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 » Tue Dec 31, 2013 9:23 pm

Today, I was browsing through Klitzing's website, and when I saw this picture of the paradiminished rhombicosidodecahedron (J80):

Image

I got the idea that perhaps partial Stott expansion can be extended to symmetries other than n-cube symmetry. Consider the above picture, for example. If you imagine the pentagonal faces being "pushed inwards" so that the triangles and squares collapse into points and edges respectively, then the result is a regular dodecahedron. So the reverse process, going from the dodecahedron to J80, is just like regular Stott expansion/contraction except that the two decagonal faces did not undergo a full expansion, but only a partial expansion in their respective planes (pentagon -> decagon). Or, another way to look at it, is that two different Stott expansions are taking place here: the pentagon->decagon expansion (which, if applied to the entire dodecahedron, would yield a truncated dodecahedron), and the {pentagon->pentagon, vertex->triangle, edge->square} expansion (which, if applied to the entire dodecahedron, would yield the (non-diminished) rhombicosidodecahedron). These two expansions just so happen to share a compatible interface, so they can take place simultaneously within different parts of the dodecahedron.

I haven't tried specific cases in 4D yet, but it does seem promising as another way to search for CRFs in the 120-cell/600-cell family of polychora. Specifically, I wonder if there's a hybrid Stott expansion that may somehow produce a polychoron with pairs of J80 conjoined at their decagonal faces -- this may turn out to be a new kind of CRF that perhaps isn't obtainable via ordinary cut-n-paste operations? E.g., if we imagine the 120-cell being partially expanded in such a way that some cells aren't fully expanded into x5o3x's, but become J80's or some of the other x5o3x diminishings. I don't know for sure whether this is actually possible (with CRF results), but it seems to be an interesting new direction to explore.
quickfur
Pentonian
 
Posts: 2848
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North

Re: Johnsonian Polytopes

Postby student91 » Thu Jan 02, 2014 1:06 pm

student91 wrote:[...]
The first two restrictions in my previous post still are true: a triangle can be altered anyhow, a square only in an xyyx-fashion (then it's the sech.prod. of a digonal prism and a x||y-polytope) the triangle in the first restriction can be replaced with any simplex. (this way you can see sechmentotopes are alternations of 1-simplices (except for the highlighted cases in my first post about this) ). those simplices keep bothering me, because I can't find a way to "tame" them (i.e. put them in a system of operations)

student91


Found the solution for this as well: it's just that simplices can be altered anyhow, and if they're altered with two polytopes (which give a valid sechmentotope(so P1||P2 exists)), it can be seen as a sechmentotopical product of those two polytopes atop of each other, and two, not necessarily equal simplices placed atop each other. (so the ortho- and magnabicupolic rings can be seen as sech. products of line||point and a cupola, whereas the gyrobicupolic ring cannot be seen as a sechmentotopical product). I will call these simplical sechmentotopes. If you want to peform a sechmentotopical product on two simplistical sechmentotopes, you have to align the "vertices" of the simplexes with each other, product the vertices, and then rebuild the simplex, with smaller edges of the base-simplex.

you might wonder, why I want to make this aditional system of sechmentotopes, because there is already a sufficient system: Klitzings
This is because, in my opinion, Klitzings way of looking at it provides a good way for sechmentotopes that are based on an (n-1)-symmetry, but it is not very good in describing sechmentotopes based on a less-than-(n-1)-symmetry. (e.g. the ortobicupolic ring: Klitzing sees it as either cube||octagon or 4-cupola||square, whereas I would see it as square square octagon in a triangle fashion, out of which Klitzings views can be derived (as either (square||square)||octagon or square||(square||octagon) ) On the other hand, it's true that klitzings way of looking at it has been studied more and better, making it more reliable.

this way of looking at it might also give easier calculations, let me give an example:
As you are looking for sechmentotera, I thought I could look for equivalents of xfo3oox&#xt||o3x.
now xfo3oox&#xt ||o3x can be seen in my new way of looking at it as x3o, f3o and o3x on a line (you could represent those by vertices, and then the distance between those vertices should correspond to the heights of the corresponding sechmentohedra (x3o||f3o and f3o||o3x) ), and on top of that line there's another "vertex"(o3x), which is connected to all three other "vertices."(again, those lines have the length of the corresponding sechmentotope). Now a new polyteron might look like xfo3oox3ooo&#xt || o3x3x. the only thing we have to calculate in order to know whether this is a possible polyteron or not, is if the vertices with given connection lenghts make a planar drawing (you can even only use heron's formula if the heights of the sechmentotopes are known). I get stuck at calculating the height of f3o3o||o3x3x.
however,in a similar way, we can see that both (xfo3oox&#xt||o3x)||o3x and (xfo3oox&#xt||o3x)||xfo3oox&#xt are valid sechmentotera (although they're quite obvious, and actually sechmentotopical products of line||point and xfo3oox&#xt||o3x)

student91
How easily one gives his confidence to persons who know how to give themselves the appearance of more knowledge, when this knowledge has been drawn from a foreign source.
-Stern/Multatuli/Eduard Douwes Dekker
student91
Tetronian
 
Posts: 328
Joined: Tue Dec 10, 2013 3:41 pm

Re: Johnsonian Polytopes

Postby Klitzing » Thu Jan 02, 2014 3:50 pm

just want to point out:
... sechmentotope ...

should be spelled: segmentotope

(there is a noun called "segment",
but no noun called "sechment" :) )
--- rk
Klitzing
Pentonian
 
Posts: 1616
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

Re: Johnsonian Polytopes

Postby Klitzing » Thu Jan 02, 2014 4:17 pm

If I get it correctly, one of student91's points in his recent post was just the rediscovering of Wendy's lace simplexes.

