ab-angle->ABCF C

Average Accuracy: 67.8% → 67.4%

Time: 18.5s

Precision: binary64

Cost: 45504

Math

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

↓

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

FPCore

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

↓

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

C

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

↓

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

Java

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

↓

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

Python

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

↓

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

Julia

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

↓

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

MATLAB

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

↓

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

Wolfram

code[a_, b_, angle_] := N[(N[Power[N[(a * N[Cos[N[(Pi * N[(angle / 180.0), $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]

↓

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

Derivation

1. Initial program 67.8%

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

2. Applied egg-rr 67.6%

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

   [Start] 67.8	\[ {\left(a \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(b \cdot \sin \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} \]
   associate-*r/ [=>] 67.8	\[ {\left(a \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(b \cdot \sin \color{blue}{\left(\frac{\pi \cdot angle}{180}\right)}\right)}^{2} \]
   associate-/l* [=>] 67.8	\[ {\left(a \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(b \cdot \sin \color{blue}{\left(\frac{\pi}{\frac{180}{angle}}\right)}\right)}^{2} \]
   add-sqr-sqrt [=>] 67.6	\[ {\left(a \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(b \cdot \sin \left(\frac{\color{blue}{\sqrt{\pi} \cdot \sqrt{\pi}}}{\frac{180}{angle}}\right)\right)}^{2} \]
   associate-/l* [=>] 67.6	\[ {\left(a \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(b \cdot \sin \color{blue}{\left(\frac{\sqrt{\pi}}{\frac{\frac{180}{angle}}{\sqrt{\pi}}}\right)}\right)}^{2} \]

3. Simplified 67.5%

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

   [Start] 67.6	\[ {\left(a \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(b \cdot \sin \left(\frac{\sqrt{\pi}}{\frac{\frac{180}{angle}}{\sqrt{\pi}}}\right)\right)}^{2} \]
   associate-/r/ [=>] 67.6	\[ {\left(a \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(b \cdot \sin \color{blue}{\left(\frac{\sqrt{\pi}}{\frac{180}{angle}} \cdot \sqrt{\pi}\right)}\right)}^{2} \]
   *-commutative [=>] 67.6	\[ {\left(a \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(b \cdot \sin \color{blue}{\left(\sqrt{\pi} \cdot \frac{\sqrt{\pi}}{\frac{180}{angle}}\right)}\right)}^{2} \]
   associate-/r/ [=>] 67.5	\[ {\left(a \cdot \cos \left(\pi \cdot \frac{angle}{180}\right)\right)}^{2} + {\left(b \cdot \sin \left(\sqrt{\pi} \cdot \color{blue}{\left(\frac{\sqrt{\pi}}{180} \cdot angle\right)}\right)\right)}^{2} \]

4. Taylor expanded in angle around 0 67.4%

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

5. Final simplification 67.4%

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

Alternatives

Alternative 1
Accuracy	67.7%
Cost	26240

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

Alternative 2
Accuracy	67.7%
Cost	26240

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

Alternative 3
Accuracy	66.4%
Cost	20552

\[\begin{array}{l} t_0 := \frac{b}{\frac{2}{b}}\\ \mathbf{if}\;angle \leq -0.0044:\\ \;\;\;\;{a}^{2} + t_0 \cdot \left(1 - \cos \left(\pi \cdot \left(angle \cdot 0.011111111111111112\right)\right)\right)\\ \mathbf{elif}\;angle \leq 1.4 \cdot 10^{-44}:\\ \;\;\;\;{a}^{2} + {\left(\pi \cdot \left(b \cdot angle\right)\right)}^{2} \cdot 3.08641975308642 \cdot 10^{-5}\\ \mathbf{else}:\\ \;\;\;\;{a}^{2} + t_0 \cdot \left(1 - \cos \left(\left(\pi \cdot 0.005555555555555556\right) \cdot \left(angle + angle\right)\right)\right)\\ \end{array} \]

