ab-angle->ABCF C

Percentage Accurate: 80.0% → 80.0%

Time: 20.1s

Alternatives: 7

Speedup: 1.5×

• 37.3% of points produce a very large (infinite) output. You may want to add a precondition.

Specification

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Initial Program: 80.0% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Alternative 1: 80.0% accurate, 1.5× speedup

Math

\[\begin{array}{l} \\ {a}^{2} + {\left(b \cdot \sin \left(\frac{1}{\frac{\frac{180}{angle}}{\pi}}\right)\right)}^{2} \end{array} \]

FPCore

(FPCore (a b angle)
 :precision binary64
 (+ (pow a 2.0) (pow (* b (sin (/ 1.0 (/ (/ 180.0 angle) PI)))) 2.0)))

C

double code(double a, double b, double angle) {
	return pow(a, 2.0) + pow((b * sin((1.0 / ((180.0 / angle) / ((double) M_PI))))), 2.0);
}

Java

public static double code(double a, double b, double angle) {
	return Math.pow(a, 2.0) + Math.pow((b * Math.sin((1.0 / ((180.0 / angle) / Math.PI)))), 2.0);
}

Python

def code(a, b, angle):
	return math.pow(a, 2.0) + math.pow((b * math.sin((1.0 / ((180.0 / angle) / math.pi)))), 2.0)

Julia

function code(a, b, angle)
	return Float64((a ^ 2.0) + (Float64(b * sin(Float64(1.0 / Float64(Float64(180.0 / angle) / pi)))) ^ 2.0))
end

MATLAB

function tmp = code(a, b, angle)
	tmp = (a ^ 2.0) + ((b * sin((1.0 / ((180.0 / angle) / pi)))) ^ 2.0);
end

Wolfram

code[a_, b_, angle_] := N[(N[Power[a, 2.0], $MachinePrecision] + N[Power[N[(b * N[Sin[N[(1.0 / N[(N[(180.0 / angle), $MachinePrecision] / Pi), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 77.2%

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

2. Taylor expanded in angle around 0 77.5%

   \[\leadsto {\left(a \cdot \color{blue}{1}\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]
3. Step-by-step derivation

   1. associate-*r/ 77.5%

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

   2. clear-num 77.5%

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

   3. *-commutative 77.5%

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

   4. associate-/r* 77.6%

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

4. Applied egg-rr 77.6%

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

5. Final simplification 77.6%

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

Alternative 2: 80.1% accurate, 1.5× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 77.2%

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

3. Simplified 77.1%

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

4. Taylor expanded in b around 0 66.2%

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

5. Taylor expanded in angle around 0 66.6%

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

   1. add-sqr-sqrt 66.6%

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

   2. unpow2 66.6%

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

7. Applied egg-rr 77.5%

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

8. Final simplification 77.5%

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

Alternative 3: 80.1% accurate, 1.5× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

code[a_, b_, angle_] := N[(N[Power[a, 2.0], $MachinePrecision] + N[Power[N[(b * N[Sin[N[(angle / N[(180.0 / Pi), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 77.2%

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

2. Taylor expanded in angle around 0 77.5%

   \[\leadsto {\left(a \cdot \color{blue}{1}\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]
3. Step-by-step derivation

   1. associate-*r/ 77.5%

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

   2. *-commutative 77.5%

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

   3. associate-/l* 77.5%

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

4. Applied egg-rr 77.5%

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

5. Final simplification 77.5%

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

Alternative 4: 80.0% accurate, 1.5× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

code[a_, b_, angle_] := N[(N[Power[a, 2.0], $MachinePrecision] + N[Power[N[(b * N[Sin[N[(Pi / N[(180.0 / angle), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 77.2%

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

2. Taylor expanded in angle around 0 77.5%

   \[\leadsto {\left(a \cdot \color{blue}{1}\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]
3. Step-by-step derivation

   1. clear-num 77.5%

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

   2. un-div-inv 77.6%

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

4. Applied egg-rr 77.6%

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

5. Final simplification 77.6%

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

Alternative 5: 67.5% accurate, 2.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;b \leq 1.4 \cdot 10^{-78}:\\ \;\;\;\;{a}^{2}\\ \mathbf{else}:\\ \;\;\;\;{a}^{2} + {\left(0.005555555555555556 \cdot \left(angle \cdot \left(b \cdot \pi\right)\right)\right)}^{2}\\ \end{array} \end{array} \]