Consider % to be any of "o" or "x", independently at any position.
Then a lace prism e.g. is %%P%%Q%%...&#x.

You could stack several lace prisms atop of each other.
Then you get lace towers: %..%P%..%Q%..%.....&#xt.

Instead you could run cyclically, getting lace rings: %..%P%..%Q%..%.....&#xr.

But you well could consider not only consecutive layers as lace prisms, but any pair of layers too.
This then is what is a lace simplex, i.e. %..%P%..%Q%..%.....&#x
(still being written as laced, but without any additional qualifier at the end).

For sure, a lace prism itself then is just the degenerate digonal simplex and therefore follows this convention correctly.

Thus the cube||octagon or 4-cupola||square, mentioned by him, clearly is nothing but the lace simplex xxx4oox&#x read in 2 different orientations.

--- rk
Klitzing
Pentonian
 
Posts: 1616
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

Re: Johnsonian Polytopes

Postby quickfur » Thu Jan 02, 2014 6:54 pm

student91 wrote:[...]
you might wonder, why I want to make this aditional system of sechmentotopes, because there is already a sufficient system: Klitzings
This is because, in my opinion, Klitzings way of looking at it provides a good way for sechmentotopes that are based on an (n-1)-symmetry, but it is not very good in describing sechmentotopes based on a less-than-(n-1)-symmetry. (e.g. the ortobicupolic ring: Klitzing sees it as either cube||octagon or 4-cupola||square, whereas I would see it as square square octagon in a triangle fashion, out of which Klitzings views can be derived (as either (square||square)||octagon or square||(square||octagon) )
[...]

Keiji would love you just for that statement alone. :) He has always argued that the bicupolic rings are better thought of as 2D extensions to a 2D polygon, rather than a 1D extension of a 3D polyhedron (segmentochoron).

This analysis could also be seen as a derivative of the duoprisms being Cartesian products of two polygons (rather than some kind of extension of a 3D object). Of course, these are just different ways of looking at the same thing, and I think both are helpful. For example, while I had heard of duoprisms long before, I had never really understood them until I came to see them as two interlocking rings of 3D facets. That led be to (re)discover the circle-polygon Cartesian products (I called them prismic cylinders, though that's probably not really a good name) and the duocylinder (originally called "double torus" until I realized it was the same thing as the duocylinder).
quickfur
Pentonian
 
Posts: 2848
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North

Re: Johnsonian Polytopes

Postby student91 » Thu Jan 02, 2014 11:01 pm

Klitzing wrote:If I get it correctly, one of student91's points in his recent post was just the rediscovering of Wendy's lace simplexes.
[...]
--- rk


those lace simplices as you describe them are indeed the same as what I had in mind :D , except for that I also add lengths to the edges of the simplex, with which you can calculate whether it's a valid polytope or not. (It's not a very big change, but it can make it easier to calculate catesian coordinates for it)

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)

student91
How easily one gives his confidence to persons who know how to give themselves the appearance of more knowledge, when this knowledge has been drawn from a foreign source.
-Stern/Multatuli/Eduard Douwes Dekker
student91
Tetronian
 
Posts: 328
Joined: Tue Dec 10, 2013 3:41 pm

Re: Johnsonian Polytopes

Postby wendy » Fri Jan 03, 2014 1:40 am

One should not really diss Klitzing's A || B notation. He went out to solve a specific problem for which this is the right notation.

On the other hand, i figured out how to hook multiple vertex nodes to the same dynkin symbol, and devised a notation for it.

It is interesting that student91 uses my notation rather heavily. You can change the lacing edge, the x in #x.. is the same x as in x3o etc. It's hardly a product, though, because it does not come to a algebraic multiply.
The dream you dream alone is only a dream
the dream we dream together is reality.

\ ( \(\LaTeX\ \) \ ) [no spaces] at https://greasyfork.org/en/users/188714-wendy-krieger
User avatar
wendy
Pentonian
 
Posts: 1988
Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia

Re: Johnsonian Polytopes

Postby student91 » Fri Jan 03, 2014 10:52 am

wendy wrote:[...]
It's hardly a product, though, because it does not come to a algebraic multiply.

When I was talking about a product, I mostly was talking about the sechmentotopical product I described a few posts ago. however, that post might be a little bit shitty, so I will explain it again:
a cartesian product of a pentagonal cupola and a square can be peformed in the following way:
you have the coordinates of a pentagonal cupola, lets say {(a,b,c),(d,e,f),(g,h,i),(j,k,l),...}
and the coordinates of a square: {(1/2,1/2),(1/2,-1/2),(-1/2,1/2),(-1/2,-1/2)}.
now you augment every coordinate of the pent.cup. with every coordinate of the square, so {(a,b,c,1/2,1/2),(a,b,c,1/2,-1/2),(a,b,c,-1/2,1/2),(a,b,c,-1/2,-1/2),(d,e,f,1/2,1/2),...}. you can see this as putting a square at every point of the pentagonal cupola, e.g. (a,b,c): it has 4 copies ( (a,b,c,1/2,1/2),(a,b,c,1/2,-1/2),(a,b,c,-1/2,1/2),(a,b,c,-1/2,-1/2) ) that make a square, so point (a,b,c) is "replaced" with a square.
wherever (p,q,r) has an edge to (s,t,v), a square||square occurs.

