Migdal et al, Equation (51)

?

Percentage Accurate: 99.4% → 99.5%
Time: 13.0s
Precision: binary64
Cost: 33024

?

\[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
↓
\[\begin{array}{l} t_0 := 2 \cdot \left(\pi \cdot n\right)\\ \frac{1}{\sqrt{k}} \cdot \frac{\sqrt{t_0}}{{t_0}^{\left(k \cdot 0.5\right)}} \end{array} \]
(FPCore (k n)
 :precision binary64
 (* (/ 1.0 (sqrt k)) (pow (* (* 2.0 PI) n) (/ (- 1.0 k) 2.0))))
↓
(FPCore (k n)
 :precision binary64
 (let* ((t_0 (* 2.0 (* PI n))))
   (* (/ 1.0 (sqrt k)) (/ (sqrt t_0) (pow t_0 (* k 0.5))))))
double code(double k, double n) {
	return (1.0 / sqrt(k)) * pow(((2.0 * ((double) M_PI)) * n), ((1.0 - k) / 2.0));
}
↓
double code(double k, double n) {
	double t_0 = 2.0 * (((double) M_PI) * n);
	return (1.0 / sqrt(k)) * (sqrt(t_0) / pow(t_0, (k * 0.5)));
}

public static double code(double k, double n) {
	return (1.0 / Math.sqrt(k)) * Math.pow(((2.0 * Math.PI) * n), ((1.0 - k) / 2.0));
}
↓
public static double code(double k, double n) {
	double t_0 = 2.0 * (Math.PI * n);
	return (1.0 / Math.sqrt(k)) * (Math.sqrt(t_0) / Math.pow(t_0, (k * 0.5)));
}

def code(k, n):
	return (1.0 / math.sqrt(k)) * math.pow(((2.0 * math.pi) * n), ((1.0 - k) / 2.0))
↓
def code(k, n):
	t_0 = 2.0 * (math.pi * n)
	return (1.0 / math.sqrt(k)) * (math.sqrt(t_0) / math.pow(t_0, (k * 0.5)))

function code(k, n)
	return Float64(Float64(1.0 / sqrt(k)) * (Float64(Float64(2.0 * pi) * n) ^ Float64(Float64(1.0 - k) / 2.0)))
end
↓
function code(k, n)
	t_0 = Float64(2.0 * Float64(pi * n))
	return Float64(Float64(1.0 / sqrt(k)) * Float64(sqrt(t_0) / (t_0 ^ Float64(k * 0.5))))
end

function tmp = code(k, n)
	tmp = (1.0 / sqrt(k)) * (((2.0 * pi) * n) ^ ((1.0 - k) / 2.0));
end
↓
function tmp = code(k, n)
	t_0 = 2.0 * (pi * n);
	tmp = (1.0 / sqrt(k)) * (sqrt(t_0) / (t_0 ^ (k * 0.5)));
end

code[k_, n_] := N[(N[(1.0 / N[Sqrt[k], $MachinePrecision]), $MachinePrecision] * N[Power[N[(N[(2.0 * Pi), $MachinePrecision] * n), $MachinePrecision], N[(N[(1.0 - k), $MachinePrecision] / 2.0), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]
↓
code[k_, n_] := Block[{t$95$0 = N[(2.0 * N[(Pi * n), $MachinePrecision]), $MachinePrecision]}, N[(N[(1.0 / N[Sqrt[k], $MachinePrecision]), $MachinePrecision] * N[(N[Sqrt[t$95$0], $MachinePrecision] / N[Power[t$95$0, N[(k * 0.5), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]

\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}
↓
\begin{array}{l}
t_0 := 2 \cdot \left(\pi \cdot n\right)\\
\frac{1}{\sqrt{k}} \cdot \frac{\sqrt{t_0}}{{t_0}^{\left(k \cdot 0.5\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.

Herbie found 12 alternatives:

AlternativeAccuracySpeedup

Accuracy vs Speed

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.

Bogosity?

Bogosity

Try it out?

Your Program's Arguments

Results

Enter valid numbers for all inputs

Derivation?

  1. Initial program 99.5%

    \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]

  2. Applied egg-rr99.7%

    \[\leadsto \frac{1}{\sqrt{k}} \cdot \color{blue}{\frac{\sqrt{2 \cdot \left(\pi \cdot n\right)}}{{\left(2 \cdot \left(\pi \cdot n\right)\right)}^{\left(k \cdot 0.5\right)}}} \]
    Step-by-step derivation

