January 2013 - mathfun@mailman.xmission.com

[math-fun] Go to Mars. Get your tickets now.
by Warren Smith 10 Jan '13

10 Jan '13

http://herocomplex.latimes.com/2013/01/09/life-on-mars-dutch-company-to-off… Sorry, this does not seem to be "math." Nor even halfway logical.

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[math-fun] samosas
by Bill Gosper 09 Jan '13

09 Jan '13

WFL>[Thread mysteriously split into two ...] There seems to be a growing latency before posts to math-fun are appearing, leading people to respond to stale posts. BobB> Does anyone have the equations for either a Reuleaux or a Meissner tetrahedron that could be plugged into Mathematica? Try hal.archives-ouvertes.fr/docs/00/38/51/13/*PDF*/*spheroforms*.*pdf* Here's my attempt at plain Reuleaux. It produces a plausible Plot, but integrates to an absurd face area. Perhaps someone will be good enough to designate my error. If it's truly egregious, I point the finger of blame squarely at Neil, whose directions I was trying to follow. (His 3D printerware intersected the spheres for him.) In[728]:= four = PolyhedronData["Tetrahedron", "Faces"][[1]] Out[728]= {{0, 0, Sqrt[2/3] - 1/(2 Sqrt[6])}, {-(1/(2 Sqrt[3])), -(1/2), -(1/( 2 Sqrt[6]))}, {-(1/(2 Sqrt[3])), 1/2, -(1/(2 Sqrt[6]))}, {1/Sqrt[ 3], 0, -(1/(2 Sqrt[6]))}} (*Centroid is {0,0,0}:*) In[729]:= Plus @@ % Out[729]= {0, 0, 0} (*Edge lengths = 1 *) In[730]:= EuclideanDistance @@ # & /@ Permutations[%%, {2}] Out[730]= {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1} In[731]:= ParametricPlot3D[ Table[{t, u, 1 - t - u}.Drop[four, {k}], {k, 4}], {t, 0, 1}, {u, 0, 1 - t}] (*plots the tetrahedron*) In[740]:= Table[# /. Solve[EuclideanDistance[four[[k]], #] == 1, v][[2]] &[ v*{t, u, 1 - t - u}.Drop[four, {k}]], {k, 4}] (*Purporting to project each flat face from {0,0,0} onto the unit sphere centered at the opposite vertex. Plot looks plausible:*) In[741]:= ParametricPlot3D[%, {t, 0, 1}, {u, 0, 1 - t}] (*Try for surface area of one face:*) In[742]:= Simplify[Cross[D[%740[[1]], t], D[%740[[1]], u]]] (*half a page of quadratic surds*) In[743]:= Simplify[Sqrt[%.%], 0 < t < 1 && 0 < u < 1] (*√ half a page of quadratic surds*) In[744]:= Integrate[%, {t, 0, 1}, {u, 0, 1 - t}] (* No progress *) In[745]:= N[%] Out[745]= 8.37513 This is way off the wall--should be <1. I'm being incredibly dumb. --rwg

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[math-fun] stupid 3d graphics problem
by Henry Baker 08 Jan '13

08 Jan '13

A usual trick in modeling 3D surfaces is to "triangulate" them with triangles that could in principle be hooked together with hinges. Q. Is it possible to "quadrangulate" a 3D surface with _only_ (planar) quadrilaterals "hinged" together ? (A quadrilateral has to have 4 non-trivial sides, so no triangles as limiting cases.)

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[math-fun] Can't you follow directions?
by Bill Gosper 06 Jan '13

06 Jan '13

You know those black and white traffic signs at stoplights that say, e.g., "U Turn from left lane only" only the only text is "ONLY" and the rest is various arrows? Can you top this <http://gosper.org/arrows.JPG>? California has legalized driverless cars. As a step toward mandating them. --rwg

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Re: [math-fun] USA Today's fake photo of forming solar system
by Phil Carmody 06 Jan '13

