## Regular polytopes

Discussion of known convex regular-faced polytopes, including the Johnson solids in 3D, and higher dimensions; and the discovery of new ones.

### Regular polytopes

Why is it that for more than 4 dimensions there are only three regular polytopes? Why can't there be more of them?
People may consider as God the beings of finite higher dimensions,
though in truth, God has infinite dimensions Prashantkrishnan
Trionian

Posts: 114
Joined: Mon Jan 13, 2014 5:37 pm
Location: Kochi, Kerala, India

### Re: Regular polytopes

Prashantkrishnan wrote:Why is it that for more than 4 dimensions there are only three regular polytopes? Why can't there be more of them?

Simply said, because the math doesn't work that way.

A n-dimensional regular polytope can be imagined as a string of n-1 numbers (the Schlafli numbers, each number >= 3). In 2D, you can take any number you wish and make into a polygon - {3}, {4}, {5}, etc.

In 3D, you get the five regular polytopes:
{3,3} - three triangles per vertex - tetrahedron
{3,4} - four triangles per vertex - octahedron
{3,5} - five triangles per vertex - icosahedron
{4,3} - three squares per vertex - cube
{5,3} - three pentagons per vertex - dodecahedron

It's fairly easy to see that there are no other solutions. {3,6} is flat and {3,7+} hyperbolic. {4,4} is flat and {4,5+} hyperbolic. {5,4+} is hyperbolic. {6,3} is flat and {6,4+} and {7+,3+} are all hyperbolic. In order for {a,b} to be a regular polyhedron, (inner angle of a-gon)*b must be below 360 degrees.

For 4D, {a,b,c} is regular polychoron if (dihedral angle of {a,b})*c is below 360 degrees. But the smallest possible angle in this formula is actually larger than smallest possible angle in 3D. The smallest dihedral angle is in tetrahedron, and it's bigger than 60 degrees of triangle. In fact, dihedral angle in any simplex is larger than the dihedral angle in the previous one; that's because if you build an infinite prism on a simplex, the dihedral angle of such prism will be equal to the dihedral angle of the base simplex. If you incline the edges to intersect into a tall simplex pyramid, the dihedral angle starts to increase and any such pyramid, including the higher-dimensional simplex itself will have larger dihedral angle than its base.
Another rule is that {a,b,c} can only be a polychoron if both {a,b} and {b,c} are valid polyhedra. That's because existence {a,b,c} implies the existence of the dual, {c,b,a}.

So, out of five polyhedra, we can create 11 theoretically valid combinations:
{3,3,3}, {3,3,4}, {3,3,5}
{3,4,3}
{3,5,3}
{4,3,3}, {4,3,4}, {4,3,5}
{5,3,3}, {5,3,4}, {5,3,5}

Six of these, {3,3,3}, {3,3,4}, {3,3,5}, {3,4,3}, {4,3,3} and {5,3,3} are regular polychora. {4,3,4} is the cubic tesselation of space. The remaining four, {3,5,3}, {4,3,5}, {5,3,4} and {5,3,5} are tesselations of hyperbolic space.

So, what about five dimensions? There, the regular polytopes will be {a,b,c,d} with {a,b,c} and {b,c,d} as valid polychora. Once again, there are 11 combinations that can be created from the six 4D polychora:
{3,3,3,3}, {3,3,3,4}, {3,3,3,5}
{3,3,4,3}
{3,4,3,3}
{4,3,3,3}, {4,3,3,4}, {4,3,3,5}
{5,3,3,3}, {5,3,3,4}, {5,3,3,5}

Exploring these, we find that only three, {3,3,3,3}, {3,3,3,4} and {4,3,3,3} work. The dihedral angle of pentachoron exceeds 72 degrees, so {3,3,3,5} no longer fits and can only exist in hyperbolic space. On the plus side, dihedral angle of 16-cell and 24-cell is exactly 120 degrees, which means that {4,3,3,4}, {3,3,4,3} and {3,4,3,3} are all valid tesselations of Euclidean 4D space. But all the five tesselations that involve number 5, {3,3,3,5}, {4,3,3,5}, {5,3,3,3}, {5,3,3,4} and {5,3,3,5} are hyperbolic.

