Migdal et al, Equation (51)

Average Error: 0.5 → 0.4

Time: 16.9s

Precision: binary64

Cost: 32960

Math

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

↓

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

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
 (let* ((t_0 (* PI (* 2.0 n))))
   (/ (* (pow t_0 (* k -0.5)) (pow t_0 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) {
	double t_0 = ((double) M_PI) * (2.0 * n);
	return (pow(t_0, (k * -0.5)) * pow(t_0, 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) {
	double t_0 = Math.PI * (2.0 * n);
	return (Math.pow(t_0, (k * -0.5)) * Math.pow(t_0, 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):
	t_0 = math.pi * (2.0 * n)
	return (math.pow(t_0, (k * -0.5)) * math.pow(t_0, 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)
	t_0 = Float64(pi * Float64(2.0 * n))
	return Float64(Float64((t_0 ^ Float64(k * -0.5)) * (t_0 ^ 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)
	t_0 = pi * (2.0 * n);
	tmp = ((t_0 ^ (k * -0.5)) * (t_0 ^ 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_] := Block[{t$95$0 = N[(Pi * N[(2.0 * n), $MachinePrecision]), $MachinePrecision]}, N[(N[(N[Power[t$95$0, N[(k * -0.5), $MachinePrecision]], $MachinePrecision] * N[Power[t$95$0, 0.5], $MachinePrecision]), $MachinePrecision] / N[Sqrt[k], $MachinePrecision]), $MachinePrecision]]

Derivation

1. Initial program 0.5

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

2. Applied egg-rr 0.5

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

3. Applied egg-rr 0.4

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

4. Final simplification 0.4

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

Alternatives

Alternative 1
Error	0.5
Cost	20036

\[\begin{array}{l} \mathbf{if}\;k \leq 4.1 \cdot 10^{-20}:\\ \;\;\;\;\frac{\sqrt{n \cdot \left(2 \cdot \pi\right)}}{\sqrt{k}}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{\sqrt{k \cdot {\left(2 \cdot \left(n \cdot \pi\right)\right)}^{\left(k + -1\right)}}}\\ \end{array} \]

Alternative 2
Error	13.9
Cost	19972

\[\begin{array}{l} t_0 := \frac{\sqrt{n \cdot \left(2 \cdot \pi\right)}}{\sqrt{k}}\\ \mathbf{if}\;n \leq 1.4 \cdot 10^{-77}:\\ \;\;\;\;t_0\\ \mathbf{else}:\\ \;\;\;\;\left(2 + t_0\right) + -2\\ \end{array} \]

Alternative 3
Error	14.0
Cost	19972

\[\begin{array}{l} \mathbf{if}\;n \leq 1.4 \cdot 10^{-77}:\\ \;\;\;\;\frac{\sqrt{n \cdot \left(2 \cdot \pi\right)}}{\sqrt{k}}\\ \mathbf{else}:\\ \;\;\;\;\left(1 + \frac{\sqrt{2 \cdot \left(n \cdot \pi\right)}}{\sqrt{k}}\right) - 1\\ \end{array} \]

Alternative 4
Error	0.5
Cost	19904

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

Alternative 5
Error	22.0
Cost	19584

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

Reproduce

herbie shell --seed 2023096 
(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))))