October 2016 - mathfun@mailman.xmission.com

[math-fun] Julian has constructed a spacefill dense with sextuple points
by Bill Gosper 26 Jan '17

26 Jan '17

using a 25-way conditional. ClearAll[piecewiserecursivefractal]; piecewiserecursivefractal[x_, f_, which_, iters_, fns_] := piecewiserecursivefractal[x, g_, which, iters, fns] = ((piecewiserecursivefractal[x, h_, which, iters, fns] := Block[{y}, y /. Solve[f[y] == h[y], y]]); Union @@ ((fns[[#]] /@piecewiserecursivefractal[iters[[#]][x], Composition[f, fns[[#]]], which, iters, fns]) & /@which[x])); (I cannot praise this function enough.) In[363]:= trifil25[t_] := piecewiserecursivefractal[t, Identity, Piecewise[{{{1}, #1 == 0}, {{Ceiling[25*#1]}, Inequality[0, Less, #1, LessEqual, 1]}}, {}] & , {25*#1 + 0 & , -25*#1 + 2 & , 25*#1 - 2 & , 25*#1 - 3 & , 25*#1 - 4 & , 25*#1 - 5 & , 25*#1 - 6 & , 25*#1 - 7 & , 25*#1 - 8 & , -25*#1 + 10 & , -25*#1 + 11 & , 25*#1 - 11 & , -25*#1 + 13 & , -25*#1 + 14 & , 25*#1 - 14 & , -25*#1 + 16 & , 25*#1 - 16 & , -25*#1 + 18 & , 25*#1 - 18 & , 25*#1 - 19 & , -25*#1 + 21 & , -25*#1 + 22 & , 25*#1 - 22 & , -25*#1 + 24 & , 25*#1 - 24 & }, {#1/5 & , (1/10)*(3 + I*Sqrt[3] + (-1 - I*Sqrt[3])* #1) & , (1/10)*(3 + I*Sqrt[3] + #1 - I*Sqrt[3]*#1) & , (2 + #1)/5 & , (1/10)*(6 + #1 + I*Sqrt[3]*#1) & , (1/10)*(7 + I*Sqrt[3] + I*(I + Sqrt[3])*#1) & , (1/5)*(3 + I*Sqrt[3] - #1) & , (1/10)*(4 + 2*I*Sqrt[3] + (-1 - I*Sqrt[3])*#1) & , (1/10)*(3 + I*Sqrt[3] - 2*#1) & , (-(1/10))*I*(-I + Sqrt[3])*(-2 + #1) & , (1/5)*(2 + I*Sqrt[3] - #1) & , (1/5)*(2 + I*Sqrt[3] + #1) & , (1/10)*(5 + 3*I*Sqrt[3] + #1 - I*Sqrt[3]*#1) & , (1/10)*(4 + 2*I*Sqrt[3] + #1 + I*Sqrt[3]*#1) & , (1/10)*(4 + 2*I*Sqrt[3] + I*(I + Sqrt[3])*#1) & , (-(1/10))*I*(-I + Sqrt[3])*(-4 + #1) & , (1/5)*(2 + 2*I*Sqrt[3] + #1) & , (1/10)*(5 + 3*I*Sqrt[3] + #1 + I*Sqrt[3]*#1) & , (1/10)*(5 + 3*I*Sqrt[3] + 2*#1) & , (1/10)*(7 + 3*I*Sqrt[3] + (-1 - I*Sqrt[3])*#1) & , (1/5)*(4 + I*Sqrt[3] - #1) & , (1/10)*(9 + I*Sqrt[3] + I*(I + Sqrt[3])*#1) & , (1/10)*(9 + I*Sqrt[3] - 2*#1) & , (1/10)*(8 + I*(I + Sqrt[3])*#1) & , (4 + #1)/5 & }][[1]] E.g., In[361]:= trifil25[2/25 + 2/75] // ComplexExpand Out[361]= {1/2 + (I Sqrt[3])/10} In[367]:= Graphics[{Hue@#, Line[ReIm@{trifil25@#, trifil25[# + 