FPCore

(FPCore (a b angle)
 :precision binary64
 (if (<= b 1.4e-78)
   (pow a 2.0)
   (+ (pow a 2.0) (pow (* 0.005555555555555556 (* angle (* b PI))) 2.0))))

C

double code(double a, double b, double angle) {
	double tmp;
	if (b <= 1.4e-78) {
		tmp = pow(a, 2.0);
	} else {
		tmp = pow(a, 2.0) + pow((0.005555555555555556 * (angle * (b * ((double) M_PI)))), 2.0);
	}
	return tmp;
}

Java

public static double code(double a, double b, double angle) {
	double tmp;
	if (b <= 1.4e-78) {
		tmp = Math.pow(a, 2.0);
	} else {
		tmp = Math.pow(a, 2.0) + Math.pow((0.005555555555555556 * (angle * (b * Math.PI))), 2.0);
	}
	return tmp;
}

Python

def code(a, b, angle):
	tmp = 0
	if b <= 1.4e-78:
		tmp = math.pow(a, 2.0)
	else:
		tmp = math.pow(a, 2.0) + math.pow((0.005555555555555556 * (angle * (b * math.pi))), 2.0)
	return tmp

Julia

function code(a, b, angle)
	tmp = 0.0
	if (b <= 1.4e-78)
		tmp = a ^ 2.0;
	else
		tmp = Float64((a ^ 2.0) + (Float64(0.005555555555555556 * Float64(angle * Float64(b * pi))) ^ 2.0));
	end
	return tmp
end

MATLAB

function tmp_2 = code(a, b, angle)
	tmp = 0.0;
	if (b <= 1.4e-78)
		tmp = a ^ 2.0;
	else
		tmp = (a ^ 2.0) + ((0.005555555555555556 * (angle * (b * pi))) ^ 2.0);
	end
	tmp_2 = tmp;
end

Wolfram

code[a_, b_, angle_] := If[LessEqual[b, 1.4e-78], N[Power[a, 2.0], $MachinePrecision], N[(N[Power[a, 2.0], $MachinePrecision] + N[Power[N[(0.005555555555555556 * N[(angle * N[(b * Pi), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if b < 1.40000000000000012e-78

   1. Initial program 75.0%

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

   3. Simplified 74.9%

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

   5. Applied egg-rr 45.7%

      \[\leadsto {\left(a \cdot \cos \left(\frac{\pi}{-180} \cdot angle\right)\right)}^{2} + \color{blue}{\frac{{b}^{2} \cdot \left(\cos \left(\left(\pi \cdot angle\right) \cdot 0\right) - \cos \left(\left(\pi \cdot angle\right) \cdot 0\right)\right)}{2}} \]
   6. Step-by-step derivation

      1. *-commutative 45.7%

         \[\leadsto {\left(a \cdot \cos \left(\frac{\pi}{-180} \cdot angle\right)\right)}^{2} + \frac{\color{blue}{\left(\cos \left(\left(\pi \cdot angle\right) \cdot 0\right) - \cos \left(\left(\pi \cdot angle\right) \cdot 0\right)\right) \cdot {b}^{2}}}{2} \]

      2. associate-/l* 45.7%

         \[\leadsto {\left(a \cdot \cos \left(\frac{\pi}{-180} \cdot angle\right)\right)}^{2} + \color{blue}{\frac{\cos \left(\left(\pi \cdot angle\right) \cdot 0\right) - \cos \left(\left(\pi \cdot angle\right) \cdot 0\right)}{\frac{2}{{b}^{2}}}} \]

      3. +-inverses 45.7%

         \[\leadsto {\left(a \cdot \cos \left(\frac{\pi}{-180} \cdot angle\right)\right)}^{2} + \frac{\color{blue}{0}}{\frac{2}{{b}^{2}}} \]

      4. div0 54.9%

         \[\leadsto {\left(a \cdot \cos \left(\frac{\pi}{-180} \cdot angle\right)\right)}^{2} + \color{blue}{0} \]

   7. Simplified 54.9%

      \[\leadsto {\left(a \cdot \cos \left(\frac{\pi}{-180} \cdot angle\right)\right)}^{2} + \color{blue}{0} \]

   8. Taylor expanded in angle around 0 55.4%

      \[\leadsto {\color{blue}{a}}^{2} + 0 \]

   if 1.40000000000000012e-78 < b

   1. Initial program 81.6%

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

   2. Taylor expanded in angle around 0 81.7%

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

   3. Taylor expanded in angle around 0 78.6%

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

4. Final simplification 63.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;b \leq 1.4 \cdot 10^{-78}:\\ \;\;\;\;{a}^{2}\\ \mathbf{else}:\\ \;\;\;\;{a}^{2} + {\left(0.005555555555555556 \cdot \left(angle \cdot \left(b \cdot \pi\right)\right)\right)}^{2}\\ \end{array} \]

