April 2014 - mathfun@mailman.xmission.com

Re: [math-fun] A knight's surprise?
by Bill Gosper 08 May '14

08 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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Re: [math-fun] square-ball non-dissectability
by Bill Gosper 30 Apr '14

30 Apr '14

On 2014-04-30 05:47, Adam P. Goucher wrote: > I think you want the *signed* total length of curvature-1 boundary, namely: > > (length of (convex) curvature-(+1) boundary) > - (length of (concave) curvature-(-1) boundary) > > In which case Warren's proof will work, assuming that you are only allowed > compact pieces with piecewise-C2 boundaries. I don't know what the definition > of scissors-congruence is, but I imagine that it would be something like > that (after all, can *you* cut out a shape that isn't compact or lacks a > piecewise-twice-continuously-differentiable boundary using scissors???). No, but I could describe one. And scissors "cheat" by making two closed edges, much reminiscent of spacefilled patches. So we ought to be able to create area with 2D cuts. --rwg > > > Sincerely, > > > Adam P. Goucher > >> ----- Original Message ----- >> From: Allan Wechsler >> Sent: 04/29/14 11:06 PM >> To: math-fun >> Subject: Re: [math-fun] square-ball non-dissectability >> >> The lemma is false, unless I have misunderstood it. Suppose you cut out a >> "biscuit" by making a cut along a circular arc of radius 1, centered on the >> circumference of the original circle. Now the amount of curvature-1 >> boundary has increased. Now glue the two pieces back together. You haven't >> made any progress, it's true, but at the second step the amount of >> curvature-1 boundary has decreased. >> >> Something like this argument may go through, but this particular >> formulation looks _too_ facile to me. >> >> >> >> On Tue, Apr 29, 2014 at 5:39 PM, Warren D Smith <warren.wds(a)gmail.com >wrote: >> >> > A square and circular disc of equal area >> > are NOT inter-dissectable (i.e. scissors congruent, finite # pieces). >> > Easy proof: >> > >> > Lemma: The total length of curvature-1 boundary can never decrease >> > as you cut up your disc and move the pieces around (gluing >> > them at common boundary). >> > >> > But it must decrease, in fact to 0, to achieve the conversion. >> > Contradiction.

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[math-fun] Golyhedra
by Adam P. Goucher 30 Apr '14

30 Apr '14

Alex, Ed, and anyone else who enjoys recreational mathematics: A _golyhedron_ is a simply-connected lattice polyhedron whose face areas form the sequence {1, 2, 3, ..., n} for some n. Joseph O'Rourke conjectured on MathOverflow that no such golyhedra exist. After some deliberation, however, I was able to find a 32-faced counter-example, given below: http://cp4space.files.wordpress.com/2014/04/labelled-golyhedron.png Here's an article introducing golyhedra and explaining how I discovered this elusive beast: http://cp4space.wordpress.com/2014/04/30/golygons-and-golyhedra/ My golyhedron has 32 faces. I include a proof that no golyhedron can have fewer than 11 faces, so the challenge is to decrease the (rather large!) gap between 11 and 32 by either providing smaller examples or stronger proofs. I have really no intuition as to what an optimal golyhedron would resemble. I have a feeling that you can probably reduce 32 quite considerably by using a similar approach with more work (I made no attempt to optimise my golyhedron), but you may need an entirely different approach to get down to 20 or fewer faces. Sincerely, Adam P. Goucher

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Re: [math-fun] square-ball non-dissectability
by Adam P. Goucher 30 Apr '14

30 Apr '14

I think you want the *signed* total length of curvature-1 boundary, namely: (length of (convex) curvature-(+1) boundary) - (length of (concave) curvature-(-1) boundary) In which case Warren's proof will work, assuming that you are only allowed compact pieces with piecewise-C2 boundaries. I don't know what the definition of scissors-congruence is, but I imagine that it would be something like that (after all, can *you* cut out a shape that isn't compact or lacks a piecewise-twice-continuously-differentiable boundary using scissors???). Sincerely, Adam P. Goucher > ----- Original Message ----- > From: Allan Wechsler > Sent: 04/29/14 11:06 PM > To: math-fun > Subject: Re: [math-fun] square-ball non-dissectability > > The lemma is false, unless I have misunderstood it. Suppose you cut out a > "biscuit" by making a cut along a circular arc of radius 1, centered on the > circumference of the original circle. Now the amount of curvature-1 > boundary has increased. Now glue the two pieces back together. You haven't > made any progress, it's true, but at the second step the amount of > curvature-1 boundary has decreased. > > Something like this argument may go through, but this particular > formulation looks _too_ facile to me. > > > > On Tue, Apr 29, 2014 at 5:39 PM, Warren D Smith <warren.wds(a)gmail.com>wrote: > > > A square and circular disc of equal area > > are NOT inter-dissectable (i.e. scissors congruent, finite # pieces). > > Easy proof: > > > > Lemma: The total length of curvature-1 boundary can never decrease > > as you cut up your disc and move the pieces around (gluing > > them at common boundary). > > > > But it must decrease, in fact to 0, to achieve the conversion. > > Contradiction. > > > > _______________________________________________ > > math-fun mailing list > > math-fun(a)mailman.xmission.com > > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

