Solving floating-point rounding issues C++

Question

I develop a scientific application (simulation of chromosomes moving in a cell nucleus). The chromosomes are divided in small fragments that rotate around a random axis using 4x4 rotation matrices.

The problem is that the simulation performs hundreds of billions of rotations, therefore the floating-point rounding errors stack up and grow exponentially, so the fragments tend to "float away" and detach from the rest of the chromosome as time passes.

I use double precision with C++. The soft runs on CPU for the moment but will be ported for CUDA, and simulations can last for 1 month at most.

I have no idea how I could somehow renormalize the chromosome, because all fragments are chained together (you can see it as a doubly linked-list), but I think that would be the best idea, if possible.

Do you have any suggestions ? I feel a bit lost.

Thank you very much,

H.

EDIT: Added a simplified sample code. You can assume all matrix math are classical implementations.

// Rotate 1000000 times
for (int i = 0; i < 1000000; ++i)
{
    // Pick a random section start
    int istart = rand() % chromosome->length;

    // Pick the end 20 segments further (cyclic)
    int iend = (istart + 20) % chromosome->length;

    // Build rotation axis
    Vector4 axis = chromosome->segments[istart].position - chromosome->segments[iend].position;
    axis.normalize();

    // Build rotation matrix and translation vector
    Matrix4 rotm(axis, rand() / float(RAND_MAX));
    Vector4 oldpos = chromosome->segments[istart].position;

    // Rotate each segment between istart and iend using rotm
    for (int j = (istart + 1) % chromosome->length; j != iend; ++j, j %= chromosome->length)
    {
        chromosome->segments[j].position -= oldpos;
        chromosome->segments[j].position.transform(rotm);
        chromosome->segments[j].position += oldpos;
    }
}

Answer 1

You need to find some constraint for your system and work to keep that within some reasonable bounds. I've done a bunch of molecular collision simulations and in those systems the total energy is conserved, so every step I double check the total energy of the system and if it varies by some threshold, then I know that my time step was poorly chosen (too big or too small) and I pick a new time step and rerun it. That way I can keep track of what's happening to the system in real time.

For this simulation, I don't know what conserved quantity you have, but if you have one, you can try to keep that constant. Remember, making your time step smaller doesn't always increase the accuracy, you need to optimize the step size with the amount of precision you have. I've had numerical simulations run for weeks of CPU time and conserved quantities were always within 1 part in 10^8, so it is possible, you just need to play around some.

Also, as Tomalak said, maybe try to always reference your system to the start time rather than to the previous step. So rather than always moving your chromosomes keep the chromosomes at their start place and store with them a transformation matrix that gets you to the current location. When you compute your new rotation, just modify the transformation matrix. It may seem silly, but sometimes this works well because the errors average out to 0.

For example, lets say I have a particle that sits at (x,y) and every step I calculate (dx, dy) and move the particle. The step-wise way would do this

t0 (x0,y0)
t1 (x0,y0) + (dx,dy) -> (x1, y1)
t2 (x1,y1) + (dx,dy) -> (x2, y2)
t3 (x2,y2) + (dx,dy) -> (x3, y3)
t4 (x3,30) + (dx,dy) -> (x4, y4)
...

If you always reference to t0, you could do this

t0 (x0, y0) (0, 0)
t1 (x0, y0) (0, 0) + (dx, dy) -> (x0, y0) (dx1, dy1)
t2 (x0, y0) (dx1, dy1) + (dx, dy) -> (x0, y0) (dx2, dy2)
t3 (x0, y0) (dx2, dy2) + (dx, dy) -> (x0, y0) (dx3, dy3)

So at any time, tn, to get your real position you have to do (x0, y0) + (dxn, dyn)

Now for simple translation like my example, you're probably not going to win very much. But for rotation, this can be a life saver. Just keep a matrix with the Euler angles associated with each chromosome and update that rather than the actual position of the chromosome. At least this way they won't float away.

Answer 2

Write your formulae so that the data for timestep T does not derive solely from the floating-point data in timestep T-1. Try to ensure that the production of floating-point errors is limited to a single timestep.

It's hard to say anything more specific here without a more specific problem to solve.