With this filling function, the triangle's centroid is a triple point: In[434]:= wolloh[1/2 (3 + I Sqrt[3])] // tim During evaluation of In[434]:= 1.413051,3 Out[434]= {3/10, 1/2, 7/10} But there is a tidier order of filling the one large + five small subtriangles: http://gosper.org/tameavoid2592.png http://gosper.org/tamedoily1728.png and Julian makes the remarkable observation that this function likewise has (≥) three preimages of the centroid, but one of them is irratonal! In[432]:= unTrifill[1/2 (3 + I Sqrt[3])] // tim During evaluation of In[432]:= 166.861545,2 Out[432]= {4/15, 2/3} The evidence for the irrational is the timing: 167 sec (vs 1.4), which would have been "$RecursionLimit exceeded" but for a safety valve tripping for denominators near a googol. Irrationality could be proven with a bound on preimage denominators, which should be feasible. Anybody need something to work on? --rwg On Tue, Nov 1, 2016 at 2:29 AM, Bill Gosper <billgosper@gmail.com> wrote:
It turns out the preimage of 3/2 + i √3/4 under the nonuniform triangle-filling function is BaseForm[401474566812783980411091190154813161/ 1403266106709659827505450695237042176., 6]
(0.14144414144141141144444441444414414444411411414...)₆, so we "know" it isn't algebraic. For lack of understanding Julian's remark, "You can invert it just fine and it will always converge, though good luck finding points with explicit inverses except by going forward.", it took me a week to figure out that there's a straightforward process for graphically determining two or three base 6 digits at a time. So we have the incautious conjecture that the preimages of the "Eisenstein rationals" are either rational, or "Cantorish", base 6. --rwg
On Sun, Oct 30, 2016 at 5:15 PM, Bill Gosper <billgosper@gmail.com> wrote:
There's an interesting difference between the uniform triangle fill (by quarters) vs the nonuniform five ninths plus a quadruple ninth version. It appears that the former maps the rationals in [0,1] onto the "Eisenstein rationals" (r₁ + r₂ i √3) in the (equilateral) triangle based on [0,1]. So a (= the) preimage of the center
Out[416]= 1/2 + I/(2 Sqrt[3])
is In[417]:= untrifil@%
Out[417]= {2/5} Check: In[418]:= trifil @@ % // Simplify
Out[418]= {1/6 (3 + I Sqrt[3])}
More convincingly, the preimage of
Out[405]= 113/355+(7 I Sqrt[3])/22
is In[406]:= untrifil@%//tim
During evaluation of In[406]:= 188.873607 secs Out[406]= {66067177350394731417555234130689770663588400360845847462258 734629314942043718190483511 /12433080910244666053884556203670521002511403769933692936011 5994223289874253133343883328} Check: In[407]:= trifil@@%//tim During evaluation of In[407]:= 42.187317 secs Out[407]= {113/355+(7 I Sqrt[3])/22}
Interestingly, the centroid of the nonuniform filler (based on [0,3] instead of [0,1]) is a triple point: In[420]:= wolloh[3 %416] (triple size triangle)
Out[420]= {3/10, 1/2, 7/10} Check: In[422]:= Simplify[hollow /@ %%]/3
Out[422]= {{1/6 (3 + I Sqrt[3])}, {1/6 (3 + I Sqrt[3])}, {1/6 (3 + I Sqrt[3])}}
Likewise, the preimage of Out[408]= 13/9 + (13 I)/(6 Sqrt[3])
is In[423]:= wolloh@%408
Out[423]= {1204367/1680912} Check: In[424]:= Expand@Simplify[hollow @@ %]
Out[424]= {13/9 + (13 I)/(6 Sqrt[3])}
But wolloh[3/2 + I 3 Sqrt[3]/4] apparently recurses forever. I don't even have a good enough approximation to guess if this preimage is algebraic. --rwg
On Wed, Oct 26, 2016 at 7:22 PM, Bill Gosper <billgosper@gmail.com> wrote:
On Fri, Oct 21, 2016 at 7:25 PM, Bill Gosper <billgosper@gmail.com> wrote:
See gosper.org/julianstrifle.png . Note the tiny self-crossings not visible in the cursive drawing. It is usually far clearer to portray a <2-dimensional curve as the boundary of a two-dimensional region. --rwg
Julian's Fourier expander takes an optional argument to arrange m copies of a curve around a regular m-gon. gosper.org/6trifil.png . Instead of spacefilling an equilateral triangle divided into quarters, I tried dividing it into five little ones plus a double size one. Arranged around a hexagon, gosper.org/ringnoring1234.png (one of the few cases where the outline looks maybe better than the filled area.) (Shades of https://en.wikipedia.org/wiki/Uniform_tilings_in_hyperbol ic_plane ?) Note that with 2468 rotors, the Fourier approximation gets nowhere near the middle. So is this a spacefill, or not? Julian: "Of course it is ..." Proof that it reaches all three vertices of each triangle: In[422]:= hollow /@ {0, 4/5, 1} // Expand
Out[422]= {{0}, {3/2 + (3 I Sqrt[3])/2}, {3}}
via (Julian's ever amazing) piecewiserecursivefractal Clear[hollow]; hollow[t_] := piecewiserecursivefractal[t, Identity, {Min[6, 1 + Floor[6*#]]} &, {6 # &, 6 # - 1 &, 3 - 6 # &, 6 # - 3 &, 6 # - 4 &, 6 - 6 # &}, {#/3 &, #/3 + 1 &, #/3/I^(2/3) + 3/2 + I Sqrt[3]/2 &, 3/2 + I Sqrt[3]/2 - #/3 &, 2 #/3 + 1/2 + I Sqrt[3]/2 &, #/3*I^(4/3) + 3 &}] A polygonal sampling: gosper.org/hollowfill150.png (Reminder: True spacefills self-contact like crazy.) --rwg
On Thu, Oct 20, 2016 at 4:49 AM, Bill Gosper <billgosper@gmail.com> wrote:
On 2016-10-19 10:11, James Propp wrote:
Very pretty. What's the math here? (Apologies if you've already answered this question.)
Jim Propp
Julian wrote a nifty Fourier expander for recursive Koch polygons. E.g., ptsgnlst2Fouriermat[{0, 1, I^(2/3), 1 + I^(2/3), 2}, {1, -1, 1, -1}] spacefills the triangle joining 0 to 2 via 1+i√3. Actually, it makes an infinite 3x3 matrix product for the coefficients a(k). Then Sum a(k+1/m) exp(2 i π (k+1/m)) repeats the fractal on the sides of an m-gon. The gif just accumulates consecutive harmonics. --rwg
On Wednesday, October 19, 2016, Bill Gosper <billgosper@gmail.com>
wrote:
> With much help from Julian, gosper.org/hellodoily.gif > --rwg >
> >>[chop]