Timeline for How can I randomly draw an ensemble of unit vectors that sum to zero?
Current License: CC BY-SA 3.0
18 events
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| Dec 10 at 1:32 | answer | added | Jason Cantarella | timeline score: 2 | |
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Oct 23, 2013 at 10:22 | history | edited | Ricardo Andrade |
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| Oct 23, 2013 at 10:17 | history | edited | Ricardo Andrade |
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| May 22, 2013 at 10:31 | comment | added | j.c. | @Joseph O'Rourke The paper can be accessed here: citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.44.6191 However, it addresses the number of components of the space of embeddings (i.e. which types of knots are possible and how they may be realized) and not the relative measure of these components (how probable each knot type would be). | |
| May 21, 2013 at 17:47 | comment | added | Joseph O'Rourke | I cannot access this paper at the moment, but its title suggests it might be relevant for the hexagonal trefoil: J. A. Calvo, The embedding space of hexagonal knots, Topolo. Appl. 112(2) (2001) 137–174. | |
| May 21, 2013 at 14:45 | comment | added | Jon Peterson | If $n$ is large, then the distribution of the path should converge to that of a $d$-dimensional Brownian bridge (essentially a Brownian motion conditioned to end back at the origin). | |
| May 21, 2013 at 13:20 | comment | added | Dustin G. Mixon | Apparently, the $k$-dimensional Hausdorff measure on a set of dimension $k$ is the "uniform" measure on that set, and my question concerns the uniform distribution on a $(d-1)(n-1)$-dimensional variety in $\mathbb{R}^{dn}$. | |
| May 21, 2013 at 12:32 | history | edited | Dustin G. Mixon | CC BY-SA 3.0 |
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| May 20, 2013 at 11:48 | answer | added | j.c. | timeline score: 9 | |
| May 20, 2013 at 2:14 | answer | added | Joseph O'Rourke | timeline score: 12 | |
| May 20, 2013 at 1:58 | comment | added | Alekk | suppose that you have computed the densities $d_2(\cdot)$ and $d_3(\cdot)$ of the law of the sum of $2$ and $3$ unit vectors. You can first simulate the sum $\sum_1^3 v_i$ conditionally on $\sum_1^6 v_i=0$ by simulating from a distribution proportional to $d_3^2$ (rejection sampling). You can then simulate $v_1$ and $\sum_1^5 v_i$ conditionally on the value of $X$ by simulating from a density on the unit sphere that is proportional to $d_2(X-v)$ (again, rejection sampling). Remains then to simulate $\sum_1^2 v_i$ and $\sum_1^4 v_i$ by the same approach. | |
| May 19, 2013 at 22:03 | comment | added | Yoav Kallus | @Carlo, since Allen says "I expect this would converge quickly," I take it to be implied that the process is iterated. | |
| May 19, 2013 at 21:30 | comment | added | Carlo Beenakker | @Allen, wouldn't the rescaling to unit length spoil the sum to zero? | |
| May 19, 2013 at 21:06 | comment | added | Aaron Meyerowitz | For your first question you have an ordered list of $6$ unit vectors in general position (seems right to assume that). In any of the $6!$ orders the sum is zero. We can start anywhere and traverse the knot in either order so there are essentially $60$ knots. I wonder how many of them could be a trefoil. Might be an easy question. The easiest non-trivial case is $5$ unit vectors in $\mathbb{R}^2.$ | |
| May 19, 2013 at 18:14 | comment | added | Allen Knutson | Idea: pick $n$ randomly, subtract $1/n \sum v_i$ from each, then rescale each back to unit. I expect this would converge quickly, but don't have any reason to believe that it gets you exactly to the uniform distribution. | |
| May 19, 2013 at 17:27 | comment | added | Yoav Kallus | With unit vectors, I think this would be very hard. But if you're willing to go with Gaussians, I think you can calculate the conditional probability distributions without much trouble. | |
| May 19, 2013 at 17:22 | history | asked | Dustin G. Mixon | CC BY-SA 3.0 |