May 2014 - mathfun@mailman.xmission.com

Re: [math-fun] A knight's surprise?
by Bill Gosper 07 May '14

07 May '14

Years ago Scott Kim told me that when shown the rules of chess as a child, he exclaimed "Ooh, a tesseract!" when they came to the knight. --rwg On 2014-04-13 09:09, Fred Lunnon wrote: > Yeah, that's much more direct than my argument! WFL > > > > On 4/13/14, Adam P. Goucher <apgoucher(a)gmx.com> wrote: >> Very nice discovery! >> >> >> Lemma: The graph is isomorphic to the 4-by-4 toroidal grid graph. >> >> Proof: draw the graph in the way you defined (which is on the 4-by-4 >> torus, but has lots of edges crossing). Then, translate all of the black >> vertices by (2,2) whilst preserving the positions of the white vertices. >> Then every vertex is connected to its four neighbours and nothing else. >> >> -------- >> >> Corollary: (2) >> >> Proof: Trivial. >> >> -------- >> >> Corollary: (1) >> >> Proof: The graph is isomorphic to the product C4 x C4 (where C4 is >> the cycle graph on four vertices), so is isomorphic to the product >> K2 x K2 x K2 x K2 (since C4 is isomorphic to K2 x K2). Consequently, the >> graph is isomorphic to the skeleton of the tesseract (four-dimensional >> hypercube). >> >> -------- >> >> The last isomorphism I used (C4 x C4 == K2 x K2 x K2 x K2) may be familiar >> to any electronic engineers amongst you, on the basis that it allows >> Karnaugh maps to work. >> >> >> Sincerely, >> >> >> Adam P. Goucher >> >> >>> ----- Original Message ----- >>> From: Fred Lunnon >>> Sent: 04/13/14 12:04 PM >>> To: math-fun >>> Subject: [math-fun] A knight's surprise? >>> >>> Consider the graph whose vertices & edges represent squares & knight >>> moves on a 4x4 chessboard tiling the torus. >>> >>> (1) Find a spatial embedding with the full symmetry of the graph. >>> >>> (2) Deduce that the graph can be embedded (without crossings) on the >>> torus. >>> >>> Fred Lunnon

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[math-fun] Where the Buffalo roam
by James Propp 07 May '14

07 May '14

Okay, so who _else_ was annoyed as a child by the ambiguity of the phrase "and the skies are not cloudy all day"? (I mean, are the skies merely not cloudy-all-day, or are they actually not-cloudy all day?) This has bothered me for as long as I can remember. That's not exactly a mathematical issue, but part of the skill-set of the mathematician is knowing how to be extremely literal (whereas part of the deficit-set of many mathematicians is not knowing how not to be!). Which reminds me of an idea I have an idea for a cartoon, captioned "Over-thinking preschooler", showing a toddler sitting in a circle playing a clapping game, thinking to himself "I'm happy, but do I know it?" Jim P.S. The juxtaposition of the words "Buffalo" and "roam" bothers me because Buffalo and Rome are both cities in New York state, but that's another thing entirely.

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[math-fun] brain injury turns schmoe into "math genius"
by Warren D Smith 06 May '14

06 May '14

http://news.discovery.com/human/life/brain-injury-turns-man-into-math-geniu… except I actually see zero evidence in this article that this guy is now a "math genius." He says he has altered perceptions, but that is all I see there.

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[math-fun] What is the period of this integer sequence?
by Bill Gosper 05 May '14

05 May '14

0, 1, 7, -2, -8, 3, 14, -4, -16, 5, 21, -6, -24, 7, 28, -7, -25, 7, 27, -7, -26, 7, 26, -6, -19, 5, 18, -4, -12, 3, 10, -2, -5, 1, 2, 0, 2, -1, -6, 3, 16, -5, -22, 7, 30, -8, -30, 8, 30, -7, -23, 6, 22, -5, -16, 4, 14, -3, -9, 2, 6, -1, -2, 1, 5, -1, -3, 1, 4, -1, ... a[0] = 0; a[1] = 1; a[k_?OddQ] := a[k] = a[k - 2] - Floor[9*a[k - 1]/17]; a[k_?EvenQ] := a[k] = a[k - 2] + Floor[15*a[k - 1]/2] This is the "Minsky Stock Index" that Corey and Julian ran fruitlessly from a[-10^14] to a[10^14] a few years ago. "If you haven't looked at a problem in the last few years, you haven't looked at it." --Ed Pegg, Jr. There is a slight chance that the period is infinite. (When it reached 18 trillion, we exclaimed AT&T!) The above recursive definition crashes Mathematica in under a million, even with memoizing. Instead use the iteration In[115]:= NestList[Function[xy, {#, xy[[2]] + Floor[15*#/2]} & [xy[[1]] - Floor[9*xy[[2]]/17]]], {1, 0}, 35] e.g., for {a[-1],a[0}, {a[1],a[2]},...{a[69],a[70]} Out[115]= {{1, 0}, {1, 7}, ..., {-1, -4}} Background: http://www.blurb.com/books/2172660-minskys-trinskys-3rd-edition Unlike Collatz or twin prime searching, this is a very specific question about very specific quantities, rather than a question about the infinitude of integers. --rwg

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[math-fun] Suitable subjects for 3D-demonstration
by Guy Haworth 04 May '14

04 May '14

Here at the University of Reading, we have an 'immersion room' giving the full 3D-experience in which one can 'walk around' 3D-models. We also have a portable 'cube' version of the same which - with much less expensive technology these days - cost about an eighth of what the fixed system cost. I was wondering what 3D mathematical models could be built, and shown and maybe animated [how?] using these 3D-room facilities. First choices should favour ease of creating the models - and be models which would be useful as learning resources for final-school-exam students. We have a work-experience 16-year-old who is into 3D-modelling, and going forward to Maths/CS, visiting us to do some modelling work in the summer. Ideas welcome: thank you. Guy Haworth, UoR "Three more men: maxMate times ten!"

