Johnsonian Polytopes

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 quickfur » Mon Jan 06, 2014 3:30 pm

Klitzing wrote:
quickfur wrote:So basically your segmentotopic product (A||B) # (C||D) is basically just (A x C)||(B x D) where x denotes the Cartesian product, and # is a placeholder symbol for your "segmentotopic product"? In other words, you take the Cartesian product of the "top" layer of the segmentotope A||B with the "top" layer of the segmentotope C||D, and likewise take the Cartesian product of the respective bottom layers, and then construct a new segmentotope out of the results. Right? And what you're claiming then is, if A||B and C||D exist, then (A x C)||(B x D) must also exist?

Seems reasonable to me, though it doesn't really add anything new. It's just a way of analysing a segmentotope by decomposition into Cartesian products (or conversely, of constructing a higher-dimensional segmentotope from two lower-dimensional ones). Or am I missing something?

Well, you just miss here the restriction about the height to be real and truely positive.

But if we already know that A||B and C||D exist, then (A x C)||(B x D) should also exist, right?

A different variation of this idea, is if we start with an even polytope E (where "even" means it can be alternated) -- say we call its two alternated forms E+ and E- -- and a segmentotope F||G, then we take the convex hull of (E+ x F) U (E- x G), where x denotes the Cartesian product and U denotes set union. Under what conditions would the result contain F||G as facets? Is it possible to make the result CRF? For example, one could start with a cube and, say, a pentagonal cupola (pentagon||decagon). Since the cube alternates into two dual tetrahedra, we form the 5D Cartesian products tetrahedron x pentagon and dual_tetrahedron x decagon, and take their convex hull. The result should contain pentagonal cupolae as surtopes. Is it possible to make the result CRF?

Union would not work, as then those would all remain within the same Hyperspace and you'd get rather a compound than a pair of stacked bases.

I think you misunderstood that (1) I was not looking for creating more segmentotopes, I was looking to create new polytopes that contain segmentotope surtopes (surface elements); (2) I defined the operation take the convex hull of the compound.

For example, I just figured out last night that the product of a cube (as convex_hull(tet, dual_tet)) with a triangle (as point||line) can produce the topological equivalent of two 16-cells sharing a cell, or a tetra-augmented tetrahedral prism (probably cannot be made CRF, though).
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Re: Johnsonian Polytopes

Postby student91 » Mon Jan 06, 2014 4:26 pm

quickfur wrote:
Klitzing wrote:
quickfur wrote:[...]

Well, you just miss here the restriction about the height to be real and truely positive.

But if we already know that A||B and C||D exist, then (A x C)||(B x D) should also exist, right?

no, the height of the "base"-polytope should be shrunken to sqrt(h1²+h2²-1), as I explained in my previous post:
student91 wrote:[...]
let's consider the same segmentotopes as you used: b1||b2 (with height h1), and b3||b4 (with height h3).
[...]
now the true segmentotopical product: b1×b3||b2×b4. I'll look at this from the viewpoint of b1||b2 with b3's on the vertices of b1, and b4's at the vertices of b2. We could color the vertices where a b3 will be placed red, and the vertices where a b4 will be placed blue. the edges of b1||b2 connection a red vertex with a red one now will yield b3||b3-polytopes, connecting blue with blue yields b4||b4, and connecting blue with red yields b3||b4.
In order to connect properly, the edges of b1||b2 that connect blue with red sould be shrunk to h3.[!!!]
Those edges are the edges connecting b1 with b2. if we change the distance between b1 and b2 (i.e. h1), those edges will get a different length as well, and thus, it might be possible to make the red-blue edges of length h3. we'll get to whether this is possible or not later.
[...]
The height of the new polytope might indeed be a problem, but it's not that hard to calculate:
if we have (again) the "base"-polytope b1||b2, the edges connecting b1 with b2 should be given lenght h3. To calculate the distance between b1 and b2 for those edges to become h3, I do the following: take v1 (a vertex of b1, see above). draw a line through v1 perpendicular to the hyperplane in which b1 lies. call the point were this line intersects the hyperplane of b2 H. now the lenght of Hv1=h1, v1v2=1 and thus Hv2=sqrt(1-h1²).
now we shrink v1v2 down to the size of h3, so we'd get v1v2=h3, Hv2=sqrt(1-h1²), and thus Hv1=height new segmentotope=sqrt(h3²-(Hv2)²)=sqrt(h3²+h1²-1). this height should of course be real for the polytope to exist.



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Re: Johnsonian Polytopes

Postby student91 » Mon Jan 06, 2014 7:42 pm

Klitzing wrote:
student91 wrote:we should also include the snubs into the wythoffians.
(the last one can be seen as a segmentotopical product of a CRF-prism and another segmentotope, so it can be discarted)
do you think this might be provable? (at least all 4d-segmentotopes are of one of these categories)

OMG, please don't intend that!
[...]
The very intend of the usage of "Wythoffian" is just to discard all that stuff beyond.
In fact, "Wythoffian" and the set of all polytopes which can be described by a decorated Coxeter-Dynkin symbol, i.e. using node symbols "o" (resp. unringed node) and "x" (resp. ringed node) only, are synonyms.

The introduction of the node symbol "s" (resp. empty ring "node") is a completely different construction device, it therefore enriches the set of polytopes which are describable by Coxeter-Dynkin symbols.

In fact it was my paper on snubs, which finally detailed how snubs generally can be constructed from Wythoffian starting figures by means of an alternated faceting wrt. suitable subelements. Up to then only vertices had been used. But it does not restrict to those only, as I've described therein.
--- rk

I did not mean "redefine the word wythoffian so it also includes snubs." I understand why you thought this, as my scentence was bad-phrased. Wat I did mean is something like: the set of wythoffians in my previous post should be augmented with snubs, so I ment more like a subsitution rather than a redefenition. when we take "polytopes derived from the same symmetry," I think I include most of what I ment.
my statement would then be something like:
every segmentotope is either:
a lace simplex of polytopes derived from the same symmetry.
a segmentotopical product of multiple segmentotopes
multiple lace simplices placed base-to-base
a diminishing of those.
an augmention of a diminishing

But actually it doesn't really matter what closely-related polytopes I use for the lace simplices part, as long It's possible to prove that every segmentotope is one of the things above.
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Re: Johnsonian Polytopes

Postby Klitzing » Mon Jan 06, 2014 10:34 pm

quickfur wrote:For example, I just figured out last night that the product of a cube (as convex_hull(tet, dual_tet)) with a triangle (as point||line) can produce the topological equivalent of two 16-cells sharing a cell, or a tetra-augmented tetrahedral prism (probably cannot be made CRF, though).


The compound, you are looking for, thus is ox xo3oo3ox.
(Or, taken in the language of segmentotopes, even when disregarding height and lacings, it would look like tet || prism of dual tet.)

Convex hull then, even so an easy idea, is a quite complicate thing, both for calculations as well as for construction.
Thus I won't recomment to use that one heavily! Esp. not as a construction device...

But for the case of your example it is easier so. The vertices of the "top layer" are (1/rt8, 1/rt8, 1/rt8, 0, H) and all even changes of sign in the first 3 coordinates, while that of the "bottom layer" are (1/rt8, 1/rt8, -1/rt8, 1/2, 0) and all even changes of sign in the first 3 coordinates plus the change of sign in the 4th. The height H is a free variable, introduced in the sense of a segmentoteron, which would be assumed to be zero for your compound. Thus the squared distance of the 2 given vertices (squared lacing length) then calculates as dist2 = 0 + 0 + 1/2 + 1/4 + H2. Thus, in order to get it with all unit edges, you would have to take H2 = 1/4 instead. - Furthermore you might like to shift the 5th coordinate in such a way that the circumcenter would be the origin. The total circumradius evaluates here simply as R = rt(5/8).

Thus, after all, your figure itself is not CRF, but it can be made so, when the bases would be separated in the sense of a segmentoteron. Then it becomes ox ox3oo3xo&#x.

We even managed to calculate the incidence matrix of that thingy:
Code: Select all
ox ox3oo3xo&#x

o. o.3o.3o.    | 4 * | 3  6 0  0 | 3  3  6 12 0 0 | 1  3  6 2  6 6 0 0 | 1 3 3 2 0
.o .o3.o3.o    | * 8 | 0  3 1  3 | 0  3  6  3 3 3 | 0  6  3 3  3 1 3 1 | 3 3 1 1 1
---------------+-----+-----------+----------------+--------------------+----------
.. .. .. x.    | 2 0 | 6  * *  * | 2  0  0  4 0 0 | 1  0  2 0  2 4 0 0 | 0 1 2 2 0
oo oo3oo3oo&#x | 1 1 | * 24 *  * | 0  1  2  2 0 0 | 0  2  2 1  2 1 0 0 | 1 2 1 1 0
.x .. .. ..    | 0 2 | *  * 4  * | 0  3  0  0 3 0 | 0  6  3 0  0 0 3 0 | 3 3 1 0 1
.. .x .. ..    | 0 2 | *  * * 12 | 0  0  2  0 1 2 | 0  2  0 2  1 0 2 1 | 2 1 0 1 1
---------------+-----+-----------+----------------+--------------------+----------
.. .. o.3x.    | 3 0 | 3  0 0  0 | 4  *  *  * * * | 1  0  0 0  0 2 0 0 | 0 0 1 2 0
ox .. .. ..&#x | 1 2 | 0  2 1  0 | * 12  *  * * * | 0  2  2 0  0 0 0 0 | 1 2 1 0 0
.. ox .. ..&#x | 1 2 | 0  2 0  1 | *  * 24  * * * | 0  1  0 1  1 0 0 0 | 1 1 0 1 0
.. .. .. xo&#x | 2 1 | 1  2 0  0 | *  *  * 24 * * | 0  0  1 0  1 1 0 0 | 0 1 1 1 0
.x .x .. ..    | 0 4 | 0  0 2  2 | *  *  *  * 6 * | 0  2  0 0  0 0 2 0 | 2 1 0 0 1
.. .x3.o ..    | 0 3 | 0  0 0  3 | *  *  *  * * 8 | 0  0  0 1  0 0 1 1 | 1 0 0 1 1
---------------+-----+-----------+----------------+--------------------+----------
.. o.3o.3x.    | 4 0 | 6  0 0  0 | 4  0  0  0 0 0 | 1  *  * *  * * * * | 0 0 0 2 0 tet
ox ox .. ..&#x | 1 4 | 0  4 2  2 | 0  2  2  0 1 0 | * 12  * *  * * * * | 1 1 0 0 0 squippy
ox .. .. xo&#x | 2 2 | 1  4 1  0 | 0  2  0  2 0 0 | *  * 12 *  * * * * | 0 1 1 0 0 tet
.. ox3oo ..&#x | 1 3 | 0  3 0  3 | 0  0  3  0 0 1 | *  *  * 8  * * * * | 1 0 0 1 0 tet
.. ox .. xo&#x | 2 2 | 1  4 0  1 | 0  0  2  2 0 0 | *  *  * * 12 * * * | 0 1 0 1 0 tet
.. .. oo3xo&#x | 3 1 | 3  3 0  0 | 1  0  0  3 0 0 | *  *  * *  * 8 * * | 0 0 1 1 0 tet
.x .x3.o ..    | 0 6 | 0  0 3  6 | 0  0  0  0 3 2 | *  *  * *  * * 4 * | 1 0 0 0 1 trip
.. .x3.o3.o    | 0 4 | 0  0 0  6 | 0  0  0  0 0 4 | *  *  * *  * * * 2 | 0 0 0 1 1 tet
---------------+-----+-----------+----------------+--------------------+----------
ox ox3oo ..&#x | 1 6 | 0  6 3  6 | 0  3  6  0 3 2 | 0  3  0 2  0 0 1 0 | 4 * * * * trippy
ox ox .. xo&#x | 2 4 | 1  8 2  2 | 0  4  4  4 1 0 | 0  2  2 0  2 0 0 0 | * 6 * * * squippypy
ox .. oo3xo&#x | 3 2 | 3  6 1  0 | 1  3  0  6 0 0 | 0  0  3 0  0 2 0 0 | * * 4 * * pen
.. ox3oo3xo&#x | 4 4 | 6 12 0  6 | 4  0 12 12 0 4 | 1  0  0 4  6 4 0 1 | * * * 2 * hex
.x .x3.o3.o    | 0 8 | 0  0 4 12 | 0  0  0  0 6 8 | 0  0  0 0  0 0 4 2 | * * * * 1 tepe

I.e. it results in a nice tiny polyteron with facet set: 4x trippy + 6x squippypy + 4x pen + 2x hex + 1x tepe.

