88 posts
• Page **2** of **3** • 1, **2**, 3

I'm sorry, but I didn't calculated it. I found more than a hundred , so I looked for an algorithm on internet.

This e-book is about icosahedral proteins, but it contains the problem of find m-tuples of icosahedral vertices.

<https://books.google.com.br/books?id=Q81Gwbd4l74C&pg=PA50&lpg=PA50&dq=Icosahedron+vertex+combination&source=bl&ots=us2ebpIf-D&sig=UJA_q1KtoapSart3uzw0tJlJODs&hl=pt-BR&sa=X&ved=0ahUKEwiMzfaNs-jLAhWKEZAKHYl9ANMQ6AEIWzAL#v=onepage&q=Icosahedron%20vertex%20combination&f=false>

Beacuse it's the dodecahedron's dual, it's essentially the same.

This e-book is about icosahedral proteins, but it contains the problem of find m-tuples of icosahedral vertices.

<https://books.google.com.br/books?id=Q81Gwbd4l74C&pg=PA50&lpg=PA50&dq=Icosahedron+vertex+combination&source=bl&ots=us2ebpIf-D&sig=UJA_q1KtoapSart3uzw0tJlJODs&hl=pt-BR&sa=X&ved=0ahUKEwiMzfaNs-jLAhWKEZAKHYl9ANMQ6AEIWzAL#v=onepage&q=Icosahedron%20vertex%20combination&f=false>

Beacuse it's the dodecahedron's dual, it's essentially the same.

- JMBR
- Mononian
**Posts:**8**Joined:**Wed Mar 30, 2016 12:58 pm

I forgot two details. The cuboctahedral prism can be augmented on each of its cubes, giving 9 crf.

And the icosidodecahedral prism, the same as dodecahedron, which i'm not sure how many are.

And about the tetrahedral corona, I was looking at the final johnson solids trying to found 4d analogues. Thanks for the segmentochora paper.

But i was searching fo a sphenocorona equivalent by its net, since I have no 4d program for now. The spheno part can be considered a layer of digonal prisms joined by its lateral faces and with digonal pyramids at its sides. To end up this arrangement, a corona of triangles (or a pseudoyramid of the hole) would be sufficient. I was trying to generate this with triangular, square or pentagonal prisms (hexagonal prism is forbidden by its pyramid), so it came with 5 possibilities (3, 4 or 5 triangular pisms, 3 cubes or 3 pentagonal prisms) around an edge capped by pyramids on its bases and square pyramids on its sides. Or even two side-by-side prisms covered with the same pyramids. The only problem is that i I don't know how to close them. I think that a structure of tetrahedra would fit, but there may be also a lost equivalent of disphenocingulum with two of these. So, this is the idea.

And the icosidodecahedral prism, the same as dodecahedron, which i'm not sure how many are.

And about the tetrahedral corona, I was looking at the final johnson solids trying to found 4d analogues. Thanks for the segmentochora paper.

But i was searching fo a sphenocorona equivalent by its net, since I have no 4d program for now. The spheno part can be considered a layer of digonal prisms joined by its lateral faces and with digonal pyramids at its sides. To end up this arrangement, a corona of triangles (or a pseudoyramid of the hole) would be sufficient. I was trying to generate this with triangular, square or pentagonal prisms (hexagonal prism is forbidden by its pyramid), so it came with 5 possibilities (3, 4 or 5 triangular pisms, 3 cubes or 3 pentagonal prisms) around an edge capped by pyramids on its bases and square pyramids on its sides. Or even two side-by-side prisms covered with the same pyramids. The only problem is that i I don't know how to close them. I think that a structure of tetrahedra would fit, but there may be also a lost equivalent of disphenocingulum with two of these. So, this is the idea.

- JMBR
- Mononian
**Posts:**8**Joined:**Wed Mar 30, 2016 12:58 pm

JMBR wrote:... And an equivalent of sphenocorona may be possible. Starting with three cubes or triangular prisms around an edge and putting pyramids on the other faces creates a luna-like shape, which I don't know if it can be completed with tetrahedral coronae.

Yep, take 3 triangular prisms around an edge, attach one tetrahedron (kind a trigonal pyramid) at either triangular face so far, and bend this such that those would adjoin as well. Then we truely have the 4D equivalent of the biluna-thingy of the Johnson solids. Now consider the remaining open (un-adjoined) faces. These are one triangle of each tetrahedron and one square of each prism. That is, it just allows us to "complete" this shape by a fourth such "lunar" part, i.e. adding a fourth elongated triangular bipyramid.

And in fact, what we just constructed indeed is nothing but the elongated tetrahedral bipyramid, i.e. o3o3o || x3o3o || x3o3o || o3o3o = oxxo3oooo3oooo&#xt.

Moreover, we even could have used n such triangular prisms around the edge instead. The same process then would have constructed similarily the tristratic tower point || n-pyramid || n-pyramid || point, i.e. the elongated version of the n-pyramidal bipyramid. Obviously 2 < n < 6. The general cell count thus is: n trips + 2n tets + 2 n-pyramids + 1 n-prism.

More explicite:

for n=3 we have: 4 trips + 8 tets,

for n=4 we have: 4 trips + 8 tets + 2 squippies + 1 cube,

and for n=5 we have: 5 trips + 10 tets + 2 peppies + 1 pip.

--- rk

- Klitzing
- Pentonian
**Posts:**1627**Joined:**Sun Aug 19, 2012 11:16 am**Location:**Heidenheim, Germany

Thought about these elongated n-pyramidal bipyramids again.

Found out that the critical dihedral angle would be that between the n-gonal prism and those base-wise attached pyramids, which join to the extremal tips.