So in 5D we only have three regular polytera. If we go to 6D, we find that there are only four ways to join them to valid combinations of five numbers:
{3,3,3,3,3}, {3,3,3,3,4}
{4,3,3,3,3}, {4,3,3,3,4}
{4,3,3,3,4} is an Euclidean tesselation, and so we are left with only three polypeta. And we get the same result in every higher dimension. Combining three lower-dimensional polytopes will only give us their three analogues, plus one Euclidean tesselation.

I recommend http://en.wikipedia.org/wiki/Coxeter%E2 ... in_diagram
Marek14
Pentonian

Posts: 1122
Joined: Sat Jul 16, 2005 6:40 pm

### Re: Regular polytopes

Well, in order to be a tiny bit more precise, the above mail is not fully correct! This is because you implicitely restrict yourself to convex shapes only. (A quite common fault indeed.)

Within 2D you could have virtually any polygram {n/d}, where n>2, 0<2d<n, gcd(n,d)=1. It always would be regular.
(The last restiction could be relaxed, but that then opens up for interpretation, and thus should always be used within some context only, where its interpretation is pre-set.)

For 3D it then follows that both the faces and the vertex figures of regular polyhedra could be stary, so you'd get something like {n/d, m/b} in general. - But, in order to remain at least somehow sphere-like (positive curvature and esp. finite extend) we need as a further restriction:
Code: Select all
`d/n + b/m > 1/2`

That one not only allows for
{3,3} (tetrahedron),
{3,4} (octahedron),
{3,5} (icosahedron),
{4,3} (cube or hexahedron),
{5,3} (dodecahedron)
, i.e. the Platonic solids,
but obviously also for
{5/2,3} (great icosahedron),
{5/2,5} (small stellated dodecahedron),
{3,5/2} (great stellated dodecahedron),
{5,5/2} (great dodecahedron)
, i.e. the Kepler-Poinsot solids.

But this is not all to be considered here! E.g. {8,16/7} also would pass the above mentioned restriction. The additional restriction being used to cast off such weird examples are not only the finiteness of extent, but also the finiteness of elements. I.e. that given example would wrap around the center for ever, without closing back. Thereby all elements would become infinitely dense.

For 4D we then would get some general regular polychoron {n/d, m/b, k/c}. But furthermore both its cells {n/d, m/b} and its vertex figures {m/b, k/c} need to be valid regular polyhedra.
Further one has still to bow to positive curvature (i.e. finiteness of extend) and to finiteness of elements.
The corresponding research then is the main topic of Coxeter's famous book on Regular Polytopes.
E.g. the finiteness of extend here is provided by the inequality
Code: Select all
`sin(pi.d/n).sin(pi.c/k) < cos(pi.b/m).`

That inequality then would pass the known convex regular polychora
{3,3,3} (pentachoron),
{3,3,5} (hexacosachoron),
{3,4,3} (icositetrachoron),
{4,3,3} (tesseract or octachoron),
{5,3,3} (hecatonicosachoron)
.
But also the Schläfli-Hess polychora
{3,3,5/2} (grand hexacosachoron),
{3,5/2,5} (great faceted hexadecachoron),
{5/2,3,3} (great grand stellated hecatonicosachoron),
{5/2,5,3} (small stellated hecatonicosachoron),
{5,5/2,3} (great grand hecatonicosachoron),
{5/2,3,5} (great stellated hecatonicosachoron),
{5,3,5/2} (grand hecatonicosachoron),
{5,5/2,5} (great hecatonicosachoron),
{5/2,5,5/2} (grand stellated hecatonicosachoron)
.
And once more there would pass some infinitely dense ones, like {3,5/2,3}, {4,3,5/2}, {5/2,3,4}, {5/2,3,5/2}.

For 5D we have the general regular polyteron setup {n/d, m/b, k/c, l/a}. The inequality for finiteness of extend here then reads like:
Code: Select all
`cos²(pi.b/m)/sin²(pi.d/n) + cos²(pi.c/k)/sin²(pi.a/l) < 1.`

This inequality surely do pass the known convex cases
{3,3,3,3} (hexateron),
{3,3,3,4} (triacontiditeron),
{4,3,3,3} (penteract or decateron).

But together with the restriction to finiteness of elements (non-dense ones) it happens to be impossible to have stary forms in this dimension any more.

