Generalizing the rotatope concept

[Marek14]

OK, since I'm lazy, I'll just repost the emails I wrote as I was sending them...

Yesterday, I was browsing through here:
http://tetraspace.alkaline.org/shapes/rotatopes.htm

Then, when I was lying in my bed trying to sleep, something connected in my head and I saw a way to entangle these shapes with graph theory... and this is what resulted:

This is a message written at 3AM:
"Good morning,

I've found page yesterday and it intrigued me. The idea of rotatopes was something I never thought about before, but the logic is sound.
After I assembled five- and six-dimensional ones in my head (which is quite easy, if you imagine 5D as 2D array of 3D shapes and 6D as 3D array of the same), I was heading to bed. On the drift towards sleep, I suddenly realized the connection between rotatopes and graph theory, plus a weird idea how to extend the notion of rotatopes to some weirder shapes.

Let's look at the three 3D rotatopes. Each one of them can be assigned to a particular graph on three nodes:

Cube can be assigned to graph without edges: O O O
Cylinder is a graph with one edge: O=O O
Sphere is a graph with three edges: =O=O=O=, where the left and right side are connected.

This correspondence has certain properties - for example, a shadow cast by a rotatope in one of the main direction is gotten by eliminating one of the nodes from the graph. So far there's nothing new - just the numbers in rotatopes descriptions were replaced by complete graphs.
Now, we can assign arbitrary numbers to the nodes, as "sizes" of the rotatope. 1 1 is a square of edge 1. 1 2 is a rectangle. 1=1 is a circle, 1=2 would be an ellipse. If we start with a rotatope, we can now derive how it will look, passing through the plane: the nodes not connected to the removed node will stay on constant value, while those connected to the removed node will start at zero, grow to their original value, then shrink to zero again. Passing a cylinder 1=1 1 by removing 1st node will produce a series of rectangles from =0 1 through =0.5 1 to square =1 1, and then back again through =0.5 1 to =0 1. The = symbolizes that there is a free edge, and therefore a dynamic size.

Now comes the trick: there is one more graph with three nodes: O=O=O. Does this graph have a rotatope associated with it like the others?

Surprisingly, it turns out it does. First, we review its properties: It has to cast two circular shadows (when we remove the end-nodes) and one square shadow (removing the middle-node). Passing it through the plane will result in two possibilities: either a square that will start from a point, grow, then shrink...
...or we start with a line. This will grow into circle (originally I thought it was through ellipse, but in fact the intermediate shapes are circles with their top and bottom cut). Then it shrinks back into line.

The best way to describe the shape I found would be this:

Start with a circle. Now, construct a second circle perpendicular to it. For example, meridians 0-180 and meridians 90-270 on Earth.
Now, cover this whole thing with tight canvas. The two circles will stretch it and form a square by the equator. All the latitudal lines will be squares too. I'm not yet sure if this would indeed form a sequence of ellipses when passed through a plane in direction perpendicular to one of the circles, but it should.
I just realized that I spent several weeks of family holidays in a tent which had shape of one half of this creature...

The point here is that there are 11 different graphs on 4 nodes. 5 of them have only complete components and correspond to 5 rotatopes you have on your page. The other six have non-complete graphs, and all of them would contain the O=O=O shape as a shadow in one or more direction (one of them, indeed, in ALL 4 directions). I think these are worth checking, in order to enlarge the zoo a bit.

Since these are obviously not true rotatopes, we will need a new name... Would "graphotopes" work? "

I still couldn't sleep. I posted a second message almost immediately:

"The shape I proposed works, but the rigid circles I envisioned are not the main directions - otherwise it would be impossible to achieve non-square rectangle as a cross-section. The true main circles pass through centers of the equator edges.

I tried to devise an equation for this shape - it should be max(a*x2,b*y2)+c*z2=1, where a,b,c are constants, or simply max(x2,y2)+z2=1 for the unit case."

And in 6AM, I sent in the finale...

"
Instead of sleeping, I developed my idea a bit further:

The equations are usually derived in a simple way. I'll show:

1 dimension - x2=1. There is no other option.
2 dimensions - square is max(x2,y2)=1, circle is x2+y2=1. The graphs of square and circle are complementary. Whenever we have complementary graphs, we can get equation of one graphotope from the other by exchanging adding and maximum operation.
In 3 dimensions, cube (max(x2,y2,z2)=1) and sphere (x2+y2+z2=1) form one such doublet. The second one is formed by cylinder (max(x2+y2,z2)=1) and the shape I found, which I came to call DOME: (max(x2+y2)+z2=1).

In cylinder, second rule can be seen: if the graph is formed by multiple components, its equation is formed by taking a maximum of all the individual components.
This means that every doublet has one connected and one disconnected graph. I can't (yet) form simple equations for graphotopes that are connected in both normal and complementary form. Equation max(x,y)=t can be approximated with "either x or y is t", and therefore (x-t)(y-t)=0, but no idea yet where this leads.

We can write the doublets in another way - the first one as 3, and the second one as 2+1, or as (1+1+1) and ((1+1)+1). Every parentheses represents either maximum or addition, and these must alternate (maximum parentheses can only contain additions inside, and vice versa)

4D graphotopes form 5 doublets:
4 doublet - formed by tesseract max(x2,y2,z2,w2)=1 and glome x2+y2+z2+w2=1
3+1 doublet - formed by spherinder max(x2+y2+z2,w2) and TRIDOME max(x2,y2,z2)+w2=1. Tridome cross-section in three dimensions is the dome, in fourth dimension it's a cube.
(2+1)+1 doublet - formed by DOMINDER max(max(x2,y2)+z2,w2)=1 and SPHERIDOME max(x2+y2,z2)+w2=1. Dominder cross-sections are cube, cylinder, cylinder and dome, spheridome cross-sections are cylinder, dome, dome, and sphere.
2+1+1 doublet - formed by cubinder max(x2+y2,z2,w2)=1 and SEMIGLOME max(x2+y2)+z2+w2=1. Semiglome cross-sections are dome, dome, sphere and sphere
2+2 doublet - formed by duocylinder max(x2+y2,z2+w2)=1 and CYCLODOME max(x2,y2)+max(z2,w2)=1. Cyclodome cross-sections are always domes.

The eleventh graph on four nodes is self-complementary O=O=O=O, which I call LONGDOME, and which has cross-sections of cylinder, cylinder, dome, and dome. It looks like it exists, but no specifics are known yet.

