ab-angle->ABCF B

Percentage Accurate: 54.3% → 65.4%

Time: 36.4s

Alternatives: 5

Speedup: 5.5×

• 32.9% 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: 54.3% 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: 65.4% accurate, 1.5× speedup

Math

\[\begin{array}{l} a = |a|\\ \\ \begin{array}{l} \mathbf{if}\;a \leq 1.45 \cdot 10^{+250}:\\ \;\;\;\;2 \cdot \left(\left(a + b\right) \cdot \left(\left(a - b\right) \cdot \sin \left(\pi \cdot \left(angle \cdot -0.005555555555555556\right)\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(2 \cdot \sin \left(\pi \cdot {\left(\sqrt[3]{angle \cdot -0.005555555555555556}\right)}^{3}\right)\right) \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\\ \end{array} \end{array} \]

FPCore

NOTE: a should be positive before calling this function
(FPCore (a b angle)
 :precision binary64
 (if (<= a 1.45e+250)
   (* 2.0 (* (+ a b) (* (- a b) (sin (* PI (* angle -0.005555555555555556))))))
   (*
    (* 2.0 (sin (* PI (pow (cbrt (* angle -0.005555555555555556)) 3.0))))
    (* (+ a b) (- a b)))))

C

a = abs(a);
double code(double a, double b, double angle) {
	double tmp;
	if (a <= 1.45e+250) {
		tmp = 2.0 * ((a + b) * ((a - b) * sin((((double) M_PI) * (angle * -0.005555555555555556)))));
	} else {
		tmp = (2.0 * sin((((double) M_PI) * pow(cbrt((angle * -0.005555555555555556)), 3.0)))) * ((a + b) * (a - b));
	}
	return tmp;
}

Java

a = Math.abs(a);
public static double code(double a, double b, double angle) {
	double tmp;
	if (a <= 1.45e+250) {
		tmp = 2.0 * ((a + b) * ((a - b) * Math.sin((Math.PI * (angle * -0.005555555555555556)))));
	} else {
		tmp = (2.0 * Math.sin((Math.PI * Math.pow(Math.cbrt((angle * -0.005555555555555556)), 3.0)))) * ((a + b) * (a - b));
	}
	return tmp;
}

Julia

a = abs(a)
function code(a, b, angle)
	tmp = 0.0
	if (a <= 1.45e+250)
		tmp = Float64(2.0 * Float64(Float64(a + b) * Float64(Float64(a - b) * sin(Float64(pi * Float64(angle * -0.005555555555555556))))));
	else
		tmp = Float64(Float64(2.0 * sin(Float64(pi * (cbrt(Float64(angle * -0.005555555555555556)) ^ 3.0)))) * Float64(Float64(a + b) * Float64(a - b)));
	end
	return tmp
end

Wolfram

NOTE: a should be positive before calling this function
code[a_, b_, angle_] := If[LessEqual[a, 1.45e+250], N[(2.0 * N[(N[(a + b), $MachinePrecision] * N[(N[(a - b), $MachinePrecision] * N[Sin[N[(Pi * N[(angle * -0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(2.0 * N[Sin[N[(Pi * N[Power[N[Power[N[(angle * -0.005555555555555556), $MachinePrecision], 1/3], $MachinePrecision], 3.0], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] * N[(N[(a + b), $MachinePrecision] * N[(a - b), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if a < 1.45000000000000004e250

   1. Initial program 53.2%

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

   3. Simplified 54.2%

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

      1. unpow2 54.2%

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

      2. unpow2 54.2%

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

      3. difference-of-squares 59.2%

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

   5. Applied egg-rr 59.2%

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

   6. Taylor expanded in angle around 0 60.5%

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

   7. Taylor expanded in angle around 0 60.8%

      \[\leadsto \left(2 \cdot \sin \color{blue}{\left(-0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)}\right) \cdot \left(1 \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right) \]
   8. Step-by-step derivation

      1. *-commutative 60.8%

         \[\leadsto \left(2 \cdot \sin \color{blue}{\left(\left(angle \cdot \pi\right) \cdot -0.005555555555555556\right)}\right) \cdot \left(1 \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right) \]

