January 2014 - mathfun@mailman.xmission.com

Re: [math-fun] Julian's dodecahedron
by Bill Gosper 20 Feb '14

20 Feb '14

On 2013-09-02 15:15, Adam P. Goucher wrote: > The legendary Bill Gosper wrote: I'd rather be young. > >> is made by replacing ten of the regular pentagons of a dodecahedron with >> concave equilateral pentagons, preserving the topology: >> http://gosper.org/dodohedron.png >> ( !#%&*@! was deleted again. Apologies to any of you who looked for it.) >> >> http://en.wikipedia.org/wiki/Dodecahedron glaringly omits it: > > Even worse, the Wikipedia article omits the much more well-known > _endo-dodecahedron_ (the Symmetries of Things, page 328), which has > concave regular pentagonal faces and is *face-transitive*. And it's equilateral! > Its symmetry group is pyritohedral (3*2), and it is defined as: > > "The spaces left over by positioning dodecahedra of maximum size in > the natural orientation, centred on the points of the FCC lattice" > > The faces are different from the concave pentagons of the > tympanohedron, having internal angles of {7pi/5, pi/5, 3pi/5, 3pi/5, > pi/5} in cyclic order. > > > Sincerely, > > > Adam P. Goucher > > http://cp4space.wordpress.com Actually the concave figure in http://en.wikipedia.org/wiki/Pyritohedron is probably your endododecahedron, but they don't say. It would appear in their stupid throbbing animation, if the author wasn't so hung up on convexiity. Here's my stupid throbbing animation: ListAnimate[ Table[Graphics3D[ Polygon[Join[#, -#] &@ Join[#, Map[RotateLeft, #, {2}], Map[RotateRight, #, {2}]] &@{{{-(1/2) + 2 z^2, -(1/2) + z, 0}, {-(1/2), -(1/2), -(1/2)}, {0, -(1/2) + 2 z^2, -(1/2) + z}, {1/2, -(1/2), -(1/2)}, {1/2 - 2 z^2, -(1/2) + z, 0}}, {{1/2 - 2 z^2, -(1/2) + z, 0}, {1/2, -(1/2), 1/ 2}, {0, -(1/2) + 2 z^2, 1/2 - z}, {-(1/2), -(1/2), 1/ 2}, {-(1/2) + 2 z^2, -(1/2) + z, 0}}}]] /. z -> (5*Sin[t] + Sin[5*t])/12, {t, -\[Pi]/2, \[Pi]/2, \[Pi]/30}], AnimationRunning -> True, AnimationDirection -> ForwardBackward] The mgif: gosper.org/dodex.gif --rwg BTW, yesterday Neil predictively sketched the key frames of this on a small piece of paper while riding in a car in the Santa Cruz Mountains.

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[math-fun] Hardin colourings & tri-colouring grids
by Fred Lunnon 05 Feb '14

05 Feb '14

Within the (exiguous) bowels of OEIS A078099 there lurks a throwaway comment to the effect that 3-colouring the nodes of a m x n grid such that edge-adjacent pairs of nodes are assigned distinct colours is equivalent to 2-colouring the nodes such that no square (4-circuit) has both diagonals assigned distinct colours. [Establishing this equivalence is not entirely trivial.] In default of any earlier attribution, I propose to dub the latter colourings "Hardin" colourings. Though I haven't sat down to prove it, it seems reasonable to assume that as n,m -> oo the density of diagonal pairs with equal colours approaches a limit d , for a `random' Hardin colouring. [We can just sum frequencies over distinct n x n colourings where n is large, then take means.] I can show d > 0.5382 , and extrapolation suggests d ~ 0.5395 +/- 0.00005 ; can anybody improve on these values, find an explicit expression, or show a non-trivial upper bound for d ? Fred Lunnon

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Re: [math-fun] Maximum number of triples of orthogonal directions
by Allan Wechsler 01 Feb '14

