Question

I’m computing the ordinate y of a point on a line at a given abscissa x. The line is defined by its two end points coordinates (x0,y0)(x1,y1). End points coordinates are floats and the computation must be done in float precision for use in GPU.

The maths, and thus the naive implementation, are trivial.

Let t = (x – x0)/(x1 – x0), then y = (1 – t) * y0 + t * y1 = y0 + t * (y1 – y0).

The problem is when x1 – x0 is small. The result will introduce cancellation error. When combined with the one of x – x0, in the division I expect a significant error in t.

The question is if there exist another way to determine y with a better accuracy ?

i.e. should I compute (x – x0)*(y1 – y0) first, and divide by (x1 – x0) after ?

The difference y1 – y0 will always be big.

Answer 1

To a large degree, your underlying problem is fundamental. When (x1-x0) is small, it means there are only a few bits in the mantissa of x1 and x0 which differ. And by extension, there are only a limted number of floats between x0 and x1. E.g. if only the lower 4 bits of the mantissa differ, there are at most 14 values between them.

In your best algorithm, the t term represents these lower bits. And to continue or example, if x0 and x1 differ by 4 bits, then t can take on only 16 values either. The calculation of these possible values is fairly robust. Whether you’re calculating 3E0/14E0 or 3E-12/14E-12, the result is going to be close to the mathematical value of 3/14.

Your formula has the additional advantage of having y0 <= y <= y1, since 0 <= t <= 1

(I’m assuming that you know enough about float representations, and therefore “(x1-x0) is small” really means “small, relative to the values of x1 and x0 themselves”. A difference of 1E-1 is small when x0=1E3 but large if x0=1E-6 )