And because subdimensional regulars have to function both as subelements and as vertex figures of the plusdimensional regulars, it becomes obvious that even beyond no regular stary forms can occur!

--- rk
Klitzing
Pentonian

Posts: 1464
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

### Re: Regular polytopes

Klitzing wrote:{3,3,5/2} (grand hexacosachoron),
{3,5,5/2} (faceted hexadecachoron), - should be faceted hexacosachoron
{3,5/2,5} (great faceted hexadecachoron), - should be great faceted hexacosachoron
{5/2,3,3} (great grand stellated hecatonicosachoron),
{5/2,5,3} (small stellated hecatonicosachoron),
{5,5/2,3} (great grand hecatonicosachoron),
{5/2,3,5} (great stellated hecatonicosachoron),
{5,3,5/2} (grand hecatonicosachoron),
{5,5/2,5} (great hecatonicosachoron),
{5/2,5,5/2} (grand stellated hecatonicosachoron)
.

Found a couple typos, corrected above.

If we really want to get a bit sinister, there are also lots of regular abstract polytopes like the abstract polyhedron with 24 heptagons as faces, AKA the Klein quartic - I imagine these abstract polytopes would make excellent mazes where all the rooms have the same shape, here heptagons, where each wall has a door that leads to another room. There seems to be an endless array of these sort of polytopes, some are non-orientable.
Whale Kumtu Dedge Ungol.
Polyhedron Dude
Trionian

Posts: 190
Joined: Sat Nov 08, 2003 7:02 am
Location: Texas

### Re: Regular polytopes

Polyhedron Dude wrote:Found a couple typos, corrected above.

If we really want to get a bit sinister, there are also lots of regular abstract polytopes like the abstract polyhedron with 24 heptagons as faces, AKA the Klein quartic - I imagine these abstract polytopes would make excellent mazes where all the rooms have the same shape, here heptagons, where each wall has a door that leads to another room. There seems to be an endless array of these sort of polytopes, some are non-orientable.

Probably not excellent mazes, since a good maze should be solvable -- unless you count the maze's success by how many of your enemies it has successfully dispatched.

So, a mental exercise: what would you do if you found yourself in heptagonal maze like this? Oh, and you don't know its topology beforehand. What strategy would you use?
Marek14
Pentonian

Posts: 1122
Joined: Sat Jul 16, 2005 6:40 pm

### Re: Regular polytopes

The list of regular complex polytopes is finite too, except for the polypisms and polytgums.

2D 3(3)3, 3(4)3, 3(5)3, #3(6)3, 4(3)4, #4(4)4, 5(3)5, #6(3)6
All p(2q)r, where p,q,r is any of 2,3,3; 2,3,4; 2,3,5; 2,2,p; #2,3,6; or #2,4,4 or #3,3,3

3D p(4)2(3)2, 2(3)2(4)p, 3(3)3(3)3, 3(3)3(4)2, 2(4)3(3)3

4D p(4)2(3)2(3)2, 2(3)2(2)2(4)p, and 3(3)3(3)3(3)3.

Thereafter only the first two carry through to infinite dimensions.

Starry polytopes also occur, eg 3(4)5 gives 3(10/3)5, 3(6/2)5, and 3(10/4)5.
The dream you dream alone is only a dream
the dream we dream together is reality.

\(Latex\) at https://greasyfork.org/en/users/188714-wendy-krieger wendy
Pentonian

Posts: 1874
Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia

### Re: Regular polytopes

Coxeter did a lot of interesting things with finite maps, such as x5o5oA5p, a polyhhedron with sixtysix faces, and every petrie polygon is five long. I have a list of them on my puter, but its down at this time.
The dream you dream alone is only a dream
the dream we dream together is reality.

\(Latex\) at https://greasyfork.org/en/users/188714-wendy-krieger wendy
Pentonian

Posts: 1874
Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia

### Re: Regular polytopes

Polyhedron Dude wrote:
Klitzing wrote:{3,5,5/2} (faceted hexadecachoron), - should be faceted hexacosachoron
{3,5/2,5} (great faceted hexadecachoron), - should be great faceted hexacosachoron
.

Found a couple typos, corrected above.

Oops, right you are, indeed.

