8 * | 8 0 0 | 12 0 0 | 6 0 0 cube vf.
* 64 | 1 3 3 | 3 6 3 | 3 3 1
-----+----------+----------+---------
1 1 | 64 * * | 3 0 0 | 3 0 0
0 2 | * 96 * | 1 2 0 | 2 1 0 ortho
0 2 | * * 96 | 0 2 2 | 1 2 1 para
-----+----------+----------+---------
1 2 | 2 1 0 | 96 * * | 2 0 0
0 4 | 0 2 2 | * 96 * | 1 1 0
0 3 | 0 0 3 | * * 64 | 0 1 1
-----+----------+----------+---------
2 8 | 8 8 4 | 8 4 0 | 24 * * esquidpy
0 6 | 0 3 6 | 0 3 2 | * 32 * trip
0 4 | 0 0 6 | 0 0 4 | * * 16 tet
64 * | 3 1 3 0 | 3 6 3 0 0 | 1 3 3 0 tet vertices
* 96 | 0 0 2 4 | 0 4 1 2 2 | 0 2 2 1 co vertices
------+---------------+-----------------+-----------
2 0 | 96 * * * | 2 2 0 0 0 | 1 1 2 0 edges of tet
2 0 | * 32 * * | 0 0 3 0 0 | 0 3 0 0 tet connectings
1 1 | * * 192 * | 0 2 1 0 0 | 0 2 1 0
0 2 | * * * 192 | 0 1 0 1 1 | 0 1 1 1
------+---------------+-----------------+-----------
3 0 | 3 0 0 0 | 64 * * * * | 1 0 1 0
2 2 | 1 0 2 1 | * 192 * * * | 0 1 1 0
2 1 | 0 1 2 0 | * * 96 * * | 0 2 0 0
0 4 | 0 0 0 4 | * * * 48 * | 0 1 0 1
0 3 | 0 0 0 3 | * * * * 64 | 0 0 1 1
------+---------------+-----------------+-----------
4 0 | 6 0 0 0 | 4 0 0 0 0 | 16 * * * tet
8 8 | 4 4 16 8 | 0 8 8 2 0 | * 24 * * squobcu
3 3 | 3 0 3 3 | 1 3 0 0 1 | * * 64 * trip
0 12 | 0 0 0 24 | 0 0 0 6 8 | * * * 8 co
Klitzing wrote:Ad 1)
Considering CRF polychora the verfs in fact might be taken orbiform. But then its faces nedn't be planar any longer.
Truely speaking, verfs generally are the intersection of the vertex surroundings with a small sphere. Therefore vertex figure polytopes are per se spherical tesselations.
--- rk
Marek14 wrote:Klitzing wrote:Ad 1)
Considering CRF polychora the verfs in fact might be taken orbiform. But then its faces nedn't be planar any longer.
Truely speaking, verfs generally are the intersection of the vertex surroundings with a small sphere. Therefore vertex figure polytopes are per se spherical tesselations.
--- rk
In this case, though, I'd prefer a bit different definition: intersection with unit hypersphere. Since the polytope is CRF, the unit hypersphere centred at a vertex will, by definition, contain all neighbouring vertices.
Then, instead of checking the edge lengths and comparing them to chords, it might be arguably easier to check angles - a triangle means that two edges from central vertex have angle of 60 degrees, square means 90 degrees, and so on. And if, say, four triangles meet at one vertex and form a quadrangle on the vertex figure, the angles corresponding to diagonals can be checked as well -- two angles of 90 degrees signify a square verf corresponding to octahedron, square pyramid, or similar, while trigonal bipyramid, for example, would have different diagonal angles.
Klitzing wrote:Marek14 wrote:Klitzing wrote:Ad 1)
Considering CRF polychora the verfs in fact might be taken orbiform. But then its faces nedn't be planar any longer.
Truely speaking, verfs generally are the intersection of the vertex surroundings with a small sphere. Therefore vertex figure polytopes are per se spherical tesselations.
--- rk
In this case, though, I'd prefer a bit different definition: intersection with unit hypersphere. Since the polytope is CRF, the unit hypersphere centred at a vertex will, by definition, contain all neighbouring vertices.
Then, instead of checking the edge lengths and comparing them to chords, it might be arguably easier to check angles - a triangle means that two edges from central vertex have angle of 60 degrees, square means 90 degrees, and so on. And if, say, four triangles meet at one vertex and form a quadrangle on the vertex figure, the angles corresponding to diagonals can be checked as well -- two angles of 90 degrees signify a square verf corresponding to octahedron, square pyramid, or similar, while trigonal bipyramid, for example, would have different diagonal angles.
different?
you just use the implicite "hyper" prefix explicitely. And the size (radius) does not matter, as long it is lower or equal to unity (edge length).
