- mathfun@mailman.xmission.com - mailman.xmission.com

RE: Re: Re: [math-fun] Choreography of large complete graphs
by Andy Latto 01 Sep '06

01 Sep '06

> -----Original Message----- > From: math-fun-bounces+andy.latto=pobox.com(a)mailman.xmission.com > [mailto:math-fun-bounces+andy.latto=pobox.com@mailman.xmission.com]On > Behalf Of Michael Kleber > Sent: Friday, September 01, 2006 12:22 PM > To: math-fun > Subject: Re: Re: Re: [math-fun] Choreography of large complete graphs > > > In my rephrasing of Otto's solution, I should have mentioned that > this method of having everyone meet everyone else is commonly > used in English Country line dances, those described by the > original source (John Playford, from 1651 on) as "longways for as > many as will". It's also what change-ringers refer to as "plain hunt on an odd number of bells", and may be even older in this context than in dancing. Andy Latto andy.latto(a)pobox.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >

1 0

[math-fun] William Shanks, historical computation of pi available on google books.
by Simon Plouffe 31 Aug '06

31 Aug '06

Hello, there are a number of significant documents that are available with Google Book service, they apparently added quite a collection of full and complete documents. One of them is of interest : The historical computation of Pi using Machin formula by William Shanks. http://books.google.com/books?vid=0A2yyrfAXNsNSwg9B-&id=axjockOKyikC&pg=PA2… and the following pages describe the digits, the formulas and results, quite interesting! and the result is : http://books.google.com/books?vid=0A2yyrfAXNsNSwg9B-&id=axjockOKyikC&pg=PA2… which contains the known error. Another document is the Differential Calculus by De Morgan: http://books.google.com/books?vid=08Wx_wKklYK4hOCS&id=95x4IrIcHrgC&printsec… These documents are normaly quite difficult to get and see for real even in a big city with lots of libraries around... Simon Plouffe

1 0

Re: [math-fun] Poincare Conjecture & Perelman
by Daniel Asimov 31 Aug '06

31 Aug '06

Now that the Poincare conjecture appears to be solidly proven, I'd like to see the known corollaries. Over the years a number of theorems have proved "Either there's a counterexample to the Poincare conjecture, or else something very strange happens in 4 or 5 dimensions" (with the something very strange being specified). Very weird stuff is already known to happen in 4 dimensions. E.g., it's the unique dimension whose Euclidean space R^4 carries more than one differentiable structure, up to equivalence (diffeomorphism). There is a continuum of them! Also, given any finitely-presented group G, there is a 4-dimensional manifold fundamental group M whose fundamental group pi_1(M) is isomorphic to G (using topological operations that mimic the presentation using glueing of 1-handles & 2-handles, then capping it off with a conractible 4-manifold with the same boundary -- I think). Since a theorem of Markov says there is no algorithm which, given two arbitrary finite group presentations, can determine if they define isomorphic groups, it follows that there is no algorithm to classify all (even compact) 4-manifolds up to homeomorphism. (Okay, okay, this is also true in dimensions > 4, but 4 is the lowest case where it happens.) --Dan

2 1

Re: [math-fun] Choreography of large complete graphs
by Daniel Asimov 31 Aug '06

31 Aug '06

Ed writes: << At a convention, 128 people all need to meet with each other, 1 on 1, for a paper exchange. Suppose a room is divided into a 8x8 square, and a clock is set up to ding 127 times, once every 30 seconds. Each person is given a map of their starting square, and a map of their route over the next hour and 4 minutes. Only two people can be in each cell. Ideally, most moves should be King moves. Not needing to move during a turn is acceptable. Simplicity is judged by the amount of text needed to explain the choreography, and the orderliness of the room. Alternately, the length of the maximum walk should be minimised. >> This is a lovely problem, even if we're still discussing its precise definition. I'm partial to the triangular lattice since it's the only symmetric 2D lattice such that any intersecting pair of Voronoi regions (in this case closed hexagons) intersect along an edge. (Though there is no necessary reason the network need be 2D; there is a natural generization in each dimension; in 3D it's the tiling by truncated octahedra). And I'm notorious in these quarters for preferring a torus over a bounded region, to avoid those pesky edge effects. Imo the most symmetric such tori of all are those arising from an iterative tiling by rosettes of 7 (of the preceding rosette, the first one just being 6 hexagons around 1 of them). Thus the number of nodes in these cases is a power of 7: 7, 49, 343, 2401, 16807, . . .. A torus of 49 seems a nice place to start. (Though 49 is clearly not as friendly a number as 120, which is my guess why Ed wanted to use that one -- its being both 5! and T_15.) Another nice feature of 7^n tori like this is that their network can be colored with 7 colors in a highly symmetric repeating pattern. A compromise would be considering the next most symmetrical such (triangular-lattice-based) tori, which imo are those which arise from an n x n rhombus; the nicest of reasonable size might be 12 x 12, since it's a both a square and a Fibonacci number, and also has more factors than any nearby number other than 120. (All these 60-deg. rhombic tori are in a strong sense "hexagonal" too, but the hexagon doesn't interact as nicely with the network when the # of nodes is not a power of 7.) But I'm interested in starting with the highly symmetric 7 x 7 hexagonal torus network of 49 nodes, each connected to 6 others. We can ask this question: Given 98 people, two on each node of this network at each stage, we want each one to have an itinerary (moving to an adjacent node, or staying put, on each step) such that each person shares a node with each of the other 49, such that the least total number of moves-to-an-adjacent-node are used. (Making no assumptions about the number of steps necessary.) This is clearly only one of many possible ways the details can be made precise. --Dan

