April 2019 - mathfun@mailman.xmission.com

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

24 Apr '19

Given two sets A = {x_j} and B = {y_j} of N points in R^2, there is of course the set of points {(x_j, y_j)} in R^4 that has A and B as its two projected images. Problem solved! (The earlier problem that Michael Kleber solved was whether there are 36 points in R^3 (not R^4) with suitable projections.) *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. —Dan ----- ..... 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? -----

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

24 Apr '19

I did google this, but I didn't see any OEIS sequences in the top hits. —Dan

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Re: [math-fun] Number of 15x15 crossword puzzle grids
by Dan Asimov 23 Apr '19

23 Apr '19

Nice! Now let's find the counts for the grids that *also* have symmetry group G, for each subgroup G of Isom([0,1]x[0,1]) = D_4, the dihedral group of order 8. (There are 10 subgroups altogether: the trivial group, the whole group, and 5 isomorphic to Z_2, 1 to Z_4, 2 to Z_2 + Z_2.) —Dan ----- The 538 Riddler from two and a half weeks ago asked for the number of 15x15 black-and-white grids that are legal for crossword puzzles. https://fivethirtyeight.com/features/how-many-crossword-puzzles-can-you-mak… Rules: Every white square must be "checked" (in both a horizontal and vertical word), words must be at least 3 letters long, the whole grid must be symmetric under 180-degree rotation, and the set of white squares must be connected. Evidently nobody solved it, so I decided to :-) Answer: there are 347804238364806 legal 15x15 grids according to those rules (347 trillion). And if you don't like the fact that those rules allow some all-black rows around the edges of the puzzle, and want them to be "really" 15x15, then that cuts down the count to 342935406100702. To convey the information to the outside world (including the 538 Riddler guy), I've posted some more about this on Twitter. Please forgive me. https://twitter.com/Log3overLog2/status/1092472516571000839 NJAS, the sequence of counts of nxn grids for n=3,5,7,9,11,13,15 is: All grids: 1, 15, 397, 35184, 17431781, 37147554097, 347804238364806 No black edges: 1, 12, 312, 31047, 16459558, 36076951460, 342935406100702 -----

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[math-fun] Packing problem
by James Propp 23 Apr '19

23 Apr '19

In helping my aging parents move from one apartment to another, I faced the problem of bundling rolls of wrapping paper together. I laid a bungee cord on a flat surface, stacked the rolls athwart the cord in what seemed to be a reasonably compact formation, fastened the cord (turning the triangular stack into something more circular), and picked up the bundle of rolls. Unfortunately, when I turned the bundle on its side, one of the middle rolls fell out. In terms of its cross section, that roll was what in disk-packing parlance is called a “rattler”, and friction wasn’t holding it in place. Are there good heuristics for avoiding the rattler problem when packing cylinders? For instance, would my chances of success have been greater if I’d put the thicker rolls in the middle of the stack and the thinner rolls near the periphery? (As it happens, the rattler was a thin roll near the middle.) And should I have tried to ensure that adjacent rolls had roughly the same diameter? If that’s too loosey-goosey for you, here’s a precise math question (though perhaps in its current form it’s merely pre-precise): Is there a set of rolls (i.e., a multiset of diameters) that cannot be bungee-bundled, in the sense that for any stable way of wrapping them circumferentially with an elastic cord, there’ll be a rattler? My notion of stability is based on the assumption that the (potential) energy of the system is the length of the cord; I assume that energy cannot increase, that there is no slippage between rolls, and that there is no slippage between the bungee cord and the rolls that it touches. At first I thought that if one diameter is tiny relative to all the others, it must be a rattler, but when I thought about it harder and remembered the no-slip condition, it no longer seemed obvious to me. Indeed, when the diameters are epsilon, 1, and 1, I think the 1,1,epsilon “shishkebab arrangement” is stable. Jim Propp PS: Maybe my attempted formalization of the notion of stability needs some improvement. I think that the 1,epsilon,1 shishkebab is technically stable because of the no-slip conditions, but this doesn’t correspond to physical reality.

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[math-fun] Neutron stars as atoms
by Warren D Smith 23 Apr '19

23 Apr '19

Whitfield Diffie I recall reading the intreaguing observation that stable neutron structures run up to around atomic wieght 300 and then their is a long gap before you get to the neutron stars with atomic weights around 10^50. Perhaps we should regard the neutron stars as isotopes of very heavy elements. The atomic numbers would be miniscule by comparison with the atomic weights but a mildely positively charged neutron star doesn't seem impossible: would adding a billion protons make a difference? --WDS: A better way to view this is: replace the word "stable" with "stable under pressure." See, I would claim that it is an illusion that there is no stable nucleus with atomic weight 1000. Actually, if you put it under enough pressure, then it would presumably become stable. The only problem is humans have no hope of sustaining pressures that large and the only known way to reach them is inside a neutron star. Anyhow, with suitable pressure ramp up, probably there is no "gap" at all, and practically the full range of atomic weights become reachable with stability. Another thing is: the usual nuclei generally are in the ground state. Excited states that last long, are rare for nuclei. But for a neutron star, it presumably never reaches ground, there are too many excited states it gets trapped in for long periods. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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Re: [math-fun] "maa.org" has been compromised?
by Steve Witham 23 Apr '19

