Interesting convex polyhedra with 1:1:√2 triangles

Discussion of tapertopes, uniform polytopes, and other shapes with flat hypercells.

Interesting convex polyhedra with 1:1:√2 triangles

For a while now, I've been wondering about an extension of the Johnson solids where 1:1:√2 isosceles triangles (I call them Q-triangles) are allowed as faces in addition to the usual regular polygons. It's a kind of reverse-dimensional-analogy from 4D, where we have (at least) 3 different kinds of triangle analogues allowed in CRF polytopes: the regular tetrahedron, the square pyramid, and pentagonal pyramid. Better yet, recently my wife bought a set of magnetic tiles for my son, having squares, triangles, and Q-triangles, so I got to actually playing with actual physical constructions of these polyhedra.

Obviously, a good number of Johnson solids can have some triangles replaced with Q-triangles and remain convex, but those are not as interesting because topologically they aren't anything new. The more interesting cases and where the Q-triangles permit new topologies not possible with equilateral triangles. Here are some cases that I found:

1) (Irregular) hexagonal bipyramid:

Thanks to the shallower angles of the Q-triangle, it actually allows 6 polygons per vertex while still remaining convex, so it's possible to construct a hexagonal pyramid out of it, albeit the hexagon is not regular so it needs to be "covered up" by a bipyramid or an elongated bipyramid in order to remain under the restriction of regular faces or Q-triangles only. Note that this particular arrangement around the apex (2 pairs of Q triangles interspersed by equilateral triangles) is only one of several options, all of which gives rise to various hexagonal bipyramids and other constructs.

The purple edges indicate where the long edges (√2-edges) are.

2) The hexo- complex from (1) is quite flexible; it can replace the corona in the sphenocorona (J86) to produce what I call a "sphenohexona":

The hexagonal cross-section is almost, but not quite, planar. The long √2 edge allows the tips of the spheno complex to meet at a common apex; whereas in J86 a slightly larger corona complex with an intervening edge is need to bridge the gap.

There's also a hebespheno variant, that I haven't constructed in digital form yet. :-)

3) Hexagonal spicamesocorona: this one is a novelty I discovered, which is quite interesting in several respects:

From the name: spica (Lat. "spike") refers to the apex pictured here, made from 6 Q-triangles with their long edges meeting at a common vertex, and closed up below from a 10-triangle corona larger than the 8-triangle corona in J86, but smaller than the 12-triangle corona in J88, so I dubbed it a mesocorona.

It has trigonal symmetry, and can be thought of as 3 pentagonal pyramids modified to have a √2 edge each, glued together around the √2 edge and closed up with a unit triangle at the bottom.

It can also be analysed as half an icosahedron (the mesocorona) glued to half of a cube: 3 square faces that have been split along their diagonals into pairs of Q-triangles so that the two pieces will close up. So kind of like a chimera of an icosahedron and a "broken" cube.

Alternatively, it can also be thought of as an icosahedron where a cluster of 4 triangles were removed, and 6 of the surrounding triangles distorted into Q-triangles and sharpened into a spike. So a "sharpened" icosahedron.

In any case, it's very pretty IMO, one of the prettiest I've found so far.

------

In addition to the above, I've also found several highly-irregular constructs with new topologies that I can't quite remember; I'll post those when I manage to reconstruct them again. (The little boy destroyed them before I could note them down, so I have to reconstruct them from memory! )
quickfur
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Re: Interesting convex polyhedra with 1:1:√2 triangles

BTW, here are the algebraic coordinates for these polyhedra:

1) Hexagonal bipyramid:

Code: Select all
<0, 0, ±H><±1, ±A, 0><±B, 0, 0>where   A = √((√17 - 1)/2)   B = (1 + √17)/2   H = √((7 - √17)/2)

Interestingly enough, these coordinates are expressible as quadratics over Q(√17). Why √17? It's anybody's guess.

2) Sphenohexona:

Code: Select all
<±1, 0, 0><±1, ±A, B><0, 0, H><±C, 0, D>where   A = 2*√(3/7)   B = 4/√7   C = 5/2   D = √7/2   H = √7

Surprisingly enough, these coordinates are simple quadratic irrationals, and C is even rational! Certifiably insane!

3) Hexagonal spicamesocorona:

Code: Select all
<0, 2/√3, 0><±1, -1/√3, 0><±A, A/√3, B><0, -2*A/√3, B><0, 2*C/√3, D><±C, -C/√3, D><0, 0, E>where   A = root of the following polynomial between 1.6 and 1.7:      15A^8 + 20A^7 - 254A^6 - 140A^5 + 1367A^4 + 144A^3 - 2768A^2 + 384A + 1472   B = 2*√((2 + A - A^2) / 3)   C = root of the following polynomial between 1 and 2:      16C^8 - 96C^7 + 24C^6 + 488C^5 - 167C^4 - 760C^3 + 96C^2 + 320C + 64   D = 2*√((-C^2 + 2*C + 2)/3)   E = D + 2*√(1 - C^2/3)Numerical approximations:   A = 1.678983956575393   B = 1.070823222398758   C = 1.396342784166336   D = 1.946933418512344   E = 3.130277186632156

(Yes, these are 8th degree irreducible polynomials. There are also minimal polynomial expressions for B, D, E, but they are 16th-degree polynomials (8th degree in the square) with large, ugly, up-to-9-digit coefficients.)
quickfur
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Re: Interesting convex polyhedra with 1:1:√2 triangles

Here's a model of the hebesphenohexona, a modification of the hebesphenomegacorona where the megacorona is replaced by a hexo complex:

The dihedral angle around the two √2 edges is quite large, 165.34°; so the Q-triangles are almost coplanar.

The coordinates are as follows:
Code: Select all
<±1, ±1, 0><±1, ±A, B><±C, 0, D><0, 0, E>where      7*A^3 - 9*A^2 - 4*A + 4                 [1 < A < 2]      49*B^6 - 430*B^4 + 1137*B^2 - 800       [1.959 < B < 1.960]      5*C^3 - 22*C^2 + 17*C + 16              [2 < C < 3]      25*D^6 - 56*D^4 + 29*D^2 - 2            [1.2 < D < 1.3]      E^6 - 22*E^4 + 120*E^2 - 32             [2 < E < 3]Approximate root values:A = 1.402663473932304B = 1.959046228847307C = 2.223041856439680D = 1.226445521577123E = 2.975183610748060

Interestingly, the polynomial for B has another root very close to B, at around 1.955, so the bounds have to be quite precise in order to select the correct root. I'm not sure why it's this way, but it was interesting to note.

It's also particularly interesting that we're "merely" dealing with 6th-degree polynomials here, 3rd degree in the squares of B, D, and E, as opposed to the polynomials for the "real" hebesphenomegacorona (J89) which are 20th degree (10th degree in the squares of several variables). So in terms of CVP, these polyhedra are definitely "simpler" in some sense than J88, J89, or J90. I'm not sure how valid the whole concept of CVP is in the first place, but there's definitely something going on with those Johnson crown jewels that seems likely to imply that non-trivial 4D CRFs that involve them as cells may not exist.

Given the number of CVP 3 polyhedra, though, like snub cube, snub dodecahedron, snub disphenoid, and a number of others, I'm wondering if there might be a way for these to somehow "complement" each other in some sense such that some combination of them might close up in a CRF fashion. It seems unlikely but I haven't ruled it out just yet. J86 is another candidate that involves "only" 4th degree polynomials, along with the augmented prisms, so there may be some hope there as well.

All this given the very big assumption, of course, that the concept of CVP actually has any validity at all, which is currently just pure conjecture.
quickfur
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Re: Interesting convex polyhedra with 1:1:√2 triangles

I think it should be easy to define CVP (or something similar) in terms of algebraic number theory, but I'm no expert on that subject.