5^-5]}]} & /@ Range[0, 1 - 5^-5, 5^-5]] gosper.org/trifil25.png (Those little white hexagons are sextuple points.) The inverse fractal, constructed more or less automatically: In[380]:= untrifil25[t_] := piecewiserecursivefractal[t, Identity, If[0 <= Re[#1] <= 1 && 0 <= Im[#1] <= Sqrt[3]/2, {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25}, {}] &, {ComplexExpand[5 #1] &, ComplexExpand[1/4 (6 - 2 I Sqrt[3] - 10 #1 + 10 I Sqrt[3] #1)] &, ComplexExpand[1/4 (-4 I Sqrt[3] + 10 #1 + 10 I Sqrt[3] #1)] &, ComplexExpand[-2 + 5 #1] &, ComplexExpand[1/4 (-6 + 6 I Sqrt[3] + 10 #1 - 10 I Sqrt[3] #1)] &, ComplexExpand[1/4 (4 + 8 I Sqrt[3] - 10 #1 - 10 I Sqrt[3] #1)] &, ComplexExpand[3 + I Sqrt[3] - 5 #1] &, ComplexExpand[1/4 (10 - 2 I Sqrt[3] - 10 #1 + 10 I Sqrt[3] #1)] &, ComplexExpand[1/2 (3 + I Sqrt[3] - 10 #1)] &, ComplexExpand[1/4 (8 - 10 #1 + 10 I Sqrt[3] #1)] &, ComplexExpand[2 + I Sqrt[3] - 5 #1] &, ComplexExpand[-2 - I Sqrt[3] + 5 #1] &, ComplexExpand[1/4 (4 - 8 I Sqrt[3] + 10 #1 + 10 I Sqrt[3] #1)] &, ComplexExpand[1/4 (-10 + 2 I Sqrt[3] + 10 #1 - 10 I Sqrt[3] #1)] &, ComplexExpand[1/4 (-2 + 6 I Sqrt[3] - 10 #1 - 10 I Sqrt[3] #1)] &, ComplexExpand[1/4 (16 - 10 #1 + 10 I Sqrt[3] #1)] &, ComplexExpand[-2 - 2 I Sqrt[3] + 5 #1] &, ComplexExpand[1/4 (-14 + 2 I Sqrt[3] + 10 #1 - 10 I Sqrt[3] #1)] &, ComplexExpand[1/2 (-3 I Sqrt[3] + 5 (-1 + 2 #1))] &, ComplexExpand[1/4 (16 - 4 I Sqrt[3] - 10 #1 + 10 I Sqrt[3] #1)] &, ComplexExpand[4 + I Sqrt[3] - 5 #1] &, ComplexExpand[1/4 (6 + 10 I Sqrt[3] - 10 #1 - 10 I Sqrt[3] #1)] &, ComplexExpand[1/2 (9 + I Sqrt[3] - 10 #1)] &, ComplexExpand[1/4 (8 + 8 I Sqrt[3] - 10 #1 - 10 I Sqrt[3] #1)] &, ComplexExpand[-4 + 5 #1] &}, {#1/25 &, (2 - #1)/25 &, (2 + #1)/ 25 &, (3 + #1)/25 &, (4 + #1)/25 &, (5 + #1)/25 &, (6 + #1)/ 25 &, (7 + #1)/25 &, (8 + #1)/25 &, (10 - #1)/25 &, (11 - #1)/ 25 &, (11 + #1)/25 &, (13 - #1)/25 &, (14 - #1)/25 &, (14 + #1)/ 25 &, (16 - #1)/25 &, (16 + #1)/25 &, (18 - #1)/25 &, (18 + #1)/ 25 &, (19 + #1)/25 &, (21 - #1)/25 &, (22 - #1)/25 &, (22 + #1)/ 25 &, (24 - #1)/25 &, (24 + #1)/25 &}] Here it finds the preimages of the simplest sextuple point: In[381]:= untrifil25[1/2 + I Sqrt[3]/10] Out[381]= {8/75, 11/75, 14/75, 17/75, 4/15, 23/75} --rwg

