Math FPCore C Java Python Julia MATLAB Wolfram TeX \[\pi \cdot \ell - \frac{1}{F \cdot F} \cdot \tan \left(\pi \cdot \ell\right)
\]
↓
\[\begin{array}{l}
\mathbf{if}\;\pi \cdot \ell \leq -5 \cdot 10^{+16} \lor \neg \left(\pi \cdot \ell \leq 2 \cdot 10^{-5}\right):\\
\;\;\;\;\pi \cdot \ell\\
\mathbf{else}:\\
\;\;\;\;\pi \cdot \ell + \frac{\tan \left(\pi \cdot \ell\right) \cdot \frac{-1}{F}}{F}\\
\end{array}
\]
(FPCore (F l)
:precision binary64
(- (* PI l) (* (/ 1.0 (* F F)) (tan (* PI l))))) ↓
(FPCore (F l)
:precision binary64
(if (or (<= (* PI l) -5e+16) (not (<= (* PI l) 2e-5)))
(* PI l)
(+ (* PI l) (/ (* (tan (* PI l)) (/ -1.0 F)) F)))) double code(double F, double l) {
return (((double) M_PI) * l) - ((1.0 / (F * F)) * tan((((double) M_PI) * l)));
}
↓
double code(double F, double l) {
double tmp;
if (((((double) M_PI) * l) <= -5e+16) || !((((double) M_PI) * l) <= 2e-5)) {
tmp = ((double) M_PI) * l;
} else {
tmp = (((double) M_PI) * l) + ((tan((((double) M_PI) * l)) * (-1.0 / F)) / F);
}
return tmp;
}
public static double code(double F, double l) {
return (Math.PI * l) - ((1.0 / (F * F)) * Math.tan((Math.PI * l)));
}
↓
public static double code(double F, double l) {
double tmp;
if (((Math.PI * l) <= -5e+16) || !((Math.PI * l) <= 2e-5)) {
tmp = Math.PI * l;
} else {
tmp = (Math.PI * l) + ((Math.tan((Math.PI * l)) * (-1.0 / F)) / F);
}
return tmp;
}
def code(F, l):
return (math.pi * l) - ((1.0 / (F * F)) * math.tan((math.pi * l)))
↓
def code(F, l):
tmp = 0
if ((math.pi * l) <= -5e+16) or not ((math.pi * l) <= 2e-5):
tmp = math.pi * l
else:
tmp = (math.pi * l) + ((math.tan((math.pi * l)) * (-1.0 / F)) / F)
return tmp
function code(F, l)
return Float64(Float64(pi * l) - Float64(Float64(1.0 / Float64(F * F)) * tan(Float64(pi * l))))
end
↓
function code(F, l)
tmp = 0.0
if ((Float64(pi * l) <= -5e+16) || !(Float64(pi * l) <= 2e-5))
tmp = Float64(pi * l);
else
tmp = Float64(Float64(pi * l) + Float64(Float64(tan(Float64(pi * l)) * Float64(-1.0 / F)) / F));
end
return tmp
end
function tmp = code(F, l)
tmp = (pi * l) - ((1.0 / (F * F)) * tan((pi * l)));
end
↓
function tmp_2 = code(F, l)
tmp = 0.0;
if (((pi * l) <= -5e+16) || ~(((pi * l) <= 2e-5)))
tmp = pi * l;
else
tmp = (pi * l) + ((tan((pi * l)) * (-1.0 / F)) / F);
end
tmp_2 = tmp;
end
code[F_, l_] := N[(N[(Pi * l), $MachinePrecision] - N[(N[(1.0 / N[(F * F), $MachinePrecision]), $MachinePrecision] * N[Tan[N[(Pi * l), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]
↓
