some mistakes

[wintersolstice]

I recently noticed that it had been mentioned in the "wiki" that there are 82 "pyramid forms" (as far as convex regular faced polychora go) I'm guessing is was from my post of "Johnson Polychora", the problem is, is that I've made a recent discovery:

a lot of them are invalid!! (some are concave, some can't be constructed with regular faces, at least two are identical) I haven't got round to re-enumerating them yet. The discovery has indicated that "seeing if the cells can fit round an edge by finding their angles, given that all the cells have regular faces", isn't enough to prove it can be constructed with regular faces!!! I admit I have suspected this for a long time but chose to ignore it. Given that a lot of polychora on my list need re-enumerating for other reasons I've scrapped the search and starting from the beginning. (I think it's safe to assume that the 92 prisms of Johnson solids are valid)


Also in the wiki I spotted some mistakes in the element counts for the "square and hexagonal octogonaltriates" (I am still figuring powertopes out!) but the these shapes are convex so the Euler Charactistic should be "0" but on these two it's not!
the other octogonaltriates seem to have an Euler Charactistic of "0" so they're probably correct.

Also it specifies the edge length of the octogonaltriates without specifying the edge length for the "base" or "the power"
I've been trying to understand powertopes by trying to figure out how much the base is stretched and squashed, in relation to the length of the edges the "orthotope" (hyper cuboid) (when forming the duoprisms for which the powertope is the convex hull)

Plus if the power must be a brick and the even polygons are all bricks the "hexagon" would be a valid power (I'm guessing the trivial powers are the rectange, rhombus and the square) and it says that the octagon is the simplest non-trivial, shouldn't it be the hexagon, (I think I might have identified the "hexagon" of a "triangle" and a "square", and I think the "hexagontriates" might be self dual!) that's if they exist!

[Keiji]

I recently noticed that it had been mentioned in the "wiki" that there are 82 "pyramid forms" (as far as convex regular faced polychora go) I'm guessing is was from my post of "Johnson Polychora"

Yes, I made a reference (the link is on the word "including") to wherever you originally posted them.

a lot of them are invalid!! (some are concave, some can't be constructed with regular faces, at least two are identical) I haven't got round to re-enumerating them yet. The discovery has indicated that "seeing if the cells can fit round an edge by finding their angles, given that all the cells have regular faces", isn't enough to prove it can be constructed with regular faces!!! I admit I have suspected this for a long time but chose to ignore it. Given that a lot of polychora on my list need re-enumerating for other reasons I've scrapped the search and starting from the beginning. (I think it's safe to assume that the 92 prisms of Johnson solids are valid)

Oh dear, this is no good

Now that you point it out I realize how it's not sufficient, since when folding up the polytope the sides might not reach each other.
Well, I'll leave it to you, feel free to edit the wiki yourself if you want, otherwise just post your findings on the forum and I'll put them in.

Also in the wiki I spotted some mistakes in the element counts for the "square and hexagonal octogonaltriates" (I am still figuring powertopes out!) but the these shapes are convex so the Euler Charactistic should be "0" but on these two it's not!
the other octogonaltriates seem to have an Euler Charactistic of "0" so they're probably correct.

Hmm - if the square/hexagonal ones are wrong then the others are probably also wrong, since the hexagonal octagoltriate was the first one I did following an email from Bowers about how powertopes are constructed, and then did the others by analogy. While I do understand their construction I have not been able to come up with a proven-correct formula for the element counts, as any attempt gives me a headache (it is a shame there are no non-trivial powertopes in 3D!). However I have been working on finding algorithms to apply construction operations to incidence matrices, from which the element counts (and much more information) can be extracted - the ultimate goal of this project is to find a general algorithm of this form for the brick product: any powertope operation can be represented by a brick product, so this would tell us for sure the element counts of the powertopes.

Also it specifies the edge length of the octogonaltriates without specifying the edge length for the "base" or "the power"
I've been trying to understand powertopes by trying to figure out how much the base is stretched and squashed, in relation to the length of the edges the "orthotope" (hyper cuboid) (when forming the duoprisms for which the powertope is the convex hull)

The wiki has them as (1) and (1+√2), but it should really have them as (a) and (a(1+√2)) - the point being that the ratio between the edge lengths is what matters, not their absolute dimensions. FWIW, square^square produces a tesseract where the edge length of the tesseract is equal to the edge length of the squares (of course, if the squares have different edge lengths, the power is not a regular tesseract, but the analog of a cuboid) - and any other 2^2 powertope where the operands fit exactly inside those squares will also fit exactly inside that tesseract.

Plus if the power must be a brick and the even polygons are all bricks the "hexagon" would be a valid power (I'm guessing the trivial powers are the rectange, rhombus and the square) and it says that the octagon is the simplest non-trivial, shouldn't it be the hexagon, (I think I might have identified the "hexagon" of a "triangle" and a "square", and I think the "hexagontriates" might be self dual!) that's if they exist!

The hexagon is indeed a brick and therefore a valid value for P in A^P, however the octagon has diagonal symmetry (by which I mean that the axes x and y can be swapped and the shape remains the same) whereas the hexagon does not, hence why I regard the octagoltriates as the simplest non-trivial even though the hexagon has fewer sides. I have certainly considered the hexagoltriates, but haven't tried to explicitly construct one.

Anyhow, thanks for your research, I am glad to see someone is reading the wiki and noting its faults, especially if we can correct them!

[wintersolstice]

I've been meaning to mention for a while, in reference to the 82 pyramid forms I can be 99.9% certain (not 100% yet) that the following are true:

of the 82

45 are valid and distinct Johnson polychora

28 are invalid

3 are concave

2 are identical (which brings the total list up to 46)

the last 4 are inconclusive (I haven't verified them yet)

(that should add up to 82 )

[Robert May]

One thing to keep in mind about pyramids.

If you have a polytope with regular faces and you want to construct a pyramid over it with the lateral faces regular, then the following two conditions must be satisfied:
1. The base polytope must be circumscribable.
2. The ratio of its circumradius to the length of its edge must be less than 1.

[wintersolstice]

One thing to keep in mind about pyramids.

If you have a polytope with regular faces and you want to construct a pyramid over it with the lateral faces regular, then the following two conditions must be satisfied:
1. The base polytope must be circumscribable.
2. The ratio of its circumradius to the length of its edge must be less than 1.

That's what I discovered that's why I knew they were invalid!

anyway welcome to the forum