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Today I was thinking about various 3D crown jewels, and how they might be derived from "simpler" shapes.

J84 (snub disphenoid) - there are various different ways to analyze the snub disphenoid, but recently I noticed that it can be derived from a pentagonal bipyramid: imagine a paper model of a pentagonal bipyramid, where you select a pair of adjacent edges on the pentagonal cross-section. Use scissors to cut it up along those edges, then push out the now-split vertex between them, until you can fit in a new edge. The result is a snub disphenoid. I'm not sure if the icosahedral bipyramid in 4D has enough degrees of freedom to allow an analogous construction, but it might be worth trying?

J85 (sphenocorona) - if we start with a pentagonal bipyramid, and this time cut four consecutive edges along the pentagonal cross-section, then we can stretch the now cut vertices apart and fill in the gaps with a triangle-square-square-triangle patch to make a sphenocorona. This does require a great deal of freedom of deformation, though; I'm not sure if it's possible in 4D?

J86 (sphenomegacorona) - it seems to be an analogous derivation to J84, applied to an icosahedron instead of a pentagonal bipyramid.

J87 (hebesphenomegacorona) - seems to be an analogous derivation to J85, applied to an icosahedron instead of a pentagonal bipyramid, and the inserted patch has 3 squares instead of just two.

J88 (disphenocingulum) - appears to be a distorted hexagonal antiprism capped by two "spheno" wedges. Not sure what the 4D analogue would be; perhaps some kind of duoprism suitably deformed in order to fit odd-shaped augments with non-flat joining surfaces? Sorta like D4.8.x in the sense of having non-flat joining surface, except simpler.

The question of interest, of course, is whether the 4D analogues of the pentagonal bipyramid / icosahedron have enough degrees of freedom to allow similar kinds of cuttings and deformations? Thanks to the Blind couple's results, we know that any crown jewels in this direction cannot consist only of tetrahedra or other regular cells; in all likelihood the "pushing apart" stage of the derivation will introduce triangular prisms and maybe other kinds of prisms. But still, these derivations are a lot simpler than the currently-known BT polychora, so perhaps there's a chance we can find something here?

Of particular interest is, suppose we cut off a piece of the 600-cell's surface to get a bunch of linked tetrahedra. What is the maximum number of tetrahedra in the patch such that it will have at least 1 degree of freedom? When does a patch containing tetrahedra, 5 around each edge, become forced to take on the exact curvature of the 600-cell? Such "flexible patches" of tetrahedra may be useful in constructing 4D crown jewels, since their extra degree(s) of freedom may permit enough deformation to insert unusual cell shapes to close them up in a CRF way.

Furthermore, suppose we have a CRF that contains two conjoined icosahedral cells, with dichoral symmetry between them. Suppose we now perform the ike -> J86 derivation on these cells, in correponding positions. Will it be possible to insert CRF cells to close up the new shape? If so, this will produce a non-trivial CRF with two J86 cells. The idea behind this derivation is that if the higher interconnectivity of 4D polychora makes it hard for direct generalization of the ike->J86 deformation, maybe we can use the 3D ike->J86 deformation, but applies to ikes within a 4D polytope, then patch up the result with new cells to close it up.

Thoughts?

J84 (snub disphenoid) - there are various different ways to analyze the snub disphenoid, but recently I noticed that it can be derived from a pentagonal bipyramid: imagine a paper model of a pentagonal bipyramid, where you select a pair of adjacent edges on the pentagonal cross-section. Use scissors to cut it up along those edges, then push out the now-split vertex between them, until you can fit in a new edge. The result is a snub disphenoid. I'm not sure if the icosahedral bipyramid in 4D has enough degrees of freedom to allow an analogous construction, but it might be worth trying?

J85 (sphenocorona) - if we start with a pentagonal bipyramid, and this time cut four consecutive edges along the pentagonal cross-section, then we can stretch the now cut vertices apart and fill in the gaps with a triangle-square-square-triangle patch to make a sphenocorona. This does require a great deal of freedom of deformation, though; I'm not sure if it's possible in 4D?

J86 (sphenomegacorona) - it seems to be an analogous derivation to J84, applied to an icosahedron instead of a pentagonal bipyramid.

J87 (hebesphenomegacorona) - seems to be an analogous derivation to J85, applied to an icosahedron instead of a pentagonal bipyramid, and the inserted patch has 3 squares instead of just two.

J88 (disphenocingulum) - appears to be a distorted hexagonal antiprism capped by two "spheno" wedges. Not sure what the 4D analogue would be; perhaps some kind of duoprism suitably deformed in order to fit odd-shaped augments with non-flat joining surfaces? Sorta like D4.8.x in the sense of having non-flat joining surface, except simpler.

The question of interest, of course, is whether the 4D analogues of the pentagonal bipyramid / icosahedron have enough degrees of freedom to allow similar kinds of cuttings and deformations? Thanks to the Blind couple's results, we know that any crown jewels in this direction cannot consist only of tetrahedra or other regular cells; in all likelihood the "pushing apart" stage of the derivation will introduce triangular prisms and maybe other kinds of prisms. But still, these derivations are a lot simpler than the currently-known BT polychora, so perhaps there's a chance we can find something here?

Of particular interest is, suppose we cut off a piece of the 600-cell's surface to get a bunch of linked tetrahedra. What is the maximum number of tetrahedra in the patch such that it will have at least 1 degree of freedom? When does a patch containing tetrahedra, 5 around each edge, become forced to take on the exact curvature of the 600-cell? Such "flexible patches" of tetrahedra may be useful in constructing 4D crown jewels, since their extra degree(s) of freedom may permit enough deformation to insert unusual cell shapes to close them up in a CRF way.

Furthermore, suppose we have a CRF that contains two conjoined icosahedral cells, with dichoral symmetry between them. Suppose we now perform the ike -> J86 derivation on these cells, in correponding positions. Will it be possible to insert CRF cells to close up the new shape? If so, this will produce a non-trivial CRF with two J86 cells. The idea behind this derivation is that if the higher interconnectivity of 4D polychora makes it hard for direct generalization of the ike->J86 deformation, maybe we can use the 3D ike->J86 deformation, but applies to ikes within a 4D polytope, then patch up the result with new cells to close it up.

Thoughts?

- quickfur
- Pentonian
**Posts:**3004**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

There is a piece of software ("Hedron" by Jim McNeill, cf. his website), which solves that problem quite nicely for polyhedra (i.e. for 3D): Provided a set of connectivities it sets up a random distribution of vertices and then uses random disturbations of those initial positions, subject to forces which represent the required incidences and coplanarities, to cool that shape down, always measuring the still contained deviations from that target figure (zero forces).

What you are asking here in your described transfer of building procedures of Johnson solid crown jewels from easier, i.e. readily constructable figures by means of cutting open and following insertions onto potential new 4D CRFs, would not only ask for ideas how to potentially construct those, but also would need for a similar piece of software which would allow to test such hypotheses.

That one could be set up completely in analogue to Hedron, I suppose. Just that one would have to use 4D coordinates instead of 3D. And one would have then 3 types of forces instead of only 2 so far:

a) Requiring unit edge lengths for the set of initially provided abstract edges, i.e. vertex pairs,

b) requiring coplanarity of 2D boundaries (including the required lengths of inner invisible "edges" / 2D-chords),

c) and now a third which requires corealmity of 3D cells (including achieving the required inner 3D-chords).

The setup of such a piece of software ought to be mostly straight forward. But the gaining of the required chords looks more difficult. Perhaps it might be useful here to have an external config file, providing all those informations for any allowed to use cells. (That externality would help to test earlier releases of that software with some restricted set of possible cells first. But not having to change its coding only for the reason of adding further allowed cells.)

- Anyone out there who would like to program such a software?

It furthermore ought to be easy to handle, running at least on any Windows engine (potentially also Linux?) - so perhaps in Java or even Web-based (e.g. PHP or JavaScript)? And, for sure, I'd like if it would become a freeware or a free accessible service.

--- rk

What you are asking here in your described transfer of building procedures of Johnson solid crown jewels from easier, i.e. readily constructable figures by means of cutting open and following insertions onto potential new 4D CRFs, would not only ask for ideas how to potentially construct those, but also would need for a similar piece of software which would allow to test such hypotheses.

That one could be set up completely in analogue to Hedron, I suppose. Just that one would have to use 4D coordinates instead of 3D. And one would have then 3 types of forces instead of only 2 so far:

a) Requiring unit edge lengths for the set of initially provided abstract edges, i.e. vertex pairs,

b) requiring coplanarity of 2D boundaries (including the required lengths of inner invisible "edges" / 2D-chords),

c) and now a third which requires corealmity of 3D cells (including achieving the required inner 3D-chords).

The setup of such a piece of software ought to be mostly straight forward. But the gaining of the required chords looks more difficult. Perhaps it might be useful here to have an external config file, providing all those informations for any allowed to use cells. (That externality would help to test earlier releases of that software with some restricted set of possible cells first. But not having to change its coding only for the reason of adding further allowed cells.)

- Anyone out there who would like to program such a software?

It furthermore ought to be easy to handle, running at least on any Windows engine (potentially also Linux?) - so perhaps in Java or even Web-based (e.g. PHP or JavaScript)? And, for sure, I'd like if it would become a freeware or a free accessible service.

--- rk

- Klitzing
- Pentonian
**Posts:**1641**Joined:**Sun Aug 19, 2012 11:16 am**Location:**Heidenheim, Germany

I haven't used Windows in any non-trivial capacity for a looong time... I use Linux both at home and at work. So any software I write is likely to be specific to Linux. But I suppose I could hook it up to a web service that you can access with a browser. If I find the time to work on this, that is. I have been extremely busy lately -- I still haven't finished this month's Polytope of the Month! (And btw, it will be spidrox... I've only just finished the renders for the 12 rings of prisms/antiprisms yesterday, and now I have to do the renders for the 20 rings of pyramids.)

I was thinking more along the lines of analytic solutions, though. Suppose we are given some CRF that has two icosahedra sharing a face, preferably with a narrow dichoral angle. Would it be possible to, say, deform the icosahedra into J86's, while preserving most of the existing CRF structure, perhaps modifying them suitably, so that the result is CRF-able? This is somewhat similar to bilbiro'ing or thawro'ing, in that we start with an existing CRF structure and modify the icosahedra into bilbiro's/thawro's; except here, unlike bilbiro'ing or thawro'ing, dichoral angles may not be preserved.

The first step to such analysis, of course, is to find out how many degrees of freedom we have in 4D, given some CRF patch of cells. Does the extra dimension of cell-connectivity increase or decrease the number of degrees of freedom in deforming a given (open) cell complex? What are the conditions for maximizing/minimizing the flexibility of a given cell complex? For example, in 3D, if we remove the pentagonal face from a pentagonal pyramid, the remaining 5 tetrahedra show a considerable amount of flexibility in their dihedral angles. In fact, if I'm not wrong, there are 2 degrees of freedom in how you can deform this net without breaking the triangles apart. In 4D, suppose we remove the icosahedron from an icosahedral pyramid. Can the remaining 20 tetrahedra vary in their dichoral angles freely? If so, how many degrees of freedom do you have in deforming this net? And if not, what are the restrictions on what kind of 4D nets confer flexibility of deformation?

I was thinking more along the lines of analytic solutions, though. Suppose we are given some CRF that has two icosahedra sharing a face, preferably with a narrow dichoral angle. Would it be possible to, say, deform the icosahedra into J86's, while preserving most of the existing CRF structure, perhaps modifying them suitably, so that the result is CRF-able? This is somewhat similar to bilbiro'ing or thawro'ing, in that we start with an existing CRF structure and modify the icosahedra into bilbiro's/thawro's; except here, unlike bilbiro'ing or thawro'ing, dichoral angles may not be preserved.

The first step to such analysis, of course, is to find out how many degrees of freedom we have in 4D, given some CRF patch of cells. Does the extra dimension of cell-connectivity increase or decrease the number of degrees of freedom in deforming a given (open) cell complex? What are the conditions for maximizing/minimizing the flexibility of a given cell complex? For example, in 3D, if we remove the pentagonal face from a pentagonal pyramid, the remaining 5 tetrahedra show a considerable amount of flexibility in their dihedral angles. In fact, if I'm not wrong, there are 2 degrees of freedom in how you can deform this net without breaking the triangles apart. In 4D, suppose we remove the icosahedron from an icosahedral pyramid. Can the remaining 20 tetrahedra vary in their dichoral angles freely? If so, how many degrees of freedom do you have in deforming this net? And if not, what are the restrictions on what kind of 4D nets confer flexibility of deformation?

- quickfur
- Pentonian
**Posts:**3004**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

Let's look at a general 5-triangle vertex.

We put the vertex in [0,0,0]. Then the endpoints of edges will lie on the unit sphere. First vertex will be at [1,0,0]. Second will be still somewhere in xy plane, at [x2,y2,0]. The position of remaining three is general: [x3,y3,z3], [x4,y4,z4] and [x5,y5,z5].

