Migdal et al, Equation (51)

Average Error: 0.5 → 1.1

Time: 15.2s

Precision: binary64

Cost: 19972

Math

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

↓

\[\begin{array}{l} \mathbf{if}\;k \leq 2.6057697274586915 \cdot 10^{-83}:\\ \;\;\;\;\frac{\sqrt{n}}{\sqrt{k \cdot \frac{0.5}{\pi}}}\\ \mathbf{else}:\\ \;\;\;\;{\left(\frac{k}{{\left(\pi \cdot \left(n + n\right)\right)}^{\left(1 - k\right)}}\right)}^{-0.5}\\ \end{array} \]

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
 (if (<= k 2.6057697274586915e-83)
   (/ (sqrt n) (sqrt (* k (/ 0.5 PI))))
   (pow (/ k (pow (* PI (+ n n)) (- 1.0 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) {
	double tmp;
	if (k <= 2.6057697274586915e-83) {
		tmp = sqrt(n) / sqrt((k * (0.5 / ((double) M_PI))));
	} else {
		tmp = pow((k / pow((((double) M_PI) * (n + n)), (1.0 - k))), -0.5);
	}
	return tmp;
}

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 tmp;
	if (k <= 2.6057697274586915e-83) {
		tmp = Math.sqrt(n) / Math.sqrt((k * (0.5 / Math.PI)));
	} else {
		tmp = Math.pow((k / Math.pow((Math.PI * (n + n)), (1.0 - k))), -0.5);
	}
	return tmp;
}

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):
	tmp = 0
	if k <= 2.6057697274586915e-83:
		tmp = math.sqrt(n) / math.sqrt((k * (0.5 / math.pi)))
	else:
		tmp = math.pow((k / math.pow((math.pi * (n + n)), (1.0 - k))), -0.5)
	return tmp

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)
	tmp = 0.0
	if (k <= 2.6057697274586915e-83)
		tmp = Float64(sqrt(n) / sqrt(Float64(k * Float64(0.5 / pi))));
	else
		tmp = Float64(k / (Float64(pi * Float64(n + n)) ^ Float64(1.0 - k))) ^ -0.5;
	end
	return tmp
end

MATLAB

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

↓

function tmp_2 = code(k, n)
	tmp = 0.0;
	if (k <= 2.6057697274586915e-83)
		tmp = sqrt(n) / sqrt((k * (0.5 / pi)));
	else
		tmp = (k / ((pi * (n + n)) ^ (1.0 - k))) ^ -0.5;
	end
	tmp_2 = tmp;
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_] := If[LessEqual[k, 2.6057697274586915e-83], N[(N[Sqrt[n], $MachinePrecision] / N[Sqrt[N[(k * N[(0.5 / Pi), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[Power[N[(k / N[Power[N[(Pi * N[(n + n), $MachinePrecision]), $MachinePrecision], N[(1.0 - k), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], -0.5], $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if k < 2.6057697274586915e-83

   1. Initial program 0.6

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

   2. Applied egg-rr 19.9

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

   3. Taylor expanded in k around 0 19.9

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

   4. Simplified 19.9

      \[\leadsto \sqrt{\color{blue}{\frac{n}{k} \cdot \left(2 \cdot \pi\right)}} \]
      Proof
      (*.f64 (/.f64 n k) (*.f64 2 (PI.f64))): 0 points increase in error, 0 points decrease in error
      (*.f64 (/.f64 n k) (Rewrite=> *-commutative_binary64 (*.f64 (PI.f64) 2))): 0 points increase in error, 0 points decrease in error
      (Rewrite<= associate-*l*_binary64 (*.f64 (*.f64 (/.f64 n k) (PI.f64)) 2)): 0 points increase in error, 0 points decrease in error
      (*.f64 (Rewrite=> associate-*l/_binary64 (/.f64 (*.f64 n (PI.f64)) k)) 2): 47 points increase in error, 40 points decrease in error
      (Rewrite<= *-commutative_binary64 (*.f64 2 (/.f64 (*.f64 n (PI.f64)) k))): 0 points increase in error, 0 points decrease in error

   5. Applied egg-rr 19.9

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

   6. Applied egg-rr 0.4

      \[\leadsto \color{blue}{\frac{\sqrt{n}}{\sqrt{k \cdot \frac{0.5}{\pi}}}} \]

   if 2.6057697274586915e-83 < k

   1. Initial program 0.4

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

   2. Applied egg-rr 3.3

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

   3. Applied egg-rr 1.7

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

   4. Applied egg-rr 1.7

      \[\leadsto \color{blue}{{\left(\frac{k}{{\left(\pi \cdot \left(n + n\right)\right)}^{\left(1 - k\right)}}\right)}^{-0.5}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 1.1

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 2.6057697274586915 \cdot 10^{-83}:\\ \;\;\;\;\frac{\sqrt{n}}{\sqrt{k \cdot \frac{0.5}{\pi}}}\\ \mathbf{else}:\\ \;\;\;\;{\left(\frac{k}{{\left(\pi \cdot \left(n + n\right)\right)}^{\left(1 - k\right)}}\right)}^{-0.5}\\ \end{array} \]

Alternatives

Alternative 1
Error	0.4
Cost	32896

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

Alternative 2
Error	0.5
Cost	20032

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

Alternative 3
Error	0.5
Cost	19968

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

Alternative 4
Error	1.3
Cost	19908

\[\begin{array}{l} \mathbf{if}\;k \leq 2.6057697274586915 \cdot 10^{-83}:\\ \;\;\;\;\frac{\sqrt{n}}{\sqrt{k \cdot \frac{0.5}{\pi}}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{{\left(2 \cdot \left(n \cdot \pi\right)\right)}^{\left(1 - k\right)}}{k}}\\ \end{array} \]

Alternative 5
Error	21.2
Cost	19844

\[\begin{array}{l} \mathbf{if}\;k \leq 1.8 \cdot 10^{+80}:\\ \;\;\;\;\frac{\sqrt{n}}{\sqrt{k \cdot \frac{0.5}{\pi}}}\\ \mathbf{else}:\\ \;\;\;\;{\left({\left(\frac{\frac{k}{2 \cdot \pi}}{n}\right)}^{1.5}\right)}^{-0.3333333333333333}\\ \end{array} \]

Alternative 6
Error	13.0
Cost	19844

\[\begin{array}{l} \mathbf{if}\;k \leq 48000000:\\ \;\;\;\;\frac{\sqrt{n}}{\sqrt{k \cdot \frac{0.5}{\pi}}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\mathsf{fma}\left(\frac{n}{k}, 2 \cdot \pi, 1\right) + -1}\\ \end{array} \]

Alternative 7
Error	21.4
Cost	19780

\[\begin{array}{l} \mathbf{if}\;k \leq 1.65 \cdot 10^{+88}:\\ \;\;\;\;\frac{\sqrt{n}}{\sqrt{k \cdot \frac{0.5}{\pi}}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt[3]{{\left(n \cdot \left(2 \cdot \frac{\pi}{k}\right)\right)}^{1.5}}\\ \end{array} \]

Alternative 8
Error	22.3
Cost	19584

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

Alternative 9
Error	22.2
Cost	19584

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

Alternative 10
Error	31.9
Cost	13248

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

Alternative 11
Error	32.5
Cost	13184

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

Alternative 12
Error	32.5
Cost	13184

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

Alternative 13
Error	32.5
Cost	13184

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

Alternative 14
Error	32.5
Cost	13184

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

Reproduce

herbie shell --seed 2022317 
(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))))