ab-angle->ABCF A

Percentage Accurate: 79.4% → 79.4%
Time: 35.1s
Alternatives: 4
Speedup: N/A×

Specification

?
\[\begin{array}{l} \\ \begin{array}{l} t_0 := \frac{angle}{180} \cdot \pi\\ {\left(a \cdot \sin t_0\right)}^{2} + {\left(b \cdot \cos t_0\right)}^{2} \end{array} \end{array} \]
(FPCore (a b angle)
 :precision binary64
 (let* ((t_0 (* (/ angle 180.0) PI)))
   (+ (pow (* a (sin t_0)) 2.0) (pow (* b (cos t_0)) 2.0))))
double code(double a, double b, double angle) {
	double t_0 = (angle / 180.0) * ((double) M_PI);
	return pow((a * sin(t_0)), 2.0) + pow((b * cos(t_0)), 2.0);
}
public static double code(double a, double b, double angle) {
	double t_0 = (angle / 180.0) * Math.PI;
	return Math.pow((a * Math.sin(t_0)), 2.0) + Math.pow((b * Math.cos(t_0)), 2.0);
}
def code(a, b, angle):
	t_0 = (angle / 180.0) * math.pi
	return math.pow((a * math.sin(t_0)), 2.0) + math.pow((b * math.cos(t_0)), 2.0)
function code(a, b, angle)
	t_0 = Float64(Float64(angle / 180.0) * pi)
	return Float64((Float64(a * sin(t_0)) ^ 2.0) + (Float64(b * cos(t_0)) ^ 2.0))
end
function tmp = code(a, b, angle)
	t_0 = (angle / 180.0) * pi;
	tmp = ((a * sin(t_0)) ^ 2.0) + ((b * cos(t_0)) ^ 2.0);
end
code[a_, b_, angle_] := Block[{t$95$0 = N[(N[(angle / 180.0), $MachinePrecision] * Pi), $MachinePrecision]}, N[(N[Power[N[(a * N[Sin[t$95$0], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[N[(b * N[Cos[t$95$0], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]]
\begin{array}{l}

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

Sampling outcomes in binary64 precision:

Local Percentage Accuracy vs ?

The average percentage accuracy by input value. Horizontal axis shows value of an input variable; the variable is choosen in the title. Vertical axis is accuracy; higher is better. Red represent the original program, while blue represents Herbie's suggestion. These can be toggled with buttons below the plot. The line is an average while dots represent individual samples.

Accuracy vs Speed?

Herbie found 4 alternatives:

AlternativeAccuracySpeedup
The accuracy (vertical axis) and speed (horizontal axis) of each alternatives. Up and to the right is better. The red square shows the initial program, and each blue circle shows an alternative.The line shows the best available speed-accuracy tradeoffs.

Initial Program: 79.4% accurate, 1.0× speedup?

\[\begin{array}{l} \\ \begin{array}{l} t_0 := \frac{angle}{180} \cdot \pi\\ {\left(a \cdot \sin t_0\right)}^{2} + {\left(b \cdot \cos t_0\right)}^{2} \end{array} \end{array} \]
(FPCore (a b angle)
 :precision binary64
 (let* ((t_0 (* (/ angle 180.0) PI)))
   (+ (pow (* a (sin t_0)) 2.0) (pow (* b (cos t_0)) 2.0))))
double code(double a, double b, double angle) {
	double t_0 = (angle / 180.0) * ((double) M_PI);
	return pow((a * sin(t_0)), 2.0) + pow((b * cos(t_0)), 2.0);
}
public static double code(double a, double b, double angle) {
	double t_0 = (angle / 180.0) * Math.PI;
	return Math.pow((a * Math.sin(t_0)), 2.0) + Math.pow((b * Math.cos(t_0)), 2.0);
}
def code(a, b, angle):
	t_0 = (angle / 180.0) * math.pi
	return math.pow((a * math.sin(t_0)), 2.0) + math.pow((b * math.cos(t_0)), 2.0)
function code(a, b, angle)
	t_0 = Float64(Float64(angle / 180.0) * pi)
	return Float64((Float64(a * sin(t_0)) ^ 2.0) + (Float64(b * cos(t_0)) ^ 2.0))
end
function tmp = code(a, b, angle)
	t_0 = (angle / 180.0) * pi;
	tmp = ((a * sin(t_0)) ^ 2.0) + ((b * cos(t_0)) ^ 2.0);
end
code[a_, b_, angle_] := Block[{t$95$0 = N[(N[(angle / 180.0), $MachinePrecision] * Pi), $MachinePrecision]}, N[(N[Power[N[(a * N[Sin[t$95$0], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[N[(b * N[Cos[t$95$0], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]]
\begin{array}{l}

