quad2m (problem 3.2.1, negative)

Average Error: 34.2 → 10.4

Time: 5.1s

Precision: 64

Math

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

↓

\[\begin{array}{l} \mathbf{if}\;b_2 \le -4.4270058556435274 \cdot 10^{-117}:\\ \;\;\;\;\frac{-1}{2} \cdot \frac{c}{b_2}\\ \mathbf{elif}\;b_2 \le 2.49922826628406174 \cdot 10^{84}:\\ \;\;\;\;\frac{\left(-b_2\right) - \sqrt{b_2 \cdot b_2 - a \cdot c}}{a}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{2} \cdot \frac{c}{b_2} - 2 \cdot \frac{b_2}{a}\\ \end{array}\]

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 <= -4.4270058556435274e-117)) {
		temp = (-0.5 * (c / b_2));
	} else {
		double temp_1;
		if ((b_2 <= 2.4992282662840617e+84)) {
			temp_1 = ((-b_2 - sqrt(((b_2 * b_2) - (a * c)))) / a);
		} else {
			temp_1 = ((0.5 * (c / b_2)) - (2.0 * (b_2 / a)));
		}
		temp = temp_1;
	}
	return temp;
}

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 < -4.4270058556435274e-117

   1. Initial program 51.5

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

   2. Taylor expanded around -inf 11.0

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

   if -4.4270058556435274e-117 < b_2 < 2.4992282662840617e+84

   1. Initial program 12.4

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

   3. Applied div-inv 12.5

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

   5. Applied un-div-inv 12.4

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

   if 2.4992282662840617e+84 < b_2

   1. Initial program 43.1

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

   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 10.4

   \[\leadsto \begin{array}{l} \mathbf{if}\;b_2 \le -4.4270058556435274 \cdot 10^{-117}:\\ \;\;\;\;\frac{-1}{2} \cdot \frac{c}{b_2}\\ \mathbf{elif}\;b_2 \le 2.49922826628406174 \cdot 10^{84}:\\ \;\;\;\;\frac{\left(-b_2\right) - \sqrt{b_2 \cdot b_2 - a \cdot c}}{a}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{2} \cdot \frac{c}{b_2} - 2 \cdot \frac{b_2}{a}\\ \end{array}\]

Reproduce

herbie shell --seed 2020056 +o rules:numerics
(FPCore (a b_2 c)
  :name "quad2m (problem 3.2.1, negative)"
  :precision binary64
  (/ (- (- b_2) (sqrt (- (* b_2 b_2) (* a c)))) a))