01 Feb '14

f(8) is at least 4, because ABC+CDE+EFG+GHA can be realized. I don't think it would be too much work to show that 5 is not possible. 1,1,2,2,3,4, unsurprisingly, has hundreds of hits on OEIS, so we need more terms. I have a suspicion that all the graphs of the class you describe (no four vertices with a "lune", AB+AC+BC+BD+CD) are "realizable" as orthogonality graphs in the projective plane. If they are different, then your function f probably won't diverge from the corresponding Ramsey Theory question (largest number of triangles in a luneless graph) for quite a while. We should think about trying to prove the realizability conjecture. I have only the very vaguest inklings about how to try that. On Thu, Jan 30, 2014 at 8:35 PM, Warren D Smith <warren.wds(a)gmail.com>wrote: > I posted this to math-fun also, but it has not been "approved" yet by RCS: > > I can prove f(6)=2 and f(7)=3. > > An upper bound on F(n) which shows this is: > the maximum number of triangles in an n-vertex graph G, where > only G which do not contain <|> > as a 4-vertex subgraph, are allowed. > > This function of n is also interesting by itself, and I do not know what > it > is. >

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Re: [math-fun] Maximum number of triples of orthogonal directions
by Warren D Smith 31 Jan '14

31 Jan '14

WDS: Allan Wechsler sent me various confused emails, but eventually unconfused himself, with the result his construction showing f(8)>=4, is now declared broken. My use of it to try to show f(n+6) >= 4+f(n) is similarly declared broken. > AW:: > Taking a deep breath and trying again. I think my eight-direction > configuration is not, in fact, realizable. The trouble is that AC, CE, EG, > GA are all orthogonal pairs. I don't think that this spherical > quadrilateral actually can flex in the way that we imagined. It can "fold", > but one of the pairs AE, CG must remain antipodal. This forces > identifications, so that the whole thing is reduced to the "bow-tie" > configuration of five points we identified earlier, PQR+RST. --WDS: Aha. The "flexible quadrilateral" ACEG... has a problem. If fix A at North pole of sphere of directions, then C,G both must be on equator no matter how you flex, which means E must be at either North or South pole no matter how you flex. Which means you lose, since A and E then represent the same direction, which is forbidden. It's kind of strange: essentially any spherical quadrilateral is flexible, but for this particular one, it has this miraculous sort of semi-rigid property. --However, to now make lemonade from lemons, this tells us that the 4-cycle is a forbidden subgraph, leading to stronger upper bounds. Specifically, an upper bound U on f(N) is: the most triangles any N-vertex graph G can have, where only graphs G not containing a 4-cycle, are allowed. On this topic, see http://arxiv.org/abs/1201.4912 which is about a different problem, but perhaps relevant.

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[math-fun] Maximum number of triples of orthogonal directions
by Warren D Smith 31 Jan '14

31 Jan '14

A "direction" is a 3-vector (x,y,z) obeying either x>0, or x=0 and y>0, or x=y=0 and z>0, and x^2+y^2+z^2=1. The point of that is (a) to forbid 0,0,0 and (b) I want to regard negative directions and rescaled directions as same thing as original directions, so only allow one. Let f(N) be the maximum possible number of (unordered) triples of mutually-orthogonal directions in a set of N distinct directions. What can be said about f(N)? Here is a table of lower bounds L and upper bounds U on f(N) for small N. Which probably are both pretty weak. N=3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 L=1 1 2 2 3 3 4 5 5 6 6 7 8 9 9 10 11 11 12 13 14 14 15 15 16 17 18 18 19 19 U=1 1 2 4 7 8 12 13 17 20 26 28 35 37 44 48 57 60 70 73 83 88 100 104 117 121 -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Tough sequence.
by David Wilson 30 Jan '14

30 Jan '14

Conjecturally, the last element of A020495 (a(21) = 21679) is the largest number which is neither a square nor square + prime. Likewise, the last element of A045911 (a(6195) = 78526384) is the conjectured largest number which is neither a (positive) cube or a cube + prime. In each case, the sequence is statistically finite, and a large range has been searched for additional elements, so there is a reasonable argument that these sequences are finite and correct. This, of course, suggests a sequence starting 21697, 78526384, ... The next element would the largest value which is neither a fourth power nor a fourth power + prime. This number will be very large. Looks like a real programming challenge.

