Fast quadratic minimizer

Question

Given a quadratic function, that is f(x) = ax^2 + bx + c, what is the fastest way to find x in [-1, 1] which minimizes f(x)?

So far this is the function I've come up with:

double QuadraticMinimizer(double a, double b, double c) {   
    double x = 1 - 2*(b > 0);
    if (a > 0) {
        x = -b/(2*a);
        if (fabs(x) > 1)
            x = Sign(x);
    }
    return x;
}

Is it possible to do better?

The minimum / maximum can be located at one of just 3 positions: one end point of the interval, or the place where f'(x) == 0, or x == -b / (2a) (may not be in the interval). Just check all of them. — Cong Ma, Sep 20 '15 at 18:26
@Cong Ma: Correct, that's what the above code snippet does. But it is possible to do it in FEWER steps? — user92382, Sep 20 '15 at 18:33
You have one step, how faster could any code be? Also, why a > 0 and not abs(a) > 0. — user1095108, Sep 20 '15 at 18:36
As in less multiplication/division, or less multiplication + division + comparison? — Cong Ma, Sep 20 '15 at 18:36
Why the heck is everyone so inexact here, OP didn't say he wanted real solutions. — user1095108, Sep 20 '15 at 18:43

score 1 · Answer 1 · 2015-09-21T13:32:31.197

There is no "fastest way" because the running time depends on the particular machine and the particular distribution of the input parameters. Also, there is not much that you can remove from the initial code.

If the location of the extremum -b/2a frequently falls outside of the interval [-1,1], you can avoid the division in those cases.

If you allow to hack the sign bit from the floating-point representation to implement fast abs, sgn and setsgn functions, you can use something like

a*= -2;
if (hack_abs(b) >= hack_abs(a))
  return hack_setsgn(1, hack_sgn(a) ^ hack_sgn(b));

return b / a;

You can also try with the more portable copysign function.

Fast quadratic minimizer

1 Answers1