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

Alternative 1: 79.4% accurate, 1.5× speedup?

\[\begin{array}{l} \\ {\left(a \cdot \sin \left(angle \cdot \left(0.005555555555555556 \cdot \pi\right)\right)\right)}^{2} + {b}^{2} \end{array} \]
(FPCore (a b angle)
 :precision binary64
 (+ (pow (* a (sin (* angle (* 0.005555555555555556 PI)))) 2.0) (pow b 2.0)))
double code(double a, double b, double angle) {
	return pow((a * sin((angle * (0.005555555555555556 * ((double) M_PI))))), 2.0) + pow(b, 2.0);
}
public static double code(double a, double b, double angle) {
	return Math.pow((a * Math.sin((angle * (0.005555555555555556 * Math.PI)))), 2.0) + Math.pow(b, 2.0);
}
def code(a, b, angle):
	return math.pow((a * math.sin((angle * (0.005555555555555556 * math.pi)))), 2.0) + math.pow(b, 2.0)
function code(a, b, angle)
	return Float64((Float64(a * sin(Float64(angle * Float64(0.005555555555555556 * pi)))) ^ 2.0) + (b ^ 2.0))
end
function tmp = code(a, b, angle)
	tmp = ((a * sin((angle * (0.005555555555555556 * pi)))) ^ 2.0) + (b ^ 2.0);
end
code[a_, b_, angle_] := N[(N[Power[N[(a * N[Sin[N[(angle * N[(0.005555555555555556 * Pi), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] + N[Power[b, 2.0], $MachinePrecision]), $MachinePrecision]
\begin{array}{l}

\\
{\left(a \cdot \sin \left(angle \cdot \left(0.005555555555555556 \cdot \pi\right)\right)\right)}^{2} + {b}^{2}
\end{array}
Derivation
  1. Initial program 81.3%

    \[{\left(a \cdot \sin \left(\frac{angle}{180} \cdot \pi\right)\right)}^{2} + {\left(b \cdot \cos \left(\frac{angle}{180} \cdot \pi\right)\right)}^{2} \]
  2. Taylor expanded in angle around 0 81.9%

    \[\leadsto {\left(a \cdot \sin \left(\frac{angle}{180} \cdot \pi\right)\right)}^{2} + {\left(b \cdot \color{blue}{1}\right)}^{2} \]
  3. Taylor expanded in angle around inf 81.8%

    \[\leadsto {\left(a \cdot \color{blue}{\sin \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)}\right)}^{2} + {\left(b \cdot 1\right)}^{2} \]
  4. Step-by-step derivation
    1. *-commutative81.8%

      \[\leadsto {\left(a \cdot \sin \color{blue}{\left(\left(angle \cdot \pi\right) \cdot 0.005555555555555556\right)}\right)}^{2} + {\left(b \cdot 1\right)}^{2} \]
    2. associate-*r*81.9%

      \[\leadsto {\left(a \cdot \sin \color{blue}{\left(angle \cdot \left(\pi \cdot 0.005555555555555556\right)\right)}\right)}^{2} + {\left(b \cdot 1\right)}^{2} \]
    3. *-commutative81.9%

      \[\leadsto {\left(a \cdot \sin \left(angle \cdot \color{blue}{\left(0.005555555555555556 \cdot \pi\right)}\right)\right)}^{2} + {\left(b \cdot 1\right)}^{2} \]
  5. Simplified81.9%

    \[\leadsto {\left(a \cdot \color{blue}{\sin \left(angle \cdot \left(0.005555555555555556 \cdot \pi\right)\right)}\right)}^{2} + {\left(b \cdot 1\right)}^{2} \]
  6. Final simplification81.9%

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

Alternative 2: 74.1% accurate, 2.0× speedup?

