Bialternatosnub polytopes

[Mercurial, the Spectre]

Disclaimer
Bialternatosnub is my own term for bilateral alternatosnub with the word alternatosnub meaning "alternate snubbing." i.e. the vertices are halved similar to a full snub but preserving bilateral symmetry instead of making it chiral. Note that this topic will focus on regular polytopal bases. I discovered the process myself while experimenting with orthogonal 3D projections of duoprisms in GeoGebra.

IMPORTANT NOTICE: Bialternatosnubbing is not a commutative operation, so s2s4x (a rectangular trapezoprism, basically a D2d-symmetric cube with 2 rectangles and 4 lateral trapezoidal faces) and s4s2x (a square prism) are not geometrically the same (although in some cases, such as these pairs, are topologically the same polyhedron). It can also be applied to tilings and honeycombs as well. They are isogonal, so their duals will have congruent facets.

Introduction
Bialternatosnub polytopes are derived from omnitruncated polytopes that have been alternated in a different way compared to full snubs preserving bilateral symmetry along the hyperplanes of the alternated facets. The base polytopes will be from regular polytopes, tilings, or honeycombs of Schläfli symbol {p,q,..,y,z} with shorthand notation s{p}s{q}...s{z}x with the last letter being an even number (z in this case). If the last letter represents an odd number, then the result is a holosnub instead of a full-fledged bialternatosnub. (example: s3s3s4x denoting a bialternatosnub 16-cell which contains 16 icosahedra, 32 triangular prisms (represented as s3s2x or bialternatosnub trigonal dihedron), 24 rectangular trapezoprisms (s2s4x or bialternatosnub square hosohedron), 8 pyritohedral rhombicuboctahedra, and 96 wedges as sectioning facets. It has no uniform realization. Fun fact: an x,u-variant of s3s3s4x occurs as a diminishing of prissi).

In 2D, the bialternatosnub of {2p} is the 2p-gon itself with p-gonal symmetry. Its uniform (and regular) realization is none other than the 2p-gon itself.
In 3D, the bialternatosnub of {p,2q} has faces of type sPs (same as the p-gon), s2Qx (mentioned before) with trapezoidal faces as sectioning facets. Its uniform realization is xPo2Qx (cantellated {p,2q}). Example is the bialternatosnub {3,4} which turns out to be a pyritohedral rhombicuboctahedron.
In 4D, the bialternatosnub of {p,q,2r} has cells of type sPsQs, sQs(2R)x, sPs2x, s2s(2R)x with wedges (triangular prisms) as sectioning facets. In general, like the 4D snubs, there is no uniform realization.

One of my favorites is the bialternatosnub 2-4 duoprism (s2s2s4x) which I discovered months ago. It is nonuniform and can be derived from the 4-8 duoprism. It has tetrahedra, rectangular trapezoprisms and wedges for cells (plus square faces that are actually derived from the cubes; these alternate with the tetrahedra). In fact the variant with squares and regular tetrahedra appears as a diminishing of rit (rectified tesseract).
I'd love to see its dual that has 16 identical bilaterally symmetric triangular prism cells (basically the dual of s2s2s4x's vertex figure, which is a bilaterally symmetric trigonal bipyramid) but I don't have Stella4D to check it out.

Best to you out there guys, as this is my first post

[Klitzing]

Welcome Mercurial!

Way back in the dawn of this millenium I invented the symbolical usage of "s" nodes (rings without nodes), "x" nodes (ringed nodes), and "o" nodes (unringed nodes) within the same Dynkin symbol. And then also outlined its general construction device. Cf. the according publication of 2010 here.

To recall the general concept, when considering any decorated Dynkin symbol, with some (possibly none) "x" nodes, some (possibly none) "o" nodes and some (at least 1) "s" node:

• Replace all s nodes by x nodes. - This would be the representation of your starting figure.
• Delete all s nodes and their incident links. - This would be the representation of the to be alternated element.
• Underneath each deleted element a sectioning facet would be used instead.
• Thus the elements of the thus obtained alternated faceting then are the maintained half of the alternated elements, the sectioning facets underneath the deleted ones, and the remainders of the former D-1 dimensional faces. For the latter the same process then applies in turn, within 1 dimension less.

