Migdal et al, Equation (51)

Average Accuracy: 99.3% → 99.4%

Time: 14.4s

Precision: binary64

Cost: 33152

Math

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

↓

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

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
 (/
  (/
   (sqrt (* n 2.0))
   (/ (pow (* n 2.0) (* 0.5 k)) (pow PI (+ 0.5 (* k -0.5)))))
  (sqrt k)))

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

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

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

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(sqrt(Float64(n * 2.0)) / Float64((Float64(n * 2.0) ^ Float64(0.5 * k)) / (pi ^ Float64(0.5 + Float64(k * -0.5))))) / sqrt(k))
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 = (sqrt((n * 2.0)) / (((n * 2.0) ^ (0.5 * k)) / (pi ^ (0.5 + (k * -0.5))))) / sqrt(k);
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[(N[Sqrt[N[(n * 2.0), $MachinePrecision]], $MachinePrecision] / N[(N[Power[N[(n * 2.0), $MachinePrecision], N[(0.5 * k), $MachinePrecision]], $MachinePrecision] / N[Power[Pi, N[(0.5 + N[(k * -0.5), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / N[Sqrt[k], $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 99.3%

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

2. Simplified 99.3%

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

   [Start] 99.3	\[ \frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
   associate-*l/ [=>] 99.3	\[ \color{blue}{\frac{1 \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}}{\sqrt{k}}} \]
   *-lft-identity [=>] 99.3	\[ \frac{\color{blue}{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}}}{\sqrt{k}} \]
   sqr-pow [=>] 99.1	\[ \frac{\color{blue}{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{\frac{1 - k}{2}}{2}\right)} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{\frac{1 - k}{2}}{2}\right)}}}{\sqrt{k}} \]
   sqr-pow [<=] 99.3	\[ \frac{\color{blue}{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}}}{\sqrt{k}} \]
   *-commutative [=>] 99.3	\[ \frac{{\left(\color{blue}{\left(\pi \cdot 2\right)} \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}}{\sqrt{k}} \]
   associate-*l* [=>] 99.3	\[ \frac{{\color{blue}{\left(\pi \cdot \left(2 \cdot n\right)\right)}}^{\left(\frac{1 - k}{2}\right)}}{\sqrt{k}} \]
   div-sub [=>] 99.3	\[ \frac{{\left(\pi \cdot \left(2 \cdot n\right)\right)}^{\color{blue}{\left(\frac{1}{2} - \frac{k}{2}\right)}}}{\sqrt{k}} \]
   metadata-eval [=>] 99.3	\[ \frac{{\left(\pi \cdot \left(2 \cdot n\right)\right)}^{\left(\color{blue}{0.5} - \frac{k}{2}\right)}}{\sqrt{k}} \]

3. Applied egg-rr 99.3%

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

   [Start] 99.3	\[ \frac{{\left(\pi \cdot \left(2 \cdot n\right)\right)}^{\left(0.5 - \frac{k}{2}\right)}}{\sqrt{k}} \]
   unpow-prod-down [=>] 99.2	\[ \frac{\color{blue}{{\pi}^{\left(0.5 - \frac{k}{2}\right)} \cdot {\left(2 \cdot n\right)}^{\left(0.5 - \frac{k}{2}\right)}}}{\sqrt{k}} \]
   *-commutative [=>] 99.2	\[ \frac{\color{blue}{{\left(2 \cdot n\right)}^{\left(0.5 - \frac{k}{2}\right)} \cdot {\pi}^{\left(0.5 - \frac{k}{2}\right)}}}{\sqrt{k}} \]
   pow-sub [=>] 99.3	\[ \frac{\color{blue}{\frac{{\left(2 \cdot n\right)}^{0.5}}{{\left(2 \cdot n\right)}^{\left(\frac{k}{2}\right)}}} \cdot {\pi}^{\left(0.5 - \frac{k}{2}\right)}}{\sqrt{k}} \]
   pow1/2 [<=] 99.3	\[ \frac{\frac{\color{blue}{\sqrt{2 \cdot n}}}{{\left(2 \cdot n\right)}^{\left(\frac{k}{2}\right)}} \cdot {\pi}^{\left(0.5 - \frac{k}{2}\right)}}{\sqrt{k}} \]
   associate-*l/ [=>] 99.3	\[ \frac{\color{blue}{\frac{\sqrt{2 \cdot n} \cdot {\pi}^{\left(0.5 - \frac{k}{2}\right)}}{{\left(2 \cdot n\right)}^{\left(\frac{k}{2}\right)}}}}{\sqrt{k}} \]
   sub-neg [=>] 99.3	\[ \frac{\frac{\sqrt{2 \cdot n} \cdot {\pi}^{\color{blue}{\left(0.5 + \left(-\frac{k}{2}\right)\right)}}}{{\left(2 \cdot n\right)}^{\left(\frac{k}{2}\right)}}}{\sqrt{k}} \]
   div-inv [=>] 99.3	\[ \frac{\frac{\sqrt{2 \cdot n} \cdot {\pi}^{\left(0.5 + \left(-\color{blue}{k \cdot \frac{1}{2}}\right)\right)}}{{\left(2 \cdot n\right)}^{\left(\frac{k}{2}\right)}}}{\sqrt{k}} \]
   metadata-eval [=>] 99.3	\[ \frac{\frac{\sqrt{2 \cdot n} \cdot {\pi}^{\left(0.5 + \left(-k \cdot \color{blue}{0.5}\right)\right)}}{{\left(2 \cdot n\right)}^{\left(\frac{k}{2}\right)}}}{\sqrt{k}} \]
   distribute-rgt-neg-in [=>] 99.3	\[ \frac{\frac{\sqrt{2 \cdot n} \cdot {\pi}^{\left(0.5 + \color{blue}{k \cdot \left(-0.5\right)}\right)}}{{\left(2 \cdot n\right)}^{\left(\frac{k}{2}\right)}}}{\sqrt{k}} \]
   metadata-eval [=>] 99.3	\[ \frac{\frac{\sqrt{2 \cdot n} \cdot {\pi}^{\left(0.5 + k \cdot \color{blue}{-0.5}\right)}}{{\left(2 \cdot n\right)}^{\left(\frac{k}{2}\right)}}}{\sqrt{k}} \]
   div-inv [=>] 99.3	\[ \frac{\frac{\sqrt{2 \cdot n} \cdot {\pi}^{\left(0.5 + k \cdot -0.5\right)}}{{\left(2 \cdot n\right)}^{\color{blue}{\left(k \cdot \frac{1}{2}\right)}}}}{\sqrt{k}} \]
   metadata-eval [=>] 99.3	\[ \frac{\frac{\sqrt{2 \cdot n} \cdot {\pi}^{\left(0.5 + k \cdot -0.5\right)}}{{\left(2 \cdot n\right)}^{\left(k \cdot \color{blue}{0.5}\right)}}}{\sqrt{k}} \]
   *-commutative [=>] 99.3	\[ \frac{\frac{\sqrt{2 \cdot n} \cdot {\pi}^{\left(0.5 + k \cdot -0.5\right)}}{{\left(2 \cdot n\right)}^{\color{blue}{\left(0.5 \cdot k\right)}}}}{\sqrt{k}} \]

