November 2013 - mathfun@mailman.xmission.com

[math-fun] ice cube finds high energy neutrinos
by Warren D Smith 21 Nov '13

21 Nov '13

http://www.cbpf.br/~icrc2013/papers/icrc2013-0650.pdf the highest energy neutrino they detected was claimed to be 1200 TeV, which is pretty absurd (equivalent to about 1.3 million proton masses). This seems far greater than could be produced in any nuclear reaction, and neutrinos since neutral cannot be accelerated. The only explanation I can think of (besides ice cube just being wrong) was some charged particle got accelerated to absurd energies, then emitted a neutrino.

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[math-fun] recent gamma ray bursts
by Warren D Smith 21 Nov '13

21 Nov '13

A.Maselli & 57 others: GRB 130427A: A Nearby Ordinary Monster, http://www.sciencemag.org/content/early/2013/11/20/science.1242279.abstract claims "long duration gamma bursts" come from large-star collapses and in this case immediately followed by a supernova. This event on 27 april 2013 was the brightest gamma flash ever seen at 3*10^46 watts. It was 4 billion light years away. Optical glow was detected by human observers 20 minutes after the gamma pulse (or perhaps less, but anyway there was a guy optically seeing it starting at 20 minutes). NASA's "swift" satellite detected the burst (as it does for 100 bursts/year) It relays a 3 arcminute position estimate to the ground within 20 seconds of the initial detection. It also then slews over to observe the burst with Xray and UV+optical telescopes, but this slewing takes 90 seconds. In this particular case it took 140 seconds to get the Xray telescope to see it and 147 for the UV/optical telescope. These then can provide more accurate position estimates on the order of 0.5 to 5 arcseconds, in this case 5. http://swift.gsfc.nasa.gov/about_swift/ Several gamma bursts seen by swift have been over 13 billion light years distant (based on optical red shifts) e.g. http://en.wikipedia.org/wiki/GRB_090423 which had gammas lasting 10 sec. Swift on this one failed to see it in the optical despite trying starting 77 sec later. But optically it was seen in the infrared 21 min later using ground based telescope in Hawaii. GRB080913 was also 13 GLyr away http://www.nasa.gov/mission_pages/swift/bursts/farthest_grb.html and was seen "in less than 2 minutes" by swift in the Xrays, and then after 1 further minute it was seen by ground based telescopes. The short gamma bursts appear due to mergers of compact objects (e.g. neutron stars). -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Primes in Intervals of Bounded Length
by Hans Havermann 20 Nov '13

20 Nov '13

"In April 2013, Yitang Zhang proved the existence of a finite bound B such that there are infinitely many pairs of distinct primes which differ by no more than B... In November, inspired by Zhang’s extraordinary breakthrough, James Maynard dramatically slashed this bound to 600, by a substantially easier method. Both Maynard, and Terry Tao who had independently developed the same idea, were able to extend their proofs to show that for any given integer m >= 1 there exists a bound B_m such that there are infinitely many intervals of length B_m containing at least m distinct primes. We will also prove this much stronger result herein, even showing that one can take B_m = e^(8m+5)." Andrew Granville has made available a pdf of this paper: http://www.dms.umontreal.ca/~andrew/CEBBrochureFinal.pdf

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Re: [math-fun] Newman = -1/Oldman
by Bill Gosper 18 Nov '13

18 Nov '13

Fred>On 11/17/13, Bill Gosper <billgosper(a)gmail.com> wrote: On Thu, Nov 14, 2013 at 3:39 AM, Bill Gosper <billgosper(a)gmail.com> wrote: This must have been found by Newman himself but I only just noticed. ... Hypothesis: the thread topic is (distantly related to) Calkin-Wilf and Stern-Brocot trees --- see eg. http://en.wikipedia.org/wiki/Calkin%E2%80%93Wilf_tree Adam>Here's one I prepared earlier: http://cp4space.wordpress.com/2013/10/24/enumerating-the-rationals/ Sincerely, Adam P. Goucher Eye gotcha: Plotting renders both functions newm[x_] := 1/(1 - x + 2 Floor[x]) oldm[x_] := 2*Ceiling[#] - 1 - # &[1/x] Plot[{newm[r], oldm[r]}, {r,-π,π}, AspectRatio -> Automatic] visually indistinguishable from odd ones, although the same could be said of Floor[x+1/2]. Plot[oldm[oldm[oldm[x]]], {x,-π,π}] reminds me of graphs in Gödel, Escher, Bach. Strangely, the minor change Clear[newm]; newm[x_?NumericQ] := 1/(1 - x + 2 Floor[x]) Plot[newm[newm[x]], {x,-π,π}] ruins the effect by trying to draw a continuous function connected by bogus "verticals". --rwg

