Migdal et al, Equation (51)

Percentage Accurate: 99.4% → 99.3%

Time: 16.2s

Precision: binary64

Cost: 19908

• could not determine a ground truth

  1. k = 2.00000000000000007e154
  2. n = 0.0

Math

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

↓

\[\begin{array}{l} \\ \begin{array}{l} \mathbf{if}\;k \leq 8.5 \cdot 10^{-51}:\\ \;\;\;\;\frac{\sqrt{2 \cdot n}}{\sqrt{\frac{k}{\pi}}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{{\left(\pi \cdot \left(2 \cdot n\right)\right)}^{\left(1 - k\right)}}{k}}\\ \end{array} \end{array} \]

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 8.5e-51)
   (/ (sqrt (* 2.0 n)) (sqrt (/ k PI)))
   (sqrt (/ (pow (* PI (* 2.0 n)) (- 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 <= 8.5e-51) {
		tmp = sqrt((2.0 * n)) / sqrt((k / ((double) M_PI)));
	} else {
		tmp = sqrt((pow((((double) M_PI) * (2.0 * n)), (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 <= 8.5e-51) {
		tmp = Math.sqrt((2.0 * n)) / Math.sqrt((k / Math.PI));
	} else {
		tmp = Math.sqrt((Math.pow((Math.PI * (2.0 * n)), (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 <= 8.5e-51:
		tmp = math.sqrt((2.0 * n)) / math.sqrt((k / math.pi))
	else:
		tmp = math.sqrt((math.pow((math.pi * (2.0 * n)), (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 <= 8.5e-51)
		tmp = Float64(sqrt(Float64(2.0 * n)) / sqrt(Float64(k / pi)));
	else
		tmp = sqrt(Float64((Float64(pi * Float64(2.0 * n)) ^ 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 <= 8.5e-51)
		tmp = sqrt((2.0 * n)) / sqrt((k / pi));
	else
		tmp = sqrt((((pi * (2.0 * n)) ^ (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, 8.5e-51], N[(N[Sqrt[N[(2.0 * n), $MachinePrecision]], $MachinePrecision] / N[Sqrt[N[(k / Pi), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], N[Sqrt[N[(N[Power[N[(Pi * N[(2.0 * n), $MachinePrecision]), $MachinePrecision], N[(1.0 - k), $MachinePrecision]], $MachinePrecision] / k), $MachinePrecision]], $MachinePrecision]]

Derivation

1. Split input into 2 regimes

2. if k < 8.50000000000000036e-51

   1. Initial program 98.6%

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

   2. Applied egg-rr 52.7%

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

      [Start] 98.6%	\[ \frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
      *-commutative [=>] 98.6%	\[ \color{blue}{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \cdot \frac{1}{\sqrt{k}}} \]
      *-commutative [=>] 98.6%	\[ {\left(\color{blue}{\left(\pi \cdot 2\right)} \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \cdot \frac{1}{\sqrt{k}} \]
      associate-*r* [<=] 98.6%	\[ {\color{blue}{\left(\pi \cdot \left(2 \cdot n\right)\right)}}^{\left(\frac{1 - k}{2}\right)} \cdot \frac{1}{\sqrt{k}} \]
      div-inv [<=] 98.6%	\[ \color{blue}{\frac{{\left(\pi \cdot \left(2 \cdot n\right)\right)}^{\left(\frac{1 - k}{2}\right)}}{\sqrt{k}}} \]
      expm1-log1p-u [=>] 92.3%	\[ \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{{\left(\pi \cdot \left(2 \cdot n\right)\right)}^{\left(\frac{1 - k}{2}\right)}}{\sqrt{k}}\right)\right)} \]
      expm1-udef [=>] 82.2%	\[ \color{blue}{e^{\mathsf{log1p}\left(\frac{{\left(\pi \cdot \left(2 \cdot n\right)\right)}^{\left(\frac{1 - k}{2}\right)}}{\sqrt{k}}\right)} - 1} \]

   3. Simplified 66.2%

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

      [Start] 52.7%	\[ e^{\mathsf{log1p}\left(\sqrt{\frac{{\left(2 \cdot \left(\pi \cdot n\right)\right)}^{\left(1 - k\right)}}{k}}\right)} - 1 \]
      expm1-def [=>] 62.7%	\[ \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\sqrt{\frac{{\left(2 \cdot \left(\pi \cdot n\right)\right)}^{\left(1 - k\right)}}{k}}\right)\right)} \]
      expm1-log1p [=>] 66.2%	\[ \color{blue}{\sqrt{\frac{{\left(2 \cdot \left(\pi \cdot n\right)\right)}^{\left(1 - k\right)}}{k}}} \]
      *-commutative [=>] 66.2%	\[ \sqrt{\frac{{\left(2 \cdot \color{blue}{\left(n \cdot \pi\right)}\right)}^{\left(1 - k\right)}}{k}} \]
      associate-*r* [=>] 66.2%	\[ \sqrt{\frac{{\color{blue}{\left(\left(2 \cdot n\right) \cdot \pi\right)}}^{\left(1 - k\right)}}{k}} \]
      *-commutative [=>] 66.2%	\[ \sqrt{\frac{{\left(\color{blue}{\left(n \cdot 2\right)} \cdot \pi\right)}^{\left(1 - k\right)}}{k}} \]

   4. Taylor expanded in k around 0 66.2%

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

   5. Simplified 66.2%

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

      [Start] 66.2%	\[ \sqrt{\frac{2 \cdot \left(n \cdot \pi\right)}{k}} \]
      *-commutative [=>] 66.2%	\[ \sqrt{\frac{2 \cdot \color{blue}{\left(\pi \cdot n\right)}}{k}} \]

