ab-angle->ABCF B

Percentage Accurate: 54.4% → 58.2%

Time: 26.1s

Alternatives: 5

Speedup: 27.9×

• 32.7% of points produce a very large (infinite) output. You may want to add a precondition.

Specification

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \pi \cdot \frac{angle}{180}\\ \left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin t\_0\right) \cdot \cos t\_0 \end{array} \end{array} \]

FPCore

(FPCore (a b angle)
 :precision binary64
 (let* ((t_0 (* PI (/ angle 180.0))))
   (* (* (* 2.0 (- (pow b 2.0) (pow a 2.0))) (sin t_0)) (cos t_0))))

C

double code(double a, double b, double angle) {
	double t_0 = ((double) M_PI) * (angle / 180.0);
	return ((2.0 * (pow(b, 2.0) - pow(a, 2.0))) * sin(t_0)) * cos(t_0);
}

Java

public static double code(double a, double b, double angle) {
	double t_0 = Math.PI * (angle / 180.0);
	return ((2.0 * (Math.pow(b, 2.0) - Math.pow(a, 2.0))) * Math.sin(t_0)) * Math.cos(t_0);
}

Python

def code(a, b, angle):
	t_0 = math.pi * (angle / 180.0)
	return ((2.0 * (math.pow(b, 2.0) - math.pow(a, 2.0))) * math.sin(t_0)) * math.cos(t_0)

Julia

function code(a, b, angle)
	t_0 = Float64(pi * Float64(angle / 180.0))
	return Float64(Float64(Float64(2.0 * Float64((b ^ 2.0) - (a ^ 2.0))) * sin(t_0)) * cos(t_0))
end

MATLAB

function tmp = code(a, b, angle)
	t_0 = pi * (angle / 180.0);
	tmp = ((2.0 * ((b ^ 2.0) - (a ^ 2.0))) * sin(t_0)) * cos(t_0);
end

Wolfram

code[a_, b_, angle_] := Block[{t$95$0 = N[(Pi * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]}, N[(N[(N[(2.0 * N[(N[Power[b, 2.0], $MachinePrecision] - N[Power[a, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[Sin[t$95$0], $MachinePrecision]), $MachinePrecision] * N[Cos[t$95$0], $MachinePrecision]), $MachinePrecision]]

Initial Program: 54.4% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \pi \cdot \frac{angle}{180}\\ \left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin t\_0\right) \cdot \cos t\_0 \end{array} \end{array} \]

FPCore

(FPCore (a b angle)
 :precision binary64
 (let* ((t_0 (* PI (/ angle 180.0))))
   (* (* (* 2.0 (- (pow b 2.0) (pow a 2.0))) (sin t_0)) (cos t_0))))

C

double code(double a, double b, double angle) {
	double t_0 = ((double) M_PI) * (angle / 180.0);
	return ((2.0 * (pow(b, 2.0) - pow(a, 2.0))) * sin(t_0)) * cos(t_0);
}

Java

public static double code(double a, double b, double angle) {
	double t_0 = Math.PI * (angle / 180.0);
	return ((2.0 * (Math.pow(b, 2.0) - Math.pow(a, 2.0))) * Math.sin(t_0)) * Math.cos(t_0);
}

Python

def code(a, b, angle):
	t_0 = math.pi * (angle / 180.0)
	return ((2.0 * (math.pow(b, 2.0) - math.pow(a, 2.0))) * math.sin(t_0)) * math.cos(t_0)

Julia

function code(a, b, angle)
	t_0 = Float64(pi * Float64(angle / 180.0))
	return Float64(Float64(Float64(2.0 * Float64((b ^ 2.0) - (a ^ 2.0))) * sin(t_0)) * cos(t_0))
end

MATLAB

function tmp = code(a, b, angle)
	t_0 = pi * (angle / 180.0);
	tmp = ((2.0 * ((b ^ 2.0) - (a ^ 2.0))) * sin(t_0)) * cos(t_0);
end

Wolfram

code[a_, b_, angle_] := Block[{t$95$0 = N[(Pi * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]}, N[(N[(N[(2.0 * N[(N[Power[b, 2.0], $MachinePrecision] - N[Power[a, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[Sin[t$95$0], $MachinePrecision]), $MachinePrecision] * N[Cos[t$95$0], $MachinePrecision]), $MachinePrecision]]

Alternative 1: 58.2% accurate, 1.9× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \pi \cdot \frac{angle}{180}\\ \cos t\_0 \cdot \left(\left(2 \cdot \left(\left(b - a\right) \cdot \left(b + a\right)\right)\right) \cdot \sin t\_0\right) \end{array} \end{array} \]

