January 2018 - mathfun@mailman.xmission.com

[math-fun] PS Re: Further dissection thoughts
by Dan Asimov 13 Jan '18

13 Jan '18

Correction to -----[WRONG STATEMENT:] WLOG up to rotation, flips, and uniform scaling we can consider just the parallelograms generated by the complex numbers 1 and tau where Im(tau) > 0. ----- I should not have included flips; a correct statement is: ------ WLOG, up to rotation and uniform scaling we can consider just the parallelograms generated by the complex numbers 1 and tau where Im(tau) > 0. ----- (This way, amphicheiral mirror-image parallelograms correspond to distinct tau in the upper half plane.) -----Original Message----- >From: Dan Asimov <dasimov(a)earthlink.net> >Sent: Jan 13, 2018 8:48 PM >To: math-fun <math-fun(a)mailman.xmission.com> >Subject: Further dissection thoughts > >Further thoughts: The total measure of the cuts needed for a dissection >might be an interesting quantity to optimize in typical dissection problems. > >Maybe the simplest plane dissection problem is > > an arbitrary rectangle of unit area —> unit square > >or more generally, > > an arbitrary parallelogram of unit area —> unit square. > >WLOG up to rotation, flips, and uniform scaling we can consider just the >parallelograms generated by the complex numbers 1 and tau where Im(tau) > 0. > >E.g., for dissecting a parallelogram of unit area into the unit square, what is >the infimum of the total lengths of the sets of cuts that suffice? > >(Simple case: an n^2 x 1/n^2 rectangle. Answer appears to be 2n(n-1).) > >Q2: Is the infimum of the total cut length necessarily realized by an actual >dissection? > >Q3: A slightly more interesting problem might be to answer the same questions >for the two tori C/L and C/L' where > > L = Z + iZ > >(the "square" torus), and > > L' = Z + tau*Z > >(a parallelogramic one) where tau is a complex number in the upper half plane (Im(tau) > 0). > >are two lattices that are subgroups of the complex numbers C (as additive groups). > >Example: C/L' = an n^2 x 1/n^2 rectangular torus. >Answer appears to be 2n^2.))) > >Example: C/L' where tau = -1/2 + i*sqrt(3)/2 is the hexagonal torus, the result of >identifying opposite sides of a regular hexagon. (But we need to first scale it to have >area = 1, which I think means we need to use tau' = (4/3)^(1/4) * tau.) The hexagonal torus >is the only shape of torus besides the square one that has "maximal symmetry": Any small >enough perturbation of the torus results in one with at most equal symmetry. > >I don't know the answer for the hexagonal torus, but it is sure to be interesting. > >Final questions: > >FQ1: What is the *supremum* over all shapes of torus, of all the infima >of the sets of total lengths of sets of cuts that suffice to dissect? > >This might turn out to be infinite. > >FQ2: If it's finite, is it realized by an actual parallelogramic torus? > >—Dan > >----- >I thought there might be some other interesting things to optimize here. > >Given a sphere, what's the largest exocube* it can be dissected into >using *any* number of pieces? We include infinitely many, so we should say: >What is the supremum of the exocubes that can be so made? > >[Aside: I was momentarily startled to see my friend's computer's autocorrect >had changed exocubes to excuses, altering the sense a just a bit.] > >Alas, I think the answer is that an arbitrarily large exocube could be built. > >(((Hmm, what if you went the other way, from a solid cube to an exosphere? >Again it looks as if there is no upper bound to the size of the exosphere.))) > > * * * > >But it might be interesting to minimize (find the infimum of) the *total area* >of *all cuts*. Given a sphere containing volume = V, clearly the exocube must >have side >= V^(1/3), so the TAC = total area of all cuts must be >= 6*V^(2/3). > >But that can't be the infimum, can it? > >—Dan >————————————————————————————————————————————————————————————————————————————— >* Definition: > ----------- >An *exocube* looks like a cube from the outside and consists of the closure of >a cube after their interiors of at most countably many disjoint closed topological >3-balls have removed from it. > > >Allan Wechsler wrote: >----- >It's obviously possible to dissect a sphere into pieces that could be >reassembled into something that looks like a cube from the outside. Inside, >it would be hollow, possibly with a bunch of spare pieces rattling around >inside. > >Feasibility sketch: build thin polyhedral planks from the inner part of the >sphere, with 45-degree bevels where necessary to let them form the cube's >edges and vertices. All the extra material could be chopped up into >manageable chunks and hidden in the interior. If there's too much extra >material, redesign with thinner planks to make the outer cube bigger. > >Intuitively, it feels like we could get away with just a few dozen pieces >or so: maybe 4 to 6 per face, and then an approximately equal number to >pack inside ... but I don't know. Can anybody provide an explicit >construction, or make lower-bound arguments about the number of pieces? >----- >-----

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[math-fun] Further dissection thoughts
by Dan Asimov 13 Jan '18

