x (of l.c.) ones
 *  *  * 6  * |  0  0  0  0  0  0  0  0  4  0  0 2 0 0 | 0  0  0  0  0  0  4  0  0  0  0 0  2  2  0  0  0  0 1 | 2  0  0 2  0 1 0  0  x3o (of l.c.) ones
 *  *  * * 12 |  0  0  0  0  0  2  0  0  0  2  2 0 1 1 | 0  0  0  2  0  0  0  0  2  2  2 0  0  0  1  1  2  4 0 | 0  0  4 1  2 0 1  2  x3F (of l.c.) ones
--------------+----------------------------------------+-------------------------------------------------------+--------------------
 2  0  0 0  0 | 12  *  *  *  *  *  *  *  *  *  * * * * | 1  1  1  0  0  0  0  0  0  0  0 0  0  0  0  0  0  0 0 | 2  1  0 0  0 0 0  0
 1  1  0 0  0 |  * 24  *  *  *  *  *  *  *  *  * * * * | 0  1  0  1  1  1  0  0  0  0  0 0  0  0  0  0  0  0 0 | 1  1  2 0  0 0 0  0
 1  0  1 0  0 |  *  * 24  *  *  *  *  *  *  *  * * * * | 0  0  1  0  1  1  0  0  0  0  0 0  0  0  0  0  0  0 0 | 1  1  1 0  0 0 0  0
 0  1  1 0  0 |  *  *  * 24  *  *  *  *  *  *  * * * * | 0  0  0  0  1  0  1  0  0  1  0 0  0  0  0  0  0  0 0 | 1  0  1 1  0 0 0  0  at single thawro
 0  1  1 0  0 |  *  *  *  * 24  *  *  *  *  *  * * * * | 0  0  0  0  0  1  0  1  1  0  0 0  0  0  0  0  0  0 0 | 0  1  1 0  1 0 0  0  between neighbouring thawros
 0  1  0 0  1 |  *  *  *  *  * 24  *  *  *  *  * * * * | 0  0  0  1  0  0  0  0  1  1  1 0  0  0  0  0  0  0 0 | 0  0  2 1  1 0 0  0
 0  0  2 0  0 |  *  *  *  *  *  * 12  *  *  *  * * * * | 0  0  1  0  0  0  0  1  0  0  0 1  1  0  0  0  1  0 0 | 1  1  0 0  1 1 1  0  at single thawro
 0  0  2 0  0 |  *  *  *  *  *  *  * 12  *  *  * * * * | 0  0  0  0  0  0  0  0  0  0  0 1  0  1  1  1  0  0 0 | 0  0  0 1  0 1 1  1  between neighbouring thawros
 0  0  1 1  0 |  *  *  *  *  *  *  *  * 24  *  * * * * | 0  0  0  0  0  0  1  0  0  0  0 0  1  1  0  0  0  0 0 | 1  0  0 1  0 1 0  0
 0  0  1 0  1 |  *  *  *  *  *  *  *  *  * 24  * * * * | 0  0  0  0  0  0  0  0  1  0  0 0  0  0  1  0  1  1 0 | 0  0  1 0  1 0 1  1  peppy-squippy-trip-tet
 0  0  1 0  1 |  *  *  *  *  *  *  *  *  *  * 24 * * * | 0  0  0  0  0  0  0  0  0  1  0 0  0  0  0  1  0  1 0 | 0  0  1 1  0 0 0  1  peppy-mibdi-tet
 0  0  0 2  0 |  *  *  *  *  *  *  *  *  *  *  * 6 * * | 0  0  0  0  0  0  2  0  0  0  0 0  0  0  0  0  0  0 1 | 2  0  0 1  0 0 0  0
 0  0  0 0  2 |  *  *  *  *  *  *  *  *  *  *  * * 6 * | 0  0  0  2  0  0  0  0  0  0  0 0  0  0  0  0  0  4 0 | 0  0  4 0  0 0 0  2  at peppies
