Hi.gher. Space

by **Klitzing** » Fri Dec 07, 2012 2:53 pm

Don't know whether you already kept track (in your CRF-research) of this one:

Bidex - (BI dex / bi24 diminished 600) - cells are 48 teddis, teddi is the tridiminished icosahedron. This object is convex and has directional tetrahedral swirlprism symmetry - its an unusual shaped dice! Bidex was discovered by Andrew Weimholt in 2004 and it got a bit of attention amongst the polychoronist. It's verf is a 6 sided chiral faceting of ike with 4 isosceles triangles and 2 trapezoids. Bidex has 72 vertices and it is orientable.

Just stumbled upon it in hedrondudes website at: Category S3: Special Scaliforms.

--- rk

by **quickfur** » Fri Dec 07, 2012 4:16 pm

Klitzing wrote:Don't know whether you already kept track (in your CRF-research) of this one:

Bidex - (BI dex / bi24 diminished 600) - cells are 48 teddis, teddi is the tridiminished icosahedron. This object is convex and has directional tetrahedral swirlprism symmetry - its an unusual shaped dice! Bidex was discovered by Andrew Weimholt in 2004 and it got a bit of attention amongst the polychoronist. It's verf is a 6 sided chiral faceting of ike with 4 isosceles triangles and 2 trapezoids. Bidex has 72 vertices and it is orientable.

Just stumbled upon it in hedrondudes website at: Category S3: Special Scaliforms.

--- rk

I already have a page for it. Complete with a mind-blowing animation at the end.

by **Klitzing** » Sat Dec 08, 2012 12:22 am

Wasn't aware, quickfur...

Well, I'd read that very thread in those days, as Andrew came up with it (back in 2004). But I had completely forgotten about it. Just found it again on hedrondude's page. - Inbetween those 2 posts I've also calculated the incidence matrix of bidex. That one shows its high degree of symmetry even more: it is a rather small matrix!

Code: Select all: 72 | 2 4 | 2 3 5 | 6 ---+--------+----------+--- 2 | 72 * | 2 1 1 | 4 (bottom or top edges of teddi) 2 | * 144 | 0 1 2 | 3 (other edges of teddi) ---+--------+----------+--- 3 | 3 0 | 48 * * | 2 (bottom or top {3} of teddi) 3 | 1 2 | * 72 * | 2 (lateral {3} of teddi) 5 | 1 4 | * * 72 | 2 ---+--------+----------+--- 9 | 6 9 | 2 3 3 | 48 (teddi)

Edit: this matrix was wrong! Corrected below.

--- rk

by **quickfur** » Sat Dec 08, 2012 12:40 am

It's a pretty cool shape. It's cell-transitive and vertex-transitive, and has a chiral version of the augmented 24-cell symmetry exhibited by o3x4x3o. It has two kinds of edges, as found by Keiji & myself in this thread, which are distributed in a chiral way round each cell. I independently (re)discovered its swirlprism-like symmetry while I was doing its projections for my website. It was a pretty exhilarating moment when it dawned on me.

by **Klitzing** » Sat Dec 08, 2012 9:50 am

quickfur wrote:It's a pretty cool shape. It's cell-transitive and vertex-transitive, and has a chiral version of the augmented 24-cell symmetry exhibited by o3x4x3o. It has two kinds of edges, as found by Keiji & myself in this thread, which are distributed in a chiral way round each cell. I independently (re)discovered its swirlprism-like symmetry while I was doing its projections for my website. It was a pretty exhilarating moment when it dawned on me.

Hy quickfur, now I see, that this was the thingy you were already discussing (in that thread) when I was entering the forum. oops. More bad, I even contributed to that thread.

Well, yesterday I was just using Andrews information from the polylist discussion, that there are 2 types of edges with 3 resp. 4 cells around. Those would be 144 and 72 in count. So this would reflect onto the teddies that their edges have to be divided into 9+6. As Andrew )and others) also pointed out that the teddies come aligned in circles of 6, adjoining top and bottom triangle, it got clear, that those 6 edges are of the same type.

Keiji then first assumed (in that thread) that all edges of teddi-triangles would be alike (giving the false dissection of 12+3). I yestreday assumed it the other way round: all non-top/non-bottom edges to be in the remaining class: this lead to the correct dissection of 6+9. But now I see, that you already have prooved by computer aid that the 2 non-top edges of the lateral triangles are neither both of type A nor both of type B, infact they make teddi itself chiral: 1 is of type A, one of type B, as in your pic:

Damn. This too makes a division of teddi edges into 6+9. Thus my yesterday incidence matrix comes out to be wrong after all. With this additional input it should read instead:

Code: Select all: 72 | 4 2 | 2 3 5 | 6 ---+--------+----------+--- 2 | 144 * | 1 1 1 | 3 : bottom or top edges + one chiral edge each of the lateral {3} of teddi 2 | * 72 | 0 1 3 | 4 : other edges of teddi ---+--------+----------+--- 3 | 3 0 | 48 * * | 2 : bottom or top {3} of teddi 3 | 2 1 | * 72 * | 2 : lateral {3} of teddi 5 | 2 3 | * * 72 | 2 ---+--------+----------+--- 9 | 9 6 | 2 3 3 | 48 : teddi

Thanks for that hint!

--- rk

by **Klitzing** » Sun Dec 09, 2012 5:11 pm

Recently I have revisited ex (x3o3o5o), sadi (s3s4o3o), and bidex (a swirlsymmetric noble scaliform, all of its cells are teddi, i.e. tridiminished icosahedra).

It is well-known, that sadi is a special diminishing of ex. In fact, into ex a tau-scaled ico (x3o4o3o) can be vertex inscribed. If those 24 vertices of ex would be chopped off, resulting in an icosahedral section each, the outcome is nothing but sadi. - The same holds true for bidex with respect to that 48 vertices of ex, which correspond to the ones of 2 vertex inscribed, tau-scaled icoes, provided these 2 icoes would be chosen such that these do not have any common vertices. (Moreover this is the reason for the its name, bidex: "b"i-"i"co-"d"iminished-"ex"). As those chopped vertices in this latter case come closer than before, the correponding sections would intersect themselves. Therefore those ikes (x3o5o) of sadi result in teddies for bidex, i.e. "t"ri-"d"iminished "i"kes.

Today I would like to have a look onto the related expanded versions each.

Ex then becomes sidpixhi ("s"mall "d"i-prismatic he"x"acosi-"h"ecaton"i"cosachoron, x3o3o5x). This operation is a Stott expansion (the one with respect to the edges of the dual hi, i.e. o3o3o5x): The cells of ex (tets, x3o3o) will be pulled apart by 1 edge length. Thus the original triangles result in trips (x x3o). The former edges would be replaced by pips (x x5o), and the vertices by the duals of the former vertex figure, i.e. by does (o3o5x).

The ikes of sadi are the sefas ("se"ctioning "fa"cets underneath) of the ex-vertices, they are the bottom faces of the chopped off vertex pyramids. For sidpixhi the icosahedral axes would not end in vertices again, but at its does. Accordingly we would have to look now for the sefas of the does of sidpith, i.e. the bottom faces of the to be chopped off doe-cupolae.

Any doe-face will be adjacent to a pip, moreover the pentagon will have the same orientation as that of doe. Inbetween 2 pips a trip can be placed, thus extending the doe-edges into squares at the farer end. Finally into the remaining pinches tets can be inserted, extending doe-vertices towards the farer end as additional triangles. Accordingly that searched for bottom face of the doe cupola is just a srid (x3o5x).

That is, the idsid pixhi ("i"co-"d"iminished "sidpixhi") would ask for 24 srids, 24+96=120 tets, 96+96+288=480 trip, 144+288=720 pip, and 96 doe.

Now for bidex there will be some bidsid pixhi ("b"i-"d"iminished "sidpixhi"). The corresponding diminished section here is the tedrid ("t"ri-"d"iminished s"rid"), an other Johnson solid. The decagons here represent a complex of 1 pip, 5 trip, and 5 tet from sidpixhi. This is similar as for bidex itself, where the pentagons of the teddies each represent a complex of 5 tetrahedra around a common orthogonal edge. Accordingly bidsid pixhi would have as set of cells: 48 tedrids (corresponding to the teddies of bidex), 48+72=120 trip (corr. to the {3} of bidex), 144+72=216 pip (corr. to the edges of bidex), and 72 does (corr. to the vertices of bidex).

All of these clearly are convex and regular faced. They even have an unique circumradius.

--- rk

by **wendy** » Mon Dec 10, 2012 7:21 am

I had a look last night at the 'bidex' or diminished snub 24choron. This is what i found by trying to make a model of it.

The polytope is read as subregimental to the (3,3,5), that is, its vertices and edges are a subset of the vertices and edges of (3,3,5). The hedra include pentagons, not found in 3,3,5.

The faces are not related to o3x4x3o. One can see this, because the o3x4x3o would create the faces in line with the simplex not connected to the icosahedra. However, the diminishing creates faces with normals pointing to a vertex. For it to contain the base chiral group of the octagonny (ie <48>), it would need to be class 4, but the thing is a class 2 structure.

The next step is to see if it contained the swirlgroup <24>, the primary group in (3,4,3) and shared with (3,3,5) (which is <120>. The compound (3,3,5)[5(3,4,3)](5,3,3) is a likely place to start: three of its 24chora would belong to vertices, and the other two are face-normals. It is possible to associate a vector with this particular face, pointing from the fat end to the single end.

The indicated vertex figures suggest that this vector does not go nose to tail with the next, which means that the faces do not fall into eight rings of six. Instead, the arrows point to one of the base triangles in the next cell. I would need to look at this.

Good work to all involved!!!

by **Klitzing** » Mon Dec 10, 2012 4:42 pm

Klitzing wrote:Recently I have revisited ex (x3o3o5o), sadi (s3s4o3o), and bidex (a swirlsymmetric noble scaliform, all of its cells are teddi, i.e. tridiminished icosahedra).

[...]

Today I would like to have a look onto the related expanded versions each.
Edit: i.e. corresponding to the decoration string xoox

[...]

Moreover we could try to use the small rhombated version (xoxo).

Ex then becomes srix ("s"mall "r"hombi-he"x"acosachoron). Having 600 coes ("c"ub-"o"ctahedra, which relate to the tets of ex), 720 pips ("p"entagonal "p"risms, corr. to the edges of ex), and 120 ids ("i"cosi"d"odecahedra, corr. to the vertices of ex). The triangles of ex here remain triangles (even so twice as many additional ones will come in here).

Similar as yesterday, any former vertex direction (of ex) now would be the one towards the ids (of srix). So, when chopping those off while diminishing, we have first to construct the corresponding cupolae, in order to get the corresponding sefas. The surroundings of each id are 12 pip and furthermore 20 half-coes. These half-coes are attached to the triangles of id. Accordingly those are tricues ("tri"gonal "cu"polae). That equatorial section gives rise to new hexagons. Thus we would get as bottom facet of this id-cupola a ti ("t"runcated "i"cosahedron).

Now let's have a look at idsrix ("i"cosatetra-"d"iminished "srix"). That one would have 600-24*20=120 coes (those would correspond to those 120 tets of sadi, which are completely surrounded by further tets), 720-24*12=432(=144+288) pips, 120-24=96 ids, 24 ties, but there will be also 24*20=480 tricues! - Probably a nice finding again, as that one is convex, regular faced (i.e. CRF), and uses Johnson solids as cells.

As to the bidex analogue, we would have to intersect 2 ties by reverse connection of a pip, which chops off a pentagon together with all 5 incident edges. But this would half the adjacent hexagons (becoming trapezia of side-lengths 1-1-1-2). And in fact, those tricues, occuring in idsrix, would likewise be cut down to thingies bounded by 1 square, 2 triangles, and 2 half-hexagons. Therefore, even so such an analogue clearly exists too, sadly it would not be CRF in that case here.

--- rk

PS @Jonathan:
None of those names, "idsid pixhi", "bidsid pixhi", and "idsrix" so far has got the status of a true OBSA ("o"ficial "B"owers-"s"tyle "a"cronym). They seem not been used so far, as to my listings of those. Would you please cross-check in your listings? (Or would you even like to respell them?)

PPS @Jonathan:
BTW, both "bidex" and "bidsid pixhi" don't make clear, whether to mean "bi"-"d"iminished or "b"i-"i"cositetra-"d"iminished. You should keep track for that! (Even so bidex now is coined such for a period of 8 years...) - Perhaps bidiminishing would ask for a further adjective (such as para- and meta-, as used within the Johnson solids, so there could be a need for some prefixing letters in front of that "bid" in these opposing cases?

by **Klitzing** » Tue Dec 11, 2012 2:27 pm

Klitzing wrote:
Klitzing wrote:Recently I have revisited ex (x3o3o5o), sadi (s3s4o3o), and bidex (a swirlsymmetric noble scaliform, all of its cells are teddi, i.e. tridiminished icosahedra).

[...]

Today I would like to have a look onto the related expanded versions each.
Edit: i.e. corresponding to the decoration string xoox

[...]

Moreover we could try to use the small rhombated version (xoxo).

{...]

Still not the end of the story!

We could try to use the truncated version (xxoo). Applied to the ex we'd get tex ("t"runcated "ex"). That one has 600 tuts (corr. to the tets of ex) and 120 ikes (corr. to its vertices). But then, when looking for according diminishings, we would have to chop off some of those ikes one edge-length deep. As any ike is completely surrounded by tuts, this results in cutting those tuts into 2, i.e. getting triangular sections of edge length 2! The corresponding ike-cap in fact is nothing but a variation of an ike-prism, the bottom base having twice the size of the top one. - Thus, even so we can apply that pattern onto ex, we can't not even on sadi (at least within CRFs).

Thus, so far we have the followings:

"xxoo" can only be applied onto ex (-> tex).
"xoxo" can be applied onto ex (-> srix) and sadi (-> idsrix).
"xoox" can be applied onto ex (-> sidpixhi), sadi (-> idsid pixhi), and bidex (-> bidsid pixhi).

Therefore we could try "xoxx" next. That one should then be applicable onto ex and sadi.

"xoxx" applied to ex results in prix ("p"rismato-"r"hombated he"x"acosachoron). Its cells are 600 coes (like for srix, corr. to the tets of ex), 1200 trips (like for sidpixhi, corr. to the triangles of ex), 720 dips ("d"ecagonal "p"risms, corr. to the edges of ex), and 120 tids ("t"runc. "d"odecahedra, corr. to the vertices of ex).

For diminishing in sadi style, we need to have a look at those caps in the icosahedral symmetry directions. Thus, what is below these tids? The decagons of tid clearly connect to dips. The triangles of tid connect to coes. And the dodecahedral edges of tid connect to further trips (used here as digonal cupolae). Accordingly the next section below is a grid (great rhombicosidodecahedron). Again this halves the coes into 2 tricues. One half being counted in those chopped off caps, the other one being contained within the remainder. - BTW. this again becomes the reason, why the bidex analogue won't work here too.

Thus we will get here for the "xoxx-sadi", i.e. the idprix ("i"cositetra-"d"iminished "prix"), a further CRF polychoron using Johnson solids within its cell set: 600-24*20=120 coes (likewise corr. to those tets of sadi, which are completely surrounded by further tets), 1200-24*30=480 trips, 720-24*12=432 dips, 120-24=96 tids, those new 24 grids, and finally the 24*20=480 tricues.

--- rk

by **Klitzing** » Tue Dec 11, 2012 3:32 pm

Klitzing wrote:... 600-24*20=120 coes (likewise corr. to those tets of sadi, which are completely surrounded by further tets) ...

Oops, sadi clearly has 120 tets in total, which fall into 24 (completely surrounded) plus 96 (pyramidal symmetric only). Accordingly the 120 coes of idsrix as well as those of idprix would fall into 24+96 likewise.

--- rk

by **Klitzing** » Tue Dec 11, 2012 8:21 pm

I've not fully got around the symmetry orbits of the respective elements for these recently found CRFs,
but at least I got the total counts each! (Provided I made no miss-counts...)

