Ian Simplification

Average Error: 59.7 → 58.3

Time: 27.9s

Precision: binary64

Cost: 20104

Math

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

↓

\[\begin{array}{l} t_0 := \frac{\pi}{2} - 2 \cdot \sin^{-1} \left(\sqrt{\frac{1 - x}{2}}\right)\\ \mathbf{if}\;x \leq -1.65 \cdot 10^{-162}:\\ \;\;\;\;t_0\\ \mathbf{elif}\;x \leq 5.5 \cdot 10^{-17}:\\ \;\;\;\;\left(\pi \cdot 0.5 + 1\right) + \left(-1 - 2 \cdot \sin^{-1} \left(\sqrt{0.5}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;t_0\\ \end{array} \]

FPCore

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

↓

(FPCore (x)
 :precision binary64
 (let* ((t_0 (- (/ PI 2.0) (* 2.0 (asin (sqrt (/ (- 1.0 x) 2.0)))))))
   (if (<= x -1.65e-162)
     t_0
     (if (<= x 5.5e-17)
       (+ (+ (* PI 0.5) 1.0) (- -1.0 (* 2.0 (asin (sqrt 0.5)))))
       t_0))))

C

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

↓

double code(double x) {
	double t_0 = (((double) M_PI) / 2.0) - (2.0 * asin(sqrt(((1.0 - x) / 2.0))));
	double tmp;
	if (x <= -1.65e-162) {
		tmp = t_0;
	} else if (x <= 5.5e-17) {
		tmp = ((((double) M_PI) * 0.5) + 1.0) + (-1.0 - (2.0 * asin(sqrt(0.5))));
	} else {
		tmp = t_0;
	}
	return tmp;
}

Java

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

↓

public static double code(double x) {
	double t_0 = (Math.PI / 2.0) - (2.0 * Math.asin(Math.sqrt(((1.0 - x) / 2.0))));
	double tmp;
	if (x <= -1.65e-162) {
		tmp = t_0;
	} else if (x <= 5.5e-17) {
		tmp = ((Math.PI * 0.5) + 1.0) + (-1.0 - (2.0 * Math.asin(Math.sqrt(0.5))));
	} else {
		tmp = t_0;
	}
	return tmp;
}

Python

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

↓

def code(x):
	t_0 = (math.pi / 2.0) - (2.0 * math.asin(math.sqrt(((1.0 - x) / 2.0))))
	tmp = 0
	if x <= -1.65e-162:
		tmp = t_0
	elif x <= 5.5e-17:
		tmp = ((math.pi * 0.5) + 1.0) + (-1.0 - (2.0 * math.asin(math.sqrt(0.5))))
	else:
		tmp = t_0
	return tmp

Julia

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

↓

function code(x)
	t_0 = Float64(Float64(pi / 2.0) - Float64(2.0 * asin(sqrt(Float64(Float64(1.0 - x) / 2.0)))))
	tmp = 0.0
	if (x <= -1.65e-162)
		tmp = t_0;
	elseif (x <= 5.5e-17)
		tmp = Float64(Float64(Float64(pi * 0.5) + 1.0) + Float64(-1.0 - Float64(2.0 * asin(sqrt(0.5)))));
	else
		tmp = t_0;
	end
	return tmp
end

MATLAB

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

↓

function tmp_2 = code(x)
	t_0 = (pi / 2.0) - (2.0 * asin(sqrt(((1.0 - x) / 2.0))));
	tmp = 0.0;
	if (x <= -1.65e-162)
		tmp = t_0;
	elseif (x <= 5.5e-17)
		tmp = ((pi * 0.5) + 1.0) + (-1.0 - (2.0 * asin(sqrt(0.5))));
	else
		tmp = t_0;
	end
	tmp_2 = tmp;
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]

↓

code[x_] := Block[{t$95$0 = 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]}, If[LessEqual[x, -1.65e-162], t$95$0, If[LessEqual[x, 5.5e-17], N[(N[(N[(Pi * 0.5), $MachinePrecision] + 1.0), $MachinePrecision] + N[(-1.0 - N[(2.0 * N[ArcSin[N[Sqrt[0.5], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], t$95$0]]]

Target

Original	59.7
Target	0
Herbie	58.3

\[\sin^{-1} x \]

Derivation

1. Split input into 2 regimes

2. if x < -1.65000000000000007e-162 or 5.50000000000000001e-17 < x

   1. Initial program 53.6

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

   if -1.65000000000000007e-162 < x < 5.50000000000000001e-17

   1. Initial program 62.1

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

   2. Applied egg-rr 60.1

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

   3. Taylor expanded in x around 0 60.1

      \[\leadsto \left(\pi \cdot 0.5 + 1\right) + \left(-1 - 2 \cdot \sin^{-1} \color{blue}{\left(\sqrt{0.5}\right)}\right) \]
3. Recombined 2 regimes into one program.

4. Final simplification 58.3

   \[\leadsto \begin{array}{l} \mathbf{if}\;x \leq -1.65 \cdot 10^{-162}:\\ \;\;\;\;\frac{\pi}{2} - 2 \cdot \sin^{-1} \left(\sqrt{\frac{1 - x}{2}}\right)\\ \mathbf{elif}\;x \leq 5.5 \cdot 10^{-17}:\\ \;\;\;\;\left(\pi \cdot 0.5 + 1\right) + \left(-1 - 2 \cdot \sin^{-1} \left(\sqrt{0.5}\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{\pi}{2} - 2 \cdot \sin^{-1} \left(\sqrt{\frac{1 - x}{2}}\right)\\ \end{array} \]

Alternatives

Alternative 1
Error	58.8
Cost	20096

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

Alternative 2
Error	60.5
Cost	19840

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

Alternative 3
Error	61.4
Cost	19584

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

Reproduce

herbie shell --seed 2023074 
(FPCore (x)
  :name "Ian Simplification"
  :precision binary64

  :herbie-target
  (asin x)

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