ab-angle->ABCF C

Percentage Accurate: 79.8% → 79.7%

Time: 26.9s

Alternatives: 9

Speedup: 1.0×

• 36.2% 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: 79.8% 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: 79.7% accurate, 0.7× speedup

Math

\[\begin{array}{l} angle = |angle|\\ \\ {\left(a \cdot \cos \left(e^{\log \left({\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}^{0.3333333333333333}\right) \cdot 3}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \end{array} \]

FPCore

NOTE: angle should be positive before calling this function
(FPCore (a b angle)
 :precision binary64
 (+
  (pow
   (*
    a
    (cos
     (exp
      (*
       (log (pow (* PI (* angle 0.005555555555555556)) 0.3333333333333333))
       3.0))))
   2.0)
  (pow (* b (sin (* PI (/ angle 180.0)))) 2.0)))

C

angle = abs(angle);
double code(double a, double b, double angle) {
	return pow((a * cos(exp((log(pow((((double) M_PI) * (angle * 0.005555555555555556)), 0.3333333333333333)) * 3.0)))), 2.0) + pow((b * sin((((double) M_PI) * (angle / 180.0)))), 2.0);
}

Java

angle = Math.abs(angle);
public static double code(double a, double b, double angle) {
	return Math.pow((a * Math.cos(Math.exp((Math.log(Math.pow((Math.PI * (angle * 0.005555555555555556)), 0.3333333333333333)) * 3.0)))), 2.0) + Math.pow((b * Math.sin((Math.PI * (angle / 180.0)))), 2.0);
}

Python

angle = abs(angle)
def code(a, b, angle):
	return math.pow((a * math.cos(math.exp((math.log(math.pow((math.pi * (angle * 0.005555555555555556)), 0.3333333333333333)) * 3.0)))), 2.0) + math.pow((b * math.sin((math.pi * (angle / 180.0)))), 2.0)

Julia

angle = abs(angle)
function code(a, b, angle)
	return Float64((Float64(a * cos(exp(Float64(log((Float64(pi * Float64(angle * 0.005555555555555556)) ^ 0.3333333333333333)) * 3.0)))) ^ 2.0) + (Float64(b * sin(Float64(pi * Float64(angle / 180.0)))) ^ 2.0))
end

MATLAB

angle = abs(angle)
function tmp = code(a, b, angle)
	tmp = ((a * cos(exp((log(((pi * (angle * 0.005555555555555556)) ^ 0.3333333333333333)) * 3.0)))) ^ 2.0) + ((b * sin((pi * (angle / 180.0)))) ^ 2.0);
end

Wolfram

NOTE: angle should be positive before calling this function
code[a_, b_, angle_] := N[(N[Power[N[(a * N[Cos[N[Exp[N[(N[Log[N[Power[N[(Pi * N[(angle * 0.005555555555555556), $MachinePrecision]), $MachinePrecision], 0.3333333333333333], $MachinePrecision]], $MachinePrecision] * 3.0), $MachinePrecision]], $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[N[(b * N[Sin[N[(Pi * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 81.5%

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

   1. add-cube-cbrt 81.4%

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

   2. pow3 81.5%

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

   3. metadata-eval 81.5%

      \[\leadsto {\left(a \cdot \cos \left({\left(\sqrt[3]{\pi \cdot \frac{angle}{180}}\right)}^{\color{blue}{\left(1 + 2\right)}}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]

   4. pow-to-exp 39.9%

      \[\leadsto {\left(a \cdot \cos \color{blue}{\left(e^{\log \left(\sqrt[3]{\pi \cdot \frac{angle}{180}}\right) \cdot \left(1 + 2\right)}\right)}\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]

   5. div-inv 39.9%

      \[\leadsto {\left(a \cdot \cos \left(e^{\log \left(\sqrt[3]{\pi \cdot \color{blue}{\left(angle \cdot \frac{1}{180}\right)}}\right) \cdot \left(1 + 2\right)}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]

   6. metadata-eval 39.9%

      \[\leadsto {\left(a \cdot \cos \left(e^{\log \left(\sqrt[3]{\pi \cdot \left(angle \cdot \color{blue}{0.005555555555555556}\right)}\right) \cdot \left(1 + 2\right)}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]

   7. metadata-eval 39.9%

      \[\leadsto {\left(a \cdot \cos \left(e^{\log \left(\sqrt[3]{\pi \cdot \left(angle \cdot 0.005555555555555556\right)}\right) \cdot \color{blue}{3}}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]

3. Applied egg-rr 39.9%

   \[\leadsto {\left(a \cdot \cos \color{blue}{\left(e^{\log \left(\sqrt[3]{\pi \cdot \left(angle \cdot 0.005555555555555556\right)}\right) \cdot 3}\right)}\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]
4. Step-by-step derivation