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[math-fun] Extending Rogers-Ramanujan Identities
by Robert Smith 29 Apr '14

29 Apr '14

"A framework of Rogers-Ramanujan identities and their arithmetic properties" http://arxiv.org/pdf/1401.7718v2.pdf Beware, there's a good amount of patent-office algebra. Cheers, Robert

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[math-fun] square-ball non-dissectability
by Warren D Smith 29 Apr '14

29 Apr '14

A square and circular disc of equal area are NOT inter-dissectable (i.e. scissors congruent, finite # pieces). Easy proof: Lemma: The total length of curvature-1 boundary can never decrease as you cut up your disc and move the pieces around (gluing them at common boundary). But it must decrease, in fact to 0, to achieve the conversion. Contradiction.

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[math-fun] trigsign
by Bill Gosper 29 Apr '14

29 Apr '14

By default (i.e., trigsign:true), Macsyma (but alas, not Maxima) pulls out the sign: (c1) sin((-x)^integer) (d1) ( - 1)^integer * sin(x^integer) (Alternatively, (declare(n,integer), sin((-x)^n)) But I can't for the life of me get Mma even to pull the sign out of Assuming[n \[Element] Integers, Simplify[FunctionExpand[Sin[(-1)^n*x]]]] Yet it knows In[375]:= FullSimplify[Sin[(-1)^n*x] == (-1)^n*Sin[x], n \[Element] Integers] Out[375]= True What am I doing wrong? --rwg ARRGGH! Maxima: (%i2) binomial(-3,-5); (%o2) 6 They must have taken DoE Macsyma and *broken* it! Similarly broken is Mma, whose doc falsely claims this gives 0, which it damwell should.

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Re: [math-fun] Banach-Tarski vs scissors
by Bill Gosper 28 Apr '14

28 Apr '14

DanA> Is there a proof that a 4-dimensional cube and 4-dimensional ball (sphere + interior) of the same volume are not scissors congruent? <DanAsimov I missed something. Is there a proof that a TWO-dimensional "cube" and TWO-dimensional "ball" (disk) of the same area are not interdissectible w/o A/C? It seems to me there's a lot of middle ground between "piece-wise smooth" (if that is indeed what scissors require) and "non-measurable". Also, I need to be educated as to how the following two conditions can be inequivalent: WDS> However, Dehn showed in one of the first Hilbert problem solves, that a regular tetrahedron and cube (same volume) are NOT scissors congruent. ----- DanA>I thought Dehn proved only that they could not each be dissected into the same finite set of polyhedra (up to isometries), analogous to the Bolyai dissections. ----- --rwg

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Re: [math-fun] Banach-Tarski vs scissors
by Adam P. Goucher 28 Apr '14

28 Apr '14

> > Nope. You can prove a set is nonempty without constructing > > any elements. > > How? For example, let S be the set of well-orderings of the reals. This is non-empty, but you can't construct any of its elements. A slightly more sophisticated example is the compact Hausdorff space betaN \ N of non-principal ultrafilters. Sincerely, Adam P. Goucher

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[math-fun] Novel ways to present proofs
by Adam P. Goucher 28 Apr '14

28 Apr '14

What novel ways have been used to present proofs or (more generally) mathematical ideas? I've encountered people answer olympiad problems with proofs written as limericks and sonnets before; there's also the Socratic dialogue, used by one student in a selection test and later to popularise the elementary measure-theoretic proof of Poncelet's porism: https://cp4space.wordpress.com/2014/04/19/poncelets-porism-the-socratic-dia… I've also heard of lecture in which the lecturer wrote down the truth table for logical conjunction on an overhead projector transparency like so: t ^ t = t t ^ f = f f ^ t = f f ^ f = f He then flipped the entire OHP sheet over, transforming every AND into an OR whilst transforming the `f's into `t's and vice-versa. Indeed, this is not the only spectacular use of an overhead projector. The late Christopher Bradley famously used an overhead projector to demonstrate projective transformations in a beautifully literal way: he tilted the projector, demonstrating the following concepts: -- Interchangeability of different types of conics; -- Interchangeability of parallel lines and convergent lines; -- Preservation of collinearity and concurrency; -- Conservation of cross-ratio. Any other examples of novel presentations of mathematical ideas? Sincerely, Adam P. Goucher

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