nonconvex regular faced polyhedron

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nonconvex regular faced polyhedron

Postby dodecahedron » Wed Oct 03, 2012 12:55 am

Would someone like to hep me find nonconvex johnson solids.
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Re: nonconvex regular faced polyhedron

Postby quickfur » Wed Oct 03, 2012 1:23 am

dodecahedron wrote:Would someone like to hep me find nonconvex johnson solids.

There are an infinite number of them. For example, start with a tetrahedron, attach more tetrahedrons to it in a chain. The result will be a non-convex deltahedron. There are an infinite number of possible combinations, since you can just make a chain that goes on and on. Then you can have branching structures, looping structures, etc.. And this is just with tetrahedra... the possibilties explode rapidly if you allow any closed assembly of triangles.

And then once you start allow polygons other than triangles... there's simply no way to enumerate all of these possibilities.
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Re: nonconvex regular faced polyhedron

Postby dodecahedron » Wed Oct 03, 2012 12:43 pm

OK but i was thinking of self-interscting.

If these are infinite how about under a set number of faces.
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Re: nonconvex regular faced polyhedron

Postby Klitzing » Wed Oct 03, 2012 2:29 pm

Have not given any thought so far, but how about this setup:

Not considering regular integral faces and non-convex solids, but the other way round. I.e.
  • allowing for polygrammal faces,
  • but still asking for local convexity at least.

Local convexity differs from global convexity. Global convexity does not allow for polygrammal faces. But local convexity does. In fact it is defined as a solid, with all of its vertex figures are globally convex - with respect to that figure of one dimension less. E.g. the sissid = {5/2, 5} is locally convex. But gad = {5, 5/2} OTOH is NOT locally convex.

Has anyone some ideas on how to estimate the potential magnitude of this set?

--- rk
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Re: nonconvex regular faced polyhedron

Postby quickfur » Wed Oct 03, 2012 2:50 pm

Hmm. What about orbiform (i.e. vertices lie on surface of a sphere) non-convex Johnsons? The orbiform requirement will eliminate a lot of probably-uninteresting combinations (like unending chains of tetrahedra, say), as well as make computations much more tractable.
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Re: nonconvex regular faced polyhedron

Postby dodecahedron » Wed Oct 03, 2012 5:14 pm

Orbiform is ok but only one johnson solis is orbiform and its locally vertex transtive

here is a possible one

"inaugment"

construction:take a tetrahedron push the triangular faces through eachother and get a pseudo-triangle.
Then take another tetrahedron take a face off and put it on fill in the gaps.

10 triangles

3 or 5 triangles meeting at a vertex

there should also be a square and pentagonal form
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Re: nonconvex regular faced polyhedron

Postby wendy » Mon Oct 08, 2012 10:06 am

One of the interesting ones is the pentagon-pentagram tegum, which is made of 25 regular tetrahedra.
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Re: nonconvex regular faced polyhedron

Postby quickfur » Mon Oct 08, 2012 2:27 pm

wendy wrote:One of the interesting ones is the pentagon-pentagram tegum, which is made of 25 regular tetrahedra.

That is very interesting. I was hunting for CRF tegums a while back but didn't find anything beyond the 16-cell. Are there any other tegums with CRF cells?
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Re: nonconvex regular faced polyhedron

Postby dodecahedron » Mon Oct 08, 2012 11:34 pm

I also suppose by pentagon-pentagram you mean pentagonal crossed cupola
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Re: nonconvex regular faced polyhedron

Postby quickfur » Tue Oct 09, 2012 12:22 am

dodecahedron wrote:I also suppose by pentagon-pentagram you mean pentagonal crossed cupola

No, she's talking about tegums, which are, loosely speaking, "diamond-like" shapes. The pentagon-pentagram tegum is what you get by placing a pentagon in the XY plane, a pentagram in the ZW plane, and taking the convex hull.
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Re: nonconvex regular faced polyhedron