\[\begin{array}{l} \\ {b}^{2} + 3.08641975308642 \cdot 10^{-5} \cdot {\left(angle \cdot \left(a \cdot \pi\right)\right)}^{2} \end{array} \]
(FPCore (a b angle)
 :precision binary64
 (+ (pow b 2.0) (* 3.08641975308642e-5 (pow (* angle (* a PI)) 2.0))))
double code(double a, double b, double angle) {
	return pow(b, 2.0) + (3.08641975308642e-5 * pow((angle * (a * ((double) M_PI))), 2.0));
}
public static double code(double a, double b, double angle) {
	return Math.pow(b, 2.0) + (3.08641975308642e-5 * Math.pow((angle * (a * Math.PI)), 2.0));
}
def code(a, b, angle):
	return math.pow(b, 2.0) + (3.08641975308642e-5 * math.pow((angle * (a * math.pi)), 2.0))
function code(a, b, angle)
	return Float64((b ^ 2.0) + Float64(3.08641975308642e-5 * (Float64(angle * Float64(a * pi)) ^ 2.0)))
end
function tmp = code(a, b, angle)
	tmp = (b ^ 2.0) + (3.08641975308642e-5 * ((angle * (a * pi)) ^ 2.0));
end
code[a_, b_, angle_] := N[(N[Power[b, 2.0], $MachinePrecision] + N[(3.08641975308642e-5 * N[Power[N[(angle * N[(a * Pi), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]
\begin{array}{l}

\\
{b}^{2} + 3.08641975308642 \cdot 10^{-5} \cdot {\left(angle \cdot \left(a \cdot \pi\right)\right)}^{2}
\end{array}
Derivation
  1. Initial program 81.3%

    \[{\left(a \cdot \sin \left(\frac{angle}{180} \cdot \pi\right)\right)}^{2} + {\left(b \cdot \cos \left(\frac{angle}{180} \cdot \pi\right)\right)}^{2} \]
  2. Taylor expanded in angle around 0 81.9%

    \[\leadsto {\left(a \cdot \sin \left(\frac{angle}{180} \cdot \pi\right)\right)}^{2} + {\left(b \cdot \color{blue}{1}\right)}^{2} \]
  3. Taylor expanded in angle around 0 75.8%

    \[\leadsto {\color{blue}{\left(0.005555555555555556 \cdot \left(angle \cdot \left(a \cdot \pi\right)\right)\right)}}^{2} + {\left(b \cdot 1\right)}^{2} \]
  4. Step-by-step derivation
    1. *-commutative75.8%

      \[\leadsto {\left(0.005555555555555556 \cdot \left(angle \cdot \color{blue}{\left(\pi \cdot a\right)}\right)\right)}^{2} + {\left(b \cdot 1\right)}^{2} \]
  5. Simplified75.8%

    \[\leadsto {\color{blue}{\left(0.005555555555555556 \cdot \left(angle \cdot \left(\pi \cdot a\right)\right)\right)}}^{2} + {\left(b \cdot 1\right)}^{2} \]
  6. Taylor expanded in angle around 0 64.8%

    \[\leadsto \color{blue}{3.08641975308642 \cdot 10^{-5} \cdot \left({angle}^{2} \cdot \left({a}^{2} \cdot {\pi}^{2}\right)\right)} + {\left(b \cdot 1\right)}^{2} \]
  7. Step-by-step derivation
    1. unpow264.8%

      \[\leadsto 3.08641975308642 \cdot 10^{-5} \cdot \left(\color{blue}{\left(angle \cdot angle\right)} \cdot \left({a}^{2} \cdot {\pi}^{2}\right)\right) + {\left(b \cdot 1\right)}^{2} \]
    2. *-commutative64.8%

      \[\leadsto 3.08641975308642 \cdot 10^{-5} \cdot \color{blue}{\left(\left({a}^{2} \cdot {\pi}^{2}\right) \cdot \left(angle \cdot angle\right)\right)} + {\left(b \cdot 1\right)}^{2} \]
    3. *-commutative64.8%

      \[\leadsto 3.08641975308642 \cdot 10^{-5} \cdot \left(\color{blue}{\left({\pi}^{2} \cdot {a}^{2}\right)} \cdot \left(angle \cdot angle\right)\right) + {\left(b \cdot 1\right)}^{2} \]
    4. unpow264.8%

      \[\leadsto 3.08641975308642 \cdot 10^{-5} \cdot \left(\left(\color{blue}{\left(\pi \cdot \pi\right)} \cdot {a}^{2}\right) \cdot \left(angle \cdot angle\right)\right) + {\left(b \cdot 1\right)}^{2} \]
    5. unpow264.8%

