[math-fun] "quasicrystal"
I'm intrigued by Jack Holloway's recursive pentagram snowflake design, Fig 31 in the Minskys & Trinskys book<http://www.blurb.com/bookstore/detail/2172660>. However, circularizing and sampling obscure the pixel structure, and how, if at all, it relates to Fig 30. As usual, Julian comes to the rescue, with a "period 5" trinsky in which orbital x+y stays constant mod 2. The "black squares" (checkerboard) subset of this bitmap strongly resembles Fig 31, except the bounding pentagons have vertex angles of π/2 and 3π/4 instead of 3π/5. So I linearly crunched it to have "rhombic pixels" and wrote http://www.tweedledum.com/rwg/rhombicp5.svg The .svg is so you can scale it up without antialiasing. Unfortunately, my Apple Firefox scales it right off the screen, and refuses to scroll it back! Safari scrolls, but won't let you zoom very far. And Preview won't load svgs at all! And, as of this AM, gosper.org delivers source text instead of graphics for it. So to mega zoom it, Neil suggested downloading inkscape<http://inkscape.org/>, which seems to work great. Can one speak of the limit of this pattern in the large? The pentagram outlines grow without bound and get "infinitely thin", yet seem to fill space. And what is a formula for their limiting shape? Julian (quoted without permission) explains the original image as follows: (d_t means the trinsky d multiplier, etc.) It's the Trinsky equivalent of p5d1, which is d=g=1, e=2*sin^2(π/5). The Minsky-Trinsky equivalence is has d_t=g_t=d_m, e_t=e_m/2, has x about twice as large to compensate. This is what causes the checkerboarded images–for a given Trinsky (x,y), the equivalent Minsky point is ((x+⸤d_t*y⸥)/2,y) (sorry about the floors I haven't found any decent floor characters to use), and the checkerboard pattern comes from whether (x+⸤d_t*y⸥)/2 is an integer or not. The "checkerboard" pattern is only when the Trinksy d=g=1, in which case x+y is constant mod 2: x'=x-y=c mod 2, y'=y+⸤e*x⸥, x''=x'-y'=c-y' mod 2. A granularity of 1/2 would not produce checkerboarded images–it would produce four normal (i.e. at coordinates (ax+b,cy+d)) images (or two, alternating in either columns or row but not both, if only one granularity=1/2). --rwg
On Thu, May 24, 2012 at 4:17 AM, Bill Gosper <billgosper@gmail.com> wrote:
I'm intrigued by Jack Holloway's recursive pentagram snowflake design, Fig 31 in the Minskys & Trinskys book<http://www.blurb.com/bookstore/detail/2172660>. However, circularizing and sampling obscure the pixel structure, and how, if at all, it relates to Fig 30. As usual, Julian comes to the rescue, with a "period 5" trinsky in which orbital x+y stays constant mod 2. The "black squares" (checkerboard) subset of this bitmap strongly resembles Fig 31, except the bounding pentagons have vertex angles of π/2 and 3π/4 instead of 3π/5. So I linearly crunched it to have "rhombic pixels" and wrote
http://www.tweedledum.com/rwg/rhombicp5.svg
The .svg is so you can scale it up without antialiasing. Unfortunately, my Apple Firefox scales it right off the screen, and refuses to scroll it back! Safari scrolls, but won't let you zoom very far. And Preview won't load svgs at all! And, as of this AM, gosper.org delivers source text instead of graphics for it. So to mega zoom it, Neil suggested downloading inkscape<http://inkscape.org/>, which seems to work great.
Can one speak of the limit of this pattern in the large? The pentagram outlines grow without bound and get "infinitely thin", yet seem to fill space. And what is a formula for their limiting shape?
Our Esteemed Moderator was so peeved by the stupidity of this question that, a couple of hours ago, he actually walked into my house, said
z[t]==Sum[((-1)^k*E^(I*(-(1/5) + k)*π*t)* Product[1 - Sqrt[5 - 2*Sqrt[5]]*Tan[((-(1/5) + k)*π)/(-2)^n],{n, ∞}])/(-(1/5) + k)^2, {k, -∞,∞}] 0≤t<10 and walked out. See http://gosper.org/fourierfracpent.png --rwg Julian (quoted without permission) explains the original image as follows:
(d_t means the trinsky d multiplier, etc.)
