?

Average Accuracy: 74.1% → 99.1%
Time: 14.1s
Precision: binary64
Cost: 32969

?

\[\pi \cdot \ell - \frac{1}{F \cdot F} \cdot \tan \left(\pi \cdot \ell\right) \]
\[\begin{array}{l} \mathbf{if}\;\pi \cdot \ell \leq -2 \cdot 10^{+18} \lor \neg \left(\pi \cdot \ell \leq 10000\right):\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \ell - \frac{\frac{\tan \left(\pi \cdot \ell\right)}{F}}{F}\\ \end{array} \]
(FPCore (F l)
 :precision binary64
 (- (* PI l) (* (/ 1.0 (* F F)) (tan (* PI l)))))
(FPCore (F l)
 :precision binary64
 (if (or (<= (* PI l) -2e+18) (not (<= (* PI l) 10000.0)))
   (* PI l)
   (- (* PI l) (/ (/ (tan (* PI l)) 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) <= -2e+18) || !((((double) M_PI) * l) <= 10000.0)) {
		tmp = ((double) M_PI) * l;
	} else {
		tmp = (((double) M_PI) * l) - ((tan((((double) M_PI) * l)) / 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) <= -2e+18) || !((Math.PI * l) <= 10000.0)) {
		tmp = Math.PI * l;
	} else {
		tmp = (Math.PI * l) - ((Math.tan((Math.PI * l)) / 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) <= -2e+18) or not ((math.pi * l) <= 10000.0):
		tmp = math.pi * l
	else:
		tmp = (math.pi * l) - ((math.tan((math.pi * l)) / 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) <= -2e+18) || !(Float64(pi * l) <= 10000.0))
		tmp = Float64(pi * l);
	else
		tmp = Float64(Float64(pi * l) - Float64(Float64(tan(Float64(pi * l)) / 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) <= -2e+18) || ~(((pi * l) <= 10000.0)))
		tmp = pi * l;
	else
		tmp = (pi * l) - ((tan((pi * l)) / 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], -2e+18], N[Not[LessEqual[N[(Pi * l), $MachinePrecision], 10000.0]], $MachinePrecision]], N[(Pi * l), $MachinePrecision], N[(N[(Pi * l), $MachinePrecision] - N[(N[(N[Tan[N[(Pi * l), $MachinePrecision]], $MachinePrecision] / F), $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 -2 \cdot 10^{+18} \lor \neg \left(\pi \cdot \ell \leq 10000\right):\\
\;\;\;\;\pi \cdot \ell\\

\mathbf{else}:\\
\;\;\;\;\pi \cdot \ell - \frac{\frac{\tan \left(\pi \cdot \ell\right)}{F}}{F}\\


\end{array}

Error?

Try it out?

Your Program's Arguments

Results

Enter valid numbers for all inputs

Derivation?

  1. Split input into 2 regimes
  2. if (*.f64 (PI.f64) l) < -2e18 or 1e4 < (*.f64 (PI.f64) l)

    1. Initial program 62.8%

      \[\pi \cdot \ell - \frac{1}{F \cdot F} \cdot \tan \left(\pi \cdot \ell\right) \]
    2. Simplified62.8%

      \[\leadsto \color{blue}{\pi \cdot \ell - \frac{\tan \left(\pi \cdot \ell\right)}{F \cdot F}} \]
      Proof

      [Start]62.8

      \[ \pi \cdot \ell - \frac{1}{F \cdot F} \cdot \tan \left(\pi \cdot \ell\right) \]

      associate-*l/ [=>]62.8

      \[ \pi \cdot \ell - \color{blue}{\frac{1 \cdot \tan \left(\pi \cdot \ell\right)}{F \cdot F}} \]