      2. *-commutative 60.8%

         \[\leadsto \left(2 \cdot \sin \left(\color{blue}{\left(\pi \cdot angle\right)} \cdot -0.005555555555555556\right)\right) \cdot \left(1 \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right) \]

      3. associate-*r* 62.6%

         \[\leadsto \left(2 \cdot \sin \color{blue}{\left(\pi \cdot \left(angle \cdot -0.005555555555555556\right)\right)}\right) \cdot \left(1 \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right) \]

   9. Simplified 62.6%

      \[\leadsto \left(2 \cdot \sin \color{blue}{\left(\pi \cdot \left(angle \cdot -0.005555555555555556\right)\right)}\right) \cdot \left(1 \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right) \]

   10. Taylor expanded in angle around inf 60.8%

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

       1. *-commutative 60.8%

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

       2. *-commutative 60.8%

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

       3. *-commutative 60.8%

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

       4. associate-*r* 62.6%

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

       5. associate-*l* 67.9%

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

   12. Simplified 67.9%

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

   if 1.45000000000000004e250 < a

   1. Initial program 58.3%

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

   3. Simplified 58.3%

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

      1. unpow2 58.3%

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

      2. unpow2 58.3%

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

      3. difference-of-squares 67.9%

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

   5. Applied egg-rr 67.9%

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

   6. Taylor expanded in angle around 0 76.2%

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

   7. Taylor expanded in angle around 0 67.9%

      \[\leadsto \left(2 \cdot \sin \color{blue}{\left(-0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)}\right) \cdot \left(1 \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right) \]
   8. Step-by-step derivation

      1. *-commutative 67.9%

         \[\leadsto \left(2 \cdot \sin \color{blue}{\left(\left(angle \cdot \pi\right) \cdot -0.005555555555555556\right)}\right) \cdot \left(1 \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right) \]

      2. *-commutative 67.9%

         \[\leadsto \left(2 \cdot \sin \left(\color{blue}{\left(\pi \cdot angle\right)} \cdot -0.005555555555555556\right)\right) \cdot \left(1 \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right) \]

      3. associate-*r* 59.6%

         \[\leadsto \left(2 \cdot \sin \color{blue}{\left(\pi \cdot \left(angle \cdot -0.005555555555555556\right)\right)}\right) \cdot \left(1 \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right) \]

   9. Simplified 59.6%

      \[\leadsto \left(2 \cdot \sin \color{blue}{\left(\pi \cdot \left(angle \cdot -0.005555555555555556\right)\right)}\right) \cdot \left(1 \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right) \]
   10. Step-by-step derivation

       1. add-cube-cbrt 76.2%

          \[\leadsto \left(2 \cdot \sin \left(\pi \cdot \color{blue}{\left(\left(\sqrt[3]{angle \cdot -0.005555555555555556} \cdot \sqrt[3]{angle \cdot -0.005555555555555556}\right) \cdot \sqrt[3]{angle \cdot -0.005555555555555556}\right)}\right)\right) \cdot \left(1 \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right) \]

       2. pow3 84.6%

          \[\leadsto \left(2 \cdot \sin \left(\pi \cdot \color{blue}{{\left(\sqrt[3]{angle \cdot -0.005555555555555556}\right)}^{3}}\right)\right) \cdot \left(1 \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right) \]

   11. Applied egg-rr 84.6%

       \[\leadsto \left(2 \cdot \sin \left(\pi \cdot \color{blue}{{\left(\sqrt[3]{angle \cdot -0.005555555555555556}\right)}^{3}}\right)\right) \cdot \left(1 \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right) \]
3. Recombined 2 regimes into one program.

4. Final simplification 68.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;a \leq 1.45 \cdot 10^{+250}:\\ \;\;\;\;2 \cdot \left(\left(a + b\right) \cdot \left(\left(a - b\right) \cdot \sin \left(\pi \cdot \left(angle \cdot -0.005555555555555556\right)\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\left(2 \cdot \sin \left(\pi \cdot {\left(\sqrt[3]{angle \cdot -0.005555555555555556}\right)}^{3}\right)\right) \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\\ \end{array} \]

