Ian Simplification

Percentage Accurate: 6.8% → 8.2%

Time: 24.3s

Alternatives: 9

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: 8.2% accurate, 0.8× speedup

Math

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

FPCore

(FPCore (x)
 :precision binary64
 (if (<= x -1e-310)
   (- (/ PI 2.0) (* 2.0 (asin (sqrt (/ (- 1.0 x) 2.0)))))
   (- (/ PI 2.0) (* 2.0 (asin (/ (sqrt (- 1.0 x)) (sqrt 2.0)))))))

C

double code(double x) {
	double tmp;
	if (x <= -1e-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((sqrt((1.0 - x)) / sqrt(2.0))));
	}
	return tmp;
}

Java

public static double code(double x) {
	double tmp;
	if (x <= -1e-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((Math.sqrt((1.0 - x)) / Math.sqrt(2.0))));
	}
	return tmp;
}

Python

def code(x):
	tmp = 0
	if x <= -1e-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((math.sqrt((1.0 - x)) / math.sqrt(2.0))))
	return tmp

Julia

function code(x)
	tmp = 0.0
	if (x <= -1e-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(sqrt(Float64(1.0 - x)) / sqrt(2.0)))));
	end
	return tmp
end

MATLAB

function tmp_2 = code(x)
	tmp = 0.0;
	if (x <= -1e-310)
		tmp = (pi / 2.0) - (2.0 * asin(sqrt(((1.0 - x) / 2.0))));
	else
		tmp = (pi / 2.0) - (2.0 * asin((sqrt((1.0 - x)) / sqrt(2.0))));
	end
	tmp_2 = tmp;
end

Wolfram

code[x_] := If[LessEqual[x, -1e-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[(N[Sqrt[N[(1.0 - x), $MachinePrecision]], $MachinePrecision] / N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if x < -9.999999999999969e-311

   1. Initial program 7.7%

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

   if -9.999999999999969e-311 < x

   1. Initial program 5.3%

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

      1. sqrt-div 8.9%

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

      2. div-inv 8.9%

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

   3. Applied egg-rr 8.9%

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

      1. associate-*r/ 8.9%

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

      2. *-rgt-identity 8.9%

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

   5. Simplified 8.9%

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

4. Final simplification 8.3%

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

Alternative 2: 8.2% accurate, 0.1× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := 0.5 - 0.5 \cdot x\\ t_1 := \sqrt{t_0}\\ t_2 := {\sin^{-1} t_1}^{2}\\ \frac{{\pi}^{2} \cdot 0.25 - 4 \cdot \sqrt[3]{t_2 \cdot \left(t_2 \cdot t_2\right)}}{\mathsf{fma}\left(2, \sin^{-1} \left(\sqrt[3]{t_0 \cdot t_1}\right), 0.5 \cdot \left(\sqrt[3]{\pi} \cdot {\left(\sqrt[3]{\pi}\right)}^{2}\right)\right)} \end{array} \end{array} \]

FPCore

(FPCore (x)
 :precision binary64
 (let* ((t_0 (- 0.5 (* 0.5 x))) (t_1 (sqrt t_0)) (t_2 (pow (asin t_1) 2.0)))
   (/
    (- (* (pow PI 2.0) 0.25) (* 4.0 (cbrt (* t_2 (* t_2 t_2)))))
    (fma
     2.0
     (asin (cbrt (* t_0 t_1)))
     (* 0.5 (* (cbrt PI) (pow (cbrt PI) 2.0)))))))

C

double code(double x) {
	double t_0 = 0.5 - (0.5 * x);
	double t_1 = sqrt(t_0);
	double t_2 = pow(asin(t_1), 2.0);
	return ((pow(((double) M_PI), 2.0) * 0.25) - (4.0 * cbrt((t_2 * (t_2 * t_2))))) / fma(2.0, asin(cbrt((t_0 * t_1))), (0.5 * (cbrt(((double) M_PI)) * pow(cbrt(((double) M_PI)), 2.0))));
}