13 Jan '18

Further thoughts: The total measure of the cuts needed for a dissection might be an interesting quantity to optimize in typical dissection problems. Maybe the simplest plane dissection problem is an arbitrary rectangle of unit area —> unit square or more generally, an arbitrary parallelogram of unit area —> unit square. WLOG up to rotation, flips, and uniform scaling we can consider just the parallelograms generated by the complex numbers 1 and tau where Im(tau) > 0. E.g., for dissecting a parallelogram of unit area into the unit square, what is the infimum of the total lengths of the sets of cuts that suffice? (Simple case: an n^2 x 1/n^2 rectangle. Answer appears to be 2n(n-1).) Q2: Is the infimum of the total cut length necessarily realized by an actual dissection? Q3: A slightly more interesting problem might be to answer the same questions for the two tori C/L and C/L' where L = Z + iZ (the "square" torus), and L' = Z + tau*Z (a parallelogramic one) where tau is a complex number in the upper half plane (Im(tau) > 0). are two lattices that are subgroups of the complex numbers C (as additive groups). Example: C/L' = an n^2 x 1/n^2 rectangular torus. Answer appears to be 2n^2.))) Example: C/L' where tau = -1/2 + i*sqrt(3)/2 is the hexagonal torus, the result of identifying opposite sides of a regular hexagon. (But we need to first scale it to have area = 1, which I think means we need to use tau' = (4/3)^(1/4) * tau.) The hexagonal torus is the only shape of torus besides the square one that has "maximal symmetry": Any small enough perturbation of the torus results in one with at most equal symmetry. I don't know the answer for the hexagonal torus, but it is sure to be interesting. Final questions: FQ1: What is the *supremum* over all shapes of torus, of all the infima of the sets of total lengths of sets of cuts that suffice to dissect? This might turn out to be infinite. FQ2: If it's finite, is it realized by an actual parallelogramic torus? —Dan ----- I thought there might be some other interesting things to optimize here. Given a sphere, what's the largest exocube* it can be dissected into using *any* number of pieces? We include infinitely many, so we should say: What is the supremum of the exocubes that can be so made? [Aside: I was momentarily startled to see my friend's computer's autocorrect had changed exocubes to excuses, altering the sense a just a bit.] Alas, I think the answer is that an arbitrarily large exocube could be built. (((Hmm, what if you went the other way, from a solid cube to an exosphere? Again it looks as if there is no upper bound to the size of the exosphere.))) * * * But it might be interesting to minimize (find the infimum of) the *total area* of *all cuts*. Given a sphere containing volume = V, clearly the exocube must have side >= V^(1/3), so the TAC = total area of all cuts must be >= 6*V^(2/3). But that can't be the infimum, can it? —Dan ————————————————————————————————————————————————————————————————————————————— * Definition: ----------- An *exocube* looks like a cube from the outside and consists of the closure of a cube after their interiors of at most countably many disjoint closed topological 3-balls have removed from it. Allan Wechsler wrote: ----- It's obviously possible to dissect a sphere into pieces that could be reassembled into something that looks like a cube from the outside. Inside, it would be hollow, possibly with a bunch of spare pieces rattling around inside. Feasibility sketch: build thin polyhedral planks from the inner part of the sphere, with 45-degree bevels where necessary to let them form the cube's edges and vertices. All the extra material could be chopped up into manageable chunks and hidden in the interior. If there's too much extra material, redesign with thinner planks to make the outer cube bigger. Intuitively, it feels like we could get away with just a few dozen pieces or so: maybe 4 to 6 per face, and then an approximately equal number to pack inside ... but I don't know. Can anybody provide an explicit construction, or make lower-bound arguments about the number of pieces? ----- -----

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Re: [math-fun] Sphere to cube dissection?
by Dan Asimov 12 Jan '18

12 Jan '18

I thought there might be some other interesting things to optimize here. Given a sphere, what's the largest exocube* it can be dissected into using *any* number of pieces? We include infinitely many, so we should say: What is the supremum of the exocubes that can be so made? [Aside: I was momentarily startled to see my friend's computer's autocorrect had changed exocubes to excuses, altering the sense a just a bit.] Alas, I think the answer is that an arbitrarily large exocube could be built. (((Hmm, what if you went the other way, from a solid cube to an exosphere? Again it looks as if there is no upper bound to the size of the exosphere.))) * * * But it might be interesting to minimize (find the infimum of) the *total area* of *all cuts*. Given a sphere containing volume = V, clearly the exocube must have side >= V^(1/3), so the TAC = total area of all cuts must be >= 6*V^(2/3). But that can't be the infimum, can it? —Dan ————————————————————————————————————————————————————————————————————————————— * Definition: ----------- An *exocube* looks like a cube from the outside and consists of the closure of a cube after their interiors of at most countably many disjoint closed topological 3-balls have removed from it. Allan Wechsler wrote: ----- It's obviously possible to dissect a sphere into pieces that could be reassembled into something that looks like a cube from the outside. Inside, it would be hollow, possibly with a bunch of spare pieces rattling around inside. Feasibility sketch: build thin polyhedral planks from the inner part of the sphere, with 45-degree bevels where necessary to let them form the cube's edges and vertices. All the extra material could be chopped up into manageable chunks and hidden in the interior. If there's too much extra material, redesign with thinner planks to make the outer cube bigger. Intuitively, it feels like we could get away with just a few dozen pieces or so: maybe 4 to 6 per face, and then an approximately equal number to pack inside ... but I don't know. Can anybody provide an explicit construction, or make lower-bound arguments about the number of pieces? -----

