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Re: [math-fun] Returns to zero when subtracting and adding squares
by Keith F. Lynch 09 Sep '19

09 Sep '19

Allan Wechsler <acwacw(a)gmail.com> wrote: > Michael Collins <mjcollins10(a)gmail.com> wrote: >> I suspect that, more generally, a sum a_1 + ... + a_n with each a_k >> being either k^2 or -k^2 can never be zero? > Ah, no, I thought of that, and there are examples! > 1 + 4 - 9 + 16 - 25 - 36 + 49 = 0 > Here is another: 1 - 4 - 9 + 16 - 25 + 36 + 49 - 64 = 0 Apparently these have to end with the square of a number that's congruent to 0 or 3 mod 4. Can anyone find a counterexample? The number of solutions, starting with 0, with the 1 positive, are: 1,0,0,0,0,0,0,1,1,0,0,1,5,0,0,43,57,0,0,239,430,0,0,2904,5419 I see that it's already in OEIS. A083527. Then I tried it with cubes. Here's the first solution: > 1 + 8 - 27 + 64 - 125 - 216 - 343 + 512 + 729 - 1000 - 1331 + 1728 Again, apparently these have to end with the cube of a number that's congruent to 0 or 3 mod 4. The numbers of solutions, starting with 12, with the 1 positive, are: 1,0,0,2,1,0,0,2,62,0,0,268,356. A113263. I also tried 4th powers. And 5th. And 1st. A058377. Again, congruent to 0 or 3 mod 4. Powers of 2? No solutions, unsurprisingly. Fibonacci series? It's past my bedtime so I'll check tomorrow.

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[math-fun] Returns to zero when subtracting and adding squares
by Allan Wechsler 08 Sep '19

08 Sep '19

A recent discussion on "Sequence Fanatics" has me thinking about a general process. Let An be any non-negative integer sequence. Derive Bn with the rule: B0 = A0 and for n > 0 Bn = B(n-1) - An if this is non-negative Bn = B(n-1) + An otherwise. If An is the non-negative integers (A001477), then Bn is A008344. If An is the perfect squares (A000290), then Bn is A076042. Question: Does A076042 ever revisit 0? If you play with it, you will see that the sequence is divided into eras where adding and subtracting strictly alternate. These eras are punctuated by an extra add when subtracting would go negative. The sequence reaches a local minimum just at the end of each era. If we define these minima as the last value in the sequence before a double add, they are: 1, 5, 5, 7, 4, 19, 104, 74, 193, 515, 725, 241 ... This sequence is not in OEIS, nor are any of its first few truncations. The indices of the minima are not present either. Intuitively, this sequence is aiming at 0 from greater and greater distances, and is unlikely to ever hit the bullseye. But I think the problem is amenable to more rigorous analysis. Can any funsters contribute additional insight?

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[math-fun] Who has written a clear summary of what research mathematicians do?
by Scott Kim 08 Sep '19

08 Sep '19

Does anyone know of a clear explanation written for nonmathematicians of the fundamental types of things research mathematicians do, and why? I'm planning to write about how the process of inventing puzzles is similar to the process fo doing mathematical research, and how we can teach kids about mathematical research by having them invent puzzles. I'd like to reference other authors, and wonder what has been written that isn't just mathematicians talking to mathematicians. The authors I know who have attempted this are Keith Devlin in Introduction to Mathematical Thinking, written to tell math grad students how "real" mathematics differs from how math is conventionally taught in K-12, and Eugenia Cheng in many of her books, starting with How to Bake Pi. These are good, but there's room for more voices.

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[math-fun] percolation is odd
by Cris Moore 05 Sep '19

05 Sep '19

My friend Stephan Mertens and I just proved something you might enjoy. Consider filling a subset of the cells of an n-by-m rectangular lattice. Call the resulting configuration “spanning" if there is a path of occupied cells from the top row to the bottom row (stepping orthogonally). Let N_even (N_odd) be the number of spanning configurations with an even (resp. odd) number of occupied cells. Show that N_even-N_odd = +1 or -1, and determine the sign as a function of n and m. As a corollary, the total number of spanning configurations is odd. For instance, for n=m=2 there are 7 spanning configurations, with N_even=3 and N_odd=4. Cris Cris Moore moore(a)santafe.edu

