Ian Simplification

Percentage Accurate: 6.8% → 8.3%

Time: 13.4s

Alternatives: 5

Speedup: 1.0×

Specification

Math

\[\begin{array}{l} \\ \frac{\pi}{2} - 2 \cdot \sin^{-1} \left(\sqrt{\frac{1 - x}{2}}\right) \end{array} \]

FPCore

(FPCore (x)
 :precision binary64
 (- (/ PI 2.0) (* 2.0 (asin (sqrt (/ (- 1.0 x) 2.0))))))

C

double code(double x) {
	return (((double) M_PI) / 2.0) - (2.0 * asin(sqrt(((1.0 - x) / 2.0))));
}

Java

public static double code(double x) {
	return (Math.PI / 2.0) - (2.0 * Math.asin(Math.sqrt(((1.0 - x) / 2.0))));
}

Python

def code(x):
	return (math.pi / 2.0) - (2.0 * math.asin(math.sqrt(((1.0 - x) / 2.0))))

Julia

function code(x)
	return Float64(Float64(pi / 2.0) - Float64(2.0 * asin(sqrt(Float64(Float64(1.0 - x) / 2.0)))))
end

MATLAB

function tmp = code(x)
	tmp = (pi / 2.0) - (2.0 * asin(sqrt(((1.0 - x) / 2.0))));
end

Wolfram

code[x_] := N[(N[(Pi / 2.0), $MachinePrecision] - N[(2.0 * N[ArcSin[N[Sqrt[N[(N[(1.0 - x), $MachinePrecision] / 2.0), $MachinePrecision]], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Initial Program: 6.8% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \frac{\pi}{2} - 2 \cdot \sin^{-1} \left(\sqrt{\frac{1 - x}{2}}\right) \end{array} \]

FPCore

(FPCore (x)
 :precision binary64
 (- (/ PI 2.0) (* 2.0 (asin (sqrt (/ (- 1.0 x) 2.0))))))

C

double code(double x) {
	return (((double) M_PI) / 2.0) - (2.0 * asin(sqrt(((1.0 - x) / 2.0))));
}

Java

public static double code(double x) {
	return (Math.PI / 2.0) - (2.0 * Math.asin(Math.sqrt(((1.0 - x) / 2.0))));
}

Python

def code(x):
	return (math.pi / 2.0) - (2.0 * math.asin(math.sqrt(((1.0 - x) / 2.0))))

Julia

function code(x)
	return Float64(Float64(pi / 2.0) - Float64(2.0 * asin(sqrt(Float64(Float64(1.0 - x) / 2.0)))))
end

MATLAB

function tmp = code(x)
	tmp = (pi / 2.0) - (2.0 * asin(sqrt(((1.0 - x) / 2.0))));
end

Wolfram

code[x_] := N[(N[(Pi / 2.0), $MachinePrecision] - N[(2.0 * N[ArcSin[N[Sqrt[N[(N[(1.0 - x), $MachinePrecision] / 2.0), $MachinePrecision]], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Alternative 1: 8.3% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;x \leq -4 \cdot 10^{-310}:\\ \;\;\;\;\frac{\pi}{2} - 2 \cdot \sin^{-1} \left(\sqrt{\frac{1 - x}{2}}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{\pi}{2} - 2 \cdot \sin^{-1} \left(\frac{1}{\sqrt{\frac{2}{1 - x}}}\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (x)
 :precision binary64
 (if (<= x -4e-310)
   (- (/ PI 2.0) (* 2.0 (asin (sqrt (/ (- 1.0 x) 2.0)))))
   (- (/ PI 2.0) (* 2.0 (asin (/ 1.0 (sqrt (/ 2.0 (- 1.0 x)))))))))

C

double code(double x) {
	double tmp;
	if (x <= -4e-310) {
		tmp = (((double) M_PI) / 2.0) - (2.0 * asin(sqrt(((1.0 - x) / 2.0))));
	} else {
		tmp = (((double) M_PI) / 2.0) - (2.0 * asin((1.0 / sqrt((2.0 / (1.0 - x))))));
	}
	return tmp;
}