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[math-fun] Sphere to cube dissection?
by Allan Wechsler 12 Jan '18

12 Jan '18

It's obviously possible to dissect a sphere into pieces that could be reassembled into something that looks like a cube from the outside. Inside, it would be hollow, possibly with a bunch of spare pieces rattling around inside. Feasibility sketch: build thin polyhedral planks from the inner part of the sphere, with 45-degree bevels where necessary to let them form the cube's edges and vertices. All the extra material could be chopped up into manageable chunks and hidden in the interior. If there's too much extra material, redesign with thinner planks to make the outer cube bigger. Intuitively, it feels like we could get away with just a few dozen pieces or so: maybe 4 to 6 per face, and then an approximately equal number to pack inside ... but I don't know. Can anybody provide an explicit construction, or make lower-bound arguments about the number of pieces?

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[math-fun] Draft of my January 2018 blog post
by James Propp 10 Jan '18

10 Jan '18

My new piece is called "On Size, Death, and Dinosaurs" and as usual I plan to publish it on the 17th. (It's a bit light on math, but it leads into next month's piece on dimensional analysis, which will be meatier.) Please leave your feedback here: https://mathenchant.wordpress.com?p=2091&shareadraft=5a5650a17d19c If you have comments, please sign them, so that I can ask followup questions. Thanks, Jim Propp Title: On Size, Death, and Dinosaurs Beginning: Here's how Roald Dahl describes what happened one morning when the Queen of England invited a large and unexpected guest ("The BFG" ) to have breakfast with her: There was a frantic scurry among the Palace servants when orders were received from the Queen that a twenty-four-foot giant must be seated... Read more: https://mathenchant.wordpress.com?p=2091&shareadraft= 5a5650a17d19c Thanks, jamespropp

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Re: [math-fun] Sampled twindragon tiles Moorish floor.
by Bill Gosper 07 Jan '18

07 Jan '18

On 2018-01-06 21:49, Tom Karzes wrote: > Nice Bill! > > This one really fooled me for a minute: When I viewed it in my > browser, I saw the image controls at the top, and I thought, "Oh, my > last browser update must have added this to my image viewer". But > when I tried to click on them, they didn't work. Finally I realized > the "controls" were actually just a part of the image - duh! > > Tom > > Bill Gosper writes: > > gosper.org/floordragons.png > > --rwg > Tom, here's what you missed: gosper.org/drankle2304.gif --rwg

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[math-fun] Sampled twindragon tiles Moorish floor.
by Bill Gosper 06 Jan '18

06 Jan '18

gosper.org/floordragons.png --rwg

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[math-fun] I think, I'm Siamese
by françois mendzina essomba2 06 Jan '18

06 Jan '18

hello... tanh(x) =-csch(2*x)+1/(-csch(4*x)+1/(-csch(8*x)+1/(-csch(16*x)+1/(-csch(32*x)+1/(-csch(64*x)+1/(-csch(128*x)+1/(…))))))); approximation. tanh(x)=-csch(2*x)+1/(-csch(4*x)+1/(-csch(8*x)+1/(-csch(16*x)+1/(-csch(32*x)+1/(-csch(64*x)+1/(-csch(128*x)+1/(1))))))); FME...

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[math-fun] Polyhedral graphs from consecutive power distances
by Ed Pegg Jr 04 Jan '18

04 Jan '18

Puzzle: use each of the edgelengths 1., 1.10487, 1.22074, 1.34877, 1.49022, 1.6465, 1.81917, 2.00995, 2.22074, 2.45364 a single time to make a polyhedral graph. These values are powers 0 to 9 of the second real root of x^8-x^2-1. So far, I've found 4 solutions for polyhedral graphs made from consecutive power distances. I posted pictures at https://math.stackexchange.com/questions/2592534/polyhedral-graphs-from-con… You won't believe the fourth polynomial! --Ed Pegg Jr

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[math-fun] new Mersenne prime, 2^77,232,917 -- 1
by Ray Tayek 04 Jan '18

04 Jan '18

https://science.slashdot.org/story/18/01/04/0531203/largest-prime-number-di… osted by msmash on Thursday January 04, 2018 @04:11PM from the breakthrough dept. chalsall writes: Persistence pays off. Jonathan Pace, a GIMPS volunteer for over 14 years, discovered the 50th known Mersenne prime, 2^77,232,917 -- 1 on December 26, 2017. The prime number is calculated by multiplying together 77,232,917 twos, and then subtracting one. It weighs in at 23,249,425 digits, becoming the largest prime number known to mankind. It bests the previous record prime, also discovered by GIMPS, by 910,807 digits. You can read a little more in the press release. -- Honesty is a very expensive gift. So, don't expect it from cheap people - Warren Buffett http://tayek.com/

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