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Re: [math-fun] Draft of short essay on Scott Kim's motley dissections
by Bill Gosper 04 Sep '19

04 Sep '19

On Thu, Jul 11, 2019 at 3:14 AM rwg <rwg(a)ma.sdf.org> wrote: > > > -------- Original Message -------- > Subject: Re: [math-fun] Draft of short essay on Scott Kim's motley > dissections > Date: 2019-07-10 12:59 > From: James Propp <jamespropp(a)gmail.com> > To: math-fun <math-fun(a)mailman.xmission.com> > Reply-To: math-fun <math-fun(a)mailman.xmission.com> > > Beautiful! > > Thanks, Fred! > > Jim > I'll second that! How hard would it be to twist the whole scene so that the front face doesn't rotate? We can change the problem so that the answer is an ancient Macsyma animation most of you should recall. Problem: Reshape the six "squares" so they can simply rotate, while four "triangles" alternately swell, then shrink away, while the other four complementarily follow suit in the opposite phase. The edges need no longer lie on great circles. Spoiler Lissajous arcs <http://gosper.org/lisspump.gif> On Wed, Jul 10, 2019 at 3:41 PM Fred Lunnon <fred.lunnon(a)gmail.com> > wrote: > > > I have posted an animation at > > https://www.dropbox.com/s/5yrmef1djjuatfc/pseudo_movie.gif > > of a spherical pseudo-cuboctahedron (SPCO) morphing continuously via > > octahedron -- big triangles -- cuboctahedron -- big squares -- cube, > > then back again via the mirror-images; Maple program is available > > on request. > > > > This should run continuously when the link is opened in a browser: > > please advise me of any problems! > > > > Time is proportional to `twist' angle t between (plane containing) > > SPCO edge arc and associated edge of scaffolding cube, modulo pi . > > SPCO vertices are given by 24 cube symmetries (with even resigned > > permutations of components) of point with Cartesian coordinate > > P(t) = (a^2 + a*b, a^2 - a*b, a^2 + b^2)/d , > > where > > b = cos t , a = (1/sqrt2) sin t , d^2 = 1 - a^4 . > > > > For octahedron t = 0 , for cube t = pi/2 ; for cuboctahedron > > t = arccos(-1/3)/2 . For Jim Propp's case, where edge plane meets > > 4 vertices of the scaffold cube, rather unexpectedly t = > > arccos(+1/3)/2 > > --- just pi/2 minus the cuboctahedron angle! > > > > While I was busily upstaging his previous version, Christian has > > quietly replaced that static CGI view with a 3D-printed solid model: > > see > > https://www.thingiverse.com/thing:3726912 > > > > Hmmm ... a mechanically functional solid morphing model, anybody? > The only hope would be van Deventer. —rwg > > > > Finally, note that a similar method should cope straightforwardly > > with morphing pseudo-octahedron and pseudo-icosidodecahedron. > > > > Fred Lunnon > > > > > > On 7/6/19, James Propp <jamespropp(a)gmail.com> wrote: > > > I've figured out one way to make a 3D realization of a > pseudo-octahedron, > > > by decorating the faces of a tetrahedron. (Maybe this was obvious to > you > > > all, but I didn't see it.) See > > > > > > http://faculty.uml.edu/jpropp/mathenchant/semi-pseudo-octahedron.JPG > > > > > > which shows how two of the four faces should be decorated, with three > > > vertices per face. If you breathe into the tetrahedron to inflate it to > > > become a sphere, the pattern of decorations on the tetrahedron should > > > become a motley dissection of the sphere with 12 vertices, 6 > > pseudo-edges, > > > and 8 three-sided pseudo-faces. > > > > > > Can Christian (or someone else) create an image of this for me? I'd > love > > to > > > use it in my next mini-essay (~ 1000 words) for the Big Internet > > Math-Off, > > > as well as in the longer Mathematical Enchantments piece on the subject > > of > > > motley dissections that I plan to write later this summer or in the > Fall. > > > > > > Thanks, > > > > > > Jim > > > > > > > > > > > > On Thu, Jul 4, 2019 at 10:39 AM ed pegg <ed(a)mathpuzzle.com> wrote: > > > > > >> Related: > > >> Moritz W. Schmitt, On Space Groups and Dirichlet-Voronoi Stereohedra > > >> https://refubium.fu-berlin.de/handle/fub188/10176 > > >> > > >> I want to see pictures of all of these space filling plesiohedra. > I've > > >> been wanting to see Engel's 38-sided spacefiller for years. > > >> --Ed Pegg Jr On Thursday, July 4, 2019, 07:58:05 AM CDT, Christian > > >> Lawson-Perfect <christianperfect(a)gmail.com> wrote: > > >> > > >> Scott, I'm interested in giving it a go. I've got a couple of student > > >> working with me over the summer on 3d printing mathematical objects, > so > > >> this'd be great. > > >> > > >> On Thu, 4 Jul 2019 at 06:46, Scott Kim <scott(a)scottkim.com> wrote: > > >> > > >> > The pseudo octahedron as drawn has bar 3 symmetry, meaning 4-fold > > >> > rotational symmetry plus top and lower hemispheres congruent but > > mirror > > >> > images (I think...not sure). The pseudo-cube needs to be redrawn to > > work > > >> > nicely