Here is the lace city for n=4:

It can be seen here, that those squippies here already become corealmic to that prism, i.e. those unite here into an esquidpy (J15). For n=3 those extremal end points would move slightly up (either mirror half then would represent a tetrahedron), whereas for n=5 those extremal vertices would move down, i.e. below the line marked by the pentagonal prism. Therefore my recent estimate 2 < n < 6 has now to be downsized to 2 < n <= 4 only.

On the other hand, we surely could add the lower bit wrt. to that lace prismatic display: the elongated octahedral bipyramid oxxo3oooo4oooo&#xt surely would exist as well as the elongated icosahedral bipyramid oxxo3oooo5oooo&#xt.

--- rk

Found out that the critical dihedral angle would be that between the n-gonal prism and those base-wise attached pyramids, which join to the extremal tips.

Here is the lace city for n=4:

- Code: Select all
`o4o o4o`

o4o x4o x4o o4o

It can be seen here, that those squippies here already become corealmic to that prism, i.e. those unite here into an esquidpy (J15). For n=3 those extremal end points would move slightly up (either mirror half then would represent a tetrahedron), whereas for n=5 those extremal vertices would move down, i.e. below the line marked by the pentagonal prism. Therefore my recent estimate 2 < n < 6 has now to be downsized to 2 < n <= 4 only.

On the other hand, we surely could add the lower bit wrt. to that lace prismatic display: the elongated octahedral bipyramid oxxo3oooo4oooo&#xt surely would exist as well as the elongated icosahedral bipyramid oxxo3oooo5oooo&#xt.

--- rk

- Klitzing
- Pentonian
**Posts:**1627**Joined:**Sun Aug 19, 2012 11:16 am**Location:**Heidenheim, Germany

Moreover diminishings of ike there would work as well, provided that the circumcenter of ike would be contained within the body of that diminishing. That is, pt||ike||ike||pt, pt||gyepip||gyepip||pt, pt||pap||pap|pt, pt||mibdi||mibdi||pt, and pt||teddi||teddi||pt all would work. (But, ín contrast, that pt||peppy||peppy|pt would not be convex.)

And further we likewise could consider oxxo4oooo3oooo&#xt = pt||cube||cube||pt, which then obviously would require square prisms (aka cubes) within the medial segment - instead of those triangular ones in the cases of pt||X||X||pt with X=tet, oct, or ike. In fact, that pt||cube||cube||pt is nothing than an external blend of a tesseract with 2 antipoldally attached cubical pyramids.

But then, it should be noted that the cubical pyramid and the tesseract both have a circumradius of 1 (in edge units). And a tesseract can be edge-inscribed into the 24-cell. Therefore it turns out, that pt||cube||cube||pt actually would be nothing but a hexadiminished 24-cell.

--- rk

And further we likewise could consider oxxo4oooo3oooo&#xt = pt||cube||cube||pt, which then obviously would require square prisms (aka cubes) within the medial segment - instead of those triangular ones in the cases of pt||X||X||pt with X=tet, oct, or ike. In fact, that pt||cube||cube||pt is nothing than an external blend of a tesseract with 2 antipoldally attached cubical pyramids.

But then, it should be noted that the cubical pyramid and the tesseract both have a circumradius of 1 (in edge units). And a tesseract can be edge-inscribed into the 24-cell. Therefore it turns out, that pt||cube||cube||pt actually would be nothing but a hexadiminished 24-cell.

--- rk

- Klitzing
- Pentonian
**Posts:**1627**Joined:**Sun Aug 19, 2012 11:16 am**Location:**Heidenheim, Germany

BTW, just as

so too

In fact, these are the first instances I'm aware of!

--- rk

- xxxx4oxxo&#xt = x3o4x and
- xxxo4oxxx&#xt = J37

so too

- pt||pap||pt and
- pt||mibdi||pt

- pt||pap||pap||pt and
- pt||mibdi||mibdi||pt

In fact, these are the first instances I'm aware of!

--- rk

- Klitzing
- Pentonian
**Posts:**1627**Joined:**Sun Aug 19, 2012 11:16 am**Location:**Heidenheim, Germany

Klitzing wrote:In fact, these are the first instances I'm aware of!

Well, not quite.

E.g. the 2 bipyramidal ones are just different possibilities of (non-intersecting) bidiminishings at equatorial vertices of pt||ike||pt.

And I surely knew already about the 7 different possibilities of (non-intersecting) vertex bidiminishings of the 600-cell (which then obviously each would show up total counts of 560 tets + 2 ikes). I just forgot about those...

Here those are (for reference):

a) layer 1 + layer 3 (adjacent):

- Code: Select all
`x3o o3f f3o o3x`

o3o o3f f3x x3f f3o

f3o o3F F3o o3f

o3x x3f F3o f3f o3F f3x x3o

f3o o3F F3o o3f

o3o o3f f3x x3f f3o o3o

x3o o3f f3o o3x

o3o

b) layer 1 + layer 7:

- Code: Select all
`x3o o3f f3o o3x`

o3o o3f f3x x3f f3o o3o

f3o o3F F3o o3f

o3x x3f F3o f3f o3F f3x x3o

f3o o3F F3o o3f

o3o o3f f3x x3f f3o

x3o o3f f3o o3x

o3o

c) layer 1 + layer 9 (antipodal):

- Code: Select all
`x3o o3f f3o o3x`

o3o o3f f3x x3f f3o o3o

f3o o3F F3o o3f

o3x x3f F3o f3f o3F f3x x3o

f3o o3F F3o o3f

o3o o3f f3x x3f f3o o3o

x3o o3f f3o o3x

(same within different symmetry):

- Code: Select all
`o5o o5o`

o5x

x5o x5o

o5o o5o

f5o

o5f o5f

o5x o5x

f5o f5o

x5x

o5f o5f

x5o x5o

f5o f5o

o5f

o5o o5o

o5x o5x

x5o

o5o o5o

(same within an other different symmetry):

