July 2011 - mathfun@mailman.xmission.com

[math-fun] Morpion Solitaire: new world record
by rcs＠xmission.com 19 Aug '11

19 Aug '11

[forwarded for Christian Boyer, cboyer(a)club-internet.fr --Rich] ----- Morpion Solitaire: new world record, by Chris Rosin. 177 moves, 5 more moves than his previous record. This record will be presented at IJCAI'11 (International Joint Conference on Artificial Intelligence), July 2011. May 2011 news of www.morpionsolitaire.com Christian.

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[math-fun] quintic question
by Bill Gosper 03 Aug '11

03 Aug '11

I believe the saying goes: A quintic with rational coeffs is solvable in radicals iff it factors or the Lagrange resolvent sextic (discussed here with JHC decades ago) has a rational root. What about if the quintic has radical coeffs? E.g., the center disk in the D_5 packing if 31 disks has radius a satisfying (43875 + 37180*I) + 74360*(-1)^(1/10) - 97460*(-1)^(3/10) - 70760*(-1)^(2/5) + 70760*(-1)^(3/5) + 23100*(-1)^(7/10) + ((47063255 + 40030096*I) + 80060192*(-1)^(1/10) - 104803824*(-1)^(3/10) - 76142960*(-1)^(2/5) + 76142960*(-1)^(3/5) + 24743632*(-1)^(7/10))*a + ((-4765010 - 4048824*I) - 8097648*(-1)^(1/10) + 10603816*(-1)^(3/10) + 7702560*(-1)^(2/5) - 7702560*(-1)^(3/5) - 2506168*(-1)^(7/10))*a^2 + ((1630 + 336*I) + 672*(-1)^(1/10) - 1584*(-1)^(3/10) - 1040*(-1)^(2/5) + 1040*(-1)^(3/5) + 912*(-1)^(7/10))*a^3 + ((-225 + 1116*I) + 2232*(-1)^(1/10) - 2964*(-1)^(3/10) + 200*(-1)^(2/5) - 200*(-1)^(3/5) + 732*(-1)^(7/10))*a^4 - 5*a^5 Can we say that such a quintic is solvable in radicals iff the resolvent has a root expressible in radicals? If not can we at least say that no resolvent solution implies no quintic solution? -rwg I have the resolventin Mma (LeafCount 3029) and Macsyma, if anyone wants. OLEO-GUM-RESIN NUMEROLOGIES

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[math-fun] flowsnake bathroom tile
by Bill Gosper 02 Aug '11

02 Aug '11

(to go with the Penrose toilet paper) Neil, Corey, and I (over two sessions) worked out how, with tiles of two solid colors, to map a fill of the flowsnake (Gosper) curve onto a hex grid,<http://gosper.org/hexflo.png> a staggered square grid <http://gosper.org/staggerflo.png>, and a square grid <http://gosper.org/gridflo.png> wherein squares are considered joined at their NE and SW corners but not the other two. The illustrations are of three flowsnakes joined in a triangle to create a closed curve. A few weeks(?) ago I had a burning desire to find this hex version in conjunction with either a math-fun thread or a project with the kids, but da*ned if I can remember what it was. Maybe it was just to emphasize to Erez Lieberman that hex grids and square grids aren't all that different for spacefilling purposes. --rwg

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[math-fun] Things they never tell you about Dupin cyclides
by Fred lunnon 01 Aug '11