Here are the total counts for the sadi relatives, based on those of the ex relatives:

Code: Select all: | ex | 1-dim | sadi ------------+-------+-------+------- vertices | 120 | -1 | 96 ------------+-------+-------+------- edges | 720 | -12 | 432 ------------+-------+-------+------- {3} | 1200 | -30 | 480 ------------+-------+-------+------- tet | 600 | -20 | 120 ike | 0 | +1 | 24

Code: Select all: | srix | 1-dim | idsrix ------------+-------+-------+------- vertices | 3600 | -30 | 2880 ------------+-------+-------+------- edges (sum) | 10800 | -120 | 7920 ------------+-------+-------+------- {3} (sum) | 3600 | -50 | 2400 {4} | 3600 | -60 | 2160 {5} | 1440 | -12 | 1152 {6} | 0 | +20 | 480 ------------+-------+-------+------- co | 600 | -20 | 120 pip | 720 | -12 | 432 id | 120 | -1 | 96 tricu | 0 | +20 | 480 ti | 0 | +1 | 24

Code: Select all: | prix | 1-dim | idprix ------------+-------+-------+------- vertices | 7200 | -60 | 5760 ------------+-------+-------+------- edges (sum) | 18000 | -210 | 12960 ------------+-------+-------+------- {3} (sum) | 4800 | -80 | 2880 {4} (sum) | 7200 | -120 | 4320 {6} | 0 | +20 | 480 {10} | 1440 | -12 | 1152 ------------+-------+-------+------- co | 600 | -20 | 120 trip | 1200 | -30 | 480 dip | 720 | -12 | 432 tid | 120 | -1 | 96 tricu | 0 | +20 | 480 grid | 0 | +1 | 24

Code: Select all: | sid- | 1-dim | idsid | pixhi | | pixhi ------------+-------+-------+------- vertices | 2400 | -20 | 1920 ------------+-------+-------+------- edges (sum) | 7200 | -90 | 5040 ------------+-------+-------+------- {3} | 2400 | -60 | 960 {4} | 3600 | -60 | 2160 {5} | 1440 | -12 | 1152 ------------+-------+-------+------- tet | 600 | -20 | 120 trip | 1200 | -30 | 480 pip | 720 | -12 | 432 doe | 120 | -1 | 96 srid | 0 | +1 | 24

By "1-dim" the contribution by any single diminishing is meant. The final column thus adds that medial values 24 times to the left values, in accordance to the icositetrachoral-symmetric diminishing.

For the sadi relatives those 1-dims are just reductions by according cupolae (resp. pyramids for sadi itself): all non-bottom-base elements will be reduced by the provided values. No intersection of 2 neighbouring diminishings does occur.

--- rk

by **Klitzing** » Tue Dec 11, 2012 8:44 pm

Now to the more complex case of the total counts for the bidex relatives (in fact, bidex itself and bidsid pixhi only). Again in the column "1-dim" the reductions for any single diminishing is counted. But here, the corresponding values will have to be multiplied by 48=2*24. Further, those diminishings would intersect oneanother. Therefore some parts are counted in units, while others are counted in halves only - as the latter would be contained within the intersection of 2 diminishings.

Code: Select all: | ex | 1-dim | bidex ------------+-------+------------+------- vertices | 120 | -1 | 72 ------------+-------+------------+------- edges | 720 | -7 -3/2 | 216 ------------+-------+------------+------- {3} | 1200 | -15 -15/2 | 120 {5} | 0 | +3/2 | 72 ------------+-------+------------+------- tet | 600 | -5 -15/2 | 0 teddi | 0 | +1 | 48

Code: Select all: | sid- | 1-dim | bidsid | pixhi | | pixhi ------------+-------+-----------+------- vertices | 2400 | -20 | 1440 ------------+-------+-----------+------- edges (sum) | 7200 | -75 -15/2 | 3240 ------------+-------+-----------+------- {3} | 2400 | -30 -30/2 | 240 {4} | 3600 | -45 -15/2 | 1080 {5} | 1440 | -12 | 864 {10} | 0 | +3/2 | 72 ------------+-------+-----------+------- tet | 600 | -5 -15/2 | 0 trip | 1200 | -15 -15/2 | 120 pip | 720 | -9 -3/2 | 216 doe | 120 | -1 | 72 tedrid | 0 | +1 | 48

Edit: edge & triangle counts formerly had been wrong.

Note that the tets will be reduced completely in either case of the bidex relatives.

--- rk

by **wendy** » Wed Dec 12, 2012 11:08 am

Incidence matrices can not distinguish between two uniform polytopes that have the same vertex figure, but different constructs.

by **Klitzing** » Wed Dec 12, 2012 1:35 pm

wendy wrote:Incidence matrices can not distinguish between two uniform polytopes that have the same vertex figure, but different constructs.

Hein? For sure they can. Incidence matrices are equivalent to abstract polytopes. Their geometric realisation is not contained, neither in the abstract description, nor in the incidence matrices. Vertex figures on the other hand are like the polytopes you'd consider, geometrical shapes. If you will differ 2 equivalent shapes by the means of their construction, then those are realisations of 2 different abstract polytopes. And those will have respective incidence matrices!

BTW. what are you after by that comment? Can't see its connection to the previous posts.

--- rk

by **Klitzing** » Thu Dec 13, 2012 2:45 pm

To derive the respective incidence matrices of these new figures described in the last few days, it supposedly would be better not to try for the whole polychora at once, but to restrict counting onto some specific regions only. What this could mean will be outlined below.

First we look for the fundamental region of sadi. Consider for that reason the vertices of sadi as a set of points on the hypersphere. Next derive the respective Voronoi cells, i.e. the regions of points which are closer to a specific set member than to any other. Applying this to sadi, which is a uniform polychoron, therefore clearly results in Voronoi cells which are all identical. Thus that cell would be the searched for fundamental region.

Now, what does that one look like? Obviously it would be the dual of the vertex figure. The latter one is teddi (the tridiminished ike). Dualizing a diminishing of some figure is equivalent to looking for a stellation of the dual figure. Here we have a diminishing that is a partial faceting, so we will get a partial stellation as well.

The dual of ike (x3o5o) is doe (o3o5x). Teddi diminishes 3 mutually non-adjacent vertices of ike. Accordingly we will have to stellate 3 non-adjacent faces of doe, kind as for the sissid (small stellated doe). The resultung figure could be understood like this:

Take a regular pentagon. Next use one such and omitt one of its sides, extending the adjacent ones until they meet. This gives a phylloidal tetragon of sides 1-1-tau-tau. Then use one such and do the same thing with one of the remaining short sides. This results in a flat triangle of sides tau-tau-(tau^2). We will need 3 of those tetragons and 6 of those triangles: we connect the tetragons at their short sides. Into each gap we insert one triangle each. Thus there will be 3 edges left to connect, all of which are of size tau^2. The remaining 3 triangles build the surface area of a triangular pyramid, base attached to the former construction.

Those pentagons are not only a building guide for those to be used faces, they likewise play an essential role in the followings. Especially their face-centers. Further those points on any of the edges of that fundamental region, which correspond to the side-centers of the original pentagons: In fact, those pentagonal face-centers correspond to the edge-midpoints of sadi, and those just marked points on the edges of the fundamental region correspond to the centers of the triangles of sadi. (Note that sadi has no other 2-dimensional face type.)

According to the vertex count of sadi, there will be 96 copies of that fundamental domain, which will make for a full representation of its surface. Obviously those tetragons can connect in just a single way (esp. as the figure has a threefold axial symmetry). But the 6 triangles fall in 2 types: those 3 (type A) directly connected to the tetragons and those 3 (type B) opposite to them. When placing 3 such fundamental domains tetragon to tetragon around one of the small edges, we see that their A-triangles (incident to one vertex of that very edge) form a dimple, which can be filled only by that vertex of the domain which is incident to the 3 B-triangles. I.e. any A-triangle will connect to a B-triangle and vice versa. (Therefore this figure will be not a mirror margin figure of Wendy's terminology).

The idea now is to derive the desired element counts within this much smaller unit, multiply the derived values by 96, and get the searched for element counts of any of those sadi derivatives. (A similar procedure later might serve for the bidex relatives.) Any point inside the fundamental domain will count for itself. Any point on its faces would count 1/2, according to dyadicness. Any point on an edge counts 1/3, as those are orthogonal to the sadi triangles through their face centers. Any domain vertex, which is one of the doe vertices, counts for 1/4, as it is the body center of a tet of sadi. And any point at 3 the peaky tips of the domain counts 1/12, as those are the body centers of sadis ikes and the corresponding directions come from the vertices of the ike. (In fact, the domain itself was build around one of the sadi vertices, which is incident to 3 ikes.)

Before applying to new shores, we'd like to test this procedure for sadi itself:
There is exactly 1 vertex of sadi in this domain. (It is the point around which the domain was constructed.) Hence sadi has 1*96=96 vertices.
There is exactly 1 half-edge of sadi running from that specific inner point to the midpoint of any of those pentagons. As the domain has 9 faces, we get 9*96/2=432 edges (division by 2, as 2 half-edges equate to a single complete one). Moreover we could divide this count into those edges, which are orthogonal to those tetragons and those to the triangles of the domain. Then we would get 3*96/2+6*96/2=144+288.
There is exactly 1 third-triangle (vertex - side-center - face-center - side-center) for any domain edge. In fact any (total) triangle is centered at those specific points on those edges. There are 3 edges of length 1, 9 edges of length tau, and 3 edges of length tau^2. Accordingly we have (3+9+3)*96/3=480 triangles in sadi, which can be grouped (according to the different edge lengths, i.e. by the inter domain connectedness) into 3*96/3+9*96/3+3*96/3=96+288+96.
There is exactly 1 quarter-tet for any doe-type vertex of the domain: (1+3+1)*96/4=120, and there is exactly 1 twelfth-ike for any sissid-type vertex of the domain: 3*96/12=24. Moreover the doe-type vertices could be grouped into subsets according to being coincident to a tip of the flat triangles, resp. not coincident. This gives 1*96/4+(3+1)*96/4=24+96.
I.e. the corresponding counts are thus correctly recovered. In fact, the incidence matrix of sadi is known to be

Code: Select all: s3s4o3o demi( . . . . ) | 96 | 3 6 | 3 9 3 | 3 1 4 ----------------+----+---------+-----------+--------- . s4o . | 2 | 144 * | 0 2 2 | 1 1 2 sefa( s3s . . ) | 2 | * 288 | 1 2 0 | 2 0 1 ----------------+----+---------+-----------+--------- s3s . . | 3 | 0 3 | 96 * * | 2 0 0 sefa( s3s4o . ) | 3 | 1 2 | * 288 * | 1 0 1 sefa( . s4o3o ) | 3 | 3 0 | * * 96 | 0 1 1 ----------------+----+---------+-----------+--------- s3s4o . | 12 | 6 24 | 8 12 0 | 24 * * (ike) . s4o3o | 4 | 6 0 | 0 0 4 | * 24 * (tet, in full symm.) sefa( s3s4o3o ) | 4 | 3 3 | 0 3 1 | * * 96 (tet, in pyr. symm.)

Next we like to project this domain onto the sidpixhi inscribed idsid pixhi.

The single vertex of ex becomes a full doe in sidpixhi. Still this doe is completely contained within the domain of consideration. Thus we get 20*96=1920 vertices for idsid pixhi. According to the overall symmetry of the domain we could devide those into (1(A)+3(B)+6(C)+6(D)+3(E)+1(F))*96=96(A)+288(B)+576(C)+576(D)+288(E)+96(F).
Next are the edges. Clearly that doe has 30 edges. It is completely contained within the domain, so these count fully. Then each ex edge in sidpixhi is substituted by a pip, i.e. gets multiplied by 5 (for the lacing ones). Those will likewise intersect the pentagonal regions of the domain (the ones directing to the domains 3 tips are replaced as srids), thus these are half-edges again: 30*96+(12-3)*5*96/2=5040. The 2880 edges of those does themselves can be devided into (3(AB)+6(BC)+3(CC)+6(CD)+3(DD)+6(DE)+3(EF))*96=288(AB)+576(BC)+288(CC)+576(CD)+288(DD)+576(DE)+288(EF). The 2160 halved edges would be divided into (3(AA')+6(BB')+6(CC')+3(BB*)+6(CC*)+6(DD*)+6(DD#)+6(EE#)+3(FF#))*96/2=144(AA')+288(BB')+288(CC')+144(BB*)+288(CC*)+288(DD*)+288(DD#)+288(EE#)+144(FF#). - With respect to those halved edges we further should apply the outer connectedness of the domains. Then (AA'), (BB'), and (CC') clearly will be reflected into themselves. (BB*) will be reflected into (FF#), (CC*) into (EE#), and (DD*) into (DD#). Therefore we could further identify accordingly (speaking of the same elements in fact), i.e. 2160=144(AA')+288(BB')+288(CC')+288(BB*&FF#)+576(CC*&EE#)+576(DD*&DD#).
Pentagons are completely contained in the domain. There thus are 12*96=1152. Those can be divided into (3(ABC)+3(BCD)+3(CDE)+3(DEF))*96=288(ABC)+288(BCD)+288(CDE)+288(DEF). Note that only the pentagons of type (CDE) connect the does to srids, all others connect the does to pips.
Squares connect edges of one doe to an other doe (along a pip). Thus those are contained only as halfes. We have in total: (12-3)*5*96/2=2160, respectively according to the domain symmetry: (6(AB')+6(BC')+3(CC')+6(BC*)+6(CD*)+3(DD*)+3(DD#)+6(DE#)+6(EF#))*96/2=288(AB')+288(BC')+144(CC')+288(BC*)+288(CD*)+144(DD*)+144(DD#)+288(DE#)+288(EF#). From inter domain connectedness we can unite (BC*) with (EF#), (CD*) with (DE#), and (DD*) with (DD#). Thus: 2160=288(AB')+288(BC')+144(CC')+576(BC*&EF#)+576(CD*&DE#)+288(DD*&DD#). Only the types (CC') and (CD*&DE#) connect pips to srids, the others pips to trips.
Triangles are a bit harder. Those still are orthogonal to and centered around the domain edges. In fact 2 each for every edge. But they count only 1/3 each. Accordingly we have: 15*2*96/3=960, which (for a single domain) would divide into (3(AB)+3(AB')+6(BC)+6(BC')+2*3(DD)+3(EF)+3(EF'))*96/3=96(AB)+96(AB')+192(BC)+192(BC')+192(DD)+96(EF)+96(EF'). From inter domain connection we further get the identification of triangle types (BC) and (EF') resp. of (BC') and (EF). Thus 960=96(AB)+96(AB')+288(BC&EF')+288(BC'&EF)+192(DD). Here (BC'&EF) and (DD) connect trip to srid, all others connect trip to tet.
Tets correspond to the doe-type vertices of the domain. Each would therefore belong for 1/4 only to the domain. Thus we have: (1(A)+3(B)+1(F))*96/4=24(A)+72(B)+24(F), i.e. a total of 120. Due to the domain connectedness we can unite (B) and (F). Thus: 120=24(A)+96(B&F).
Trips are situated between those triangle pairs, i.e. around those domain edges each. Like for the triangle they count also by 1/3: (3(AB)+6(BC)+3(DD)+3(EF))*96/3=96(AB)+192(BC)+96(DD)+96(EF), or 480 in total. Again (BC) and (EF) could be united: 480=96(AB)+288(BC&EF)+96(DD).
Pips run through those pentagonal regions of the domain faces. They contribute 1/2 each: (3(ABC)+3(BCD)+3(DEF))*96/2=144(ABC)+144(BCD)+144(DEF), or 432 in total. Here again (BCD) and (DEF) can be unified: 432=144(ABC)+288(BCD&DEF).
Does are completely contained within the fundamental cell. There is 1 such each, fully contributing: 1*96=96.
Srids correspond to the ikes of sadi. Each region uses parts of 3 of them (at the sissid type vertices). They contribute 1/12 each. Thus: 3*96/12=24. Here any pentagon of srid is adjacent to a (CDE)-pentagon of doe. Accordingly the srid vertices divide into 60=12*(2(C)+2(D)+1(E))=24(C)+24(D)+12(E). From within the region it further follows that the triangles of srid belong by 2/5 to (BC'), by 2/5 to (DD), and by 1/5 to (EF). As the first and last were unified we get 20=4(2(BC')+2(DD)+1(EF))=12(BC'&EF)+8(DD) - just as for the division of the triangles of ike within sadi. Likewise from within the region we get for the squares of srid: 1/5 belongs to (CC'), 2/5 to (CD*), and 2/5 to (DE#). Here the latter ones were unified, thus 30=6(1(CC')+2(CD*)+2(DE#))=6(CC')+24(CD*&DE#).