- Code: Select all
`o2x f2o x2f f2o o2x`

x2o f2f o2F F2x o2F f2f x2o

o2f F2o f2F F2o o2f

o2o f2x x2F F2f Vo2oV F2f x2F f2x o2o

o2f F2o f2F F2o o2f

x2o f2f o2F F2x o2F f2f x2o

o2x f2o x2f f2o o2x

d) layer 1 + layer 4 (tip-to-tip):

- Code: Select all
`o5o`

o5x

x5o x5o

o5o o5o

f5o

o5f o5f

o5x o5x

f5o f5o

x5x o5o

o5f o5f

x5o x5o

f5o f5o

o5f

o5o o5o

o5x o5x

x5o

o5o o5o

e) layer 1 + layer 6:

- Code: Select all
`o5o`

o5x

x5o x5o

o5o o5o

f5o

o5f o5f

o5x o5x

f5o f5o

x5x o5o

o5f o5f

x5o x5o

f5o f5o

o5f

o5o o5o

o5x o5x

x5o

o5o o5o

f) layer 1 + layer 8:

- Code: Select all
`o5o o5o`

o5x

x5o x5o

o5o

f5o

o5f o5f

o5x o5x

f5o f5o

x5x o5o

o5f o5f

x5o x5o

f5o f5o

o5f

o5o o5o

o5x o5x

x5o

o5o o5o

g) layer 1 + layer 5 (orthogonally):

- Code: Select all
`o2x f2o x2f f2o o2x`

x2o f2f o2F F2x o2F f2f x2o

o2f F2o f2F F2o o2f

o2o f2x x2F F2f Vo2oV F2f x2F f2x

o2f F2o f2F F2o o2f

x2o f2f o2F F2x o2F f2f x2o

o2x f2o x2f f2o o2x

o2o

--- rk

- Klitzing
- Pentonian
**Posts:**1627**Joined:**Sun Aug 19, 2012 11:16 am**Location:**Heidenheim, Germany

Back to the sphenocorona analogy, let's say we start with the 3 triangular prisms around an edge, and attach 6 tetrahedra to the ends to make the 4D analog of the spheno complex. Instead of adding a fourth triangular prism + 2 tetrahedra complex to close it up, what if we attach 3 square pyramids to the exposed square faces of the prisms instead? Then, possibly, we could insert tetrahedra between each pair of square pyramids to make a ring of alternating square pyramids + tetrahedra, leaving 2*6 = 12 unattached triangles on the outside. 6 of these share an edge with the tetrahedra on the ends of the triangular prisms, so we could conceivably link them up by adding tetrahedra. Hmm... not sure where I'm going with this, but I vaguely have in mind exploiting the triangular symmetry of the construction so that we reduce the complexity of computing coordinates by 1 dimension, and perhaps somehow reducing the rest of the complex to a system of equations similar to that of the sphenocorona? Would it be possible to close up this polychoron?

- quickfur
- Pentonian
**Posts:**2873**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

So you'd say you'll take here my etedpy (elongated tetrahedral dipyramid, i.e. pen + tepe + pen) and make an external blend with traf (o x3o || x o3x) at one of the trips (and try to complete that initial structure somehow). But when considering the dihedral angles of the trips within etedpy, then it's clear that that one equates to the dihedral angle of the triangles within tet, i.e. is arccos(1/3) = 70.528779°. OTOH the dihedral angle between trip and squippy within traf is arccos(-1/sqrt(6)) = 114.094843°. Thus those 2 angles would already sum up to more than 180°, i.e. would make up an non-convex angle... (if I'm not wrong).

--- rk

--- rk

- Klitzing
- Pentonian
**Posts:**1627**Joined:**Sun Aug 19, 2012 11:16 am**Location:**Heidenheim, Germany

Hmm you're right. I failed to take into account that wrapping 3 trigonal prisms around an edge already fixes all dichoral angles. I suppose for my construction to work, one would have to start with 4 trigonal prisms around an edge, not necessarily in tetragonal symmetry (there being 1 degree of freedom of deformation in the 4-fold edge), attach tets to the ends and 5pyr's to the sides, and see if there's a way to close it.

On further thought, though, this may not lead to anything new, because having tets at the ends of the trigonal prisms leaves a rather narrow gap between adjacent tets, too narrow for another tet to fit in (since 3 adjacent tets around an edge already requires >180 degrees), so the only solution is to fold the tets inwards and adjoin to each other. But that means we will have tetragonal symmetry after all, i.e., it will just be a (partial) elongated octahedral dipyramid. So it's unlikely to have enough degrees of freedom to allow for a sphenocorona-like construction that isn't something we already can build by gluing currently-known CRFs together.

So it seems that to have anything new, we cannot start with more than 2 trigonal prisms, attached at a square face, and capped with tets, i.e., a "true" 4D analog of a spheno complex. This allows us 1 degree of freedom in folding the spheno complex together (assuming the adjacent tets are joined together by a face -- nothing else would fit between them at this angle). We can then add 4 square pyramids to the unattached square faces of the trigonal prisms, to produce a network of triangles. Is it possible to close up this network by adding more tets (and possibly some other kind of pyramidal cells)?

On further thought, though, this may not lead to anything new, because having tets at the ends of the trigonal prisms leaves a rather narrow gap between adjacent tets, too narrow for another tet to fit in (since 3 adjacent tets around an edge already requires >180 degrees), so the only solution is to fold the tets inwards and adjoin to each other. But that means we will have tetragonal symmetry after all, i.e., it will just be a (partial) elongated octahedral dipyramid. So it's unlikely to have enough degrees of freedom to allow for a sphenocorona-like construction that isn't something we already can build by gluing currently-known CRFs together.