Alternative 6: 67.6% accurate, 2.0× speedup

Math

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;b \leq 5 \cdot 10^{-80}:\\ \;\;\;\;{a}^{2}\\ \mathbf{else}:\\ \;\;\;\;{a}^{2} + {\left(0.005555555555555556 \cdot \left(\pi \cdot \left(b \cdot angle\right)\right)\right)}^{2}\\ \end{array} \end{array} \]

FPCore

(FPCore (a b angle)
 :precision binary64
 (if (<= b 5e-80)
   (pow a 2.0)
   (+ (pow a 2.0) (pow (* 0.005555555555555556 (* PI (* b angle))) 2.0))))

C

double code(double a, double b, double angle) {
	double tmp;
	if (b <= 5e-80) {
		tmp = pow(a, 2.0);
	} else {
		tmp = pow(a, 2.0) + pow((0.005555555555555556 * (((double) M_PI) * (b * angle))), 2.0);
	}
	return tmp;
}

Java

public static double code(double a, double b, double angle) {
	double tmp;
	if (b <= 5e-80) {
		tmp = Math.pow(a, 2.0);
	} else {
		tmp = Math.pow(a, 2.0) + Math.pow((0.005555555555555556 * (Math.PI * (b * angle))), 2.0);
	}
	return tmp;
}

Python

def code(a, b, angle):
	tmp = 0
	if b <= 5e-80:
		tmp = math.pow(a, 2.0)
	else:
		tmp = math.pow(a, 2.0) + math.pow((0.005555555555555556 * (math.pi * (b * angle))), 2.0)
	return tmp

Julia

function code(a, b, angle)
	tmp = 0.0
	if (b <= 5e-80)
		tmp = a ^ 2.0;
	else
		tmp = Float64((a ^ 2.0) + (Float64(0.005555555555555556 * Float64(pi * Float64(b * angle))) ^ 2.0));
	end
	return tmp
end

MATLAB

function tmp_2 = code(a, b, angle)
	tmp = 0.0;
	if (b <= 5e-80)
		tmp = a ^ 2.0;
	else
		tmp = (a ^ 2.0) + ((0.005555555555555556 * (pi * (b * angle))) ^ 2.0);
	end
	tmp_2 = tmp;
end

Wolfram

code[a_, b_, angle_] := If[LessEqual[b, 5e-80], N[Power[a, 2.0], $MachinePrecision], N[(N[Power[a, 2.0], $MachinePrecision] + N[Power[N[(0.005555555555555556 * N[(Pi * N[(b * angle), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if b < 5e-80

   1. Initial program 75.0%

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

   3. Simplified 74.9%

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

   5. Applied egg-rr 45.7%

      \[\leadsto {\left(a \cdot \cos \left(\frac{\pi}{-180} \cdot angle\right)\right)}^{2} + \color{blue}{\frac{{b}^{2} \cdot \left(\cos \left(\left(\pi \cdot angle\right) \cdot 0\right) - \cos \left(\left(\pi \cdot angle\right) \cdot 0\right)\right)}{2}} \]
   6. Step-by-step derivation

      1. *-commutative 45.7%

         \[\leadsto {\left(a \cdot \cos \left(\frac{\pi}{-180} \cdot angle\right)\right)}^{2} + \frac{\color{blue}{\left(\cos \left(\left(\pi \cdot angle\right) \cdot 0\right) - \cos \left(\left(\pi \cdot angle\right) \cdot 0\right)\right) \cdot {b}^{2}}}{2} \]

      2. associate-/l* 45.7%

         \[\leadsto {\left(a \cdot \cos \left(\frac{\pi}{-180} \cdot angle\right)\right)}^{2} + \color{blue}{\frac{\cos \left(\left(\pi \cdot angle\right) \cdot 0\right) - \cos \left(\left(\pi \cdot angle\right) \cdot 0\right)}{\frac{2}{{b}^{2}}}} \]

      3. +-inverses 45.7%

         \[\leadsto {\left(a \cdot \cos \left(\frac{\pi}{-180} \cdot angle\right)\right)}^{2} + \frac{\color{blue}{0}}{\frac{2}{{b}^{2}}} \]

