[math-fun] Dragon functional equations
A couple of weeks ago, Julian found Drag[(m + 1/2 + 1/4*(-1)^m)/2^n] == 1/2 (Drag[m/2^n] + Drag[(m + 1)/2^n]), for positive integers m, n, where Drag(t) is the usual Dragon-filling map of [0,1] into the bounding box [-1/3 - i/3, 7/6 + 2i/3]. This describes where along the Dragon is the *m*th vertex of the (self-avoiding) *n*th order "median curve". Using Julian's remarkable piecewiserecursivefractal interpolator-inverter applied to the Dragon's skin, as defined by http://larryriddle.agnesscott.org/ifs/heighway/heighwayBoundary.htm, I just noticed Out[161]= ConditionalExpression[dragskin[1/3 - t] ==3/2 + I/2 - dragskin[1/3 + t], Abs[t] <= 1/6] In[125]:= ConditionalExpression[I dragskin[1 - 3 t] == dragskin@t, 0 <= t <= 1/12] In[126]:= ConditionalExpression[I ( dragskin[1/2 - t] - 1) -> dragskin[1/2 + t] - 1, Abs@t <= 1/4] Plus an intriguing symmetry I haven't yet quantified, smooth scaling? <http://gosper.org/skin 012o3.png> The whole boundary is dragskin([0,1]). Unfortunately, Riddle's (really Mandelbrot's?) clever construction draws the skin at an unboundedly nonuniform rate. I.e., it has dimension D (an unlovely cubic surd between 1 (length) and 2 (area)), but its fill rate, measure(dragskin(t),t1<t<t2)/(t2-t1)^D, gets arbitrarily small. (And arbitrarily large?) But these functional relations raise my hope of finding some continuous, monotone u(t) to give dragskin(u) a uniform fill rate. —rwg Mandelbrot once challenged me to find the skin dimension. I disbelieved the irreducible cubic from a 5⨉5 determinant. Mandelbrot wrote only terse expressions. So I found a 6⨉6. Same damn cubic. I called him up. "Yes, that's the cubic." "How the heck did you get that? it's not your style!" "I just looked at how the wiggles fit into each other." It just occurs to me that I must have had two different, uniformly filling skin descriptions to write those determinants! Oh, to be young again.
For some reason I find myself wondering: Consider the set B = {t in [0,1] | Drag(t) is on the boundary}. I would assume that B is a Cantor-dust-like set. Does it have any visible structure? Does it have an obvious construction? Is the fractal dimension of B exactly one less than the dimension of the actual boundary? On Wed, Oct 9, 2019 at 1:00 PM Bill Gosper <billgosper@gmail.com> wrote:
A couple of weeks ago, Julian found
Drag[(m + 1/2 + 1/4*(-1)^m)/2^n] == 1/2 (Drag[m/2^n] + Drag[(m + 1)/2^n]),
for positive integers m, n, where Drag(t) is the usual Dragon-filling map of [0,1] into the bounding box [-1/3 - i/3, 7/6 + 2i/3]. This describes where along the Dragon is the *m*th vertex of the (self-avoiding) *n*th order "median curve".
Using Julian's remarkable piecewiserecursivefractal interpolator-inverter applied to the Dragon's skin, as defined by http://larryriddle.agnesscott.org/ifs/heighway/heighwayBoundary.htm, I just noticed
Out[161]= ConditionalExpression[dragskin[1/3 - t] ==3/2 + I/2 - dragskin[1/3 + t], Abs[t] <= 1/6]
In[125]:= ConditionalExpression[I dragskin[1 - 3 t] == dragskin@t, 0 <= t <= 1/12]
In[126]:= ConditionalExpression[I ( dragskin[1/2 - t] - 1) -> dragskin[1/2 + t] - 1, Abs@t <= 1/4]
Plus an intriguing symmetry I haven't yet quantified, smooth scaling? <http://gosper.org/skin 012o3.png>
The whole boundary is dragskin([0,1]).
Unfortunately, Riddle's (really Mandelbrot's?) clever construction draws the skin at an unboundedly nonuniform rate. I.e., it has dimension D (an unlovely cubic surd between 1 (length) and 2 (area)), but its fill rate, measure(dragskin(t),t1<t<t2)/(t2-t1)^D, gets arbitrarily small. (And arbitrarily large?)
But these functional relations raise my hope of finding some continuous, monotone u(t) to give dragskin(u) a uniform fill rate. —rwg
Mandelbrot once challenged me to find the skin dimension. I disbelieved the irreducible cubic from a 5⨉5 determinant. Mandelbrot wrote only terse expressions. So I found a 6⨉6. Same damn cubic. I called him up. "Yes, that's the cubic." "How the heck did you get that? it's not your style!" "I just looked at how the wiggles fit into each other."