Alternative 4
Accuracy	67.4%
Cost	20425

\[\begin{array}{l} \mathbf{if}\;angle \leq -0.0044 \lor \neg \left(angle \leq 65000000000\right):\\ \;\;\;\;{a}^{2} + \frac{b}{\frac{2}{b}} \cdot \left(1 - \cos \left(\pi \cdot \left(angle \cdot 0.011111111111111112\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;{a}^{2} + {\left(\pi \cdot \left(b \cdot angle\right)\right)}^{2} \cdot 3.08641975308642 \cdot 10^{-5}\\ \end{array} \]

Alternative 5
Accuracy	66.4%
Cost	20424

\[\begin{array}{l} \mathbf{if}\;angle \leq -0.0044:\\ \;\;\;\;{a}^{2} + \frac{b}{\frac{2}{b}} \cdot \left(1 - \cos \left(\pi \cdot \left(angle \cdot 0.011111111111111112\right)\right)\right)\\ \mathbf{elif}\;angle \leq 1.4 \cdot 10^{-44}:\\ \;\;\;\;{a}^{2} + {\left(\pi \cdot \left(b \cdot angle\right)\right)}^{2} \cdot 3.08641975308642 \cdot 10^{-5}\\ \mathbf{else}:\\ \;\;\;\;{a}^{2} + \frac{b \cdot b}{2} \cdot \left(1 - \cos \left(angle \cdot \left(\pi \cdot 0.011111111111111112\right)\right)\right)\\ \end{array} \]

Alternative 6
Accuracy	62.7%
Cost	20360

\[\begin{array}{l} \mathbf{if}\;b \leq -4 \cdot 10^{+73}:\\ \;\;\;\;{a}^{2} + {\left(\pi \cdot \left(b \cdot angle\right)\right)}^{2} \cdot 3.08641975308642 \cdot 10^{-5}\\ \mathbf{elif}\;b \leq 10^{-10}:\\ \;\;\;\;{a}^{2} + 3.08641975308642 \cdot 10^{-5} \cdot \left({\pi}^{2} \cdot \left(angle \cdot \left(angle \cdot \left(b \cdot b\right)\right)\right)\right)\\ \mathbf{else}:\\ \;\;\;\;{a}^{2} + 3.08641975308642 \cdot 10^{-5} \cdot {\left(b \cdot \left(\pi \cdot angle\right)\right)}^{2}\\ \end{array} \]

Alternative 7
Accuracy	59.2%
Cost	20232

\[\begin{array}{l} \mathbf{if}\;b \leq -1.15 \cdot 10^{-161}:\\ \;\;\;\;{a}^{2} + {\left(\pi \cdot \left(b \cdot angle\right)\right)}^{2} \cdot 3.08641975308642 \cdot 10^{-5}\\ \mathbf{elif}\;b \leq 1.58 \cdot 10^{-162}:\\ \;\;\;\;{a}^{2} + 3.08641975308642 \cdot 10^{-5} \cdot \left(\left(b \cdot b\right) \cdot {\left(\pi \cdot angle\right)}^{2}\right)\\ \mathbf{else}:\\ \;\;\;\;{a}^{2} + 3.08641975308642 \cdot 10^{-5} \cdot {\left(b \cdot \left(\pi \cdot angle\right)\right)}^{2}\\ \end{array} \]

Alternative 8
Accuracy	59.3%
Cost	20232

\[\begin{array}{l} \mathbf{if}\;b \leq -1.15 \cdot 10^{-161}:\\ \;\;\;\;{a}^{2} + {\left(\pi \cdot \left(b \cdot angle\right)\right)}^{2} \cdot 3.08641975308642 \cdot 10^{-5}\\ \mathbf{elif}\;b \leq 1.55 \cdot 10^{-162}:\\ \;\;\;\;{a}^{2} + \left(b \cdot b\right) \cdot {\left(0.005555555555555556 \cdot \left(\pi \cdot angle\right)\right)}^{2}\\ \mathbf{else}:\\ \;\;\;\;{a}^{2} + 3.08641975308642 \cdot 10^{-5} \cdot {\left(b \cdot \left(\pi \cdot angle\right)\right)}^{2}\\ \end{array} \]

Alternative 9
Accuracy	58.9%
Cost	19840

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

Alternative 10
Accuracy	58.9%
Cost	19840

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

Alternative 11
Accuracy	58.9%
Cost	19840

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

Reproduce

herbie shell --seed 2023130 
(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)))