And sure "tesselation" likewise was meant to be the intersection of those vertex incident elements with the surface of that "sphere".
--- rk
o-infin-o o-infin-o o-infin-o = point
o-infin-o o-infin-o o-infin-x = aze (integer line)
o-infin-o o-infin-x o-infin-x = squat (integer plane, square grid)
o-infin-x o-infin-x o-infin-x = chon (cubical honeycomb)
o4o3o4o = point
? : aze
? : squat
o4o3o4x = chon
x-infin-o x-infin-o x-infin-o
x-infin-o x-infin-o x-infin-x
x-infin-o x-infin-x x-infin-x
x-infin-x x-infin-x x-infin-x
x4o3o4o
?
?
x4o3o4x
o4x3o4o = rich
? : pexrich (partially expanded rich)
? : pacsrich (partially contracted srich)
o4x3o4x = srich
rich pexrich pacsrich srich
N oct esquidpy squobcu sirco
N co co co co
N {4} {4} {4} cube
N {4} {4} cube cube
N {4} cube cube cube
x4x3o4o = tich
? : pextich (partially expanded tich)
? : pacprich (partially contracted prich)
x4x3o4x = prich
tich pextich pacprich prich
N oct esquidpy squobcu sirco
N tic tic tic tic
N {8} {8} {8} op
N {8} {8} op op
N {8} op op op
N edge edge {4} cube
N edge {4} {4} cube
N edge {4} cube cube
s4o3o4o = octet ( = o3o3x *4o )
? : pextoh (partially extended tetrahedral-octahedral honeycomb)
? : pacratoh (partially contracted ratoh)
s4o3o4x = ratoh ( = o3o3x *b4x )
octet pextoh pacratoh ratoh
N tet tet tet tet
N tet tet tet tet
N oct esquidpy squobcu sirco
N point edge {4} cube
s4x3o4o = o4x3o4o
?
?
s4x3o4x = o4x3o4x
| o4o3a patex-o4o3a o4x3a :: tetrahedral
---+----------------------------
its 1st type of cells (in the right case for any "a", in the left only for "a" = "x"):
4 | . o3a . o3a -> . x3a :: triangular
4 | . o3a -> . x3a . x3a
---+----------------------------
its 2nd type of cells (only for "a" = "x"):
12 | o . a o . a o . a :: none
---+----------------------------
its 3rd type of cells (in the right case ever, in the left one never):
6 | o4o . -> pex-o4o -> o4x . :: axial
| a3o4o3b pox-a3o4o3b poc-a3o4x3b a3o4x3b :: hexadecachoral
---+-------------------------------------------------
its 1st type of cells (in the right case for any "b", in the left only for "b" = "x"):
8 | . o4o3b -> . o4o3b patex-o4o3b . o4x3b :: tetrahedral
8 | . o4o3b patex-o4o3b -> patex-o4o3b . o4x3b
8 | . o4o3b patex-o4o3b . o4x3b -> . o4x3b
---+-------------------------------------------------
its 2nd type of cells (in the right case for any "b" and "a" = "x", in the left only for "a" = "b" = "x"):
32 | a . o3b a . o3b a . o3b -> a . x3b :: triangular
32 | a . o3b a . o3b -> a . x3b a . x3b
32 | a . o3b -> a . x3b a . x3b a . x3b
---+-------------------------------------------------
its 3rd type of cells (only for "a" = "b" = "x"):
96 | a3o . b a3o . b a3o . b a3o . b :: none
---+-------------------------------------------------
its 4th type of cells (in the right case ever, in the left only for "a" = "x"):
24 | a3o4o . -> pex-a3o4o -> pac-a3o4x -> a3o4x . :: axial
| a3b3o4o3c phextex-a3b3o4o3c pabhextex-a3b3o4o3c phextco-a3b3o4x3c a3b3o4x3c :: hexadecachoric-tetracombal
----+----------------------------------------------------------------------------------
its 1st type of cells (in the right case for any "b" and "c", in the left only for "b" = "x" or "c" = "x"):
N | . b3o4o3c -> . b3o4o3c pox-b3o4o3c poc-b3o4x3c . b3o4x3c :: hexadecachoral
N | . b3o4o3c pox-b3o4o3c -> pox-b3o4o3c poc-b3o4x3c . b3o4x3c
N | . b3o4o3c pox-b3o4o3c poc-b3o4x3c -> poc-b3o4x3c . b3o4x3c
N | . b3o4o3c pox-b3o4o3c poc-b3o4x3c . b3o4x3c -> . b3o4x3c
----+----------------------------------------------------------------------------------
its 2nd type of cells (in the right case only for "a" = "x", in the left only for "a" = "c" = "x"):
8N | a . o4o3c a . o4o3c a . o4o3c -> patex-o4o3a-p -> a . o4x3c :: tetrahedral