on a sphere, as does the pseudo-dodecahedron. The 3d models I > > >> really > > >> > want to have 3d printed are the pseudo 5-cell (easy) and pseudo > > 16-cell > > >> > (hard), which have exteriors that are concave curivlinear > tetrahedra. > > I > > >> can > > >> > provide drawings if anyone's interested in giving them a try. — > Scott > > >> > > > >> > On Wed, Jul 3, 2019 at 9:07 PM James Propp <jamespropp(a)gmail.com> > > wrote: > > >> > > > >> > > And it would be good to have a picture of the pseudo-cube as an > > actual > > >> > > motley dissection of the sphere, but I have no idea how to realize > > it. > > >> > I've > > >> > > put Scott's sketch of the combinatorial structure of the > pseudo-cube > > >> > > at > > >> > > http://mathenchant.org/pseudo-cube.png; in both this picture and > > the > > >> > > pseudo-octahedron picture, it's important to keep in mind that the > > >> > > noncompact region in the picture is also a pseudo-face. In the > > >> > symmetrical > > >> > > pseudo-cube, the six pseudo-faces would be centered at the > vertices > > of > > >> an > > >> > > octahedron, but I don't see how the pseudo-faces fit together. Can > > any > > >> of > > >> > > you figure this out? > > >> > > > > >> > > Jim > > >> > > > > >> > > On Wed, Jul 3, 2019 at 9:28 PM James Propp <jamespropp(a)gmail.com> > > >> wrote: > > >> > > > > >> > > > It would also be good to have a picture of the > pseudo-octahedron. > > >> I've > > >> > > put > > >> > > > Scott Kim's picture of it at > > >> > > http://mathenchant.org/pseudo-octahedron.png; > > >> > > > there should be a realization of it as a motley dissection of > the > > >> > surface > > >> > > > of the sphere that has all the rotational symmetries of the > > >> octahedron > > >> > > (but > > >> > > > none of the reflection symmetries). Can Christian (or anyone > else) > > >> > > > construct a virtual model? In some ways it might be harder to > > >> visually > > >> > > > parse the pseudo-octahedral motley dissection than the > > >> > > pseudo-cuboctahedral > > >> > > > motley dissection, even though the former is simpler, because > the > > >> > visible > > >> > > > hemisphere offers less information. But an interactive rotatable > > >> > version > > >> > > > (or just a GIF that shows the object rotating) would solve that > > >> > problem. > > >> > > > > > >> > > > Jim > > >> > > > > > >> > > > On Wed, Jul 3, 2019 at 2:49 PM James Propp < > jamespropp(a)gmail.com> > > >> > wrote: > > >> > > > > > >> > > >> Wow; this is great! > > >> > > >> > > >> > > >> Is there a way to rotate the view? > > >> > > >> > > >> > > >> Jim > > >> > > >> > > >> > > >> On Wed, Jul 3, 2019 at 3:38 AM Christian Lawson-Perfect < > > >> > > >> christianperfect(a)gmail.com> wrote: > > >> > > >> > > >> > > >>> I've made a 3d model in OpenSCAD and uploaded it to > thingiverse: > > >> > > >>> https://www.thingiverse.com/thing:3726912. I've put a simple > > >> > rendering > > >> > > >>> of > > >> > > >>> the shape on that page; I'm going to both 3d print it and do a > > >> nicer > > >> > > CGI > > >> > > >>> rendering now. > > >> > > >>> > > >> > > >>> On Tue, 2 Jul 2019 at 19:24, James Propp < > jamespropp(a)gmail.com> > > >> > wrote: > > >> > > >>> > > >> > > >>> > The link to the video apparently didn't survive the > conversion > > >> from > > >> > > >>> HTML to > > >> > > >>> > PDF, so here's the URL for Scott Kim's video: > > >> > > >>> > > > >> > > >>> > https://youtu.be/xK1QA0Oi7iE > > >> > > >>> > > > >> > > >>> > Jim > > >> > > >>> > > > >> > > >>> > > > >> > > >>> > > > >> > > >>> > > > >> > > >>> > On Tue, Jul 2, 2019 at 12:03 PM James Propp < > > >> jamespropp(a)gmail.com> > > >> > > >>> wrote: > > >> > > >>> > > > >> > > >>> > > At > > http://faculty.uml.edu/jpropp/mathenchant/motley-draft3.pdf > > >> > > >>> you'll > > >> > > >>> > > find a draft of an essay that will be published on July > > 11th. > > >> > > >>> Comments > > >> > > >>> > are > > >> > > >>> > > welcome, but even more welcome would be a good picture of > > the > > >> > > >>> "spherical > > >> > > >>> > > pseudo-cuboctahedron"! In principle I know how to make > such > > a > > >> > > >>> picture, > > >> > > >>> > sort > > >> > > >>> > > of: take the motley dissection of the surface of the cube > > >> (shown > > >> > in > > >> > > >>> the > > >> > > >>> > > essay), push all the vertices out onto a sphere, and draw > > >> > spherical > > >> > > >>> arcs > > >> > > >>> > > joining up those vertices. This seems like a job for > > >> Mathematica, > > >> > > >>> but my > > >> > > >>> > > mastery of 3D graphics isn't up to the job. I'll > gratefully > > >> > > >>> acknowledge > > >> > > >>> > > anyone who can create a picture I can use! > > >> > > >>> > > > > >> > > >>> > > Thanks, > > >> > > >>> > > > > >> > > >>> > > Jim Propp > > >> > > >>> > > > > >> > > >>> > _______________________________________________ >

