April 2019 - mathfun@mailman.xmission.com

[math-fun] The Lost Code
by Fred Lunnon 26 Apr '19

26 Apr '19

Seated one day at an analogue of the aperiodic Penrose tiling, I struck one 10-dimensional rank-4 null space of beauty with the (redundant) weight-4 basis: [ 1, 0, 0, 0, -1, -1, -1, 0, 0, 0] [ 0, 1, 0, 0, -1, 0, 0, 0, -1, 1] [ 0, 0, 1, 0, 0, -1, 0, 1, 0, -1] [ 0, 0, 0, 1, 0, 0, -1, -1, 1, 0] [ 1, -1, -1, -1, 0, 0, 0, 0, 0, 0] ; and on dropping signs it struck me (back!) that this was case n,m = 5,2 of an elementary binary block code with: number = n words, weight = (n-1)_C_(m-1) , length = n_C_m , distance = 2 (n-2)_C_(m-1) , where n_C_m denotes binomial coefficient. I have sought, but I seek it vainly --- along with its name and rank. Can somebody please jog my memory? Fred Lunnon

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Re: [math-fun] Hot tea sucks.
by Dan Asimov 26 Apr '19

26 Apr '19

Drop the bottle on the sidewalk from 3 ft. above. The tea will come out immediately. —Dan ----- I brew it in a 46oz pickle jar that easily tips over, and leaks if the lid isn't snug. But if I forget to momentarily loosen the lid within a few minutes, a powerful vacuum develops that makes it virtually impossible to unscrew. What is the trivial solution? -----

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Re: [math-fun] Square hex numbers?
by Dan Asimov 26 Apr '19

26 Apr '19

Thanks, Adam! I thought there should be some general method for solving P(n) = Q(K) where P, Q are quadratic polynomials over the integers, but didn't know where to go from there. * * * When there is an integer like 169 that belongs to two figurate numbers (H_7 = 169 = 13^2), I wonder if there is a set of 169 points in R^4 = R^2 x R^2 whose projection by p_12 onto one R^2 factor is the first figure and onto the other by p_34 is the second figure. I seem to recall asking this here some years ago for T_8 = 6^2, and that Michael Kleber showed that there's no set X of 36 points in R^4 with p_12(X) = triangle of side 8 and p_34(X) = square of side 6. Maybe there's a general statement about when such a thing can exist? —Dan —Dan Adam Goucher wrote: ----- Your equation is of the form: (quadratic in n) = (quadratic in K) ... ... -----

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Re: [math-fun] Square hex numbers?
by Dan Asimov 25 Apr '19

25 Apr '19

This suggests the question: ----- P and Q be integer polynomials of degrees >= 2. Find the solution set {(N,K) in Z^2 | P(N) = Q(K)}. ----- What's known about the general case when max(deg(P), deg(Q)} = 2, 3, 4, or >= 5 ??? I would try googling this but I'm not sure how to search for it. —Dan Adam Goucher wrote: ----- Your equation is of the form: (quadratic in n) = (quadratic in K) so you can complete the square on both sides and ..... -----

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Re: [math-fun] Packing problem
by Dan Asimov 25 Apr '19

25 Apr '19

Rigorous posts don't get no props. My previous attemptedly rigorous post on the subject was just to say, I guess opaquely, that it seems to capture the essence of a stable arrangement — like one forced by an elastic collar — to merely require: any sufficiently small perturbation of the arrangement (modify the centers but not the radii) cannot decrease the *diameter* of the arrangement (= the diameter of the smallest disk containing all disks of the arrangement). It seems clear that a rosette of 7 equal disks, but with the central one removed, is stable in this sense (despite the hole). —Dan ----- On 4/23/19, Allan Wechsler <acwacw(a)gmail.com> wrote: > I don't mind any particular model, but I think I just don't understand the > no-slip condition. Can you rigorize it? -----

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Re: [math-fun] "Non-orientable gyroid" question from John Graham
by Dan Asimov 25 Apr '19

25 Apr '19

Apologies for sending a private e-mail to math-fun under this subject. —Dan

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Re: [math-fun] "Non-orientable gyroid" question from John Graham
by Dan Asimov 25 Apr '19