Now, we have 11 unknowns and we can set constraints:

Unit sphere constraints:

1. x2^2 + y2^2 = 1

2. x3^2 + y3^2 + z3^2 = 1

3. x4^2 + y4^2 + z4^2 = 1

4. x5^2 + y5^2 + z5^2 = 1

Edge constraints:

5. (x2 - 1)^2 + y2^2 = 1

6. (x3 - x2)^2 + (y3 - y2)^2 + z3^2 = 1

7. (x4 - x3)^2 + (y4 - y3)^2 + (z4 - z3)^2 = 1

8. (x5 - x4)^2 + (y5 - y4)^2 + (z5 - z4)^2 = 1

9. (x5 - 1) ^2 + y5^2 + z5^2 = 1

So we have 9 equation for 11 unknowns, and so we see there are two degrees of freedom. Adding coplanarity condition would add two more equations (you make an equation of plane passing through first three points and then force 4th and 5th point to lie in it as well), so we'd get no degrees of freedom then.

Now look at 4D case. In 4D, vertices typically have significantly more faces than in 3D. A tetrahedral vertex has 6 faces -- that is 6 chords whose lengths must be precisely set to create a CRF figure.

Let's look at an icosahedral vertex. It has 12 edges and 30 faces. Each of the adjacent vertices will have 4 coordinates in 4D:

[1,0,0,0]

[x2,y2,0,0]

[x3,y3,z3,0]

[x4,y4,z4,w4]

...

[x12,y12,z12,w12]

So we have 41 unknowns altogether.

Constraints are:

11 equations enforcing unit hypersphere

30 equations enforcing chord lengths (corresponding to 30 faces at the central vertex).

So unless some of these equations are dependent (and I don't know whether they are], it would mean that vertex of icosahedral pyramid actually has no degrees of freedom at all.

Let's look at the general case. In 3D, you generally have an n-gon with 3*n coordinates. 4 of these are fixed ([1,0,0] for first vertex and 0 z coordinate for second), so you have 3n - 4. Then you have n - 1 unit sphere constraints and n chords. This translates in n-3 degrees of freedom (0 for triangles, 1 for quadrangles and 2 for pentagons -- no higher polygons can occur as vertex figures). In 4D, the number of coordinates to find is 4n - 7, n - 1 unit sphere constraints, and then the number of chords varies.

What's the maximum number of chords? Well, that would be reached if the vertex polyhedron was deltahedron, i.e. composed solely of triangles. If it's not a deltahedron, then you could split a non-triangular face by a diagonal.

So, how many edges does a deltahedron have? If it has n triangles, then it has 3n/2 edges, as every triangle has 3 and every edge belongs to 2 triangles. Tetrahedron (4 triangles) has 6 edges, triangular bipyramid (6 triangles) has 9 edges, icosahedron (20 triangles) has 30 edges, etc.

How many vertices does a deltahedron have? That can be obtained from Euler's formula V + F - E = 2. If F is n and E is 3n/2, we get

V + n - 3n/2 = 2

V - n/2 = 2

V = 2 + n/2

So tetrahedron has 4 (2 + 4/2) vertices, triangular bipyramid has 5 (2 + 6/2) and icosahedron has 12 (2 + 20/2).

We'll use 2n for number of faces from now on to make it simpler (deltahedra must have even number of faces, of course).

So, for a deltahedron-like vertex figure with 2n triangles, you have 2 + n vertices. These have 4n + 1 coordinates to determine (4*(2 + n) - 7).

Now, we have n + 1 unit hypersphere constraints and 3n chords.

See the problem? Together, they add to 4n + 1 -- in other words, we get 0 degrees of freedom!

Tetrahedral vertex -- 2*2 triangles, 9 coordinates, 3 + 6 constraints = 0 degrees of freedom.

Triangular bipyramidal vertex -- 2*3 triangles, 13 coordinates, 4 + 9 constraints = 0 degrees of freedom.

Octahedral vertex -- 2*4 triangles, 17 coordinates, 5 + 12 cponstraints = 0 degrees of freedom.

Icosahedral vertex -- 2*10 triangles, 41 coordinates, 11 + 30 constraints = 0 degrees of freedom!

So, what about vertex figures that are NOT deltahedra? Actually, there's no such thing! If the vertex figure contains a polygon with more than three sides, that polygon must correspond to one of the finite list of valid polyhedra. That means that its chords CANNOT vary freely, they must take one of a discrete set of values. For example, you can have a quadrangle with unit edges corresponging to a octahedron's square, a quadrangle corresponding to equatorial vertex of triangular bipyramid or quadrangle corresponding to a 4-triangle vertex of snub disphenoid, but not a generic, flexible quadrangle! So every time you have a non-deltahedron vertex figure, you still have to add additional chords to ensure the big polygon will work out correctly, and that will, in effect, turn it into a deltahedron!

So... unless there are prevalent dependencies between these equations, I think that this shows that ANY 4D vertex of CRF polychoron must have degree of freedom 0, and therefore is rigid.

We put the vertex in [0,0,0]. Then the endpoints of edges will lie on the unit sphere. First vertex will be at [1,0,0]. Second will be still somewhere in xy plane, at [x2,y2,0]. The position of remaining three is general: [x3,y3,z3], [x4,y4,z4] and [x5,y5,z5].

Now, we have 11 unknowns and we can set constraints:

Unit sphere constraints:

1. x2^2 + y2^2 = 1

2. x3^2 + y3^2 + z3^2 = 1

3. x4^2 + y4^2 + z4^2 = 1

4. x5^2 + y5^2 + z5^2 = 1

Edge constraints:

5. (x2 - 1)^2 + y2^2 = 1

6. (x3 - x2)^2 + (y3 - y2)^2 + z3^2 = 1

7. (x4 - x3)^2 + (y4 - y3)^2 + (z4 - z3)^2 = 1

8. (x5 - x4)^2 + (y5 - y4)^2 + (z5 - z4)^2 = 1

9. (x5 - 1) ^2 + y5^2 + z5^2 = 1

So we have 9 equation for 11 unknowns, and so we see there are two degrees of freedom. Adding coplanarity condition would add two more equations (you make an equation of plane passing through first three points and then force 4th and 5th point to lie in it as well), so we'd get no degrees of freedom then.

Now look at 4D case. In 4D, vertices typically have significantly more faces than in 3D. A tetrahedral vertex has 6 faces -- that is 6 chords whose lengths must be precisely set to create a CRF figure.

Let's look at an icosahedral vertex. It has 12 edges and 30 faces. Each of the adjacent vertices will have 4 coordinates in 4D:

[1,0,0,0]

[x2,y2,0,0]

[x3,y3,z3,0]

[x4,y4,z4,w4]

...

[x12,y12,z12,w12]

So we have 41 unknowns altogether.

Constraints are:

11 equations enforcing unit hypersphere

30 equations enforcing chord lengths (corresponding to 30 faces at the central vertex).

So unless some of these equations are dependent (and I don't know whether they are], it would mean that vertex of icosahedral pyramid actually has no degrees of freedom at all.

Let's look at the general case. In 3D, you generally have an n-gon with 3*n coordinates. 4 of these are fixed ([1,0,0] for first vertex and 0 z coordinate for second), so you have 3n - 4. Then you have n - 1 unit sphere constraints and n chords. This translates in n-3 degrees of freedom (0 for triangles, 1 for quadrangles and 2 for pentagons -- no higher polygons can occur as vertex figures). In 4D, the number of coordinates to find is 4n - 7, n - 1 unit sphere constraints, and then the number of chords varies.

What's the maximum number of chords? Well, that would be reached if the vertex polyhedron was deltahedron, i.e. composed solely of triangles. If it's not a deltahedron, then you could split a non-triangular face by a diagonal.

So, how many edges does a deltahedron have? If it has n triangles, then it has 3n/2 edges, as every triangle has 3 and every edge belongs to 2 triangles. Tetrahedron (4 triangles) has 6 edges, triangular bipyramid (6 triangles) has 9 edges, icosahedron (20 triangles) has 30 edges, etc.

How many vertices does a deltahedron have? That can be obtained from Euler's formula V + F - E = 2. If F is n and E is 3n/2, we get

V + n - 3n/2 = 2

V - n/2 = 2

V = 2 + n/2

So tetrahedron has 4 (2 + 4/2) vertices, triangular bipyramid has 5 (2 + 6/2) and icosahedron has 12 (2 + 20/2).

We'll use 2n for number of faces from now on to make it simpler (deltahedra must have even number of faces, of course).

So, for a deltahedron-like vertex figure with 2n triangles, you have 2 + n vertices. These have 4n + 1 coordinates to determine (4*(2 + n) - 7).

Now, we have n + 1 unit hypersphere constraints and 3n chords.

See the problem? Together, they add to 4n + 1 -- in other words, we get 0 degrees of freedom!

Tetrahedral vertex -- 2*2 triangles, 9 coordinates, 3 + 6 constraints = 0 degrees of freedom.

Triangular bipyramidal vertex -- 2*3 triangles, 13 coordinates, 4 + 9 constraints = 0 degrees of freedom.

Octahedral vertex -- 2*4 triangles, 17 coordinates, 5 + 12 cponstraints = 0 degrees of freedom.

Icosahedral vertex -- 2*10 triangles, 41 coordinates, 11 + 30 constraints = 0 degrees of freedom!

So, what about vertex figures that are NOT deltahedra? Actually, there's no such thing! If the vertex figure contains a polygon with more than three sides, that polygon must correspond to one of the finite list of valid polyhedra. That means that its chords CANNOT vary freely, they must take one of a discrete set of values. For example, you can have a quadrangle with unit edges corresponging to a octahedron's square, a quadrangle corresponding to equatorial vertex of triangular bipyramid or quadrangle corresponding to a 4-triangle vertex of snub disphenoid, but not a generic, flexible quadrangle! So every time you have a non-deltahedron vertex figure, you still have to add additional chords to ensure the big polygon will work out correctly, and that will, in effect, turn it into a deltahedron!

So... unless there are prevalent dependencies between these equations, I think that this shows that ANY 4D vertex of CRF polychoron must have degree of freedom 0, and therefore is rigid.

- Marek14
- Pentonian
**Posts:**1191**Joined:**Sat Jul 16, 2005 6:40 pm

But if 4D vertices are always rigid, then doesn't that mean we can just brute-force enumerate all possible vertex configurations, and thereby find all 4D CRFs? We don't even need to precompute any verfs; just enumerate all possible combinations of 3D CRFs around a vertex, and discard those that don't close up locally.

- quickfur
- Pentonian
**Posts:**3004**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

quickfur wrote:But if 4D vertices are always rigid, then doesn't that mean we can just brute-force enumerate all possible vertex configurations, and thereby find all 4D CRFs? We don't even need to precompute any verfs; just enumerate all possible combinations of 3D CRFs around a vertex, and discard those that don't close up locally.

Haven't I been mentioning the possibility of something like that for quite some time? But it was just a hunch, only today I actually tried to calculate the conditions. The result was surprising

Of course, it IS possible that for some vertices the equations will turn out to be dependent and they won't be rigid, but it should be the exception, not the rule.

- Marek14
- Pentonian
**Posts:**1191**Joined:**Sat Jul 16, 2005 6:40 pm

Hmm. So it appears that while edges are flexible, vertices are not (at least, not in the general case). So it seems that the condition for a 4D cell complex (non-closed net) to be flexible, all of its vertices must be open (i.e., lie on the unconnected boundary of the net).

This is a very interesting result. In 3D, this doesn't hold -- a closed vertex (completely surrounded by faces) may still be flexible, e.g., the vertex surrounded by 4 or 5 triangles. Because of this, a large-enough 3D net may have transitive flexibility. For example, if you cut a series of edges on an icosahedron, the closed vertices (not adjacent to any of the cut edges) are still flexible, and they can deform enough to accomodate the cut edges to widen into a gap where more faces can be filled in, thus producing the sphenocoronas. It's because of this transitive flexibility that you can end up with some complicated dependencies in dihedral angles: as you widen the gap where the cut was made, the deformation propagates transitively throughout the remainder of the polyhedron, modifying dihedral angles accordingly, so in the general case, to compute the coordinates of the deformed shape you need to solve a whole series of interdependent equations.

But in 4D, if this is indeed true that all vertices are rigid, then there cannot be any transitive flexibility; the only flexible parts of a partially-constructed 4D CRF must lie on its unconnected boundary! Which means that once a vertex in the net becomes closed, it forces that part of the net to become rigid, and so the closed portion of the net is always rigid, and any dependencies of dichoral angles must only lie on the open (unconnected) boundary of the net.

This seems to suggest that maybe the only 4D CVP3 CRFs are those that contain CVP3 polyhedra as cells... because the boundary of the net is, at most, a 3-manifold, so if there are any CVP3 relationships between vertices (which arise from flexibility, since otherwise they can be solved with just independent quadratic equations and so can't be CVP3), it must arise from the 3D boundary of the net at some point during its construction. Which in turn suggests that such relationships must come from a CVP3 cell somewhere on the boundary of the net. So without a CVP3 cell somewhere in the net during the construction of the polytope, its coordinates will not be CVP3.

This is a very interesting result. In 3D, this doesn't hold -- a closed vertex (completely surrounded by faces) may still be flexible, e.g., the vertex surrounded by 4 or 5 triangles. Because of this, a large-enough 3D net may have transitive flexibility. For example, if you cut a series of edges on an icosahedron, the closed vertices (not adjacent to any of the cut edges) are still flexible, and they can deform enough to accomodate the cut edges to widen into a gap where more faces can be filled in, thus producing the sphenocoronas. It's because of this transitive flexibility that you can end up with some complicated dependencies in dihedral angles: as you widen the gap where the cut was made, the deformation propagates transitively throughout the remainder of the polyhedron, modifying dihedral angles accordingly, so in the general case, to compute the coordinates of the deformed shape you need to solve a whole series of interdependent equations.