Alternative 2: 56.3% accurate, 2.9× speedup

Math

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

FPCore

NOTE: a should be positive before calling this function
(FPCore (a b angle)
 :precision binary64
 (* 2.0 (* (* (+ a b) (- a b)) (sin (* -0.005555555555555556 (* PI angle))))))

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

NOTE: a should be positive before calling this function
code[a_, b_, angle_] := N[(2.0 * N[(N[(N[(a + b), $MachinePrecision] * N[(a - b), $MachinePrecision]), $MachinePrecision] * N[Sin[N[(-0.005555555555555556 * N[(Pi * angle), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 53.4%

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

3. Simplified 54.4%

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

   1. unpow2 54.4%

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

   2. unpow2 54.4%

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

   3. difference-of-squares 59.6%

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

5. Applied egg-rr 59.6%

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

6. Taylor expanded in angle around 0 61.3%

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

7. Taylor expanded in angle around inf 61.2%

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

8. Final simplification 61.2%

   \[\leadsto 2 \cdot \left(\left(\left(a + b\right) \cdot \left(a - b\right)\right) \cdot \sin \left(-0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)\right) \]

Alternative 3: 66.5% accurate, 2.9× speedup

Math

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

FPCore

NOTE: a should be positive before calling this function
(FPCore (a b angle)
 :precision binary64
 (* 2.0 (* (+ a b) (* (- a b) (sin (* PI (* angle -0.005555555555555556)))))))

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

NOTE: a should be positive before calling this function
code[a_, b_, angle_] := N[(2.0 * N[(N[(a + b), $MachinePrecision] * N[(N[(a - b), $MachinePrecision] * N[Sin[N[(Pi * N[(angle * -0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 53.4%

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

3. Simplified 54.4%

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

   1. unpow2 54.4%

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

   2. unpow2 54.4%

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

   3. difference-of-squares 59.6%

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

5. Applied egg-rr 59.6%

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

6. Taylor expanded in angle around 0 61.3%

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

7. Taylor expanded in angle around 0 61.2%

   \[\leadsto \left(2 \cdot \sin \color{blue}{\left(-0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)}\right) \cdot \left(1 \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right) \]
8. Step-by-step derivation

   1. *-commutative 61.2%

      \[\leadsto \left(2 \cdot \sin \color{blue}{\left(\left(angle \cdot \pi\right) \cdot -0.005555555555555556\right)}\right) \cdot \left(1 \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right) \]

   2. *-commutative 61.2%

      \[\leadsto \left(2 \cdot \sin \left(\color{blue}{\left(\pi \cdot angle\right)} \cdot -0.005555555555555556\right)\right) \cdot \left(1 \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right) \]

   3. associate-*r* 62.5%

      \[\leadsto \left(2 \cdot \sin \color{blue}{\left(\pi \cdot \left(angle \cdot -0.005555555555555556\right)\right)}\right) \cdot \left(1 \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right) \]

9. Simplified 62.5%

   \[\leadsto \left(2 \cdot \sin \color{blue}{\left(\pi \cdot \left(angle \cdot -0.005555555555555556\right)\right)}\right) \cdot \left(1 \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right) \]

10. Taylor expanded in angle around inf 61.2%

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

    1. *-commutative 61.2%

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

    2. *-commutative 61.2%

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

    3. *-commutative 61.2%

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

    4. associate-*r* 62.5%

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

    5. associate-*l* 67.9%

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

12. Simplified 67.9%

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

13. Final simplification 67.9%

    \[\leadsto 2 \cdot \left(\left(a + b\right) \cdot \left(\left(a - b\right) \cdot \sin \left(\pi \cdot \left(angle \cdot -0.005555555555555556\right)\right)\right)\right) \]

Alternative 4: 54.4% accurate, 5.5× speedup

Math

\[\begin{array}{l} a = |a|\\ \\ -0.011111111111111112 \cdot \left(angle \cdot \left(\pi \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right)\right) \end{array} \]

FPCore

NOTE: a should be positive before calling this function
(FPCore (a b angle)
 :precision binary64
 (* -0.011111111111111112 (* angle (* PI (* (+ a b) (- a b))))))

C

a = abs(a);
double code(double a, double b, double angle) {
	return -0.011111111111111112 * (angle * (((double) M_PI) * ((a + b) * (a - b))));
}