It's the Trinsky equivalent of p5d1, which is d=g=1, e=2*sin^2(π/5). The Minsky-Trinsky equivalence is has d_t=g_t=d_m, e_t=e_m/2, has x about twice as large to compensate. This is what causes the checkerboarded images–for a given Trinsky (x,y), the equivalent Minsky point is ((x+⸤d_t*y⸥)/2,y) (sorry about the floors I haven't found any decent floor characters to use), and the checkerboard pattern comes from whether (x+⸤d_t*y⸥)/2 is an integer or not. The "checkerboard" pattern is only when the Trinksy d=g=1, in which case x+y is constant mod 2: x'=x-y=c mod 2, y'=y+⸤e*x⸥, x''=x'-y'=c-y' mod 2. A granularity of 1/2 would not produce checkerboarded images–it would produce four normal (i.e. at coordinates (ax+b,cy+d)) images (or two, alternating in either columns or row but not both, if only one granularity=1/2).
--rwg
On Thu, May 24, 2012 at 10:54 PM, Bill Gosper <billgosper@gmail.com> wrote:
On Thu, May 24, 2012 at 4:17 AM, Bill Gosper <billgosper@gmail.com> wrote:
I'm intrigued by Jack Holloway's recursive pentagram snowflake design, Fig 31 in the Minskys & Trinskys book<http://www.blurb.com/bookstore/detail/2172660>. However, circularizing and sampling obscure the pixel structure, and how, if at all, it relates to Fig 30. As usual, Julian comes to the rescue, with a "period 5" trinsky in which orbital x+y stays constant mod 2. The "black squares" (checkerboard) subset of this bitmap strongly resembles Fig 31, except the bounding pentagons have vertex angles of π/2 and 3π/4 instead of 3π/5. So I linearly crunched it to have "rhombic pixels" and wrote
http://www.tweedledum.com/rwg/rhombicp5.svg
The .svg is so you can scale it up without antialiasing. Unfortunately, my Apple Firefox scales it right off the screen, and refuses to scroll it back! Safari scrolls, but won't let you zoom very far. And Preview won't load svgs at all! And, as of this AM, gosper.org delivers source text instead of graphics for it. So to mega zoom it, Neil suggested downloading inkscape<http://inkscape.org/>, which seems to work great.
Can one speak of the limit of this pattern in the large? The pentagram outlines grow without bound and get "infinitely thin", yet seem to fill space. And what is a formula for their limiting shape?
Our Esteemed Moderator was so peeved by the stupidity of this question that, a couple of hours ago, he actually walked into my house, said
z[t]==Sum[((-1)^k*E^(I*(-(1/5) + k)*π*t)* Product[1 - Sqrt[5 - 2*Sqrt[5]]*Tan[((-(1/5) + k)*π)/(-2)^n],{n, ∞}])/(-(1/5) + k)^2, {k, -∞,∞}] 0≤t<10
and walked out. See http://gosper.org/fourierfracpent.png
(Now uprezzed. Notice that, even though enlarged, the higher res (3511 harmonics, 10240 points) curve looks darker. Since the fractal it's illustrating has dimension < 2, both curves should actually be invisible, unless revealed as the boundary between 2D regions. However, there is a virtue to painting the points of a fractal curve with microscopic disks, because the darkness increases with dimension and order of approximation. And pixels approximate microscopic disks.)
--rwg
What I was really seeking was a dissection of the interior of this figure into finitely many similar figures, like Mandelbrot's dissections of the snowflake interior into 7 and 13 snowflakes. Or failing that, into a finite set of pieces, each of which dissects into two or more pieces similar to those in the set. And what about similar treatments for the two heptagrams? --rwg PS, Henry has kindly "stapled" the 7 scanned pages of the Hashlife paper, which I'll forward to any ancient history fans who request it. Presumably the quality of the Elsevier version is far higher, in case you want to buy it, or wait for them to give it away. I actually have an ancient Symbolics TeX source file, but not the quaint Lisp Machine screenshot illustrations. Since the paper is utterly obsolete, I see no reason to reconstruct it with modern illustrations. Julian (quoted without permission) explains the original image as follows:
(d_t means the trinsky d multiplier, etc.)