Julia

function code(x)
	t_0 = Float64(0.5 - Float64(0.5 * x))
	t_1 = sqrt(t_0)
	t_2 = asin(t_1) ^ 2.0
	return Float64(Float64(Float64((pi ^ 2.0) * 0.25) - Float64(4.0 * cbrt(Float64(t_2 * Float64(t_2 * t_2))))) / fma(2.0, asin(cbrt(Float64(t_0 * t_1))), Float64(0.5 * Float64(cbrt(pi) * (cbrt(pi) ^ 2.0)))))
end

Wolfram

code[x_] := Block[{t$95$0 = N[(0.5 - N[(0.5 * x), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[Sqrt[t$95$0], $MachinePrecision]}, Block[{t$95$2 = N[Power[N[ArcSin[t$95$1], $MachinePrecision], 2.0], $MachinePrecision]}, N[(N[(N[(N[Power[Pi, 2.0], $MachinePrecision] * 0.25), $MachinePrecision] - N[(4.0 * N[Power[N[(t$95$2 * N[(t$95$2 * t$95$2), $MachinePrecision]), $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / N[(2.0 * N[ArcSin[N[Power[N[(t$95$0 * t$95$1), $MachinePrecision], 1/3], $MachinePrecision]], $MachinePrecision] + N[(0.5 * N[(N[Power[Pi, 1/3], $MachinePrecision] * N[Power[N[Power[Pi, 1/3], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]]

Derivation

1. Initial program 6.5%

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

   1. flip-- 6.4%

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

3. Applied egg-rr 6.4%

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

   1. add-cbrt-cube 7.6%

      \[\leadsto \frac{{\pi}^{2} \cdot 0.25 - 4 \cdot \color{blue}{\sqrt[3]{\left({\sin^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right)}^{2} \cdot {\sin^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right)}^{2}\right) \cdot {\sin^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right)}^{2}}}}{\mathsf{fma}\left(2, \sin^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right), \pi \cdot 0.5\right)} \]

   2. *-commutative 7.6%

      \[\leadsto \frac{{\pi}^{2} \cdot 0.25 - 4 \cdot \sqrt[3]{\left({\sin^{-1} \left(\sqrt{0.5 - \color{blue}{0.5 \cdot x}}\right)}^{2} \cdot {\sin^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right)}^{2}\right) \cdot {\sin^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right)}^{2}}}{\mathsf{fma}\left(2, \sin^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right), \pi \cdot 0.5\right)} \]

   3. *-commutative 7.6%

      \[\leadsto \frac{{\pi}^{2} \cdot 0.25 - 4 \cdot \sqrt[3]{\left({\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2} \cdot {\sin^{-1} \left(\sqrt{0.5 - \color{blue}{0.5 \cdot x}}\right)}^{2}\right) \cdot {\sin^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right)}^{2}}}{\mathsf{fma}\left(2, \sin^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right), \pi \cdot 0.5\right)} \]

   4. *-commutative 7.6%

      \[\leadsto \frac{{\pi}^{2} \cdot 0.25 - 4 \cdot \sqrt[3]{\left({\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2} \cdot {\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2}\right) \cdot {\sin^{-1} \left(\sqrt{0.5 - \color{blue}{0.5 \cdot x}}\right)}^{2}}}{\mathsf{fma}\left(2, \sin^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right), \pi \cdot 0.5\right)} \]

5. Applied egg-rr 7.6%

   \[\leadsto \frac{{\pi}^{2} \cdot 0.25 - 4 \cdot \color{blue}{\sqrt[3]{\left({\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2} \cdot {\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2}\right) \cdot {\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2}}}}{\mathsf{fma}\left(2, \sin^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right), \pi \cdot 0.5\right)} \]
6. Step-by-step derivation

   1. add-cbrt-cube 7.6%

      \[\leadsto \frac{{\pi}^{2} \cdot 0.25 - 4 \cdot \sqrt[3]{\left({\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2} \cdot {\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2}\right) \cdot {\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2}}}{\mathsf{fma}\left(2, \sin^{-1} \color{blue}{\left(\sqrt[3]{\left(\sqrt{0.5 - x \cdot 0.5} \cdot \sqrt{0.5 - x \cdot 0.5}\right) \cdot \sqrt{0.5 - x \cdot 0.5}}\right)}, \pi \cdot 0.5\right)} \]

