?

Average Accuracy: 74.0% → 99.0%
Time: 12.6s
Precision: binary64
Cost: 32969

?

\[\pi \cdot \ell - \frac{1}{F \cdot F} \cdot \tan \left(\pi \cdot \ell\right) \]
\[\begin{array}{l} \mathbf{if}\;\pi \cdot \ell \leq -1 \cdot 10^{+22} \lor \neg \left(\pi \cdot \ell \leq 200000000\right):\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \ell - \frac{\frac{\tan \left(\pi \cdot \ell\right)}{F}}{F}\\ \end{array} \]
(FPCore (F l)
 :precision binary64
 (- (* PI l) (* (/ 1.0 (* F F)) (tan (* PI l)))))
(FPCore (F l)
 :precision binary64
 (if (or (<= (* PI l) -1e+22) (not (<= (* PI l) 200000000.0)))
   (* 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) <= -1e+22) || !((((double) M_PI) * l) <= 200000000.0)) {
		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) <= -1e+22) || !((Math.PI * l) <= 200000000.0)) {
		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) <= -1e+22) or not ((math.pi * l) <= 200000000.0):
		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) <= -1e+22) || !(Float64(pi * l) <= 200000000.0))
		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) <= -1e+22) || ~(((pi * l) <= 200000000.0)))
		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], -1e+22], N[Not[LessEqual[N[(Pi * l), $MachinePrecision], 200000000.0]], $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 -1 \cdot 10^{+22} \lor \neg \left(\pi \cdot \ell \leq 200000000\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) < -1e22 or 2e8 < (*.f64 (PI.f64) l)

    1. Initial program 63.1%

      \[\pi \cdot \ell - \frac{1}{F \cdot F} \cdot \tan \left(\pi \cdot \ell\right) \]
    2. Simplified63.1%

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

      [Start]63.1

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

      associate-*l/ [=>]63.1

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

      *-lft-identity [=>]63.1

      \[ \pi \cdot \ell - \frac{\color{blue}{\tan \left(\pi \cdot \ell\right)}}{F \cdot F} \]
    3. Taylor expanded in l around inf 99.4%

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

    if -1e22 < (*.f64 (PI.f64) l) < 2e8

    1. Initial program 85.4%

      \[\pi \cdot \ell - \frac{1}{F \cdot F} \cdot \tan \left(\pi \cdot \ell\right) \]
    2. Applied egg-rr98.7%

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

      [Start]85.4

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

      associate-*l/ [=>]86.1

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

      *-un-lft-identity [<=]86.1

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

      associate-/r* [=>]98.7

      \[ \pi \cdot \ell - \color{blue}{\frac{\frac{\tan \left(\pi \cdot \ell\right)}{F}}{F}} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification99.0%

    \[\leadsto \begin{array}{l} \mathbf{if}\;\pi \cdot \ell \leq -1 \cdot 10^{+22} \lor \neg \left(\pi \cdot \ell \leq 200000000\right):\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \ell - \frac{\frac{\tan \left(\pi \cdot \ell\right)}{F}}{F}\\ \end{array} \]

Alternatives

Alternative 1
Accuracy98.4%
Cost26569
\[\begin{array}{l} \mathbf{if}\;\pi \cdot \ell \leq -1 \cdot 10^{+22} \lor \neg \left(\pi \cdot \ell \leq 0.001\right):\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \ell - \frac{\pi \cdot \frac{\ell}{F}}{F}\\ \end{array} \]
Alternative 2
Accuracy92.5%
Cost13641
\[\begin{array}{l} \mathbf{if}\;\ell \leq -5 \cdot 10^{+15} \lor \neg \left(\ell \leq 3.5\right):\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \ell - \pi \cdot \frac{\ell}{F \cdot F}\\ \end{array} \]
Alternative 3
Accuracy98.7%
Cost13641
\[\begin{array}{l} \mathbf{if}\;\ell \leq -3.5 \cdot 10^{+14} \lor \neg \left(\ell \leq 3.5\right):\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \ell - \frac{\ell}{F} \cdot \frac{\pi}{F}\\ \end{array} \]
Alternative 4
Accuracy92.2%
Cost13513
\[\begin{array}{l} \mathbf{if}\;\ell \leq -1.5 \cdot 10^{+15} \lor \neg \left(\ell \leq 3.5\right):\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{else}:\\ \;\;\;\;\ell \cdot \left(\pi - \frac{\pi}{F \cdot F}\right)\\ \end{array} \]
Alternative 5
Accuracy92.2%
Cost13513
\[\begin{array}{l} \mathbf{if}\;\ell \leq -5 \cdot 10^{+15} \lor \neg \left(\ell \leq 3.5\right):\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \left(\ell - \ell \cdot {F}^{-2}\right)\\ \end{array} \]
Alternative 6
Accuracy79.1%
Cost13448
\[\begin{array}{l} \mathbf{if}\;F \leq -2.8 \cdot 10^{-58}:\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{elif}\;F \leq -2.2 \cdot 10^{-74}:\\ \;\;\;\;\pi \cdot \left({F}^{-2} \cdot \left(-\ell\right)\right)\\ \mathbf{elif}\;F \leq 3.5 \cdot 10^{-157}:\\ \;\;\;\;1 + \left(\pi \cdot \ell + -1\right)\\ \mathbf{elif}\;F \leq 6.5 \cdot 10^{-10}:\\ \;\;\;\;\frac{\pi \cdot \left(-\ell\right)}{F \cdot F}\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \ell\\ \end{array} \]
Alternative 7
Accuracy79.0%
Cost7376
\[\begin{array}{l} t_0 := \frac{\pi \cdot \left(-\ell\right)}{F \cdot F}\\ \mathbf{if}\;F \leq -1.55 \cdot 10^{-58}:\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{elif}\;F \leq -2.1 \cdot 10^{-74}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;F \leq 1.1 \cdot 10^{-159}:\\ \;\;\;\;1 + \left(\pi \cdot \ell + -1\right)\\ \mathbf{elif}\;F \leq 1.2 \cdot 10^{-8}:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \ell\\ \end{array} \]
Alternative 8
Accuracy78.9%
Cost6528
\[\pi \cdot \ell \]

Error

Reproduce?

herbie shell --seed 2023144 
(FPCore (F l)
  :name "VandenBroeck and Keller, Equation (6)"
  :precision binary64
  (- (* PI l) (* (/ 1.0 (* F F)) (tan (* PI l)))))