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Re: [math-fun] Returns to zero when subtracting and adding squares
by Bill Gosper 02 Sep '19

02 Sep '19

(This Subject reminds me of Hofstadter's yields FALSE when preceded by its quotation.) Just an aside: Mathematica makes it ridiculously easy to "bisect" formal sums of numbers: haf[L_List] := Block[{t = Total@L/2}, If[True, Select[Subsets@L, Total@# == t &], {}]] (To speed up our integer case, replace True with IntegerQ@t .) E.g., instead of Allan's seven squares, In[439]:=√haf[Range@7^2] Out[439]= {{3, 5, 6}, {1, 2, 4, 7}} (one solution), you need at least 24 5th powers: {{1, 2, 4, 10, 11, 13, 14, 15, 16, 18, 22, 24}, {1, 2, 5, 6, 10, 14, 15, 16, 18, 19, 20, 24}, {1, 2, 5, 7, 10, 13, 14, 15, 19, 20, 21, 23}, {3, 4, 6, 8, 9, 11, 12, 16, 17, 18, 22, 24} {3, 4, 7, 8, 9, 11, 12, 13, 17, 21, 22, 23}, {3, 5, 6, 7, 8, 9, 12, 17, 19, 20, 21, 23}} (three solutions, book-matched) Next come eight solutions with 28 5th powers: {{1, 4, 7, 14, 15, 19, 21, 22, 27, 28}, {1, 4, 11, 12, 15, 17, 19, 24, 27, 28}, {2, 4, 5, 12, 13, 15, 18, 21, 22, 24, 25, 27}, {2, 5, 8, 11, 12, 16, 18, 19, 20, 23, 26, 28}, {3, 5, 6, 12, 13, 14, 18, 20, 21, 22, 26, 28}, {3, 6, 8, 10, 14, 15, 16, 17, 22, 23, 26, 28}, {1, 2, 3, 7, 8, 9, 10, 15, 16, 18, 22, 23, 26, 28}, {1, 5, 7, 8, 12, 14, 17, 18, 19, 20, 22, 24, 25, 26}, ...} (half-shown; obvious DIY). —rwg -------- Original Message -------- Date: 2019-08-29 11:30 From: Tomas Rokicki <rokicki(a)gmail.com> To: math-fun <math-fun(a)mailman.xmission.com> Reply-To: math-fun <math-fun(a)mailman.xmission.com> Since there are 2^n ways to assign signs, and since the maximum sum is O(n^3), I expect the number of ways to assign signs that sum to zero to increase exponentially with n. On Thu, Aug 29, 2019 at 11:26 AM Allan Wechsler <acwacw(a)gmail.com> wrote: > Here is another: 1 - 4 - 9 + 16 - 25 + 36 + 49 - 64 = 0 > > On Thu, Aug 29, 2019 at 2:21 PM Allan Wechsler <acwacw(a)gmail.com> wrote: > > > Ah, no, I thought of that, and there are examples! > > > > 1 + 4 - 9 + 16 - 25 - 36 + 49 = 0 > > > > On Thu, Aug 29, 2019 at 2:17 PM Michael Collins <mjcollins10(a)gmail.com> > > wrote: > > > >> I suspect that, more generally, a sum a_1 + ... + a_n with each a_k > being > >> either k^2 or -k^2 can never be zero? > >> > >> > >> On Thu, Aug 29, 2019 at 12:10 PM Tomas Rokicki <rokicki(a)gmail.com> > wrote: > >> > >> > Sure, I'll do that. It's easy to write a Python program that walks > the > >> > sequence, calculating the lengths of the epochs and doing each in > >> > one computational step. Since the length of the epochs increases > >> > exponentially, the resulting program gives you a polynomial algorithm > >> > to calculate a given term (polynomial in the bit length of the input). > >> > > >> > I will say further that it is rare for the b(n) to be less than n; > this > >> > only > >> > happens at the end of epochs. Indeed, b(n) does not appear to get > >> > below 1e9 between about n=2^32 .. n^5000. > >> > > >> > I'm fairly certain a simple probabilistic argument can be made that > >> > this sequence returns to zero with