   2. add-sqr-sqrt 7.6%

      \[\leadsto \frac{{\pi}^{2} \cdot 0.25 - 4 \cdot \sqrt[3]{\left({\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2} \cdot {\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2}\right) \cdot {\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2}}}{\mathsf{fma}\left(2, \sin^{-1} \left(\sqrt[3]{\color{blue}{\left(0.5 - x \cdot 0.5\right)} \cdot \sqrt{0.5 - x \cdot 0.5}}\right), \pi \cdot 0.5\right)} \]

   3. *-commutative 7.6%

      \[\leadsto \frac{{\pi}^{2} \cdot 0.25 - 4 \cdot \sqrt[3]{\left({\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2} \cdot {\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2}\right) \cdot {\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2}}}{\mathsf{fma}\left(2, \sin^{-1} \left(\sqrt[3]{\left(0.5 - \color{blue}{0.5 \cdot x}\right) \cdot \sqrt{0.5 - x \cdot 0.5}}\right), \pi \cdot 0.5\right)} \]

   4. *-commutative 7.6%

      \[\leadsto \frac{{\pi}^{2} \cdot 0.25 - 4 \cdot \sqrt[3]{\left({\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2} \cdot {\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2}\right) \cdot {\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2}}}{\mathsf{fma}\left(2, \sin^{-1} \left(\sqrt[3]{\left(0.5 - 0.5 \cdot x\right) \cdot \sqrt{0.5 - \color{blue}{0.5 \cdot x}}}\right), \pi \cdot 0.5\right)} \]

7. Applied egg-rr 7.6%

   \[\leadsto \frac{{\pi}^{2} \cdot 0.25 - 4 \cdot \sqrt[3]{\left({\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2} \cdot {\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2}\right) \cdot {\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2}}}{\mathsf{fma}\left(2, \sin^{-1} \color{blue}{\left(\sqrt[3]{\left(0.5 - 0.5 \cdot x\right) \cdot \sqrt{0.5 - 0.5 \cdot x}}\right)}, \pi \cdot 0.5\right)} \]
8. Step-by-step derivation

   1. add-cube-cbrt 7.6%

      \[\leadsto \frac{{\pi}^{2} \cdot 0.25 - 4 \cdot \sqrt[3]{\left({\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2} \cdot {\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2}\right) \cdot {\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2}}}{\mathsf{fma}\left(2, \sin^{-1} \left(\sqrt[3]{\left(0.5 - 0.5 \cdot x\right) \cdot \sqrt{0.5 - 0.5 \cdot x}}\right), \color{blue}{\left(\left(\sqrt[3]{\pi} \cdot \sqrt[3]{\pi}\right) \cdot \sqrt[3]{\pi}\right)} \cdot 0.5\right)} \]

   2. pow2 7.6%

      \[\leadsto \frac{{\pi}^{2} \cdot 0.25 - 4 \cdot \sqrt[3]{\left({\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2} \cdot {\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2}\right) \cdot {\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2}}}{\mathsf{fma}\left(2, \sin^{-1} \left(\sqrt[3]{\left(0.5 - 0.5 \cdot x\right) \cdot \sqrt{0.5 - 0.5 \cdot x}}\right), \left(\color{blue}{{\left(\sqrt[3]{\pi}\right)}^{2}} \cdot \sqrt[3]{\pi}\right) \cdot 0.5\right)} \]

9. Applied egg-rr 7.6%

   \[\leadsto \frac{{\pi}^{2} \cdot 0.25 - 4 \cdot \sqrt[3]{\left({\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2} \cdot {\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2}\right) \cdot {\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2}}}{\mathsf{fma}\left(2, \sin^{-1} \left(\sqrt[3]{\left(0.5 - 0.5 \cdot x\right) \cdot \sqrt{0.5 - 0.5 \cdot x}}\right), \color{blue}{\left({\left(\sqrt[3]{\pi}\right)}^{2} \cdot \sqrt[3]{\pi}\right)} \cdot 0.5\right)} \]

