Also in Bill/Julian's calculation, you can see the double, triple, and quadruple points on the "peanobert" Z-function with steps 1/72. So this appears consistent with Wunderlich F.5, nice! As for nomenclature, the argument against the name "peanobert" is that Hilbert's curve has linear inflation factor 2, incommensurate with 3. If a "peanobert" curve were to exist, a more likely setting would be a replacement grid of one square to 6-by-6. The argument for the name "peanobert"? I don't know. --Brad On Tue, Sep 24, 2019 at 10:24 AM Brad Klee <bradklee@gmail.com> wrote:
Isn't the valency of a multi-point limited by the valency of vertices in the unit cell tiling? Square unit cells seem to imply that quadruple points have the most possible degeneracy. Or am I wrong?
Considering only tiling topology, it would also be possible to have triple points, but it looks like the symmetry of the unit pattern forbids this. The curve snakes across the unit cell between opposite corners, so divides any Z^2/3^n lattice into subsets 2*Z^2/3^n and 2*Z^2/3^n + 1. One associates to quadruple points and the other to double points. This is already visible in the nice "tetraskelion" pic.
From the perspective of multipoints, Wunderlich "Figur 5" is the most interesting. It looks to have double, triple, and quadruple points.
--Brad
On Tue, Sep 24, 2019 at 7:38 AM Bill Gosper <billgosper@gmail.com> wrote:
Here's a raw sandbox with Peano and Wunderlich samplings: https://www.wolframcloud.com/obj/1fa6d29d-7451-4e68-ae7b-dded77390040 The "tetraskelion"s are dense with quadruple points. I need to figure out how, with Julian's tools, to make inversepeano and see if there are quintuple or sextuple points (which would be surprising). —rwg