But in 4D, if this is indeed true that all vertices are rigid, then there cannot be any transitive flexibility; the only flexible parts of a partially-constructed 4D CRF must lie on its unconnected boundary! Which means that once a vertex in the net becomes closed, it forces that part of the net to become rigid, and so the closed portion of the net is always rigid, and any dependencies of dichoral angles must only lie on the open (unconnected) boundary of the net.

This seems to suggest that maybe the only 4D CVP3 CRFs are those that contain CVP3 polyhedra as cells... because the boundary of the net is, at most, a 3-manifold, so if there are any CVP3 relationships between vertices (which arise from flexibility, since otherwise they can be solved with just independent quadratic equations and so can't be CVP3), it must arise from the 3D boundary of the net at some point during its construction. Which in turn suggests that such relationships must come from a CVP3 cell somewhere on the boundary of the net. So without a CVP3 cell somewhere in the net during the construction of the polytope, its coordinates will not be CVP3.

- quickfur
- Pentonian
**Posts:**3004**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

Indeed. Formerly, the absence of such calculations made that I didn't consider such an approach as easy. However, now that you have made these, I think we can finally investigate some CVP3-polytopes (or prove their non-existence)Marek14 wrote:quickfur wrote:But if 4D vertices are always rigid, then doesn't that mean we can just brute-force enumerate all possible vertex configurations, and thereby find all 4D CRFs? We don't even need to precompute any verfs; just enumerate all possible combinations of 3D CRFs around a vertex, and discard those that don't close up locally.

Haven't I been mentioning the possibility of something like that for quite some time? But it was just a hunch, only today I actually tried to calculate the conditions. The result was surprising

I think you don't have to worry about that much. let's take a central vertex. now we add some vertices. let's say these vertices are all part of a surtope. This means the added vertices should make the verf of the central vertex. This verf can be seen in matrix context: vertex 1 has distance x to vertex 2, vertex 2 has distance y to vertex 3, vertex 1 has distance z to vertex 3, vertex 3 has distance w to vertex 2, etc, nicely placed in a matrix.Of course, it IS possible that for some vertices the equations will turn out to be dependent and they won't be rigid, but it should be the exception, not the rule.

This verf is (clearly) rigid (though you can turn it around in space, but it stays congruent all the time). If we place some more vertices now, of a surtope which shares vertices 1 and 2 and the central vertex, these vertices make another verf, and thus should have another matrix. Of course, these verfs are independent, though they have an overlapping distance between 1 and 2. When we build the rest of the verf, all surtopal verfs stay independent, with occasional overlap. This means all distances between vertices are independent, if the surtopal verfs don't have internal independencies themselves. You shouldn't worry about those either, as they only occur with verfs with more than 3 vertices, and you already proved these differently.

Now this result is very useful for determining CVP-3 polytopes, as this means dichoral angles are fixed as well. As quickfur already noted, vertices are only CVP3 if they have CVP3-surtopes themselves. Now what I'm hoping is that these CVP3-surtopal verfs make the dichoral angles horrible, so that we can only combine CVP3 verfs with CVP3 verfs. This means the number of CVP3-polytopes stays conveniently small, and thus they won't combinatorically explode, allowing a computer search. of course we first have to proof my first hypothesis.

How easily one gives his confidence to persons who know how to give themselves the appearance of more knowledge, when this knowledge has been drawn from a foreign source.

-Stern/Multatuli/Eduard Douwes Dekker

-Stern/Multatuli/Eduard Douwes Dekker

- student91
- Tetronian
**Posts:**328**Joined:**Tue Dec 10, 2013 3:41 pm

First step here might be enumeration of dichoral angles. I enumerated those about two years ago:

From these, it should be possible to enumerate 4D edges.

- Code: Select all
`So, let's look at dihedral angles. Stella is a big help here since its net mode computes dihedral angles automatically.`

I include all prisms and antiprisms up to 10, for completeness.

BTW, you shouldn't automatically assume that there can be only 5 faces to an edge -- dihedral angles of some Johnson solids can get quite small.

Now, let's talk "possibilities". Basically, in some Johnson solids, the same dihedral angle exists at several nonequivalent places. I note this. Some of those places might be also assymetrical (i.e. joining two identical polygons, but without an axis of symmetry passing through the edge, so you have to try to fit it in both orientations).

The exact number of possiblities for CRF purposes will have to be checked, some possibilities might be chiral.

For the weird solids (sphenocorona and up) I've given up on describing the edges and counting possibilities...

31.7175 (dpc4)

4-10 in pentagonal cupola

37.3774 (dpp)

3-5 in pentagonal pyramid

3-10 in pentagonal cupola

45

4-8 in square cupola

54.7356 (90 - d4/2)

3-4 in square pyramid

4-6 in triangular cupola

3-8 in square cupola

60

4-4 in triangular prism

4-4 in elongated triangular pyramid (asymmetrical)

4-4 in elongated triangular dipyramid

4-4 in augmented triangular prism

63.4349 (2*dpc4)

5-10 in pentagonal rotunda

4-4 in pentagonal orthobicupola

5-5 in metabidiminished icosahedron

5-5 in tridiminished icosahedron (asymmetrical)

5-5 in augmented tridiminished icosahedron (asymmetrical)

5-5 in bilunabirotunda

69.0948 (dpc4 + dpp = d20/2)

3-4 in pentagonal gyrobicupola (equatorial)

70.5288 (let's call it d4 because it's an important number)

3-3 in tetrahedron

6-6 in truncated tetrahedron

3-6 in triangular cupola

3-3 in elongated triangular pyramid (asymmetrical)

3-3 in triangular dipyramid (apex) (asymmetrical)

3-3 in elongated triangular dipyramid (asymmetrical)

3-3 in augmented tridiminished icosahedron (augment) (asymmetrical)

6-6 in augmented truncated tetrahedron

72.973

4-4 in sphenomegacorona

74.7547 (2*dpp)

3-3 in pentagonal dipyramid (equatorial)

3-3 in pentagonal orthobicupola

79.1877 (dpr3; dpr3 + 2*dpc4 = did)

3-10 in pentagonal rotunda

86.7268

3-3 in sphenomegacorona

90

4-4 in cube

8-8 in truncated cube

3-4 in triangular prism

4-5 in pentagonal prism

4-6 in hexagonal prism

4-7 in heptagonal prism

4-8 in octagonal prism

4-9 in enneagonal prism

4-10 in decagonal prism

3-4 in elongated triangular pyramid (base)

4-4 in elongated square pyramid (2 possibilities) (both asymmetrical)

4-5 in elongated pentagonal pyramid

4-4 in elongated square dipyramid

4-6 in elongated triangular cupola (2 possibilities)

4-8 in elongated square cupola (2 possibilities)

4-10 in elongated pentagonal cupola (2 possibilities)

4-10 in elongated pentagonal rotunda (2 possibilities)

3-4 in gyrobifastigium (individual prisms)

4-4 in square orthobicupola (equatorial)

3-4 in augmented triangular prism (base/side)

3-4 in biaugmented triangular prism (base/side)

4-5 in augmented pentagonal prism (2 possibilities)

4-5 in biaugmented pentagonal prism (2 possibilities)

4-6 in augmented hexagonal prism (3 possibilities)

4-6 in parabiaugmented hexagonal prism

4-6 in metabiaugmented hexagonal prism (3 possibilities)

4-6 in triaugmented hexagonal prism

8-8 in augmented truncated cube (2 possibilities) (both asymmetrical)

8-8 in biaugmented truncated cube

95.1524 (3*dpc4)

4-5 in pentagonal gyrocupolarotunda (equatorial)

95.2466 (da10)

3-10 in decagonal antiprism

3-10 in gyroelongated pentagonal cupola

3-10 in gyroelongated pentagonal rotunda

95.843

3-9 in enneagonal antiprism

96.1983

3-3 in snub disphenoid (type 1) (2 possibilities) (1 asymmetrical)

96.5945 (da8)

3-8 in octagonal antiprism

3-8 in gyroelongated square cupola

97.4555

3-4 in sphenocorona

3-4 in augmented sphenocorona

97.5723

3-7 in heptagonal antiprism

98.8994 (da6)

3-6 in hexagonal antiprism

3-6 in gyroelongated triangular cupola

99.7356 (135 - d4/2)

3-4 in square gyrobicupola (equatorial)

100.194

4-4 in disphenocingulum

100.812 (d20 - dpp = dpp + 2*dpc4)

3-5 in pentagonal antiprism

3-5 in gyroelongated pentagonal pyramid

3-5 in pentagonal orthocupolarotunda (equatorial)

3-5 in metabidiminished icosahedron (2 possibilities)

3-5 in tridiminished icosahedron (2 possibilities)

3-5 in augmented tridiminished icosahedron (main body)

3-5 in bilunabirotunda

3-5 in triangular hebesphenorotunda

102.524

4-4 in hebesphenomegacorona

103.836 (da4)

3-4 in square antiprism

3-4 in gyroelongated square pyramid

108

4-4 in pentagonal prism

4-4 in elongated pentagonal pyramid (asymmetrical)

4-4 in elongated pentagonal dipyramid

4-4 in augmented pentagonal prism (2 possibilities) (1 asymmetrical)

4-4 in biaugmented pentagonal prism

109.471 (180 - d4)

3-3 in octahedron

3-6 in truncated tetrahedron

6-6 in truncated octahedron

3-3 in square pyramid

3-3 in elongated square pyramid (asymmetrical)

3-3 in gyroelongated square pyramid (apex) (asymmetrical)

3-3 in elongated square dipyramid (asymmetrical)

3-3 in gyroelongated square dipyramid (apex) (asymmetrical)

4-4 in triangular orthobicupola

3-3 in square orthobicupola

3-3 in augmented triangular prism (augment) (asymmetrical)

3-3 in biaugmented triangular prism (augment) (2 possibilities) (both asymmetrical)

3-3 in triaugmented triangular prism (augment) (asymmetrical)

3-3 in augmented pentagonal prism (asymmetrical)

3-3 in biaugmented pentagonal prism (2 possibilities) (both asymmetrical)

3-3 in augmented hexagonal prism (asymmetrical)

3-3 in parabiaugmented hexagonal prism (asymmetrical)

3-3 in metabiaugmented hexagonal prism (2 possibilities) (both asymmetrical)

3-3 in triaugmented hexagonal prism (asymmetrical)

3-6 in augmented truncated tetrahedron (main body) (2 possibilities)

3-3 in augmented sphenocorona

109.524

3-4 in sphenocorona

3-4 in augmented sphenocorona

110.905 (dpc4 + dpr3)

3-4 in pentagonal orthocupolarotunda (equatorial)

3-4 in bilunabirotunda

4-6 in triangular hebesphenorotunda

3-4 in triangular hebesphenorotunda

111.735

3-3 in hebesphenomegacorona

114.645

3-3 in snub square antiprism (middle edges) (2 possibilities)

114.736 (150 - d4/2)

3-4 in augmented triangular prism (augment/side)

3-4 in biaugmented triangular prism (augment/side)

116.565 (d12 = dpp + dpr3)

5-5 in dodecahedron

10-10 in truncated dodecahedron

3-3 in pentagonal gyrocupolarotunda

5-5 in augmented dodecahedron (4 possibilities) (all asymmetrical)

5-5 in parabiaugmented dodecahedron (2 possibilities) (1 asymmetrical)

5-5 in metabiaugmented dodecahedron (7 possibilities) (5 asymmetrical)

5-5 in triaugmented dodecahedron (4 possibilities) (all asymmetrical)

10-10 in augmented truncated dodecahedron (4 possibilities) (all asymmetrical)

10-10 in parabiaugmented truncated dodecahedron (2 possibilities) (1 asymmetrical)

10-10 in metabiaugmented truncated dodecahedron (7 possibilities) (5 asymmetrical)

10-10 in triaugmented truncated dodecahedron (4 possibilities) (all asymmetrical)

5-10 in diminished rhombicosidodecahedron

5-10 in paragyrate diminished rhombicosidodecahedron

5-10 in metagyrate diminished rhombicosidodecahedron (3 possibilities)

5-10 in bigyrate diminished rhombicosidodecahedron (3 possibilities)

5-10 in parabidiminished rhombicosidodecahedron

5-10 in metabidiminished rhombicosidodecahedron (3 possibilities)

5-10 in gyrate bidiminished rhombicosidodecahedron (5 possibilities)

5-10 in tridiminished rhombicosidodecahedron (3 possibilities)

117.019 (dsp)

4-4 in sphenocorona

117.356

3-3 in sphenomegacorona

118.892

3-3 in sphenocorona

3-3 in augmented sphenocorona

120

4-4 in hexagonal prism

4-4 in elongated triangular cupola (prism) (asymmetrical)

4-4 in elongated triangular orthobicupola (prism) (asymmetrical)

4-4 in elongated triangular gyrobicupola (prism)

4-4 in augmented hexagonal prism (2 possibilities) (both asymmetrical)

4-4 in parabiaugmented hexagonal prism

4-4 in metabiaugmented hexagonal prism (asymmetrical)

121.717 (90 + dpc4)

4-4 in elongated pentagonal cupola (cupola/prism) (asymmetrical)

4-4 in elongated pentagonal orthobicupola (cupola/prism) (asymmetrical)