Java

public static double code(double x) {
	double tmp;
	if (x <= -4e-310) {
		tmp = (Math.PI / 2.0) - (2.0 * Math.asin(Math.sqrt(((1.0 - x) / 2.0))));
	} else {
		tmp = (Math.PI / 2.0) - (2.0 * Math.asin((1.0 / Math.sqrt((2.0 / (1.0 - x))))));
	}
	return tmp;
}

Python

def code(x):
	tmp = 0
	if x <= -4e-310:
		tmp = (math.pi / 2.0) - (2.0 * math.asin(math.sqrt(((1.0 - x) / 2.0))))
	else:
		tmp = (math.pi / 2.0) - (2.0 * math.asin((1.0 / math.sqrt((2.0 / (1.0 - x))))))
	return tmp

Julia

function code(x)
	tmp = 0.0
	if (x <= -4e-310)
		tmp = Float64(Float64(pi / 2.0) - Float64(2.0 * asin(sqrt(Float64(Float64(1.0 - x) / 2.0)))));
	else
		tmp = Float64(Float64(pi / 2.0) - Float64(2.0 * asin(Float64(1.0 / sqrt(Float64(2.0 / Float64(1.0 - x)))))));
	end
	return tmp
end

MATLAB

function tmp_2 = code(x)
	tmp = 0.0;
	if (x <= -4e-310)
		tmp = (pi / 2.0) - (2.0 * asin(sqrt(((1.0 - x) / 2.0))));
	else
		tmp = (pi / 2.0) - (2.0 * asin((1.0 / sqrt((2.0 / (1.0 - x))))));
	end
	tmp_2 = tmp;
end

Wolfram

code[x_] := If[LessEqual[x, -4e-310], N[(N[(Pi / 2.0), $MachinePrecision] - N[(2.0 * N[ArcSin[N[Sqrt[N[(N[(1.0 - x), $MachinePrecision] / 2.0), $MachinePrecision]], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(Pi / 2.0), $MachinePrecision] - N[(2.0 * N[ArcSin[N[(1.0 / N[Sqrt[N[(2.0 / N[(1.0 - x), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if x < -3.999999999999988e-310

   1. Initial program 10.2%

      \[\frac{\pi}{2} - 2 \cdot \sin^{-1} \left(\sqrt{\frac{1 - x}{2}}\right) \]
   2. Add Preprocessing

   if -3.999999999999988e-310 < x

   1. Initial program 3.5%

      \[\frac{\pi}{2} - 2 \cdot \sin^{-1} \left(\sqrt{\frac{1 - x}{2}}\right) \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. clear-num 3.5%

         \[\leadsto \frac{\pi}{2} - 2 \cdot \sin^{-1} \left(\sqrt{\color{blue}{\frac{1}{\frac{2}{1 - x}}}}\right) \]

      2. sqrt-div 6.7%

         \[\leadsto \frac{\pi}{2} - 2 \cdot \sin^{-1} \color{blue}{\left(\frac{\sqrt{1}}{\sqrt{\frac{2}{1 - x}}}\right)} \]

      3. metadata-eval 6.7%

         \[\leadsto \frac{\pi}{2} - 2 \cdot \sin^{-1} \left(\frac{\color{blue}{1}}{\sqrt{\frac{2}{1 - x}}}\right) \]

   4. Applied egg-rr 6.7%

      \[\leadsto \frac{\pi}{2} - 2 \cdot \sin^{-1} \color{blue}{\left(\frac{1}{\sqrt{\frac{2}{1 - x}}}\right)} \]
3. Recombined 2 regimes into one program.
4. Add Preprocessing

Alternative 2: 8.2% accurate, 0.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \sqrt{0.5 - 0.5 \cdot x}\\ \frac{0.25 \cdot \left(\pi \cdot \pi\right) - 4 \cdot {\left(\pi \cdot 0.5 - \cos^{-1} t\_0\right)}^{2}}{\pi \cdot 0.5 + 2 \cdot \sin^{-1} t\_0} \end{array} \end{array} \]