We should probably focus on the distances between vertices, or equivalently the dot products between edge/chord vectors*, rather than the non-unique coordinates. Already that defies the term "coordinate vertex polynomial", so whatever this is should have a different name. Yet, given the distances, we can easily set up a coordinate system (perhaps with the x-axis along an edge, and the xy-plane containing an incident face; think of the Gram-Schmidt process) such that all coordinates are derivable from the distances using only addition, subtraction, multiplication, division, and square roots.

*(When I say "vector", I usually mean a geometric arrow, not a list of numbers. Of course they're equivalent after a coordinate system is chosen. But, for example, the list (8,12,6), for the cube's numbers of vertices, edges, and faces, I would not call a vector. And the list (1,sqrt2,sqrt3), for the cube's chord lengths, I would not call a vector.)

The set of distances generates an algebraic field K, which contains a bunch of smaller sub-fields, the smallest of which is the rational numbers Q. We can organize these into a partially ordered set, where F > G if F is an extension of G, meaning F contains G as a sub-field. F>G also implies that F is a linear space (often called a vector space) over G, so it has a dimension n, which is called the degree of the extension, and denoted [F : G] = n. Now consider the set of all degrees [F : G], where F and G are any sub-fields of K (possibly equal to K), F>G, and there is no field H between them F>H>G. The maximum of this set of degrees will be called the CVP of the polytope.

Do you think this is the right approach?
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mr_e_man
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Re: Interesting convex polyhedra with 1:1:√2 triangles

I can imagine a situation with, say, 4 fields, K>F>Q, K>G>Q, F not comparable to G, [K:F]=4, [F:Q]=4, [K:G]=2, [G:Q]=8. According to the above definition, the CVP would be 8, but I think it ought to be 4. Or maybe this situation can't happen anyway.

Perhaps instead we should take the maximum degree in a chain K>F>G>...>Q, then take the minimum over all such chains. In other words, we want to get from Q up to K using the smallest steps possible. That step size is the CVP.
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mr_e_man
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Re: Interesting convex polyhedra with 1:1:√2 triangles

Actually, I think your idea of using algebraic number theory and vectors may be much more useful than CVP, which I'm not even sure is actually a useful concept in the first place.

Consider this: every time you take two polytopes A, B and fold them into part of a higher-dimensional polytope P, you're essentially performing an algebraic operation on the vectors that define A and B. This operation depends on the chords of the vertex figures of the vertices of the facet at which the join happens (because the difacetal angle is determined, at least in part, by these chords). Furthermore, the join will create new distances in P, both as chords in the resulting vertex figures, and more generally as distances between vertices. Obviously, these distances will also be derived from the distances in A and B.

Now consider the algebraic number theory side of this. Suppose the chords and lengths between vertices of A can be expressed as elements of some number field F, and the chords and lengths between vertices of B can be expressed as elements of some number field G. Then the chords and lengths found in P must be expressible in some number field H, which will be some combination of F and G. (I'm not 100% sure this is entirely a linear combination, so I'm refraining from assuming that at the moment.) Let's say we continue the process of building P by adding more cells to it, until it is almost ready to be closed. The distances in the remaining gaps must therefore be expressible as elements of H. If there is a candidate polytope C whose internal measurements are elements of some other field J which contains elements outside of H, then C cannot possibly fit into the remaining gap in a way that will close up the polytope.

So let's consider a concrete examples. Let's say we take a bunch of cubes and square cupolae and triangular prisms, say, the internal measurements of which are expressible as some extension E of the rationals, something like Q(√2,√3) or some combination thereof (maybe with one or two more quadratic elements, I'm not 100% sure). E has some even degree (there are no cube roots in the extension elements). In the process of folding up a polychoron from these cells, we're performing some operations on E that may increase its degree by some integer factor n (I didn't check, but I suspect n=2 or n=4 or some such). Now consider the snub cube, which has elements that belong to some extension field F of degree 3 (or some multiple thereof), due to its connection with the tribonacci constant in some of its proportions. Since the degree of E is even, and the act of building up the polychoron only multiplies its degree by an even number, that means no combination of E can possibly represent distances in a gap that might fit a snub cube as a cell. I.e., there cannot exist a polychoron that contains 1 snub cube and the rest of the cells are cubes, square cupolae, prisms, etc..

Now consider something with even higher-degree proportions, like J89. The coordinates require 10th degree polynomials to express, and while I agree with you that the exact coordinates themselves are not unique, and therefore on their own don't mean very much, they nevertheless do form an important ratio with the chosen integer edge length (I chose edge length 2 in this case), meaning that the internal ratios of J89 will necessarily involve algebraic numbers of degree 10 (or possibly higher). This means that any polychoron built with J89 cells will have internal proportions that involve algebraic numbers of degree 10, so any resulting gaps will likely also have proportions of similar algebraic degree, so you would not be able to close it up with polyhedra that don't have such internal proportions.

There is one catch to all of this though. And that is non-rigid vertices that we find in 3D, which allows a fully-surrounded vertex to deform continuously, and thereby be able to generate arbitrary proportions in spite of the proportions in the constituent polygons. J89 itself, for example, is built from squares and equilateral triangles, but because of this continuous deformation of vertices of degree 4 and higher, it can form proportions of degree 10 in spite of squares and triangles being fully expressible in Q(√2,√3). In 4D, as far as we know, vertices are rigid, as in, once all the cells around a vertex are determined, the vertex can no longer deform continuously. If this is true (and it very likely is, even though we haven't formally proved it yet), then in 4D and higher basically we're "locked in" to the proportions of the lower-dimensional elements; we can no longer generate arbitrary proportions but can only derive proportions that belong to extension fields of the constituent subpolytopes. Then the above arguments ought to apply.
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Re: Interesting convex polyhedra with 1:1:√2 triangles

The number-systems that define the sorts of polytope seen at this site are the standard unitary-integer ones. This equates to having algebraic-integer coordinates, when the edge is set to some algebraic-integer. Getting general fractions (Q, ie F), is not something in the geometric construction by polygons. General fractions lead to non-rational angles.

A general polytope, like the cell of {8, Wa, 8/3}, where Wa is a polygon with a vertex-figure 1:1:1/sqrt(2), ie a=1/2, leads to an integer-equation with larger than unit. The general span of these includes Bb (where x is the lead coefficient) in the set-product, here B2Z4 (base-2 numbers over [1, sqrt 2]). The vertex-angle here is irrational, but the girth of polygons is precise (it's an octagon).

Rational angles have a chord (2 cos pi*m/n), that is an algebraic integer producing 1 as the determinate, or for where n is a power of p, p^{2/P^{n-1}Ø(p) } where Ø(n) is the euler totative, the polygon having p^n sides, and the value declared, is the weight of the prime in the algebraic integer, that is one power of p for each of (P^n-P^(n-1))/2 rows of the matrix.

Note that the weight of a number is independent of the size of the number-system it falls in, or whether the prime is direct or not. (A direct prime, is such that some power of it is a multiple of p, such as 1+i. An indirect prime never gives a multiple of the prime, even though it has that prime as its weight (2+i ~ 5^½ even though no power of 2+i is a multiple of 5). (2+i)^2 = 3+4i, so it solves 0 = x^2 -4x - 5.

Various nested sets do occur in the ordinary polygons. [class x can be supposed to be the span over x variables, of a prime equation]

Z5 is Z(1,ø). Z10 is Z(1,ø)(1,g), Z15 is Z(1,ø)(1,hg). are both 4th ordinal equations, are supersets of Z5, and a subset of Z(1,ø)(1,h, g,gh ) [a class-8 system]

g = sqrt(ø sqrt(5)), h = sqrt(3). A third class-4 system occurs as Z(1,f)(1,h), occurs with the complex-polytope 3(10)2 and the group o3o5o3o6z.