4. Simplified 99.4%

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

   [Start] 99.3	\[ \frac{\frac{\sqrt{2 \cdot n} \cdot {\pi}^{\left(0.5 + k \cdot -0.5\right)}}{{\left(2 \cdot n\right)}^{\left(0.5 \cdot k\right)}}}{\sqrt{k}} \]
   associate-/l* [=>] 99.4	\[ \frac{\color{blue}{\frac{\sqrt{2 \cdot n}}{\frac{{\left(2 \cdot n\right)}^{\left(0.5 \cdot k\right)}}{{\pi}^{\left(0.5 + k \cdot -0.5\right)}}}}}{\sqrt{k}} \]
   *-commutative [=>] 99.4	\[ \frac{\frac{\sqrt{\color{blue}{n \cdot 2}}}{\frac{{\left(2 \cdot n\right)}^{\left(0.5 \cdot k\right)}}{{\pi}^{\left(0.5 + k \cdot -0.5\right)}}}}{\sqrt{k}} \]
   *-commutative [=>] 99.4	\[ \frac{\frac{\sqrt{n \cdot 2}}{\frac{{\color{blue}{\left(n \cdot 2\right)}}^{\left(0.5 \cdot k\right)}}{{\pi}^{\left(0.5 + k \cdot -0.5\right)}}}}{\sqrt{k}} \]

5. Final simplification 99.4%

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

Alternatives

Alternative 1
Accuracy	99.4%
Cost	32960

\[\begin{array}{l} t_0 := \left(n \cdot 2\right) \cdot \pi\\ \sqrt{t_0} \cdot \left({k}^{-0.5} \cdot {t_0}^{\left(k \cdot -0.5\right)}\right) \end{array} \]

Alternative 2
Accuracy	99.4%
Cost	32960

\[\begin{array}{l} t_0 := n \cdot \left(2 \cdot \pi\right)\\ \frac{\sqrt{t_0}}{{t_0}^{\left(0.5 \cdot k\right)}} \cdot {k}^{-0.5} \end{array} \]

Alternative 3
Accuracy	99.4%
Cost	32896

\[\begin{array}{l} t_0 := \left(n \cdot 2\right) \cdot \pi\\ \frac{\frac{\sqrt{t_0}}{{t_0}^{\left(0.5 \cdot k\right)}}}{\sqrt{k}} \end{array} \]

Alternative 4
Accuracy	99.2%
Cost	26624

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

Alternative 5
Accuracy	99.3%
Cost	19968

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

Alternative 6
Accuracy	99.0%
Cost	19908

\[\begin{array}{l} \mathbf{if}\;k \leq 5.6 \cdot 10^{-43}:\\ \;\;\;\;\sqrt{n \cdot 2} \cdot \sqrt{\frac{\pi}{k}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{{\left(n \cdot \left(2 \cdot \pi\right)\right)}^{\left(1 - k\right)}}{k}}\\ \end{array} \]

Alternative 7
Accuracy	99.3%
Cost	19904

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

Alternative 8
Accuracy	65.6%
Cost	19584

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

Alternative 9
Accuracy	65.6%
Cost	19584

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

Alternative 10
Accuracy	48.9%
Cost	13184

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

Alternative 11
Accuracy	48.9%
Cost	13184

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

Reproduce

herbie shell --seed 2023137 
(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))))