--- rk
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Re: Johnsonian Polytopes

Postby quickfur » Tue Jan 07, 2014 1:31 am

Klitzing wrote:
quickfur wrote:For example, I just figured out last night that the product of a cube (as convex_hull(tet, dual_tet)) with a triangle (as point||line) can produce the topological equivalent of two 16-cells sharing a cell, or a tetra-augmented tetrahedral prism (probably cannot be made CRF, though).


The compound, you are looking for, thus is ox xo3oo3ox.
(Or, taken in the language of segmentotopes, even when disregarding height and lacings, it would look like tet || prism of dual tet.)

I think you are still misunderstanding my construction. The result is a 4D polychoron, the convex hull of a tetrahedral prism and a dual tetrahedron that intersects the midplane of the prism.

The original motivation for student91's segmentotopic product is a modified Cartesian product in the following sense: the traditional Cartesian product A x B may be understood as "substituting" each point in A with a copy of B in an orthogonal space. His idea was to modify this construction so that the points in the top layer C of a segmentotope C||D would be "substituted" with copies of one polytope, E, whereas the points in the bottom layer D would be "substituted" with copies of another polytope F. This modified Cartesian product would that produce E||F as a surtope due to the original C||D having lacing edges between points in C and D, which have now been replaced with copies of E and F, respectively, in the orthogonal space.

My construction is a different development of this idea: instead of choosing all points in one layer to be "substituted" with copies of the first operand E, and choosing all the points in another layer to be "substituted" with copies of the second operand F, first I start with an even polytope, and then mark each vertex as red or black (i.e., consider the alternations of the polytope). Then I replace red vertices with copies of E in the orthogonal space, and black vertices with copies of F in the orthogonal space. The convex hull operation is really just a simple way to interpolate what shape gets substituted in points between red and black vertices. The idea behind this is that since red and black vertices alternate, the resulting polytope would contain E||F surtopes (not necessarily (n-1)-dimensional) produced by the adjacency of E's (substituting red vertices) and F's (substituting black vertices).

More formally, given 3 polytopes A of dimension p, B of dimension q, and C of dimension r, where A is alternable, mark the vertices of A as red/black such that every red vertex is adjacent to a black vertex, and vice versa. Assume without loss of generality that q<r. Then construct the result R in (p+r)-dimensional space by placing copies of B at the points (x1,x2,...xp,0,0,0,...0), for each red vertex (x1,x2,...xp) of A, and copies of C at the analogous points for each black vertex of A, with the B's and C's oriented in the r-dimensional hyperplane orthogonal to A; then interpolate between these "expanded" vertices by taking the convex hull of the result.

In other words, the result R is the convex hull of the union of (A x B) and (A x C) in (p+r)-dimensional space.
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Re: Johnsonian Polytopes

Postby Klitzing » Tue Jan 07, 2014 2:31 pm

quickfur wrote:I think you are still misunderstanding my construction. The result is a 4D polychoron, ...

Not at all, quickfur.

The mere 4D compound (without that convex hull stuff) just is "ox ox3oo3xo". Note that we have 4 node positions, and there is no trailing "&#x" added, that is, it truely is meant to be read as a coplanar compound - without any further shift in orthogonal direction.

But, what I was pointing out in my mailing, that the lacing edges, which then will occur in the hull procedure, come out to be shorter than unity. (Where unity is the edge length of both components.) Therefore it will be allowed to add some shift into 5D, separating these components, and construct thereby a true (likewise convex) segmentoteron, i.e. a 5D figure.

The thus obtained 5D figure was then detailed in the remainder of my post: a tiny polyteron with only 12 vertices and 17 facets.

--- rk
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Re: Johnsonian Polytopes

Postby Klitzing » Tue Jan 07, 2014 3:22 pm

quickfur wrote:More formally, given 3 polytopes A of dimension p, B of dimension q, and C of dimension r, where A is alternable, mark the vertices of A as red/black such that every red vertex is adjacent to a black vertex, and vice versa. Assume without loss of generality that q<r. Then construct the result R in (p+r)-dimensional space by placing copies of B at the points (x1,x2,...xp,0,0,0,...0), for each red vertex (x1,x2,...xp) of A, and copies of C at the analogous points for each black vertex of A, with the B's and C's oriented in the r-dimensional hyperplane orthogonal to A; then interpolate between these "expanded" vertices by taking the convex hull of the result.

In other words, the result R is the convex hull of the union of (A x B) and (A x C) in (p+r)-dimensional space.


Two minor errors in that part.
  • without loss of generality you'd have only q<=r.
  • you'd mean the convex hull of the union of (A+ x B) and (A- x C).


In order to be CRF you'd have to rescale both A+ and A- back to unit edges, that is, for convex A this would ask for factors k with 0<k<=1 (as the shortchords of the A-polygons then are generally larger than unity). This requirement is because most of the edges of A+ x B and of A- x C tend to survive the hull operation. Accordingly the height of those to be used segmentotopes B||C (between a red and a black vertex of A) then would have to be k as well.

But the height of a segmentotope (i.e. having all unit edges) is predefined by the choice of B and C. Therefore your construction, in order to result in a CRF, cannot vary A, B, and C independently!

Rather it requires
  • A shall be alternatable into A+ and A-.
  • A shall have only one type of polygonal faces. (Else the corresponding shortchords would vary and thus no uniform scaling could be defined.)
  • Then only those pairs of B and C could be allowed to be placed alternatingly in perp space, which yield a segmentotopal height which equates to the inverse of that mentioned shortchord.

--- rk
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Re: Johnsonian Polytopes

Postby student91 » Tue Jan 07, 2014 5:20 pm

just stumbled upon K4.109. It is a counterexample to my previous assumption that every ring could be dissected to a prism, cartesian product or lace simplex:
student91 wrote:But those rings (%..%P%..%Q%..%...&#xr), do they occur with more than 4 %'s? Because I think those are either impossible or derived from either a cartesian producted polygon, a prism or a lace simplex. (look, for example, to the triaugmented trigonal prism. it can be seen as oxoxox&#r, but the o's can be cut of, and hence it's not a "new" thing. the same with the triaugmented hexagonal prism oxxoxxoxx&#xr). I think this, because if the ring is bigger than 4, and it exists of more than 1 %P%Q%.., (and it's not a prism) you have to connect a diagonal. if diagonals cross (like oxox4xxxx&#xr in my previous post), you get a degenerate higher-dimensional polytope. If diagonals don't cross, you could cut the shape in two at this diagonal. if you keep cutting in this way, you always get either a cartesian producted polygon, a prism or a lace simplex (because something else could be cut again)

K4.109 has a interesting structure though. first of all it can be written as xxoo4oxxo&#xr. This means, that there is a quadiliteral with on its vertices a o4o, a x4o, a o4x and a x4x possible, in which everything connects properly.
first of all we know the distances x4o||x4x, x4o||o4o, o4x||x4x and o4x||o4o. These are all equal to sqrt(1/2), so the "base"-polytope has to be a rhomb with edge-length sqrt(1/2). Furthermore, the vertices of the x4x and the x4o are connected to the o4o and respectively the o4x with an edge of length sqrt(2)=q (the diagonals of the connecting squares). The height of ox4ox&#q is sqrt((2-sqrt(2))/2), and the height of xo4ox&#q is sqrt((2+sqrt(2))/2). If we give our rhomb a diagonal of one of those lengths, the other length will automatically appear at the other diagonal, so this "base"-polytope is planar.
Code: Select all
x4o    o4o           x4o-x-o4o
                     |\   / |
              =      x  q   x
                     |/   \ |
x4x    o4x          x4x-x-o4x      x=sqrt(1/2), q=sqrt((2+sqrt(2))/2) or sqrt((2-sqrt(2))/2)

This gave me a new idea: maybe rings are possible, if the height of lacings with a diagonal edge all inserted make a planar drawing.
e.g. take a pentagonal ring with polytopes ABCDE (so %%%%%P%%%%%&#xr where the %P%'s are A,B,C etc.) Now suggest this ring is undiminishable. It should be clear that the edges of the "base"-polytope connecting A with B should get lenght height(A||B), and so with B||C, C||D, D||E and E||A. Now there are some diagonals, namely AC, BD, CE, DA and EB, that should also get a length in the "base"-polytope. These lengths should correspond to the heights of %%P%%&#f of all pairs of polytopes. Now if the resulting "base"-polytope is planar, the connections going from A to B to C and so on should generate pentagons, and hence the ring ABCDE should be possible.
when you see rings this way, cartesian products of polygons and something else also generates (trivial) valid rings: lets say we have a ring of length P consisting of identical elements A. (i.e. the cartesian product of a P-sided polygon and A). Now the edges of the "base"-polytope should get the length of height AA&#x, so 1. Furthermore, a shordchord of P with length p should be drawn by a diagonal with length of the height of AA&#p, so p. this means the "base"-polytope becomes exactly polygon P itself. I think it to be probable that a similar thing applies for any subsitution of the vertices of a polytope with other polytopes. (and then the segmentotopical products are a specific case that always works). it needs some working out however.

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Re: Johnsonian Polytopes

Postby quickfur » Tue Jan 07, 2014 8:15 pm

Klitzing wrote:
quickfur wrote:I think you are still misunderstanding my construction. The result is a 4D polychoron, ...

Not at all, quickfur.

The mere 4D compound (without that convex hull stuff) just is "ox ox3oo3xo". Note that we have 4 node positions, and there is no trailing "&#x" added, that is, it truely is meant to be read as a coplanar compound - without any further shift in orthogonal direction.

But, what I was pointing out in my mailing, that the lacing edges, which then will occur in the hull procedure, come out to be shorter than unity. (Where unity is the edge length of both components.) Therefore it will be allowed to add some shift into 5D, separating these components, and construct thereby a true (likewise convex) segmentoteron, i.e. a 5D figure.

The thus obtained 5D figure was then detailed in the remainder of my post: a tiny polyteron with only 12 vertices and 17 facets.

--- rk

Ah, I see what you're getting at now. Sorry. :oops:

Klitzing wrote:
quickfur wrote:More formally, given 3 polytopes A of dimension p, B of dimension q, and C of dimension r, where A is alternable, mark the vertices of A as red/black such that every red vertex is adjacent to a black vertex, and vice versa. Assume without loss of generality that q<r. Then construct the result R in (p+r)-dimensional space by placing copies of B at the points (x1,x2,...xp,0,0,0,...0), for each red vertex (x1,x2,...xp) of A, and copies of C at the analogous points for each black vertex of A, with the B's and C's oriented in the r-dimensional hyperplane orthogonal to A; then interpolate between these "expanded" vertices by taking the convex hull of the result.

In other words, the result R is the convex hull of the union of (A x B) and (A x C) in (p+r)-dimensional space.


Two minor errors in that part.
  • without loss of generality you'd have only q<=r.
  • you'd mean the convex hull of the union of (A+ x B) and (A- x C).

You're right, it should be q≤r, not q<r. And also on the second point. :) I was in a hurry and didn't check what I wrote. :oops:

In order to be CRF you'd have to rescale both A+ and A- back to unit edges, that is, for convex A this would ask for factors k with 0<k<=1 (as the shortchords of the A-polygons then are generally larger than unity). This requirement is because most of the edges of A+ x B and of A- x C tend to survive the hull operation. Accordingly the height of those to be used segmentotopes B||C (between a red and a black vertex of A) then would have to be k as well.