This all put together results in the incidence matrix of idsid pixhi:

Code: Select all: A | 96 * * * * * | 3 0 0 0 0 0 0 3 0 0 0 0 0 | 3 0 0 0 0 6 0 0 0 0 0 3 0 0 0 | 1 0 3 0 0 3 0 1 0 B | * 288 * * * * | 1 2 0 0 0 0 0 0 2 0 1 0 0 | 0 1 2 0 0 2 2 0 2 0 0 2 1 0 0 | 0 1 1 2 0 2 1 1 0 C | * * 576 * * * | 0 1 1 1 0 0 0 0 0 1 0 1 0 | 0 0 0 1 0 0 1 1 1 1 0 1 1 1 0 | 0 0 0 1 0 1 1 1 1 D | * * * 576 * * | 0 0 0 1 1 1 0 0 0 0 0 0 2 | 0 0 0 0 1 0 0 0 0 2 2 0 1 1 1 | 0 0 0 0 1 0 2 1 1 E | * * * * 288 * | 0 0 0 0 0 2 1 0 0 0 0 2 0 | 0 0 0 1 0 0 0 0 2 2 0 0 0 1 2 | 0 0 0 1 0 0 2 1 1 F | * * * * * 96 | 0 0 0 0 0 0 3 0 0 0 3 0 0 | 0 0 3 0 0 0 0 0 6 0 0 0 0 0 3 | 0 1 0 3 0 0 3 1 0 --------+-----------------------+-----------------------------------------------------+-----------------------------------------------------------+------------------------------ AB | 1 1 0 0 0 0 | 288 * * * * * * * * * * * * | 0 0 0 0 0 2 0 0 0 0 0 2 0 0 0 | 0 0 1 0 0 2 0 1 0 BC | 0 1 1 0 0 0 | * 576 * * * * * * * * * * * | 0 0 0 0 0 0 1 0 1 0 0 1 1 0 0 | 0 0 0 1 0 1 1 1 0 CC | 0 0 2 0 0 0 | * * 288 * * * * * * * * * * | 0 0 0 0 0 0 0 1 0 0 0 1 0 1 0 | 0 0 0 0 0 1 0 1 1 CD | 0 0 1 1 0 0 | * * * 576 * * * * * * * * * | 0 0 0 0 0 0 0 0 0 1 0 0 1 1 0 | 0 0 0 0 0 0 1 1 1 DD | 0 0 0 2 0 0 | * * * * 288 * * * * * * * * | 0 0 0 0 0 0 0 0 0 0 2 0 1 0 1 | 0 0 0 0 1 0 2 1 0 DE | 0 0 0 1 1 0 | * * * * * 576 * * * * * * * | 0 0 0 0 0 0 0 0 0 1 0 0 0 1 1 | 0 0 0 0 0 0 1 1 1 EF | 0 0 0 0 1 1 | * * * * * * 288 * * * * * * | 0 0 0 0 0 0 0 0 2 0 0 0 0 0 2 | 0 0 0 1 0 0 2 1 0 AA' | 2 0 0 0 0 0 | * * * * * * * 144 * * * * * | 2 0 0 0 0 2 0 0 0 0 0 0 0 0 0 | 1 0 2 0 0 1 0 0 0 BB' | 0 2 0 0 0 0 | * * * * * * * * 288 * * * * | 0 1 1 0 0 1 1 0 0 0 0 0 0 0 0 | 0 1 1 1 0 1 0 0 0 CC' | 0 0 2 0 0 0 | * * * * * * * * * 288 * * * | 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 | 0 0 0 1 0 1 0 0 1 BB*&FF# | 0 1 0 0 0 1 | * * * * * * * * * * 288 * * | 0 0 2 0 0 0 0 0 2 0 0 0 0 0 0 | 0 1 0 2 0 0 1 0 0 CC*&EE# | 0 0 1 0 1 0 | * * * * * * * * * * * 576 * | 0 0 0 1 0 0 0 0 1 1 0 0 0 0 0 | 0 0 0 1 0 0 1 0 1 DD*&DD# | 0 0 0 2 0 0 | * * * * * * * * * * * * 576 | 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 | 0 0 0 0 1 0 1 0 1 --------+-----------------------+-----------------------------------------------------+-----------------------------------------------------------+------------------------------ AB | 3 0 0 0 0 0 | 0 0 0 0 0 0 0 3 0 0 0 0 0 | 96 * * * * * * * * * * * * * * | 1 0 1 0 0 0 0 0 0 AB' | 0 3 0 0 0 0 | 0 0 0 0 0 0 0 0 3 0 0 0 0 | * 96 * * * * * * * * * * * * * | 0 1 1 0 0 0 0 0 0 BC&EF' | 0 2 0 0 0 1 | 0 0 0 0 0 0 0 0 1 0 2 0 0 | * * 288 * * * * * * * * * * * * | 0 1 0 1 0 0 0 0 0 BC'&EF | 0 0 2 0 1 0 | 0 0 0 0 0 0 0 0 0 1 0 2 0 | * * * 288 * * * * * * * * * * * | 0 0 0 1 0 0 0 0 1 DD | 0 0 0 3 0 0 | 0 0 0 0 0 0 0 0 0 0 0 0 3 | * * * * 192 * * * * * * * * * * | 0 0 0 0 1 0 0 0 1 AB' | 2 2 0 0 0 0 | 2 0 0 0 0 0 0 1 1 0 0 0 0 | * * * * * 288 * * * * * * * * * | 0 0 1 0 0 1 0 0 0 BC' | 0 2 2 0 0 0 | 0 2 0 0 0 0 0 0 1 1 0 0 0 | * * * * * * 288 * * * * * * * * | 0 0 0 1 0 1 0 0 0 CC' | 0 0 4 0 0 0 | 0 0 2 0 0 0 0 0 0 2 0 0 0 | * * * * * * * 144 * * * * * * * | 0 0 0 0 0 1 0 0 1 BC*&EF# | 0 1 1 0 1 1 | 0 1 0 0 0 0 1 0 0 0 1 1 0 | * * * * * * * * 576 * * * * * * | 0 0 0 1 0 0 1 0 0 CD*&DE# | 0 0 1 2 1 0 | 0 0 0 1 0 1 0 0 0 0 0 1 1 | * * * * * * * * * 576 * * * * * | 0 0 0 0 0 0 1 0 1 DD*&DD# | 0 0 0 4 0 0 | 0 0 0 0 2 0 0 0 0 0 0 0 2 | * * * * * * * * * * 288 * * * * | 0 0 0 0 1 0 1 0 0 ABC | 1 2 2 0 0 0 | 2 2 1 0 0 0 0 0 0 0 0 0 0 | * * * * * * * * * * * 288 * * * | 0 0 0 0 0 1 0 1 0 BCD | 0 1 2 2 0 0 | 0 2 0 2 1 0 0 0 0 0 0 0 0 | * * * * * * * * * * * * 288 * * | 0 0 0 0 0 0 1 1 0 CDE | 0 0 2 2 1 0 | 0 0 1 2 0 2 0 0 0 0 0 0 0 | * * * * * * * * * * * * * 288 * | 0 0 0 0 0 0 0 1 1 DEF | 0 0 0 2 2 1 | 0 0 0 0 1 2 2 0 0 0 0 0 0 | * * * * * * * * * * * * * * 288 | 0 0 0 0 0 0 1 1 0 --------+-----------------------+-----------------------------------------------------+-----------------------------------------------------------+------------------------------ A | 4 0 0 0 0 0 | 0 0 0 0 0 0 0 6 0 0 0 0 0 | 4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 | 24 * * * * * * * * B&F | 0 3 0 0 0 1 | 0 0 0 0 0 0 0 0 3 0 3 0 0 | 0 1 3 0 0 0 0 0 0 0 0 0 0 0 0 | * 96 * * * * * * * AB | 3 3 0 0 0 0 | 3 0 0 0 0 0 0 3 3 0 0 0 0 | 1 1 0 0 0 3 0 0 0 0 0 0 0 0 0 | * * 96 * * * * * * BC&EF | 0 2 2 0 1 1 | 0 2 0 0 0 0 1 0 1 1 2 2 0 | 0 0 1 1 0 0 1 0 2 0 0 0 0 0 0 | * * * 288 * * * * * DD | 0 0 0 6 0 0 | 0 0 0 0 3 0 0 0 0 0 0 0 6 | 0 0 0 0 2 0 0 0 0 0 3 0 0 0 0 | * * * * 96 * * * * ABC | 2 4 4 0 0 0 | 4 4 2 0 0 0 0 1 2 2 0 0 0 | 0 0 0 0 0 2 2 1 0 0 0 2 0 0 0 | * * * * * 144 * * * BCD&DEF | 0 1 2 4 2 1 | 0 2 0 2 2 2 2 0 0 0 1 2 2 | 0 0 0 0 0 0 0 0 2 2 1 0 1 0 1 | * * * * * * 288 * * doe | 1 3 6 6 3 1 | 3 6 3 6 3 6 3 0 0 0 0 0 0 | 0 0 0 0 0 0 0 0 0 0 0 3 3 3 3 | * * * * * * * 96 * srid | 0 0 24 24 12 0 | 0 0 12 24 0 24 0 0 0 12 0 24 24 | 0 0 0 12 8 0 0 6 0 24 0 0 0 12 0 | * * * * * * * * 24

(Damn! would there any possibilities to have a horizontal slidebar as well? - Else you should copy that matrix into an other text editor, to get the long lines either un-wrapped or not so tiny... :angry:

)

(Others will follow as time permits... :-) )

--- rk

by **Klitzing** » Fri Dec 14, 2012 11:01 am

Today we want to apply the help from that (recently derived) fundamental domain of sadi onto idsrix

The single vertex of ex here becomes a full id (o3x5o), which is completely contained within that domain. Accordingly idsrix would have 30*96=2880 vertices. Those fall with respect to the symmetry of that fundamenal domain into (3(A)+6(B)+3(C)+6(D)+3(E)+6(F)+3(G))*96=288(A)+576(B)+288(C)+576(D)+288(E)+576(F)+288(G).
For edges idsrix on the one hand clearly has all the edges of these ids. Those again are completely contained within the domain and thus have to be counted fully. So we get, already divided accordingly: (3(AA)+6(AB)+3(BB)+6(BC)+6(BD)+6(CD)+6(DE)+6(DF)+6(EF)+3(FF)+6(FG)+3(GG))*96=288(AA)+576(AB)+288(BB)+576(BC)+576(BD)+576(CD)+576(DE)+576(DF)+576(EF)+288(FF)+576(FG)+288(GG), a total of 5760. On the other hand there will be additionally the lacing edges of all those pips, which remain from srix in the actual diminishing. The remaining pips are adjacent to the ids and do cross the faces of that fundamental region exactly in the pentagonal base parts of those faces. Thus we have for the corresponding half-edges: (6(AA')+6(BB')+3(CC')+6(BB*)+6(DD*)+3(EE*)+3(EE#)+6(FF#)+6(GG#))*96/2=288(AA')+288(BB')+144(CC')+288(BB*)+288(DD*)+144(EE*)+144(EE#)+288(FF#)+288(GG#), a further total of 2160 - thus we get 5760+2160=7920 edges all together. We just half to take account for the appropriate connectedness of these half-edges across the domain margins. As the flat triangles of those domains are connected to the same shapes (for sure), but to the other type each (while the tetragons connect to themselves), we will have to unite the classes (BB*) and (GG#), (DD*) and (FF#), resp. (EE*) and EE#). Resulting in: 2160=288(AA')+288(BB')+144(CC')+576(BB*&GG#)+576(DD*&FF#)+288(EE*&EE#).
Next there are triangles. Again of 2 types: On the one hand those of id; these are contained completely within the fundamental region. So: (1(AAA)+3(ABB)+6(BCD)+6(DEF)+3(FFG)+1(GGG))*96=96(AAA)+288(ABB)+576(BCD)+576(DEF)+288(FFG)+96(GGG), in total 1920. On the other hand there are triangles of srix, where coes connect to coes. These will be located orthogonally at the centers of the pentagon sides contained within the domain edges. These will account by 1/3 each, i.e. (3(A)+6(B)+3(E)+3(G))*96/3=96(A)+192(B)+96(E)+96(G). That division has to be unified according to the outer domain connection into 480=96(A)+288(B&G)+96(E).
Followed up by the squares, which occur in srix at the coes only. - With respect to the id they will emanate from its edges. For the diminishing under consideration this holds true only for those edges of id, which are not coincident to the bases of diminishing, the ties. Further those count by 1/2 each for any fundamental domain. Thus: (3(AA)+6(AB)+3(BB)+6(BC)+6(BD)+6(DE)+6(EF)+6(FG)+3(GG))*96/2=144(AA)+288(AB)+144(BB)+288(BC)+288(BD)+288(DE)+288(EF)+288(FG)+144(GG). We have to unify moreover (BB) and (GG), (BD) and (FG), resp. (DE) and (EF). Resulting in 2160=144(AA)+288(AB)+288(BB&GG)+288(BC)+576(BD&FG)+576(DE&EF).
Pentagons occure only at the ids (either connecting to pips or to ties). Those are contained completely in the fundamental domain, so the pentagons can be counted fully. (3(ABC)+3(BDE)+3(CDF)+3(EFG))*96=288(ABC)+288(BDE)+288(CDF)+288(EFG), together 1152. Only the (CDF) ones connect to ties.
Hexagons occur as sections of coes, i.e. as new faces in idsrix at the ties and tricues only. Those are incident at the id edges only, which connect to ties. At the fundamental domain they occur onla perpendicular to the 2 types of longer edges; in fact one at any elongation of the former pentagonal sides. They contirbute 1/3 each. (6(CD)+6(DF)+3(FF))*96/3=192(CD)+192(DF)+96(FF). By exterior symmetry (CD) and (FF) are to be unified: 480=288(CD&FF)+192(DF).
The coes are situated at the positions of the tets of sadi. Thus they are at the dodecahedral vertices of the domain. They are to be counted as 1/4. Getting (1(AAA)+3(ABB)+1(GGG))*96/4=24(AAA)+72(ABB)+24(GGG). Outer symmetry further unites the latter 2 classes: 120=24(AAA)+96(ABB&GGG).
Pips run through those pentagonal regions of the domain faces. They contribute 1/2 each: (3(ABC)+3(BDE)+3(EFG))*96/2=144(ABC)+144(BDE)+144(EFG). Connectedness unifies the last 2 classes: 432=144(ABC)+288(BDE&EFG).
The id is completely contained in the domain. It thus contributes fully: 1*96=96.
Tricues occur at the domain at any inscribed dodecahedral vertex, which not becomes a vertex of the domain. They contribute 1/3 each into the domain. (6(BCD)+6(DEF)+3(FFG))*96/3=192(BCD)+192(DEF)+96(FFG). But (BCD) and (FFG) have to be unified: 480=288(BCD&FFG)+192(DEF).
Ties occur around the sissid-type vertices of the domain. They contribute 1/12 each. 3*96/12=24.

