\[ \begin{array}{c}[x, y] = \mathsf{sort}([x, y])\\ \end{array} \]
\[\frac{\frac{1}{x}}{y \cdot \left(1 + z \cdot z\right)}
\]
↓
\[\begin{array}{l}
t_0 := \sqrt{y} \cdot \mathsf{hypot}\left(1, z\right)\\
\mathbf{if}\;y \cdot \left(1 + z \cdot z\right) \leq 5 \cdot 10^{+305}:\\
\;\;\;\;\frac{\frac{1}{x}}{y + z \cdot \left(y \cdot z\right)}\\
\mathbf{else}:\\
\;\;\;\;\frac{1}{t_0 \cdot \left(x \cdot t_0\right)}\\
\end{array}
\]
(FPCore (x y z) :precision binary64 (/ (/ 1.0 x) (* y (+ 1.0 (* z z)))))
↓
(FPCore (x y z)
:precision binary64
(let* ((t_0 (* (sqrt y) (hypot 1.0 z))))
(if (<= (* y (+ 1.0 (* z z))) 5e+305)
(/ (/ 1.0 x) (+ y (* z (* y z))))
(/ 1.0 (* t_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 = sqrt(y) * hypot(1.0, z);
double tmp;
if ((y * (1.0 + (z * z))) <= 5e+305) {
tmp = (1.0 / x) / (y + (z * (y * z)));
} else {
tmp = 1.0 / (t_0 * (x * t_0));
}
return tmp;
}
public static double code(double x, double y, double z) {
return (1.0 / x) / (y * (1.0 + (z * z)));
}
↓
public static double code(double x, double y, double z) {
double t_0 = Math.sqrt(y) * Math.hypot(1.0, z);
double tmp;
if ((y * (1.0 + (z * z))) <= 5e+305) {
tmp = (1.0 / x) / (y + (z * (y * z)));
} else {
tmp = 1.0 / (t_0 * (x * t_0));
}
return tmp;
}
def code(x, y, z):
return (1.0 / x) / (y * (1.0 + (z * z)))
↓
def code(x, y, z):
t_0 = math.sqrt(y) * math.hypot(1.0, z)
tmp = 0
if (y * (1.0 + (z * z))) <= 5e+305:
tmp = (1.0 / x) / (y + (z * (y * z)))
else:
tmp = 1.0 / (t_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(sqrt(y) * hypot(1.0, z))
tmp = 0.0
if (Float64(y * Float64(1.0 + Float64(z * z))) <= 5e+305)
tmp = Float64(Float64(1.0 / x) / Float64(y + Float64(z * Float64(y * z))));
else
tmp = Float64(1.0 / Float64(t_0 * Float64(x * t_0)));
end
return tmp
end
function tmp = code(x, y, z)
tmp = (1.0 / x) / (y * (1.0 + (z * z)));
end
↓
function tmp_2 = code(x, y, z)
t_0 = sqrt(y) * hypot(1.0, z);
tmp = 0.0;
if ((y * (1.0 + (z * z))) <= 5e+305)
tmp = (1.0 / x) / (y + (z * (y * z)));
else
tmp = 1.0 / (t_0 * (x * t_0));
end
tmp_2 = 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[y], $MachinePrecision] * N[Sqrt[1.0 ^ 2 + z ^ 2], $MachinePrecision]), $MachinePrecision]}, If[LessEqual[N[(y * N[(1.0 + N[(z * z), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 5e+305], N[(N[(1.0 / x), $MachinePrecision] / N[(y + N[(z * N[(y * z), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(1.0 / N[(t$95$0 * N[(x * t$95$0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]
\frac{\frac{1}{x}}{y \cdot \left(1 + z \cdot z\right)}
↓
\begin{array}{l}
t_0 := \sqrt{y} \cdot \mathsf{hypot}\left(1, z\right)\\