If we really want to get a bit sinister, there are also lots of regular abstract polytopes like the abstract polyhedron with 24 heptagons as faces, AKA the Klein quartic - I imagine these abstract polytopes would make excellent mazes where all the rooms have the same shape, here heptagons, where each wall has a door that leads to another room. There seems to be an endless array of these sort of polytopes, some are non-orientable.

Hmmm, well, I was taking up the tune of the first post in this thread, i.e. taking over that setup, which indeed leads to 3 regular polytopes above 4D. - OTOH, there are others, not only within abstract spaces, but also quite normal regular polytopes within euclidean or hyperbolic spaces. It is just that those have not to be considered as tesselations only but that those ought to bear some body "underneath". - This in fact also is how euclidean "tesselations" are re-used for cells within non-compact hyperbolic tesselations. Or even hyperbolic ones of some lower absolut value of curvature within ones of a higher one.

But when you come up with abstract ones, then we could mention likewise the Grünbaum-Coxeter polytopes, esp. the 11-cell and the 57-cell.

--- rk
Klitzing
Pentonian

Posts: 1464
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

### Re: Regular polytopes

Wendy, what is that # symbol wrt. complex polytopes?
And what means that p in x5o5oA5p ?
--- rk
Klitzing
Pentonian

Posts: 1464
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

### Re: Regular polytopes

# is just meant to point to a non-existant foot note, saying these things are tilings. So x4o4o is a tiling, as is 3(4)3(4)3. But since tilings are polyhedra, i included the polygonal tilings (ie horogons) here.

p in x5o5oA5p is a petrie node. It means that the petrie polygon of the figure in question is 5. Coxeter's regular maps can be constructed as three arms radiating from a single node, one the vertex, one the cell, and the third one is the petrie polygon. A given map like x3o7oA13p, has a petrie polygon of 13 edges (and a symmetry of 1092 = 9.12).
The dream you dream alone is only a dream
the dream we dream together is reality.

\(Latex\) at https://greasyfork.org/en/users/188714-wendy-krieger wendy
Pentonian

Posts: 1874
Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia

### Re: Regular polytopes

I'm not too used to Petrie polygons. That h ought to be generally calcuable from {p,q,...}, I suppose.
But the only thing I found (in a quick Google search) so far was just
Code: Select all
`cos²(pi/h) = cos²(pi/p) + cos²(pi/q)`
for 2D fabrics with positive curvature (i.e. polyhedra). None for higher D nor for zero or negative curvature.

The Klein quartic case, mentioned by Hedrondude, can be derived from the incmat of x7o3o
Code: Select all
`o3o7x   (N → ∞). . . | 14N |   3 |  3------+-----+-----+---. . x |   2 | 21N |  2------+-----+-----+---. o7x |   7 |   7 | 6N`
just by taking N=4. I.e. it occurs here as a rather low approximant. And wrt. x7o3o it conversely is a rather high mod-wrap, corresponding to a 24-coloring thereof.

If I knew how to derive h from such an incmat, I might transport that understanding to the matrix of x5o5o
Code: Select all
`x5o5o   (N → ∞). . . | 2N |  5 |  5------+----+----+---x . . |  2 | 5N |  2------+----+----+---x5o . |  5 |  5 | 2N`
as well. And then could derive which approximant (i.e. which choice of N) you'd need for that x5o5oA5p = x5o5o *b5p ...

--- rk
Klitzing
Pentonian

Posts: 1464
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

### Re: Regular polytopes

Marek14 wrote:
Polyhedron Dude wrote:Found a couple typos, corrected above.

If we really want to get a bit sinister, there are also lots of regular abstract polytopes like the abstract polyhedron with 24 heptagons as faces, AKA the Klein quartic - I imagine these abstract polytopes would make excellent mazes where all the rooms have the same shape, here heptagons, where each wall has a door that leads to another room. There seems to be an endless array of these sort of polytopes, some are non-orientable.

Probably not excellent mazes, since a good maze should be solvable -- unless you count the maze's success by how many of your enemies it has successfully dispatched.

So, a mental exercise: what would you do if you found yourself in heptagonal maze like this? Oh, and you don't know its topology beforehand. What strategy would you use?

