October 2015 - mathfun@mailman.xmission.com

Re: [math-fun] Totally unsupported polygons and polyhedra
by Warren D Smith 14 Oct '15

14 Oct '15

A D-dimensional totally unsupported polytope, necessarily has at least D+1 convex vertices (i.e. which are on the convex hull). Say it has C, where C>=D+1. * If D=2, it must also have C concave vertices for a simple polygon otherwise a supported edge would occur. * If D=3 and C=4, then it must also have at least 2 nonconvex vertices since it it had 0 it'd be convex, and if it had 1 then by considering all possible combinators for the edge graph you can see there would have to be a support face. And if D=3, C>=5 at least 1 nonconvex vertex would be required. Then, IF the faces are triangles Euler formula allows you to see at least 8 faces would be needed. So I claim the above argument PROVES that the "alternating hexagon" and the "Asimov pringle" are both minimal examples, in the sense the former has the min # edges, and the latter the min # triangle faces (provided all faces triangles and interior is topologically a ball). I tend to doubt allowing nontriangle faces would help, but have not proven it (and even my proof above of what I did prove. feels a bit handwavy)... if so then Asimov pringle is minimal-faced for all polyhedra whose interior has the topology of a ball.

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Re: [math-fun] Totally unsupported polygons and polyhedra
by Dan Asimov 13 Oct '15

13 Oct '15

Small correction: Another math-fun member has kindly pointed out that the intersection of AB with BC in the first paragraph below is not necessarily in the interior of both of them. If the intersection of AB and CD occurs at one of the 4 endpoints (but necessarily not at a common endpoint), the difference with what I wrote below is that one of the triangles ACD or BCD (but not both) will be degenerate. Fortunately, the conclusion still follows. —Dan > On Oct 13, 2015, at 3:55 PM, rcs(a)xmission.com wrote: > > Quoting Dan Asimov <asimov(a)msri.org <mailto:asimov@msri.org>>: > >> Assume A, B, C, D are distinct points in the plane such that >> segments AB and CD intersect (necessarily in their interiors). >> >> Then ACD and BCD are triangles sharing the edge CD >> but otherwise disjoint, since A and B are on opposite sides of >> the line extending CD. >> >> Thus BC and AD are disjoint. >> >> ?Dan >> >> >>> On Oct 13, 2015, at 8:02 AM, James Propp <jamespropp(a)gmail.com <mailto:jamespropp@gmail.com>> wrote: >>> >>> 3) Although this is more relevant to my mistake from last night than to the >>> questions I'm raising here, I'd still like to know the right way to see >>> that, given points A, B, C, and D in the plane, it isn't possible for line >>> segment AB to intersect line segment CD *and* for line segment BC to >>> intersect line segment AD unless two or more of the points coincide.

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Re: [math-fun] Totally unsupported polygons and polyhedra
by Leo Broukhis 13 Oct '15

13 Oct '15

A friend of mine has proposed to name the saddle-like figure "atcohedron". Leo

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Re: [math-fun] math-fun Digest, Vol 152, Issue 16
by Leo Broukhis 13 Oct '15

13 Oct '15

On Tue, 13 Oct 2015 13:59:56 -0500, James Buddenhagen <jbuddenh(a)gmail.com> wrote: > Here is a pic: > https://www.dropbox.com/s/0nizg6iy93t03eu/pringle1-3.png?dl=0 Hello James, Could your software save it as .STL for 3D-printing? Thanks, Leo

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[math-fun] Totally unsupported polygons and polyhedra
by James Propp 13 Oct '15

13 Oct '15

I was thinking sloppily when I wrote (as part of the math-fun thread "5 and 7 outcome fair dice") that the pentagram is optimal. For purposes of this problem, the pentagram should be viewed as a non-convex decagon. And 10 sides isn't optimal (see below). So here are my questions: 1) Call a side of a polygon "supporting" if it's part of a support line of the polygon (that is, the polygon lies in one of the closed half-planes determined by the line) and "unsupporting" otherwise, and call a polygon "totally unsupported" if it has no supporting sides. What is the smallest number of sides a totally unsupported polygon can have? My record is 6 (a hexagon whose sides alternate between concave angles and reflex angles). 2) Call a face of a polyhedron "supporting" if it's part of a support plane of the polyhedron (that is, the polyhedron lies in one of the closed half-spaces determined by the plane) and "unsupporting" otherwise, and call a polyhedron "totally unsupported" if it has no supporting faces. What is the smallest number of faces a totally unsupported polyhedron can have? My record is 12 (see below). Also: 3) Although this is more relevant to my mistake from last night than to the questions I'm raising here, I'd still like to know the right way to see that, given points A, B, C, and D in the plane, it isn't possible for line segment AB to intersect line segment CD *and* for line segment BC to intersect line segment AD unless two or more of the points coincide. Jim On Tue, Oct 13, 2015 at 12:07 AM, James Propp <jamespropp(a)gmail.com> wrote: > I like the idea of a concave polyhedron that cannot rest on any of > its faces (stably or unstably), i.e., a polyhedron none of whose faces > belong to supporting planes. (Sort of like a holeyhedron, but a whole lot > easier to construct.) > > How few faces can such a polyhedron have? > > Take a solid tetrahedron and excavate triangular-pyramidal holes on each > of its four faces; we get a polyhedron with twelve faces that can't rest > (stably or unstably) on any of them. > > Or: stick two tetrahedra together along a face, but give one of the > tetrahedra a 180-degree twist, so that the two abutting triangular > faces form a star of David. We again get a polyhedron with twelve faces > that can't rest on any of them. > > Can anyone do better? > > (I can prove by brute force that the pentagram is the optimal solution to > the 2D version of the question. Is there a slick way to see that a > tetragram can't have the property? This last question smells like an > Olympiad problem to me.) > > Jim Propp > > On Monday, October 12, 2015, Bill Gosper <billgosper(a)gmail.com> wrote: > >> Odd barrel dice have the problem that two faces are always uppermost, >> so how do you label them? Alternative to face-transitivity, there are >> fair dice which cannot rest on any face, and yet have one face >> uppermost. Spoiler: gosper.org/5&7dice.png >> --rwg >> I assume these are old ideas. Check out Wikipedia if you haven't >> seen spherical, 6 outcome dice. >> _______________________________________________ >> math-fun mailing list >> math-fun(a)mailman.xmission.com >> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >> >

