?

Average Accuracy: 76.6% → 99.2%
Time: 16.5s
Precision: binary64
Cost: 32969

?

\[\pi \cdot \ell - \frac{1}{F \cdot F} \cdot \tan \left(\pi \cdot \ell\right) \]
↓
\[\begin{array}{l} \mathbf{if}\;\pi \cdot \ell \leq -5 \cdot 10^{+18} \lor \neg \left(\pi \cdot \ell \leq 1000000000000\right):\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \ell - \frac{\frac{\tan \left(\pi \cdot \ell\right)}{F}}{F}\\ \end{array} \]
(FPCore (F l)
 :precision binary64
 (- (* PI l) (* (/ 1.0 (* F F)) (tan (* PI l)))))
↓
(FPCore (F l)
 :precision binary64
 (if (or (<= (* PI l) -5e+18) (not (<= (* PI l) 1000000000000.0)))
   (* 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) <= -5e+18) || !((((double) M_PI) * l) <= 1000000000000.0)) {
		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) <= -5e+18) || !((Math.PI * l) <= 1000000000000.0)) {
		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) <= -5e+18) or not ((math.pi * l) <= 1000000000000.0):
		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) <= -5e+18) || !(Float64(pi * l) <= 1000000000000.0))
		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) <= -5e+18) || ~(((pi * l) <= 1000000000000.0)))
		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], -5e+18], N[Not[LessEqual[N[(Pi * l), $MachinePrecision], 1000000000000.0]], $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 -5 \cdot 10^{+18} \lor \neg \left(\pi \cdot \ell \leq 1000000000000\right):\\
\;\;\;\;\pi \cdot \ell\\

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


\end{array}

Error?

Bogosity?

Bogosity

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) < -5e18 or 1e12 < (*.f64 (PI.f64) l)

    1. Initial program 68.2%

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

    2. Taylor expanded in l around 0 52.8%

      \[\leadsto \pi \cdot \ell - \color{blue}{\frac{\ell \cdot \pi}{{F}^{2}}} \]

    3. Simplified52.8%

      \[\leadsto \pi \cdot \ell - \color{blue}{\frac{\ell}{F \cdot F} \cdot \pi} \]
      Proof

      [Start]52.8

      \[ \pi \cdot \ell - \frac{\ell \cdot \pi}{{F}^{2}} \]

      associate-/l* [=>]52.8

      \[ \pi \cdot \ell - \color{blue}{\frac{\ell}{\frac{{F}^{2}}{\pi}}} \]

      associate-/r/ [=>]52.8

      \[ \pi \cdot \ell - \color{blue}{\frac{\ell}{{F}^{2}} \cdot \pi} \]

      unpow2 [=>]52.8

      \[ \pi \cdot \ell - \frac{\ell}{\color{blue}{F \cdot F}} \cdot \pi \]

    4. Taylor expanded in F around inf 99.6%

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

    if -5e18 < (*.f64 (PI.f64) l) < 1e12

    1. Initial program 87.8%

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

    2. Applied egg-rr99.5%

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

      [Start]87.8

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

      associate-*l/ [=>]89.2

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

      *-un-lft-identity [<=]89.2

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

      associate-/r* [=>]99.5

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

  4. Final simplification99.6%

    \[\leadsto \begin{array}{l} \mathbf{if}\;\pi \cdot \ell \leq -5 \cdot 10^{+18} \lor \neg \left(\pi \cdot \ell \leq 1000000000000\right):\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \ell - \frac{\frac{\tan \left(\pi \cdot \ell\right)}{F}}{F}\\ \end{array} \]

Alternatives

Alternative 1
Accuracy98.4%
Cost26568
\[\begin{array}{l} \mathbf{if}\;\pi \cdot \ell \leq -0.6:\\ \;\;\;\;1 + \left(\pi \cdot \ell + -1\right)\\ \mathbf{elif}\;\pi \cdot \ell \leq 1000000000000:\\ \;\;\;\;\pi \cdot \ell - \frac{\frac{\pi \cdot \ell}{F}}{F}\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \ell\\ \end{array} \]
Alternative 2
Accuracy98.4%
Cost13640
\[\begin{array}{l} \mathbf{if}\;\ell \leq -0.28:\\ \;\;\;\;1 + \left(\pi \cdot \ell + -1\right)\\ \mathbf{elif}\;\ell \leq 5 \cdot 10^{+14}:\\ \;\;\;\;\pi \cdot \ell - \frac{\ell \cdot \frac{\pi}{F}}{F}\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \ell\\ \end{array} \]
Alternative 3
Accuracy74.5%
Cost7640
\[\begin{array}{l} t_0 := 1 + \left(\pi \cdot \ell + -1\right)\\ t_1 := \frac{\ell}{F} \cdot \frac{\pi}{-F}\\ \mathbf{if}\;F \leq -2 \cdot 10^{-138}:\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{elif}\;F \leq -3.5 \cdot 10^{-255}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;F \leq -4.9 \cdot 10^{-290}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;F \leq 1.7 \cdot 10^{-136}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;F \leq 4.1 \cdot 10^{-75}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;F \leq 3.65 \cdot 10^{-9}:\\ \;\;\;\;t_1\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \ell\\ \end{array} \]
Alternative 4
Accuracy74.3%
Cost7640
\[\begin{array}{l} t_0 := 1 + \left(\pi \cdot \ell + -1\right)\\ t_1 := \frac{\ell}{F} \cdot \frac{\pi}{-F}\\ \mathbf{if}\;F \leq -1.9 \cdot 10^{-138}:\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{elif}\;F \leq -6.5 \cdot 10^{-248}:\\ \;\;\;\;\pi \cdot \left(-\frac{\frac{\ell}{F}}{F}\right)\\ \mathbf{elif}\;F \leq -3.05 \cdot 10^{-291}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;F \leq 4.2 \cdot 10^{-137}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;F \leq 4.5 \cdot 10^{-68}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;F \leq 1.02 \cdot 10^{-9}:\\ \;\;\;\;t_1\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \ell\\ \end{array} \]
Alternative 5
Accuracy98.4%
Cost7176
\[\begin{array}{l} \mathbf{if}\;\ell \leq -0.28:\\ \;\;\;\;1 + \left(\pi \cdot \ell + -1\right)\\ \mathbf{elif}\;\ell \leq 5 \cdot 10^{+14}:\\ \;\;\;\;\pi \cdot \left(\ell - \frac{\frac{\ell}{F}}{F}\right)\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \ell\\ \end{array} \]
Alternative 6
Accuracy73.0%
Cost6528
\[\pi \cdot \ell \]

Error

Reproduce?

herbie shell --seed 2023160 
(FPCore (F l)
  :name "VandenBroeck and Keller, Equation (6)"
  :precision binary64
  (- (* PI l) (* (/ 1.0 (* F F)) (tan (* PI l)))))