probably less than 2^5000 or > >> > some such if you make some fairly broad and tenuous assumptions > >> > about the distribution of values seen. > >> > > >> > On Thu, Aug 29, 2019 at 10:44 AM Neil Sloane <njasloane(a)gmail.com> > >> wrote: > >> > > >> > > Tomas, You said > >> > > > >> > > (quote) > >> > > Empirically, [A076042] does not appear to return to zero within > >> > > at least 2^4000 terms. I can easily increase the exponent > >> > > (perhaps up to 40,000). > >> > > (end quote) > >> > > > >> > > Very interesting! Could you explain what you mean? And could you > >> add it > >> > > as a comment to the sequence? > >> > > > >> > > Best regards > >> > > Neil > >> > > > >> > > Neil J. A. Sloane, President, OEIS Foundation. > >> > > 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. > >> > > Also Visiting Scientist, Math. Dept., Rutgers University, > Piscataway, > >> NJ. > >> > > Phone: 732 828 6098; home page: http://NeilSloane.com > >> > > Email: njasloane(a)gmail.com > >> > > > >> > > > >> > > > >> > > On Thu, Aug 29, 2019 at 1:26 PM Tomas Rokicki <rokicki(a)gmail.com> > >> wrote: > >> > > > >> > > > Empirically, this does not appear to return to zero within > >> > > > at least 2^4000 terms. I can easily increase the exponent > >> > > > (perhaps up to 40,000). > >> > > > > >> > > > On Thu, Aug 29, 2019 at 7:29 AM Allan Wechsler <acwacw(a)gmail.com> > >> > wrote: > >> > > > > >> > > > > A recent discussion on "Sequence Fanatics" has me thinking > about a > >> > > > general > >> > > > > process. Let An be any non-negative integer sequence. Derive Bn > >> with > >> > > the > >> > > > > rule: > >> > > > > > >> > > > > B0 = A0 > >> > > > > > >> > > > > and for n > 0 > >> > > > > > >> > > > > Bn = B(n-1) - An if this is non-negative > >> > > > > Bn = B(n-1) + An otherwise. > >> > > > > > >> > > > > If An is the non-negative integers (A001477), then Bn is > A008344. > >> > > > > > >> > > > > If An is the perfect squares (A000290), then Bn is A076042. > >> > > > > > >> > > > > Question: Does A076042 ever revisit 0? If you play with it, you > >> will > >> > > see > >> > > > > that the sequence is divided into eras where adding and > >> subtracting > >> > > > > strictly alternate. These eras are punctuated by an extra add > when > >> > > > > subtracting would go negative. The sequence reaches a local > >> minimum > >> > > just > >> > > > at > >> > > > > the end of each era. If we define these minima as the last value > >> in > >> > the > >> > > > > sequence before a double add, they are: 1, 5, 5, 7, 4, 19, 104, > >> 74, > >> > > 193, > >> > > > > 515, 725, 241 ... This sequence is not in OEIS, nor are any of > its > >> > > first > >> > > > > few truncations. The indices of the minima are not present > either. > >> > > > > > >> > > > > Intuitively, this sequence is aiming at 0 from greater and > greater > >> > > > > distances, and is unlikely to ever hit the bullseye. But I think > >> the > >> > > > > problem is amenable to more rigorous analysis. > >> > > > > > >> > > > > Can any funsters contribute additional insight? > >> > > > > _______________________________________________