25 Apr '19

Joe, Aha! Speaking to you suggested a revised and much more reasonable condition on A \int B \int (∂M x ∂N) than the one we spoke about: that it must be empty. Checking the new one out now. If the proof works that way, then we're probably OK. I thought about Berkeley, and we might or might not be able to make it. But meanwhile, I look forward to the next time we can talk about math unhurriedly in person. —Dan

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[math-fun] Square hex numbers?
by Dan Asimov 24 Apr '19

24 Apr '19

Just happened to notice that in the centered hexagonal numbers H_n = 1 + 6 T_n = 3n^2 + 3n + 1 we have H_7 = 13^2. Also H_0 = 1^2. So I tried to find all such cases of H_n = K^2, but am not sure of the best way to proceed. Suggestions? —Dan

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[math-fun] "Non-orientable gyroid" question from John Graham
by Bill Gosper 24 Apr '19

24 Apr '19

(He knows about Scherk surfaces.) ---------- Forwarded message --------- From: john graham <ikojag(a)yahoo.com> Date: Wed, Apr 24, 2019 at 10:12 AM Subject: Re: Infinite Mobius To: Bill Gosper <billgosper(a)gmail.com> Hello Bill. The Gyroid is indeed a wondrous fascinating shape, it is evidently non self-intersecting and has the potential for infinite growth through the repeated addition of identical forms. However I see that it is a surface of two opposite faces that divide space into two opposite but mutually exclusive and interlocking labyrinths. In its developed state it is edgeless. My object, on the other hand, is monospatial and is clearly a single surface as can be demonstrated locally and elementally, such as to infer that the single surface could be repeated ad infinitum. The object consists of twisted octagons interconnected four ways along a three dimensional cubic lattice grid, with the result that the edges are related helical lines of a fixed and regular period. I'd be interested to find out if the object is related to any of the known mathematical surfaces. John.. On ‎Tuesday‎, ‎23‎ ‎April‎ ‎2019‎ ‎18‎:‎12‎:‎24‎ ‎BST, Bill Gosper < billgosper(a)gmail.com> wrote: Can you explain your object compared to https://en.wikipedia.org/wiki/Gyroid ? —Bill On Tue, Apr 23, 2019 at 5:23 AM john graham <ikojag(a)yahoo.com> wrote: Hello Bill, I browsed through Shapeways catalogue, and homed in especially on complex surfaces. It was really very interesting. In fact I've ordered a Scherks First surface, a fascinating shape in itself and one that comes pretty close to what I've created/discovered? but clearly is not. I am sending you some images of the two models that I have made by hand one of which I've resolved into equilateral triangles. Inevitably any embodiment or manifestation of the infinite Mobius will be a fragment, as is shown here, but under certain geometrical conditions the loose ends can be tied up to unite the edges into a single entitiy, so that the mathematical integrity of the Mobius (singe edge/single surface) is maintained. I apologise if the images take some working out. My workmanship is less than perfect, and I do not, at present. have the IT skills to produce CGIs, but nevertheless I look forward to hearing from you. John.

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Re: [math-fun] Square hex numbers?
by Dan Asimov 24 Apr '19

24 Apr '19

More generally yet: Let A={a_j} and B={b_j} be sets of N points in R^a and R^b, respectively, and ask whether there is a set X of points in R^n, n < a + b, and subspaces P = R^a and Q = R^b of R^n such that the orthogonal projections of X onto P and Q are congruent to A and B, respectively. ----- *This* is the possibly interesting question: ----- Given two sets A and B of N points in R^2, when is there a set X in R^3 and 2-planes P, Q in R^3 with the orthogonal projections of X onto P, Q having images congruent to A and B, respectively? More generally, replace R^2 with R^k, and R^3 with R^n for n < 2k. ----- E.g.: ----- Given any two triangles T_1, T_2, when does there exist a set X of size 3 in R^3, and two 2-planes P_1, P_2 in R^3 with the orthogonal projection of X onto P_j congruent to T_j ??? —Dan

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