Average Error: 6.5 → 5.3
Time: 4.7s
Precision: binary64
\[\frac{\frac{1}{x}}{y \cdot \left(1 + z \cdot z\right)}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -1.39422003712020579 \cdot 10^{-79} \lor \neg \left(x \le 1.39993456316826 \cdot 10^{-84}\right):\\ \;\;\;\;\frac{\frac{1}{x}}{y \cdot \left(1 + z \cdot z\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{y} \cdot \frac{1}{x \cdot \left(1 + z \cdot z\right)}\\ \end{array}\]
\frac{\frac{1}{x}}{y \cdot \left(1 + z \cdot z\right)}
↓
\begin{array}{l}
\mathbf{if}\;x \le -1.39422003712020579 \cdot 10^{-79} \lor \neg \left(x \le 1.39993456316826 \cdot 10^{-84}\right):\\
\;\;\;\;\frac{\frac{1}{x}}{y \cdot \left(1 + z \cdot z\right)}\\

\mathbf{else}:\\
\;\;\;\;\frac{1}{y} \cdot \frac{1}{x \cdot \left(1 + z \cdot z\right)}\\

\end{array}
double code(double x, double y, double z) {
	return ((double) (((double) (1.0 / x)) / ((double) (y * ((double) (1.0 + ((double) (z * z))))))));
}
↓
double code(double x, double y, double z) {
	double VAR;
	if (((x <= -1.3942200371202058e-79) || !(x <= 1.3999345631682573e-84))) {
		VAR = ((double) (((double) (1.0 / x)) / ((double) (y * ((double) (1.0 + ((double) (z * z))))))));
	} else {
		VAR = ((double) (((double) (1.0 / y)) * ((double) (1.0 / ((double) (x * ((double) (1.0 + ((double) (z * z))))))))));
	}
	return VAR;
}

Error

Bits error versus x

Bits error versus y

Bits error versus z

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original6.5
Target5.8
Herbie5.3
\[\begin{array}{l} \mathbf{if}\;y \cdot \left(1 + z \cdot z\right) \lt -inf.0:\\ \;\;\;\;\frac{\frac{1}{y}}{\left(1 + z \cdot z\right) \cdot x}\\ \mathbf{elif}\;y \cdot \left(1 + z \cdot z\right) \lt 8.68074325056725162 \cdot 10^{305}:\\ \;\;\;\;\frac{\frac{1}{x}}{\left(1 + z \cdot z\right) \cdot y}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{1}{y}}{\left(1 + z \cdot z\right) \cdot x}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if x < -1.39422003712020579e-79 or 1.39993456316826e-84 < x

    1. Initial program 2.5

      \[\frac{\frac{1}{x}}{y \cdot \left(1 + z \cdot z\right)}\]

    if -1.39422003712020579e-79 < x < 1.39993456316826e-84

    1. Initial program 15.0

      \[\frac{\frac{1}{x}}{y \cdot \left(1 + z \cdot z\right)}\]
    2. Using strategy rm

    3. Applied *-un-lft-identity15.0

      \[\leadsto \frac{\frac{1}{\color{blue}{1 \cdot x}}}{y \cdot \left(1 + z \cdot z\right)}\]

    4. Applied *-un-lft-identity15.0

      \[\leadsto \frac{\frac{\color{blue}{1 \cdot 1}}{1 \cdot x}}{y \cdot \left(1 + z \cdot z\right)}\]

    5. Applied times-frac15.0

      \[\leadsto \frac{\color{blue}{\frac{1}{1} \cdot \frac{1}{x}}}{y \cdot \left(1 + z \cdot z\right)}\]

    6. Applied times-frac11.2

      \[\leadsto \color{blue}{\frac{\frac{1}{1}}{y} \cdot \frac{\frac{1}{x}}{1 + z \cdot z}}\]

    7. Simplified11.2

      \[\leadsto \color{blue}{\frac{1}{y}} \cdot \frac{\frac{1}{x}}{1 + z \cdot z}\]

    8. Simplified11.2

      \[\leadsto \frac{1}{y} \cdot \color{blue}{\frac{1}{x \cdot \left(1 + z \cdot z\right)}}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification5.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -1.39422003712020579 \cdot 10^{-79} \lor \neg \left(x \le 1.39993456316826 \cdot 10^{-84}\right):\\ \;\;\;\;\frac{\frac{1}{x}}{y \cdot \left(1 + z \cdot z\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{y} \cdot \frac{1}{x \cdot \left(1 + z \cdot z\right)}\\ \end{array}\]

Reproduce

herbie shell --seed 2020184 
(FPCore (x y z)
  :name "Statistics.Distribution.CauchyLorentz:$cdensity from math-functions-0.1.5.2"
  :precision binary64

  :herbie-target
  (if (< (* y (+ 1.0 (* z z))) (- INFINITY)) (/ (/ 1.0 y) (* (+ 1.0 (* z z)) x)) (if (< (* y (+ 1.0 (* z z))) 8.680743250567252e+305) (/ (/ 1.0 x) (* (+ 1.0 (* z z)) y)) (/ (/ 1.0 y) (* (+ 1.0 (* z z)) x))))

  (/ (/ 1.0 x) (* y (+ 1.0 (* z z)))))