June 2013 - mathfun@mailman.xmission.com

Re: [math-fun] Naturally-occurring banana shaped like an integral sign
by Phil Carmody 08 Jun '13

08 Jun '13

From: "Adam P. Goucher" <apgoucher(a)gmx.com> > The title is pretty self-explanatory. > > http://cp4space.wordpress.com/2013/06/05/integral-banana/ "It is renowned for having a constant positive curvature" ?!??! If you've not noticed the saddle, you've clearly never ridden on a giant inflatable one! Phil -- () ASCII ribbon campaign () Hopeless ribbon campaign /\ against HTML mail /\ against gratuitous bloodshed [stolen with permission from Daniel B. Cristofani]

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[math-fun] (Generalized) cross products as 'caches'
by Henry Baker 06 Jun '13

06 Jun '13

The vector cross product in 3D is a very curious construct. Gibbs found quaternions to be too limiting for his applications, and so he split their product into two pieces -- the dot product & the cross product: If p,q are 'vector' quaternions (whose scalar component is zero), then pq = pxq - p.q Cross products are a kind of 'incomplete' determinant: | i j k| |ux uy uz| = uxv |vx vy vz| where i, j, k are formal variables to be substituted later to form the complete 'triple product' via the dot product: |wx wy wz| |ux uy uz| = w.(uxv) |vx vy vz| But suppose we didn't know (or care) what the _interpretation_ of uxv was; suppose we only wanted to compute a bunch of 3x3 determinants whose 2nd & 3rd rows were always the constants [ux,uy,uz] and [vx,vy,vz]. Then the computation of these determinants is speeded up by the concept of cross products, because we've already done most of the work by pre-computing the determinant minors. So, in effect, the cross product has acted like a 'cache' for these pre-computed results. Since 3x3 determinants compute _signed_ areas, they are extremely useful in computer graphics to compute whether some point falls 'above' or 'below' some line segment. The cross product effectively pre-computes the information from the line segment so that we can test many other points against this line segment. This also works for higher dimensions. We can test whether a point p is 'above' or 'below' the plane determined by the triangle u,v,w with the following determinant: |px py pz 1| |ux uy uz 1| |vx vy vz 1| = D |wx wy wz 1| But we often have to perform this test with a large number of different p's. This means that if we could _precompute_ the appropriate minors and gather this information up into a 4D 'vector', we could do each test with a simple 4D dot product. D = |uy uz 1| |vy vz 1| * px - |wy wz 1| |ux uz 1| |vx vz 1| * py + |wx wz 1| |ux uy 1| |vx vy 1| * pz - |wx wy 1| |ux uy uz| |vx vy vz| = |wx wy wz| [px,py,pz,1].Dxyz1 where Dxyz1 is the appropriate 4D vector of pre-computed minors. Note that we don't need to know or care about any interpretation of this vector, other than that it is an 'incomplete' determinant. Once we have the notion of an 'incomplete' determinant, we can come up with macro code for incomplete determinants of more general kinds, by including formal variables for various arbitrary entries in a determinant. We leave it as an exercise for some CS major to automatically 'compile' such an incomplete determinant by expanding-by-minors so as to minimize the number of operations left to be performed once the formal variables become bound.

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[math-fun] 3D polyhedra with thousands of holes (big genus)
by Henry Baker 06 Jun '13

06 Jun '13

I've been hacking some code for building 'networks' in 3D out of 'nodes' and 'links'. Here's the problem. I have a very large undirected graph -- e.g., O(10^6) vertices/faces/edges -- that I want to embed in 3D. I make an assignment of vertices to 3D points, and would then like to 'link' these points together with 'links'/'struts' which are right regular cylindrical polyhedra. Thus, given points P1=[x1,y1,z1], P2=[x2,y2,z2], I want to 'link' P1 to P2 with a 'strut' whose cross section is a regular n-gon (e.g., equilateral triangle) of some fixed size (e.g., side length = 10 millimeters). For the moment, I will completely ignore the problem of choosing the 3D locations of the vertices so that the struts do not interfere/intersect -- except where two (or more) struts meet at a vertex node. (This will be easy if I construct the graph embedding in the first place with no such interferences/intersections.) We thus assume that we have a 'good' embedding of vertices/nodes in 3D such that the struts don't interfere/intersect except at the vertices/nodes themselves. Note that when we have a 'strut' from P1 to P2, the line down the exact center of this strut goes exactly from P1 to P2. However, the angles/faces where two such struts meet can be quite complex. Thus, the only degree of freedom for the strut from P1 to P2 is the _rotation angle_ of the strut around the line from P1 to P2 as the axis of rotation. I'd like to minimize the complexity (in terms of the number of faces/facets) of the overall structure, which means minimizing the number of faces where the struts come together at the vertices. Consider one of the simplest examples: a complete graph of 3 vertices joined by 3 edges/links, which I embed in 3D as an equilateral triangle. I want the links to have cross-sections which are regular n-gons -- e.g., triangles. I could choose to have the 'flat' part of all three struts on the _outside_ of the big triangle linking the 3 nodes. Each pair of struts would meet at a 60 degree angle and generate no additional faces. Or I could choose to have the 'edge' part of all three struts on the _ outside_ of the big triangle linking the 3 nodes. Each pair of struts would still meet at a 60 degree angle and generate no additional faces. But I could choose _any angle_ for one strut around its long axis, and -- so long as I choose the _same angle_ for the other two struts -- I can still arrange for the struts to meet at the vertices/nodes with no additional faces. I.e., the complete graph of 3 vertices and 3 edges has only 1 degree of freedom. Q: what happens if the location assignment of the graph to 3D is no longer equilateral? Am I now forced to have additional faces? A: No, we can still have simple joins. Proof: look at the 3 planes which are the angle bisectors of the embedded triangle. When 2 struts meet at such an angle bisector plane, they intersect the angle bisector plane at the same triangle if their rotation angle about their long axis is the same. Thus, _any_ embedded triangle still has exactly 1 degree of freedom: if we choose the rotation angle of any 1 of the 3 struts, the rotation angles of the other two struts are constrained to be the same. Q: what happens with more complex objects -- e.g., what happens if I start with a _planar_ graph & use a location assignment that preserves its planarity. What exactly do the nodes look like where the struts come together at random angles? Q: what happens with non-planar embeddings -- e.g., a graph that looks like a regular (or perhaps non-regular) tetrahedron?

