The solution for the 3D case is similar. Each pair of points defines a great circle across which the order of the points changes, so we need to count the number of disjoint regions on the surface of the sphere defined by all n(n-1)/2 great circles. That is, if we consider the graph whose vertices and edges are defined by the great circles, we need to count the faces. Again, it's easy to arrange for all the great circles to be distinct. The tricky part is that for every group of three points, all three of the great circles defined by pairs of those points intersect in just two vertices (each of degree six). However, we can arrange the points so that this is the only case in which more than two great circles intersect at a vertex. There are n(n-1)(n-2)/6 groups of three points, each of which defines two vertices of degree 6. There are n(n-1)(n-2)(n-3)/8 pairs of pairs of points, each of which defines two vertices of degree 4. So: V = n(n-1)(n-2)/3 + n(n-1)(n-2)(n-3)/4 = n(n-1)(n-2)(3n-5)/12 E = [6 * n(n-1)(n-2)/3 + 4 * n(n-1)(n-2)(n-3)/4] / 2 = n(n-1)(n-2)(n-1)/2 F = 2 + E - V = 2 + n(n-1)(n-2)(3n-1)/12 J.P. On Fri, Oct 17, 2014 at 9:28 PM, Tom Rokicki <rokicki@gmail.com> wrote:
I think the generalization to the marble (3D) case is actually pretty easy, and I think the upper bound is n(n-1)(n-2) for n>2. That's what I'm putting my money on.
-tom
On Fri, Oct 17, 2014 at 5:34 PM, Tom Rokicki <rokicki@gmail.com> wrote:
For the circle case: There are at most n(n-1)^2 points at which the order changes (when the line connecting two dots is horizontal). It is easy to arrange points so these all have distinct angles (with small perturbations). So for the 2D case, it's clearly n(n-1) distinct orders. It's easy to show if all the angles are distinct, the n(n-1) distinct orders are also.
I do not yet see how to generalize this to the 3D case.
On Fri, Oct 17, 2014 at 5:04 PM, Neil Sloane <njasloane@gmail.com> wrote:
For the "circle" version, a person who might know the answer is William (Tom) Trotter, trotter@math.gatech.edu - suggest you ask him. It's just his kind of problem - and he is, or was, the editor of the journal "Order"!
Best regards Neil
Neil J. A. Sloane, President, OEIS Foundation. 11 South Adelaide Avenue, Highland Park, NJ 08904, USA. Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ. Phone: 732 828 6098; home page: http://NeilSloane.com Email: njasloane@gmail.com
On Fri, Oct 17, 2014 at 7:00 PM, Charles Greathouse < charles.greathouse@case.edu> wrote:
I agree, great question. The 1-dimensiononal version is trivial, so the simplest (possibly-)nontrivial version is dimension 2 with a circle rolling on a 1-dimensional floor (I think we always want to assume codimension 1 here). So 1, 2, and 3 bubbles can be arranged in all ways by taking symmetrical positions, leaving 4 the first nontrivial case.The best I can do offhand is to put one in the center and three evenly spaced along the edge, which I think gives 12 orders.
But there are over 2000 entries in the OEIS starting 1, 2, 6, 12-24, = 12, ... so this isn't even enough to see if the sequence has been studied before.
Charles Greathouse Analyst/Programmer Case Western Reserve University
On Fri, Oct 17, 2014 at 6:02 PM, Dan Asimov <dasimov@earthlink.net> wrote:
What a great question!
--Dan
On Oct 17, 2014, at 2:54 PM, David Wilson <davidwwilson@comcast.net> wrote:
A clear marble has N tiny bubbles in it, numbered 1 through N.
Roll the marble on the floor, then list the bubbles in order of their distance from the floor (ignore situations where two or more bubbles are at the same height).
Given an optimal distribution of bubbles, what is the largest possible number of bubble orders you could record?
What if you measure distance from the contact point of the marble and floor?
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