?

Average Accuracy: 73.5% → 98.6%
Time: 13.4s
Precision: binary64
Cost: 33097.00

?

\[\pi \cdot \ell - \frac{1}{F \cdot F} \cdot \tan \left(\pi \cdot \ell\right) \]
\[\begin{array}{l} \mathbf{if}\;\pi \cdot \ell \leq -5 \cdot 10^{+24} \lor \neg \left(\pi \cdot \ell \leq 5 \cdot 10^{-7}\right):\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \ell + \frac{\frac{-1}{F}}{\frac{F}{\tan \left(\pi \cdot \ell\right)}}\\ \end{array} \]
(FPCore (F l)
 :precision binary64
 (- (* PI l) (* (/ 1.0 (* F F)) (tan (* PI l)))))
(FPCore (F l)
 :precision binary64
 (if (or (<= (* PI l) -5e+24) (not (<= (* PI l) 5e-7)))
   (* PI l)
   (+ (* PI l) (/ (/ -1.0 F) (/ F (tan (* PI l)))))))
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+24) || !((((double) M_PI) * l) <= 5e-7)) {
		tmp = ((double) M_PI) * l;
	} else {
		tmp = (((double) M_PI) * l) + ((-1.0 / F) / (F / tan((((double) M_PI) * l))));
	}
	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+24) || !((Math.PI * l) <= 5e-7)) {
		tmp = Math.PI * l;
	} else {
		tmp = (Math.PI * l) + ((-1.0 / F) / (F / Math.tan((Math.PI * l))));
	}
	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+24) or not ((math.pi * l) <= 5e-7):
		tmp = math.pi * l
	else:
		tmp = (math.pi * l) + ((-1.0 / F) / (F / math.tan((math.pi * l))))
	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+24) || !(Float64(pi * l) <= 5e-7))
		tmp = Float64(pi * l);
	else
		tmp = Float64(Float64(pi * l) + Float64(Float64(-1.0 / F) / Float64(F / tan(Float64(pi * l)))));
	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+24) || ~(((pi * l) <= 5e-7)))
		tmp = pi * l;
	else
		tmp = (pi * l) + ((-1.0 / F) / (F / tan((pi * l))));
	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+24], N[Not[LessEqual[N[(Pi * l), $MachinePrecision], 5e-7]], $MachinePrecision]], N[(Pi * l), $MachinePrecision], N[(N[(Pi * l), $MachinePrecision] + N[(N[(-1.0 / F), $MachinePrecision] / N[(F / N[Tan[N[(Pi * l), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $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^{+24} \lor \neg \left(\pi \cdot \ell \leq 5 \cdot 10^{-7}\right):\\
\;\;\;\;\pi \cdot \ell\\

\mathbf{else}:\\
\;\;\;\;\pi \cdot \ell + \frac{\frac{-1}{F}}{\frac{F}{\tan \left(\pi \cdot \ell\right)}}\\


\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) < -5.00000000000000045e24 or 4.99999999999999977e-7 < (*.f64 (PI.f64) l)

    1. Initial program 63.2%

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

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

      [Start]63.2

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

      associate-*l/ [=>]63.2

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

      *-lft-identity [=>]63.2

      \[ \pi \cdot \ell - \frac{\color{blue}{\tan \left(\pi \cdot \ell\right)}}{F \cdot F} \]
    3. Taylor expanded in l around inf 98.7%

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

    if -5.00000000000000045e24 < (*.f64 (PI.f64) l) < 4.99999999999999977e-7

    1. Initial program 84.8%

      \[\pi \cdot \ell - \frac{1}{F \cdot F} \cdot \tan \left(\pi \cdot \ell\right) \]
    2. Applied egg-rr98.4%

      \[\leadsto \pi \cdot \ell - \color{blue}{\frac{\frac{1}{F}}{\frac{F}{\tan \left(\pi \cdot \ell\right)}}} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification98.6%

    \[\leadsto \begin{array}{l} \mathbf{if}\;\pi \cdot \ell \leq -5 \cdot 10^{+24} \lor \neg \left(\pi \cdot \ell \leq 5 \cdot 10^{-7}\right):\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \ell + \frac{\frac{-1}{F}}{\frac{F}{\tan \left(\pi \cdot \ell\right)}}\\ \end{array} \]