Of course, I am a hobbyist and physicist, not a mathematician. It's only after interacting with the web and mail-lists, that I discover that i am pretty much with the bleeding edge in what i do. So much of the terminology and symbols are different.
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wendy
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Re: Interesting convex polyhedra with 1:1:√2 triangles

More interesting polyhedra with Q-triangles:

5) Dihexocingulum: a modification of J90 by substituting the spheno complexes with hexo complexes.

It's notable that the dihedral angle between the pair of lateral triangles in front of the image, that bridge the side of the top Q-triangle to the bottom Q-triangle, is 179.342784°, which is less than 1° from being coplanar(!). There are 3 such pairs in this polyhedron.

Coordinates:
Code: Select all
<0, 0, ±H><±1, ±A, B><±A, ±1, -B><±C, 0, D><0, ±C, -D>where   A^12 - 44*A^10 + 64*A^9 + 366*A^8 - 784*A^7 - 580*A^6 + 2464*A^5 - 887*A^4 - 2256*A^3 + 2104*A^2 - 192*A - 368 = 0 [1 < A < 2]   B = √((1 + 2*A - A^2) / 2)   15*C^12 - 4*C^11 - 798*C^10 + 1228*C^9 + 13959*C^8 - 35776*C^7 - 71552*C^6 + 273920*C^5 - 5632*C^4 - 458752*C^3 - 65536*C^2 + 393216*C + 196608 = 0 [2.3 < C < 2.4]   H = B + √(3 - A^2)   D = (H^2 - B*H + C - 4) / √(3 - A^2)Numerically:   A = 1.375551957559394   B = 0.964095619524664   C = 2.361503491386901   D = 0.459947769190601   E = 2.016643391388026

There are polynomial expressions for B, D, and H, but they are pretty nasty-looking, especially the one for D, which is degree 24 (degree 12 in D^2), with 10-digit coefficients.

6) Dispicatrisphenoid (AKA trihexona):

This one is formed by two "spikes" formed by 3 pairs of Q-triangles each, joined together at the ends of the long edges, with 3 pairs of equilateral triangles filling the gaps.

It can also be analysed as 3 hexo complexes folded together in trigonal symmetry. I can't think of any direct analogue with a Johnson solid, though. This is also the last in the series of multi-hexo complexes that is still convex; 4 hexo complexes around a vertex would give it an angle defect of 0, forcing it to be flat.

Coordinates:
Code: Select all
<0, 0, ±H><0, 2*A/√3, ±1><±A, -A/√3, ±1><0, -2*C/√3, 0><±C, C/√3, 0>where   4*A^4 + 4*A^3 + A^2 - 12*A - 12         [1.4 < A < 1.5]   4*C^4 + 8*C^3 - 17*C^2 - 24*C + 24      [1.5 < C < 1.6]   3*H^4 - 6*H^3 - 17*H^2 + 16*H + 40      [2.1 < H < 2.2]Numerically:   A = 1.408466207136316   C = 1.577254986178371   H = 2.164029176813618

Interestingly enough, only degree 4 polynomials are necessary here to express the coordinates. Though perhaps that's largely irrelevant, since the concept of CVP is questionable anyway.
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Re: Interesting convex polyhedra with 1:1:√2 triangles

Here's an interesting fact about the dispicatrisphenoid AKA trihexona that I just discovered this morning: we can replace one of the hexona complexes with a spheno group, and it remains convex, but the remaining Q-triangles become coplanar and merge into squares, yielding a polyhedron with regular triangles and squares. It turns out that this is none other than the triangular orthobicupola, a Johnson solid (J27).

So now we have a direct connection with a Johnson solid: the dispicatrisphenoid is basically the result of taking J27, replacing one of its spheno groups with a hexo group, and splitting the remaining square faces into Q-triangles so that they can be bent to fit the hexo group and close up the polyhedron. Fascinating!

P.S. There is a similar connection with the hebesphenohexona, besides its relation to J89: if you replace the middle lune of the hebespheno complex with a hexo group, all the Q-triangles will become coplanar and merge into squares; the result is the uniform cuboctahedron. :-)
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Re: Interesting convex polyhedra with 1:1:√2 triangles

OK, so today I had the idea of a modification of J90 where only 1 spheno group was replaced by a hexona group. It looks like it approximately ought to close up, but when I fed the system of polynomials to the solver, it came back with polynomials of degrees 76 and 152, having absolutely gigantic coefficients (up to 88+ digits!). The solver software I'm using, Singular, was able to handle these monsters just fine, albeit with some performance hits. Another helper CAD program, Maxima, was also able to extract roots from these beasts. However, the rest of my pipeline was defeated by the huge coefficients... I'd have to rework several key components to use BigInt's and BigFloat's in order to be able to get useful coordinates out of this beast.

So I'm currently unable to confirm whether this beast actually exists, even though it looks like it ought to. It's hard to tell, though. Just because the solver found a solution does not guarantee it will be convex, and there are too many roots per polynomial to be able to sort through things by hand. In any case, even if this monster exists, its direct algebraic representation will be extremely nasty, quite unpresentable.

Makes me wonder if 4D CRFs that contain some of the rarer Johnson crown jewels as cells, if they exist, would exhibit the same tractibility issues. I usually dislike working only numerically because of the chance of roundoff errors producing misleading results, but in cases such as these it seems there are not many other alternatives.
quickfur
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Re: Interesting convex polyhedra with 1:1:√2 triangles

quickfur wrote:the act of building up the polychoron only multiplies its degree by an even number

Why is that?

quickfur wrote:There is one catch to all of this though. And that is non-rigid vertices that we find in 3D, which allows a fully-surrounded vertex to deform continuously, and thereby be able to generate arbitrary proportions in spite of the proportions in the constituent polygons. J89 itself, for example, is built from squares and equilateral triangles, but because of this continuous deformation of vertices of degree 4 and higher, it can form proportions of degree 10 in spite of squares and triangles being fully expressible in Q(√2,√3). In 4D, as far as we know, vertices are rigid, as in, once all the cells around a vertex are determined, the vertex can no longer deform continuously. If this is true (and it very likely is, even though we haven't formally proved it yet), then in 4D and higher basically we're "locked in" to the proportions of the lower-dimensional elements; we can no longer generate arbitrary proportions but can only derive proportions that belong to extension fields of the constituent subpolytopes. Then the above arguments ought to apply.

I think even if vertices are rigid, the degree could be multiplied by an odd number.

And I think there is some theorem, analogous to Cauchy's theorem for 3D Euclidean space, that convex polyhedra in 3D hyperbolic or spherical space are rigid. This would apply to our polychora vertex figures. See this, and the comments under this.
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Re: Interesting convex polyhedra with 1:1:√2 triangles

wendy wrote:The number-systems that define the sorts of polytope seen at this site are the standard unitary-integer ones. This equates to having algebraic-integer coordinates, when the edge is set to some algebraic-integer. Getting general fractions (Q, ie F), is not something in the geometric construction by polygons. General fractions lead to non-rational angles.

I went through the Platonic solids, prisms, antiprisms, and half of the Archimedean solids (I didn't check the ike/doe family), and found that the chords are indeed all algebraic-integers. Along the way, I discovered that the truncated octahedron's squared chord lengths are exactly the integers from one to ten!