   6. Taylor expanded in n around 0 66.2%

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

   7. Simplified 66.2%

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

      [Start] 66.2%	\[ \sqrt{2 \cdot \frac{n \cdot \pi}{k}} \]
      associate-/l* [=>] 66.2%	\[ \sqrt{2 \cdot \color{blue}{\frac{n}{\frac{k}{\pi}}}} \]

   8. Applied egg-rr 99.3%

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

      [Start] 66.2%	\[ \sqrt{2 \cdot \frac{n}{\frac{k}{\pi}}} \]
      associate-*r/ [=>] 66.2%	\[ \sqrt{\color{blue}{\frac{2 \cdot n}{\frac{k}{\pi}}}} \]
      sqrt-div [=>] 99.3%	\[ \color{blue}{\frac{\sqrt{2 \cdot n}}{\sqrt{\frac{k}{\pi}}}} \]

   9. Simplified 99.3%

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

      [Start] 99.3%	\[ \frac{\sqrt{2 \cdot n}}{\sqrt{\frac{k}{\pi}}} \]
      *-commutative [=>] 99.3%	\[ \frac{\sqrt{\color{blue}{n \cdot 2}}}{\sqrt{\frac{k}{\pi}}} \]

   if 8.50000000000000036e-51 < k

   1. Initial program 99.5%

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

   2. Applied egg-rr 90.8%

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

      [Start] 99.5%	\[ \frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
      *-commutative [=>] 99.5%	\[ \color{blue}{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \cdot \frac{1}{\sqrt{k}}} \]
      *-commutative [=>] 99.5%	\[ {\left(\color{blue}{\left(\pi \cdot 2\right)} \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \cdot \frac{1}{\sqrt{k}} \]
      associate-*r* [<=] 99.5%	\[ {\color{blue}{\left(\pi \cdot \left(2 \cdot n\right)\right)}}^{\left(\frac{1 - k}{2}\right)} \cdot \frac{1}{\sqrt{k}} \]
      div-inv [<=] 99.5%	\[ \color{blue}{\frac{{\left(\pi \cdot \left(2 \cdot n\right)\right)}^{\left(\frac{1 - k}{2}\right)}}{\sqrt{k}}} \]
      expm1-log1p-u [=>] 99.2%	\[ \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{{\left(\pi \cdot \left(2 \cdot n\right)\right)}^{\left(\frac{1 - k}{2}\right)}}{\sqrt{k}}\right)\right)} \]
      expm1-udef [=>] 90.8%	\[ \color{blue}{e^{\mathsf{log1p}\left(\frac{{\left(\pi \cdot \left(2 \cdot n\right)\right)}^{\left(\frac{1 - k}{2}\right)}}{\sqrt{k}}\right)} - 1} \]

   3. Simplified 99.5%

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

      [Start] 90.8%	\[ e^{\mathsf{log1p}\left(\sqrt{\frac{{\left(2 \cdot \left(\pi \cdot n\right)\right)}^{\left(1 - k\right)}}{k}}\right)} - 1 \]
      expm1-def [=>] 99.2%	\[ \color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\sqrt{\frac{{\left(2 \cdot \left(\pi \cdot n\right)\right)}^{\left(1 - k\right)}}{k}}\right)\right)} \]
      expm1-log1p [=>] 99.5%	\[ \color{blue}{\sqrt{\frac{{\left(2 \cdot \left(\pi \cdot n\right)\right)}^{\left(1 - k\right)}}{k}}} \]
      *-commutative [=>] 99.5%	\[ \sqrt{\frac{{\left(2 \cdot \color{blue}{\left(n \cdot \pi\right)}\right)}^{\left(1 - k\right)}}{k}} \]
      associate-*r* [=>] 99.5%	\[ \sqrt{\frac{{\color{blue}{\left(\left(2 \cdot n\right) \cdot \pi\right)}}^{\left(1 - k\right)}}{k}} \]
      *-commutative [=>] 99.5%	\[ \sqrt{\frac{{\left(\color{blue}{\left(n \cdot 2\right)} \cdot \pi\right)}^{\left(1 - k\right)}}{k}} \]

3. Recombined 2 regimes into one program.

4. Final simplification 99.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 8.5 \cdot 10^{-51}:\\ \;\;\;\;\frac{\sqrt{2 \cdot n}}{\sqrt{\frac{k}{\pi}}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{{\left(\pi \cdot \left(2 \cdot n\right)\right)}^{\left(1 - k\right)}}{k}}\\ \end{array} \]

Alternatives

Alternative 1
Accuracy	99.3%
Cost	19908

\[\begin{array}{l} \mathbf{if}\;k \leq 8.5 \cdot 10^{-51}:\\ \;\;\;\;\frac{\sqrt{2 \cdot n}}{\sqrt{\frac{k}{\pi}}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\frac{{\left(\pi \cdot \left(2 \cdot n\right)\right)}^{\left(1 - k\right)}}{k}}\\ \end{array} \]

Alternative 2
Accuracy	99.4%
Cost	26304

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

Alternative 3
Accuracy	99.4%
Cost	20032

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

Alternative 4
Accuracy	99.5%
Cost	19968

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

Alternative 5
Accuracy	99.5%
Cost	19904

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

Alternative 6
Accuracy	37.6%
Cost	19584

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

Alternative 7
Accuracy	48.9%
Cost	19584

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

Alternative 8
Accuracy	37.6%
Cost	13312

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

Alternative 9
Accuracy	37.1%
Cost	13184

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

Reproduce

herbie shell --seed 2023167 
(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))))