\mathbf{if}\;y \cdot \left(1 + z \cdot z\right) \leq 5 \cdot 10^{+305}:\\
\;\;\;\;\frac{\frac{1}{x}}{y + z \cdot \left(y \cdot z\right)}\\
\mathbf{else}:\\
\;\;\;\;\frac{1}{t_0 \cdot \left(x \cdot t_0\right)}\\
\end{array}
Alternatives
| Alternative 1 |
|---|
| Error | 1.1 |
|---|
| Cost | 13636 |
|---|
\[\begin{array}{l}
\mathbf{if}\;x \leq -40000000000000:\\
\;\;\;\;\frac{\frac{1}{x}}{y + z \cdot \left(y \cdot z\right)}\\
\mathbf{else}:\\
\;\;\;\;\frac{1}{\left(y \cdot \mathsf{hypot}\left(1, z\right)\right) \cdot \left(x \cdot \mathsf{hypot}\left(1, z\right)\right)}\\
\end{array}
\]
| Alternative 2 |
|---|
| Error | 3.7 |
|---|
| Cost | 968 |
|---|
\[\begin{array}{l}
\mathbf{if}\;z \leq -1 \cdot 10^{+17}:\\
\;\;\;\;\frac{\frac{1}{z \cdot \left(y \cdot z\right)}}{x}\\
\mathbf{elif}\;z \leq 1.7 \cdot 10^{+51}:\\
\;\;\;\;\frac{\frac{1}{x}}{y \cdot \left(1 + z \cdot z\right)}\\
\mathbf{else}:\\
\;\;\;\;\frac{\frac{1}{y}}{z \cdot \left(z \cdot x\right)}\\
\end{array}
\]
| Alternative 3 |
|---|
| 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 4 |
|---|
| Error | 4.3 |
|---|
| Cost | 840 |
|---|
\[\begin{array}{l}
\mathbf{if}\;z \leq -1:\\
\;\;\;\;\frac{1}{x \cdot \left(z \cdot \left(y \cdot z\right)\right)}\\
\mathbf{elif}\;z \leq 1:\\
\;\;\;\;\frac{\frac{1}{x}}{y}\\
\mathbf{else}:\\
\;\;\;\;\frac{1}{y \cdot \left(z \cdot \left(z \cdot x\right)\right)}\\
\end{array}
\]
| Alternative 5 |
|---|
| Error | 4.3 |
|---|
| Cost | 840 |
|---|
\[\begin{array}{l}
\mathbf{if}\;z \leq -0.86:\\
\;\;\;\;\frac{1}{x \cdot \left(z \cdot \left(y \cdot z\right)\right)}\\
\mathbf{elif}\;z \leq 0.86:\\
\;\;\;\;\frac{1 - z \cdot z}{y \cdot x}\\
\mathbf{else}:\\
\;\;\;\;\frac{1}{y \cdot \left(z \cdot \left(z \cdot x\right)\right)}\\
\end{array}
\]
| Alternative 6 |
|---|
| Error | 4.3 |
|---|
| Cost | 840 |
|---|
\[\begin{array}{l}
\mathbf{if}\;z \leq -0.86:\\
\;\;\;\;\frac{1}{x \cdot \left(z \cdot \left(y \cdot z\right)\right)}\\
\mathbf{elif}\;z \leq 0.86:\\
\;\;\;\;\frac{1 - z \cdot z}{y \cdot x}\\
\mathbf{else}:\\
\;\;\;\;\frac{\frac{1}{y}}{z \cdot \left(z \cdot x\right)}\\
\end{array}
\]
| Alternative 7 |
|---|
| Error | 2.4 |
|---|
| Cost | 836 |
|---|
\[\begin{array}{l}
\mathbf{if}\;y \leq 5 \cdot 10^{-38}:\\
\;\;\;\;\frac{\frac{1}{x}}{y + z \cdot \left(y \cdot z\right)}\\
\mathbf{else}:\\
\;\;\;\;\frac{\frac{\frac{1}{x}}{y}}{1 + z \cdot z}\\
\end{array}
\]
| Alternative 8 |
|---|
| Error | 3.7 |
|---|
| Cost | 704 |
|---|
\[\frac{\frac{1}{x}}{y + z \cdot \left(y \cdot z\right)}
\]
| Alternative 9 |
|---|
| Error | 29.5 |
|---|
| Cost | 320 |
|---|
\[\frac{1}{y \cdot x}
\]