Java

a = Math.abs(a);
public static double code(double a, double b, double angle) {
	return -0.011111111111111112 * (angle * (Math.PI * ((a + b) * (a - b))));
}

Python

a = abs(a)
def code(a, b, angle):
	return -0.011111111111111112 * (angle * (math.pi * ((a + b) * (a - b))))

Julia

a = abs(a)
function code(a, b, angle)
	return Float64(-0.011111111111111112 * Float64(angle * Float64(pi * Float64(Float64(a + b) * Float64(a - b)))))
end

MATLAB

a = abs(a)
function tmp = code(a, b, angle)
	tmp = -0.011111111111111112 * (angle * (pi * ((a + b) * (a - b))));
end

Wolfram

NOTE: a should be positive before calling this function
code[a_, b_, angle_] := N[(-0.011111111111111112 * N[(angle * N[(Pi * N[(N[(a + b), $MachinePrecision] * N[(a - b), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 53.4%

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

3. Simplified 54.4%

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

   1. unpow2 54.4%

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

   2. unpow2 54.4%

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

   3. difference-of-squares 59.6%

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

5. Applied egg-rr 59.6%

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

6. Taylor expanded in angle around 0 56.2%

   \[\leadsto \color{blue}{-0.011111111111111112 \cdot \left(angle \cdot \left(\pi \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right)\right)} \]

7. Final simplification 56.2%

   \[\leadsto -0.011111111111111112 \cdot \left(angle \cdot \left(\pi \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right)\right) \]

Alternative 5: 54.4% accurate, 5.5× speedup

Math

\[\begin{array}{l} a = |a|\\ \\ -0.011111111111111112 \cdot \left(\left(\left(a + b\right) \cdot \left(a - b\right)\right) \cdot \left(\pi \cdot angle\right)\right) \end{array} \]

FPCore

NOTE: a should be positive before calling this function
(FPCore (a b angle)
 :precision binary64
 (* -0.011111111111111112 (* (* (+ a b) (- a b)) (* PI angle))))

C

a = abs(a);
double code(double a, double b, double angle) {
	return -0.011111111111111112 * (((a + b) * (a - b)) * (((double) M_PI) * angle));
}

Java

a = Math.abs(a);
public static double code(double a, double b, double angle) {
	return -0.011111111111111112 * (((a + b) * (a - b)) * (Math.PI * angle));
}

Python

a = abs(a)
def code(a, b, angle):
	return -0.011111111111111112 * (((a + b) * (a - b)) * (math.pi * angle))

Julia

a = abs(a)
function code(a, b, angle)
	return Float64(-0.011111111111111112 * Float64(Float64(Float64(a + b) * Float64(a - b)) * Float64(pi * angle)))
end

MATLAB

a = abs(a)
function tmp = code(a, b, angle)
	tmp = -0.011111111111111112 * (((a + b) * (a - b)) * (pi * angle));
end

Wolfram

NOTE: a should be positive before calling this function
code[a_, b_, angle_] := N[(-0.011111111111111112 * N[(N[(N[(a + b), $MachinePrecision] * N[(a - b), $MachinePrecision]), $MachinePrecision] * N[(Pi * angle), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 53.4%

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

3. Simplified 54.4%

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

   1. unpow2 54.4%

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

   2. unpow2 54.4%

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

   3. difference-of-squares 59.6%

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

5. Applied egg-rr 59.6%

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

6. Taylor expanded in angle around 0 56.2%

   \[\leadsto \color{blue}{-0.011111111111111112 \cdot \left(angle \cdot \left(\pi \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right)\right)} \]
7. Step-by-step derivation

   1. associate-*r* 56.2%

      \[\leadsto -0.011111111111111112 \cdot \color{blue}{\left(\left(angle \cdot \pi\right) \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right)} \]

8. Simplified 56.2%

   \[\leadsto \color{blue}{-0.011111111111111112 \cdot \left(\left(angle \cdot \pi\right) \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right)} \]

9. Final simplification 56.2%

   \[\leadsto -0.011111111111111112 \cdot \left(\left(\left(a + b\right) \cdot \left(a - b\right)\right) \cdot \left(\pi \cdot angle\right)\right) \]

Reproduce

herbie shell --seed 2023308 
(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)))))