Migdal et al, Equation (51)

Percentage Accurate: 99.3% → 99.3%

Time: 9.4s

Precision: binary64

Cost: 19908

• could not determine a ground truth

  1. k = 2.00000000000000007e154
  2. n = 0.0

Math

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

↓

\[\begin{array}{l} \mathbf{if}\;k \leq 3.8 \cdot 10^{-22}:\\ \;\;\;\;\frac{\sqrt{\pi \cdot \left(2 \cdot n\right)}}{\sqrt{k}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{{\left(n \cdot \left(\pi \cdot 2\right)\right)}^{\left(1 - k\right)}}{k}}\\ \end{array} \]

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
 (if (<= k 3.8e-22)
   (/ (sqrt (* PI (* 2.0 n))) (sqrt k))
   (sqrt (/ (pow (* n (* PI 2.0)) (- 1.0 k)) 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 tmp;
	if (k <= 3.8e-22) {
		tmp = sqrt((((double) M_PI) * (2.0 * n))) / sqrt(k);
	} else {
		tmp = sqrt((pow((n * (((double) M_PI) * 2.0)), (1.0 - k)) / k));
	}
	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 <= 3.8e-22) {
		tmp = Math.sqrt((Math.PI * (2.0 * n))) / Math.sqrt(k);
	} else {
		tmp = Math.sqrt((Math.pow((n * (Math.PI * 2.0)), (1.0 - k)) / k));
	}
	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 <= 3.8e-22:
		tmp = math.sqrt((math.pi * (2.0 * n))) / math.sqrt(k)
	else:
		tmp = math.sqrt((math.pow((n * (math.pi * 2.0)), (1.0 - k)) / k))
	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 <= 3.8e-22)
		tmp = Float64(sqrt(Float64(pi * Float64(2.0 * n))) / sqrt(k));
	else
		tmp = sqrt(Float64((Float64(n * Float64(pi * 2.0)) ^ Float64(1.0 - k)) / k));
	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 <= 3.8e-22)
		tmp = sqrt((pi * (2.0 * n))) / sqrt(k);
	else
		tmp = sqrt((((n * (pi * 2.0)) ^ (1.0 - k)) / k));
	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, 3.8e-22], N[(N[Sqrt[N[(Pi * N[(2.0 * n), $MachinePrecision]), $MachinePrecision]], $MachinePrecision] / N[Sqrt[k], $MachinePrecision]), $MachinePrecision], N[Sqrt[N[(N[Power[N[(n * N[(Pi * 2.0), $MachinePrecision]), $MachinePrecision], N[(1.0 - k), $MachinePrecision]], $MachinePrecision] / k), $MachinePrecision]], $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if k < 3.80000000000000023e-22

   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. Taylor expanded in k around 0 99.2%

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

   3. Applied egg-rr 74.3%

      \[\leadsto \color{blue}{e^{\mathsf{log1p}\left(\frac{\sqrt{\pi \cdot \left(2 \cdot n\right)}}{\sqrt{k}}\right)} - 1} \]
      Step-by-step derivation

      [Start] 99.2%	\[ \frac{1}{\sqrt{k}} \cdot \left(\sqrt{2} \cdot \sqrt{n \cdot \pi}\right) \]
      expm1-log1p-u [=>] 93.7%	\[ \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{1}{\sqrt{k}} \cdot \left(\sqrt{2} \cdot \sqrt{n \cdot \pi}\right)\right)\right)} \]
      expm1-udef [=>] 74.3%	\[ \color{blue}{e^{\mathsf{log1p}\left(\frac{1}{\sqrt{k}} \cdot \left(\sqrt{2} \cdot \sqrt{n \cdot \pi}\right)\right)} - 1} \]
      associate-*l/ [=>] 74.3%	\[ e^{\mathsf{log1p}\left(\color{blue}{\frac{1 \cdot \left(\sqrt{2} \cdot \sqrt{n \cdot \pi}\right)}{\sqrt{k}}}\right)} - 1 \]
      *-un-lft-identity [<=] 74.3%	\[ e^{\mathsf{log1p}\left(\frac{\color{blue}{\sqrt{2} \cdot \sqrt{n \cdot \pi}}}{\sqrt{k}}\right)} - 1 \]
      *-commutative [=>] 74.3%	\[ e^{\mathsf{log1p}\left(\frac{\color{blue}{\sqrt{n \cdot \pi} \cdot \sqrt{2}}}{\sqrt{k}}\right)} - 1 \]
      pow1/2 [=>] 74.3%	\[ e^{\mathsf{log1p}\left(\frac{\color{blue}{{\left(n \cdot \pi\right)}^{0.5}} \cdot \sqrt{2}}{\sqrt{k}}\right)} - 1 \]
      pow1/2 [=>] 74.3%	\[ e^{\mathsf{log1p}\left(\frac{{\left(n \cdot \pi\right)}^{0.5} \cdot \color{blue}{{2}^{0.5}}}{\sqrt{k}}\right)} - 1 \]
      pow-prod-down [=>] 74.3%	\[ e^{\mathsf{log1p}\left(\frac{\color{blue}{{\left(\left(n \cdot \pi\right) \cdot 2\right)}^{0.5}}}{\sqrt{k}}\right)} - 1 \]
      *-commutative [=>] 74.3%	\[ e^{\mathsf{log1p}\left(\frac{{\left(\color{blue}{\left(\pi \cdot n\right)} \cdot 2\right)}^{0.5}}{\sqrt{k}}\right)} - 1 \]
      associate-*r* [<=] 74.3%	\[ e^{\mathsf{log1p}\left(\frac{{\color{blue}{\left(\pi \cdot \left(n \cdot 2\right)\right)}}^{0.5}}{\sqrt{k}}\right)} - 1 \]
      pow1/2 [<=] 74.3%	\[ e^{\mathsf{log1p}\left(\frac{\color{blue}{\sqrt{\pi \cdot \left(n \cdot 2\right)}}}{\sqrt{k}}\right)} - 1 \]
      *-commutative [=>] 74.3%	\[ e^{\mathsf{log1p}\left(\frac{\sqrt{\pi \cdot \color{blue}{\left(2 \cdot n\right)}}}{\sqrt{k}}\right)} - 1 \]