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[math-fun] gray code and hilbert curve connection
by Warren D Smith 18 Nov '13

18 Nov '13

The D-hypercube has 2^D vertices that are the D-bit binary words. A "gray code" is a hamiltonian tour on those 2^D vertices using edges to nearest neighbors only, i.e. each gray word differs from predecessor at a single bit. There are many gray codes when D is large, but some are nicer and easier to deal with than others. A Hilbert curve is a space filling curve filling the D-hypercube at unit density i.e. maps the real interval [0,1] into the cube continuously. To do it, partition [0,1] into 2^D equal subintervals, each is mapped into one of the 2^D corner half-size subcubes via a re-scaled Hilbert curve. The subcubes are ordered in gray code order. The key is to rotate and/or flip each subcube appropriately (by permuting and/or negating the D coordinates) so all the little half hilberts stitch together to make a full hilbert. OK, understand now? Now to make good elegant efficient program for this, it would if we had a fast simple routine to compute the Nth gray-word and which bit changes to reach its successor (or predecessor). I'm rather ignorant about grayness, but google found the following brilliant: uint gray(uint N){ return( N ^ (N >> 1) ); } will return the Nth gray code word! To translate that C code into english: unsigned integer gray(unsigned integer N){ return( N XOR RightShift(N) ); } I believe in a recent mathfun post it was explained how to produce the inverse of this function in O(log(D)) runtime... The goal is good hilbert code that fits on 1 page or less. It is desirable to have inverse routine too, i.e. inputs D coordinates and outputs t. The Peano-Smith curve seems way simpler than Hilbert and also basically unique (unlike hilbert which because gray codes are very non-unique, are too). But Peano-Smith naturally lives in a hyper-rectangle with edge lengths 3^(k/D) for k=1..D, not a hypercube. Needless to say, the appropriate name for D-dimensional hilbert is "Dilbert." -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] multi-D hilbert curve
by Bill Gosper 18 Nov '13