It's the Trinsky equivalent of p5d1, which is d=g=1, e=2*sin^2(π/5). The Minsky-Trinsky equivalence is has d_t=g_t=d_m, e_t=e_m/2, has x about twice as large to compensate. This is what causes the checkerboarded images–for a given Trinsky (x,y), the equivalent Minsky point is ((x+⸤d_t*y⸥)/2,y) (sorry about the floors I haven't found any decent floor characters to use), and the checkerboard pattern comes from whether (x+⸤d_t*y⸥)/2 is an integer or not. The "checkerboard" pattern is only when the Trinksy d=g=1, in which case x+y is constant mod 2: x'=x-y=c mod 2, y'=y+⸤e*x⸥, x''=x'-y'=c-y' mod 2. A granularity of 1/2 would not produce checkerboarded images–it would produce four normal (i.e. at coordinates (ax+b,cy+d)) images (or two, alternating in either columns or row but not both, if only one granularity=1/2).
--rwg
The Fourier series mentioned below is a specialization of the one in gosper.org/fst.pdf, which alternatively specializes to the Koch's Snowflake, Peano's spacefiller, etc., according to the parameter s. For convergence, the geometry of the recursion underlying the construction of the Fourier coefficients requires |s-i| < 2, but at s= √3, the infinite product telescopes, revealing that series convergence also extends to this boundary point. A couple of days ago I thought I noticed convergence beyond |s-i| = 2, but common sense and recent experiments at Julian's house convinced me I was hallucinating. In the process, we produced some pleasantly Picassoid squiggle <http://gosper.org/snowfourier.pdf>s. (Or maybe just Dr. Seuss.) The function futst(t,s,m,cutoff) traces the outline of the curve repeated around a regular m-gon as 0≤t≤2m, low-pass (t=time) filtered to the cutoff frequency (default ±69 harmonics). Thus lacking evidence of anomalous convergence, I was puzzled to receive just now (again, quoting without permission): You may have figured this out already, but just in case: the reason that it can converge while the fractal doesn't is that you found the series from an integral, but you don't actually compute the integral. Instead you compute a sequence of Riemann sums (intervals of size 2^-n), but this only necessarily computes the integral when the function in question is continuous. The fractal curve is not defined, and thus is not continuous, so the Fourier series is just some (necessarily discontinuous on a dense set, as otherwise you could find an interval where it would be continuous and match the "fractal" at dyadic rational values) function. It may or may not have to pass through the same points as the "fractal" at dyadic rational values. I don't know why it matches the predictions for 1/3, but the problem seems approachable. Julian He's apparently right! Out[219]= {futst[1/3, s, m, 9999], (I + s)/(3 + I s) + Cot[π/m]} In[227]:= %219 /. {9999 -> 99999, m -> 2., s -> 7/4.} Out[227]= {0.586944 - 0.00655471 I, 0.580311 - 0.00518135 I} where the Cot expression is the theoretical t=1/3 value for cutoff→∞. Trying 2 million terms, In[228]:= %219 /. {9999 -> 999999, m -> 2., s -> 7/4.} NEVER RETURNS, because somewhere between 99999 and 999999, Sum arbitrarily, silently, and PERVERSELY changes 1/2. from .5 to 1/2 ! This is infuriating. This is evil. --rwg Apropos Wikipedia math, I note that en.wikipedia.org/wiki/Generalized_continued_fraction is chatty, avuncular, and opinionated, and contains a URL apparently to the author's private web page, yet is free of all the usuak editorial unworthiness stigmata, as if it approached the pristine ideal of the breathlessly inconsequential http://en.wikipedia.org/wiki/MS_Smyril . Warren: For long range battleship type gunnery, the Ray solution would be useless because the atmosphere gets exponentially thinner at higher altitude which should be a major effect that he can't handle. In fact, to maximize range, they aim *above* 45°, the opposite of what you'd do with uniform drag [citation needed]. Henry: <minimum publishable quantum, fun facts> These are more than fun facts--they're valuable optimizations. We should gather up our various complex plane computational geometry slick tricks (circumcenter, circumradius, incenter, inradius, polygon area, polygon-point winding number, line-segment intersection, ...) and publish them somehow. Henry: (This is also true in the arts; take a look at Beethoven's manuscripts -- not everyone wrote music like Mozart, which were perfect the first time.) This is probably because Mozart wrote almost nothing in C#. Even less in C++. On Thu, May 24, 2012 at 10:54 PM, Bill Gosper <billgosper@gmail.com> wrote:
On Thu, May 24, 2012 at 4:17 AM, Bill Gosper <billgosper@gmail.com> wrote:
I'm intrigued by Jack Holloway's recursive pentagram snowflake design, Fig 31 in the Minskys & Trinskys book<http://www.blurb.com/bookstore/detail/2172660>. However, circularizing and sampling obscure the pixel structure, and how, if at all, it relates to Fig 30. As usual, Julian comes to the rescue, with a "period 5" trinsky in which orbital x+y stays constant mod 2. The "black squares" (checkerboard) subset of this bitmap strongly resembles Fig 31, except the bounding pentagons have vertex angles of π/2 and 3π/4 instead of 3π/5. So I linearly crunched it to have "rhombic pixels" and wrote
http://www.tweedledum.com/rwg/rhombicp5.svg
The .svg is so you can scale it up without antialiasing. Unfortunately, my Apple Firefox scales it right off the screen, and refuses to scroll it back! Safari scrolls, but won't let you zoom very far. And Preview won't load svgs at all! And, as of this AM, gosper.org delivers source text instead of graphics for it. So to mega zoom it, Neil suggested downloading inkscape<http://inkscape.org/>, which seems to work great.
