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[math-fun] Help with Mathematica desired
by Dan Asimov 23 Oct '20

23 Oct '20

I tried to submit this to Wolfram Alpha in correct syntax, but it never gave any indication that it had the vaguest idea that this was an equation to solve, no less the correct equation to solve for real numbers u, v, w. Solve[Exp[I(3u+v)] + Exp[I(3v+w)] + Exp[I(3w+u)] == 0, {u,v,w}, Reals] What am I doing wrong? Thanks, Dan

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Re: [math-fun] Help with Mathematica desired
by Dan Asimov 23 Oct '20

23 Oct '20

Jim Propp wrote << Did you try putting the equation in curly braces? I didn't, because I couldn't see how that would make a difference. (But I just tried it now, and it didn't.) —Dan

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Re: [math-fun] Help with Mathematica desired
by Dan Asimov 23 Oct '20

23 Oct '20

This time I tried a simplified, all-real command: Solve[Cos[3 x+y] + Cos[3 y+z] + Cos[3 z+x] == 0, {x,y,z}] (verbatim). Wolfram Alpha had absolutely no idea that it was even being asked to solve an equation. —Dan

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Re: [math-fun] Help with Mathematica desired
by Dan Asimov 23 Oct '20

23 Oct '20

In fact I already tried both 1. and 2. right after posting to math-fun, and there was no improvement! —Dan Gareth McCaughan wrote: ----- On 23/10/2020 22:36, Dan Asimov wrote: > I tried to submit this to Wolfram Alpha in correct syntax, > but it never gave any indication that it had the vaguest idea > that this was an equation to solve, no less the correct equation > to solve for real numbers u, v, w. > > Solve[Exp[I(3u+v)] + Exp[I(3v+w)] + Exp[I(3w+u)] == 0, {u,v,w}, Reals] > > What am I doing wrong? I don't know how helpful the following is, but: 1. If I put exactly that into Mathematica, it complains about the instances of I, on the grounds that if you're trying to solve an equation over the reals then all the constants involved must be real. 2. If I replace it with Cos[...]+...==0 && Sin[...]+...==0, then instead I get "The system cannot be solved with the methods available to Solve." -----

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Re: [math-fun] Z^2 scale/rotate/translate puzzle
by Steve Witham 23 Oct '20

23 Oct '20

As the Sesame Street Martians say, Yyyyyyyep. Yyyyep. Yep-yep-yep-yep-yep-yep. GCD is the next best thing to factoring. GCD of Gaussian integers looks slightly trickier than the ordinary version. My brain says "I don't want to learn a whole new skill!" and I have to tell it, "You put your shoes on, stop hiding in your comfort zone, go outside and have some math-fun young man." --Steve > From: Allan Wechsler <acwacw(a)gmail.com> > Date: 10/22/20, 11:32 AM > > Yes. Then take the GCD of all the remaining points (considered as Gaussian > integers), and divide out by the GCD to obtain the solution. > > On Thu, Oct 22, 2020 at 1:19 AM Steve Witham <sw(a)tiac.net> wrote: > >> Not only does reflection not matter to the problem, >> but translate the problem so that one original point >> is (0, 0) and let the same point in the solution be (0, 0). >> >>> From: Henry Baker <hbaker1(a)pipeline.com> >>> Date: 10/21/20, 10:27 AM >>> >>> I'm guessing factorization of Gaussian integers is involved... >>> >>> At 12:51 AM 10/21/2020, Tom Karzes wrote: >>>> Here's a puzzle some of you might like: (fin)

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Re: [math-fun] Zonohedral surfaces
by Dan Asimov 22 Oct '20

22 Oct '20

Not at all. My answer was the result of very deep thought! —Dan Jim Propp wrote: ----- I was just overlooking the obvious! -----

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Re: [math-fun] Zonohedral surfaces
by Dan Asimov 22 Oct '20

22 Oct '20

For some reason this is reminding me of a very pretty polyhedral surface of genus 3 that does not exist as a subset of R^3 but instead as a subset of the torus R^3/Z^3. Truncate the corners of a unit cube so that what remains is a truncated octahedron (the corners become 8 hexagons and the faces become 6 smaller squares at 45º angles). Now just considering the 14 faces, delete the 6 squares and identify the opposite faces of the original cube so that it becomes the torus T^3 = R^3/Z^3. The truncated octahedron is now a polyhedral surface M_3 of genus 3 inside the torus T^3, tiled by 8 hexagons, 4 per vertex. There is an isometry of the whole M_3 to itself taking a given hexagon to any hexagon in any of 12 dihedral ways, for a total of 96 isometries altogether. —Dan Tomas Rokicki wrote: ----- Or take the eight cube solution and skew two opposite cubes so the visible parts are not rectangles. This introduces the necessary separating edges. -----