    [Start]99.5%

    \[ \frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]

    div-sub [=>]99.5%

    \[ \frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\color{blue}{\left(\frac{1}{2} - \frac{k}{2}\right)}} \]

    metadata-eval [=>]99.5%

    \[ \frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\color{blue}{0.5} - \frac{k}{2}\right)} \]

    pow-sub [=>]99.7%

    \[ \frac{1}{\sqrt{k}} \cdot \color{blue}{\frac{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{0.5}}{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{k}{2}\right)}}} \]

    pow1/2 [<=]99.7%

    \[ \frac{1}{\sqrt{k}} \cdot \frac{\color{blue}{\sqrt{\left(2 \cdot \pi\right) \cdot n}}}{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{k}{2}\right)}} \]

    associate-*l* [=>]99.7%

    \[ \frac{1}{\sqrt{k}} \cdot \frac{\sqrt{\color{blue}{2 \cdot \left(\pi \cdot n\right)}}}{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{k}{2}\right)}} \]

    associate-*l* [=>]99.7%

    \[ \frac{1}{\sqrt{k}} \cdot \frac{\sqrt{2 \cdot \left(\pi \cdot n\right)}}{{\color{blue}{\left(2 \cdot \left(\pi \cdot n\right)\right)}}^{\left(\frac{k}{2}\right)}} \]

    div-inv [=>]99.7%

    \[ \frac{1}{\sqrt{k}} \cdot \frac{\sqrt{2 \cdot \left(\pi \cdot n\right)}}{{\left(2 \cdot \left(\pi \cdot n\right)\right)}^{\color{blue}{\left(k \cdot \frac{1}{2}\right)}}} \]

    metadata-eval [=>]99.7%

    \[ \frac{1}{\sqrt{k}} \cdot \frac{\sqrt{2 \cdot \left(\pi \cdot n\right)}}{{\left(2 \cdot \left(\pi \cdot n\right)\right)}^{\left(k \cdot \color{blue}{0.5}\right)}} \]

  3. Final simplification99.7%

    \[\leadsto \frac{1}{\sqrt{k}} \cdot \frac{\sqrt{2 \cdot \left(\pi \cdot n\right)}}{{\left(2 \cdot \left(\pi \cdot n\right)\right)}^{\left(k \cdot 0.5\right)}} \]

Alternatives

Alternative 1
Accuracy99.5%
Cost33024
\[\begin{array}{l} t_0 := 2 \cdot \left(\pi \cdot n\right)\\ \frac{1}{\sqrt{k}} \cdot \frac{\sqrt{t_0}}{{t_0}^{\left(k \cdot 0.5\right)}} \end{array} \]
Alternative 2
Accuracy99.5%
Cost19968
\[{k}^{-0.5} \cdot {\left(n \cdot \left(2 \cdot \pi\right)\right)}^{\left(\frac{1 - k}{2}\right)} \]
Alternative 3
Accuracy98.9%
Cost19908
\[\begin{array}{l} \mathbf{if}\;k \leq 7 \cdot 10^{-67}:\\ \;\;\;\;\frac{\sqrt{\pi \cdot \left(2 \cdot n\right)}}{\sqrt{k}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{{\left(2 \cdot \left(\pi \cdot n\right)\right)}^{\left(1 - k\right)}}{k}}\\ \end{array} \]
Alternative 4
Accuracy99.5%
Cost19904
\[\frac{{\left(\pi \cdot \left(2 \cdot n\right)\right)}^{\left(\frac{1 - k}{2}\right)}}{\sqrt{k}} \]
Alternative 5
Accuracy50.1%
Cost19780
\[\begin{array}{l} \mathbf{if}\;k \leq 5.1 \cdot 10^{+271}:\\ \;\;\;\;\frac{\sqrt{\pi \cdot \left(2 \cdot n\right)}}{\sqrt{k}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt[3]{{\left(\pi \cdot \left(n \cdot \frac{2}{k}\right)\right)}^{1.5}}\\ \end{array} \]
Alternative 6
Accuracy50.0%
Cost19780
\[\begin{array}{l} \mathbf{if}\;k \leq 4.4 \cdot 10^{+274}:\\ \;\;\;\;{k}^{-0.5} \cdot \sqrt{\pi \cdot \left(2 \cdot n\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt[3]{{\left(\pi \cdot \left(n \cdot \frac{2}{k}\right)\right)}^{1.5}}\\ \end{array} \]
Alternative 7
Accuracy49.1%
Cost19584
\[\frac{\sqrt{\pi \cdot \left(2 \cdot n\right)}}{\sqrt{k}} \]
Alternative 8
Accuracy37.7%
Cost13184
\[\sqrt{2 \cdot \left(n \cdot \frac{\pi}{k}\right)} \]
Alternative 9
Accuracy37.7%
Cost13184
\[\sqrt{2 \cdot \left(\pi \cdot \frac{n}{k}\right)} \]
Alternative 10
Accuracy37.7%
Cost13184
\[\sqrt{2 \cdot \frac{n}{\frac{k}{\pi}}} \]
Alternative 11
Accuracy37.7%
Cost13184
\[\sqrt{\frac{\pi}{\frac{\frac{k}{2}}{n}}} \]
Alternative 12
Accuracy37.7%
Cost13184
\[\sqrt{\frac{\pi \cdot \left(2 \cdot n\right)}{k}} \]

Reproduce?

herbie shell --seed 2023166 
(FPCore (k n)
  :name "Migdal et al, Equation (51)"
  :precision binary64
  (* (/ 1.0 (sqrt k)) (pow (* (* 2.0 PI) n) (/ (- 1.0 k) 2.0))))