10. Final simplification 7.6%

    \[\leadsto \frac{{\pi}^{2} \cdot 0.25 - 4 \cdot \sqrt[3]{{\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2} \cdot \left({\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2} \cdot {\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2}\right)}}{\mathsf{fma}\left(2, \sin^{-1} \left(\sqrt[3]{\left(0.5 - 0.5 \cdot x\right) \cdot \sqrt{0.5 - 0.5 \cdot x}}\right), 0.5 \cdot \left(\sqrt[3]{\pi} \cdot {\left(\sqrt[3]{\pi}\right)}^{2}\right)\right)} \]

Alternative 3: 8.2% accurate, 0.2× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := 0.5 - 0.5 \cdot x\\ t_1 := \sqrt{t_0}\\ \frac{{\pi}^{2} \cdot 0.25 - 4 \cdot \sqrt[3]{{\sin^{-1} t_1}^{2} \cdot {\left({\sin^{-1} \left(\sqrt{0.5 + x \cdot -0.5}\right)}^{2}\right)}^{2}}}{\mathsf{fma}\left(2, \sin^{-1} \left(\sqrt[3]{t_0 \cdot t_1}\right), \pi \cdot 0.5\right)} \end{array} \end{array} \]

FPCore

(FPCore (x)
 :precision binary64
 (let* ((t_0 (- 0.5 (* 0.5 x))) (t_1 (sqrt t_0)))
   (/
    (-
     (* (pow PI 2.0) 0.25)
     (*
      4.0
      (cbrt
       (*
        (pow (asin t_1) 2.0)
        (pow (pow (asin (sqrt (+ 0.5 (* x -0.5)))) 2.0) 2.0)))))
    (fma 2.0 (asin (cbrt (* t_0 t_1))) (* PI 0.5)))))

C

double code(double x) {
	double t_0 = 0.5 - (0.5 * x);
	double t_1 = sqrt(t_0);
	return ((pow(((double) M_PI), 2.0) * 0.25) - (4.0 * cbrt((pow(asin(t_1), 2.0) * pow(pow(asin(sqrt((0.5 + (x * -0.5)))), 2.0), 2.0))))) / fma(2.0, asin(cbrt((t_0 * t_1))), (((double) M_PI) * 0.5));
}

Julia

function code(x)
	t_0 = Float64(0.5 - Float64(0.5 * x))
	t_1 = sqrt(t_0)
	return Float64(Float64(Float64((pi ^ 2.0) * 0.25) - Float64(4.0 * cbrt(Float64((asin(t_1) ^ 2.0) * ((asin(sqrt(Float64(0.5 + Float64(x * -0.5)))) ^ 2.0) ^ 2.0))))) / fma(2.0, asin(cbrt(Float64(t_0 * t_1))), Float64(pi * 0.5)))
end

Wolfram

code[x_] := Block[{t$95$0 = N[(0.5 - N[(0.5 * x), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[Sqrt[t$95$0], $MachinePrecision]}, N[(N[(N[(N[Power[Pi, 2.0], $MachinePrecision] * 0.25), $MachinePrecision] - N[(4.0 * N[Power[N[(N[Power[N[ArcSin[t$95$1], $MachinePrecision], 2.0], $MachinePrecision] * N[Power[N[Power[N[ArcSin[N[Sqrt[N[(0.5 + N[(x * -0.5), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]], $MachinePrecision], 2.0], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / N[(2.0 * N[ArcSin[N[Power[N[(t$95$0 * t$95$1), $MachinePrecision], 1/3], $MachinePrecision]], $MachinePrecision] + N[(Pi * 0.5), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Initial program 6.5%

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

   1. flip-- 6.4%

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

3. Applied egg-rr 6.4%

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

   1. add-cbrt-cube 7.6%

      \[\leadsto \frac{{\pi}^{2} \cdot 0.25 - 4 \cdot \color{blue}{\sqrt[3]{\left({\sin^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right)}^{2} \cdot {\sin^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right)}^{2}\right) \cdot {\sin^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right)}^{2}}}}{\mathsf{fma}\left(2, \sin^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right), \pi \cdot 0.5\right)} \]

