?

Average Accuracy: 74.5% → 98.4%
Time: 10.9s
Precision: binary64
Cost: 26569

?

\[\pi \cdot \ell - \frac{1}{F \cdot F} \cdot \tan \left(\pi \cdot \ell\right) \]
\[\begin{array}{l} \mathbf{if}\;\pi \cdot \ell \leq -5 \cdot 10^{+16} \lor \neg \left(\pi \cdot \ell \leq 10^{-19}\right):\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \ell - \frac{\frac{\pi \cdot \ell}{F}}{F}\\ \end{array} \]
(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) 1e-19)))
   (* PI l)
   (- (* PI l) (/ (/ (* 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+16) || !((((double) M_PI) * l) <= 1e-19)) {
		tmp = ((double) M_PI) * l;
	} else {
		tmp = (((double) M_PI) * l) - (((((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+16) || !((Math.PI * l) <= 1e-19)) {
		tmp = Math.PI * l;
	} else {
		tmp = (Math.PI * l) - (((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+16) or not ((math.pi * l) <= 1e-19):
		tmp = math.pi * l
	else:
		tmp = (math.pi * l) - (((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+16) || !(Float64(pi * l) <= 1e-19))
		tmp = Float64(pi * l);
	else
		tmp = Float64(Float64(pi * l) - Float64(Float64(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+16) || ~(((pi * l) <= 1e-19)))
		tmp = pi * l;
	else
		tmp = (pi * l) - (((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+16], N[Not[LessEqual[N[(Pi * l), $MachinePrecision], 1e-19]], $MachinePrecision]], N[(Pi * l), $MachinePrecision], N[(N[(Pi * l), $MachinePrecision] - N[(N[(N[(Pi * l), $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^{+16} \lor \neg \left(\pi \cdot \ell \leq 10^{-19}\right):\\
\;\;\;\;\pi \cdot \ell\\

\mathbf{else}:\\
\;\;\;\;\pi \cdot \ell - \frac{\frac{\pi \cdot \ell}{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) < -5e16 or 9.9999999999999998e-20 < (*.f64 (PI.f64) l)

    1. Initial program 65.4%

      \[\pi \cdot \ell - \frac{1}{F \cdot F} \cdot \tan \left(\pi \cdot \ell\right) \]
    2. Simplified65.4%

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

      [Start]65.4

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

      associate-*l/ [=>]65.4

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

      *-lft-identity [=>]65.4

      \[ \pi \cdot \ell - \frac{\color{blue}{\tan \left(\pi \cdot \ell\right)}}{F \cdot F} \]
    3. Taylor expanded in l around inf 98.2%

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

    if -5e16 < (*.f64 (PI.f64) l) < 9.9999999999999998e-20

    1. Initial program 84.7%

      \[\pi \cdot \ell - \frac{1}{F \cdot F} \cdot \tan \left(\pi \cdot \ell\right) \]
    2. Applied egg-rr99.0%

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

      [Start]84.7

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

      associate-*l/ [=>]85.7

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

      *-un-lft-identity [<=]85.7

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

      associate-/r* [=>]99.0

      \[ \pi \cdot \ell - \color{blue}{\frac{\frac{\tan \left(\pi \cdot \ell\right)}{F}}{F}} \]
    3. Taylor expanded in l around 0 98.6%

      \[\leadsto \pi \cdot \ell - \frac{\color{blue}{\frac{\ell \cdot \pi}{F}}}{F} \]
  3. Recombined 2 regimes into one program.
  4. Final simplification98.4%

    \[\leadsto \begin{array}{l} \mathbf{if}\;\pi \cdot \ell \leq -5 \cdot 10^{+16} \lor \neg \left(\pi \cdot \ell \leq 10^{-19}\right):\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \ell - \frac{\frac{\pi \cdot \ell}{F}}{F}\\ \end{array} \]

Alternatives

Alternative 1
Accuracy98.6%
Cost13641
\[\begin{array}{l} \mathbf{if}\;\ell \leq -2 \cdot 10^{+16} \lor \neg \left(\ell \leq 3.1 \cdot 10^{-9}\right):\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \ell - \frac{\pi}{F} \cdot \frac{\ell}{F}\\ \end{array} \]
Alternative 2
Accuracy98.6%
Cost13641
\[\begin{array}{l} \mathbf{if}\;\ell \leq -2 \cdot 10^{+16} \lor \neg \left(\ell \leq 3.1 \cdot 10^{-9}\right):\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \ell - \frac{\frac{\pi}{\frac{F}{\ell}}}{F}\\ \end{array} \]
Alternative 3
Accuracy91.8%
Cost13513
\[\begin{array}{l} \mathbf{if}\;\ell \leq -2 \cdot 10^{+16} \lor \neg \left(\ell \leq 3.1 \cdot 10^{-9}\right):\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{else}:\\ \;\;\;\;\ell \cdot \left(\pi - \frac{\pi}{F \cdot F}\right)\\ \end{array} \]
Alternative 4
Accuracy79.2%
Cost7888
\[\begin{array}{l} t_0 := \ell \cdot \frac{-\pi}{F \cdot F}\\ t_1 := \left(\pi \cdot \ell + 1\right) + -1\\ \mathbf{if}\;F \cdot F \leq 5 \cdot 10^{-239}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;F \cdot F \leq 5 \cdot 10^{-127}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;F \cdot F \leq 2 \cdot 10^{-71}:\\ \;\;\;\;t_1\\ \mathbf{elif}\;F \cdot F \leq 5 \cdot 10^{-42}:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \ell\\ \end{array} \]
Alternative 5
Accuracy78.8%
Cost6528
\[\pi \cdot \ell \]

Error

Reproduce?

herbie shell --seed 2023137 
(FPCore (F l)
  :name "VandenBroeck and Keller, Equation (6)"
  :precision binary64
  (- (* PI l) (* (/ 1.0 (* F F)) (tan (* PI l)))))