November 2010 - mathfun@mailman.xmission.com

[math-fun] Words, integers and loops
by Eric Angelini 22 Nov '10

22 Nov '10

Hello Math-Fun, Infinite loop: a) 12 - 12 - 12 - 12 - 12 ... Shortest loops so far: b) 6 - 10 - 15 - 14 - 11 - 8 - (6) c) 6 - 10 - 16 - 17 - 13 - 8 - (6) A longer loop: d) 6 - 10 - 15 - 14 - 12 - 11 - 8 - (6) A combination of (c) and (d): e) 6 - 10 - 16 - 17 - 13 - 8 - 6 - 10 - 15 - 14 - 12 - 11 - 8 - (6) This loop (e) shows ten different "words". The rule is: i) Replace all integers by their English spelling ii) a(n) is the count of the letters used by a(n)+a(n+1) Example for (d): d) 6 - 10 - 15 - 14 - 12 - 11 - 8 - (6) d) six-ten-fifteen-fourteen-twelve-eleven-eight-(six) Letters: 3 3 7 8 6 6 5 3 Letter-sum: 6 10 15 14 12 11 8 ... this being the sequence itself. Is it possible to find more loops? To insert more words not already used above? Best, É.

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Re: [math-fun] equilateral elliptical n-gons
by Henry Baker 21 Nov '10

21 Nov '10

Andy is right. The recursive subdivision doesn't work correctly -- it produces pairs of equal sides, but those pairs don't necessarily equal pairs futher away. Back to the drawing board... At 02:37 PM 11/18/2010, Andy Latto wrote: >On Thu, Nov 18, 2010 at 5:10 PM, Henry Baker <hbaker1(a)pipeline.com> wrote: >> [My statement below about "notchy" is *wrong*. Given an ellipse, you can start my recursive perpendicular bisector subdivision from _any_ line through the center of the ellipse. > > But as someone pointed out, the recursive perpendicular bisector >subdivsion doesn't work. You replace side AB with sides AC and CB, >where |AC| = |CB|, and you replace side XY with sides XZ and ZY, with >|XZ| = |ZY|, but there's no reason to expect that just because |AB| = >|XY|, that |AC| should equal |XZ|. > >Andy

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[math-fun] E. Coli Bacteria Solve Sudoku Puzzle
by Henry Baker 20 Nov '10

20 Nov '10

FYI -- Now I can solve sudoku's by "gut instinct" ??? E. Coli Bacteria Solve Sudoku Puzzle, Programmed By University Of Tokyo Students http://www.huffingtonpost.com/2010/11/18/e-coli-bacteria-solve-sud_n_785494… Problem-solving bacteria crack sudoku http://www.newscientist.com/article/dn19733-problemsolving-bacteria-crack-s…

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Re: [math-fun] equilateral elliptical n-gons
by Henry Baker 19 Nov '10

19 Nov '10

Perhaps "sliding" may not be the best word here. The situation is most dramatic with 2,3,4 sided n-gons. In the case of the 2-gon, the length of the sides varies between 2a and 2b, where a,b are the major,minor radii. Similarly, with a rhombus, the length of a side appears to be greatest [sqrt(a^2+b^2)] when it makes a diamond formation, and smallest when it has to be an upright square within the ellipse. As the number of sides increases, the difference in length between a side and the segment of the ellipse it cuts off becomes vanishingly small, so the variation in length during sliding also becomes vanishingly small. --- I still don't know how to construct a non-power of 2 number of vertices -- e.g., an equilateral pentagon -- within an ellipse. At 03:50 PM 11/18/2010, Dan Asimov wrote: >This statement puzzles me, since what I meant by being able to slide a polygon around a curve is that the vertices remain on the curve and the edges are rigid segments. (So during such a sliding each edge, and so the total perimeter, remains constant.) > >PUZZLE: Specify a case where a finite number of unit balls are each tangent to two others in a unit-radius tunnel* through R^3 (and not otherwise intersect) . . . but cannot be slid (i.e., so that each ball moves to the position of an adjacent one, through a continuous family of such cyclically tangent arrangements) -- and prove this. > >--Dan >__________________________________________ >* Meaning: there is a C^1 curve C in R^3 such that the tunnel is the union of all the unit-radius disks -- required to be disjoint -- that are each centered at a point of C and perpendicular to C. > ><< >. . . Given an ellipse, you can start my recursive perpendicular bisector subdivision from _any_ line through the center of the ellipse. So, it _is_ possible to "slide" the vertices around the perimeter of the ellipse, at least for equilateral 2^n-gons. Note, however, that the total perimeter of the 2^n-gon does vary slightly as it slides around the ellipse. >>>

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Re: [math-fun] equilateral elliptical n-gons
by Dan Asimov 18 Nov '10

