Falkner and Boettcher, Equation (22+)

Percentage Accurate: 98.5% → 98.9%
Time: 1.2s
Alternatives: 1
Speedup: 3.2×

Specification

?
\[\begin{array}{l} \\ \frac{4}{\left(\left(3 \cdot \pi\right) \cdot \left(1 - v \cdot v\right)\right) \cdot \sqrt{2 - 6 \cdot \left(v \cdot v\right)}} \end{array} \]
(FPCore (v)
 :precision binary64
 (/ 4.0 (* (* (* 3.0 PI) (- 1.0 (* v v))) (sqrt (- 2.0 (* 6.0 (* v v)))))))
double code(double v) {
	return 4.0 / (((3.0 * ((double) M_PI)) * (1.0 - (v * v))) * sqrt((2.0 - (6.0 * (v * v)))));
}

public static double code(double v) {
	return 4.0 / (((3.0 * Math.PI) * (1.0 - (v * v))) * Math.sqrt((2.0 - (6.0 * (v * v)))));
}

def code(v):
	return 4.0 / (((3.0 * math.pi) * (1.0 - (v * v))) * math.sqrt((2.0 - (6.0 * (v * v)))))

function code(v)
	return Float64(4.0 / Float64(Float64(Float64(3.0 * pi) * Float64(1.0 - Float64(v * v))) * sqrt(Float64(2.0 - Float64(6.0 * Float64(v * v))))))
end

function tmp = code(v)
	tmp = 4.0 / (((3.0 * pi) * (1.0 - (v * v))) * sqrt((2.0 - (6.0 * (v * v)))));
end

code[v_] := N[(4.0 / N[(N[(N[(3.0 * Pi), $MachinePrecision] * N[(1.0 - N[(v * v), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[Sqrt[N[(2.0 - N[(6.0 * N[(v * v), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

\begin{array}{l}

\\
\frac{4}{\left(\left(3 \cdot \pi\right) \cdot \left(1 - v \cdot v\right)\right) \cdot \sqrt{2 - 6 \cdot \left(v \cdot v\right)}}
\end{array}

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 1 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: 98.5% accurate, 1.0× speedup?

\[\begin{array}{l} \\ \frac{4}{\left(\left(3 \cdot \pi\right) \cdot \left(1 - v \cdot v\right)\right) \cdot \sqrt{2 - 6 \cdot \left(v \cdot v\right)}} \end{array} \]
(FPCore (v)
 :precision binary64
 (/ 4.0 (* (* (* 3.0 PI) (- 1.0 (* v v))) (sqrt (- 2.0 (* 6.0 (* v v)))))))
double code(double v) {
	return 4.0 / (((3.0 * ((double) M_PI)) * (1.0 - (v * v))) * sqrt((2.0 - (6.0 * (v * v)))));
}

public static double code(double v) {
	return 4.0 / (((3.0 * Math.PI) * (1.0 - (v * v))) * Math.sqrt((2.0 - (6.0 * (v * v)))));
}

def code(v):
	return 4.0 / (((3.0 * math.pi) * (1.0 - (v * v))) * math.sqrt((2.0 - (6.0 * (v * v)))))

function code(v)
	return Float64(4.0 / Float64(Float64(Float64(3.0 * pi) * Float64(1.0 - Float64(v * v))) * sqrt(Float64(2.0 - Float64(6.0 * Float64(v * v))))))
end

function tmp = code(v)
	tmp = 4.0 / (((3.0 * pi) * (1.0 - (v * v))) * sqrt((2.0 - (6.0 * (v * v)))));
end

code[v_] := N[(4.0 / N[(N[(N[(3.0 * Pi), $MachinePrecision] * N[(1.0 - N[(v * v), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[Sqrt[N[(2.0 - N[(6.0 * N[(v * v), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

\begin{array}{l}

\\
\frac{4}{\left(\left(3 \cdot \pi\right) \cdot \left(1 - v \cdot v\right)\right) \cdot \sqrt{2 - 6 \cdot \left(v \cdot v\right)}}
\end{array}

Alternative 1: 98.9% accurate, 3.2× speedup?

\[\begin{array}{l} \\ \frac{1.3333333333333333}{\pi \cdot \sqrt{2}} \end{array} \]
(FPCore (v) :precision binary64 (/ 1.3333333333333333 (* PI (sqrt 2.0))))
double code(double v) {
	return 1.3333333333333333 / (((double) M_PI) * sqrt(2.0));
}

public static double code(double v) {
	return 1.3333333333333333 / (Math.PI * Math.sqrt(2.0));
}

def code(v):
	return 1.3333333333333333 / (math.pi * math.sqrt(2.0))

function code(v)
	return Float64(1.3333333333333333 / Float64(pi * sqrt(2.0)))
end

function tmp = code(v)
	tmp = 1.3333333333333333 / (pi * sqrt(2.0));
end

code[v_] := N[(1.3333333333333333 / N[(Pi * N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

\begin{array}{l}

\\
\frac{1.3333333333333333}{\pi \cdot \sqrt{2}}
\end{array}
Derivation

  1. Initial program 98.5%

    \[\frac{4}{\left(\left(3 \cdot \pi\right) \cdot \left(1 - v \cdot v\right)\right) \cdot \sqrt{2 - 6 \cdot \left(v \cdot v\right)}} \]

  2. Taylor expanded in v around 0

    \[\leadsto \color{blue}{\frac{\frac{4}{3}}{\mathsf{PI}\left(\right) \cdot \sqrt{2}}} \]
  3. Step-by-step derivation

    1. lower-/.f64N/A

      \[\leadsto \frac{\frac{4}{3}}{\color{blue}{\mathsf{PI}\left(\right) \cdot \sqrt{2}}} \]

    2. lower-*.f64N/A

      \[\leadsto \frac{\frac{4}{3}}{\mathsf{PI}\left(\right) \cdot \color{blue}{\sqrt{2}}} \]

    3. lift-PI.f64N/A

      \[\leadsto \frac{\frac{4}{3}}{\pi \cdot \sqrt{\color{blue}{2}}} \]

    4. lower-sqrt.f6498.9

      \[\leadsto \frac{1.3333333333333333}{\pi \cdot \sqrt{2}} \]

  4. Applied rewrites98.9%

    \[\leadsto \color{blue}{\frac{1.3333333333333333}{\pi \cdot \sqrt{2}}} \]
  5. Add Preprocessing

Reproduce

?
herbie shell --seed 2025135 
(FPCore (v)
  :name "Falkner and Boettcher, Equation (22+)"
  :precision binary64
  (/ 4.0 (* (* (* 3.0 PI) (- 1.0 (* v v))) (sqrt (- 2.0 (* 6.0 (* v v)))))))