This all put together results in the incidence matrix of idsrix:

Code: Select all: A | 288 * * * * * * | 2 2 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 | 1 1 0 0 0 0 1 0 0 2 2 0 0 0 0 2 0 0 0 0 0 | 1 1 2 0 1 0 0 0 B | * 576 * * * * * | 0 1 1 1 1 0 0 0 0 0 0 0 0 1 0 1 0 0 | 0 1 1 0 0 0 0 1 0 0 1 1 1 1 0 1 1 0 0 0 0 | 0 1 1 1 1 1 0 0 C | * * 288 * * * * | 0 0 0 2 0 2 0 0 0 0 0 0 0 0 1 0 0 0 | 0 0 2 0 0 0 0 0 0 0 0 0 2 0 0 1 0 1 0 2 0 | 0 0 1 0 1 2 0 1 D | * * * 576 * * * | 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 1 0 | 0 0 1 1 0 0 0 0 0 0 0 0 0 1 1 0 1 1 0 1 1 | 0 0 0 1 1 1 1 1 E | * * * * 288 * * | 0 0 0 0 0 0 2 0 2 0 0 0 0 0 0 0 0 2 | 0 0 0 2 0 0 0 0 1 0 0 0 0 0 4 0 1 0 1 0 0 | 0 0 0 2 1 0 2 0 F | * * * * * 576 * | 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 1 0 | 0 0 0 1 1 0 0 0 0 0 0 0 0 1 1 0 0 1 1 1 1 | 0 0 0 1 1 1 1 1 G | * * * * * * 288 | 0 0 0 0 0 0 0 0 0 0 2 2 0 0 0 2 0 0 | 0 0 0 0 1 1 0 1 0 0 0 2 0 2 0 0 0 0 2 0 0 | 0 1 0 2 1 1 0 0 --------+-----------------------------+-------------------------------------------------------------------------+---------------------------------------------------------------------------------+---------------------------- AA | 2 0 0 0 0 0 0 | 288 * * * * * * * * * * * * * * * * * | 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 | 1 0 1 0 1 0 0 0 AB | 1 1 0 0 0 0 0 | * 576 * * * * * * * * * * * * * * * * | 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 | 0 1 1 0 1 0 0 0 BB | 0 2 0 0 0 0 0 | * * 288 * * * * * * * * * * * * * * * | 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 | 0 1 0 1 1 0 0 0 BC | 0 1 1 0 0 0 0 | * * * 576 * * * * * * * * * * * * * * | 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 | 0 0 1 0 1 1 0 0 BD | 0 1 0 1 0 0 0 | * * * * 576 * * * * * * * * * * * * * | 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 | 0 0 0 1 1 1 0 0 CD | 0 0 1 1 0 0 0 | * * * * * 576 * * * * * * * * * * * * | 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 | 0 0 0 0 1 1 0 1 DE | 0 0 0 1 1 0 0 | * * * * * * 576 * * * * * * * * * * * | 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 | 0 0 0 1 1 0 1 0 DF | 0 0 0 1 0 1 0 | * * * * * * * 576 * * * * * * * * * * | 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 | 0 0 0 0 1 0 1 1 EF | 0 0 0 0 1 1 0 | * * * * * * * * 576 * * * * * * * * * | 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 | 0 0 0 1 1 0 1 0 FF | 0 0 0 0 0 2 0 | * * * * * * * * * 288 * * * * * * * * | 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 1 0 | 0 0 0 0 1 1 0 1 FG | 0 0 0 0 0 1 1 | * * * * * * * * * * 576 * * * * * * * | 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 | 0 0 0 1 1 1 0 0 GG | 0 0 0 0 0 0 2 | * * * * * * * * * * * 288 * * * * * * | 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 | 0 1 0 1 1 0 0 0 AA' | 2 0 0 0 0 0 0 | * * * * * * * * * * * * 288 * * * * * | 0 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 | 1 1 1 0 0 0 0 0 BB' | 0 2 0 0 0 0 0 | * * * * * * * * * * * * * 288 * * * * | 0 0 0 0 0 0 0 1 0 0 1 0 1 0 0 0 0 0 0 0 0 | 0 1 1 0 0 1 0 0 CC' | 0 0 2 0 0 0 0 | * * * * * * * * * * * * * * 144 * * * | 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 2 0 | 0 0 1 0 0 2 0 1 BB*&GG# | 0 1 0 0 0 0 1 | * * * * * * * * * * * * * * * 576 * * | 0 0 0 0 0 0 0 1 0 0 0 1 0 1 0 0 0 0 0 0 0 | 0 1 0 1 0 1 0 0 DD*&FF# | 0 0 0 1 0 1 0 | * * * * * * * * * * * * * * * * 576 * | 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 1 1 | 0 0 0 1 0 1 1 1 EE*&EE# | 0 0 0 0 2 0 0 | * * * * * * * * * * * * * * * * * 288 | 0 0 0 0 0 0 0 0 1 0 0 0 0 0 2 0 0 0 0 0 0 | 0 0 0 1 0 0 2 0 --------+-----------------------------+-------------------------------------------------------------------------+---------------------------------------------------------------------------------+---------------------------- AAA | 3 0 0 0 0 0 0 | 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 | 96 * * * * * * * * * * * * * * * * * * * * | 1 0 0 0 1 0 0 0 ABB | 1 2 0 0 0 0 0 | 0 2 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 | * 288 * * * * * * * * * * * * * * * * * * * | 0 1 0 0 1 0 0 0 BCD | 0 1 1 1 0 0 0 | 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 | * * 576 * * * * * * * * * * * * * * * * * * | 0 0 0 0 1 1 0 0 DEF | 0 0 0 1 1 1 0 | 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 | * * * 576 * * * * * * * * * * * * * * * * * | 0 0 0 0 1 0 1 0 FFG | 0 0 0 0 0 2 1 | 0 0 0 0 0 0 0 0 0 1 2 0 0 0 0 0 0 0 | * * * * 288 * * * * * * * * * * * * * * * * | 0 0 0 0 1 1 0 0 GGG | 0 0 0 0 0 0 3 | 0 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 | * * * * * 96 * * * * * * * * * * * * * * * | 0 1 0 0 1 0 0 0 A | 3 0 0 0 0 0 0 | 0 0 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 | * * * * * * 96 * * * * * * * * * * * * * * | 1 1 0 0 0 0 0 0 B&G | 0 2 0 0 0 0 1 | 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 2 0 0 | * * * * * * * 288 * * * * * * * * * * * * * | 0 1 0 0 0 1 0 0 E | 0 0 0 0 3 0 0 | 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 | * * * * * * * * 96 * * * * * * * * * * * * | 0 0 0 0 0 0 2 0 AA | 4 0 0 0 0 0 0 | 2 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 | * * * * * * * * * 144 * * * * * * * * * * * | 1 0 1 0 0 0 0 0 AB | 2 2 0 0 0 0 0 | 0 2 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 | * * * * * * * * * * 288 * * * * * * * * * * | 0 1 1 0 0 0 0 0 BB&GG | 0 2 0 0 0 0 2 | 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 2 0 0 | * * * * * * * * * * * 288 * * * * * * * * * | 0 1 0 1 0 0 0 0 BC | 0 2 2 0 0 0 0 | 0 0 0 2 0 0 0 0 0 0 0 0 0 1 1 0 0 0 | * * * * * * * * * * * * 288 * * * * * * * * | 0 0 1 0 0 1 0 0 BD&FG | 0 1 0 1 0 1 1 | 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 1 1 0 | * * * * * * * * * * * * * 576 * * * * * * * | 0 0 0 1 0 1 0 0 DE&EF | 0 0 0 1 2 1 0 | 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 1 1 | * * * * * * * * * * * * * * 576 * * * * * * | 0 0 0 1 0 0 1 0 ABC | 2 2 1 0 0 0 0 | 1 2 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 | * * * * * * * * * * * * * * * 288 * * * * * | 0 0 1 0 1 0 0 0 BDE | 0 2 0 2 1 0 0 | 0 0 1 0 2 0 2 0 0 0 0 0 0 0 0 0 0 0 | * * * * * * * * * * * * * * * * 288 * * * * | 0 0 0 1 1 0 0 0 CDF | 0 0 1 2 0 2 0 | 0 0 0 0 0 2 0 2 0 1 0 0 0 0 0 0 0 0 | * * * * * * * * * * * * * * * * * 288 * * * | 0 0 0 0 1 0 0 1 EFG | 0 0 0 0 1 2 2 | 0 0 0 0 0 0 0 0 2 0 2 1 0 0 0 0 0 0 | * * * * * * * * * * * * * * * * * * 288 * * | 0 0 0 1 1 0 0 0 CD&FF | 0 0 2 2 0 2 0 | 0 0 0 0 0 2 0 0 0 1 0 0 0 0 1 0 2 0 | * * * * * * * * * * * * * * * * * * * 288 * | 0 0 0 0 0 1 0 1 DF | 0 0 0 3 0 3 0 | 0 0 0 0 0 0 0 3 0 0 0 0 0 0 0 0 3 0 | * * * * * * * * * * * * * * * * * * * * 192 | 0 0 0 0 0 0 1 1 --------+-----------------------------+-------------------------------------------------------------------------+---------------------------------------------------------------------------------+---------------------------- AAA | 12 0 0 0 0 0 0 | 12 0 0 0 0 0 0 0 0 0 0 0 12 0 0 0 0 0 | 4 0 0 0 0 0 4 0 0 6 0 0 0 0 0 0 0 0 0 0 0 | 24 * * * * * * * ABB&GGG | 3 6 0 0 0 0 3 | 0 6 3 0 0 0 0 0 0 0 0 3 3 3 0 6 0 0 | 0 3 0 0 0 1 1 3 0 0 3 3 0 0 0 0 0 0 0 0 0 | * 96 * * * * * * ABC | 4 4 2 0 0 0 0 | 2 4 0 4 0 0 0 0 0 0 0 0 2 2 1 0 0 0 | 0 0 0 0 0 0 0 0 0 1 2 0 2 0 0 2 0 0 0 0 0 | * * 144 * * * * * BDE&EFG | 0 2 0 2 2 2 2 | 0 0 1 0 2 0 2 0 2 0 2 1 0 0 0 2 2 1 | 0 0 0 0 0 0 0 0 0 0 0 1 0 2 2 0 1 0 1 0 0 | * * * 288 * * * * id | 3 6 3 6 3 6 3 | 3 6 3 6 6 6 6 6 6 3 6 3 0 0 0 0 0 0 | 1 3 6 6 3 1 0 0 0 0 0 0 0 0 0 3 3 3 3 0 0 | * * * * 96 * * * BCD&FFG | 0 2 2 2 0 2 1 | 0 0 0 2 2 2 0 0 0 1 2 0 0 1 1 2 2 0 | 0 0 2 0 1 0 0 1 0 0 0 0 1 2 0 0 0 0 0 1 0 | * * * * * 288 * * DEF | 0 0 0 3 3 3 0 | 0 0 0 0 0 0 3 3 3 0 0 0 0 0 0 0 3 3 | 0 0 0 3 0 0 0 0 1 0 0 0 0 0 3 0 0 0 0 0 1 | * * * * * * 192 * ti | 0 0 12 24 0 24 0 | 0 0 0 0 0 24 0 24 0 12 0 0 0 0 6 0 24 0 | 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 12 0 12 8 | * * * * * * * 24

Sure, like the Johnson solids themselves, these icositetra-diminishings are generally no longer uniform. Sadi was an exception. The incidence matrix of idsid pixhi from yesterday and that one of idsrix above clearly show up several different classes of vertices. I.e. those are not contained under a single orbit of symmetry. - None the less they all are regular faced and convex (which was the reason of consideration).

(Stay tuned for idprix ...)

--- rk

by **Polyhedron Dude** » Fri Dec 14, 2012 1:08 pm

The names seem to be available. Although idsrix would be a good name for the 7-D id-srix duoprism, we could give the 4-D one dibs on the name. Bidiminished ex, srix, etc will have several varieties and less symmetry than the bi24diminished cases, so they will need an extra prefix in the name.

by **Klitzing** » Sat Dec 15, 2012 11:09 am

Polyhedron Dude wrote:The names seem to be available. Although idsrix would be a good name for the 7-D id-srix duoprism, we could give the 4-D one dibs on the name. Bidiminished ex, srix, etc will have several varieties and less symmetry than the bi24diminished cases, so they will need an extra prefix in the name.

Good for having them free. Sure, you might them use for duoprisms too, as you mentioned. But in that direction your previous opinion, pointed out several times, was: if ambiguous, then the lower dimensional reading wins. The higher one shall get a different name. - One the other hand, those are not uniform, so more "regular" figures could get precedence anyway. - So what would be your preferred names, both for those under consideration, and the alternate meanings, you mentioned (applies for all sadi relatives: idsrix, idprix, idsid pixhi)?

Not clear what you meant to say by "we could give the 4-D one dibs on the name". Were you refering to bids instead? - The remainder of your reply was exactly how I felt too. :-)

--- rk

by **wendy** » Sun Dec 16, 2012 7:12 am

'dibs' means something like claim. Give something first dibs means to give it a first claim on it.

by **Klitzing** » Sun Dec 16, 2012 5:07 pm

wendy wrote:'dibs' means something like claim. Give something first dibs means to give it a first claim on it.

Oh thanx, Wendy! - So after all, it thus seems that Jonathan felt okay with those names proposed by me. That's great, as they occured now for a while in this thread, and renaming always makes some trouble. - Which, btw. is one of your favourit themes, Wendy, isn't it?

Its just that Jonathan has to come up with alternate names for the "dips".

--- rk

by **Klitzing** » Sun Dec 16, 2012 5:24 pm

Application of the fundamental domain of sadi onto idprix:

Idprix is the icositetra-diminished version of prix (x3o3x5x). Sadi itself is a likewise diminishing of ex (x3o3o5o). Thus this domain represents the localized structure of either polychoron, as both belong to the same global symmetry. The domain itself is derived as Voronoi cell of one (and thus any) vertex of sadi (with respect to the set of its vertex points).
The vertex of sadi/ex is represented by a complete tid (o3x5x) in idprix/prix. This tid lies completely within that domain, thus has to be fully counted. (3(A)+3(B)+6(C)+6(D)+6(E)+6(F)+6(G)+6(H)+6(I)+6(J)+3(K)+3(L))*96=288(A)+288(B)+576(C)+576(D)+576(E)+576(F)+576(G)+576(H)+576(I)+576(J)+288(K)+288(L), i.e. a total of 5760.
As in the cases considered before we have 2 types of edges: the ones completely contained in the region (i.e. the ones of tid, and those connecting one region with the adjacent. Any completely contained edge is counted by 1, any connecting one is counted by 1/2. The full edges are (3(AA)+3(AB)+6(BC)+3(CC)+6(CD)+6(DE)+6(DF)+3(EE)+6(EF)+6(FG)+6(GH)+6(GI)+3(HH)+6(HI)+6(IJ)+3(JJ)+6(JK)+3(KL)+3(LL))*96=288(AA)+288(AB)+576(BC)+288(CC)+576(CD)+576(DE)+576(DF)+288(EE)+576(EF)+576(FG)+576(GH)+576(GI)+288(HH)+576(HI)+576(IJ)+288(JJ)+576(JK)+288(KL)+288(LL), i.e. 8640 of those. The half-edges are (6(AA')+6(BB')+6(CC')+6(DD')+6(EE')+6(CC*)+6(DD*)+6(FF*)+6(GG*)+6(HH*)+6(HH#)+6(II#)+6(JJ#)+6(KK#)+6(LL#))*96/2=288(AA')+288(BB')+288(CC')+288(DD')+288(EE')+288(CC*)+288(DD*)+288(FF*)+288(GG*)+288(HH*)+288(HH#)+288(II#)+288(JJ#)+288(KK#)+288(LL#). The outer symmetry, i.e. the connectedness of those domains, implies the unification of classes (CC*) and (LL#), (DD*) and (KK#), (FF*) and (JJ#), (GG*) and (II#), resp. (HH*) and (HH#). Thus: 4320=288(AA')+288(BB')+288(CC')+288(DD')+288(EE')+576(CC*&LL#)+576(DD*&KK#)+576(FF*&JJ#)+576(GG*&II#)+576(HH*&HH#). In total this results in 8640+4320=12960 idprix edges.
Next there are triangles. Again of 2 types: On the one hand those of tid; these are contained completely within the fundamental region. So: (1(AAA)+3(BCC)+6(DEF)+6(GHI)+3(JJK)+1(LLL))*96=96(AAA)+288(BCC)+576(DEF)+576(GHI)+288(JJK)+96(LLL), in total 1920. On the other hand there are the triangles of the trips. These will be located orthogonally at the pentagon sides contained within the domain edges. These will account by 1/3 each, i.e. (3(A)+3(B)+6(C)+6(D)+6(H)+3(K)+3(L))*96/3=96(A)+96(B)+192(C)+192(D)+192(H)+96(K)+96(L). That division has to be unified according to the outer domain connection as (C) with (L), resp. (D) with (K). Thus 960=96(A)+96(B)+288(C&L)+288(D&K)+192(H). - Alltogether there are 1920+960=2880 triangles in idprix.
Then there are the squares. Those occur in prix as latterals of the dips: either connecting to coes or to trips. Within idprix these dips (the remaining ones) would connect by their squares to trips, coes, tricues, and grids. Still those take acount for all the squares. Dips cross the pentagonal regions of the domains faces. Accordingly the squares too would intersect those, contributing 1/2 each to each domain. Thus we have: (3(AA')+2*3(AB')+6(BC')+6(CD')+6(DE')+3(EE')+3(CC*)+6(CD*)+6(DF*)+6(FG*)+6(GH*)+3(HH*)+3(HH#)+6(HI#)+6(IJ#)+6(JK#)+2*3(KL#)+3(LL#))*96/2=144(AA')+288(AB')+288(BC')+288(CD')+288(DE')+144(EE')+144(CC*)+288(CD*)+288(DF*)+288(FG*)+288(GH*)+144(HH*)+144(HH#)+288(HI#)+288(IJ#)+288(JK#)+288(KL#)+144(LL#). From outer symmetry we have to unify (CC*) with (LL#), (CD*) with (KL#), (DF*) with (JK#), (FG*) with (IJ#), (GH*) with (HI#), and (HH*) with (HH#). Resulting in: 1440=144(AA')+288(AB')+288(BC')+288(CD')+288(DE')+144(EE')+288(CC*&LL#)+576(CD*&KL#)+576(DF*&JK#)+576(FG*&IJ#)+576(GH*&HI#)+288(HH*&HH#).
Hexagons occur as sections of coes, i.e. as new faces in idprix at the grids and tricues only. Those are incident at the tid edges only, which connect to grids and not are incident to squares. At the fundamental domain they occur only perpendicular to the 2 types of longer edges; in fact one at any elongation of the former pentagonal sides. They contirbute 1/3 each. (6(EF)+6(GI)+3(JJ))*96/3=192(EF)+192(GI)+96(JJ). By exterior symmetry (EF) and (JJ) are to be unified. Thus: 480=288(EF&JJ)+192(GI).
All decagons are faces of the tids, both for prix and for idprix. So they are contained completely within the (corresponding) domain and count by 1. Those are (3(ABCDE)+3(CDFGH)+3(EFGIJ)+3(HIJKL))*96=288(ABCDE)+288(CDFGH)+288(EFGIJ)+288(HIJKL), i.e. a total of 1152.
The coes are situated at the positions of the tets of sadi. Thus they are at the dodecahedral vertices of the domain. They are to be counted as 1/4. Getting (1(AAA)+3(BCC)+1(LLL))*96/4=24(AAA)+72(BCC)+24(LLL). Here from outer symmetry the latter ones have to be unified: 120=24(AAA)+96(BCC&LLL).
Trips occur around those domain edges each. Like for the triangle they count also by 1/3: (3(AB)+6(CD)+3(HH)+3(KL))*96/3=96(AB)+192(CD)+96(HH)+96(KL). (CD) and (KL) are to be unified again, thus: 480=96(AB)+288(CD&KL)+96(HH).
Dips are situated at those decagons of tid which correspond to the pentagonal regions of the domain faces. In fact they intersect those, thus they are counted by 1/2 each. (3(ABCDE)+3(CDFGH)+3(HIJKL))*96/2=144(ABCDE)+144(CDFGH)+144(HIJKL). By symmetry the latter ones are to be unified, resulting in 432=144(ABCDE)+288(CDFGH&HIJKL).
The tids themselves are completely contained within the domain, so they count by 1: 1*96=96.
Tricues will be situated around the domain edges again, one for each elongation of the former pentagonal side. They contribute 1/3 each. (6(D)+6(H)+3(K))*96/3=192(D)+192(H)+96(K). First and last are to be unified here: 480=288(D&K)+192(H).
Grids occur at the sissid-type vertices of the domain. They contribute 1/12 each: 3*96/12=24.