18 Nov '13

3D (e.g.) Hilbert is easy with the finite state machine. Three bits of t in, one bit of x, y, and z out, plus state change. --rwg --well... actually figure out the state machine for arbitrary D, and do it in a way that we do not need no stinking huge 2^D entry table to store the state-rules, and then I'm sold. There's some disgusting computer programs out there I found with google to do multi-D hilbert curves, but there really ought to be a nice way that takes under 1 page of code. And they haven't found it. --wdsmith OK, but first, a vid of Gareth Conway <http://daftmusings.com/wp-content/uploads/2011/06/G4G9-052.jpg> bouncing on a steel one: http://blog.makezine.com/archive/2010/11/math_monday_3d_hilbert_curve_in_st… Here's some relevant old funmail. The three pngs at the end have as usual vanished from gosper.org, so I wrote a three page pdf of the 208 KB notebook containing them. 78.8MB!! gosper.org/Trilbertloopscenterscorners.pdf If Adobe Reader ever finishes "optimizing" it, I'll rewrite it. I could write a png, but then you couldn't copy the text. On 2011-01-05 14:39, Bill Gosper wrote: > Last year I sent one named Treano and posted some graphs. It had > three problems: > 1) As Neil Bickford points out, it should have been named Trilbert. > 2) It spanned a face diagonal instead of an edge, creating > inconsistent textures > when sampled at fixed frequencies. > 3) It was so obscure that when I found the edge-spanning construction, I > couldn't decipher and modify my own code. > > Fortunately, Corey Ziegler Hunts was up to both tasks: > I3 = IdentityMatrix[3]; > > In[720]:= Vects = {{0, 0, 0}, {0, 0, 1}, {0, 1, 1}, {0, 2, 1}, > {1, 2, 1}, {2, 2, 1}, {2, 1, 1}, {2, 0, 1}, {2, 0, 0}}/2; > > Mats = {{{0, 0, 1}, {0, 1, 0}, {-1, 0, 0}}.{{-1, 0, 0}, {0, 1, 0}, {0, > 0, 1}}, {{0, -1, 0}, {1, 0, 0}, {0, 0, 1}}.{{1, 0, 0}, {0, > 0, -1}, {0, 1, 0}}, {{0, -1, 0}, {1, 0, 0}, {0, 0, 1}}.{{1, 0, > 0}, {0, 0, -1}, {0, 1, 0}}, {{1, 0, 0}, {0, 0, -1}, {0, 1, > 0}}.{{1, 0, 0}, {0, -1, 0}, {0, 0, 1}}, {{1, 0, 0}, {0, > 0, -1}, {0, 1, 0}}.{{1, 0, 0}, {0, -1, 0}, {0, 0, 1}}, {{0, 1, > 0}, {-1, 0, 0}, {0, 0, 1}}.{{1, 0, 0}, {0, 0, -1}, {0, 1, > 0}}, {{0, 1, 0}, {-1, 0, 0}, {0, 0, 1}}.{{1, 0, 0}, {0, > 0, -1}, {0, 1, 0}}, {{0, 0, -1}, {0, 1, 0}, {1, 0, 0}}.{{-1, 0, > 0}, {0, 1, 0}, {0, 0, 1}}, > {{1, 0, 0}, {0, 1, 0}, {0, 0, 1}}}/2; > > > In[722]:= Trilbert[t_, mat_: I3, vect_: {0, 0, 0}] := > (Trilbert[t, b1_: I3, b0_: {0, 0, 0}] = > (Trilbert[t, matend_: I3, vectend_: {0, 0, 0}] := > Inverse[matend - mat].(vect - vectend); > Module[{t8 = 8*t, n}, > n = Floor[t8] + 1; > Vects[[n]] + > Mats[[n]].Trilbert[FractionalPart[t8], mat.Mats[[n]], > vect + mat.Vects[[n]]]])) > > (style-tweaked in ways Corey might disavow). > > The inverse images of the eight vertices are > > In[724]:= Trilbert /@ {0, 5/28, 9/28, 3/7, 4/7, 19/28, 23/28, 1} > > Out[724]= {{0, 0, 0}, {0, 0, 1}, {0, 1, 1}, {0, 1, 0}, {1, 1, 0}, > {1, 1, 1}, {1, 0, 1}, {1, 0, 0}} > > (which you couldn't directly test without the exact rational trick). > There are eight(!) inverse[s] of the center point: > > In[729]:= {19/28, 1 + 23/28, 2 + 5/28, 3 + 23/28, 4 + 5/28, 5 + 23/28, > 6 + 5/28, 7 + 9/28}/8 > > Out[729]= > 19 51 61 107 117 163 173 205 > {---, ---, ---, ---, ---, ---, ---, ---} > 224 224 224 224 224 224 224 224 > > Trilbert /@ % > Out[728]= > 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 > {{-, -, -}, {-, -, -}, {-, -, -}, {-, -, -}, {-, -, -}, {-, -, -}, {-, -, -}, {-, -, -}} > 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 > > Every interior triple of dyadic rationals has eight inverses. > > Plots: > http://gosper.org/Trilbertloops.png > http://gosper.org/Trilbertcenters.png > http://gosper.org/Trilbertcorners.png > --rwg > (While I was typing this, a popup said "Mathematica quit unexpectedly". > When I clicked Restart, it said my license had expired, demanding a > password, which I requested upon the first warning Christmas Eve. Why > does WRI synchronize this with everyone's vacation?) > It turns out the unexpected quit was due to a 7.0 bug eating the entire > Apple VM while trying to convert to a .pdf the small .nb containing these > three plots. > This version of Trilbert used the exact rational trick. Corey's faster but lengthier dyadic rational definition: ClearAll[Trilbert]; Trilbert[x_, funcs_: Table[{0 &, 1, 1}, {i, 1, 9, 1}]] := Block[{vects, mats, reflect, vect = {0, 0, 0}, mat = IdentityMatrix[3], vectr, matr, digits, digit, nonrep = 0, rep, special, period, i}, vects = {{0, 0, 0}, {0, 0, 1/2}, {0, 1/2, 1/2}, {0, 1, 1/2}, {1/2, 1, 1/2}, {1, 1, 1/2}, {1, 1/2, 1/2}, {1, 0, 1/2}, {1, 0, 0}}; mats = Table[0, {i, 1, 9, 1}]; mats[[1]] = (1/ 2)*{{0, 1, 0}, {-1, 0, 0}, {0, 0, 1}}.