 0  0  0 0  2 |  *  *  *  *  *  *  *  *  *  *  * * * 6 | 0  0  0  0  0  0  0  0  0  0  2 0  0  0  0  0  2  0 0 | 0  0  0 1  2 0 1  0  at mibdies resp. trips
--------------+----------------------------------------+-------------------------------------------------------+--------------------
 6  0  0 0  0 |  6  0  0  0  0  0  0  0  0  0  0 0 0 0 | 2  *  *  *  *  *  *  *  *  *  * *  *  *  *  *  *  * * | 2  0  0 0  0 0 0  0  {6}
 2  1  0 0  0 |  1  2  0  0  0  0  0  0  0  0  0 0 0 0 | * 12  *  *  *  *  *  *  *  *  * *  *  *  *  *  *  * * | 1  1  0 0  0 0 0  0  {3} thawro S-polar
 2  0  2 0  0 |  1  0  2  0  0  0  1  0  0  0  0 0 0 0 | *  * 12  *  *  *  *  *  *  *  * *  *  *  *  *  *  * * | 1  1  0 0  0 0 0  0  {4} thawro S-tropal
 1  2  0 0  2 |  0  2  0  0  0  2  0  0  0  0  0 0 1 0 | *  *  * 12  *  *  *  *  *  *  * *  *  *  *  *  *  * * | 0  0  2 0  0 0 0  0  {5} peppy base
 1  1  1 0  0 |  0  1  1  1  0  0  0  0  0  0  0 0 0 0 | *  *  *  * 24  *  *  *  *  *  * *  *  *  *  *  *  * * | 1  0  1 0  0 0 0  0  {3} thawro S-tropal
 1  1  1 0  0 |  0  1  1  0  1  0  0  0  0  0  0 0 0 0 | *  *  *  *  * 24  *  *  *  *  * *  *  *  *  *  *  * * | 0  1  1 0  0 0 0  0  {3} between neighbouring thawros
 0  1  2 2  0 |  0  0  0  2  0  0  0  0  2  0  0 1 0 0 | *  *  *  *  *  * 12  *  *  *  * *  *  *  *  *  *  * * | 1  0  0 1  0 0 0  0  {5} thawro N-tropal
 0  1  2 0  0 |  0  0  0  0  2  0  1  0  0  0  0 0 0 0 | *  *  *  *  *  *  * 12  *  *  * *  *  *  *  *  *  * * | 0  1  0 0  1 0 0  0  {3} between neighbouring thawros
 0  1  1 0  1 |  0  0  0  0  1  1  0  0  0  1  0 0 0 0 | *  *  *  *  *  *  *  * 24  *  * *  *  *  *  *  *  * * | 0  0  1 0  1 0 0  0  {3} peppy-squippy
 0  1  1 0  1 |  0  0  0  1  0  1  0  0  0  0  1 0 0 0 | *  *  *  *  *  *  *  *  * 24  * *  *  *  *  *  *  * * | 0  0  1 1  0 0 0  0  {3} peppy-mibdi
 0  1  0 0  2 |  0  0  0  0  0  2  0  0  0  0  0 0 0 1 | *  *  *  *  *  *  *  *  *  * 12 *  *  *  *  *  *  * * | 0  0  0 1  1 0 0  0  {3}
 0  0  4 0  0 |  0  0  0  0  0  0  2  2  0  0  0 0 0 0 | *  *  *  *  *  *  *  *  *  *  * 6  *  *  *  *  *  * * | 0  0  0 0  0 1 1  0  {4}
 0  0  2 1  0 |  0  0  0  0  0  0  1  0  2  0  0 0 0 0 | *  *  *  *  *  *  *  *  *  *  * * 12  *  *  *  *  * * | 1  0  0 0  0 1 0  0  {3} thawro N-tropal