In five dimensions, there are following doublets (first name is always the disconnected graph)
5: Penteract (all CS are tesseracts) and pentaglome (all CS are glomes)
4+1: Glominder (4 spherinders + glome) and tetradome (four tridomes and tesseract)
(3+1)+1: Tridominder (tridome, 3 dominders and tesseract) and glomidome (glome, 3 spheridomes and spherinder)
((2+1)+1)+1: Spheridominder (spheridome, spherinder, 2 dominders and cubinder) and bispheridome (semiglome, 2 spheridomes, tridome and dominder)
(2+1+1)+1: Semiglominder (semiglome, 2 spherinders and 2 dominders) and spheritridome (2 spheridomes, 2 tridomes and 1 cubinder)
(2+2)+1: Cyclodominder (cyclodome and 4 dominders) and duospheridome (4 spheridomes and duocylinder)
3+1+1: Spherisquare (2 spherinders and 3 cubinders) and sesquiglome (3 semiglomes and 2 tridomes)
(2+1)+1+1: Glomosquare (2 glominders, 2 cubinders and tesseract) and spheriglome (glome, 2 semiglomes and 2 spheridomes)
3+2: Sphericircle (2 spherinders and 3 duocylinders) and tricycle (3 cyclodomes and 2 tridomes)
(2+1)+2: Domicircle (2 dominders, 2 duocylinders and cubinder) and siamese spheridome (semiglome, 2 spheridomes and 2 cyclodomes)
2+1+1+1: Cubicircle (3 cubinders and 2 tesseracts) and semipentaglome (2 glomes and 3 semiglomes)
2+2+1: Dual cylinder (duocylinder and 4 cubinders) and pyraglome (4 semiglomes and cyclodome)

Plus, there should be ten more graphotopes with more complex equations. All of these involve longdomes as cross-sections.

Longdominder (longdome, 2 dominders and 2 cubinders) and trialong (2 semiglomes, 2 spheridomes and longdome)
Branchdome (tridome, 2 longdomes, dominder and cubinder) and sesquispheridome (semiglome, spheridome, spherinder and 2 longdomes)
Superdome (2 longdomes, 2 dominders and duocylinder) and triquad (2 spheridomes, cyclodome and 2 longdomes)
self-complementary bicesphere (2 spheridomes, longdome and 2 dominders)
Longspheridome (spheridome, spherinder, 2 longdomes and cubinder) and branchcyclodome (tridome, cyclodome, 2 longdomes and dominder)
self-complementary cyclongdome (5 longdomes)

...All the names are my own and the graphs corresponding to each one are left as an exercise for the reader, so there.

[jinydu]

Whoa, that's an impressive first post!

[Marek14]

Well, I'm interested in multidimensional geometry for a long time - I just discovered this place a short while ago.

[wendy]

Very strange indeed. Can't help but to think it's a kind of tegmic product. Have not figured out whether it's associative as yet, though!

W

[Marek14]

What it boils down to is finding a shape with a certain combination of squares and circles as it's cross-sections in 2D coordinate planes.

If one would use a diamond abs(x)+abs(y)=1, instead of either circle or square, what would result, I wonder? Or if you could use all three shapes?

[wendy]

Suppose one treats a polytope as an angular function, such that the surface is f(a)=1. For any given ray from inside, the ray crosses the surface, and this defines 1 in that angle.

The prismic product, then is "max[f(a),g(a),....]" for orthogonal products. This makes the cube out of crossing lines.

The tegmic product is "sum[f(a),g(a),...]" This makes a rhombus out of crossing lines.

We have also rss[f(a),g(a),...], where rss is root-sum-squares, gives a spheric product.

When i look at this spheric product, it is a tegmic-like product, based on the rss function. But i continue to be pussled by the differentiation between, eg ((2,1),1) and (2,1,1), since both the tegmic and spheric functions are open ones. So also is the prismic one, ie

max [ max[ list1 ] max[ list2 ]] = max[list1, list2]
sum [ sum[list1] sum[list2] ] = sum[list1, list2]

In practice, the three functions are "open-bag" functions: One can join bags of the same function, and draw in any order.

The general observation i make on the rototopes, is that they are essentially prisms of equal-in-diameter spheres, and not, say the result of an rss function.

On the other hand, there _is_ a more rotary function. This is what you are trying to invoke with the dome. Here, the function is rss[square, circle],

For example, suppose we did an rss[ line, pentagon]. We would end up with something that has five gores, or quarts like an orange. The five lines of longitude woild gradually disappear to the polar region.

We now might want to rss[line, rss[] ]. This figure is essentially the same as a rss[ line, line, pentagon], or rss[circle, pentagon].

The whole general class of rototopes can then be described as a combination of rss[] and max[] applied over the equalateral, centred 1d line segment.

You ask about the tegmic product. The interesting thing is that the dual of a figure made out of rss[] and max[] functions is the corresponding rss[] and sum[] function.

For example, the dual of the duo-cylinder or bi-circular prism is the bi-curcular tegum.

When one uses non-equal lines, the rss-function gives rise to ellipses, the sum to elongated tegums, and the max[] to rectangles. You can make these all eccentric, giving egg-shaped figures, and strombii. [One of JHC's terms].

[Marek14]

Here's the puzzle, though: there is only limited number of ways how to combine max and rss functions in this way, while the number of possible graphs starts to exceed them eventually. I'm not sure whether things like what I call longdome exist - but if it does, I am very interested in its equation.

The longdome sliced through 3D space would start as a circle - dome cross-section through one of its vertical planes - and then expand to both sides until it becomes a complete dome, then shrink again.
The second possibility is that it starts as a line, which would be a diameter of cylinder positioned in a mid-height. This line would expand into full cylinder, and then shrink, etc.

So, I think one could try to find an equation for these two slices, then try to unify is - real longdome should produce first sequence in 2 cartesian directions, and the second one in other two, plus some extra conditions, basically that the height of cylinder in "second slice" directions must be always one of the "first slice" ones.

An idea, though: does this mean that as soon as we have an equation for ANY polytope OR another shape in any dimension, we can define the appropriate product? The limitation is that the product qould require EXACTLY as much terms as there are dimension of polytopes. This way, the prism product is only a general "measure polytope" product, comprising of "line product" with 1 term, "square product" with 2, "cube product" with 3... Whereas tegmic product is general "cross polytope product" comprising of line product, dual square abs(x)+abs(y)=1 product, octahedral product, hexadecachoral product etc...
And spheric product is an infinite line of line product, circle product, sphere product, glome product...

I know Jonathan Bowers was talking about general polytopic products a while back, and I'm not sure if this is the same concept, but still...

This, of course, requires not only a polytope to determine a product, but a SPECIFIC polytope, with equation, since square can give prism or tegnimc product based on its orientation. I'm not sure if it's mandatory that the polytope contains all [+-1,0,0...] permutations.

[wendy]

The formula given for the rototopes is simply the rss and max functions in any case. While both create products, one can not "filter out" the two. That is,

max(rss(w,x),rss(y,z)) = duocylinder.