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Re: [math-fun] Space-filling curves again
by Bill Gosper 21 Nov '16

21 Nov '16

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(a)gmail.com> wrote: > >> With much help from Julian, gosper.org/hellodoily.gif >> --rwg >> >> On Fri, Sep 30, 2016 at 10:13 AM, Bill Gosper <billgosper(a)gmail.com >> <javascript:;>> wrote: >> >> > On Thu, Sep 29, 2016 at 8:33 PM, Bill Gosper <billgosper(a)gmail.com >> <javascript:;>> wrote: >> > >> >> The animation I intended was perhaps too computationally ambitious. >> >> Meanwhile, gosper.org/FHilbert.gif is just conventional plots of the >> >> first 288 >> >> approximations. I dislike such plots due to the finite line thickness, >> >> which someone might imagine contributed to spacefilling. >> >> If you read all the frames into Preview, say, you can see that the >> texture >> >> of frame n repeats at frame 4n. This gives us a way to interpolate >> >> the nonexistent frame n+1/4 from frame 4n+1. >> >> >> >> Also uploaded: gosper.org/outcirc.gif made by flipping a sign in >> >> incirc.gif. --rwg >> >> Stay tuned for a Ptolemaic Hilbert sweep from a multi-hour computation. >> >> >> > >> > Three hours. Foo, the Symbolics machine could do this nearly in real >> time. >> > gosper.org/hilbert286.gif is the 287 rotor sweep. (288 would >> > have been too gross.) It was to have been a single frame of the >> frequency >> > expansion animation, analogous to gosper.org/FHilbert.gif , but the >> > project >> > would not have contributed much insight into the final "catastrophic" >> > dimension bump. >> > >> > I haven't tried very hard to telescope Julian's 3⨉3 matrix product for >> the >> > Fourier coefficients. Success would provide a flood of strange >> identities >> > like (d247) and (d246) in http://www.tweedledum.com/rwg/idents.htm , >> > which come from the family that includes the Sierpinski Curve and >> > Koch's Snowflake (http://gosper.org/fst.pdf pp137, 138) rather than >> > Hilbert's curve. Failure to telescope would encumber the infinite >> > products in the identities with matrices instead of scalars. --rwg >> > The infinite Fourier series faithfully represents Hilbert's function, >> > except it misses the top two corners (the endpoints of Hilbert's curve) >> > and moves them instead to top center (where the animation begins and >> > ends). This is a quadruple point if you think of the Fourier function on >> > [0,1], but only a triple point if you regard it as a periodic function, >> > since >> > a period contains only one of its endpoints. >> > >> >> >> >> On 2016-09-26 18:42, James Propp wrote: >> >> > I hope Bill will write a lively document (or create a lively video) >> that >> >> > explains what's wrong with so many accounts of spacefilling curves. >> >> > >> >> > (I myself wish that more accounts started in a "Not Knot"-ish vein, >> >> > explaining why the "kindergartner's space-filling curve" --- >> scribbling >> >> > one's crayon back and forth until the square is filled --- isn't a >> >> solution >> >> > to the mathematician's problem.) >> >> > >> >> >> >> >> >> > Is there a way to relax an approximation to a space-filling curve >> in >> >> >> > continuous time so that it works out its kinks and regresses to >> >> simpler >> >> >> > approximations? >> >> >> > >> >> >> > (No interim self-intersections please!) >> >> >> > >> >> >> > Jim Propp >> >> >> >> >> >> http://gosper.org/FDrags128.mp4 >> >> >> >> >> >> If people really want to see it, Julian's recent Fourier matrix >> >> product can >> >> >> produce >> >> >> the analogous animation for Hilbert's "curve". >> >> > >> >> > >> >> > I would like to see it. >> >> > >> >> > But it is a seductive thought crime to view a spacefilling function as >> >> some >> >> >> kind of >> >> >> limit of "spacefilling curves"! Those curves Henry sketches are mere >> >> >> schematics, >> >> >> of no mathematical consequence. This leads inevitably to embarrassed >> >> >> hemming >> >> >> and hawing about how the area jumps from 0 to 1 at the very last >> >> moment, >> >> >> when >> >> >> both interior and exterior suddenly disappear and become boundary. >> >> >> Successive partial sums of the Fourier series are even more >> >> seductive. But >> >> >> no matter how many terms you take, you're still infinitely far from >> the >> >> >> end. >> >> >> --rwg >> >> > >> >> > >> >> > That's one way to look at it. But it depends on the notion of >> path-space >> >> > you use and what metric you put on it, doesn't it? In particular, it's >> >> > crucial to look at paths equipped with a parametrization. Then you >> >> really >> >> > can get convergence to a limit. And the limiting object is a >> continuous >> >> > function from a line segment onto a square. If you just look at the >> >> range >> >> > of a path and not the parametrization, you can't describe the square >> as >> >> the >> >> > limit in any meaningful way. Which I gather is part of Bill's point. >> >> > >> >> > Jim Propp

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[math-fun] Pairwise sums/differences are semiprimes.
by Zak Seidov 28 Oct '16

28 Oct '16

Set of integers that all pairwise sums are semiprimes? E.g., s = {1,3,32,54} with pairwise sums 4,33,55, 35,57, 86 all semiprimes Set of integers that all pairwise differences are semiprimes? E.g., s={ 1, 5, 11, 15, 134, 220} with pairwise differences 4,10,14,133,221, 6,10,129,215, 4,123,209 119,205 86 all semiprimes. -- Zak Seidov

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[math-fun] Golomb rulers and 492116
by Tomas Rokicki 27 Oct '16