18 Nov '10

This statement puzzles me, since what I meant by being able to slide a polygon around a curve is that the vertices remain on the curve and the edges are rigid segments. (So during such a sliding each edge, and so the total perimeter, remains constant.) PUZZLE: Specify a case where a finite number of unit balls are each tangent to two others in a unit-radius tunnel* through R^3 (and not otherwise intersect) . . . but cannot be slid (i.e., so that each ball moves to the position of an adjacent one, through a continuous family of such cyclically tangent arrangements) -- and prove this. --Dan __________________________________________ * Meaning: there is a C^1 curve C in R^3 such that the tunnel is the union of all the unit-radius disks -- required to be disjoint -- that are each centered at a point of C and perpendicular to C. << . . . Given an ellipse, you can start my recursive perpendicular bisector subdivision from _any_ line through the center of the ellipse. So, it _is_ possible to "slide" the vertices around the perimeter of the ellipse, at least for equilateral 2^n-gons. Note, however, that the total perimeter of the 2^n-gon does vary slightly as it slides around the ellipse. >> Those who sleep faster get more rest.

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Re: [math-fun] regular _elliptical_ n-gons?
by Bill Gosper 18 Nov '10

18 Nov '10

Victor Miller: >Is what you want related to Poncelet's Porism? >http://mathworld.wolfram.com/PonceletsPorism.html This subject seems to foster misnomers. Eric consistently has Poncelet traNSverse vs traverse. --rwg

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Re: [math-fun] equilateral elliptical n-gons
by Henry Baker 18 Nov '10

18 Nov '10

[My statement below about "notchy" is *wrong*. Given an ellipse, you can start my recursive perpendicular bisector subdivision from _any_ line through the center of the ellipse. So, it _is_ possible to "slide" the vertices around the perimeter of the ellipse, at least for equilateral 2^n-gons. Note, however, that the total perimeter of the 2^n-gon does vary slightly as it slides around the ellipse.] I don't see any way to put an equilateral quadrilateral inside an ellipse, except for the diamond and the square. So the rotation thing is definitely "notchy". At 10:55 AM 11/18/2010, James Buddenhagen wrote: >I think it may be more complicated than you describe below, but it >will still turn out to be algebraic, in fact if you have one vertex of >the equilateral polygon at an ellipse vertex the coordinates of the >polygon vertices will be algebraic for any algebraic a,b and any n, >because you have to solve a system of polynomial equations (arising >from the distance formula and the equation of the ellipse, and forcing >closure). > >Also, I think the solution will be isolated, no slipping of the >vertices around the ellipse as Dan mentioned, unless it is a circle, >but I would have to look more closely to be sure > >On Thu, Nov 18, 2010 at 12:12 PM, Henry Baker <hbaker1(a)pipeline.com> wrote: >> After a deep discussion with Macsyma, M convinced me that this recursive procedure works, and does indeed produce points whose coordinates are the solutions of polynomials with coefficients in the rational field extended by a & b. Furthermore, these polynomials are Galois-solvable by towers of square roots. >> >> Computing these points is really no more difficult than computing the coordinates of the 2^n-th roots of unity, except that a & b keep entering into the computation. If a=b=1, then the computations are exactly the same as those for calculating the coordinates of the 2^n-th roots of unity. >> >> Basically, given 2 points already on the ellipse, compute the equation of the perpendicular bisector line through those two points. This line intersects the ellipse in two places, _both_ of whom become new additional points (by symmetry). The equation set can be produced by resultant(ellipseeqn,bisectoreqn,y) [a quadratic in x] for the two x coordinates, and resultant(ellipseeqn,bisectoreqn,x) [a quadratic in y] for the two y coordinates. A slightly different approach can produce the points w/o any ambiguity about which x coordinate goes with which y coordinate.

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Re: [math-fun] equilateral elliptical n-gons
by Henry Baker 18 Nov '10

18 Nov '10

I don't see any way to put an equilateral quadrilateral inside an ellipse, except for the diamond and the square. So the rotation thing is definitely "notchy". At 10:55 AM 11/18/2010, James Buddenhagen wrote: >I think it may be more complicated than you describe below, but it >will still turn out to be algebraic, in fact if you have one vertex of >the equilateral polygon at an ellipse vertex the coordinates of the >polygon vertices will be algebraic for any algebraic a,b and any n, >because you have to solve a system of polynomial equations (arising >from the distance formula and the equation of the ellipse, and forcing >closure). > >Also, I think the solution will be isolated, no slipping of the >vertices around the ellipse as Dan mentioned, unless it is a circle, >but I would have to look more closely to be sure > >On Thu, Nov 18, 2010 at 12:12 PM, Henry Baker <hbaker1(a)pipeline.com> wrote: >> After a deep discussion with Macsyma, M convinced me that this recursive procedure works, and does indeed produce points whose coordinates are the solutions of polynomials with coefficients in the rational field extended by a & b. Furthermore, these polynomials are Galois-solvable by towers of square roots. >> >> Computing these points is really no more difficult than computing the coordinates of the 2^n-th roots of unity, except that a & b keep entering into the computation. If a=b=1, then the computations are exactly the same as those for calculating the coordinates of the 2^n-th roots of unity. >> >> Basically, given 2 points already on the ellipse, compute the equation of the perpendicular bisector line through those two points. This line intersects the ellipse in two places, _both_ of whom become new additional points (by symmetry). The equation set can be produced by resultant(ellipseeqn,bisectoreqn,y) [a quadratic in x] for the two x coordinates, and resultant(ellipseeqn,bisectoreqn,x) [a quadratic in y] for the two y coordinates. A slightly different approach can produce the points w/o any ambiguity about which x coordinate goes with which y coordinate.