06 Jan '13

> From: Warren Smith <warren.wds(a)gmail.com> > >From: Phil Carmody <thefatphil(a)yahoo.co.uk> > >Anyway, back to: > > > http://www.usatoday.com/story/tech/sciencefair/2013/01/02/planet-formation-… > > It's a lot less remarkable than most paintings I see in the local art > galleries near here. You didn't believe USAToday when they credited it ^^^^^^^^^^^^^^^^^ > "(Photo: M. Kornmesser, ALMA (ESO/NAOJ/NRAO))", did you? It's just an ^^^^^^^^ > artist's impression, that's all. > > The actual *photo* is here: > http://www.eso.org/public/archives/images/screen/eso1301b.jpg > > --hmm. The USA Today I got this from said "photo" not "artist impression" Ack, see above. That was the bit that they, and others like them, did wrong. > and, what really sealed the deal to my gullible mind, was they also > gave a VIDEO not just still, which I foolishly thought actually meant > it was, like, real data from ESO. It was real *PR* from ESO. That's what (IMHO) they did wrong. > Not just an attempt to be even more artistic. > But when you think about it, you realize there is no way in hell that > star-in-process-of-forming is going to be flying across the background > field of fixed stars at that high a rate, so this video was actually a complete > fabrication!! That's really really obnoxious to a degree I would have > thought unbelievable. Lower my opinion of the press to new lows. The video was ESO's PR too. It was gag-makingly bad. Heads, or PR-agency contracts, should roll. (Again, IMHO.) I know some in the field who are very much pro this kind of "give them pretty pictures to hook them when they're young" approach, so I know that My Honest Opinion is just that and nothing more. I am hoping to have a chat about this very example with one of them real soon now... The press' involvement in the scam was just a small (one word wrong) bit of sloppiness. (The same magnitude of error that can turn "rumours" into "reports", for example.) > The real "photo" which by the way was from radio telescope array, not optical, > is not nearly as nice and convincing as the fake. Is it legitimate to > draw the conclusion expressed by the fake? Not from a single photo. I'm sure they have many, and have studied them closely together before arriving at their conclusion. For example, the smears which are claimed to be bars of dust presumably rotate. I'm sure the scientists are doing their job responsibly. Alas they don't seem to share all their source (or processed) data. Phil -- () ASCII ribbon campaign () Hopeless ribbon campaign /\ against HTML mail /\ against gratuitous bloodshed [stolen with permission from Daniel B. Cristofani]

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Re: [math-fun] poisoned wine bottles
by Veit Elser 06 Jan '13

06 Jan '13

After three hours wax-&-wane found the following 50/4 code: {225, 593, 92, 198, 394, 46, 553, 291, 816, 674, 538, 320, 780, 114, \ 53, 147, 75, 660, 897, 281, 141, 104, 278, 69, 208, 712, 420, 612, \ 578, 64, 160, 24, 6, 513, 400, 261, 48, 136, 516, 3, 640, 36, 296, 9, \ 770, 18, 528, 130, 33, 12} I used d=6 for a slightly deeper, less greedy search. This code has the symmetry of the 8-element quaternion group: i = (2) (4) (1 6 9 3) (5 8 7 10) j = (2) (4) (1 7 9 5) (3 10 6 8) k = (2) (4) (1 10 9 8) (3 5 6 7) (These are the permutations on the codewords as bit strings.) I am now intrigued to see what other symmetries might turn up -- recall that wax-&-wane makes no symmetry restrictions whatsoever. So now we can evenly divide up the 1000 bottles into 50 sets of 20 from which at most 80 must be rejected. -Veit On Jan 2, 2013, at 9:19 PM, Veit Elser <ve10(a)cornell.edu> wrote: > Thanks for pointing out the rounding issue; it's gratifying to know I > helped in a small way to salvage another bottle of wine! > > I spent a couple of hours turning my Mma program into C code, in case > anyone wants to play with it. It generates 49/4 codes at a rate of about > 1 per minute. Here's a brief description of the algorithm: > > There are two operations, wax and wane, in each cycle. As the names > suggest, the code grows in size during wax and shrinks during wane. > > Wax is the simpler of the two: unused codewords are added as long as > the "cap" on the code (4 in the case of 49/4) is not exceeded. This introduces > randomness, because the algorithm tries the unused codewords in a > random (not lexicographic) order. > > During wane, codewords are deleted that have an overabundant > representation in the unions that are at their capped values. The rationale > is that growth of the code is more likely when fewer unions are capped, > and if the rule is to always delete the same number d of codewords, one > should preferentially delete those that "un-cap" the greatest number of > capped unions. So the algorithm assigns a cost to each codeword equal > to the number of capped unions it contributes to and then deletes the > d codewords having the highest cost. > > Starting from the empty code the algorithm performs many wax and wane > cycles and outputs the code whenever an increased size is found. During wax > there is no limit to the growth of the code, while the size always shrinks by > d during wane. I get good results for the 49/4 code when d=5. > > -Veit > > On Jan 2, 2013, at 7:46 PM, Michael Kleber <michael.kleber(a)gmail.com> > wrote: > >> Oh hey, my Z2xZ2 symmetry search just turned up a 49/4 also: >> >> 5,160,640,20,6,192,384,12,10,320,33,528,34,65,272,520,40,257,80,514,51,609,816,537,60,897,240,519,66,264,77,418,712,278,90,834,360,267,102,195,408,780,132,145,548,169,293,596,658 >> >> (Note the presence of 132 = 0010000100, a symmetry orbit of size 1, >> allowing the code to be of odd size.) >> >> Still eager to see Veit's approach, though; I remain afraid that this >> symmetry thing is ultimately a distraction. >> >> --Michael > > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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Re: [math-fun] quadrangulation
by Henry Baker 06 Jan '13