   1. pow1/3 39.9%

      \[\leadsto {\left(a \cdot \cos \left(e^{\log \color{blue}{\left({\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}^{0.3333333333333333}\right)} \cdot 3}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]

5. Applied egg-rr 39.9%

   \[\leadsto {\left(a \cdot \cos \left(e^{\log \color{blue}{\left({\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}^{0.3333333333333333}\right)} \cdot 3}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]

6. Final simplification 39.9%

   \[\leadsto {\left(a \cdot \cos \left(e^{\log \left({\left(\pi \cdot \left(angle \cdot 0.005555555555555556\right)\right)}^{0.3333333333333333}\right) \cdot 3}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]

Alternative 2: 79.7% accurate, 0.7× speedup

Math

\[\begin{array}{l} angle = |angle|\\ \\ {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(a \cdot \cos \left(e^{3 \cdot \log \left(\sqrt[3]{\pi \cdot \left(angle \cdot 0.005555555555555556\right)}\right)}\right)\right)}^{2} \end{array} \]

FPCore

NOTE: angle should be positive before calling this function
(FPCore (a b angle)
 :precision binary64
 (+
  (pow (* b (sin (* PI (/ angle 180.0)))) 2.0)
  (pow
   (* a (cos (exp (* 3.0 (log (cbrt (* PI (* angle 0.005555555555555556))))))))
   2.0)))

C

angle = abs(angle);
double code(double a, double b, double angle) {
	return pow((b * sin((((double) M_PI) * (angle / 180.0)))), 2.0) + pow((a * cos(exp((3.0 * log(cbrt((((double) M_PI) * (angle * 0.005555555555555556)))))))), 2.0);
}

Java

angle = Math.abs(angle);
public static double code(double a, double b, double angle) {
	return Math.pow((b * Math.sin((Math.PI * (angle / 180.0)))), 2.0) + Math.pow((a * Math.cos(Math.exp((3.0 * Math.log(Math.cbrt((Math.PI * (angle * 0.005555555555555556)))))))), 2.0);
}

Julia

angle = abs(angle)
function code(a, b, angle)
	return Float64((Float64(b * sin(Float64(pi * Float64(angle / 180.0)))) ^ 2.0) + (Float64(a * cos(exp(Float64(3.0 * log(cbrt(Float64(pi * Float64(angle * 0.005555555555555556)))))))) ^ 2.0))
end

Wolfram

NOTE: angle should be positive before calling this function
code[a_, b_, angle_] := N[(N[Power[N[(b * N[Sin[N[(Pi * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[N[(a * N[Cos[N[Exp[N[(3.0 * N[Log[N[Power[N[(Pi * N[(angle * 0.005555555555555556), $MachinePrecision]), $MachinePrecision], 1/3], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 81.5%

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

   1. add-cube-cbrt 81.4%

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

   2. pow3 81.5%

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

   3. metadata-eval 81.5%

      \[\leadsto {\left(a \cdot \cos \left({\left(\sqrt[3]{\pi \cdot \frac{angle}{180}}\right)}^{\color{blue}{\left(1 + 2\right)}}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]

   4. pow-to-exp 39.9%

      \[\leadsto {\left(a \cdot \cos \color{blue}{\left(e^{\log \left(\sqrt[3]{\pi \cdot \frac{angle}{180}}\right) \cdot \left(1 + 2\right)}\right)}\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]

   5. div-inv 39.9%

      \[\leadsto {\left(a \cdot \cos \left(e^{\log \left(\sqrt[3]{\pi \cdot \color{blue}{\left(angle \cdot \frac{1}{180}\right)}}\right) \cdot \left(1 + 2\right)}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]

   6. metadata-eval 39.9%

      \[\leadsto {\left(a \cdot \cos \left(e^{\log \left(\sqrt[3]{\pi \cdot \left(angle \cdot \color{blue}{0.005555555555555556}\right)}\right) \cdot \left(1 + 2\right)}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]

   7. metadata-eval 39.9%

      \[\leadsto {\left(a \cdot \cos \left(e^{\log \left(\sqrt[3]{\pi \cdot \left(angle \cdot 0.005555555555555556\right)}\right) \cdot \color{blue}{3}}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]

3. Applied egg-rr 39.9%

   \[\leadsto {\left(a \cdot \cos \color{blue}{\left(e^{\log \left(\sqrt[3]{\pi \cdot \left(angle \cdot 0.005555555555555556\right)}\right) \cdot 3}\right)}\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]

4. Final simplification 39.9%

   \[\leadsto {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(a \cdot \cos \left(e^{3 \cdot \log \left(\sqrt[3]{\pi \cdot \left(angle \cdot 0.005555555555555556\right)}\right)}\right)\right)}^{2} \]