So it seems that to have anything new, we cannot start with more than 2 trigonal prisms, attached at a square face, and capped with tets, i.e., a "true" 4D analog of a spheno complex. This allows us 1 degree of freedom in folding the spheno complex together (assuming the adjacent tets are joined together by a face -- nothing else would fit between them at this angle). We can then add 4 square pyramids to the unattached square faces of the trigonal prisms, to produce a network of triangles. Is it possible to close up this network by adding more tets (and possibly some other kind of pyramidal cells)?

- quickfur
- Pentonian
**Posts:**2873**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

Just considered autip = J49. Its dihedral angle between the augmentation triangles and the remaining squares is 114.735610 degrees.

When building a prism with autip bases, then the same angle is between the respectively adjoined lacing trip and cube.

Knowing on the other hand that the dihedral angle at the base of a cubical pyramid is just 45 degrees, this proves that either cube (or even both) of the autip-prism could be augmented by such pyramids!

Similarily the single cube of the bautip-prism (bautip = J50) could be augmanted that way!

--- rk

When building a prism with autip bases, then the same angle is between the respectively adjoined lacing trip and cube.

Knowing on the other hand that the dihedral angle at the base of a cubical pyramid is just 45 degrees, this proves that either cube (or even both) of the autip-prism could be augmented by such pyramids!

Similarily the single cube of the bautip-prism (bautip = J50) could be augmanted that way!

--- rk

- Klitzing
- Pentonian
**Posts:**1627**Joined:**Sun Aug 19, 2012 11:16 am**Location:**Heidenheim, Germany

The J49 prism is equivalent to a 3,4-duoprism augmented with a square pyramid prism. The 3,4-duoprism can also be augmented with cubical pyramids, too. So we can consider the set of all possible combinations of these two augments on the 3,4-duoprism.

Some years ago I wrote a program to calculate the number of n-prism pyramid augmentations of m,n-duoprisms, which includes augmentations of the 3,4-duoprism with cubical pyramids, but not square pyramid prisms. So that's something new.

In fact, another thought just occurred to me: what about augmentations with square pyramid prisms that are in the "wrong" orientation, so that the square pyramid cells don't merge with the triangular prism into J49's? AFAICT the dichoral angle between the square pyramid and adjacent cube would be pretty large, so you probably can't fit another augment after that. But perhaps the cubical pyramid might still fit?

Some years ago I wrote a program to calculate the number of n-prism pyramid augmentations of m,n-duoprisms, which includes augmentations of the 3,4-duoprism with cubical pyramids, but not square pyramid prisms. So that's something new.

In fact, another thought just occurred to me: what about augmentations with square pyramid prisms that are in the "wrong" orientation, so that the square pyramid cells don't merge with the triangular prism into J49's? AFAICT the dichoral angle between the square pyramid and adjacent cube would be pretty large, so you probably can't fit another augment after that. But perhaps the cubical pyramid might still fit?

- quickfur
- Pentonian
**Posts:**2873**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

Here's a render of the square pyramid skew-augmented 3,4-duoprism:

It should be relatively easy to imagine if we rotate the augment (mostly outlined by the blue edges) 90° from the vertical, the top and bottom square pyramids would be brought into the hyperplane of a pair of trigonal prisms, and being corealmar would merge into J49's.

Coordinates for the square pyramid skew-augmented 3,4-duoprism:

Edit: cell counts: 8 trigonal prisms, 2 square pyramids, 2 cubes. Total 12 cells.

It should be relatively easy to imagine if we rotate the augment (mostly outlined by the blue edges) 90° from the vertical, the top and bottom square pyramids would be brought into the hyperplane of a pair of trigonal prisms, and being corealmar would merge into J49's.

Coordinates for the square pyramid skew-augmented 3,4-duoprism:

- Code: Select all
`# 3,4-duoprism`

<±1, 0, ±1, ±1>

< 0, √3, ±1, ±1>

# Augmentation with skew orientation.

<±1, -√2, 0, 0>

Edit: cell counts: 8 trigonal prisms, 2 square pyramids, 2 cubes. Total 12 cells.

- quickfur
- Pentonian
**Posts:**2873**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

Provided that your gyautisdip (gyro-augmented triangle-square-duoprism) works and is truely convex, we even could go beyond:

In fact, it will be describable as xxo oxx4ooo&#xt .

Then we could well expand it at the last node position to xxo oxx4xxx&#xt !

Edit: that latter then is nothing but a stack of the 3-8-duoprism (todip) and the J4-prism (squacupe). In fact it places a cubical "tip" atop one of 8-prisms (op) of todip. As such it thus is also an augmented todip - or "autodip" for short.

Cells then are: 1+4=5 cubes, 4+4+4=12 trips, 2 squacues, 2 ops.

--- rk

In fact, it will be describable as xxo oxx4ooo&#xt .

Then we could well expand it at the last node position to xxo oxx4xxx&#xt !

Edit: that latter then is nothing but a stack of the 3-8-duoprism (todip) and the J4-prism (squacupe). In fact it places a cubical "tip" atop one of 8-prisms (op) of todip. As such it thus is also an augmented todip - or "autodip" for short.

Cells then are: 1+4=5 cubes, 4+4+4=12 trips, 2 squacues, 2 ops.

--- rk

- Klitzing
- Pentonian
**Posts:**1627**Joined:**Sun Aug 19, 2012 11:16 am**Location:**Heidenheim, Germany

I think it's pretty clear from the coordinates that it is convex, no?

Also, our current naming scheme is incomplete because it is ambiguous what the shape of the augment is. It is possible to augment a 3,4-duoprism with a cubical pyramid or a gyro square pyramid prism, and indeed, due to the shape of the 3,4-duoprism it can also attach to a tesseract, another 3,4-duoprism (in various orientations), etc., but the name does not unambiguously indicate which augmentation is meant.

