VandenBroeck and Keller, Equation (6)

?

Percentage Accurate: 76.4% → 98.3%
Time: 12.5s
Precision: binary64
Cost: 33097

?

\[\pi \cdot \ell - \frac{1}{F \cdot F} \cdot \tan \left(\pi \cdot \ell\right) \]
↓
\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;\pi \cdot \ell \leq -5 \cdot 10^{+37} \lor \neg \left(\pi \cdot \ell \leq 100000000000\right):\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \ell + \frac{\tan \left(\pi \cdot \ell\right) \cdot \frac{-1}{F}}{F}\\ \end{array} \end{array} \]
(FPCore (F l)
 :precision binary64
 (- (* PI l) (* (/ 1.0 (* F F)) (tan (* PI l)))))
↓
(FPCore (F l)
 :precision binary64
 (if (or (<= (* PI l) -5e+37) (not (<= (* PI l) 100000000000.0)))
   (* PI l)
   (+ (* PI l) (/ (* (tan (* PI l)) (/ -1.0 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+37) || !((((double) M_PI) * l) <= 100000000000.0)) {
		tmp = ((double) M_PI) * l;
	} else {
		tmp = (((double) M_PI) * l) + ((tan((((double) M_PI) * l)) * (-1.0 / 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+37) || !((Math.PI * l) <= 100000000000.0)) {
		tmp = Math.PI * l;
	} else {
		tmp = (Math.PI * l) + ((Math.tan((Math.PI * l)) * (-1.0 / 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+37) or not ((math.pi * l) <= 100000000000.0):
		tmp = math.pi * l
	else:
		tmp = (math.pi * l) + ((math.tan((math.pi * l)) * (-1.0 / 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+37) || !(Float64(pi * l) <= 100000000000.0))
		tmp = Float64(pi * l);
	else
		tmp = Float64(Float64(pi * l) + Float64(Float64(tan(Float64(pi * l)) * Float64(-1.0 / 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+37) || ~(((pi * l) <= 100000000000.0)))
		tmp = pi * l;
	else
		tmp = (pi * l) + ((tan((pi * l)) * (-1.0 / 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+37], N[Not[LessEqual[N[(Pi * l), $MachinePrecision], 100000000000.0]], $MachinePrecision]], N[(Pi * l), $MachinePrecision], N[(N[(Pi * l), $MachinePrecision] + N[(N[(N[Tan[N[(Pi * l), $MachinePrecision]], $MachinePrecision] * N[(-1.0 / F), $MachinePrecision]), $MachinePrecision] / F), $MachinePrecision]), $MachinePrecision]]

\pi \cdot \ell - \frac{1}{F \cdot F} \cdot \tan \left(\pi \cdot \ell\right)
↓
\begin{array}{l}

\\
\begin{array}{l}
\mathbf{if}\;\pi \cdot \ell \leq -5 \cdot 10^{+37} \lor \neg \left(\pi \cdot \ell \leq 100000000000\right):\\
\;\;\;\;\pi \cdot \ell\\

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


\end{array}
\end{array}

Local Percentage Accuracy vs ?

The average percentage accuracy by input value. Horizontal axis shows value of an input variable; the variable is choosen in the title. Vertical axis is accuracy; higher is better. Red represent the original program, while blue represents Herbie's suggestion. These can be toggled with buttons below the plot. The line is an average while dots represent individual samples.

Herbie found 8 alternatives:

AlternativeAccuracySpeedup

Accuracy vs Speed

The accuracy (vertical axis) and speed (horizontal axis) of each alternatives. Up and to the right is better. The red square shows the initial program, and each blue circle shows an alternative.The line shows the best available speed-accuracy tradeoffs.

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) < -4.99999999999999989e37 or 1e11 < (*.f64 (PI.f64) l)

    1. Initial program 61.4%

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

    2. Simplified61.4%

      \[\leadsto \color{blue}{\pi \cdot \ell - \frac{\tan \left(\pi \cdot \ell\right)}{F \cdot F}} \]
      Step-by-step derivation

      [Start]61.4%

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

      associate-*l/ [=>]61.4%

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

      *-lft-identity [=>]61.4%

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

    3. Taylor expanded in l around 0 43.8%

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

    4. Simplified43.8%

      \[\leadsto \color{blue}{\ell \cdot \left(\pi - \frac{\pi}{F \cdot F}\right)} \]
      Step-by-step derivation