But the Johnson solids provide plenty of counter-examples. The squared height of the triangular bipyramid is 8/3. In the (elongated or not) square gyrobicupola, the squared distance between vertices in the top square and the bottom square involves sqrt2/2. These are not algebraic-integers. So we are dealing with fields in general.
ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎
ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
mr_e_man
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Re: Interesting convex polyhedra with 1:1:√2 triangles

The edge is usually, but not always, one. Figures from the same symmetry can have an edge of sqrt(n+1), which is the usual role for simplexes.

The bi-triangular tegum in the equiangle form would have edges of r2 and r3, the middle axis is r2. It is a feature of symmetry to propagate a number-axis from place to place, and if this involves fractions, it comes to be infinitely dense (specifically a further C2 element, because the general integer-polynomial with term 1 in B, gives rise to a Bb (base-b) number.
The dream you dream alone is only a dream
the dream we dream together is reality.

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wendy
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Posts: 1937
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Location: Brisbane, Australia

Re: Interesting convex polyhedra with 1:1:√2 triangles

mr_e_man wrote:[...] Along the way, I discovered that the truncated octahedron's squared chord lengths are exactly the integers from one to ten!

This is not a surprise, because the truncated octahedron happens to be the same as the omnitruncated tetrahedron (3-simplex), which in turn is the same as the 3-dimensional permutohedron. A permutohedron is the convex hull of all permutations of <0, 1, 2, ... n>. In general, the n-permutohedron is the same as the (n-1)-dimensional omnitruncated simplex. To see this, note that the omnitruncated n-cube's coordinates are <1, 1+√2, 1+2√2, ... 1+n√2>. The corner facet of the omnitruncated n-cube (i.e, the facet in the direction of <1,1,1,...1>) is an omnitruncated (n-1)-simplex (all of its facets are omnitruncated (n-1)-polytopes), so its coordinates are all permutations of <1, 1+√2, 1+2√2, ... 1+n√2>, which, subtracting <1,1,1,...> and dividing by √2 (translate and scale the facet, which does not change its shape), yields the n-permutohedron.
quickfur
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Re: Interesting convex polyhedra with 1:1:√2 triangles

The surprise is not that the chord-squares are integers, but that they're consecutive integers, without any gaps. (I'm using edge length 1.) Is that a general property of permutohedra/permutotopes?
ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎
ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
mr_e_man
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Re: Interesting convex polyhedra with 1:1:√2 triangles

For higher dimensions, it is fairly common to fill in the gaps. Even the pentagonal ones are pretty good at filling in the gaps.

For the pentagonal numbers, you use base ø² = $$1\frac 12 + \frac 12\sqrt{5}$$. The digits are 0, 1, f.

For the finite polytopes, the last non-digit zero of the r² value can not be 'f'. So there is no chord f, f0, 10f, f00, f0f, etc. These numbers are alt-negative.

There is no dodecahedral chord {5}, {5,3}, {5,3,3}, {5,3,3,3}, ... that ends in 11.

The icosahedral chords then are 1, 10, 11, f1, 100, 101, 110, 111, f01, f10, f11, ff1, 1000, 1001, 1010, 1011, 10f1, 1100, 1101, 1110, 1111, f001, f010, f011, f0f1, f100, etc. The dodecahedral ones are the same, except the last two digits are not 11. This is because the alt form of the dodecahedron has a circumdiam of 1.1, and any larger number is too big for it.

Note that in {5,3,3}, you get the first eight multiples of phi^4, viz 100, f10, 1010, 1110, f0f1, ff01, 10001, and 10101 (which is the diam, edge 1).

Of course, it's hardly a supprise.

The hyperbolic chords of those tilings, relative to the edges, follow the same plan, but the actual lengths are positive-altnegatives, ie the last non-zero digit is a f.
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wendy
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Location: Brisbane, Australia

Re: Interesting convex polyhedra with 1:1:√2 triangles

quickfur wrote:OK, so today I had the idea of a modification of J90 where only 1 spheno group was replaced by a hexona group.[...]

I've solved the system!!! As it turns out, there were several bugs in my program, and also a good number of silly inefficiencies and redundant computations, all of which contributed to the failed attempt last time. This time, I managed to squeeze out usable numerical coordinates from the system. Polyview confirmed that edge lengths are as expected. So here we have it:

6) Sphenohexonacingulum:

The two blue edges are the long edges (√2-edges), showing where the 4 Q-triangles are. This is basically the intermediate between J90 and (5). As it turns out, the sphenocingulum complex is flexible enough to be able to accomodate a hexona complex, in spite of the lack of vertical symmetry.

Here are the coordinates:
Code: Select all
<0, 0, A><±1, ±B, 0><±C, 0, D><0, ±E, F><±G, ±1, H><0, ±1, J>where   A = 1.08060673360778   B = 1.35362073243635   C = 2.42268468625875   D = -0.379050387420082   E = 2.18728234995118   F = -1.51822538099011   G = 1.56710352628884   H = -1.88503208674599   J = -3.1276851106803

For completeness' sake, here are the polynomials from which these numerical values are extracted, along with the root brackets. Beware that they are hilariously high-degree and sport ludicrously huge coefficients. As described before, they are 76th degree in x or 152nd degree in x^2, and are therefore not a sight for the faint of heart. Some of the coefficients go up to 97 digits(!).