Postby wendy » Tue Oct 09, 2012 7:15 am

The polytope i mentioned is a tegum-product, a cover or drawing of surfaces of its bases, tegums are duals to prisms.
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Re: nonconvex regular faced polyhedron

Postby Klitzing » Tue Oct 09, 2012 9:03 am

quickfur wrote:
dodecahedron wrote:I also suppose by pentagon-pentagram you mean pentagonal crossed cupola

No, she's talking about tegums, which are, loosely speaking, "diamond-like" shapes. The pentagon-pentagram tegum is what you get by placing a pentagon in the XY plane, a pentagram in the ZW plane, and taking the convex hull.


Hehe, speaking of convex hull in this case seems wrong, as you are using a non-convex spanning element within the ZW plane. :)

Even so, the point (towards "dodecahedron") is correct. You will have to connect every vertex of the pentagon with every vertex of the pentagram in order to get the lacing edges (those in addition to the 2x5 ones of those 2 polygons). Faces would be wone by making all the lace prisms of type pentagon-edge || pentagram-vertex, resp. pentagon-vertex || pentagram-edge. And finally the cells would be derived as (any) pentagon-edge || (any, thus orthogonal) pentagram-edge. All these cells are tetrahedra.

The same holds true for the other tegum you mentioned, that of 2 completely orthogonal squares. That figure turns out to be nothing but the hexadecachoron.

That those 2 tegums do work can be seen from their circumradii: the unit square has radius 1/sqrt(2). Accordingly the lacing edge would have length (applying Pythagoras):
Code: Select all
sqrt[(1/sqrt(2))^2 + (1/sqrt(2))^2] = sqrt[1/2 + 1/2] = 1 (i.e. unity again)

In case of Wendys pentagon-pentagram-tegum we have the circumradius of the pentagon being sqrt[(5+sqrt(5))/10] while that of the pentagram equals sqrt[(5-sqrt(5))/10]. Accordingly Pythagoras would give there:
Code: Select all
sqrt[(sqrt[(5+sqrt(5))/10])^2 + (sqrt[(5-sqrt(5))/10])^2] = sqrt[(5+sqrt(5))/10 + (5-sqrt(5))/10] = sqrt[(5+sqrt(5)+5-sqrt(5))/10] = sqrt[10/10] = 1


The same principle would apply more generally: if you would consider a tegum with an unit-edged {n/d} within the XY plane (centered at the origin) and an unit-edged {m/b} in the ZW plane (also centered at the origin) and would considere the tegum therefrom derived, you could calculate the lacing edge lengths right from the respective circumradii:
Code: Select all
sqrt([1/(2*sin(pi*d/n))]^2 + [1/(2*sin(pi*b/m))]^2)

In order to have only unit edges throughout, i.e. the resulting tegum being bounded by regular tetrahedra only, we get a diophantine equation: that formula for the lacing edge length has to equal 1, subject to the further restriction that its ingrediants, i.e. n, d, m, and b, all have to (positive) integers. - We already have two solutions of that problem: n=m=4, d=b=1 and n=m=5, d=1, b=2. - My conjecture would be that those are the only ones; but I have no proof.

Someone would like to come in, providing one?

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Re: nonconvex regular faced polyhedron

Postby wendy » Tue Oct 09, 2012 10:47 am

If one takes d2(5)+d2(5/2), one gets 4, as one can see from the vf of the discrete lattice o5o3x3o5/2o.

This means the corresponding tegum, with a unit-edge 5, and a unit-edge 5/2, has, as a distance between the vertices, also a unit edge. This makes it bounded throughout by regular tetrahedra. I'm pretty sure i know what i am talking about. The other polygon-tegum that is bounded by regular tetrahedra also comes from a discrete tiling o4o3x3o4o.

It is not, however, the tiling cell of the discrete lattice o5o3x3o5/2o, since i am pretty sure that its vertex-angles are not integer.

