Average Error: 19.5 → 13.5
Time: 5.6s
Precision: binary64
\[c0 \cdot \sqrt{\frac{A}{V \cdot \ell}}\]
↓
\[\begin{array}{l} \mathbf{if}\;V \cdot \ell \le -3.9279438898199403 \cdot 10^{-218}:\\ \;\;\;\;c0 \cdot \sqrt{A \cdot \frac{1}{V \cdot \ell}}\\ \mathbf{elif}\;V \cdot \ell \le 0.0:\\ \;\;\;\;c0 \cdot \sqrt{\frac{1}{V} \cdot \frac{A}{\ell}}\\ \mathbf{else}:\\ \;\;\;\;c0 \cdot \left(\sqrt{A} \cdot \sqrt{\frac{1}{V \cdot \ell}}\right)\\ \end{array}\]
c0 \cdot \sqrt{\frac{A}{V \cdot \ell}}
↓
\begin{array}{l}
\mathbf{if}\;V \cdot \ell \le -3.9279438898199403 \cdot 10^{-218}:\\
\;\;\;\;c0 \cdot \sqrt{A \cdot \frac{1}{V \cdot \ell}}\\

\mathbf{elif}\;V \cdot \ell \le 0.0:\\
\;\;\;\;c0 \cdot \sqrt{\frac{1}{V} \cdot \frac{A}{\ell}}\\

\mathbf{else}:\\
\;\;\;\;c0 \cdot \left(\sqrt{A} \cdot \sqrt{\frac{1}{V \cdot \ell}}\right)\\

\end{array}
double code(double c0, double A, double V, double l) {
	return ((double) (c0 * ((double) sqrt((A / ((double) (V * l)))))));
}
↓
double code(double c0, double A, double V, double l) {
	double VAR;
	if ((((double) (V * l)) <= -3.9279438898199403e-218)) {
		VAR = ((double) (c0 * ((double) sqrt(((double) (A * (1.0 / ((double) (V * l)))))))));
	} else {
		double VAR_1;
		if ((((double) (V * l)) <= 0.0)) {
			VAR_1 = ((double) (c0 * ((double) sqrt(((double) ((1.0 / V) * (A / l)))))));
		} else {
			VAR_1 = ((double) (c0 * ((double) (((double) sqrt(A)) * ((double) sqrt((1.0 / ((double) (V * l)))))))));
		}
		VAR = VAR_1;
	}
	return VAR;
}

Error

Bits error versus c0

Bits error versus A

Bits error versus V

Bits error versus l

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Derivation

  1. Split input into 3 regimes

  2. if (* V l) < -3.9279438898199403e-218

    1. Initial program 14.4

      \[c0 \cdot \sqrt{\frac{A}{V \cdot \ell}}\]
    2. Using strategy rm

    3. Applied div-inv14.4

      \[\leadsto c0 \cdot \sqrt{\color{blue}{A \cdot \frac{1}{V \cdot \ell}}}\]

    if -3.9279438898199403e-218 < (* V l) < 0.0

    1. Initial program 51.6

      \[c0 \cdot \sqrt{\frac{A}{V \cdot \ell}}\]
    2. Using strategy rm

    3. Applied *-un-lft-identity51.6

      \[\leadsto c0 \cdot \sqrt{\frac{\color{blue}{1 \cdot A}}{V \cdot \ell}}\]

    4. Applied times-frac33.5

      \[\leadsto c0 \cdot \sqrt{\color{blue}{\frac{1}{V} \cdot \frac{A}{\ell}}}\]

    if 0.0 < (* V l)

    1. Initial program 15.5

      \[c0 \cdot \sqrt{\frac{A}{V \cdot \ell}}\]
    2. Using strategy rm

    3. Applied div-inv15.6

      \[\leadsto c0 \cdot \sqrt{\color{blue}{A \cdot \frac{1}{V \cdot \ell}}}\]

    4. Applied sqrt-prod7.3

      \[\leadsto c0 \cdot \color{blue}{\left(\sqrt{A} \cdot \sqrt{\frac{1}{V \cdot \ell}}\right)}\]
  3. Recombined 3 regimes into one program.

  4. Final simplification13.5

    \[\leadsto \begin{array}{l} \mathbf{if}\;V \cdot \ell \le -3.9279438898199403 \cdot 10^{-218}:\\ \;\;\;\;c0 \cdot \sqrt{A \cdot \frac{1}{V \cdot \ell}}\\ \mathbf{elif}\;V \cdot \ell \le 0.0:\\ \;\;\;\;c0 \cdot \sqrt{\frac{1}{V} \cdot \frac{A}{\ell}}\\ \mathbf{else}:\\ \;\;\;\;c0 \cdot \left(\sqrt{A} \cdot \sqrt{\frac{1}{V \cdot \ell}}\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020182 
(FPCore (c0 A V l)
  :name "Henrywood and Agarwal, Equation (3)"
  :precision binary64
  (* c0 (sqrt (/ A (* V l)))))