April 2019 - mathfun@mailman.xmission.com

[math-fun] "maa.org" has been compromised?
by Henry Baker 19 Apr '19

19 Apr '19

I just received a "phishing" email whose header said it came from webwork.maa.org. This was a very good imitation of a Chase Bank email. There are some real pros out there. Be careful!

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[math-fun] Antipodal New Zealand
by Hans Havermann 19 Apr '19

19 Apr '19

2001 vs 2019, in Mathematica. http://gladhoboexpress.blogspot.com/2019/04/antipodal-new-zealand.html

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[math-fun] One base, two digit sets
by Bill Gosper 19 Apr '19

19 Apr '19

On 2019-04-17 16:27, Adam P. Goucher wrote: > Interesting! Are there any reasons to prefer quater-imaginary over > base i-1? Rectangular vs fractal pure fractions region? Julian's magic piecewiserecursivefractal function, seems to solve membership questions. E.g., it can give the exact real preimages of any complex value of the Heighway Dragon function.Heighway triple point <http://gosper.org/dragtrip!.png>> > As mentioned before, base 3 + omega (with digits 0 and the 6 roots > of unity) is my favourite positional number system for the complex > numbers. The reason being that: > > (a) the sum of any 3 one-digit numbers is a two-digit number; > (b) the product of any 2 one-digit numbers is a one-digit number; > > which together make it possible to build a ripple carry adder out > of 'full adders'. (The same is, of course, true for binary.) > > But balanced ternary, which Knuth himself describes as the most > beautiful of all positional number systems (which I agree with), > satisfies a strengthened version, where '3' is replaced with '4' > in (a). > > So a balanced ternary adder can sum *three* numbers together, > instead of (as is the case in binary) requiring sums to be > performed pairwise. (That's actually quite cute: in binary, the > adder is a binary function; in balanced ternary, it's a ternary > function.) > The magicd function in gasket Fourier <http://gosper.org/gaskettalk.pdf> is a fairly exotic application of balanced ternary. balter[x_Integer] := Sign[x]* (IntegerDigits[Abs[x] + (1/2)* (3^Ceiling[Log[3, 1 + Abs[2*x]]] - 1), 3] - 1) That paper also illustrates an interesting alternative to convergent sums, products, and continued fractions: Infinite products of idempotent matrices with vanishing perturbations. —rwg > > Best wishes, > > > Adam P. Goucher > > >> Sent: Thursday, April 18, 2019 at 12:14 AM >> From: "Bill Gosper" <billgosper(a)gmail.com> >> To: math-fun(a)mailman.xmission.com >> Subject: [math-fun] One base, two digit sets >> >> Knuth has proposed two positional base + digits schemes for representing >> complex numbers. >> Base i-1, digits {0,1}: >> https://en.wikipedia.org/wiki/Complex-base_system#/media/File:ComplexTwindr… >> , >> gosper.org/basei-1.png >> and Base 2i, digits {0,1,2,3}: >> https://en.wikipedia.org/wiki/Quater-imaginary_base >> These systems are closely related, because (i-1)² = -2i . >> So in base 2i, you can get the mirror image of the twindragon system by >> using the digits {0, 1, - i - 1, -i} instead of {0,1,2,3}! —rwg >> ______

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[math-fun] Double normal puzzle + question
by Dan Asimov 17 Apr '19

17 Apr '19

Some time ago I wondered if there could be a smooth simple closed curve in the plane without a double normal — a chord of the curve that is perpendicular at both ends. This is a fun puzzle which I won't spoil for anyone who hasn't seen it. But today I wondered about the same question for a smooth surface in R^3 topologically equivalent to the 2-sphere S^2. Now for many smooth S^2 surfaces M — like convex ones — following the internal normal N(p) at point p of M until it first hits the surface, say at point Q(p) — the map Q : M —> M taking p to Q(p) is a continuous. If so, then having obtained Q(p) from p we now follow the internal normal N(Q(p)) at Q(p) until it first intersects the tangent plane T_p(M) of M at p. Let this point be f(p). Assuming no double normal, we know f has no fixed points: for all p in M, f(p) ≠ p. But then f(p) - p provides a vector field on S^2 continuous in p, known to be impossible (the hairy ball theorem), contradiction. But if that map Q above is *not* continuous, this argument doesn't work. So, Question: Does there exist a smooth S^2 in R^3 without a double normal? —Dan

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[math-fun] Mathematica CDF player
by Fred Lunnon 17 Apr '19

17 Apr '19

I'm considering implementing a solid honeycomb / 3-D tiling demo for the CDF player, and soliciting feedback concerning how well it works for author and user. I also wonder how in practice casual users might feel about going to the trouble of downloading this player to view interactive video material. Fred Lunnon

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[math-fun] One base, two digit sets
by Bill Gosper 17 Apr '19