No other pairing of polygons leads to a uniform-surhedron polytope, as did quickfur ask.

Klitzing wrote:We already have two solutions of that problem: n=m=4, d=b=1 and n=m=5, d=1, b=2. - My conjecture would be that those are the only ones; but I have no proof.

Someone would like to come in, providing one?


Proof exists in the ZZ number theory, which is why one does not forrage too far, but rk found my introduction to it a little hard to fathom. In essence, one uses various isomorphisms to demonstrate that a pair of class-2 polygon is the largest that support this, and running through the cases, one finds that of (5,5/2), (8, 8/3), (10,10/3), (12, 12/5), and (4,4), (4,6) and (6,6) are the only solutions for which the diameter-squares of two polygons gives rise to an integer. Since we require that the value be 4 to give a quarter-diameter equal to the edge, we are left with (4,4) and (5,5/2). Higher-order classes leave too many isomorphs open and therefore do not close over two values.

Three polygons, can be solved by augmenting (4) or (6) to the above, along with the likes of (3,3,3), (7,7/2,7/3), (9,9/2,9/4) and (14, 14/5, 14/7) and (18, 18/5, 18/7). However, there's not a lot of call for the diameter of a prism of three polygons to come to call, save in a group where these are the three arms. All of these are hyperbolic, and rather dense, except the first (which is 33A).

The only solutions involving three non-zero short-chords adding to 4 are (3,3,4) and (3,5,5/2). This is well known, the proof in geometry is hard, although not impossible (coxeter spends several pages on it, by first supposing that a group whose mirrors cross the fundemental cell (like 5/2, 5), must give a second group which doesn't (like (3,5)), and then show that these two solutions are the only ones that pass. In number theory, one notes that any solution of three shortchords adding to an integer must be class-3 or less, which cuts down the search to Z2,Z3,Z4,Z5,Z6,Z7,Z9.
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Re: nonconvex regular faced polyhedron

Postby dodecahedron » Tue Oct 09, 2012 1:16 pm

ok, i think the pentagon-pentagram tegum is interesting.however look at the topic it's polyhedron.
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Re: nonconvex regular faced polyhedron

Postby wendy » Thu Oct 11, 2012 10:08 am

Klitzing wrote: We already have two solutions of that problem: n=m=4, d=b=1 and n=m=5, d=1, b=2. - My conjecture would be that those are the only ones; but I have no proof.

Someone would like to come in, providing one?


The actual proof runs along the line that if some p/d is involved, then there is a matching solution using p. This is the principle of isomporphism, which transforms any solution of an equation through all roots. Since the required lacing is 1, which is automorphic, we simply need to search for solutions with at least one is a polygon, p/1, This means that we look for complements to {3} (fails), {4} and {5} (succeeds).
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Re: nonconvex regular faced polyhedron

Postby quickfur » Thu Oct 11, 2012 2:31 pm

dodecahedron wrote:ok, i think the pentagon-pentagram tegum is interesting.however look at the topic it's polyhedron.

Take a look at: http://www.orchidpalms.com/polyhedra/index.html

There are a lot of non-convex polyhedra described there, as well as families of them.

Like I said, the possibilities are infinite. It comes down to what you consider as "interesting".
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Re: nonconvex regular faced polyhedron

Postby dodecahedron » Fri Oct 12, 2012 2:01 pm

i already suggested self intersection.
i have been on orchid palms,i got there looking for "near misses".
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Re: nonconvex regular faced polyhedron

Postby Klitzing » Mon Oct 15, 2012 7:46 pm

Klitzing wrote:Have not given any thought so far, but how about this setup:

Not considering regular integral faces and non-convex solids, but the other way round. I.e.
  • allowing for polygrammal faces,
  • but still asking for local convexity at least.