FPCore

(FPCore (a b angle)
 :precision binary64
 (let* ((t_0 (* PI (/ angle 180.0))))
   (* (cos t_0) (* (* 2.0 (* (- b a) (+ b a))) (sin t_0)))))

C

double code(double a, double b, double angle) {
	double t_0 = ((double) M_PI) * (angle / 180.0);
	return cos(t_0) * ((2.0 * ((b - a) * (b + a))) * sin(t_0));
}

Java

public static double code(double a, double b, double angle) {
	double t_0 = Math.PI * (angle / 180.0);
	return Math.cos(t_0) * ((2.0 * ((b - a) * (b + a))) * Math.sin(t_0));
}

Python

def code(a, b, angle):
	t_0 = math.pi * (angle / 180.0)
	return math.cos(t_0) * ((2.0 * ((b - a) * (b + a))) * math.sin(t_0))

Julia

function code(a, b, angle)
	t_0 = Float64(pi * Float64(angle / 180.0))
	return Float64(cos(t_0) * Float64(Float64(2.0 * Float64(Float64(b - a) * Float64(b + a))) * sin(t_0)))
end

MATLAB

function tmp = code(a, b, angle)
	t_0 = pi * (angle / 180.0);
	tmp = cos(t_0) * ((2.0 * ((b - a) * (b + a))) * sin(t_0));
end

Wolfram

code[a_, b_, angle_] := Block[{t$95$0 = N[(Pi * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]}, N[(N[Cos[t$95$0], $MachinePrecision] * N[(N[(2.0 * N[(N[(b - a), $MachinePrecision] * N[(b + a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[Sin[t$95$0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Initial program 58.9%

   \[\left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
2. Add Preprocessing
3. Step-by-step derivation

   1. unpow2 58.9%

      \[\leadsto \left(\left(2 \cdot \left(\color{blue}{b \cdot b} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   2. unpow2 58.9%

      \[\leadsto \left(\left(2 \cdot \left(b \cdot b - \color{blue}{a \cdot a}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   3. difference-of-squares 63.3%

      \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

4. Applied egg-rr 63.3%

   \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

5. Final simplification 63.3%

   \[\leadsto \cos \left(\pi \cdot \frac{angle}{180}\right) \cdot \left(\left(2 \cdot \left(\left(b - a\right) \cdot \left(b + a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \]
6. Add Preprocessing

Alternative 2: 58.1% accurate, 1.9× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := angle \cdot \frac{\pi}{-180}\\ 2 \cdot \left(\cos t\_0 \cdot \left(\sin t\_0 \cdot \left(\left(b + a\right) \cdot \left(a - b\right)\right)\right)\right) \end{array} \end{array} \]

FPCore

(FPCore (a b angle)
 :precision binary64
 (let* ((t_0 (* angle (/ PI -180.0))))
   (* 2.0 (* (cos t_0) (* (sin t_0) (* (+ b a) (- a b)))))))

C

double code(double a, double b, double angle) {
	double t_0 = angle * (((double) M_PI) / -180.0);
	return 2.0 * (cos(t_0) * (sin(t_0) * ((b + a) * (a - b))));
}

Java

public static double code(double a, double b, double angle) {
	double t_0 = angle * (Math.PI / -180.0);
	return 2.0 * (Math.cos(t_0) * (Math.sin(t_0) * ((b + a) * (a - b))));
}

Python

def code(a, b, angle):
	t_0 = angle * (math.pi / -180.0)
	return 2.0 * (math.cos(t_0) * (math.sin(t_0) * ((b + a) * (a - b))))

Julia

function code(a, b, angle)
	t_0 = Float64(angle * Float64(pi / -180.0))
	return Float64(2.0 * Float64(cos(t_0) * Float64(sin(t_0) * Float64(Float64(b + a) * Float64(a - b)))))
end

MATLAB

function tmp = code(a, b, angle)
	t_0 = angle * (pi / -180.0);
	tmp = 2.0 * (cos(t_0) * (sin(t_0) * ((b + a) * (a - b))));
end