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CIA Releases Analysts Tale of Cracking the Kryptos Sculpture
by Henry Baker 05 Jun '13

05 Jun '13

FYI -- http://www.wired.com/threatlevel/2013/06/analyst-who-cracked-kryptos/

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[math-fun] Naturally-occurring banana shaped like an integral sign
by Adam P. Goucher 05 Jun '13

05 Jun '13

The title is pretty self-explanatory. http://cp4space.wordpress.com/2013/06/05/integral-banana/ Sincerely, Adam P. Goucher

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[math-fun] squared squares
by Bill Gosper 04 Jun '13

04 Jun '13

http://www.squaring.net/sq/ss/spss/o22/spsso22.pdf has largest/smallest = 30. Is this minimal? What about squared rectangles? --rwg

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[math-fun] Some binomial double sums
by Warren D Smith 04 Jun '13

04 Jun '13

The r-power "binomial moments" are f(k,r) = sum_p binomial(2*k, k+p) * |p|^r and for any fixed r=0,1,2,... can be evaluated in closed form, see http://arxiv.org/abs/math/0606080 (and also MAPLE9 is capable of doing these for r=0,1,2,...,9 with some human help). I instead want to discuss some DOUBLE sums defining a different kind of r-power moment. Let f(k,s,r) = sum_p sum_q binomial(2*k, k+p)*binomial(2*k, k+q)*|p^2+s*q^2|^r where I have in mind that s=+1 or s=-1, and r=0,1,2,3... The sums are over all integers p,q, or if you want to be pedestrian then for -k <= p <= k, -k <= q <= k. If s=+1 or if r=even then the absolute value sign is obviated whereupon the double sums are expressible in terms of the above single-sums, no problemo. Life only is difficult if r=odd and s=-1. But in those cases Richard Brent & I surprised ourselves by discovering various closed forms: f(k, -1, 1) = 2 * k^2 * binomial(2*k,k)^2 f(k, -1, 3) = 8*binomial(2*k-2,k-1)^2 * k * (2*k-1) * (8*k^2-12*k+5) f(k, -1, 5) = 8*binomial(2*k-2,k-1)^2 * k * (256*k^5-1152*k^4+2112*k^34-1936*k^2+874*k-153) ...except I do not know how to prove these. (They are empirically correct.) Brent thinks he can prove the first, but I'll believe his proof when I see it :) So, what is the general formula and how to prove it?

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Re: [math-fun] squared squares
by Stuart Anderson 04 Jun '13

04 Jun '13

Armin Singer has written a program to search for 'nice' squared squares http://www.squaring.net/downloads/squaredance-0.10.tgz He has a couple of 'nice' files in the tarball Also James Williams produced over 15 million new perfect squared squares earlier this year. Most of them havnt been added to the website yet. There are bound to be a whole lot of record breakers in that lot. There are some 'nice' squared rectangles here http://www.squaring.net/sq/sr/spsr/spsr_minmax.html Stuart

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Re: [math-fun] Norman Routledge (1928--2013) - FRS or die?
by Fred lunnon 04 Jun '13

04 Jun '13

Douglas Rogers cites the following obit. WFL <http://www.independent.co.uk/news/obituaries/norman-routledge-inspirational…>

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[math-fun] artistically-optimal squared squares
by Warren D Smith 04 Jun '13

04 Jun '13

I am also interested in the length of the minimal line segment not just the minimal square tile side. For example in http://www.squaring.net/sq/ss/spss/o22/spsso22.pdf there are two adjacent squares of sides 6 & 7 causing a line segment of length=1, and by that reckoning the ratio of max/min is 60/1.

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