Local convexity differs from global convexity. Global convexity does not allow for polygrammal faces. But local convexity does. In fact it is defined as a solid, with all of its vertex figures are globally convex - with respect to that figure of one dimension less. E.g. the sissid = {5/2, 5} is locally convex. But gad = {5, 5/2} OTOH is NOT locally convex.

Has anyone some ideas on how to estimate the potential magnitude of this set?

--- rk


Just were roaming: looking for locally convex polyhedra with regular faces. - Sure we would have the inclusion sequence: Platonics < Archimedeans and Prisms < Johnsonians (inclusive set) < locally convex ones. But just as the Johnsonians are usually given as an exclusive set, we could look at the locally convex ones too exclusively, i.e. the locally convex ones which are non-convex.

We would have some uniforms:
  • sissid: 12(5/2)^5
  • gissid: 20(5/2)^3
  • tigid: 60(5/2,10,10)
  • quith: 24(8/3,8/3,3)
  • quit sissid: 60(10/3,10/3,5)
  • quit gissid: 60(3,10/3,10/3)
  • did: 30(5/2,5)^2
  • gid: 30(5/2,3)^2
  • sidtid: 20(5/2,3)^3
  • raded: 60(5/2,4,5,4)
  • gocco: 24(8/3,3,8/3,4)
  • gidditdid: 60(3,10/3,5,10/3)
  • gaddid: 60(5/2,10/3,3,10/3)
  • siid: 60(5/2,6,3,6)
  • quitco: 48(8/3,4,6)
  • quitdid: 120(10/3,4,10)
  • gaquatid: 120(10/3,4,6)
  • cotco: 48(8/3,6,8)
  • iddid: 120(10/3,6,10)
  • siddid: 60(5/2,3,3,5,3) - 2 enantiomorphs!
  • gosid: 60(5/2,3^4)
  • seside: 60(5/2,3^5)
  • n/d-p (any n/d > 2, d > 1): 2n(n/d,4,4)
  • n/d-ap (any n/d > 2, d > 1): 2n(n/d,3^3)

Then I looked through my list of edge-facetings of uniforms (cf. http://www.polyedergarten.de/polyhedrix/e_klintro.htm). There are rather few:

Furthermore there clearly are cupolae and their relatives:
  • n/d-cu (2 < n/d < 6, d > 1 odd): n(n/d,4,3,4)+2n(2n/d,3,4)
  • ortho bi-n/d-cu (2 < n/d < 6, d > 1 odd): 2n(n/d,4,3,4)+2n(3,3,4,4)
  • gyro bi-n/d-cu (2 < n/d < 6, d > 1 odd): 2n(n/d,4,3,4)+2n(3,4)^2
  • elong n/d-cu (2 < n/d < 6, d > 1 odd): n(n/d,4,3,4)+2n(3,4^3)+2n(2n/d,4,4)
  • elong ortho bi-n/d-cu (2 < n/d < 6, d > 1 odd): 2n(n/d,4,3,4)+4n(3,4^3)
  • elong gyro bi-n/d-cu (2 < n/d < 6, d > 1 odd): 2n(n/d,4,3,4)+4n(3,4^3)
  • gyro-elong n/d-cu (2 < ? < n/d < 6, d > 1 odd): n(n/d,4,3,4)+2n(3^4,4)+2n(2n/d,3^3)
  • gyro-elong bi-n/d-cu (2 < ? < n/d < 6, d > 1 odd): 2n(n/d,4,3,4)+4n(3^4,4) - 2 enantiomorphs!

Having spoken of Johnsonians, there clearly also is a relative to 2 of them:

Sure, there will be more. This listing by no means is complete.

All what I've found does include one polygram at least.
So I wonder whether there are also localy convex, non-convex, regular faced polyhedra which have convex faces only?

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Re: nonconvex regular faced polyhedron

Postby dodecahedron » Thu Oct 18, 2012 9:02 pm

how about non prismatic/uniform ncrf's
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