 0  0  2 1  0 |  0  0  0  0  0  0  0  1  2  0  0 0 0 0 | *  *  *  *  *  *  *  *  *  *  * *  * 12  *  *  *  * * | 0  0  0 1  0 1 0  0  {3} between neighbouring thawros
 0  0  2 0  1 |  0  0  0  0  0  0  0  1  0  2  0 0 0 0 | *  *  *  *  *  *  *  *  *  *  * *  *  * 12  *  *  * * | 0  0  0 0  0 0 1  1  {3} at trip
 0  0  2 0  1 |  0  0  0  0  0  0  0  1  0  0  2 0 0 0 | *  *  *  *  *  *  *  *  *  *  * *  *  *  * 12  *  * * | 0  0  0 1  0 0 0  1  {3} at mibdi
 0  0  2 0  2 |  0  0  0  0  0  0  1  0  0  2  0 0 0 1 | *  *  *  *  *  *  *  *  *  *  * *  *  *  *  * 12  * * | 0  0  0 0  1 0 1  0  {4}
 0  0  1 0  2 |  0  0  0  0  0  0  0  0  0  1  1 0 1 0 | *  *  *  *  *  *  *  *  *  *  * *  *  *  *  *  * 24 * | 0  0  1 0  0 0 0  1  {3}
 0  0  0 3  0 |  0  0  0  0  0  0  0  0  0  0  0 3 0 0 | *  *  *  *  *  *  *  *  *  *  * *  *  *  *  *  *  * 2 | 2  0  0 0  0 0 0  0  {3} thawro top
--------------+----------------------------------------+-------------------------------------------------------+--------------------
 6  3  6 3  0 |  6  6  6  6  0  0  3  0  6  0  0 3 0 0 | 1  3  3  0  6  0  3  0  0  0  0 0  3  0  0  0  0  0 1 | 4  *  * *  * * *  *  thawro
 2  1  2 0  0 |  1  2  2  0  2  0  1  0  0  0  0 0 0 0 | 0  1  1  0  0  2  0  1  0  0  0 0  0  0  0  0  0  0 0 | * 12  * *  * * *  *  squippy (at thawro's 4s)
 1  2  1 0  2 |  0  2  1  1  1  2  0  0  0  1  1 0 1 0 | 0  0  0  1  1  1  0  0  1  1  0 0  0  0  0  0  0  1 0 | *  * 24 *  * * *  *  peppy
 0  2  4 2  2 |  0  0  0  4  0  4  0  2  4  0  4 1 0 1 | 0  0  0  0  0  0  2  0  0  4  2 0  0  2  0  2  0  0 0 | *  *  * 6  * * *  *  mibdi
 0  1  2 0  2 |  0  0  0  0  2  2  1  0  0  2  0 0 0 1 | 0  0  0  0  0  0  0  1  2  0  1 0  0  0  0  0  1  0 0 | *  *  * * 12 * *  *  squippy (at trip's 2-cup-lacing-4s)
 0  0  4 1  0 |  0  0  0  0  0  0  2  2  4  0  0 0 0 0 | 0  0  0  0  0  0  0  0  0  0  0 1  2  2  0  0  0  0 0 | *  *  * *  * 6 *  *  squippy (at trip's 2-cup-base-4)
 0  0  4 0  2 |  0  0  0  0  0  0  2  2  0  4  0 0 0 1 | 0  0  0  0  0  0  0  0  0  0  0 1  0  0  2  0  2  0 0 | *  *  * *  * * 6  *  trip
 0  0  2 0  2 |  0  0  0  0  0  0  0  1  0  2  2 0 1 0 | 0  0  0  0  0  0  0  0  0  0  0 0  0  0  1  1  0  2 0 | *  *  * *  * * * 12  tet