This all put together results in the incidence matrix of idprix:

Code: Select all: A | 288 * * * * * * * * * * * | 2 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 | 1 0 0 0 0 0 1 0 0 0 0 2 2 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 | 1 0 1 0 0 2 0 1 0 0 0 B | * 288 * * * * * * * * * * | 0 1 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 | 0 1 0 0 0 0 0 1 0 0 0 0 2 2 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 | 0 1 1 0 0 2 0 1 0 0 0 C | * * 576 * * * * * * * * * | 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 | 0 1 0 0 0 0 0 0 1 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 0 1 1 0 0 | 0 1 0 1 0 1 1 1 0 0 0 D | * * * 576 * * * * * * * * | 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 | 0 0 1 0 0 0 0 0 0 1 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 1 1 0 0 | 0 0 0 1 0 1 1 1 1 0 0 E | * * * * 576 * * * * * * * | 0 0 0 0 0 1 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 | 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 0 1 0 1 0 | 0 0 0 0 0 1 0 1 1 0 1 F | * * * * * 576 * * * * * * | 0 0 0 0 0 0 1 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 | 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 1 1 0 | 0 0 0 0 0 0 1 1 1 0 1 G | * * * * * * 576 * * * * * | 0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 | 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 1 1 0 | 0 0 0 0 0 0 1 1 0 1 1 H | * * * * * * * 576 * * * * | 0 0 0 0 0 0 0 0 0 0 1 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 | 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 2 2 0 0 0 1 0 1 | 0 0 0 0 1 0 2 1 0 1 0 I | * * * * * * * * 576 * * * | 0 0 0 0 0 0 0 0 0 0 0 1 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 | 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 1 1 | 0 0 0 0 0 0 1 1 0 1 1 J | * * * * * * * * * 576 * * | 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 1 0 0 | 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 1 1 | 0 0 0 0 0 0 1 1 1 0 1 K | * * * * * * * * * * 288 * | 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 1 0 0 0 0 0 0 0 2 0 0 0 | 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 2 2 0 0 0 0 0 0 0 0 2 | 0 0 0 1 0 0 2 1 1 0 0 L | * * * * * * * * * * * 288 | 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 2 0 0 0 0 0 2 0 0 0 0 | 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 2 2 0 0 0 0 0 0 0 0 0 2 | 0 1 0 1 0 0 2 1 0 0 0 ------------+-------------------------------------------------+---------------------------------------------------------------------------------------------------------------------+-----------------------------------------------------------------------------------------------------------------+-------------------------------------- AA | 2 0 0 0 0 0 0 0 0 0 0 0 | 288 * * * * * * * * * * * * * * * * * * * * * * * * * * * * | 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 | 1 0 0 0 0 1 0 1 0 0 0 AB | 1 1 0 0 0 0 0 0 0 0 0 0 | * 288 * * * * * * * * * * * * * * * * * * * * * * * * * * * | 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 | 0 0 1 0 0 2 0 1 0 0 0 BC | 0 1 1 0 0 0 0 0 0 0 0 0 | * * 576 * * * * * * * * * * * * * * * * * * * * * * * * * * | 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 | 0 1 0 0 0 1 0 1 0 0 0 CC | 0 0 2 0 0 0 0 0 0 0 0 0 | * * * 288 * * * * * * * * * * * * * * * * * * * * * * * * * | 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 | 0 1 0 0 0 0 1 1 0 0 0 CD | 0 0 1 1 0 0 0 0 0 0 0 0 | * * * * 576 * * * * * * * * * * * * * * * * * * * * * * * * | 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 1 1 0 0 | 0 0 0 1 0 1 1 1 0 0 0 DE | 0 0 0 1 1 0 0 0 0 0 0 0 | * * * * * 576 * * * * * * * * * * * * * * * * * * * * * * * | 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 | 0 0 0 0 0 1 0 1 1 0 0 DF | 0 0 0 1 0 1 0 0 0 0 0 0 | * * * * * * 576 * * * * * * * * * * * * * * * * * * * * * * | 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 | 0 0 0 0 0 0 1 1 1 0 0 EE | 0 0 0 0 2 0 0 0 0 0 0 0 | * * * * * * * 288 * * * * * * * * * * * * * * * * * * * * * | 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 1 0 | 0 0 0 0 0 1 0 1 0 0 1 EF | 0 0 0 0 1 1 0 0 0 0 0 0 | * * * * * * * * 576 * * * * * * * * * * * * * * * * * * * * | 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 | 0 0 0 0 0 0 0 1 1 0 1 FG | 0 0 0 0 0 1 1 0 0 0 0 0 | * * * * * * * * * 576 * * * * * * * * * * * * * * * * * * * | 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 1 0 | 0 0 0 0 0 0 1 1 0 0 1 GH | 0 0 0 0 0 0 1 1 0 0 0 0 | * * * * * * * * * * 576 * * * * * * * * * * * * * * * * * * | 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 | 0 0 0 0 0 0 1 1 0 1 0 GI | 0 0 0 0 0 0 1 0 1 0 0 0 | * * * * * * * * * * * 576 * * * * * * * * * * * * * * * * * | 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 | 0 0 0 0 0 0 0 1 0 1 1 HH | 0 0 0 0 0 0 0 2 0 0 0 0 | * * * * * * * * * * * * 288 * * * * * * * * * * * * * * * * | 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 1 0 1 | 0 0 0 0 1 0 2 1 0 0 0 HI | 0 0 0 0 0 0 0 1 1 0 0 0 | * * * * * * * * * * * * * 576 * * * * * * * * * * * * * * * | 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 | 0 0 0 0 0 0 1 1 0 1 0 IJ | 0 0 0 0 0 0 0 0 1 1 0 0 | * * * * * * * * * * * * * * 576 * * * * * * * * * * * * * * | 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 1 | 0 0 0 0 0 0 1 1 0 0 1 JJ | 0 0 0 0 0 0 0 0 0 2 0 0 | * * * * * * * * * * * * * * * 288 * * * * * * * * * * * * * | 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 | 0 0 0 0 0 0 0 1 1 0 1 JK | 0 0 0 0 0 0 0 0 0 1 1 0 | * * * * * * * * * * * * * * * * 576 * * * * * * * * * * * * | 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 | 0 0 0 0 0 0 1 1 1 0 0 KL | 0 0 0 0 0 0 0 0 0 0 1 1 | * * * * * * * * * * * * * * * * * 288 * * * * * * * * * * * | 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 2 | 0 0 0 1 0 0 2 1 0 0 0 LL | 0 0 0 0 0 0 0 0 0 0 0 2 | * * * * * * * * * * * * * * * * * * 288 * * * * * * * * * * | 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 | 0 1 0 0 0 0 1 1 0 0 0 AA' | 2 0 0 0 0 0 0 0 0 0 0 0 | * * * * * * * * * * * * * * * * * * * 288 * * * * * * * * * | 0 0 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 | 1 0 1 0 0 1 0 0 0 0 0 BB' | 0 2 0 0 0 0 0 0 0 0 0 0 | * * * * * * * * * * * * * * * * * * * * 288 * * * * * * * * | 0 0 0 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 | 0 1 1 0 0 1 0 0 0 0 0 CC' | 0 0 2 0 0 0 0 0 0 0 0 0 | * * * * * * * * * * * * * * * * * * * * * 288 * * * * * * * | 0 0 0 0 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 | 0 1 0 1 0 1 0 0 0 0 0 DD' | 0 0 0 2 0 0 0 0 0 0 0 0 | * * * * * * * * * * * * * * * * * * * * * * 288 * * * * * * | 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 | 0 0 0 1 0 1 0 0 1 0 0 EE' | 0 0 0 0 2 0 0 0 0 0 0 0 | * * * * * * * * * * * * * * * * * * * * * * * 288 * * * * * | 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 1 0 0 0 0 0 | 0 0 0 0 0 1 0 0 1 0 1 CC*&LL# | 0 0 1 0 0 0 0 0 0 0 0 1 | * * * * * * * * * * * * * * * * * * * * * * * * 576 * * * * | 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 | 0 1 0 1 0 0 1 0 0 0 0 DD*&KK# | 0 0 0 1 0 0 0 0 0 0 1 0 | * * * * * * * * * * * * * * * * * * * * * * * * * 576 * * * | 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 | 0 0 0 1 0 0 1 0 1 0 0 FF*&JJ# | 0 0 0 0 0 1 0 0 0 1 0 0 | * * * * * * * * * * * * * * * * * * * * * * * * * * 576 * * | 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0 | 0 0 0 0 0 0 1 0 1 0 1 GG*&II# | 0 0 0 0 0 0 1 0 1 0 0 0 | * * * * * * * * * * * * * * * * * * * * * * * * * * * 576 * | 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 | 0 0 0 0 0 0 1 0 0 1 1 HH*&HH# | 0 0 0 0 0 0 0 2 0 0 0 0 | * * * * * * * * * * * * * * * * * * * * * * * * * * * * 576 | 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 | 0 0 0 0 1 0 1 0 0 1 0 ------------+-------------------------------------------------+---------------------------------------------------------------------------------------------------------------------+-----------------------------------------------------------------------------------------------------------------+-------------------------------------- AAA | 3 0 0 0 0 0 0 0 0 0 0 0 | 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 | 96 * * * * * * * * * * * * * * * * * * * * * * * * * * * * | 1 0 0 0 0 0 0 1 0 0 0 BCC | 0 1 2 0 0 0 0 0 0 0 0 0 | 0 0 2 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 | * 288 * * * * * * * * * * * * * * * * * * * * * * * * * * * | 0 1 0 0 0 0 0 1 0 0 0 DEF | 0 0 0 1 1 1 0 0 0 0 0 0 | 0 0 0 0 0 1 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 | * * 576 * * * * * * * * * * * * * * * * * * * * * * * * * * | 0 0 0 0 0 0 0 1 1 0 0 GHI | 0 0 0 0 0 0 1 1 1 0 0 0 | 0 0 0 0 0 0 0 0 0 0 1 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 | * * * 576 * * * * * * * * * * * * * * * * * * * * * * * * * | 0 0 0 0 0 0 0 1 0 1 0 JJK | 0 0 0 0 0 0 0 0 0 2 1 0 | 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 2 0 0 0 0 0 0 0 0 0 0 0 0 | * * * * 288 * * * * * * * * * * * * * * * * * * * * * * * * | 0 0 0 0 0 0 0 1 1 0 0 LLL | 0 0 0 0 0 0 0 0 0 0 0 3 | 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 0 0 0 0 | * * * * * 96 * * * * * * * * * * * * * * * * * * * * * * * | 0 1 0 0 0 0 0 1 0 0 0 A | 3 0 0 0 0 0 0 0 0 0 0 0 | 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 0 0 0 | * * * * * * 96 * * * * * * * * * * * * * * * * * * * * * * | 1 0 1 0 0 0 0 0 0 0 0 B | 0 3 0 0 0 0 0 0 0 0 0 0 | 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 0 0 | * * * * * * * 96 * * * * * * * * * * * * * * * * * * * * * | 0 1 1 0 0 0 0 0 0 0 0 C&L | 0 0 2 0 0 0 0 0 0 0 0 1 | 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 2 0 0 0 0 | * * * * * * * * 288 * * * * * * * * * * * * * * * * * * * * | 0 1 0 1 0 0 0 0 0 0 0 D&K | 0 0 0 2 0 0 0 0 0 0 1 0 | 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 2 0 0 0 | * * * * * * * * * 288 * * * * * * * * * * * * * * * * * * * | 0 0 0 1 0 0 0 0 1 0 0 H | 0 0 0 0 0 0 0 3 0 0 0 0 | 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 | * * * * * * * * * * 192 * * * * * * * * * * * * * * * * * * | 0 0 0 0 1 0 0 0 0 1 0 AA' | 4 0 0 0 0 0 0 0 0 0 0 0 | 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 | * * * * * * * * * * * 144 * * * * * * * * * * * * * * * * * | 1 0 0 0 0 1 0 0 0 0 0 AB' | 2 2 0 0 0 0 0 0 0 0 0 0 | 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 | * * * * * * * * * * * * 288 * * * * * * * * * * * * * * * * | 0 0 1 0 0 1 0 0 0 0 0 BC' | 0 2 2 0 0 0 0 0 0 0 0 0 | 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 | * * * * * * * * * * * * * 288 * * * * * * * * * * * * * * * | 0 1 0 0 0 1 0 0 0 0 0 CD' | 0 0 2 2 0 0 0 0 0 0 0 0 | 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 | * * * * * * * * * * * * * * 288 * * * * * * * * * * * * * * | 0 0 0 1 0 1 0 0 0 0 0 DE' | 0 0 0 2 2 0 0 0 0 0 0 0 | 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 | * * * * * * * * * * * * * * * 288 * * * * * * * * * * * * * | 0 0 0 0 0 1 0 0 1 0 0 EE' | 0 0 0 0 4 0 0 0 0 0 0 0 | 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 | * * * * * * * * * * * * * * * * 144 * * * * * * * * * * * * | 0 0 0 0 0 1 0 0 0 0 1 CC*&LL# | 0 0 2 0 0 0 0 0 0 0 0 2 | 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 2 0 0 0 0 | * * * * * * * * * * * * * * * * * 288 * * * * * * * * * * * | 0 1 0 0 0 0 1 0 0 0 0 CD*&KL# | 0 0 1 1 0 0 0 0 0 0 1 1 | 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 1 0 0 0 | * * * * * * * * * * * * * * * * * * 576 * * * * * * * * * * | 0 0 0 1 0 0 1 0 0 0 0 DF*&JK# | 0 0 0 1 0 1 0 0 0 1 1 0 | 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 1 0 0 | * * * * * * * * * * * * * * * * * * * 576 * * * * * * * * * | 0 0 0 0 0 0 1 0 1 0 0 FG*&IJ# | 0 0 0 0 0 1 1 0 1 1 0 0 | 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 1 0 | * * * * * * * * * * * * * * * * * * * * 576 * * * * * * * * | 0 0 0 0 0 0 1 0 0 0 1 GH*&HI# | 0 0 0 0 0 0 1 2 1 0 0 0 | 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 | * * * * * * * * * * * * * * * * * * * * * 576 * * * * * * * | 0 0 0 0 0 0 1 0 0 1 0 HH*&HH# | 0 0 0 0 0 0 0 4 0 0 0 0 | 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 | * * * * * * * * * * * * * * * * * * * * * * 288 * * * * * * | 0 0 0 0 1 0 1 0 0 0 0 EF&JJ | 0 0 0 0 2 2 0 0 0 2 0 0 | 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 2 0 0 | * * * * * * * * * * * * * * * * * * * * * * * 288 * * * * * | 0 0 0 0 0 0 0 0 1 0 1 GI | 0 0 0 0 0 0 3 0 3 0 0 0 | 0 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 0 | * * * * * * * * * * * * * * * * * * * * * * * * 192 * * * * | 0 0 0 0 0 0 0 0 0 1 1 ABCDE | 2 2 2 2 2 0 0 0 0 0 0 0 | 1 2 2 0 2 2 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 | * * * * * * * * * * * * * * * * * * * * * * * * * 288 * * * | 0 0 0 0 0 1 0 1 0 0 0 CDFGH | 0 0 2 2 0 2 2 2 0 0 0 0 | 0 0 0 1 2 0 2 0 0 2 2 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 | * * * * * * * * * * * * * * * * * * * * * * * * * * 288 * * | 0 0 0 0 0 0 1 1 0 0 0 EFGIJ | 0 0 0 0 2 2 2 0 2 2 0 0 | 0 0 0 0 0 0 0 1 2 2 0 2 0 0 2 1 0 0 0 0 0 0 0 0 0 0 0 0 0 | * * * * * * * * * * * * * * * * * * * * * * * * * * * 288 * | 0 0 0 0 0 0 0 1 0 0 1 HIJKL | 0 0 0 0 0 0 0 2 2 2 2 2 | 0 0 0 0 0 0 0 0 0 0 0 0 1 2 2 0 2 2 1 0 0 0 0 0 0 0 0 0 0 | * * * * * * * * * * * * * * * * * * * * * * * * * * * * 288 | 0 0 0 0 0 0 1 1 0 0 0 ------------+-------------------------------------------------+---------------------------------------------------------------------------------------------------------------------+-----------------------------------------------------------------------------------------------------------------+-------------------------------------- AAA | 12 0 0 0 0 0 0 0 0 0 0 0 | 12 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 12 0 0 0 0 0 0 0 0 0 | 4 0 0 0 0 0 4 0 0 0 0 6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 | 24 * * * * * * * * * * BCC&LLL | 0 3 6 0 0 0 0 0 0 0 0 3 | 0 0 6 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 0 3 3 0 0 6 0 0 0 0 | 0 3 0 0 0 1 0 1 3 0 0 0 0 3 0 0 0 3 0 0 0 0 0 0 0 0 0 0 0 | * 96 * * * * * * * * * AB | 3 3 0 0 0 0 0 0 0 0 0 0 | 0 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 3 0 0 0 0 0 0 0 0 | 0 0 0 0 0 0 1 1 0 0 0 0 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 | * * 96 * * * * * * * * CD&KL | 0 0 2 2 0 0 0 0 0 0 1 1 | 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 1 0 2 2 0 0 0 | 0 0 0 0 0 0 0 0 1 1 0 0 0 0 1 0 0 0 2 0 0 0 0 0 0 0 0 0 0 | * * * 288 * * * * * * * HH | 0 0 0 0 0 0 0 6 0 0 0 0 | 0 0 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 6 | 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 0 0 | * * * * 96 * * * * * * ABCDE | 4 4 4 4 4 0 0 0 0 0 0 0 | 2 4 4 0 4 4 0 2 0 0 0 0 0 0 0 0 0 0 0 2 2 2 2 2 0 0 0 0 0 | 0 0 0 0 0 0 0 0 0 0 0 1 2 2 2 2 1 0 0 0 0 0 0 0 0 2 0 0 0 | * * * * * 144 * * * * * CDFGH&HIJKL | 0 0 2 2 0 2 2 4 2 2 2 2 | 0 0 0 1 2 0 2 0 0 2 2 0 2 2 2 0 2 2 1 0 0 0 0 0 2 2 2 2 2 | 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 2 2 2 2 1 0 0 0 1 0 1 | * * * * * * 288 * * * * tid | 3 3 6 6 6 6 6 6 6 6 3 3 | 3 3 6 3 6 6 6 3 6 6 6 6 3 6 6 3 6 3 3 0 0 0 0 0 0 0 0 0 0 | 1 3 6 6 3 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 3 3 3 | * * * * * * * 96 * * * D&K | 0 0 0 2 2 2 0 0 0 2 1 0 | 0 0 0 0 0 2 2 0 2 0 0 0 0 0 0 1 2 0 0 0 0 0 1 1 0 2 2 0 0 | 0 0 2 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 0 2 0 0 0 1 0 0 0 0 0 | * * * * * * * * 288 * * H | 0 0 0 0 0 0 3 3 3 0 0 0 | 0 0 0 0 0 0 0 0 0 0 3 3 0 3 0 0 0 0 0 0 0 0 0 0 0 0 0 3 3 | 0 0 0 3 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 3 0 0 1 0 0 0 0 | * * * * * * * * * 192 * grid | 0 0 0 0 24 24 24 0 24 24 0 0 | 0 0 0 0 0 0 0 12 24 24 0 24 0 0 24 12 0 0 0 0 0 0 0 12 0 0 24 24 0 | 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 6 0 0 0 24 0 0 12 8 0 0 12 0 | * * * * * * * * * * 24

