quad2m (problem 3.2.1, negative)

Average Error: 34.7 → 10.2

Time: 11.6s

Precision: 64

Math

\[\frac{\left(-b_2\right) - \sqrt{b_2 \cdot b_2 - a \cdot c}}{a}\]

↓

\[\begin{array}{l} \mathbf{if}\;b_2 \le -4.6537017569063518 \cdot 10^{-82}:\\ \;\;\;\;\frac{-1}{2} \cdot \frac{c}{b_2}\\ \mathbf{elif}\;b_2 \le 1.0479007947857462 \cdot 10^{99}:\\ \;\;\;\;\frac{1}{\frac{a}{\left(-b_2\right) - \sqrt{b_2 \cdot b_2 - a \cdot c}}}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{2} \cdot \frac{c}{b_2} - 2 \cdot \frac{b_2}{a}\\ \end{array}\]

C

double f(double a, double b_2, double c) {
        double r45538 = b_2;
        double r45539 = -r45538;
        double r45540 = r45538 * r45538;
        double r45541 = a;
        double r45542 = c;
        double r45543 = r45541 * r45542;
        double r45544 = r45540 - r45543;
        double r45545 = sqrt(r45544);
        double r45546 = r45539 - r45545;
        double r45547 = r45546 / r45541;
        return r45547;
}

↓

double f(double a, double b_2, double c) {
        double r45548 = b_2;
        double r45549 = -4.653701756906352e-82;
        bool r45550 = r45548 <= r45549;
        double r45551 = -0.5;
        double r45552 = c;
        double r45553 = r45552 / r45548;
        double r45554 = r45551 * r45553;
        double r45555 = 1.0479007947857462e+99;
        bool r45556 = r45548 <= r45555;
        double r45557 = 1.0;
        double r45558 = a;
        double r45559 = -r45548;
        double r45560 = r45548 * r45548;
        double r45561 = r45558 * r45552;
        double r45562 = r45560 - r45561;
        double r45563 = sqrt(r45562);
        double r45564 = r45559 - r45563;
        double r45565 = r45558 / r45564;
        double r45566 = r45557 / r45565;
        double r45567 = 0.5;
        double r45568 = r45567 * r45553;
        double r45569 = 2.0;
        double r45570 = r45548 / r45558;
        double r45571 = r45569 * r45570;
        double r45572 = r45568 - r45571;
        double r45573 = r45556 ? r45566 : r45572;
        double r45574 = r45550 ? r45554 : r45573;
        return r45574;
}

Error

Bits error versus a

Bits error versus b_2

Bits error versus c

Derivation

1. Split input into 3 regimes

2. if b_2 < -4.653701756906352e-82

   1. Initial program 52.8

      \[\frac{\left(-b_2\right) - \sqrt{b_2 \cdot b_2 - a \cdot c}}{a}\]

   2. Taylor expanded around -inf 9.1

      \[\leadsto \color{blue}{\frac{-1}{2} \cdot \frac{c}{b_2}}\]

   if -4.653701756906352e-82 < b_2 < 1.0479007947857462e+99

   1. Initial program 13.4

      \[\frac{\left(-b_2\right) - \sqrt{b_2 \cdot b_2 - a \cdot c}}{a}\]
   2. Using strategy rm

   3. Applied clear-num 13.5

      \[\leadsto \color{blue}{\frac{1}{\frac{a}{\left(-b_2\right) - \sqrt{b_2 \cdot b_2 - a \cdot c}}}}\]

   if 1.0479007947857462e+99 < b_2

   1. Initial program 47.6

      \[\frac{\left(-b_2\right) - \sqrt{b_2 \cdot b_2 - a \cdot c}}{a}\]

   2. Taylor expanded around inf 4.1

      \[\leadsto \color{blue}{\frac{1}{2} \cdot \frac{c}{b_2} - 2 \cdot \frac{b_2}{a}}\]
3. Recombined 3 regimes into one program.

4. Final simplification 10.2

   \[\leadsto \begin{array}{l} \mathbf{if}\;b_2 \le -4.6537017569063518 \cdot 10^{-82}:\\ \;\;\;\;\frac{-1}{2} \cdot \frac{c}{b_2}\\ \mathbf{elif}\;b_2 \le 1.0479007947857462 \cdot 10^{99}:\\ \;\;\;\;\frac{1}{\frac{a}{\left(-b_2\right) - \sqrt{b_2 \cdot b_2 - a \cdot c}}}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{2} \cdot \frac{c}{b_2} - 2 \cdot \frac{b_2}{a}\\ \end{array}\]

Reproduce

herbie shell --seed 2020065 
(FPCore (a b_2 c)
  :name "quad2m (problem 3.2.1, negative)"
  :precision binary64
  (/ (- (- b_2) (sqrt (- (* b_2 b_2) (* a c)))) a))