ab-angle->ABCF B

Percentage Accurate: 54.4% → 58.2%

Time: 53.7s

Alternatives: 6

Speedup: 32.2×

• 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

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

FPCore

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

Python

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

MATLAB

function tmp = code(a, b, angle)
	t_0 = pi * (angle / 180.0);
	tmp = ((2.0 * ((b - a) * (b + a))) * 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[(b - a), $MachinePrecision] * N[(b + a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[Sin[t$95$0], $MachinePrecision]), $MachinePrecision] * N[Cos[t$95$0], $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%

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

Alternative 2: 57.9% accurate, 1.9× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} t_0 := -0.005555555555555556 \cdot \left(\pi \cdot angle\right)\\ \mathbf{if}\;b \leq 1.05 \cdot 10^{+217}:\\ \;\;\;\;2 \cdot \left(\cos t\_0 \cdot \left(\left(\left(b + a\right) \cdot \left(a - b\right)\right) \cdot \sin t\_0\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\cos \left(\pi \cdot \frac{angle}{180}\right) \cdot \left(0.011111111111111112 \cdot \left(angle \cdot \left(\pi \cdot \left(\left(b - a\right) \cdot \left(b + a\right)\right)\right)\right)\right)\\ \end{array} \end{array} \]

FPCore

(FPCore (a b angle)
 :precision binary64
 (let* ((t_0 (* -0.005555555555555556 (* PI angle))))
   (if (<= b 1.05e+217)
     (* 2.0 (* (cos t_0) (* (* (+ b a) (- a b)) (sin t_0))))
     (*
      (cos (* PI (/ angle 180.0)))
      (* 0.011111111111111112 (* angle (* PI (* (- b a) (+ b a)))))))))

C

double code(double a, double b, double angle) {
	double t_0 = -0.005555555555555556 * (((double) M_PI) * angle);
	double tmp;
	if (b <= 1.05e+217) {
		tmp = 2.0 * (cos(t_0) * (((b + a) * (a - b)) * sin(t_0)));
	} else {
		tmp = cos((((double) M_PI) * (angle / 180.0))) * (0.011111111111111112 * (angle * (((double) M_PI) * ((b - a) * (b + a)))));
	}
	return tmp;
}

Java

public static double code(double a, double b, double angle) {
	double t_0 = -0.005555555555555556 * (Math.PI * angle);
	double tmp;
	if (b <= 1.05e+217) {
		tmp = 2.0 * (Math.cos(t_0) * (((b + a) * (a - b)) * Math.sin(t_0)));
	} else {
		tmp = Math.cos((Math.PI * (angle / 180.0))) * (0.011111111111111112 * (angle * (Math.PI * ((b - a) * (b + a)))));
	}
	return tmp;
}

Python

def code(a, b, angle):
	t_0 = -0.005555555555555556 * (math.pi * angle)
	tmp = 0
	if b <= 1.05e+217:
		tmp = 2.0 * (math.cos(t_0) * (((b + a) * (a - b)) * math.sin(t_0)))
	else:
		tmp = math.cos((math.pi * (angle / 180.0))) * (0.011111111111111112 * (angle * (math.pi * ((b - a) * (b + a)))))
	return tmp

Julia

function code(a, b, angle)
	t_0 = Float64(-0.005555555555555556 * Float64(pi * angle))
	tmp = 0.0
	if (b <= 1.05e+217)
		tmp = Float64(2.0 * Float64(cos(t_0) * Float64(Float64(Float64(b + a) * Float64(a - b)) * sin(t_0))));
	else
		tmp = Float64(cos(Float64(pi * Float64(angle / 180.0))) * Float64(0.011111111111111112 * Float64(angle * Float64(pi * Float64(Float64(b - a) * Float64(b + a))))));
	end
	return tmp
end

MATLAB

function tmp_2 = code(a, b, angle)
	t_0 = -0.005555555555555556 * (pi * angle);
	tmp = 0.0;
	if (b <= 1.05e+217)
		tmp = 2.0 * (cos(t_0) * (((b + a) * (a - b)) * sin(t_0)));
	else
		tmp = cos((pi * (angle / 180.0))) * (0.011111111111111112 * (angle * (pi * ((b - a) * (b + a)))));
	end
	tmp_2 = tmp;
end