The 2 figures described by Mercurial then are: a) x2s4s


resp. b) s2s4x


The pictures are color-coded: red are the rejected alternating elements, black outline is the starting figure, yellow are the sectioning facets underneath, and the remainder of the former octagons is shown here in purple.

Note that those Pictures, as well as the above description, only represent the mere sectioning part. That one is clearly applicable in general. But often one wants to get the snubs to be uniform again. This then would require some further process of variation of the edge lengths. That additional process might succeed within 3D in several cases, but beyond it usually does not. This is just a matter of degrees of freedom.

The intended symbols of Mercurial always use linear Dynkin symbols only. He applies onle s nodes throughout, except of a single x node at one of its ends. Thus he restricts himself to omnitruncated starting figures only. Further he only consideres the alternation of a very specific edge type only (any one of the ends of the symbol). - So the above described method of alternated faceting applies generally. - So I think there is no need for a further naming. None the less my congratulations for your (partial) redescovery, Mercurial!

--- rk

[username5243]

These might have been discovered, but they haven't have names before. In particular that "s3s3s4x" (as given on your website) doesn't yet seem to have a name (and probably can't be made uniform either), but it should still be convex...

[Klitzing]

Just as an addition to the former:

here is a display of a different application of the alternated faceting method, which does not conform with Mercurial's Intention.
This time I'm alternating the triangles of the small rhombicuboctahedron, s4o3x,

(again using the same color-coding as explained above), which (as mere sectioning) results in a variant of the truncated tetrahedron.

--- rk

[Mercurial, the Spectre]

Just as an addition to the former:

here is a display of a different application of the alternated faceting method, which does not conform with Mercurial's Intention.
This time I'm alternating the triangles of the small rhombicuboctahedron, s4o3x,

(again using the same color-coding as explained above), which (as mere sectioning) results in a variant of the truncated tetrahedron.

--- rk

Nice to see you here Klitzing! I love reading your incmats since it offers me a way to explore new polytopes. Actually, this is a part of my isogonal polychora project (which includes step prisms, double symmetry compounds of 5-cell and 24-cell polychora, and lots more!) which are duals of dice (congruent facets or sides).

Tomruen (possibly) concepted the idea of omnisnubs (full snubs) which has all snub nodes. But he has only shown its vertex figures (fun fact: for an alternated prism the vertex figure is a simplex|verf of snubbed base cupola, so s3s3s2s (pyritohedral icosahedral antiprism or omnisnub tetrahedral antiprism) has a verf of a triangle|pentagon cupola, the triangle being a simplex and the pentagon being the verf of the alternated truncated octahedron (icosahedron)).

Besides, I was concerned of coincidental possibilities of uniformity among rank 4 omnisnubs {p,q,r} where each of p, q, and r is greater than 3 (maybe include fractional values such as 5/2 and 5/3). This excludes the trivial s2s2s2s (16-cell) which, along with gudap, has uniform realizations. Their verfs would be similar to the omnisnub 5-cell (it has C2 symmetry, so it is chiral. To picture this topologically, imagine the 3.4.3.4 configuration of faces. Now add opposite triangles at each of the tetragonal faces (making sure that the projection viewed from the 3.4.3.4 vertex retains 2-fold symmetry), then close the figure by adding an edge between the new vertices of the two triangles (one that doesn't share its vertex with the tetragonal faces) so that it creates two additional pentagons).

Anyways, back to the topic:
The polytope you described is x(1)s(3){3,4} which represents a ringed node at the first, and a snub node at the third using the octahedron as the base. In reference
bialternatosnubs would be s(1,2,3,...,n-2,n-1)x(n){p,q,...y,2z} which describes snub nodes at all except the last, which instead is ringed. That would also be called a cantic cube, which is not a full snub (or bialternatosnub) by proper definition. It is properly classified as a cantic (in fact, the wiki doesn't even list it as a snub) since it alternates the rhombicuboctahedron based on demicubic symmetry (which incidentally is full tetrahedral symmetry). Consequently it has half the vertices of the original figure. These can be extended to any dimension and to any geometry while retaining uniformity (since there are an infinite number of regular polytopes with even faces and even in one specific dimension if we allow different geometries such as hyperbolic).

