\[ \begin{array}{c}[x, y] = \mathsf{sort}([x, y])\\ \end{array} \]
\[\frac{\frac{1}{x}}{y \cdot \left(1 + z \cdot z\right)}
\]
↓
\[\begin{array}{l}
t_0 := \mathsf{hypot}\left(1, z\right) \cdot \sqrt{y}\\
\mathbf{if}\;y \cdot \left(1 + z \cdot z\right) \leq 5 \cdot 10^{+303}:\\
\;\;\;\;\frac{\frac{1}{x}}{\mathsf{fma}\left(y \cdot z, z, y\right)}\\
\mathbf{else}:\\
\;\;\;\;\frac{1}{t_0} \cdot \frac{\frac{1}{x}}{t_0}\\
\end{array}
\]
(FPCore (x y z) :precision binary64 (/ (/ 1.0 x) (* y (+ 1.0 (* z z)))))
↓
(FPCore (x y z)
:precision binary64
(let* ((t_0 (* (hypot 1.0 z) (sqrt y))))
(if (<= (* y (+ 1.0 (* z z))) 5e+303)
(/ (/ 1.0 x) (fma (* y z) z y))
(* (/ 1.0 t_0) (/ (/ 1.0 x) t_0)))))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 t_0 = hypot(1.0, z) * sqrt(y);
double tmp;
if ((y * (1.0 + (z * z))) <= 5e+303) {
tmp = (1.0 / x) / fma((y * z), z, y);
} else {
tmp = (1.0 / t_0) * ((1.0 / x) / t_0);
}
return tmp;
}
function code(x, y, z)
return Float64(Float64(1.0 / x) / Float64(y * Float64(1.0 + Float64(z * z))))
end
↓
function code(x, y, z)
t_0 = Float64(hypot(1.0, z) * sqrt(y))
tmp = 0.0
if (Float64(y * Float64(1.0 + Float64(z * z))) <= 5e+303)
tmp = Float64(Float64(1.0 / x) / fma(Float64(y * z), z, y));
else
tmp = Float64(Float64(1.0 / t_0) * Float64(Float64(1.0 / x) / t_0));
end
return tmp
end
code[x_, y_, z_] := N[(N[(1.0 / x), $MachinePrecision] / N[(y * N[(1.0 + N[(z * z), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]
↓
code[x_, y_, z_] := Block[{t$95$0 = N[(N[Sqrt[1.0 ^ 2 + z ^ 2], $MachinePrecision] * N[Sqrt[y], $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[(y * N[(1.0 + N[(z * z), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 5e+303], N[(N[(1.0 / x), $MachinePrecision] / N[(N[(y * z), $MachinePrecision] * z + y), $MachinePrecision]), $MachinePrecision], N[(N[(1.0 / t$95$0), $MachinePrecision] * N[(N[(1.0 / x), $MachinePrecision] / t$95$0), $MachinePrecision]), $MachinePrecision]]]
\frac{\frac{1}{x}}{y \cdot \left(1 + z \cdot z\right)}
↓
\begin{array}{l}
t_0 := \mathsf{hypot}\left(1, z\right) \cdot \sqrt{y}\\
\mathbf{if}\;y \cdot \left(1 + z \cdot z\right) \leq 5 \cdot 10^{+303}:\\
\;\;\;\;\frac{\frac{1}{x}}{\mathsf{fma}\left(y \cdot z, z, y\right)}\\
\mathbf{else}:\\
\;\;\;\;\frac{1}{t_0} \cdot \frac{\frac{1}{x}}{t_0}\\
\end{array}
Alternatives
| Alternative 1 |
|---|
| Error | 1.3 |
|---|