   2. *-commutative 7.6%

      \[\leadsto \frac{{\pi}^{2} \cdot 0.25 - 4 \cdot \sqrt[3]{\left({\sin^{-1} \left(\sqrt{0.5 - \color{blue}{0.5 \cdot x}}\right)}^{2} \cdot {\sin^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right)}^{2}\right) \cdot {\sin^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right)}^{2}}}{\mathsf{fma}\left(2, \sin^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right), \pi \cdot 0.5\right)} \]

   3. *-commutative 7.6%

      \[\leadsto \frac{{\pi}^{2} \cdot 0.25 - 4 \cdot \sqrt[3]{\left({\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2} \cdot {\sin^{-1} \left(\sqrt{0.5 - \color{blue}{0.5 \cdot x}}\right)}^{2}\right) \cdot {\sin^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right)}^{2}}}{\mathsf{fma}\left(2, \sin^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right), \pi \cdot 0.5\right)} \]

   4. *-commutative 7.6%

      \[\leadsto \frac{{\pi}^{2} \cdot 0.25 - 4 \cdot \sqrt[3]{\left({\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2} \cdot {\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2}\right) \cdot {\sin^{-1} \left(\sqrt{0.5 - \color{blue}{0.5 \cdot x}}\right)}^{2}}}{\mathsf{fma}\left(2, \sin^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right), \pi \cdot 0.5\right)} \]

5. Applied egg-rr 7.6%

   \[\leadsto \frac{{\pi}^{2} \cdot 0.25 - 4 \cdot \color{blue}{\sqrt[3]{\left({\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2} \cdot {\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2}\right) \cdot {\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2}}}}{\mathsf{fma}\left(2, \sin^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right), \pi \cdot 0.5\right)} \]
6. Step-by-step derivation

   1. add-cbrt-cube 7.6%

      \[\leadsto \frac{{\pi}^{2} \cdot 0.25 - 4 \cdot \sqrt[3]{\left({\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2} \cdot {\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2}\right) \cdot {\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2}}}{\mathsf{fma}\left(2, \sin^{-1} \color{blue}{\left(\sqrt[3]{\left(\sqrt{0.5 - x \cdot 0.5} \cdot \sqrt{0.5 - x \cdot 0.5}\right) \cdot \sqrt{0.5 - x \cdot 0.5}}\right)}, \pi \cdot 0.5\right)} \]

   2. add-sqr-sqrt 7.6%

      \[\leadsto \frac{{\pi}^{2} \cdot 0.25 - 4 \cdot \sqrt[3]{\left({\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2} \cdot {\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2}\right) \cdot {\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2}}}{\mathsf{fma}\left(2, \sin^{-1} \left(\sqrt[3]{\color{blue}{\left(0.5 - x \cdot 0.5\right)} \cdot \sqrt{0.5 - x \cdot 0.5}}\right), \pi \cdot 0.5\right)} \]

   3. *-commutative 7.6%

      \[\leadsto \frac{{\pi}^{2} \cdot 0.25 - 4 \cdot \sqrt[3]{\left({\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2} \cdot {\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2}\right) \cdot {\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2}}}{\mathsf{fma}\left(2, \sin^{-1} \left(\sqrt[3]{\left(0.5 - \color{blue}{0.5 \cdot x}\right) \cdot \sqrt{0.5 - x \cdot 0.5}}\right), \pi \cdot 0.5\right)} \]

   4. *-commutative 7.6%

      \[\leadsto \frac{{\pi}^{2} \cdot 0.25 - 4 \cdot \sqrt[3]{\left({\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2} \cdot {\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2}\right) \cdot {\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2}}}{\mathsf{fma}\left(2, \sin^{-1} \left(\sqrt[3]{\left(0.5 - 0.5 \cdot x\right) \cdot \sqrt{0.5 - \color{blue}{0.5 \cdot x}}}\right), \pi \cdot 0.5\right)} \]