FPCore

(FPCore (x)
 :precision binary64
 (let* ((t_0 (sqrt (- 0.5 (* 0.5 x)))))
   (/
    (- (* 0.25 (* PI PI)) (* 4.0 (pow (- (* PI 0.5) (acos t_0)) 2.0)))
    (+ (* PI 0.5) (* 2.0 (asin t_0))))))

C

double code(double x) {
	double t_0 = sqrt((0.5 - (0.5 * x)));
	return ((0.25 * (((double) M_PI) * ((double) M_PI))) - (4.0 * pow(((((double) M_PI) * 0.5) - acos(t_0)), 2.0))) / ((((double) M_PI) * 0.5) + (2.0 * asin(t_0)));
}

Java

public static double code(double x) {
	double t_0 = Math.sqrt((0.5 - (0.5 * x)));
	return ((0.25 * (Math.PI * Math.PI)) - (4.0 * Math.pow(((Math.PI * 0.5) - Math.acos(t_0)), 2.0))) / ((Math.PI * 0.5) + (2.0 * Math.asin(t_0)));
}

Python

def code(x):
	t_0 = math.sqrt((0.5 - (0.5 * x)))
	return ((0.25 * (math.pi * math.pi)) - (4.0 * math.pow(((math.pi * 0.5) - math.acos(t_0)), 2.0))) / ((math.pi * 0.5) + (2.0 * math.asin(t_0)))

Julia

function code(x)
	t_0 = sqrt(Float64(0.5 - Float64(0.5 * x)))
	return Float64(Float64(Float64(0.25 * Float64(pi * pi)) - Float64(4.0 * (Float64(Float64(pi * 0.5) - acos(t_0)) ^ 2.0))) / Float64(Float64(pi * 0.5) + Float64(2.0 * asin(t_0))))
end

MATLAB

function tmp = code(x)
	t_0 = sqrt((0.5 - (0.5 * x)));
	tmp = ((0.25 * (pi * pi)) - (4.0 * (((pi * 0.5) - acos(t_0)) ^ 2.0))) / ((pi * 0.5) + (2.0 * asin(t_0)));
end

Wolfram

code[x_] := Block[{t$95$0 = N[Sqrt[N[(0.5 - N[(0.5 * x), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]}, N[(N[(N[(0.25 * N[(Pi * Pi), $MachinePrecision]), $MachinePrecision] - N[(4.0 * N[Power[N[(N[(Pi * 0.5), $MachinePrecision] - N[ArcCos[t$95$0], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / N[(N[(Pi * 0.5), $MachinePrecision] + N[(2.0 * N[ArcSin[t$95$0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Initial program 6.9%

   \[\frac{\pi}{2} - 2 \cdot \sin^{-1} \left(\sqrt{\frac{1 - x}{2}}\right) \]
2. Add Preprocessing
3. Step-by-step derivation

   1. flip-- 6.9%

      \[\leadsto \color{blue}{\frac{\frac{\pi}{2} \cdot \frac{\pi}{2} - \left(2 \cdot \sin^{-1} \left(\sqrt{\frac{1 - x}{2}}\right)\right) \cdot \left(2 \cdot \sin^{-1} \left(\sqrt{\frac{1 - x}{2}}\right)\right)}{\frac{\pi}{2} + 2 \cdot \sin^{-1} \left(\sqrt{\frac{1 - x}{2}}\right)}} \]

   2. div-inv 6.9%

      \[\leadsto \color{blue}{\left(\frac{\pi}{2} \cdot \frac{\pi}{2} - \left(2 \cdot \sin^{-1} \left(\sqrt{\frac{1 - x}{2}}\right)\right) \cdot \left(2 \cdot \sin^{-1} \left(\sqrt{\frac{1 - x}{2}}\right)\right)\right) \cdot \frac{1}{\frac{\pi}{2} + 2 \cdot \sin^{-1} \left(\sqrt{\frac{1 - x}{2}}\right)}} \]