4-4 in elongated pentagonal gyrobicupola (cupola/prism) (asymmetrical)

4-4 in elongated pentagonal orthocupolarotunda (cupola/prism) (asymmetrical)

4-4 in elongated pentagonal gyrocupolarotunda (cupola/prism) (asymmetrical)

4-10 in diminished rhombicosidodecahedron

4-10 in paragyrate diminished rhombicosidodecahedron

4-10 in metagyrate diminished rhombicosidodecahedron (3 possibilities)

4-10 in bigyrate diminished rhombicosidodecahedron (3 possibilities)

4-10 in parabidiminished rhombicosidodecahedron

4-10 in metabidiminished rhombicosidodecahedron (3 possibilities)

4-10 in gyrate bidiminished rhombicosidodecahedron (5 possibilities)

4-10 in tridiminished rhombicosidodecahedron (3 possibilities)

121.743

3-3 in snub disphenoid (type 2) (2 possibilities) (both asymmetrical)

124.702

3-3 in disphenocingulum

125.264 (90 + d4/2)

3-4 in cuboctahedron

4-6 in truncated octahedron

3-8 in truncated cube

6-8 in truncated cuboctahedron

3-4 in triangular cupola (2 possibilities)

3-4 in elongated triangular cupola (cupola) (2 possibilities)

3-4 in gyroelongated triangular cupola (cupola) (2 possibilities)

3-4 in triangular orthobicupola (2 possibilities)

3-4 in elongated triangular orthobicupola (cupola) (2 possibilities)

3-4 in elongated triangular gyrobicupola (cupola) (2 possibilities)

3-4 in gyroelongated triangular bicupola (cupola) (2 possibilities)

3-4 in augmented truncated tetrahedron (augment) (2 possibilities)

3-8 in augmented truncated cube (3 possibilities)

3-8 in biaugmented truncated cube

126.87 (4*dpc4)

5-5 in pentagonal orthobirotunda

126.964 (da10 + dpc)

3-4 in gyroelongated pentagonal cupola (cupola/antiprism)

3-4 in gyroelongated pentagonal bicupola (cupola/antiprism)

3-4 in gyroelongated pentagonal cupolarotunda (cupola/antiprism)

127.377 (90 + dpp)

3-4 in elongated pentagonal pyramid

3-4 in elongated pentagonal dipyramid

3-4 in elongated pentagonal cupola (cupola/prism)

3-4 in elongated pentagonal orthobicupola (cupola/prism)

3-4 in elongated pentagonal gyrobicupola (cupola/prism)

3-4 in elongated pentagonal orthocupolarotunda (cupola/prism)

3-4 in elongated pentagonal gyrocupolarotunda (cupola/prism)

127.552

3-3 in square antiprism

3-3 in gyroelongated square pyramid (antiprism band) (asymmetrical)

3-3 in gyroelongated square dipyramid (antiprism band)

128.496

3-3 in hebesphenomegacorona

128.571

4-4 in heptagonal prism

129.445

3-3 in sphenomegacorona

131.442

3-3 in augmented sphenocorona

132.624 (da10 + dpp)

3-3 in gyroelongated pentagonal cupola (cupola/antiprism) (asymmetrical)

3-3 in gyroelongated pentagonal bicupola (cupola/antiprism) (asymmetrical)

3-3 in gyroelongated pentagonal cupolarotunda (cupola/antiprism) (asymmetrical)

133.591

3-3 in disphenocingulum

133.973

3-4 in hebesphenomegacorona

135

4-4 in rhombicuboctahedron (asymmetrical)

4-8 in truncated cuboctahedron

4-4 in octagonal prism

4-4 in square cupola

4-4 in elongated square cupola (3 possibilities) (all asymmetrical)

4-4 in gyroelongated square cupola (asymmetrical)

4-4 in square orthobicupola (cupola) (asymmetrical)

4-4 in square gyrobicupola (asymmetrical)

4-4 in elongated square gyrobicupola (3 possibilities) (2 of them asymmetrical)

4-4 in gyroelongated square bicupola (asymmetrical)

4-4 in augmented truncated cube (asymmetrical)

4-4 in biaugmented truncated cube (asymmetrical)

135.992

3-3 in sphenocorona

3-3 in augmented sphenocorona

136.336

3-4 in disphenocingulum

137.24

3-4 in sphenomegacorona

138.19 (d20 = 2*(dpc4 + dpp)

3-3 in icosahedron

6-6 in truncated icosahedron

3-3 in pentagonal antiprism

3-3 in pentagonal pyramid

3-3 in elongated pentagonal pyramid (asymmetrical)

3-3 in gyroelongated pentagonal pyramid (3 possibilities) (all asymmetrical)

3-3 in pentagonal dipyramid (apex) (asymmetrical)

3-3 in elongated pentagonal dipyramid (asymmetrical)

3-3 in augmented dodecahedron (asymmetrical)

3-3 in parabiaugmented dodecahedron (asymmetrical)

3-3 in metabiaugmented dodecahedron (3 possibilities) (all asymmetrical)

3-3 in triaugmented dodecahedron (3 possibilities) (all asymmetrical)

3-3 in metabidiminished icosahedron (4 possibilities) (3 asymmetrical)

3-3 in tridiminished icosahedron (asymmetrical)

3-3 in augmented tridiminished icosahedron (main body) (asymmetrical)

3-6 in triangular hebesphenorotunda

3-3 in triangular hebesphenorotunda

140

4-4 in enneagonal prism

141.058 (2*d4)

3-3 in triangular dipyramid (equatorial)

3-3 in triangular orthobicupola

3-6 in augmented truncated tetrahedron (augment/main body)

141.31

3-3 in hebesphenomegacorona

141.595 (45 + da8)

3-4 in gyroelongated square cupola (cupola/antiprism)

3-4 in gyroelongated square bicupola (cupola/antiprism)

142.623 (did)

3-5 in icosidodecahedron

5-6 in truncated icosahedron

3-10 in truncated dodecahedron

6-10 in truncated icosidodecahedron

3-5 in pentagonal rotunda (3 possibilities)

3-5 in elongated pentagonal rotunda (3 possibilities)

3-5 in gyroelongated pentagonal cupola (rotunda) (3 possibilities)

3-5 in pentagonal orthocupolarotunda (rotunda) (3 possibilities)

3-5 in pentagonal gyrocupolarotunda (3 possibilities)

3-5 in pentagonal orthobirotunda (3 possibilities)

3-5 in elongated pentagonal orthocupolarotunda (3 possibilities)

3-5 in elongated pentagonal gyrocupolarotunda (3 possibilities)

3-5 in elongated pentagonal orthobirotunda (3 possibilities)

3-5 in elongated pentagonal gyrobirotunda (3 possibilities)

3-5 in gyroelongated pentagonal cupolarotunda (3 possibilities)

3-5 in gyroelongated pentagonal birotunda (3 possibilities)

3-10 in augmented truncated dodecahedron (main body) (7 possibilities)

3-10 in parabiaugmented truncated dodecahedron (main body) (3 possibilities)

3-10 in metabiaugmented truncated dodecahedron (main body) (14 possibilities)

3-10 in triaugmented truncated dodecahedron (main body) (9 possibilities)

3-5 in bilunabirotunda

3-5 in triangular hebesphenorotunda

142.983

3-4 in snub cube

143.479

3-3 in sphenocorona

3-3 in augmented sphenocorona

143.738

3-3 in sphenomegacorona

144

4-4 in decagonal prism

4-4 in elongated pentagonal cupola (prism) (asymmetrical)

4-4 in elongated pentagonal rotunda (asymmetrical)

4-4 in elongated pentagonal orthobicupola (prism) (asymmetrical)

4-4 in elongated pentagonal gyrobicupola (prism)

4-4 in elongated pentagonal orthocupolarotunda (prism) (asymmetrical)

4-4 in elongated pentagonal gyrocupolarotunda (prism) (asymmetrical)

4-4 in elongated pentagonal orthobirotunda (asymmetrical)

4-4 in elongated pentagonal gyrobirotunda

144.144

3-3 in snub square antiprism (other edges of triangles adjacent to squares) (2 possibilities) (both asymmetrical)

144.736 (180 - d4/2)

3-4 in rhombicuboctahedron

4-6 truncated cuboctahedron

3-4 in square cupola

3-4 in elongated square pyramid

3-4 in elongated square dipyramid

4-4 in elongated triangular cupola (cupola/prism) (asymmetrical)

3-4 in elongated square cupola (2 possibilities)

3-4 in gyroelongated square cupola (cupola)

3-4 in square orthobicupola

3-4 in square gyrobicupola (cupola)

4-4 in elongated triangular orthobicupola (cupola/prism) (asymmetrical)

4-4 in elongated triangular gyrobicupola (cupola/prism) (asymmetrical)

3-4 in elongated square gyrobicupola (2 possibilities)

3-4 in gyroelongated square bicupola (cupola)

3-3 in augmented triangular prism (augment/base) (asymmetrical)

3-3 in biaugmented triangular prism (augment/base) (asymmetrical)

3-3 in triaugmented triangular prism (augment/base) (asymmetrical)

3-5 in augmented pentagonal prism

3-5 in biaugmented pentagonal prism

3-6 in augmented hexagonal prism

3-6 in parabiaugmented hexagonal prism

3-6 in metabiaugmented hexagonal prism

3-6 in triaugmented hexagonal prism

3-4 in augmented truncated cube (augment)

3-8 in augmented truncated cube (augment/main body)

3-4 in biaugmented truncated cube (augment)

3-8 in biaugmented truncated cube (augment/main body)

145.222

3-3 in hexagonal antiprism

3-3 in gyroelongated triangular cupola (antiprism) (2 possibilities) (both asymmetrical)

3-3 in in gyroelongated triangular bicupola (antiprism) (3 possibilities) (1 asymmetrical)

145.441

3-4 in snub square antiprism

148.283

4-5 in rhombicosidodecahedron

4-10 in truncated icosidodecahedron

4-5 in pentagonal cupola

4-5 in elongated pentagonal cupola

4-5 in gyroelongated pentagonal cupola

4-5 in pentagonal orthobicupola

4-5 in pentagonal gyrobicupola

4-5 in pentagonal orthocupolarotunda

4-5 in pentagonal gyrocupolarotunda (cupola)

4-5 in elongated pentagonal orthobicupola

4-5 in elongated pentagonal gyrobicupola

4-5 in elongated pentagonal orthocupolarotunda (cupola)

4-5 in elongated pentagonal gyrocupolarotunda (cupola)

4-5 in gyroelongated pentagonal bicupola

4-5 in augmented truncated dodecahedron

4-5 in parabiaugmented truncated dodecahedron

4-5 in metabiaugmented truncated dodecahedron (3 possibilities)

4-5 in triaugmented truncated dodecahedron (3 possibilities)

4-5 in gyrate rhombicosidodecahedron (7 possibilities)

4-5 in parabigyrate rhombicosidodecahedron (3 possibilities)

4-5 in metabigyrate rhombicosidodecahedron (15 possibilities)

4-5 in trigyrate rhombicosidodecahedron (10 possibilities)

4-5 in diminished rhombicosidodecahedron (6 possibilities)

4-5 in paragyrate diminished rhombicosidodecahedron (5 possibilities)

4-5 in metagyrate diminished rhombicosidodecahedron (22 possibilities)

4-5 in bigyrate diminished rhombicosidodecahedron (21 possibilities)

4-5 in parabidiminished rhombicosidodecahedron (2 possibilities)

4-5 in metabidiminished rhombicosidodecahedron (11 possibilities)

4-5 in gyrate bidiminished rhombicosidodecahedron (20 possibilities)

4-5 in tridiminished rhombicosidodecahedron (6 possibilities)

148.434

3-3 in disphenocingulum

149.565

3-3 in hebesphenomegacorona

150

3-4 in gyrobifastigium (blend)

150.222

3-3 in heptagonal antiprism

151.33 (90 + da8 - d4/2)

3-3 in gyroelongated square cupola (cupola/antiprism) (asymmetrical)

3-3 in gyroelongated square bicupola (cupola/antiprism) (asymmetrical)

152.191

3-3 in augmented sphenocorona

152.93

3-5 in snub dodecahedron

152.976

3-4 in hebesphenomegacorona

153.235

3-3 in snub cube (2 possibilities) (1 asymmetrical)

153.435 (90 + 2*dpc4)

4-5 in elongated pentagonal rotunda

4-5 in elongated pentagonal orthocupolarotunda (rotunda/prism)

4-5 in elongated pentagonal gyrocupolarotunda (rotunda/prism)

4-5 in elongated pentagonal orthobirotunda

4-5 in elongated pentagonal gyrobirotunda

4-4 in gyrate rhombicosidodecahedron (asymmetrical)

4-4 in parabigyrate rhombicosidodecahedron (asymmetrical)

4-4 in metabigyrate rhombicosidodecahedron (3 possibilities) (all asymetrical)

4-4 in trigyrate rhombicosidodecahedron (3 possibilities) (all asymmetrical)

4-4 in paragyrate diminished rhombicosidodecahedron (asymmetrical)

4-4 in metagyrate diminished rhombicosidodecahedron (3 possibilities) (all asymmetrical)

4-4 in bigyrate diminished rhombicosidodecahedron (5 possibilities) (all asymmetrical)

4-4 in gyrate bidiminished rhombicosidodecahedron (3 possibilities) (all asymmetrical)