BTW, capirsit is my favorite honeycomb, practially due to coincidental uniformity that is derived from gicaricot. It has links to the golden ratio and cypit despite having the same symmetry as sadit, although it is neither an omnisnub nor bialternatosnub. One of its facets though is a topological bialternatosnub tetrahedral dihedron (s3s3s2x), which is nothing more than a tetrahedral-symmetric icosahedral prism.)

Thanks for reading!

[username5243]

I'd be interested in hearing about your project, especially the duals of the "gyrochoron" type dice (see Hedrondude's dice pages)...

I've meanwhile made a list of (full) snubs (including honeycombs) in 4 and 5 dimensions. Note that facets are not assumed to be generally uniform. I've only included prismatic groups in the 4D spherical cases.

Code: Select all

4D spherical:
s3s3s3s - snip (snub pentachoron). Cells: 10 ikes, 20 octs, 60 tets.
s3s3s4s - snet (snub tesseract). Cells: 8 snics, 16 ikes, 24 squaps, 32 octs, 192 tets.
s3s4s3s - snico (snub icositetrachoron). Cells: 48 snics, 192 octs, 576 tets.
s3s3s5s - snahi (snub hecatonicosachoron. Cells: 120 snids, 600 ikes, 720 paps, 1200 octs, 7200 tets.
s3s3s *b3s = s3s3s4o = s3s4o3o - sadi (snub disicositetrachoron). Cells: 24 ikes, 24+96 tets. This one can be made uniform.
s2s3s3s - snittap (snub tetrahedral antiprism). Cells: 2 ikes, 8 octs, 6+24 tets.
s2s3s4s - sniccap (snub cubic antiprism). Cells: 2 snics, 6 squaps, 8 tets, 12+48 tets.
s2s3s5s - sniddap (snub dodecahedral antiprism). cells: 2 snids, 12 paps, 20 octs, 30+120 tets.
sps2sqs (p,q>2): p,q-dap (p,q-duoantiprism). Cells: 2q p-aps, 2p q-aps, 2pq tets.
s2s2sps (p > 2): 2,p-dap (2,p-duoantiprism). cells: 4 n-aps, 2p+4p tets.
s2s2s2s - hex (hexadecachoron). Cells: 16 tets) This one is uniform.

Honeycombs:
s3s4s3s - snich (snub cubic honeycomb). Cells, snics, squaps, tets.
s3s3s *b4s = s4s3s4o - ???. Cells: snics, ikes, tets (both main and snub).
s3s3s3s3*a = s3s3s *b4o = o4s3s4o - ????. Cells: ikes, tets.

5D spherical:
s3s3s3s3s - snix (snub hexateron). Facetes: 12 snips, 30 snittaps, 20 triddaps (3-3 duoantirprisms), 360 pens.
s3s3s3s4s - snan (snub penteract). Facets: 10 snets, 32 snips, 40 sniccaps, 80 snittaps, 80 tisdaps (3-4 duoantiprisms), 1920 pens.
s3s3s *b3s3s = s3s3s3s4o - ?????. Cells: 10 sadis, 32 snips, 40 snittaps, 80 ditdaps (2-3 duoantiprisms), 960 pens.

Tetracombs:
s4s3s3s4s - snettit (snub tesseractic tetracomb). Cells: snets, sniccaps, squiddaps (4-4 duoantiprisms), pens.
s3s3s *b3s4s = s4s3s3s4o - ??????. Facets: snets, sadis, snittaps, disdaps (2-4 duoantiprisms), pens.
s3s3s *b3s *b3s = s3s3s *b3s4o = o4s3s3s4o = s3s3s4o3o = o3o3o4s3s - sadit (snub disicositetrachoric tetracomb). Cells: sadis, hexes, pens. This one is uniform.
s3s3s4s3s = snicot (snub icositetrachoric tetracomb). Cells: snicoes, snets, sniccaps, snittaps, triddaps, pens.
s3s3s3s3s3*a - ?????. Cells, snips, pens.


[Mercurial, the Spectre]

I'd be interested in hearing about your project, especially the duals of the "gyrochoron" type dice (see Hedrondude's dice pages)...