But anyway, yes the Stott-expanded equivalent certainly should exist too. I have realized some years ago that for every augmentation of an m,n-duoprism with m-prism pyramids, there is a corresponding augmentation of a 2m,n-duoprism with m-gon||2m-prisms. I.e., a Stott expansion of the original augmented m,n-duoprism. I'm not clear exactly what the conditions are for augments other than m-prism pyramids, though.

In any case, even just within the confines of m-gon||2m-prism augmentations of 2m,n-duoprisms, there is still research to be done on counting exactly how many such augmentations (modulo 2m,n-duoprism symmetry) exist as CRFs. Marek & myself have already counted the n-prism pyramid / m-prism pyramid augmentations of m,n-duoprisms; there are 1633 such augmentations. I wrote a program for counting all augmentations with n-prism pyramids / n-gon||2n-prisms, and got somewhere around 1.6 million (most of which are caused by the combinatorial explosion of augmentations of the 20,20-duoprism, which are possible thanks to the very shallow height of 5gon||10-prism). However, this count was never independently verified.

If we include other augment shapes such as n-pyramid prisms, the counts will have to be adjusted accordingly. Though I doubt it will add significantly more augmentations to the total, because the 90° dichoral angles of the n-pyramid prisms would mean that only 3,n-duoprisms and 4,n-duoprisms are augmentable with them, so they would not experience the same kind of combinatorial explosion the augmented 10,10-duoprism and 20,20-duoprism exhibits.

Also, our current naming scheme is incomplete because it is ambiguous what the shape of the augment is. It is possible to augment a 3,4-duoprism with a cubical pyramid or a gyro square pyramid prism, and indeed, due to the shape of the 3,4-duoprism it can also attach to a tesseract, another 3,4-duoprism (in various orientations), etc., but the name does not unambiguously indicate which augmentation is meant.

But anyway, yes the Stott-expanded equivalent certainly should exist too. I have realized some years ago that for every augmentation of an m,n-duoprism with m-prism pyramids, there is a corresponding augmentation of a 2m,n-duoprism with m-gon||2m-prisms. I.e., a Stott expansion of the original augmented m,n-duoprism. I'm not clear exactly what the conditions are for augments other than m-prism pyramids, though.

In any case, even just within the confines of m-gon||2m-prism augmentations of 2m,n-duoprisms, there is still research to be done on counting exactly how many such augmentations (modulo 2m,n-duoprism symmetry) exist as CRFs. Marek & myself have already counted the n-prism pyramid / m-prism pyramid augmentations of m,n-duoprisms; there are 1633 such augmentations. I wrote a program for counting all augmentations with n-prism pyramids / n-gon||2n-prisms, and got somewhere around 1.6 million (most of which are caused by the combinatorial explosion of augmentations of the 20,20-duoprism, which are possible thanks to the very shallow height of 5gon||10-prism). However, this count was never independently verified.

If we include other augment shapes such as n-pyramid prisms, the counts will have to be adjusted accordingly. Though I doubt it will add significantly more augmentations to the total, because the 90° dichoral angles of the n-pyramid prisms would mean that only 3,n-duoprisms and 4,n-duoprisms are augmentable with them, so they would not experience the same kind of combinatorial explosion the augmented 10,10-duoprism and 20,20-duoprism exhibits.

- quickfur
- Pentonian
**Posts:**2873**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

P.S. I had forgotten the actual count of duoprism augmentations; according to this post of mine it is not the 20,20-duoprism that produces the most augmentations, because in fact the 20,20-duoprism has no augmentations (there is no CRF augment I know of that has a 20-gon prism cell that can augment a duoprism in a convex way). The largest count actually comes from the 10,10-duoprism, which, due to various coinciding factors, permits simultaneous augmentations on both rings and at the same time allows gyrated augments, causing a combinatorial explosion that leads to 11 million distinct augmentations (modulo 10,10-duoprism symmetry). The second runner-up in terms of number of augmentations is the 10,20-duoprism, numbering about 12 thousand augmentations.

Even though the 10,20-duoprism has more possible positions for augmentation, only one ring can be augmented (again because the 20-gonal prism has no convex augments that I know of), which greatly reduces the number of possible combinations. Even though adjacent augments are not allowed on the one ring (they would be concave), it still includes analogues of single-ring 10,10-duoprism augmentations, and adds a few more too (e.g., augments spaced an odd number of positions apart, which have no counterpart in the 10,10-duoprism augmentations). Nevertheless, these do not make up for the reduction of combinations due to only one ring being augmentable. So it numbers only in the thousands, rather than the millions.

Even though the 10,20-duoprism has more possible positions for augmentation, only one ring can be augmented (again because the 20-gonal prism has no convex augments that I know of), which greatly reduces the number of possible combinations. Even though adjacent augments are not allowed on the one ring (they would be concave), it still includes analogues of single-ring 10,10-duoprism augmentations, and adds a few more too (e.g., augments spaced an odd number of positions apart, which have no counterpart in the 10,10-duoprism augmentations). Nevertheless, these do not make up for the reduction of combinations due to only one ring being augmentable. So it numbers only in the thousands, rather than the millions.

- quickfur
- Pentonian
**Posts:**2873**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

Thought about constructing a CRF which is not too rigid by construction already, just as the Johnson solids with 5 polygons per vertex.

That is to use lot of solids around an edge here.

I started to consider the cycle (4-trip-4-squippy-3-squippy-4-trip-4-squippy-3-squippy-4). Found that it might be possible to add a further squippy at the top resp. bottom (in the sense of a cube-pyramid) and that we there then will have the open faces cycle (-3(trip)-4(squippy)-3(trip)-4(squippy)-), which then could be joined to a co each. Thus we are back to cope, with all ist cubes being augmented by cube-pyramids.