Esp. his provided cell counts this time come out the same here. :P

--- rk
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Re: Johnsonian Polytopes

Postby quickfur » Sun Feb 09, 2014 1:28 am

student91 wrote:[...]
Maybe the CVP as we have it doesn't tell you how easy it was to construct a polytope, but I think this algebraic twist to polytopes might be a way to say what polytopes can be used in CRF's. After all, it is quite perculiar that things with CVP=3 seemesly only occur in cartesian products, so I think, although it's not what you wanted, the CVP is a very potential invention.

You're right, even though heptagons and 7,7-duoprisms are easy to construct abstractly, the fact that the CVP of a heptagon is 3 probably has something to do with the fact that none of the Johnson solids have a heptagon in it! And the only 3D CRFs containing heptagons are the infinite families of prisms and antiprisms. So maybe the CVP does tell us something about CRF-ability.

Having said that, though, there are some non-convex Johnson solid generalizations that do sport heptagons; I remember seeing on orchidpalms one of the hebespheno-thingies analyzed as the only convex member of a larger family, and some of them do sport heptagons.

Now, it'd be interesting to think of a polytope (not necessarily CRF) that requires solving a quintic equation to find its coordinates...
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Re: Johnsonian Polytopes

Postby quickfur » Sun Feb 09, 2014 1:36 am

Klitzing wrote:As up to the name of quickfurs J92 beasty, my working title would go along of the Bowers style namings, generally. His acronym for J92 is thawro = trigonal hebe spheno (=w) rotunda. And your polychoron uses 4 such thingies in a 60 degree rhomb. That is, up to some better name, I'd call it frankly a "thawro rhomb". - But we might ask HedronDude directly about his usually keen sense for namings ...[...]

Rhombus! I think you hit the nail on the head! The J92 beasty is a 4D rhombus! The 60°-120°-60°-120° dichoral angles are very rhombus-like. So maybe I'll call it the J92-rhombus. Or J92-rhombochoron, to avoid confusion with 2D rhombuses. :D

And on that note, it seems everyone is jumping on to the other Johnson crown jewels already, but I think we may not be quite done with J92 yet. I'm still considering the possibility that perhaps a 120° dichoral angle at the hexagonal face between two J92's may be CRF-able as well. After all, there are other J92-J92 possibilities that haven't been fully explored yet! Perhaps a hexagonal arrangement of J92's might work (120° at both hexagons and triangles between J92 cells)?
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Re: Johnsonian Polytopes

Postby Polyhedron Dude » Sun Feb 09, 2014 6:23 am

J92 - rhombochoron - love it! We can call it thawrorh for short.
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Re: Johnsonian Polytopes

Postby Keiji » Sun Feb 09, 2014 11:20 am

quickfur wrote:J92-rhombochoron


So, in full it would be the triangular hebesphenorotundaeic rhombochoron?
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Re: Johnsonian Polytopes

Postby Klitzing » Sun Feb 09, 2014 11:36 am

Keiji wrote:
quickfur wrote:J92-rhombochoron


So, in full it would be the triangular hebesphenorotundaeic rhombochoron?


The problem of such three termed names is the respective adjectivation used, i.e. the intended way of placing parantheses.
Should it be read as an A (B C)
or rather an (A B) C?

This is why I usually place hyphens between the closer to be connected ones:

    small-rhombicuboctahedron prism
    triangular-prism pyramid
    ...
    triangular-hebesphenorotunda rhombohedron

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Re: Johnsonian Polytopes

Postby Keiji » Sun Feb 09, 2014 10:21 pm

That's not a problem, since I will always read (and write) them as (A B) C.
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Re: Johnsonian Polytopes

Postby Marek14 » Mon Feb 10, 2014 10:41 am

Ok, let's try if J91 can be generalized in 4D:

First, let's take the icosidodecahedral rotunda and the dodecahedral semipericupola x5o3x||x5x3o.

Both of these are cuts with truncated dodecahedron on bottom. Now...

Dodecahedral semipericupola has following cells:
1 rhombicosidodecahedron
1 truncated dodecahedron
12 pentagonal cupolae
30 triangular prisms
20 octahedra

It's a cut from x5o3x3o. Triangular prisms and octahedra are normal parts of that polytope. The pentagonal cupolae, though, are cut rhombicosidodecahedra.
Now, what will happen if we cut it again, so the rhombicosidodecahedron we DO have will be ALSO reduced to a pentagonal cupola?

The bottom truncated dodecahedron will become something strange, but that's ok -- we'll correct that later.