8N | a . o4o3c a . o4o3c -> patex-o4o3a-p patex-o4o3a-p a . o4x3c
8N | a . o4o3c a . o4o3c patex-o4o3a-p patex-o4o3a-p a . o4x3c
8N | a . o4o3c -> patex-o4o3a-p patex-o4o3a-p -> a . o4x3c a . o4x3c
8N | a . o4o3c patex-o4o3a-p patex-o4o3a-p a . o4x3c a . o4x3c
8N | a . o4o3c patex-o4o3a-p -> a . o4x3c a . o4x3c a . o4x3c
----+----------------------------------------------------------------------------------
its 3rd type of cells (in the right case only for "a" = "x" or "b" = "x", in the left only for ("a" = "x" or "b" = "x") and "c" = "x"):
32N | a3b . o3c a3b . o3c a3b . o3c a3b . o3c -> a3b . x3c :: triangular
32N | a3b . o3c a3b . o3c a3b . o3c -> a3b . x3c a3b . x3c
32N | a3b . o3c a3b . o3c -> a3b . x3c a3b . x3c a3b . x3c
32N | a3b . o3c -> a3b . x3c a3b . x3c a3b . x3c a3b . x3c
----+----------------------------------------------------------------------------------
its 4th type of cells (only for ("a" = "x" or "b" = "x") and "c" = "x"):
96N | a3b3o . c a3b3o . c a3b3o . c a3b3o . c a3b3o . c :: none
----+----------------------------------------------------------------------------------
its 5th type of cells (in the right case ever, in the left only for "a" = "x" or "b" = "x"):
12N | a3b3o4o . -> pex-a3b3o4o -> pabex-a3b3o4o -> pac-a3b3o4x -> a3b3o4x . :: axial
Marek14 wrote:Would that mean that 4-3-3-5 hyperbolic tetracomb could be partially reduced according to the branched subsymmetry as well?
o3o...o3o4o with o o ... o o o
o4o3o...o3o4o with o-infin-o o-infin-o ... o-infin-o o-infin-o
o3o...o3o4o3o with o3o3o *b3o ... *b3o
Marek14 wrote:Would that mean that 4-3-3-5 hyperbolic tetracomb could be partially reduced according to the branched subsymmetry as well?
Would that mean that 4-3-... could be partially reduced according to the branched subsymmetry as well?
o4y3z.. = y3z3y *b...
o4o3z.. = o3z3o *b.. <-> x3z3o *b.. = pex-o4o3z.. = pac-o4x3z.. <-> o4x3z.. = x3z3x *b..
(1)...o4o...(2) <-> (several partial expansion/contraction steps) <-> (1)...o4x...(2)
octahedron <-> elongated square dipyramid <-> square orthobicupola <-> small rhombicuboctahedron
wendy wrote:I think i see your game.
What you are trying to do, is to deconstruct a stott-expansion over the whole symmetry into several bits.
For example, in going from x3o4o to x3o4x, one adds the X axis, and then the Y axis, and then the Z axis.
If you stop working with bollo-apeirotopes (tilings), and work with real polytopes (eg {3,8} in 3D) then yes, it will work.
The resulting polytopes won't be co-radial (that is a bollosphere through the vertices), but the will pass by the other rules. It should work over any value of 2P, not just 4, but some of the intermediate steps might not give regular polygons,
now thinking of x3o6o to x3o6x.
I'm trying to think of the {5,3,4}. If you look at the dodecahedron, by way of example, there are six edges which are parallel/perpendicular. These are the six edges that do not connect to an inscribed cube. In the {5,3,4}, one can colour these in three different colours (eg red, blue, green), which gives one some room to work with. (see 'Not Knot' on youtube, part 2, stimly).
If you are looking at a polytope, say a cell of {5,3,4,3}, then it should be possible to cut the planes that contain red lines, and insert a prismatic layer there. It won't be concentric on the vertices as such, but it is possible, because you can insert struts between the fragments to keep the general curvature. This would eventually introduce pentagonal prisms, and square prisms along the red edges. Yes, that would work. The second progression would introduce pentagonal prisms along the the second. Yes, it would be possible to convert x5o3o4o into x5o3o4x in three steps, akin to x3o3o4x and x4o3o4x.
