stolfi@jumbo.DEC.COM (Jorge Stolfi) (11/01/86)
Larry Bruns asked: > > If three distinct points are chosen randomly anywhere in space, > what is the probability that the resulting triangle is acute? > (i.e. that none of the 3 angles is greater than 90 degrees). As others pointed out, "choosing a point randomly anywhere in space" is an ill-defined notion: there is no probability distribution that is uniform over the whole space. This is not by itself a fatal flaw. For some problems of this type one can get a reasonable answer by a limiting process, as suggested by David Moews: 1. Let P(x) be a generic probability distribution on three-space, and d a generic "scale factor" 2. Compute the answer for the distribution P(x) "spread out" by a scale factor d (that is, for the distribution (1/d**3) P(x/d) ), as a function of d and P 3. Find the limit as d goes to infinity If the limit does not depend on P, then you may reasonably take that as the answer for an "uniform distribution over all space". Unfortunately, this does not work for the proposed problem. Scaling a triangle by a factor of d does not change its angles; therefore, the answer computed in step 2 will depend on P but NOT on d, and the same would be true of the limit. By varying P, you can get any probability between 0% and 100%. It would still be interesting to compute the probability for some common distributions (Gaussian, uniform inside a sphere, in a cube, on the surface of a sphere, etc.). Note that we can assume zero mean and unit variance, without loss of generality. It would be interesting also to compute the maximum and minimum probability for some restricted classes of distributions (say, spherically symmetric, or symmetric and unimodal). Mark Leeper and Spiros Boucouris concluded the probability is zero by reasoning roughly as follows: > > It really doesn't make much difference how you pick the first > two points, call them A and B. Take the sphere whose axis is > the segment from A to B and the two parallel planes tangent to > the sphere at A and B. Elementary geometry tells us the > triangle will be acute if the third point, C, is in the region > Z(A,B) between the two planes but outside the sphere. > That is an infinitesimally small fraction of the whole space, > hence the probability that C is between the two planes goes to > zero. Needless to say, this argument does not work for any meaningful (non-uniform) distribution P. The positions of A and B DO matter (some pairs are more likely than others), and the probability of C being in the stated region is NOT zero (it depends on A and B). Someone also said that since the triangle is planar, we can as well solve the same problem on the plane. This too is incorrect. Even if we take "similar" distributions on two and three dimensions (for example, uniform over a disk vs. uniform inside a ball), most attributes of a random triangle will have different distributions. For example, the vertices are much more likely to be near the boundary of the region in three dimensions than in two. The correct answer is of course given by the integral / / / | | | | dA | dB | dC P(A)P(B)P(C) | | | / / / R^3 R^3 Z(A,B) where Z(A,B) is the region described above. Any volunteers? jorge ------------------------------------------------------------ DISCLAIMER: The opinions above are not those of my computer.
cc@locus.ucla.edu (Mitch) (11/04/86)
Maybe I should have caught up on this newsgroup before rushing off that last reply. I seem to be the only one to interpret "space" in the original posting as referring to the final frontier. If we must think of it in terms of pure geometry, why this silly preference for a space with zero curvature? If we assume positive curvature, as for Riemannian spherical geometry, we lend an entirely new twist to the problem. Seriously, consider three points chosen in such a space - I think the problem is a little more interesting. Mitch Gunzler --- and then I said, "alright, what IS the 3n+1 problem?"
colonel@sunybcs.UUCP (Col. G. L. Sicherman) (11/05/86)
> If we must think of it in terms of pure geometry, why this silly > preference for a space with zero curvature? If we assume positive > curvature, as for Riemannian spherical geometry, we lend an entirely > new twist to the problem. Why consider 3-space at all? I would guess that the original problem was meant for 2-space. So how about 3 random vertices on a sphere? That's meaningful. Shouldn't be too hard to figure out the integral. -- Col. G. L. Sicherman UU: ...{rocksvax|decvax}!sunybcs!colonel CS: colonel@buffalo-cs BI: colonel@sunybcs, csdsiche@sunyabvc
sewilco@mecc.UUCP (Scot E. Wilcoxon) (11/05/86)
In article <486@jumbo.DEC.COM> stolfi@jumbo.UUCP (Jorge Stolfi) writes: >Larry Bruns asked: >> >> If three distinct points are chosen randomly anywhere in space, >> what is the probability that the resulting triangle is acute? >> (i.e. that none of the 3 angles is greater than 90 degrees). > I missed the original posting, but isn't this the same as "what is the ratio between { the range of three angles which are in acute triangle } and { the range of three angles which are in triangles which are not acute } ?" Ignore the distance and just look at the possible angles. -- Scot E. Wilcoxon Minn Ed Comp Corp {quest,dayton,meccts}!mecc!sewilco 45 03 N 93 08 W (612)481-3507 ihnp4!meccts!mecc!sewilco Laws are society's common sense, recorded for the stupid. The alert question everything, and most laws are obvious to them.
colonel@sunybcs.UUCP (Col. G. L. Sicherman) (11/06/86)
> So how about 3 random vertices on a sphere? That's meaningful. Shouldn't > be too hard to figure out the integral. And I'll bet it comes out less than 1/4. -- Col. G. L. Sicherman UU: ...{rocksvax|decvax}!sunybcs!colonel CS: colonel@buffalo-cs BI: colonel@sunybcs, csdsiche@sunyabvc