      *-lft-identity [=>]62.8

      \[ \pi \cdot \ell - \frac{\color{blue}{\tan \left(\pi \cdot \ell\right)}}{F \cdot F} \]
    3. Taylor expanded in l around inf 99.2%

      \[\leadsto \color{blue}{\ell \cdot \pi} \]

    if -2e18 < (*.f64 (PI.f64) l) < 1e4

    1. Initial program 86.2%

      \[\pi \cdot \ell - \frac{1}{F \cdot F} \cdot \tan \left(\pi \cdot \ell\right) \]
    2. Applied egg-rr99.1%

      \[\leadsto \pi \cdot \ell - \color{blue}{\frac{\frac{\tan \left(\pi \cdot \ell\right)}{F}}{F}} \]
      Proof

      [Start]86.2

      \[ \pi \cdot \ell - \frac{1}{F \cdot F} \cdot \tan \left(\pi \cdot \ell\right) \]

      associate-*l/ [=>]86.9

      \[ \pi \cdot \ell - \color{blue}{\frac{1 \cdot \tan \left(\pi \cdot \ell\right)}{F \cdot F}} \]

      *-un-lft-identity [<=]86.9

      \[ \pi \cdot \ell - \frac{\color{blue}{\tan \left(\pi \cdot \ell\right)}}{F \cdot F} \]

      associate-/r* [=>]99.1

      \[ \pi \cdot \ell - \color{blue}{\frac{\frac{\tan \left(\pi \cdot \ell\right)}{F}}{F}} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification99.1%

    \[\leadsto \begin{array}{l} \mathbf{if}\;\pi \cdot \ell \leq -2 \cdot 10^{+18} \lor \neg \left(\pi \cdot \ell \leq 10000\right):\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \ell - \frac{\frac{\tan \left(\pi \cdot \ell\right)}{F}}{F}\\ \end{array} \]

Alternatives

Alternative 1
Accuracy98.5%
Cost26569
\[\begin{array}{l} \mathbf{if}\;\pi \cdot \ell \leq -2 \cdot 10^{+18} \lor \neg \left(\pi \cdot \ell \leq 10^{-15}\right):\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \ell - \frac{\frac{\pi \cdot \ell}{F}}{F}\\ \end{array} \]
Alternative 2
Accuracy98.8%
Cost13641
\[\begin{array}{l} \mathbf{if}\;\ell \leq -5.5 \lor \neg \left(\ell \leq 1.1 \cdot 10^{-10}\right):\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \ell - \frac{\pi}{F} \cdot \frac{\ell}{F}\\ \end{array} \]
Alternative 3
Accuracy78.9%
Cost7376
\[\begin{array}{l} t_0 := \frac{\pi}{F} \cdot \frac{-\ell}{F}\\ \mathbf{if}\;F \leq -0.35:\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{elif}\;F \leq -6 \cdot 10^{-184}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;F \leq 1.15 \cdot 10^{-60}:\\ \;\;\;\;\left(\pi \cdot \ell + 1\right) + -1\\ \mathbf{elif}\;F \leq 0.00375:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \ell\\ \end{array} \]
Alternative 4
Accuracy98.8%
Cost7241
\[\begin{array}{l} \mathbf{if}\;\ell \leq -5.5 \lor \neg \left(\ell \leq 1.1 \cdot 10^{-10}\right):\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \left(\ell + \frac{\frac{\ell}{F}}{-F}\right)\\ \end{array} \]
Alternative 5
Accuracy92.9%
Cost7177
\[\begin{array}{l} \mathbf{if}\;\ell \leq -5.5 \lor \neg \left(\ell \leq 1.1 \cdot 10^{-10}\right):\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \left(\ell - \frac{\ell}{F \cdot F}\right)\\ \end{array} \]
Alternative 6
Accuracy80.2%
Cost6528
\[\pi \cdot \ell \]
Alternative 7
Accuracy3.2%
Cost64
\[0 \]

Error

Reproduce?

herbie shell --seed 2023133 
(FPCore (F l)
  :name "VandenBroeck and Keller, Equation (6)"
  :precision binary64
  (- (* PI l) (* (/ 1.0 (* F F)) (tan (* PI l)))))