23 Apr '19

> From: Henry Baker <hbaker1(a)pipeline.com> > Date: 4/19/19, 11:49 AM > > I just received a "phishing" email whose header said it came from webwork.maa.org. > From: Tomas Rokicki <rokicki(a)gmail.com> > Date: 4/19/19, 11:57 AM > > SMTP is fundamentally insecure. Headers are trivial to spoof. A recent favorite spam I've received comes "From" my own email address. That's an old one, but the fun thing is that it says "Look at the 'From' address of this email. It is your own address! This way I prove that I have hacked your computer," and that I should take seriously that the sender really has encrypted all my data and will erase it unless I send xxxxx bitcoin to yyyyyyyy. --Steve

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Re: [math-fun] Packing problem
by Dan Asimov 22 Apr '19

22 Apr '19

These are very interesting questions! There's a useful concept of stability that's based on considering what happens under an arbitrarily small perturbation. Omitting mention of physics, we can define * for any set R = {r_j | j in J} of radii r_j > 0 (where J is some index set) with finite sum Sum_{j in J} r_j, let disk D_j in the plane have radius r_j. * Define an arrangement A = {D_j} in the plane of non-overlapping disks D_j in the plane, where the radius of D_j is r_j. This means that for each D_j, A is a choice of *center* = c_j in R^2. Non-overlapping means that for each pair of indices j ≠ k, we have ||c_j - c_k|| >= r_j + r_k. * Let D(A) be the smallest disk in R^2 containing each D_j in its interior. Then define diam(A) = diameter(D). * For an arrangement A and eps > 0, define an *eps-close perturbation* of A to be an arrangement A' with the same radii r_j but with each center c_j replaced by a new center c'_j, such that the sum of the center discrepancies ||c_j - c'_j|| is bounded above by eps: Sum_{j in J} ||c_j - c'_j|| < eps. * Finally, say an arrangement A is *stable* if there exists eps > 0 such that every eps-close perturbation A' of A satisfies diam(A') >= diam(A). In other words, small perturbations to A cannot decrease diam(A). —Dan ----- ... ... Is there a ... multiset of diameters ... that cannot be bungee-bundled, in the sense that for any stable way of wrapping them circumferentially with an elastic cord, there’ll be a rattler? My notion of stability is based on the assumption that the (potential) energy of the system is the length of the cord; I assume that energy cannot increase, that there is no slippage between rolls, and that there is no slippage between the bungee cord and the rolls that it touches. ... PS: Maybe my attempted formalization of the notion of stability needs some improvement. I think that the 1,epsilon,1 shishkebab is technically stable because of the no-slip conditions, but this doesn’t correspond to physical reality. -----

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[math-fun] Numerical fluke (?)
by Bill Gosper 22 Apr '19

22 Apr '19

(balter:=balanced ternary) Nine trits of even balanced ternary of π: In[329]:= 2 balter@Floor[3^9 π/2] Out[329]= {2, -2, -2, -2, 0, 2, 2, -2, 0, 2, -2} Check: In[330]:= FromDigits[{%, 2}, 3] // N Out[330]= 3.14149265863944 But In[337]:= Sum[4 (-1)^k/(1 - 2 k), {k, 10^4}] // N Out[337]= 3.14149265359004 ! —rwg

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[math-fun] Berlekamp
by rcs＠xmission.com 20 Apr '19

20 Apr '19

Elwyn Berlekamp died April 9, age 78. I heard this from a G4G update. Obituary at https://news.berkeley.edu/2019/04/18/elwyn-berlekamp-game-theorist-and-codi… His web page https://math.berkeley.edu/~berlek/index.html Wikipedia https://en.wikipedia.org/wiki/Elwyn_Berlekamp --- Rich

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[math-fun] Even balanced ternary
by Bill Gosper 20 Apr '19

20 Apr '19

Base 3, digits -2, 0, 2 would seem silly, representing only the even integers. But it is momentarily startling, and then trivially obvious that the pure fractions in this system completely cover [-1,1]. Digits {-1,0,1} only cover [-1/2,1/2]: In[306]:= Table[FromDigits[{{d1, d2, d3}, 0}, 3], {d1, -1, 1}, {d2, -1, 1}, {d3, -1, 1}] // Flatten Out[306]= {-(13/27), -(4/9), -(11/27), -(10/27), -(1/3), -(8/27), -(7/27), -(2/9), -(5/27), -(4/27), -(1/9), -(2/27), -(1/27), 0, 1/27, 2/27, 1/9, 4/27, 5/27, 2/9, 7/27, 8/27, 1/3, 10/27, 11/27, 4/9, 13/27} (stupid parentheses) But with {-2,0,2} In[307]:= Table[FromDigits[{{d1, d2, d3}, 0}, 3], {d1, -2, 2, 2}, {d2, -2, 2, 2}, {d3, -2, 2, 2}] // Flatten Out[307]= {-(26/27), -(8/9), -(22/27), -(20/27), -(2/3), -(16/27), -(14/27), -(4/9), -(10/27), -(8/27), -(2/9), -(4/27), -(2/27), 0, 2/27, 4/27, 2/9, 8/27, 10/27, 4/9, 14/27, 16/27, 2/3, 20/27, 22/27, 8/9, 26/27} You may object that there are no odd numerators. But .2222... is obviously 1, .2➋2➋2➋... = ½, and .20➋020➋020➋0... = ⅗, etc. And then 3 = 2.2222... = 2➋.➋➋➋... etc, and {-2,0,2} (more than) covers all the reals!—rwg

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