Alternative 3: 79.8% accurate, 1.0× speedup

Math

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

FPCore

NOTE: angle should be positive before calling this function
(FPCore (a b angle)
 :precision binary64
 (let* ((t_0 (* PI (/ angle 180.0))))
   (+ (pow (* b (sin t_0)) 2.0) (pow (* a (cos t_0)) 2.0))))

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 81.5%

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

2. Final simplification 81.5%

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

Alternative 4: 79.8% accurate, 1.0× speedup

Math

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

FPCore

NOTE: angle should be positive before calling this function
(FPCore (a b angle)
 :precision binary64
 (+
  (pow (* b (sin (* PI (/ angle 180.0)))) 2.0)
  (pow (* a (cos (/ PI (/ 180.0 angle)))) 2.0)))

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

NOTE: angle should be positive before calling this function
code[a_, b_, angle_] := N[(N[Power[N[(b * N[Sin[N[(Pi * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[N[(a * N[Cos[N[(Pi / N[(180.0 / angle), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 81.5%

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

   1. clear-num 81.5%

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

   2. un-div-inv 81.5%

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

3. Applied egg-rr 81.5%

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

4. Final simplification 81.5%

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

Alternative 5: 79.7% accurate, 1.5× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 81.5%

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

2. Taylor expanded in angle around 0 81.1%

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

3. Taylor expanded in b around 0 80.7%

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

4. Final simplification 80.7%

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

Alternative 6: 79.7% accurate, 1.5× speedup

Math

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

FPCore

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

C

angle = abs(angle);
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

angle = Math.abs(angle);
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

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

Julia

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

MATLAB

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

Wolfram

NOTE: angle should be positive before calling this function
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 81.5%

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

2. Taylor expanded in angle around 0 81.1%

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

3. Taylor expanded in b around 0 80.7%

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

   1. *-commutative 80.7%

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

   2. *-commutative 80.7%

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

   3. associate-*r* 81.1%

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

   4. *-commutative 81.1%

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

5. Simplified 81.1%

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

6. Final simplification 81.1%

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

Alternative 7: 79.7% accurate, 1.5× speedup

Math

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

FPCore

NOTE: angle should be positive before calling this function
(FPCore (a b angle)
 :precision binary64
 (+ (pow (* b (sin (* PI (/ angle 180.0)))) 2.0) (pow a 2.0)))

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

NOTE: angle should be positive before calling this function
code[a_, b_, angle_] := N[(N[Power[N[(b * N[Sin[N[(Pi * N[(angle / 180.0), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[a, 2.0], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 81.5%

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

2. Taylor expanded in angle around 0 81.1%

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

3. Final simplification 81.1%

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

Alternative 8: 74.7% accurate, 1.9× speedup

Math

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

FPCore

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

C

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 81.5%

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

2. Taylor expanded in angle around 0 81.1%

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

3. Taylor expanded in angle around 0 76.9%

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

   1. associate-*r* 76.8%

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

   2. *-commutative 76.8%

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

   3. associate-*l* 76.9%

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

   4. associate-*r* 76.9%

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

   5. metadata-eval 76.9%

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

   6. div-inv 76.9%

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

   7. *-commutative 76.9%

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

   8. unpow2 76.9%

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

   9. associate-*r* 76.9%

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

   10. associate-*r* 76.9%

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

5. Applied egg-rr 76.9%

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

6. Final simplification 76.9%

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

Alternative 9: 74.6% accurate, 2.0× speedup

Math

\[\begin{array}{l} angle = |angle|\\ \\ {a}^{2} + 3.08641975308642 \cdot 10^{-5} \cdot {\left(angle \cdot \left(\pi \cdot b\right)\right)}^{2} \end{array} \]

FPCore

NOTE: angle should be positive before calling this function
(FPCore (a b angle)
 :precision binary64
 (+ (pow a 2.0) (* 3.08641975308642e-5 (pow (* angle (* PI b)) 2.0))))

C

angle = abs(angle);
double code(double a, double b, double angle) {
	return pow(a, 2.0) + (3.08641975308642e-5 * pow((angle * (((double) M_PI) * b)), 2.0));
}

Java

angle = Math.abs(angle);
public static double code(double a, double b, double angle) {
	return Math.pow(a, 2.0) + (3.08641975308642e-5 * Math.pow((angle * (Math.PI * b)), 2.0));
}

Python

angle = abs(angle)
def code(a, b, angle):
	return math.pow(a, 2.0) + (3.08641975308642e-5 * math.pow((angle * (math.pi * b)), 2.0))

Julia

angle = abs(angle)
function code(a, b, angle)
	return Float64((a ^ 2.0) + Float64(3.08641975308642e-5 * (Float64(angle * Float64(pi * b)) ^ 2.0)))
end

MATLAB

angle = abs(angle)
function tmp = code(a, b, angle)
	tmp = (a ^ 2.0) + (3.08641975308642e-5 * ((angle * (pi * b)) ^ 2.0));
end