The bag concept only applies when you have the same function at two levels, eg

rss(rss(w,x),rss(y,z)) = rss(w,x,y,z) = glome

rss(rss(w,x),max(y,z)) = rss(w,x,max(y,z))

So what ye end up with is a nested tree, of alternate layers of rss and max, in the form, eg

rss(a,b,max(c,d,e),max(f,g,rss(i,j)),max(k,l,rss(m,n,max(o,p)))

Visually, you get a tree of alternating max and rss nodes like the one below, and the variables connect to nodes. Each node must have at least two inputs to prevent collapse.

Code: Select all


                        / max
     max --- rss  = max - rss
          \                   
             rss  --- max   




rss(a, max(rss(b,c)) = rss(a, rss(b,c)) = rss(a,b,c)

So there is still quite a lot of room there for the rototopes.

P.S.

Jonathan's polytope-product is the same idea as my 'cuboctahedral product'. Even the simple examples are frought with some complexity, because even with, say tegums, moving the centre makes a different shape. (Only "max" is free from this).

The idea is that one treats the elements as a sphere (ie a surface R=1), and then take several bases, where there exists R1, R2, R3. This makes three dimensions. One can place an octant of some figure into that octant, and make a figure.

One then has

rss (root-sum-square) = spheric product [ie spheres]
max (maximum value) = prism product [ie cubes, squares]
sum (sum of values) = tegum product [ie cross polytopes]

The max and sum products define a set of coherent units. That is, one can define the prismic or tegmic unit in terms of the max() or sum() of unit orthogonals. All other volumes are compared against these. Then the volume of the max() is the product of the prism-volumes, while the volume of the sum() is the product of the tegum-volumes.

The thing with, say the cuboctahedral or icosahedral products, is that these do not define a coheret unit. Using the icosahedral octant, one sees that the list it governs is now order-dependent: the ico(a,b,c) != ico(a,c,b).

Some products, like the comb, torus and pyramid, have no expression in terms of this tegmic space.

W

[wendy]

I had a closer look at Marek's rototopes.

While the simpler ones may be described in terms of the max() and rss()
functions, the thing may be indeed more complex than is suggested.
The idea is that we can reduce nodes in his graph, to P or S nodes,
being prism and sphere, if they are everywhere connected to any other
node by the same order.

Code: Select all

We have this representation:

     ()  defines a spheric or rss product
     []  defines a prismic or max product
     <>  defines a tegmic or sum product

    unconnected nodes project in 2d as squares
    connected nodes project in 2d as circles.

  ----------------------------------------------------

      o---o        S2      circle   (1,2)

      o   o        P2      prism    [1,2]

    o   o   o      P3      cube     [ 1,2,3]

    o---o---o--z   S3      sphere   (1,2,3)

    o---o   o              cylinder  [(1,2),3]

    o---o---o              dome      ([1,3],2]



Marek then goes to list 11 examples of rototopes in four dimensions, of which 10 give products. But if the graph model is correct, then we should see that even more is permitted.

Code: Select all


    1---2     <  glome      (1,2,3,4)       1   2
    | X |
    3---4        tesseract   [1,2,3,4] >    3   4


    1   4    <  spherinder  [(1,2,3),4 ]    1 -- 4
    | \                                        / |
    2 - 3       tridome   ( [1,2,3], 4]   >  2   3


    1---2     < duocylinder [(1,2),(3,4)]     1---4
                                              |   |
    3---4       cyclodome  ( [1,2],[3,4]) >   3---2


    1---2      <- cubinder  [(1,2),3,4]      1   2
                                             | X |
    3   4       semiglome   ([1,2],3,4) >    3 -- 4

    1 -- 2    < -- spheridome (2,[1,(3,4)])     1   2
      /  |                                      | \
    3 ---4        dominder [2,(1,[3,4])]    >   3   4

     1---2     <--- longdome
         |
     4---3



The longdome has no projection into products, because no nodes are of identical connexions. This means that it can not be reduced into a product of lesser figures. In the main, this does not prevent the figure from existing.

The 2d sections of this are then circles for 1-2, 2-3, and 3-4, and squares for 3-1, 1-4 and 4-2 hedroaxies. The real question then is does this project into a single terid [4d solid]?

Need to think over this!

[wendy]

We now look at the possible zoo for the 5d rototopes. We see that that there are these examples, of which 10 can not be constructed as a product.

Marek's bracket code corresponds to the nesting of products. The innermost product must contain two elements.

Code: Select all


   all branches alike

     o o o o o
      P      [1,2,3,4,5]    peneteract
      S      (1,2,3,4,5)    pentaglome

   one branch different

     o--o o o o
             ([1,2],3,4,5)   semipentaglome
             [(1,2),3,4,5]   cubicircle

    two branches different

     o---o  o---o  o
           [(1,2),(3,4),5]   dual cylinder
           ([1,2],[3,4],5)   pyraglome

     o---o---o   o   o
           [(2,[1,3]),4,5]    cubicircle
           ([2,(1,3)],4,5]    semipentaglome

     three branches different

      o---o---o-:  o   o
           [(1,2,3),4,5]     spherisquare
           ([1,2,3),4,5)     sesquiglome

       o---o---o   o---o
           [(2,[1,3]),(4,5)]   dormicircle
           ([2,(1,3)],[4,5])   siamese spheridome

      o---o---o- A-o   o
           [(2,[1,3,4]),5]    tridominder
           ([2,(1,3,4)],5)    glomidome

       o---o---o---o   o
            *  1
            *  2

      four branches different

      o---o---o-:   o---o
           [(1,2,3),(4,5)]     sphericircle
           ([1,2,3],[4,5])     tricycle

      o---o---o---o--:   o
           [([1,2],[3,4]),5]    cyclodominder
           ([(1,2),(3,4)],5)    duospheridome

      o---o---o-:--o    o
           [([(1,2),4],3),5]   spheridominder
           ([([1,2],4),3],5)   spheritridome

      o---o---o---o---o
           * 3
           * 4

      o---o---o---o -A-o
           * 5
           * 6

      o---o---o -A-o -B-o
           (2,[1,3,4,5])   tetradome
           [2,(1,3,4,5)]   glominder

      five different branches

      o---o---o---o---o--:
            * 7

      o---o---o-: --o -A-o
           (3,[(1,2),4,5])     glomosquare
           [3,([1,2],4,5)]     spheriglome

       o---o---o---o-: -o
       o---o---o-: -o---o
            * 8
            * 9

       o---o---o-: --o -C-o
             * 10




At least our enumeration of graphs are identical.