Alternatives

Alternative 1
Accuracy98.6%
Cost32969.00
\[\begin{array}{l} \mathbf{if}\;\pi \cdot \ell \leq -5 \cdot 10^{+24} \lor \neg \left(\pi \cdot \ell \leq 5 \cdot 10^{-7}\right):\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \ell - \frac{\frac{\tan \left(\pi \cdot \ell\right)}{F}}{F}\\ \end{array} \]
Alternative 2
Accuracy98.3%
Cost26697.00
\[\begin{array}{l} \mathbf{if}\;\pi \cdot \ell \leq -5 \cdot 10^{+24} \lor \neg \left(\pi \cdot \ell \leq 5 \cdot 10^{-7}\right):\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \ell + \frac{\frac{-1}{F}}{\frac{F}{\pi \cdot \ell}}\\ \end{array} \]
Alternative 3
Accuracy98.7%
Cost13769.00
\[\begin{array}{l} \mathbf{if}\;\ell \leq -2.4 \cdot 10^{+16} \lor \neg \left(\ell \leq 1.22 \cdot 10^{-5}\right):\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \ell + \frac{\ell}{\frac{F}{\pi}} \cdot \frac{-1}{F}\\ \end{array} \]
Alternative 4
Accuracy98.7%
Cost13641.00
\[\begin{array}{l} \mathbf{if}\;\ell \leq -2.4 \cdot 10^{+16} \lor \neg \left(\ell \leq 2 \cdot 10^{-6}\right):\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \ell - \frac{\pi}{F} \cdot \frac{\ell}{F}\\ \end{array} \]
Alternative 5
Accuracy79.6%
Cost7376.00
\[\begin{array}{l} t_0 := \frac{\pi}{F \cdot F} \cdot \left(-\ell\right)\\ \mathbf{if}\;F \leq -8.6 \cdot 10^{-14}:\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{elif}\;F \leq -4.2 \cdot 10^{-95}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;F \leq 4.6 \cdot 10^{-85}:\\ \;\;\;\;\left(\pi \cdot \ell + 1\right) + -1\\ \mathbf{elif}\;F \leq 2.5 \cdot 10^{-13}:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \ell\\ \end{array} \]
Alternative 6
Accuracy79.6%
Cost7376.00
\[\begin{array}{l} \mathbf{if}\;F \leq -9.4 \cdot 10^{-14}:\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{elif}\;F \leq -1.9 \cdot 10^{-101}:\\ \;\;\;\;\frac{\ell}{F \cdot \frac{-F}{\pi}}\\ \mathbf{elif}\;F \leq 3.9 \cdot 10^{-85}:\\ \;\;\;\;\left(\pi \cdot \ell + 1\right) + -1\\ \mathbf{elif}\;F \leq 3.4 \cdot 10^{-5}:\\ \;\;\;\;\frac{\pi}{F \cdot F} \cdot \left(-\ell\right)\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \ell\\ \end{array} \]
Alternative 7
Accuracy98.7%
Cost7305.00
\[\begin{array}{l} \mathbf{if}\;\ell \leq -2.4 \cdot 10^{+16} \lor \neg \left(\ell \leq 1.22 \cdot 10^{-5}\right):\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \left(\ell + \frac{\ell}{F} \cdot \frac{-1}{F}\right)\\ \end{array} \]
Alternative 8
Accuracy91.8%
Cost7177.00
\[\begin{array}{l} \mathbf{if}\;\ell \leq -1.75 \cdot 10^{-39} \lor \neg \left(\ell \leq 1.22 \cdot 10^{-5}\right):\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \left(\ell - \frac{\ell}{F \cdot F}\right)\\ \end{array} \]
Alternative 9
Accuracy79.5%
Cost6528.00
\[\pi \cdot \ell \]

Error

Reproduce?

herbie shell --seed 2023096 
(FPCore (F l)
  :name "VandenBroeck and Keller, Equation (6)"
  :precision binary64
  (- (* PI l) (* (/ 1.0 (* F F)) (tan (* PI l)))))