From: Fred W. Helenius Subject: Re: geometry/combinatorics/whatever problem Date: Tue, 18 Jul 2000 18:33:35 -0400 Newsgroups: sci.math Summary: [missing] Jan Kristian Haugland wrote: >Let d(x, y) denote the distance between two points >x and y. >Is there an *odd* integer n >= 7 for which it is >possible to find n points p_0, p_1, ..., p_(n-1) >in Euclidean 3-space such that > d(p_i, p_(i+1)) = 1, d(p_i, p_(i+2)) = sqrt(3) >for each i = 0, 1, ..., n - 1 (the indices being >reduced modulo n) ? In other words, you have a bunch of unit lengths of pipe and 120-degree pipe elbows, and you want to know if an odd number of them can be joined in a ring. I looked for a solution with n = 7 by fixing three of the vertices at (-sqrt(3)/2,1/2,0), (0,0,0) and (sqrt(3)/2,1/2,0), assuming the remaining four were located in symmetric pairs with respect to the y-z plane, and solving numerically. One solution for the remaining points is (a,b,c), (1/2,d,e), (-1/2,d,e), (-a,b,c), where a = 1, b = 1.267949243... = sqrt(6)*sqrt(2 - sqrt(3)), c = 0.626342434... = 2/sqrt(5 + sqrt(27)), d = 1.300425797... , e = 1.491758675... . b and c were identified by the Inverse Symbolic Calculator, http://www.cecm.sfu.ca/projects/ISC/ ; d and e can in principle be determined in closed terms by solving a pair of simultaneous quadratics with coefficients involving b and c. -- Fred W. Helenius ============================================================================== From: haoyuep@aol.com (Dan Hoey) Subject: Re: geometry/combinatorics/whatever problem Date: 25 Jul 2000 03:10:22 GMT Newsgroups: sci.math Fred W. Helenius wrote: >In other words, you have a bunch of unit lengths of pipe and >120-degree pipe elbows, and you want to know if an odd number >of them can be joined in a ring. > >I looked for a solution with n = 7 by fixing three of the >vertices at (-sqrt(3)/2,1/2,0), (0,0,0) and (sqrt(3)/2,1/2,0), >assuming the remaining four were located in symmetric pairs with >respect to the y-z plane, and solving numerically. I did the solution symbolically with Mathematica, and got four solutions, approximately > {{0, 0, 0}, {.5, .86602, 0}, {1.26795, 1, .62634}, > {1.30042, .5, 1.49176}, {1.30042, -.5, 1.49176}, > {1.26795, -1, .62634}, {.5, -.86602, 0}}, > {{0, 0, 0}, {.5, .86602, 0}, {1.26795, 1, .62634}, > {2.12222, .5, .48416}, {2.12222, -.5, .48416}, > {1.26795, -1, .62634}, {.5, -.86602, 0}}, and the same two reflected in the x-y plane. (There are also four imaginary solutions). I'm interested, though, in what happens if we drop the requirement of symmetry. After fixing three vertices, we have a system of eleven equations in twelve unknowns, so there should be a set of one-parameter solutions. I haven't been able to get Mathematica to solve such a large system, though. I'm curious about what the topology of the configuration space is. For instance, is it connected? Dan Hoey Posted and e-mailed. [A previous version of this message was sent by mistake. I hope I cancelled it successfully.] ============================================================================== From: dtd@world.std.com (Don Davis) Subject: Re: geometry/combinatorics/whatever problem Date: Wed, 19 Jul 2000 01:52:56 GMT Newsgroups: sci.math In article , Jan Kristian Haugland wrote: > Is there an *odd* integer n >= 7 for which it is > possible to find n points p_0, p_1, ..., p_(n-1) > in Euclidean 3-space such that > > d(p_i, p_(i+1)) = 1, d(p_i, p_(i+2)) = sqrt(3) ? arrange your points along the edge of a m�bius strip, imbedded in the 3-space. for example, for 5 points: p_0 *====> .\ . . \ . . 1\ . . \. . * p_1 . /. . 1/ . . / . ./ . p_2 * . sqrt3 .\ . . \ . . 1\ . . \. . * p_3 . /. . 1/ . . / . ./ . p_4 * . sqrt3 .\ . . \ . . 1\ . . \. <====* p_0 it should be _possible_, but hard, to vary the points' spacings and the strip's width, so as to get exactly the distances you want. but it's straightforward to build a model strip from cardbard that does what you want. - don davis, boston ============================================================================== From: Jan Kristian Haugland Subject: Re: geometry/combinatorics/whatever problem Date: Wed, 19 Jul 2000 12:34:32 -0400 Newsgroups: sci.math On Wed, 19 Jul 2000, Don Davis wrote: > In article , Jan > Kristian Haugland wrote: > > > Is there an *odd* integer n >= 7 for which it is > > possible to find n points p_0, p_1, ..., p_(n-1) > > in Euclidean 3-space such that > > > > d(p_i, p_(i+1)) = 1, d(p_i, p_(i+2)) = sqrt(3) ? > > arrange your points along the edge of a m=F6bius strip, > imbedded in the 3-space. for example, for 5 points: From the requirements it is easy to prove that d(p_0, p_3) >= 2, and it is therefore impossible to have n <= 5. [quote of remainder of previous message deleted --djr] ============================================================================== From: dtd@world.std.com (Don Davis) Subject: Re: geometry/combinatorics/whatever problem Date: Thu, 20 Jul 2000 03:33:49 GMT Newsgroups: sci.math In article , Jan Kristian Haugland wrote: >> > [can one] find n points p_0, p_1, ..., p_(n-1) in [R3] >> > such that d(p_i, p_(i+1)) =1, d(p_i, p_(i+2)) =sqrt(3) ? >> >> arrange your points along the edge of a m�bius strip, >> imbedded in the 3-space. for example, for 5 points: ... > > ... it is easy to prove that d(p_0, p_3) >= 2, > and it is therefore impossible to have n <= 5. i drew 5 points just to keep the diagram small, but i failed to check whether the 5-point band will close. for 7 points, we have to check whether folding brings the band's ends together in the right orientation, so as to allow the band to close. from the diagram below, the 7-point band seems to close easily. - don davis, boston p_3 ,*---------------------*. p_4 // ,_-" \ ,'/ ,_-" `. / / ,_-"_, \ ,' | ,_-" "-_, `. / / ,_-" "-_, \ ,' |_-" "-_, `. / / |"-_, \ ,' | \ "-_,`. p_2 * / | * p_5 :\ | \ : :\ / | : : \ | \ : : \ / | : : \ \ : : \ | : : \ /\ : : \ / | : : \ / \ : : \ / : p_1 *.---___ / * p_6 `. """"---O O--"""/ .' `. | | / .' `. | | / .' `. | | / .' `. | | / .' `. | |/.' `* p_0 p_0 *' -