quickfur wrote:student91 wrote:And this one can be found by thawro-ing the o5o3x3o too. (the bilbiro-pseudopyramid can be found by bilbiro-ing the o5o3x3o).

a part of the o5o3x3o looks like xoxFof.....3oxoofo.....5ooxoox.....&#xt. Now a "thawro-ing" is done by deleting the o3f5o-layer of vertices, and scaling down the f-hexagons that can be found in the f3o5x-layer (three f's, together with three diagonals of o5x make an f-hexagon). This is done by changing f3o5x to x3x3o. This gives us oxFx3xoox5oxox&#xt.

I think you have a typo there, it should be oxFx3xoox5oxoo&#xt. What you wrote is non-CRF because the bottom x3x5x is too big, and requires either non-CRF lacing edges or self-intersection with the other layers.

I liked this idea. But sadly both of your descriptions look wrong in that bottom figure.

You start with f3o5x. Then you select 8 of those f3o . triangles (in fact those coplanar to a circumscribing large oct). These are connected to f . x rectangles, respectively . o5x Pentagons. Therefore these triangles indeed can be replaced by regular f3f hexagons.

But then, what would be the remainder of this still unscaled figure? YOu'll have 6 rectangles being left (in fact those coplanar to a circumscribing large cube). And the small edges of those will be connected to acute golden triangles x-f-f, while the large ones would be connected to regular triangles f-f-f (the remaining 12, not being replaced by hexagons).

Thus the figure thus described would be a variation of the rectification of the truncated octahedron. I.e. having 8 hexagons, 6 tetragons, and 24 triangles. But this figure supposedly cannot be made CRF. For sure not uniform. And esp. neither x3x3o, nor x3x5x, nor x3x5o does describe this figure!

--- rk