Intersection probability for 'N' fixed-length rods in one- or two-dimensions - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T11:50:44Z http://mathoverflow.net/feeds/question/61404 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/61404/intersection-probability-for-n-fixed-length-rods-in-one-or-two-dimensions Intersection probability for 'N' fixed-length rods in one- or two-dimensions Rob Grey 2011-04-12T14:02:36Z 2011-04-12T16:39:54Z <p>Please consider the case where I have 'N' rods of length L (and width W) placed on a one- or two-dimensional surface with dimensions [0, A] in 1D, and [ [0, A], [0, B] ] in 2D. For the two-dimensional case, L and W are &lt;&lt; A or B. </p> <p>As a function of the number of rods N, and the relative dimensions of the rods and the surface on which they are placed, is there a reasonably easy derivation for the number of expected intersections between rods / the probability of an intersection occurring? I feel like this should have been solved somewhere in the literature, but I was unable to find anything.</p> <p>Edit - When I mentioned the rods should be 'placed', I failed to clarify that my treatment has been to assume that one end of each rod is placed with uniform random probability somewhere inside the specified dimensions of the one- or two-dimensional surface, the angle of the rod should be random, and any rod sections outside the bounded surface should either be treated or ignored depending on how easy it makes treating boundary conditions.</p> http://mathoverflow.net/questions/61404/intersection-probability-for-n-fixed-length-rods-in-one-or-two-dimensions/61418#61418 Answer by Anthony Quas for Intersection probability for 'N' fixed-length rods in one- or two-dimensions Anthony Quas 2011-04-12T16:20:16Z 2011-04-12T16:39:54Z <p>To compute the expected number of intersections, use the fact that expectation is additive. The expected number of intersections is just $\binom{N}{2}p$ where $p$ is the probability of an intersection. </p> <p>To compute the probability of an intersection you can make your life easier (with essentially no cost to the accuracy) by assuming your surface/line wraps around. This means there are no special places on your surface. Now you need to compute the probability that two rods intersect. </p> <p>In one dimension imagine the first rod is placed somewhere. What is the probability that the second rod intersects it? $2L/A$ (either the second rod is placed so its left end lies inside the first rod; or that its right end lies inside the first rod).</p> <p>In two dimensions, fix the position of the first rod. Then suppose the second rod is inclined at angle $\theta$ to the first rod (we can assume that $0&lt;\theta&lt;\pi/2$). We now need the area of the set of positions of the "top left" vertex of the second rod such that there is an intersection with the first rod. This set of positions is an octagon (whose area (as long as my geometry is correct) is $2L^2\sin\theta+2W^2\sin\theta+2WL(1+\cos\theta)$). (In the case where $W\ll L$ it reduces to a rhombus for which the area is just $L^2\sin\theta$). The intersection probability is then just $(2/\pi)$ times the integral of this area over $\theta$ divided by $AB$. i.e. $(4(L^2+W^2)/\pi+LW(2+4/\pi))/AB$</p>