Average Error: 6.6 → 5.1
Time: 6.7s
Precision: binary64
\[[x, y]=\mathsf{sort}([x, y])\]
\[\frac{\frac{1}{x}}{y \cdot \left(1 + z \cdot z\right)}\]
↓
\[\begin{array}{l} \mathbf{if}\;y \cdot \left(1 + z \cdot z\right) \leq 1.4633473160021232 \cdot 10^{+200}:\\ \;\;\;\;\frac{\frac{1}{x}}{y \cdot \left(1 + z \cdot z\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{1}{y}}{x + \left(z \cdot z\right) \cdot x}\\ \end{array}\]
\frac{\frac{1}{x}}{y \cdot \left(1 + z \cdot z\right)}
↓
\begin{array}{l}
\mathbf{if}\;y \cdot \left(1 + z \cdot z\right) \leq 1.4633473160021232 \cdot 10^{+200}:\\
\;\;\;\;\frac{\frac{1}{x}}{y \cdot \left(1 + z \cdot z\right)}\\

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

\end{array}
(FPCore (x y z) :precision binary64 (/ (/ 1.0 x) (* y (+ 1.0 (* z z)))))
↓
(FPCore (x y z)
 :precision binary64
 (if (<= (* y (+ 1.0 (* z z))) 1.4633473160021232e+200)
   (/ (/ 1.0 x) (* y (+ 1.0 (* z z))))
   (/ (/ 1.0 y) (+ x (* (* z z) x)))))
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 * (1.0 + (z * z))) <= 1.4633473160021232e+200) {
		tmp = (1.0 / x) / (y * (1.0 + (z * z)));
	} else {
		tmp = (1.0 / y) / (x + ((z * z) * x));
	}
	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

Original6.6
Target5.0
Herbie5.1
\[\begin{array}{l} \mathbf{if}\;y \cdot \left(1 + z \cdot z\right) < -\infty:\\ \;\;\;\;\frac{\frac{1}{y}}{\left(1 + z \cdot z\right) \cdot x}\\ \mathbf{elif}\;y \cdot \left(1 + z \cdot z\right) < 8.680743250567252 \cdot 10^{+305}:\\ \;\;\;\;\frac{\frac{1}{x}}{\left(1 + z \cdot z\right) \cdot y}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{1}{y}}{\left(1 + z \cdot z\right) \cdot x}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if (*.f64 y (+.f64 1 (*.f64 z z))) < 1.4633473160021232e200

    1. Initial program 2.1

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

    if 1.4633473160021232e200 < (*.f64 y (+.f64 1 (*.f64 z z)))

    1. Initial program 13.3

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

    3. Applied associate-/r*_binary64_92299.7

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

    4. Simplified9.9

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

    6. Applied *-un-lft-identity_binary64_92859.9

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

    7. Applied times-frac_binary64_92919.8

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

    8. Applied associate-/l*_binary64_92309.6

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

    9. Simplified9.6

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

  4. Final simplification5.1

    \[\leadsto \begin{array}{l} \mathbf{if}\;y \cdot \left(1 + z \cdot z\right) \leq 1.4633473160021232 \cdot 10^{+200}:\\ \;\;\;\;\frac{\frac{1}{x}}{y \cdot \left(1 + z \cdot z\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{\frac{1}{y}}{x + \left(z \cdot z\right) \cdot x}\\ \end{array}\]

Reproduce

herbie shell --seed 2021040 
(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)))))