I've meanwhile made a list of (full) snubs (including honeycombs) in 4 and 5 dimensions. Note that facets are not assumed to be generally uniform. I've only included prismatic groups in the 4D spherical cases.

Code: Select all

4D spherical:
s3s3s3s - snip (snub pentachoron). Cells: 10 ikes, 20 octs, 60 tets.
s3s3s4s - snet (snub tesseract). Cells: 8 snics, 16 ikes, 24 squaps, 32 octs, 192 tets.
s3s4s3s - snico (snub icositetrachoron). Cells: 48 snics, 192 octs, 576 tets.
s3s3s5s - snahi (snub hecatonicosachoron. Cells: 120 snids, 600 ikes, 720 paps, 1200 octs, 7200 tets.
s3s3s *b3s = s3s3s4o = s3s4o3o - sadi (snub disicositetrachoron). Cells: 24 ikes, 24+96 tets. This one can be made uniform.
s2s3s3s - snittap (snub tetrahedral antiprism). Cells: 2 ikes, 8 octs, 6+24 tets.
s2s3s4s - sniccap (snub cubic antiprism). Cells: 2 snics, 6 squaps, 8 tets, 12+48 tets.
s2s3s5s - sniddap (snub dodecahedral antiprism). cells: 2 snids, 12 paps, 20 octs, 30+120 tets.
sps2sqs (p,q>2): p,q-dap (p,q-duoantiprism). Cells: 2q p-aps, 2p q-aps, 2pq tets.
s2s2sps (p > 2): 2,p-dap (2,p-duoantiprism). cells: 4 n-aps, 2p+4p tets.
s2s2s2s - hex (hexadecachoron). Cells: 16 tets) This one is uniform.

Honeycombs:
s3s4s3s - snich (snub cubic honeycomb). Cells, snics, squaps, tets.
s3s3s *b4s = s4s3s4o - ???. Cells: snics, ikes, tets (both main and snub).
s3s3s3s3*a = s3s3s *b4o = o4s3s4o - ????. Cells: ikes, tets.

5D spherical:
s3s3s3s3s - snix (snub hexateron). Facetes: 12 snips, 30 snittaps, 20 triddaps (3-3 duoantirprisms), 360 pens.
s3s3s3s4s - snan (snub penteract). Facets: 10 snets, 32 snips, 40 sniccaps, 80 snittaps, 80 tisdaps (3-4 duoantiprisms), 1920 pens.
s3s3s *b3s3s = s3s3s3s4o - ?????. Cells: 10 sadis, 32 snips, 40 snittaps, 80 ditdaps (2-3 duoantiprisms), 960 pens.

Tetracombs:
s4s3s3s4s - snettit (snub tesseractic tetracomb). Cells: snets, sniccaps, squiddaps (4-4 duoantiprisms), pens.
s3s3s *b3s4s = s4s3s3s4o - ??????. Facets: snets, sadis, snittaps, disdaps (2-4 duoantiprisms), pens.
s3s3s *b3s *b3s = s3s3s *b3s4o = o4s3s3s4o = s3s3s4o3o = o3o3o4s3s - sadit (snub disicositetrachoric tetracomb). Cells: sadis, hexes, pens. This one is uniform.
s3s3s4s3s = snicot (snub icositetrachoric tetracomb). Cells: snicoes, snets, sniccaps, snittaps, triddaps, pens.
s3s3s3s3s3*a - ?????. Cells, snips, pens.


Any news on the verfs of all of these, especially the 5D and 4D honeycomb examples?

Anyways, I worked on the conjecture of uniform omnisnubs. Turns out that full fledged examples don't include those of type {p,q,p} when p and q are both greater than 2. Their omnitruncated variants have 2p-prism cells and omnitruncated {p,q} cells. My proof is that using extended symmetry, the topological rectangles of the omnitruncated {p,q} cells have to be squares because of the transitivity of the {p,q,p}. These in turn connect to the 2p-prisms by their square faces. The 2p-prisms then have alternating square and rectangle faces. Clearly the omnisnubs must ideally contain uniform p-antiprism cells derived from 2p-prisms with identical rectangular sides in general. If we consider square sides from prisms then it must be a 4-prism, or a cube. That means p has to be 2 to even have this coinciding feature that makes the antiprism uniform, which is always a degenerate tetrahedron. Then the ones of the form {2,p,2} are 2-p duoantiprisms, of which the only solution is the 16-cell, which is trivial. QED.

