- mathfun@mailman.xmission.com - mailman.xmission.com

Re: [math-fun] Tammes problem for the square and cubical torus
by Dan Asimov 11 Sep '20

11 Sep '20

Good point. —Dan Allan Wechsler wrote: ----- I would think that for big enough N = a^2 + b^2, the moderately obvious "Pythagorean" solution would not be optimal, and that you could do better with big patches of points arranged in a triangular lattice. -----

1 0

Re: [math-fun] Tammes problem for the square and cubical torus
by Dan Asimov 11 Sep '20

11 Sep '20

I think that for an approximate incremental algorithm, probably Steve's idea is optimal (in some precise sense). (Years ago, I used something like that to get geodesic segments S(L) of length L on the n-torus T^n = R^n/Z^n that were intended to minimize the radius of the largest open ball B(r), r=r(L), in the complement T^n - S(L).) For the exact question, I think that for N = a^2 + b^2 points in T^2, where gcd(a,b) = 1, you can tile T^2 by N squares (usually tilted). Then the vertices of the tiling (or equally the centers of the tiles) form an optimal arrangement. Giving the max-min distance as sqrt(1/N). (I think this is closely related to finite factor rings of the Gaussian integers.) Checking N = 5, N = 10, N = 13, this looks an awful lot like exactly what Steve got for numbers of the form N = a^2 + b^2, gcd(a,b) = 1. I'm not sure about exact solutions on T^3. But another method for approximating the exact answer is to consider the function M : (T^3)^N —> R on arrangements of N points on T^3 (or any-dimensional torus), that gives the max-min distance, and try to climb the gradient. One problem is that this function isn't always differentiable. Another is that, a priori, there might be local maxes that aren't global maxes, though I don't know if this can occur. —Dan Steve Witham wrote: ----- You can get a pretty nice packing on a square torus using the golden ratio. For the unit square, pick one of these spacings (Yes, they are > 1.) h = floor(N phi) / N or h = ceil(N phi) / N, put the points at (k h (mod 1), k / N) for k = 1 to N, and pick the better of the two packings. They are uniform, which makes it < O(N) to measure the closest spacing. > But what would seem to be the best arrangements > for 1 <= N <= 10 on T^2 and T^3 ??? Here are my numbers. The sequence is not monotonic, which means you can sometimes do better by using the next spacing and leaving out a point! Using the best of `floor()` or `ceil()` (on a unit square): N min distance 2 0.7071067811865476 3 0.4714045207910316 4 0.5 5 0.44721359549995765 6 0.3726779962499649 7 0.3194382824999699 8 0.3535533905932738 9 0.3333333333333333 10 0.3162277660168369 11 0.2874797872880344 12 0.25 13 0.27735009811261385 14 0.22587697572631218 I don't know an analogous method for the cube torus. But you *could* "pack" points on a unit interval: k phi (mod 1) for k = 1 to N. ...Why would you want to do that!? you ask. Because you can keep adding points to the interval incrementally, and the neighbor distances (mod 1) are always within some constant factor of 1/N. I have searched but not found a similarly simple way to add points incrementally to the square torus that gets remotely good packing. (I call this the peppering problem.) --Steve > From: Dan Asimov <dasimov(a)earthlink.net> > Date: 9/10/20, 6:26 PM > > The Tammes problem is the well-known problem to find the arrangement > of N points on the unit sphere S^2 that maximizes the minimum distance > between any pair of them. > > What about the same question for N points on the square torus > > T^2 = R^2/Z^2 > > or the cubical torus > > T^3 = R^3/Z^3 > > ??? -----

2 1

Re: [math-fun] Tammes problem for the square and cubical torus
by Steve Witham 11 Sep '20