01 Aug '11

Following my recent ignominious extraction from the Dupin node quicksands with the assistance of a flying-squad who do know some algebraic geometry, and muster a computer algebra system which can solve polynomial equations, I remounted my cyclide and pedalled off shakily to further adventures in neglected corners of this subfusc mathematical suburb. Among the customary extensive selection of technical lamposts, potholes and manhole-covers lying subsequently in wait, there lurked the following conundrum. To recap a recent post by Ralph, a Dupin cyclide in canonical pose may be specified by three parameters (due apparently to Maxwell, rather than to Cayley as earlier hazarded). The radius "a" specifies a central circle at the origin; with centre on this, a perpendicular cross-sectional circle sweeps out a tube with mean radius "m"; while in the course of a single revolution, the actual tube radius is offset from m-c through m+c and back to m-c again. In terms of c,a,m, the implicit equation is then a Cartesian quartic (x^2 + y^2 + z^2 - m^2 - a^2 + c^2)^2 - 4(a x - c m)^2 - 4(a^2 + c^2)y^2 = 0 . The general shape of the surface will be "horned" when 0 < m < c < a ; "ring" when 0 < c < m < a ; "spindle" when 0 < c < a < m ; with various special cases at boundaries between. For example, when the differential offset vanishes c = 0 we have a torus; when in addition the tube radius equals the central radius m = a we have a double sphere centred at the origin (and giving the average graphics surface plotter a furry tongue). Now consider the special case where offset equals central radius c = a. It's easily established that the equation factors as the product of (x - 2a)^2 + y^2 + z^2 - (m - 2a)^2 , with (x + 2a)^2 + y^2 + z^2 - (m + 2a)^2 ; the surface comprises two tangent spheres with centres on the x-axis (by the way, as oriented spheres properly tangent, not anti-tangent). 'Ang abaht tho' --- one definition of a Dupin cyclide is the envelope (unique when spheres are oriented) swept out by a sphere moving tangent to 3 fixed spheres. For what 3 spheres (including planes and points) could such an envelope possibly be a pair of tangent spheres? Fred Lunnon

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Re: [math-fun] fractal optimizer
by Dan Asimov 31 Jul '11

31 Jul '11

I love the fractal I call the map-of-France (but which RWG can probably correct my terminology as he may have done in the past): Start with a hexagon (stage 0) centered at 0 in C, and with vertical sides 1 unit apart. Now surround that with six copies of itself and normalize this rosette by dividing the whole thing by sqrt(7) to get stage 1. Repeat ad infinitum. The thing would converge to a fractal shape with 6-fold symmetry except that the stages keep rotating a bit. So the normalizations should divide not just by sqrt(7) but by a complex number of length sqrt(7) that counteracts this tendency to rotate. (Does it work if that number is always 2 + exp(2pi*i/6), or does the rotation angle need to vary with the stages?). The result is a fractal-boundary hexagon, a rosette of 7 copies of which have the identical shape. So, the boundary B_n of stage n converges to a fractal curve B of Hausdorff dimension d(B) = log_7(9) = 1.129.... It seems as if no matter how this is done, B will not have mirror symmetry, just rotational. BUT there is a lot of choice in how 6 copies of each rosette are placed around itself, so this method can give a plethora of different curves in the limit. (Probably continuum many, given a countable number of discrete choices.) QUESTION: Among all these choices, is there some optimization problem that will result in a much smaller collection of maps-of-France, or ideally only essentially one? Perhaps the Hausdorff dimension log_7(9) *measure* of the limiting shape can be minimized? I'm fuzzy about just how to compute Hausdorff *measure* from its usual definition. --Dan Veit asks: << Are fractal sets ever the solution of an optimization problem? Sometimes the brain has a mind of its own.

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[math-fun] fractal optimizer
by Veit Elser 31 Jul '11

31 Jul '11

The Mandelbrot posts reminded me of something: Are fractal sets ever the solution of an optimization problem? Here's one instance I discovered a couple of years ago: Take an infinite square grid of red points and an identical grid of blue points. Now take the blue grid and rotate it by 45 degrees with respect to the red grid. The problem is to form a perfect matching of red and blue points such that the matching minimizes the sum of the squared distances between pairs. If the infinite value of the thing being minimized bothers you, do this: Use a continued fraction approximation for sqrt(2) to approximate the blue grid so that it becomes commensurate with the red grid (at periods that increase with the order of the continued fraction approximation). Now solve the optimization problem on just one (finite) period. To get the fractal, translate all the matched (red, blue) pairs so the red point is at the origin, and observe how the set of blue points evolves with order of approximation. Veit

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Re: [math-fun] Mandelbrot set history (Byte magazine September 1984) (was Re: Mandelbrot set question)
by Dan Asimov 31 Jul '11

31 Jul '11

I just don't remember, but was the Mandelbrot set not in either of his two books "Fractals: Form, Chance, and Dimension" (1977), or "The Fractal Geometry of Nature" (1982) ??? --Dan Robert M. wrote: << For those who are interested, I found those first Mandelbrot set images from Byte magazine September 1984. . . . >> Sometimes the brain has a mind of its own.