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[math-fun] Add 1 and reverse
by James Propp 13 Oct '15

13 Oct '15

I've been playing with the compound operation on bit-strings of length m in which you (a) add 1 mod 2^m and (b) reverse the order of the bits. Has anyone seen this before? It seems sufficiently simple that I doubt I'm the first person to have played with it. Jim Propp

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[math-fun] deterring evolution of antibiotic resistance
by Warren D Smith 12 Oct '15

12 Oct '15

it finally occurred to somebody to use mathematics / modelling: Daniel Nichol, Peter Jeavons, Alexander G. Fletcher, Robert A. Bonomo, Philip K. Maini, Jerome L. Paul, Robert A. Gatenby, Alexander R.A. Anderson, Jacob G. Scott. Steering Evolution with Sequential Therapy to Prevent the Emergence of Bacterial Antibiotic Resistance. PLOS Computational Biology, 2015; 11 (9): e1004493 http://journals.plos.org/ploscompbiol/article?id=10.1371/journal.pcbi.10044… will it work? -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Draft of October essay
by James Propp 12 Oct '15

12 Oct '15

I'd welcome input on an essay on Polya's urn model that I'll be posting at Mathematical Enchantments on Saturday. (It's still a bit rough; I plan to re-do all the figures as well as the embedded video, and if time permits I'll add some extra video content as well.) See https://mathenchant.wordpress.com/?p=203&shareadraft=561bd3e655f2b Title: Pólya's Urn Beginning: "Life has a way of magnifying chance effects. Bad luck in romance can make you embittered and wary, making you less attractive to potential partners. If the post office delays delivery of a big check you were counting on, so that a check you wrote ends up bouncing, you'll instantly get charged overdraft..." I'd especially welcome references and links to accessible articles, applets, and apps that are relevant to the Polya urn process, or to what I call the "non-Polya urn process" (which is just a nonce-name for generating a binomial random variable from a Bernoulli process), or to quasi-random simulation of random processes via chip-firing. The best way to reply is via email (directly to me or, if you think your remarks will be of general interest, to math-fun). If you do use the feedback link at Wordpress, keep in mind that I have no way of knowing who you are unless you sign your feedback. I prefer non-anonymous comments, so that if I have follow-up questions I know whom to ask. (Did Martin Gardner ever write about the Polya urn model? If so, I'd like to include a reference. But I just searched the CD containing all of Gardner's columns and couldn't find anything.) Jim Propp

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[math-fun] time
by Warren D Smith 12 Oct '15

12 Oct '15

Krzysztof Lorek, Jorma Louko, Andrzej Dragan: Ideal clocks—a convenient fiction. Classical and Quantum Gravity 32,17 (2015) 175003 sounds interesting. According to Einstein, time passes at rates that depend on the observer, but I (at rest) could compute your rate by knowing your velocity. However -- might the rate of passage of time depend not only on velocity, but also on acceleration? Well, the "Unruh effect" says an accelerating observer will perceive more thermal noise. And this noise will cause your velocity to fluctuate as thermal background photons hit you and push you around. That in turn will alter your perception of time. So, yes. And actually, just sitting on Earth I am accelerating to counteract gravity all the time (in the general relativistic view) and am indeed experiencing a slightly different rate of time versus some observer far from Earth -- although for a different reason than I just said, this is because of the non-flatness of spacetime. The Unruhish effect just mentioned goes beyond merely general relativity. It should make quantum gravity nastier.

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Re: [math-fun] infinite comic strip proof of knot non-cancellation?
by Warren D Smith 09 Oct '15

09 Oct '15

I'm worried about the validity of that proof. The issue I have is that perhaps this would "prove" that unknot+unknot does not equal unknot? See, if we had a finite sum of unknots, then they ought to be untangleable in a finite number of steps to make it obvious that the result is the unknot. But it perhaps is not obvious that an infinite sum can thus be untangled.

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