Wolfram

code[a_, b_, angle_] := Block[{t$95$0 = N[(angle * N[(Pi / -180.0), $MachinePrecision]), $MachinePrecision]}, N[(2.0 * N[(N[Cos[t$95$0], $MachinePrecision] * N[(N[Sin[t$95$0], $MachinePrecision] * N[(N[(b + a), $MachinePrecision] * N[(a - b), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Initial program 58.9%

   \[\left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

3. Simplified 58.4%

   \[\leadsto \color{blue}{2 \cdot \left(\cos \left(angle \cdot \frac{\pi}{-180}\right) \cdot \left(\sin \left(angle \cdot \frac{\pi}{-180}\right) \cdot \left({a}^{2} - {b}^{2}\right)\right)\right)} \]
4. Add Preprocessing
5. Step-by-step derivation

   1. unpow2 58.4%

      \[\leadsto 2 \cdot \left(\cos \left(angle \cdot \frac{\pi}{-180}\right) \cdot \left(\sin \left(angle \cdot \frac{\pi}{-180}\right) \cdot \left(\color{blue}{a \cdot a} - {b}^{2}\right)\right)\right) \]

   2. unpow2 58.4%

      \[\leadsto 2 \cdot \left(\cos \left(angle \cdot \frac{\pi}{-180}\right) \cdot \left(\sin \left(angle \cdot \frac{\pi}{-180}\right) \cdot \left(a \cdot a - \color{blue}{b \cdot b}\right)\right)\right) \]

   3. difference-of-squares 62.4%

      \[\leadsto 2 \cdot \left(\cos \left(angle \cdot \frac{\pi}{-180}\right) \cdot \left(\sin \left(angle \cdot \frac{\pi}{-180}\right) \cdot \color{blue}{\left(\left(a + b\right) \cdot \left(a - b\right)\right)}\right)\right) \]

6. Applied egg-rr 62.4%

   \[\leadsto 2 \cdot \left(\cos \left(angle \cdot \frac{\pi}{-180}\right) \cdot \left(\sin \left(angle \cdot \frac{\pi}{-180}\right) \cdot \color{blue}{\left(\left(a + b\right) \cdot \left(a - b\right)\right)}\right)\right) \]

7. Final simplification 62.4%

   \[\leadsto 2 \cdot \left(\cos \left(angle \cdot \frac{\pi}{-180}\right) \cdot \left(\sin \left(angle \cdot \frac{\pi}{-180}\right) \cdot \left(\left(b + a\right) \cdot \left(a - b\right)\right)\right)\right) \]
8. Add Preprocessing

Alternative 3: 56.6% accurate, 3.4× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \left(b - a\right) \cdot \left(b + a\right)\\ t_1 := \pi \cdot \frac{angle}{180}\\ \mathbf{if}\;a \leq 2 \cdot 10^{+219}:\\ \;\;\;\;\left(2 \cdot t\_0\right) \cdot \sin t\_1\\ \mathbf{else}:\\ \;\;\;\;\cos t\_1 \cdot \left(0.011111111111111112 \cdot \left(angle \cdot \left(t\_0 \cdot \pi\right)\right)\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (a b angle)
 :precision binary64
 (let* ((t_0 (* (- b a) (+ b a))) (t_1 (* PI (/ angle 180.0))))
   (if (<= a 2e+219)
     (* (* 2.0 t_0) (sin t_1))
     (* (cos t_1) (* 0.011111111111111112 (* angle (* t_0 PI)))))))

C

double code(double a, double b, double angle) {
	double t_0 = (b - a) * (b + a);
	double t_1 = ((double) M_PI) * (angle / 180.0);
	double tmp;
	if (a <= 2e+219) {
		tmp = (2.0 * t_0) * sin(t_1);
	} else {
		tmp = cos(t_1) * (0.011111111111111112 * (angle * (t_0 * ((double) M_PI))));
	}
	return tmp;
}

Java

public static double code(double a, double b, double angle) {
	double t_0 = (b - a) * (b + a);
	double t_1 = Math.PI * (angle / 180.0);
	double tmp;
	if (a <= 2e+219) {
		tmp = (2.0 * t_0) * Math.sin(t_1);
	} else {
		tmp = Math.cos(t_1) * (0.011111111111111112 * (angle * (t_0 * Math.PI)));
	}
	return tmp;
}

Python

def code(a, b, angle):
	t_0 = (b - a) * (b + a)
	t_1 = math.pi * (angle / 180.0)
	tmp = 0
	if a <= 2e+219:
		tmp = (2.0 * t_0) * math.sin(t_1)
	else:
		tmp = math.cos(t_1) * (0.011111111111111112 * (angle * (t_0 * math.pi)))
	return tmp