What I had in mind would go something like this:

You start off in a heptagon shaped area around 80 to 100 feet across. The area can be filled with all sorts of terrain, objects, rooms, pools, etc to make each of the 24 areas unique and interesting. Each of the seven walls of the heptagon has a locked door or gate, and you must solve riddles and puzzles as well as search for keys and levers to open the various gates and doors. So the first goal would be to unlock all 24 heptagonal areas to explore. Next would be to unlock every gate and door, once all seven gates/doors in an area is unlocked, it gives you access to one of the 24 key items. Once all of the 24 items are found, then you put them together like a puzzle and use this as a key to gain access to a transporter to let you exit the entire maze - and then you enter the next maze - the 57-choron .
Whale Kumtu Dedge Ungol.
Polyhedron Dude
Trionian

Posts: 190
Joined: Sat Nov 08, 2003 7:02 am
Location: Texas

### Re: Regular polytopes

Polyhedron Dude wrote:
Marek14 wrote:
Polyhedron Dude wrote:Found a couple typos, corrected above.

If we really want to get a bit sinister, there are also lots of regular abstract polytopes like the abstract polyhedron with 24 heptagons as faces, AKA the Klein quartic - I imagine these abstract polytopes would make excellent mazes where all the rooms have the same shape, here heptagons, where each wall has a door that leads to another room. There seems to be an endless array of these sort of polytopes, some are non-orientable.

Probably not excellent mazes, since a good maze should be solvable -- unless you count the maze's success by how many of your enemies it has successfully dispatched.

So, a mental exercise: what would you do if you found yourself in heptagonal maze like this? Oh, and you don't know its topology beforehand. What strategy would you use?

What I had in mind would go something like this:

You start off in a heptagon shaped area around 80 to 100 feet across. The area can be filled with all sorts of terrain, objects, rooms, pools, etc to make each of the 24 areas unique and interesting. Each of the seven walls of the heptagon has a locked door or gate, and you must solve riddles and puzzles as well as search for keys and levers to open the various gates and doors. So the first goal would be to unlock all 24 heptagonal areas to explore. Next would be to unlock every gate and door, once all seven gates/doors in an area is unlocked, it gives you access to one of the 24 key items. Once all of the 24 items are found, then you put them together like a puzzle and use this as a key to gain access to a transporter to let you exit the entire maze - and then you enter the next maze - the 57-choron .

Incidentally, that is not that far from a game I've found recently, based on hyperbolic geometry. Except that it has unlimited areas -- you collect treasures and avoid or kill monsters. The final puzzle in the game is that you're sent to retrieve a key from distant region, and then you have to find your way back -- which is pretty difficult!
Marek14
Pentonian

Posts: 1122
Joined: Sat Jul 16, 2005 6:40 pm

### Re: Regular polytopes

All regular polytopes have a petrie polygon. This is the movement of a vertex when the reflections are taken in (any) cyclic order. For the simplex, all vertices are visited, for the cross and tegum, h = 2n, for 3,5 it's 10, and for 3,3,5, it's 30. It exists for the gosset groups E6 has 12, E7 has 18, and E8 has 30.

For polyhedra, it is Wx, where the polyhedron is Wa, Wb, then x=a+b. This is specifically why the notation W is invented!

A polytope described in such a group, can not be any larger than a polygon xh, where x is the number of marked nodes.

The number of mirrors is exactly 2m = nh, where m is the mirrors, n is the dimension, and h is the petrie polygon. This is because the petrie polygon crosses each mirror exactly twice. So, for E7, n=7, m=63 (half the verticies of 2_31), thence h = 63*2/7 = 18.
The dream you dream alone is only a dream
the dream we dream together is reality.

\(Latex\) at https://greasyfork.org/en/users/188714-wendy-krieger wendy
Pentonian

Posts: 1874
Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia

### Re: Regular polytopes

It might be well-known, but it only occured to me these days.
I was reconsidering the cross polytopes in general, throughout all dimensions.

Just as the measure polytope of dimension d is the cartesian product of any codimensional pair of boundary elements (which are measure polytopes again, i.e. having dimensions n and d-n), so too the cross-polytope of dimension d is the tegum sum of any codimensional pair of lower-dimensional cross-polytopes, i.e. having dimensions n an d-n.

Further it is known, that any cross-polytope can be represented as a lace prism, when being based on one of ist facets. Thus we have the d-1-dimensional simplex atop its dual d-1-dimensional simplex.