Wolfram

code[a_, b_, angle_] := Block[{t$95$0 = N[(-0.005555555555555556 * N[(Pi * angle), $MachinePrecision]), $MachinePrecision]}, If[LessEqual[b, 1.05e+217], N[(2.0 * N[(N[Cos[t$95$0], $MachinePrecision] * N[(N[(N[(b + a), $MachinePrecision] * N[(a - b), $MachinePrecision]), $MachinePrecision] * N[Sin[t$95$0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[Cos[N[(Pi * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(0.011111111111111112 * N[(angle * N[(Pi * N[(N[(b - a), $MachinePrecision] * N[(b + a), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]

Derivation

1. Split input into 2 regimes

2. if b < 1.05e217

   1. Initial program 59.9%

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

   3. Simplified 59.5%

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

      1. unpow2 53.5%

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

      2. unpow2 53.5%

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

      3. difference-of-squares 56.5%

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

   6. Applied egg-rr 62.0%

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

   7. Taylor expanded in angle around inf 60.7%

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

      1. *-commutative 60.7%

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

   9. Simplified 60.7%

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

   if 1.05e217 < b

   1. Initial program 45.2%

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

      1. unpow2 45.2%

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

      2. unpow2 45.2%

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

      3. difference-of-squares 73.5%

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

   4. Applied egg-rr 73.5%

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

   5. Taylor expanded in angle around 0 79.1%

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

4. Final simplification 61.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;b \leq 1.05 \cdot 10^{+217}:\\ \;\;\;\;2 \cdot \left(\cos \left(-0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right) \cdot \left(\left(\left(b + a\right) \cdot \left(a - b\right)\right) \cdot \sin \left(-0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;\cos \left(\pi \cdot \frac{angle}{180}\right) \cdot \left(0.011111111111111112 \cdot \left(angle \cdot \left(\pi \cdot \left(\left(b - a\right) \cdot \left(b + a\right)\right)\right)\right)\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 3: 58.2% accurate, 1.9× speedup

Math

\[\begin{array}{l} \\ 2 \cdot \left(\sin \left(\pi \cdot \left(angle \cdot -0.005555555555555556\right)\right) \cdot \left(\cos \left(angle \cdot \frac{\pi}{-180}\right) \cdot \left(\left(b + a\right) \cdot \left(a - b\right)\right)\right)\right) \end{array} \]

FPCore

(FPCore (a b angle)
 :precision binary64
 (*
  2.0
  (*
   (sin (* PI (* angle -0.005555555555555556)))
   (* (cos (* angle (/ PI -180.0))) (* (+ b a) (- a b))))))

C

double code(double a, double b, double angle) {
	return 2.0 * (sin((((double) M_PI) * (angle * -0.005555555555555556))) * (cos((angle * (((double) M_PI) / -180.0))) * ((b + a) * (a - b))));
}

Java

public static double code(double a, double b, double angle) {
	return 2.0 * (Math.sin((Math.PI * (angle * -0.005555555555555556))) * (Math.cos((angle * (Math.PI / -180.0))) * ((b + a) * (a - b))));
}

Python

def code(a, b, angle):
	return 2.0 * (math.sin((math.pi * (angle * -0.005555555555555556))) * (math.cos((angle * (math.pi / -180.0))) * ((b + a) * (a - b))))

Julia

function code(a, b, angle)
	return Float64(2.0 * Float64(sin(Float64(pi * Float64(angle * -0.005555555555555556))) * Float64(cos(Float64(angle * Float64(pi / -180.0))) * Float64(Float64(b + a) * Float64(a - b)))))
end

MATLAB

function tmp = code(a, b, angle)
	tmp = 2.0 * (sin((pi * (angle * -0.005555555555555556))) * (cos((angle * (pi / -180.0))) * ((b + a) * (a - b))));
end

Wolfram

code[a_, b_, angle_] := N[(2.0 * N[(N[Sin[N[(Pi * N[(angle * -0.005555555555555556), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(N[Cos[N[(angle * N[(Pi / -180.0), $MachinePrecision]), $MachinePrecision]], $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 58.5%

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

   1. unpow2 52.9%

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

   2. unpow2 52.9%

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

   3. difference-of-squares 56.5%

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

6. Applied egg-rr 62.5%

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

7. Taylor expanded in angle around 0 63.2%

   \[\leadsto 2 \cdot \left(\sin \color{blue}{\left(-0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)} \cdot \left(\cos \left(\frac{\pi}{-180} \cdot angle\right) \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right)\right) \]
8. Step-by-step derivation

   1. associate-*r* 62.9%

      \[\leadsto 2 \cdot \left(\sin \color{blue}{\left(\left(-0.005555555555555556 \cdot angle\right) \cdot \pi\right)} \cdot \left(\cos \left(\frac{\pi}{-180} \cdot angle\right) \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right)\right) \]