Julia

function code(a, b, angle)
	t_0 = Float64(Float64(b - a) * Float64(b + a))
	t_1 = Float64(pi * Float64(angle / 180.0))
	tmp = 0.0
	if (a <= 2e+219)
		tmp = Float64(Float64(2.0 * t_0) * sin(t_1));
	else
		tmp = Float64(cos(t_1) * Float64(0.011111111111111112 * Float64(angle * Float64(t_0 * pi))));
	end
	return tmp
end

MATLAB

function tmp_2 = code(a, b, angle)
	t_0 = (b - a) * (b + a);
	t_1 = pi * (angle / 180.0);
	tmp = 0.0;
	if (a <= 2e+219)
		tmp = (2.0 * t_0) * sin(t_1);
	else
		tmp = cos(t_1) * (0.011111111111111112 * (angle * (t_0 * pi)));
	end
	tmp_2 = tmp;
end

Wolfram

code[a_, b_, angle_] := Block[{t$95$0 = N[(N[(b - a), $MachinePrecision] * N[(b + a), $MachinePrecision]), $MachinePrecision]}, Block[{t$95$1 = N[(Pi * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[a, 2e+219], N[(N[(2.0 * t$95$0), $MachinePrecision] * N[Sin[t$95$1], $MachinePrecision]), $MachinePrecision], N[(N[Cos[t$95$1], $MachinePrecision] * N[(0.011111111111111112 * N[(angle * N[(t$95$0 * Pi), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]]

Derivation

1. Split input into 2 regimes

2. if a < 1.99999999999999993e219

   1. Initial program 59.6%

      \[\left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. unpow2 59.6%

         \[\leadsto \left(\left(2 \cdot \left(\color{blue}{b \cdot b} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      2. unpow2 59.6%

         \[\leadsto \left(\left(2 \cdot \left(b \cdot b - \color{blue}{a \cdot a}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      3. difference-of-squares 62.3%

         \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   4. Applied egg-rr 62.3%

      \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   5. Step-by-step derivation

      1. add-cbrt-cube 60.2%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\color{blue}{\sqrt[3]{\left(\pi \cdot \pi\right) \cdot \pi}} \cdot \frac{angle}{180}\right) \]

      2. pow3 60.2%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\sqrt[3]{\color{blue}{{\pi}^{3}}} \cdot \frac{angle}{180}\right) \]

   6. Applied egg-rr 60.2%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\color{blue}{\sqrt[3]{{\pi}^{3}}} \cdot \frac{angle}{180}\right) \]

   7. Taylor expanded in angle around 0 56.5%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \color{blue}{1} \]

   if 1.99999999999999993e219 < a

   1. Initial program 50.8%

      \[\left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. unpow2 50.8%

         \[\leadsto \left(\left(2 \cdot \left(\color{blue}{b \cdot b} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      2. unpow2 50.8%

         \[\leadsto \left(\left(2 \cdot \left(b \cdot b - \color{blue}{a \cdot a}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      3. difference-of-squares 73.9%

         \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   4. Applied egg-rr 73.9%

      \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   5. Taylor expanded in angle around 0 64.8%

      \[\leadsto \color{blue}{\left(0.011111111111111112 \cdot \left(angle \cdot \left(\pi \cdot \left(\left(a + b\right) \cdot \left(b - a\right)\right)\right)\right)\right)} \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
3. Recombined 2 regimes into one program.

4. Final simplification 57.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;a \leq 2 \cdot 10^{+219}:\\ \;\;\;\;\left(2 \cdot \left(\left(b - a\right) \cdot \left(b + a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\\ \mathbf{else}:\\ \;\;\;\;\cos \left(\pi \cdot \frac{angle}{180}\right) \cdot \left(0.011111111111111112 \cdot \left(angle \cdot \left(\left(\left(b - a\right) \cdot \left(b + a\right)\right) \cdot \pi\right)\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 4: 56.4% accurate, 3.5× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \left(b - a\right) \cdot \left(b + a\right)\\ \mathbf{if}\;a \leq 10^{+187}:\\ \;\;\;\;\left(2 \cdot t\_0\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\\ \mathbf{else}:\\ \;\;\;\;2 \cdot \left(0.005555555555555556 \cdot \left(angle \cdot \left(t\_0 \cdot \pi\right)\right)\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (a b angle)
 :precision binary64
 (let* ((t_0 (* (- b a) (+ b a))))
   (if (<= a 1e+187)
     (* (* 2.0 t_0) (sin (* PI (/ angle 180.0))))
     (* 2.0 (* 0.005555555555555556 (* angle (* t_0 PI)))))))