3 3

Re: [math-fun] Choreography of large complete graphs
by Daniel Asimov 30 Aug '06

30 Aug '06

<< I'm interested in starting with the highly symmetric 7 x 7 hexagonal torus network of 49 nodes, each connected to 6 others. We can ask this question: Given 98 people, two on each node of this network at each stage, we want each one to have an itinerary (moving to an adjacent node, or staying put, on each step) such that each person shares a node with each of the other ***49***, such that the least total number of moves-to-an-adjacent-node are used. (Making no assumptions about the number of steps necessary.) >> (asterisks added for emphasis) Of course, silly me, this should read 97, not 49. Gareth: Your solution looks promising, but I'm not sure exactly what you mean by n. --Dan

1 0

OOPS Re: [math-fun] Choreography of large complete graphs
by Daniel Asimov 30 Aug '06

30 Aug '06

I see from the Subject line that Ed intended for one step to be to any other of the 64 squares, so I retract that question. --Dan

1 0

Re: [math-fun] Choreography of large complete graphs
by Daniel Asimov 30 Aug '06

30 Aug '06

Ed asks: << At a convention, 120 people all need to meet with each other, 1 on 1. Suppose a room is divided into a 8x8 square, and a clock is set up to ring 60 times, once per minute. Each person is given a map of their starting square, and a map of their route over the next hour. What is the simplest set of maps? What is the length of the minimal walks people need to take? >> This is an intriguing question! One thing I'm not sure I understand is what kind of step one is allowed to take. Must it be N,S,E,W ? Or does it also include NE, SE, NW, SW ? Another question is how would one meet 119 other people in only 60 steps? I must be misunderstanding something. (Or is the real question how each person could meet 60 others in 60 steps?) A third question I have is: Is it obvious that (whatever the goal) there is necessarily some way to achieve it in the allotted time? ------------------------------------------------------------------------------------ I think that at any stage there are supposed to be two people in each of 60 squares, with 4 empty squares. (Is that right, Ed?) (Another possible avenue is to use a symmetrtic network with exactly 120 nodes, like T_15.) --Dan

1 0

Re: [math-fun] Choreography of large complete graphs
by Jpr2718＠aol.com 30 Aug '06

30 Aug '06

Ed Pegg Jr. wrote: > At a convention, 120 people all need to meet with each other, 1 on 1. > Suppose a room is divided into a 8x8 square, and a clock is set > up to ring 60 times, once per minute. Each person is given a > map of their starting square, and a map of their route over the next > hour. Wouldn't each person need to have 119 meetings? Is it required that there are 60 meetings at a time (maybe this is optimal anyway)? John Robertson

1 0

Re: [math-fun] Possibly naive question about algebraic numbers
by Daniel Asimov 29 Aug '06

29 Aug '06

Thanks a lot, Franklin and Gareth. I suspected that conditions (*) and (**) together would result in algebraic coordinates, mainlly because I'd expect to have heard of these numbers if they defined a larger class. The theorem Gareth proves seems very basic, so I wonder if it has a name and attribution. Also, given such an integer-polynomial function P: C^n -> C^n such that its Jacobian at each of its roots Z in C^n is non-singular: QUESTION: What can be said about the size of the (necessarily) 0-dimensional set of such roots? (Where of course a root of P means a Z in C^n where P(Z) is the origin in C^n.) --Dan

2 1

RE: [math-fun] Possibly naive question about algebraic numbers
by Andy Latto 29 Aug '06

29 Aug '06

> -----Original Message----- > From: math-fun-bounces+andy.latto=pobox.com(a)mailman.xmission.com > [mailto:math-fun-bounces+andy.latto=pobox.com@mailman.xmission.com]On > Behalf Of Daniel Asimov > Sent: Monday, August 28, 2006 11:48 PM > To: math-fun > Subject: Re: [math-fun] Possibly naive question about > algebraic numbers > > Also, given such an integer-polynomial function > P: C^n -> C^n such that its Jacobian at each of its > roots Z in C^n is non-singular: > > QUESTION: What can be said about the size of the > (necessarily) 0-dimensional set of such roots? If you ask the question in projective space, rather than in affine space, you get the answer you want, which is that provided all the roots are simple (that is, the Jacobian is non-singular), then the number of roots is exactly equal to the product of the degrees of the polynomials. If you ask the question in affine space, the best you can say is that the number of roots is <= this product, because you have left out the "roots at infinity". In two dimensions, this goes by the name of Bezout's theorem. Andy Latto andy.latto(a)pobox.com > > (Where of course a root of P means a Z in C^n where > P(Z) is the origin in C^n.) > > --Dan > > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >

1 0