153.635 (180 + da6 - d4/2)

3-4 in gyroelongated triangular cupola (cupola/antiprism)

3-4 in gyroelongated triangular bicupola (cupola/antiprism)

153.942 (dpp + d12)

3-5 in augmented dodecahedron

3-5 in parabiaugmented dodecahedron

3-5 in metabiaugmented dodecahedron (3 possibilities)

3-5 in triaugmented dodecahedron (3 possibilities)

3-10 in augmented truncated dodecahedron (augment/main body)

3-10 in parabiaugmented truncated dodecahedron (augment/main body)

3-10 in metabiaugmented truncated dodecahedron (augment/main body) (3 possibilities)

3-10 in triaugmented truncated dodecahedron (augment/main body) (3 possibilities)

3-5 in gyrate rhombicosidodecahedron

3-5 in parabigyrate rhombicosidodecahedron

3-5 in metabigyrate rhombicosidodecahedron (3 possibilities)

3-5 in trigyrate rhombicosidodecahedron (3 possibilities)

3-5 in paragyrate diminished rhombicosidodecahedron

3-5 in metagyrate diminished rhombicosidodecahedron (3 possibilities)

3-5 in bigyrate diminished rhombicosidodecahedron (5 possibilities)

3-5 in gyrate bidiminished rhombicosidodecahedron (3 possibilities)

153.962

3-3 in octagonal antiprism

3-3 in gyroelongated square cupola (antiprism) (2 possibilities) (both asymmetrical)

3-3 in gyroelongated square bicupola (antiprism) (3 possibilities) (1 asymmetrical)

154.419

3-4 in disphenocingulum

154.722

3-4 in sphenomegacorona

156.866

3-3 in enneagonal antiprism

157.148

3-3 in hebesphenomegacorona

158.375 (2*dpr3)

3-3 in pentagonal orthobirotunda

158.572 (90 + da4 - d4/2)

3-3 in gyroelongated square pyramid (apex/antiprism) (asymmetrical)

3-3 in in gyroelongated square dipyramid (apex/antiprism) (asymmetrical)

158.682 (da10 + 2*dpc4)

3-5 in gyroelongated pentagonal rotunda (rotunda/antiprism)

3-5 in gyroelongated pentagonal cupolarotunda (rotunda/antiprism)

3-5 in gyroelongated pentagonal birotunda (rotunda/antiprism)

159.095

3-4 in rhombicosidodecahedron

4-6 in truncated icosidodecahedron

3-4 in pentagonal cupola

3-4 in elongated pentagonal cupola (cupola)

3-4 in gyroelongated pentagonal cupola (cupola)

3-4 in pentagonal orthobicupola

3-4 in pentagonal gyrobicupola (cupola)

3-4 in pentagonal orthocupolarotunda (cupola)

3-4 in pentagonal gyrocupolarotunda

3-4 in elongated pentagonal orthobicupola (cupola)

3-4 in elongated pentagonal gyrobicupola (cupola)

3-4 in elongated pentagonal orthocupolarotunda (cupola)

3-4 in elongated pentagonal gyrocupolarotunda (cupola)

3-4 in gyroelongated pentagonal bicupola (cupola)

3-4 in gyroelongated pentagonal cupolarotunda (cupola)

3-4 in augmented truncated dodecahedron (augment)

3-4 in parabiaugmented truncated dodecahedron (augment)

3-4 in metabiaugmented truncated dodecahedron (augment) (5 possibilities)

3-4 in triaugmented truncated dodecahedron (augment) (5 possibilities)

3-4 in gyrate rhombicosidodecahedron (7 possibilities)

3-4 in parabigyrate rhombicosidodecahedron (3 possibilities)

3-4 in metabigyrate rhombicosidodecahedron (14 possibilities)

3-4 in trigyrate rhombicosidodecahedron (9 possibilities)

3-4 in diminished rhombicosidodecahedron (6 possibilities)

3-4 in paragyrate diminished rhombicosidodecahedron (5 possibilities)

3-4 in metagyrate diminished rhombicosidodecahedron (21? possibilities -- I'm honestly not quite sure here)

3-4 in bigyrate diminished rhombicosidodecahedron (17? possibilities)

3-4 in parabidiminished rhombicosidodecahedron (2 possibilities)

3-4 in metabidiminished rhombicosidodecahedron (9 possibilities)

3-4 in gyrate bidiminished rhombicosidodecahedron (13 possibilities)

3-4 in tridiminished rhombicosidodecahedron (4 possibilities)

3-4 in bilunabirotunda

3-4 in triangular hebesphenorotunda

159.187

3-3 in decagonal antiprism

3-3 in gyroelongated pentagonal cupola (antiprism) (2 possibilities) (both asymmetrical)

3-3 in gyroelongated pentagonal rotunda (antiprism) (2 possibilities) (both asymmetrical)

3-3 in gyroelongated pentagonal bicupola (antiprism) (3 possibilities) (1 asymmetrical)

3-3 in gyroelongated pentagonal cupolarotunda (antiprism) (4 possibilities) (all asymmetrical)

3-3 in gyroelongated pentagonal birotunda (3 possibilities) (1 asymmetrical)

159.892

3-3 in sphenocorona

3-3 in augmented sphenocorona

160.529 (90 + d4)

3-4 in elongated triangular pyramid (apex)

3-4 in elongated triangular dipyramid

3-4 in elongated triangular cupola (cupola/prism)

3-4 in elongated triangular orthobicupola (cupola/prism)

3-4 in elongated triangular gyrobicupola (cupola/prism)

161.483

3-3 in sphenomegacorona

162.736 (198 - d4/2)

3-4 in augmented pentagonal prism (augment/side)

3-4 in biaugmented pentagonal prism (2 possibilities)

164.172

3-3 in snub dodecahedron (2 possibilities) (1 asymmetrical)

164.207 (270 - 3*d4/2)

3-4 in augmented truncated tetrahedron (augment/main body)

164.257

3-3 in snub square antiprism (vertical edges) (asymmetrical)

166.441

3-3 in snub disphenoid (type 3) (2 possibilities)

166.811

3-3 in disphenocingulum

169.188 (90 + dpr3)

3-4 in elongated pentagonal rotunda

3-4 in elongated pentagonal orthocupolarotunda (rotunda/prism)

3-4 in elongated pentagonal gyrocupolarotunda (rotunda/prism)

3-4 in elongated pentagonal orthobirotunda

3-4 in elongated pentagonal gyrobirotunda

169.428 (d4 + da6)

3-3 in gyroelongated triangular cupola (cupola/antiprism) (asymmetrical)

3-3 in gyroelongated triangular bicupola (cupola/antiprism) (asymmetrical)

169.471 (240 - d4)

3-3 in biaugmented triangular prism (augment/augment)

3-3 in triaugmented triangular prism (augment/augment)

170.264 (135 + d4/2)

3-4 in augmented truncated cube (augment/main body)

3-4 in biaugmented truncated cube (augment/main body)

171.341 (dpp + 2*dpc4 + d4)

3-5 in augmented tridiminished icosahedron (augment/main body)

171.646

3-3 in sphenomegacorona

171.755 (90 + dsp - d4/2)

3-4 in augmented sphenocorona

174.34 (dpc4 + did)

3-4 in augmented truncated dodecahedron (augment/main body)

3-4 in parabiaugmented truncated dodecahedron (augment/main body)

3-4 in metabiaugmented truncated dodecahedron (augment/main body) (3 possibilities)

3-4 in triaugmented truncated dodecahedron (augment/main body) (3 possibilities)

174.434 (da10 + dpr3)

3-3 in gyroelongated pentagonal rotunda (rotunda/antiprism) (asymmetrical)

3-3 in gyroelongated pentagonal cupolarotunda (rotunda/antiprism)

3-3 in gyroelongated pentagonal birotunda (rotunda/antiprism)

174.736 (210 - d4/2)

3-4 in augmented hexagonal prism

3-4 in parabiaugmented hexagonal prism

3-4 in metabiaugmented hexagonal prism (2 possibilities)

3-4 in triaugmented hexagonal prism

Of course, you must also take into account chiral polyhedra:

snub cube, snub dodecahedron, gyroelongated triangular bicupola, gyroelongated square bicupola, gyroelongated pentagonal bicupola, gyroelongated pentagonal cupolarotunda and gyroelongated pentagonal birotunda. Those might count twice in all cases.

From these, it should be possible to enumerate 4D edges.

- Marek14
- Pentonian
**Posts:**1191**Joined:**Sat Jul 16, 2005 6:40 pm

Recently I looked over Marek's proof again. It seems pretty solid, in proving that 4D vertices must be rigid, once the surrounding cells have been determined. There are simply too many constraints on vertices for there to be any additional degrees of freedom to deform the vertex.

However, I'm not so sure that this proof alone gives us a feasible computer brute force search algorithm. One significant oversight in Marek's proof is in the following statement:

The conclusion is sound, but the statement that the polygon must correspond to a finite list of valid polyhedra is inaccurate, because there are an infinite number of CRFs in 3D! Namely, the prisms and antiprisms, which are surely CRF, even though they do not belong to the set of Johnson solids. What this means it that the set of possible chord lengths is countable, but not necessarily finite. This means it's not so simple for a brute-force algorithm to merely try to enumerate all possible vertices, because such an algorithm might not terminate!

So it's true that you won't have the continuously-deformable vertices you have in 3D, but there are still potentially an infinite (albeit countable) number of possibilities. Unless you're willing to make the assumption of excluding polygons past a certain size (e.g., no polygons greater than 20 vertices will be checked), that means the number of verfs will still be infinite, and will need careful representation in order to avoid an infinite loop in the algorithm.

And we already have proof that we must at least consider polygons up to degree 20, because there exist non-trivial CRF augmentations of the 10,20-duoprism that contain elongated pentagonal bipyramid cells. They may not be crown jewels per se, but they also suggest that non-trivial CRF constructions may exist at least up to a 20-gon. Norman Johnson included prism augmentations among the Johnson solids, so arguably duoprism augmentations should not be excluded from a theoretical list of all 4D CRFs, even if we were to exclude the infinite families of CRFs. The fact that they are possible at least up to 10,20-duoprisms suggests that we may miss some actual crown jewels if we were to cut off the search for CRFs with polygons anywhere less than 20-gons. That's a lot of polygons to take into consideration, which also exponentially increases the number of possibilities a brute-force algorithm would have to check.

OTOH, this means there is the slim but exciting possibility that there may exist 4D CRF crown jewels that contain more unusual polygons, perhaps like CVP-3 heptagons, bridging 3D crown jewel cells like snub disphenoids. I suspect this is unlikely to be the case, but Marek's proof does not exclude this possibility, and the potentially infinite list of polygons (resp. prisms and antiprisms) we could choose from in order to bridge the CVP ≥3 vertices of the 3D crown jewels means there's a remote possibility that such things might actually exist.

However, I'm not so sure that this proof alone gives us a feasible computer brute force search algorithm. One significant oversight in Marek's proof is in the following statement:

So, what about vertex figures that are NOT deltahedra? Actually, there's no such thing! If the vertex figure contains a polygon with more than three sides, that polygon must correspond to one of the finite list of valid polyhedra. That means that its chords CANNOT vary freely, they must take one of a discrete set of values.

The conclusion is sound, but the statement that the polygon must correspond to a finite list of valid polyhedra is inaccurate, because there are an infinite number of CRFs in 3D! Namely, the prisms and antiprisms, which are surely CRF, even though they do not belong to the set of Johnson solids. What this means it that the set of possible chord lengths is countable, but not necessarily finite. This means it's not so simple for a brute-force algorithm to merely try to enumerate all possible vertices, because such an algorithm might not terminate!

So it's true that you won't have the continuously-deformable vertices you have in 3D, but there are still potentially an infinite (albeit countable) number of possibilities. Unless you're willing to make the assumption of excluding polygons past a certain size (e.g., no polygons greater than 20 vertices will be checked), that means the number of verfs will still be infinite, and will need careful representation in order to avoid an infinite loop in the algorithm.

And we already have proof that we must at least consider polygons up to degree 20, because there exist non-trivial CRF augmentations of the 10,20-duoprism that contain elongated pentagonal bipyramid cells. They may not be crown jewels per se, but they also suggest that non-trivial CRF constructions may exist at least up to a 20-gon. Norman Johnson included prism augmentations among the Johnson solids, so arguably duoprism augmentations should not be excluded from a theoretical list of all 4D CRFs, even if we were to exclude the infinite families of CRFs. The fact that they are possible at least up to 10,20-duoprisms suggests that we may miss some actual crown jewels if we were to cut off the search for CRFs with polygons anywhere less than 20-gons. That's a lot of polygons to take into consideration, which also exponentially increases the number of possibilities a brute-force algorithm would have to check.

OTOH, this means there is the slim but exciting possibility that there may exist 4D CRF crown jewels that contain more unusual polygons, perhaps like CVP-3 heptagons, bridging 3D crown jewel cells like snub disphenoids. I suspect this is unlikely to be the case, but Marek's proof does not exclude this possibility, and the potentially infinite list of polygons (resp. prisms and antiprisms) we could choose from in order to bridge the CVP ≥3 vertices of the 3D crown jewels means there's a remote possibility that such things might actually exist.