Seems to be one of the largest matrices, I've ever worked out so far... :o_o:

I'd agree that the font size chosen is way too small to read anything out. So please copy that and paste it into any text editor. - My main problem here is just that this forum does not allow for horizontal slide bars, at it seems.

Finally just bidsid pixhi still is missing. So stay tuned!

Well, that one is no sadi relative any longer. Those are done now. Rather it will be a bidex relative. Thus we will need to apply this domain-idea onto bidex first, to see what the domain of that one would look like. And then work out the application of that to be derived domain onto bidsid pixhi. ...

--- rk

by **Klitzing** » Tue Dec 18, 2012 5:46 pm

The incidence matrix of the vertex figure of bidex reads like this:

Code: Select all: A | 2 * * | 1 1 1 1 0 0 | 2 1 1 B | * 2 * | 0 1 1 0 1 0 | 1 1 1 C | * * 2 | 0 0 0 1 1 1 | 0 1 2 -----+-------+-------------+------ AA | 2 0 0 | 1 * * * * * | 2 0 0 (tau-edge) AB | 1 1 0 | * 2 * * * * | 1 0 1 (1-edges) AB | 1 1 0 | * * 2 * * * | 1 1 0 (tau-edges) AC | 1 0 1 | * * * 2 * * | 0 1 1 (tau-edges) BC | 0 1 1 | * * * * 2 * | 0 1 1 (1-edges) CC | 0 0 2 | * * * * * 1 | 0 0 2 (1-edge) -----+-------+-------------+------ AAB | 2 1 0 | 1 1 1 0 0 0 | 2 * * ABC | 1 1 1 | 0 0 1 1 1 0 | * 2 * ABCC | 1 1 2 | 0 1 0 1 1 1 | * * 2

From the seeming rotation symmetry of 180 degrees (within the writing plane) bidex' verf not only looks like being self-dual, it is self-dual (as can be verified directly by geometric construction). Moreover the handedness of that figure thereby will be maintained: the dual comes out to be nothing but a rotated copy.

Like the domain of sadi resulted a faceting of sissid (x5/2o5o), the domain of bidex comes out to be a faceting of gad (x5o5/2o). In fact, within the building phase of sadi's domain the maintained vertices of teddi (here considered as an ike faceting) are dualized to pentagon-planes of the dodecahedron, which either use 3 or 4 sides of that very pentagon itself for borders (in fact y-y-y-y-n resp. y-y-n-y-n). For bidex' domain there are 2 or even only 1 being re-used (in fact y-n-n-n-n resp. y-n-y-n-n). Therefore here any re-used side will be elongated into both directions by tau (= golden ratio, 1.618...) times the side length, thus becoming a new side of 1+2*tau=tau^3. Any pentagon vertex between 2 non-maintained sides would thereby be replaced by an side of length tau^2 times the former pentagonal side. - Just as gad can be considered as the 2nd doe stellation, thus having larger pantagons which are anti-oriented with the small one of the kernel dodecahedron and in size are scaled by tau^2, this holds true here as well: the pentagonal faces of that irregular stellation kernel get extended into facetings of those larger pentagons, where the smaller sides of those are in dual position with respect to those of th kernel. Those needed pentagonal facetings are again golden acute triangles resp. golden trapezia.

Therefore in total, the fundamental domain of bidex looks exactly the same as its vertex figure!

Next we have to look for potential outer symmetries of that fundamental domains. Obviously the tetragons can connect to tetragons only. (The inner symmetry would not allow to distinguish those.) Allowing for a single handedness of these domains only (corresponding to the overall swirl symmetry), this very connection forces around the aligned longer edges (AC) of either domain the further incidence of that specific single longer edge (AA) of the domain which uses anti-aligned incident acute triangles. Therefrom we get for free, that here too those triangles fall into 2 disjunct sets (those incident to (AA), resp. those being not incident to (AA) - edge type namings again with respect to each single domain only!), and that these 2 types of triangles always do cross-connect only.

Bidex has 72 vertices. Accordingly we need 72 such fundamental domains, each being nothing but a Voronoi cell of the vertex set of bidex.

Now, just as was already outlined for the sadi domain, before similarily applying to new shores, we'd again like to test this procedure for bidex itself:

There is exactly 1 vertex of bidex in this domain. (It is the point around which the domain was constructed.) Hence bidex has 1*72=72 vertices.
There is exactly 1 half-edge of bidex running from that specific inner point to the midpoint of any of those pentagons. As the domain has 6 faces, we get 6*72/2=216 edges (division by 2, as 2 half-edges equate to a single complete one). Moreover we could divide this count into those edges, which are orthogonal to those tetragons and those to the triangles of the domain. Then we would get 4*72/2+2*72/2=144+72.
There is exactly 1 third-triangle for any of the longer edges of the domain. Thus 5*72/3=120. From the outer symmetries we got that the AA- and AC-edges of the domain will be placed together, while the (longer) AB-edges will be kept apart. Accordingly those divide into 3*72/3+2*72/3=72+48.
The shorter edges of the domain take account for 1/5 of an pentagon each. Thus 5*72/5=72.
The teddies are situated at the vertices of the domain. As those cells are not uniform themselves, they contribute with different fractions each. Even so we can calculate like this: 6 vertices of the domain, 72 domains in total, and 9 vertices of teddi results in 6*72/9=48.

Therefore the corresponding counts are thus all correctly recovered. In fact, the incidence matrix of bidex was previously (and independantly) derived as:

Code: Select all: 72 | 4 2 | 2 3 5 | 6 ---+--------+----------+--- 2 | 144 * | 1 1 1 | 3 : bottom or top edges + 1 chiral edge each of the lateral {3} of teddi 2 | * 72 | 0 1 3 | 4 : other edges of teddi ---+--------+----------+--- 3 | 3 0 | 48 * * | 2 : bottom or top {3} of teddi 3 | 2 1 | * 72 * | 2 : lateral {3} of teddi 5 | 2 3 | * * 72 | 2 ---+--------+----------+--- 9 | 9 6 | 2 3 3 | 48 : teddi

Now, stay tuned for the final application of this domain onto bidsid pixhi.

--- rk

by **wendy** » Wed Dec 19, 2012 11:51 am

Klitzing wrote:... and renaming always makes some trouble. - Which, btw. is one of your favourit themes, Wendy, isn't it?

--- rk

Most of my renamings have been to restore words to their original meanings. Words have several senses and extended meanings, which leads to drift of meaning. cf English fathom and german faden, both derive from the same root, related to ladin petal (to open, embrase), to gothic fathom (enclosure), germanic fathom (arm-span, 6 feet-measure). But the 6 ft measure leads to different meanings. English fathoms were used to measure the depths of water, and then thought, while german fathoms were used to measure threads, and thence a thread itself (as well as fathom-thin = "thread-thin" 'stick-thin')

When you restore a kind of gramatical gender to the words (body vs point), and think of words as being body-sense or point-sense, then the point-sense do not change dimensions as all-space increases, whereas body-sense does. Most of my 'changes in meaning' have been to identify and apply this division, largely to bring into line what is used elsewhere.

7 yes, it is my favourite sense. (for 7, see unicode 8266 = x204A - it's an old english 'and'.)

by **Klitzing** » Wed Dec 19, 2012 1:41 pm

wendy wrote:...
7 yes, it is my favourite sense. (for 7, see unicode 8266 = x204A - it's an old english 'and'.)

Well, I'd suppose that this gaelic "et"-sign (in the form of a "7") most probably has the same derivation history as you were just describing for words and meanings. Start with latin "et" for "and". Then write the "e" more like a greek epsilon and the "t" like a greek tau, and it becomes the origin of what is now known as the ampersent "&". Finally drop even that "e" in total and do not prolong the top bar of that tau beyond its support, and you get that gaelic sign. - Alternatively it could be thought of being derived from the plus sign, where the two lines don't cross properly. (Or even th other way round: this could likewise be the derivation of the plus sign!) - In fact, all these "newer letters" derive from some often used and thus slightly deformed hand-written short-cuts.

