Average Error: 5.8 → 2.0
Time: 1.8s
Precision: binary64
\[\frac{x \cdot y}{z}\]
\[\begin{array}{l} \mathbf{if}\;\frac{x \cdot y}{z} = -inf.0:\\ \;\;\;\;x \cdot \frac{y}{z}\\ \mathbf{elif}\;\frac{x \cdot y}{z} \le -1.7134703487508276 \cdot 10^{-293}:\\ \;\;\;\;\frac{x \cdot y}{z}\\ \mathbf{elif}\;\frac{x \cdot y}{z} \le 0.0:\\ \;\;\;\;\frac{x}{\frac{z}{y}}\\ \mathbf{else}:\\ \;\;\;\;\frac{x \cdot y}{z}\\ \end{array}\]
\frac{x \cdot y}{z}
\begin{array}{l}
\mathbf{if}\;\frac{x \cdot y}{z} = -inf.0:\\
\;\;\;\;x \cdot \frac{y}{z}\\

\mathbf{elif}\;\frac{x \cdot y}{z} \le -1.7134703487508276 \cdot 10^{-293}:\\
\;\;\;\;\frac{x \cdot y}{z}\\

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

\mathbf{else}:\\
\;\;\;\;\frac{x \cdot y}{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) (((double) (x * y)) / z)) <= -inf.0)) {
		VAR = ((double) (x * ((double) (y / z))));
	} else {
		double VAR_1;
		if ((((double) (((double) (x * y)) / z)) <= -1.7134703487508276e-293)) {
			VAR_1 = ((double) (((double) (x * y)) / z));
		} else {
			double VAR_2;
			if ((((double) (((double) (x * y)) / z)) <= 0.0)) {
				VAR_2 = ((double) (x / ((double) (z / y))));
			} else {
				VAR_2 = ((double) (((double) (x * y)) / z));
			}
			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

Original5.8
Target6.1
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 3 regimes
  2. if (/ (* x y) z) < -inf.0

    1. Initial program 64.0

      \[\frac{x \cdot y}{z}\]
    2. Using strategy rm
    3. Applied *-un-lft-identity64.0

      \[\leadsto \frac{x \cdot y}{\color{blue}{1 \cdot z}}\]
    4. Applied times-frac0.3

      \[\leadsto \color{blue}{\frac{x}{1} \cdot \frac{y}{z}}\]
    5. Simplified0.3

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

    if -inf.0 < (/ (* x y) z) < -1.7134703487508276e-293 or 0.0 < (/ (* x y) z)

    1. Initial program 2.5

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

    if -1.7134703487508276e-293 < (/ (* x y) z) < 0.0

    1. Initial program 10.7

      \[\frac{x \cdot y}{z}\]
    2. Using strategy rm
    3. Applied associate-/l*0.7

      \[\leadsto \color{blue}{\frac{x}{\frac{z}{y}}}\]
  3. Recombined 3 regimes into one program.
  4. Final simplification2.0

    \[\leadsto \begin{array}{l} \mathbf{if}\;\frac{x \cdot y}{z} = -inf.0:\\ \;\;\;\;x \cdot \frac{y}{z}\\ \mathbf{elif}\;\frac{x \cdot y}{z} \le -1.7134703487508276 \cdot 10^{-293}:\\ \;\;\;\;\frac{x \cdot y}{z}\\ \mathbf{elif}\;\frac{x \cdot y}{z} \le 0.0:\\ \;\;\;\;\frac{x}{\frac{z}{y}}\\ \mathbf{else}:\\ \;\;\;\;\frac{x \cdot y}{z}\\ \end{array}\]

Reproduce

herbie shell --seed 2020161 
(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))