- quickfur
- Pentonian
**Posts:**3004**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

quickfur wrote:Recently I looked over Marek's proof again. It seems pretty solid, in proving that 4D vertices must be rigid, once the surrounding cells have been determined. There are simply too many constraints on vertices for there to be any additional degrees of freedom to deform the vertex.

However, I'm not so sure that this proof alone gives us a feasible computer brute force search algorithm. One significant oversight in Marek's proof is in the following statement:So, what about vertex figures that are NOT deltahedra? Actually, there's no such thing! If the vertex figure contains a polygon with more than three sides, that polygon must correspond to one of the finite list of valid polyhedra. That means that its chords CANNOT vary freely, they must take one of a discrete set of values.

The conclusion is sound, but the statement that the polygon must correspond to a finite list of valid polyhedra is inaccurate, because there are an infinite number of CRFs in 3D! Namely, the prisms and antiprisms, which are surely CRF, even though they do not belong to the set of Johnson solids. What this means it that the set of possible chord lengths is countable, but not necessarily finite. This means it's not so simple for a brute-force algorithm to merely try to enumerate all possible vertices, because such an algorithm might not terminate!

So it's true that you won't have the continuously-deformable vertices you have in 3D, but there are still potentially an infinite (albeit countable) number of possibilities. Unless you're willing to make the assumption of excluding polygons past a certain size (e.g., no polygons greater than 20 vertices will be checked), that means the number of verfs will still be infinite, and will need careful representation in order to avoid an infinite loop in the algorithm.

And we already have proof that we must at least consider polygons up to degree 20, because there exist non-trivial CRF augmentations of the 10,20-duoprism that contain elongated pentagonal bipyramid cells. They may not be crown jewels per se, but they also suggest that non-trivial CRF constructions may exist at least up to a 20-gon. Norman Johnson included prism augmentations among the Johnson solids, so arguably duoprism augmentations should not be excluded from a theoretical list of all 4D CRFs, even if we were to exclude the infinite families of CRFs. The fact that they are possible at least up to 10,20-duoprisms suggests that we may miss some actual crown jewels if we were to cut off the search for CRFs with polygons anywhere less than 20-gons. That's a lot of polygons to take into consideration, which also exponentially increases the number of possibilities a brute-force algorithm would have to check.

OTOH, this means there is the slim but exciting possibility that there may exist 4D CRF crown jewels that contain more unusual polygons, perhaps like CVP-3 heptagons, bridging 3D crown jewel cells like snub disphenoids. I suspect this is unlikely to be the case, but Marek's proof does not exclude this possibility, and the potentially infinite list of polygons (resp. prisms and antiprisms) we could choose from in order to bridge the CVP ≥3 vertices of the 3D crown jewels means there's a remote possibility that such things might actually exist.

I think that the part about finite list was referring to the fact that each specific vertex figure of a polygon can be deformed only in finitely many CRF ways. For large prisms and antiprisms, there is no variation at all -- they must be planar.

- Marek14
- Pentonian
**Posts:**1191**Joined:**Sat Jul 16, 2005 6:40 pm

Ah, I think I understand what you were trying to say.

Still, it's an important detail that the number of types of faces that can appear in a vertex figure (of a prospective 4D CRF's vertex) is not necessarily finite. Take an n-gonal prism, for example: for each n, you'd get a vertex figure that's a triangle whose side length depends on n. Since there's theoretically no upper bound on n, that means there's no limit on the number of different types of faces that may appear in a verf. Well, no limit as far as constraints on verf faces are concerned; of course there are limits imposed by other factors, e.g. perhaps verfs that contain triangles derived from a 17-gonal antiprism, say, may not have any way of closing up in a CRF way. And in spite of being possibly unbounded, the number of possible face types must be discrete; there's no possibility for the continuous deformation that we see in 3D verfs.

Nevertheless, this means that unless we make further assumptions, we cannot impose a finite bound on the number of different verfs that a 4D CRF may have. Which complicates any potential brute-force search algorithm, because that in turn means you can't just build a finite list of possible verfs, and just iterate over every combination of them to find CRFs. You'll have to do it in a way that won't get stuck in an infinite loop. And there may not be a way of terminating the entire algorithm: we would not know when to stop, because the algorithm wouldn't know whether there are more CRFs that may be produced by verfs of higher-order prisms / antiprisms. We would need some way of identifying, at an early stage, whether a particular combination of verfs corresponds with a CRF of a known infinite family, and we would need some way of telling whether all remaining verfs to be checked cannot belong to anything outside the known infinite families. I'm not sure, at this point, whether it's possible to programmatically determine this without human intervention (which would be infeasible, since the point is to let the computer do the work). So this is a significant obstacle to a feasible brute-force CRF search algorithm.

OTOH, perhaps the right approach isn't to try to enumerate all CRFs, but to try to search specifically for crown jewels. By your proof, I think it's pretty safe to assume that CVP 3 vertices in any 4D CRF can only come from a CVP 3 Johnson solid, so perhaps the way to go is to specifically enumerate verfs that contain one or more of the chords of a CVP 3 Johnson solid. That would cut out most of the "uninteresting" 4D CRFs from consideration, and would lead to a direct search for CRFs that contain at least one 3D crown jewel. Having said that, though, the non-finiteness problem still applies to the rest of the polytope, since we can't tell beforehand whether a starting cell complex containing a 3D crown jewel might need an unusual prism or antiprism in order to close up in a CRF way. So we would still need some way of proving that a particular combination is not possible.

In both cases, perhaps a first stab at the problem might be a carefully-chosen search order, such that we try verfs of lower-order prisms/antiprisms first, and run through combinations of those, before trying verfs containing higher-order polygons. That would allow us to temporarily terminate the algorithm, say, after exhausting verfs up to 20-gons (or maybe even 7-gons, for an initial run), and be able to say "no CRF crown jewels exist that contain polygons of degree ≤ n" for some value of n, even if we cannot say for certain whether no CRF crown jewels exist for all n.

Still, it's an important detail that the number of types of faces that can appear in a vertex figure (of a prospective 4D CRF's vertex) is not necessarily finite. Take an n-gonal prism, for example: for each n, you'd get a vertex figure that's a triangle whose side length depends on n. Since there's theoretically no upper bound on n, that means there's no limit on the number of different types of faces that may appear in a verf. Well, no limit as far as constraints on verf faces are concerned; of course there are limits imposed by other factors, e.g. perhaps verfs that contain triangles derived from a 17-gonal antiprism, say, may not have any way of closing up in a CRF way. And in spite of being possibly unbounded, the number of possible face types must be discrete; there's no possibility for the continuous deformation that we see in 3D verfs.

Nevertheless, this means that unless we make further assumptions, we cannot impose a finite bound on the number of different verfs that a 4D CRF may have. Which complicates any potential brute-force search algorithm, because that in turn means you can't just build a finite list of possible verfs, and just iterate over every combination of them to find CRFs. You'll have to do it in a way that won't get stuck in an infinite loop. And there may not be a way of terminating the entire algorithm: we would not know when to stop, because the algorithm wouldn't know whether there are more CRFs that may be produced by verfs of higher-order prisms / antiprisms. We would need some way of identifying, at an early stage, whether a particular combination of verfs corresponds with a CRF of a known infinite family, and we would need some way of telling whether all remaining verfs to be checked cannot belong to anything outside the known infinite families. I'm not sure, at this point, whether it's possible to programmatically determine this without human intervention (which would be infeasible, since the point is to let the computer do the work). So this is a significant obstacle to a feasible brute-force CRF search algorithm.

OTOH, perhaps the right approach isn't to try to enumerate all CRFs, but to try to search specifically for crown jewels. By your proof, I think it's pretty safe to assume that CVP 3 vertices in any 4D CRF can only come from a CVP 3 Johnson solid, so perhaps the way to go is to specifically enumerate verfs that contain one or more of the chords of a CVP 3 Johnson solid. That would cut out most of the "uninteresting" 4D CRFs from consideration, and would lead to a direct search for CRFs that contain at least one 3D crown jewel. Having said that, though, the non-finiteness problem still applies to the rest of the polytope, since we can't tell beforehand whether a starting cell complex containing a 3D crown jewel might need an unusual prism or antiprism in order to close up in a CRF way. So we would still need some way of proving that a particular combination is not possible.

In both cases, perhaps a first stab at the problem might be a carefully-chosen search order, such that we try verfs of lower-order prisms/antiprisms first, and run through combinations of those, before trying verfs containing higher-order polygons. That would allow us to temporarily terminate the algorithm, say, after exhausting verfs up to 20-gons (or maybe even 7-gons, for an initial run), and be able to say "no CRF crown jewels exist that contain polygons of degree ≤ n" for some value of n, even if we cannot say for certain whether no CRF crown jewels exist for all n.

- quickfur
- Pentonian
**Posts:**3004**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

In fact, expanding upon my last paragraph, perhaps a nice initial step is to set n=3, i.e., search for CRF crown jewels that contain only triangles as surtopes. This would exclude things like the snub square antiprism from consideration, but would include the snub disphenoid. The search space would be much smaller (and more feasible for brute-force search!), and would be able to tell us whether non-triangular faces are a necessary condition for 4D CRF crown jewels -- by seeing whether the algorithm finds the snub disphenoid prism as the only possibility.

After we've done that, we could do the search for n=4, which would include a much larger set of possibilities. That may turn up some more useful information before we step it up to higher values of n.

Another way of conducting the search is to limit the set of 3D cells allowed, then we would have an actual, finite set of possible verfs that can be brute-force searched. Say if we only allow tetrahedra, square pyramids, and pentagonal pyramids, then we would have quite a small number of possible verfs, so we'd be able to run through the search in a relatively short time to see if anything interesting turns up. Or if we allow said pyramids, plus snub disphenoids thrown into the mix, then we might be able to find out whether it's possible to construct a CRF containing only snub disphenoids and pyramids.

After we've done that, we could do the search for n=4, which would include a much larger set of possibilities. That may turn up some more useful information before we step it up to higher values of n.

Another way of conducting the search is to limit the set of 3D cells allowed, then we would have an actual, finite set of possible verfs that can be brute-force searched. Say if we only allow tetrahedra, square pyramids, and pentagonal pyramids, then we would have quite a small number of possible verfs, so we'd be able to run through the search in a relatively short time to see if anything interesting turns up. Or if we allow said pyramids, plus snub disphenoids thrown into the mix, then we might be able to find out whether it's possible to construct a CRF containing only snub disphenoids and pyramids.

- quickfur
- Pentonian
**Posts:**3004**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

In fact, in my search for convex segmentochora I already made a dissection into N=3, 4, 5, 6, 8, 10 at the one hand side and all other N's on the other.

--- rk

--- rk

- Klitzing
- Pentonian
**Posts:**1641**Joined:**Sun Aug 19, 2012 11:16 am**Location:**Heidenheim, Germany

Side note: I wonder if we can include the snub cube and snub dodecahedron among the 3D crown jewels, because they are CVP 3 in spite of being uniform. I would really love to see what kinds of CRFs could be constructed that contained these two polyhedra as cells (besides their obvious prisms, of course). So far, all my attempts to build a CRF out of the snub cube besides its prism have failed.

- quickfur
- Pentonian
**Posts:**3004**Joined:**Thu Sep 02, 2004 11:20 pm**Location:**The Great White North

I'm writing a program for calculating dihedral angles given face angles. I tested it on the sphenocorona, and compared the results with Marek's list; they mostly agreed. But the angle at the edge directly opposite to the two squares is 131.442°; this is in the sphenocorona itself, not only the augmented sphenocorona. And Marek's list is missing 164.2596°, for one of the three angles produced by the augmentation.

Later, while checking the hebesphenomegacorona, I got 141.3411° in place of the list's 141.31°.

Later, while checking the hebesphenomegacorona, I got 141.3411° in place of the list's 141.31°.

ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎

ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

- mr_e_man
- Tetronian
**Posts:**531**Joined:**Tue Sep 18, 2018 4:10 am

Here are pictures of the crown jewels, showing exactly which edges have which dihedral angles. This should make it easier to calculate possible 4D verfs.

Snub disphenoid:

Snub square antiprism:

Sphenocorona:

Augmented sphenocorona: Just add the square pyramid's angle arccos(1/√3)=54.7356° to the angles around a square in waco (edges 0, 1, 5) to get 171.7546446866206°, 164.2596439739962°, 152.1911326249878°. And of course include the octahedron's angle arccos(-1/3)=109.4712°. I know we probably don't need so many digits, but it doesn't hurt; there's enough space on this page.

Sphenomegacorona:

Hebesphenomegacorona:

Disphenocingulum:

Bilunabirotunda and triangular hebesphenorotunda: I didn't bother with these, since their angles are easily expressible in terms of uniform polyhedra or trivalent vertices. For example, the angle θ between the hexagon and a square in thawro is the same as if a pentagon (rather than two triangles) completed the vertex:

cos θ = (cos 108° - cos 90° cos 120°)/(sin 90° sin 120°) = -φ⁻¹/√3;

θ = 110.9052°.

Snub disphenoid:

Snub square antiprism:

Sphenocorona:

Augmented sphenocorona: Just add the square pyramid's angle arccos(1/√3)=54.7356° to the angles around a square in waco (edges 0, 1, 5) to get 171.7546446866206°, 164.2596439739962°, 152.1911326249878°. And of course include the octahedron's angle arccos(-1/3)=109.4712°. I know we probably don't need so many digits, but it doesn't hurt; there's enough space on this page.