now the segmentotopical product is wat you get if you'd place different polytopes (lets say squares and octagons) on the vertices of the "base"-polytope (the pentagonal cupola). the only way to do this properly is by placing one polytope on the bottom of the "base", and the other on the top (so let's say we put the octagon on the decagon("bottom") and the square on the pentagon("top")) now some edges of the base-polytope (those going from bottom to top) connect a square with an octagon. these edges will give square||octagon-polytopes, and thus these edges should get the length of the height of square||octagon. If we place the squares and the octagons now, they'd perfectly connect. Of course the height of the base-polytope would get smaller (to be more precise: it would be of a length of sqrt([height pentagonal cupola]²+[height square||octagon]²-1). this formula is derived as follows: take an edge connecting the top with the bottom. draw the height of one point of this edge. the edge and the height form a right triangle. the third edge of this triangle won't change if you'd change the height. the edge itself would get the length of square||octagon. the third edge can be calculated as (3'd)=sqrt(1-[height pent.cup.]²) the new height can be calculated as sqrt([height square||octagon]²-(3'd)²)=sqrt([height square||octagon]²+[height pentagonal cupola]²-1). so our new polytope is pentagon×square||decagon×octagon. it is derived from 2 symmetries (.4. and .5.), so a notation can be ox5xx[some multiplication sign]ox4xx&#x
EDIT: it's height evaluates to sqrt(1/2+(5-sqrt(5))/10-1)=sqrt(-sqrt(5)/10), so this one doesn't exist, and hence it's a stupid example.

student91
Last edited by student91 on Sat Jan 04, 2014 11:20 am, edited 3 times in total.
How easily one gives his confidence to persons who know how to give themselves the appearance of more knowledge, when this knowledge has been drawn from a foreign source.
-Stern/Multatuli/Eduard Douwes Dekker
student91
Tetronian
 
Posts: 328
Joined: Tue Dec 10, 2013 3:41 pm

Re: Johnsonian Polytopes

Postby student91 » Fri Jan 03, 2014 12:25 pm

wendy wrote:One should not really diss Klitzing's A || B notation. He went out to solve a specific problem for which this is the right notation.

On the other hand, i figured out how to hook multiple vertex nodes to the same dynkin symbol, and devised a notation for it.

It is interesting that student91 uses my notation rather heavily. You can change the lacing edge, the x in #x.. is the same x as in x3o etc.
[...]


I did not intend to diss Klitzings notation, and I agree that it is useful for a lot of cases, and that the use of words instead of xo3oo3ox&#x is much more comprehensible for people who don't understand CD-diagrams. (like me when I was exploring 4-space). And it is also useful for shapes we don't have a good notation for yet (e.g. K4.13: it can be written as x2o||(x2x||x2x)||o2x or (x2o||x2x)||(x2x||o2x), but every way to write it must use a ||), or shapes that are based on a (n-1)-dimensional symmetry. However, some cases have better notations (like the lace simplices, they can all be written in some way as a segmentotope, but that doesn't show it's true simplectic structure), and that's what I thought I had invented. I like your way of writing, because it's the structure of the polytope that is described by it, rather than a construction.
I don't understand how something with a different lacing edge (say f) can yield a valid CRF. do you have an example, or is it just a unused feature that might be used one day?

student91
How easily one gives his confidence to persons who know how to give themselves the appearance of more knowledge, when this knowledge has been drawn from a foreign source.
-Stern/Multatuli/Eduard Douwes Dekker
student91
Tetronian
 
Posts: 328
Joined: Tue Dec 10, 2013 3:41 pm

Re: Johnsonian Polytopes

Postby wendy » Sat Jan 04, 2014 3:13 am

Lace notation was actually invented to describe the vertex figures of the uniform polyhedra. These do not have in general, lacings of x, but usually q, h, and longer ones.

It's a modification made on Klitzing's behest that uses x to mean a lacing of 1, rather than 1.73205080757.
The dream you dream alone is only a dream
the dream we dream together is reality.

\ ( \(\LaTeX\ \) \ ) [no spaces] at https://greasyfork.org/en/users/188714-wendy-krieger
User avatar
wendy
Pentonian
 
Posts: 1988
Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia

Re: Johnsonian Polytopes

Postby student91 » Sat Jan 04, 2014 3:00 pm

ahh, so it is used, but not for CRF's, and in the same way rings are used, but not for CRF's. :D

besides, found some new polytera: ooxx2oxoo&#x (can be either triangle||tetahedron or triangle||orthogonal square :) , consists of only 5-cells and square pyramid pyramids)
point||(3,3)-duoprism, point||(3,4)-duoprism, line||(3,3)-duoprism, line||(3,4)-duoprism, triangle||orthogonal triangular prism, triangle||orthogonal cube, cube||orthogonal triangular prism, square||(3,4)-duprism
(3,3)-duoprism||(6,6)-duoprism, (3,6)-duoprism||(6,3)-duoprism, (3,8)-duoprism||(6,4)-duoprism, (3,4)-duoprism||(6,8)-duoprism,
hexagon||(3,3)-duoprism, triangle||(6,3)-duoprism, hexagon||(3,4)-duoprism, triangle||(6,4)-duoprism, octagon||(4,3)-duoprism, square||(8,3)-duoprism,
triangular prism||(6,3)-duoprism, hexagonal prism||(3,3)-duoprism, cube||(8,3)-duoprism, octagonal prism||(4,3)-duoprism, triangular prism||(4,6)-duoprism, hexagonal prism||(4,3)-duoprism, cube||(4,8)-duoprism, octagonal prism||tesseract, pentagonal prism||(4,10)-duoprism, decagonal prism||(4,5)-duoprism, line||(3,4)-duoprism, triangular prism||square, line||tesseract, cube||square, line||(4,5)-duoprism, pentagonal prism||square.
you might have noticed that line||(3,4)-duoprism occured twice. these are different, as in one the line is parallel to an edge of the triangles in the (3,4)-duoprism, whereas in the other the line is parallel to two edges of every square of the (3,4)-duoprism