Note that this doesn't prove the nonexistence of uniform omnisnubs of type {p,q,r}, p ≠ r. I still have to work on these cases.

[username5243]

Snich's verf should be similar to snip's verf - it should have 2 isosceles pentagons, 2 trapezoids, and 4 triangles. The next ones verf will be some stretched version of a teddi (sadi verf), with 3 pentagons and 5 triangles. The last one's (s3s3s3s3*a)'s verf will have 4 pentagons and 4 triangles, but I'm not quite sure what it's like.

Snix and snan (as well as snettit and snicot) have verfs with 2 snip-verf-types, 2 snittap-verf-types, 1 duoantiprism-verf-type (which I think is a blend of two wedges at their bases) and 5 tets. s3s3s *b3s3s and s3s3s *b3s4s will have 2 snip-verf-types, 1 teddi, 1 snittap-verf-type, 1 2,n-dap type (similar to the general duoantiprism verf, but the top edge of one of the wedges has collapsed to a point), and 5 tets. Sadit's verf has 4 teddis, 5 tets, and 1 oct, and in the uniform version it has all regular faces (Klitzing has its incmat on his site). Finally, s3s3s3s3s3*a's verf should have 5 snip-verfs and 5 tets.

I have no clue how the verfs actually can be constructed, but that's just based on their facets.

[Mercurial, the Spectre]

Snich's verf should be similar to snip's verf - it should have 2 isosceles pentagons, 2 trapezoids, and 4 triangles. The next ones verf will be some stretched version of a teddi (sadi verf), with 3 pentagons and 5 triangles. The last one's (s3s3s3s3*a)'s verf will have 4 pentagons and 4 triangles, but I'm not quite sure what it's like.

Snix and snan (as well as snettit and snicot) have verfs with 2 snip-verf-types, 2 snittap-verf-types, 1 duoantiprism-verf-type (which I think is a blend of two wedges at their bases) and 5 tets. s3s3s *b3s3s and s3s3s *b3s4s will have 2 snip-verf-types, 1 teddi, 1 snittap-verf-type, 1 2,n-dap type (similar to the general duoantiprism verf, but the top edge of one of the wedges has collapsed to a point), and 5 tets. Sadit's verf has 4 teddis, 5 tets, and 1 oct, and in the uniform version it has all regular faces (Klitzing has its incmat on his site). Finally, s3s3s3s3s3*a's verf should have 5 snip-verfs and 5 tets.

I have no clue how the verfs actually can be constructed, but that's just based on their facets.

I see. Fun fact: s3s3s3s3s3*a's verf has 5-2 step prism symmetry, order 10. The placement of tets and snip-verfs are those of two topological pentachora. Its convex hull, when adjusted so it has two verfs (one tet, one dual-snip cell), is the cell of s3s3s3s3s3*a's dual.
Also, s3s3s3s3*a's verf is a tut, with two opposite edges reduced into points so that the hexagons become pentagons. Its dual is the ten-of-diamonds decahedron which happens to fill space.

Anyways, back to the topic, it has drifted too far...

I'll list the verfs of the bialternatosnub polychora:
s2s2s2x (bialternatosnub 2-2 duoprism) - triangular pyramid. This is actually the tetrahedral prism.
s3s2s2x (bialternatosnub 3-2 duoprism) - square pyramid. This is actually the octahedral prism.
In general, the bialternatosnub p-2 duoprism is nothing more than the p-gonal antiprismatic prism. Verfs take on trapezoidal pyramids.

s2sPs2x is the same as those mentioned above.

s3s3s2x (bialternatosnub tetrahedral dichoron) - pentagonal pyramid. This is actually the icosahedral prism.
s3s4s2x (bialternatosnub octahedral dichoron) - isosceles-pentagonal pyramid. This is actually the snub cube prism.
s3s5s2x (bialternatosnub icosahedral dichoron) - isosceles-pentagonal pyramid. This is actually the snub dodecahedron prism.