Sphenomegacorona:

Hebesphenomegacorona:

Disphenocingulum:

Bilunabirotunda and triangular hebesphenorotunda: I didn't bother with these, since their angles are easily expressible in terms of uniform polyhedra or trivalent vertices. For example, the angle θ between the hexagon and a square in thawro is the same as if a pentagon (rather than two triangles) completed the vertex:

cos θ = (cos 108° - cos 90° cos 120°)/(sin 90° sin 120°) = -φ⁻¹/√3;

θ = 110.9052°.

ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎

ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

- mr_e_man
- Tetronian
**Posts:**531**Joined:**Tue Sep 18, 2018 4:10 am

My program is based on the following fact: If a vertex has (for example) four faces 0,1,2,3, and corresponding face angles and (exterior) dihedral angles, then this quaternion equation must be satisfied:

exp(½ iθ₀) exp(½ jθ₀₁) exp(½ iθ₁) exp(½ jθ₁₂) exp(½ iθ₂) exp(½ jθ₂₃) exp(½ iθ₃) exp(½ jθ₃₀) = ±1.

(Recall Euler's formula: exp(½ uθ) = cos ½θ + u sin ½θ, where u is anything with u² = -1, and θ is any real number (or an angle expressed in radians).)

The left side of this equation is a function of the eight angles, whose output is a quaternion. Setting this equal to ±1 is the same as setting its i,j,k components equal to 0; since all of the factors are unit quaternions, the product is also a unit quaternion, so if three of its components are 0, the fourth component must be ±1.

We can interpret quaternions as rotation operators, so this equation is saying that the composition of rotations by these angles, going around the vertex, should result in the identity transformation.

This is essentially the same situation I described here. In the Euclidean plane, the sum of edge vectors of a polygon must be 0, which gives 2 equations; and the sum of exterior angles must be 2π, which gives 1 equation; in total, we get 3 equations for the n edge lengths and n angles. On the surface of a sphere, where we have a verf of a polyhedron, we also get 3 equations relating the n edge lengths (corresponding to face angles) and spherical polygon angles (corresponding to dihedral angles). Indeed there should be (n - 3) degrees of freedom.

So, if the polyhedron has V vertices, then we get 3V equations that need to be satisfied. My program tries to minimize the sum of squares of these functions.

One advantage to this approach is that it works just as well in non-Euclidean spaces. All that's required is that the tangent space at each vertex is a 3D Euclidean space (that's basically the definition of a Riemannian 3-manifold), so quaternions can be applied; and that the dihedral angle is the same at both ends of an edge.

It should also work for non-convex or irregular-faced polyhedra.

In Euclidean space, after we've solved for the dihedral angles, we can use quaternions to calculate the orientations of edges at a vertex, and thus travel to nearby vertices, carrying orientation quaternions along, and eventually get coordinates for all vertices relative to one vertex. That's just what I did to generate the pictures. I think this can be modified to work in 3D spherical space, using 4D rotors (or pairs of quaternions).

exp(½ iθ₀) exp(½ jθ₀₁) exp(½ iθ₁) exp(½ jθ₁₂) exp(½ iθ₂) exp(½ jθ₂₃) exp(½ iθ₃) exp(½ jθ₃₀) = ±1.

(Recall Euler's formula: exp(½ uθ) = cos ½θ + u sin ½θ, where u is anything with u² = -1, and θ is any real number (or an angle expressed in radians).)

The left side of this equation is a function of the eight angles, whose output is a quaternion. Setting this equal to ±1 is the same as setting its i,j,k components equal to 0; since all of the factors are unit quaternions, the product is also a unit quaternion, so if three of its components are 0, the fourth component must be ±1.

We can interpret quaternions as rotation operators, so this equation is saying that the composition of rotations by these angles, going around the vertex, should result in the identity transformation.

This is essentially the same situation I described here. In the Euclidean plane, the sum of edge vectors of a polygon must be 0, which gives 2 equations; and the sum of exterior angles must be 2π, which gives 1 equation; in total, we get 3 equations for the n edge lengths and n angles. On the surface of a sphere, where we have a verf of a polyhedron, we also get 3 equations relating the n edge lengths (corresponding to face angles) and spherical polygon angles (corresponding to dihedral angles). Indeed there should be (n - 3) degrees of freedom.

So, if the polyhedron has V vertices, then we get 3V equations that need to be satisfied. My program tries to minimize the sum of squares of these functions.

One advantage to this approach is that it works just as well in non-Euclidean spaces. All that's required is that the tangent space at each vertex is a 3D Euclidean space (that's basically the definition of a Riemannian 3-manifold), so quaternions can be applied; and that the dihedral angle is the same at both ends of an edge.

It should also work for non-convex or irregular-faced polyhedra.

In Euclidean space, after we've solved for the dihedral angles, we can use quaternions to calculate the orientations of edges at a vertex, and thus travel to nearby vertices, carrying orientation quaternions along, and eventually get coordinates for all vertices relative to one vertex. That's just what I did to generate the pictures. I think this can be modified to work in 3D spherical space, using 4D rotors (or pairs of quaternions).

ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎

ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

- mr_e_man
- Tetronian
**Posts:**531**Joined:**Tue Sep 18, 2018 4:10 am

In the approach in the previous post, each dihedral angle adds a degree of freedom.

Here's a different way to determine the angles, using just one degree of freedom.

In this post I'll give a low-level description for two polyhedra, using trigonometric functions.

In the next post I'll give a higher-level description for all six polyhedra.

(You might want to open another copy of this page, so you can look at the pictures and the text at the same time.)

We'll start with the snub disphenoid, and a variable dihedral angle a0 at edge 0. Consider the angle a1,2 between edges 1 and 2, cutting through the polyhedron. By the law of cosines (applied to the verf),

a1,2 = arccos(cos 60° cos 60° + sin 60° sin 60° cos a0) = arccos((1 + 3 cos a0)/4);

and we could instead get a1,2 by starting with a5 (the dihedral angle at edge 5), so the two equations imply a5 = a0. The dihedral angle at edge 1 is (by considering the two halves of the verf)

a1 = 2 arccos((cos 60° - cos 60° cos a1,2)/(sin 60° sin a1,2)) = 2 arccos(1/√3 tan(1/2 a1,2)).

Going to the other end of edge 5, we see that a7,8 = a1,2. Also by symmetry a3 = a1. Now looking at the front vertex,

a3,7 = arccos(cos 60° cos 60° + sin 60° sin 60° cos a1) = arccos((1 + 3 cos a1)/4).

Introducing the notation a1,3,7 = a7,3,1 for the dihedral angle at edge 3 of the tetrahedron formed by edges 1, 3, and 7, we have

a1,3,7 = arccos((cos 60° - cos 60° cos a3,7)/(sin 60° sin a3,7)) = arccos(1/√3 tan(1/2 a3,7)),

a7,3,9 = a3 - a1,3,7,

a9,7 = arccos(cos 60° cos a3,7 + sin 60° sin a3,7 cos a7,3,9).

Now we've fully determined the shape formed by the six triangular faces on the bottom half of the polyhedron, as a function of a0. Notice that the bottom half and the top half are equivalent by a 90° rotoreflection, and for the two halves to fit together we need a9,7 = a7,8. As these two angles depend on a0, this is an equation that we can numerically solve for a0.

There's still a "loose end": the dihedral angle a7, or equivalently a9. That is gotten from

a3,9,7 = arccos((cos a3,7 - cos 60° cos a9,7)/(sin 60° sin a9,7)),

a7,9,11 = a8,7,5 = arccos((cos 60° - cos 60° cos a7,8)/(sin 60° sin a7,8)) = a2,1,5 = 1/2 a1,

a9 = a3,9,11 = a3,9,7 + a7,9,11.

That's it for the snub disphenoid.

Next is the snub square antiprism. (Sorry about the picture; the 1,5,6 triangle is hard to see.) The top half, consisting of 1 square and 12 triangles, is equivalent to the bottom half by a 45° rotoreflection.

a1 = a0

a0,5 = arccos(cos 90° cos 60° + sin 90° sin 60° cos a1) = arccos(√3/2 cos a1)

a1,0,5 = arccos((cos 60° - cos 90° cos a0,5)/(sin 90° sin a0,5)) = arccos(1/(2 sin a0,5))

a5,0,4 = a0 - a1,0,5

a4,5 = arccos(cos 60° cos a0,5 + sin 60° sin a0,5 cos a5,0,4)

a12 = arccos((cos a4,5 - cos 60° cos 60°)/(sin 60° sin 60°)) = arccos((4 cos a4,5 - 1)/3)

a16,17 = arccos(cos 60° cos 60° + sin 60° sin 60° cos a12) = a4,5

a0,4,5 = arccos((cos a0,5 - cos 60° cos a4,5)/(sin 60° sin a4,5))

a5,4,12 = arccos((cos 60° - cos 60° cos a4,5)/(sin 60° sin a4,5)) = arccos(1/√3 tan(1/2 a4,5))

a4 = a0,4,12 = a0,4,5 + a5,4,12

a11,16 = arccos(cos 60° cos 60° + sin 60° sin 60° cos a4) = arccos((1 + 3 cos a4)/4)

a11 = a4

a4,11,16 = arccos((cos 60° - cos 60° cos a11,16)/(sin 60° sin a11,16)) = arccos(1/√3 tan(1/2 a11,16))

a16,11,23 = a11 - a4,11,16

a23,16 = arccos(cos 60° cos a11,16 + sin 60° sin a11,16 cos a16,11,23)

a23,16 ≟ a16,17 (solve this for a0)

a11,23,16 = arccos((cos a11,16 - cos 60° cos a23,16)/(sin 60° sin a23,16))

a16,23,24 = a17,16,12 = arccos((cos 60° - cos 60° cos a16,17)/(sin 60° sin a16,17)) = a5,4,12

a16 = a23 = a11,23,24 = a11,23,16 + a16,23,24

Here's a different way to determine the angles, using just one degree of freedom.

In this post I'll give a low-level description for two polyhedra, using trigonometric functions.

In the next post I'll give a higher-level description for all six polyhedra.

(You might want to open another copy of this page, so you can look at the pictures and the text at the same time.)

We'll start with the snub disphenoid, and a variable dihedral angle a0 at edge 0. Consider the angle a1,2 between edges 1 and 2, cutting through the polyhedron. By the law of cosines (applied to the verf),

a1,2 = arccos(cos 60° cos 60° + sin 60° sin 60° cos a0) = arccos((1 + 3 cos a0)/4);

and we could instead get a1,2 by starting with a5 (the dihedral angle at edge 5), so the two equations imply a5 = a0. The dihedral angle at edge 1 is (by considering the two halves of the verf)

a1 = 2 arccos((cos 60° - cos 60° cos a1,2)/(sin 60° sin a1,2)) = 2 arccos(1/√3 tan(1/2 a1,2)).

Going to the other end of edge 5, we see that a7,8 = a1,2. Also by symmetry a3 = a1. Now looking at the front vertex,

a3,7 = arccos(cos 60° cos 60° + sin 60° sin 60° cos a1) = arccos((1 + 3 cos a1)/4).

Introducing the notation a1,3,7 = a7,3,1 for the dihedral angle at edge 3 of the tetrahedron formed by edges 1, 3, and 7, we have

a1,3,7 = arccos((cos 60° - cos 60° cos a3,7)/(sin 60° sin a3,7)) = arccos(1/√3 tan(1/2 a3,7)),

a7,3,9 = a3 - a1,3,7,

a9,7 = arccos(cos 60° cos a3,7 + sin 60° sin a3,7 cos a7,3,9).

Now we've fully determined the shape formed by the six triangular faces on the bottom half of the polyhedron, as a function of a0. Notice that the bottom half and the top half are equivalent by a 90° rotoreflection, and for the two halves to fit together we need a9,7 = a7,8. As these two angles depend on a0, this is an equation that we can numerically solve for a0.

There's still a "loose end": the dihedral angle a7, or equivalently a9. That is gotten from

a3,9,7 = arccos((cos a3,7 - cos 60° cos a9,7)/(sin 60° sin a9,7)),

a7,9,11 = a8,7,5 = arccos((cos 60° - cos 60° cos a7,8)/(sin 60° sin a7,8)) = a2,1,5 = 1/2 a1,

a9 = a3,9,11 = a3,9,7 + a7,9,11.

That's it for the snub disphenoid.