Code: Select all
#    79228162514264337593543950336*A^152 + 713053462628379038341895553024*A^150 - 112229168081534004961804803899392*A^148 + 650566891720400362241018633912320*A^146 + 52137697597736261164176421668519936*A^144 - 827383910948620976022188823307878400*A^142 - 6683981213750077788906902397015556096*A^140 + 262943735908044260293328233671549779968*A^138 - 1241655426123378369645699768841029025792*A^136 - 31235691657571220078333849336007704444928*A^134 + 453106474678886456398235855874531300212736*A^132 - 273936174159445445864673365245306018463744*A^130 - 43367223817547908591590655019237954486796288*A^128 + 369459483359141253100163121914700601124454400*A^126 + 495249855909812408931511073674336076048629760*A^124 - 28500338774209688538775842000638599350473195520*A^122 + 170401595751901340001764896845520476978460753920*A^120 + 371748555260994616505680790786997593358391574528*A^118 - 10596403451775151446223606834566498103396564729856*A^116 + 50532824131266923120167689348372033752864726712320*A^114 + 108824725449902668896297031279655676479344918659072*A^112 - 2454492672133928507342954629845980320499951338520576*A^110 + 10581593303715014626046177352043043768668389963202560*A^108 + 14069284172166282897573012743523671325156566955458560*A^106 - 367406146727290496816953829841972705390886680331419648*A^104 + 1608309088887384573588526779000128571683347730750242816*A^102 + 124856796249861037452398764909640124386626896860839936*A^100 - 34946436058545143884897226704662972539063303705850789888*A^98 + 171370647416807125487361686130873010655881874909280164864*A^96 - 198348574612038306161961897041275298842511551439447077888*A^94 - 1932337141150198543134930216913110741543206002562730259584*A^92 + 12034057435844289331546557406160875798141687542952716828992*A^90 - 26724218011947112966104059720293191805824784653084159010303*A^88 - 42289652793299020829969412734216511321687920657167970412984*A^86 + 515225830825036574557741585254539118970943866680743342291320*A^84 - 1699203534938648161457680771806302729330030195239012274491904*A^82 + 1511590936516138455303698842097480360655742461935760610157032*A^80 + 11027771064130042338562357816398591965611406079303232041507680*A^78 - 59118581091155146438060958460557691294562618184517303648660336*A^76 + 138815418867290099330006682789214886436064214897683805893308256*A^74 - 57692635524564413430734531684211699401334013009191771516619236*A^72 - 856728058804386495652622736498403346463865468333786090796715264*A^70 + 3771005805892384763548538834849938309975315060969699590940622288*A^68 - 8884811309184740945275317601380775082949347442077209676323606048*A^66 + 11086688017453739086911597804580502383410203766662485303966948872*A^64 + 6007676196049277343144015819628662678905822277840402879126506592*A^62 - 69255626826018348661373636146914007762248205435815016347965395824*A^60 + 193065043323381685449719164412869404764111563792895300735642670176*A^58 - 326139664699523241942217942133997137741777977183925653209968747194*A^56 + 294434185586383768887134217363616333801729715644960800895584418384*A^54 + 183845440091814170856931813138400810563844656934594348616485766784*A^52 - 1327790560667678757483428110963246933820791451080782470747401382304*A^50 + 2982062710260772104679920041444760340691011337294875060630048374376*A^48 - 4360770291743953108780156503442376272934716176919505896230084798944*A^46 + 4154262731269107670445204910931609903251970337805097124174108642096*A^44 - 1152176930042652166395408286075445546412448247032071975977594743264*A^42 - 4877553060072343902880292343238715170746530677902278162641693458660*A^40 + 12670399840397938284062164680171606926887757475653906355046244608832*A^38 - 19824447552056384075457762134181355296667662314529919504774888860496*A^36 + 23951683960286545959773552559594705323199032026196872261059781186464*A^34 - 23898269441231595104160160792220224154927503576305711728411337389000*A^32 + 20221180591963033628357926404530781595962026814453091955949004539168*A^30 - 14672946479100509676916706077011731945481760868145128658109998531280*A^28 + 9161525837052793693576180573350171286286208247151133570654080375008*A^26 - 4913594895426519803772407944743868441421058824583162098522807964415*A^24 + 2250036098904711166305614589398212679975592474237019396402208732456*A^22 - 870494203378047068927328519345575077269332163212995543281822373496*A^20 + 279958642440178344922484510017053817842813159784360018244653641248*A^18 - 73029061022957754296020248816041085194561213782232682454619438864*A^16 + 14859308727809273081793331450916522295931647931469599299726017792*A^14 - 2199878718854027964299260760914655753391988991240695697114634496*A^12 + 202275900512669842714959537954044703260249327950803182288696320*A^10 - 5293093173402125725339137843833699793837747700008932367323392*A^8 - 918990117746728195376184711432192208731049254006315465537536*A^6 + 56187637729168759149804876214524238426009136612681799010304*A^4 + 3404814066099039110048484757823552322386318713084688244736*A^2 + 7626362531205726622944487194005446285942361313180258304   [1.080 < A < 1.081]#    281474976710656*B^76 - 5629499534213120*B^75 + 22940210601918464*B^74 + 267401227875123200*B^73 - 2587287199598772224*B^72 - 45467004831793152*B^71 + 83122804731714469888*B^70 - 260156927290241974272*B^69 - 1071234598126124793856*B^68 + 7623062968767004803072*B^67 - 1713071695074837921792*B^66 - 101922657606678815440896*B^65 + 249716395644941557039104*B^64 + 577006544838522826653696*B^63 - 3693275096910174261084160*B^62 + 2361812729439789567180800*B^61 + 26015361525218704741171200*B^60 - 68805187232574201951682560*B^59 - 56202666271726552639602688*B^58 + 561645238115334053380489216*B^57 - 646147885625691839583174656*B^56 - 2160951970847205904221143040*B^55 + 7135862307292942127935922176*B^54 - 469513992118409082661912576*B^53 - 33834780687989018142485002752*B^52 + 54121432768419052189277829120*B^51 + 65841544801139772291189310976*B^50 - 318631764186455147677471521792*B^49 + 195213881460749164906654450624*B^48 + 914991698784129609766756544768*B^47 - 1937378195143073900751575919008*B^46 - 532003073301189604668196149440*B^45 + 7045581775727649084933281873057*B^44 - 7359883253373184195268415089388*B^43 - 12394754282662507745185921777470*B^42 + 37036094605344423985493926291908*B^41 - 7652234442488264006759598518737*B^40 - 95449215681630516173091510829016*B^39 + 122713308823162411556384773248956*B^38 + 127628383845188862702311493741928*B^37 - 428966332607212342293886143252849*B^36 + 60497216266074559019841959631044*B^35 + 938003348563474545432495010638210*B^34 - 798105841252578274457559259352684*B^33 - 1407855845419166376657203442327743*B^32 + 2337516949389277750303042986411776*B^31 + 1318032008017967969460153666688960*B^30 - 4494895272255354652069456903629312*B^29 - 192577799878753701893741374869632*B^28 + 6496857367434099372888469252087808*B^27 - 1875341565937573597491996075787264*B^26 - 7363462414446847467236069452861440*B^25 + 4038613971611800660046742091586560*B^24 + 6630765443692010220575981298188288*B^23 - 5212747234434537018760886546743296*B^22 - 4733333647269406969209039684501504*B^21 + 4902670078997763221247802636140544*B^20 + 2637620585242631433897242027098112*B^19 - 3526683465300663499923182142357504*B^18 - 1110352033758567790445345607516160*B^17 + 1956598453741323435069654721495040*B^16 + 330506336236132070277667415392256*B^15 - 825593870721194331843828629110784*B^14 - 59238095739378391345015700848640*B^13 + 254942270656693817932430643822592*B^12 + 2938167275913464679767153311744*B^11 - 52780876603024478656855405494272*B^10 + 565655307802876779527984381952*B^9 + 5553662420847886563084249071616*B^8 + 308047935710300870597901549568*B^7 + 266042554199788357709270614016*B^6 - 194573680505296139697036197888*B^5 - 155936576211820054875534786560*B^4 + 36878616801497919916961955840*B^3 + 15134837619591487105592721408*B^2 - 2493504132905103826651971584*B - 242845922085090878887559168   [1.35 < B < 1.36]#    2288725400390625*C^76 - 91374636937500000*C^75 + 1309228154043750000*C^74 - 3686436598747500000*C^73 - 114686513001613125000*C^72 + 1365014595979392300000*C^71 - 1884004149397676670000*C^70 - 72084798944374677372000*C^69 + 532353857139290345478300*C^68 + 372848385999654736328160*C^67 - 22996795611197961660521232*C^66 + 93948164329033342074195360*C^65 + 305078308050619541847658824*C^64 - 3849923604212856863476475808*C^63 + 7283543577257292805889209680*C^62 + 59621753526937454273796688672*C^61 - 356106676376898778368288674170*C^60 + 91451049554390909664638332576*C^59 + 5675917981490201890485132248656*C^58 - 18868941807033329147293728286624*C^57 - 23375588783744831924734913778872*C^56 + 308352549589555328501825950306720*C^55 - 578769968739897253807515600586512*C^54 - 1773535335624598989781805998978848*C^53 + 10343844955890646654784455703943836*C^52 - 9873851834025050359544593621390176*C^51 - 62906273055551552813776349140575280*C^50 + 226236018104434409409931543369519072*C^49 - 71079993338635105134953444860771720*C^48 - 1340589370904742953909823842372044256*C^47 + 3352537140730627308926901870763798000*C^46 + 516563359523198697171355232065925216*C^45 - 18559727731682004010983121518547746079*C^44 + 34146677119327820780964903450761358400*C^43 + 16795154428365077685610876425456444032*C^42 - 168470281000523238117529442016106000384*C^41 + 237492069173356795094482844611001398784*C^40 + 148154093245273828678086090393742229504*C^39 - 947352979282057910861831256959653388288*C^38 + 1168798747566770363657853799384906334208*C^37 + 243783538796791562859325499313312219136*C^36 - 2808042239529091440828533974339531309056*C^35 + 4957322913320351634558503999740741943296*C^34 - 5049823900349181519954426544795028553728*C^33 - 1554774178913554952318346451616681426944*C^32 + 20724683597992536951337518711541382250496*C^31 - 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1522952507636112428208688365091880331677620821935361000538957582472405760*J^58 - 4933738287509712972458981528964479132465842362948423420636571192904060544*J^56 + 9393717951791379475513368966277928931367355618157555200666050279108137984*J^54 + 31348177493065736732464782137563884137186808152080372778859142502895803392*J^52 - 41180772611141662271975284828414968483215499402550503995277557540524572672*J^50 - 182278488061671291370114976963512982974375934844973437683116052322182743040*J^48 + 105055333688117010059755612523860321842704889282673802253201845950325800960*J^46 + 888845111696911128444653426135966387888404610731095136577317256665741312000*J^44 + 161424950252140778434323607227087280077379215853275293997314697336746999808*J^42 - 3162928716762824955013203057255666612751948304889885639687916850098094702592*J^40 - 3275502339664748566513291489492283268740849393784792294156940475258563198976*J^38 + 5349774046326111838934283898629591912323040607416208680969210284970303553536*J^36 + 12822357905233373401285157643195443943120119426557106118987224169672923938816*J^34 + 5124491455344715528961535654825962616250265771394805904387676163332630773760*J^32 - 9994921020172236527186328168951634586674288468192685590039454043609306759168*J^30 - 13561462708253868918523159358113545896427391457092040327734710656952562614272*J^28 - 4271029740958885345851483309296445880842085786923869765738971413992244772864*J^26 + 4719025155398711451719622974622010030623919115567297728752803788564084031488*J^24 + 6231097477101206192807082265126098237153841416570995534158836350728091467776*J^22 + 3620536496427089743465365074649127872199863380880304506873146520046426652672*J^20 + 1317026318341117423933247913034896268904102665244245726504778729622620602368*J^18 + 332986933686923799049248245711689283440505242076293215790314243102473191424*J^16 + 54733254769942030402368964911319661923167202430797298587806817506516008960*J^14 - 352231893236691163681822616312134178401988626885086707620603523290365952*J^12 - 3416861871089229370315880999226358424888708607087397870601718317157711872*J^10 - 674827655006035921176091174273539712695607162677217521130099287263281152*J^8 + 66360812079865509029300851268926343725429251514546932510532187405156352*J^6 + 683064815820596109330436278803974403403130372751033014640825459539968*J^4 - 73523340591661249405200124709722097445434526813651427329420191334400*J^2 + 562354198139683355573410799245520114347260164865269094129174839296   [-3.13 < J < -3.12]