It just occurs to me that I must have had two different, uniformly filling skin descriptions to write those determinants! Oh, to be young again. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
For Triangle space-filler, another neat example, see also [1]. It is not 2x2 per se, but has the same inflation factor, as does half-hex substitution system [1]. It is possible to make many half-hex space fillers using at most 4 symbols in 4 reflection classes. I don't think this area has been explored yet. It looks like JZH's PRF function is easier to call when the replacement has one or two symbols, but input arguments become cumbersome when adding so many rules (cf. [1] "trifil25"). JZH probably does have the helper functions for producing arguments from, say, Lindenmayer rules, but documentation so far is scarce. Does he plan to write up a paper, user manual, or demonstration? Here are a few more objectives, not too difficult: * Complete Q-functions for 96 2x2 Z-functions on four letters and compare multi-points. More "functional equations" / identities / addition rules? * Calculate Z functions for half-hex substitution system (too many to count?). * Analysis of Z-function scrambles such as [3]? When is the output easy or difficult to understand? --Brad [1] https://community.wolfram.com/groups/-/m/t/912279 [2] https://tilings.math.uni-bielefeld.de/substitution/half-hex/ ( and also: https://tilings.math.uni-bielefeld.de/substitution/hexagonal-aperiodic-monot... ) [3] https://www.youtube.com/watch?v=Wdd9JFD_N0c (blame youtube encoding for blurry vid. argggg.) On Wed, Oct 9, 2019 at 12:00 PM Bill Gosper <billgosper@gmail.com> wrote:
Oh, to be young again.
On Wed, Oct 9, 2019 at 9:59 AM Bill Gosper <billgosper@gmail.com> wrote:
A couple of weeks ago, Julian found
Drag[(m + 1/2 + 1/4*(-1)^m)/2^n] == 1/2 (Drag[m/2^n] + Drag[(m + 1)/2^n]),
for positive integers m, n, where Drag(t) is the usual Dragon-filling map of [0,1] into the bounding box [-1/3 - i/3, 7/6 + 2i/3]. This describes where along the Dragon is the *m*th vertex of the (self-avoiding) *n*th order "median curve".
Using Julian's remarkable piecewiserecursivefractal interpolator-inverter applied to the Dragon's skin, as defined by http://larryriddle.agnesscott.org/ifs/heighway/heighwayBoundary.htm, I just noticed
Out[161]= ConditionalExpression[dragskin[1/3 - t] ==3/2 + I/2 - dragskin[1/3 + t], Abs[t] <= 1/6]
In[125]:= ConditionalExpression[I dragskin[1 - 3 t] == dragskin@t, 0 <= t <= 1/12]
In[126]:= ConditionalExpression[I ( dragskin[1/2 - t] - 1) -> dragskin[1/2 + t] - 1, Abs@t <= 1/4]
Plus an intriguing symmetry I haven't yet quantified, smooth scaling? <http://gosper.org/skin%20012o3.png>
Nothing so exotic. Just ConditionalExpression[dragskin[1/6 + t] == (1/2 + I/2) - 1/2 I dragskin[t], 0 <= t <= 1/4] It is remarkable that there are identities which scale the dependent variable (dragskin) by 2 and others that scale the independent variable t by 3. This is somewhat obfuscated (but not caused) by Riddle's(?) (wildly) nonuniform sampling density.
The whole boundary is dragskin([0,1]).
Unfortunately, Riddle's (really Mandelbrot's?) clever construction draws the skin at an unboundedly nonuniform rate. I.e., it has dimension D (an unlovely cubic surd between 1 (length) and 2 (area)), but its fill rate, measure(dragskin(t),t1<t<t2)/(t2-t1)^D, gets arbitrarily small. (And arbitrarily large?)
But these functional relations raise my hope of finding some continuous, monotone u(t) to give dragskin(u) a uniform fill rate.
—rwg
Mandelbrot once challenged me to find the skin dimension. I disbelieved the irreducible cubic from a 5⨉5 determinant. Mandelbrot wrote only terse expressions. So I found a 6⨉6. Same damn cubic. I called him up. "Yes, that's the cubic." "How the heck did you get that? it's not your style!" "I just looked at how the wiggles fit into each other."
It just occurs to me that I must have had two different, uniformly filling skin descriptions to write those determinants! Oh, to be young again.
I just remembered an ancient, Mil-Spec looseleaf notebook of XGP output including a uniformly sampled, hollow Twindragon, the source of at least one of those determinants. I may even have scanned it.
participants (3)
-
Allan Wechsler -
Bill Gosper -
Brad Klee