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[math-fun] Cognitive Dissonance: The Homology Concept
by Brad Klee 01 Sep '19

01 Sep '19

Subtitle: A Curious Case of Batesian Mimicry. During late August in Arkansas it is often possible to glimpse a black-bodied insect darting through the tree tops. Looking closer, the dorsal side of the hind wings reflects a blue-green hue. You might naturally wonder, what species is that? On the running trail: https://0x0.st/z4Wy.JPG On U of A campus: https://0x0.st/z4Wt.JPG Or At Devil's Den State Park: https://0x0.st/z4Wv.JPG https://0x0.st/z4Ww.JPG It could easily be confused for a swallowtail, either pipevine or spicebush, when in fact the creature is from family Nymphalidae rather than Papilionidae. During a flight, it is impossible to closely observe the fore-wing venation, so humans nor birds can be blamed for falling fool to an effective mimic (at least at first). Even upon a closer look at the ventral side, the poorly-named red-spoted purple admiral ( why not orange-spotted blue? ) appears very similar to the pipevine swallowtail. Both butterflies have seven orange ocelli along the border of the hind wing. The ocelli are homologous under biological definitions, but they certainly are not mathematical homologs. Compare data: https://www.butterfliesandmoths.org/species/Battus-philenor https://www.butterfliesandmoths.org/species/Limenitis-arthemis Roughly speaking, if evolution acts as an anatomical homotopy, then two organisms along separate branches of a family tree should end up biological homologs. With Batesian mimicry, survival depends on homology, because predators need to see similar biological features to confuse a palatable species (here arthemis) with an unpalatable species (here philenor). However, biological homology is a looser concept than mathematical homology, and it is very possible that evolution introduces additional structures that disrupt mathematical homology. Comparing arthemis and philenor, the idea of a common ancestor is easy to accept. As it so happens, evolution has produced mathematically distinct ocelli. With arthemis, the eyespots are usually sole orange disks, each possibly connected to another black disk. Philenor has more complicated geometry involving an orange disk with black and white surroundings. To characterize the homology of two typical patterns for Philenor eyespots, I made the following disk-edge-vertex decomposition graphs: https://0x0.st/z4WA.png A next step would be to collect lots of data and measure area of disks, length of edges, and angles around each vertex. This would associate each graph vertex to a metric, thus would give a decent idea as to what an average philenor eyespot looks like. Of course, the same could be done for Arthemis. It would be an interesting exercise of bio-statistics to compare the orange-disk area metric between species. It looks that Arthemis has smaller spots, but survival of the species implies that the spots are not too small to ruin the mimicry effect. Another interesting question: is there a tolerance limit for pattern agreement, which bounds the ability of one species to effectively mimic another? Cheers --Brad

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Re: [math-fun] Sugar + water paradox:
by Bill Gosper 01 Sep '19