W

[Marek14]

Our enumeration of graphs is indeed identical. Which of my name is corresponding to which of the graphs can be actually deduced too (as I provided the cross-sections for all of them):

*1 - longdominder
*2 - trialong
*3 - superdome
*4 - triquad
*5 - branchdome
*6 - sesquispheridome
*7 - cyclongdome
*8 - branchcyclodome
*9 - longspheridome
*10 - bicesphere

There is one thing I don't really understand, Wendy - I agree that icosahedral product wouldn't be commutative, but cuboctahedral? If you aim the three axes to pass through the centers of the square faces, the three-member product absed on this would be still commutative. The same thing should hold for all products based on a polytope from measure/cross groups.

Let's look at the simplest non-trivial cuboctahedral product: n-gon/line/line cuboctaproduct. The cuboctahedron here is "standard" one, with vertices at [1,1,0] with all sign changes and permutations.

Slicing this thing through 3D space, we start with n-gonal prism of edge 1. This transforms in n-gonal bipyramid (tegum) dual to prism with edge sqrt(2). Then it goes back into its original size and shape. I have a nagging suspicion that the shapes it passes "between" correspond to "morphing duals with rectangles" feature of Great Stella. Square/line/line cuboctaproduct would (on its way from cube to octahedron) pass through rhombicuboctahedron.

In 5D, there would be polyhedron-line-line cuboctaproducts formed in similar way, and also polygon-polygon-line ones which would pass from duoprism to duotegum and back. It's only 6D with its polygon-polygon-polygon cuboctaproducts that puzzles me so far...

[wendy]

The CO generalises into the vertex figure of the semi-cubic tiling, that is, the rectate of the cross polytope. This has the general coordinates (1,1,0,0,0,...).

While the product CO( line) ^n gives the general CO, the CO product over cuboctahedra do not give a cuboctahedra, but rather a subset of the vertices of the tri-rectate (ie a subset of four units, and the rest zeros).

This is why the CO was included here: it is not a power-product in the manner that the tegum (sum), pyramid (xor), prism (max) or spheric (rss) are. [XOR is exclusive or, like radio-buttons, not PARity, like even=0.]

The general construction in terms of radius-space, allows us to make all sorts of interesting polytopes. All we do is to map on each axis, a radius of a polytope, and draw the figure therein.

Let's look at the CO of three polygons, in six dimensions.

We pick three polygons, CO(x,y,z).

In the positive octant, the CO has four faces, the triangle, and quarter-squares.

The q-square faces have the vertex (1,1,0) to (1,0,1). This can be represented as, eg as max(x, sum(y,z)). This is a prism product of a product of x and the tegum over y and z, or say [x,<y,z>].

The remaining face is best described as a lace prism, of the kind

oxxXooo & xoxYooo & xxoZooo &#x.

We can then evaluate the surtopes in the usual way for lace-prisms.

The three edges, represent, eg

ox.Xoo. & xo.Yoo. & xx.Zoo. &#x

This is a prism, bases Z and the pyramid product of X and Y. This margin is shared between the two kinds of face.

The other margin, is shared between different faces of the triangle-type, are the three faces of the type

... ... & xoxYooo & xxoZooo &#x

This is a bisare thing, which i can largely feel, but have not figured out what it might be yet. The diagram says that it is, in RK's || notation, the triangle-array {y}{z} || {y} || {z}. In four dimensions, you could construct this as {2}{2} || {2} || {2}, this is a triangle prism, lieing on its back. The square is in the wx plane, and we have the lines parallel to the w, x axies, displaced in the y and z spaces respectively. Between the two lines stretches a tetrahedron. So it's tetrahedron || square.

It's all terribly simple when you understand what kind of space the product is sketched out in. Each point (x,y,z) represents a prism-product of the polytopes X,Y,Z, scaled to sizes x,y,z. Kind of like the stott-vector space.

A line, is little more than a progression from one polytope to another.

A lace-prism, then is little more than a simplex stretched out in this kind of space. For this purpose, a simplex equates to a vertex-count, so the CO3 product itself leads to a simplex, but higher ones add extra vertices. It is in general the rectate-cube that gives a general simplex [ie all 1, except one zero].

For something like the rCO product (general vertex here is (2,1,1), you have to sit down and feel out each of the faces of the rCO itself: faces correspond to faces, and margins are either in the same plane as faces (ie connecting two faces of the same derivation), or margins of the general projection.

So in the rCO product, the faces of the rCO give the facets of the result, but the ridges occur in both the rCO faces [between two faces that project onto the same rCO face], or on the rCO ridges [between two different kinds].

Generally, element-wise, one can use lace-prism calculations to explore it bit by bit: that's how i do it. I just do it a tad faster than most.

W

[Marek14]

We pick three polygons, CO(x,y,z).

In the positive octant, the CO has four faces, the triangle, and quarter-squares.

The q-square faces have the vertex (1,1,0) to (1,0,1). This can be represented as, eg as max(x, sum(y,z)). This is a prism product of a product of x and the tegum over y and z, or say [x,<y,z>].

Correct me if I'm wrong, but wouldn't [x,<y,z>] still have six dimensions, like the whole thing? One dimension has to be dropped somewhere, I think.

That being said, I thought about the way these various rotatopes would actually roll. The results are interesting. It seems that "rolling" is a way to transform one "shadow" in a (n-1)-dimensional hyperplane into another.

For example, cylinder can be dropped into plane in three cardinal orientations. In one of them, it projects as a circle - this is a 2D object, and so the cylinder can't roll. In two orientations, however, it projects as a line - which is 1D, so the cylinder is able to roll along the perpendicular direction. If you roll the cylinder 90 degrees, however, you arrive into the other "rolling" orientation from the one where you started. These two orientations are undistinguishable, however.

Try rolling a dome, though: When a dome is positioned on one of its poles, it can roll in two perpendicular directions. Once you start to roll it, though, it will start to touch the plane in a line instead of point, and the direction of roll can be no longer changed. After 90-degree rotation, the dome is positioned on one of its equatorial "edges", and can be still only rolled in one direction. Only after you roll it back to polar orientation, you can change the direction of roll.

In the graph representation, all possible rolls are represented with sequences of edges. Continual rolling in one direction corresponds to endless going back and forth over the same edge. The pole corresponds with the node of degree 2, so two perpendicular directions of rolling are availaible from it. The end-nodes have degree 1, so you can only roll in 1 direction from them.

How do the 4-dimensional rotatopes roll?

TESSERACT always projects as a cube, and so it can never roll.

GLOME projects as a point and can be rolled in either of three perpendicular directions.

SPHERINDER projects either as a sphere (which can't roll) or as line which can roll in two perpendicular directions

TRIDOME projects as either a square or a point. From the "point" form it can be rolled in three perpendicular directions, in all cases becoming a square. Only after you roll it back into point, you can change the direction.