Just occured to me that that one already was found within this very thread 13 months ago. Thus I just rediscovered it.

But then I considered to write that very one in student91's tegum sum notation. Obviously it is the sum of cope and some scaled oct. Thus we have oa3xo4oo2xo&#zx. And that "a" calculates to be a = w/q = 1.707. (Cells here are: 2 coes, 8 trips, 12+24=36 squippies.)

But then it ought be possible to do the Stott expansions thereof as well:

First we have oa3xo4xx2xo&#zx. Cells then are: 2tics, 8+24=32 trips, 24 squippies, 12 squacues.

But also the other "free" node (i.e. w/o true edges) could be expanded independently, resulting in xb3xo4oo2xo&#zx, where b=a+x. Cells then are: 2toe, 8 hip, 12+24=36 squippies

Or both expansions being applied simultanuously, i.e. xb3xo4xx2xo&#zx. Then the cells are: 2 gircoes, 8 hips, 12 cubes, 24 trips, 24 squippies, 12 squacues.

So, at least, we got 3 more CRFs from that known one for free.

--- rk

That is to use lot of solids around an edge here.

I started to consider the cycle (4-trip-4-squippy-3-squippy-4-trip-4-squippy-3-squippy-4). Found that it might be possible to add a further squippy at the top resp. bottom (in the sense of a cube-pyramid) and that we there then will have the open faces cycle (-3(trip)-4(squippy)-3(trip)-4(squippy)-), which then could be joined to a co each. Thus we are back to cope, with all ist cubes being augmented by cube-pyramids.

Just occured to me that that one already was found within this very thread 13 months ago. Thus I just rediscovered it.

But then I considered to write that very one in student91's tegum sum notation. Obviously it is the sum of cope and some scaled oct. Thus we have oa3xo4oo2xo&#zx. And that "a" calculates to be a = w/q = 1.707. (Cells here are: 2 coes, 8 trips, 12+24=36 squippies.)

But then it ought be possible to do the Stott expansions thereof as well:

First we have oa3xo4xx2xo&#zx. Cells then are: 2tics, 8+24=32 trips, 24 squippies, 12 squacues.

But also the other "free" node (i.e. w/o true edges) could be expanded independently, resulting in xb3xo4oo2xo&#zx, where b=a+x. Cells then are: 2toe, 8 hip, 12+24=36 squippies

Or both expansions being applied simultanuously, i.e. xb3xo4xx2xo&#zx. Then the cells are: 2 gircoes, 8 hips, 12 cubes, 24 trips, 24 squippies, 12 squacues.

So, at least, we got 3 more CRFs from that known one for free.

--- rk

- Klitzing
- Pentonian
**Posts:**1627**Joined:**Sun Aug 19, 2012 11:16 am**Location:**Heidenheim, Germany

Having mentioned these, I just went on in generalizing them.

So I considered the cases

oa3xoPoo2xo&#zx

oa3xoPxx2xo&#zx

xb3xoPoo2xo&#zx

xb3xoPxx2xo&#zx

where a=a(P) to be evaluated individually and b=a+x.

Here P = 2,3,4,5. - The case P=4 was considered yesterday. The case P=2 becomes degenerate (remind the part ...2oo2... therein) and happens to be 3D, or (for the ...2xx2... cases) to be a mere prism thereof. Obviously oa3xo(2oo)2xo&#zx just is the Johnson solid J51 (= trip with 3 squippies attached) and xb3xo(2oo)2xo&#zx then is just J57 (= hip with 3 squippies attached).

So we are left with the cases P=3 and P=5.

oa3xo5oo2xo&#zx then is the tegum sum of iddip and an a-scaled ike, or, worded differently, iddip with all lateral pips augmented by pippies.

Here one derives a = f = (1+sqrt(5))/2 and thus b = F = (3+sqrt(5))/2.

Thus we get of3xo5oo2xo&#zx, of3xo5xx2xo&#zx, xF3xo5oo2xo&#zx, and xF3xo5xx2xo&#zx.

oa3xo3oo2xo&#zx OTOH is the tegum sum of ope and an a-scaled tet, or, worded differently, ope with all alternate trips augmented by trippies.

Here one derives a = (2+sqrt(10))/3 = 1.720759 and thus b = (5+sqrt(10))/3 = 2.720759.

Thus we get oa3xo3oo2xo&#zx, oa3xo3xx2xo&#zx, xb3xo3oo2xo&#zx, and xb3xo3xx2xo&#zx.

All these cases again are CRF, i.e. are regular faced by construction and happen to be convex as well.

--- rk

So I considered the cases

oa3xoPoo2xo&#zx

oa3xoPxx2xo&#zx

xb3xoPoo2xo&#zx

xb3xoPxx2xo&#zx

where a=a(P) to be evaluated individually and b=a+x.

Here P = 2,3,4,5. - The case P=4 was considered yesterday. The case P=2 becomes degenerate (remind the part ...2oo2... therein) and happens to be 3D, or (for the ...2xx2... cases) to be a mere prism thereof. Obviously oa3xo(2oo)2xo&#zx just is the Johnson solid J51 (= trip with 3 squippies attached) and xb3xo(2oo)2xo&#zx then is just J57 (= hip with 3 squippies attached).

So we are left with the cases P=3 and P=5.

oa3xo5oo2xo&#zx then is the tegum sum of iddip and an a-scaled ike, or, worded differently, iddip with all lateral pips augmented by pippies.

Here one derives a = f = (1+sqrt(5))/2 and thus b = F = (3+sqrt(5))/2.

