March 2019 - mathfun@mailman.xmission.com

Re: [math-fun] Why isn’t PentakisDodecahedron Johnson 93??
by Bill Gosper 12 Mar '19

12 Mar '19

On 2019-03-11 13:56, Allan Wechsler wrote: > The missing fact, I think, is that if you augment two adjacent pentagons of > a dodecahedron, the edge between the two pyramids will be a valley and not > a ridge, so the resulting polyhedron won't be convex. > Alan, if you would just visit https://www.wolframcloud.com/objects/810fbe92-6ab3-480b-9f86-e4869990cf46 (and maybe refresh a couple of times) you can tumble the graphic and see that it is manifestly convex. And, were it not, just augmenting one face of a dodecahedron could serve as J93. And if you don't believe your eyes, why not believe In[29]:= PolyhedronData@60 Out[29]= {"CubeTenCompound", "CumulatedDodecahedron", \ "DeltoidalHexecontahedron", "DodecahedronFiveCompound", \ "MathematicaPolyhedron", "PentagonalHexecontahedron", \ "PentakisDodecahedron", "RhombicHexecontahedron", \ "SmallTriambicIcosahedron", "TriakisIcosahedron"} In[30]:= Intersection[%, PolyhedronData@"Johnson"] Out[30]= {} In[2]:= Max@PolyhedronData["PentakisDodecahedron", "DihedralAngles"] Out[2]= \[Pi] - ArcCos[1/109 (80 + 9 Sqrt[5])] In[3]:= %/Degree // N Out[3]= 156.718553726459 This is getting unreal. —rwg

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Re: [math-fun] Venn diagram
by Dan Asimov 11 Mar '19

11 Mar '19

I certainly agree the xkcd Venn diagram is very nice. (But: Define awkward.) This article is also nice: <http://www.ams.org/notices/200611/fea-wagon.pdf>. —Dan Rich Schroeppel écrit: ----- Today's XKCD has the best 5-variable Venn diagram drawing that I've seen. It's possible to draw VDs with any number of variables, but the diagrams are tortured, with the regions having odd shapes that make the visualization useless. It's even possible to make the single-variable regions convex, but the intersections are mostly sliver-shaped, and the dimensions vary widely. The XKCD layout is symmetric in each variable, and has reasonable size & shape for every region. There are lots of concavities, but they aren't awkward. https://xkcd.com/2122/ -----

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[math-fun] Why isn’t PentakisDodecahedron Johnson 93??
by Bill Gosper 11 Mar '19

11 Mar '19

Johnson 93? <https://www.wolframcloud.com/objects/810fbe92-6ab3-480b-9f86-e4869990cf46> —rwg And why can't we tumble that figure?

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Re: [math-fun] Venn diagram
by Stuart Anderson 11 Mar '19

11 Mar '19

software for producing nice-looking Venn diagrams https://stackoverflow.com/questions/32440128/nice-looking-five-sets-venn-di… > ------------------------------ > > Message: 2 > Date: Mon, 11 Mar 2019 11:18:18 -0600 > From: rcs(a)xmission.com > To: math-fun(a)mailman.xmission.com > Cc: rcs(a)xmission.com > Subject: [math-fun] Venn diagram > Message-ID: > <20190311111818.d0lrksi0sgs4cs8c(a)webmail-backup.xmission.com> > Content-Type: text/plain; charset=ISO-8859-1; DelSp="Yes"; > format="flowed" > > Today's XKCD has the best 5-variable Venn diagram > drawing that I've seen. It's possible to draw VDs > with any number of variables, but the diagrams are > tortured, with the regions having odd shapes that > make the visualization useless. It's even possible > to make the single-variable regions convex, but the > intersections are mostly sliver-shaped, and the > dimensions vary widely. > The XKCD layout is symmetric in each variable, and > has reasonable size & shape for every region. There > are lots of concavities, but they aren't awkward. > > https://xkcd.com/2122/ > > Rich > > > >

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[math-fun] Fwd: Statistics for a class of random solids
by Lucas, Stephen K - lucassk 11 Mar '19