C

double code(double a, double b, double angle) {
	double t_0 = (b - a) * (b + a);
	double tmp;
	if (a <= 1e+187) {
		tmp = (2.0 * t_0) * sin((((double) M_PI) * (angle / 180.0)));
	} else {
		tmp = 2.0 * (0.005555555555555556 * (angle * (t_0 * ((double) M_PI))));
	}
	return tmp;
}

Java

public static double code(double a, double b, double angle) {
	double t_0 = (b - a) * (b + a);
	double tmp;
	if (a <= 1e+187) {
		tmp = (2.0 * t_0) * Math.sin((Math.PI * (angle / 180.0)));
	} else {
		tmp = 2.0 * (0.005555555555555556 * (angle * (t_0 * Math.PI)));
	}
	return tmp;
}

Python

def code(a, b, angle):
	t_0 = (b - a) * (b + a)
	tmp = 0
	if a <= 1e+187:
		tmp = (2.0 * t_0) * math.sin((math.pi * (angle / 180.0)))
	else:
		tmp = 2.0 * (0.005555555555555556 * (angle * (t_0 * math.pi)))
	return tmp

Julia

function code(a, b, angle)
	t_0 = Float64(Float64(b - a) * Float64(b + a))
	tmp = 0.0
	if (a <= 1e+187)
		tmp = Float64(Float64(2.0 * t_0) * sin(Float64(pi * Float64(angle / 180.0))));
	else
		tmp = Float64(2.0 * Float64(0.005555555555555556 * Float64(angle * Float64(t_0 * pi))));
	end
	return tmp
end

MATLAB

function tmp_2 = code(a, b, angle)
	t_0 = (b - a) * (b + a);
	tmp = 0.0;
	if (a <= 1e+187)
		tmp = (2.0 * t_0) * sin((pi * (angle / 180.0)));
	else
		tmp = 2.0 * (0.005555555555555556 * (angle * (t_0 * pi)));
	end
	tmp_2 = tmp;
end

Wolfram

code[a_, b_, angle_] := Block[{t$95$0 = N[(N[(b - a), $MachinePrecision] * N[(b + a), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[a, 1e+187], N[(N[(2.0 * t$95$0), $MachinePrecision] * N[Sin[N[(Pi * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[(2.0 * N[(0.005555555555555556 * N[(angle * N[(t$95$0 * Pi), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if a < 9.99999999999999907e186

   1. Initial program 60.6%

      \[\left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. unpow2 60.6%

         \[\leadsto \left(\left(2 \cdot \left(\color{blue}{b \cdot b} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      2. unpow2 60.6%

         \[\leadsto \left(\left(2 \cdot \left(b \cdot b - \color{blue}{a \cdot a}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      3. difference-of-squares 62.9%

         \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   4. Applied egg-rr 62.9%

      \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]
   5. Step-by-step derivation

      1. add-cbrt-cube 60.8%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\color{blue}{\sqrt[3]{\left(\pi \cdot \pi\right) \cdot \pi}} \cdot \frac{angle}{180}\right) \]

      2. pow3 60.8%

         \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\sqrt[3]{\color{blue}{{\pi}^{3}}} \cdot \frac{angle}{180}\right) \]

   6. Applied egg-rr 60.8%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\color{blue}{\sqrt[3]{{\pi}^{3}}} \cdot \frac{angle}{180}\right) \]

   7. Taylor expanded in angle around 0 56.4%

      \[\leadsto \left(\left(2 \cdot \left(\left(b + a\right) \cdot \left(b - a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \color{blue}{1} \]

   if 9.99999999999999907e186 < a

   1. Initial program 46.4%

      \[\left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   3. Simplified 46.4%

      \[\leadsto \color{blue}{2 \cdot \left(\left({b}^{2} - {a}^{2}\right) \cdot \left(\sin \left(\pi \cdot \frac{angle}{180}\right) \cdot \cos \left(\frac{\pi \cdot angle}{-180}\right)\right)\right)} \]
   4. Add Preprocessing

   5. Taylor expanded in angle around 0 49.6%

      \[\leadsto 2 \cdot \color{blue}{\left(0.005555555555555556 \cdot \left(angle \cdot \left(\pi \cdot \left({b}^{2} - {a}^{2}\right)\right)\right)\right)} \]
   6. Step-by-step derivation

      1. unpow2 46.4%

         \[\leadsto \left(\left(2 \cdot \left(\color{blue}{b \cdot b} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      2. unpow2 46.4%

         \[\leadsto \left(\left(2 \cdot \left(b \cdot b - \color{blue}{a \cdot a}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

      3. difference-of-squares 65.9%

         \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   7. Applied egg-rr 65.9%

      \[\leadsto 2 \cdot \left(0.005555555555555556 \cdot \left(angle \cdot \left(\pi \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right)\right)\right) \]
3. Recombined 2 regimes into one program.

