February 2018 - mathfun@mailman.xmission.com

[math-fun] Unlucky thirteen
by James Propp 24 Mar '18

24 Mar '18

#1. Puzzle: Show that for every positive integer n less than thirteen, there exists a Laurent polynomial p(x) with coefficients in {0,1,2,3,4} satisfying p(1/2) = p(1/3) = n. (Hint: Use an iterative construction n -> n+1.) #2. Problem: Show that this NOT the case for thirteen or any larger positive integer. (I don't know how to prove this.) If #1 seems tricky, here's a simpler version: Show that for every positive integer n there exists a Laurent polynomial p(x) with coefficients in {0,1,2} satisfying p(1) = p(2) = n. Equivalently, show that every positive integer n can be written as the sum of n powers of 2, each of which may appear at most twice (where "power of 2" means "2^k for some k in Z"). Jim Propp

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[math-fun] Curved space paradox
by Allan Wechsler 23 Mar '18

23 Mar '18

The following intuitive line of reasoning leads to a contradictory conclusion, and I don't know what's wrong with it. Obviously one of my assumptions or beliefs about spaces with constant Gaussian curvature is wrong, but I can't figure out which. Suppose we have a 3-dimensional space of constant, negative Gaussian curvature. For some large enough R, construct the locus of points at distance R from a fixed center C. This forms a 2-dimensional surface which is topologically an ordinary sphere. Select a point P on this sphere, and begin drawing circles of increasing radius r on the sphere centered at P. I mean r to be measured along the surface of the sphere, not straight through the hyperbolic 3-space in which the sphere is embedded. Now, this sphere must itself have constant Gaussian curvature; I don't have a fast proof but surely there is a symmetry argument that can establish it. Watch how the circumference of a circle, centered at P, changes with its radius r. For small r, it looks like 2 pi r. But as r gets larger, the negative curvature of the ambient hyperbolic space begins to make itself felt, and the circumference goes above 2 pi r, by some factor related to sinh. (If it doesn't, you didn't make R big enough -- that is exactly what I meant by "large enough R" above.) But later, as r becomes commensurate with R, the circumference rises to a maximum and begins to shrink again, because the sphere itself has finite circumference. On a surface of constant curvature, the circumference should either fall below 2 pi r, or it should exceed it forever. It should not do both. Surely I am not the first person to think these confusing thoughts. What is wrong with my reasoning?

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[math-fun] packing polyominoes?
by Mike Stay 02 Mar '18

02 Mar '18

My family recently gave me some packing puzzles for my birthday. One is a rectangle packing puzzle, where the rectangles' edges remain parallel to the sides of the tray; there are some pretty obvious algorithms for solving that one. The other is "Stewart's Coffin", which Martin Gardner pronounced "...the finest dissection puzzle of all time. It looks easy but is fiendishly difficult." It involves four laser-cut polyominoes and a tray that is just slightly too small in either direction for easy placement of the polynomials. Clearly the puzzle involves placing the pieces at an angle, but exactly which angle and how to arrange the pieces is the hard problem. Are there algorithms for solving a puzzle like this other than brute force? -- Mike Stay - metaweta(a)gmail.com http://www.math.ucr.edu/~mike http://reperiendi.wordpress.com

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[math-fun] Math camps...help!
by Marnie Kanarek 28 Feb '18

28 Feb '18

Hi all, I don't know if you can help me but does anyone have or know anyone that has any experience with CTY summer camps? Gabe really wants to do a math camp and CTY has a campus like 10 mins away (as opposed to Epsilon or C&amp which are very far away). They are all so expensive so we want to know that they are good before we commit. Also, if anyone has any summer camp options that would be...appropriate for a super advanced 8 year old that AREN'T super expensive, I'm all ears! (I tried to convince Gray to bring Gabe into work for a week at STSCI and just let him bug the astronomers...but that is probably not a realistic option :P) Thanks in advance! Marnie

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Re: [math-fun] packing polyominoes?
by Dan Asimov 27 Feb '18