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Re: [math-fun] equilateral elliptical n-gons
by Henry Baker 18 Nov '10

18 Nov '10

After a deep discussion with Macsyma, M convinced me that this recursive procedure works, and does indeed produce points whose coordinates are the solutions of polynomials with coefficients in the rational field extended by a & b. Furthermore, these polynomials are Galois-solvable by towers of square roots. Computing these points is really no more difficult than computing the coordinates of the 2^n-th roots of unity, except that a & b keep entering into the computation. If a=b=1, then the computations are exactly the same as those for calculating the coordinates of the 2^n-th roots of unity. Basically, given 2 points already on the ellipse, compute the equation of the perpendicular bisector line through those two points. This line intersects the ellipse in two places, _both_ of whom become new additional points (by symmetry). The equation set can be produced by resultant(ellipseeqn,bisectoreqn,y) [a quadratic in x] for the two x coordinates, and resultant(ellipseeqn,bisectoreqn,x) [a quadratic in y] for the two y coordinates. A slightly different approach can produce the points w/o any ambiguity about which x coordinate goes with which y coordinate. >Date: Thu, 18 Nov 2010 07:29:26 -0800 >From: Henry Baker <hbaker1(a)pipeline.com> > >I guess one way to inscribe an equilateral n-gon in an ellipse of semimajor axis a and semiminor axis b would be to start with the 4 points > >(a,0), (0,b), (-a,0), (0,-b) > >forming a diamond inscribed in the ellipse, and then recursively constructing a little triangle on each of these diagonals which intersects the ellipse. In this way, we can construct an equilateral 2^n-gon within the ellipse. I assume that calculating the point of intersection of the perpendicular bisector of an existing side with the ellipse is the solution of a polynomial equation ?? > >At 06:47 AM 11/18/2010, Henry Baker wrote: >>I was looking for an equi _lateral_ n-gon inscribed in an ellipse. It is more important that the (straight line segment) sides be equal, than that the vertices occur at equal curve segments around the edge. In an ellipse, it won't be possible to do both at the same time. >> >>At 06:40 AM 11/18/2010, Michael Kleber wrote: >>>Wait, Henry, do you really mean you want "equal spacing along the perimeter >>>of the ellipse"? I thought you wanted equal side lengths, and didn't care >>>about the length of the ellipse arc they subtended. >>> >>>--Michael >>> >>>On Thu, Nov 18, 2010 at 9:30 AM, Henry Baker <hbaker1(a)pipeline.com> wrote: >>> >>>> I think that by using the term "regular" instead of "equilateral", I >>>> confused everyone. >>>> >>>> I'm trying to find a simple way to generate n-vertex equilateral ellipses. >>>> It is particularly important for the application I have in mind that the >>>> vertices are at equal spacing along the perimeter of the ellipse, and most >>>> importantly, that the sequence closes after one trip around -- i.e., there >>>> can't be any gap. >>>> >>>> The reason is that I want to use these little vectors as accelerations, and >>>> if they don't sum up exactly to zero, then the velocities & positions don't >>>> have any hope of closing up.

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Re: [math-fun] equilateral elliptical n-gons
by Henry Baker 18 Nov '10

18 Nov '10

I was looking for an equi _lateral_ n-gon inscribed in an ellipse. It is more important that the (straight line segment) sides be equal, than that the vertices occur at equal curve segments around the edge. In an ellipse, it won't be possible to do both at the same time. At 06:40 AM 11/18/2010, Michael Kleber wrote: >Wait, Henry, do you really mean you want "equal spacing along the perimeter >of the ellipse"? I thought you wanted equal side lengths, and didn't care >about the length of the ellipse arc they subtended. > >--Michael > >On Thu, Nov 18, 2010 at 9:30 AM, Henry Baker <hbaker1(a)pipeline.com> wrote: > >> I think that by using the term "regular" instead of "equilateral", I >> confused everyone. >> >> I'm trying to find a simple way to generate n-vertex equilateral ellipses. >> It is particularly important for the application I have in mind that the >> vertices are at equal spacing along the perimeter of the ellipse, and most >> importantly, that the sequence closes after one trip around -- i.e., there >> can't be any gap. >> >> The reason is that I want to use these little vectors as accelerations, and >> if they don't sum up exactly to zero, then the velocities & positions don't >> have any hope of closing up.

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