{{0, 0, -1}, {0, 1, 0}, {1, 0, 0}}; mats[[2]] = (1/ 2)*{{0, -1, 0}, {1, 0, 0}, {0, 0, 1}}.{{1, 0, 0}, {0, 0, -1}, {0, 1, 0}}; mats[[3]] = (1/ 2)*{{0, -1, 0}, {1, 0, 0}, {0, 0, 1}}.{{1, 0, 0}, {0, 0, -1}, {0, 1, 0}}; mats[[4]] = (1/ 2)*{{1, 0, 0}, {0, 0, -1}, {0, 1, 0}}.{{1, 0, 0}, {0, 0, -1}, {0, 1, 0}}; mats[[5]] = (1/ 2)*{{1, 0, 0}, {0, 0, -1}, {0, 1, 0}}.{{1, 0, 0}, {0, 0, -1}, {0, 1, 0}}; mats[[6]] = (1/ 2)*{{0, 1, 0}, {-1, 0, 0}, {0, 0, 1}}.{{1, 0, 0}, {0, 0, -1}, {0, 1, 0}}; mats[[7]] = (1/ 2)*{{0, 1, 0}, {-1, 0, 0}, {0, 0, 1}}.{{1, 0, 0}, {0, 0, -1}, {0, 1, 0}}; mats[[8]] = (1/ 2)*{{0, -1, 0}, {1, 0, 0}, {0, 0, 1}}.{{0, 0, 1}, {0, 1, 0}, {-1, 0, 0}}; mats[[9]] = (1/2)*{{1, 0, 0}, {0, 1, 0}, {0, 0, 1}}; reflect = {{1, 0, 0}, {0, 0, 1}, {0, 1, 0}}; If[And[0 < x, x < 1], digits = RealDigits[x + 1, 8][[1, 2 ;; -1]] + 1, If[x == 0, digits = {1}, digits = {9} ] ]; mat = MatrixPower[reflect, funcs[[1, 1]][0]]; For[{i = 1}, And[i <= Length[digits], ! ListQ[digits[[i]]]], i++, digit = digits[[i]]; vect = vect + mat.vects[[digit]]; mat = mat.mats[[digit]].MatrixPower[reflect, funcs[[digit, 1]][i]]; nonrep++ ]; If[ListQ[digits[[-1]]], digits = digits[[-1]]; rep = Length[digits]; special = Max[funcs[[Union[digits], 2]]] - nonrep; For[{i = 1}, i <= special, i++, digit = digits[[Mod[i - 1, rep] + 1]]; vect = vect + mat.vects[[digit]]; mat = mat.mats[[digit]].MatrixPower[reflect, funcs[[digit, 1]][i + nonrep]] ]; If[special < 0, special = 0 ]; vectr = {0, 0, 0}; matr = IdentityMatrix[3]; period = Apply[LCM, Append[funcs[[Union[digits], 3]], Length[digits]]]; For[{i = 1}, i <= period, i++, digit = digits[[Mod[i + special - 1, rep] + 1]]; vectr = vectr + matr.vects[[digit]]; matr = matr.mats[[digit]].MatrixPower[reflect, funcs[[digit, 1]][i + nonrep + special]] ]; vect + mat.Inverse[IdentityMatrix[3] - matr].vectr, vect ] ] Block[{order = 2, max = 85/126, funcs, data, str, cases, i}, funcs = Table[{0 &, 0, 1}, {i, 1, 9, 1}]; funcs[[1]] = {Piecewise[{{0, # == 0}, {1, # < order - 1}, {0, # == order - 1}}, 0] &, 2}; funcs[[2]] = {Piecewise[{{1, # < order - 1}, {0, # == order - 1}}, 0] &, order, 2}; funcs[[3]] = {Piecewise[{{1, # < order - 1}, {0, # == order - 1}}, 0] &, order, 2}; funcs[[4]] = {Piecewise[{{0, # < order - 1}, {1, # == order - 1}}, 0] &, order, 2}; funcs[[5]] = {Piecewise[{{0, # < order - 1}, {1, # == order - 1}}, 0] &, order, 2}; funcs[[6]] = {Piecewise[{{1, # < order - 1}, {0, # == order - 1}}, 0] &, order, 2}; funcs[[7]] = {Piecewise[{{1, # < order - 1}, {0, # == order - 1}}, 0] &, order, 2}; funcs[[8]] = {Piecewise[{{1, # < order - 1}, {0, # == order - 1}}, 0] &, order, 2}; data = Table[ Trilbert[ Block[{num = (i + max/8)/(8^order)}, If[And[0 <= num, num <= 1], num, Mod[num, 1]]], funcs], {i, 0, 8^order - 1}]; str = StringJoin @@ (Differences[data]*(2^order) /. {{1, 0, 0} -> "x", {0, 1, 0} -> "y", {0, 0, 1} -> "z", {-1, 0, 0} -> "a", {0, -1, 0} -> "b", {0, 0, -1} -> "c"}); cases = DeleteDuplicates[StringCases[str, a__ ~~ a_]]; Print[ If[Length[cases] < 1, cases, Table[{cases[[i]], StringPosition[str, cases[[i]]]}, {i, 1, Length[cases], 1}] // MatrixForm ] ]; str = StringJoin @@ (Abs[ Differences[data]*(2^order)] /. {{1, 0, 0} -> "x", {0, 1, 0} -> "y", {0, 0, 1} -> "z"}); cases = DeleteDuplicates[StringCases[str, a__ ~~ a_]]; Print[ If[Length[cases] < 1, cases, Table[{cases[[i]], StringPosition[str, cases[[i]]]}, {i, 1, Length[cases], 1}] // MatrixForm ] ]; Graphics3D[Tube[data, 1/50], Axes -> True, AxesLabel -> {"x", "y", "z"}, PlotRange -> {{-1/20, 21/20}, {-1/20, 21/20}, {-1/20, 21/20}}] ] That same notebook contains an isomorphic pair of statements defining and invoking "Spirilbert", which sweeps the cube in a sort of spiral, instead of by octants. --rwg