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[math-fun] Mandelbrot set history (Byte magazine September 1984) (was Re: Mandelbrot set question)
by Robert Munafo 30 Jul '11

30 Jul '11

For those who are interested, I found those first Mandelbrot set images from Byte magazine September 1984. I added the picture to my Lambda map page ( http://www.mrob.com/pub/muency/lambdamap.html) and the figure itself is here: http://www.mrob.com/images/0-muency/byte-198409-p158.jpg Image by David Turk, the computer graphics researcher and professor at Georgia Tech, when he was still an undergrad. Based on the article text, the images were probably created on a VAX 11/750 at West Coast University. - Robert Munafo On Fri, Jul 29, 2011 at 02:14, Robert Munafo <mrob27(a)gmail.com> wrote: > The only "az^2+bz+c" variation I've seen is the "lambda map", f(z) = c z > (1-z) = c z - c z^2. Here is an image: > > mrob.com/images/0-muency/lambda.jpg > > Notice the shape is different from the normal Mandelbrot shape, and they > are related by "squaring" one to get the other. > > This version of the Mandelbrot set was actually published in Byte > magazine[1] some months earlier than the Scientific American[2] that is > normally thought to be the popular introduction of the Mandelbrot set. The > Byte article was about fractals and simply showed a few pictures of the > Lambda fractal, without a formula or other explanation. The "interior" was > white, as in my image above, not black as the Mandelbrot set is more > commonly drawn. I immediately recognized the importance of this new fractal > and wished I could find out how it was calculated. > [...] > -- Robert Munafo -- mrob.com Follow me at: gplus.to/mrob - fb.com/mrob27 - twitter.com/mrob_27 - mrob27.wordpress.com - youtube.com/user/mrob143 - rilybot.blogspot.com

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[math-fun] Mandelbrot set question
by Dan Asimov 29 Jul '11

29 Jul '11

Speaking of the Mandelbrot set, denoted by M: The functions z^2 + c, c in C, contain just one examplar of each linear equivalence class of quadratic maps. (I.e., in the set of maps Q = {f(z) = a z^2 + b z + c with a,b,c in C and a != 0} let f,g in Q be equivalent if there exists a linear map L(z) = d z + e with d,e in C and d != 0, such that Linv o f o L == g, where Linv is the inverse function of L.) But {z^2 + c} is only one of many ways to parametrize this set of equivalence classes, and the choice of parametrization will drastically affect the appearance of the Mandelbrot set. And so I'm wondering if it's a lot more symmetrical than it looks. Some ways to define its symmetry group might be 1) All maps f:M -> M that are restrictions of a conformal bijection f: U -> V between two open neighborhoods U, V of M, with f(M) = M. 2) All self-homeomorphisms h:M -> M, factored out by the equivalence relation of isotopy. (I.e., homeos h,k: M -> M are equivalent if there is a continuous mapping H: M x [0,1] -> M such that for all z in M and t in [0,1] we have * H(0,z) = h(z), * H(1,z) = k(z), and * the map H( ,t): M -> M is a self-homeo of M. ------------------------------------------------ Maybe 2) is more interesting, since it would show only the fundamentally different self-homeos of M. Does anyone know if there are results on this question? --Dan Sometimes the brain has a mind of its own.

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Re: [math-fun] Mandelbrot's Conjecture (as mentioned in this video interview)
by Dan Asimov 29 Jul '11

29 Jul '11

Suddenly two definitions of the Mandelbrot set come to mind, the first one I learned (I), and the much more common one (II): For any c in C, define f_c(z) as z^2 + c. I. The set of c in C for which the Julia set of f_c is connected. (See < http://en.wikipedia.org/wiki/Julia_set >.) II. the set of c in C for which the orbit of 0, under (forward) iteration of f_c, is bounded. I've never seen a proof that these definitions are equivalent, though I haven't looked very hard, either. (In II, it seems to me that considering the orbits of 0 rather than of any other point is somewhat arbitrary.) --Dan Sometimes the brain has a mind of its own.

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