7. Applied egg-rr 7.6%

   \[\leadsto \frac{{\pi}^{2} \cdot 0.25 - 4 \cdot \sqrt[3]{\left({\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2} \cdot {\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2}\right) \cdot {\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2}}}{\mathsf{fma}\left(2, \sin^{-1} \color{blue}{\left(\sqrt[3]{\left(0.5 - 0.5 \cdot x\right) \cdot \sqrt{0.5 - 0.5 \cdot x}}\right)}, \pi \cdot 0.5\right)} \]
8. Step-by-step derivation

   1. pow2 7.6%

      \[\leadsto \frac{{\pi}^{2} \cdot 0.25 - 4 \cdot \sqrt[3]{\color{blue}{{\left({\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2}\right)}^{2}} \cdot {\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2}}}{\mathsf{fma}\left(2, \sin^{-1} \left(\sqrt[3]{\left(0.5 - 0.5 \cdot x\right) \cdot \sqrt{0.5 - 0.5 \cdot x}}\right), \pi \cdot 0.5\right)} \]

   2. cancel-sign-sub-inv 7.6%

      \[\leadsto \frac{{\pi}^{2} \cdot 0.25 - 4 \cdot \sqrt[3]{{\left({\sin^{-1} \left(\sqrt{\color{blue}{0.5 + \left(-0.5\right) \cdot x}}\right)}^{2}\right)}^{2} \cdot {\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2}}}{\mathsf{fma}\left(2, \sin^{-1} \left(\sqrt[3]{\left(0.5 - 0.5 \cdot x\right) \cdot \sqrt{0.5 - 0.5 \cdot x}}\right), \pi \cdot 0.5\right)} \]

   3. metadata-eval 7.6%

      \[\leadsto \frac{{\pi}^{2} \cdot 0.25 - 4 \cdot \sqrt[3]{{\left({\sin^{-1} \left(\sqrt{0.5 + \color{blue}{-0.5} \cdot x}\right)}^{2}\right)}^{2} \cdot {\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2}}}{\mathsf{fma}\left(2, \sin^{-1} \left(\sqrt[3]{\left(0.5 - 0.5 \cdot x\right) \cdot \sqrt{0.5 - 0.5 \cdot x}}\right), \pi \cdot 0.5\right)} \]

9. Applied egg-rr 7.6%

   \[\leadsto \frac{{\pi}^{2} \cdot 0.25 - 4 \cdot \sqrt[3]{\color{blue}{{\left({\sin^{-1} \left(\sqrt{0.5 + -0.5 \cdot x}\right)}^{2}\right)}^{2}} \cdot {\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2}}}{\mathsf{fma}\left(2, \sin^{-1} \left(\sqrt[3]{\left(0.5 - 0.5 \cdot x\right) \cdot \sqrt{0.5 - 0.5 \cdot x}}\right), \pi \cdot 0.5\right)} \]

10. Final simplification 7.6%

    \[\leadsto \frac{{\pi}^{2} \cdot 0.25 - 4 \cdot \sqrt[3]{{\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2} \cdot {\left({\sin^{-1} \left(\sqrt{0.5 + x \cdot -0.5}\right)}^{2}\right)}^{2}}}{\mathsf{fma}\left(2, \sin^{-1} \left(\sqrt[3]{\left(0.5 - 0.5 \cdot x\right) \cdot \sqrt{0.5 - 0.5 \cdot x}}\right), \pi \cdot 0.5\right)} \]

Alternative 4: 8.2% accurate, 0.2× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)\\ \frac{{\pi}^{2} \cdot 0.25 - 4 \cdot \sqrt[3]{{t_0}^{2} \cdot {t_0}^{4}}}{\mathsf{fma}\left(2, t_0, \pi \cdot 0.5\right)} \end{array} \end{array} \]

