August 2017 - mathfun@mailman.xmission.com

Re: [math-fun] Is there a name for permutations that preserve no consecutive pairs?
by Dan Asimov 18 Aug '17

18 Aug '17

Don't know a name for this, other than "well-mixed anagrams". megachiropteran cinematographer —Dan Bill Gosper wrote: ----- I find them surprisingly tricky to eyeball correctly. E.g., unshuf[a_String, b_String] := StringCases[a ~~ "¢" ~~ b, ___ ~~ x_ ~~ y_ ~~ ___ ~~ "¢" ~~ ___ ~~ x_ ~~ y_ ~~ ___ :> x ~~ y] TableForm[{#, unshuf @@ #} & /@ {{"aniseed oils", "loin disease"}, {"disanimate", "mediastina"}, {"pericleans", "linespacer"}, {"scalyanteaters", "transacetylase"}, {"laborintensive", "inventoriables"}, {"pleonastic", "pointlaces"}, {"celeritous", "reticulose"}, {"linesofcredit", "selfdirection"}, {"nitrosamines", "nestorianism"}, {"crucible steel", "electric blues"}, {"blueshirt", "thuribles"}, {"cornmeals", "macrolens"}, {"cathartine", "anthracite"}, {"treelawns", "waterlens"}, {"dipstream", "predatism"}}, TableSpacing -> {1, 49}] aniseed oils loin disease oi disanimate mediastina di pericleans linespacer er scalyanteaters transacetylase an laborintensive inventoriables ve pleonastic pointlaces celeritous reticulose linesofcredit selfdirection di nitrosamines nestorianism es crucible steel electric blues el blueshirt thuribles es cornmeals macrolens cathartine anthracite th treelawns waterlens ns dipstream predatism re redtapism spermatid ammonolytic commonality on partleading plantigrade ad astonished dositheans he alicesfern craneflies es discrepate pederastic forcertain refraction camelspider premedicals ls bestarred redbreast ed apostlebird pilotsbread -----

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[math-fun] wrongish looking elementary identity
by Bill Gosper 18 Aug '17

18 Aug '17

ArcTan[(Tan[π/√2] - Tanh[π/√2])/( Tan[π/√2] + Tanh[π/√2])] == Arg[Sinh[π √I]] - π/4 FullSimplify fails. FullSimplify[Tan/@%] fails. But then FullSimplify@TrigExpand@% says True. Imagine if FullSimplify[ ArcTan[(Tan[π/√2] - Tanh[π/√2])/(Tan[π/√2] + Tanh[π/√2])] + π/4] gave Arg[Sinh[π √I]] because it has a (much) lower ByteCount! Who could possibly object?-) --rwg

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[math-fun] Odd-sided equilateral integer polygons
by Bill Gosper 16 Aug '17

16 Aug '17

For a little animation of Conway's(?) extreme proof, gosper.org/homeplate.html --rwg -------- Original Message -------- Date: 2017-08-14 12:37 From: Thane Plambeck <tplambeck(a)gmail.com> To: math-fun <math-fun(a)mailman.xmission.com> Cc: "Schroeppel, Richard" <rschroe(a)sandia.gov> Reply-To: math-fun <math-fun(a)mailman.xmission.com> suppose P is a convex polygon with vertices at the usual integer (2d) lattice points (Z2) Is P necessarily similar to a polygon T obtained by (1) deleting finitely many points from Z2, and then (2) computing the Voronoi cells of the remaining points? if the answer is yes, then up to similarity, every convex polygon with Z2 vertices has (up to similarity) a Voronoi deletion birthday. for example, the square has birthday zero. a "picket" shaped polygon of area 1.25 is the unique polygon at birthday one. three more are born at day two, including a pentagon that is obviously the "correct" resolution of the Major League Baseball impossible home plate problem, described here http://mathworld.wolfram.com/HomePlate.html another example: what is the birthday of the 45-45-90 triangle? i worked with a high school student, Michael Nizsenson, on related problems recently. [megachop]

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[math-fun] Is there a name for permutations that preserve no consecutive pairs?
by Bill Gosper 16 Aug '17

16 Aug '17

I find them surprisingly tricky to eyeball correctly. E.g., unshuf[a_String, b_String] := StringCases[a ~~ "¢" ~~ b, ___ ~~ x_ ~~ y_ ~~ ___ ~~ "¢" ~~ ___ ~~ x_ ~~ y_ ~~ ___ :> x ~~ y] TableForm[{#, unshuf @@ #} & /@ {{"aniseed oils", "loin disease"}, {"disanimate", "mediastina"}, {"pericleans", "linespacer"}, {"scalyanteaters", "transacetylase"}, {"laborintensive", "inventoriables"}, {"pleonastic", "pointlaces"}, {"celeritous", "reticulose"}, {"linesofcredit", "selfdirection"}, {"nitrosamines", "nestorianism"}, {"crucible steel", "electric blues"}, {"blueshirt", "thuribles"}, {"cornmeals", "macrolens"}, {"cathartine", "anthracite"}, {"treelawns", "waterlens"}, {"dipstream", "predatism"}}, TableSpacing -> {1, 49}] aniseed oils loin disease oi disanimate mediastina di pericleans linespacer er scalyanteaters transacetylase an laborintensive inventoriables ve pleonastic pointlaces celeritous reticulose linesofcredit selfdirection di nitrosamines nestorianism es crucible steel electric blues el blueshirt thuribles es cornmeals macrolens cathartine anthracite th treelawns waterlens ns dipstream predatism re redtapism spermatid ammonolytic commonality on partleading plantigrade ad astonished dositheans he alicesfern craneflies es discrepate pederastic forcertain refraction camelspider premedicals ls bestarred redbreast ed apostlebird pilotsbread --rwg