student91
Last edited by student91 on Sat Jan 04, 2014 3:51 pm, edited 1 time in total.
How easily one gives his confidence to persons who know how to give themselves the appearance of more knowledge, when this knowledge has been drawn from a foreign source.
-Stern/Multatuli/Eduard Douwes Dekker
student91
Tetronian
 
Posts: 328
Joined: Tue Dec 10, 2013 3:41 pm

Re: Johnsonian Polytopes

Postby Klitzing » Sat Jan 04, 2014 3:50 pm

No, you are wrong. f-nodes etc. are well used in CRFs.

First of all:
x denotes an edge of length 1
f denotes an edge of length 1.618 (tau)
v denotes an edge of length 0.618 (1/tau = tau - 1)
q denotes an edge of length 1.414 (sqrt2)
h denotes an edge of length 1.732 (sqrt3)
u denotes an edge of length 2
w denotes an edge of length 2.414 (1 + sqrt2)
...
x occures as vertex figure of x3o, f occures as vertex figure of x5o, v occures as vertex figure of x5/2o, q occures as vertex figure of x4o, h occures as vertex figure of x6o, and u occurs as vertex figure of the apeirogon (x-infin-o).

It is the invention of lace towers, which makes all these figures occur. Note that then you will consider vertex layers, resp. the structures described by them. But not all edges of the dissected structure will fall in those layers only. In the contrary, those sections would often need for vertex distances not being described by edge length distances only.

A regular pentagon e.g. then can be described as ofx&#xt, i.e. a tower of 3 layers with a single point in the uppermost, 2 points in a medial layer, which are spaced by tau, and finally 2 points within the 3rd layer, spaced by unity. Vertices of the different layers then are connected by unit lacing edges (...&#xt). Similarily a regular hexagon can be given either as xux&#xt or as ohho&#xt. And a regular octagon as xwwx&#xt. You even could write a square not only as xx&#x, but likewise as oqo&#xt. - All these figures clearly use unit distances for their edges only, but there are other vertex distances within these layers, which not qualify as true edges, which rather require other lengths.

A unit-edged cuboctahedron e.g. can be given as xox4oqo&#xt (i.e. as x4o || pseudo o4q || x4o). A unit-edged truncated cube as xwwx4xoox&#xt. The unit-edged great rhombicuboctahedron becomes xxwwxx4xuxxux&#xt. Etc.

--- rk
Last edited by Klitzing on Sat Jan 04, 2014 4:27 pm, edited 3 times in total.
Klitzing
Pentonian
 
Posts: 1616
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

Re: Johnsonian Polytopes

Postby Klitzing » Sat Jan 04, 2014 4:07 pm

Wrt lace rings, those clearly are useful descriptions for CRFs too.
In fact a regular n-gon can be given as oo...oo&#xr (n unringed nodes, i.e. cyclically consecutive vertices).
Likewise the semiregular n-gonal prism then would be xx...xx&#xr (n ringed nodes, i.e. cyclically consecutive orthogonal edges).
The tetrahedron could be given as oox&#xr, the square pyramid as oxx&#xr, and a triangle prism could be given as ooxx&#xr.

But note, that those are just quite simple, low dimensional examples. And therefore might look superfluous.
Consider e.g. the hexadecachoron x3o3o4o. It can be written as segmentochoron tet || dual tet, and therefore as lace prism xo3oo3ox&#x. But as the tetrahedron itself can be given as a segmentohedron o3o || x3o (and the dual one then as o3x || o3o), the whole thing thus could be rewritten as oxoo3ooox&#xr.

--- rk
Klitzing
Pentonian
 
Posts: 1616
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

Re: Johnsonian Polytopes

Postby student91 » Sat Jan 04, 2014 4:29 pm

Klitzing wrote:No, you are wrong. f-edges etc. are well used in CRFs.

First of all:
x denotes an edge of length 1
f denotes an edge of length 1.618 (tau)
v denotes an edge of length 0.618 (1/tau = tau - 1)
q denotes an edge of length 1.414 (sqrt2)
h denotes an edge of length 1.732 (sqrt3)
u denotes an edge of length 2
w denotes an edge of length 2.414 (1 + sqrt2)
...
x occures as vertex figure of x3o, f occures as vertex figure of x5o, v occures as vertex figure of x5/2o, q occures as vertex figure of x4o, h occures as vertex figure of x6o, and u occurs as vertex figure of the apeirogon (x-infin-o).

It is the invention of lace towers, which makes all these figures occur. Note that then you will consider vertex layers, resp. the structures described by them. But not all edges of the dissected structure will fall in those layers only. In the contrary, those sections would often Need for vertex distances not being described by edge length distances only.

A regular pentagon e.g. then can be described as ofx&#xt, i.e. a tower of 3 layers with a single point in the uppermost, 2 points in a medial layer, which are spaced by tau, and finally 2 points within the 3rd layer, spaced by unity. Vertices of the different layers then are connected by unit lacing edges (...&#xt). Similarily a regular hexagon can be given either as xux&#xt or as ohho&#xt. And a regular octagon as xwwx&#xt. You even could write a square not only as xx&#x, but likewise as oqo&#xt. - All these figures clearly use unit distances for their edges only, but there are other vertex distances within these layers, which not qualify as true edges, which rather require other lengths.