11 Sep '20

You can get a pretty nice packing on a square torus using the golden ratio. For the unit square, pick one of these spacings (Yes, they are > 1.) h = floor(N phi) / N or h = ceil(N phi) / N, put the points at (k h (mod 1), k / N) for k = 1 to N, and pick the better of the two packings. They are uniform, which makes it < O(N) to measure the closest spacing. > But what would seem to be the best arrangements > for 1 <= N <= 10 on T^2 and T^3 ??? Here are my numbers. The sequence is not monotonic, which means you can sometimes do better by using the next spacing and leaving out a point! Using the best of `floor()` or `ceil()` (on a unit square): N min distance 2 0.7071067811865476 3 0.4714045207910316 4 0.5 5 0.44721359549995765 6 0.3726779962499649 7 0.3194382824999699 8 0.3535533905932738 9 0.3333333333333333 10 0.3162277660168369 11 0.2874797872880344 12 0.25 13 0.27735009811261385 14 0.22587697572631218 I don't know an analogous method for the cube torus. But you *could* "pack" points on a unit interval: k phi (mod 1) for k = 1 to N. ...Why would you want to do that!? you ask. Because you can keep adding points to the interval incrementally, and the neighbor distances (mod 1) are always within some constant factor of 1/N. I have searched but not found a similarly simple way to add points incrementally to the square torus that gets remotely good packing. (I call this the peppering problem.) --Steve > From: Dan Asimov <dasimov(a)earthlink.net> > Date: 9/10/20, 6:26 PM > > The Tammes problem is the well-known problem to find the arrangement > of N points on the unit sphere S^2 that maximizes the minimum distance > between any pair of them. > > What about the same question for N points on the square torus > > T^2 = R^2/Z^2 > > or the cubical torus > > T^3 = R^3/Z^3 > > ???

1 0

[math-fun] stochastic analysis
by Bernie Cosell 11 Sep '20

11 Sep '20

https://www.theguardian.com/science/2020/sep/10/uk-mathematician-martin-hai… /Bernie\ Bernie Cosell bernie(a)fantasyfarm.com -- Too many people; too few sheep --

1 0

Re: [math-fun] Tammes problem for the square and cubical torus
by Dan Asimov 10 Sep '20

10 Sep '20

Thanks, but actually I have no trouble looking up relevant papers. I posted the problem because it seemed as if it would be fun to discuss on math-fun. —Dan ----- https://arxiv.org/abs/1809.10525 seems relevant.. -----

1 0

[math-fun] Ball sorting puzzle
by Allan Wechsler 10 Sep '20

10 Sep '20

I have played a little bit with a new-to-me puzzle type. The "board" consists of some number of stacks of colored balls. (Each stack is depicted as a test tube, making it more plausible that you can arrange smooth balls in a stack.) A legal move is to remove the top ball from some stack, and then place it on top of any other stack whose top ball is the same color as the one you're moving. An empty stack is a "wild card" -- you can move any ball to an empty stack. The stacks have a maximum capacity, typically 4 balls. The initial position has some number of full stacks -- typically 3 or more of them -- and two empty stacks. There are 4 balls of each color, so the number of colors matches the number of initial stacks. The object of the game is to rearrange all the balls into stacks of completely matching colors. Moves are not in general reversible (although some implementations offer an "undo" option), so it is definitely possible to get stuck and have to start over. The app never shows you a puzzle that can't be solved. Obvious parameters are the number of colors (C), the number of balls of each color (B), which is usually the same as the maximum stack height (H), and the number of empty stacks at the outset (E). One question is, for given values of C, B, and H, what is the smallest E that guarantees that any initial position can be solved? What is a good algorithm for generating puzzles? Some of the puzzles are very hard and require a lot of lookahead (or false starts). Many implementations I have found either cost money, or have extremely annoying advertising, or offer only a tiny selection of teaser puzzles. But https://www.crazygames.com/game/bubble-sorting looks like a good place to try it out.