But the height of a segmentotope (i.e. having all unit edges) is predefined by the choice of B and C. Therefore your construction, in order to result in a CRF, cannot vary A, B, and C independently!

Right, I wasn't expecting that the result can be made CRF in the general case. It was just a way to produce topological polytopes that might suggest CRF possibilities, but obviously many cases won't be CRF-able.

Rather it requires
  • A shall be alternatable into A+ and A-.
  • A shall have only one type of polygonal faces. (Else the corresponding shortchords would vary and thus no uniform scaling could be defined.)
  • Then only those pairs of B and C could be allowed to be placed alternatingly in perp space, which yield a segmentotopal height which equates to the inverse of that mentioned shortchord.

--- rk

Thanks for working out these conditions for me. :lol: I was going to study this a little more but you beat me to it. :D

In any case, this does suggest some interesting possibilties -- A can be a snub disphenoid, for example -- but I'm not sure if it's possible to choose B and C such that a CRF would be produced (other than setting B = C = line segment, in which case you just get the snub disphenoid prism, which isn't that interesting).

Anyway, I've been exploring these ideas as ways of attacking the CRF crown jewel problem, but so far it seems that none of them have yielded any promising leads. :\

On the other hand, though, I've recently noticed that o5o3x3x appears to admit a bathodiminishing with a x5o3x cell, which could mean that the bipentacyclodiminishing might also be possible (with parabidiminished x5o3x cells)? Have you considered this case when we last discussed the 120-cell family diminishings with 5,5-duoprism symmetry? I admit I haven't been able to keep up with the number of incmats you've been posting, :oops: so I may have missed it.
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Re: Johnsonian Polytopes

Postby Klitzing » Tue Jan 07, 2014 8:36 pm

Dear student91,

you probably should consider to take a look at Wendy's lace cities. Those in fact are exactly what are you looking for all the way...

A lace prism can be stacked to provide lace towers. But then several such lace towers in turn can be stacked in a different direction to provide what is a lace city.


I'll provide here some easy to visualize 3D examples. But the same holds true for higher dimensions as well. Then the orthogonal space becomes D-2 dimensional.
Code: Select all
x x    <- square
x x    <- square

^ ^    : cube
| |
s s
q q
u u
a a
r r
e e


Code: Select all
  x x     <- square
x w w x   <- octagon
x w w x   <- octagon
  x x     <- square

^ ^ ^ ^   : sirco
| | | |
s o o s
q c c q
u t t u
a a a a
r g g r
e o o e
  n n 


Code: Select all
x w   w x   <- octagon
w       w   <- w-square
         
w       w   <- w-square
x w   w x   <- octagon

^ ^   ^ ^   : tic
| |   | |
o w   w o
c -   - c
t s   s t
a q   q a
g u   u g
o a   a o
n r   r n
  e   e


And to provide a 4D example: ico could be given either according to its orthogonal 3-fold symmetry as
Code: Select all
   o3x   x3o   
               
x3o   x3x   o3x
               
   o3x   x3o   
or according to its orthogonal 4-fold symmetry as
Code: Select all
o4o  x4o  o4o
             
x4o  o4q  x4o
             
o4o  x4o  o4o


Even the 6D Gosset polytope jak = 2_1,2 could be done this way:
Code: Select all
o
    B
A       o
    C
o
where A, B, C all are hexadecachora, given in the bifurcated Dynkin graph display, i.e. x3o3o *b3o, each time an other end being ringed. (And o represents a single vertex.)


In essence, those all mime exactly what you are trying: to attach different shapes of perp space onto diferent positions of para space!

Thus your "segmentotopal product" not only was known far before your attempt, but moreover extends way beyond your application to segmentotopes (or lace prisms) only!

--- rk
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Re: Johnsonian Polytopes

Postby Klitzing » Tue Jan 07, 2014 8:53 pm

quickfur wrote:On the other hand, though, I've recently noticed that o5o3x3x appears to admit a bathodiminishing with a x5o3x cell, which could mean that the bipentacyclodiminishing might also be possible (with parabidiminished x5o3x cells)? Have you considered this case when we last discussed the 120-cell family diminishings with 5,5-duoprism symmetry? I admit I haven't been able to keep up with the number of incmats you've been posting, :oops: so I may have missed it.


Hmm, suppose that I lack some understandings of your namings. Bathodiminishing = ? Bipentacyclodiminishing = ? - The latter one at least is descriptive, but still neither selfexplaning nor unique I fear.

In tex = x3x3o5o I don't find a x3o5x section. At least I do not see any. The 2 next parallel sections underneath the ike = x3o5 then would be a u-ike = u3o5o (|u| = 2|x|) and then a ti = x3x5o. Any deeper I haven't digged so far in this case.

--- rk
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Re: Johnsonian Polytopes

Postby student91 » Tue Jan 07, 2014 9:26 pm

quickfur wrote:
Rather it requires
  • A shall be alternatable into A+ and A-.
  • A shall have only one type of polygonal faces. (Else the corresponding shortchords would vary and thus no uniform scaling could be defined.)
  • Then only those pairs of B and C could be allowed to be placed alternatingly in perp space, which yield a segmentotopal height which equates to the inverse of that mentioned shortchord.

--- rk

Thanks for working out these conditions for me. :lol: I was going to study this a little more but you beat me to it. :D

I just got a new idea that states those conditions don't have to be true (most of the time they are, though), although, I think I might be able to prove that rings, cartesian products and segmentotopical products are the only "subsitute polytopes", or at least I might restrict the possible "substitute polytopes" drastically. My previous post tells you something about this idea.
In any case, this does suggest some interesting possibilties -- A can be a snub disphenoid, for example -- but I'm not sure if it's possible to choose B and C such that a CRF would be produced (other than setting B = C = line segment, in which case you just get the snub disphenoid prism, which isn't that interesting).

Note that you could also set B=line, C=orthogonal line, yielding a (n+1)-dimensional demicube, although that's not that special either.

just read Klitzings new post about lace cities.
I think lace cities are not exactly the same as "subsitute polytopes". "subsitute polytopes" use two different symmetries to gain a polytope that also has those two symmetries.
lace cities, as far as I get them, are a way of displaying how parts of a polytope with the same symmetry are connected. those lace cities consist most of the time of "subsituted" simplices, polygons or hypercubes connected edge to edge (the 3-representatation of the 24-cell can be seen as 6× xxo3xox&#x joined at their edges). My segmetotopical products, on the contrary, don't consist of simplices, polygons and hypercubes, but have much complexer structures. look, for example at the segmentotopical product xo42xx¤xx5ox&#x (so pentagon×square||decagon×line). As a lace city this would be represented as (i guess):
Code: Select all
                   o2x
         o2x                 o2x
                     x2x
o2x        x2x                      o2x
                            x2x
o2x        x2x                      o2x
                     x2x
       o2x                    o2x
                 o2x

as a lace city this would be seen as a x2x×o5x augmented with (x2x||x2x)||(o2x||o2x), and closed by (o2x||o2x)||x2x. This would be a 4D-polytope. My polytope would be 5D, and also have a o2x×x5x and a lot of polytopes with .5.-symmetry.
Last edited by student91 on Thu Jan 09, 2014 6:54 pm, edited 1 time in total.
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Re: Johnsonian Polytopes

Postby student91 » Tue Jan 07, 2014 11:30 pm

student91 wrote:just stumbled upon K4.109. [...]
I think it to be probable that a similar thing applies for any subsitution of the vertices of a polytope with other polytopes. (and then the segmentotopical products are a specific case that always works). it needs some working out however.


I think I will just post what I've got so far:
Consider a "base"-polytope B with vertices v1, v2, v3,... and a "symmetry group" .P.Q.R. that will be shorthandedly called S
now first of all I will introduce some new specific notations:
h(xoPoxQoxRxx&#x) means the height of segmentotope xPoQoRx||oPxQxRx.
l(v1,v5) means the length of the line connecting v1 with v5.
S(v1) means the polytope with symmetry S with which v1 should be subsituted
S(v2,v4)&#x means the polytope of symmetry group S with what v2 should be subsituted, atop the polytope with what v4 should be subsituted
so h(S(v3,v6)&#l(v3,v6)) means the height of the "segmentotope:" the polytope on v3, atop the polytope on v6, with a lacing edge equal to the length of the line connecting v3 with v6.

now if we want to subsitute multiple vertices of B with different polytopes of S, I think we want copies of B to occur. (something else would probably not yield CRF's). Now B can be defined by it's set of vertices, and the length of the connections between those. (so {v1,v2,...} and {l(v1,v2),l(v1,v3),l(v1,v4),...,l(v2,v3),l(v2,v4),...} define B). actually the lengths imply the vertices, so we can take only the set {l(v1,v2),l(v1,v3),l(v1,v4),...,l(v2,v3),l(v2,v4),...} to define B.

now if we are going to subsitute every vertex vn of B with S(vn), we have to modify the "base"-polytope B so that the connecting lines will be of length h(S(vn,vm)&#l(vn,vm)). If we do this for every two vertices vn and vm, and then place the polytopes S(v1),S(v2),..., the new polytope should be (C)RF, and have lots of B's as facets (because a lot of sets {l(v1,v2),l(v1,v3),l(v1,v4),...,l(v2,v3),l(v2,v4),...} will occur}).

Note that the request of every diagonal to be h(S(vn,vm)&#l(vn,vm)) is very demanding, and most polytopes B will not be able to undergo such an operation. (unless all S(vn)'s are equal, this would make every diagonal be the same length as they were previously, and construct a cartesian product.)
EDIT: furthermore, h(S(vn,vm)&#l(vn,vm)) is determined by h(S(vn,vm)&#x),through the formula: sqrt(l(vn,vm)² + h(S(vn,vm)&#x)² - 1). this will restrict the possible deformations of B even more.

the operations mentioned above are not violated by any known (at least to me) examples.
I hope my cryptix has been clear enough

student91
Last edited by student91 on Tue Jan 07, 2014 11:47 pm, edited 2 times in total.
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Re: Johnsonian Polytopes

Postby quickfur » Tue Jan 07, 2014 11:32 pm

Klitzing wrote:
quickfur wrote:On the other hand, though, I've recently noticed that o5o3x3x appears to admit a bathodiminishing with a x5o3x cell, which could mean that the bipentacyclodiminishing might also be possible (with parabidiminished x5o3x cells)? Have you considered this case when we last discussed the 120-cell family diminishings with 5,5-duoprism symmetry? I admit I haven't been able to keep up with the number of incmats you've been posting, :oops: so I may have missed it.


Hmm, suppose that I lack some understandings of your namings. Bathodiminishing = ? Bipentacyclodiminishing = ? - The latter one at least is descriptive, but still neither selfexplaning nor unique I fear.

Sorry, I should have explained my terminology instead of assuming people will understand. "Bathodiminishing" is a term I coined from Greek "batho-" meaning "deep", + diminishing, which I use to indicate a diminishing where the cutting hyperplane intersects the polytope deeper than a (presumably obvious) shallower diminishing.

In this case, the most obvious diminishing of o5o3x3x is the one that cuts the polytope 1 edge length below an icosahedral cell, so the cut-off piece consists of 1 icosahedron, 20 octahedra, and 12 pentagonal pyramids, plus a o5x3o that lies on the cutting plane.

However, I notice that there is a deeper cut possible (hence bathodiminishing), in which the cutting hyperplane truncates the layer of 12 icosahedra beneath the top-most one (ie., the one parallel to the cutting hyperplane), such that they get cut into truncated icosahedra (aka gyroelongated pentagonal pyramids). This cutting also splits 30 of the octahedral cells into square pyramids, so the cut-off piece consists of 1 icosahedron, some number of octahedra (I didn't count them exactly), 12 truncated icosahedra, and 30 square pyramids, plus a x5o3x lying on the cutting plane. So the remaining part of the o5o3x3x would have a x5o3x where the cut was made, plus 30 square pyramids (the other halves of the octahedra that got cut off) and 12 pentagonal pyramids.