11 Mar '19

Let me add that there is a paper on asymptotics of the number of vertices, perimeter, and area of a convex hull for standard normally distributed independent points in the plane: Irene Hueter, The Convex Hull of a Normal Sample, Adv. App. Prob. 26 855-875 (1994). It is available at JSTOR. She proves that the number of vertices for the convex hull N satisfies: (N-2\sqrt(2*pi*ln n)) —> sqrt(2\sqrt(2*pi*ln n)*(1+pi*c)) times N(0,1), c_2 about 0.1 So average number of vertices grows like square root of logarithm of n. Steve

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Re: [math-fun] Statistics for a class of random solids
by Dan Asimov 11 Mar '19

11 Mar '19

Wait — you mean that isn't the actual origin of the name "Box-Muller" ? —Dan ----- Note that the probability density function on R^2 where each coordinate is independently uniform [0, 1] is shaped like a box. As such, it seems reasonable to refer to the shape of the probability density function of a bivariate standard Gaussian as a 'muller', so that the Box-Muller transform lives up to its name. -----

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[math-fun] Venn diagram
by rcs＠xmission.com 11 Mar '19

11 Mar '19

Today's XKCD has the best 5-variable Venn diagram drawing that I've seen. It's possible to draw VDs with any number of variables, but the diagrams are tortured, with the regions having odd shapes that make the visualization useless. It's even possible to make the single-variable regions convex, but the intersections are mostly sliver-shaped, and the dimensions vary widely. The XKCD layout is symmetric in each variable, and has reasonable size & shape for every region. There are lots of concavities, but they aren't awkward. https://xkcd.com/2122/ Rich

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Re: [math-fun] Statistics for a class of random solids
by Dan Asimov 10 Mar '19

10 Mar '19

It's not like there's some particular bad interaction between pseudo-random number generators and high-dimensional geometry. Isn't it really that a) one often needs many samples to get accurate estimates in the vastness of high-dimensional space, and b) with too many samples, a pseudorandom number generator isn't. —Dan Fred Lunnon écrit: ----- Where higher dimensional geometry is involved, bear in mind that even a well-designed pseudo-random number generator is only independent to some fixed precision. Once dimension x (user-demanded precision) exceeds this quantity, the randomness becomes compromised: heed the awful warning in G. Marsaglia "Random numbers fall mainly in the planes" https://www.ncbi.nlm.nih.gov/pmc/articles/PMC285899/pdf/pnas00123-0038.pdf This effective precision is invariably omitted from a PRNG specification, and must instead be deduced from a detailed inspection of the algorithm. -----

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Re: [math-fun] 33 as sum of 3 integer cubes
by Keith F. Lynch 10 Mar '19

10 Mar '19

I can't find any description of the search algorithm that found it. Is Timothy Browning keeping it a secret? Obviously, he didn't simply try every pair of 16-digit numbers to see if their difference, plus 33, was a cube. There isn't enough computer power in the world for such a search. Maybe something to do with modular arithmetic? It looks like, modulo any prime in the form 6k+1, cubes have only about a third of the possible residues (e.g. cubes have 35 residues mod 103). And they're of course all independent, so by combining a bunch of them you can shrink the search space by as much as desired. Unless that just triggers a different combinatorial explosion. It's cool that the next unknown is 42. I already presented the problem to a Douglas Adams fan as the 42 problem years ago, skipping all mention that 33 was also, at the time, unknown.

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Re: [math-fun] 33 as sum of 3 integer cubes
by Dan Asimov 10 Mar '19

10 Mar '19

What about other residue class restrictions besides mod 9 ? Googling, I found this discussion, <https://mathoverflow.net/questions/138886/which-integers-can-be-expressed-a…>, where someone states no other such restrictions are known. Most interesting of all, it is apparently unsolved whether for *every* N ≠ ±4 (mod 9) the equation x^3 + y^3 + z^3 = N has infinitely many integer solutions in x, y, z. This is known for N = a perfect cube or twice a perfect cube, but it seems in no other case. Another cute tidbit: N = 3 has two known solutions mod permutations: 1 + 1 + 1 = 3 and 4^3 + 4^3 + (-5)^3 = 3. Are there any more? Not known. —Dan Allan Wechsler wrote: ----- Aside from n = 4 or 5 mod 9, which are all impossible, the next unresolved case is x^3 + y^3 + z^3 = 42. -----

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