4. Applied egg-rr 6.9%

   \[\leadsto \color{blue}{\left({\left(\pi \cdot 0.5\right)}^{2} - {\left(2 \cdot \sin^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right)\right)}^{2}\right) \cdot \frac{1}{\mathsf{fma}\left(2, \sin^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right), \pi \cdot 0.5\right)}} \]
5. Step-by-step derivation

   1. asin-acos 8.1%

      \[\leadsto \left({\left(\pi \cdot 0.5\right)}^{2} - {\left(2 \cdot \color{blue}{\left(\frac{\pi}{2} - \cos^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right)\right)}\right)}^{2}\right) \cdot \frac{1}{\mathsf{fma}\left(2, \sin^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right), \pi \cdot 0.5\right)} \]

   2. div-inv 8.1%

      \[\leadsto \left({\left(\pi \cdot 0.5\right)}^{2} - {\left(2 \cdot \left(\color{blue}{\pi \cdot \frac{1}{2}} - \cos^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right)\right)\right)}^{2}\right) \cdot \frac{1}{\mathsf{fma}\left(2, \sin^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right), \pi \cdot 0.5\right)} \]

   3. metadata-eval 8.1%

      \[\leadsto \left({\left(\pi \cdot 0.5\right)}^{2} - {\left(2 \cdot \left(\pi \cdot \color{blue}{0.5} - \cos^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right)\right)\right)}^{2}\right) \cdot \frac{1}{\mathsf{fma}\left(2, \sin^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right), \pi \cdot 0.5\right)} \]

   4. *-commutative 8.1%

      \[\leadsto \left({\left(\pi \cdot 0.5\right)}^{2} - {\left(2 \cdot \left(\pi \cdot 0.5 - \cos^{-1} \left(\sqrt{0.5 - \color{blue}{0.5 \cdot x}}\right)\right)\right)}^{2}\right) \cdot \frac{1}{\mathsf{fma}\left(2, \sin^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right), \pi \cdot 0.5\right)} \]

6. Applied egg-rr 8.1%

   \[\leadsto \left({\left(\pi \cdot 0.5\right)}^{2} - {\left(2 \cdot \color{blue}{\left(\pi \cdot 0.5 - \cos^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)\right)}\right)}^{2}\right) \cdot \frac{1}{\mathsf{fma}\left(2, \sin^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right), \pi \cdot 0.5\right)} \]

7. Taylor expanded in x around 0 8.1%

   \[\leadsto \color{blue}{\frac{0.25 \cdot {\pi}^{2} - 4 \cdot {\left(0.5 \cdot \pi - \cos^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)\right)}^{2}}{0.5 \cdot \pi + 2 \cdot \sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}} \]
8. Step-by-step derivation

   1. unpow2 8.1%

      \[\leadsto \frac{0.25 \cdot \color{blue}{\left(\pi \cdot \pi\right)} - 4 \cdot {\left(0.5 \cdot \pi - \cos^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)\right)}^{2}}{0.5 \cdot \pi + 2 \cdot \sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)} \]

9. Applied egg-rr 8.1%

   \[\leadsto \frac{0.25 \cdot \color{blue}{\left(\pi \cdot \pi\right)} - 4 \cdot {\left(0.5 \cdot \pi - \cos^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)\right)}^{2}}{0.5 \cdot \pi + 2 \cdot \sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)} \]

10. Final simplification 8.1%

    \[\leadsto \frac{0.25 \cdot \left(\pi \cdot \pi\right) - 4 \cdot {\left(\pi \cdot 0.5 - \cos^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)\right)}^{2}}{\pi \cdot 0.5 + 2 \cdot \sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)} \]
11. Add Preprocessing

Alternative 3: 8.2% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \frac{\pi}{2} + 2 \cdot \left(\cos^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right) - \pi \cdot 0.5\right) \end{array} \]

FPCore

(FPCore (x)
 :precision binary64
 (+ (/ PI 2.0) (* 2.0 (- (acos (sqrt (- 0.5 (* 0.5 x)))) (* PI 0.5)))))