Putting these ideas together, we now get for lace city display generally a rhomb (sometimes becoming a square) of the following form:
Code: Select all
`A    B         D    C`

where the perp-dimensional space is represented as An x Ad-n-2 for any 0 <= n <= (d-2)/2 according to:
• A = x3o...o3o o3o...o3o
• B = o3o...o3o x3o...o3o
• C = o3o...o3x o3o...o3o
• D = o3o...o3o o3o...o3x
In fact, for n > (d-2)/2 the rhomb just gets slanted into the other direction. And therefore, for even dimensions d one gets for n = (d-2)/2 the square case for that lace city display.

And, for sure, any of those lace city displays then allows for an own incidence matrix representation of that cross-polytope:
xooo3oooo...oooo3ooxo oxoo3oooo...oooo3ooox &#xr   (where that trailing r reminds to being a lace ring symbol).

--- rk
Klitzing
Pentonian

Posts: 1464
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

### Re: Regular polytopes

The diagram that Richard shows, is a general illistration that

"the tegum-product of several antiprisms, is the anti-prism of the pyramid-product of the antiprism-bases."

The same holds true for the general tegum<>prism exchange. so

"the prism-product of several antitegums, is the anti-tegum of the pyramid-product of the antitegum-bases."

For the example like Richard shows, we have

A B

dB dA

A || dA is an antiprism, and B || dB ditto. A and B are completely orthogonal, so A+B is a pyramid-product, as are any of the four sides.

Any of the four sides are dual to the opposite side, and thus form an antiprism, eg A+dB is dual to dA+B, etc.

The special case is the tegum powers, which form the orthotopes. Here, A and B are simplexes, the vertex count makes the dimension. dA and dB are inverted simplexs. In this case, A-dA and B-dB cross at right angles, but the lengths are different if the dimension-count is different.
The dream you dream alone is only a dream
the dream we dream together is reality.

\(Latex\) at https://greasyfork.org/en/users/188714-wendy-krieger wendy
Pentonian

Posts: 1874
Joined: Tue Jan 18, 2005 12:42 pm
Location: Brisbane, Australia

### Re: Regular polytopes

Klitzing wrote:Putting these ideas together, we now get for lace city display generally a rhomb (sometimes becoming a square) of the following form:
Code: Select all
`A    B         D    C`

where the perp-dimensional space is represented as An x Ad-n-2 for any 0 <= n <= (d-2)/2 according to:
• A = x3o...o3o o3o...o3o
• B = o3o...o3o x3o...o3o
• C = o3o...o3x o3o...o3o
• D = o3o...o3o o3o...o3x

Just to be even a bit more explicite, we have:

3D-cross-polytope = oct =
Code: Select all
`o x x o`

4D-cross-polytope = hex =
Code: Select all
`o3o   x3o                              o3x   o3o`
or
Code: Select all
`x o   o x                           o x   x o`

5D-cross-polytope = tac =
Code: Select all
`o3o3o   x3o3o                                                 o3o3x   o3o3o`
or
Code: Select all
`x o3o   o x3o                                                          o o3x   x o3o`

6D-cross-polytope = gee =
Code: Select all
`o3o3o3o   x3o3o3o                                                                                      o3o3o3x   o3o3o3o`
or
Code: Select all
`x o3o3o   o x3o3o                                                                                                   o o3o3x   x o3o3o`
or
Code: Select all
`x3o o3o   o3o x3o                                                                                                      o3o o3x   o3x o3o`

7D-cross-polytope = zee =
Code: Select all
`o3o3o3o3o   x3o3o3o3o                                                                                                                              o3o3o3o3x   o3o3o3o3o`
or
Code: Select all
`x o3o3o3o   o x3o3o3o                                                                                                                                              o o3o3o3x   x o3o3o3o`
or
Code: Select all
`x3o o3o3o   o3o x3o3o                                                                                                                                                            o3o o3o3x   o3x o3o3o`

etc.

The first of those each most probably is common knowledge, because that one just refers to the bipyramid description of the cross-polytope.
The additional ones had been encountered already in one or two cases, when using such cross-polytopes as slanted faces within larger polytopes.
But the general behaviour of this series seems none the less surprising.
Even so, in view of what I'd outlined above and what Wendy supported then, it becomes obvious, that those are valide descriptions indeed. --- rk
Klitzing
Pentonian

Posts: 1464
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany 