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[math-fun] Custom number crunching programs --- expectations?
by Andres Valloud 17 Apr '19

17 Apr '19

Hello, I've used a GNFS factoring package, and it seemed to have been modeled after a C program targeting POSIX, and emitting output to stdout. I also wrote similar programs with the same mode of operation. Have you had experience with custom number crunching programs like that? How did you interacting with them? What would you expect? Thanks in advance. Andres.

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Re: [math-fun] That curve I regrettably named flowsnake
by Bill Gosper 17 Apr '19

17 Apr '19

On 2019-04-10 06:10, Joerg Arndt wrote: > Pics or it didn't happen! > Agreed: https://www.wolframcloud.com/objects/797a1110-f0cd-47a5-90f6-f0b3d1d8ce4b > Joking aside: I'd like to understand, but don't. > > Best regards, jj > > > * Bill Gosper <billgosper(a)gmail.com> [Apr 08. 2019 15:50]: >> On Sat, Apr 6, 2019 at 9:20 AM Bill Gosper <billgosper(a)gmail.com> wrote: >> >> > Mandelbrot detested the name. Then he discovered a spacefill of Koch's >> > Snowflake. >> > The worthy claimant to the Flowsnake name! Which he still detested. >> > I recently realized how little I know about my own curve, which fills the >> > FranceFlake, the fundamental region for the positional number system >> > Base: 2 + i^(2/3) = √7 e^(i arccos(11/14)/2) >> > and the (necessarily) seven digits 0 and the 6th roots of 1. >> > >> >> SPOILERS >> >> > If FranceFill(0)= 0+0i, FranceFill(1) = 1 + 0i: >> > 1) What is the area of a FranceFlake? >> > >> Only √3/2. >> >> > 2) Where is its centroid? >> > >> 1/2+i/2/√3 >> >> > 3) Is its circumradius √(52/147) ? >> > >> No. ≥ √(6140317/3)/2401. (What the heck is this approaching?) >> The same as the FranceFlake, duh. >> >> > 4) Ironically, it was popular partly for how completely its canonical >> > sampling (at 0/7, 1/7,...,1) self-avoided. But like all true spacefills, >> > it is *dense* with *triple* points. >> > E.g., what are the preimages of (9 + i √3)/21? >> > >> {5/42, 11/42, 17/42} >> Are there quadruple points? How can we prove "No."? >> >> > —rwg

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[math-fun] Substitution Tilings
by Ed Pegg Jr 17 Apr '19

17 Apr '19

I've recently found a lot of new substitution tilings. They are listed in the program at http://demonstrations.wolfram.com/SubstitutionTilings/ -Ed Pegg Jr

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[math-fun] The St. Ives tessellation
by Dan Asimov 14 Apr '19

14 Apr '19

Let the map-of-France shape M be the outer boundary of a (limiting) flowsnake: M_0 = a 2D regular hexagon, including its boundary curve. M_(n+1) = a rosette of 7 copies of M_n, after the rosette has been shrunk to total volume = 1 (linear factor of 1/sqrt(7)) and a slight rotation ... ... such that lim M_n = M. Since at each stage the M_n tessellate R^2, so does M. Call this tessellation by T_0. Then each tile of T_0 (a copy of M) is a rosette of 7 smaller ones. Call the tessellation by *those* by T_1, lather, rinse, repeat. Since a tessellation is nothing more than the set whose members are its tiles: Finally let T denote the set of all tiles of any T_n: T = T_0 u T_1 u ... u T_n u ... Whew! Now consider the *symmetry group* of all similarity mappings taking *any tile in T* to *any tile (possibly the same one) in T*, in *any rotational position* (there are 6). (Flips are excluded, of course, since all of M's isometries with itself are orientation preserving.) This is a very interesting group that I know little about. * * * What is the 1-dimensional version of this? That looks like a pretty interesting group G as well. A member of this group G takes any interval (K/2^n, (K+1)/2^n) to any interval of the same form, i.e., (L/2^m, (L+1)/2^m): (K/2^n, (K+1)/2^n) —> (L/2^m, (L+1)/2^m) by either preserving or reversing orientation and expanding length by a constant factor, i..e., by the map x |—> ax + b determined by (y - (L+1)/2^m) / (x - L/2^m) = ((L+1)/2^m - (K+1)/2^n) / (L/2^m - K/2^n). —Dan

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Re: [math-fun] Restaurant placemat mazes
by Keith F. Lynch 14 Apr '19

14 Apr '19

Cris Moore <moore(a)santafe.edu> wrote: > Site percolation on the square lattice isn?t self-dual since there > can be ?checkerboard? states where neither path is possible, I could be wrong, but I don't think that's what anyone is talking about. The square lattice, or maybe I should say grid, is like a fly screen, consisting entirely of lots of equally spaced horizontal and vertical lines. Some proportion of the segments of those horizontal lines between two consecutive vertical lines, and the same proportion of the segments of the vertical lines between two consecutive horizontal lines, are randomly removed, to create a random maze. "Checkerboard" implies that at least some of the small squares are filled in. They're all hollow. The dual of the maze consists of the remaining lines. Think of them as wires, which can conduct current from one side of the maze to the opposite side only if there's continuity, i.e. an unbroken wire path. To me it's obvious that either the empty spaces connect two opposite faces or the solid lines connect the other two faces. It's impossible for both to be true or for neither to be true. > and the threshold is 0.592746? > 1/2. Where does that number come from? Allan Wechsler <acwacw(a)gmail.com> wrote: > Keith's confusion, which I share, is raised by the wording in the > Bollobas & Riordan paper which he quotes upthread. It seems to > Keith and me that pH = 1/2 follows instantly from self-duality, but > Bollobas & Riordan say that the proof is deep and non-trivial. I > know Bollobas's name -- these are no lightweights. It follows that > Keith and I are missing some complication to the question. All I can think of is that maybe what I describe above as obvious is one of those things that's very difficult to formally prove. I certainly wouldn't know how to begin to do so. It's one of those things that would seem to not need a proof. I guess that means I'll never be a real mathematician. See https://xkcd.com/2042/

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[math-fun] Three old "threequences" from Gosper
by Allan Wechsler 14 Apr '19

14 Apr '19

OEIS has three sequences, A019989, A019990, and A019991, entered by Neil Sloane in 1996 but credited to Bill Gosper. The wording is quite terse, but reasonably clear. There is little commentary, and only about 60 terms are given. Recently on Sequence Fanatics, Sean Irvine reported that he tried to reproduce the data from the given definitions and failed. I was skeptical, but then I tried the experiment myself and also failed. Obviously Sean and I are misunderstanding the wording in the same way, and I'd like to reconstruct what was meant. Below, I'm going to give my reconstruction, which must be wrong at some point! I hope somebody (hint, hint) can clear up my misunderstanding. The underlying structure is a set of six boolean sequences, named a, b, c, A, B, C. At n=0, they are initialized with a(0) true, and the rest false. For any future n, the values are calculated from one particular set of six old values, the ones for k = floor((n+1)/3). For example, the value at 117 are calculated from those at 39. The value of each of the sequences at n is derived from the values at k by OR-ing together three of the values at k. The rule is quite symmetrical: the new value for each sequence is derived from the old value of the same sequence, OR-ed together with the values of the two sequences that differ from the given one in both case and letter. For example, B(n) = B(k) OR a(k) OR c(k). The three sequences in OEIS are defined in terms of a(2n), b(2n), and c(2n). A number n is in the list A019989 if a(2n) is true, and the other two sequences are built from b and c respectively. The problem that Sean and I encountered is that, because OR is monotonic, all six sequences seem to be quickly driven to true, and stay there. My reconstruction is as follows: n=0: abc/ABC = TFF/FFF n=1: TFF/FTT n=2, 3, 4: TTT/FTT all n>=5: TTT/TTT Sean tried exclusive OR, and some other options, and it failed in different ways that he did not share. What was the intent here?

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Re: [math-fun] Possibly unsolved problem about continuous functions f : [0,1]^3 —> R^2
by Dan Asimov 14 Apr '19

14 Apr '19

That Andy! Always with the counterexamples! I had been reading about maps from R^n to R^q and talking about (n-q)-dimensional area, but tripped up on 3-2 = 1. Ah, the glories of old age. Anyhow by 2-dimensional area I meant 1-dimensional length. So: Given a continuous g : [0, 1]^3 —> R^2, the problem is: Prove there exists p in R^2 with the *length* of the inverse image g^(-1)(p) of p is >= 1. ( Or more generally, for any continuous map g : [0, 1]^n —> R^q show there exists a point p of R^q such that (the (n-q)-dimensional "area" of g^(-1)(p)) >= 1 ) —Dan ----- > > Say > > f : [0,1]^3 —> R^2 > > is a continuous map from the unit cube to the plane. > > Unsolved problem (I think): > > ----- > Prove there is some point p in R^2 whose inverse image f^(-1)(p) has area >= 1. Take f(x,y, z) = (x,y). The inverse image of any point is a line segment, which has area 0. Andy > ----- > > Here "area" means 2-dimensional Hausdorff measure.* > > This should properly be called a conjecture. It is related to an apparently > difficult — but proven — theorem in geometry, the "waist inequality" of Gromov: > > Let S^3 be the unit sphere in 4-space, and let > > g : S^3 —> R^2 > > be any continuous map, where D^3 is the closed unit ball in R^3. > > ----- > Then there exists some point p of R^2 whose inverse image g^(-1)(p) has area > at least the area of the equator of S^3. That is, >= 4π^2, since the equator > is a unit 2-sphere). > ----- > > —Dan > > ————— > * Hausdorff r-measure is an ingenious way to assign a reasonable meaning to > the notion of r-dimensional measure of a compact subset K of Euclidean space > R^N, for any positive real number r. When r is an integer it works like this: > > Let alph(N) > 0 be the constant for which the volume of an N-dimensional > ball D_R of radius R in R^n is equal to > > vol(D_R) = alpha(N) R^N. > > Now in R^N, let C be any covering of K by open balls of radii <= eps, > for some eps > 0. > > Define > > |C|_(r, eps) = alph(N) * (the sum of the r'th power of each radius of a ball in C). > > Finally, define > > |C|_r = inf |C|_(r, eps) > eps > 0 > > It's a bit complicated, but it's exactly the right definition. -----

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[math-fun] Possibly unsolved problem about continuous functions f : [0,1]^3 —> R^2
by Dan Asimov 13 Apr '19

13 Apr '19

Say f : [0,1]^3 —> R^2 is a continuous map from the unit cube to the plane. Unsolved problem (I think): ----- Prove there is some point p in R^2 whose inverse image f^(-1)(p) has area >= 1. ----- Here "area" means 2-dimensional Hausdorff measure.* This should properly be called a conjecture. It is related to an apparently difficult — but proven — theorem in geometry, the "waist inequality" of Gromov: Let S^3 be the unit sphere in 4-space, and let g : S^3 —> R^2 be any continuous map, where D^3 is the closed unit ball in R^3. ----- Then there exists some point p of R^2 whose inverse image g^(-1)(p) has area at least the area of the equator of S^3. That is, >= 4π^2, since the equator is a unit 2-sphere). ----- —Dan ————— * Hausdorff r-measure is an ingenious way to assign a reasonable meaning to the notion of r-dimensional measure of a compact subset K of Euclidean space R^N, for any positive real number r. When r is an integer it works like this: Let alph(N) > 0 be the constant for which the volume of an N-dimensional ball D_R of radius R in R^n is equal to vol(D_R) = alpha(N) R^N. Now in R^N, let C be any covering of K by open balls of radii <= eps, for some eps > 0. Define |C|_(r, eps) = alph(N) * (the sum of the r'th power of each radius of a ball in C). Finally, define |C|_r = inf |C|_(r, eps) eps > 0 It's a bit complicated, but it's exactly the right definition.

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Re: [math-fun] Restaurant placemat mazes
by Keith F. Lynch 13 Apr '19

13 Apr '19

Allan Wechsler <acwacw(a)gmail.com> wrote: > Just in case anybody doesn't know: questions like this have been > studied at least since the 1950s, under the rubric of "percolation > theory". The seminal work appears to be Broadbent & Hammersley > 1957. See the Wikipedia article on percolation theory for the > complete reference. Thanks. Wikipedia led me, indirectly, to https://arxiv.org/pdf/math/0410359.pdf which says: The problem of investigating pH in a variety of contexts was posed by Broadbent and Hammersley [7] in 1957. Hammersley [11, 12, 13] proved general results implying in particular that 0.35 < pH(Z2) < 0.65. The first major progress was due to Harris [14], who proved in 1960 that pH(Z2) >= 1/2. His proof makes use of the "self-duality" of Z^2, and is highly non-trivial. For many years it was believed that pH = 1/2 for bond percolation in Z^2; see, for example, Sykes and Essam [22]. However, it was only in 1980, twenty years after Harris proved that pH >= 1/2, that Kesten [16] proved this conjecture, following significant progress by Russo [20] and Seymour and Welsh [21]. I'm puzzled by this. I would think that knowing that it was self dual and that the odds of a solution were greater than or equal to one half would prove that the dual also has odds of a solution greater than or equal to one half. And since the dual obviously has a solution if and only if the original does not, it must follow that both are exactly equal to one half. What am I missing?

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[math-fun] Draft of my April 17, 2019 essay
by James Propp 12 Apr '19

12 Apr '19

I've posted a draft of the essay I plan to post on the morning of the 17th (or maybe the night before) at http://mathenchant.org/047-draft1.pdf I don't expect to get comments from many of you, but all comments will be appreciated and acknowledged. One thing that's missing from the essay is a good reference for readers who want to learn more about modern mazes. For that matter, do any of you know of any good references about how to create proofs that explicitly advocate the work-from-the-outside-in tactic? I've never read "How to Read and Do Proofs" or the many similar books that are out there, but if you think readers of my blog would like a particular book in this genre, I'd be glad to mention it. Thanks, Jim

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