Thus we get of3xo5oo2xo&#zx, of3xo5xx2xo&#zx, xF3xo5oo2xo&#zx, and xF3xo5xx2xo&#zx.

oa3xo3oo2xo&#zx OTOH is the tegum sum of ope and an a-scaled tet, or, worded differently, ope with all alternate trips augmented by trippies.

Here one derives a = (2+sqrt(10))/3 = 1.720759 and thus b = (5+sqrt(10))/3 = 2.720759.

Thus we get oa3xo3oo2xo&#zx, oa3xo3xx2xo&#zx, xb3xo3oo2xo&#zx, and xb3xo3xx2xo&#zx.

All these cases again are CRF, i.e. are regular faced by construction and happen to be convex as well.

--- rk

- Klitzing
- Pentonian
**Posts:**1627**Joined:**Sun Aug 19, 2012 11:16 am**Location:**Heidenheim, Germany

Recently, I was thinking about some of the simpler polyhedral prism augmentations. Specifically, i waas wondering about the tepe (tet prism) augmentations, and if a tetra-augmentation would be possible.

Turns out it isn't. The dihedral angle between trips (triangle prisms) in the tepe is the same as that of tet, 70.53º. The relevant dihedral angles in the trippies (trippy = 3-prism pyramid) are the trip-squippy ones (squippy = square pyramid). tHose, at least from Klitzing's incmats site, seem to be 65.90º. Adding double that to 70.53º results in >180º, so the tetra-augmented tepe isn't convex.

The octa-augmented ope and 20-augmented ipe would similarly not be convex, as in those the trip-trip dihedral angles are even larger. Only the hexa-augmented cube prism (which is just a parabidiminished ico - here the squippies will combine to octahedra) and the dodeca-augmented dope (which was the first polytope to be mentioned in this thread) would work.

Turns out it isn't. The dihedral angle between trips (triangle prisms) in the tepe is the same as that of tet, 70.53º. The relevant dihedral angles in the trippies (trippy = 3-prism pyramid) are the trip-squippy ones (squippy = square pyramid). tHose, at least from Klitzing's incmats site, seem to be 65.90º. Adding double that to 70.53º results in >180º, so the tetra-augmented tepe isn't convex.

The octa-augmented ope and 20-augmented ipe would similarly not be convex, as in those the trip-trip dihedral angles are even larger. Only the hexa-augmented cube prism (which is just a parabidiminished ico - here the squippies will combine to octahedra) and the dodeca-augmented dope (which was the first polytope to be mentioned in this thread) would work.

- username5243
- Trionian
**Posts:**125**Joined:**Sat Mar 18, 2017 1:42 pm

username5243 wrote:The octa-augmented ope ... would similarly not be convex, as in those the trip-trip dihedral angles are even larger.

But, as I've just mentioned before, the tetra-augmented ope is !

--- rk

- Klitzing
- Pentonian
**Posts:**1627**Joined:**Sun Aug 19, 2012 11:16 am**Location:**Heidenheim, Germany

Putting it all together, we have 2 old known seed CRFs, the dodecahedral prism and the icosidodecahedral prism, each with all pentagonal prisms being blended with pentagonal prism pyramids, thereby producing according augmentations oa3oo5xo2xo&#zx and oa3xo5oo2xo&#zx respecively ("a" to be calculated accordingly). Thereon we can do several variations: first we can do several Stott expansions, second we could change the "5" by other numbers P.

The second type of variation is limited by P=4 for the first set, where cells become corealmic (e.g. pairs of squippies merge into octs, etc.), and by P=2 for the second, where the polychora become dimensionally degenerate (becoming J51 and J57, resp. their prisms). Accordingly we have here the following bunch of valide CRFs:

oq3oo4xo xo&#zx = tes + 6 (equatorial) cubpies

oq3xx4xo xo&#zx = ticcup + 6 (gyro) squipufs

xw3oo4xo xo&#zx = sircope + 6 cubpies

xw3xx4xo xo&#zx = gircope + 6 (gyro) squipufs

oa3oo5xo xo&#zx = dope + 12 pippies

oa3xx5xo xo&#zx = tiddip + 12 (gyro) pepufs

xb3oo5xo xo&#zx = sriddip + 12 pippies

xb3xx5xo xo&#zx = griddip + 12 (gyro) pepufs

(where a = 3/sqrt(5) = 1.341641, b = a+x = 2.341641)

oa3xo3oo xo&#zx = ope + 4 (alternate) trippies

oa3xo3xx xo&#zx = tuttip + 4 tripufs

xb3xo3oo xo&#zx = tuttip + 4 trippies

xb3xo3xx xo&#zx = tope + 4 (alternate) tripufs

(where a = (2+sqrt(10))/3 = 1.720759, b = a+x = 2.720759)

oa3xo4oo xo&#zx = cope + 6 cubpies

oa3xo4xx xo&#zx = ticcup + 6 (ortho) squipufs

xb3xo4oo xo&#zx = tope + 6 cubpies

xb3xo4xx xo&#zx = gircope + 6 (ortho) squipufs

(where a = w/q = 1.707107, b = a+x = 2.707107)

of3xo5oo xo&#zx = iddip + 12 peppies

of3xo5xx xo&#zx = tiddip + 12 (ortho) pepufs

xF3xo5oo xo&#zx = tipe + 12 peppies

xF3xo5xx xo&#zx = griddip + 12 (ortho) pepufs

(where F = ff = f+x)

And, of course, one might use less than the mentioned totals of possible augmentations.