DOMINDER projects either as a dome (it can't be rolled in this orientation), or as a square or as a line. From "line" you can roll it just like dome, in two directions. It transforms into rectangle, then square, and you can't change direction until you roll it back into line.

SPHERIDOME starts to get interesting. It has three kinds of projections: circle, line and point.
If you start rolling it from the circle, it will shrink into point, at which point you can choose either of three perpendicular directions.
If you start rolling from a line, you can take two directions with varying results - in one of them the line shrinks into a point, in the other, it will stay the same - all orientations in this plane are, in fact, equivalent.
If you start rolling from the point, then you get to circle if you go in one direction, and to line if you go in other two.
Since the spheridome graph is connected, you can put it into any orientation without lifting it from the conceptual 3D paper.

CUBINDER projects either as a cylinder (which can't roll) or as a square which can roll in the direction of its normal.

SEMIGLOME is, once again connected. It projects as either line or point. From the line, you always roll it to point, no matter which of the two directions you take, but from a point, only two roll directions take you to line - the third will keep semiglome projection a single point (or a small dome if you press real hard on it).

DUOCYLINDER will always project as a circle and can be rolled in a single direction.

CYCLODOME will always project as a line. However, as you roll it in one of the two allowed directions, it will go through rectangle, square and rectangle into perpendicular line. It can roll all over the 3D paper, conceptually.

LONGDOME is, once again, interesting. It projects as either line or circle. When you roll it from a circle, it will always shrink into line and allow you to change direction. Rolling it from line will either expand it into circle or transform it into perpendicular line.

[wendy]

Sorry. Had a lend of you - must had been very cold + tired.

Even though i have found a way of describing the figure in total, the area is not one i looked into a great deal, save to explore the three products. But you are correct regarding the dimensionality of the figure [x,<y,z>] as six dimensions.

Also the lace tower oxx X ooo & xox Y ooo & xxo Z ooo &#x has nine nodes, which makes it eight-dimensional. The figure in question is the flat form (no altitude), rather like a compound of three polychora in six dimensions. A similar array exists in the icosa = 3 crossing golden rectangles.

Consider first the pentagonal prism. In terms of the axies {5} and {}, it is the point (1,1), which makes perpendiculars to these walls. The faces of the pentagonal prism are then represented by the lines (1,1) - (0,1) and (1,1) to (0,1). Letting S(X) stand for the surface of X, we have the faces as

[S{5},{}] a ring of five squares (surface of {5} = five edges)

[{5},S{}] two pentagons (surface of a line = its end points)

We now turn to the CO product, of three polygons, X,Y,Z.

The square face (1,1,0) to (1,0,1) belongs also to a prism [X,<Y,Z>]. By themselves, these squares define a prism, which represents this. The square faces are actually rhombii, representing the face [S(X),<y,z>].

The three squares together, represent three rings of prisms, the height being a line, and the base a y-z tegum. What the triangle-facet symbolises is then used to fill in the gaps between these three faces.

The vertices of these squares (ie 0,1,1), represent the product of surfaces, ie [S(Y),S(Z)]. For Y=Z=polygons, this makes a terid (solid in 4d) sheet of squares, rather like the plane that appears in the duocylinder.

The triangle-facet is a lace-prism, we can analyse it as follows:

oxx & xox & xxo &#x

{the understanding of the simplex product was done in exactly the same way, zB oox & oxo & xoo &#x = lace-prism of orthogonal bases.

This particular lace tegum has six nodes (three mirrors, three vertex-nodes), and thus is five-dimensional. We see that the three vertices give rise to squares, ie o.. & x.. & x.. &#x

The opposite 4d face is .xx & .ox & xo &#x, is a line * tetrahedron prism. One takes these two figures and makes a segmentotope from it, to get the required face.

The section of this figure can be likewise derived, although the most obvious are the hedric sections.

The intercept of the product with the plane X=0 gives the square marked by the point (0,1,1). This is simply [Y,Z]. The faces of [Y,Z] are [S(Y),Z] and [Y,S(Z)] appear as the two edges of the square (which is not a surtope), and the vertex becomes [S(Y),S(Z)], the prism-product of the surfaces of Y,Z.

Although the hedric outline makes a [Y,Z] shaddow, the faces of this shaddow makes eg [Y,line] prism. This prism is also a section of the tegum-prism [Y,<line, X>], where the hedric projection has reduced the tegum to its axis. (rather like projecting a sphere to its diameter).

The latric section (ie projection down a line), needs to be more carefully plotted, but in every case, it turns out to be a CO product.

Consider that the CO-space has axies that represent the radius of a surface function, that is, the x axis is any radius of the polytope X. That is, the point (x) on the x axis represents xX, an X of size x.

Because we are not disturbing the overall product figure, the section down any axis (where the axis is part of a larger-dimensional axis), does not disturb the product. It's only when you start projecting down lines that run across the product.

This then applies in general to any product drawn on that space, eg spheric, tegum, prism, pyramid.

So, we have the sections of CO{X,Y,Z} = CO{X", Y, Z}, where X" is a section of X. In the case of polygons, this gives CO{line, Y, Z}. For odd polygons, the resulting section requires that the sectioning line X" be off-centre, and so also the figure in total.

Consider what happens when do a section of a triangle-triangle tegum, into three dimensions. We have <3", 3>, where 3" is the section of the triangle. For this, we note that the centre of the line is not at 0:60, the centre, but at 0:40, or 1/3. So it gives not a balanced triangle-bipyramid, but one where one pyramid is twice the height of the other.

None the same, it _is_ still a tegum, off-centre relative to the height, rather than the triangle-base.

[wendy]

ROLLING THINGS

The trouble with the rototope notion is that there are other things that can roll as well, such as cones. One has also to consider the eccentric and bisarre figure <O,O> , or <(x,x),(x,x)>, the tegum-product of two circles. In practice this can provide a rather interesting mechanism for steering a thing that rolls on a line.

The notation i use for spheres, ellipsoids etc, is to treat the spghere as regular figures {O}, {O,O}, etc. When nodes are marked with x-like nodes, (eg x, q, ...), it gives a progressively increasing set of axies, ie xOoOo = sphere, xOoOq = 1:1:r2 or prolate ellipsoid, while xOqOo = 1:r2:r2 or oblate ellipsoid. When the slash-notation is used, the use of slashes give increasing size, eg /OO = sphere, /O/O = x<y=z = oblate ellipsoid, while /OO/ = prolate ellipsoid.

One then can consider the great class of ellipsoids in 4d. An ellipsoid will roll if the second node is unmarked. If it is marked, the thing will end up rolling to a stop, like a squashed sphere does.