Now, we know that the icosahedral cells of o5o3x3x can be grouped into rings, connected by their top and bottom vertices, such that they form great circles of 10 icosahedra each. If we look at a single ring of icosahedra, then we see that the bathodiminishing removes one icosahedron (let's call that the "first" one) and truncates the two adjacent to it in the ring (i.e., the 2nd and 10th icosahedra). This suggests that maybe it's possible to perform a second bathodiminishing, centered on the 3rd icosahedron in the ring, which would completely remove the pentagonal pyramid remaining from the 2nd icosahedron, and also truncate the 4th icosahedron. Since there's originally a ring of 10 icosahedra and multiple bathodiminishings seem to be permissible as long as they remove non-adjacent icosahedra in the ring, we can perform a maximum of 5 bathodiminishings in a single ring. The x5o3x's resulting from each cut would, of course, overlap with each other; but since x5o3x's can be bidiminished, it seems to suggest that after these 5 bathodiminishings (I think it's clear by now that this is the cyclo-pentadiminishing I was referring to), we should end up with a polytope containing a ring of 5 parabidiminished x5o3x's.

This is only a single cyclopentadiminishing, of course; we know that there's another ring of 10 icosahedra lying on the plane orthogonal to the plane of the first ring. These two rings seem to be separated far enough that it should be possible to also perform a cyclopentadiminishing on the second ring. So this would be a bi-penta-cyclodiminished o5o3x3x, which, if my conjecture about the possibility of the cyclo-pentadiminishing is correct, should contain two orthogonal rings of 5 parabidiminished x5o3x's with 5,5-duoprism symmetry. Furthermore, if my assumptions are correct, then this should also be CRF, and probably will contain some number of pentagonal pyramids and square pyramids in addition to the two rings of parabidiminished x5o3x's.
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Re: Johnsonian Polytopes

Postby Klitzing » Tue Jan 07, 2014 11:50 pm

student91 wrote:for example at the segmentotopical product xo4xx¤xx5ox&#x (so pentagon×square||decagon×line).


xo4xx&#x # xx5ox&#x := (x4x X x5o) || (o4x X x5x) = xo4xx2xx5ox&#x
(Esp. neither square nor line, rather octagon and square, :P )

Note that the rightmost display shows that there is no need for any funny product sign whatsoever. In fact it is just a further link mark 2 of the total Dynkin diagram.

Finally, this very choice of yours, xo4xx2xx5ox&#x, would calculate a purely imaginary height if all edges should be taken unity. Thus it could be implemented in hyperbolic geometry only. (Else one should replace "&#x" by "&#y", i.e. with some different (longer) lacing edges.)


as a lace city ... . This would be a 4D-polytope. My polytope would be 5D,

The lace city stuff indeed is more meant in the product sense (placing different objects in perp space). The additional height of segmentotopes there indeed is not yet included.

--- rk
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Re: Johnsonian Polytopes

Postby student91 » Tue Jan 07, 2014 11:59 pm

Klitzing wrote:
student91 wrote:for example at the segmentotopical product xo4xx¤xx5ox&#x (so pentagon×square||decagon×line).


xo4xx&#x # xx5ox&#x := (x4x X x5o) || (o4x X x5x) = xo4xx2xx5ox&#x
(Esp. neither square nor line, rather octagon and square, :P )

oops, ment a trigonal prism (xo2xx&#x) :oops: , then the height would be real. ( :oops: :oops: )
Note that the rightmost display shows that there is no need for any funny product sign whatsoever. In fact it is just a further link mark 2 of the total Dynkin diagram.

I see, hadn't thought of the symmetry of a duoprism before.
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Re: Johnsonian Polytopes

Postby Klitzing » Wed Jan 08, 2014 12:46 am

Ah, quickfur, you tricked me!
You always write x3x3o5o = tex,
but you mean instead o3x3o5o = rox

The ike-first series of vertex layers of rox indeed starts with: x3o5o || o3x5o || x3o5x || ...
That is, you speak of a bistratic diminishing here ("bathodiminishing").

Btw. here is the lace city of rox:
Code: Select all
                                            o5o                                           
                                    x5o             x5o                                   
                                            o5f                                           
                        o5x                                     o5x                       
                o5o             x5x                     x5x             o5o               
                                                                                          
                    f5o                     x5f                     f5o                          F=ff=x+f=2x+v,
            o5x                     F5o             F5o                     o5x                  V=F+v=2f=2x+2v
                                o5F                     o5F                               
                        f5x                                     f5x                       
                                                                                          
            x5x                     f5f             f5f                     x5x           
    x5o             o5F                     V5o                     o5F             x5o   
                                                                                          
                F5o             o5V                     o5V             F5o               
o5o                     f5f                                     f5f                     o5o
    o5f                                     x5F                                     o5f   
            x5f                     F5x             F5x                     x5f           
                                                                                          
x5o                     V5o                                     V5o                     x5o
                                x5F                     x5F                               
            F5o                                                             F5o           
    x5x             f5f                     F5F                     f5f             x5x   
            o5F                                                             o5F           
                                F5x                     F5x                               
o5x                     o5V                                     o5V                     o5x
                                                                                          
            f5x                     x5F             x5F                     f5x           
    f5o                                     F5x                                     f5o   
o5o                     f5f                                     f5f                     o5o
                o5F             V5o                     V5o             o5F               
                                                                                          
    o5x             F5o                     o5V                     F5o             o5x   
            x5x                     f5f             f5f                     x5x           
                                                                                          
                        x5f                                     x5f                       
                                F5o                     F5o                               
            x5o                     o5F             o5F                     x5o           
                    o5f                     f5x                     o5f                   
                                                                                          
                o5o             x5x                     x5x             o5o               
                        x5o                                     x5o                       
                                            f5o                                           
                                    o5x             o5x                                   
                                            o5o                                           


And yes, 5 such bistratic diminishings could be chosen in a pentagonal fashion, i.e. cycling around. Then adjacent cuts will be that deep, so they would intersect. Therefore indeed parabidiminished rhombicosidodecahedra would result as those facets. - Below the resulting cypdrox (cyclically penta-diminished rox) is displayed as lace city:
Code: Select all
                                            x5x                           
                                                                          
                                x5f                                              F=ff=x+f=2x+v,
                        F5o             F5o                                      V=F+v=2f=2x+2v
                    o5F                     o5F                           
            f5x                                     f5x                   
                                                                          
x5x                     f5f             f5f                               
        o5F                     V5o                     o5F               
                                                                          
    F5o             o5V                     o5V             F5o           
            f5f                                     f5f                   
                                x5F                                       
x5f                     F5x             F5x                     x5f       
                                                                          
            V5o                                     V5o                   
                    x5F                     x5F                           
F5o                                                             F5o       
        f5f                     F5F                     f5f             x5x
o5F                                                             o5F       
                    F5x                     F5x                           
            o5V                                     o5V                   
                                                                          
f5x                     x5F             x5F                     f5x       
                                F5x                                       
            f5f                                     f5f                   
    o5F             V5o                     V5o             o5F           
                                                                          
        F5o                     o5V                     F5o               
x5x                     f5f             f5f                               
                                                                          
            x5f                                     x5f                   
                    F5o                     F5o                           
                        o5F             o5F                               
                                f5x                                       
                                                                          
                                            x5x                           

Whether so a bi-cyclical penta-diminishing, i.e. a corresponding diminishing at 2 orthogonal rings of 5, would exist, resp. how those would or not would intersect each other, eludes me.

Yes I had considered formerly the bicyclical pentadiminishings for rahi (bistratic) resp. for srix (tristratic). Those were just touching, no intersections. But rox so far has not been considered so far. - Thank you for that one.

--- rk
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Re: Johnsonian Polytopes

Postby Klitzing » Wed Jan 08, 2014 12:58 am

Okay, h( xo2xx2xx5ox&#x ) = sqrt[(5-2 sqrt(5))/20] = 0.162460.
Two 2's look not too interesting so.

But you could consider instead:
h( xo3xx2xx4ox&#x ) = 1/sqrt(6) = 0.408248.

Or even:
h( xo3xx2xx3ox&#x ) = 1/sqrt(3) = 0.577350.

--- rk
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Re: Johnsonian Polytopes

Postby quickfur » Wed Jan 08, 2014 1:05 am

Klitzing wrote:Ah, quickfur, you tricked me!
You always write x3x3o5o = tex,
but you mean instead o3x3o5o = rox

:o_o: You're right, I made a mistake!! It's supposed to be o5o3x3o, not o5o3x3x. :oops: :oops: :oops:

The ike-first series of vertex layers of rox indeed starts with: x3o5o || o3x5o || x3o5x || ...
That is, you speak of a bistratic diminishing here ("bathodiminishing").

Btw. here is the lace city of rox:
Code: Select all
                                            o5o                                           
                                    x5o             x5o                                   
                                            o5f                                           
                        o5x                                     o5x                       
                o5o             x5x                     x5x             o5o               
                                                                                          
                    f5o                     x5f                     f5o                          F=ff=x+f=2x+v,
            o5x                     F5o             F5o                     o5x                  V=F+v=2f=2x+2v
                                o5F                     o5F                               
                        f5x                                     f5x                       
                                                                                          
            x5x                     f5f             f5f                     x5x           
    x5o             o5F                     V5o                     o5F             x5o   
                                                                                          
                F5o             o5V                     o5V             F5o               
o5o                     f5f                                     f5f                     o5o
    o5f                                     x5F                                     o5f   
            x5f                     F5x             F5x                     x5f           
                                                                                          
x5o                     V5o                                     V5o                     x5o
                                x5F                     x5F                               
            F5o                                                             F5o           
    x5x             f5f                     F5F                     f5f             x5x   
            o5F                                                             o5F           
                                F5x                     F5x                               
o5x                     o5V                                     o5V                     o5x
                                                                                          
            f5x                     x5F             x5F                     f5x           
    f5o                                     F5x                                     f5o   
o5o                     f5f                                     f5f                     o5o
                o5F             V5o                     V5o             o5F               
                                                                                          
    o5x             F5o                     o5V                     F5o             o5x   
            x5x                     f5f             f5f                     x5x           
                                                                                          
                        x5f                                     x5f                       
                                F5o                     F5o                               
            x5o                     o5F             o5F                     x5o           
                    o5f                     f5x                     o5f                   
                                                                                          
                o5o             x5x                     x5x             o5o               
                        x5o                                     x5o                       
                                            f5o                                           
                                    o5x             o5x                                   
                                            o5o                                           


And yes, 5 such bistratic diminishings could be chosen in a pentagonal fashion, i.e. cycling around. Then adjacent cuts will be that deep, so they would intersect. Therefore indeed parabidiminished rhombicosidodecahedra would result as those facets. - Below the resulting cypdrox (cyclically penta-diminished rox) is displayed as lace city:
Code: Select all
                                            x5x                           
                                                                          
                                x5f                                              F=ff=x+f=2x+v,
                        F5o             F5o                                      V=F+v=2f=2x+2v
                    o5F                     o5F                           
            f5x                                     f5x                   
                                                                          
x5x                     f5f             f5f                               
        o5F                     V5o                     o5F               
                                                                          
    F5o             o5V                     o5V             F5o           
            f5f                                     f5f                   
                                x5F                                       
x5f                     F5x             F5x                     x5f       
                                                                          
            V5o                                     V5o                   
                    x5F                     x5F                           
F5o                                                             F5o       
        f5f                     F5F                     f5f             x5x
o5F                                                             o5F       
                    F5x                     F5x                           
            o5V                                     o5V                   
                                                                          
f5x                     x5F             x5F                     f5x       
                                F5x                                       
            f5f                                     f5f                   
    o5F             V5o                     V5o             o5F           
                                                                          
        F5o                     o5V                     F5o               
x5x                     f5f             f5f                               
                                                                          
            x5f                                     x5f                   
                    F5o                     F5o                           
                        o5F             o5F                               
                                f5x                                       
                                                                          
                                            x5x                           

Whether so a bi-cyclical penta-diminishing, i.e. a corresponding diminishing at 2 orthogonal rings of 5, would exist, resp. how those would or not would intersect each other, eludes me.