Can one speak of the limit of this pattern in the large? The pentagram outlines grow without bound and get "infinitely thin", yet seem to fill space. And what is a formula for their limiting shape?
Our Esteemed Moderator was so peeved by the stupidity of this question that, a couple of hours ago, he actually walked into my house, said
z[t]==Sum[((-1)^k*E^(I*(-(1/5) + k)*π*t)* Product[1 - Sqrt[5 - 2*Sqrt[5]]*Tan[((-(1/5) + k)*π)/(-2)^n],{n, ∞}])/(-(1/5) + k)^2, {k, -∞,∞}] 0≤t<10
and walked out. See http://gosper.org/fourierfracpent.png --rwg
Julian (quoted without permission) explains the original image as follows:
(d_t means the trinsky d multiplier, etc.)
It's the Trinsky equivalent of p5d1, which is d=g=1, e=2*sin^2(π/5). The Minsky-Trinsky equivalence is has d_t=g_t=d_m, e_t=e_m/2, has x about twice as large to compensate. This is what causes the checkerboarded images–for a given Trinsky (x,y), the equivalent Minsky point is ((x+⸤d_t*y⸥)/2,y) (sorry about the floors I haven't found any decent floor characters to use), and the checkerboard pattern comes from whether (x+⸤d_t*y⸥)/2 is an integer or not. The "checkerboard" pattern is only when the Trinksy d=g=1, in which case x+y is constant mod 2: x'=x-y=c mod 2, y'=y+⸤e*x⸥, x''=x'-y'=c-y' mod 2. A granularity of 1/2 would not produce checkerboarded images–it would produce four normal (i.e. at coordinates (ax+b,cy+d)) images (or two, alternating in either columns or row but not both, if only one granularity=1/2).
--rwg
On Mon, Jun 4, 2012 at 6:44 PM, Bill Gosper <billgosper@gmail.com> wrote:
The Fourier series mentioned below is a specialization of the one in gosper.org/fst.pdf, which alternatively specializes to the Koch's Snowflake, Peano's spacefiller, etc., according to the parameter s. For convergence, the geometry of the recursion underlying the construction of the Fourier coefficients requires |s-i| < 2, but at s= √3, the infinite product telescopes, revealing that series convergence also extends to this boundary point. A couple of days ago I thought I noticed convergence beyond |s-i| = 2, but common sense and recent experiments at Julian's house convinced me I was hallucinating. In the process, we produced some pleasantly Picassoid squiggle <http://gosper.org/snowfourier.pdf>s. (Or maybe just Dr. Seuss.) The function futst(t,s,m,cutoff) traces the outline of the curve repeated around a regular m-gon as 0≤t≤2m, low-pass (t=time) filtered to the cutoff frequency (default ±69 harmonics).
Thus lacking evidence of anomalous convergence, I was puzzled to receive just now (again, quoting without permission):
You may have figured this out already, but just in case: the reason that it can converge while the fractal doesn't is that you found the series from an integral, but you don't actually compute the integral. Instead you compute a sequence of Riemann sums (intervals of size 2^-n), but this only necessarily computes the integral when the function in question is continuous. The fractal curve is not defined, and thus is not continuous, so the Fourier series is just some (necessarily discontinuous on a dense set, as otherwise you could find an interval where it would be continuous and match the "fractal" at dyadic rational values) function. It may or may not have to pass through the same points as the "fractal" at dyadic rational values. I don't know why it matches the predictions for 1/3, but the problem seems approachable.
