Quadratic Equation and the Evil of Floating Point

April 29, 2018

Today I came across an interesting discussion of floating point arithmetic by John D Cook: The quadratic formula and low-precision arithmetic. The take away message is: “when the linear coefficient b is large relative to the other coefficients, the quadratic formula can give wrong results when implemented in floating point arithmetic.”

It made me think that floating-point arithmetic is hard, really hard, and that we must take care of very fine details, even when working with simple arithmetic expression like quadratic equation. I had to search my (very old) archives of Pascal and C code to find an illustration of the above trap (substracting nearly equal numbers) and how to alleviate it.

The following snippet (full C code, trinome.c) computes the two real or complex roots of a second degree polynomial. All variables are declared as float and EPS is simply defined as static double EPS = 1E-10. Note that we only compute one of the two roots and use Vieta’s formulas to compute the second real root. Given the quadratic equation $ax^2 + bx + c = 0$, the roots $x_1$ and $x_2$ satisfy the following two relations: $x_1 + x_2 = -\tfrac{b}{a}$ and $ x_1x_2 = \tfrac{c}{a}$. The first root is computed according to the sign of $b$:

delta = b*b - 4*a*c;
if (delta >= 0) {
  if (b > 0)
    r1 = -(b+sqrt(delta))/(2*a);
    r1 = (-b+sqrt(delta))/(2*a);
  r2 = abs(r1) < EPS ? 0 : c/(a*r1);
  i1 = i2 = 0;
else {
  r1 = r2 = -b/(2*a);
  i1 = sqrt(-delta)/(2*a);
  i2 = -i1;

As a I said, floating-point arithmetic is hard, even when it is visually explained (HN thread).

Other ressources of potential interest:

♪ Bill Callahan • Sometimes I Wish We Were an Eagle

See Also

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