quad2p (problem 3.2.1, positive)

Average Error: 34.0 → 10.1

Time: 6.3s

Precision: 64

Math

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

↓

\[\begin{array}{l} \mathbf{if}\;b_2 \le -1.257476678127677856918278287038350045718 \cdot 10^{107}:\\ \;\;\;\;\frac{1}{2} \cdot \frac{c}{b_2} - 2 \cdot \frac{b_2}{a}\\ \mathbf{elif}\;b_2 \le 4.90028183252923720758757892253110653773 \cdot 10^{-79}:\\ \;\;\;\;\frac{1}{\frac{a}{\sqrt{b_2 \cdot b_2 - a \cdot c} - b_2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{-1}{2} \cdot \frac{c}{b_2}\\ \end{array}\]

C

double f(double a, double b_2, double c) {
        double r21149 = b_2;
        double r21150 = -r21149;
        double r21151 = r21149 * r21149;
        double r21152 = a;
        double r21153 = c;
        double r21154 = r21152 * r21153;
        double r21155 = r21151 - r21154;
        double r21156 = sqrt(r21155);
        double r21157 = r21150 + r21156;
        double r21158 = r21157 / r21152;
        return r21158;
}

↓

double f(double a, double b_2, double c) {
        double r21159 = b_2;
        double r21160 = -1.2574766781276779e+107;
        bool r21161 = r21159 <= r21160;
        double r21162 = 0.5;
        double r21163 = c;
        double r21164 = r21163 / r21159;
        double r21165 = r21162 * r21164;
        double r21166 = 2.0;
        double r21167 = a;
        double r21168 = r21159 / r21167;
        double r21169 = r21166 * r21168;
        double r21170 = r21165 - r21169;
        double r21171 = 4.900281832529237e-79;
        bool r21172 = r21159 <= r21171;
        double r21173 = 1.0;
        double r21174 = r21159 * r21159;
        double r21175 = r21167 * r21163;
        double r21176 = r21174 - r21175;
        double r21177 = sqrt(r21176);
        double r21178 = r21177 - r21159;
        double r21179 = r21167 / r21178;
        double r21180 = r21173 / r21179;
        double r21181 = -0.5;
        double r21182 = r21181 * r21164;
        double r21183 = r21172 ? r21180 : r21182;
        double r21184 = r21161 ? r21170 : r21183;
        return r21184;
}

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 < -1.2574766781276779e+107

   1. Initial program 48.1

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

   2. Taylor expanded around -inf 3.2

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

   if -1.2574766781276779e+107 < b_2 < 4.900281832529237e-79

   1. Initial program 12.8

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

   3. Applied clear-num 12.9

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

   4. Simplified 12.9

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

   if 4.900281832529237e-79 < b_2

   1. Initial program 53.1

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

   2. Taylor expanded around inf 9.2

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

4. Final simplification 10.1

   \[\leadsto \begin{array}{l} \mathbf{if}\;b_2 \le -1.257476678127677856918278287038350045718 \cdot 10^{107}:\\ \;\;\;\;\frac{1}{2} \cdot \frac{c}{b_2} - 2 \cdot \frac{b_2}{a}\\ \mathbf{elif}\;b_2 \le 4.90028183252923720758757892253110653773 \cdot 10^{-79}:\\ \;\;\;\;\frac{1}{\frac{a}{\sqrt{b_2 \cdot b_2 - a \cdot c} - b_2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{-1}{2} \cdot \frac{c}{b_2}\\ \end{array}\]

Reproduce

herbie shell --seed 2020001 +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))