So, /OOO glome, and the ellipsoids /OO/O, /OOO/, and /OO/O/ all roll, while the ellipsoids /O/OOO, /O/O/O, /O/OO/ and /O/O/O/ do not. The degree of freedom of rolling is how many unmarked nodes follow the first slash. So /OOO can roll over a 3d flat, /OOO/ is over 2d, /OO/O and /OO/O/ over 1d. None the same, these are not physically restricted by extensive flat surfaces from being steered: all ellipsoids stand on a point.

In terms of your diagram of nodes and chords, it may well be the case that one might be able to have a third kind of branch, forming tegum-like products (ie having sum along rss and max branches).

On the main, i have very poor ability to rotate things in my mind. I'm one who turns the map around when i'm navigating.

Your reading on how things roll is how i read it too, so we agree there.

On the other hand, some circular products do not all that easily roll. For example, the (x,[x,x]) or dome, does have circular projections. But a device made of this, would tend to roll on the 1:sqrt(2) elliptical sections through the faces, and then would tend to lie there haplessly trapped by the solitary point near the centre.

On the other hand, one can have a figure like [(x,x),(x,[x,x])], a five dimensional figure, that has a glome as a cross-section. In practice, this would tend to fall into one of four ruts, where the cross-section is the 4d oblate ellipsoid xOqOqOq, and it would tend to roll only in the deeper x axis, and generally eschew the three q axies.

[Marek14]

Basically, I only considered "ideal" rollings, in cardinal directions. I found that if two projections can be rolled one into another without changing the "trace" in any way, that the thing has a circular symmetry along that plane and any point along the roll can be considered as one of these two orientations.

Your musings about ellipsoids got me thinking of another thing, though - what kinds of quadrics exist in 4D?

I can understand a glome or 4D ellipsoid x^2+y^2+z^2+w^2=1, three kinds of hyperboloids, but I'm puzzled on parabolic creatures.

[quickfur]

[...]Then, when I was lying in my bed trying to sleep, something connected in my head and I saw a way to entangle these shapes with graph theory... and this is what resulted:

Wow. Impressive.

[...]Now comes the trick: there is one more graph with three nodes: O=O=O. Does this graph have a rotatope associated with it like the others?

Surprisingly, it turns out it does. First, we review its properties: It has to cast two circular shadows (when we remove the end-nodes) and one square shadow (removing the middle-node). Passing it through the plane will result in two possibilities: either a square that will start from a point, grow, then shrink...
...or we start with a line. This will grow into circle (originally I thought it was through ellipse, but in fact the intermediate shapes are circles with their top and bottom cut). Then it shrinks back into line.

I believe this shape is called a "crind", and is mentioned by alkaline on the rotatopes page as a (rather crude) analogue of the duocylinder. Using alkaline's definition of rotatope, this shape doesn't qualify as one because it can't be generated by rotating or extruding a 2D figure. However, I think it well deserves to be called a rotatope, since it does have interesting rotational properties. :-)

[...]The point here is that there are 11 different graphs on 4 nodes. 5 of them have only complete components and correspond to 5 rotatopes you have on your page. The other six have non-complete graphs, and all of them would contain the O=O=O shape as a shadow in one or more direction (one of them, indeed, in ALL 4 directions). I think these are worth checking, in order to enlarge the zoo a bit.

Cool. I'm curious to find out what 4D shape has a crind-shaped shadow in all 4 directions! :-)

Since these are obviously not true rotatopes, we will need a new name... Would "graphotopes" work? "[...]

I'm not sure about that. Your description essentially covers shapes composed of line segments and circles, and neatly encapsulates them as graphs; however, I don't see why graphs in general can't be extended to cover other types of shapes (e.g., if you have a second kind of edge that is based on a triangular symmetry, or some such.) Maybe "circulotopes"? :-)

You got me thinking, though. I'm sure I'll be suffering from insomnia tonight as well, trying to visualize that 4D crind shape! :-)

(I'll reply to the rest of your discussion in another post.)

[quickfur]

Here's the puzzle, though: there is only limited number of ways how to combine max and rss functions in this way, while the number of possible graphs starts to exceed them eventually. I'm not sure whether things like what I call longdome exist - but if it does, I am very interested in its equation.

I've thought about this a bit more, and I don't see why your longdome shouldn't exist. As far as I can tell, its projection into 3-space will have a crind (dome) shaped envelope. If you assume two of its vertices (where the circles intersect) lie along the Y axis, then the other two may lie either on the Z axis or the X axis. If two of the circles project to the XY and YZ planes, the other would project on the ZW plane or XW plane, respectively. Assuming the former, this projection would correspond with an edge-first view.

Now, while the longdome is quite interesting in itself, I find far more fascinating the tridome. According to my visualization methods, the tridome would project into 3-space also with a crind (dome) shaped envelope, like the longdome, but with a different internal structure. The tridome has 4 perpendicular circles that intersect at 8 vertices: this is interesting, since these 8 vertices coincide with those of the 16-cell! Two of these circles project onto the XZ and YZ planes, and the other two project onto perpendicular line segments along the X and Y axes, respectively. The nearest and farthest vertices from the 4D viewpoint project onto the origin.

[Marek14]

Yesterday I came up with an idea to picture the rotatopes as "shadows with trace inside". That's a bit hard-to-understand concept, so here goes:

in 3D, a cube has square shadow in two dimensions. Dropped onto a plane, it leaves square imprint. So, we will picture it as a square with square inside, and since both squares are the same size, it looks like an EMPTY SQUARE.
Empty shapes can't roll. It's as simple as that.

Cylinder can be pictured as an EMPTY CIRCLE. But when we drop it on its side, so it can roll, it's shadow is square, but it touches the paper in a line. We can represent it as SQUARE WITH A LINE:

+-----+
| |
+-----+
| |
+-----+

It can roll in direction PERPENDICULAR to the line

Sphere will be pictured as CIRCLE WITH A POINT INSIDE. Since the point has zero dimension, two less than the circle, the sphere has two degrees of freedom, and can roll in two perpendicular directions.

We have two possible figures left: SQUARE WITH A POINT, and CIRCLE WITH A LINE, and these correspond to "dome" or "crind" - I will stick with "dome", since that is what I made my terminology for. If we put a dome in the origin, in the square-with-a-point orientation, we can roll it up or down (along y) one unit, and it becomes circle with horizontal line. If we roll it left or right instead, it becomes a circle with vertical line instead. If we roll it two units in any direction, it returns into original form.

Now, in 4D shapes, the shadow will be always 3D. Moreover, the "trace" will be always "cut" through the shadow, of some dimension.