--- rk

ah, I see, so they're frequently used. Are they also used for the lacing distance in CRF's? (so &#ut for example?)
Klitzing wrote:Wrt lace rings, those clearly are useful descriptions for CRFs too.
In fact a regular n-gon can be given as oo...oo&#xr (n unringed nodes, i.e. cyclically consecutive vertices).
Likewise the semiregular n-gonal prism then would be xx...xx&#xr (n ringed nodes, i.e. cyclically consecutive orthogonal edges).
The tetrahedron could be given as oox&#xr, the square pyramid as oxx&#xr, and a triangle prism could be given as ooxx&#xr.

But note, that those are just quite simple, low dimensional examples. And therefore might look superfluous.
Consider e.g. the hexadecachoron x3o3o4o. It can be written as segmentochoron tet || dual tet, and therefore as lace prism xo3oo3ox&#x. But as the tetrahedron itself can be given as a segmentohedron o3o || x3o (and the dual one then as o3x || o3o), the whole thing thus could be rewritten as oxoo3ooox&#xr.

--- rk

I was referring to my previous post:
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)

the thing you've written (oxoo3ooox&#xr) can be decomposed in this way to oxo3xoo&#xt and xoo3oxo&#xr, joined at their xo3ox&#x-edge, and the ooo3ooo&#xr and xxx3xxx&#xr are just cartesian products (and a little superfluous, as you said)

besides the segmentotera I mentioned above there are some infinite families of segmentotera:
gyrated N-gonal prism||(3,N)-duoprism (N>1), gyrated N-gonal prism||(4,N)-duoprism (N>1),
gyrated N-gon||(3,N)-duoprism (N>1), gyrated N-gon||(4,N)-duoprism (N>2), gyrated N-gon||(5,N)-duoprism (N>3)
gyrated (3,N)-duoprism||(6,N)-duoprism (N>1), gyrated (4,N)-duoprism||(8,N)-duoprism (N>2), gyrated (5,N)-duoprism||(10,N)-duoprism (N>3)
(gyrated N,gyrated M)-duoprism||(N,M)-duoprism (N>1,M>1)
together with multiplications of segmentohedra with a polygon

student91
How easily one gives his confidence to persons who know how to give themselves the appearance of more knowledge, when this knowledge has been drawn from a foreign source.
-Stern/Multatuli/Eduard Douwes Dekker
student91
Tetronian
 
Posts: 328
Joined: Tue Dec 10, 2013 3:41 pm

Re: Johnsonian Polytopes

Postby Klitzing » Sat Jan 04, 2014 5:21 pm

student91 wrote:ah, I see, so they're frequently used. Are they also used for the lacing distance in CRF's? (so &#ut for example?)

No. Please reread my (meanwhile extended) post. Sure you could use longer lacings too. But usually those are true edges and thus would be of unit size in CRFs. But cross sections at vertex layers generally cut some lacing faces into pieces such that the cut becomes a pseudo edge which then is longer than unity.

Cf.
Klitzing wrote:A unit-edged cuboctahedron e.g. can be given as xox4oqo&#xt (i.e. as x4o || pseudo o4q || x4o). A unit-edged truncated cube as xwwx4xoox&#xt. The unit-edged great rhombicuboctahedron becomes xxwwxx4xuxxux&#xt. Etc.


--- rk
Klitzing
Pentonian
 
Posts: 1616
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

Re: Johnsonian Polytopes

Postby Klitzing » Sat Jan 04, 2014 5:39 pm

student91 wrote:the thing you've written (oxoo3ooox&#xr) can be decomposed in this way to oxo3xoo&#xt and xoo3oxo&#xr, joined at their xo3ox&#x-edge, and the ooo3ooo&#xr and xxx3xxx&#xr are just cartesian products (and a little superfluous, as you said)

Probably a typo: you'd mean oxo3xoo&#xr and xoo3oxo&#xr.

oxoo3ooox&#xr is just a different notation (with the same meaning) as the lace city:
Code: Select all
o3o   x3o


o3x   o3o

i.e. a cyclical arrangement of Dynkin symbols. The left tower (in fact a place prism only) is point || triangle, the right is dual triangle || point. - Yes the diagonal from bottom-left to top-right will be a further lace prism section (here not a facet). In fact the octahedron. I.e. it uses true unit lacing edges too. Thus one can dissect the hexadecachoron in 2 octahedral pyramds.

But: the other diagonal, from top-left to bottom-right has not a unit lacing. In fact that distance between those opposite points would be of size q. And thus is a pseudo "edge" only. And a corresponding split is likewise impossible (as CRF).

ooo3ooo&#xr is not at all a cartesian product. It is just a complicated description of a triangle:
Code: Select all
o3o   o3o

   o3o

i.e. a lace city from 3 points.

xxx3xxx&#xr accordingly would be a triangular arrangement of hexagons (layers x3x). Thus it is nothing but thiddip, the trigonal-hexagonal-duoprism.

--- rk
Klitzing
Pentonian
 
Posts: 1616
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

Re: Johnsonian Polytopes

Postby student91 » Sat Jan 04, 2014 7:41 pm

Klitzing wrote:Probably a typo: you'd mean oxo3xoo&#xr and xoo3oxo&#xr.
indeed

ooo3ooo&#xr is not at all a cartesian product. It is just a complicated description of a triangle:
Code: Select all
o3o   o3o

   o3o

i.e. a lace city from 3 points.