FPCore

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

C

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

Julia

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

Wolfram

code[x_] := Block[{t$95$0 = N[ArcSin[N[Sqrt[N[(0.5 - N[(0.5 * x), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]], $MachinePrecision]}, N[(N[(N[(N[Power[Pi, 2.0], $MachinePrecision] * 0.25), $MachinePrecision] - N[(4.0 * N[Power[N[(N[Power[t$95$0, 2.0], $MachinePrecision] * N[Power[t$95$0, 4.0], $MachinePrecision]), $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / N[(2.0 * t$95$0 + N[(Pi * 0.5), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Initial program 6.5%

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

   1. flip-- 6.4%

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

3. Applied egg-rr 6.4%

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

   1. add-cbrt-cube 7.6%

      \[\leadsto \frac{{\pi}^{2} \cdot 0.25 - 4 \cdot \color{blue}{\sqrt[3]{\left({\sin^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right)}^{2} \cdot {\sin^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right)}^{2}\right) \cdot {\sin^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right)}^{2}}}}{\mathsf{fma}\left(2, \sin^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right), \pi \cdot 0.5\right)} \]

   2. *-commutative 7.6%

      \[\leadsto \frac{{\pi}^{2} \cdot 0.25 - 4 \cdot \sqrt[3]{\left({\sin^{-1} \left(\sqrt{0.5 - \color{blue}{0.5 \cdot x}}\right)}^{2} \cdot {\sin^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right)}^{2}\right) \cdot {\sin^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right)}^{2}}}{\mathsf{fma}\left(2, \sin^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right), \pi \cdot 0.5\right)} \]

   3. *-commutative 7.6%

      \[\leadsto \frac{{\pi}^{2} \cdot 0.25 - 4 \cdot \sqrt[3]{\left({\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2} \cdot {\sin^{-1} \left(\sqrt{0.5 - \color{blue}{0.5 \cdot x}}\right)}^{2}\right) \cdot {\sin^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right)}^{2}}}{\mathsf{fma}\left(2, \sin^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right), \pi \cdot 0.5\right)} \]

   4. *-commutative 7.6%

      \[\leadsto \frac{{\pi}^{2} \cdot 0.25 - 4 \cdot \sqrt[3]{\left({\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2} \cdot {\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2}\right) \cdot {\sin^{-1} \left(\sqrt{0.5 - \color{blue}{0.5 \cdot x}}\right)}^{2}}}{\mathsf{fma}\left(2, \sin^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right), \pi \cdot 0.5\right)} \]

5. Applied egg-rr 7.6%

   \[\leadsto \frac{{\pi}^{2} \cdot 0.25 - 4 \cdot \color{blue}{\sqrt[3]{\left({\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2} \cdot {\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2}\right) \cdot {\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2}}}}{\mathsf{fma}\left(2, \sin^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right), \pi \cdot 0.5\right)} \]

6. Taylor expanded in x around 0 7.6%

   \[\leadsto \frac{{\pi}^{2} \cdot 0.25 - 4 \cdot \sqrt[3]{\color{blue}{{\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{4}} \cdot {\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2}}}{\mathsf{fma}\left(2, \sin^{-1} \left(\sqrt{0.5 - x \cdot 0.5}\right), \pi \cdot 0.5\right)} \]

7. Final simplification 7.6%

   \[\leadsto \frac{{\pi}^{2} \cdot 0.25 - 4 \cdot \sqrt[3]{{\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{2} \cdot {\sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right)}^{4}}}{\mathsf{fma}\left(2, \sin^{-1} \left(\sqrt{0.5 - 0.5 \cdot x}\right), \pi \cdot 0.5\right)} \]

Alternative 5: 8.3% accurate, 0.8× 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 6.5%

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

   1. asin-acos 7.6%

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

   2. div-inv 7.6%

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

   3. metadata-eval 7.6%

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

   4. div-sub 7.6%

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

   5. metadata-eval 7.6%

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

   6. div-inv 7.6%

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

   7. metadata-eval 7.6%

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

3. Applied egg-rr 7.6%

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

4. Final simplification 7.6%

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

Alternative 6: 6.7% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

code[x_] := 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. Initial program 6.5%