27 Oct '16

I thought I would share that Gil Dogon and I, inspired by an Al Zimmermann programming contest, went ahead and calculated subquadratic Golomb rulers as reasonably far as we could. We were able to find subquadratic rulers through 492115 marks. At 492116 marks, none of the known (to us) constructions led to a subquadratic ruler. Is it possible that no subquadratic Golomb ruler exists at 492116 marks? Or is it more likely that one exists, and if so, is it something that we will ever explicitly know? -- -- http://cube20.org/ -- [ <http://golly.sf.net/>Golly link suppressed; ask me why] --

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[math-fun] Cooper/Zudilin: Hypergeometric modular equations
by Joerg Arndt 26 Oct '16

26 Oct '16

Those of us who appreciate that kind of stuff might find the following entertaining. Shaun Cooper, Wadim Zudilin: Hypergeometric modular equations, http://arxiv.org/abs/1609.07276 Best regards, jj

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[math-fun] Mandelbrot''s (frac-7) snowflake fill
by Bill Gosper 25 Oct '16

25 Oct '16

Howard Cannon used Julian's fractal Fourier expander to make gosper.org/mandelfou.png illustrating that it fills some areas unboundedly faster than others. (dArea/dt is interesting.) The function has no symmetry, i.e., for no (nontrivial) a, b does Z(t) = a(Z(b(t)), and there is no symmetry in the filled areas of winding number 1 of the Fourier approximations, such as illustrated. (Except for the fundamental circle and ellipse. gosper.org/mandelfou-1-2-3.png) Yet for infinitely many discrete samplings, the filled (polygonal) area does have symmetry, e.g, gosper.org/blusno.gif . --rwg

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[math-fun] physicist click bait
by Bill Gosper 23 Oct '16

23 Oct '16

http://www.symmetrymagazine.org/article/july-2015/lhc-physicists-discover-f… trumpets for screenfuls the discovery of two pentaquarks. But a URL at the bottom of the page says Oops, never mind. So why don't they warn readers at the top of the article? --rwg

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[math-fun] 50% gray
by Bill Gosper 23 Oct '16

23 Oct '16

On 2016-10-22 15:07, Eugene Salamin via math-fun wrote: > Let's ask for such a set S as a subset of the unit interval instead of > the entire real line. Then S = S intersect [0,1], and so S must have > measure 1/2. So if S can't be measurable, that completes the proof. > But why is S nonmeasurable? > > -- Gene Julian (quoted without permission): >rwg>I thought I read somewhere that there was no subset of the Reals whose > >intersection with every interval has measure density ½. > >Correct. Lebesgue measure is defined (when using Carathéodory) as the inf of the measures of all countable unions of intervals containing the set. Taking a union of intervals with measure ≤1.01 times the measure of the set, one of the intervals >must have measure ≤1.01 times the measure of its intersection with the set, i.e. in that interval the set has density >99%. rwg>Yet I think I see a nifty one. What gives? >Depends on your proposed construction. Most likely, it either gives a non-measurable set, or gives a set of density 0 or 1 (most recursive constructions do this—almost all points will alternate between being in and out), or is not well-defined. > From: James Propp <jamespropp(a)gmail.com> > To: math-fun <math-fun(a)mailman.xmission.com> > Sent: Saturday, October 22, 2016 12:46 PM > Subject: Re: [math-fun] 50% gray > > Are you using the axiom of choice? Or is your set constructible? (Clearly > it's not measurable.) > > Jim Propp I am using hallucinogens in conjunction with the axiom of stupidity. My "solution" was to partition the reals according to the relation of Khinchin's constant to the geometric mean of their partial quotients! (Sorry.) --rwg > > .On Saturday, October 22, 2016, Bill Gosper <billgosper(a)gmail.com> wrote: > >> I thought I read somewhere that there was no subset of the Reals whose >> intersection with every interval has measure density ½. Yet I think I see >> a nifty one. What gives? --rwg >

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[math-fun] 50% gray
by Bill Gosper 22 Oct '16

22 Oct '16

I thought I read somewhere that there was no subset of the Reals whose intersection with every interval has measure density ½. Yet I think I see a nifty one. What gives? --rwg

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[math-fun] https://upload.wikimedia.org/wikipedia/commons/a/ae/Ishihara_19.PNG
by Bill Gosper 22 Oct '16

22 Oct '16

claims to be a red-green deficiency "trick question". You only see the numeral if you're colorblind. But after pasting the image into Mathematica, Rohan and I were unable to find a mapping, e.g., Image[(ImageData@%702) /. {r_, g_, b_, a_} :> {r', g',b',a'}] to reveal the hidden numeral. What's the trick? --rwg

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