Average Error: 6.3 → 2.0
Time: 2.5s
Precision: binary64
\[\frac{x \cdot y}{z}\]
↓
\[\begin{array}{l} \mathbf{if}\;x \cdot y \le -2.04235723035023199 \cdot 10^{221} \lor \neg \left(x \cdot y \le -4.3330948919630908 \cdot 10^{-163} \lor \neg \left(x \cdot y \le 2.87538483372854104 \cdot 10^{-162} \lor \neg \left(x \cdot y \le 19606962.26358\right)\right)\right):\\ \;\;\;\;x \cdot \frac{y}{z}\\ \mathbf{else}:\\ \;\;\;\;\left(x \cdot y\right) \cdot \frac{1}{z}\\ \end{array}\]
\frac{x \cdot y}{z}
↓
\begin{array}{l}
\mathbf{if}\;x \cdot y \le -2.04235723035023199 \cdot 10^{221} \lor \neg \left(x \cdot y \le -4.3330948919630908 \cdot 10^{-163} \lor \neg \left(x \cdot y \le 2.87538483372854104 \cdot 10^{-162} \lor \neg \left(x \cdot y \le 19606962.26358\right)\right)\right):\\
\;\;\;\;x \cdot \frac{y}{z}\\

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

\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)) <= -2.042357230350232e+221) || !((((double) (x * y)) <= -4.333094891963091e-163) || !((((double) (x * y)) <= 2.875384833728541e-162) || !(((double) (x * y)) <= 19606962.263580322))))) {
		VAR = ((double) (x * ((double) (y / z))));
	} else {
		VAR = ((double) (((double) (x * y)) * ((double) (1.0 / 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

Original6.3
Target6.6
Herbie2.0
\[\begin{array}{l} \mathbf{if}\;z \lt -4.262230790519429 \cdot 10^{-138}:\\ \;\;\;\;\frac{x \cdot y}{z}\\ \mathbf{elif}\;z \lt 1.70421306606504721 \cdot 10^{-164}:\\ \;\;\;\;\frac{x}{\frac{z}{y}}\\ \mathbf{else}:\\ \;\;\;\;\frac{x}{z} \cdot y\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if (* x y) < -2.04235723035023199e221 or -4.3330948919630908e-163 < (* x y) < 2.87538483372854104e-162 or 19606962.26358 < (* x y)

    1. Initial program 10.7

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

    3. Applied *-un-lft-identity10.7

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

    4. Applied times-frac3.2

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

    5. Simplified3.2

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

    if -2.04235723035023199e221 < (* x y) < -4.3330948919630908e-163 or 2.87538483372854104e-162 < (* x y) < 19606962.26358

    1. Initial program 0.2

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

    3. Applied div-inv0.3

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

  4. Final simplification2.0

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \cdot y \le -2.04235723035023199 \cdot 10^{221} \lor \neg \left(x \cdot y \le -4.3330948919630908 \cdot 10^{-163} \lor \neg \left(x \cdot y \le 2.87538483372854104 \cdot 10^{-162} \lor \neg \left(x \cdot y \le 19606962.26358\right)\right)\right):\\ \;\;\;\;x \cdot \frac{y}{z}\\ \mathbf{else}:\\ \;\;\;\;\left(x \cdot y\right) \cdot \frac{1}{z}\\ \end{array}\]

Reproduce

herbie shell --seed 2020162 
(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))