Wolfram

NOTE: angle should be positive before calling this function
code[a_, b_, angle_] := N[(N[Power[a, 2.0], $MachinePrecision] + N[(3.08641975308642e-5 * N[Power[N[(angle * N[(Pi * b), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 81.5%

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

2. Taylor expanded in angle around 0 81.1%

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

3. Taylor expanded in angle around 0 76.9%

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

4. Taylor expanded in angle around 0 68.8%

   \[\leadsto {\left(a \cdot 1\right)}^{2} + \color{blue}{3.08641975308642 \cdot 10^{-5} \cdot \left({angle}^{2} \cdot \left({b}^{2} \cdot {\pi}^{2}\right)\right)} \]
5. Step-by-step derivation

   1. unpow2 68.8%

      \[\leadsto {\left(a \cdot 1\right)}^{2} + 3.08641975308642 \cdot 10^{-5} \cdot \left(\color{blue}{\left(angle \cdot angle\right)} \cdot \left({b}^{2} \cdot {\pi}^{2}\right)\right) \]

   2. unpow2 68.8%

      \[\leadsto {\left(a \cdot 1\right)}^{2} + 3.08641975308642 \cdot 10^{-5} \cdot \left(\left(angle \cdot angle\right) \cdot \left(\color{blue}{\left(b \cdot b\right)} \cdot {\pi}^{2}\right)\right) \]

   3. associate-*r* 68.8%

      \[\leadsto {\left(a \cdot 1\right)}^{2} + 3.08641975308642 \cdot 10^{-5} \cdot \color{blue}{\left(\left(\left(angle \cdot angle\right) \cdot \left(b \cdot b\right)\right) \cdot {\pi}^{2}\right)} \]

   4. swap-sqr 76.9%

      \[\leadsto {\left(a \cdot 1\right)}^{2} + 3.08641975308642 \cdot 10^{-5} \cdot \left(\color{blue}{\left(\left(angle \cdot b\right) \cdot \left(angle \cdot b\right)\right)} \cdot {\pi}^{2}\right) \]

   5. unpow2 76.9%

      \[\leadsto {\left(a \cdot 1\right)}^{2} + 3.08641975308642 \cdot 10^{-5} \cdot \left(\left(\left(angle \cdot b\right) \cdot \left(angle \cdot b\right)\right) \cdot \color{blue}{\left(\pi \cdot \pi\right)}\right) \]

   6. swap-sqr 76.9%

      \[\leadsto {\left(a \cdot 1\right)}^{2} + 3.08641975308642 \cdot 10^{-5} \cdot \color{blue}{\left(\left(\left(angle \cdot b\right) \cdot \pi\right) \cdot \left(\left(angle \cdot b\right) \cdot \pi\right)\right)} \]

   7. *-commutative 76.9%

      \[\leadsto {\left(a \cdot 1\right)}^{2} + 3.08641975308642 \cdot 10^{-5} \cdot \left(\color{blue}{\left(\pi \cdot \left(angle \cdot b\right)\right)} \cdot \left(\left(angle \cdot b\right) \cdot \pi\right)\right) \]

   8. *-commutative 76.9%

      \[\leadsto {\left(a \cdot 1\right)}^{2} + 3.08641975308642 \cdot 10^{-5} \cdot \left(\left(\pi \cdot \left(angle \cdot b\right)\right) \cdot \color{blue}{\left(\pi \cdot \left(angle \cdot b\right)\right)}\right) \]

   9. unpow2 76.9%

      \[\leadsto {\left(a \cdot 1\right)}^{2} + 3.08641975308642 \cdot 10^{-5} \cdot \color{blue}{{\left(\pi \cdot \left(angle \cdot b\right)\right)}^{2}} \]

   10. *-commutative 76.9%

       \[\leadsto {\left(a \cdot 1\right)}^{2} + 3.08641975308642 \cdot 10^{-5} \cdot {\color{blue}{\left(\left(angle \cdot b\right) \cdot \pi\right)}}^{2} \]

   11. associate-*r* 76.9%

       \[\leadsto {\left(a \cdot 1\right)}^{2} + 3.08641975308642 \cdot 10^{-5} \cdot {\color{blue}{\left(angle \cdot \left(b \cdot \pi\right)\right)}}^{2} \]

6. Simplified 76.9%

   \[\leadsto {\left(a \cdot 1\right)}^{2} + \color{blue}{3.08641975308642 \cdot 10^{-5} \cdot {\left(angle \cdot \left(b \cdot \pi\right)\right)}^{2}} \]

7. Final simplification 76.9%

   \[\leadsto {a}^{2} + 3.08641975308642 \cdot 10^{-5} \cdot {\left(angle \cdot \left(\pi \cdot b\right)\right)}^{2} \]

Reproduce

herbie shell --seed 2023200 
(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)))