4. Final simplification 57.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;a \leq 10^{+187}:\\ \;\;\;\;\left(2 \cdot \left(\left(b - a\right) \cdot \left(b + a\right)\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\\ \mathbf{else}:\\ \;\;\;\;2 \cdot \left(0.005555555555555556 \cdot \left(angle \cdot \left(\left(\left(b - a\right) \cdot \left(b + a\right)\right) \cdot \pi\right)\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 5: 54.5% accurate, 27.9× speedup

Math

\[\begin{array}{l} \\ 2 \cdot \left(0.005555555555555556 \cdot \left(angle \cdot \left(\left(\left(b - a\right) \cdot \left(b + a\right)\right) \cdot \pi\right)\right)\right) \end{array} \]

FPCore

(FPCore (a b angle)
 :precision binary64
 (* 2.0 (* 0.005555555555555556 (* angle (* (* (- b a) (+ b a)) PI)))))

C

double code(double a, double b, double angle) {
	return 2.0 * (0.005555555555555556 * (angle * (((b - a) * (b + a)) * ((double) M_PI))));
}

Java

public static double code(double a, double b, double angle) {
	return 2.0 * (0.005555555555555556 * (angle * (((b - a) * (b + a)) * Math.PI)));
}

Python

def code(a, b, angle):
	return 2.0 * (0.005555555555555556 * (angle * (((b - a) * (b + a)) * math.pi)))

Julia

function code(a, b, angle)
	return Float64(2.0 * Float64(0.005555555555555556 * Float64(angle * Float64(Float64(Float64(b - a) * Float64(b + a)) * pi))))
end

MATLAB

function tmp = code(a, b, angle)
	tmp = 2.0 * (0.005555555555555556 * (angle * (((b - a) * (b + a)) * pi)));
end

Wolfram

code[a_, b_, angle_] := N[(2.0 * N[(0.005555555555555556 * N[(angle * N[(N[(N[(b - a), $MachinePrecision] * N[(b + a), $MachinePrecision]), $MachinePrecision] * Pi), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 58.9%

   \[\left(\left(2 \cdot \left({b}^{2} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

3. Simplified 58.4%

   \[\leadsto \color{blue}{2 \cdot \left(\left({b}^{2} - {a}^{2}\right) \cdot \left(\sin \left(\pi \cdot \frac{angle}{180}\right) \cdot \cos \left(\frac{\pi \cdot angle}{-180}\right)\right)\right)} \]
4. Add Preprocessing

5. Taylor expanded in angle around 0 52.9%

   \[\leadsto 2 \cdot \color{blue}{\left(0.005555555555555556 \cdot \left(angle \cdot \left(\pi \cdot \left({b}^{2} - {a}^{2}\right)\right)\right)\right)} \]
6. Step-by-step derivation

   1. unpow2 58.9%

      \[\leadsto \left(\left(2 \cdot \left(\color{blue}{b \cdot b} - {a}^{2}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   2. unpow2 58.9%

      \[\leadsto \left(\left(2 \cdot \left(b \cdot b - \color{blue}{a \cdot a}\right)\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

   3. difference-of-squares 63.3%

      \[\leadsto \left(\left(2 \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right) \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right) \cdot \cos \left(\pi \cdot \frac{angle}{180}\right) \]

7. Applied egg-rr 56.5%

   \[\leadsto 2 \cdot \left(0.005555555555555556 \cdot \left(angle \cdot \left(\pi \cdot \color{blue}{\left(\left(b + a\right) \cdot \left(b - a\right)\right)}\right)\right)\right) \]

8. Final simplification 56.5%

   \[\leadsto 2 \cdot \left(0.005555555555555556 \cdot \left(angle \cdot \left(\left(\left(b - a\right) \cdot \left(b + a\right)\right) \cdot \pi\right)\right)\right) \]
9. Add Preprocessing

Reproduce

herbie shell --seed 2024039 
(FPCore (a b angle)
  :name "ab-angle->ABCF B"
  :precision binary64
  (* (* (* 2.0 (- (pow b 2.0) (pow a 2.0))) (sin (* PI (/ angle 180.0)))) (cos (* PI (/ angle 180.0)))))