      \[\leadsto 3.08641975308642 \cdot 10^{-5} \cdot \left(\left(\left(\pi \cdot \pi\right) \cdot \color{blue}{\left(a \cdot a\right)}\right) \cdot \left(angle \cdot angle\right)\right) + {\left(b \cdot 1\right)}^{2} \]
    6. swap-sqr64.8%

      \[\leadsto 3.08641975308642 \cdot 10^{-5} \cdot \left(\color{blue}{\left(\left(\pi \cdot a\right) \cdot \left(\pi \cdot a\right)\right)} \cdot \left(angle \cdot angle\right)\right) + {\left(b \cdot 1\right)}^{2} \]
    7. swap-sqr75.8%

      \[\leadsto 3.08641975308642 \cdot 10^{-5} \cdot \color{blue}{\left(\left(\left(\pi \cdot a\right) \cdot angle\right) \cdot \left(\left(\pi \cdot a\right) \cdot angle\right)\right)} + {\left(b \cdot 1\right)}^{2} \]
    8. associate-*r*75.8%

      \[\leadsto 3.08641975308642 \cdot 10^{-5} \cdot \left(\color{blue}{\left(\pi \cdot \left(a \cdot angle\right)\right)} \cdot \left(\left(\pi \cdot a\right) \cdot angle\right)\right) + {\left(b \cdot 1\right)}^{2} \]
    9. associate-*r*75.8%

      \[\leadsto 3.08641975308642 \cdot 10^{-5} \cdot \left(\left(\pi \cdot \left(a \cdot angle\right)\right) \cdot \color{blue}{\left(\pi \cdot \left(a \cdot angle\right)\right)}\right) + {\left(b \cdot 1\right)}^{2} \]
    10. unpow275.8%

      \[\leadsto 3.08641975308642 \cdot 10^{-5} \cdot \color{blue}{{\left(\pi \cdot \left(a \cdot angle\right)\right)}^{2}} + {\left(b \cdot 1\right)}^{2} \]
    11. associate-*r*75.8%

      \[\leadsto 3.08641975308642 \cdot 10^{-5} \cdot {\color{blue}{\left(\left(\pi \cdot a\right) \cdot angle\right)}}^{2} + {\left(b \cdot 1\right)}^{2} \]
    12. *-commutative75.8%

      \[\leadsto 3.08641975308642 \cdot 10^{-5} \cdot {\color{blue}{\left(angle \cdot \left(\pi \cdot a\right)\right)}}^{2} + {\left(b \cdot 1\right)}^{2} \]
  8. Simplified75.8%

    \[\leadsto \color{blue}{3.08641975308642 \cdot 10^{-5} \cdot {\left(angle \cdot \left(\pi \cdot a\right)\right)}^{2}} + {\left(b \cdot 1\right)}^{2} \]
  9. Final simplification75.8%

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

Alternative 3: 74.1% accurate, 2.0× speedup?

\[\begin{array}{l} \\ {b}^{2} + {\left(a \cdot \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)\right)}^{2} \end{array} \]
(FPCore (a b angle)
 :precision binary64
 (+ (pow b 2.0) (pow (* a (* 0.005555555555555556 (* angle PI))) 2.0)))
double code(double a, double b, double angle) {
	return pow(b, 2.0) + pow((a * (0.005555555555555556 * (angle * ((double) M_PI)))), 2.0);
}
public static double code(double a, double b, double angle) {
	return Math.pow(b, 2.0) + Math.pow((a * (0.005555555555555556 * (angle * Math.PI))), 2.0);
}
def code(a, b, angle):
	return math.pow(b, 2.0) + math.pow((a * (0.005555555555555556 * (angle * math.pi))), 2.0)
function code(a, b, angle)
	return Float64((b ^ 2.0) + (Float64(a * Float64(0.005555555555555556 * Float64(angle * pi))) ^ 2.0))
end
function tmp = code(a, b, angle)
	tmp = (b ^ 2.0) + ((a * (0.005555555555555556 * (angle * pi))) ^ 2.0);
end
code[a_, b_, angle_] := N[(N[Power[b, 2.0], $MachinePrecision] + N[Power[N[(a * N[(0.005555555555555556 * N[(angle * Pi), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]
\begin{array}{l}

