quad2p (problem 3.2.1, positive)

Average Error: 34.1 → 9.8

Time: 22.6s

Precision: 64

Math

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

↓

\[\begin{array}{l} \mathbf{if}\;b_2 \le -1.008402374157600307221015587992064225208 \cdot 10^{154}:\\ \;\;\;\;\mathsf{fma}\left(\frac{c}{b_2}, \frac{1}{2}, \frac{b_2}{a} \cdot -2\right)\\ \mathbf{elif}\;b_2 \le 1.61145084478121505718169973575148582501 \cdot 10^{-34}:\\ \;\;\;\;\frac{\sqrt{b_2 \cdot b_2 - a \cdot c}}{a} - \frac{b_2}{a}\\ \mathbf{else}:\\ \;\;\;\;\frac{-1}{2} \cdot \frac{c}{b_2}\\ \end{array}\]

C

double f(double a, double b_2, double c) {
        double r24529 = b_2;
        double r24530 = -r24529;
        double r24531 = r24529 * r24529;
        double r24532 = a;
        double r24533 = c;
        double r24534 = r24532 * r24533;
        double r24535 = r24531 - r24534;
        double r24536 = sqrt(r24535);
        double r24537 = r24530 + r24536;
        double r24538 = r24537 / r24532;
        return r24538;
}

↓

double f(double a, double b_2, double c) {
        double r24539 = b_2;
        double r24540 = -1.0084023741576003e+154;
        bool r24541 = r24539 <= r24540;
        double r24542 = c;
        double r24543 = r24542 / r24539;
        double r24544 = 0.5;
        double r24545 = a;
        double r24546 = r24539 / r24545;
        double r24547 = -2.0;
        double r24548 = r24546 * r24547;
        double r24549 = fma(r24543, r24544, r24548);
        double r24550 = 1.611450844781215e-34;
        bool r24551 = r24539 <= r24550;
        double r24552 = r24539 * r24539;
        double r24553 = r24545 * r24542;
        double r24554 = r24552 - r24553;
        double r24555 = sqrt(r24554);
        double r24556 = r24555 / r24545;
        double r24557 = r24556 - r24546;
        double r24558 = -0.5;
        double r24559 = r24558 * r24543;
        double r24560 = r24551 ? r24557 : r24559;
        double r24561 = r24541 ? r24549 : r24560;
        return r24561;
}

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 < -1.0084023741576003e+154

   1. Initial program 64.0

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

   2. Simplified 64.0

      \[\leadsto \color{blue}{\frac{\sqrt{b_2 \cdot b_2 - a \cdot c} - b_2}{a}}\]

   3. Taylor expanded around -inf 1.7

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

   4. Simplified 1.7

      \[\leadsto \color{blue}{\mathsf{fma}\left(\frac{c}{b_2}, \frac{1}{2}, \frac{b_2}{a} \cdot -2\right)}\]

   if -1.0084023741576003e+154 < b_2 < 1.611450844781215e-34

   1. Initial program 13.5

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

   2. Simplified 13.5

      \[\leadsto \color{blue}{\frac{\sqrt{b_2 \cdot b_2 - a \cdot c} - b_2}{a}}\]
   3. Using strategy rm

   4. Applied div-sub 13.5

      \[\leadsto \color{blue}{\frac{\sqrt{b_2 \cdot b_2 - a \cdot c}}{a} - \frac{b_2}{a}}\]

   if 1.611450844781215e-34 < b_2

   1. Initial program 55.0

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

   2. Simplified 55.0

      \[\leadsto \color{blue}{\frac{\sqrt{b_2 \cdot b_2 - a \cdot c} - b_2}{a}}\]

   3. Taylor expanded around inf 7.0

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

4. Final simplification 9.8

   \[\leadsto \begin{array}{l} \mathbf{if}\;b_2 \le -1.008402374157600307221015587992064225208 \cdot 10^{154}:\\ \;\;\;\;\mathsf{fma}\left(\frac{c}{b_2}, \frac{1}{2}, \frac{b_2}{a} \cdot -2\right)\\ \mathbf{elif}\;b_2 \le 1.61145084478121505718169973575148582501 \cdot 10^{-34}:\\ \;\;\;\;\frac{\sqrt{b_2 \cdot b_2 - a \cdot c}}{a} - \frac{b_2}{a}\\ \mathbf{else}:\\ \;\;\;\;\frac{-1}{2} \cdot \frac{c}{b_2}\\ \end{array}\]

Reproduce

herbie shell --seed 2019325 +o rules:numerics
(FPCore (a b_2 c)
  :name "quad2p (problem 3.2.1, positive)"
  :precision binary64
  (/ (+ (- b_2) (sqrt (- (* b_2 b_2) (* a c)))) a))