--- rk

by **Klitzing** » Thu Dec 20, 2012 10:14 am

Application of the fundamental domain of bidex onto bidsid pixhi:

Bidsid pixhi is the bi-icositetra-diminished version of sidpixhi (x3o3o5x). Bidex itself is a likewise diminishing of ex (x3o3o5o). Thus this domain represents the localized structure of either polychoron, as both belong to the same global symmetry. The domain itself is derived as Voronoi cell of one (and thus any) vertex of bidex (with respect to the set of its vertex points).
The vertex of bidex/ex here now is represented by a complete doe (o3o5x) in bidsid pixhi/sidpixhi. This doe lies completely within that domain, thus has to be fully counted. The domain itself has 2-fold rotation symmetry only. Accordingly we get (2(A)+2(B)+2(C)+2(D)+2(E)+2(F)+2(G)+2(H)+2(I)+2(J))*72=144(A)+144(B)+144(C)+144(D)+144(E)+144(F)+144(G)+144(H)+144(I)+144(J).
Edges again fall into 2 classes: those incident to the doe (which count fully), and those intersecting the faces of the domain (orthogonally, hence being counted 1/2 each). In the first class we have (1(AA)+2(AB)+2(AC)+2(BD)+2(BE)+2(CD)+2(CF)+2(DG)+2(EF)+2(EH)+2(FI)+2(GH)+2(GI)+2(HJ)+2(IJ)+1(JJ))*72=72(AA)+144(AB)+144(AC)+144(BD)+144(BE)+144(CD)+144(CF)+144(DG)+144(EF)+144(EH)+144(FI)+144(GH)+144(GI)+144(HJ)+144(IJ)+72(JJ), i.e. 2160 of those. The other ones are (4(A')+2(B')+2(C')+2(D')+2(C*)+2(D*)+2(F*)+2(G*)+2(I*)+2(E#)+2(F#)+2(H#)+2(I#)+2(J#))*72/2=144(A')+72(B')+72(C')+72(D')+72(C*)+72(D*)+72(F*)+72(G*)+72(I*)+72(E#)+72(F#)+72(H#)+72(I#)+72(J#). From outer symmetry we get here the following unifications: that (A'), which is closer to (C), with (F*), that (A'), which is closer to (B), with (I*) - thus the (A') after all have to be divided further! -, (B') with (G*), (C') with (C*), (D') with (D*), (E#) with (J#), and (F#) with (I#): 1080=144(Ac'&F*)+144(Ab'&I*)+144(B'&G*)+144(C'&C*)+144(D'&D*)+144(E#&J#)+144(F#&I#)+72(H#)
There are exactly 2 third-triangles for any of the longer edges of the domain. Thus (2(A)+2(C)+2(D)+2(F)+2(I))*72/3=48(A)+48(C)+48(D)+48(F)+48(I). Outer symmetry implies unification of (A) with (F) and (I): 240=144(A&F&I)+48(C)+48(D).
Squares should emanate from the edges of the doe (2 each, if not to be omitted in this diminishing), and are orthogonal to the domain faces, i.e. they count 1/2 each. (2(AA')+2(AB')+2(AC')+2(BD')+2(CD')+2(CD*)+2(CF*)+2(DG*)+2(FI*)+2(GI*)+2(EF#)+2(EH#)+2(FI#)+2(HJ#)+2(IJ#))*72/2=72(AA')+72(AB')+72(AC')+72(BD')+72(CD')+72(CD*)+72(CF*)+72(DG*)+72(FI*)+72(GI*)+72(EF#)+72(EH#)+72(FI#)+72(HJ#)+72(IJ#). Here from outer symmetry we have to unify (AA') with (FI*), (AB') with (GI*), (AC') with (CF*), (BD') with (DG*), (CD') with (CD*), (EF#) with (IJ#), and (EH#) with (HJ#). Becoming 1080=144(AA'&FI*)+144(AB'&GI*)+144(AC'&CF*)+144(BD'&DG*)+144(CD'&CD*)+144(EF#&IJ#)+144(EH#&HJ#)+72(FI#).
The pentagons of bidsid pixhi are exactly those of the doe in any domain. Thus (2(AABDC)+2(ABEFC)+2(BDGHE)+2(CDGIF)+2(EFIJH)+2(GHJJI))*72=144(AABDC)+144(ABEFC)+144(BDGHE)+144(CDGIF)+144(EFIJH)+144(GHJJI), a total of 864. Pentagons (AABDC), (CDGIF), and (EFIJH) will attach to pips, while (ABEFC), (BDGHE), and (GHJJI) attach to tedrids.
Decagons occur orthogonal to the shorter edges of the domain, in fact at their centers. They contribute 1/5 in total. (2(BE)+2(GH)+1(JJ))*72/5. From outer symmetry we derive the individual side cyclus to be (BE)-(E#&J#)-(JJ)-(J#&E#)-(EB)-(B'&G*)-(GH)-(H#&H#)-(HG)-(G*&B'), that is, all decagons become equivalent, they contribute within each domain just in different orientations.
As tets would occur in sidpixhi at each doe vertex in opposite position, those would remain only in bidsid pixhi, if there would be a doe vertex, which is not incident to a tedrid, as the corresponding diminishing would chop them off. In fact this is not the case, any vertex of doe is incident to 1 or 2 tedrids. So the total count of tets is simply 0.
The trips are incident in opposite position at those edges of doe, which are not incident to tedrids. Each trip counts 1/3. (1(AA)+2(CD)+2(FI))*72/3=24(AA)+48(CD)+48(FI). Outer symmetry unifies (A)) and (FI), thus: 120=72(AA&FI)+48(CD). As no tets remain in bidsid pixhi, all triangles connect trips to tedrids only.
Pips are attached to those doe pentagons mentioned already above. They count 1/2 each. (2(AABDC)+2(CDGIF)+2(EFIJH))*72/2=72(AABDC)+72(CDGIF)+72(EFIJH). Symmetry unifies (AABDC) with (CDGIF), thus 216=144(AABDC&CDGIF)+72(EFIJH).
Clearly there is 1 doe per domain: 1*72=72.
Tedrids occur at those pentagons mentioned already above. Those contribute with different fractions. There are 48 in total (according to the count of diminishings). The tedrids all belong to the same orbit of symmetry.

Here is the geometry (application of letters to vertices), I use for the doe within the fundamental domain:

Code: Select all: C--A__ / -__ ---A__ / . -D------B F b | . \ |\ . . | . c \ E \ __G__ . E | I-- d - __ _ - | | | . H F | | / / H_|. g . / / J-- __ /_I J AA long edge of domain CD long edge of domain FI long edge of domain AABDC inc. to pip (attach to type CDGIF): ' ABEFC inc. to tedrid BDGHE inc. to tedrid CDGIF inc. to pip (attach to type AABDC): * EFIJH inc. to pip (attach to type EFIJH): # GHJJI inc. to tedrid BE inc. to 10 GH inc. to 10 JJ inc. to 10

This all put together results in the incidence matrix of bidsid pixhi:

Code: Select all: A | 144 * * * * * * * * * | 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 | 1 0 0 2 1 1 0 0 0 0 0 2 1 0 0 0 0 0 | 1 0 2 0 1 1 B | * 144 * * * * * * * * | 0 1 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 | 0 0 0 0 1 0 1 0 0 0 0 1 1 1 0 0 0 0 | 0 0 1 0 1 2 C | * * 144 * * * * * * * | 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 | 0 1 0 0 0 2 0 2 0 0 0 1 1 0 1 0 0 0 | 0 1 2 0 1 1 D | * * * 144 * * * * * * | 0 0 0 1 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 | 0 0 1 0 0 0 2 2 0 0 0 1 0 1 1 0 0 0 | 0 1 2 0 1 1 E | * * * * 144 * * * * * | 0 0 0 0 1 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 | 0 0 0 0 0 0 0 0 1 1 0 0 1 1 0 1 0 1 | 0 0 0 1 1 2 F | * * * * * 144 * * * * | 0 0 0 0 0 0 1 0 1 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 | 1 0 0 1 0 1 0 0 1 0 1 0 1 0 1 1 0 0 | 1 0 1 1 1 1 G | * * * * * * 144 * * * | 0 0 0 0 0 0 0 1 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 | 0 0 0 0 1 0 1 0 0 0 0 0 0 1 1 0 1 1 | 0 0 1 0 1 2 H | * * * * * * * 144 * * | 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0 0 0 0 0 0 0 0 0 1 | 0 0 0 0 0 0 0 0 0 2 0 0 0 1 0 1 1 1 | 0 0 0 1 1 2 I | * * * * * * * * 144 * | 0 0 0 0 0 0 0 0 0 0 1 0 1 0 1 0 0 1 0 0 0 0 1 0 | 1 0 0 1 1 0 0 0 1 0 1 0 0 0 1 1 1 0 | 1 0 1 1 1 1 J | * * * * * * * * * 144 | 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 1 0 0 | 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 1 2 1 | 0 0 0 1 1 2 ------------+-----------------------------------------+----------------------------------------------------------------------------------------------+---------------------------------------------------------------------+------------------- AA | 2 0 0 0 0 0 0 0 0 0 | 72 * * * * * * * * * * * * * * * * * * * * * * * | 0 0 0 2 0 0 0 0 0 0 0 2 0 0 0 0 0 0 | 1 0 2 0 1 0 AB | 1 1 0 0 0 0 0 0 0 0 | * 144 * * * * * * * * * * * * * * * * * * * * * * | 0 0 0 0 1 0 0 0 0 0 0 1 1 0 0 0 0 0 | 0 0 1 0 1 1 AC | 1 0 1 0 0 0 0 0 0 0 | * * 144 * * * * * * * * * * * * * * * * * * * * * | 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 0 | 0 0 1 0 1 1 BD | 0 1 0 1 0 0 0 0 0 0 | * * * 144 * * * * * * * * * * * * * * * * * * * * | 0 0 0 0 0 0 1 0 0 0 0 1 0 1 0 0 0 0 | 0 0 1 0 1 1 BE | 0 1 0 0 1 0 0 0 0 0 | * * * * 144 * * * * * * * * * * * * * * * * * * * | 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 1 | 0 0 0 0 1 2 CD | 0 0 1 1 0 0 0 0 0 0 | * * * * * 144 * * * * * * * * * * * * * * * * * * | 0 0 0 0 0 0 0 2 0 0 0 1 0 0 1 0 0 0 | 0 1 2 0 1 0 CF | 0 0 1 0 0 1 0 0 0 0 | * * * * * * 144 * * * * * * * * * * * * * * * * * | 0 0 0 0 0 1 0 0 0 0 0 0 1 0 1 0 0 0 | 0 0 1 0 1 1 DG | 0 0 0 1 0 0 1 0 0 0 | * * * * * * * 144 * * * * * * * * * * * * * * * * | 0 0 0 0 0 0 1 0 0 0 0 0 0 1 1 0 0 0 | 0 0 1 0 1 1 EF | 0 0 0 0 1 1 0 0 0 0 | * * * * * * * * 144 * * * * * * * * * * * * * * * | 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 1 0 0 | 0 0 0 1 1 1 EH | 0 0 0 0 1 0 0 1 0 0 | * * * * * * * * * 144 * * * * * * * * * * * * * * | 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 1 0 0 | 0 0 0 1 1 1 FI | 0 0 0 0 0 1 0 0 1 0 | * * * * * * * * * * 144 * * * * * * * * * * * * * | 0 0 0 1 0 0 0 0 0 0 1 0 0 0 1 1 0 0 | 1 0 1 1 1 0 GH | 0 0 0 0 0 0 1 1 0 0 | * * * * * * * * * * * 144 * * * * * * * * * * * * | 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1 | 0 0 0 0 1 2 GI | 0 0 0 0 0 0 1 0 1 0 | * * * * * * * * * * * * 144 * * * * * * * * * * * | 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 1 0 | 0 0 1 0 1 1 HJ | 0 0 0 0 0 0 0 1 0 1 | * * * * * * * * * * * * * 144 * * * * * * * * * * | 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 1 0 | 0 0 0 1 1 1 IJ | 0 0 0 0 0 0 0 0 1 1 | * * * * * * * * * * * * * * 144 * * * * * * * * * | 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 1 0 | 0 0 0 1 1 1 JJ | 0 0 0 0 0 0 0 0 0 2 | * * * * * * * * * * * * * * * 72 * * * * * * * * | 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 1 | 0 0 0 0 1 2 Ac'&F* | 1 0 0 0 0 1 0 0 0 0 | * * * * * * * * * * * * * * * * 144 * * * * * * * | 1 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 | 1 0 1 0 0 1 Ab'&I* | 1 0 0 0 0 0 0 0 1 0 | * * * * * * * * * * * * * * * * * 144 * * * * * * | 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 | 1 0 1 0 0 1 B'&G* | 0 1 0 0 0 0 1 0 0 0 | * * * * * * * * * * * * * * * * * * 144 * * * * * | 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 1 | 0 0 1 0 0 2 C'&C* | 0 0 2 0 0 0 0 0 0 0 | * * * * * * * * * * * * * * * * * * * 144 * * * * | 0 1 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 | 0 1 1 0 0 1 D'&D* | 0 0 0 2 0 0 0 0 0 0 | * * * * * * * * * * * * * * * * * * * * 144 * * * | 0 0 1 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 | 0 1 1 0 0 1 E#&J# | 0 0 0 0 1 0 0 0 0 1 | * * * * * * * * * * * * * * * * * * * * * 144 * * | 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 1 | 0 0 0 1 0 2 F#&I# | 0 0 0 0 0 1 0 0 1 0 | * * * * * * * * * * * * * * * * * * * * * * 144 * | 1 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 | 1 0 0 1 0 1 H# | 0 0 0 0 0 0 0 2 0 0 | * * * * * * * * * * * * * * * * * * * * * * * 72 | 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 1 | 0 0 0 1 0 2 ------------+-----------------------------------------+----------------------------------------------------------------------------------------------+---------------------------------------------------------------------+------------------- A&F&I | 1 0 0 0 0 1 0 0 1 0 | 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 1 0 | 144 * * * * * * * * * * * * * * * * * | 1 0 0 0 0 1 C | 0 0 3 0 0 0 0 0 0 0 | 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 0 0 0 0 | * 48 * * * * * * * * * * * * * * * * | 0 1 0 0 0 1 D | 0 0 0 3 0 0 0 0 0 0 | 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 0 0 0 | * * 48 * * * * * * * * * * * * * * * | 0 1 0 0 0 1 AA'&FI* | 2 0 0 0 0 1 0 0 1 0 | 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 | * * * 144 * * * * * * * * * * * * * * | 1 0 1 0 0 0 AB'&GI* | 1 1 0 0 0 0 1 0 1 0 | 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0 0 | * * * * 144 * * * * * * * * * * * * * | 0 0 1 0 0 1 AC'&CF* | 1 0 2 0 0 1 0 0 0 0 | 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 | * * * * * 144 * * * * * * * * * * * * | 0 0 1 0 0 1 BD'&DG* | 0 1 0 2 0 0 1 0 0 0 | 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 1 0 0 0 | * * * * * * 144 * * * * * * * * * * * | 0 0 1 0 0 1 CD'&CD* | 0 0 2 2 0 0 0 0 0 0 | 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 | * * * * * * * 144 * * * * * * * * * * | 0 1 1 0 0 0 EF#&IJ# | 0 0 0 0 1 1 0 0 1 1 | 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 1 1 0 | * * * * * * * * 144 * * * * * * * * * | 0 0 0 1 0 1 EH#&HJ# | 0 0 0 0 1 0 0 2 0 1 | 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 1 0 1 | * * * * * * * * * 144 * * * * * * * * | 0 0 0 1 0 1 FI# | 0 0 0 0 0 2 0 0 2 0 | 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 0 0 0 2 0 | * * * * * * * * * * 72 * * * * * * * | 1 0 0 1 0 0 AABDC | 2 1 1 1 0 0 0 0 0 0 | 1 1 1 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 | * * * * * * * * * * * 144 * * * * * * | 0 0 1 0 1 0 ABEFC | 1 1 1 0 1 1 0 0 0 0 | 0 1 1 0 1 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 | * * * * * * * * * * * * 144 * * * * * | 0 0 0 0 1 1 BDGHE | 0 1 0 1 1 0 1 1 0 0 | 0 0 0 1 1 0 0 1 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 | * * * * * * * * * * * * * 144 * * * * | 0 0 0 0 1 1 CDGIF | 0 0 1 1 0 1 1 0 1 0 | 0 0 0 0 0 1 1 1 0 0 1 0 1 0 0 0 0 0 0 0 0 0 0 0 | * * * * * * * * * * * * * * 144 * * * | 0 0 1 0 1 0 EFIJH | 0 0 0 0 1 1 0 1 1 1 | 0 0 0 0 0 0 0 0 1 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 | * * * * * * * * * * * * * * * 144 * * | 0 0 0 1 1 0 GHJJI | 0 0 0 0 0 0 1 1 1 2 | 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0 0 0 | * * * * * * * * * * * * * * * * 144 * | 0 0 0 0 1 1 decagon | 0 2 0 0 2 0 2 2 0 2 | 0 0 0 0 2 0 0 0 0 0 0 2 0 0 0 1 0 0 2 0 0 2 0 1 | * * * * * * * * * * * * * * * * * 72 | 0 0 0 0 0 2 ------------+-----------------------------------------+----------------------------------------------------------------------------------------------+---------------------------------------------------------------------+------------------- AA&FI | 2 0 0 0 0 2 0 0 2 0 | 1 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 2 2 0 0 0 0 2 0 | 2 0 0 2 0 0 0 0 0 0 1 0 0 0 0 0 0 0 | 72 * * * * * CD | 0 0 3 3 0 0 0 0 0 0 | 0 0 0 0 0 3 0 0 0 0 0 0 0 0 0 0 0 0 0 3 3 0 0 0 | 0 1 1 0 0 0 0 3 0 0 0 0 0 0 0 0 0 0 | * 48 * * * * AABDC&CDGIF | 2 1 2 2 0 1 1 0 1 0 | 1 1 1 1 0 2 1 1 0 0 1 0 1 0 0 0 1 1 1 1 1 0 0 0 | 0 0 0 1 1 1 1 1 0 0 0 1 0 0 1 0 0 0 | * * 144 * * * EFIJH | 0 0 0 0 2 2 0 2 2 2 | 0 0 0 0 0 0 0 0 2 2 2 0 0 2 2 0 0 0 0 0 0 2 2 1 | 0 0 0 0 0 0 0 0 2 2 1 0 0 0 0 2 0 0 | * * * 72 * * doe | 2 2 2 2 2 2 2 2 2 2 | 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 0 0 0 0 0 0 0 0 | 0 0 0 0 0 0 0 0 0 0 0 2 2 2 2 2 2 0 | * * * * 72 * tedrid | 3 6 3 3 6 3 6 6 3 6 | 0 3 3 3 6 0 3 3 3 3 0 6 3 3 3 3 3 3 6 3 3 6 3 3 | 3 1 1 0 3 3 3 0 3 3 0 0 3 3 0 0 3 3 | * * * * * 48

This post finally completes this small series of sadi- and bidex relatives!