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[math-fun] Quantum Resistant Ledger
by Ray Tayek 16 Aug '17

16 Aug '17

https://github.com/theQRL/QRL/blob/master/QRL_whitepaper.pdf anyone know if this is rational or are these guys blowing smoke? thanks -- Honesty is a very expensive gift. So, don't expect it from cheap people - Warren Buffett http://tayek.com/

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[math-fun] space is big,
by Bill Gosper 16 Aug '17

16 Aug '17

space is dark. you'll always find a place to park. --Myanmar Shave Two Conway Life puffers fly out the x and y axes, firing spaceships back toward to origin. During a brief inflationary period, seven (invisibly) small fixed objects form south of the x-axis. In the ensuing 41G generations, exactly one more fixed object (π heptomino ash cloud) arises there, the white pixel in the green (8M x 8M) box, gosper.org/isolation.png, 314M east and 27M south. Scale 2^19 means each screen pixel = half a million x half a million Life cells. --rwg

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[math-fun] Diophantine puzzle
by Dan Asimov 15 Aug '17

15 Aug '17

Puzzle: Solve this equation in positive integers U, V, W: U/(V + W) + V/(W + U) + W/(U + V) = 4 (But don't spend *too* much time on it; it's hard.) When you give up, the answer can be found here: http://tinyurl.com/y9neq7p7. —Dan

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[math-fun] Draft of my August 2017 blog post
by James Propp 15 Aug '17

15 Aug '17

I started writing a new draft titled "Prof. Engel’s Marvelously Improbable Machines" and would love to get your feedback. I plan on publishing it on Thursday; I apologize for the shortness of notice. (Though one the nice thing about publishing in WordPress is that there's no real deadline, so comments received after August 17 will still be acted upon and acknowledged.) I also apologize for the crude illustrations. Over the next couple of days I hope to replace the photos of hand-written sketches by something less sloppy. Please leave your feedback here: https://mathenchant.wordpress.com?p=1802&shareadraft=5992885a0dab8 (keeping in mind that if you don't sign your comment I won't know who you, with the consequence that I won't be able to ask follow-up questions that your comments may elicit). Title: Prof. Engel’s Marvelously Improbable Machines Beginning: When a simple question has a simple answer, there's something unsatisfying about getting from the question to the answer via a route that's more complicated than either of them. You can just shrug off your dissatisfaction, or you can think harder and try to find a different route. This is a story of... Read more: https://mathenchant.wordpress.com?p=1802&shareadraft=5992885 a0dab8 Thanks, jamespropp

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[math-fun] P vs NP result?
by Michael Kleber 14 Aug '17

14 Aug '17

Whoa. Anyone have opinions on this? https://arxiv.org/pdf/1708.03486.pdf "A Solution of the P versus NP Problem" Norbert Blum Institut f¨ur Informatik, Universit¨at Bonn Friedrich-Ebert-Allee 144, D-53113 Bonn, Germany email: blum(a)cs.uni-bonn.de August 14, 2017 *Abstract* Berg and Ulfberg [4] and Amano and Maruoka [2] have used CNFDNF-approximators to prove exponential lower bounds for the monotone network complexity of the clique function and of Andreev’s function. We show that these approximators can be used to prove the same lower bound for their non-monotone network complexity. This implies P != NP. --Michael -- Forewarned is worth an octopus in the bush.

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Re: [math-fun] Odd-sided equilateral integer polygons
by Fred Lunnon 14 Aug '17

14 Aug '17

Fred Lunnon <fred.lunnon(a)gmail.com> wrote: > Adam P. Goucher <apgoucher(a)gmx.com> wrote: >> Now, if L is even, then for any pair of adjacent vertices v, w in >> Z[i], we have that (v - w) is divisible by 1 + i. > C'mon, you young hotshots --- spare a thought for the feeble-minded > pensioners among us! Why is that? Adam beat me to finding a proof that there are no equilateral odd-N-gons in the square lattice. (At least I somehow realized that there weren't any, though I didn't have a proof.) I'll try to redeem myself by explaining. My apologies if I over-explain, since I'm not sure quite what your question is. Gaussian integers are complex numbers in the form A+Bi where A and B are integers. Gaussian primes are gaussian integers that are not divisible by any gaussian integers whose absolute magnitude isn't 1 or their own. The "first" gaussian prime is i+1. It's a lot like 2 is to regular primes, in that half of all gaussian integers are divisible by it (roughly speaking). These "even" gaussian integers form a checkerboard pattern. The "odd" gaussian integers, i.e. those not divisible by 1+i, form the complementary checkerboard pattern. (2 itself is divisible by 1+i, hence is not a gaussian prime.) Any line which contains more than one gaussian integer will contain infinitely many, and they will either all be "odd," all be "even," or they will strictly alternate. Any line segment that begins and ends on gaussian integers will have either an "even" length, meaning that the end points are of the same parity, or an "odd" length, meaning that the end points are of the opposite parity. Any other line segment of the same length will share the same parity. If a set of line segments (e.g. a polygon) are all of "even" length, and terminate on "even" gaussian integers, you can divide everything by 1+i, and the whole figure will rotate by 45 degrees and shrink by a factor of sqrt(2), but wherever there were gaussian integers on the lines, there still will be, but the parities will in general be different. Of course if the line segments were originally all the same length, they'll still all have the same (smaller) length, hence will have the same parity as each other, though not necessarily the same parity as before the shrinkage. Adam's proof could be recast entirely in terms of a checkerboard, never mentioning complex numbers or gaussian integers or primes.

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