quad2p (problem 3.2.1, positive)

Average Error: 34.1 → 9.2

Time: 4.5s

Precision: 64

Math

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

↓

\[\begin{array}{l} \mathbf{if}\;b_2 \le -2.37749702272254886 \cdot 10^{101}:\\ \;\;\;\;\frac{1}{2} \cdot \frac{c}{b_2} - 2 \cdot \frac{b_2}{a}\\ \mathbf{elif}\;b_2 \le 1.9238883452280037 \cdot 10^{-130}:\\ \;\;\;\;\frac{\sqrt{b_2 \cdot b_2 - a \cdot c} + \left(-b_2\right)}{a}\\ \mathbf{elif}\;b_2 \le 4.01993084419163312 \cdot 10^{109}:\\ \;\;\;\;\frac{\frac{0 + a \cdot c}{\left(-b_2\right) - \sqrt{b_2 \cdot b_2 - a \cdot c}}}{a}\\ \mathbf{else}:\\ \;\;\;\;\frac{-1}{2} \cdot \frac{c}{b_2}\\ \end{array}\]

C

double code(double a, double b_2, double c) {
	return ((-b_2 + sqrt(((b_2 * b_2) - (a * c)))) / a);
}

↓

double code(double a, double b_2, double c) {
	double temp;
	if ((b_2 <= -2.377497022722549e+101)) {
		temp = ((0.5 * (c / b_2)) - (2.0 * (b_2 / a)));
	} else {
		double temp_1;
		if ((b_2 <= 1.9238883452280037e-130)) {
			temp_1 = ((sqrt(((b_2 * b_2) - (a * c))) + -b_2) / a);
		} else {
			double temp_2;
			if ((b_2 <= 4.019930844191633e+109)) {
				temp_2 = (((0.0 + (a * c)) / (-b_2 - sqrt(((b_2 * b_2) - (a * c))))) / a);
			} else {
				temp_2 = (-0.5 * (c / b_2));
			}
			temp_1 = temp_2;
		}
		temp = temp_1;
	}
	return temp;
}

Error

Bits error versus a

Bits error versus b_2

Bits error versus c

Derivation

1. Split input into 4 regimes

2. if b_2 < -2.377497022722549e+101

   1. Initial program 47.1

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

   2. Taylor expanded around -inf 3.6

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

   if -2.377497022722549e+101 < b_2 < 1.9238883452280037e-130

   1. Initial program 11.9

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

   3. Applied +-commutative 11.9

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

   if 1.9238883452280037e-130 < b_2 < 4.019930844191633e+109

   1. Initial program 40.3

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

   3. Applied flip-+ 40.3

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

   4. Simplified 15.5

      \[\leadsto \frac{\frac{\color{blue}{0 + a \cdot c}}{\left(-b_2\right) - \sqrt{b_2 \cdot b_2 - a \cdot c}}}{a}\]

   if 4.019930844191633e+109 < b_2

   1. Initial program 59.9

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

   2. Taylor expanded around inf 2.4

      \[\leadsto \color{blue}{\frac{-1}{2} \cdot \frac{c}{b_2}}\]
3. Recombined 4 regimes into one program.

4. Final simplification 9.2

   \[\leadsto \begin{array}{l} \mathbf{if}\;b_2 \le -2.37749702272254886 \cdot 10^{101}:\\ \;\;\;\;\frac{1}{2} \cdot \frac{c}{b_2} - 2 \cdot \frac{b_2}{a}\\ \mathbf{elif}\;b_2 \le 1.9238883452280037 \cdot 10^{-130}:\\ \;\;\;\;\frac{\sqrt{b_2 \cdot b_2 - a \cdot c} + \left(-b_2\right)}{a}\\ \mathbf{elif}\;b_2 \le 4.01993084419163312 \cdot 10^{109}:\\ \;\;\;\;\frac{\frac{0 + a \cdot c}{\left(-b_2\right) - \sqrt{b_2 \cdot b_2 - a \cdot c}}}{a}\\ \mathbf{else}:\\ \;\;\;\;\frac{-1}{2} \cdot \frac{c}{b_2}\\ \end{array}\]

Reproduce

herbie shell --seed 2020060 +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))