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[math-fun] Zonohedral surfaces
by James Propp 22 Oct '20

22 Oct '20

George Hart’s Celebration of Mind talk (just concluded) makes me wonder: is there a non-self-intersecting torus surface made of parallelograms? Jim Propp

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Re: [math-fun] Z^2 scale/rotate/translate puzzle
by Dan Asimov 22 Oct '20

22 Oct '20

Tom Karzes wrote: ----- The minimal solution is unique up to multiplication by a unit (1, i, -1, -i). ----- This depends on the point set. If say it's X = {(0,0), (5,0), (10,0), (15,0), (20,0)} Then there are 12 rotations about the origin that carry X into a set of integer points. (By angles N*90º and N*90º ± atan2(3,4), N = 0, 1, 2, 3.) —Dan

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Re: [math-fun] Z^2 scale/rotate/translate puzzle
by Henry Baker 22 Oct '20

22 Oct '20

I can imagine that 3D is easier, since we only have to solve for *one* rotation. In 4D, there are 2 independent rotations. Perhaps the 4D answer is related to the fact that there are *two* gcd algorithms: a 'left' gcd and a 'right' gcd. So perhaps you have to find both gcd's for the final result? At 10:48 AM 10/22/2020, Fred Lunnon wrote: > Hurwitz quaternions work in 3-space --- see for example my write-up for when >this algorithm last appeared on math-fun, > https://arxiv.org/abs/1202.3198 > >I never needed to consider 4-space, since nobody has managed to discover a >Heronian pentatope! > >WFL > >On 10/22/20, Henry Baker <hbaker1(a)pipeline.com> wrote: >> Hurwitz quaternions might work for 4D. >> >> At 09:32 AM 10/22/2020, Tom Karzes wrote: >>>Good job! That's the solution I found when I first came up >>>with this problem: >>> >>> First translate so that one of the points is at the origin. >>> Then divide all the points by the GCD. >>> >>>That's all there is to it. It doesn't matter which point ends >>>up at the origin, as long as one of them does. The GCD will be >>>the same no matter which one you choose. You can think of the >>>translated set of points as displacement vectors (relative to >>>the point at the origin). >>> >>>Treating the points as Gaussian integers is a key insight. >>>So is translating one of them to the origin, and of course >>>dividing by their GCD. I went out of my way not to mention >>>Gaussian integers in the problem description, just to make >>>the solution a little less obvious. I also made sure my >>>example didn't have any of the points at the origin, again >>>to make it less obvious. >>> >>>The minimal solution is unique up to multiplication by a >>>unit (1, i, -1, -i). >>> >>>Once you have a minimal solution, multiplication by any >>>nonzero Gaussian integer will give you a new scaled/rotated >>>set of integer points with the same shape. If you don't want >>>to rotate, you can just multiply by a positive integer. And >>>of course, you can reflect by taking the complex conjugate. >>> >>>To compute the Gaussian GCD, the standard Euclidean Algorithm >>>works perfectly well. It just needs to operate on Gaussian >>>integers. >>> >>>I actually used this for some raster graphics I was doing, where >>>I had sets of points that I wanted to scale up or down without >>>introducing any distortion. This solved the problem nicely. >>> >>>Any idea how to scale this up to 3 (or more) dimensions? The >>>Gaussian integer approach works great for 2D, but obviously >>>not for higher dimensions. Is there a way to represent higher- >>>dimensional points as, say, orthogonal matrices, or polynomials, >>>or something along those lines, then obtain some sort of GCD and >>>somehow factor it out? I guess part of the difficulty is that >>>rotations don't commute in 3 or more dimensions. >>> >>>Still, the basic problem is perfectly well-defined for higher >>>dimensions, so some solution must be possible. >>> >>>Tom >>> >>>Allan Wechsler writes: >>> > Yes. Then take the GCD of all the remaining points (considered as >>> > Gaussian >>> > integers), and divide out by the GCD to obtain the solution. >>> > >>> > On Thu, Oct 22, 2020 at 1:19 AM Steve Witham <sw(a)tiac.net> wrote: >>> > > Not only does reflection not matter to the problem, >>> > > but translate the problem so that one original point >>> > > is (0, 0) and let the same point in the solution be (0, 0). >>> > > >>> > > Without loss of generality, >>> > > --Steve >>> > > >>> > > > From: Henry Baker <hbaker1(a)pipeline.com> >>> > > > Date: 10/21/20, 10:27 AM >>> > > > To: >>> > > > >>> > > > I'm guessing factorization of Gaussian integers is involved... >>> > > > >>> > > > At 12:51 AM 10/21/2020, Tom Karzes wrote: >>> > > >> Here's a puzzle some of you might like:

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