Average Error: 11.3 → 5.7
Time: 3.5s
Precision: binary64
\[\frac{a1 \cdot a2}{b1 \cdot b2}\]
↓
\[\begin{array}{l} \mathbf{if}\;a1 \cdot a2 \le -3.3079264831196082 \cdot 10^{194} \lor \neg \left(a1 \cdot a2 \le -5.632465143843106 \cdot 10^{-172} \lor \neg \left(a1 \cdot a2 \le 3.7907289754606943 \cdot 10^{-297}\right) \land a1 \cdot a2 \le 4.11512080060718968 \cdot 10^{280}\right):\\ \;\;\;\;a2 \cdot \left(\frac{1}{b1} \cdot \frac{a1}{b2}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{b2 \cdot \frac{b1}{a1 \cdot a2}}\\ \end{array}\]
\frac{a1 \cdot a2}{b1 \cdot b2}
↓
\begin{array}{l}
\mathbf{if}\;a1 \cdot a2 \le -3.3079264831196082 \cdot 10^{194} \lor \neg \left(a1 \cdot a2 \le -5.632465143843106 \cdot 10^{-172} \lor \neg \left(a1 \cdot a2 \le 3.7907289754606943 \cdot 10^{-297}\right) \land a1 \cdot a2 \le 4.11512080060718968 \cdot 10^{280}\right):\\
\;\;\;\;a2 \cdot \left(\frac{1}{b1} \cdot \frac{a1}{b2}\right)\\

\mathbf{else}:\\
\;\;\;\;\frac{1}{b2 \cdot \frac{b1}{a1 \cdot a2}}\\

\end{array}
double code(double a1, double a2, double b1, double b2) {
	return ((double) (((double) (a1 * a2)) / ((double) (b1 * b2))));
}
↓
double code(double a1, double a2, double b1, double b2) {
	double VAR;
	if (((((double) (a1 * a2)) <= -3.307926483119608e+194) || !((((double) (a1 * a2)) <= -5.632465143843106e-172) || (!(((double) (a1 * a2)) <= 3.790728975460694e-297) && (((double) (a1 * a2)) <= 4.11512080060719e+280))))) {
		VAR = ((double) (a2 * ((double) (((double) (1.0 / b1)) * ((double) (a1 / b2))))));
	} else {
		VAR = ((double) (1.0 / ((double) (b2 * ((double) (b1 / ((double) (a1 * a2))))))));
	}
	return VAR;
}

Error

Bits error versus a1

Bits error versus a2

Bits error versus b1

Bits error versus b2

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original11.3
Target11.4
Herbie5.7
\[\frac{a1}{b1} \cdot \frac{a2}{b2}\]

Derivation

  1. Split input into 2 regimes

  2. if (* a1 a2) < -3.3079264831196082e194 or -5.632465143843106e-172 < (* a1 a2) < 3.7907289754606943e-297 or 4.11512080060718968e280 < (* a1 a2)

    1. Initial program 23.0

      \[\frac{a1 \cdot a2}{b1 \cdot b2}\]

    2. Taylor expanded around 0 23.0

      \[\leadsto \color{blue}{\frac{a1 \cdot a2}{b1 \cdot b2}}\]

    3. Simplified11.2

      \[\leadsto \color{blue}{a2 \cdot \frac{a1}{b1 \cdot b2}}\]
    4. Using strategy rm

    5. Applied *-un-lft-identity11.2

      \[\leadsto a2 \cdot \frac{\color{blue}{1 \cdot a1}}{b1 \cdot b2}\]

    6. Applied times-frac6.1

      \[\leadsto a2 \cdot \color{blue}{\left(\frac{1}{b1} \cdot \frac{a1}{b2}\right)}\]

    if -3.3079264831196082e194 < (* a1 a2) < -5.632465143843106e-172 or 3.7907289754606943e-297 < (* a1 a2) < 4.11512080060718968e280

    1. Initial program 4.9

      \[\frac{a1 \cdot a2}{b1 \cdot b2}\]
    2. Using strategy rm

    3. Applied clear-num5.2

      \[\leadsto \color{blue}{\frac{1}{\frac{b1 \cdot b2}{a1 \cdot a2}}}\]

    4. Simplified5.5

      \[\leadsto \frac{1}{\color{blue}{b2 \cdot \frac{b1}{a1 \cdot a2}}}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification5.7

    \[\leadsto \begin{array}{l} \mathbf{if}\;a1 \cdot a2 \le -3.3079264831196082 \cdot 10^{194} \lor \neg \left(a1 \cdot a2 \le -5.632465143843106 \cdot 10^{-172} \lor \neg \left(a1 \cdot a2 \le 3.7907289754606943 \cdot 10^{-297}\right) \land a1 \cdot a2 \le 4.11512080060718968 \cdot 10^{280}\right):\\ \;\;\;\;a2 \cdot \left(\frac{1}{b1} \cdot \frac{a1}{b2}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{b2 \cdot \frac{b1}{a1 \cdot a2}}\\ \end{array}\]

Reproduce

herbie shell --seed 2020185 
(FPCore (a1 a2 b1 b2)
  :name "Quotient of products"
  :precision binary64

  :herbie-target
  (* (/ a1 b1) (/ a2 b2))

  (/ (* a1 a2) (* b1 b2)))