01 Sep '19

On Wed, Mar 20, 2019 at 11:37 AM <Shy respondent> wrote: > [just us] > > I think you may just need to wait longer. > Yes. > If the water isn't stirred (either manually or by boiling) > then the liquid near the sugar pile will be highly saturated > and you'll have to wait for random thermal stirring to transport it away > But for several minutes, at least, the amount of dissolved sugar actually declines. On 2019-03-20 12:17, Allan Wechsler wrote: > The obvious question, for me, is how are you measuring this process? I > imagine that you could extract a cc of solution each minute, then evaporate > them and weigh the amount of sugar in each. But I think you're probably not > doing that, but rather relying on some kind of visual measure. What is the > observed dynamic? Does it dissolve for a while until the concentration hits > a plateau, and then sugar starts to precipitate out? Exactly! There's way more sugar in the bottom after it cools for a while. Huge density gradient from lack of stirring. You should have no trouble observing this. Unless I decontrolled it with a teabag of Stash Double Bergamot. —rwg > > On Wed, Mar 20, 2019 at 1:57 PM Bill Gosper <billgosper(a)gmail.com> wrote: > >> Dump a couple of Tbsp of sugar into a quart or two jar of tepid water. >> Don't stir. >> 𝜇wave a couple minutes, hot but not boiling. Let it sit. Experimental >> claim: >> The total dissolved sugar is not an increasing function of time, even >> though a >> couple of Tbsp is << enough to saturate even cold water. >> —rwg > > > > On Mar 20, 2019, at 10:57 AM, Bill Gosper <billgosper(a)gmail.com> wrote: > > > > Dump a couple of Tbsp of sugar into a quart or two jar of tepid water. > > Don't stir. > > 𝜇wave a couple minutes, hot but not boiling. Let it sit. Experimental > > claim: > > The total dissolved sugar is not an increasing function of time, even > > though a > > couple of Tbsp is << enough to saturate even cold water. > > —rwg > > _______________________________________________ > >

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[math-fun] Online bipartite matching question
by Richard Chatwin 30 Aug '19

30 Aug '19

Hi all, I wonder if anyone can provide some insight into the following problem. The idea is to analyze the performance of the Ranking algorithm for online bipartite matching with random arrivals as applied to a specific graph. Background: - Given a bipartite graph; call the two sides L and R - Assume a perfect matching exists - The nodes from R arrive in an online manner - The order of arrival is a random permutation of the nodes in R selected uniformly from among all such permutations - When node j in R arrives, its edges to nodes in L are revealed - The algorithm then selects one of the neighbor nodes in L that has not previously been matched to match to j - The Ranking algorithm selects a random permutation of the nodes of L uniformly from all such permutations - When node j in R arrives, Ranking matches j to the highest ranking unmatched neighbor to j in L Goal: - Given a bipartite graph, compute the competitive ratio achieved by Ranking on the graph - The competitive ratio is the expected size of the matching achieved by Ranking divided by the size of the perfect matching Specific graph G: - L consists of 2n nodes, labeled 1, …, 2n. Nodes 1, …, n are in L1; nodes n+1, …, 2n are in L2 - R consists of 2n nodes, labeled 1, …, 2n. Nodes 1, …, n are in R1; nodes n+1, …, 2n are in R2 - The adjacency matrix is given by: - A(i, j) = 1 if i = j - A(i, j) = 1 if i <= n and j > n - A(i, j) = 0 otherwise - In summary: - each node in L1 is adjacent to a unique neighbor in R1 - each node in L2 is adjacent to a unique neighbor in R2 - all nodes in L1 are adjacent to all nodes in R2 - Ranking will match all nodes in L1 and all nodes in R2 but whenever it includes edge (i, j) with i in L1 and j in R2, then the neighbor to i in R1 and the neighbor to j in L2 remain unmatched Specific goal: - Show that the competitive ratio of Ranking on G is a decreasing function of n with limit 0.75 Context: - This graph has been proposed in Karande, Mehta and Tripathi, Online Bipartite Matching with Unknown Distributions, to provide an upper bound on the performance of Ranking on online bipartite graphs with random arrivals. They provide a theoretical analysis to establish the 0.75 bound but I don’t believe their analysis. - I believe the bound can be established empirically via simulation. - I don’t see an obvious way to approach this from a theoretical perspective so, as it’s a cute problem that seems like it should be amenable to some form of probabilistic analysis, I thought I’d see if anyone here has any insight. Cheers, Richard