17 Apr '19

Knuth has proposed two positional base + digits schemes for representing complex numbers. Base i-1, digits {0,1}: https://en.wikipedia.org/wiki/Complex-base_system#/media/File:ComplexTwindr… , gosper.org/basei-1.png and Base 2i, digits {0,1,2,3}: https://en.wikipedia.org/wiki/Quater-imaginary_base These systems are closely related, because (i-1)² = -2i . So in base 2i, you can get the mirror image of the twindragon system by using the digits {0, 1, - i - 1, -i} instead of {0,1,2,3}! —rwg

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[math-fun] Custom number crunching programs --- expectations?
by Andres Valloud 17 Apr '19

17 Apr '19

Hello, I've used a GNFS factoring package, and it seemed to have been modeled after a C program targeting POSIX, and emitting output to stdout. I also wrote similar programs with the same mode of operation. Have you had experience with custom number crunching programs like that? How did you interacting with them? What would you expect? Thanks in advance. Andres.

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Re: [math-fun] That curve I regrettably named flowsnake
by Bill Gosper 17 Apr '19

17 Apr '19

On 2019-04-10 06:10, Joerg Arndt wrote: > Pics or it didn't happen! > Agreed: https://www.wolframcloud.com/objects/797a1110-f0cd-47a5-90f6-f0b3d1d8ce4b > Joking aside: I'd like to understand, but don't. > > Best regards, jj > > > * Bill Gosper <billgosper(a)gmail.com> [Apr 08. 2019 15:50]: >> On Sat, Apr 6, 2019 at 9:20 AM Bill Gosper <billgosper(a)gmail.com> wrote: >> >> > Mandelbrot detested the name. Then he discovered a spacefill of Koch's >> > Snowflake. >> > The worthy claimant to the Flowsnake name! Which he still detested. >> > I recently realized how little I know about my own curve, which fills the >> > FranceFlake, the fundamental region for the positional number system >> > Base: 2 + i^(2/3) = √7 e^(i arccos(11/14)/2) >> > and the (necessarily) seven digits 0 and the 6th roots of 1. >> > >> >> SPOILERS >> >> > If FranceFill(0)= 0+0i, FranceFill(1) = 1 + 0i: >> > 1) What is the area of a FranceFlake? >> > >> Only √3/2. >> >> > 2) Where is its centroid? >> > >> 1/2+i/2/√3 >> >> > 3) Is its circumradius √(52/147) ? >> > >> No. ≥ √(6140317/3)/2401. (What the heck is this approaching?) >> The same as the FranceFlake, duh. >> >> > 4) Ironically, it was popular partly for how completely its canonical >> > sampling (at 0/7, 1/7,...,1) self-avoided. But like all true spacefills, >> > it is *dense* with *triple* points. >> > E.g., what are the preimages of (9 + i √3)/21? >> > >> {5/42, 11/42, 17/42} >> Are there quadruple points? How can we prove "No."? >> >> > —rwg

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[math-fun] Substitution Tilings
by Ed Pegg Jr 17 Apr '19

17 Apr '19

I've recently found a lot of new substitution tilings. They are listed in the program at http://demonstrations.wolfram.com/SubstitutionTilings/ -Ed Pegg Jr

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[math-fun] The St. Ives tessellation
by Dan Asimov 14 Apr '19

14 Apr '19

Let the map-of-France shape M be the outer boundary of a (limiting) flowsnake: M_0 = a 2D regular hexagon, including its boundary curve. M_(n+1) = a rosette of 7 copies of M_n, after the rosette has been shrunk to total volume = 1 (linear factor of 1/sqrt(7)) and a slight rotation ... ... such that lim M_n = M. Since at each stage the M_n tessellate R^2, so does M. Call this tessellation by T_0. Then each tile of T_0 (a copy of M) is a rosette of 7 smaller ones. Call the tessellation by *those* by T_1, lather, rinse, repeat. Since a tessellation is nothing more than the set whose members are its tiles: Finally let T denote the set of all tiles of any T_n: T = T_0 u T_1 u ... u T_n u ... Whew! Now consider the *symmetry group* of all similarity mappings taking *any tile in T* to *any tile (possibly the same one) in T*, in *any rotational position* (there are 6). (Flips are excluded, of course, since all of M's isometries with itself are orientation preserving.) This is a very interesting group that I know little about. * * * What is the 1-dimensional version of this? That looks like a pretty interesting group G as well. A member of this group G takes any interval (K/2^n, (K+1)/2^n) to any interval of the same form, i.e., (L/2^m, (L+1)/2^m): (K/2^n, (K+1)/2^n) —> (L/2^m, (L+1)/2^m) by either preserving or reversing orientation and expanding length by a constant factor, i..e., by the map x |—> ax + b determined by (y - (L+1)/2^m) / (x - L/2^m) = ((L+1)/2^m - (K+1)/2^n) / (L/2^m - K/2^n). —Dan

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