I saw it as multiplication of the "base"-polytope (a triangle) with it's things (o3o). this is indeed not a real multiplication, more like multiplying with 1.

besides, I think I can conjecture (without proof):
every segmentotope is either:
a lace simplex of wythoffian polytopes with 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
A cartesian product of a segmentotope and another CRF-polytope


student91
How easily one gives his confidence to persons who know how to give themselves the appearance of more knowledge, when this knowledge has been drawn from a foreign source.
-Stern/Multatuli/Eduard Douwes Dekker
student91
Tetronian
 
Posts: 328
Joined: Tue Dec 10, 2013 3:41 pm

Re: Johnsonian Polytopes

Postby wendy » Sun Jan 05, 2014 4:53 am

The idiom behind a lace figure is that of a pyramid. You have an altitude with little symmetry, and bases perpendicular to the altitude, with more symmetry. Although the early lace prisms had simplex-altitudes, the notion was extended, and 'lace-simplex' is largely depreciated. Lace prism is the correct term. The dual of a lace prism is a lace tegum.

A lace tower is a string of bases on a linear altitude. A lace-city is an array of bases on a 2space (hedrix). They're useful for stratic polytopes. A lace ring is a kind of lace city that can be turned into a loop and thus enumerated.

Wythoff-mirror-edge polytopes refer to the use of the symmetry group as an axial coordinate. That is, a form like x2f is actually a true coordinate on a generally oblique axis-system, and a valid WME. You can use lengths of any value, the ones i specify are the shortchords of the polygons, because that's the ones that came up most often. In other geometries, there are lots of non WME uniforms. 's' is not a valid coordinate, so the representations using it are not valid WME'. So so3so4ox&#x is not a valid WLP. Nor is o3o4x || x3o5o.

Nothing stops you from having non-unit lacing in a lace prism. In practice, the ones i used to deal with, have q and h and longer measures as lacing, rarely x. These are the sort of edges one gets in vertex-figures, such as oxo8ooo&p8t, or oxqxo8ooooo&xqt. p8 is a lacing equal to the shortchord of an octagon, viz sqrt(3.41421356238).

I'm not sure what is meant by 'segmentopic product'. Richard Klitzing's segmentotopes are 'outcomes', not 'processes'. In other words, any shape that passes a 'segmentotope' is one, whereas a WLP is a pre-defined construction that may or may not work.

I think you're confusing 'segmentotope' (which is an equal-edge bistratic polytope), with CRH (ie CRF, except that it's the hedra, not the faces, that are regular). Something like oxx4ooo3ooo&#xt is a CRH, but not a segmentotope. Many of Johnson's polytopes are polystratic, eg oxo5oox&#xt 'diminished icosahedron'. It may be that segmentotopes are a subset of CRH, but not the other way around.

One normally does not count 'cartesian products' into the set of CRH. CRH is a residue set, excluding higher-order members. So even the diminished icosahedral prism is not a CRH. However, other products are fair game, eg the tegum-product of a line and an icosahedron (ie oxo3ooo5ooo&#xt), is fair game.

Augmenting is fair game, but you have to watch the margin-angles don't exceed c60. Otherwise the thing is no longer convex. On the other hand, it is possible to join large numbers of cells of x5o3o5o, into a convex yickloid region, as long as there is at least one hedron of x5o3o between each join.
The dream you dream alone is only a dream
the dream we dream together is reality.

\ ( \(\LaTeX\ \) \ ) [no spaces] at https://greasyfork.org/en/users/188714-wendy-krieger
User avatar
wendy
Pentonian
 
Posts: 1988
Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia

Re: Johnsonian Polytopes

Postby Klitzing » Sun Jan 05, 2014 10:02 am

wendy wrote:So so3so4ox&#x is not a valid WLP. Nor is o3o4x || x3o5o.


Well, the latter is out of scope of lace prisms because the are not based on the same symmetry. The former would be based on that, but it is the the snubbing process which conflicts here.

Note that so3so4ox&#x could be read as s3s4o || o3o4x. But it is meant instead as
alternated faceting {starting figure = xo3xo4ox&#y; alternated elements = ..3..4o.; subsequent relaxing = true}


--- rk
Klitzing
Pentonian
 
Posts: 1616
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

Re: Johnsonian Polytopes

Postby Klitzing » Sun Jan 05, 2014 11:24 am

wendy wrote:I'm not sure what is meant by 'segmentopic product'.

Me neither fully.

But as far as I got it, he is aiming just for a reversion of operations.

Consider 2 segmentotopes b1 || b2 and b3 || b4.
Then he calls (b1||b2) x (b3||b4) := (b1xb3) || (b2xb4).

But I already argued that then it would be essential to know in advance the preconditions under which those results would be valid or not. (In the sense of kind a theorem.) Not only that all asked for subelements would exist (i.e. connect correctly) but that moreover the required heights do come out positive, resp. taken the other way round, that the lacing edges can be chosen as unity too.

@student91: please verify.

--- rk
Klitzing
Pentonian
 
Posts: 1616
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

Re: Johnsonian Polytopes

Postby student91 » Sun Jan 05, 2014 11:50 pm

Klitzing wrote:
wendy wrote:I'm not sure what is meant by 'segmentopic product'.

Me neither fully.

But as far as I got it, he is aiming just for a reversion of operations.