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Re: [math-fun] Newman = -1/Oldman
by Adam P. Goucher 18 Nov '13

18 Nov '13

> APG: thankyou for that. For "Newnham’s" read "Newman's", perhaps? WFL Oops, Freudian slip... http://en.wikipedia.org/wiki/Newnham_College,_Cambridge Sincerely, Adam P. Goucher > On 11/17/13, Adam P. Goucher <apgoucher(a)gmx.com> wrote: > >> Fred Lunnon wrote: > >> > >> I should be much obliged if somebody would confirm or refute this > >> conjecture, > >> if possible attaching relevant references. > > > > > > Here's one I prepared earlier: > > > > http://cp4space.wordpress.com/2013/10/24/enumerating-the-rationals/ > > > > > > > > Sincerely, > > > > > > Adam P. Goucher > > > > _______________________________________________ > > math-fun mailing list > > math-fun(a)mailman.xmission.com > > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > >

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[math-fun] The Giphoscope turns your favorite GIF into a hand-cranked work of art
by Henry Baker 18 Nov '13

18 Nov '13

FYI -- Perhaps some wonderful animated math gifs would look good on this 19th century device? Perhaps a math museum? http://www.theverge.com/2013/11/17/5112314/the-giphoscope-turns-your-favori… The Giphoscope turns your favorite GIF into a hand-cranked work of art By Bryan Bishop on November 17, 2013 06:01 am @bcbishop 26 If you were to accuse us of being fans of GIFs we wouldn't disagree, but what happens when you want to share that perfect, singular moment away from the confines of a computer screens? One particularly beautiful solution is called The Giphoscope. The work of Marco Calabrese and Alessandro Scali, it's made up of an aluminum frame mounted on a base of italian walnut. Your GIF is then broken down into individual images, which are mounted onto individual cards around a central core. Cranking through the images relies on persistence of vision to create the looped movement of the original GIF, mashing up your imagery with the hand-crafted charm of some of the earliest motion picture devices (the creators point to the Mutoscope as their inspiration). http://en.wikipedia.org/wiki/Mutoscope Each individual unit is put together by hand, with the type of aluminum and the color of the wood and screws all specifically chosen to work in concert with the GIF itself. It's a breathtaking piece of design and craftsmanship, though with prices starting at Â299 (around $403) you'll certainly be paying for that attention to detail. Of course, when it comes to creating the perfect GIF Â or finding the perfect way to show it off Â is money really a factor?