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Re: [math-fun] https://en.wikipedia.org/wiki/Dragon_curve
by Bill Gosper 28 Aug '19

28 Aug '19

On 2019-08-28 09:34, Brad Klee wrote: > The article should also say something about limit-periodicity. > For example try: > > SetDragon[Root_, BinaryBranching_] := Append[ > Flatten[Reap[FoldList[(Sow[ > {x_Integer /; IntegerQ[ > Divide[x - #1 - (2^#2[[2]])*BitXor[0, #2[[1]]], > 2^(#2[[2]] + 1)]] :> 1, x_Integer /; IntegerQ[ > Divide[x - #1 - (2^#2[[2]])*BitXor[1, #2[[1]]], > 2^(#2[[2]] + 1)]] :> 0}]; #1 + (2^#2[[2]])* > BitXor[0, #2[[1]]] + 2^(#2[[2]] - 1)) &, > Root, Transpose[{BinaryBranching, > Range[Length@BinaryBranching]}]] > ][[2]]], _Integer :> "?"] > > ArrayPlot[ > Map[Range[0, 2^6] /. SetDragon[0, #] &, > NestList[Append[#, RandomInteger[{0, 1}]] &, > RandomInteger[{0, 1}, 1], 6]], > Mesh -> True, ImageSize -> 500] > Wow, for the first time on Planet Earth, an actual brillathong keyboard. Brad, this is too dense for me. For starters, is a Fourier transform pure point the same as a periodic point? I can see why the Wikipedist dodged this can of worms. The simplest Fourier transform I can find for Heighway's Dragon has for its coefficients infinite products of non-triangular 3⨉3 matrices. first few dragon harmonics <http://gosper.org/FDrags128.gif> —rwg > This definition clearly shows relation of the tiling > space to a binary tree, and also explains why the > Fourier transform should have pure points. It also > helps to explain why the dragon sequence shows > up as a factor in Socolar-Taylor tiling, and in other > hexagonal or square quadtree tilings. > > --Brad > > > On Wed, Aug 28, 2019 at 10:30 AM Bill Gosper <billgosper(a)gmail.com> wrote: >> >> is the usual mix of good and bad. >> Good: Illustration that 4 Dragons joined at the nose >> < https://en.wikipedia.org/wiki/Dragon_curve#/media/File:Dragon_tiling1.svg> >> fill a tile different >> from four joined at the tail >> < https://en.wikipedia.org/wiki/Dragon_curve#/media/File:Dragon_tiling2.svg>. >> Oops: Fails to note that twin1 >> < https://en.wikipedia.org/wiki/Dragon_curve#/media/File:Dragon_tiling3.svg> >> and twin2 >> < https://en.wikipedia.org/wiki/Dragon_curve#/media/File:Dragon_tiling5.svg> are >> *both* twindragons*.* >> (Svgs, but painted with a horrible fat brush that *ruins* the symmetry.) >> Howler: (Re: the bounding box) >> "Note that the dimensions 1, and 1.5 are limits >> <https://en.wikipedia.org/wiki/Limit_(mathematics)> and not actual values." >> There remains almost universal ignorance of Dragon curves being merely >> lousy plots of the (almighty) Dragon Function. Axis-aligned >> tangents to the Dragon image intersect it in (uncountable) Cantor sets! >> E.g., some points on the bottom edge (Im(z) = -1/3) have spacings >> {1, 3, 1, 11, 1, 3, 1, 43, 1, 3, 1, 11, 1, 3, 1, 171, >> 1, 3, 1, 11, 1, 3, 1, 43, 1, 3, 1, 11, 1, 3, 1, . . .}/1023. Next term: 4⨉171 - 1 = 683 >> This is *a276391**,* the spacings of *preimages* of quadruple points of the >> *Hilbert *curve! >> More incredibly, another subset of the tangent points along x-i/3 has >> differences >> {1, 3, 1, 11, 1, 3, 1, 43, 1, 3, 1, 11, 1, 3, 1, 170, . . .}/4080, with 170 >> instead of 171! Next terms: 1, 3, 1, 11, 1, 3, 1, 43, 1, 3, 1, 11, 1, 3, 1, 680, . . . >> —rwg

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