Yes I had considered formerly the bicyclical pentadiminishings for rahi (bistratic) resp. for srix (tristratic). Those were just touching, no intersections. But rox so far has not been considered so far. - Thank you for that one.

--- rk

In fact, I have just found out that o5o3x3o (probably!) admits an even deeper cut (at least tristratic, I'm not sure but it may be even more than that) -- this cut removes 33 icosahedral cells (1+12+20), cuts gyroelongated pentagonal pyramids off another 12 icosahedra, and leaves behind an almost-bisected o5o3x3o containing a truncated dodecahedron, 12 pentagonal cupolae, and a bunch of square & pentagonal pyramids.

The cut-off piece looks to be a Stott expansion of the hemi-600-cell ("bisected" 600-cell with pentagonal pyramids inserted in place of the florets of tetrahedra that got bisected by the cutting hyperplane); it contains 33 icosahedral cells and 12 gyroelongated pentagonal pyramids, with a bunch of square pyramids, octahedra, and a truncated dodecahedron.

If we perform this very-deep cut on both sides of the o5o3x3o, then what's left is a kind of "meatball sandwich" made of two truncated dodecahedra lying in parallel hyperplanes, with 30 icosahedra ("meatballs") between them along with a bunch of square pyramids, pentagonal cupolae, and octahedra.

Can you confirm if this is actually CRF? (Hopefully I didn't make a horrible mistake again this time. :sweatdrop: )
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Re: Johnsonian Polytopes

Postby Klitzing » Wed Jan 08, 2014 1:26 am

Yes there is a single very deep cut too, which again has only unit edges, that tid.

The according paradiminishing then looks like this:
Code: Select all
            o5o           
    x5o             x5o   
            o5f           
                           
x5x                     x5x
                           
            x5f                  F=ff=x+f=2x+v,
    F5o             F5o          V=F+v=2f=2x+2v
o5F                     o5F
                           
                           
    f5f             f5f   
            V5o           
                           
o5V                     o5V
                           
            x5F           
    F5x             F5x   
                           
                           
x5F                     x5F
                           
            F5F           
                           
F5x                     F5x
                           
                           
    x5F             x5F   
            F5x           
                           
V5o                     V5o
                           
            o5V           
    f5f             f5f   
                           
                           
F5o                     F5o
    o5F             o5F   
            f5x           
                           
x5x                     x5x
                           
            f5o           
    o5x             o5x   
            o5o           

or just: o3x5x || f3x5o || (V3o5o + x3o5f) || f3x5o || o3x5x.
For sure it is CRF.

--- rk
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Re: Johnsonian Polytopes

Postby quickfur » Wed Jan 08, 2014 6:25 am

Klitzing wrote:Yes there is a single very deep cut too, which again has only unit edges, that tid.

The according paradiminishing then looks like this:
Code: Select all
            o5o           
    x5o             x5o   
            o5f           
                           
x5x                     x5x
                           
            x5f                  F=ff=x+f=2x+v,
    F5o             F5o          V=F+v=2f=2x+2v
o5F                     o5F
                           
                           
    f5f             f5f   
            V5o           
                           
o5V                     o5V
                           
            x5F           
    F5x             F5x   
                           
                           
x5F                     x5F
                           
            F5F           
                           
F5x                     F5x
                           
                           
    x5F             x5F   
            F5x           
                           
V5o                     V5o
                           
            o5V           
    f5f             f5f   
                           
                           
F5o                     F5o
    o5F             o5F   
            f5x           
                           
x5x                     x5x
                           
            f5o           
    o5x             o5x   
            o5o           

or just: o3x5x || f3x5o || (V3o5o + x3o5f) || f3x5o || o3x5x.
For sure it is CRF.

--- rk

Thanks, Klitzing!

So this means we can stratify o5oxo into a few layers: the first one cut off by the first diminishing is o5ox||o5xo, then there's the layer between the 1st cutting hyperplane and the 2nd cutting hyperplane (o5xo||x5ox, I think), then between the 2nd cutting hyperplane and the 3rd: not sure about the exact structure, it starts with x5ox then there are a bunch of vertex layers before it gets to x5xo. Looks like a new polystratic? This polystratic then is the layer before the "sandwich" / parabidiminishing layer that you posted above, and after that we have the polystratic again (in reverse), then x5ox||o5xo, then o5xo||o5ox at the antipode.
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Re: Johnsonian Polytopes

Postby quickfur » Wed Jan 08, 2014 6:34 am

P.S. Since we established there are 3 different depths at which o5oxo may be diminished, I'm thinking perhaps what I previously called a bathodiminishing should be renamed mesodiminishing, since it's not really the deepest diminishing anymore. So we have:

- diminishing -- cuts off o5ox||o5xo (maybe micro-diminishing for being the smallest diminishing?)
- mesodiminishing -- cuts off o5ox||o5xo||x5ox
- bathodiminishing -- cuts up to a x5xo cross-section.

I'm not sure if there are any other diminishings between meso- and batho-; from a cursory examination, doesn't look like there are, because after the x5ox cross-section, there are 20 icosahedra in a triangle-first orientation, which means there will be no CRF cutting until the opposite triangle face, which is where the bathodiminishing's cutting hyperplane lies.
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Re: Johnsonian Polytopes

Postby Klitzing » Wed Jan 08, 2014 9:01 am

Rox = o3x3o5o = x3o5o || o3x5o || x3o5x || F3o5o || o3f5o || f3o5x || o3x5x || f3x5o || (V3o5o + x3o5f) || f3x5o || o3x5x || f3o5x || o3f5o || F3o5o || x3o5x || o3x5o || x3o5o
(Here I used: f = tau = 1.618, v = 1/tau, F=ff=x+f=2x+v, V=F+v=2f=2x+2v)

--- rk
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Re: Johnsonian Polytopes

Postby student91 » Wed Jan 08, 2014 9:28 am

Note that there's another (quite obvious) diminishing of o3x3o5o: just cut out a single vertex. The resulting gap will be a pentagonal prism. Maybe we can call this a micro-diminishing? :D

There might be also some octahedron-centered diminishings. The octahedron should be seen as a tetatetrahedron, as it's faces are alternated. so it should be seen as o3x3o.
now a diminishing would look like something as o3x3o||f3o3x||....||something cut off. The something should have tetahedral symmetry as well (I guess). and furthermore, it should have only unit-edges, so it should be one of the tetahedral uniforms. therefore, I think we could also look for diminishings in the following way: take a instance of the CD-diagram (o3x3o5o), e.g. o3x3o.. now this is a tetratetrahedron, so we could look for tetrahedral uniforms in the octahedral stacking of the polytope. these should occur on the hypersphere of o3x3o5o, so we could just look if they occur at the place where there is a sphere with radius corresponding to the diameter of the supposed tetrahedral uniform. Now the o3x3o5o exceeds the circumradius of a x3x3x quite quickly, without having shown a tetahedral uniform, so a octahedral diminishing is impolssible. The other surtope o...o5o is exceeded almost immediately after the pentagonal prism.
Is this way of looking at it useful? :D

student91
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Re: Johnsonian Polytopes

Postby student91 » Wed Jan 08, 2014 9:37 am

furthermore, I found a (new?) diminishing of the cantellated 600-cell (o3x3x5o x3o3x5o). it can be cut off at a x5x3x, cutting a cuboctahedron and an icosidodecahedron in half. (used http://eusebeia.dyndns.org/4d/cant600cell, the x5x3x can be seen in the second picture of the third layer. (your renders are very helpfull, quickfur! :D :D )
EDIT: It is known already, as the expanded icosidodecahedral cupola is known
Last edited by student91 on Wed Jan 08, 2014 11:36 pm, edited 2 times in total.
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Re: Johnsonian Polytopes

Postby student91 » Wed Jan 08, 2014 9:49 am

student91 wrote:
Klitzing wrote:Note that the rightmost display shows that there is no need for any funny product sign whatsoever. In fact it is just a further link mark 2 of the total Dynkin diagram.

I see, hadn't thought of the symmetry of a duoprism before.

thinking of this, I think I should get back on my previous post:
student91 wrote:
quickfur wrote:[...]
So basically your segmentotopic product (A||B) # (C||D) is basically just (A x C)||(B x D) where x denotes the Cartesian product, and # is a placeholder symbol for your "segmentotopic product"? In other words, you take the Cartesian product of the "top" layer of the segmentotope A||B with the "top" layer of the segmentotope C||D, and likewise take the Cartesian product of the respective bottom layers, and then construct a new segmentotope out of the results. Right? And what you're claiming then is, if A||B and C||D exist, then (A x C)||(B x D) must also exist?

Seems reasonable to me, though it doesn't really add anything new. It's just a way of analysing a segmentotope by decomposition into Cartesian products (or conversely, of constructing a higher-dimensional segmentotope from two lower-dimensional ones). Or am I missing something?

You're not missing something. The segmentotopical product is indeed a way to construct segmentotopes from lower-dimensional segmentotopes. However, where those didn't exist in 4d, they will exist in >4d, and thus they are "something new" in the sense that they are a "new" class of segmentotopes that occur in >4d. Furthermore, they're the first segmentotopes I know of that can have multiple symmetry-groups. All the 4d-segmentotopes are 3d-thing ||3d-thing. a 3d-thing can only be gained through cartesian production if you mutiply a line with a polygon (or a point with a polygon, you'd get 2d-things), so the segmentotopical products in 4d are multiplications of a 2d-segmentotope and a 3d-segmentotope. the 2d-segmentotopes are point||line and line||line, so the segmentotopical products do not differ from other segmentotopes in 4d (they're some of the cupolic rings, the (3,N)-duoprisms, and the prisms of the cupola's and prisms). In >5d, however, I think they will be quite different from others.

Because I didn't think of duoprism-symmetries to be existent, I thought my segmentotopical product was something completely new. However, as these do exist, you were right that they're just another way of analyzing duoprism-segmentochora.
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Re: Johnsonian Polytopes

Postby Klitzing » Wed Jan 08, 2014 10:08 am

student91 wrote:Note that there's another (quite obvious) diminishing of o3x3o5o: just cut out a single vertex. The resulting gap will be a pentagonal prism. Maybe we can call this a micro-diminishing? :D

There might be also some octahedron-centered diminishings. The octahedron should be seen as a tetatetrahedron, as it's faces are alternated. so it should be seen as o3x3o.
now a diminishing would look like something as o3x3o||f3o3x||....||something cut off. The something should have tetahedral symmetry as well (I guess). and furthermore, it should have only unit-edges, so it should be one of the tetahedral uniforms. therefore, I think we could also look for diminishings in the following way: take a instance of the CD-diagram (o3x3o5o), e.g. o3x3o.. now this is a tetratetrahedron, so we could look for tetrahedral uniforms in the octahedral stacking of the polytope. these should occur on the hypersphere of o3x3o5o, so we could just look if they occur at the place where there is a sphere with radius corresponding to the diameter of the supposed tetrahedral uniform. Now the o3x3o5o exceeds the circumradius of a x3x3x quite quickly, without having shown a tetahedral uniform, so a octahedral diminishing is impolssible. The other surtope o...o5o is exceeded almost immediately after the pentagonal prism.
Is this way of looking at it useful? :D

student91

Exactly this Investigation has been done already, e.g. cf. http://bendwavy.org/klitzing/incmats/rox.htm, section "Vertex layers": not only the possible uniform polytopes within the section hyperplane are given, in fact all, i.e. using other edge lengths, and for any relevant subsymmetry. Note also the leftmost numbering of layers: within several instances those polytopes cannot given directly, rather they tend to be compounds of several coincident ones.

The same analysis is provided for lots of other polytopes as well, so not all :)
(Feel free to mail me further additions.)