--- rk

The second type of variation is limited by P=4 for the first set, where cells become corealmic (e.g. pairs of squippies merge into octs, etc.), and by P=2 for the second, where the polychora become dimensionally degenerate (becoming J51 and J57, resp. their prisms). Accordingly we have here the following bunch of valide CRFs:

oq3oo4xo xo&#zx = tes + 6 (equatorial) cubpies

oq3xx4xo xo&#zx = ticcup + 6 (gyro) squipufs

xw3oo4xo xo&#zx = sircope + 6 cubpies

xw3xx4xo xo&#zx = gircope + 6 (gyro) squipufs

oa3oo5xo xo&#zx = dope + 12 pippies

oa3xx5xo xo&#zx = tiddip + 12 (gyro) pepufs

xb3oo5xo xo&#zx = sriddip + 12 pippies

xb3xx5xo xo&#zx = griddip + 12 (gyro) pepufs

(where a = 3/sqrt(5) = 1.341641, b = a+x = 2.341641)

oa3xo3oo xo&#zx = ope + 4 (alternate) trippies

oa3xo3xx xo&#zx = tuttip + 4 tripufs

xb3xo3oo xo&#zx = tuttip + 4 trippies

xb3xo3xx xo&#zx = tope + 4 (alternate) tripufs

(where a = (2+sqrt(10))/3 = 1.720759, b = a+x = 2.720759)

oa3xo4oo xo&#zx = cope + 6 cubpies

oa3xo4xx xo&#zx = ticcup + 6 (ortho) squipufs

xb3xo4oo xo&#zx = tope + 6 cubpies

xb3xo4xx xo&#zx = gircope + 6 (ortho) squipufs

(where a = w/q = 1.707107, b = a+x = 2.707107)

of3xo5oo xo&#zx = iddip + 12 peppies

of3xo5xx xo&#zx = tiddip + 12 (ortho) pepufs

xF3xo5oo xo&#zx = tipe + 12 peppies

xF3xo5xx xo&#zx = griddip + 12 (ortho) pepufs

(where F = ff = f+x)

And, of course, one might use less than the mentioned totals of possible augmentations.

--- rk

- Klitzing
- Pentonian
**Posts:**1627**Joined:**Sun Aug 19, 2012 11:16 am**Location:**Heidenheim, Germany

Hmm. Is it possible to make a CRF out of a 3prism and a line perpendicular to the axis of the prism (i.e., parallel to one of the edges of a triangular face)? Topologically, at least, I can see it as having these cells: 2 triangular prisms, 1 tetrahedron, 2+2=4 square pyramids. But I'm not sure if it can be made CRF.

The reason I ask, is because this could potentially serve as an augment of a different kind that could be used for augmenting various polyhedral prisms, if it turns out to be CRF and have suitable dichoral angles for augmentation.

The reason I ask, is because this could potentially serve as an augment of a different kind that could be used for augmenting various polyhedral prisms, if it turns out to be CRF and have suitable dichoral angles for augmentation.

- quickfur
- Pentonian
**Posts:**2873**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

quickfur wrote:Hmm. Is it possible to make a CRF out of a 3prism and a line perpendicular to the axis of the prism (i.e., parallel to one of the edges of a triangular face)? Topologically, at least, I can see it as having these cells: 2 triangular prisms, 1 tetrahedron, 2+2=4 square pyramids. But I'm not sure if it can be made CRF.

The reason I ask, is because this could potentially serve as an augment of a different kind that could be used for augmenting various polyhedral prisms, if it turns out to be CRF and have suitable dichoral angles for augmentation.

Are you talking about K4.8?

That's what it sounds like you are describing, and the provided cells (1 tet + 4 squippies + 2 trips) match your post.

- username5243
- Trionian
**Posts:**125**Joined:**Sat Mar 18, 2017 1:42 pm

Hmm. This would seem to be equivalent to tetrahedron || square. Is this one in your list, Klitzing?

- quickfur
- Pentonian
**Posts:**2873**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

Ah yes, looks like it is K4.8 indeed! Thanks!

- quickfur
- Pentonian
**Posts:**2873**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

The next question is, can K4.8 be used for augmenting polyhedral prisms?

- quickfur
- Pentonian
**Posts:**2873**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

Unfortunately, the relevant dichoral angles are not provided there (unlike other polychora on that site). Anyone know how to derive them?

- username5243
- Trionian
**Posts:**125**Joined:**Sat Mar 18, 2017 1:42 pm

quickfur wrote:The next question is, can K4.8 be used for augmenting polyhedral prisms?

Well, depends for sure on where you want it to be attached to.

The dihedrals of K-4.8 = bidrap (yes, username5243 got it right!) directly can be deduced from those of rap and the 2 chopped off trippies.

Cf. also the lace cities of rap and bidrap respectively:

- Code: Select all
`x o o x x o o x`

o o x x o o x x

- at {4} between squippy and trip: arccos(-1/sqrt(6)) = 114.094843°
- at {3} between squippy and tet: arccos(-1/4) = 104.477512°
- at {3} between squippy and squippy: arccos(1/4) = 75.522488°
- at {3} between squippy and trip: arccos(sqrt[1/6]) = 65.905157°
- at {4} between trip and trip: arccos(2/3) = 48.189685°

- Klitzing
- Pentonian
**Posts:**1627**Joined:**Sun Aug 19, 2012 11:16 am**Location:**Heidenheim, Germany

Hmm. Some of those angles are rather large, so that might limit where we can attach it. Since we're augmenting polyhedral prisms, we'd be using the 3prism as the attachment cell; but there's one large angle there (114.09°). So it would have to fit where there's lots of extra room, ergo, at sharp angles in the base polyhedron.

Do you think it might fit on a snub disphenoid prism? There's a sharp angle between two pairs of triangles, where we might be able to accomodate the 114.09° angle, perhaps?

Do you think it might fit on a snub disphenoid prism? There's a sharp angle between two pairs of triangles, where we might be able to accomodate the 114.09° angle, perhaps?

- quickfur
- Pentonian
**Posts:**2873**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

88 posts
• Page **2** of **3** • 1, **2**, 3

Users browsing this forum: No registered users and 1 guest