\\
{b}^{2} + {\left(a \cdot \left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)\right)}^{2}
\end{array}
Derivation
  1. Initial program 81.3%

    \[{\left(a \cdot \sin \left(\frac{angle}{180} \cdot \pi\right)\right)}^{2} + {\left(b \cdot \cos \left(\frac{angle}{180} \cdot \pi\right)\right)}^{2} \]
  2. Taylor expanded in angle around 0 81.9%

    \[\leadsto {\left(a \cdot \sin \left(\frac{angle}{180} \cdot \pi\right)\right)}^{2} + {\left(b \cdot \color{blue}{1}\right)}^{2} \]
  3. Taylor expanded in angle around 0 75.8%

    \[\leadsto {\left(a \cdot \color{blue}{\left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)}\right)}^{2} + {\left(b \cdot 1\right)}^{2} \]
  4. Final simplification75.8%

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

Alternative 4: 74.1% accurate, 2.0× speedup?

\[\begin{array}{l} \\ {b}^{2} + {\left(a \cdot \left(angle \cdot \left(0.005555555555555556 \cdot \pi\right)\right)\right)}^{2} \end{array} \]
(FPCore (a b angle)
 :precision binary64
 (+ (pow b 2.0) (pow (* a (* angle (* 0.005555555555555556 PI))) 2.0)))
double code(double a, double b, double angle) {
	return pow(b, 2.0) + pow((a * (angle * (0.005555555555555556 * ((double) M_PI)))), 2.0);
}
public static double code(double a, double b, double angle) {
	return Math.pow(b, 2.0) + Math.pow((a * (angle * (0.005555555555555556 * Math.PI))), 2.0);
}
def code(a, b, angle):
	return math.pow(b, 2.0) + math.pow((a * (angle * (0.005555555555555556 * math.pi))), 2.0)
function code(a, b, angle)
	return Float64((b ^ 2.0) + (Float64(a * Float64(angle * Float64(0.005555555555555556 * pi))) ^ 2.0))
end
function tmp = code(a, b, angle)
	tmp = (b ^ 2.0) + ((a * (angle * (0.005555555555555556 * pi))) ^ 2.0);
end
code[a_, b_, angle_] := N[(N[Power[b, 2.0], $MachinePrecision] + N[Power[N[(a * N[(angle * N[(0.005555555555555556 * Pi), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]
\begin{array}{l}

\\
{b}^{2} + {\left(a \cdot \left(angle \cdot \left(0.005555555555555556 \cdot \pi\right)\right)\right)}^{2}
\end{array}
Derivation
  1. Initial program 81.3%

    \[{\left(a \cdot \sin \left(\frac{angle}{180} \cdot \pi\right)\right)}^{2} + {\left(b \cdot \cos \left(\frac{angle}{180} \cdot \pi\right)\right)}^{2} \]
  2. Taylor expanded in angle around 0 81.9%

    \[\leadsto {\left(a \cdot \sin \left(\frac{angle}{180} \cdot \pi\right)\right)}^{2} + {\left(b \cdot \color{blue}{1}\right)}^{2} \]
  3. Taylor expanded in angle around 0 75.8%

    \[\leadsto {\left(a \cdot \color{blue}{\left(0.005555555555555556 \cdot \left(angle \cdot \pi\right)\right)}\right)}^{2} + {\left(b \cdot 1\right)}^{2} \]
  4. Step-by-step derivation
    1. *-commutative75.8%

      \[\leadsto {\left(a \cdot \color{blue}{\left(\left(angle \cdot \pi\right) \cdot 0.005555555555555556\right)}\right)}^{2} + {\left(b \cdot 1\right)}^{2} \]
    2. associate-*r*75.8%

      \[\leadsto {\left(a \cdot \color{blue}{\left(angle \cdot \left(\pi \cdot 0.005555555555555556\right)\right)}\right)}^{2} + {\left(b \cdot 1\right)}^{2} \]
    3. *-commutative75.8%

      \[\leadsto {\left(a \cdot \left(angle \cdot \color{blue}{\left(0.005555555555555556 \cdot \pi\right)}\right)\right)}^{2} + {\left(b \cdot 1\right)}^{2} \]
  5. Simplified75.8%

    \[\leadsto {\left(a \cdot \color{blue}{\left(angle \cdot \left(0.005555555555555556 \cdot \pi\right)\right)}\right)}^{2} + {\left(b \cdot 1\right)}^{2} \]
  6. Final simplification75.8%

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

Reproduce

?
herbie shell --seed 2023243 
(FPCore (a b angle)
  :name "ab-angle->ABCF A"
  :precision binary64
  (+ (pow (* a (sin (* (/ angle 180.0) PI))) 2.0) (pow (* b (cos (* (/ angle 180.0) PI))) 2.0)))