(Just in time for the Xmas break...)

--- rk

by **Klitzing** » Thu Jan 03, 2013 3:01 pm

Recently I applied icositetra-diminishings onto the hexacosachoron (ex) and its Wythoffian relatives. That one was related to the chiral compound of 5 icositetrachora (chi), which could be vertex-inscribed into ex (up to a scaling of tau). A similar compound-inscription is available for ico itself: the compound of 3 hexadecachora.

Thus we here likewise could consider the relations:

Code: Select all: -----------+--------------+-------------+ decoration | non-dim. | 8-dim. | -----------+--------------+-------------+ xooo | ico | tes | | (24 octs) | (8 cubes) | -----------+--------------+-------------+ xoox | spic | srit | | (48 octs, | (8 sircoes, | | 192 trips) | 32 trips, | | | 16 octs) | -----------+--------------+-------------+ xoxo | srico | proh | | (24 sircoes, | (8 tics, | | 96 trips, | 24 ops, | | 24 coes) | 32 trips, | | | 16 coes) | -----------+--------------+-------------+ xoxx | prico | gidpith | | (24 sircoes, | (8 gircoes, | | 96 trips, | 24 ops, | | 96 hips, | 32 hips, | | 24 toes) | 16 toes) | -----------+--------------+-------------+ xxoo | tico | - | -----------+--------------+-------------+ xxxo | grico | - | -----------+--------------+-------------+ xxox | prico | - | -----------+--------------+-------------+ xxxx | gippic | - | -----------+--------------+-------------+

The "-" each denotes cases where the diminishing would result in half-hexagons, i.e. the diminished figure would no longer be CRF.

In total we have:

ico can have chopped off 8 cube-pyramidal parts to become a tes.
spic can have chopped off 8 oct || sirco segmentochora to become a srit.
srico can have chopped off 8 co || tic segmentochora to become a proh.
prico can have chopped off 8 toe || girco segmentochora to become a gidpith.

(Respectively the other way round by corresponding augmentations, which each result in a polychoron of higher symmetry!) - Well, none does result in a non-Wythoffian figure here. So this thread might seem wrong after all. But it relates to the former mails by concept.

--- rk

by **quickfur** » Tue Jan 08, 2013 1:28 am

Whoa. I've been away from this forum for several months, and now I feel like I'm falling behind already.

Would someone kindly summarize what's been going on here? I've to confess I haven't been able to keep up with all the detailed posts here.

Anyway, I'm kinda back in the fray with 4D CRFs at the moment -- i just finished the first draft of a new page for my website that describes in detail the structure of the paratetragyrated cantellated tesseract. That is, x4o3x3o with the four 8prism||square segments corresponding to the para- positions on a 24-cell's 2-faces rotated by 45°. This breaks up 12 of the octahedral cells into square pyramids, and turns all of the x4o3x's into J37's (that is, Johnson solid 37, elongated square gyrobicupola). Four of the x4o3x3o's octahedra survive untouched; in place of the rest, there are now pairs of square pyramid/triangular prism or triangular prism/triangular prism. (Due to the dihedral angle between them, the pairs don't merge into augmented tri-prisms or gyrobifastigiums, though, which would have been cool.) Half of the original triangular prisms remain in place; the others have been replaced by square pyramids. I haven't determined whether this shape is chiral -- most likely, it is not, since it appears to have a symmetry related to the 8,8-duoprism. (I've already posted about this shape in the past, but now I finally got some time to do detailed renders of its structure. I'll post the renders in February.)

Another variation of this, is the orthotetragyrated cantellated tesseract -- that is, rotate by 45° four of the 8prism||square segmentotopes that lie along a single great circle. This one will have all octahedral cells broken up into square pyramids.

It's possible that gyrating the segments corresponding with one of the meximal 24-cell metadiminishings may yield highly-irregular CRFs that nevertheless still have some kind of symmetric structure. I'll have to see.

I still haven't made any progress on finding new CRFs produced by diminishings, though: my last study was the diminishings of o3x4o3x, but it's still incomplete. Anybody discovered new CRFs recently?

by **Klitzing** » Tue Jan 08, 2013 9:46 am

quickfur wrote:Whoa. I've been away from this forum for several months, and now I feel like I'm falling behind already.
...

Wellcome back - and all the best for 2013!

... Would someone kindly summarize what's been going on here? I've to confess I haven't been able to keep up with all the detailed posts here.
...

Well, the idea of my researches was as follows:
Both, sadi and bidex are based on ex (x3o3o5o). In fact they are the 24-diminished ex and the bi-24-diminished ex. I wondered if the same diminishings would apply as well to the higher Wythoffian analogues as well (i.e. on tex (x3x3o5o), srix (x3o3x5o), sidpith (x3o3o5x), etc.). - The outcome, in detail described in those posts, is that the neighbouring node can be ringed only (when considering CRFs only) if we don't apply such diminishings (thus remaining within the mere Wythoffians). That the one but next node could be ringed if we want to apply 24-diminishing only onto it. And that the opposite node might be ringed if we want to apply bi-24-diminishing onto it.
This results in:

Code: Select all: -----------+-------------+---------------+---------------+ decoration | non-dim. | 24-dim. | bi-24-dim. | -----------+-------------+---------------+---------------+ xooo | ex | sadi | bidex | | (600 tets) | (120 tets, | (48 teddies) | | | 600 ikes) | | -----------+-------------+---------------+---------------+ xoox | sidpixhi | idsid pixhi | bidsid pixhi | | (600 tets, | (120 tets, | (120 trips, | | 1200 trips, | 480 trips, | 216 pips, | | 720 pips, | 432 pips, | 72 does, | | 120 does) | 96 does, | 48 tedrids) | | | 24 srids) | | -----------+-------------+---------------+---------------+ xoxo | srix | idsrix | - | | (600 coes, | (120 coes, | | | 720 pips, | 432 pips, | | | 120 ids) | 96 ids, | | | | 480 tricues, | | | | 24 ties) | | -----------+-------------+---------------+---------------+ xoxx | prix | idprix | - | | (600 coes, | (120 coes, | | | 1200 trips, | 480 trips, | | | 720 dips, | 432 dips, | | | 120 tids) | 96 tids, | | | | 480 tricues, | | | | 24 grids) | | -----------+-------------+---------------+---------------+ xxoo | tex | - | - | -----------+-------------+---------------+---------------+ xxxo | grix | - | - | -----------+-------------+---------------+---------------+ xxox | prahi | - | - | -----------+-------------+---------------+---------------+ xxxx | gidpixhi | - | - | -----------+-------------+---------------+---------------+

And, most recently, I thought about some parallels. Sadi, and all these, were produced by an inscribed compound of 5 (tau scaled) icositetrachora (= ico) into the hexacosachoron (= ex). (In fact, the vertices of any of those icoes gives rise to a further 24-diminishing.) So we could see, what could be derived by the compound of 3 hexadecachora inscribed into an ico. So I investigated the application of 8-diminishings onto ico (x3o4o3o) and its higher Wythoffian relatives (i.e. tico (x3x4o3o), srico (x3o4x3o), spic (x3o4o3x), etc.). - The outcame was given in my last post of this thread.

... Anyway, I'm kinda back in the fray with 4D CRFs at the moment -- i just finished the first draft of a new page for my website that describes in detail the structure of the paratetragyrated cantellated tesseract.
...

Have to think about the remainder of your post for a while.

(Btw., the greek words para and ortho seem not to apply here correctly. If I get your procedure correctly, you rather are investigating an 8-gyration respectively a 4-gyration, right?)

--- rk

by **quickfur** » Tue Jan 08, 2013 3:06 pm

Klitzing wrote:[...]
Wellcome back - and all the best for 2013!

Thanks!

[...]
Code: Select all
-----------+-------------+---------------+---------------+ decoration | non-dim. | 24-dim. | bi-24-dim. | -----------+-------------+---------------+---------------+ xooo | ex | sadi | bidex | | (600 tets) | (120 tets, | (48 teddies) | | | 600 ikes) | | -----------+-------------+---------------+---------------+ xoox | sidpixhi | idsid pixhi | bidsid pixhi | | (600 tets, | (120 tets, | (120 trips, | | 1200 trips, | 480 trips, | 216 pips, | | 720 pips, | 432 pips, | 72 does, | | 120 does) | 96 does, | 48 tedrids) | | | 24 srids) | |

These ones (24-dim and bi-24-dim of xoox) looks interesting! Maybe I'll make some renders of them! (If I can figure out the coordinates

) Is the bi-24-dim-xoo5x chiral too?

Code: Select all
-----------+-------------+---------------+---------------+ xoxo | srix | idsrix | - | | (600 coes, | (120 coes, | | | 720 pips, | 432 pips, | | | 120 ids) | 96 ids, | | | | 480 tricues, | | | | 24 ties) | | -----------+-------------+---------------+---------------+ xoxx | prix | idprix | - | | (600 coes, | (120 coes, | | | 1200 trips, | 480 trips, | | | 720 dips, | 432 dips, | | | 120 tids) | 96 tids, | | | | 480 tricues, | | | | 24 grids) | |

These ones look interesting too. But the bi-24-dims are non-CRF?

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-----------+-------------+---------------+---------------+ xxoo | tex | - | - | -----------+-------------+---------------+---------------+ xxxo | grix | - | - | -----------+-------------+---------------+---------------+ xxox | prahi | - | - | -----------+-------------+---------------+---------------+ xxxx | gidpixhi | - | - | -----------+-------------+---------------+---------------+

So these don't produce CRF diminishings. Hmm.

And, most recently, I thought about some parallels. Sadi, and all these, were produced by an inscribed compound of 5 (tau scaled) icositetrachora (= ico) into the hexacosachoron (= ex). (In fact, the vertices of any of those icoes gives rise to a further 24-diminishing.) So we could see, what could be derived by the compound of 3 hexadecachora inscribed into an ico. So I investigated the application of 8-diminishings onto ico (x3o4o3o) and its higher Wythoffian relatives (i.e. tico (x3x4o3o), srico (x3o4x3o), spic (x3o4o3x), etc.). - The outcame was given in my last post of this thread.

Interesting! I'll have to go back and read it more carefully.

... Anyway, I'm kinda back in the fray with 4D CRFs at the moment -- i just finished the first draft of a new page for my website that describes in detail the structure of the paratetragyrated cantellated tesseract.
...

Have to think about the remainder of your post for a while.
(Btw., the greek words para and ortho seem not to apply here correctly. If I get your procedure correctly, you rather are investigating an 8-gyration respectively a 4-gyration, right?)

OK, I guess I'm really just borrowing the terminology from the 24-cell diminishings, because of the correspondence between the vertices of the 24-cell and the square faces of the tesseract. The diminishings of the 24-cell correspond with subsets of non-adjacent vertices of the 24-cell. Now since x4o3x3o can be gyrated as long as the rotated 8-prism||square segments are non-adjacent, and these segments correspond with the 2-faces of the tesseract (and therefore the vertices of the 24-cell), I thought it was convenient to borrow the terminology from the 24-cell diminishings to indicate which subset of 8-prism||square segments are rotated.

So I'm using para here in the sense of being antipodes (i.e., opposite vertices of the 24-cell, or antipodal 2-faces of the tesseract, and therefore the 8-prism||square segments corresponding to antipodal 2-faces of the tesseract). There is only one possible combination of non-adjacent vertices of the 24-cell that consists of two pairs of antipodal vertices -- that is, they correspond with the vertices of an inscribed 16-cell, and so "paratetra" uniquely identifies this combination of vertices on the 24-cell, and by extension, the particular combination of two pairs of antipodal 8-prism||square segments on the x4o3x3o.

With ortho, it means a pair of vertices of the 24-cell such that their respective vectors to the center of the 24-cell are 90° to each other. And I think I made a mistake in my usage in my post; a tetraortho- would mean 4 vertices of a tetrahedron in a 16-cell inscribed in the 24-cell, whereas I meant a cyclotetragyration (i.e., the 4 non-adjacent 8-prism||square segments that lie on a great circle, corresponding with the 4 vertices of 16-cell inscribed in a 24-cell that lie on a single great circle).

With meta, on the 24-cell it means two vertices A and B such that B is adjacent to A's antipode. The various metadiminishings of the 24-cell happen to be precisely those that do not correspond with any augmentation of the tesseract. It also happens that there are only two maximal diminishings of the 24-cell besides the tesseract itself (maximal means no more diminishing of non-adjacent vertices is possible), and both are metadiminishings. So they represent a kind of interesting configuration of vertices on the 24-cell where there are no remaining vertices that are non-adjacent to the rest. By extension, they also represent maximal subsets of non-adjacent 8-prism||square on the x4o3x3o that do not correspond with augmentations of an 8,8-duoprism, so they are a special kind of diminishing of the x4o3x3o.

by **quickfur** » Tue Jan 08, 2013 3:37 pm

Klitzing wrote:Recently I applied icositetra-diminishings onto the hexacosachoron (ex) and its Wythoffian relatives. That one was related to the chiral compound of 5 icositetrachora (chi), which could be vertex-inscribed into ex (up to a scaling of tau). A similar compound-inscription is available for ico itself: the compound of 3 hexadecachora.

Thus we here likewise could consider the relations:

Code: Select all
-----------+--------------+-------------+ decoration | non-dim. | 8-dim. | -----------+--------------+-------------+ xooo | ico | tes | | (24 octs) | (8 cubes) | -----------+--------------+-------------+ xoox | spic | srit | | (48 octs, | (8 sircoes, | | 192 trips) | 32 trips, | | | 16 octs) | -----------+--------------+-------------+ xoxo | srico | proh | | (24 sircoes, | (8 tics, | | 96 trips, | 24 ops, | | 24 coes) | 32 trips, | | | 16 coes) | -----------+--------------+-------------+ xoxx | prico | gidpith | | (24 sircoes, | (8 gircoes, | | 96 trips, | 24 ops, | | 96 hips, | 32 hips, | | 24 toes) | 16 toes) | -----------+--------------+-------------+ xxoo | tico | - | -----------+--------------+-------------+ xxxo | grico | - | -----------+--------------+-------------+ xxox | prico | - | -----------+--------------+-------------+ xxxx | gippic | - | -----------+--------------+-------------+

The "-" each denotes cases where the diminishing would result in half-hexagons, i.e. the diminished figure would no longer be CRF.

In total we have:
ico can have chopped off 8 cube-pyramidal parts to become a tes.
spic can have chopped off 8 oct || sirco segmentochora to become a srit.
srico can have chopped off 8 co || tic segmentochora to become a proh.
prico can have chopped off 8 toe || girco segmentochora to become a gidpith.
(Respectively the other way round by corresponding augmentations, which each result in a polychoron of higher symmetry!) - Well, none does result in a non-Wythoffian figure here. So this thread might seem wrong after all. But it relates to the former mails by concept.

--- rk

You can also do a kind of partial Stott expansion of x3o4o3x, such that instead of expanding all 24 elements, you expand only 8. This produces a lower-symmetry polychoron (it has tesseractic symmetry, no longer 24-cell symmetry) that contains elongated square bipyramids (J15), which I've posted before:

It's the same thing as a x4o3x3x augmented with eight x4o3x||o4o3x's. I've tried to find a CRF that contains J15's, that also has 24-cell symmetry, but I haven't found anything yet.

Hi.gher. Space

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