| Cost | 7492 |
|---|
\[\begin{array}{l}
\mathbf{if}\;y \cdot \left(1 + z \cdot z\right) \leq 5 \cdot 10^{+303}:\\
\;\;\;\;\frac{\frac{1}{x}}{\mathsf{fma}\left(y \cdot z, z, y\right)}\\
\mathbf{else}:\\
\;\;\;\;\frac{\frac{\frac{1}{z}}{x}}{y \cdot z}\\
\end{array}
\]
| Alternative 2 |
|---|
| Error | 1.4 |
|---|
| Cost | 1736 |
|---|
\[\begin{array}{l}
t_0 := y \cdot \left(1 + z \cdot z\right)\\
\mathbf{if}\;t_0 \leq -\infty:\\
\;\;\;\;\frac{1}{z} \cdot \frac{\frac{1}{y}}{z \cdot x}\\
\mathbf{elif}\;t_0 \leq 5 \cdot 10^{+303}:\\
\;\;\;\;\frac{\frac{1}{x}}{t_0}\\
\mathbf{else}:\\
\;\;\;\;\frac{\frac{\frac{1}{z}}{x}}{y \cdot z}\\
\end{array}
\]
| Alternative 3 |
|---|
| Error | 1.4 |
|---|
| Cost | 1736 |
|---|
\[\begin{array}{l}
t_0 := y \cdot \left(1 + z \cdot z\right)\\
\mathbf{if}\;t_0 \leq -\infty:\\
\;\;\;\;\frac{1}{z} \cdot \frac{\frac{1}{y}}{z \cdot x}\\
\mathbf{elif}\;t_0 \leq 5 \cdot 10^{+303}:\\
\;\;\;\;\frac{\frac{1}{x}}{y + y \cdot \left(z \cdot z\right)}\\
\mathbf{else}:\\
\;\;\;\;\frac{\frac{\frac{1}{z}}{x}}{y \cdot z}\\
\end{array}
\]
| Alternative 4 |
|---|
| Error | 4.3 |
|---|
| Cost | 841 |
|---|
\[\begin{array}{l}
\mathbf{if}\;z \leq -1 \lor \neg \left(z \leq 1\right):\\
\;\;\;\;\frac{1}{x \cdot \left(z \cdot \left(y \cdot z\right)\right)}\\
\mathbf{else}:\\
\;\;\;\;\frac{\frac{1}{x}}{y}\\
\end{array}
\]
| Alternative 5 |
|---|
| Error | 4.3 |
|---|
| Cost | 840 |
|---|
\[\begin{array}{l}
\mathbf{if}\;z \leq -1:\\
\;\;\;\;\frac{1}{z \cdot \left(z \cdot \left(y \cdot x\right)\right)}\\
\mathbf{elif}\;z \leq 1:\\
\;\;\;\;\frac{\frac{1}{x}}{y}\\
\mathbf{else}:\\
\;\;\;\;\frac{1}{x \cdot \left(z \cdot \left(y \cdot z\right)\right)}\\
\end{array}
\]
| Alternative 6 |
|---|
| Error | 2.5 |
|---|
| Cost | 836 |
|---|
\[\begin{array}{l}
\mathbf{if}\;z \cdot z \leq 1:\\
\;\;\;\;\frac{\frac{1}{x}}{y}\\
\mathbf{else}:\\
\;\;\;\;\frac{1}{\left(y \cdot z\right) \cdot \left(z \cdot x\right)}\\
\end{array}
\]
| Alternative 7 |
|---|
| Error | 2.2 |
|---|
| Cost | 836 |
|---|
\[\begin{array}{l}
\mathbf{if}\;z \cdot z \leq 1:\\
\;\;\;\;\frac{\frac{1}{x}}{y}\\
\mathbf{else}:\\
\;\;\;\;\frac{\frac{\frac{1}{z}}{x}}{y \cdot z}\\
\end{array}
\]
| Alternative 8 |
|---|
| Error | 28.4 |
|---|
| Cost | 320 |
|---|
\[\frac{1}{y \cdot x}
\]
| Alternative 9 |
|---|
| Error | 28.5 |
|---|
| Cost | 320 |
|---|
\[\frac{\frac{1}{x}}{y}
\]