Julian
He's apparently right! Out[219]= {futst[1/3, s, m, 9999], (I + s)/(3 + I s) + Cot[π/m]}
In[227]:= %219 /. {9999 -> 99999, m -> 2., s -> 7/4.}
Out[227]= {0.586944 - 0.00655471 I, 0.580311 - 0.00518135 I}
where the Cot expression is the theoretical t=1/3 value for cutoff→∞. Trying 2 million terms, In[228]:= %219 /. {9999 -> 999999, m -> 2., s -> 7/4.}
NEVER RETURNS, because somewhere between 99999 and 999999, Sum arbitrarily, silently, and PERVERSELY changes 1/2. from .5 to 1/2 !
This is infuriating. This is evil. --rwg
So I forced the term calculation (fut) to use machine precision, conceding the ability for the series (futst) to control the precision. Trying 400000 terms, In[287]:= (Print[#1]; #2) & @@AbsoluteTiming[%219 /. {9999 -> 199999, m -> 2., s -> 7/4.}] During evaluation of In[287]:= 122.436966 Out[287]= {0.575792 - 0.0102995 I, 0.580311 - 0.00518135 I} And then 600000: In[288]:= (Print[#1]; #2) & @@ AbsoluteTiming[%219 /. {9999 -> 299999, m -> 2., s -> 7/4.}] During evaluation of In[288]:= 126.319336 Out[288]= {0.578973 - 0.0108453 I, 0.580311 - 0.00518135 I} And then 1.4 million: In[283]:= (Print[#1]; #2) & @@ AbsoluteTiming[%219 /. {9999 -> 699999, m -> 2., s -> 7/4.}] During evaluation of In[286]:= 2722.704824 Out[283]= {0.101321 ((8.87115 + 1.71512 I) - Limit[(-1)^k E^(1/3 I (0.5 + k) \[Pi]) fut[0.5 + k, 1.75], k -> 0, Assumptions -> True]), 0.580311 - 0.00518135 I} Somebody should be shot. Despite these efforts, Mathematica failed to conceal a surprise in the actual math: The series seems to get close and then simply dither, never actually converging! Upping the precision for 600000: In[290]:= (Print[#1]; #2) & @@ AbsoluteTiming[%219 /. {9999 -> 299999, m -> 2.`25, s -> 7/4.`25}] During evaluation of In[292]:= 1672.388773 Out[290]= {0.5789734583798222659864 - 0.0108452753013866670336 I, 0.580310880829015544041451 - 0.0051813471502590673575130 I} in perfect agreement with machine precision. Let's make some plots<http://gosper.org/snowfourier+.pdf> . %264 is the apparently convergent s=√3 case on an (m=2)gon. . %267 is part of the questionable s=7/4 case on an (m=3)gon. (Anatomists, not to mention trigonometrists, actually speak of trigons.) %268 is 288 terms of s=7/4. A cross between a bowling lane and a cricket pitch! Upping to 999 terms, %277 grows another row of "circles" and switch to pool. The One Ball was amputated by PlotRange->Automatic. %278 is the same, with finer plotting. Why fifteen loops? And what are those little tripods? Still with s=7/4, %279 takes 20000 series terms, grows another row of loops, and becomes UNDERsampled, while the tripods become indented triangles. %280 takes 200000 terms (after many hours), concentrating on 2<t<2.1, a short, off center subinterval. And it's still undersampled. Each loop is orbited hundreds of times. Perfunctory experiments with s=1.742 show very similar behavior. This is very peculiar non-convergence. This was a case where pictures actually clarified misleading numbers (unlike Mike Stay's recent 8^2=5*13 mention, or my old dysplotsia<http://www.tweedledum.com/rwg/dysplotsia.htm>post.) --rwg "It turns out that the purpose of computing was numbers after all. sorry for the confusion." --"W.R. Hamming" (Who, really?) (Which reminded Neil of "Computers are useless--they can only give you answers." --Picasso.) Too bad Mathworld is also hard to update.
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Bill Gosper