We'll get a shape with 2 pentagonal cupolas, 5 triangular prisms and 5 octahedra. It's also bordered by 2 truncated-dodecahedral cuts which intersect each other.

Now for icosidodecahedral rotunda. This is made of
1 icosidodecahedron
1 truncated dodecahedron (this cell is missing from wiki)
12 pentagonal rotundae
40 tetrahedra (in 2 layers)

It's a cut from o5x3o3o. The pentagonal rotundae are cuts of icosidodecahedra. So, we'll do the same thing: cut the top icosidodecahedron in half to get a pentagonal rotunda.
We'll get a shape with 7 pentagonal rotundae and 20 tetrahedra. It's bordered, once again, by 2 truncated-dodecahedral cuts.

Now, my question is this: are the dichoral angles between these truncated-dodecahedral cuts such that the both of these semi-shapes can be combined together, thus creating a 4D analogue of J91?

EDIT: This technique might be much more productive in 4D than in 3D; just realized, though, that elongated square bipyramid can be also thought of in this manner (as four "lunae" from rhombicuboctahedron put together).

Also, I realize that the rotundaic part of J91 is not a full wedge of pentagonal rotunda, but a "cut" wedge with two end triangles removed -- not sure which/how many parts would be needed to remove from the icosidodecahedral rotunda wedge to make it fit with the wedge created from dodecahedral semipericupola.
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Re: Johnsonian Polytopes

Postby Keiji » Mon Feb 10, 2014 6:05 pm

Marek14 wrote:(this cell is missing from wiki)


If you find such things are missing, why not add them? ;)
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Re: Johnsonian Polytopes

Postby quickfur » Mon Feb 10, 2014 7:26 pm

Y'know, during the calculations that eventually led to the J92 rhombochoron, I got the impression that the 600-cell family may have a LOT of "patchwork" CRFs, that is, CRFs made of different patches of surfaces taken from different uniforms in that family.

Consider, for example, the fact that you can cut a 600-cell into several strata: point||icosahedron, icosahedron||dodecahedron, dodecahedron||icosidodecahedron. Each cut causes some non-CRFs to appear because of various tetrahedra getting bisected at odd angles, but those can be deleted altogether and replaced with pentagonal pyramids. Let's take the easiest case: the hemi-600-cell, which contains an icosidodecahedron at the bottom. This icosidodecahedron can be bisected at various angles to give the 600-cell lunae, but you can also consider this icosidodecahedron as part of the larger surface of a rectified 120-cell. You can then insert patches from the rectified 120-cell in place of some of the tetrahedra, and borrow other patches to make things convex again by joining back up to the remaining 600-cell tetrahedra. Conversely, you can start with a patch of the rectified 120-cell's surface, stop it at some point, and insert 600-cell tetrahedra on top of the icosidodecahedral cells, then close the thing up with various other bits of pentagonal CRF polyhedra.

I think many of these cases should be CRF-able, if we would only look.
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Re: Johnsonian Polytopes

Postby quickfur » Mon Feb 10, 2014 7:29 pm

Keiji wrote:I would propose numbering them as such:

CJ4.1 snubdis antiprism
CJ4.2.1 tetrahedral ursachoron
CJ4.2.2 expanded tetrahedral ursachoron
CJ4.2.3 octahedral ursachoron
CJ4.2.4 expanded octahedral ursachoron
CJ4.2.5 icosahedral ursachoron
CJ4.2.6 expanded icosahedral ursachoron
CJ4.3.1 castellated rhodoperihedral prism (what is H101, by the way?)
CJ4.3.2 castellated rhodopantohedral prism (the expanded version of CJ4.3.1)
CJ4.4 hebesphenochoron (or whatever we are calling it)
[...]

Belated reply: I second this proposal. Maybe we should use the crown jewel wiki page as the canonical authority on these numberings, then each new crown jewel we discover can be assigned a unique numbering that can also be used as an identifier until we decide on a name for it.

P.S. About H101: sorry, it's my own notation for x5o3x ((small) rhombicosidodecahedron).
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