I really don't know what to say about this. The closely-related dihexocingulum involves "merely" 12th degree polynomials in x (24th degree in x^2), and the Johnson solid J90 also "merely" involves 12th degree polynomials in x (also 24th degree in x^2). But this seemingly-innocuous in-between hybrid sees an explosion of polynomial degree and coefficient sizes unlike anything I've seen so far. The lack of symmetry really launched the complexity of the system into outer space.

Note especially the tight bounds on the value of A; this is because the polynomial has roots that are very close by, so at least 3 digits after the decimal point are necessary to disambiguate the correct root. This shows the trickiness of working with these beasts; just a teeny bit off, and you're in a different (and wrong) local minima for the system.

P.S. And just in case you were wondering, these polynomials are all irreducible, and furthermore were factored from even higher-degree polynomials output by the polynomial solver I'm using (Singular). So yeah. There's some nasty inherent complexity going on here that cannot be simplified.
quickfur
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Re: Interesting convex polyhedra with 1:1:√2 triangles

Here's a more irregular polyhedron with Q-triangles:

7) Hexiscindosphen (9 vertices, 21 edges, 14 faces):

It can be considered to be the result of pasting a hexi- complex (a hexagonal pyramid with 2 Q-triangles) with a spheno complex whose square faces have been split into pairs of Q-triangles (hence, scindo- (Lat. "split"), referring to the split square faces, + -sphen). The dihedral angle between the Q-triangles of a split square face is about 167.5°, so about 12.5° from being coplanar. This splitting is necessary to accomodate the non-symmetric shape of the hexi complex, which requires one end of the spheno complex to dip downwards and the other up to rise upwards in order to close up the polyhedron. The hexi complex is slightly "fatter" on the side away from the Q-triangles, so the lateral edges of the split spheno complex are not parallel (otherwise the right angles would force its respective Q-triangles to be coplanar and merge into squares).

As far as naming is concerned: there are 4 possibilities of a degree-6 vertex involving Q-triangles and equilateral triangles: (1) 1 pair of Q-triangles, which I'm calling hexi-; (2) 2 pairs of opposite Q-triangles, which I'm calling hexo- or hexona; (3) 2 adjacent pairs of Q-triangles, which I'm tentatively calling hexa-; (4) 3 pairs of Q-triangles around a vertex, forming a spike-like configuration, which I call spica-.

Anyway, here are the coordinates:
Code: Select all
<A, 0, B><C, 0, D><E, ±F, 0><G, ±H, 0><J, 0, K><L, 0, M><0, 0, N>where   A = -0.981600318420614   B = 1.22637059157515   C = 1.00542822560813   D = 1.45378555417659   E = -0.841242484580385   F = 1.57363110372136   G = 1.14795967730473   H = 1.36608650814826   J = -1.96227359712285   K = -0.516695649403492   L = 2.58994610583885   M = 0.233415492584973   N = -0.903325097369556

Yeah, there were 13 unknowns in the polynomial system I used to find these coordinates. It took Singular >3 hours to compute the solution polynomials. Well, actually, technically it "only" takes ~15 mins to solve the system; the 3 hours is because the helper program I wrote does things the stupid way by recomputing the system once per unknown, so that's 15 mins multiplied by 13. I really should fix this... but in the meantime, this is what I've got to work with.