--- rk
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Re: Johnsonian Polytopes

Postby Klitzing » Wed Jan 08, 2014 10:26 am

student91 wrote:furthermore, I found a (new?) diminishing of the cantellated 600-cell (o3x3x5o). it can be cut off at a x5x3x, cutting a cuboctahedron and an icosidodecahedron in half. (used http://eusebeia.dyndns.org/4d/cant600cell, the x5x3x can be seen in the second picture of the third layer. (your renders are very helpfull, quickfur! :D :D )
EDIT: It is known already, as the expanded icosidodecahedral cupola is known


xhi = o3x3x5o is not what you are refering to.
It rather is srix = x3o3x5o.

And yes, along the icosahedral axis there you'd have a sequence starting with:
o3x5o || x3x5o || x3o5f || x3x5x || ..

--- rk
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Re: Johnsonian Polytopes

Postby quickfur » Wed Jan 08, 2014 6:14 pm

student91 wrote:furthermore, I found a (new?) diminishing of the cantellated 600-cell (o3x3x5o). it can be cut off at a x5x3x, cutting a cuboctahedron and an icosidodecahedron in half. (used http://eusebeia.dyndns.org/4d/cant600cell, the x5x3x can be seen in the second picture of the third layer. (your renders are very helpfull, quickfur! :D :D )
EDIT: It is known already, as the expanded icosidodecahedral cupola is known

It should be o5x3o3x (or x3o3x5o as Klitzing said), not o3x3x5o.

But yeah, I derived the 3 diminishings of o5o3x3o that I described in my previous posts just by looking at my own renders. :P It's not always obvious at first, but once you understand what the images represent, it's actually quite easy to visually pick out diminishing possibilities, esp. CRF diminishings where the edge outline already suggests the shape of the cut, then it's just a matter of checking that the cells that protrude through the edge outline will still remain CRF after being partially cut off.
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Re: Johnsonian Polytopes

Postby quickfur » Sat Jan 11, 2014 7:49 am

Klitzing wrote:[...]
And yes, 5 such bistratic diminishings could be chosen in a pentagonal fashion, i.e. cycling around. Then adjacent cuts will be that deep, so they would intersect. Therefore indeed parabidiminished rhombicosidodecahedra would result as those facets. - Below the resulting cypdrox (cyclically penta-diminished rox) is displayed as lace city:
[...snipped...]
Whether so a bi-cyclical penta-diminishing, i.e. a corresponding diminishing at 2 orthogonal rings of 5, would exist, resp. how those would or not would intersect each other, eludes me.

Yes I had considered formerly the bicyclical pentadiminishings for rahi (bistratic) resp. for srix (tristratic). Those were just touching, no intersections. But rox so far has not been considered so far. - Thank you for that one.

--- rk

I have successfully constructed the bi-cyclo-penta-mesodiminished o5o3x3o. Basically, I identified two mutually orthogonal rings of 10 icosahedra each, then picked out 5 non-adjacent icosahedra from each ring. Then for each selected icosahedron, I cut off a o5ox||o5xo||x5ox from the polychoron at that point. These cuttings overlap, so the x5ox's get truncated into parabidiminished x5ox's (Johnson solid #80, J80) in two rings of 5, joined by their decagonal faces. The J80's from each ring also touches a face of a J80 from the other ring, so the rings are quite closely packed.

The resulting CRF polychoron has exactly 300 vertices, 985 faces (10 decagons, 100 pentagons, 175 squares, 700 triangles), exactly 1000 edges, and 285 cells (10 J80's, 100 pentagonal pyramids, 25 octahedra, and 150 square pyramids; no icosahedra are left intact). My polytope viewer program confirms that all edge lengths are equal. Here's a preliminary rendering of it:

Image

Only the parabidiminished x5o3x (J80)'s are shown here; I omitted everything else because I haven't had the time to sit down and do a more careful render yet. :) Part of its 5,5-duoprism symmetry is evident here, in the 5 J80's that wrap around the 3 other J80's visible from this viewpoint. These 3 J80's are part of the second ring of 5 J80's; the other two are on the far side of the polytope so they are not included here (I left clipping on since otherwise the image would be too tangled). Maybe tomorrow morning I'll do a nicer render.

I have had the suspicion that these bi-cyclo-penta-diminished uniform polychora from the 120-cell may exhibit distinct gyrated variants, in which one of the rings of 5 cells may be rotated relative to the other ring to yield two distinct polychora, perhaps enantiomers. But this is yet to be confirmed. It could be the case that the gyrated variants are identical to a rotated copy of the original due to the inherited 120-cell symmetry, then they would not be distinct.

