quad2m (problem 3.2.1, negative)

Average Error: 32.8 → 10.2

Time: 30.9s

Precision: 64

Math

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

↓

\[\begin{array}{l} \mathbf{if}\;b_2 \le -3.031575300615258 \cdot 10^{-39}:\\ \;\;\;\;\frac{-1}{2} \cdot \frac{c}{b_2}\\ \mathbf{elif}\;b_2 \le 1.9089378078751267 \cdot 10^{+122}:\\ \;\;\;\;\frac{-b_2}{a} - \frac{\sqrt{b_2 \cdot b_2 - c \cdot a}}{a}\\ \mathbf{else}:\\ \;\;\;\;\frac{c}{b_2} \cdot \frac{1}{2} - 2 \cdot \frac{b_2}{a}\\ \end{array}\]

C

double f(double a, double b_2, double c) {
        double r676162 = b_2;
        double r676163 = -r676162;
        double r676164 = r676162 * r676162;
        double r676165 = a;
        double r676166 = c;
        double r676167 = r676165 * r676166;
        double r676168 = r676164 - r676167;
        double r676169 = sqrt(r676168);
        double r676170 = r676163 - r676169;
        double r676171 = r676170 / r676165;
        return r676171;
}

↓

double f(double a, double b_2, double c) {
        double r676172 = b_2;
        double r676173 = -3.031575300615258e-39;
        bool r676174 = r676172 <= r676173;
        double r676175 = -0.5;
        double r676176 = c;
        double r676177 = r676176 / r676172;
        double r676178 = r676175 * r676177;
        double r676179 = 1.9089378078751267e+122;
        bool r676180 = r676172 <= r676179;
        double r676181 = -r676172;
        double r676182 = a;
        double r676183 = r676181 / r676182;
        double r676184 = r676172 * r676172;
        double r676185 = r676176 * r676182;
        double r676186 = r676184 - r676185;
        double r676187 = sqrt(r676186);
        double r676188 = r676187 / r676182;
        double r676189 = r676183 - r676188;
        double r676190 = 0.5;
        double r676191 = r676177 * r676190;
        double r676192 = 2.0;
        double r676193 = r676172 / r676182;
        double r676194 = r676192 * r676193;
        double r676195 = r676191 - r676194;
        double r676196 = r676180 ? r676189 : r676195;
        double r676197 = r676174 ? r676178 : r676196;
        return r676197;
}

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 < -3.031575300615258e-39

   1. Initial program 53.3

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

   2. Taylor expanded around -inf 7.8

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

   if -3.031575300615258e-39 < b_2 < 1.9089378078751267e+122

   1. Initial program 13.7

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

   3. Applied div-sub 13.7

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

   if 1.9089378078751267e+122 < b_2

   1. Initial program 50.0

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

   2. Taylor expanded around inf 3.4

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

4. Final simplification 10.2

   \[\leadsto \begin{array}{l} \mathbf{if}\;b_2 \le -3.031575300615258 \cdot 10^{-39}:\\ \;\;\;\;\frac{-1}{2} \cdot \frac{c}{b_2}\\ \mathbf{elif}\;b_2 \le 1.9089378078751267 \cdot 10^{+122}:\\ \;\;\;\;\frac{-b_2}{a} - \frac{\sqrt{b_2 \cdot b_2 - c \cdot a}}{a}\\ \mathbf{else}:\\ \;\;\;\;\frac{c}{b_2} \cdot \frac{1}{2} - 2 \cdot \frac{b_2}{a}\\ \end{array}\]

Reproduce

herbie shell --seed 2019128 +o rules:numerics
(FPCore (a b_2 c)
  :name "quad2m (problem 3.2.1, negative)"
  (/ (- (- b_2) (sqrt (- (* b_2 b_2) (* a c)))) a))