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Re: [math-fun] Some superficial remarks on http://arxiv.org/pdf/1401.7718v2.pdf
by Bill Gosper 03 May '14

03 May '14

DanA> Am curious to know how "simpler" is defined. --Dan On May 2, 2014, at 6:41 PM, Bill Gosper <billgosper(a)gmail.com> wrote: . . . a radical denester which simplifies (1 + 3*Sqrt[5] - 2*Sqrt[10 + 2*Sqrt[5]])^(1/4) . . . to √(Sqrt[2] 5^(1/4) - Sqrt[1 + Sqrt[5]]) |. . .] Good question. Corey defines simpler as smaller denominator of outermost power, even if the radicand lengthens. Here, it shortened, so there's no contest, assuming that you got a square root sign in front of the 2nd expression. Rich complains that he and presumably other funsters are getting ?s instead of square root, cube root, Phi, pi, and subscript characters, and quoted my message back so damaged, even though my copy from math-fun looks fine. (Except for !@#$%^&* gratuitous linebreaks, for which I blame GMail.) Anyway, since it was short, here it is again in Old High Cuneiform: [Goaded by Mike Hirschhorn] A couple of years ago, a young friend, Corey Ziegler Hunts, wrote a radical denester which simplifies (1 + 3*Sqrt[5] - 2*Sqrt[10 + 2*Sqrt[5]])^(1/4), the fourth root of a trinomial in your expression for G(e^-(2Pi)), to Sqrt[Sqrt[2] 5^(1/4) - Sqrt[1 + Sqrt[5]]], the square root of a binomial. Similarly for H. Three lines later on p2 are two typos: "surprizingly" and "were were". Corey's younger brother Julian and I wrote a sextic solver which led to (6 Cos[Pi/18] - 6 Cos[Pi/9] + 6 Cos[2Pi/9] - 6 Sin[2Pi/9] - 1)^(1/3) ~ 0.2170953869504416 as the value of Phi_1b(2,2;i/3) on p10. --rwg If you mean how "simpler" is defined to Mathematica, I wonder if anyone knows. It seems to be based on LeafCount and a capricious tree search, influenced, in the case of FullSimplify, by an optionally supplied ComplexityFunction. Simplicity, like Intelligence, is a multidimensional quality that confounds ordering.

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[math-fun] Some superficial remarks on http://arxiv.org/pdf/1401.7718v2.pdf
by Bill Gosper 03 May '14

03 May '14

[Goaded by Mike Hirschhorn] A couple of years ago, a young friend, Corey Ziegler Hunts, wrote a radical denester which simplifies (1 + 3*Sqrt[5] - 2*Sqrt[10 + 2*Sqrt[5]])^(1/4), the fourth root of a trinomial in your expression for G(e^-(2π)), to √(Sqrt[2] 5^(1/4) - Sqrt[1 + Sqrt[5]]), the square root of a binomial. Similarly for H. Three lines later on p2 are two typos: "surprizingly" and "were were". Corey's younger brother Julian and I wrote a sextic solver which led to ∛(6 Cos[π/18] - 6 Cos[π/9] + 6 Cos[2π/9] - 6 Sin[2π/9] - 1) ~ 0.2170953869504416 as the value of Φ₁b(2,2;i/3) on p10. --Bill Gosper

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Re: [math-fun] Math belt
by Ray Tayek 02 May '14

02 May '14

At 10:37 AM 5/2/2014, you wrote: >I have an old belt that I'd like to replace with a new one of the same >design. I call it my "mathematician belt" ... >Has anyone seen such a belt, or know who makes them? ... >at http://jamespropp.org/belt.jpg . i had one many years ago. i used it for backpacking. googling for: belt with many holes will get some pics and links. thanks --- co-chair http://ocjug.org/

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Re: [math-fun] Math belt
by Bill Gosper 02 May '14

02 May '14

W.Edwin Clark>There are apparently lots of belts with lots of holes:< https://www.google.com/search?site=imghp&tbm=isch&source=hp&biw=1280&bih=89…> Found at Google Images. On Fri, May 2, 2014 at 1:37 PM, James Propp <jamespropp(a)gmail.com> wrote: I have an old belt that I'd like to replace with a new one of the same design. I call it my "mathematician belt" because it has two tongues instead of just one and they can be fit into a 2-by-38 grid of holes. (One could also call it a person-of-variable-girth belt, but I don't.) Has anyone seen such a belt, or know who makes them? You can see a photo at http://jamespropp.org/belt.jpg . I bought it at Old Navy a decade ago, but nobody there recognizes them, and I haven't been able to find them on the web. Jim Propp Due to a seriously dyslexic manufacturer, my extra holes wound up in the *buckle*. http://gosper.org/IMG_0111.JPG --rwg

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[math-fun] Math belt
by James Propp 02 May '14

02 May '14

I have an old belt that I'd like to replace with a new one of the same design. I call it my "mathematician belt" because it has two tongues instead of just one and they can be fit into a 2-by-38 grid of holes. (One could also call it a person-of-variable-girth belt, but I don't.) Has anyone seen such a belt, or know who makes them? You can see a photo at http://jamespropp.org/belt.jpg . I bought it at Old Navy a decade ago, but nobody there recognizes them, and I haven't been able to find them on the web. Jim Propp

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