Anyway, the solution polynomials involve 19-digit coefficients, so I decided to just go with the numerical solution instead. But regardless, the polynomials are posted below for those who do not fear for their eyes. Interestingly enough, the polynomials have degree 28 in x^2 (oddly, there are none with degree 14 in x), and there are 14 faces (7 if you discount the bilateral symmetry). This is probably just pure coincidence, but I can't help thinking there's something more to it, since this is the first time I've encountered solution polynomials whose degrees are divisible by 7. But then again, the solution for sphenohexonacingulum involves polynomials of degree 152 // 76 == 2^n*19, and 19 doesn't appear to have any direct connection with surface element counts, so perhaps I'm just reading too much into this. That aside, here are the polynomials in all their g(l)ory:
Code: Select all
#   6946371*A^28 - 620628696*A^26 + 7165196248*A^24 + 37828191094*A^22 + 508232506170*A^20 - 655671626230*A^18 - 2874019751637*A^16 - 90652092262*A^14 + 10277611312945*A^12 - 5802293017386*A^10 - 5737753237865*A^8 + 5445125953376*A^6 - 1205092086912*A^4 + 87493478400*A^2 - 1866240000   [-0.99 < A < -0.98]#   48624597*B^28 + 4532322906*B^26 + 18678729181*B^24 - 748883023456*B^22 + 5357203184256*B^20 - 19624321715328*B^18 + 44298179428864*B^16 - 66328586590208*B^14 + 68012437147648*B^12 - 48086301048832*B^10 + 23102657593344*B^8 - 7253379252224*B^6 + 1381025972224*B^4 - 138680467456*B^2 + 5435817984   [1.2 < B < 1.3]#   62517339*C^28 - 462527226*C^26 - 4856723423*C^24 + 29423236866*C^22 - 6036759219*C^20 - 391133436094*C^18 + 2413988507114*C^16 - 9324065757012*C^14 + 22445398009791*C^12 - 32193312447328*C^10 + 25113105587968*C^8 - 6362158268416*C^6 - 5041303846912*C^4 + 4287719538688*C^2 - 966367641600   [1.005 < C < 1.006]#   437621373*D^28 - 17365092768*D^26 + 300870197536*D^24 - 2959727171712*D^22 + 17798504304384*D^20 - 63205240428544*D^18 + 91719054651392*D^16 + 257634704654336*D^14 - 1764727613227008*D^12 + 4635185885741056*D^10 - 6911347543506944*D^8 + 5904450992472064*D^6 - 2493171875446784*D^4 + 269822730436608*D^2 + 37950331027456   [1.45 < D < 1.46]#   173659275*E^28 + 8467134582*E^26 + 41845781361*E^24 + 4651995410*E^22 - 339647503699*E^20 + 20123688538*E^18 + 979171021354*E^16 - 516014697220*E^14 - 1320763244377*E^12 + 2035179659016*E^10 - 1252171770088*E^8 + 399411439072*E^6 - 65330028432*E^4 + 4413844224*E^2 - 18662400   [-0.85 < E < -0.84]#   1575*F^28 - 125696*F^26 + 3859112*F^24 - 64620000*F^22 + 681552624*F^20 - 4885857408*F^18 + 24867263232*F^16 - 92066064384*F^14 + 250557112320*F^12 - 500538933248*F^10 + 723756449792*F^8 - 734557896704*F^6 + 493782827008*F^4 - 195957882880*F^2 + 34359738368   [1.5 < F < 1.6]#   809310879744*G^28 - 6254667497472*G^26 - 96861921714176*G^24 + 1560844937729024*G^22 - 9253237638360400*G^20 + 30242165864199080*G^18 - 62342737807167733*G^16 + 96152350352838118*G^14 - 156399285396499199*G^12 + 303749521653489490*G^10 - 512315048955128635*G^8 + 602914350881321082*G^6 - 450666762644180486*G^4 + 191105650786183452*G^2 - 34571631968477289   [1.14 < G < 1.15]#   7340032*H^28 - 171180032*H^26 + 1681383424*H^24 - 9188598784*H^22 + 31107320944*H^20 - 68645848040*H^18 + 101017560071*H^16 - 99714390072*H^14 + 65806807696*H^12 - 28960546304*H^10 + 8551297280*H^8 - 1683855360*H^6 + 213610496*H^4 - 16056320*H^2 + 524288   [1.36 < H < 1.37]#   771819*J^28 - 10985904*J^26 + 48806758*J^24 - 112223690*J^22 + 1000981605*J^20 - 9798493102*J^18 + 60619660149*J^16 - 256506698390*J^14 + 677749911731*J^12 - 950861542512*J^10 + 434423312280*J^8 + 361079412736*J^6 - 399558083792*J^4 + 118699868800*J^2 - 10899360000   [-1.97 < J < -1.96]#   1800911*K^28 + 32639622*K^26 + 80794361*K^24 - 438436182*K^22 + 549157723*K^20 - 969388402*K^18 + 1504993498*K^16 - 1003184754*K^14 + 307136864*K^12 - 108523308*K^10 + 68806979*K^8 - 24951000*K^6 + 4118232*K^4 - 270432*K^2 + 3888   [-0.52 < K < -0.51]#   3161370624*L^28 - 188809893888*L^26 + 4836805254400*L^24 - 69183270040832*L^22 + 598309100102224*L^20 - 3047143628314200*L^18 + 5778096552165395*L^16 + 44876281203292400*L^14 - 495672940747609028*L^12 + 2436826645106101760*L^10 - 6480595526215811072*L^8 + 8340886307106455552*L^6 - 7294897311820283904*L^4 + 15038868256557367296*L^2 - 746542006980837376   [2.5 < L < 2.6]#   7376531456*M^28 - 19316680704*M^26 + 10033117440*M^24 - 30206178560*M^22 + 138339289872*M^20 - 197166000360*M^18 - 96251973225*M^16 + 525540172518*M^14 - 388840126293*M^12 + 86351682088*M^10 + 650046672*M^8 - 1103251200*M^6 + 74036480*M^4 - 1677312*M^2 + 12288   [0.23 < M < 0.24]#   5402733*N^28 - 134799186*N^26 + 870934351*N^24 + 3512821234*N^22 - 72890031501*N^20 + 358063409738*N^18 - 435177472130*N^16 - 2897970166308*N^14 + 15712625619017*N^12 - 37636736759416*N^10 + 52908903816944*N^8 - 45862477866240*N^6 + 24021154551552*N^4 - 6947440392192*N^2 + 848521007104[-1.0 < N < -0.9]

Definitely not for the fainthearted, though a lot tamer in comparison with those insane 152nd degree polynomials with 98-digit coefficients in sphenohexonacingulum!
quickfur
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Re: Interesting convex polyhedra with 1:1:√2 triangles

After the irregular hexiscindosphen, I decided to try something tamer:

(8) Dihexoluna (10 vertices, 23 edges, 15 faces):

This one is made by two hexo complexes and a luna, pasted together in a triangular configuration. The dihedral angle at the long edges is approx 154.88°, not too extreme, but getting close to being coplanar. Inserting another luna between the hexo complexes on the opposite side does make it coplanar, resulting in the uniform cuboctahedron.

Due to the comparatively high degree of symmetry, the coordinates are much tamer than 6) or 7):
Code: Select all
<±1, ±1, 0><±A, 0, B><0, ±C, D><±1, 0, E>where   5*A^3 - 19*A^2 + 40             [2.3 < A < 2.4]   25*B^6 + 21*B^4 - 64*B^2 + 4    [1.0 < B < 1.1]   4*C^3 - 15*C^2 + 8*C + 8        [1 < C < 2]   16*D^6 - 55*D^4 + 28*D^2 + 4    [1.6 < D < 1.7]   E^6 - E^4 - 41*E^2 + 25         [2 < E < 3]Numerically:   A = 2.34418544136513   B = 1.09232115205283   C = 1.47517871197982   D = 1.66559454600488   E = 2.57325514761847

Yep, only degree 3 in x and degree 6 in x^2. In spite of being CVP 3, it's still very tame compared to the degree 152 insanity of sphenohexonacingulum.
quickfur
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Re: Interesting convex polyhedra with 1:1:√2 triangles

quickfur wrote:
quickfur wrote:OK, so today I had the idea of a modification of J90 where only 1 spheno group was replaced by a hexona group.[...]

I've solved the system!!! As it turns out, there were several bugs in my program, and also a good number of silly inefficiencies and redundant computations, all of which contributed to the failed attempt last time. This time, I managed to squeeze out usable numerical coordinates from the system. Polyview confirmed that edge lengths are as expected. So here we have it:

6) Sphenohexonacingulum:

It's interesting (but perhaps expected?) that the polynomials bulge outward rather smoothly, with smaller coefficients at the ends, and smallest at the highest-degree term.

Note that the coordinates B,C,E,G are actually chords, and they have simpler polynomials than A,D,F,H,J.
ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎
ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
mr_e_man
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Re: Interesting convex polyhedra with 1:1:√2 triangles

Do you know of accurate numerical algorithms for solving multi-variable systems directly, without reducing them to single-variable equations and exploding the polynomial sizes? I can see a few possible ways to generalize Newton's method.