Klitzing wrote:Ey, you are right indeed! So far I did not succeed, with anything having a 5 in its symbol, but you are right indeed.
wendy wrote:There's quite a few of these partial stott expansions. The sufficient condition is the segment ..o4o... to ..o4x..
One can find the number of independent types of mirrors in a symmetry by removing all of the even branches. So, in the group o5o3o4o, there are two kinds of mirror, those internal to the dodecahedron, and the walls between the dodecahedron. Each of these groups stands freely. What it does not tell you, is that the resulting group r (those mirrors o5o3o4r), also degenerates into free subsets, each subset or combination of subset stands freely.
In the case of o3oo3o4r, the group r is the rectangularoid group, each combination of W, X, Y, Z, ... stands freely. This is why there are four steps in this group.
In the case of o4o3o4r, there are six mirrors, which can be represented by eX, oX, eY, oY, eZ, oZ, ..., and this can be expanded in six steps.
In the case of o5o3o4r, the group r divides into no fewer than 12 separate mirrors. It's not apparent as yet, but consider marking the faces of a dodecahedral cell with numbers from 1 to 12. Now, just reflecting these in the walls of the dodecahedron, one sees that there are planes of pentagons marked '1', and different planes marked '2', usw. None of thes are images of each other, so we have deconstructed this 'rectododecahedral' group into the intersection of 12 separate laminatopic (polytopes bounded by unbounded faces) groups, each one presenting just a single pentagon to a given cell.
Likewise o8o4r, gives an rectioctagonal group, divided into eight laminaohedra. And o7o4r gives a rectoheptagonal group, which divides into seven laminatotopic groups. You get the idea.
Now, we see that R Klitzing's various partial stott expansions, then consist of raising stilts, or mirror edges to free combinations of these laminatopic groups, so, zB
1. octahedron has no stilts.
2. elongated square bipyramid has stilts in X axis
3. bi-square cupola has stilts in X, Y axis
4. rhombo-cuboctahedron has stilts in X, Y, Z axis.
The x5o3o4o then does the same sort of thing, but in place of three steps, twelve. You erect stilts on what's inside the dodecahedron, one face at a time, in any order. There is no need for faces to be adjacent, or side by side, or whatever. So, there are three different arrangements of 'two faces', and several more for 'three faces'.
The progression from x8o4o to x8o4x, takes eight steps, and from xPo4o to xPo4x takes no fewer than P steps. The x5o3o3o4o has 120 steps and many more intermediates, while the x3o4o3o4o has 24 steps to x3o4o3o4r.
wendy wrote:It goes further.
If one considers the group x5o3o4r, for example, then the 'r' divides to 12 separate nodes, ...
... i..xii, each of which can be marked separately, in much the same way that x3o3o4r corresponds to dividing r to four separate nodes w,x,y,z. But each of i-xii and w-z correspond to faces of the dodecahedron and tetrahedron separately, ...
... the number of unique figures is much less than 2^12 = 4096, or 2^4 = 16. In the latter, there are just 5 combinations, the former gives at least
1 for 1 and 11, 3 for 2 and 10, and various different numbers for 3,9, and 4,8 and 5,7 and 6.
In short, the partial stott expansion has a wythoff-mirror-edge explaination, to the defragment of the recto-(face of xPo4xRo) group.
wendy wrote:The progression from x8o4o to x8o4x, takes eight steps, and from xPo4o to xPo4x takes no fewer than P steps. The x5o3o3o4o has 120 steps and many more intermediates, while the x3o4o3o4o has 24 steps to x3o4o3o4r.
Klitzing wrote:wendy wrote:now thinking of x3o6o to x3o6x.
Well there a 60° rhomb would be needed... - In other cases, when considering the octagon, a longish hexagon would occure. But all these odd things I brushed away so far.
Klitzing wrote:Klitzing wrote:wendy wrote:now thinking of x3o6o to x3o6x.
[...]
Re-thought that one: [...] - But then, what about tripesic reductions at the hexagonal centers, i.e applying kind a hexagonal lattice on top of those hexagon centers of x3o6x? Then 2/3 of these hexagons would be reduced to triangles (and the corresponding squares between would be reduced to mere edges), but 1/3 of the hexagons would remain in place.
[...] - I.e. we just have described a partial Stott contraction from x3o6x down to x3o6o, using 3 independent steps! [...]
--- rk
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