Consider 2 segmentotopes b1 || b2 and b3 || b4.
Then he calls (b1||b2) x (b3||b4) := (b1xb3) || (b2xb4).

But I already argued that then it would be essential to know in advance the preconditions under which those results would be valid or not. (In the sense of kind a theorem.) Not only that all asked for subelements would exist (i.e. connect correctly) but that moreover the required heights do come out positive, resp. taken the other way round, that the lacing edges can be chosen as unity too.

@student91: please verify.

--- rk


I think I can make such a theorem, although i've never made a theorem before.

let's consider the same segmentotopes as you used: b1||b2 (with height h1), and b3||b4 (with height h3).
furthermore there's a vertex v1 that belongs to b1, and that has a connection (in b1||b2) with v2, a vertex of b2.
now if we product both b1 and b2 with b3, we'd get b1×b3||b2×b3 = b3×(b1||b2).
I see this the following way: you place a b3 at every vertex of b1||b2.
All the edges of b1||b2 (the "base"-polytope) have length 1. Because of this, all the b3's will connect properly into b3-prisms where there's an edge connecting two b3's.
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.
Now when we've shrunken h1 to the required size, all the b3's should connect correctly to the b4's, and thus, everything connects properly

now i'll give an inductive proof that all the surtopes exist:
take b1×b3||b2×b4 again, with b1||b2 as "base"-polytope.
let's take the facets of b1||b2 that are in between b1 and b2 (thus those connecting b1 and b2). those are segmentotopes as well, and those will be segmentotopically producted with b3||b4, so the (n-1)-dimensional segmentotopical product should exist in order for these facets to exist. there is another set of facets in this segmentotopical product: take a surtope of b3, say s3, that makes a segmentotope with a surtope of b4, s4.
In b1×b3, every copy of b3 has a s3, so these give b1×s3. furthermore, every copy of b4 in b2×b4 has a s4, giving b2×s4. those two are connected, so this gives us b1×s3||b2×s4.
the last two facets of b1×b3||b2×b4 are b1×b3 and b2×b4 themselves. in summary we have: the surtopes of b1×b3||b2×b4 exist, if s1×b3||s2×b4 and b1×s3||b2×s4 exist. this proces can be repeated, until the surtopes become points, so q1×point||q2×point or q3×point||q4×point should exist.(q1, q2, q3 and q4 are any surtopes of b1, b2, b3 and b4) these do, as they are simply the segmentotope q1||q2 or q3||q4 themselves.

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.

I hope I have been clear (although I suspect I wasn't), feel free to ask for more explanation.
student91
Last edited by student91 on Mon Jan 06, 2014 11:08 am, edited 1 time in total.
How easily one gives his confidence to persons who know how to give themselves the appearance of more knowledge, when this knowledge has been drawn from a foreign source.
-Stern/Multatuli/Eduard Douwes Dekker
student91
Tetronian
 
Posts: 328
Joined: Tue Dec 10, 2013 3:41 pm

Re: Johnsonian Polytopes

Postby quickfur » Mon Jan 06, 2014 2:04 am

student91 wrote:[...]
let's consider the same sechmentotopes as you used: b1||b2 (with height h1), and b3||b4 (with height h3).
furthermore there's a vertex v1 that belongs to b1, and that has a connection (in b1||b2) with v2, a vertex of b2.
now if we product both b1 and b2 with b3, we'd get b1×b3||b2×b3 = b3×(b1||b2).
I see this the following way: you place a b3 at every vertex of b1||b2.
All the edges of b1||b2 (the "base"-polytope) have length 1. Because of this, all the b3's will connect properly into b3-prisms where there's an edge connecting two b3's.
now the true sechmentotopical 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.
Now when we've shrunken h1 to the required size, all the b3's should connect correctly to the b4's, and thus, everything connects properly

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?

---

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?
quickfur
Pentonian
 
Posts: 2848
Joined: Thu Sep 02, 2004 11:20 pm
Location: The Great White North

Re: Johnsonian Polytopes

Postby student91 » Mon Jan 06, 2014 9:20 am

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.
Last edited by student91 on Mon Jan 06, 2014 11:45 am, edited 1 time in total.
How easily one gives his confidence to persons who know how to give themselves the appearance of more knowledge, when this knowledge has been drawn from a foreign source.
-Stern/Multatuli/Eduard Douwes Dekker
student91
Tetronian
 
Posts: 328
Joined: Tue Dec 10, 2013 3:41 pm

Re: Johnsonian Polytopes

Postby Klitzing » Mon Jan 06, 2014 9:53 am

student91 wrote:let's consider the same sechmentotopes as you used: b1||b2 (with height h1), and b3||b4 (with height h3).

Once more: its a segmentotope! :roll:

--- rk
Klitzing
Pentonian
 
Posts: 1616
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

Re: Johnsonian Polytopes

Postby Klitzing » Mon Jan 06, 2014 10:59 am

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.

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. But you could like to consider that in the sense of student91's segmentotopal product instead:
E @ (F||G) := (E+||E-) # (F||G) = (E+xF) || (E-xG).

But be aware then: the exisatance of such a new quickfur-operation ( :D ) would depend on the existance of E+||E-! (And on that height restriction too.) E.g. tet || dual tet (in the sense of alternated cube, i.e. s2s2s) does exist (it is hex = x3o3o4o). But neither r-snic || l-snic nor r-snid || l-snid do exist! - Thus this "@" seems to be kind of rather restricted after all...

