Migdal et al, Equation (51)

Average Accuracy: 99.2% → 99.2%

Time: 14.9s

Precision: binary64

Cost: 33152

Math

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

↓

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

FPCore

(FPCore (k n)
 :precision binary64
 (* (/ 1.0 (sqrt k)) (pow (* (* 2.0 PI) n) (/ (- 1.0 k) 2.0))))

↓

(FPCore (k n)
 :precision binary64
 (*
  (/ 1.0 (sqrt k))
  (/ (* (sqrt n) (pow (* 2.0 PI) (+ 0.5 (* k -0.5)))) (pow n (* k 0.5)))))

C

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) {
	return (1.0 / sqrt(k)) * ((sqrt(n) * pow((2.0 * ((double) M_PI)), (0.5 + (k * -0.5)))) / pow(n, (k * 0.5)));
}

Java

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) {
	return (1.0 / Math.sqrt(k)) * ((Math.sqrt(n) * Math.pow((2.0 * Math.PI), (0.5 + (k * -0.5)))) / Math.pow(n, (k * 0.5)));
}

Python

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):
	return (1.0 / math.sqrt(k)) * ((math.sqrt(n) * math.pow((2.0 * math.pi), (0.5 + (k * -0.5)))) / math.pow(n, (k * 0.5)))

Julia

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)
	return Float64(Float64(1.0 / sqrt(k)) * Float64(Float64(sqrt(n) * (Float64(2.0 * pi) ^ Float64(0.5 + Float64(k * -0.5)))) / (n ^ Float64(k * 0.5))))
end

MATLAB

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)
	tmp = (1.0 / sqrt(k)) * ((sqrt(n) * ((2.0 * pi) ^ (0.5 + (k * -0.5)))) / (n ^ (k * 0.5)));
end

Wolfram

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_] := N[(N[(1.0 / N[Sqrt[k], $MachinePrecision]), $MachinePrecision] * N[(N[(N[Sqrt[n], $MachinePrecision] * N[Power[N[(2.0 * Pi), $MachinePrecision], N[(0.5 + N[(k * -0.5), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision] / N[Power[n, N[(k * 0.5), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 99.2%

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

2. Applied egg-rr 99.2%

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

   [Start] 99.2	\[ \frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
   unpow-prod-down [=>] 99.0	\[ \frac{1}{\sqrt{k}} \cdot \color{blue}{\left({\left(2 \cdot \pi\right)}^{\left(\frac{1 - k}{2}\right)} \cdot {n}^{\left(\frac{1 - k}{2}\right)}\right)} \]
   *-commutative [=>] 99.0	\[ \frac{1}{\sqrt{k}} \cdot \color{blue}{\left({n}^{\left(\frac{1 - k}{2}\right)} \cdot {\left(2 \cdot \pi\right)}^{\left(\frac{1 - k}{2}\right)}\right)} \]
   div-sub [=>] 99.0	\[ \frac{1}{\sqrt{k}} \cdot \left({n}^{\color{blue}{\left(\frac{1}{2} - \frac{k}{2}\right)}} \cdot {\left(2 \cdot \pi\right)}^{\left(\frac{1 - k}{2}\right)}\right) \]
   pow-sub [=>] 99.2	\[ \frac{1}{\sqrt{k}} \cdot \left(\color{blue}{\frac{{n}^{\left(\frac{1}{2}\right)}}{{n}^{\left(\frac{k}{2}\right)}}} \cdot {\left(2 \cdot \pi\right)}^{\left(\frac{1 - k}{2}\right)}\right) \]
   sqrt-pow1 [<=] 99.2	\[ \frac{1}{\sqrt{k}} \cdot \left(\frac{\color{blue}{\sqrt{{n}^{1}}}}{{n}^{\left(\frac{k}{2}\right)}} \cdot {\left(2 \cdot \pi\right)}^{\left(\frac{1 - k}{2}\right)}\right) \]
   pow1 [<=] 99.2	\[ \frac{1}{\sqrt{k}} \cdot \left(\frac{\sqrt{\color{blue}{n}}}{{n}^{\left(\frac{k}{2}\right)}} \cdot {\left(2 \cdot \pi\right)}^{\left(\frac{1 - k}{2}\right)}\right) \]
   associate-*l/ [=>] 99.2	\[ \frac{1}{\sqrt{k}} \cdot \color{blue}{\frac{\sqrt{n} \cdot {\left(2 \cdot \pi\right)}^{\left(\frac{1 - k}{2}\right)}}{{n}^{\left(\frac{k}{2}\right)}}} \]
   div-sub [=>] 99.2	\[ \frac{1}{\sqrt{k}} \cdot \frac{\sqrt{n} \cdot {\left(2 \cdot \pi\right)}^{\color{blue}{\left(\frac{1}{2} - \frac{k}{2}\right)}}}{{n}^{\left(\frac{k}{2}\right)}} \]
   metadata-eval [=>] 99.2	\[ \frac{1}{\sqrt{k}} \cdot \frac{\sqrt{n} \cdot {\left(2 \cdot \pi\right)}^{\left(\color{blue}{0.5} - \frac{k}{2}\right)}}{{n}^{\left(\frac{k}{2}\right)}} \]
   div-inv [=>] 99.2	\[ \frac{1}{\sqrt{k}} \cdot \frac{\sqrt{n} \cdot {\left(2 \cdot \pi\right)}^{\left(0.5 - \color{blue}{k \cdot \frac{1}{2}}\right)}}{{n}^{\left(\frac{k}{2}\right)}} \]
   metadata-eval [=>] 99.2	\[ \frac{1}{\sqrt{k}} \cdot \frac{\sqrt{n} \cdot {\left(2 \cdot \pi\right)}^{\left(0.5 - k \cdot \color{blue}{0.5}\right)}}{{n}^{\left(\frac{k}{2}\right)}} \]
   div-inv [=>] 99.2	\[ \frac{1}{\sqrt{k}} \cdot \frac{\sqrt{n} \cdot {\left(2 \cdot \pi\right)}^{\left(0.5 - k \cdot 0.5\right)}}{{n}^{\color{blue}{\left(k \cdot \frac{1}{2}\right)}}} \]
   metadata-eval [=>] 99.2	\[ \frac{1}{\sqrt{k}} \cdot \frac{\sqrt{n} \cdot {\left(2 \cdot \pi\right)}^{\left(0.5 - k \cdot 0.5\right)}}{{n}^{\left(k \cdot \color{blue}{0.5}\right)}} \]