27 Feb '18

I have a puzzle of 4 purple plastic pieces that sounds vaguely like what you're describing. They are arranged in a square, and each piece interlocks the next one around in a keyhole-shaped hole. (http://www.cs.brandeis.edu/~storer/JimPuzzles/SHAPE/JigSawFourPiece/JigSawF…) That's not it, is it? It's available in pretty much all game stores. —Dan Mike Stay wrote: ----- My family recently gave me some packing puzzles for my birthday. One is a rectangle packing puzzle, where the rectangles' edges remain parallel to the sides of the tray; there are some pretty obvious algorithms for solving that one. The other is "Stewart's Coffin", which Martin Gardner pronounced "...the finest dissection puzzle of all time. It looks easy but is fiendishly difficult." It involves four laser-cut polyominoes and a tray that is just slightly too small in either direction for easy placement of the polynomials. Clearly the puzzle involves placing the pieces at an angle, but exactly which angle and how to arrange the pieces is the hard problem. >Are there algorithms for solving a puzzle like this other than brute force? -----

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Re: [math-fun] packing polyominoes?
by Bill Gosper 27 Feb '18

27 Feb '18

Veit, isn't this your cue? https://en.wikipedia.org/wiki/Difference-map_algorithm I wouldn't be surprised if it could _design_ the puzzle (a.k.a Four Fit, Martin's Menace): Given the pentominoes, find the minimal box. --rwg Quibble: I wouldn't call this a dissection puzzle. On 2018-02-26 22:31, Mike Stay wrote: > My family recently gave me some packing puzzles for my birthday. One is a > rectangle packing puzzle, where the rectangles' edges remain parallel to > the sides of the tray; there are some pretty obvious algorithms for solving > that one. > > The other is "Stewart's Coffin", which Martin Gardner pronounced "...the > finest dissection puzzle of all time. It looks easy but is fiendishly > difficult." It involves four laser-cut polyominoes and a tray that is just > slightly too small in either direction for easy placement of the > poly[ominoe]s. Clearly the puzzle involves placing the pieces at an angle, > but exactly which angle and how to arrange the pieces is the hard problem. > Are there algorithms for solving a puzzle like this other than brute force?

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[math-fun] student faces expulsion for joking that square-root symbol looks like a gun
by Bill Gosper 27 Feb '18

27 Feb '18

https://www.truthrevolt.org/news/student-faces-expulsion-joking-square-root… --rwg

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[math-fun] dragun[2x]/dragun[x] = {1 - I}
by Bill Gosper 26 Feb '18

26 Feb '18

(0≤x≤½) What about dragun[5x]? In[347]:= dragun[5/6]/dragun[1/6] Out[347]= {-1 - 2 I} In[348]:= dragun[5/12]/dragun[1/12] Out[348]= {-1 - 2 I} In[349]:= dragun[5/8]/dragun[1/8] Out[349]= {-1 - 2 I} In[350]:= dragun[5/7]/dragun[1/7] Out[350]= {-I} Dang. In[351]:= dragun[2/7]/dragun[1/7] Out[351]= {1 - I} --rwg

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[math-fun] One fractal, two impressions
by Bill Gosper 25 Feb '18

25 Feb '18

Julian's curvaceous Fourierization of the usual 4-for-1 triangle fill gosper.org/trifil125%25.png looks improbable when polygonally sampled instead gosper.org/polygonal trifil0.png gosper.org/polygonal trifil.png (just standard connect-the-dots plots.) --rwg It appears that the actual spacefill has no triple points-- only singles, doubles, and quadruples! {1/2 + i^(2/3), 1 + i^(2/3)/2, 1/2 + i^(2/3)/2, 3/4 + i^(2/3)} In[257]:= lifirt /@ (% /. i -> I) {{3/8, 11/24, 9/16, 17/24}, {5/16, 5/12, 13/16, 11/12}, {1/12, 3/16, 7/24, 7/16}, {11/32, 35/96, 23/32, 71/96}} The centroid is not even double: In[258]:= lifirt[1 + I/√3] Out[258]= {2/5}

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Re: [math-fun] student faces expulsion for joking that square-rootsymbol looks like a gun
by Hilarie Orman 24 Feb '18

24 Feb '18

And now teachers will get bonuses for being radicals. Hilarie

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