C

double code(double x) {
	return (((double) M_PI) / 2.0) + (2.0 * (acos(sqrt((0.5 - (0.5 * x)))) - (((double) M_PI) * 0.5)));
}

Java

public static double code(double x) {
	return (Math.PI / 2.0) + (2.0 * (Math.acos(Math.sqrt((0.5 - (0.5 * x)))) - (Math.PI * 0.5)));
}

Python

def code(x):
	return (math.pi / 2.0) + (2.0 * (math.acos(math.sqrt((0.5 - (0.5 * x)))) - (math.pi * 0.5)))

Julia

function code(x)
	return Float64(Float64(pi / 2.0) + Float64(2.0 * Float64(acos(sqrt(Float64(0.5 - Float64(0.5 * x)))) - Float64(pi * 0.5))))
end

MATLAB

function tmp = code(x)
	tmp = (pi / 2.0) + (2.0 * (acos(sqrt((0.5 - (0.5 * x)))) - (pi * 0.5)));
end

Wolfram

code[x_] := N[(N[(Pi / 2.0), $MachinePrecision] + N[(2.0 * N[(N[ArcCos[N[Sqrt[N[(0.5 - N[(0.5 * x), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]], $MachinePrecision] - N[(Pi * 0.5), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 6.9%

   \[\frac{\pi}{2} - 2 \cdot \sin^{-1} \left(\sqrt{\frac{1 - x}{2}}\right) \]
2. Add Preprocessing
3. Step-by-step derivation

   1. asin-acos 8.1%

      \[\leadsto \frac{\pi}{2} - 2 \cdot \color{blue}{\left(\frac{\pi}{2} - \cos^{-1} \left(\sqrt{\frac{1 - x}{2}}\right)\right)} \]

   2. div-inv 8.1%

      \[\leadsto \frac{\pi}{2} - 2 \cdot \left(\color{blue}{\pi \cdot \frac{1}{2}} - \cos^{-1} \left(\sqrt{\frac{1 - x}{2}}\right)\right) \]

   3. metadata-eval 8.1%

      \[\leadsto \frac{\pi}{2} - 2 \cdot \left(\pi \cdot \color{blue}{0.5} - \cos^{-1} \left(\sqrt{\frac{1 - x}{2}}\right)\right) \]

   4. div-sub 8.1%

      \[\leadsto \frac{\pi}{2} - 2 \cdot \left(\pi \cdot 0.5 - \cos^{-1} \left(\sqrt{\color{blue}{\frac{1}{2} - \frac{x}{2}}}\right)\right) \]

   5. metadata-eval 8.1%

      \[\leadsto \frac{\pi}{2} - 2 \cdot \left(\pi \cdot 0.5 - \cos^{-1} \left(\sqrt{\color{blue}{0.5} - \frac{x}{2}}\right)\right) \]

   6. div-inv 8.1%

      \[\leadsto \frac{\pi}{2} - 2 \cdot \left(\pi \cdot 0.5 - \cos^{-1} \left(\sqrt{0.5 - \color{blue}{x \cdot \frac{1}{2}}}\right)\right) \]

   7. metadata-eval 8.1%

      \[\leadsto \frac{\pi}{2} - 2 \cdot \left(\pi \cdot 0.5 - \cos^{-1} \left(\sqrt{0.5 - x \cdot \color{blue}{0.5}}\right)\right) \]

4. Applied egg-rr 8.1%

   \[\leadsto \frac{\pi}{2} - 2 \cdot \color{blue}{\left(\pi \cdot 0.5 - \cos^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right)\right)} \]

5. Final simplification 8.1%

   \[\leadsto \frac{\pi}{2} + 2 \cdot \left(\cos^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right) - \pi \cdot 0.5\right) \]
6. Add Preprocessing

Alternative 4: 6.8% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \frac{\pi}{2} - 2 \cdot \sin^{-1} \left(\sqrt{\frac{1 - x}{2}}\right) \end{array} \]