   4. Simplified 99.5%

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

      [Start] 74.3%	\[ e^{\mathsf{log1p}\left(\frac{\sqrt{\pi \cdot \left(2 \cdot n\right)}}{\sqrt{k}}\right)} - 1 \]
      expm1-def [=>] 93.8%	\[ \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{\sqrt{\pi \cdot \left(2 \cdot n\right)}}{\sqrt{k}}\right)\right)} \]
      expm1-log1p [=>] 99.5%	\[ \color{blue}{\frac{\sqrt{\pi \cdot \left(2 \cdot n\right)}}{\sqrt{k}}} \]

   if 3.80000000000000023e-22 < k

   1. Initial program 99.9%

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

   2. Applied egg-rr 95.8%

      \[\leadsto \color{blue}{e^{\mathsf{log1p}\left(\sqrt{\frac{{\left(2 \cdot \left(\pi \cdot n\right)\right)}^{\left(1 - k\right)}}{k}}\right)} - 1} \]
      Step-by-step derivation

      [Start] 99.9%	\[ \frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
      *-commutative [=>] 99.9%	\[ \color{blue}{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \cdot \frac{1}{\sqrt{k}}} \]
      *-commutative [=>] 99.9%	\[ {\left(\color{blue}{\left(\pi \cdot 2\right)} \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \cdot \frac{1}{\sqrt{k}} \]
      associate-*r* [<=] 99.9%	\[ {\color{blue}{\left(\pi \cdot \left(2 \cdot n\right)\right)}}^{\left(\frac{1 - k}{2}\right)} \cdot \frac{1}{\sqrt{k}} \]
      div-inv [<=] 99.9%	\[ \color{blue}{\frac{{\left(\pi \cdot \left(2 \cdot n\right)\right)}^{\left(\frac{1 - k}{2}\right)}}{\sqrt{k}}} \]
      expm1-log1p-u [=>] 99.8%	\[ \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{{\left(\pi \cdot \left(2 \cdot n\right)\right)}^{\left(\frac{1 - k}{2}\right)}}{\sqrt{k}}\right)\right)} \]
      expm1-udef [=>] 95.8%	\[ \color{blue}{e^{\mathsf{log1p}\left(\frac{{\left(\pi \cdot \left(2 \cdot n\right)\right)}^{\left(\frac{1 - k}{2}\right)}}{\sqrt{k}}\right)} - 1} \]

   3. Simplified 99.9%

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

      [Start] 95.8%	\[ e^{\mathsf{log1p}\left(\sqrt{\frac{{\left(2 \cdot \left(\pi \cdot n\right)\right)}^{\left(1 - k\right)}}{k}}\right)} - 1 \]
      expm1-def [=>] 99.8%	\[ \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\sqrt{\frac{{\left(2 \cdot \left(\pi \cdot n\right)\right)}^{\left(1 - k\right)}}{k}}\right)\right)} \]
      expm1-log1p [=>] 99.9%	\[ \color{blue}{\sqrt{\frac{{\left(2 \cdot \left(\pi \cdot n\right)\right)}^{\left(1 - k\right)}}{k}}} \]
      associate-*r* [=>] 99.9%	\[ \sqrt{\frac{{\color{blue}{\left(\left(2 \cdot \pi\right) \cdot n\right)}}^{\left(1 - k\right)}}{k}} \]

3. Recombined 2 regimes into one program.

4. Final simplification 99.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 3.8 \cdot 10^{-22}:\\ \;\;\;\;\frac{\sqrt{\pi \cdot \left(2 \cdot n\right)}}{\sqrt{k}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{{\left(n \cdot \left(\pi \cdot 2\right)\right)}^{\left(1 - k\right)}}{k}}\\ \end{array} \]

Alternatives

Alternative 1
Accuracy	99.3%
Cost	19908

\[\begin{array}{l} \mathbf{if}\;k \leq 3.8 \cdot 10^{-22}:\\ \;\;\;\;\frac{\sqrt{\pi \cdot \left(2 \cdot n\right)}}{\sqrt{k}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{{\left(n \cdot \left(\pi \cdot 2\right)\right)}^{\left(1 - k\right)}}{k}}\\ \end{array} \]

Alternative 2
Accuracy	99.4%
Cost	19904

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

Alternative 3
Accuracy	49.0%
Cost	19584

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

Alternative 4
Accuracy	37.8%
Cost	13184

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

Alternative 5
Accuracy	37.8%
Cost	13184

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

Alternative 6
Accuracy	37.8%
Cost	13184

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

Reproduce

herbie shell --seed 2023165 
(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))))