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Perl: the Markov chain saw

Re^2: computational efficiency

by thelycaeum (Initiate)
on Oct 10, 2011 at 19:00 UTC ( #930693=note: print w/replies, xml ) Need Help??

in reply to Re: computational efficiency
in thread computational efficiency

I've read the article a couple of times by now. Let me comment on that.

I've been planning to randomize my rotation subroutine by doing three subsequent arbitrary rotations about each axis in cartesian space, i.e. arbitrary rotation with independent randomization for each axis x, y and z. These however have to be randomized by two randomization processes each, as the sign of the cosine corresponding to the randomized sine has to randomized, too (if I'm not mistaken here, cf. my reply to moritz comment on correctness). Hence, my (thanks to your comments) improved idea would involve six randomizations for each arbitrary rotation, while the method posted on wolfram would involve only two (cf. equations 9 to 11 in the article), correct?

What I'm asking myself right now is the following: Moritz is absolutely correct, as a single rotation about one axis (which was my original question) results in omitting half of the angles, consequently incompleteness and bias towards rotation angles below 180. Would this still be true, if the routine is applied about each axis xyz? So, if Marsaglia (1972) uses two randomizations to produce an arbitrary point in three dimensional space in the polar coordinate system, would three randomizations in cartesian space still result in incompleteness? I think so, because equations 13 to 15 imply, that at least four randomizations are needed in cartesian space... ?

P.S.: MidLifeXis: Thank you, I found  use Benchmark qw(:all) ;on

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Re^3: computational efficiency
by Lotus1 (Curate) on Oct 12, 2011 at 18:53 UTC

    The approach I like for this problem is what Perlbotics suggested. Make a table. I suggest a hash of arrays where the keys are angles and each value is an array of sin and cos values. Then you choose a random angle (0-360) and get both sin and cos from a fast hash with the negative signs included. This takes care of all four quadrands.

    Do this for each axis of rotation as you proposed. The article I posted was about picking uniformly distributed points on the surface of a sphere which I believe is more difficult than your problem. The only point I was making with my previous post is that if you pick random sine values the corresponding angles are not uniformly distributed. I'm not sure about proving this approach of three rotations but it seems good to me. I'll think about it more and maybe I'll come up with a proof.

    use warnings; use strict; my $ref_trig = make_trigtable(); print "\nangle sine cosine\n"; foreach (1..15) { my $angle = sprintf "%.1f", rand(3600)/10; printf "%5.1f %8.5f %8.5f\n", $angle, $ref_trig->{$angle}[0], $ref +_trig->{$angle}[0]; } sub make_trigtable { use constant PI => 4 * atan2(1, 1); use constant DEG_TO_RAD => PI / 180; my %trig; foreach (0..3600) { my $angle = sprintf "%.1f", $_/10; $trig{$angle}=[sin $angle * DEG_TO_RAD, cos $angle * DEG_TO_RA +D]; } return \%trig; } __END__ C:\perlmonks>perl angle sine cosine 216.2 -0.59061 -0.59061 268.6 -0.99970 -0.99970 166.8 0.22835 0.22835 276.8 -0.99297 -0.99297 81.6 0.98927 0.98927 355.4 -0.08020 -0.08020 11.9 0.20620 0.20620 82.3 0.99098 0.99098 70.1 0.94029 0.94029 173.1 0.12014 0.12014 14.5 0.25038 0.25038 251.3 -0.94721 -0.94721 161.6 0.31565 0.31565 117.7 0.88539 0.88539 319.7 -0.64679 -0.64679

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