Average Error: 6.2 → 2.7
Time: 2.3s
Precision: binary64
\[\frac{x \cdot y}{z}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \cdot y \le -1.39299103375843131 \cdot 10^{-89}:\\ \;\;\;\;\frac{x \cdot y}{z}\\ \mathbf{elif}\;x \cdot y \le 0.0:\\ \;\;\;\;x \cdot \frac{y}{z}\\ \mathbf{elif}\;x \cdot y \le 3.9220340091600002 \cdot 10^{156}:\\ \;\;\;\;\frac{x \cdot y}{z}\\ \mathbf{else}:\\ \;\;\;\;\frac{x}{z} \cdot y\\ \end{array}\]
\frac{x \cdot y}{z}
↓
\begin{array}{l}
\mathbf{if}\;x \cdot y \le -1.39299103375843131 \cdot 10^{-89}:\\
\;\;\;\;\frac{x \cdot y}{z}\\

\mathbf{elif}\;x \cdot y \le 0.0:\\
\;\;\;\;x \cdot \frac{y}{z}\\

\mathbf{elif}\;x \cdot y \le 3.9220340091600002 \cdot 10^{156}:\\
\;\;\;\;\frac{x \cdot y}{z}\\

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

\end{array}
double code(double x, double y, double z) {
	return ((double) (((double) (x * y)) / z));
}
↓
double code(double x, double y, double z) {
	double VAR;
	if ((((double) (x * y)) <= -1.3929910337584313e-89)) {
		VAR = ((double) (((double) (x * y)) / z));
	} else {
		double VAR_1;
		if ((((double) (x * y)) <= 0.0)) {
			VAR_1 = ((double) (x * ((double) (y / z))));
		} else {
			double VAR_2;
			if ((((double) (x * y)) <= 3.92203400916e+156)) {
				VAR_2 = ((double) (((double) (x * y)) / z));
			} else {
				VAR_2 = ((double) (((double) (x / z)) * y));
			}
			VAR_1 = VAR_2;
		}
		VAR = VAR_1;
	}
	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

Original6.2
Target6.5
Herbie2.7
\[\begin{array}{l} \mathbf{if}\;z \lt -4.262230790519429 \cdot 10^{-138}:\\ \;\;\;\;\frac{x \cdot y}{z}\\ \mathbf{elif}\;z \lt 1.70421306606504721 \cdot 10^{-164}:\\ \;\;\;\;\frac{x}{\frac{z}{y}}\\ \mathbf{else}:\\ \;\;\;\;\frac{x}{z} \cdot y\\ \end{array}\]

Derivation

  1. Split input into 3 regimes

  2. if (* x y) < -1.39299103375843131e-89 or 0.0 < (* x y) < 3.9220340091600002e156

    1. Initial program 3.0

      \[\frac{x \cdot y}{z}\]

    if -1.39299103375843131e-89 < (* x y) < 0.0

    1. Initial program 9.8

      \[\frac{x \cdot y}{z}\]
    2. Using strategy rm

    3. Applied *-un-lft-identity9.8

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

    4. Applied times-frac1.9

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

    5. Simplified1.9

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

    if 3.9220340091600002e156 < (* x y)

    1. Initial program 18.5

      \[\frac{x \cdot y}{z}\]
    2. Using strategy rm

    3. Applied associate-/l*1.9

      \[\leadsto \color{blue}{\frac{x}{\frac{z}{y}}}\]
    4. Using strategy rm

    5. Applied associate-/r/3.7

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

  4. Final simplification2.7

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \cdot y \le -1.39299103375843131 \cdot 10^{-89}:\\ \;\;\;\;\frac{x \cdot y}{z}\\ \mathbf{elif}\;x \cdot y \le 0.0:\\ \;\;\;\;x \cdot \frac{y}{z}\\ \mathbf{elif}\;x \cdot y \le 3.9220340091600002 \cdot 10^{156}:\\ \;\;\;\;\frac{x \cdot y}{z}\\ \mathbf{else}:\\ \;\;\;\;\frac{x}{z} \cdot y\\ \end{array}\]

Reproduce

herbie shell --seed 2020155 
(FPCore (x y z)
  :name "Diagrams.Solve.Tridiagonal:solveCyclicTriDiagonal from diagrams-solve-0.1, A"
  :precision binary64

  :herbie-target
  (if (< z -4.262230790519429e-138) (/ (* x y) z) (if (< z 1.7042130660650472e-164) (/ x (/ z y)) (* (/ x z) y)))

  (/ (* x y) z))