s2s2s4x (bialternatosnub 2-4 duoprism) - isosceles-triangular bipyramid
s3s2s4x (bialternatosnub 3-4 duoprism) - a square pyramid with one of its triangles augmented
For the bialternatosnub p-q duoprism with p > 2, q > 2 and even, it should have a verf similar to s3s2s4x, otherwise if p is 2 then it has a verf similar to s2s2s4x.

s2s3s4x (bialternatosnub octahedral hosochoron) - tetragonal antiwedge, which is topologically a triangular prism with one square divided into two triangles.
s3s3s4x (bialternatosnub 16-cell) - unknown, but it has 1 pentagon, 1 trapezoid, and 5 triangles.

These are all that exist in 4-space. I'm not sure how sectioning facets of bialternatosnubs generalize - maybe a tetrahedral wedge for the 5 dimensional case?

[username5243]

What do you mean by "tetrahedral wedge"?

My guess would be some kind of variant of the tet prism for the sectioning facets oof the 5D cases.

[Mercurial, the Spectre]

What do you mean by "tetrahedral wedge"?

My guess would be some kind of variant of the tet prism for the sectioning facets oof the 5D cases.

Yes. If it can be extrapolated, of course.
In 3d, we have trapezoids and in 4d, we have trips.

[Mercurial, the Spectre]

Possible Bowers short names for the bialternatosnub polychora:
s2s2s4x - badisdip (bialternatosnub digon-square duoprism)
s3s2s4x - batisdip (bialternatosnub trigon-square duoprism)
s4s2s4x - basquiddip (bialternatosnub square-square duoprism)
s5s2s4x - bapisdip (bialternatosnub pentagon-square duoprism)
s6s2s4x - bahisdip (bialternatosnub hexagon-square duoprism)

s2s2s6x - badihdip (bialternatosnub digon-hexagon duoprism)
s3s2s6x - bathdip (bialternatosnub trigon-hexagon duoprism)
s4s2s6x - bashiddip (bialternatosnub square-hexagon duoprism)
s5s2s6x - baphiddip (bialternatosnub pentagon-hexagon duoprism)
s6s2s6x - bahiddip (bialternatosnub hexagon-hexagon duoprism)

s2s3s4x - baocho (bialternatosnub octahedral hosohedron)

s3s3s4x - bahex (bialternatosnub 16-cell)

[Mercurial, the Spectre]

One thing about these polytopes, is that in 4D, the only uniform bialternatosnubs are:
b{p,q,2} = s{p,q} x {2} or s{p,q} prism. This includes the b{p,2,2}/b{2,x,2} = s{2,p} x {2} or p-gonal antiprismatic prism.

To see, look at b{P,Q,2R} or sPsQs(2R)x. Cells then are:
sPsQs - snub {P,Q}.
sQs(2R)x - topologically identical to xQo(2R)x or cantellated {Q,R}.
sPs2x - topologically identical to the P-gonal prism (with rectangular lateral sides in general). These have P-gonal prismatic symmetry.
s2s(2R)x - topologically identical to the 2R-gonal prism (with trapezoidal sides in general). These have R-gonal antiprismatic symmetry.
sefa - triangular prism (basically a wedge). In general it has C2v symmetry.

In general it is possible for sPsQs and sPs2x to be made uniform by choosing the right parameters. But in doing so, you have to account for the distortion of xPxQx that when alternated, produces the uniform sPsQs. xQx must always be a 2Q-gon, but the uniform solution of sQs(2R)x requires that it must have xQo(2R)x symmetry (meaning that it cannot be derived from xQx2Rx; to see, look at the polygon x(2R)o. Then if it is actually derived from x(2R)x then half of x(2R)o's edges must be diagonals, which must force degeneracy of x(2R)x back into x(2R)o, since a diagonal of a regular polygon is not the same length as its edge length. This forces the omnitruncate to become a degenerate cantellation, which means xQx becomes xQo (i.e. a Q-gon). Exceptions exist for sQs2x (since xQo2x and xQx2x are the same), so the possible choices for uniformity are bialternatosnubs of {p,q,2} (mentioned earlier).

Cheers!
Mercurial