9. Simplified 62.9%

   \[\leadsto 2 \cdot \left(\sin \color{blue}{\left(\left(-0.005555555555555556 \cdot angle\right) \cdot \pi\right)} \cdot \left(\cos \left(\frac{\pi}{-180} \cdot angle\right) \cdot \left(\left(a + b\right) \cdot \left(a - b\right)\right)\right)\right) \]

10. Final simplification 62.9%

    \[\leadsto 2 \cdot \left(\sin \left(\pi \cdot \left(angle \cdot -0.005555555555555556\right)\right) \cdot \left(\cos \left(angle \cdot \frac{\pi}{-180}\right) \cdot \left(\left(b + a\right) \cdot \left(a - b\right)\right)\right)\right) \]
11. Add Preprocessing

Alternative 4: 54.6% accurate, 27.9× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

code[a_, b_, angle_] := N[(2.0 * N[(angle * N[(-0.005555555555555556 * N[(Pi * N[(N[(b + a), $MachinePrecision] * N[(a - b), $MachinePrecision]), $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 58.5%

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

5. Taylor expanded in angle around 0 52.9%

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

   1. unpow2 52.9%

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

   2. unpow2 52.9%

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

   3. difference-of-squares 56.5%

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

7. Applied egg-rr 56.5%

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

   1. pow1 56.5%

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

   2. *-commutative 56.5%

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

   3. associate-*r* 56.5%

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

   4. +-commutative 56.5%

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

9. Applied egg-rr 56.5%

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

    1. unpow1 56.5%

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

    2. associate-*l* 56.5%

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

    3. associate-*l* 56.5%

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

    4. +-commutative 56.5%

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

11. Simplified 56.5%

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

12. Final simplification 56.5%

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

Alternative 5: 54.5% accurate, 32.2× speedup

Math

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

FPCore

(FPCore (a b angle)
 :precision binary64
 (* -0.011111111111111112 (* angle (* PI (* (+ b a) (- a b))))))

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

code[a_, b_, angle_] := N[(-0.011111111111111112 * N[(angle * N[(Pi * 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 58.5%

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

5. Taylor expanded in angle around 0 52.9%

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

   1. unpow2 52.9%

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

   2. unpow2 52.9%

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

   3. difference-of-squares 56.5%

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

7. Applied egg-rr 56.5%

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

   1. pow1 56.5%

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

   2. *-commutative 56.5%

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

   3. associate-*r* 56.5%

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

   4. +-commutative 56.5%

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

9. Applied egg-rr 56.5%

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

    1. unpow1 56.5%

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

    2. associate-*l* 56.5%

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

    3. associate-*l* 56.5%

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

    4. +-commutative 56.5%

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

11. Simplified 56.5%

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

12. Taylor expanded in angle around 0 56.5%

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

    1. *-commutative 56.5%

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

14. Simplified 56.5%

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

15. Final simplification 56.5%

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

Alternative 6: 54.5% accurate, 32.2× speedup

Math

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

FPCore

(FPCore (a b angle)
 :precision binary64
 (* (* (* (+ b a) (- a b)) (* PI angle)) -0.011111111111111112))

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

code[a_, b_, angle_] := N[(N[(N[(N[(b + a), $MachinePrecision] * N[(a - b), $MachinePrecision]), $MachinePrecision] * N[(Pi * angle), $MachinePrecision]), $MachinePrecision] * -0.011111111111111112), $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 58.5%

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

5. Taylor expanded in angle around 0 52.9%

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

   1. unpow2 52.9%

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

   2. unpow2 52.9%

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

   3. difference-of-squares 56.5%

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

7. Applied egg-rr 56.5%

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

   1. pow1 56.5%

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

   2. *-commutative 56.5%

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

   3. associate-*r* 56.5%

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

   4. +-commutative 56.5%

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

9. Applied egg-rr 56.5%

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

    1. unpow1 56.5%

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

    2. associate-*l* 56.5%

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

    3. associate-*l* 56.5%

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

    4. +-commutative 56.5%

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

11. Simplified 56.5%

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

12. Taylor expanded in angle around 0 56.5%

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

    1. *-commutative 56.5%

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

    2. associate-*r* 56.5%

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

14. Simplified 56.5%

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

15. Final simplification 56.5%

    \[\leadsto \left(\left(\left(b + a\right) \cdot \left(a - b\right)\right) \cdot \left(\pi \cdot angle\right)\right) \cdot -0.011111111111111112 \]
16. 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)))))