Anyway, here are the full coordinates, in case somebody finds it useful:
Code: Select all
<-(3+sqrt(5)), -(1+sqrt(5)), 0, 0>
<-(3+sqrt(5)), -(3+sqrt(5))/2, -1, (1+sqrt(5))/2>
<-(3+sqrt(5)), -(3+sqrt(5))/2, 1, -(1+sqrt(5))/2>
<-(3+sqrt(5)), -1, -(1+sqrt(5))/2, (3+sqrt(5))/2>
<-(3+sqrt(5)), -1, (1+sqrt(5))/2, -(3+sqrt(5))/2>
<-(3+sqrt(5)), 1, -(1+sqrt(5))/2, (3+sqrt(5))/2>
<-(3+sqrt(5)), 1, (1+sqrt(5))/2, -(3+sqrt(5))/2>
<-(3+sqrt(5)), (3+sqrt(5))/2, -1, (1+sqrt(5))/2>
<-(3+sqrt(5)), (3+sqrt(5))/2, 1, -(1+sqrt(5))/2>
<-(3+sqrt(5)), 1+sqrt(5), 0, 0>
<-(2+sqrt(5)), -(2+sqrt(5)), -1, -1>
<-(2+sqrt(5)), -(2+sqrt(5)), -1, 1>
<-(2+sqrt(5)), -(2+sqrt(5)), 1, -1>
<-(2+sqrt(5)), -(2+sqrt(5)), 1, 1>
<-(2+sqrt(5)), -(5+sqrt(5))/2, 0, -(3+sqrt(5))/2>
<-(2+sqrt(5)), -(5+sqrt(5))/2, 0, (3+sqrt(5))/2>
<-(2+sqrt(5)), -(1+sqrt(5)), -(3+sqrt(5))/2, (1+sqrt(5))/2>
<-(2+sqrt(5)), -(1+sqrt(5)), (3+sqrt(5))/2, -(1+sqrt(5))/2>
<-(2+sqrt(5)), -(3+sqrt(5))/2, -(1+sqrt(5))/2, 1+sqrt(5)>
<-(2+sqrt(5)), -(3+sqrt(5))/2, (1+sqrt(5))/2, -(1+sqrt(5))>
<-(2+sqrt(5)), -(1+sqrt(5))/2, -(1+sqrt(5)), (3+sqrt(5))/2>
<-(2+sqrt(5)), -(1+sqrt(5))/2, 1+sqrt(5), -(3+sqrt(5))/2>
<-(2+sqrt(5)), -1, -1, 2+sqrt(5)>
<-(2+sqrt(5)), -1, 1, -(2+sqrt(5))>
<-(2+sqrt(5)), 0, -(3+sqrt(5))/2, (5+sqrt(5))/2>
<-(2+sqrt(5)), 0, (3+sqrt(5))/2, -(5+sqrt(5))/2>
<-(2+sqrt(5)), 1, -1, 2+sqrt(5)>
<-(2+sqrt(5)), 1, 1, -(2+sqrt(5))>
<-(2+sqrt(5)), (1+sqrt(5))/2, -(1+sqrt(5)), (3+sqrt(5))/2>
<-(2+sqrt(5)), (1+sqrt(5))/2, 1+sqrt(5), -(3+sqrt(5))/2>
<-(2+sqrt(5)), (3+sqrt(5))/2, -(1+sqrt(5))/2, 1+sqrt(5)>
<-(2+sqrt(5)), (3+sqrt(5))/2, (1+sqrt(5))/2, -(1+sqrt(5))>
<-(2+sqrt(5)), 1+sqrt(5), -(3+sqrt(5))/2, (1+sqrt(5))/2>
<-(2+sqrt(5)), 1+sqrt(5), (3+sqrt(5))/2, -(1+sqrt(5))/2>
<-(2+sqrt(5)), (5+sqrt(5))/2, 0, -(3+sqrt(5))/2>
<-(2+sqrt(5)), (5+sqrt(5))/2, 0, (3+sqrt(5))/2>
<-(2+sqrt(5)), 2+sqrt(5), -1, -1>
<-(2+sqrt(5)), 2+sqrt(5), -1, 1>
<-(2+sqrt(5)), 2+sqrt(5), 1, -1>
<-(2+sqrt(5)), 2+sqrt(5), 1, 1>
<-(5+sqrt(5))/2, -(2+sqrt(5)), -(3+sqrt(5))/2, 0>
<-(5+sqrt(5))/2, -(2+sqrt(5)), (3+sqrt(5))/2, 0>
<-(5+sqrt(5))/2, -(3+sqrt(5))/2, 0, -(2+sqrt(5))>
<-(5+sqrt(5))/2, -(3+sqrt(5))/2, 0, 2+sqrt(5)>
<-(5+sqrt(5))/2, 0, -(2+sqrt(5)), (3+sqrt(5))/2>
<-(5+sqrt(5))/2, 0, 2+sqrt(5), -(3+sqrt(5))/2>
<-(5+sqrt(5))/2, (3+sqrt(5))/2, 0, -(2+sqrt(5))>
<-(5+sqrt(5))/2, (3+sqrt(5))/2, 0, 2+sqrt(5)>
<-(5+sqrt(5))/2, 2+sqrt(5), -(3+sqrt(5))/2, 0>
<-(5+sqrt(5))/2, 2+sqrt(5), (3+sqrt(5))/2, 0>
<-(1+sqrt(5)), -(3+sqrt(5)), 0, 0>
<-(1+sqrt(5)), -(2+sqrt(5)), -(1+sqrt(5))/2, -(3+sqrt(5))/2>
<-(1+sqrt(5)), -(2+sqrt(5)), (1+sqrt(5))/2, (3+sqrt(5))/2>
<-(1+sqrt(5)), -(3+sqrt(5))/2, -(2+sqrt(5)), (1+sqrt(5))/2>
<-(1+sqrt(5)), -(3+sqrt(5))/2, 2+sqrt(5), -(1+sqrt(5))/2>
<-(1+sqrt(5)), -(1+sqrt(5))/2, -(3+sqrt(5))/2, 2+sqrt(5)>
<-(1+sqrt(5)), -(1+sqrt(5))/2, (3+sqrt(5))/2, -(2+sqrt(5))>
<-(1+sqrt(5)), 0, 0, -(3+sqrt(5))>
<-(1+sqrt(5)), 0, 0, 3+sqrt(5)>
<-(1+sqrt(5)), (3+sqrt(5))/2, -(2+sqrt(5)), (1+sqrt(5))/2>
<-(1+sqrt(5)), (3+sqrt(5))/2, 2+sqrt(5), -(1+sqrt(5))/2>
<-(1+sqrt(5)), 2+sqrt(5), -(1+sqrt(5))/2, -(3+sqrt(5))/2>
<-(1+sqrt(5)), 2+sqrt(5), -(1+sqrt(5))/2, (3+sqrt(5))/2>
<-(1+sqrt(5)), 2+sqrt(5), (1+sqrt(5))/2, -(3+sqrt(5))/2>
<-(1+sqrt(5)), 2+sqrt(5), (1+sqrt(5))/2, (3+sqrt(5))/2>
<-(3+sqrt(5))/2, -(3+sqrt(5)), -(1+sqrt(5))/2, -1>
<-(3+sqrt(5))/2, -(3+sqrt(5)), (1+sqrt(5))/2, 1>
<-(3+sqrt(5))/2, -(2+sqrt(5)), -(1+sqrt(5)), -(1+sqrt(5))/2>
<-(3+sqrt(5))/2, -(2+sqrt(5)), 1+sqrt(5), (1+sqrt(5))/2>
<-(3+sqrt(5))/2, -(5+sqrt(5))/2, -(2+sqrt(5)), 0>
<-(3+sqrt(5))/2, -(5+sqrt(5))/2, 2+sqrt(5), 0>
<-(3+sqrt(5))/2, -(1+sqrt(5)), -(1+sqrt(5))/2, -(2+sqrt(5))>
<-(3+sqrt(5))/2, -(1+sqrt(5)), (1+sqrt(5))/2, 2+sqrt(5)>
<-(3+sqrt(5))/2, -(1+sqrt(5))/2, -(2+sqrt(5)), 1+sqrt(5)>
<-(3+sqrt(5))/2, -(1+sqrt(5))/2, -1, -(3+sqrt(5))>
<-(3+sqrt(5))/2, -(1+sqrt(5))/2, -1, 3+sqrt(5)>
<-(3+sqrt(5))/2, -(1+sqrt(5))/2, 1, -(3+sqrt(5))>
<-(3+sqrt(5))/2, -(1+sqrt(5))/2, 1, 3+sqrt(5)>
<-(3+sqrt(5))/2, -(1+sqrt(5))/2, 2+sqrt(5), -(1+sqrt(5))>
<-(3+sqrt(5))/2, -1, -(3+sqrt(5)), (1+sqrt(5))/2>
<-(3+sqrt(5))/2, -1, 3+sqrt(5), -(1+sqrt(5))/2>
<-(3+sqrt(5))/2, 1, -(3+sqrt(5)), (1+sqrt(5))/2>
<-(3+sqrt(5))/2, 1, 3+sqrt(5), -(1+sqrt(5))/2>
<-(3+sqrt(5))/2, (1+sqrt(5))/2, -1, -(3+sqrt(5))>
<-(3+sqrt(5))/2, (1+sqrt(5))/2, 1, 3+sqrt(5)>
<-(3+sqrt(5))/2, 1+sqrt(5), -(1+sqrt(5))/2, -(2+sqrt(5))>
<-(3+sqrt(5))/2, 1+sqrt(5), (1+sqrt(5))/2, 2+sqrt(5)>
<-(3+sqrt(5))/2, (5+sqrt(5))/2, -(2+sqrt(5)), 0>
<-(3+sqrt(5))/2, (5+sqrt(5))/2, 2+sqrt(5), 0>
<-(3+sqrt(5))/2, 2+sqrt(5), -(1+sqrt(5)), -(1+sqrt(5))/2>
<-(3+sqrt(5))/2, 2+sqrt(5), -(1+sqrt(5)), (1+sqrt(5))/2>
<-(3+sqrt(5))/2, 2+sqrt(5), 0, -(5+sqrt(5))/2>
<-(3+sqrt(5))/2, 2+sqrt(5), 0, (5+sqrt(5))/2>
<-(3+sqrt(5))/2, 2+sqrt(5), 1+sqrt(5), -(1+sqrt(5))/2>
<-(3+sqrt(5))/2, 2+sqrt(5), 1+sqrt(5), (1+sqrt(5))/2>
<-(1+sqrt(5))/2, -(2+sqrt(5)), -(3+sqrt(5))/2, -(1+sqrt(5))>
<-(1+sqrt(5))/2, -(2+sqrt(5)), (3+sqrt(5))/2, 1+sqrt(5)>
<-(1+sqrt(5))/2, -(1+sqrt(5)), -(2+sqrt(5)), -(3+sqrt(5))/2>
<-(1+sqrt(5))/2, -(1+sqrt(5)), 2+sqrt(5), (3+sqrt(5))/2>
<-(1+sqrt(5))/2, -(3+sqrt(5))/2, -(3+sqrt(5)), -1>
<-(1+sqrt(5))/2, -(3+sqrt(5))/2, -(3+sqrt(5)), 1>
<-(1+sqrt(5))/2, -(3+sqrt(5))/2, -(1+sqrt(5)), -(2+sqrt(5))>
<-(1+sqrt(5))/2, -(3+sqrt(5))/2, 1+sqrt(5), 2+sqrt(5)>
<-(1+sqrt(5))/2, -(3+sqrt(5))/2, 3+sqrt(5), -1>
<-(1+sqrt(5))/2, -(3+sqrt(5))/2, 3+sqrt(5), 1>
<-(1+sqrt(5))/2, -1, -(5+3*sqrt(5))/2, 0>
<-(1+sqrt(5))/2, -1, -(3+sqrt(5))/2, -(3+sqrt(5))>
<-(1+sqrt(5))/2, -1, (3+sqrt(5))/2, 3+sqrt(5)>
<-(1+sqrt(5))/2, -1, (5+3*sqrt(5))/2, 0>
<-(1+sqrt(5))/2, 0, -1, -(5+3*sqrt(5))/2>
<-(1+sqrt(5))/2, 0, 1, (5+3*sqrt(5))/2>
<-(1+sqrt(5))/2, 1, -(5+3*sqrt(5))/2, 0>
<-(1+sqrt(5))/2, 1, -(3+sqrt(5))/2, -(3+sqrt(5))>
<-(1+sqrt(5))/2, 1, (3+sqrt(5))/2, 3+sqrt(5)>
<-(1+sqrt(5))/2, 1, (5+3*sqrt(5))/2, 0>
<-(1+sqrt(5))/2, (3+sqrt(5))/2, -(3+sqrt(5)), -1>
<-(1+sqrt(5))/2, (3+sqrt(5))/2, -(3+sqrt(5)), 1>
<-(1+sqrt(5))/2, (3+sqrt(5))/2, -(1+sqrt(5)), -(2+sqrt(5))>
<-(1+sqrt(5))/2, (3+sqrt(5))/2, 1+sqrt(5), 2+sqrt(5)>
<-(1+sqrt(5))/2, (3+sqrt(5))/2, 3+sqrt(5), -1>
<-(1+sqrt(5))/2, (3+sqrt(5))/2, 3+sqrt(5), 1>
<-(1+sqrt(5))/2, 1+sqrt(5), -(2+sqrt(5)), -(3+sqrt(5))/2>
<-(1+sqrt(5))/2, 1+sqrt(5), 2+sqrt(5), (3+sqrt(5))/2>
<-(1+sqrt(5))/2, 2+sqrt(5), -(3+sqrt(5))/2, -(1+sqrt(5))>
<-(1+sqrt(5))/2, 2+sqrt(5), (3+sqrt(5))/2, 1+sqrt(5)>
<-1, -(3+sqrt(5)), -(3+sqrt(5))/2, -(1+sqrt(5))/2>
<-1, -(3+sqrt(5)), (3+sqrt(5))/2, (1+sqrt(5))/2>
<-1, -(2+sqrt(5)), -(2+sqrt(5)), -1>
<-1, -(2+sqrt(5)), 2+sqrt(5), 1>
<-1, -(3+sqrt(5))/2, -(1+sqrt(5))/2, -(3+sqrt(5))>
<-1, -(3+sqrt(5))/2, (1+sqrt(5))/2, 3+sqrt(5)>
<-1, -(1+sqrt(5))/2, -(3+sqrt(5)), (3+sqrt(5))/2>
<-1, -(1+sqrt(5))/2, 0, -(5+3*sqrt(5))/2>
<-1, -(1+sqrt(5))/2, 0, (5+3*sqrt(5))/2>
<-1, -(1+sqrt(5))/2, 3+sqrt(5), -(3+sqrt(5))/2>
<-1, 0, -(5+3*sqrt(5))/2, (1+sqrt(5))/2>
<-1, 0, (5+3*sqrt(5))/2, -(1+sqrt(5))/2>
<-1, (3+sqrt(5))/2, -(1+sqrt(5))/2, -(3+sqrt(5))>
<-1, (3+sqrt(5))/2, (1+sqrt(5))/2, 3+sqrt(5)>
<-1, 2+sqrt(5), -(2+sqrt(5)), -1>
<-1, 2+sqrt(5), -(2+sqrt(5)), 1>
<-1, 2+sqrt(5), -1, -(2+sqrt(5))>
<-1, 2+sqrt(5), 1, 2+sqrt(5)>
<-1, 2+sqrt(5), 2+sqrt(5), -1>
<-1, 2+sqrt(5), 2+sqrt(5), 1>