06 Jan '13

Thanks, Warren! That gives me a way to incrementally add new faces! Suppose we have a face which is a square. Embed a new square the middle of the given square, where the new square is half as big as the given square, and aligned with the given square. Link each vertex of the new square with the corresponding vertex of the given square. Now "push out" the new square from the plane of the given square. Once this little square is "pushed out", I can now deform it to better approximate the underlying surface. ASCII art: -------------- |\ /| | \--------/ | | | | | | | | | | | | | | | | | | /--------\ | |/ \| -------------- At 09:22 AM 1/6/2013, Warren Smith wrote: >> "Henry Baker" <hbaker1(a)pipeline.com> >> Q. Is it possible to "quadrangulate" a 3D surface with _only_ (planar) >> quadrilaterals "hinged" together ? >> (A quadrilateral has to have 4 non-trivial sides, so no triangles as >> limiting cases.) > >--First answer: >A triangle can be divided into 3 quads using 4 extra vertices at edge midpoints >and some central point. >And any convex polyhedron can be regarded as having triangle faces only >(happens automatically by perturbing all vertices by epsilon randomly). > >--Additional remark: >Let F=#quads, E=#edges, V=#vertices. >F-E+V=2 (Euler) >and >E=2*F >for a quadrangle-faced polyhedron with same topology as a sphere. >So >V = 2+F.

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Re: [math-fun] math-fun Digest, Vol 119, Issue 4
by Steve Witham 06 Jan '13

06 Jan '13

>From: Bill Gosper <billgosper(a)gmail.com> > >Carolyn Bickford snapped >https://mail-attachment.googleusercontent.com/attachment/?ui=2&ik=97b2a014c… >in the Lawrence Hall of Science gift shop. >Any chance it's a legitimate formula for something? One thing I don't understand is the form of the second line with the... is that a double contour integral? Integral over a closed surface? Integral over a "torus" in 2 dimensions, each of which is complex? I don't see the variable for that funky integral...? Going to Google images with "double contour integral" is pretty entertaining. --Steve

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[math-fun] quadrangulation
by Warren Smith 06 Jan '13

06 Jan '13

> "Henry Baker" <hbaker1(a)pipeline.com> > Q. Is it possible to "quadrangulate" a 3D surface with _only_ (planar) > quadrilaterals "hinged" together ? > (A quadrilateral has to have 4 non-trivial sides, so no triangles as > limiting cases.) --First answer: A triangle can be divided into 3 quads using 4 extra vertices at edge midpoints and some central point. And any convex polyhedron can be regarded as having triangle faces only (happens automatically by perturbing all vertices by epsilon randomly). --Additional remark: Let F=#quads, E=#edges, V=#vertices. F-E+V=2 (Euler) and E=2*F for a quadrangle-faced polyhedron with same topology as a sphere. So V = 2+F.

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Re: [math-fun] math-fun Digest, Vol 119, Issue 6
by Mike Stay 06 Jan '13

06 Jan '13

> Date: Sun, 06 Jan 2013 08:53:21 -0800 > From: Marc LeBrun <mlb(a)well.com> > To: math-fun <math-fun(a)mailman.xmission.com> > Subject: Re: [math-fun] stupid 3d graphics problem > Message-ID: <CD0EEC01.E717%mlb(a)well.com> > Content-Type: text/plain; charset="US-ASCII" > >>="Henry Baker" <hbaker1(a)pipeline.com> >> A usual trick in modeling 3D surfaces is to "triangulate" them with triangles >> that could in principle be hooked together with hinges. >> >> Q. Is it possible to "quadrangulate" a 3D surface with _only_ (planar) >> quadrilaterals "hinged" together ? >> >> (A quadrilateral has to have 4 non-trivial sides, so no triangles as limiting >> cases.) > > How would we quadrangulate a tetrahedron? Quadrangulate each triangular face separately. To quadrangulate a face, pick a point in the center and draw a line from it to each side. -- Mike Stay - metaweta(a)gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com

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