code[F_, l_] := If[Or[LessEqual[N[(Pi * l), $MachinePrecision], -5e+16], N[Not[LessEqual[N[(Pi * l), $MachinePrecision], 2e-5]], $MachinePrecision]], N[(Pi * l), $MachinePrecision], N[(N[(Pi * l), $MachinePrecision] + N[(N[(N[Tan[N[(Pi * l), $MachinePrecision]], $MachinePrecision] * N[(-1.0 / F), $MachinePrecision]), $MachinePrecision] / F), $MachinePrecision]), $MachinePrecision]]
\pi \cdot \ell - \frac{1}{F \cdot F} \cdot \tan \left(\pi \cdot \ell\right)
↓
\begin{array}{l}
\mathbf{if}\;\pi \cdot \ell \leq -5 \cdot 10^{+16} \lor \neg \left(\pi \cdot \ell \leq 2 \cdot 10^{-5}\right):\\
\;\;\;\;\pi \cdot \ell\\
\mathbf{else}:\\
\;\;\;\;\pi \cdot \ell + \frac{\tan \left(\pi \cdot \ell\right) \cdot \frac{-1}{F}}{F}\\
\end{array}
Alternatives Alternative 1 Accuracy 98.8% Cost 32969
\[\begin{array}{l}
\mathbf{if}\;\pi \cdot \ell \leq -5 \cdot 10^{+16} \lor \neg \left(\pi \cdot \ell \leq 2 \cdot 10^{-5}\right):\\
\;\;\;\;\pi \cdot \ell\\
\mathbf{else}:\\
\;\;\;\;\pi \cdot \ell - \frac{\frac{\tan \left(\pi \cdot \ell\right)}{F}}{F}\\
\end{array}
\]
Alternative 2 Accuracy 98.5% Cost 26569
\[\begin{array}{l}
\mathbf{if}\;\pi \cdot \ell \leq -5 \cdot 10^{+16} \lor \neg \left(\pi \cdot \ell \leq 2 \cdot 10^{-5}\right):\\
\;\;\;\;\pi \cdot \ell\\
\mathbf{else}:\\
\;\;\;\;\pi \cdot \ell - \frac{\pi}{F} \cdot \frac{\ell}{F}\\
\end{array}
\]
Alternative 3 Accuracy 74.1% Cost 7888
\[\begin{array}{l}
t_0 := \pi \cdot \frac{-\ell}{F \cdot F}\\
t_1 := 1 + \left(\pi \cdot \ell + -1\right)\\
\mathbf{if}\;F \cdot F \leq 5 \cdot 10^{-318}:\\
\;\;\;\;t_1\\
\mathbf{elif}\;F \cdot F \leq 6.9 \cdot 10^{-216}:\\
\;\;\;\;t_0\\
\mathbf{elif}\;F \cdot F \leq 2.5 \cdot 10^{-184}:\\
\;\;\;\;t_1\\
\mathbf{elif}\;F \cdot F \leq 5.3 \cdot 10^{-126}:\\
\;\;\;\;t_0\\
\mathbf{else}:\\
\;\;\;\;\pi \cdot \ell\\
\end{array}
\]
Alternative 4 Accuracy 74.9% Cost 7378
\[\begin{array}{l}
\mathbf{if}\;\ell \leq -4.8 \cdot 10^{-18} \lor \neg \left(\ell \leq -3 \cdot 10^{-162}\right) \land \left(\ell \leq 6.2 \cdot 10^{-217} \lor \neg \left(\ell \leq 0.0065\right)\right):\\
\;\;\;\;\pi \cdot \ell\\
\mathbf{else}:\\
\;\;\;\;\frac{\pi}{F} \cdot \frac{-\ell}{F}\\
\end{array}
\]
Alternative 5 Accuracy 98.6% Cost 7177
\[\begin{array}{l}
\mathbf{if}\;\ell \leq -1450000000000 \lor \neg \left(\ell \leq 115000\right):\\
\;\;\;\;\pi \cdot \ell\\
\mathbf{else}:\\
\;\;\;\;\pi \cdot \left(\ell - \frac{\frac{\ell}{F}}{F}\right)\\
\end{array}
\]
Alternative 6 Accuracy 74.0% Cost 6528
\[\pi \cdot \ell
\]