1 0

[math-fun] Tammes problem for the square and cubical torus
by Dan Asimov 10 Sep '20

10 Sep '20

The Tammes problem is the well-known problem to find the arrangement of N points on the unit sphere S^2 that maximizes the minimum distance between any pair of them. What about the same question for N points on the square torus T^2 = R^2/Z^2 or the cubical torus T^3 = R^3/Z^3 ??? (Note that the maximum distance between any two points on T^2 is sqrt(2)/2 and between any two points on T^3 is sqrt(3)/2.) This gets very difficult on S^2 even for slightly large N, so I don't expect it will be easy to solve on T^2 or T^3. But what would seem to be the best arrangements for 1 <= N <= 10 on T^2 and T^3 ??? —Dan

2 1

Re: [math-fun] Tantalizing Mandelflake fill
by Bill Gosper 08 Sep '20

08 Sep '20

On Thu, Sep 3, 2020 at 4:56 PM Bill Gosper <billgosper(a)gmail.com> wrote: > What the heck is going on here? Alex's (fill rate) smoother idea on > Mandelbrot's > Snowflake filler: [Picture <http://gosper.org/smoothman9colors.png>, worth a thousand words, censored for bagbitting math-fun.] > somehow manages to fill both a Snowflake and one √3 times larger with the > exact same > pattern. Mandelbrot missed it, along with everybody else. I never > suspected it was even > possible. What's the rule? Note that successive sizes are "texturally > everted". —Bill > Julian to the rescue. <http://gosper.org/smoothflake.png> Oh what I'd give for BBM's reaction to this! Speaking of bounding boxes|circumradii|convex hulls, recall APG's heroic extraction of a gross of digits of the circumradius of the France ("Flowsnake") Fractal. For which we never found an expression. Do we even have an accurate bounding box for the terDragon? I just approximated (minus) the left bound of Alex Roodman's i √7 spacefiller <http://gosper.org/smoothman.png> as FromContinuedFraction{0, 18, 29, 2, 11, 1, 1, 1, 3, 2, 5, 2, 94, 1, 1, 1, 42, 1, 2, 1, 2, 1, 1, 19, 23, 1, 2, 1, 2, 16, 55, 1, 14, 93, 1, 23, 1, 2}. Octupling the sampling frequency gave {0, 18, 29, 2, 11, 1, 1, 1, 3, 2, 27, 2, 1, 2, 2, 2, 1, 2, 34, 12, 1, 8, 1, 4, 3, 1, 1, 2, 1, 1, 1, 3, 24, 1, 3, 10, 2, 1, 1, 10, 1, 1, 1, 1, 1, 1, 18, 1, 1, 1, 3, 3} I see no sign of √7. I always rave about piecewiserecursivefractal's inversion capability, but I > should caution that > fractal functions like the (optionally) smooth flakefill have inverses in > finite terms only on > arguments of the form a + i b √3, where a and b are rational. > —rwg

2 5

[math-fun] Splinter surprise
by Fred Lunnon 07 Sep '20

07 Sep '20

An oriental prestidigitation caught my eye yesterday in Guardian News: it will hardly detain anyone on math-fun for more than a few minutes. I was more puzzled to learn that it remained unsolved in competition. Note that the triangle in question is planar and Euclidean: perhaps contestants were misled by the implicit reference to a sphere? << Show that there is a triangle, the sum of whose three heights is less than 1mm, that has an area greater than the surface of the Earth (510M km^2). ... It was originally set in the final round of a Vietnamese TV gameshow for smart teenagers in 1998. (None of them solved it.) >> https://www.theguardian.com/science/2020/sep/07/can-you-solve-it-the-smalle… Fred Lunnon

1 0

Re: [math-fun] Interesting limit question (sort of spoiler alert)
by Dan Asimov 06 Sep '20

06 Sep '20

Nice analysis, George! Yup, Hans named it — the Foias constant. Apparently it was discovered because of a misprint in a copy of a question that originally asked whether the (much simpler) recurrence x_(n+1) = (1 + 1/x_n)^x_n (where x_1 > 0) could possibly approach oo. —Dan Hans Havermann wrote: ----- GH: "f(1) = 1.1874523511" Looks like Foias' constant. https://mathworld.wolfram.com/FoiasConstant.html -----

5 8