Re: [OpenJDK 2D-Dev] X11 uniform scaled wide lines and dashed lines; STROKE_CONTROL in Pisces
Correction... On 12/28/2010 3:00 PM, Jim Graham wrote: math. (Though it begs the question - is "-q/sqrt(-p^3)" more accurate than "-q/(p*sqrt(-p)"? If p is < 1 then the cube is an even smaller number, does that matter?) Make that "-q/(-p*sqrt(-p))", or "q/(p*sqrt(-p))"... ...jim
Re: [OpenJDK 2D-Dev] X11 uniform scaled wide lines and dashed lines; STROKE_CONTROL in Pisces
Hi Denis, I'm attaching a test program I wrote that compares the old and new algorithms. Obviously the old one missed a bunch of solutions because it classified all solutions as 1 or 3, but the new one also sometimes misses a solution. You might want to turn this into an automated test for the bug (and maybe use it as a stress test with a random number generator). I think one problem might be that you use "is close to zero" to check if you should use special processing. I think any tests which say "do it this way and get fewer roots" should be conservative and if we are on the borderline and we can do the code that generates more solutions then we should generate more and them maybe refine the roots and eliminate duplicates. That way we can be (more) sure not to leave any roots unsolved. The test as it is has a test case (I just chose random numbers to check and got lucky - d'oh!) that generates 1 solution from the new code even though the equation had 2 distinct solutions that weren't even near each other... ...jim CubicSolver.java Description: java/
Re: [OpenJDK 2D-Dev] X11 uniform scaled wide lines and dashed lines; STROKE_CONTROL in Pisces
Aha! I finally figured out what I was missing. On 12/24/2010 4:43 PM, Denis Lila wrote: Line 1133 - I don't understand why that term has -q in it. The above link and the original code both computed essentially the arccos of this Basically, the negative comes in when you push all of the p terms back inside the sqrt(). They had cos(3phi) = (3Q/2P)sqrt(-3/P) = (q/p)*sqrt(-1/p) = (-q/-p)*sqrt(-1/p) = -q*(-1/p)*sqrt(-1/p) = -q/(-p*sqrt(-p)) = -q/(sqrt(-p^3) which is what you calculate, so we are not calculating the negative of the angle after all - we are calculating the same angle using different math. (Though it begs the question - is "-q/sqrt(-p^3)" more accurate than "-q/(p*sqrt(-p)"? If p is < 1 then the cube is an even smaller number, does that matter?) Let X = (3q/2p)*sqrt(-3/p) where p and q are the ones from the wikipedia article, not our code. So, when we do ret[0] = t * cos(phi), that's: = t * cos(acos(-X)) actually, it really was t*cos(acos(X)), not -X. = t * cos(pi/3 - acos(X)) = t * cos(acos(X) - pi/3) = t * cos(acos(X) - pi/3 - pi) There was an error here in this step as well, this line should read: = t * cos(acos(X) - pi/3 - 2pi) = t * cos(acos(X) -7pi/3) = nothing because this isn't our math... ;-) I unfortunately don't have access to the icedtea servers at this moment, so I attached a patch. I hope that's ok. Have you updated the webrev yet? ...jim