Next is the snub square antiprism. (Sorry about the picture; the 1,5,6 triangle is hard to see.) The top half, consisting of 1 square and 12 triangles, is equivalent to the bottom half by a 45° rotoreflection.

a1 = a0

a0,5 = arccos(cos 90° cos 60° + sin 90° sin 60° cos a1) = arccos(√3/2 cos a1)

a1,0,5 = arccos((cos 60° - cos 90° cos a0,5)/(sin 90° sin a0,5)) = arccos(1/(2 sin a0,5))

a5,0,4 = a0 - a1,0,5

a4,5 = arccos(cos 60° cos a0,5 + sin 60° sin a0,5 cos a5,0,4)

a12 = arccos((cos a4,5 - cos 60° cos 60°)/(sin 60° sin 60°)) = arccos((4 cos a4,5 - 1)/3)

a16,17 = arccos(cos 60° cos 60° + sin 60° sin 60° cos a12) = a4,5

a0,4,5 = arccos((cos a0,5 - cos 60° cos a4,5)/(sin 60° sin a4,5))

a5,4,12 = arccos((cos 60° - cos 60° cos a4,5)/(sin 60° sin a4,5)) = arccos(1/√3 tan(1/2 a4,5))

a4 = a0,4,12 = a0,4,5 + a5,4,12

a11,16 = arccos(cos 60° cos 60° + sin 60° sin 60° cos a4) = arccos((1 + 3 cos a4)/4)

a11 = a4

a4,11,16 = arccos((cos 60° - cos 60° cos a11,16)/(sin 60° sin a11,16)) = arccos(1/√3 tan(1/2 a11,16))

a16,11,23 = a11 - a4,11,16

a23,16 = arccos(cos 60° cos a11,16 + sin 60° sin a11,16 cos a16,11,23)

a23,16 ≟ a16,17 (solve this for a0)

a11,23,16 = arccos((cos a11,16 - cos 60° cos a23,16)/(sin 60° sin a23,16))

a16,23,24 = a17,16,12 = arccos((cos 60° - cos 60° cos a16,17)/(sin 60° sin a16,17)) = a5,4,12

a16 = a23 = a11,23,24 = a11,23,16 + a16,23,24

ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

- mr_e_man
- Tetronian
**Posts:**531**Joined:**Tue Sep 18, 2018 4:10 am

The law of cosines is a quantitative form of the SAS (side-angle-side) triangle congruence theorem. In spherical geometry, it states

cos(s02) = cos(s01) cos(s12) + sin(s01) sin(s12) cos(a1),

where the vertices are labelled 0,1,2, and the angles and sides are labelled accordingly.

A larger polygon can be decomposed into triangles, to which the law of cosines can be applied repeatedly. Let us define functions that give the results of such an analysis, for convex spherical 3-gons, 4-gons, and 5-gons. The input is some subset of the angles and side lengths, and the output is all the other angles and side lengths.

SAS(s01, a1, s12) = (a2, s20, a0)

SSS(s01, s12, s20) = (a1, a2, a0)

SASAS(s01, a1, s12, a2, s23) = (a3, s30, a0)

SASSS(s01, a1, s12, s23, s30) = (a2, a3, a0)

SASASAS(s01, a1, s12, a2, s23, a3, s34) = (a4, s40, a0)

SASASSS(s01, a1, s12, a2, s23, s34, s40) = (a3, a4, a0)

See also Formula for the fourth side of a spherical quadrilateral

Now let's apply this to the crown jewels.

Snub disphenoid: This has 90° rotoreflection symmetry. Starting with a0, find all the other angles in the bottom half of the polyhedron. For the two halves to fit together we need a9,7 = a7,8 ; solve this equation for a0. (Note that a9,7 is not e.g. an abbreviation for a pair of dihedral angles (a9, a7); it is a single angle, between edges 9 and 7.)

(a1, a5, a2) = SASSS(60°, a0, 60°, 60°, 60°)

(a5,7,8 , a7,8 , a7,8,5) = SAS(60°, a5, 60°)

a3 = a1

(a3,9,7 , a9,7 , a9,7,1) = SASAS(60°, a1, 60°, a3, 60°)

a9,7 ≟ a7,8

(a9, a11, a7) = SASASSS(60°, a1, 60°, a3, 60°, 60°, 60°)

Snub square antiprism: This has 45° rotoreflection symmetry. Starting with a0, find all the other angles in the bottom half of the polyhedron. For the two halves to fit together we need a23,16 = a16,17 ; solve this equation for a0.

a1 = a0

(a5, a12, a4) = SASASSS(60°, a0, 90°, a1, 60°, 60°, 60°)

(a12,16,17 , a16,17 , a16,17,12) = SAS(60°, a12, 60°)

a11 = a4

(a11,23,16 , a23,16 , a23,16,4) = SASAS(60°, a4, 60°, a11, 60°)

a23,16 ≟ a16,17

(a23, a24, a16) = SASASSS(60°, a4, 60°, a11, 60°, 60°, 60°)

Sphenocorona: Starting with a0, find all the other angles in the polyhedron minus the top two triangles (i.e. omit edge 21). Calculate a17,19 , and set this equal to 60° to solve for a0.

(a1, a7, a3) = SASSS(90°, a0, 90°, 60°, 60°)

(a9, a13, a5) = SASSS(90°, a1, 60°, 60°, 60°)

a11 = a9

(a19 , a19,17 , a17) = SASASAS(60°, a9, 60°, a7, 60°, a11, 60°)

a19,17 ≟ 60°

a14 = a13

(a18, a21, a17) = SASASSS(60°, a13, 60°, a14, 60°, 60°, 60°)

Sphenomegacorona: Starting with a0, find all the other angles in the polyhedron minus the top two triangles (i.e. omit edge 27). Calculate a23,24 , and set this equal to 60° to solve for a0.

(a1, a5, a2) = SASSS(90°, a0, 90°, 60°, 60°)

(a7, a13, a8) = SASSS(60°, a5, 60°, 60°, 60°)

(a15, a19, a11) = SASASSS(90°, a1, 60°, a7, 60°, 60°, 60°)

a16 = a15

(a24 , a24,23 , a23) = SASASAS(60°, a15, 60°, a13, 60°, a16, 60°)

a24,23 ≟ 60°

a21 = a19

(a25, a27, a23) = SASASSS(60°, a19, 60°, a21, 60°, 60°, 60°)

Hebesphenomegacorona: Starting with a0, find all the other angles in the polyhedron minus the top two triangles (i.e. omit edge 32). Calculate a28,29 , and set this equal to 60° to solve for a0.

(a2, a4, a8) = SASSS(90°, a0, 90°, 60°, 60°)

a5 = a4

(a13, a18, a12) = SASASSS(60°, a4, 60°, a5, 60°, 60°, 60°)

(a20, a24, a16) = SASASSS(90°, a8, 60°, a12, 60°, 60°, 60°)

a21 = a20

(a29 , a29,28 , a28) = SASASAS(60°, a20, 60°, a18, 60°, a21, 60°)

a29,28 ≟ 60°

a26 = a24

(a30, a32, a28) = SASASSS(60°, a24, 60°, a26, 60°, 60°, 60°)

Disphenocingulum: This has 90° rotoreflection symmetry. Starting with a21 = a23, find a9 and a5 and the rest of the angles in the bottom spheno complex, and take a10 = a12 = a9 to fix those five triangles in the back. This determines half of the polyhedron. For the two halves to fit together we need a19,29 = a17,7 ; solve this equation for a21.

a23 = a21

(a23,19,29 , a19,29 , a19,29,23) = SAS(60°, a23, 60°)

(a11, a5, a9) = SASASSS(60°, a21, 60°, a23, 60°, 60°, 60°)

(a1, a0, a3) = SASSS(60°, a5, 60°, 90°, 90°)

(a9,17,7 , a17,7 , a17,7,1) = SASAS(90°, a1, 60°, a9, 60°)

a17,7 ≟ a19,29

(a17, a13, a7) = SASASSS(90°, a1, 60°, a9, 60°, 60°, 60°)

cos(s02) = cos(s01) cos(s12) + sin(s01) sin(s12) cos(a1),

where the vertices are labelled 0,1,2, and the angles and sides are labelled accordingly.

A larger polygon can be decomposed into triangles, to which the law of cosines can be applied repeatedly. Let us define functions that give the results of such an analysis, for convex spherical 3-gons, 4-gons, and 5-gons. The input is some subset of the angles and side lengths, and the output is all the other angles and side lengths.

SAS(s01, a1, s12) = (a2, s20, a0)

SSS(s01, s12, s20) = (a1, a2, a0)

SASAS(s01, a1, s12, a2, s23) = (a3, s30, a0)

SASSS(s01, a1, s12, s23, s30) = (a2, a3, a0)

SASASAS(s01, a1, s12, a2, s23, a3, s34) = (a4, s40, a0)

SASASSS(s01, a1, s12, a2, s23, s34, s40) = (a3, a4, a0)

See also Formula for the fourth side of a spherical quadrilateral

Now let's apply this to the crown jewels.

Snub disphenoid: This has 90° rotoreflection symmetry. Starting with a0, find all the other angles in the bottom half of the polyhedron. For the two halves to fit together we need a9,7 = a7,8 ; solve this equation for a0. (Note that a9,7 is not e.g. an abbreviation for a pair of dihedral angles (a9, a7); it is a single angle, between edges 9 and 7.)

(a1, a5, a2) = SASSS(60°, a0, 60°, 60°, 60°)

(a5,7,8 , a7,8 , a7,8,5) = SAS(60°, a5, 60°)

a3 = a1

(a3,9,7 , a9,7 , a9,7,1) = SASAS(60°, a1, 60°, a3, 60°)

a9,7 ≟ a7,8

(a9, a11, a7) = SASASSS(60°, a1, 60°, a3, 60°, 60°, 60°)

Snub square antiprism: This has 45° rotoreflection symmetry. Starting with a0, find all the other angles in the bottom half of the polyhedron. For the two halves to fit together we need a23,16 = a16,17 ; solve this equation for a0.

a1 = a0

(a5, a12, a4) = SASASSS(60°, a0, 90°, a1, 60°, 60°, 60°)

(a12,16,17 , a16,17 , a16,17,12) = SAS(60°, a12, 60°)

a11 = a4

(a11,23,16 , a23,16 , a23,16,4) = SASAS(60°, a4, 60°, a11, 60°)

a23,16 ≟ a16,17

(a23, a24, a16) = SASASSS(60°, a4, 60°, a11, 60°, 60°, 60°)

Sphenocorona: Starting with a0, find all the other angles in the polyhedron minus the top two triangles (i.e. omit edge 21). Calculate a17,19 , and set this equal to 60° to solve for a0.

(a1, a7, a3) = SASSS(90°, a0, 90°, 60°, 60°)

(a9, a13, a5) = SASSS(90°, a1, 60°, 60°, 60°)

a11 = a9

(a19 , a19,17 , a17) = SASASAS(60°, a9, 60°, a7, 60°, a11, 60°)

a19,17 ≟ 60°

a14 = a13

(a18, a21, a17) = SASASSS(60°, a13, 60°, a14, 60°, 60°, 60°)

Sphenomegacorona: Starting with a0, find all the other angles in the polyhedron minus the top two triangles (i.e. omit edge 27). Calculate a23,24 , and set this equal to 60° to solve for a0.

(a1, a5, a2) = SASSS(90°, a0, 90°, 60°, 60°)

(a7, a13, a8) = SASSS(60°, a5, 60°, 60°, 60°)

(a15, a19, a11) = SASASSS(90°, a1, 60°, a7, 60°, 60°, 60°)

a16 = a15

(a24 , a24,23 , a23) = SASASAS(60°, a15, 60°, a13, 60°, a16, 60°)

a24,23 ≟ 60°

a21 = a19

(a25, a27, a23) = SASASSS(60°, a19, 60°, a21, 60°, 60°, 60°)

Hebesphenomegacorona: Starting with a0, find all the other angles in the polyhedron minus the top two triangles (i.e. omit edge 32). Calculate a28,29 , and set this equal to 60° to solve for a0.

(a2, a4, a8) = SASSS(90°, a0, 90°, 60°, 60°)

a5 = a4

(a13, a18, a12) = SASASSS(60°, a4, 60°, a5, 60°, 60°, 60°)

(a20, a24, a16) = SASASSS(90°, a8, 60°, a12, 60°, 60°, 60°)

a21 = a20

(a29 , a29,28 , a28) = SASASAS(60°, a20, 60°, a18, 60°, a21, 60°)

a29,28 ≟ 60°

a26 = a24

(a30, a32, a28) = SASASSS(60°, a24, 60°, a26, 60°, 60°, 60°)

Disphenocingulum: This has 90° rotoreflection symmetry. Starting with a21 = a23, find a9 and a5 and the rest of the angles in the bottom spheno complex, and take a10 = a12 = a9 to fix those five triangles in the back. This determines half of the polyhedron. For the two halves to fit together we need a19,29 = a17,7 ; solve this equation for a21.

a23 = a21

(a23,19,29 , a19,29 , a19,29,23) = SAS(60°, a23, 60°)

(a11, a5, a9) = SASASSS(60°, a21, 60°, a23, 60°, 60°, 60°)

(a1, a0, a3) = SASSS(60°, a5, 60°, 90°, 90°)

(a9,17,7 , a17,7 , a17,7,1) = SASAS(90°, a1, 60°, a9, 60°)

a17,7 ≟ a19,29

(a17, a13, a7) = SASASSS(90°, a1, 60°, a9, 60°, 60°, 60°)

ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

- mr_e_man
- Tetronian
**Posts:**531**Joined:**Tue Sep 18, 2018 4:10 am

quickfur wrote:Side note: I wonder if we can include the snub cube and snub dodecahedron among the 3D crown jewels, because they are CVP 3 in spite of being uniform. I would really love to see what kinds of CRFs could be constructed that contained these two polyhedra as cells (besides their obvious prisms, of course). So far, all my attempts to build a CRF out of the snub cube besides its prism have failed.

For the snub cube or snub dodecahedron, solve the cubic equation s^3 - 2s = √2 or φ, and then analyze the tetrahedra formed by the chords, to get the dihedral angles.

Thanks to Wendy for pointing out these chord relations. viewtopic.php?f=25&t=2583#p28343

ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩

- mr_e_man
- Tetronian
**Posts:**531**Joined:**Tue Sep 18, 2018 4:10 am

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