3. Final simplification 99.2%

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

Alternatives

Alternative 1
Accuracy	99.3%
Cost	20032

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

Alternative 2
Accuracy	99.3%
Cost	19968

\[{\left(\pi \cdot \left(n \cdot 2\right)\right)}^{\left(0.5 + k \cdot -0.5\right)} \cdot {k}^{-0.5} \]

Alternative 3
Accuracy	97.6%
Cost	19908

\[\begin{array}{l} \mathbf{if}\;k \leq 3.1 \cdot 10^{-41}:\\ \;\;\;\;\frac{\sqrt{n \cdot 2}}{\sqrt{\frac{k}{\pi}}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{n}{k} \cdot \frac{2 \cdot \pi}{{n}^{k}}}\\ \end{array} \]

Alternative 4
Accuracy	98.9%
Cost	19908

\[\begin{array}{l} \mathbf{if}\;k \leq 10^{-41}:\\ \;\;\;\;\frac{\sqrt{n \cdot 2}}{\sqrt{\frac{k}{\pi}}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{{\left(n \cdot \left(2 \cdot \pi\right)\right)}^{\left(1 - k\right)}}{k}}\\ \end{array} \]

Alternative 5
Accuracy	99.3%
Cost	19904

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

Alternative 6
Accuracy	67.5%
Cost	19780

\[\begin{array}{l} \mathbf{if}\;k \leq 10^{+165}:\\ \;\;\;\;\frac{\sqrt{n \cdot 2}}{\sqrt{\frac{k}{\pi}}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt[3]{{\left(n \cdot \frac{2}{\frac{k}{\pi}}\right)}^{1.5}}\\ \end{array} \]

Alternative 7
Accuracy	67.5%
Cost	19780

\[\begin{array}{l} \mathbf{if}\;k \leq 10^{+114}:\\ \;\;\;\;\frac{\sqrt{n \cdot 2}}{\sqrt{\frac{k}{\pi}}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt[3]{{\left(\pi \cdot \frac{2}{\frac{k}{n}}\right)}^{1.5}}\\ \end{array} \]

Alternative 8
Accuracy	66.1%
Cost	19584

\[\sqrt{\frac{2}{k}} \cdot \sqrt{n \cdot \pi} \]

Alternative 9
Accuracy	66.1%
Cost	19584

\[\sqrt{n \cdot 2} \cdot \sqrt{\frac{\pi}{k}} \]

Alternative 10
Accuracy	66.1%
Cost	19584

\[\frac{\sqrt{n \cdot 2}}{\sqrt{\frac{k}{\pi}}} \]

Alternative 11
Accuracy	50.9%
Cost	13312

\[\frac{1}{\sqrt{0.5 \cdot \frac{\frac{k}{n}}{\pi}}} \]

Alternative 12
Accuracy	50.9%
Cost	13312

\[\frac{1}{\sqrt{\frac{k}{\pi} \cdot \frac{0.5}{n}}} \]

Alternative 13
Accuracy	49.8%
Cost	13184

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

Alternative 14
Accuracy	49.8%
Cost	13184

\[\sqrt{2 \cdot \frac{n}{\frac{k}{\pi}}} \]

Reproduce

herbie shell --seed 2023138 
(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))))