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[math-fun] Why twisted ropes don't untwist under load
by Henry Baker 17 Nov '13

17 Nov '13

[This explanation _begs_ for some pretty animations.] Consider a _single-level_ rope made of 2 individual continuous/unbroken fiber strands. We assume that these individual fiber strands are completely limp, with no resistance to bending, _no resistance to twisting_, and a typical linear spring constant stiffness to longitudinal stretching. If both untwisted fibers are of length L>>1 and radius r, then twisting them to achieve a helical angle of alpha, where alpha=0 means zero twist, will shorten the overall rope, and the twisting torque can be calculated by how much the rope length will change with a given change in twisting angle. Let us consider a unit disk in the XY plane, and a parallel unit disk around (0,0,L). We attach fiber #1 at (1,0,0) and at (1,0,L) and fiber #2 at (-1,0,0) and at (-1,0,L). We assume r<1, and both fibers are straight and parallel. If we then rotate the disk at Z=L by a certain angle alpha, then not only will the entire rope be twisted by alpha, but each fiber strand will also be twisted by alpha, but since we have assumed that torquing an individual fiber encounters no resistance and does not change its length, the overall behavior of a one-level rope of 2 strands is to untwist (unravel) under load. Let us now consider a _two-level_ rope, twisted from 2 strands, each of which is itself twisted from 2 strands of individual fibers. Let us start with all 4 fibers straight. Fiber #1 goes from (1,r,0) to (1,r,L); Fiber #2 goes from (1,-r,0) to (1,-r,L); Fiber #3 goes from (-1,r,0) to (-1,r,L); Fiber #4 goes from (-1,-r,0) to (-1,-r,L). If we now rotate the disk at Z=L by angle alpha, then all 4 fibers will wrap around each other in a _single-level_ rope with 4 strands, and since all the fibers are wrapped in the same direction, untwisting it by angle alpha will lengthen the rope, and thereby impart a torque tending to unravel the rope. But consider now what happens if we simultaneously twist the 2 lower (smaller) level strands by +alpha and twist the 2 higher (larger) level strands by -beta, such that the untwisting torque of -alpha is matched by the twisting torque of +beta. The overall tendency of the entire 2-level rope is to change length (with some sort of spring constant k) under load, but to have _no net twisting torque_. (We note that fixing the ends of the rope so that the inner strands cannot rotate against one another is essential, else the inner rotation will allow the outer torque to become non-zero, and the rope will then unravel itself catastrophically.) If we have a 2-level rope made up of m inner strands and n outer strands, and the inner strands are round fibers of radius r, then we can calculate the overall properties of the rope. (A slightly more accurate model would have the radius r of each fiber change slightly under longitudinal load, so that the _volume_ of a particular piece of fiber remains constant when it is stretched.) We can also relax the constraint on continuous/unbroken fibers. Since loading a 2-level rope increases the fiber-to-fiber friction (as in our previous untwisted WDSmith model), under normal operating conditions this friction is enough to keep the individual fibers from slipping against one another, so that a break in one individual fiber will only reduce the carrying capacity of the rope by a factor of 1/(m*n). There are still some interesting questions about what different "tiling patterns" occur with different integers m & n, where "tiling patterns" means how the various strands at level 0 interact with the strands at level 1 where they meet (and possibly nest in indentations), with depending on the different angles of twisting alpha & beta.

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[math-fun] space filling me, triple horror, and WDS should use "reply"
by Joerg Arndt 17 Nov '13

17 Nov '13

Triggered by the sheer horror of attempting to use RWG's triple-moerder-henker-OMG.pdf I produced a pdf containing 238 images (often in color) each one a grid-filling curve, together with the L-system that generates them. http://jjj.de/tmp-snap/ (there snap-2013.11.17.pdf) The file snap-2013.11.17-toc.txt gives the names of the pdfs for the individual pages, in http://jjj.de/3frac/ Happy color printing! To stick to the horror theme, here is a make-target I used for each of the pdfs: --------------------- .PHONY: z z: eval $$(xcb -p 0) --------------------- If this is not powerful (and bloody dangerous), then what is? ... and WDS should really use "reply" more often. Beer o'clock, jj

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