--- rk
Klitzing
Pentonian
 
Posts: 1616
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

Re: Johnsonian Polytopes

Postby Klitzing » Mon Jan 06, 2014 11:13 am

Some rather easy observations on student91's segmentotopal product "#":

Definition:
(A||B) # (C||D) := (AxC) || (BxD)

Observations:
if D = C then     (A||B) # (C||C) = (AxC) || (BxC) = (A||B) x C
esp. C = D = point then     (A||B) # line = (A||B) # (pt||pt) = ... = (A||B) x pt = A||B
and C = D = line then     (A||B) # square = (A||B) # (line||line) = ... = (A||B) x line = (A||B)-prism
Further:
because of PxQ = QxP (up to orientation in coodinate space) we too have P#Q = Q#P

--- rk
Klitzing
Pentonian
 
Posts: 1616
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

Re: Johnsonian Polytopes

Postby student91 » Mon Jan 06, 2014 11:34 am

Klitzing wrote:
student91 wrote:let's consider the same sechmentotopes as you used: b1||b2 (with height h1), and b3||b4 (with height h3).

Once more: its a segmentotope! :roll:

I'm sorry. When I'd treat segmentotope as a dutch word, it would be written as sechmentotope according to the dutch spelling, so it's a little against my nature to write segmentotope :\ . I'll try to pay attention to it in the future though.

Klitzing wrote:
quickfur wrote:[...]

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. But you could like to consider that in the sense of student91's segmentotopal product instead:
E @ (F||G) := (E+||E-) # (F||G) = (E+xF) || (E-xG).

But be aware then: the exisatance of such a new quickfur-operation ( :D ) would depend on the existance of E+||E-! (And on that height restriction too.) E.g. tet || dual tet (in the sense of alternated cube, i.e. s2s2s) does exist (it is hex = x3o3o4o). But neither r-snic || l-snic nor r-snid || l-snid do exist! - Thus this "@" seems to be kind of rather restricted after all...

--- rk

Although I have proven that A×C||B×D will work if A||B and C||D exist, I have not proven the opposite, so it might be possible that A×C||B×D exists if C||D doesn't exist, although I think it to be very unlikely, and maybe even provably impossible. I think I can prove it if I think a little more, but I'm not able to prove it yet. (I am very bad at proving things don't exist).
besides, I think my segmentotopical product should be implemented in Wendy's lacing notation. I was thinking about something like
%%P%%Q%%...¤%%R%%S%%..&#x. We could of course use some other multiplication sign, but both x and # are already used in the lacing notation. (and I thought winterstolice also used *). (the ¤-sign is ctrl+alt+4 or (Alt Gr)+4)
How easily one gives his confidence to persons who know how to give themselves the appearance of more knowledge, when this knowledge has been drawn from a foreign source.
-Stern/Multatuli/Eduard Douwes Dekker
student91
Tetronian
 
Posts: 328
Joined: Tue Dec 10, 2013 3:41 pm

Re: Johnsonian Polytopes

Postby student91 » Mon Jan 06, 2014 11:54 am

but let's get back to my previous statement:
student91 wrote:[...]
every segmentotope is either:
a lace simplex of wythoffian polytopes with 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
A cartesian product of a segmentotope and another CRF-polytope


student91

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)
Last edited by student91 on Mon Jan 06, 2014 12:02 pm, edited 1 time in total.
How easily one gives his confidence to persons who know how to give themselves the appearance of more knowledge, when this knowledge has been drawn from a foreign source.
-Stern/Multatuli/Eduard Douwes Dekker
student91
Tetronian
 
Posts: 328
Joined: Tue Dec 10, 2013 3:41 pm

Re: Johnsonian Polytopes

Postby student91 » Mon Jan 06, 2014 11:56 am

Klitzing wrote:Some rather easy observations on student91's segmentotopal product "#":

Definition:
(A||B) # (C||D) := (AxC) || (BxD)

Observations:
if D = C then     (A||B) # (C||C) = (AxC) || (BxC) = (A||B) x C
esp. C = D = point then     (A||B) # line = (A||B) # (pt||pt) = ... = (A||B) x pt = A||B
and C = D = line then     (A||B) # square = (A||B) # (line||line) = ... = (A||B) x line = (A||B)-prism
Further:
because of PxQ = QxP (up to orientation in coodinate space) we too have P#Q = Q#P

--- rk

I like those!!
furthermore, It's (obviously) true that if (A||B)#(C||D) is possible, (B||A)#(C||D) is also possible.
How easily one gives his confidence to persons who know how to give themselves the appearance of more knowledge, when this knowledge has been drawn from a foreign source.
-Stern/Multatuli/Eduard Douwes Dekker
student91
Tetronian
 
Posts: 328
Joined: Tue Dec 10, 2013 3:41 pm

Re: Johnsonian Polytopes

Postby Klitzing » Mon Jan 06, 2014 12:50 pm

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 term Wythoffian is made up to reflect the kaleidoscopical construction of polytopes, outlined by Wythoff.
The construction of snubs in the sense of an alternation e.g. was done by Stott.
Snubs in her sense usually do not lend for kaleidoscopical construction.
Nearly all known uniform polytopes known either qualify as Wythoffian or as snubs.

But snubs (in general) definitely are not Wythoffian. (Exceptions here are s2s2s = tet, s3s4o = ike.)
The only known convex uniform (up to my knowledge), which neither is Wythoffian nor a snub, is the grand antiprism (4D), found by Johnson.

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
Klitzing
Pentonian
 
Posts: 1616
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

PreviousNext

Return to CRF Polytopes

Who is online

Users browsing this forum: No registered users and 2 guests

cron