So we have:

Empty shapes - cube, cylinder, dome, and sphere
2D traces: cube with a square, cylinder with a square, cylinder with a circle, dome with a square, dome with a circle, sphere with a circle.
1D traces: cube with a line, cylinder with a line (two possibilities, if we put the circle in xy plane, then the line can be vertical or horizontal), dome with a line (again, two possibilities, vertical or horizontal - we put the square in xy plane), sphere with a line.
0D traces: cube with a point, cylinder with a point, dome with a point, sphere with a point.

Now, we can assign each 4D shape its various forms:

tesseract - cube
cubinder - cylinder and cube/square
dominder - dome, cylinder/square, cube/line
spherinder - sphere, cylinder/vertical line
duocylinder - cylinder/circle
tridome - dome/square, cube/point
longdome - dome/circle, cylinder/horizontal line
spheridome - sphere/circle, dome/horizontal line, cylinder/point
cyclodome - dome/vertical line
semiglome - sphere/line, dome/point
glome - sphere/point

This makes it easier to see some things - for example that spherinder in cylinder/line form can be freely rolled in any direction in the plane, as can semiglome in shape of sphere/line, because they have circular symmetry, while dominder as cube/line and longdome as cylinder/horizontal line can only roll in two speficic directions.

[wendy]

Here is one for you...

The intersection of three cylinders, in three dimensions, is not a sphere, but a thing that looks like a rhombic dodecahedron, where the inscribed octahedron is glomid (ie inscribed on a sphere). The thing then has 12 curved faces.

This still has the three perpendicular sections as circular, but is not a sphere.

One can of course do this with any surtegmate (dual of rectate). Such figures are formed by the convex hull of dual figures (eg octahedron and cube), set so that for some surtope (edge), it and the dual's matching cross. If now the dual is made a spheric tiling, rather than a polytope, one gets a spherated form.

Note the spherated form can go either way, so one can have, eg

spherated bisurtegmated {3,3,5}

the faces of this would be pentagonal tegums (bipyramids), where the pentagon is part of a sphere surface.

So, in four dimensions, we have something like five or six new rollies.


Wendy the os2 fan 2

[wendy]

Another thing i play with is the possibility that the spheric unit may be coherent to the spheric product. That is, one can find the volume of the spheric product by multiplying the s-volumes of the bases.

For the units, one takes the spheric product of a unit line, to get:

(writing something like * number after an operator makes it a power operator, eg 5 ** 3 = 5 * 5 * 5, 5 +* 3 = 5 + 5 + 5.

inch base unit

circular inch = inch xx inch

spheric inch = inch xx inch xx inch = inch xx circular inch

glomic inch = inch xx* 4 = circular inch xx* 2 = spheric inch xx inch.

It's pretty exciting... (although my maths is not up to it )


Wendy

[Marek14]

...and domic inch would be (inch x inch) xx inch?

[wendy]

I do use cylinder inches (inch X (inch XX inch)) quite a bit, actually, because many capacity units are defined in terms of cylinder contents. Cylinders are easier to render on a lathe, than cubes are to make in practice.

The US bushel, for example, is 2738 cyl in, being a cylinder 8 inches high and 18.5 inches diameter. To make a heaped bushel, one places a cone on top, the base-diameter is 19.5 inches, the height 6 inches.

The thing about the graph-notation, is that the all-marked graph does not give exclusively a sphere. In three dimensions, one can also get a rhombic dodecahedron, being the convex hull of a cube and an edge-tangent sphere. This is the intersection of three cylinderic pipes. Haven't figured out the formula for it yet, apart from:

s = max( rss(x,y), rss(y,z), rss(z,x))

The surface-function makes the surface when s=1.

Note the sphere is s = rss(x,y,z).

Note that the value of s at 1,1,1 is for the first, r2 (ie the surface lies at (1/r2, 1/r2, 1/r2), while the second lies at s=r3 (ie the point is 1/r3, 1/r3, 1/r3)

So even though the sphere gives the form of o=o=o=:, it's not the only shape to do so.

In 4d, you have many such shapes

s1 = max(rss(w,x), rss(w,y), rss(w,z), rss(x,y), rss(x,z), rss(y,z)

which has 24 faces

s2 = max(rss(w,x,y), rss(w,x,z), rss(w,y,z), rss(x,y,z)

which has 8 faces

s3 = rss(x,y,z)

which has one face

The values of the point 1,1,1,1 is then s1 = 2, s2 = r3, s3 = 2, which means that the surface point is (1,1,1,1)/s, different places for different figures.

Only the first one is expressable in terms of the spheric and prismic products.

[quickfur]

[...]The thing about the graph-notation, is that the all-marked graph does not give exclusively a sphere. In three dimensions, one can also get a rhombic dodecahedron, being the convex hull of a cube and an edge-tangent sphere. This is the intersection of three cylinderic pipes.

You're right, albeit a "puffed" rhombdode.

So even though the sphere gives the form of o=o=o=:, it's not the only shape to do so.

Yeah, I thought it was a bit fishy that other graph combinations map to crind-like shapes, but this one magically produces spheres.

[Marek14]

Hm. I wonder if there is any way to distinguish between these...

That being said, I have tried my hand in rolling rotatopes yesterday, and I found a surprising connection between the way they roll, and (of all things) Sam Lloyd's Fifteen puzzle.

Stay with me.

First of all, we will limit ourselves to rotatopes described with a connected graph. These can theoretically roll all over (n-1)-dimensional surface.

What I realized is that we can treat its graph, when rolling, just like we would treat the Fifteen.

Every node of the graph is a space, and all these spaces but one are covered with a symbol of one of the axes of the (n-1)-dimensional paper. When we do a roll, we simply move one of the symbols onto the unoccupied space.

Let's see it with a tridome.

Tridome has a graph O=O=O A=O, and I will show the configuration as a;bcd where a is the central node.

If we start from 0;xyz, how will the tridome look? It's cross-section is what we get when we remove the unmarked node from the graph - the cube xyz. We get its trace - what we get on the paper - by removing the unmarked node, AND all nodes connected to it. In this case, we remove all nodes to get a point trace.

There are edges going from the central node to all three symbols, x, y, and z. This means that tridome has three degrees of freedom (as also observed by the fact that its trace has three dimensions less than the cross-section).

When we take one of these routes, the symbol will move to the centre, leading to these configurations:
x;0yz - y;0xz - z;0xy

We will look at x;0yz, since the others are just cyclic permutations.
In this position, tridome's cross-section is a dome. More specifically, it's a dome with height in x axis (the direction we rolled it), with y and z as its cross dimensions. And its trace will be the yz square.

From this, only one direction of roll is availaible, of course, so we return back to 0;xyz.

Now, this is a direct analogy with Fifteen - just take 16-dimensional rotatope with nodes connected in square array, mark every two antipodal points on the axes with a different color, draw a 15-dimensional cube on a 15-dimensional paper, and try to roll the thing in the cube so all sides would match. What a puzzle for 16-dimensional kids

In the next part, we will look at the other rotachora and some rotaterons, and we will see what happens when some dimensions are completely equivalent.