And what do you think of the rigidity of vertices in 4D? Do you agree that it's proven now? viewtopic.php?f=25&t=2475#p27387
ΓΔΘΛΞΠΣΦΨΩ αβγδεζηθϑικλμνξοπρϱσςτυϕφχψωϖ °±∓½⅓⅔¼¾×÷†‡• ⁰¹²³⁴⁵⁶⁷⁸⁹⁺⁻⁼⁽⁾₀₁₂₃₄₅₆₇₈₉₊₋₌₍₎
ℕℤℚℝℂ∂¬∀∃∅∆∇∈∉∋∌∏∑ ∗∘∙√∛∜∝∞∧∨∩∪∫≅≈≟≠≡≤≥⊂⊃⊆⊇ ⊕⊖⊗⊘⊙⌈⌉⌊⌋⌜⌝⌞⌟〈〉⟨⟩
mr_e_man
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Re: Interesting convex polyhedra with 1:1:√2 triangles

mr_e_man wrote:[...]
It's interesting (but perhaps expected?) that the polynomials bulge outward rather smoothly, with smaller coefficients at the ends, and smallest at the highest-degree term.
[...]

This isn't always the case, but generally yes, that's expected. It's a similar phenomenon to binomial expansions.

mr_e_man wrote:Do you know of accurate numerical algorithms for solving multi-variable systems directly, without reducing them to single-variable equations and exploding the polynomial sizes? I can see a few possible ways to generalize Newton's method.

Actually, reduction to univariate polynomials is just a matter of presentation. Note that multi-variable linear systems are "trivial" to solve with Gaussian elimination, for which there are plenty of implementations out there. The trouble with many polyhedral / polytopal constructions is that they involve non-linear systems. Mostly quadratic systems, but mathematically speaking, quadratic systems are equivalent to polynomial systems of arbitrary degree (because you can always introduce a new variable y and add the equation y = x^2, so quadratic systems can encode polynomial equations of arbitrary degree).

Polynomial systems are in general very difficult to solve, even when the starting equations appear to be "simple quadratics" (which is the case with most polytopal constructions). Unlike linear systems, where the complexity of the solution generally is proportional to the complexity of the input system, polynomial systems can have solutions that are exponential (or worse) in complexity relative to the input system.

Generally, solution proceeds by first computing a so-called Gröbner basis, which is the polynomial equivalent of row-echelon form for linear systems, then extracting solutions from there using various known transformations. Here, "solution" can mean different things; usually it means a presentation from which it is relatively simple to compute numerical approximations to any desired accuracy. I chose univariate polynomials because in simple cases, it's the most concise way of presenting the solution that allows extraction of numerical values to arbitrary precision. But as you can see, sometimes the univariate representation is not very useful, in that the size of the solution is impractically large relative to the size of a numerical approximation! Generally, coefficient sizes can be exponential in the complexity of the input. It's not the only representation, however. There's something called a rational univariate representation, in which you introduce an extra variable that has a polynomial relationship with the input variables, and express the solution in terms of this variable. Generally, the coefficient sizes are linear in the complexity of the solution, rather than exponential. I've yet to find an easy implementation of this, however, because choosing the right variable to introduce is not a trivial task, and not easily automated; and the size of the solution does depend very much on the choice of this variable.

There is also the approach of purely numerical solutions, where you start by inputting a set of guessed solution values, which is then iteratively refined using the defining equations. This approach has drawbacks, however. Polynomial systems generally are not "well-behaved" like linear systems; polynomials can have multiple solutions, and if you start with the wrong set of initial values, it may "miss" the solution, diverge, or if there are multiple solutions it may find any arbitrary one without any indication of the existence or number of other solutions. Polynomial systems also tend to be numerically unstable, meaning an iterative system may accumulate roundoff errors that either overwhelm the algorithm (in the extreme case, produces nonsensical values), or else converge too slowly to be of practical use. These problems apply not just to polynomial systems; even pure Newton's method on a function of a single variable may fail to converge, or hard to control where it converges in the case of multiple solutions, depending on what initial values you give it, and how the function behaves around the initial value.

One possible hybrid approach is to start with the Gröbner basis computation, but instead of outputting the solution polynomials as-is, use a polynomial root solver to extract numerical values to the desired precision and output that instead. That's what I've chosen with sphenohexonacingulum. This way, you let the computer deal with the crazy huge coefficients behind the scenes, and give you nice decimal approximations to any desired accuracy.

Of course, the Gröbner basis method / polynomial presentation presumes a reliable polynomial root-finding algorithm. Since polynomials generally have multiple roots, after the Gröbner basis is computed you still have the question of which polynomial root represents a solution. For this, you need a reliable algorithm for separating the roots and any potential multiplicities so that you can apply a numerical algorithm to reliable extract the numerical values (otherwise you run into the multiple root problem where something like Newton's method may fail to converge, or converge to the wrong root). Fortunately there are standard methods here that are widely used. One approach I know of is to compute a Sturm sequence to bracket the polynomial's roots, then use a numerical root-finder within each bracket to extract the value of the root.

For my purposes, I use Singular for the Gröbner basis computation (for 0-dimensional polynomial systems, it implements the super-fast FGLM algorithm, which first computes the Gröbner basis using an "easy" monomial ordering, then transforms it into lexicographic ordering as a second step -- computing Gröbner basis in lex ordering is generally very expensive, very slow, and often super-exponential in complexity). Then I use Maxima to compute polynomial roots -- it supports arbitrary-precision floats; although the default settings only give about 5-6 digits' accuracy, with correct configuration it can compute the roots to any number of digits. AFAIK Maxima uses Sturm sequences to bracket roots, then some kind of root-finder to compute the values, probably Newton's method or one of the hybrid methods like Brett's method. Funnily enough, Maxima internally works with rational approximations, but that's not a problem since with the right configuration it will automatically convert that to a decimal approximation up to the desired precision.

So yeah, polynomial systems are not easy to work with. Even a seemingly simple-looking system can explode in complexity, like we see here with sphenohexonacingulum.

And what do you think of the rigidity of vertices in 4D? Do you agree that it's proven now? viewtopic.php?f=25&t=2475#p27387

I only glanced at the posts, didn't follow through in detail. But it does look like rigidity in all likelihood does apply to 4D and above. (But probably only in the convex case -- see below.)

Which means in 4D and above, you won't find the kind of arbitrary flexibility with degree ≥4 vertices that you find in 3D. So any CRF crown jewels in 4D would have to come from other directions, or be based on 3D crown jewels. Which also means that the set of possibilities of 4D CRF crown jewels is discrete, and may be amenable to brute-force search. Though I'm skeptical about the feasibility of brute-force set without further refinement / restrictions; given the 300+ million CRFs generated by non-adjacent diminishings of the 600-cell alone, some restriction of scope will be necessary otherwise the algorithm may just get stuck enumerating 600-cell diminishings. One approach, given the rigidity theorem, is to start with a 3D crown jewel and then try to disprove the possibility of closing up in a CRF way (except for the trivial prism). Failure to disprove it may reveal some new avenues of construction that might potentially lead to undiscovered 4D crown jewels.

Somebody in this forum brought up years ago that we can use the rigidity theorem to pre-compute the set of all vertex figures that can be made from the 3D CRFs. Once we exclude certain known infinite families, this set should be finite. Once we have this set, we can brute-force all CRFs simply by enumerating all possible combinations of vertex figures, or use it as a filter to prune parts of the search tree when a partially-constructed vertex sports a combination of cells that isn't a subset of any member of this set. We can also start with a vertex figure that contains a 3D crown jewel, and iteratively try to find adjacent vertex figures that might eventually close up the polytope.

On an interesting side note, the rigidity theorem I think only applies to convex polytopes. In 3D, there are certain non-convex polyhedra that have rigid faces but the polyhedron itself can deform continuously. Although the first such example was self-intersecting, it's possible to construct a similar polyhedron in a non-self-intersecting way. Now imagine building a 4D polytope out of these polyhedra as cells. That may lead to 4D continuously-deformable vertices that can produce crown-jewel-like constructions -- albeit this would no longer be CRF since (some of) the cells would be non-convex.
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