FPCore

(FPCore (x)
 :precision binary64
 (- (/ PI 2.0) (* 2.0 (asin (sqrt (/ (- 1.0 x) 2.0))))))

C

double code(double x) {
	return (((double) M_PI) / 2.0) - (2.0 * asin(sqrt(((1.0 - x) / 2.0))));
}

Java

public static double code(double x) {
	return (Math.PI / 2.0) - (2.0 * Math.asin(Math.sqrt(((1.0 - x) / 2.0))));
}

Python

def code(x):
	return (math.pi / 2.0) - (2.0 * math.asin(math.sqrt(((1.0 - x) / 2.0))))

Julia

function code(x)
	return Float64(Float64(pi / 2.0) - Float64(2.0 * asin(sqrt(Float64(Float64(1.0 - x) / 2.0)))))
end

MATLAB

function tmp = code(x)
	tmp = (pi / 2.0) - (2.0 * asin(sqrt(((1.0 - x) / 2.0))));
end

Wolfram

code[x_] := N[(N[(Pi / 2.0), $MachinePrecision] - N[(2.0 * N[ArcSin[N[Sqrt[N[(N[(1.0 - x), $MachinePrecision] / 2.0), $MachinePrecision]], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 6.9%

   \[\frac{\pi}{2} - 2 \cdot \sin^{-1} \left(\sqrt{\frac{1 - x}{2}}\right) \]
2. Add Preprocessing
3. Add Preprocessing

Alternative 5: 4.1% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \frac{\pi}{2} - 2 \cdot \sin^{-1} \left(\sqrt{0.5}\right) \end{array} \]

FPCore

(FPCore (x) :precision binary64 (- (/ PI 2.0) (* 2.0 (asin (sqrt 0.5)))))

C

double code(double x) {
	return (((double) M_PI) / 2.0) - (2.0 * asin(sqrt(0.5)));
}

Java

public static double code(double x) {
	return (Math.PI / 2.0) - (2.0 * Math.asin(Math.sqrt(0.5)));
}

Python

def code(x):
	return (math.pi / 2.0) - (2.0 * math.asin(math.sqrt(0.5)))

Julia

function code(x)
	return Float64(Float64(pi / 2.0) - Float64(2.0 * asin(sqrt(0.5))))
end

MATLAB

function tmp = code(x)
	tmp = (pi / 2.0) - (2.0 * asin(sqrt(0.5)));
end

Wolfram

code[x_] := N[(N[(Pi / 2.0), $MachinePrecision] - N[(2.0 * N[ArcSin[N[Sqrt[0.5], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 6.9%

   \[\frac{\pi}{2} - 2 \cdot \sin^{-1} \left(\sqrt{\frac{1 - x}{2}}\right) \]
2. Add Preprocessing

3. Taylor expanded in x around 0 4.2%

   \[\leadsto \frac{\pi}{2} - 2 \cdot \sin^{-1} \color{blue}{\left(\sqrt{0.5}\right)} \]
4. Add Preprocessing

Developer Target 1: 100.0% accurate, 2.1× speedup

Math

\[\begin{array}{l} \\ \sin^{-1} x \end{array} \]

FPCore

(FPCore (x) :precision binary64 (asin x))

C

double code(double x) {
	return asin(x);
}

Fortran

real(8) function code(x)
    real(8), intent (in) :: x
    code = asin(x)
end function

Java

public static double code(double x) {
	return Math.asin(x);
}

Python

def code(x):
	return math.asin(x)

Julia

function code(x)
	return asin(x)
end

MATLAB

function tmp = code(x)
	tmp = asin(x);
end

Wolfram

code[x_] := N[ArcSin[x], $MachinePrecision]

Reproduce

herbie shell --seed 2024185 
(FPCore (x)
  :name "Ian Simplification"
  :precision binary64

  :alt
  (! :herbie-platform default (asin x))

  (- (/ PI 2.0) (* 2.0 (asin (sqrt (/ (- 1.0 x) 2.0))))))