As our second example, we will take a longdome O=O=O=O. The best way to express the configuration is, probably, to take it as abcd chain.

Let's start in 0xyz position. Here, the cross-section (CS) is a dome with height in y direction and horizontal dimensions of xz. The trace is circle in yz plane. We can roll it in x direction, to... x0yz. Here, the CS is cylinder with height in x and circle in yz plane, and the trace is line in z direction.
Rolling this line in x direction takes us back, but rolling in y direction results in xy0z, which has CS of (xy)z cylinder, and trace of x line. Rolling in y takes us back, while rolling in z results in the final configuration, xyz0, with CS of (xz)y dome and trace of xy circle.

What's the total result? Unlike tridome, longdome has preferred orientations. Whenever the longdome rolls into a dome CS, the height of the dome will be always oriented the same way.

Spheridome
In Wendy's inline notation, spheridome can be written as (O=O=O=)=O. It's a triangle with fourth node attached to one of its vertices.
Spheridome is pretty interesting, as you will see.
I will write the spheridome configuration as a;b;cd, where a is the degree-1 node, b is the degree-3 node and c and d are the two degree-2 nodes.

Let's start from 0;x;yz.

The CS is a sphere - the trace is yz circle. Rolling in x direction takes us to x;0;yz, a kind of most basic orientation of the spheridome which has CS of (yz)x cylinder and trace of a point.

Rolling in y direction takes us to x;y;0z while rolling in z direction takes us to x;z;0y. Both of these have a dome as CS (either (xz)y or (xy)z), and x line as the trace. Note that in either of these orientations, NOTHING WILL CHANGE if we roll along the line that connects both degree-2 nodes. The resulting graph will be isomorphic to the original one even after taking the labels into account. This means that these two nodes are, in fact, interchangeable. Rolling among them doesn't have to come in discrete "steps" - you can roll it by an arbitrary angle.

As you do, the dome CS will rotate, its height changing from y to z direction and back. The horizontal x dimension, though, will always stay horizontal.

This means that from the x;0;yz position, we can roll not only in x, y, and z directions, but also in any other direction that lies in yz plane. The notation would be x;y/z;0(z/y). When we roll far enough to get the trace as x line, the CS will be a dome with heigth parallel to the direction we rolled the spheridome in.
We might as well label the point-trace situation as x;0;(y/z)(z/y) to express that the two degree-2 nodes can be thought of as belonging to any two perpendicular axes in yz plane.

Cyclodome
As all 4 nodes of the cyclodome are equivalent, it's rolling is very simple - it has only three possible configurations: 0xyz, 0yxz, and 0xzy. Each of these configurations has a CS of dome with ((xz)y, (yz)x, and (xy)z, respectively) and trace of line (y, x, and z). In each position, it has two directions to roll in (the two horizontal dimensions of the dome CS), which will exchange the height of the dome with the other horizontal dimension, and transform the trace line in a perpendicular line. Easy!

Semiglome
Semiglome is a bit confusing. I will label the configuration ab;cd, where ab are the degree-3 nodes, and cd are degree-2 nodes.
So, from 0x;yz, we can roll to 0x;yz, xy;0z, and xz;0y

0x;yz has (yz)x dome as CS and point trace. Rolling in x direction will keep the trace as point, and not even the CS will change in any way. The configuration xy;0z has a sphere as CS and z line as trace while the xz;0y has CS also spheric, but trace is a y line. Since the two degree-3 nodes are equivalent, however, we can write these configurations as (x/y)(y/x);0z and (x/z)(z/x);0y. We can roll arbitrarily from them, as long as we stay in the plane. By repeated rolling, we can get to (x/y)0;(y/x)z, and then to (x/y)z;0(y/x) which is, in fact, ((x/y)/z)(z/(x/y))/0(y/x) - simply said, it seems that semiglome can be rolled into a completely arbitrary 3D orientation! And, unlike full glome, it can be SEEN as having arbitrary orientation.

Ah, the glome? No need to speak about it - the four nodes are all equivalent, so it can be rolled in 3 dimensions any way you want. Its CS is sphere, and its trace is a point.

[wendy]

The 15-puzzle is a thing where one slides blocks numbered 1 to 15 around the board, there being 16!/2 = 360 5048 0000 states. However, the only transitions involve adjacent swappings of the empty square and one of its adjacents, eg 0 with X.

So there is only two, three or four connected states to any given state, the ratios stand here as 1 2 1.

I really can't see how a rototope is so limited.

[Marek14]

You can't? The point is that each node in the rotatope's graph is connected to at most 4 others - so it can never roll in more than 4 perpendicular directions out of 15 possible.

[wendy]

I see it now. i'm pretty slow sometimes

[quickfur]

Finally (after 14 years!) I got around to really reading Marek's initial post in detail.

Mainly, my interest was driven by trying to understand what exactly "longdome" is. Before we get there, however, I found the interpretation of O=O=O as "dome" (or, in more usual terminology, "crind", or according to Wikipedia, "bicylinder") a little hard to follow. If O=O=O is to be interpreted as a dome aka crind, then what stops us from interpreting =O=O=O= as a tricylinder (i.e., a Steinmetz solid) rather than a sphere? A tricylinder would also have circular projections along each axis.

Or perhaps I'm misunderstanding what the graph nodes signify. But in either case, can we express a tricylinder in any existing notation? It's no less of an interesting shape that the other more conventional rotatopes / toratopes, having a connection with the rhombic dodecahedron, which makes one wonder if a 4D analogue might have connections with the 24-cell(!).

Anyway, I'm still trying to understanding exactly what the longdome is. So more questions may follow.

[ndl]

Mainly, my interest was driven by trying to understand what exactly "longdome" is.

I also got interested in crinds recently and rediscovered this "longdome". I would call it a "Crindal Cylinder" maybe in following with the names on the wiki. I created vertices for an approximate of the shape and loaded it into Stella4d. I posted a picture here. If you want I can send you the .off or .stel file.

Or perhaps I'm misunderstanding what the graph nodes signify. But in either case, can we express a tricylinder in any existing notation? It's no less of an interesting shape that the other more conventional rotatopes / toratopes, having a connection with the rhombic dodecahedron, which makes one wonder if a 4D analogue might have connections with the 24-cell(!).

The tricylinder is very interesting because it brings out the following question: What type of operation are you looking for? A crindal operation, so to speak, takes an existing n-dimensional shape and reduces it to a point in the n+1 dimension on both sides. But there are 2 ways of doing that reduction, it can keep the entirety of the shape and scale down the size, or you can truncate it on all sides. As you pointed at those are both interesting shapes.