<0, -(2+sqrt(5)), -(5+sqrt(5))/2, -(3+sqrt(5))/2>
<0, -(2+sqrt(5)), (5+sqrt(5))/2, (3+sqrt(5))/2>
<0, -(5+sqrt(5))/2, -(3+sqrt(5))/2, -(2+sqrt(5))>
<0, -(5+sqrt(5))/2, (3+sqrt(5))/2, 2+sqrt(5)>
<0, -(1+sqrt(5)), -(3+sqrt(5)), 0>
<0, -(1+sqrt(5)), 3+sqrt(5), 0>
<0, -(1+sqrt(5))/2, -(5+3*sqrt(5))/2, 1>
<0, -(1+sqrt(5))/2, (5+3*sqrt(5))/2, -1>
<0, -1, -(1+sqrt(5))/2, -(5+3*sqrt(5))/2>
<0, -1, (1+sqrt(5))/2, (5+3*sqrt(5))/2>
<0, 1, -(1+sqrt(5))/2, -(5+3*sqrt(5))/2>
<0, 1, (1+sqrt(5))/2, (5+3*sqrt(5))/2>
<0, (1+sqrt(5))/2, -(5+3*sqrt(5))/2, 1>
<0, (1+sqrt(5))/2, (5+3*sqrt(5))/2, -1>
<0, 1+sqrt(5), -(3+sqrt(5)), 0>
<0, 1+sqrt(5), 3+sqrt(5), 0>
<0, (5+sqrt(5))/2, -(3+sqrt(5))/2, -(2+sqrt(5))>
<0, (5+sqrt(5))/2, (3+sqrt(5))/2, 2+sqrt(5)>
<0, 2+sqrt(5), -(5+sqrt(5))/2, -(3+sqrt(5))/2>
<0, 2+sqrt(5), (5+sqrt(5))/2, (3+sqrt(5))/2>
<1, -(3+sqrt(5)), -(3+sqrt(5))/2, -(1+sqrt(5))/2>
<1, -(3+sqrt(5)), (3+sqrt(5))/2, (1+sqrt(5))/2>
<1, -(2+sqrt(5)), -(2+sqrt(5)), -1>
<1, -(2+sqrt(5)), 2+sqrt(5), 1>
<1, -(3+sqrt(5))/2, -(1+sqrt(5))/2, -(3+sqrt(5))>
<1, -(3+sqrt(5))/2, (1+sqrt(5))/2, 3+sqrt(5)>
<1, -(1+sqrt(5))/2, -(3+sqrt(5)), (3+sqrt(5))/2>
<1, -(1+sqrt(5))/2, 0, -(5+3*sqrt(5))/2>
<1, -(1+sqrt(5))/2, 0, (5+3*sqrt(5))/2>
<1, -(1+sqrt(5))/2, 3+sqrt(5), -(3+sqrt(5))/2>
<1, 0, -(5+3*sqrt(5))/2, (1+sqrt(5))/2>
<1, 0, (5+3*sqrt(5))/2, -(1+sqrt(5))/2>
<1, (3+sqrt(5))/2, -(1+sqrt(5))/2, -(3+sqrt(5))>
<1, (3+sqrt(5))/2, (1+sqrt(5))/2, 3+sqrt(5)>
<1, 2+sqrt(5), -(2+sqrt(5)), -1>
<1, 2+sqrt(5), -(2+sqrt(5)), 1>
<1, 2+sqrt(5), -1, -(2+sqrt(5))>
<1, 2+sqrt(5), 1, 2+sqrt(5)>
<1, 2+sqrt(5), 2+sqrt(5), -1>
<1, 2+sqrt(5), 2+sqrt(5), 1>
<(1+sqrt(5))/2, -(2+sqrt(5)), -(3+sqrt(5))/2, -(1+sqrt(5))>
<(1+sqrt(5))/2, -(2+sqrt(5)), (3+sqrt(5))/2, 1+sqrt(5)>
<(1+sqrt(5))/2, -(3+sqrt(5))/2, -(3+sqrt(5)), 1>
<(1+sqrt(5))/2, -(3+sqrt(5))/2, 3+sqrt(5), -1>
<(1+sqrt(5))/2, 0, -1, -(5+3*sqrt(5))/2>
<(1+sqrt(5))/2, 0, 1, (5+3*sqrt(5))/2>
<(1+sqrt(5))/2, (3+sqrt(5))/2, -(3+sqrt(5)), 1>
<(1+sqrt(5))/2, (3+sqrt(5))/2, 3+sqrt(5), -1>
<(1+sqrt(5))/2, 2+sqrt(5), -(3+sqrt(5))/2, -(1+sqrt(5))>
<(1+sqrt(5))/2, 2+sqrt(5), (3+sqrt(5))/2, 1+sqrt(5)>
<(3+sqrt(5))/2, -(3+sqrt(5)), -(1+sqrt(5))/2, -1>
<(3+sqrt(5))/2, -(3+sqrt(5)), (1+sqrt(5))/2, 1>
<(3+sqrt(5))/2, -(2+sqrt(5)), -(1+sqrt(5)), -(1+sqrt(5))/2>
<(3+sqrt(5))/2, -(2+sqrt(5)), 1+sqrt(5), (1+sqrt(5))/2>
<(3+sqrt(5))/2, -(5+sqrt(5))/2, -(2+sqrt(5)), 0>
<(3+sqrt(5))/2, -(5+sqrt(5))/2, 2+sqrt(5), 0>
<(3+sqrt(5))/2, -(1+sqrt(5)), -(1+sqrt(5))/2, -(2+sqrt(5))>
<(3+sqrt(5))/2, -(1+sqrt(5)), (1+sqrt(5))/2, 2+sqrt(5)>
<(3+sqrt(5))/2, -(1+sqrt(5))/2, -(2+sqrt(5)), 1+sqrt(5)>
<(3+sqrt(5))/2, -(1+sqrt(5))/2, -1, -(3+sqrt(5))>
<(3+sqrt(5))/2, -(1+sqrt(5))/2, -1, 3+sqrt(5)>
<(3+sqrt(5))/2, -(1+sqrt(5))/2, 1, -(3+sqrt(5))>
<(3+sqrt(5))/2, -(1+sqrt(5))/2, 1, 3+sqrt(5)>
<(3+sqrt(5))/2, -(1+sqrt(5))/2, 2+sqrt(5), -(1+sqrt(5))>
<(3+sqrt(5))/2, -1, -(3+sqrt(5)), (1+sqrt(5))/2>
<(3+sqrt(5))/2, -1, 3+sqrt(5), -(1+sqrt(5))/2>
<(3+sqrt(5))/2, 1, -(3+sqrt(5)), (1+sqrt(5))/2>
<(3+sqrt(5))/2, 1, 3+sqrt(5), -(1+sqrt(5))/2>
<(3+sqrt(5))/2, (1+sqrt(5))/2, -1, -(3+sqrt(5))>
<(3+sqrt(5))/2, (1+sqrt(5))/2, 1, 3+sqrt(5)>
<(3+sqrt(5))/2, 1+sqrt(5), -(1+sqrt(5))/2, -(2+sqrt(5))>
<(3+sqrt(5))/2, 1+sqrt(5), (1+sqrt(5))/2, 2+sqrt(5)>
<(3+sqrt(5))/2, (5+sqrt(5))/2, -(2+sqrt(5)), 0>
<(3+sqrt(5))/2, (5+sqrt(5))/2, 2+sqrt(5), 0>
<(3+sqrt(5))/2, 2+sqrt(5), -(1+sqrt(5)), -(1+sqrt(5))/2>
<(3+sqrt(5))/2, 2+sqrt(5), -(1+sqrt(5)), (1+sqrt(5))/2>
<(3+sqrt(5))/2, 2+sqrt(5), 0, -(5+sqrt(5))/2>
<(3+sqrt(5))/2, 2+sqrt(5), 0, (5+sqrt(5))/2>
<(3+sqrt(5))/2, 2+sqrt(5), 1+sqrt(5), -(1+sqrt(5))/2>
<(3+sqrt(5))/2, 2+sqrt(5), 1+sqrt(5), (1+sqrt(5))/2>
<1+sqrt(5), -(3+sqrt(5)), 0, 0>
<1+sqrt(5), -(2+sqrt(5)), -(1+sqrt(5))/2, -(3+sqrt(5))/2>
<1+sqrt(5), -(2+sqrt(5)), (1+sqrt(5))/2, (3+sqrt(5))/2>
<1+sqrt(5), -(3+sqrt(5))/2, -(2+sqrt(5)), (1+sqrt(5))/2>
<1+sqrt(5), -(3+sqrt(5))/2, 2+sqrt(5), -(1+sqrt(5))/2>
<1+sqrt(5), -(1+sqrt(5))/2, -(3+sqrt(5))/2, 2+sqrt(5)>
<1+sqrt(5), -(1+sqrt(5))/2, (3+sqrt(5))/2, -(2+sqrt(5))>
<1+sqrt(5), 0, 0, -(3+sqrt(5))>
<1+sqrt(5), 0, 0, 3+sqrt(5)>
<1+sqrt(5), (3+sqrt(5))/2, -(2+sqrt(5)), (1+sqrt(5))/2>
<1+sqrt(5), (3+sqrt(5))/2, 2+sqrt(5), -(1+sqrt(5))/2>
<1+sqrt(5), 2+sqrt(5), -(1+sqrt(5))/2, -(3+sqrt(5))/2>
<1+sqrt(5), 2+sqrt(5), -(1+sqrt(5))/2, (3+sqrt(5))/2>
<1+sqrt(5), 2+sqrt(5), (1+sqrt(5))/2, -(3+sqrt(5))/2>
<1+sqrt(5), 2+sqrt(5), (1+sqrt(5))/2, (3+sqrt(5))/2>
<(5+sqrt(5))/2, -(2+sqrt(5)), -(3+sqrt(5))/2, 0>
<(5+sqrt(5))/2, -(2+sqrt(5)), (3+sqrt(5))/2, 0>
<(5+sqrt(5))/2, -(3+sqrt(5))/2, 0, -(2+sqrt(5))>
<(5+sqrt(5))/2, -(3+sqrt(5))/2, 0, 2+sqrt(5)>
<(5+sqrt(5))/2, 0, -(2+sqrt(5)), (3+sqrt(5))/2>
<(5+sqrt(5))/2, 0, 2+sqrt(5), -(3+sqrt(5))/2>
<(5+sqrt(5))/2, (3+sqrt(5))/2, 0, -(2+sqrt(5))>
<(5+sqrt(5))/2, (3+sqrt(5))/2, 0, 2+sqrt(5)>
<(5+sqrt(5))/2, 2+sqrt(5), -(3+sqrt(5))/2, 0>
<(5+sqrt(5))/2, 2+sqrt(5), (3+sqrt(5))/2, 0>
<2+sqrt(5), -(2+sqrt(5)), -1, -1>
<2+sqrt(5), -(2+sqrt(5)), -1, 1>
<2+sqrt(5), -(2+sqrt(5)), 1, -1>
<2+sqrt(5), -(2+sqrt(5)), 1, 1>
<2+sqrt(5), -(5+sqrt(5))/2, 0, -(3+sqrt(5))/2>
<2+sqrt(5), -(5+sqrt(5))/2, 0, (3+sqrt(5))/2>
<2+sqrt(5), -(1+sqrt(5)), -(3+sqrt(5))/2, -(1+sqrt(5))/2>
<2+sqrt(5), -(1+sqrt(5)), -(3+sqrt(5))/2, (1+sqrt(5))/2>
<2+sqrt(5), -(1+sqrt(5)), (3+sqrt(5))/2, -(1+sqrt(5))/2>
<2+sqrt(5), -(1+sqrt(5)), (3+sqrt(5))/2, (1+sqrt(5))/2>
<2+sqrt(5), -(3+sqrt(5))/2, -(5+sqrt(5))/2, 0>
<2+sqrt(5), -(3+sqrt(5))/2, -(1+sqrt(5))/2, -(1+sqrt(5))>
<2+sqrt(5), -(3+sqrt(5))/2, -(1+sqrt(5))/2, 1+sqrt(5)>
<2+sqrt(5), -(3+sqrt(5))/2, (1+sqrt(5))/2, -(1+sqrt(5))>
<2+sqrt(5), -(3+sqrt(5))/2, (1+sqrt(5))/2, 1+sqrt(5)>
<2+sqrt(5), -(3+sqrt(5))/2, (5+sqrt(5))/2, 0>
<2+sqrt(5), -(1+sqrt(5))/2, -(1+sqrt(5)), (3+sqrt(5))/2>
<2+sqrt(5), -(1+sqrt(5))/2, 1+sqrt(5), -(3+sqrt(5))/2>
<2+sqrt(5), -1, -(2+sqrt(5)), 1>
<2+sqrt(5), -1, -1, -(2+sqrt(5))>
<2+sqrt(5), -1, -1, 2+sqrt(5)>
<2+sqrt(5), -1, 1, -(2+sqrt(5))>
<2+sqrt(5), -1, 1, 2+sqrt(5)>
<2+sqrt(5), -1, 2+sqrt(5), -1>
<2+sqrt(5), 0, -(3+sqrt(5))/2, (5+sqrt(5))/2>
<2+sqrt(5), 0, (3+sqrt(5))/2, -(5+sqrt(5))/2>
<2+sqrt(5), 1, -(2+sqrt(5)), 1>
<2+sqrt(5), 1, -1, -(2+sqrt(5))>
<2+sqrt(5), 1, -1, 2+sqrt(5)>
<2+sqrt(5), 1, 1, -(2+sqrt(5))>
<2+sqrt(5), 1, 1, 2+sqrt(5)>
<2+sqrt(5), 1, 2+sqrt(5), -1>
<2+sqrt(5), (1+sqrt(5))/2, -(1+sqrt(5)), (3+sqrt(5))/2>
<2+sqrt(5), (1+sqrt(5))/2, 1+sqrt(5), -(3+sqrt(5))/2>
<2+sqrt(5), (3+sqrt(5))/2, -(5+sqrt(5))/2, 0>
<2+sqrt(5), (3+sqrt(5))/2, -(1+sqrt(5))/2, -(1+sqrt(5))>
<2+sqrt(5), (3+sqrt(5))/2, -(1+sqrt(5))/2, 1+sqrt(5)>
<2+sqrt(5), (3+sqrt(5))/2, (1+sqrt(5))/2, -(1+sqrt(5))>
<2+sqrt(5), (3+sqrt(5))/2, (1+sqrt(5))/2, 1+sqrt(5)>
<2+sqrt(5), (3+sqrt(5))/2, (5+sqrt(5))/2, 0>
<2+sqrt(5), 1+sqrt(5), -(3+sqrt(5))/2, -(1+sqrt(5))/2>
<2+sqrt(5), 1+sqrt(5), -(3+sqrt(5))/2, (1+sqrt(5))/2>
<2+sqrt(5), 1+sqrt(5), (3+sqrt(5))/2, -(1+sqrt(5))/2>
<2+sqrt(5), 1+sqrt(5), (3+sqrt(5))/2, (1+sqrt(5))/2>
<2+sqrt(5), (5+sqrt(5))/2, 0, -(3+sqrt(5))/2>
<2+sqrt(5), (5+sqrt(5))/2, 0, (3+sqrt(5))/2>
<2+sqrt(5), 2+sqrt(5), -1, -1>
<2+sqrt(5), 2+sqrt(5), -1, 1>
<2+sqrt(5), 2+sqrt(5), 1, -1>
<2+sqrt(5), 2+sqrt(5), 1, 1>
quickfur
Pentonian
 
Posts: 2848
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