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Re: [math-fun] Jumping square points
by Adam P. Goucher 30 Jan '14

30 Jan '14

That's a very beautiful problem. I've appended a short solution to the end of this e-mail. > ----- Original Message ----- > From: Eric Angelini > Sent: 01/29/14 10:30 AM > To: math-fun > Subject: [math-fun] Jumping square points > > Hello Math-Fun, > Four points shape a square of size "s". > Every minute a point randomly jumps over another point > and lands symmetrically behind it. > Is it possible that, at some stage, the 4 points shape > a new square having size "t" > "s"? > Best, > É. > > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun ***** spoiler alert ***** . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . No. Operate in the ring Z[i] of Gaussian integers, and let {a,b,c,d} be the Gaussian integers corresponding to the vertices of the square. When you apply the process to obtain {a,2a-b,c,d}, the greatest common divisor (again a Gaussian integer, up to multiplication by the ring units) remains invariant (similar to Euclid's algorithm). Hence, if they form a square at any time, it must be the same size and orientation as the original square. Sincerely, Adam P. Goucher

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[math-fun] Cleaning out a jar
by Bill Gosper 29 Jan '14

29 Jan '14

Jörg> Somebody wrote: (The problem is impregnation of the rubbery film of sealant inside the lid. Simmering several hours in Clorox helped, but insufficiently. Put in into NaOH solution (which is to be treated with due respect). That will help much better. NaOH + grease --> soap And the human nose is pathetic compared to a bloodhound's. Or a gila monster's egg sense.) _______________________________________________ I now demand a problem where I can point out that citric acid is almost always to be preferred over vinegar. Best, jj Can we hope that essence of sauerkraut is an oil? The old standby, Red Devil lye, was put out of business by The War on Meth Labs. http://candleandsoap.about.com/od/coldprocesssoapmaking/a/lyesources.htm But hardware stores in Latino areas still have http://www.ebay.com/itm/Red-Devil-Lye-by-Rooto-100-Household-Lye-16oz-/3205…, which Mexicans use to home-make tortillas!! Perhaps even La Rosa brand, "Casi hecha a mano". Almost made by hand. --rwg An eavesdropper wonders how to lose the Clorox smell. Leave it out in the sun? Actually, there is some slight acidity in tea. Leaving used bags in the lid seems to help.

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[math-fun] Cleaning out a jar
by James Propp 29 Jan '14

29 Jan '14

What's the optimum amount of water to put in a jar if you want to clean the jar by putting some water in it and shaking it for a bit? I grant that the problem is ill-posed in all kinds of ways, but maybe some not-too-farfetched model gives results in line with experimental data conducted on jars of ordinary shape containing food residues of an ordinary kind. I seem to recall that number theorist David Solomon, back when we were housemates in Berkeley in the 1980s, came up with an idealization of the problem for which the optimum volume of water was 1/e times the volume of the jar. I'll ask him if he remembers any details. Jim Propp

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[math-fun] Cleaning out a jar
by Bill Gosper 29 Jan '14

29 Jan '14

JPropp> What's the optimum amount of water to put in a jar if you want to clean the jar by putting some water in it and shaking it for a bit? I grant that the problem is ill-posed in all kinds of ways, but maybe some not-too-farfetched model gives results in line with experimental data conducted on jars of ordinary shape containing food residues of an ordinary kind. I seem to recall that number theorist David Solomon, back when we were housemates in Berkeley in the 1980s, came up with an idealization of the problem for which the optimum volume of water was 1/e times the volume of the jar. I'll ask him if he remembers any details. Jim Propp -- If, as in my note to Randi, you plan to brew tea in a jar sold full of sauerkraut, it's not clear that the answer is finite. --rwg (The problem is impregnation of the rubbery film of sealant inside the lid. Simmering several hours in Clorox helped, but insufficiently. And the human nose is pathetic compared to a bloodhound's. Or a gila monster's egg sense.)

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