Average Error: 6.4 → 5.3
Time: 8.3s
Precision: binary64
\[[x, y]=\mathsf{sort}([x, y])\]
\[\frac{\frac{1}{x}}{y \cdot \left(1 + z \cdot z\right)}\]
↓
\[\begin{array}{l} \mathbf{if}\;y \leq 8.07827379454449 \cdot 10^{+129}:\\ \;\;\;\;\frac{\frac{1}{y \cdot \left(1 + z \cdot z\right)}}{x}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{y} \cdot \frac{\frac{1}{x}}{1 + z \cdot z}\\ \end{array}\]
\frac{\frac{1}{x}}{y \cdot \left(1 + z \cdot z\right)}
↓
\begin{array}{l}
\mathbf{if}\;y \leq 8.07827379454449 \cdot 10^{+129}:\\
\;\;\;\;\frac{\frac{1}{y \cdot \left(1 + z \cdot z\right)}}{x}\\

\mathbf{else}:\\
\;\;\;\;\frac{1}{y} \cdot \frac{\frac{1}{x}}{1 + z \cdot z}\\

\end{array}
(FPCore (x y z) :precision binary64 (/ (/ 1.0 x) (* y (+ 1.0 (* z z)))))
↓
(FPCore (x y z)
 :precision binary64
 (if (<= y 8.07827379454449e+129)
   (/ (/ 1.0 (* y (+ 1.0 (* z z)))) x)
   (* (/ 1.0 y) (/ (/ 1.0 x) (+ 1.0 (* z z))))))
double code(double x, double y, double z) {
	return (1.0 / x) / (y * (1.0 + (z * z)));
}
↓
double code(double x, double y, double z) {
	double tmp;
	if (y <= 8.07827379454449e+129) {
		tmp = (1.0 / (y * (1.0 + (z * z)))) / x;
	} else {
		tmp = (1.0 / y) * ((1.0 / x) / (1.0 + (z * z)));
	}
	return tmp;
}

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

Original6.4
Target5.0
Herbie5.3
\[\begin{array}{l} \mathbf{if}\;y \cdot \left(1 + z \cdot z\right) < -\infty:\\ \;\;\;\;\frac{\frac{1}{y}}{\left(1 + z \cdot z\right) \cdot x}\\ \mathbf{elif}\;y \cdot \left(1 + z \cdot z\right) < 8.680743250567252 \cdot 10^{+305}:\\ \;\;\;\;\frac{\frac{1}{x}}{\left(1 + z \cdot z\right) \cdot y}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{1}{y}}{\left(1 + z \cdot z\right) \cdot x}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if y < 8.07827379454448951e129

    1. Initial program 7.2

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

    3. Applied div-inv_binary64_99647.3

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

    4. Simplified7.3

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

    6. Applied associate-*l/_binary64_99107.2

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

    if 8.07827379454448951e129 < y

    1. Initial program 4.4

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

    3. Applied *-un-lft-identity_binary64_99674.4

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

    4. Applied *-un-lft-identity_binary64_99674.4

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

    5. Applied times-frac_binary64_99734.4

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

    6. Applied times-frac_binary64_99730.6

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

  4. Final simplification5.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;y \leq 8.07827379454449 \cdot 10^{+129}:\\ \;\;\;\;\frac{\frac{1}{y \cdot \left(1 + z \cdot z\right)}}{x}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{y} \cdot \frac{\frac{1}{x}}{1 + z \cdot z}\\ \end{array}\]

Reproduce

herbie shell --seed 2021098 
(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)))))