      [Start]43.8%

      \[ \ell \cdot \left(\pi - \frac{\pi}{{F}^{2}}\right) \]

      unpow2 [=>]43.8%

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

    5. Taylor expanded in F around inf 99.4%

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

    if -4.99999999999999989e37 < (*.f64 (PI.f64) l) < 1e11

    1. Initial program 89.2%

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

    2. Applied egg-rr99.6%

      \[\leadsto \pi \cdot \ell - \color{blue}{\frac{\frac{1}{F} \cdot \tan \left(\pi \cdot \ell\right)}{F}} \]
      Step-by-step derivation

      [Start]89.2%

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

      associate-/r* [=>]89.2%

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

      metadata-eval [<=]89.2%

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

      add-sqr-sqrt [=>]44.7%

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

      sqrt-prod [<=]74.3%

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

      sqrt-div [<=]74.3%

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

      associate-*l/ [=>]74.3%

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

      sqrt-div [=>]75.4%

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

      metadata-eval [=>]75.4%

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

      sqrt-prod [=>]52.1%

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

      add-sqr-sqrt [<=]99.6%

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

  4. Final simplification99.5%

    \[\leadsto \begin{array}{l} \mathbf{if}\;\pi \cdot \ell \leq -5 \cdot 10^{+37} \lor \neg \left(\pi \cdot \ell \leq 100000000000\right):\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \ell + \frac{\tan \left(\pi \cdot \ell\right) \cdot \frac{-1}{F}}{F}\\ \end{array} \]

Alternatives

Alternative 1
Accuracy98.3%
Cost33097
\[\begin{array}{l} \mathbf{if}\;\pi \cdot \ell \leq -5 \cdot 10^{+37} \lor \neg \left(\pi \cdot \ell \leq 100000000000\right):\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \ell + \frac{\tan \left(\pi \cdot \ell\right) \cdot \frac{-1}{F}}{F}\\ \end{array} \]
Alternative 2
Accuracy98.3%
Cost33097
\[\begin{array}{l} \mathbf{if}\;\pi \cdot \ell \leq -5 \cdot 10^{+37} \lor \neg \left(\pi \cdot \ell \leq 100000000000\right):\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \ell + \frac{\tan \left(\pi \cdot \ell\right)}{F} \cdot \frac{-1}{F}\\ \end{array} \]
Alternative 3
Accuracy98.3%
Cost32969
\[\begin{array}{l} \mathbf{if}\;\pi \cdot \ell \leq -5 \cdot 10^{+37} \lor \neg \left(\pi \cdot \ell \leq 100000000000\right):\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \ell - \frac{\frac{\tan \left(\pi \cdot \ell\right)}{F}}{F}\\ \end{array} \]
Alternative 4
Accuracy97.6%
Cost26569
\[\begin{array}{l} \mathbf{if}\;\pi \cdot \ell \leq -5 \cdot 10^{+37} \lor \neg \left(\pi \cdot \ell \leq 100000000000\right):\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \ell - \frac{\frac{\pi}{\frac{F}{\ell}}}{F}\\ \end{array} \]
Alternative 5
Accuracy92.1%
Cost13641
\[\begin{array}{l} \mathbf{if}\;\ell \leq -2.6 \cdot 10^{+36} \lor \neg \left(\ell \leq 45000000000\right):\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \ell - \pi \cdot \frac{\ell}{F \cdot F}\\ \end{array} \]
Alternative 6
Accuracy92.1%
Cost13641
\[\begin{array}{l} \mathbf{if}\;\ell \leq -2.6 \cdot 10^{+36} \lor \neg \left(\ell \leq 45000000000\right):\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{else}:\\ \;\;\;\;\pi \cdot \ell - \frac{\pi \cdot \ell}{F \cdot F}\\ \end{array} \]
Alternative 7
Accuracy91.7%
Cost7305
\[\begin{array}{l} \mathbf{if}\;\ell \leq -2.6 \cdot 10^{+36} \lor \neg \left(\ell \leq 45000000000\right):\\ \;\;\;\;\pi \cdot \ell\\ \mathbf{else}:\\ \;\;\;\;\ell \cdot \left(\pi \cdot \left(1 + \frac{\frac{-1}{F}}{F}\right)\right)\\ \end{array} \]
Alternative 8
Accuracy73.9%
Cost6528
\[\pi \cdot \ell \]

Reproduce?

herbie shell --seed 2023167 
(FPCore (F l)
  :name "VandenBroeck and Keller, Equation (6)"
  :precision binary64
  (- (* PI l) (* (/ 1.0 (* F F)) (tan (* PI l)))))