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

   1. clear-num 6.4%

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

   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%

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

3. Applied egg-rr 6.7%

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

4. Final simplification 6.7%

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

Alternative 7: 5.8% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;x \leq -1 \cdot 10^{-310}:\\ \;\;\;\;\frac{\pi}{2} - 2 \cdot \sin^{-1} \left(\sqrt{0.5}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{\pi}{2} - 2 \cdot \sin^{-1} \left(\frac{1}{\sqrt{2}}\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (x)
 :precision binary64
 (if (<= x -1e-310)
   (- (/ PI 2.0) (* 2.0 (asin (sqrt 0.5))))
   (- (/ PI 2.0) (* 2.0 (asin (/ 1.0 (sqrt 2.0)))))))

C

double code(double x) {
	double tmp;
	if (x <= -1e-310) {
		tmp = (((double) M_PI) / 2.0) - (2.0 * asin(sqrt(0.5)));
	} else {
		tmp = (((double) M_PI) / 2.0) - (2.0 * asin((1.0 / sqrt(2.0))));
	}
	return tmp;
}

Java

public static double code(double x) {
	double tmp;
	if (x <= -1e-310) {
		tmp = (Math.PI / 2.0) - (2.0 * Math.asin(Math.sqrt(0.5)));
	} else {
		tmp = (Math.PI / 2.0) - (2.0 * Math.asin((1.0 / Math.sqrt(2.0))));
	}
	return tmp;
}

Python

def code(x):
	tmp = 0
	if x <= -1e-310:
		tmp = (math.pi / 2.0) - (2.0 * math.asin(math.sqrt(0.5)))
	else:
		tmp = (math.pi / 2.0) - (2.0 * math.asin((1.0 / math.sqrt(2.0))))
	return tmp

Julia

function code(x)
	tmp = 0.0
	if (x <= -1e-310)
		tmp = Float64(Float64(pi / 2.0) - Float64(2.0 * asin(sqrt(0.5))));
	else
		tmp = Float64(Float64(pi / 2.0) - Float64(2.0 * asin(Float64(1.0 / sqrt(2.0)))));
	end
	return tmp
end

MATLAB

function tmp_2 = code(x)
	tmp = 0.0;
	if (x <= -1e-310)
		tmp = (pi / 2.0) - (2.0 * asin(sqrt(0.5)));
	else
		tmp = (pi / 2.0) - (2.0 * asin((1.0 / sqrt(2.0))));
	end
	tmp_2 = tmp;
end

Wolfram

code[x_] := If[LessEqual[x, -1e-310], N[(N[(Pi / 2.0), $MachinePrecision] - N[(2.0 * N[ArcSin[N[Sqrt[0.5], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(Pi / 2.0), $MachinePrecision] - N[(2.0 * N[ArcSin[N[(1.0 / N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if x < -9.999999999999969e-311

   1. Initial program 7.7%

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

   2. Taylor expanded in x around 0 6.1%

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

   if -9.999999999999969e-311 < x

   1. Initial program 5.3%

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

      1. clear-num 5.3%

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

      2. sqrt-div 8.8%

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

      3. metadata-eval 8.8%

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

   3. Applied egg-rr 8.8%

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

   4. Taylor expanded in x around 0 6.5%

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

4. Final simplification 6.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq -1 \cdot 10^{-310}:\\ \;\;\;\;\frac{\pi}{2} - 2 \cdot \sin^{-1} \left(\sqrt{0.5}\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{\pi}{2} - 2 \cdot \sin^{-1} \left(\frac{1}{\sqrt{2}}\right)\\ \end{array} \]

Alternative 8: 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 6.5%

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

2. Final simplification 6.5%

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

Alternative 9: 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 6.5%

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

2. Taylor expanded in x around 0 4.1%

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

3. Final simplification 4.1%

   \[\leadsto \frac{\pi}{2} - 2 \cdot \sin^{-1} \left(\sqrt{0.5}\right) \]

Developer target: 100.0% accurate, 3.1× 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 2023193 
(FPCore (x)
  :name "Ian Simplification"
  :precision binary64

  :herbie-target
  (asin x)

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