The first three levels of Hilbert quadruple points. <http://gosper.org/hilbquads.png> (Connected per ordering of the unit interval preimage,) I expected it to be even more boring. But this isn't: Define hilbert:[0,1] ➝ [0,1]⨉[0i,i]. Then for integers n>0 and 0≤k<4^n, hilbert[(2 k + 1/3)/2^(2 n + 1)] == hilbert[(2 k + 1)/2^(2 n + 1)] == hilbert[(2 k + 5/3)/2^(2 n + 1)] a two-parameter family of triple points. Because of the ⅓ and 5/3, a proof might require "hand simulating" hilbert[t_?NumericQ] := piecewiserecursivefractal[t, Identity, {Min[4, 1 + Floor[4*#]]} &, {1 - 4*# &, 4*# - 1 &, 4*# - 2 &, 4 - 4*# &}, {I*(1 - #)/2 &, (I + #)/2 &, (I + 1 + #)/2 &, 1 + #*I/2 &}] with piecewiserecursivefractal defined at http://oeis.org/A260482 . Hairsplitting: It looks like the complete set of quad points will be [0,1]⨉[0i,i] minus a bunch of interior open line segments. So it's not dense. ? On Tue, May 28, 2019 at 9:32 PM Bill Gosper <billgosper@gmail.com> wrote:

to accompany should've been "sampHilbert" <http://www.tweedledum.com/rwg/sampeano.htm> Hilbert's [spacefilling] function is dense with quadruple points

??

, i.e. points in [0,*i*] with four distinct preimage points in [0,1]: gosper.org/hilbquad.png . E.g., the preimage of 1/2 + *i*/4 is {5/48, 7/48, 41/48, 43/48}. This picture connects, in the order they were swept, all the points in [0,*i*] with preimages having denominators ≤ 3⨉4096. (All quadruple point preimages have denominators 3⨉2ⁿ.) E.g., the lower-leftmost vertex is 1/32 + *i*/64, the image of 5/12288, 7/12288, 41/12288, and 43/12288. The white segments are retraced boundary, retroflexed in the middle, (making quadruple points). (Spacefilling functions map closed intervals to closed sets.)

The Heighway Dragon is dense with mere triple points. Here's one: Hi res Heighway <http://gosper.org/dragtrip!.png> . —rwg

Here are several more from the two families <http://gosper.org/heighwaytrips.png>. (The line segments just indicate adjacency in the preimage.)