      4. div0 54.9%

         \[\leadsto {\left(a \cdot \cos \left(\frac{\pi}{-180} \cdot angle\right)\right)}^{2} + \color{blue}{0} \]

   7. Simplified 54.9%

      \[\leadsto {\left(a \cdot \cos \left(\frac{\pi}{-180} \cdot angle\right)\right)}^{2} + \color{blue}{0} \]

   8. Taylor expanded in angle around 0 55.4%

      \[\leadsto {\color{blue}{a}}^{2} + 0 \]

   if 5e-80 < b

   1. Initial program 81.6%

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

   2. Taylor expanded in angle around 0 81.7%

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

   3. Taylor expanded in angle around 0 78.6%

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

      1. associate-*r* 78.6%

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

   5. Simplified 78.6%

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

4. Final simplification 63.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;b \leq 5 \cdot 10^{-80}:\\ \;\;\;\;{a}^{2}\\ \mathbf{else}:\\ \;\;\;\;{a}^{2} + {\left(0.005555555555555556 \cdot \left(\pi \cdot \left(b \cdot angle\right)\right)\right)}^{2}\\ \end{array} \]

Alternative 7: 57.3% accurate, 6.0× speedup

Math

\[\begin{array}{l} \\ {a}^{2} \end{array} \]

FPCore

(FPCore (a b angle) :precision binary64 (pow a 2.0))

C

double code(double a, double b, double angle) {
	return pow(a, 2.0);
}

Fortran

real(8) function code(a, b, angle)
    real(8), intent (in) :: a
    real(8), intent (in) :: b
    real(8), intent (in) :: angle
    code = a ** 2.0d0
end function

Java

public static double code(double a, double b, double angle) {
	return Math.pow(a, 2.0);
}

Python

def code(a, b, angle):
	return math.pow(a, 2.0)

Julia

function code(a, b, angle)
	return a ^ 2.0
end

MATLAB

function tmp = code(a, b, angle)
	tmp = a ^ 2.0;
end

Wolfram

code[a_, b_, angle_] := N[Power[a, 2.0], $MachinePrecision]

Derivation

1. Initial program 77.2%

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

3. Simplified 77.1%

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

5. Applied egg-rr 38.8%

   \[\leadsto {\left(a \cdot \cos \left(\frac{\pi}{-180} \cdot angle\right)\right)}^{2} + \color{blue}{\frac{{b}^{2} \cdot \left(\cos \left(\left(\pi \cdot angle\right) \cdot 0\right) - \cos \left(\left(\pi \cdot angle\right) \cdot 0\right)\right)}{2}} \]
6. Step-by-step derivation

   1. *-commutative 38.8%

      \[\leadsto {\left(a \cdot \cos \left(\frac{\pi}{-180} \cdot angle\right)\right)}^{2} + \frac{\color{blue}{\left(\cos \left(\left(\pi \cdot angle\right) \cdot 0\right) - \cos \left(\left(\pi \cdot angle\right) \cdot 0\right)\right) \cdot {b}^{2}}}{2} \]

   2. associate-/l* 38.8%

      \[\leadsto {\left(a \cdot \cos \left(\frac{\pi}{-180} \cdot angle\right)\right)}^{2} + \color{blue}{\frac{\cos \left(\left(\pi \cdot angle\right) \cdot 0\right) - \cos \left(\left(\pi \cdot angle\right) \cdot 0\right)}{\frac{2}{{b}^{2}}}} \]

   3. +-inverses 38.8%

      \[\leadsto {\left(a \cdot \cos \left(\frac{\pi}{-180} \cdot angle\right)\right)}^{2} + \frac{\color{blue}{0}}{\frac{2}{{b}^{2}}} \]

   4. div0 53.4%

      \[\leadsto {\left(a \cdot \cos \left(\frac{\pi}{-180} \cdot angle\right)\right)}^{2} + \color{blue}{0} \]

7. Simplified 53.4%

   \[\leadsto {\left(a \cdot \cos \left(\frac{\pi}{-180} \cdot angle\right)\right)}^{2} + \color{blue}{0} \]

8. Taylor expanded in angle around 0 53.8%

   \[\leadsto {\color{blue}{a}}^{2} + 0 \]

9. Final simplification 53.8%

   \[\leadsto {a}^{2} \]

Reproduce

herbie shell --seed 2023305 
(FPCore (a b angle)
  :name "ab-angle->ABCF C"
  :precision binary64
  (+ (pow (* a (cos (* PI (/ angle 180.0)))) 2.0) (pow (* b (sin (* PI (/ angle 180.0)))) 2.0)))